SAT Preparation Classes

SAT Quick Challenge - Week 9
Fall 2022

Saturday, November 19, 2022

Providence Baptist Church is concerned about you, your family, your friends, and the community. We are following the recommendations to limit the number of people in our in-person gatherings. For this reason, all SAT Preparation classes at the church are cancelled until further notice.

Although the current COVID-19 pandemic abruptly ended the Spring 2020 SAT classes, technology offers an opportunity for students to refresh, retain, and/or acquire SAT knowledge and skills essential for answering various kinds of questions often found on the SAT. Therefore, the Providence Baptist Church SAT Preparation Program has provided this “SAT Quick Challenge” website.

Remember the 3 Ws - Wear a mask, Wash your hands, and Watch your distance. God bless you and yours!

Merry Christmas!

So that the SAT Quick Challenge Team can spend time with family and friends during the holidays, our November 19 lesson was the final lesson of the Fall 2022 season. Our initial Spring 2023 lesson will be posted in January 2023. We hope that you and yours have a Merry Christmas, and a Happy New Year. Enjoy this time with family and friends and stay safe!

For unto you is born this day in the city of David a Saviour, which is Christ the Lord.
                                                                                                                       Luke 2: (KJV)

SAT Math - Questions from Fall 2022, Week 9  (November 19, 2022)

Absolute Value Equations

The absolute value of a number is the distance of that number from 0 on the number line. The symbol |x| is used to designate the absolute value of x. Since distance is nonnegative, the absolute value of a number is always positive or zero. Thus, |7| = 7 because 7 is seven units away from 0 on the number line. Also, |– 6| = 6 because 6 is six units away from 0 on the number line.

Some equations contain an absolute value expression, which take the form |ax + b| = c. To solve an absolute value equation, it is necessary to consider the possibility that “ax + b” may be positive and “ax + b” may be negative. To solve for x in the equation, we solve both for “ax + b = c” and “ax + b = – c”. Thus there may be two solutions for the equation. It is always necessary to check the solutions in the original equation to verify the solutions.

Examine Examples 1, 2, 3, and 4 below; solve each of the equations.

Example 1.                   |2x + 3| = 15
                                       2x + 3 = 15                  2x + 3 = – 15
                                             2x = 12                        2x = – 18
                                               x = 6                            x = – 9

We have two answers, 6 and – 9. Now we can check our answers for each value of x in the original equation.
                x = 6                                                            x = – 9
      |2x + 3| = 15                                               |2x + 3| = 15
    |2(6) + 3| = 15                                         |2(– 9) + 3| = 15
      |12 + 3| = 15                                           |– 18 + 3| = 15
            |15| = 15                                                 |– 15| = 15
              15 = 15                                                       15 = 15                      Our answers check successfully.


Example 2.          4|6 – 2x| – 5 = 27

        To solve the absolute value equation, the absolute value term must be alone on one side of the equation. Thus we need to add 5 to both sides of the equation and then divide both sides by 4.

                                  4|6 – 2x| = 32
                                    |6 – 2x| = 8
                                      6 – 2x = 8                       6 – 2x = – 8
                                         – 2x = 2                          – 2x = – 14
                                              x = – 1                            x = 7

We have two answers, – 1 and 7. We can check our answers for each value of x in the original equation.
                                               x = – 1                           x = 7

                             4|6 – 2x| – 5 = 27        4|6 – 2x| – 5 = 27
                         4|6 – 2(–1)| – 5 = 27     4|6 – 2(7)| – 5 = 27
                                4|6 +2| – 5 = 27       4|6 – 14| – 5 = 27
                                     4|8| – 5 = 27             4|– 8| – 5 = 27
                                     4(8) – 5 = 27                4(8) – 5 = 27
                                       32 – 5 = 27                  32 – 5 = 27
                                              27 = 27                        27 = 27                                                   
Our answers check successfully.


Example 3.             |3x + 9| + 4 = 0
                                     |3x + 9| = – 4

This equation has no solution. Since the absolute value of a number is always positive or zero, there is no value of x that would make this sentence true.


Example 4.                     |x – 2| = 2x – 10
                                       |x – 2| = 2x – 10             |x – 2| = 2x – 10
                                         x – 2 = 2x – 10               x – 2 = – (2x – 10)
                                            – 2 = x – 10                 x – 2 = – 2x + 10
                                               8 = x                             3x = 12
                                                                                      x = 4

We have two answers, 8 and 4. We can check our answers for each value of x in the original equation.
            x = 8                          x = 4

     |x – 2| = 2x – 10      |x – 2| = 2x – 10
     |8 – 2| = 2(8) – 10   |4 – 2| = 2(4) – 10
           |6| = 16 – 10           |2| = 8 – 10
             6 = 6                        2 = – 2         NO

In this case, only one value of x works; x = 8 works, but x = 4 does not. The only solution is x = 8.

Keeping in mind the information above, solve the following problems.
  1. |x – 3| = 17
  2. |x + 11| = 42
  3. 11|x – 9| = 121
  4. |2x + 7| = x – 4
  5. 8|x – 3| = 88
  6. 3|x + 6| = 9x – 6

SAT Math - Answers to Questions from Fall 2022, Week 9  (November 19, 2022)

  1. |x – 3| = 17

    x – 3 = 17                                                x – 3 = – 17
    x = 20                                                            x = – 14
    We have two answers, 20 and – 14. We can check our answers for each value of x in the original equation.

     x = 20                                                          x = – 14

    |x – 3| = 17                                               |x – 3| = 17

    |20 – 3| = 17                                           | – 14 – 3| = 17
          |17| = 17                                                 | – 17| = 17              Our answers check successfully.
            17 = 17

  2. |x + 11| = 42

    x + 11 = 42                                   x + 11 = – 42
             x = 31                                           x = – 53
    We have two answers, 31 and – 53. We can check our answers for each value of x in the original equation.

              x = 31                                          x = – 53

       |x + 11| = 42                              |x + 11| = 42
    |31 + 11| = 42                        | – 53 + 11| = 42
            |42| = 42                                 | – 42| = 42
              42 = 42                                       42 = 42                      Our answers check successfully.

  3. 11|x – 9| = 121

         |x – 9| = 11
           x – 9 = 11                                  x – 9 = – 11
                 x = 20                                        x = – 2

    We have two answers, 20 and – 2. We can check our answers for each value of x in the original equation.

                   x = 20                                     x = – 2

      11|x – 9| = 121                      11|x – 9| = 121
    11|20 – 9| = 121                 11| – 2 – 9| = 121
    11|11| = 121                           11| – 11| = 121
    11(11)| = 121                              11(11) = 121
    121 = 121                                        121 = 121                   Our answers check successfully.

  4. |2x + 7| = x – 4

    There is no solution for this equation. There is no value of x that will make the two sides of the
    equation equal. The absolute value of 2x + 7 will always be greater than x – 4.

  5. 8|x – 3| = 88

     |x – 3| = 11
       x – 3 = 11                                   x – 3 = – 11
             x = 14                                         x = – 8

    We have two answers, 14 and – 8. We can check our answers for each value of x in the original equation.

               x = 14                                        x = – 8

      8|x – 3| = 88                            8|x – 3| = 88
    8|14 – 3| = 88                       8| – 8 – 3| = 88
          8|11| = 88                           8| – 11| = 88
          8(11) = 88                                8(11) = 88
              88 = 88                                    88 = 88                   Our answers check successfully.

  6. 3|x + 6| = 9x – 6

      |x + 6| = 3x – 2
        x + 6 = 3x – 2                              x + 6 = – (3x – 2)
              8 = 2x                                    x + 6 = – 3x + 2
              4 = x                                          4x = – 4
                                                                  x = – 1

    We have two answers, 4 and – 1. We can check our answers for each value of x in the original equation.

              x = 4                                           x = – 1

    3|x + 6| = 9x – 6                        3|x + 6| = 9x – 6
    3|4 + 6| = 9(4) – 6                3| – 1 + 6| = 9(– 1) – 6
    3|10| = 36 – 6                                3|5| = – 9 – 6
    3(10) = 30                                       3(5) = – 15
        30 = 30                                         15 = – 15         NO


    In this case, only one value of x works; x = 4 works, but x = – 1 does not. The only solution is x = 4.

SAT Verbal - Questions from Fall 2022, Week 9 (November 19, 2022)

SAT QUICK CHALLENGE EXERCISE M22
Subject-Verb Agreement (SVA)

The Prepositional Phrase(s) Between the Subject and Verb. On an SAT subject/verb agreement
(sva) question, prepositional phrases placed between a subject and its verb can make selecting the right
verb a bit challenging, as in the following sentence: The houses with the lovely red roses in the yard
is/are going to be sold to the highest bidder next Monday. Since the subject of a sentence will never
be the object of the preposition, draw a line through each prepositional phrase that comes before the verb
in question, and then identify the subject – the closest noun to that verb. (The houses with the lovely red
roses in the yard is/are) In this sentence, that word is the plural noun “houses,” (not the singular object
of the preposition, “yard.”). Therefore, we need the plural verb “are,” and the sentence should read as
follows: The houses with the lovely red roses in the yard are going to be sold to the highest bidder next
Monday.

Verb Before the Subject. Usually, the subject comes before the verb in a sentence, but some SAT sva
questions place the verb before the subject. Such sentences may begin with a prepositional phrase that
is followed by the verb and then the subject, as follows: On the other side of the building awaits
miniature golf, kiddie pools, and a snack bar. To determine the correct verb form, draw a line through
each prepositional phrase. Then, identify the subject, and select the correct verb (singular or plural),
as follows: On the other side of the building awaits miniature golf, kiddie pools, and a snack bar.
Since the subject names three things, the subject is compound and needs the plural verb – “await.

Compound Subject Joined with “and”. A compound subject contains two or more separate nouns
joined by the word “and” or some other conjunction. A compound subject joined by “and” is always
plural. This kind of subject can be easy to overlook in longer sentences, as well as in sentences in
which the verb comes before the subject. Therefore, you must be sure to read carefully, and pay
attention to all the nouns – not just the one beside the verb. Note the sentence which follows: When
Mom drives on long trips, my brother’s silly jokes and GPA information on his phone helps her stay
awake. That sentence is incorrect because even though a singular noun, “information,” is close to the
verb, we have a compound subject – jokes and information. Therefore, we need the plural verb –
help.

Now, keeping in mind the information above, complete Exercise M22 below.

SAT QUICK CHALLENGE EXERCISE M22
Subject-Verb Agreement

Directions. Fill in the blank in each sentence below with the correct answer choice. When you have finished the exercise, use the answer key to check your work. 

1.  Along the driveway in the front of the garden ______ the newly planted roses that Mom has always wanted.

  1. blooms
  2. bloom

2. The band and their coach ______ to sit in this section of the bleachers.

  1. need
  2. needs

3. The notes and practice problems on the screen of Teddy’s computer ______ to be very good study information for the quiz.

  1. is
  2. are

SAT Verbal - Answers to Questions from Fall 2022, Week  9 (November 19, 2022)

  1. B
  2. A
  3. B

SAT Math - Questions from Fall 2022, Week 8  (November 12, 2022)

INEQUALITIES: Word Problems

The following symbols are used in inequalities:

≠ is not equal to
> is greater than
< is less than
≥ is greater than or equal to
≤ is less than or equal to

Just as some word problems can be approached by solving equations, some other word problems are solved by inequalities.

The following steps can help you solve inequality word problems:

  1. Read the problem carefully and be sure that you understand key words and concepts.
  2. Determine how the key words and concepts are related mathematically by
    converting the words to math; write the inequality in mathematical terms.
    1. It is important to determine what the variable is; express the unknown as a
      mathematical symbol (x or any other latter).
    2. It is also important to determine accurately the direction of the inequality sign:
      is it greater than or is it less than?
    3. Determine what goes on either side of the inequality sign.
  3. Solve the problem and interpret the solution.
  4. Be sure that the answer is reasonable and answers the question that was asked.

Now examine examples 1, 2, and 3.

Example 1.
Six more than twice a number is at least 60. What is the minimum value of the number?

Our unknown, our variable, is a number. Let’s call it x. Twice x plus 6 is greater than 60; it could be 80, or 300, or 20,000, or any number greater than 60. But we want the smallest possible value that meets our conditions. Thus the direction of the inequality sign is greater than or equal to since we want a value that is at least 60.

         2x + 6 ≥ 60
               2x ≥ 54
                 x ≥ 27            We want the smallest number that is greater than or equal to 27; x is any number that is greater than or equal to 27. The smallest of these is 27. Thus our answer is 27.

Example 2.
The Petty family is considering renting a boat for a fun ride on the Dan River. Booker Boats charges $260 per week, plus $2 per hour, for use of its boats. A competitor, Saulter Ships, charges $40 per week, plus $8 per hour, for use of its boats. Boats of both companies are similar. How much would the Petty family have to use the boat in a week in order for Booker Boats to be the better deal?

Because of the small weekly cost, $40 vs $260, Saulter Ships is the better deal if we use the boat just a few hours per week. But if we use the boat for many hours during the week, Booker Boats is the better choice because of its lower per hour charge. Our question, then, is when does Booker Boats become the better deal?

Our variable is the number of hours per week; let’s call it x.
We want to find x when the total cost for Booker Boats is less than the total cost for Saulter Ships. Thus the direction for the inequality is less than.

        260 + 2x < 40 + 8x
                220 < 6x
           36.667 < x Therefore the Petty family would have to use the boat for 37 or more hours per week for Booker Boats to be the better deal.

To check our work, calculate the total cost when x is greater than 37 (Booker Boats should cost less) and when x is less than 37 (Saulter Ships should cost less).

Let x = 40:  Saulter Ships: cost = 40 + 8x = 40 + 8(40) = 40 + 320 = 360
                   Booker Boats: cost = 260 + 2x = 260 + 2(40) = 260 + 80 = 340
                   Booker Boats is better

Let x = 30:  Saulter Ships: cost = 40 + 8x = 40 + 8(30) = 40 + 240 = 280
                   Booker Boats: cost = 260 + 2x = 260 + 2(30) = 260 + 60 = 320
                   Saulter Ships costs less

Example 3.
Tickets for A&T’s winter concert cost $4.00 for students and $10 for adults. If Mr. Harrison spends at least $42 but no more than $70 on x student tickets and 3 adult tickets, what is one possible value of x?

We have a compound inequality: an amount is between two values. For example, if x is
an integer and we have 5 < x ≤ 9 then x could be 6, 7, 8, or 9.

In our ticket example, we have a lower amount, $42, and a higher amount, $70, and the variable is the number of student tickets.

      42 ≤ 4x + 3(10) ≤ 70
      42 ≤ 4x + 30 ≤ 70
      12 ≤ 4x ≤ 40
        3 ≤ x ≤ 10

Thus the number of student tickets could be any number between 3 and 10, including 3 and 10.

Keeping in mind the information above, answer the following questions.


  1. Dave and Ron are the top scorers on their high school basketball team. If the team is to have any chance of winning its next game, Dave and Ron together must score at least 50 points. What is the minimum number of points that Dave must score if Ron scores 4 points less than twice the number of Dave’s points?

  2. I have 200 shares of IBM stock. My uncle, who recently retired, has 2000 shares
    of IBM stock. He has substantial investments in other companies, and wants to give the IBM stock to his five nephews. Starting tomorrow, he will give each of his five nephews, including me, one share of his IMB stock every day. How many shares of IBM stock will I have on the first day that I have more shares of the stock than my uncle has?

  3. Marcus rented a motor cycle. The rental cost $15 per hour, and he also had to pay
    for a helmet that costs $20. In total, he spent more than $90 for the rental and helmet. If the motor cycle was available for only a whole number of hours, what was the minimum number of hours that Marcus could have rented the motor cycle?

  4. C = 25x + 8y

    The formula above gives the monthly cost C, in dollars, of operating a printer when a technician works a total of x hours and when y reams of paper are used. If, in July it costs no more than $5,800 to operate the printer and 125 reams of paper were used, what is the maximum number of hours the technician could have worked?

    1. 144
    2. 160
    3. 192
    4. 240

  5. A restaurant is making its weekly purchase of beef and chicken from its supplier. The supplier will deliver no more than 250 pounds in a shipment. Each package of beef weighs 20 pounds and each package of chicken weighs 15 pounds. The restaurant wants to purchase 2 times as many packages of chicken as packages of beef, and the restaurant wants to purchase at least 100 pounds of meat. What is the minimum number of packages of beef that should be purchased?

  6. The average monthly cable television cost for the Adams family is $120. The family plans to spend $3,000 to install a satellite television system. The family estimates that the average annual satellite television cost will be $1,100. How many years after installation of the satellite system will the total amount of television cost savings exceed the installation cost?

SAT Math - Answers to Questions from Fall 2022, Week 8  (November 12, 2022)


  1. Dave and Ron are the top scorers on their high school basketball team. If the team is to have any chance of winning its next game, Dave and Ron together must score at least 50 points. What is the minimum number of points that Dave must score if Ron scores 4 points less than twice the number of Dave’s points?

    The variable is the number of points that Dave must score. The direction of the
    inequality sign is greater than. Dave’s score is x and Ron’s score is 2x – 4. Thus we have
                          x + 2x – 4 ≥ 50
                                      3x ≥ 54
                                        x ≥ 18 Thus Dave must score at least 18 points.

  2. I have 200 shares of IBM stock. My uncle, who recently retired, has 2000 shares
    of IBM stock. He has substantial investments in other companies, and wants to give the IBM stock to his five nephews. Starting tomorrow, he will give each of his five nephews, including me, one share of his IMB stock every day. How many shares of IBM stock will I have on the first day that I have more shares of the stock than my uncle has?

    Each day my uncle gives away share of stock, I gain one share. After he has given away stock on x days, I have 200 + x shares of stock. Each day my uncle gives away stock, he loses 5 shares; thus he has given away 5x shares after x days. This leaves him with 2000 – 5x shares of IBM stock. We want to know when I will have more shares than he has, so we want to know when 200 + x > 2000 – 5x

    We now solve for x.                          200 + x > 2000 – 5x
                                                                      6x > 1800
                                                                        x > 300

    The first time I will have more shares of IBM stock than my uncle is just after he has given away shares of stock 301 times (301 is the smallest number that is greater than 300). At that point, I will have 200 + 301 = 501 shares, and he will have 2000 – 5(301) = 2000 – 1505 = 495 shares. Therefore I will have 501 shares of stock the first time I have more
    shares than my uncle has.

  3. Marcus rented a motor cycle. The rental cost $15 per hour, and he also had to pay
    for a helmet that costs $20. In total, he spent more than $90 for the rental and helmet. If the motor cycle was available for only a whole number of hours, what was the minimum number of hours that Marcus could have rented the motor cycle?

    The variable, x, is the number of hours Marcus rents the motor cycle.

    There is a cost of $12 per hour; if he rents the motor cycle of x hours, he will spend 12x dollars plus an initial one time payment of $20 for the helmet. Thus the total cost of the rental and helmet is 12x + 20.

    The amount he will spend is more than 90; thus we have this inequality:

                   12x + 20 > 90

    Now solve for x:
                   12x + 20 > 90
                           12x > 70
                               x > 70/12 = 5 10/12 = 5 5/6

    Since he must rent for a whole number of hours, the minimum number of hours is 6.

  4. C = 25x + 8y

    The formula above gives the monthly cost C, in dollars, of operating a printer when a technician works a total of x hours and when y reams of paper are used. If, in July it costs no more than $5,800 to operate the printer and 125 reams of paper were used, what is the maximum number of hours the technician could have worked?

      1. 144
      2. 160
      3. 192
      4. 240


    The variable is the number of hours the technician could have worked; the direction of the inequality is less than. Thus we have the following inequality:
                                                               25x + 8y ≤ 5,800
                                                        25x + 8(125) ≤ 5,800
                                                         25x + 1,000 ≤ 5,800
                                                                      25x ≤ 4,800
                                                                          x ≤ 192 ------- C

  5. A restaurant is making its weekly purchase of beef and chicken from its supplier. The supplier will deliver no more than 250 pounds in a shipment. Each package of beef weighs 20 pounds and each package of chicken weighs 15 pounds. The restaurant wants to purchase 2 times as many packages of chicken as packages of beef, and the restaurant wants to purchase at least 100 pounds of meat. What is the minimum number of packages of beef that should be purchased?

    We have a compound inequality: an amount is between two values, 100 and 250.

    Our variable, x, is the number of packages of beef; the number of packages of chicken is 2x.

                                    100 ≤ 20x + 15(2x) ≤ 250
                                                            100 ≤ 20x + 30x ≤ 250
                                                            100 ≤ 50x ≤ 250
                                                                2 ≤ x ≤ 5

    The minimum number of packages of beef is 2.

  6. The average monthly cable television cost for the Adams family is $120. The family plans to spend $3,000 to install a satellite television system. The family estimates that the average annual satellite television cost will be $1,100. How many years after installation of the satellite system will the total amount of television cost savings exceed the installation cost?

    Our variable, x, is the number of years it takes for the savings to exceed the installation cost. There will be annual savings since the annual cost after installation, $1,100, is less than the family’s current annual cost of $1,440 ($120(12) = $1,440). The direction of the inequality sign is greater than since the total savings will exceed the installation cost. Thus the inequality is:
                        (1440 – 1100)x > 3,000
                                         340x > 3,000
                                                x > 8.82

    It will take 9 years of savings for the family to recoup the installation cost.

SAT Verbal - Questions from Fall 2022, Week 8 (November 12, 2022)

The Non-Essential Clause (NEC) Between the Subject and Verb. Subjects and their verbs are often side by side in sentences, but on SAT subject/verb agreement (sva) questions, various words, phrases, or clauses placed between a subject and its verb make matching the two correctly a bit difficult. Note the sentence which follows. Pickle ball, which is now very popular with many people, were almost unheard of until recently. When this type of sentence is followed by answer choices with various sva combinations, mostly – if not all -- in the same tense, you can be fairly certain that the question is testing whether you can match the subject and verb correctly. For such a question, draw a line completely through the nec, or place parentheses around it, as follows: Pickle ball, which is now very popular with many students, were mostly unheard of just a few years ago. That step will (1) help you ignore the nec
and (2) make it easier for you to select the correct verb for the subject. Hence, you will see that the subject (“ball” or “pickle ball”) is singular and needs the singular verb -- “was.”

The Essential Clause Beginning with “That.” In some SAT questions, a clause beginning with
“that,” separates the subject and verb, as follows: The problems that we had with the computer has
been solved. The job of that essential clause (that we had with the computer) is simply to modify the
noun that it follows – problems. That noun, by the way, is the subject of the sentence. (Note that
neither the subject nor its verb will ever be within the clause that modifies the subject.) Again,
placing parentheses around the clause or drawing a line through it will help keep you from being
distracted by it as you look for the subject and verb in order to be sure that they agree, as in the
following sentence: The problems that we had with the computer have been solved. Note that you
must treat essential clauses beginning with “which” or “who” the same way you treat essential
clauses beginning with “that.”

The Gerund. You may recall that a gerund is a word that looks like an “ing” verb but functions as a
noun. Note that gerunds are always singular. Therefore, a sentence might say, “Mom says that
baking cookies is [not are] a very relaxing way to spend the afternoon.” Do not become confused
about whether to use a singular or plural verb when a gerund is followed by a plural noun. A gerund is
always singular and must always have a singular verb. Therefore, the sentence must say that baking
cookies is Marla’s favorite Christmas tradition.

Keeping in mind the information above, complete the exercise below.

SUBJECT-VERB AGREEMENT

Directions.  Replace the underlined word(s) in each sentence below with the answer that corrects the error in the sentence. If there is no error, pick choice A – NO CHANGE.

1.  Football, which results in injuries to many players, are watched by millions of viewers every day.

  1. NO CHANGE
  2. is watched
  3. were watched
  4. were not watched

2. Norman says that using chopsticks are a fun way to eat spaghetti.

  1. NO CHANGE
  2. chopsticks were
  3. chopsticks am
  4. chopsticks is

3. The grapes that we shared with our neighbor is rich in vitamin C.

  1. NO CHANGE
  2. was rich
  3. was not rich
  4. are rich

SAT Verbal - Answers to Questions from Fall 2022, Week  8  (November 12, 2022)

  1. B
  2. D
  3. D

SAT Math - Questions from Fall 2022, Week 7  (November 5, 2022)

INEQUALITIES: No Solution; All Real Numbers Solution

As with linear equations, it is possible for inequalities to have no solution.

And it is also possible that the solution to an inequality can include all real numbers; that is, rather than x > 9 (where the solution includes only numbers greater than 9) or x < 19 (where the solution includes only numbers less than 19), the solution includes all real numbers.

The following symbols are used in inequalities:

≠ is not equal to
> is greater than
< is less than
≥ is greater than or equal to
≤ is less than or equal to

An inequality will have no solution if the “solution” states something false or gives a contradiction. For example, if we work through an inequality and arrive either at the following as our solution, there is no solution:

                                                        4 < -4
                                                        6 < x < -6

Statements similar to these are false and each inequality has no solution.

We may have other inequalities where all real numbers are solutions to an inequality. This is the case when the inequality solution is true for any value of the variable, whether it is 0, less than 0, or greater than 0. For example, if we work through an inequality and arrive at something similar to either of the following as our solution, the inequality is true for all real numbers:>
                                                       -3x + 7 ≥ 7 -3x
                                                       -1 ≤ x2 -1

Now examine examples 1, 2, and 3.
  1. What values of x satisfy 5x + 2 < 6x < 5x - 2
    Subtract 5x from each section: 5x + 6x < 5x – 2      becomes      2 < x < - 2

    We cannot pick a value of x that is greater than 2 and less than -2. Thus this inequality has
    no solution.

  2. What values of x satisfy 7x -11x + 3 ≥ 3 -4x
    Combine like terms: 7x -11x + 3 ≥ 3 -4x      becomes      -4x + 3 ≥ 3 -4x
    Add 4x to both sides: -4x + 3 ≥ 3 -4x      becomes      3 ≥ 3

    This statement is true for all real numbers.

  3. What values of x satisfy 2(3x – 4) < 4 + 6x - 15
    Remove parentheses: 2(3x – 4) < 4 + 6x – 15 becomes 6x – 8 < 4 + 6x - 15
    Combine like terms: 6x – 8 < 4 + 6x - 15 becomes 6x – 8 < 6x - 11
    Subtract 6x from both sides: 6x – 8 < 6x - 11 becomes – 8 < - 11

    This statement is false because – 8 is not less than –11. Therefore the inequality has no
    solution.

Keeping in mind the information above, solve the following problems. (In these problems, there are three possible answers: a solution, no solution, or all real numbers satisfy the inequality.)

  1. What values of x satisfy 9 + 2x – 5x ≥ –x + 12 – 2x ?

  2. What values of x satisfy 10(x -2) < 6x?

  3. What values of c satisfy 2 – 3c ≥ 6c – 3 – 9c?

  4. What values of m satisfy 9m + 5 – 12m ≥ 7 + 3m +10?

  5. What values of y satisfy 9 + 6(y + 1) ≥ 18 + 3(3y – 1) – 3y?

  6. What values of x satisfy 5(x – 1) +7 ≤ 2(x – 4) +3x + 1?

SAT Math - Answers from Fall 2022, Week 7  (November 5, 2022)

  1. What values of x satisfy 9 + 2x – 5x ≥ –x + 12 – 2x?
    Combine like terms: 9 + 2x – 5x ≥ –x + 12 – 2x becomes 9 – 3x ≥ 12 – 3x
    Add 3x to both sides: 9 – 3x ≥ 12 – 3x becomes 9 ≥ 12

    This statement is false because 9 is not greater than or equal to 12. Therefore the inequality
    has no solution.

  2. What values of x satisfy 10(x -2) < 6x?
    Remove parentheses: 10(x -2) < 6x becomes 10x - 20 < 6x
    Subtract 6x from both sides and add 20 to both sides: 10x -20 < 6x becomes 4x < 20
    Solve for x: x < 5

    This is a true statement. Here x could be any value less than 5, not including 5. We have a
    solution.

  3. What values of c satisfy 2 – 3c ≥ 6c – 3 – 9c?
    Combine like terms: 2 – 3c ≥ 6c – 3 – 9c becomes 2 – 3c ≥ – 3 – 3c
    Add 3c to both sides: 2 – 3c ≥ – 3 – 3c becomes 2 ≥ – 3

    This statement is true for all real numbers of c. Here c could be any real number to solve the
    inequality.

  4. What values of m satisfy 9m + 5 – 12m ≥ 7 + 3m +10?
    Combine like terms: 9m + 5 – 12m ≥ 7 + 3m +10 becomes 5 – 3m ≥ 17 + 3m

    Subtract 17 from both sides and add 3m to both sides: 5 – 3m ≥ 17 + 3m becomes – 12 ≥ 6m
    Solve for m: – 2 ≥ m

    This is a true statement. Here m could be any value less than or equal to 2, including 2. We
    have a solution.

  5. What values of y satisfy 9 + 6(y + 1) ≥ 18 + 3(3y – 1) – 3y?
    Remove parentheses: 9 + 6(y + 1) ≥ 18 + 3(3y – 1) – 3y      becomes      9 + 6y +6 ≥ 18 + 9y –3 – 3y
    Combine like terms: 9 + 6y +6 ≥ 18 + 9y –3 – 3y      becomes      15 + 6y ≥ 15 + 6y
    Subtract 6y from both sides: 15 + 6y ≥ 15 + 6y      becomes      15 ≥ 15

    This statement is true for all real numbers of y. Here y could be any real number to solve the
    inequality.

  6. What values of x satisfy 5(x – 1) +7 ≤ 2(x – 4) +3x + 1?
    Remove parentheses: 5(x – 1) +7 ≤ 2(x – 4) +3x + 1      becomes      5x – 5 +7 ≤ 2x – 8+3x +1
    Combine like terms: 5x – 5 +7 ≤ 2x – 8+3x +1      becomes      5x +2 ≤ 5x –7
    Subtract 5x from both sides: 5x +2 ≤ 5x –7      becomes      2 ≤ –7

    This statement is false because 2 is not less than or equal to -7. Therefore the inequality
    has no solution.

SAT Verbal - Questions from Fall 2022, Week 7  (November 5, 2022)

TIPS FOR USING TRANSITIONS CORRECTLY 
TRANSITION TYPE, JOB IT DOES, AND EXAMPLES

SAMPLE SENTENCE
AND TRANSITION SELECTION TIPS

Type: Reason, Cause/Effect 
Job: Shows why something did or did not
        happen.
Examples: consequently, as a result, hence,
                  since, therefore, accordingly

Mom was up all night with a sick baby; consequently, she overslept and was late for work.

The first part of the sentence gives the reason for what happened in the second part. Therefore, a “reason” transition is needed, and “consequently” is correct.

Type: Addition/Example
Job: Points out additional information that
        explains further a point already made.
Examples: moreover, in fact, for example,
                  furthermore, in addition,
                  likewise, similarly, additionally
We like Mama Leona’s pizza because the ingredients are always fresh; moreover, the crust is always crispy.

The second part of the sentence provides added information that explains further the point made in the first part of the sentence. Therefore, we need an “addition” transition, and “moreover” is correct.

Mom was up all night with a sick baby; consequently, she overslept and was late for work. 
Type: Contrast
Job: Indicates contrasting/conflicting ideas.
Examples: although, even so, however,                              despite, nevertheless, otherwise,                       on the other hand, alternatively
Cindy gets too much homework to bake treats on weeknights; on the other hand, she bakes lots of delicious goodies on Saturdays.

The second part of the sentence is in contrast to what was said in the first part. Therefore, a “contrast” transition is needed, and “on the other hand” is correct.


Now, keeping in mind the information above, complete Exercise K22 below.

Using Transitions Correctly

Directions.  Replace the underlined word(s) in each sentence below with the answer that corrects the error in the sentence. If there is no error, pick choice A – NO CHANGE.

1.  Rick was in a very minor auto accident last night; on the other hand, getting his car repaired should not be extremely expensive. 

  1. NO CHANGE
  2. additionally
  3. consequently
  4. nevertheless

2. Hawaii is a beautiful, warm place to spend your vacation; consequently, Antarctica’s icy temperatures make that place a poor choice for vacation fun.  

  1. NO CHANGE
  2. however
  3. furthermore
  4. likewise

3. Nutritious, well-balanced meals every day will help keep you healthy; moreover, appropriate daily exercise will also help you stay in good physical condition.

  1. NO CHANGE
  2. similarly
  3. nevertheless
  4. consequently

SAT Verbal - Answers from Fall 2022, Week 7  (November 5, 2022)

  1. C
  2. B
  3. A

SAT Math - Questions from Fall 2022, Week 6  (October 22, 2022)

Inequalities

We have reviewed linear equations, systems of equations, quadratic equations, and systems of equations with linear and quadratic equations. Another type of similar problem that appears on the SAT is an inequality.

In an equation, one side of an equation equals the other side. In an inequality, the two sides are not equal.

The following symbols are used in inequalities:

≠ is not equal to
> is greater than
< is less than
≥ is greater than or equal to
≤ is less than or equal to

You can usually work with inequalities in exactly the same way you work with equations.

  1. You can collect similar terms, and you can simplify by doing the same thing to both sides: adding, subtracting, multiplying, dividing, raising to a power, or taking a root.
  2. An important caution: multiplying both sides of an inequality by a negative number
    reverses the direction of the inequality.

Examine examples 1, 2, and 3.

  1. x > y
    1. Add 2 to both sides
    2. Multiply both sides by 10
    3. Multiply both sides by -2

  2. If 2x < 3 and 3x > 4, what is one possible value of x?

    The best approach here is to solve for x and then express each fraction as a decimal.

    2x < 3                                           3x > 4
    x < 3/2                                            x > 4/3
    x < 1.5                                            x > 1.33333

    Thus x is between 1.33333 and 1.5, not including 1.33333 and 1.5. Any value in this range would be acceptable.

  3. What values of x satisfy 7 + 2x – 5x ≥ – x + 13 – 4x ?

    Solve for x by adding, subtracting, multiplying, dividing, raising to a power, or taking a root.

    -3x + 7 ≥ –5x + 13
    2x ≥ 6
    x ≥ 3

    Thus x is any value greater than or equal to 3, including 3.

Keeping in mind the information above, solve the following problems.

  1. 6x – 9y > 12
    Which of the following inequalities is equivalent to the inequality above?

    1. x – y > 2
    2. 2x – 3y > 4
    3. 3x – 2y > 4
    4. 3y – 2x > 2
  2. If -4 < x < -2, which of the following could be the value of 3x?

    1. -2.5
    2. -3.5
    3. -4.5
    4. -7.5
    5. -14.5

  3. What values of x satisfy 7x + 3 – 10x ≥ 4 + 2x + 14?

  4. If x + 6 > 0 and 1 – 2x > - 1, then x could equal each of the following EXCEPT

    1. -6
    2. -4
    3. 0
    4. 1/2
  5. If  -6 < -4x + 10 ≤ 2, what is the least possible value of 4x + 3?

    1. 2
    2. 5
    3. 8
    4. 11

  6. If -8 < -(3/5)r + 1 ≤ -(16/5) , what is one possible value of r?

    1. 3
    2. 5
    3. 8
    4. 16

SAT Math - Answers from Fall 2022, Week 6  (October 22, 2022)

  1. 6x – 9y > 12
    Which of the following inequalities is equivalent to the inequality above?

      1. x – y > 2
      2. 2x – 3y > 4
      3. 3x – 2y > 4
      4. 3y – 2x > 2

    Divide each term by 3: 6x - 9x > 2 becomes 2x - 3y > 4.
    The answer is B.
  2. If -4 < x < -2, which of the following could be the value of 3x?

    1. -2.5
    2. -3.5
    3. -4.5
    4. -7.5
    5. -14.5

    Multiply each term by 3: -4 < x < -2 becomes -12 < 3x < -6 It could be -7.5.
    The answer is D.
  3. What values of x satisfy 7x + 3 – 10x ≥ 4 + 2x + 14?

    Solve for x: 7x + 3 – 10x ≥ 4 + 2x + 14
                                 3 – 3x ≥ 2x + 18
                                    – 5x ≥ 15
                                      – x ≥ 3
                                         x ≤ -3

  4. If x + 6 > 0 and 1 – 2x > - 1, then x could equal each of the following EXCEPT

      1. -6
      2. -4
      3. 0
      4. ½
    Solve each inequality and determine the range of values for x.

    x + 6 > 0                 1 – 2x > - 1
    x > -6                         – 2x > - 2
                                           x < 1

    Here x could be any value between -6 and 1, not including -6 and 1. Thus x could be -4, 0, or ½. It could not be -6. The answer is A.
  5. If  -6 < -4x + 10 ≤ 2, what is the least possible value of 4x + 3?

    1. 2
    2. 5
    3. 8
    4. 11

    What do we do here? We want to add, subtract, multiply, or divide so that -4x + 10 becomes 4x + 3.

    The first step is to change the -4x to 4x. We can make this change by multiplying each section by -1: -6 < -4x + 10 ≤ 2 becomes 6 > 4x - 10 ≥ - 2

    Next we need to change the -10 to +3. We can make this change by adding 13 to each section:

    6 > 4x - 10 ≥ - 2 becomes 19 > 4x +3 ≥ 11

    Thus 4x + 3 is between 11 and 19, not including 19. The lowest value is 11.
    The answer is D.

  6. If -8 < -(3/5)r + 1 ≤ -(16/5) , what is one possible value of r?

    1. 3
    2. 5
    3. 8
    4. 16

    We will solve for r. We will begin by multiplying each section by -1:
    -8 < -(3/5)r + 1 ≤ -(16/5) becomes 8 > (3/5)r - 1 ≥ (16/5)

    Now we multiply each section by 5 (to get rid of the 5 in the denominator).
    8 > (3/5)r - 1 ≥ (16/5) becomes 40 > 3r - 5 ≥ 16

    Now we add 5 to each section.
    40 > 3r - 5 ≥ 16 becomes 45 > 3r ≥ 21

    Finally we divide each section by 3. 45 > 3r ≥ 21 becomes 15 > r ≥ 7
    Thus r is any value between 7 and 15, not including 15.

SAT Verbal - Questions from Fall 2022, Week 6  (October 22, 2022)

Recognizing the Participial Phrase. You may recall that a participial phrase describes a noun or pronoun and must be placed as close as possible to the word it modifies. That kind of phrase 1) begins with a participle and (2) includes any objects, modifiers, or other words needed to complete the main idea expressed in the sentence. Present participles, which are fairly easy to identify because they end in “ing,” include words such as “playing,” “working,” and “eating.” Past participles, made from “regular” verbs, are also fairly easy to identify because they end in “ed,” as in “walked,” “helped,” and “wanted.” Although regular present and past participles generally appear more frequently on the SAT than do “irregular” ones, you should note that irregular past participle endings you might see on the SAT, along with sample words with those endings, include the following: en/taken, d/saved, t/dealt, n/seen, and ne/gone.

Punctuating the Participial Phrase. Often, the participial phrase is separated from the main idea of a sentence by commas, but not always. Note the underlined participial phrase in Sentence A, which follows:

Sentence A. Mrs. Lawson will be surprised to find out that her costumed server is her son Mike, who has been overseas in the army for 18 months. The participial phrase (costumed server) is not separated from the main idea by commas because the phrase is actually a part of the main idea. Therefore, it must not be separated from the rest of the main idea. Now, note Sentence B, which follows:

Sentence B. Thrilled about her new driver’s license, Melissa volunteered to do errands for everyone in the family. Note that a comma follows the participial phrase because when a participial phrase begins a sentence but is not a part of the main idea, the phrase must be separated from the main idea by a comma.

Now, note Sentence C, which follows.
Sentence C. The frisky little puppy, excited about seeing the children come home, barked and jumped gleefully as they headed for the yard. In this situation, the participial phrase is within the sentence, but it is between two commas. The reason is that when a participial phrase is in that position, but is not part of the main idea, commas must separate the participial phrase from the rest of the sentence.

Keeping in mind the information above, identify the letter of the answer choice which punctuates each sentence below correctly. If you think that the sentence is already punctuated correctly, select choice A – NO CHANGE. Then use the answer key to check your answers.

  1. Please give raking the leaves this message to the gardener.
    1. NO CHANGE
    2. After the word “gardener”
    3. After the word “please”
    4. After the word “this”

  2. Several of the boxes are for the new recreation center unloaded this afternoon.
    1. NO CHANGE
    2. After the word “several”
    3. After the word “are”
    4. After the word “boxes”

  3. The bus driver will take this jacket, left on the bus this morning, to the principal’s office.
    1. NO CHANGE
    2. Eliminate the commas
    3. After the word “driver”
    4. After the word “office"

SAT Verbal - Answers from Fall 2022, Week 6  (October 22, 2022)

  1. B
  2. D
  3. A

SAT Math - Questions from Fall 2022, Week 5  (October 15, 2022)

Quadratic Equations: System of Equations: Linear and Quadratic

One of the types of questions you will see on the SAT is solving systems of equations where one of the equations is linear and the other one is quadratic. In an earlier lesson, we pointed out that a system of linear equations (both equations are linear) can be solved in three ways: by substitution, by elimination, and by using the answer choices (when answer choices are given). When the alternative of using the answer choices is used, we substitute an answer choice in each of the two equations and determine which one works. The correct answer choice must work in both equations. When substitution is used, we substitute one equation in the other one and solve for the remaining variable. When we have systems of equations where one of the equations is linear and the other one is quadratic, we can solve by substitution or using the answer choices.

Consider the following example:

y = x2 - 4x + 4
           y = 4 - x

If the ordered pair (x, y) satisfies the system of equations above, what is one possible value of x?
In this example we would substitute the second equation in the first one, put the resulting quadratic equation in standard form, and solve the equation.
          4 – x = x2 - 4x + 4
               0 = x2 - 3x
               0 = x(x – 3)
               x = 0                                         x – 3 = 0
                                                                      x = 3

The value of x can be 0 or 3. We can check our work
If x = 0, y = 4 – x = 4 – 0 = 4 and                 y = x2 - 4x + 4 = 02 -4(0) + 4 = 4
If x = 3, y = 4 – x = 4 – 3 = 1 and                 y = x2 - 4x + 4 = 32 -4(3) + 4 = 9 – 12 + 4 = 9 – 12 + 4 = 1

Here is another example:

In the xy-plane, which of the following is a point of intersection between the graphs of y = x + 2 and y = x2 + x – 2?

  1. (0, -2)
  2. (0, 2)
  3. (1, 0)
  4. (2, 4)


In this problem, we have a point where the two lines cross. When we find the value of x, we would substitute it in both equations and we should get the value of y in both equations. For example, if A is the correct answer, substituting the value of x, 0, in each equation should give us -2 for y in each equation. We can solve the problem by substituting the second equation in the first equation and solving the quadratic equation, or we can solve by using the answer choices. Let’s use the answer choices. We test first by using the first equation, y = x + 2, since it is easier to work with.

  1. x = 0, y = -2: y = x + 2 = 0 + 2 = 2 This does not work since y = 2, not -2. Answer choice A is not correct.

  2. x = 0, y = 2: y = x + 2 = 0 + 2 = 2 This does work since y = 2. But now we must determine whether it works in the second equation, y = x2 + x – 2. We are looking for y = 2 when x = 0.
    y = x2 + x – 2 = 02 + 0 -2 = -2. This does not work since y = -2, not 2. Answer choice B is not correct.

  3. x = 1, y = 0: y = x + 2 = 1 + 2 = 3 This does not work since y = 3, not 0. Answer choice C is not correct.
    Thus our answer must be D. We can show it as follows:

  4. x = 2, y = 4: y = x + 2 = 2 + 2 = 4 This does work since y = 4. But now we must determine whether it works in the second equation, y = x2 + x – 2. We are looking for y = 4 when x = 2.
    y = x2 + x – 2 = 22 + 2 -2 = 4 +2 -2 = 4 This does work since y = 4. Answer choice D is correct.

Now let’s solve by substituting the second equation in the first equation and solving the quadratic equation.
y = x + 2           and           y = x2 + x – 2
x + 2 = x2 + x – 2
0 = x2 – 4 = (x – 2)(x + 2)
x – 2 = 0                                               x + 2 = 0
       x = 2                                              x = -2
Using the second equation, y = x + 2, we get y = x + 2 = 2 + 2 = 4. Thus choice D is correct.

What about the other solution to the quadratic equation, x = -2?
When x = -2, y = x + 2 = -2 + 2 = 0. This is a point (-2,0) on the graph of the quadratic, but it is not the point where the two lines cross. And note that this point was not one of the answer choices.

Keeping in mind the information above, solve the following problems.

  1. y = x2 - 6x - 9
    y = -9 - x

    If the ordered pair (x,y) satisfies the system of equations above, what is one possible value of x?

  2. y = x2 + 3
    y = 15x – 33

    The system of equations shown above is grafted into the (x,y) plane. If the system has two solutions, what is the product of the x-coordinates of the two solutions?
    1. 36
    2. 4
    3. -4
    4. -36

  3. y = x2
    2y + 6 = 2(x + 3)

    If (x,y) is a solution of the system of equations above and x > 0, what is the value of xy?
    1. 1
    2. 2
    3. 3
    4. 9
  4. x = 2y + 5
    y = (2x – 3)(x + 9)

    How many ordered pairs (x,y) satisfy the system of equations shown above?

    1. 0
    2. 1
    3. 2
    4. infinitely many

  5. y = x- 12x + 12
    y = 12 - x

    If the ordered pair (x,y) satisfies the system of equations above, what is one possible value of x?

  6. x+ y  = 16
    x - y = 4

    Which value is a y-coordinate of a solution to the system of equations above?

    1. -9
    2. -8
    3. -5
    4. -4

SAT Math - Answers from Fall 2022, Week 5  (October 15, 2022)

    1. y = x2 - 6x - 9
      y = -9 - x

      If the ordered pair (x,y) satisfies the system of equations above, what is one possible value of x?

      -9 -x = x2 - 6x - 9
            0 = x2 - 5x = x(x – 5)
      x = 0                                                      x – 5 = 0
                                                                          x = 5
      The possible values of x are 0 and 5.

    2. y = x2 + 3
      y = 15x – 33

      The system of equations shown above is grafted into the (x,y) plane. If the system has two solutions, what is the product of the x-coordinates of the two solutions?
      1. 36
      2. 4
      3. -4
      4. -36

      15x - 33 = x2 + 3
                  0 = x2 – 15x + 36 = (x – 12)(x – 3)

      x – 12 = 0                                                            x – 3 = 0
              x = 12                                                                x = 3

      The product of the x coordinates = 12(3) = 36 The answer is A

      There is another way to get the product when the quadratic equation is expressed in standard form:
      ax2 + bx + c = 0

      Recall from a previous lesson that when the quadratic equation is expressed in standard form the sum of the x coordinates = -b/a and the product of the x coordinates = c/a.

      For this problem the product = 36/1 = 36

    3. y = x2
      2y + 6 = 2(x + 3)

      If (x,y) is a solution of the system of equations above and x > 0, what is the value of xy?
      1. 1
      2. 2
      3. 3
      4. 9

      2x2 + 6 = 2(x+3) = 2x + 6
      2x2 + 6 = 2(x+3) = 2x + 6
                0 = -2x2 + 2x = 2x(-x + 1)

             2x = 0                                -x +1 = 0
              x = 0                                      -x = -1
                                                              x = 1

      Now solve for y: y = x2 = 12 = 1              Thus xy = 1(1) = 1. The answer is A.

    4. x = 2y + 5
      y = (2x – 3)(x + 9)

      How many ordered pairs (x,y) satisfy the system of equations shown above?

        1. 0
        2. 1
        3. 2
        4. infinitely many

        Simplify the second equation:
        y = (2x – 3)(x + 9) = 2x2 + 15x – 27

        Substitute the second equation into the first equation:
        x = 2y + 5 = 2(2x2 + 15x – 27) + 5 = 4x2 + 30x – 54 + 5 = 4x2 + 30x – 49

        This is the quadratic equation:
        x = 4x2 + 30x – 49

        State the quadratic equation in standard form:
        0 = 4x2 + 29x – 49

        In an earlier lesson we learned that the discriminant will tell us whether we have one solution, two
        solutions, or no real solution. We have one solution if the discriminant = 0; we have two solutions if the discriminant is positive; and we have no real solution if the discriminant is negative.

        The discriminant = b2 – 4ac where a, b, and c are the constants of the quadratic equation expressed in
        standard form: ax2 + bx + c = 0
        In this case we have b2 – 4ac = 292 – 4(4)(49) = 841 - 784 = 57. The discriminant is positive;
        thus there are two solutions. The answer is C.

        1. y = x- 12x + 12
          y = 12 - x

          If the ordered pair (x,y) satisfies the system of equations above, what is one possible value of x?

                       12 -x = x2 - 12x + 12
                             0 = x2 - 11x = x(x – 11)
                          x = 0                                               x – 11 = 0
                                                                                          x = 11
          The possible values of x are 0 and 11.

        2. x+ y  = 16
          x - y = 4

          Which value is a y-coordinate of a solution to the system of equations above?

          1. -9
          2. -8
          3. -5
          4. -4

                          Solve the second equation for y and substitute it into the first equation.
                          x - y = 4
                          x - 4 = y
                          x2 + y = 16
                         x2 + x - 4 = 16
                         x2 + x – 20 = 0
                         (x + 5)(x – 4) = 0
                         x + 5 = 0                                                            x – 4 = 0
                         x = -5                                                                       x = 4

                        We have found the two solutions, -5 and 4, and one of them is one of the answer choices, C. A                          student who is not very careful will choose C, but C is a trick answer. The question asks for the y-                    coordinate of a solution, not the solution. We have to find the value of y based on our values of x.                    Using the second equation, y = x – 4, we will obtain two values of y:

                        Using x = -5: y = x – 4 = -5 -4 = -9
                        Using x = 4: y = x – 4 = 4 – 4 = 0
             
                        Thus our answer is A since -9 is one of the answer choices.

      SAT Verbal - Questions from Fall 2022, Week 5  (October 15, 2022)

      The Participle. A participle looks like a verb, but functions as an adjective by describing a noun or pronoun. Note Examples A and B, which follow.

      Example A. Liz wrecked her car in a bad accident.
      Discussion A. In this sentence, “wrecked” is a verb because it names the action done by Liz.

      Example B. Liz’s car, wrecked when she slammed it into a huge truck, will be towed away.
      Discussion B. In this sentence, “wrecked” is a participle because it functions as an adjective by describing the car.

      The Participial Phrase. A participial phrase describes a noun or pronoun in a sentence. This kind of phrase (1) begins with a participle and (2) includes any objects, modifiers, or other words needed to complete the thought being expressed in the sentence. Note Examples C and D, which follow.

      Example C. The water in the bathtub drained slowly because hair was clogging the drain.
      Discussion C. Example C is fairly straightforward and short because its point is made clearly without the need for a participial phrase or any other additional information.

      Example D. The water in the bathtub drained slowly because hair falling into the sink when Bella would bathe her dog was clogging the drain.
      Discussion D: Unlike Example C, Example D contains a participial phrase, which begins with the participle “falling” and ends with the noun “dog.” Note that the entire participial phrase is needed to complete the thought that the sentence expresses. The participial phrase itself modifies the noun “hair.”

      Placement of the Participle. When possible, place the participle next to the word it modifies, as in “The eggs boiled this morning had to be replaced because they burned.” Note that the participle “boiled” is beside the word it modifies – eggs. When placing a modifier next to the word it modifies is not possible, place it as close as possible to that word. For instance, note Example E, which follows.

      Example E. The huge picture window in our living room, broken when a baseball smashed through it, will be very expensive to replace. Discussion E. In this sentence, the entire participial phrase, broken when a baseball smashed through it, modifies “picture window,” but so does the adjective phrase “in our living room.” However, the adjective phrase is needed to indicate which picture window the sentence is talking about. Therefore, the adjective phrase must go next to “picture window.” and the participial phrase must come afterwards. Even with that change, the sentence is as concise as possible, and it is written correctly.

      Keeping in mind the information above, complete the Quick Challenge Exercise I22 below.

      Exercise I22 The Participle Phrase

      Directions.  For each sentence below, select the answer choice that places a participle or participial phrase in the sentence correctly. If you think that the sentence already does that job, select choice A.

      1.  Kellie is boiling water so that she can make tea.

      1. NO CHANGE
      2. needs boiling water
      3. was boiling water
      4. needs to boil the water

      2. The window that broke when icy water had been splashed on it on a 105 degree morning will be replaced this afternoon.  

      1. NO CHANGE
      2. that Joe threw icy water on and broke it
      3. that was hit and shattered by Joe’s icy water
      4. broken when icy water hit it

      3. The people who live in the house that a tree fell onto and damaged the roof of must' move out
      until after the roof has been repaired properly.  

      1. NO CHANGE
      2. that a tree fell onto and damaged the roof of
      3. damaged when a tree fell onto its roof
      4. that got damaged when a tree fell onto its roof

      SAT Verbal - Answers from Fall 2022, Week 5  (October 15, 2022)

      1. B
      2. D
      3. C

      SAT Math - Questions from Fall 2022, Week 4  (October 1, 2022)

      Quadratic Equations: Word Problems

      Word problems have become more numerous in recent administrations of the SAT. The majority of word problems involve linear relationships (equations with one variable and systems of equations). But there are a significant number of word problems that require an understanding of quadratic equations. Some word problems will require that you create a quadratic equation from the information given. Sometimes you will be asked to solve the equation, but often you will just be asked to develop the equation. It is essential, then, that students be able to find and translate the information in the problem into an equation. The problem must be read carefully to ensure that the equation is consistent with the information in the problem.

      Let’s look at the following example:

      h = -4.9t + 25t

      The equation above represents the approximate height, h, in feet, of a ball t seconds after it is launched vertically upward from the ground with an initial velocity of 25 feet per second. After approximately how many seconds will the ball hit the ground?

      1. 3.4
      2. 4.0
      3. 4.5
      4. 5.0

      The height is 0 feet when the ball hits the ground. Substituting 0 for h in the equation, we get:

      h = -4.9t2 + 25t
      0 = -4.9t2 + 25t



      Put in standard form, we have     -4.9t2 +  25t  =  0
      The standard form of a quadratic equation is ax2 +  bx  +  c  =  0
      In this problem, the c term is missing; this means that  c = 0.  The c term is the y intercept; it is the value of y when x = 0.  In this case, it is the initial height; the height is 0 when we start; the ball is on the ground.

      Factoring our equation of    -4.9t2 +  25t  =  0, we get   t(-4.9t  +  25)  =  0

      Solving for t, we get: t = 0 -4.9t + 25 = 0
                                                                               -4.9t = -25
                                                                                      t = -25/-4.9
                                                                                      t = 5.1

      Thus the ball hits the ground approximately 5.0 seconds after it is launched.

      Here is another example.

      A car is travelling at x feet per second. The driver sees a red light ahead, and after 2 seconds
      reaction time, the driver applies the brake. After the brake is applied, the car takes x/20 seconds to stop, during which time the average speed of the car is x/3 feet per second. If the car travels 150 feet from the time the driver saw the red light and the time it comes to a complete stop, which of the following equations can be used to find the value of x?                                                      

      1. x2 + 60x – 9,000 = 0
      2. x2 + 120x – 9,000 = 0
      3. x2 + 60x - 6,000 = 0
      4. x2 + 120x - 6,000 = 0

      To begin, we recognize that this is a distance problem, where distance = speed times time. For example, if we are travelling at 60 miles per hour for 2 hours, our distance would be 60 x 2 = 120 miles. In this case, we have two speeds: x feet per second and x/3 feet per second. We travel at a speed of x feet per second for 2 seconds, and we travel at a speed of x/3 feet per seconds for x/20 seconds. The sum of these two gives us our total distance of 150 feet. Thus we have: (x/3)(x/20) + (x)(2) = 150


      To get rid of the 60 in the denominator, we multiply everything by 60: x2 + 120x = 9,000

      Putting it in standard form we get: x2 + 120x - 9,000 = 0. Thus the answer is B.

      Keeping in mind the information above, solve the following word problems.
      1. h = -16t2 +110t + 72
        The equation above models the height h, in feet, of an object above ground t seconds after being launched straight up in the air. What does the number 72 represent?

        1. The initial height, in feet, of the object
        2. The maximum height, in feet, of the object
        3. The initial speed, in feet per second, of the object
        4. The maximum speed, in feet per second, of the object

      2. h = -5.8t + 50t
        The equation above represents the approximate height, h, in feet, of a model rocket t seconds after it is launched vertically upward from the ground with an initial velocity of 50 feet per second. After approximately how many seconds will the rocket hit the ground?

        1. 5.8
        2. 6.5
        3. 7.6
        4. 8.6

      3. A homeowner is designing a rectangular deck in his back yard. The length of the deck is to be 8 feet longer than the with. If the area of the deck will be 240 square feet, what will be the length, in feet, of the deck?

        1. 12
        2. 15
        3. 20
        4. 25

      4. h = 1/((x-5)2 + 4(x – 5) + 4)
        For what value of x is the equation above undefined?

      5. If (ax + 2)(bx + 7) = 15x2 + cx +14 for all values of x, and a + b = 8, what are the two possible
        values for c?

        1. 3 and 5
        2. 6 and 35
        3. 10 and 21
        4. 31 and 41

      6. The product of the ages of Sue and James now is 175 more than the product of their ages 5 years prior. If Sue is 20 years older than James, how old is James now?

        1. 8
        2. 10
        3. 12
        4. 15

      SAT Math - Answers to Questions from Fall 2022, Week 4  (October 1, 2022)

          1. h = -16t2 +110t + 72
            The equation above models the height h, in feet, of an object above ground t seconds after being launched straight up in the air. What does the number 72 represent?

            1. The initial height, in feet, of the object
            2. The maximum height, in feet, of the object
            3. The initial speed, in feet per second, of the object
            4. The maximum speed, in feet per second, of the object

            In the standard form of a quadratic equation, ax2 + bx + c = 0, the c term is the y intercept; it
            is the value of y when x = 0. In this case, it is the initial height; the height is 72 when we start;
            it is 72 feet above ground 0 seconds after being launched. The answer is A.

          2. h = -5.8t + 50t
            The equation above represents the approximate height, h, in feet, of a model rocket t seconds after it is launched vertically upward from the ground with an initial velocity of 50 feet per second. After approximately how many seconds will the rocket hit the ground?

            1. 5.8
            2. 6.5
            3. 7.6
            4. 8.6

            Put in standard form, we have -5.8t2 + 50t = 0
            Factoring our equation t(-5.8t + 50) = 0

            Solving for t, we get:                                                  t = 0                                 -5.8t + 50 = 0
                                                                                                                                                -5.8t = -50
                                                                                                                                                       t = -50/-5.8
                                                                                                                                                       t = 8.62

            Thus the ball hits the ground approximately 8.6 seconds after it is launched.

          3. A homeowner is designing a rectangular deck in his back yard. The length of the deck is to be 8 feet longer than the with. If the area of the deck will be 240 square feet, what will be the length, in feet, of the deck?

            1. 12
            2. 15
            3. 20
            4. 25

            The area = length times width = 240
             Width = w; length = w + 8
            Thus the area = w(w + 8) = 240
                                       w2 + 8w = 240
                                       w2 + 8w – 240 = 0      This is a quadratic equation in standard form. We will solve
                                                                           by factoring.
                                      (w – 12)(w + 20) = 0
                                      w – 12 = 0                                   w + 20 = 0
                                              w = 12                                         w = -20

            We can discard the value w = -20 since the width cannot be negative; the width is 12.
            Therefore, the length = 12 + 8 = 20. The answer is C.

          4.                            h = 1/((x-5)2 + 4(x – 5) + 4)
            For what value of x is the equation above undefined?

            A fraction is undefined if the denominator is 0. We find the value of x that makes the
            denominator = 0.
            (x-5)2 + 4(x – 5) + 4 = 0
            x2 – 10x + 25 + 4x – 20 + 4 = 0
            x2 – 6x + 9 = 0
            (x – 3)(x – 3) = 0
            x – 3 = 0                                                                                          x – 3 = 0
                  x = 3                                                                                                x = 3                                        The answer is 3.

          5. If (ax + 2)(bx + 7) = 15x2 + cx +14 for all values of x, and a + b = 8, what are the two possible
            values for c?

            1. 3 and 5
            2. 6 and 35
            3. 10 and 21
            4. 31 and 41

            (ax + 2)(bx + 7) = 15x2 + cx +14

            Using F O I L, expand (ax + 2)(bx + 7)
            (ax + 2)(bx + 7) = abx2 + 7ax + 2bx + 14

            Combine similar terms
            abx2 + 7ax + 2bx + 14 = abx2 + (7a + 2b)x + 14

            Now we have:
            abx2 + (7a + 2b)x + 14 = 15x2 + cx +14

            Next we can equate the coefficients :
            abx2 = 15x2; therefore ab = 15

            (7a + 2b)x = cx; therefore c = 7a + 2b

            We were initially given that a + b = 8, and now we know that ab = 15. Next we must determine what are the values of a & b, which when added together = 8 and when multiplied together = 15.
            They are 3 and 5: a = 3, b = 5 and a = 5, b = 3

            Since c = 7a + 2b, there are two possible values for c:
                         c = 7a + 2b = 7(3) + 2(5) = 21 + 10 = 31
                         c = 7a + 2b = 7(5) + 2(3) = 35 + 6 = 41

            Thus the two possible values for c are 31 and 41. The answer is D.

          6. The product of the ages of Sue and James now is 175 more than the product of their ages 5 years prior. If Sue is 20 years older than James, how old is James now?

            1. 8
            2. 10
            3. 12
            4. 15

      Be sure that you incorporate each piece of information.

      Choose a variable for each age now.
      Specify “their ages 5 years prior”.
      Incorporate properly the “175 more” and the “20 years older.”

      The age of Sue is s; the age of James is j. What were their ages 5 years prior?

      The age of Sue 5 years prior was s - 5; the age of James 5 years prior was j – 5.

      The product = sj = (s-5)(j-5) + 175.

      Sue is 20 years older than James. We can write this as s = j + 20 or j = s – 20. Since we are

      looking for j, we will use s = j + 20 and substitute for s so that j will be left in the equation.

      (j + 20)j = (j + 20 – 5)(j – 5) + 175 = (j + 15)(j-5) + 175 = j2 + 10j -75 + 175= j2 + 10j + 100

      j2 + 20j = j2 + 10j + 100
       
           10j = 100
               j = 10                             The answer is B

      Note that we had a quadratic equation in this problem, but we did not have to solve the quadratic equation since the j2 terms cancelled out.

      SAT Verbal - Questions from Fall 2022, Week 4 (October 1, 2022)

      SAT QUICK CHALLENGE EXERCISE H22
      Conciseness B

      Conciseness is the ability to write clearly without repeating ideas unnecessarily. For instance, if two words mean the same thing, you must not use them both in the same sentence because the unnecessary repetition destroys conciseness and makes the sentence redundant. Some SAT questions test your ability to write concisely by assessing how well you handle (1) synonymous adverb errors and (2) implied description errors. Both types are explained below.

      Implied Description Errors. An implied description is a word that does two grammatical jobs: (1) it names something, and (2) it gives a "definition-related" description of the thing it names, as in the following sentence: "Carla needs the tan four-sided square." This sentence tells us that Carla needs a square and that it has four sides. However, the word "square" itself implies a figure with four equal sides. Therefore, mentioning the four sides makes the sentence redundant.

      Since the color of the square does not help define the figure, mentioning the color is fine. Moreover, if the color had been needed in order to identify which square Carla needs, mentioning the color would have been necessary, and the sentence would have needed to say "Carla needs the tan square."

      Passive Verb Errors. Both active and passive verbs tell what happened (is happening, etc.). However, an active verb follows the word that tells who did the action, as in the following: “Angela planted the vegetables in Grandma’s garden.” However, a passive verb comes before the word that tells who did the action, as in the following: “The vegetables in Grandma’s garden were planted by Angela.” Note that the sentence with the active verb is shorter than the sentence with the passive verb. Therefore, sentences with active verbs are more concise than their counterparts with passive verbs.

      Keeping in mind the information above, complete SAT Quick Challenge Exercise H22 below.

      Exercise H22 Conciseness B

      Directions.  On the line that follows each statement below, place the letter of the answer choice which corrects the underlined part of that statement. If you believe that the underlined part is already correct, select Choice A -- NO CHANGE. After completing this activity, use the answer key to check your work.

      1.  At Monroe Recreation Center on Saturday afternoons, children’s recreational activities are being provided by college students.

      1. NO CHANGE
      2. a variety of children’s recreational activities are being provided by college students
      3. college students are providing a variety of children's recreational activities
      4. children’s recreational activities of which college students led are being provided

      2. That stop sign which has eight sides means that you must stop and be sure that traffic is clear before you continue your trip.

      1. NO CHANGE
      2. Delete the underlined
      3. which is eight sided
      4. which has a total of eight sides

      3. The vase broke and shattered to pieces when it fell from the table.

      1. NO CHANGE
      2. broke and shattered
      3. Delete the underlined
      4. broke

      SAT Verbal - Answers to Questions from Fall 2022, Week 4  (October 1, 2022)

      1. C
      2. B
      3. D

      SAT Math - Questions from Fall 2022, Week 3  (September 24, 2022)

      Quadratic Equations: Completing the Square

      In previous lessons we have reviewed two methods for solving quadratic equations, factoring and using the quadratic formula. In this lesson we will review the third method, completing the square.
      We begin by noting three relationships that must be recognized by SAT test takers:

      1. Square of a binomial into a trinomial: (x + y)2 = x2 + 2xy + y2
        (x + 3)2 = x2 + 2x(3) + 32 = x2 + 6x + 9
        (x + 7)2 = x2 + 2x(7) + 72 = x2 + 14x + 49

      2. Square of a binomial into a trinomial: (x - y)2 = x2 - 2xy + y2
        (x - 3)2 = x2 - 2x(3) + 32 = x2 - 6x + 9
        (x - 7)2 = x2 - 2x(7) + 72 = x2 - 14x + 49

      3. Difference o two squares: x2 - y2 = (x + y)(x - y)
        x2 - 25 = (x + 5)(x - 5)
        x2 - 64 = (x + 8)(x - 8)

      These should also be recognized in reverse.
      1. A trinomial into the square of a binomial:  x2 + 2xy + y2 = (x + y)2
               x2 + 6x + 9 = (x + 3)2
               x2 + 14x + 49 = (x + 7)2

      2.  A trinomial into the square of a binomial:  x2 - 2xy + y= (x - y)2
                x2 - 6x + 9 = (x - 3)2 
                x2 - 14x + 49 = (x - 7)2

      3. Difference of two squares: (x + y)(x - y) = ( x2 - y2)
                                                   (x + 5)(x - 5) =  x2 - 25
                                                   (x + 8)(x - 8) = x2 - 64

        Expressing a trinomial as a square of a binomial (or factoring the trinomial) is the key to solving quadratic equations by completing the square. There is a six step process in using this method.

        Step 1. Begin with the equation in standard form: ax2 + bx + c = 0
        Step 2. Divide each term in the equation by the coefficient of the x2 term.
        Step 3. Move the constant term (the term without a variable) to the opposite side of the equation
                    (by adding or subtracting)
        Step 4. Take half of the coefficient of the x term, square it, and add the squared term to both sides of
                     the equation. This step will create a perfect square trinomial.
        Step 5. Factor the side of the equation that is a perfect square trinomial (we just created it) into the
                    square of a binomial, which will be in the form of (x + h)2 or (x – h)2, where h is half of the
                    coefficient of the x term.
        Step 6. Find the square root of each side of the equation and solve for x.


        Let’s look at an example where the quadratic equation is solved by completing the square.
        Solve for x:   2x2 + 6x  =  8

        Step 1.  Begin with the equation in standard form:  ax2 + bx + c = 0             
                     2x2 + 6x - 8  =  0

        Step 2.  Divide each term in the equation by the coefficient of the x2 term.                                                                
                     x2 + 3x - 4  =  0

        Step 3.  Move the constant term (the term without a variable) to the opposite side of the equation             
                      (by adding or subtracting).  In this case we add 4 to both sides.                                                                                                      x2 + 3x  =  4

        Step 4.  Take half of the coefficient of the x term, square it, and add the squared term to both sides of              
                      the equation.  This step will create a perfect square trinomial.                                                                                                        x2 + 3x + 9/4  =   4 + 9/4                 
                      (the coefficient of the x term is 3; half of 3 is 3/2; when we square 3/2 we get 9/4; 9/4 is              
                      added to both sides of the equation)             
                      Before continuing, we will simplify 4 + 9/4: 
                      4 + 9/4  =  16/4 + 9/4  =  25/4             
                      Thus we have                x2 + 3x + 9/4  =  25/4     
         
            Step 5.  Factor the side of the equation that is a perfect square trinomial (we just created it) into the              
                          square of a binomial, which will be in the form of (x + h)2 or (x – h)2, where h is half of the              coefficient of                    the x term in Step 4.
                          (x + 3/2)2 = 25/4

          Step 6. Find the square root of each side of the equation and solve for x.
                       x + 3/2 = ± 5/2
                       x + 3/2 = 5/2                                  x + 3/2 = -5/2
                      x = 2/2 = 1                                      x = -8/2 = -4


        Let’s look at another example that is solved by completing the square.
        x2 – 8 = -2x

        Step 1. x2 + 2x – 8 = 0
        Step 2. x2 + 2x – 8 = 0
        Step 3. x2 + 2x = 8
        Step 4. x2 + 2x + 1 = 8 + 1
                    x2 + 2x + 1 = 9

        Step 5. (x + 1)2 = 9

        Step 6. x + 1 = ± 3

                    x + 1 = 3                                 x + 1 = -3
                          x = 2                                        x = -4

      Keeping in mind the information above, solve the following quadratic equations by completing the square.

      1. Solve the equation:  x2 + 12x + 32 = 0.

      2. Solve the equation:  2x2 + 12x + 16 = 0.

      3. Solve the equation:  x2 - 8x  =  9.

      4. Solve the equation:   2x2 - 4x  =  3.

      5. Solve the equation:  x2 - 20x + 96 = 60.

      6. Solve the equation:  3x2 - 14x + 8 = 0.

      SAT Math - Answers to Questions from Fall 2022, Week 3  (September 24, 2022)

      1. Solve the equation:  x2 + 12x + 32 = 0.
        Step 1. x2 + 12x + 32 = 0
        Step 2. x2 + 12x + 32 = 0
        Step 3. x2 + 12x = - 32
        Step 4. x2 + 12x + 36 = -32 + 36
                   x2 + 12x + 36 = 4
        Step 5. (x + 6)2 = 4
        Step 6. x + 6 = ±2

                     x + 6 = 2                                    x + 6 = -2
                           x = -4                                        x = -8

      2. Solve the equation:  2x2 + 12x + 16 = 0.

        Step 1.  2x2 + 12x + 16 = 0
        Step 2.  x2 + 6x + 8 = 0
        Step 3.  x2 + 6x = - 8
        Step 4.  x2 + 6x + 9 = -8 + 9
                     x2 + 6x + 9 = 1
        Step 5.  (x + 3)2 = 1

        Step 6.  x + 3 = ± 1

                      x + 3 = 1                                   x + 3 = -1           
                            x = -2                                       x = -4



      3. Solve the equation:  x2 - 8x  =  9.

        Step 1.  x2 - 8x - 9 = 0
        Step 2.  x2 - 8x - 9 = 0
        Step 3.  x2 - 8x = 9
        Step 4.  x2 - 8x + 16 = 9 + 16
                     x2 - 8x + 16 = 25
        Step 5.  (x - 4)2 = 25

        Step 6.  x - 4 = ± 5

                     x - 4 = 5                                  x - 4 = -5               
                           x = 9                                        x = -1

      4. Solve the equation:   2x2 - 4x  =  3.

        Step 1.  2x2 - 4x - 3 = 0
        Step 2.  x2 - 2x – 3/2 = 0
        Step 3.  x2 – 2x = 3/2
        Step 4.  x2 - 2x + 1 = 3/2 + 1
                     x2 - 2x + 1 = 3/2 + 2/2 = 5/2 = 2.5
                     x2 - 2x + 1 = 5/2

        Step 5.  (x - 1)2 = 2.5

        Step 6.   x - 1 = ± √2.5  

                      x - 1 = √2.5                                      x - 1 = - √2.5  
                            x = 1.5811 + 1                                 x = -1.5811 + 1
                            x = 2.5811                                       x = -0.5811

      5. Solve the equation:  x2 - 20x + 96 = 60.

        Step 1. x2 - 20x + 36 = 0
        Step 2. x2 - 20x + 36 = 0
        Step 3. x2 - 20x = -36
        Step 4.  x2 - 20x + 100 = -36 + 100
                     x2 - 20x + 100 = 64
        Step 5.  (x - 10)2 = 64
        Step 6.  x - 10 = ± 8

                        x - 10 = 8                                          x - 10 = -8
                                x = 18                                                x = 2

      6. Solve the equation:  3x2 - 14x + 8 = 0.

        Step 1.  3x2 - 14x + 8 = 0.
        Step 2.  x2 - (14/3)x + 8/3 = 0
        Step 3.  x2 - (14/3)x = - 8/3
        Step 4.  x2 - (14/3)x + 49/9 = -8/3 + 49/9
                     x2 - (14/3)x + 49/9 = -24/9 + 49/9
                     x2 - (14/3)x + 49/9 = 25/9
        Step 5.  (x – 7/3)2 = 25/9
        Step 6.  x – 7/3 = ± 5/3

                      x - 7/3 = 5/3                                        x - 7/3 = -5/3
                               x = 5/3 + 7/3                                        x = - 5/3 + 7/3
                               x = 12/3 = 4                                         x = 2/3

      SAT Verbal - Questions from Fall 2022, Week 3 (September 24, 2022)

      SAT QUICK CHALLENGE EXERCISE G22
      Conciseness A

      Conciseness. Concise writing gets the message across clearly without unneeded extra words. Such writing would not have two words or phrases that mean the same thing in the same sentence. Also, it would not have extra information that is not needed to get the message across clearly. Some SAT questions ask you to select from a group of similar sentences the one that is the most concise. The information below will help you understand how to answer those kinds of questions correctly.

      Redundancy. Redundancy occurs when two or more words or phrases that mean the same thing are used in a sentence, as in the following: Hastily, Gloria ran to the bus stop quickly, hoping that she would not miss the bus. Since hastily and quickly mean the same thing, only one of the two words should be used in that sentence. Accordingly, the corrected sentence could say eitherHastily, Gloria ran to the bus stop, hoping that she would not miss the bus.” or Gloria ran to the bus stop quickly, hoping that she would not miss the bus.

      Wordiness. When a sentence contains additional words that are not needed to get the message across, those extra words make the sentence wordy, as in the following: The reason that Mom could not be present at the concert with us is because of the fact that she had to go to work on account of an unexpected emergency that she had to handle. Removing the extra words would eliminate the wordiness, as in the following: Mom couldn’t go to the concert with us because she had to handle an emergency at work.

      Keeping in mind the information above, complete SAT Quick Challenge Exercise G22 below.

      Exercise G22 Conciseness

      Directions.  Identify the letter of the answer choice which corrects the underlined part of each statement below. If you believe that the underlined part is already correct, select Choice A -- NO CHANGE. After completing this activity, use the answer key to check your work.

      1.  Now, we will discuss at this time suggestions for upgrading our computer technology.

      1. NO CHANGE
      2. we will discuss at this time ideas and suggestions for upgrading
      3. we will discuss suggestions for upgrading
      4. we will discuss ideas and suggestions for upgrading and improving

      2. On account of the weather situation, which showed that tornados could strike our area, officials said that the game had to be canceled.

      1. NO CHANGE
      2. Because of the fact that tornadoes could develop,
      3. Prompted and caused by the possibility of strong tornadoes,
      4. Because of tornado threats,

      3. When we got back from our 20-mile walk, we went right to bed because we were extremely tired and quite exhausted.

      1. NO CHANGE
      2. exhausted and tired
      3. extremely tired
      4. exhausted and very tired

      SAT Verbal - Answers to Questions from Fall 2022, Week 3  (September 24, 2022)

      1. C
      2. D
      3. C

      SAT Math - Questions from Fall 2022, Week 2 (September 17, 2022)

      Quadratic Equations: One Solution, Two Solutions, No Real Solution

      Most of the quadratic equation problems on the SAT will have two solutions. Occasionally, however, there will be a problem where there is one solution or there is no real solution. For test takers it is important to be able to determine which situation exists. The approach to make this determination is to find the value of what is known as the discriminant. The discriminant is a portion of the quadratic formula, a formula that must be memorized by test takers. You will recall that to solve a quadratic equation by using the quadratic formula, we first put the quadratic equation in standard form: ax2 + bx + c = 0. Then the value of x is given by:

              -b  ±  √b2 - 4ac 
        X =   ----------------- 
                           2a

      where a is the coefficient of the x2 term, b is the coefficient of the x term, and c is the constant.

      The discriminant is the value of the expression under the radical sign: b2 – 4ac

      If the determinant is positive, there are two solutions to the quadratic equation.

      If it is equal to zero, there is one solution to the quadratic equation.

      If it is negative, there is no real solution to the quadratic equation.

      In the following example, there is one solution:

      9x2 + 4 = 12x

      Restate in standard form: 9x2 - 12x + 4 = 0
      Then a = 9, b = -12, and c = 4

      Thus the discriminant = b2 – 4ac = (-12)2 – 4(9)(4) = 144 – 4(36) = 144 – 144 = 0
      Therefore, there is one solution since the discriminant = 0.

                                                           -b  ±  √b2 - 4ac -(-12)  ±  √(-12)2 - 4(9)(4) 12  ±  √144 - 144 
               If we solve for x, we get    X = ------------------ = ----------------------------- = ------------------------ 
                                                             2a                                               2(9)        18

      12  ±  √0                12                 2
      X = ----------     = -----      =     ---                       This is the one solution.
       18      18                 3


      In the following example, there are two solutions:

      5x + x2  = -6

      Restate in standard form: x2 + 5x + 6 = 0
      Then a = 1, b = 5, and c = 6

      Thus the discriminant = b2 – 4ac = (5)2 – 4(1)(6) = 25 – 24 = 1
      Therefore, there are two solutions since the discriminant is positive.

                                                           -b  ±  √b2 - 4ac -(5)  ±  √(5)2 - 4(1)(6) -5  ±  √25 - 24 
               If we solve for x, we get    X = ------------------ = ----------------------------- = -------------  = 
                                                             2a                                               2(1)        2

      -5  ±  √1                -5  ±  1
                ----------     = ---------
         2           2


                 -5 + 1               -4                                                                   -5 - 1               -6                                                                                                         
       X   =   -------    =   --------   =  -2                                        X   =   -------    =    --------   =     -3             These are the two solutions. 
                       2                   2                                                                       2                    2        


      In the following example, there is no real solution:

      x2 - x + 8 = 0

      The equation is in standard form.
      Then a = 1, b = -1, and c = 8

      Thus the discriminant = b2 – 4ac = (-1)2 – 4(1)(8) = 1 - 32 = -1
      Therefore, there is no real solution since the discriminant is negative. 

      There is no value of x that makes this equation true.

      Keeping in mind the information above, in each of the following equations, determine which situation exists:  one solution, two solutions, or no real solution.

      1. -5x2 + 9x = 2

      2. x2 + 8x + 16 = 0

      3. 2x2 - 5x +1 = 0

      4. 5x + 4 = -6x2
               
      5. 3 + 3x2 = -6x

      6. x2 - 12x + 5 = 0

      SAT Math - Answers to Questions from Fall 2022, Week 2  (September 17, 2022)

      1. -5x2 + 9x = 2
        Restate in standard form: -5x2 + 9x -2 = 0
        a = -5, b = 9, c = -2
        Thus the discriminant = b2 – 4ac = (9)2 – 4(-5)(-2) = 81 - 4(10) = 81 - 40 = 41
        Therefore, there are two solutions since the discriminant is positive.

      2. x2 + 8x + 16 = 0
        a = 1, b = 8, c = 16
        Thus the discriminant = b2 – 4ac = (8)2 – 4(1)(16) = 64 - 4(16) = 64 - 64 = 0     
        Therefore, there is one solution since the discriminant equals 0.
      3. 2x2 - 5x +1 = 0

        a = 2, b = - 5, c = 1
        Thus the discriminant = b2 – 4ac = (-5)2 – 4(2)(1) = 25 - 8 = 17
        Therefore, there are two solutions since the discriminant is positive.

      4. 5x + 4 = - 6x2
        Restate in standard form:  6x2 + 5x + 4 = 0a = 6, b = 5, c =4

        Thus the discriminant = b2 – 4ac = (5)2 – 4(6)(4) = 25 - 4(24) = 25 - 96 = -71     
        Therefore, there is no real solution since the discriminant is negative.

      5. 3 + 3x2 = - 6x
        Restate in standard form: 3x2 + 6x + 3 = 0
        a = 3, b = 6, c = 3
        Thus the discriminant = b2 – 4ac = (6)2 – 4(3)(3) = 36 - 4(9) = 36 - 36 = 0
        Therefore, there is one solution since the discriminant equals 0.

      6. x2 - 12x +5 = 0
        a = 1, b = - 12, c = 5
        Thus the discriminant = b2 – 4ac = (- 12)2 – 4(1)(5) = 144 - 20 = 124
        Therefore, there are two solutions since the discriminant is positive.

      SAT Verbal - Questions from Fall 2022, Week 2 (September 17, 2022)

      SAT QUICK CHALLENGE EXERCISE F
      The Relative Clause

       

      A relative clause is a dependent clause that (1) begins with a relative pronoun (such as who, whom, that, which, whose, when, where, or whereby) and (2) modifies (or provides extra information about) a noun in the independent clause of the sentence. A relative clause is a dependent clause that (1) begins with a relative pronoun (such as who, whom, that, which, whose, when, where, or whereby) and (2) modifies (or provides extra information about) a noun in the independent clause of the sentence.

      Where” or “When.”  The word where refers to placesDo not use it to refer to times, books, films, or other things that are not places.  The word when refers to times and events.  Note Examples A and B, which follow. 
      Example A.  Washington, DC is the city when we plan to spend our summer vacation. 
      Comments.  The word when is incorrect because Washington, DC is a place, and we need a word that refers to a place. 
      Correction.  Washington, DC is the city where we plan to spend our summer vacation. 
      Example B.  We need to find out the dates where we can take the new COVID- 19 booster.   
      Comments.  The word where is incorrect because this example refers to dates, which are times.  Therefore, we need a word that refers to times, and when does that job. 
      Correction. We need to find out the dates when we can take the new COVID-19 booster.    

      Using a Preposition (in, during, to, etc.) and “Which” Instead of Using “Where.  Besides using the word “ where,” you can refer to places by using an appropriate preposition (in, during, to, etc.) followed by the word “which.”  Note the following:  Washington, DC is a city to which many people like to go for their vacations. 

      Using a Preposition (in, during, to, etc.) and “Which” Instead of “When.” Besides using “when,” you can refer to time by using a preposition that helps relate to time (on, during), followed by the word “which.” Note the following: (1) Do you know the dates on which we can take the new COVID- 19 booster? (2) Summer is the season during which many families travel the most.

      Using Whereby. “Whereby” is roughly translated to mean “by which,” “according to which,” “in which,” or “during which.” The word is used to introduce a description of a process, system, or method, as in the following sentence: Our researchers have created an amazing machine whereby thoughts are translated into spoken words.

      Now, keeping in mind the information above, complete the exercise below.

      Exercise O21 - The Relative Pronoun

      Directions.  Identify the letter of the answer choice that corrects the underlined part of each sentence below. If you think a sentence is already correct, select choice A – NO CHANGE. Use the answer key to check your work.

      1.  Harvard is the college of which Steve would like to go.

      1. NO CHANGE
      2. to which
      3. which
      4. when

      2.  Researchers are testing a process where polluted water can be sterilized and then used safely.

      1. NO CHANGE
      2. to where
      3. which
      4. whereby

      3. Do you know the dates for when the COVID boosters will be given?

      1. NO CHANGE
      2. where
      3. when
      4. which

      SAT Verbal - Answers to Questions from Fall 2022, Week 2  (September 17, 2022)

      1. B
      2. D
      3. C

      SAT Math - Questions from Fall 2022, Week 1 (September 10, 2022)

      Quadratic Equations: Solving by Factoring and by Using the Quadratic Formula

      It was noted in earlier classes this summer that there are three ways that quadratic equations can be solved:  by factoring, by using the quadratic formula, and by completing the square.  In the previous two lessons, we concentrated on factoring and using the quadratic formula.  In this lesson, we will review these two methods.It was noted in earlier classes this summer  that there are three ways that quadratic equations can be solved:  by factoring, by using the quadratic formula, and by completing the square.  In the previous two lessons, we concentrated on factoring and using the quadratic formula.  In this lesson, we will review these two methods.

      Solve by Factoring

      There are three steps in solving a quadratic equation by factoring:

      1. Set the equation equal to zero (put the equation in standard form);  ax2 + bx + c  =  0 is the standard form, where a is the coefficient of the x2 term, b is the coefficient of the x term, and c is the constant. 
      2. Factor the expression.
      3. Set each factor equal to zero and solve for the variable in each factor.  There will usually be two solutions.  If an expression cannot be factored, then use the quadratic formula. 

      Using this approach, solve the following equation: x2 = 24 – 5x

      1. Set the equation equal to zero (put the equation in standard form)
        x2 = 24 – 5x
        x2 + 5x - 24 = 0

      2. Factor the expression.

        x2 + 5x - 24 = 0
        We need to find two factors of -24 which when multiplied together will equal -24 and when added together will equal +5. Consider the factors of 24: 1 x 24, 2 x 12, 3 x 8, and 4 x 6.
        Which will work? +8 and -3 will work, since -3 x 8 = -24 and -3 + 8 = + 5

        Thus, the factored expression is (x + 8)(x - 3) = 0

      3. Set each factor equal to zero and solve for the variable in each factor.
        (x + 8)(x - 3) = 0

        x + 8 = 0 x - 3 = 0
        x = -8 x = 3

        To check our work, substitute each value of x in the original equation.
        For x = -8:       x2 = 24 – 5x
                           (-8)2 = 24 – 5(-8)
                              64 = 24 + 40 = 64 The value x = -8 checks.

        For x = 3:          x2 = 24 – 5x
                               (3)2 = 24 – 5(3)
                                   9 = 24 – 15 = 9 The value x = 3 checks.

      Here is another example: Solve the equation: x2 + 3x = 28
      x2 + 3x – 28 = 0

      (x – 4)(x + 7) = 0
      (x – 4) = 0                                      (x + 7) = 0
      x = 4                                              x = -7

      Solve by Using the Quadratic Formula

      The quadratic formula is an alternative to factoring, and it can be used whether or not an expression can be factored. We begin by putting the equation in standard form (ax2 + bx + c). Then the value of x is given by:

              -b  ±  √b2 - 4ac 
        X =   ----------------- 
                           2a 

      where a is the coefficient of the x2 term, b is the coefficient of the x term, and c is the constant.

      Using this approach, solve the following equation (which was solved by factoring above):
      x2 = 24 – 5x

      Put the equation in standard form.

      x2 = 24 – 5x
      x2 + 5x – 24 = 0


      Now apply the quadratic formula: 

      -b  ±  √b2 - 4ac
      X = -------------------- a = 1, b = 5, c = -24
      2a

      -5  ±  √52 - 4(1)(-24) -5  ±  √25 + 96 -5  ±  √121 -5  ±  11
      X = ------------------------- = ------------------- = -------------- = --------------
      2(1) 2 2 2

      -5-11            -16 -5+11            6
      X = -------  = ----- = -8 X  =  --------  =      ---  =  3
      2   2 2   2

      Thus, as expected, the solution is x = -8 and x = 3.


      Here is another example: Solve the equation by using the quadratic formula: x2 + 3x = 28


      Put the equation in standard form.
      x2 + 3x = 28
      x2 + 3x -28 = 0

      Now apply the quadratic formula

      Now apply the quadratic formula: 

      -b  ±  √b2 - 4ac
      X = -------------------- a = 1, b = 3, c = -28
      2a

      -3  ±  √32 - 4(1)(-28) -3  ±  √9 + 112 -3  ±  √121 -3  ±  11
      X = ------------------------- = ------------------- = -------------- = --------------
      2(1) 2 2 2

      -3-11            -14 -3+11            8
      X = -------  = ----- = -7 X  =  --------  =      ---  =  4
      2   2 2   2

      Keeping in mind the information above, solve each of the following equations by factoring and by using the quadratic formula.

      1. Solve the equation: x2 - 8x = 9.

      2. Solve the equation: 2x2 + 12x + 16 = 0.

      3. Solve the equation: 3x2 + 8 = 14x.

      SAT Math - Answers to Questions from Fall 2022, Week 1 (September 10, 2022)

      1. Solve the equation:     x2 - 8x = 9.

        Solve by factoring:
        Put the equation in standard form:  x2 - 8x -9 = 0.
        Factor the expression:
        x2 - 8x - 9 = (x + 1)(x - 9)

        Set each factor equal to zero and solve for the variable in each factor.
        x + 1 = 0 x - 9 = 0
        x = -1 x = 9

        Solve by using the quadratic formula:

        x2 - 8x = 9.

        Put the equation in standard form:  x2 - 8x - 9 = 0        a = 1, b = -8, c = -9

        -b  ±  √b2 - 4ac
        X = -------------------- a = 1, b = 3, c = -28
        2a

        -b  ±  √b2 - 4ac -(-8)  ±  √(-8)2 -4(1)(-9)  8  ±  √64+ 36  8  ±  √100  8  ±  10
        X = ----------------- = ------------------------ = --------------- = --------------     =  ------------
        2(1) 2 2 2 2

        8 + 10            18 8 - 10            -2
        X = -------  = ----- = 9 X  =  --------  =      ---  =  -1
        2   2 2   2




      2. Solve the equation:     2x2 + 12x + 16 = 0.
        Solve by factoring:

        Factor the expression:
        2x2 + 12x + 16 = 2(x2 + 6x + 8) = 0
        x2 + 6x + 8 = 0
        x2 + 6x + 8 = (x + 2)(x + 4)

        Set each factor equal to zero and solve for the variable in each factor.
        x + 2 = 0 x + 4 = 0
        x = -2 x = -4

        Solve by using the quadratic formula:

        2x2 + 12x + 16 = 0.

        Simplify the expression:
        2x2 + 12x + 16 = 2(x2 + 6x + 8) = 0
        x2 + 6x + 8 = 0         a = 1, b = 6, c = 8

        -b  ±  √b2 - 4ac -6  ±  √62 -4(1)(8)  -6  ±  √36 - 32   -6 ±  √4   -6  ±  2
        X = ----------------- = ------------------ = ----------------- = -----------     =  ---------
        2(1) 2 2

        -6 + 2              -4 -6 - 2            -8
        X = -------  = ----- = -2 X  =  --------  =      ---  =  -4
        2   2 2   2

      3. Solve the equation:     3x2 + 8 = 14x.

        Solve by factoring:

        Put the equation in standard form:   3x2 - 14x +8 = 0

        Factor the expression:
        3x2 - 14x + 8 = (3x - 2)(x - 4)

        Set each factor equal to zero and solve for the variable in each factor.
        3x - 2 = 0 x - 4 = 0
        3x = 2 x = 4
        x = 2/3 

        Solve by using the quadratic formula: 

        Put the equation in standard form:  3x2 - 14x + 8 = 0.              a = 3, b = -14, c = 8

        -b  ±  √b2 - 4ac -(-14) ±  √(-14)2 -4(3)(8)  14  ±  √196 - 96   14 ±  √100   14  ±  10
        X = ----------------- = ------------------------- = ----------------- = ------------     =  ---------
        2a 2(3) 6

        14 + 10            24 14 - 10           4
        X = --------  = ----- = 4 X   =   --------  =      ---  =  2/3
        6   6 6   6

      SAT Verbal - Questions from Fall 2022, Week 1 (September 10, 2022)

      SAT QUICK CHALLENGE EXERCISE E
      Punctuating Nonessential Information

       

      Using the Comma for Nonessential Information. A nonessential (nes) word/word group gives extra information about another word/word group in a sentence, but the nes information is not needed in the sentence. Since the sentence will still make sense If the nes word/word group is removed, nes information must be enclosed in commas. However, information that is needed for expressing the main idea of the sentence must not be separated from the rest of the sentence. Note Examples A and B below.

      Example A: Lisa’s home-baked cookies, which she makes from a very old family recipe, taste even better than the ones we buy at the bakery.
      Comments. Example A tells how Lisa’s cookies are “even better than the ones we buy at the bakery.” The clause right after the word “cookies,” “which she makes from a very old family recipe,” gives additional information about Lisa’s cookies, but we understand the sentence’s main idea clearly without that additional information. Therefore, the clause (1) is not essential, and (2) must be enclosed in commas.

      Example B: The cookies that Lisa baked won the bake-off trophy at the county fair.
      Comments. This sentence tells us that cookies won a trophy, and the clause right after the word “cookies” tells which cookies won that trophy: the cookies that Lisa baked. Since knowing which cookies won the trophy is important, the clause giving that information is essential and must not be enclosed in commas.

      “Which” or “That”? You can determine whether to use “which” or “that” by remembering that “which” follows a comma and begins a nonessential clause, but “that” does not follow a comma and begins an essential clause. In Example A, the word “which” begins a clause that contains extra information that is not an essential part of the meaning of the sentence. Therefore, commas separate the clause from the rest of the sentence. In Example B, the word “that” is used to begin a clause that is essential to the meaning of the sentence because the clause tells which cookies we are talking about. Since the clause is essential, it is not separated from the rest of the sentence.

      Keeping in mind the information above, complete the exercise below.

      Punctuating Essential and Nonessential Information

      Directions.  Identify the letter of the answer choice that corrects each sentence below. If you think a statement is already correct, select choice A – NO CHANGE. Use the answer key to check your work.

      1.  The apple pie recipe of which Grandma gave me twenty years ago is still my favorite.

      1. NO CHANGE
      2. on which
      3. that
      4. that which

      2.  The Artemis rocket launch, on which had been scheduled for September 3, was postponed because of a fuel leak.

      1. NO CHANGE
      2. launch, which
      3. launch of which
      4. launch, that which

      3. The fox in which was trying to get into the hen house has been captured.

      1. NO CHANGE
      2. fox which
      3. fox which that
      4. fox that

      SAT Verbal - Answers to Questions from Fall 2022, Week 1  (September 10, 2022)

      1. C
      2. B
      3. D

      SAT Math - Questions fom Summer 2022, Week 3 (July 16, 2022)

      Quadratic Equations: Solve by Using the Quadratic Formula

      What is the quadratic formula? As an alternative to factoring, it is a formula that can be used to solve a quadratic equation. In the previous lesson, when we solved the quadratic equation by factoring, we put the equation in standard form (ax2 + bx + c), factored the expression, set each factor equal to zero, and solved for the variable in each factor. As noted, there were usually two solutions.

      When we solve the equation by using the quadratic formula, we also will usually have two solutions. What is the formula? Putting the equation in standard form (ax2 + bx + c), the value of x is given by:


              -b  ± √b2 - 4ac 
       X =   ----------------- 
                           2a 

      where a is the coefficient of the x2 term, b is the coefficient of the x term, and c is the constant.

      This is a formula that must be memorized for the SAT. Some formulas are listed on the SAT, but this is not one of them. It must be memorized.

      Let’s look at an example that is solved by factoring and by using the quadratic formula. Note that in each case, the equation must first be put in standard form.

      Solve this equation:   x2 + 3x = 4

      Put in standard form: x2 + 3x – 4 = 0

      Solve by factoring:    (x + 4)(x – 1) = 0
      x + 4 = 0                                                                          x – 1 = 0
            x = -4                                                                               x = 1



                                                                                                  -b  ± √b2 - 4ac   
      Solve by using the quadratic formula:       X =  --------------------             a = 1, b = 3, c = -4 
                                                                                                                2a            




              -3  ± √32 - 4(1)(-4)            -3  ± √19 + 16                -3 ± √25               3 ± 5                        
       X =   -----------------------   =  ----------------------   =      -----------   =       --------
                               2(1)                                     2                                   2                        2                            
        


               -3 - 5            -8                                             -3  + 5               2
      X =    --------   =   ---   = -4                       X  =    --------    =     ----   =  1       
                      2              2                                                   2                   2     


      Thus as expected, the solution is x = -4 and x = 1.

      This is another example: 
      Solve this equation:
      2x2 - 4x – 3 = 0 by factoring and by using the quadratic formula.


      Actually, this expression cannot be factored. When the equation cannot be factored, we must use the quadratic formula. In this case, a = 2, b = -4, and c = -3.

      Solve the equation: x2 + 12x + 32 = 0.                  a = 1, b = 12, c = 32

              -b  ± √b2 - 4ac            -(-4)  ± √(-4)2 - 4(2)(-3)          4 ± √16 + 24 
       X =   -----------------   =  ----------------------   =    -----------------   =
                           2a                                  2(2)                                         4

        4  ±  √40            4  ±  √(4)(10)           4  ±  √4   √10             4  ± 2 √10           
       ---------     =     -------------    =    --------------    =      ----------    =   1  ±  (1/2) √10
              4                            4                                4                              4

      X = 1 + (1/2)  √10  or 1 + (1/2)(3.1632) = 1 + 1.5811 = 2.5811 = 2.58 (rounded to two places) 
      X = 1 - (1/2)   √10  or 1 - (1/2)(3.1632) = 1 - 1.5811 = - 0.5811= -0.58 (rounded to two places)

      Solve this equation by using the quadratic formula.
       x2 -4x + 4 = 0          (a = 1, b = -4, c = 4)  


                  -b  ± √b2 - 4ac            -(-4)  ± √(-4)2 - 4(1)(4)          4 ± √16 - 16            4  ±  √0          4 
      X    =   -----------------     =  -----------------------    =  -------------    =    ---------  =  ---    =  2
                               2a                                    2(1)                                    2                         2               2

      Keeping in mind the information above, solve the following equations by using the quadratic equation.                    .

      1. Solve the equation:   x2 + 12x + 32 = 0.

      2. Solve the equation:   2x2 +12x + 16 = 0.

      3. Solve the equation:   x2 -8x = 9.

      4. Solve the equation:   x2 -20x + 96 = 60.

      5. Solve the equation:   3x2 -14x + 8 = 0.

      6. Solve the equation:   (2x - 1)2  = (x + 2)2.

      SAT Math - Answers to Questions from Summer 2022, Week 4 (July 16, 2022)

      1. Solve the equation: x2 + 12x + 32 = 0.                  a = 1, b = 12, c = 32

                -b  ± √b2 - 4ac            -12  ± √122 - 4(1)(32)          -12 ± √144 - 128 
         X =   -----------------   =  ----------------------   =    -----------------   =
                             2a                                  2(1)                                         2

          -12  ±  √16            -12 ± 4
        --------------     =     -------
                  2                            2
           
                 -12 + 4         -8                                          -12 - 4              -16
        X =    --------   =   ---   = -4                       X  =    --------    =     ----   =  -8       
                        2             2                                                  2                   2     

      2. Solve the equation:   2x2 + 12x + 16 = 0.
        Simplify the expression:
        2x2 + 12x + 16 = 2(x2 + 6x + 8) = 0
        x2 + 6x + 8 = 0       a = 1, b = 6, c = 8

                -b  ± √b2 - 4ac            -6  ± √62 - 4(1)(8)                 -6 ± √36 - 32          -6 ±  √4
         X =   -----------------   =  ------------------     =       ----------------   =    --------    =    
                             2a                                  2(1)                                    2                          2

          -6 ± 2        
        --------
              2                            2
           
                 -6 + 2        -4                                                              -6 - 2        -8                     
        X =    -------   =  ---   = -2                                    X =          -------   =   ---   = -4                  
                      2            2                                                                    2           2         
                                       
      3. Solve the equation:   x2 - 8x = 9.
        Put the equation in standard form:   x2 - 8x - 9 = 0        a = 1, b = -8, c = -9

                -b  ± √b2 - 4ac            -(-8)  ± √(-8)2 - 4(1)(-9)       8 ± √64 + 36          8 ±  √100
         X =   -----------------   =  ------------------     =       ----------------   =    --------    =    
                             2a                                  2(1)                                    2                          2

          8 ± 10        
        --------
              2                            2
           
                 8 + 10       18                                                              8 - 10        -2                     
        X =    -------   =  ---   = 9                                    X =          -------   =   ---   = -1                  
                      2            2                                                                    2           2                                         

      4. Solve the equation:   x2 - 20x + 96 = 60.
        Put the equation in standard form:   x2 - 20x + 36 = 0        a = 1, b = -20, c = 36

                -b  ± √b2 - 4ac            -(-20)  ± √(-20)2 - 4(1)(36)              20 ± √400 - 144          20 ±  √256
         X =   -----------------   =  --------------------------     =       ----------------   =    ------------    =    
                             2a                                  2(1)                                                  2                          2

          20 ± 16        
        --------
              2                            2
           
                 20 + 16       36                                                               20 - 16       4                     
        X =    -------   =   ---   = 18                                    X =          -------   =   ---   =  2                  
                      2              2                                                                    2              2         
                                       
      5. Solve the equation:  3x2 - 14x + 8 = 0..                                a = 3, b = -14, c = 8

                -b  ± √b2 - 4ac            -(-14)  ± √(-14)2 - 4(3)(8)                   14± √196- 96          14 ±  √100
         X =   -----------------   =  --------------------------     =       ----------------   =    ------------    =    
                             2a                                  2(3)                                                  6                          6

          14 ± 10        
        --------
              6                            
           
                 14 + 10       24                                                               14 - 10       4                     
        X =    -------   =   ---   = 4                                    X =          -------   =   ---   =  2/3                  
                      6              6                                                                    6              6         
                                       
      6. Solve the equation.    (2x - 1)2 = (x + 2)2
        Put the equation in standard form.

        (2x - 1)2 = (x + 2)2

        4x2 - 4x _ 1 = x2 + 4x + 4
        3x2 - 8x - 3 = 0                  This is the equation in standard form. Now we apply the quadratic formula.
                                                   a = 3, b = -8, c = -3


                -b  ± √b2 - 4ac               -(-8)  ± √(-8)2 - 4(3)(-3)                    8± √64-+36              8 ±  √100
         X =   -----------------   =  --------------------------     =       ----------------   =    ------------    =    
                             2a                                  2(3)                                                  6                          6

          8 ± 10        
        --------
              6                            
           
                 8 + 10         18                                                               8 - 10        -2                     
        X =    -------   =   ---   =  3                                    X =          -------   =   ---   =  -1/3                  
                      6              6                                                                   6              6         
                                       

      SAT Verbal - Questions from Summer 2022, Week 4 (July 16, 2022)

      SAT QUICK CHALLENGE EXERCISE T21
      Achieving Conciseness

      Conciseness Concise writing makes points clearly without using more words than necessary. For this reason, correct Verbal SAT answers generally are the shortest choices. Redundancy, wordiness, and passive verbs are errors that destroy conciseness because they contain words that are not needed in order to get the message across clearly. Avoiding them will help you get higher SAT scores.  

      Wordiness. A wordy sentence contains unnecessary words. Note Sentences 1 and 2, which follow. (1) Due to the fact that Kathy was tired, she went to her room, got in the bed, and took a nap. (2) Kathy took a nap because she was tired. Comments. The main idea of each sentence is that Kathy took a nap because she was tired. Sentence 1 uses many unneeded words to make its point, but Sentence 2 makes its point concisely without unnecessary words. Therefore, only Sentence 2 is written correctly.

      Redundancy. A redundant sentence contains multiple words that mean the same thing. Those errors add nothing important to a sentence, and they make it longer unnecessarily. Avoiding redundancies can help you maintain concise writing. Note Sentences 3 and 4, which follow. (3) At the present time, the delegates will now discuss options for improvements. (4) The delegates will now discuss options for improvements.
      Comments. The bolded words in Sentence 3 make it redundant because they mean the same thing. Sentence 4 has no redundancies; it is concise, and it is correct.

      Active and Passive Verbs. An active verb tells us what someone does, did, or will do. A passive verb does the same thing, but it uses more words. See Sentences 5 and 6, which follow. (5) Roland picked fresh strawberries (active verb, as in x did y). (6) Fresh strawberries were picked by Roland, (passive verb, as in y was done by x). The passive pattern can also indicate what was done without telling who did it, as in y was done. That is, "Fresh strawberries were picked today." Note that since active verbs use fewer words than passive verbs, a sentence created with active verbs is more concise than the same sentence created with passive verbs.

      Now, keeping in mind the information above, complete the exercise below.

       

      SAT Quick Challenge T21
      Achieving Conciseness

      Directions.  Identify the letter of the answer choice that corrects the underlined part of each sentence below. If you think a statement is already                          correct, select choice A – NO CHANGE. Use the answer key to check your work.

      1.  At Speedy Serve packages are delivered within 24 hours after you place your order.

      1. NO CHANGE
      2. Orders placed at Speedy Serve are delivered
      3. Speedy Serve delivers orders
      4. Your Speedy Serve order is delivered

      2.  Tanya will be our valedictorian due to the fact that she has the highest GPA in our class.

      1. NO CHANGE
      2. the reason being that
      3. on account of
      4. because

      3.  Mya loves big, round, gold earrings that are circular in shape.

      1. NO CHANGE
      2. Delete the underlined words.
      3. of circular shape
      4. shaped like circles

      SAT Verbal - Answers to Questions from Summer 2022, Week 4 (July 16, 2022)

      1. C
      2. D
      3. B

      SAT Math - Questions fom Summer 2022, Week 3 (July 9, 2022)

      1. Quadratic Equations: Solve by Factoring

        It has been noted that there are three ways that quadratic equations can be solved: by factoring, by using the quadratic equation, and by completing the square. In the previous lesson, we concentrated on the process of factoring an expression. It is important to have a thorough understanding of how to factor an expression when a quadratic equation is solved by factoring.

        This lesson will explain how to solve quadratic equations by factoring. There are three steps in this process:

        1. Set the equation equal to zero. The basic format is ax2 + bx + c = 0, where a is the coefficient of the x2 term, b is the coefficient of the x term, and c is the constant.
        2. Factor the expression.
        3. Set each factor equal to zero and solve for the variable in each factor. There will usually be two solutions. If an expression cannot be factored, then use the quadratic formula.

        Using this approach, solve the following equation: x2 = 13 – 12x

        1. Set the equation equal to zero. The basic format is ax2 + bx + c = 0, where a is the coefficient of the x2 term, b is the coefficient of the x term, and c is the constant.
          x2 = 13 – 12x

          x2 + 12x - 13 = 0

        2. Factor the expression.
          x2 + 12x - 13 = 0
          (x + 13)(x - 1) = 0

        3. Set each factor equal to zero and solve for the variable in each factor.
          (x + 13)(x - 1) = 0

          x + 13 = 0 x - 1 = 0
          x = -13

          To check our work, substitute each value of x in the original equation.
          For x = -13:     x2 = 13 – 12x
                           (-13)2 = 13 – 12(-13)
                               169 = 13 + 156 = 169

          For x = 1:        x2 = 13 – 12x
                              (1)2 = 13 – 12(1)
                                  1 = 13 – 12 = 1

          Here is another example: Solve the equation: x2 + 3x = 28
          x2 + 3x – 28 = 0

          (x – 4)(x + 7) = 0
          (x – 4) = 0                                                (x + 7) = 0
                   x = 4                                                       x = -7

          Be careful with this one:

          If x>0 and 2x2 + 3x – 2 = 0, What is the value of x?
          We factor the expression: 2x2 + 3x – 2 = (2x + ?)(x + ?)
          What are the factors of -2? -2, 1 and 2, -1
          Which one gives us a value of 3 for the coefficient of the x term?
          -2, 1 does not work
          2, -1 will work since 2(2) + 1(-1)) = 4 -1 = 3
          2x2 + 3x – 2 = (2x -1)(x + 2)

          Now set each factor equal to zero and then solve for x.
          2x -1 = 0                                                      x + 2 = 0
                2x = 1                                                           x = -2
          x = ½

          What is the answer? The answer is x = ½ since x>0. On the SAT, it would be incorrect to include x = -2 as an answer.

      SAT Math - Answers to Questions from Summer 2022, Week 3 (July 9, 2022)

      1. Solve the equation:   x2 + 12x + 32 = 0.
        Factor the expression:
        x2 + 12x + 32 = (x + 4)(x + 8)

        Set each factor equal to zero and solve for the variable in each factor.
        x + 4 = 0 x + 8 = 0
        x = -4

      2. Solve the equation: 2x2 + 12x + 16 = 0.
        Factor the expression:
        2x2 + 12x + 16 = 2(x2 + 6x + 8) = 0
        x2 + 6x + 8 = 0
        x2 + 6x + 8 = (x + 2)(x + 4)
        Set each factor equal to zero and solve for the variable in each factor.
        x + 2 = 0                                     x + 4 = 0
              x = -2                                          x = -4
      3. Solve the equation: x2 - 8x = 9.

        Put the equation in standard form:  x2 - 8x - 9 = 0
        Factor the expression:
        x2 – 8x – 9 = (x + 1)(x - 9)

        Set each factor equal to zero and solve for the variable in each factor.
        x + 1 = 0                                                          x - 9 = 0
             x = - 1                                                               x = 9

      4. Solve the equation: x2 - 20x + 96 = 60.

        Put the equation in standard form: x2 - 20x + 36 = 0
        Factor the expression:
        x2 – 20x + 36 = (x - 18)(x - 2)

        Set each factor equal to zero and solve for the variable in each factor.
        x - 18 = 0                                                            x - 2 = 0
                x = 18                                                                x = 2

      5. Solve the equation: 3x2 - 14x + 8 = 0.
        Factor the expression:
        3x2 – 14x + 8 = (3x - 2)(x - 4)

        Set each factor equal to zero and solve for the variable in each factor.
        3x - 2 = 0                                                            x - 4 = 0
        3x = 2                                                                        x = 4
          x = 2/3

      6. What is the sum of the solutions of (2x – 1)2 = ( x + 2)2 ?
        Get the two values of x and then add them together.
        We first must simplify the expressions and then put the equation in standard form.
        (2x – 1)2 = ( x + 2)2
        4x2 - 4x + 1 = x2 + 4x + 4
        3x2 - 8x - 3 = 0                                     This is the equation in standard form. Now we factor the expression,
                                                                       set each factor equal to zero, and solve for the variable in each                                                                         factor.     
        (3x + 1)(x – 3) = 0

        3x + 1 = 0                                                              x - 3 = 0
              3x = -1                                                                  x = 3
                x = -1/3
        The sum = 3 - 1/3 = 2 2/3 or 8/3

        Note for your information: there is a quicker way of finding the sum or product of the solutions once we have the equation in standard form. Remember as noted at the beginning of this lesson, in standard form the basic format is ax2 + bx + c = 0, where a is the coefficient of the x2 term, b is the coefficient of the x term, and c is the constant.

        The sum of the solutions is -b/a and the product of the solutions is c/a.
        In this case, a = 3, b = -8 and c = -3.
        Thus the sum = -b/a = -(-8)/3 = 8/3, which is the same result we got earlier.

      .

      SAT Verbal - Questions from Summer 2022, Week 3 (July 9, 2022)

      SAT QUICK CHALLENGE EXERCISE S21
      Words Often Confused - The Pronoun

      The Possessive Pronoun. This review lesson is a reminder that while the apostrophe followed by an “s” (’s) shows possession for nouns (as in Carl’s shoes or men’s jeans), the apostrophe is never used to show ownership for pronouns. Instead, possessive pronouns such as my, our, your, his, her, its, and their do that job. Thus, “Carl’s shoes” would become “his shoes,” and “men’s jeans” would become “their jeans.”

      The Pronoun Contraction. When an apostrophe is used with a pronoun, that apostrophe, combined with a shortened version of the verb which goes with that pronoun, creates a contraction that means the same thing as the original pronoun-verb combination. Hence, an apostrophe with an “s” turns “She is” into “She’s.” Thus, “She is my cousin.” becomes “She’s my cousin.” Similarly, “they’re” turns “They are my cousins.” into “They’re my cousins.”

      Pronouns Often Confused. Keeping the pronoun contraction in mind can help you answer SAT questions involving pronouns that are often confused with one another. For example, note the following sentence: _______ car is parked in the driveway? You must determine whether choice A or B, which both follow, completes the sentence correctly. (A) Who’s (B) Whose Explanation. Since the sentence is asking about the owner of a car, a possessive pronoun is needed. The apostrophe in choice A makes the sentence say “Who is” car is parked in the driveway?” That statement makes no sense. However, the possessive pronoun “Whose” in choice B makes the sentence say “Whose car is parked in the driveway?” That choice is correct.

      Sometimes, there, they’re, and their are confused with one another. Use “there,” which is the opposite of “here,” to indicate the place where something is located. If you need to, review The Possessive Pronoun and The Pronoun Contraction above to refresh your memory about how to use “their” and “they’re” correctly,

      Now, keeping in mind the information above, complete Exercise S21 below.

       

      SAT Quick Challenge S21
      The Pronoun and The Apostrophe

      Directions.  Identify the answer choice that corrects the underlined section of each statement below. If a statement has no error, mark choice “A.” When you have finished, use the answer key to check your work. 

      1.  They’ers the man who rescued the baby from that hot car.

      1. NO CHANGE
      2. Theyre’s
      3. Theirs
      4. There’s

      2.  This dessert is delicious; I think that we’re going to win the pie-baking contest.

      1. NO CHANGE
      2. we’er
      3. were
      4. we'ere

      3.  Do you know who’s baseball glove this is?

      1. NO CHANGE
      2. who is
      3. whose
      4. who’se

      SAT Verbal - Answers to Questions from Summer 2022, Week 3 (July 9, 2022)

      1. D
      2. A
      3. C

      SAT Math - Questions from Summer 2022, Week 2 (June 25, 2022)

      Quadratic Equations: Factoring

      In the last lesson we indicated that there are three ways that quadratic equations can be solved: by factoring, by using the quadratic equation, and by completing the square. It is important to have a thorough understanding of how to factor an expression when a quadratic equation is solved by factoring.
      This lesson will concentrate on factoring an expression.

      1. A factor of a number is defined as a positive integer that can be divided evenly unto the number – without a remainder. For example, the factors of 12 are 1, 2, 3, 4, 6, and 12.
      2. A variable can also be a factor. In the expression 12xy, x is a factor and y is a factor.
      3. Negative numbers can also be factors. -1, -2, -3, -4, -6, and -12 are factors of 12.


      We may be asked to factor a binomial or a trinomial. To factor a binomial, we factor out the greatest common factor. For example,
      Factor 4x + 16.         2 is a common factor but 4 is the greatest common factor. Thus, 4x + 16 = 4(x + 4).
      Factor 9a2 – 21a.     3a is the greatest common factor. Thus, 9a2 – 21a = 3a(3a - 7).

      To factor a trinomial into the product of two binomials, we write the expression in the standard format:
      ax2 + bx + c, and then we find factors that meet two criteria when the coefficient of the x2 term is 1:

      1. The product of the factors = the last term
      2. The sum of the factors = the coefficient of the x term

      For example, factor x2 + 5x + 4.                x2 + 5x + 4 = (x + ?)(x + ?) In place of the question marks, try various factor pairs so that the product of two factors = 4 and the sum of the factors = 5. The factor pairs to consider are 2, 2 and 1, 4. (2)(2) = 4 but 2 + 2 does not equal 5. Thus the factor pair 2, 2 will not work. The factor pair 1, 4 will work since (1)(4) = 4 and 1 + 4 = 5. Thus, x2 + 5x + 4 = (x + 1)(x + 4).

      Factor the expression x2 - 6x + 8.             x2 - 6x + 8 = (x + ?)(x + ?) In place of the question marks, try various factor pairs so that the product of two factors = 8 and the sum of the factors = -6. The factor pairs must be negative in this case; we can consider -1, -8 and -2, -4. (-1)(-8) = 8 but (-1) + (-8) does not equal -6. Thus, the factor pair -1, -8 will not work. The factor pair -2, -4 will work since (-2)(-4) = 8 and (-2) +(- 4) = -6. Thus, x2 - 6x + 8 = (x -2)(x - 4).

      Another example: Factor the expression 3x2 - 9x -54. Check for a greatest common factor; there is one: 3. First we factor out the 3: 3x2 - 9x -54 = 3(x2 - 3x – 18) = 3(x - ?)(x - ?) What are the factors of -18 that will give us a product of -18 and a middle term of -3? There are several to consider: 1 and -18, -1 and 18, 2 and -9, -2 and 9, 3 and -6, -3 and 6. Which one works? 3 and -6 will work. Thus 3x2 - 9x -54 = 3(x2 - 3x – 18) = 3(x + 3)(x -6).

      It is a little bit more complicated when the coefficient of the x2 term is not equal to 1 (after the greatest common factor has been factored out). In this case, we find factors that meet the following two criteria:

      1. The product of the factors = the last term
      2. To find the coefficient of the x term, we multiple one of the factors by the coefficient of the x2 term, and add that product to the other multiple.

      Factor the expression 2x2 + 7x + 6.    2x2 + 7x + 6 = (2x + ?)(x + ?) For a product of 6 for the last term, we have (1)(6), (6)(1), (2)(3), or (3)(2). The difficulty is getting 7 for the coefficient of the x term. To get 7 for the coefficient of the x term, which multiple when multiplied by 2, plus the other multiple, equals 7? In this case, (2)(2) + 3 = 7. Thus 2x2 + 7x + 6 = (2x + 3)(x + 2).

      Remember the following three important relationships:
      a2 + 2ab + b2 = (a+b)(a+b) or (a+b)2
      a2 -2ab +b2 = (a-b)(a-b) or (a-b)2
      a2 – b2 = (a+b)(a-b)

      You must memorize these: a2 + 2ab + b2 , a2 -2ab +b2 , and a2 – b2 . They appear again and again on the SAT.

      a2 + 2ab + b2 = what?      (a+b)(a+b) or (a+b)2
      a2 -2ab +b2 = what?        (a-b)(a-b) or (a-b)2
      a2 – b2 = what?                (a+b)(a-b)

      Factor the expression     x4 – y4 .                      x4 – y4 = (x2 + y2)(x2 – y2) =  (x2 + y2)(x + y)(x - y)

      Which of the following is not a factor of x5 – x3?                             

      1. x3
      2. x - 1
      3. x2 - 1
      4. x3 - 1

      x5 – x3 = x5 – x3(x + 1)(x - 1)

      Therefore, x3 is a factor, (x-1) is a factor, and (x2 – 1) is a factor; (x3 – 1) is not a factor.

      Keeping in mind the information above, ansswer the following questions.

      1. Factor the following expression: x2 – 5x + 6.

      2. (7x + 10) – (4x – 8) is equivalent to
        1. 3(x + 6)
        2. 3x + 2
        3. 3(x + 2)
        4. 3(x + 18)

        (7x + 10) – (4x – 8) = 7x + 10 – 4x + 8 = 3x + 18 = 3(x + 6) -------------- A

      3. Factor the expression 2x2 – 20x – 22

      4. Factor the expression: 3x2 + 2x - 5

      5. Where             (x + 3)         is defined, it is equivalent to which of the following expressions?
                       ----------------
                          (x2 - 2x - 15)

        1. 3 / (x - 15)
        2. 1 / (x - 5)
        3. 3 / (x + 5)
        4. 1 / (x + 5)

      6. If (a + b)2 = 169 and (a - b)2 = 25, then ab = ?

      SAT Math - Answers to Questions from Summer 2022, Week 2 (June 25, 2022)

      1. Factor the following expression: x2 – 5x + 6.

        x2 – 5x + 6 = (x + ?)(x + ?) 
        In place of the question marks, try various factor pairs so that the product of two factors = 6 and the sum of the factors = -5. The factor pairs must be negative in this case; we can consider -1, -6 and -2, -3. (-1)(-6) = 6 but (-1) + (-6) does not equal -5. Thus, the factor pair -1, -6 will not work. The factor pair -2, -3 will work since (-2)(-3) = 6 and (-2) +(- 3) = -5. Thus, x2 - 5x + 6 = (x -2)(x - 3).

      2. (7x + 10) – (4x – 8) is equivalent to
        1. 3(x + 6)
        2. 3x + 2
        3. 3(x + 2)
        4. 3(x + 18)

        (7x + 10) – (4x – 8) = 7x + 10 – 4x + 8 = 3x + 18 = 3(x + 6) -------------- A

      3. Factor the expression 2x2 – 20x – 22
        Check for a greatest common factor; there is one; it is 2. Factor out the 2:
        2x2 – 20x – 22 = 2(x2 – 10x – 11) = 2(x + ?)(x + ?) Again, in place of the question marks, try various factor pairs so that the product of two factors = -11 and the sum of the factors = -10.
        The factor pairs are (1)(-11) and (-1)(11). In this case (1)(-11) works. Thus,
        2x2 – 20x – 22 = 2(x2 – 10x – 11) = 2(x + 1)(x -11).

      4. Factor the expression: 3x2 + 2x - 5

        There is no common factor; thus 3x2 + 2x – 5 = (3x + ?)(x + ?) For product of -5 for the last term, we have (1)(-5) or (-1)(5). The difficulty is getting 2 for the coefficient of the x term. To get 2 for the coefficient of the x term, which of the two multiples, when multiplied by 3, plus the other multiple, equals 2? In this case, (3)(-1) + 5 = 2.
        Thus 3x2 + 2x -5 = (3x + 5)(x -1).

      5. Where             (x + 3)         is defined, it is equivalent to which of the following expressions?
                       ----------------
                          (x2 - 2x - 15)

        1. 3 / (x - 15)
        2. 1 / (x - 5)
        3. 3 / (x + 5)
        4. 1 / (x + 5)

        To solve this problem, we must factor the denominator. (x2 – 2x -15) = (x + ?)(x + ?) What are the factors of -15 that will give us a product of -15 and a middle term of -2? There are several to consider: 1 and -15, -1 and 15; and 3 and -5, -3 and 5. Which one works? 3 and -5 will work since (3)(-5) = -15 and 3 +(-5) = -2. Thus x2 - 2x -15 = (x + 3)(x -5).

               x + 3               =               =            x + 3                =                         1
        -------------                                 -------------                         -------------
          x2 - 2x -15                                  (x + 3) (x - 5)                                  (x - 5)

        The answer is B.

      6. If (a + b)2 = 169 and (a – b)2 = 25, then ab = ?

        We start by factoring (a + b)2 and (a-b)2.

        (a + b)2 = a2 + 2ab + b2 = 169
        (a - b)2 = a2 - 2ab + b2 = 25

        To solve for ab, multiply both sides of the second equation by -1, and then add the two equations.

         a2 + 2ab + b2 = 169
        -a2 + 2ab - b2 = -25
        --------------------
                 4ab        = 144
                   ab        = 36

        Note that we did not solve for a and b separately, we solved for ab.

      SAT Verbal - Questions from Summer 2022, Week 2 (June 25, 2022)

      SAT QUICK CHALLENGE EXERCISE R21
      The Apostrophe and The Pronoun

      The Possessive Pronoun. You will recall that we add an apostrophe and the letter “s” to a noun to show what that noun owns. For example, we write the boy’s wagon or the girl’s bike. However, we do not use apostrophes to show ownership for pronouns. Instead, we use possessive pronouns. Hence, we would say “his” wagon or “her” bike. The “Possessive Pronouns” chart below illustrates how to show ownership for pronouns.
      .
      The Pronoun Contraction. Although we never use the apostrophe to show pronoun ownership, we must use the apostrophe to make pronoun contractions. For instance, for I am, we write I’m, or for “You are,” we write “You’re.” Remembering how to use the apostrophe correctly for pronouns will enable you to earn points each time an SAT question requires you to show that you have mastered that skill. The “Pronoun Contractions” chart below shows how to write pronoun contractions correctly. After studying the charts, complete Exercise R21 (beneath the chart). Then, use the Answer Key to check your work.

      POSSESSIVE PRONOUNS PRONOUN CONTRACTIONS
      (for the verb "to be")

      Singular Plural Singular

      Plural

      My Our I am               I'm

      We are               We're

      Your Your You are          You're

      You are              You're

      His, Her, Its Their He is            She is            It is
      He's             She's             It's

      They are           They're

       

      SAT Quick Challenge R21
      The Pronoun and The Apostrophe

      Directions.  Identify the answer choice that corrects the underlined section of each statement below. If a statement has no error, mark choice “A.”                        When you have finished, use the answer key to check your work.

      1.  Some social media critics say that its harmful to children because it can place them in dangerous situations.

      1. NO CHANGE
      2. where
      3. which
      4. during which

      2.  It was four days before Christmas where Matt Lucas’s three-year-old son unwrapped all the presents that were under the Christmas tree.

      1. NO CHANGE
      2. its’
      3. I’ts
      4. it’s

      3.  The phone isn’t working because you’re battery needs to be charged.

      1. NO CHANGE
      2. you’er
      3. you
      4. your

      SAT Verbal - Answers to Questions from Summer 2022, Week 2 (June 25, 2022)

      1. D
      2. B
      3. D

      SAT Math - Questions from Summer 2022, Week 1 (June 11, 2022)

      Quadratic Equations: An Introduction

      We have looked at linear equations and systems of linear equations. We will next study a third type of equation, quadratic equations. These are equations that have an x2 term, such as x2 + 3x + 2 = 0. There are three ways that quadratic equations can be solved: by factoring, by using the quadratic equation, and by completing the square.

      We will review the factoring process. A factor of a number is defined as a positive integer that can be divided evenly unto the number – without a remainder. For example, the factors of 18 are 1,2,3,6,9, and 18.

      On the SAT you will have problems dealing with common factors and the greatest common factor. A common factor is a factor that appears in two or more terms. The greatest common factor is the largest of the common factors.

      The common factors in 10 and 30 are 1, 2, 5, and 10.

      The greatest common factor in 12 and 30 is 6; the common factors are 1, 2, 3, and 6, and the greatest of these is 6.

      What is the greatest common factor in 24x3 – 16x2 + 8x? It is 8x; 8 is the greatest common factor in 24, 16, and 8, and x is a factor that appears in all three terms.

      Factoring a polynomial is the reverse of multiplying (or distributing).

      For example, find the product: (3x + 1)(x + 2)

      To multiply two binomials, we find the sum of the products by using the F O I L method (First terms, Outer terms, Inner terms, Last terms)

                F              multiply the First terms (3x)(x) = 3x2
                O             multiply the Outer terms (3x)(2) = 6x
                I              multiply the Inner terms (1)(x) = x
                L             multiply the Last terms (1)(2) = 2

      Now we add the terms: 3x2 + 6x + x + 2 = 3x2 + 7x + 2

      Here is another example: Find the product of (3a – b)(2a + 4b).
      (3a – b)(2a + 4b) = (3a)(2a) + (3a)(4b) + (-b)(2a) + (-b)(4b) = 6a2 + 12ab – 2ab – 4b2 = 6a2 + 10ab – 4b2
                                           F              O               I                L

      Review the following three important relationships:

      1. (a+b)(a+b) or (a+b)2 = a2 + 2ab + b2
        the sum of two variables squared = first variable squared plus 2 times the product of the two
                                                                    variables plus the second variable squared

      2. (a-b)(a-b) or (a-b)2 = a2 -2ab +b2
        The difference between two variables squared = first variable squared minus 2 times the product of
        the two variables plus the second variable squared
      3. (a+b)(a-b) = a2 – b2                           the difference between two variables times the sum of the two                                                                     variables = the difference between the squares of the two variables                                                             (sometimes stated as the difference between two squares)
              

      You must memorize these: (a+b)2 , (a-b)2 , (a+b)(a-b). They appear again and again on the SAT.

      (a+b)2 = what?               a2 + 2ab + b2
      (a-b)2 = what?               a2 - 2ab + b2
      (a+b)(a-b) = what?        a2 – b2

      Keeping in mind the information above, answer the following questions.

      1. What is the value of -7x(2x3 – 5x + 3)?                

      2. What is the value of (x-4)(3x + 1)?
         
      3. What is the value of (5a – 7b2)(5a + 7b2)?

      4. What is the value of (5x – 3)2?

      5. What is the value of 3a(2a – b)2?

      6. What is the value of (3x + y)(x2 – y)?


      SAT Math - Answers to Questions from Summer 2022, Week 1 (June 11, 2022)

      1. -7x(2x3 – 5x + 3) = ?                           Here we distribute
        -7x(2x3 – 5x + 3) = -14x4 + 35 x2 – 21x

      2. (x-4)(3x + 1) = ?                                    Use F O I L   
        (x-4)(3x + 1) = (x)(3x) + (x)(1) + (-4)(3x) + (-4)(1) = 3x2 + x - 12x - 4 = 3x2 - 11x – 4
                                    F           O             I               L         
         
      3. (5a – 7b2)(5a + 7b2) = ?                     Difference of two squares
        (5a – 7b2)(5a + 7b2) = 25a2 – 49b4

      4. (5x – 3)2 = ?                                          The difference between two variables squared
        (5x – 3)2 = 25x2 – 30x + 9

      5. 3a(2a – b)2 = ?                                      The difference between two variables squared; then distribute
        3a(2a – b)2 = 3a(4a2 – 4ab + b2) = 12a3 – 12a2b + 3ab2

      6. (3x + y)(x2 – y) = ?                                  Use F O I L
        (3x + y)(x2 – y) = 3x3 – 3xy + x2y – y2
                                     F        O        I      L

      SAT Verbal - Questions from Summer 2022, Week 1 (June 11, 2022)

      SAT QUICK CHALLENGE EXERCISE Q21
      The Dangling Modifier

      The Introductory Modifying Phrase (IMP) and the Subject of a Sentence. You can think of a sentence with an introductory modifying phrase (IMP) as a two-part sentence. The first part is the IMP itself, which modifies the subject of the sentence. The second part is the main clause, which (1) follows the comma after the IMP, (2) begins with the subject, and (3) expresses the main idea. The comma after the IMP connects the IMP to the subject. Note that in Sentence A, which follows, all the sentence elements are where they should be, and the sentence is written correctly. Sentence A. Having forgotten her homework, Marla was afraid that her teacher would think that she had not done it.

      Identifying a Dangling Modifier. When the subject an IMP should modify is not included in the sentence, the IMP is sometimes described as “just hanging there” because it is not connected to that subject. For this reason, an IMP in a sentence which does not have the subject the IMP should modify is referred to as a dangling modifier. Note Sentence B, which follows. Sentence B. When driving along Highway 37, smoke from the forest fire nearby made breathing very difficult. Explanation B. As written, the IMP modifies the word “smoke” because “smoke” is the subject of the clause that follows the IMP. The subject that should be in the sentence (a noun that could drive along the highway) is not in the sentence at all, so the IMP is a dangling modifier. You can correct this error by rewriting either the IMP or the main clause, as explained below.

      Correcting a Dangling Modifier: Revising Either the IMP or the Main Clause. One way to correct Sentence B’s dangling modifier is (1) to leave the IMP as it is, (2) place the missing subject right after the comma that follows the IMP, and (3) use that subject to express the main idea of the sentence, as in Sentence C, which follows. Sentence C. When driving along Highway 37, Mrs. Cooper found that smoke from the forest fire nearby made breathing very difficult.

      Another way to correct the dangling modifier is (1) to add the subject to the IMP, and (2) leave the main clause as it is, as in Sentence D, which follows. Sentence D. When Mrs. Cooper was driving along Highway 37, smoke from the forest fire nearby made breathing very difficult.

      Now, keeping in mind the information above, complete the exercise below.

       

      SAT Quick Challenge Q21
      The Dangling Modifier

      Directions.  Select the letter of the answer choice which corrects the underlined error. If you think that the underlined words are already correct,                          select choice A -- NO CHANGE. Then check your work.

      1.  Tremendously thankful for finishing the test on time, tears of joy flowed from Amy’s eyes.

      1. NO CHANGE
      2. Amy shed heartfelt tears of gratitude
      3. Amy’s eyes filled with tears
      4. Grateful tears calmed Amy’s stress

      2.  Driving home from the fair, a traffic accident delayed traffic for over three hours.

      1. NO CHANGE
      2. When driving home from the fair
      3. Driving the children home from the fair
      4. When we were driving home from the fair

      3.  After injuring his knee at football practice, it was hard for Todd to use the stairs.

      1. NO CHANGE
      2. Todd’s knee wouldn’t bend on the stairs
      3. Todd had trouble using the stairs
      4. the stairs had to be avoided by Todd

      SAT Verbal - Answers to Questions from Summer 2022, Week 1 (June 11, 2022)

      1. B
      2. D
      3. C

      SAT Math - Questions from Spring 2022, Week 16 (June 4, 2022)

      Systems of Linear Equations, Part 6: Word Problems  

      In recent years word problems have become more numerous on the SAT. Some word problems will require that you create a system of equations from the information given. Sometimes you will be asked to solve the system of equations, but often you will just be asked to develop the two equations. It is essential, then, that you be able to find and translate the information in the problem into the equations. The following procedure is useful in determining the required information.

      Step 1: Read the problem carefully and identify these important items for each equation:

      1. The variables
      2. The coefficients of the variables
      3.  The total

      Remember that each of the two equations is in the format: Ax +By = C where x and y are variables,
      A and B are coefficients, and C is the total.

      The coefficients can be negative or positive;
      the coefficients can equal 1;
      each equation has a total on the right side of the equation;
      and the total can be negative or positive

      Step 2. Write the appropriate coefficients and total for each equation.

      Step 3. Review the problem to ensure that your equations are consistent with the information in the problem.

      Let’s look at the following example:

      Albert, Brad, and Charles are selling hamburgers and hot dogs. The hamburgers are $2.50 each and the hot dogs are $1.75 each. On yesterday they had 58 customers each buying only one item and they made $128.50. Which of the following equations will let them know how many hamburgers and how many hot dogs they sold?

      1. x + y = 58
        1.75x + 2.50y = 128.50

      2. x + y = 58
        250x + 175y = 128.50

      3. x + y = 58
        2.50x + 1.75y = 128.50

      4. x + y = 128.5
        2.50x + 1.75y = 58

      Let's use the steps outlined above:

      Step 1: Read the problem carefully and identify these important items for each equation:

      1. The variables:  The variables are the number of hamburgers and the number of hot dogs.  Let x = the number of hamburgers and y = the number of hot dogs1. The variables:  The variables are the number of hamburgers and the number of hot dogs.  Let x = the number of hamburgers and y = the number of hot dogs
      2. The coefficients of the variables:  in one equation the coefficient of x is 1 and the coefficient of y is 1; in the other equation the coefficient of x is 2.50 and the coefficient of y is 1.75.
      3. The total:  the total for one equation is 58 the total for the other equation is 128.50

      Step 2. Write the appropriate coefficients and total for each equation.
      x - y = 1
      1
      x + y = 59

      Step 3. Review the problem to ensure that your equations are consistent with the information in the problem.

      The equations are consistent.

      Keeping in mind the information above, answer the following system of equations questions.

      1. Delta State University has separate tuition rates for in-state students and out-of-state students.  In-state students are charged $421 per semester and out-of-state students are charged $879 per semester.  The university’s junior class of 1,980 students paid a total of $1,170,210 in tuition fees for the most recent semester. Which of the following systems of equations represents the number of in-state (x) and out-of-state  (y) juniors and the amount of tuition fees the two groups paid?
        1. x + y  =  1,170,210
          421x + 879y  =  1,980

        2. x + y  =  1,980
          879x + 421y  =  1,170,210

        3. x + y  =  1,980
          421x + 879y  =  1,170,210

        4. x + y  =  1,170,210
          879x + 421y  =  1,980

      2. An online bookstore sells novels and magazines.  Each novel sells for $4, and each magazine sells for $1.  If Jackie purchased a total of 11 novels and magazines that have a combined selling price of $20, how many novels did she purchase?
        1. 2
        2. 3
        3. 4
        4. 5

      3. A bead shop sells wooden beads for $0.20 each and crystal beads for $0.50 each.   If a jewelry artist buys 127 beads total and pays $41 for them, how much more did she spend on crystal beads than wooden beads?
        1. $11
        2. $15
        3. $23
        4. $26

      SAT Math - Answers to Questions from Spring 2022, Week 16 (June 4, 2022)

        1. Delta State University has separate tuition rates for in-state students and out-of-state students.  In-state students are charged $421 per semester and out-of-state students are charged $879 per semester.  The university’s junior class of 1,980 students paid a total of $1,170,210 in tuition fees for the most recent semester. Which of the following systems of equations represents the number of in-state (x) and out-of-state  (y) juniors and the amount of tuition fees the two groups paid?

          1.  x + y  =  1,170,210
            421x + 879y  =  1,980

          2.  x + y  =  1,980
            879x + 421y  =  1,170,210

          3. x + y  =  1,980
            421x + 879y  =  1,170,210

          4. x + y  =  1,170,210
            879x + 421y  =  1,980

          In this case we are not asked to solve the system of equations; just determine what they are. Use the process described earlier:


          Step 1: Read the problem carefully and identify these important items for each equation:

          - The variables:  The variables are the number of in-state juniors and the number of
             out- of-state juniors.  Let x = the number of in-state juniors and y = the number of
             out-of-state juniors. 
          - The coefficients of the variables:  in one equation the coefficient of x is 1 and the
             coefficient of y is 1; in the other equation the coefficient of x is 421 and the
             coefficient of y is 879.
          - The total:  the total for one equation is 1,980 and the total for the other equation is
             1,170,210.

          Step 2. Write the appropriate coefficients and total for each equation.

            x + y = 1.980
            421x + 879y = 1,170,210             The answer is C.

        2. An online bookstore sells novels and magazines.  Each novel sells for $4, and each magazine sells for $1.  If Jackie purchased a total of 11 novels and magazines that have a combined selling price of $20, how many novels did she purchase?

          1. 2
          2. 3
          3. 4
          4. 5

          In this case we are asked to determine the system of equations and then solve the problem. Use the process described earlier to determine the system of equations,

          Step 1:  Read the problem carefully and identify these important items for each equation:
          -  The variables:  The variables are the number of novels and the number of         
              magazines.  Let x = the number of novels  and  y = the number of magazines. 
          -  The coefficients of the variables:  in one equation the coefficient of x is $4 and the
              coefficient of y is $1; in the other equation the coefficient of x is 1 and the
              coefficient of y is 1.
            - The total:  the total for one equation is $20 and the total for the other equation is
               11.

          Step 2. Write the appropriate coefficients and total for each equation.
          4x + y = 20
          x + y = 11

          Now we can solve the system of equations; let’s use the elimination approach. Multiply the second equation by -1: -x - y = -11. Then
          4x + y = 20
          -x - y = -11
          3x      = 9
                 x = 3                The answer is B.

        3. A bead shop sells wooden beads for $0.20 each and crystal beads for $0.50 each.   If a jewelry artist buys 127 beads total and pays $41 for them, how much more did she spend on crystal beads than wooden beads?           

          1. $11
          2. $15
          3. $23
          4. $26

          In this case we are asked to determine the system of equations and then solve the problem. Use the process described earlier to determine the system of equations,

          Read the problem carefully and identify these important items for each equation:

          1. The variables:  The variables are the number of wooden beads and the number of crystal beads.     Let x = the number of  wooden beads  and  y = the number of crystal beads.
          2. The coefficients of the variables:  in one equation the coefficient of x is $0.20 and the coefficient of y is $0.50; in the other equation the coefficient of x is 1 and the coefficient of y is 1.2. The coefficients of the variables:  in one equation the coefficient of x is $0.20 and the coefficient of y is $0.50; in the other equation the coefficient of x is 1 and the coefficient of y is 1.
          3. The total:  the total for one equation is $41 and the total for the other equation is 127.

          Step 2. Write the appropriate coefficients and total for each equation.

          0.2x + 0.5 y = 41
          x + y = 127

          Now we can solve the system of equations; let’s use the elimination approach. Multiply the second equation by -0.5: -0.5x - 0.5 y = -63.5. Then
            0.2x + 0.5y =   41
          -0.5x – 0.5y = -63.5
          --------------------
          -0.3x            = -22.5
                            x = 75               = number of wooden beads

            x + y = 127
          75 + y = 127
                   y = 127 – 75 = 52 = number of crystal beads

          Cost of crystal beads = $0.5 (52)   = $26
          Cost of wooden beads = $0.2 (75) = $15
                                                                      ----
                                                                      $11      The answer is A

      SAT Verbal - Questions from Spring 2022, Week 16 (June 4, 2022)

      SAT QUICK CHALLENGE EXERCISE P21
      Placing Modifiers Correctly

      The Introductory Modifying Phrase (IMP). An introductory modifying phrase (IMP) (1) comes at the beginning of a sentence, (2) is followed by a comma, and (3) describes the subject of the sentence, which comes right after the IMP (or as close as possible to it). Note Sentence A, which follows. Sentence A. Feeling exhausted after the marathon, Carla just wanted to lie down. Note the bold, underlined IMP at the beginning of the sentence, the comma that follows the IMP, and the bold, italicized subject right after the comma -- Carla. The IMP modifies that subject.

      The Modifier That Comes After the Word It Modifies. Unlike the IMP, some modifiers follow the word they modify. At times, a “follower” modifier comes after a comma. At other times, a “follower” modifier is enclosed in parentheses. Some students think that information in parentheses is unimportant and ignore it, but such students taking the SAT have lost points because of that mistake. Do not let that happen to you! Remember that whether it follows a comma or is enclosed in parentheses, a modifier that is placed as close as possible to the word it modifies is always placed correctly. Note Sentences B and C below.

      Sentence B.. Marti says that she wants to be an aesthetician, a specialist trained in skin beautification.
      Sentence C. Marti says that she wants to be an aesthetician (a specialist trained in skin beautification).

      Comments. Although the modifier in Sentence B follows a comma, and the modifier in Sentence C is in parentheses, (1) each modifier modifies the noun that it follows ("aesthetician"), (2) each modifier is in the right place, and (3) both sentences are written correctly.

      Now, keeping in mind the information above, complete the exercise below.

       

      SAT Quick Challenge P21
      Placing Modifiers Correctly

      Directions.  For each statement below, select the letter of the answer choice which places the underlined modifier in the right place. If you think                          that the modifier is already in the right place, select choice A -- NO CHANGE. Then use the answer key to check your work.

      1.  Cassie will have her birthday celebration at Grandma’s Kitchen next Friday (the new restaurant downtown).

      1. NO CHANGE
      2. after "have"
      3. before "celebration"
      4. after "Grandma's Kitchen"

      2.  During homecoming, current and former students, faculty, staff, and other supporters gather to share fantastic old memories and make new           ones a festive occasion featuring great fun, food, and fellowship,

      1. NO CHANGE
      2. before "share"
      3. after "homecoming"
      4. after "supporters"

      3.  Sean says that he will donate his old car to a charity that repairs vehicles and gives them to people who can’t afford to buy their own eager to         help a family that needs transportation, after he purchases his new car.

      1. NO CHANGE
      2. after "Hope"
      3. before "Sean"
      4. after "car"

      SAT Verbal - Answers to Questions from Spring 2022, Week 16 (June 4, 2022)

      1. D
      2. C
      3. B

      SAT Math - Questions fom Spring 2022, Week 15 (May 21, 2022)

      Systems of Linear Equations, Part 5: One Solution No Solution, An Infinite Number of Solutions

      Most of the system of linear equations problem on the SAT will have one solution. Occasionally, however, there will be a problem where there is no solution or there is an infinite number of solutions. For test takers it is important to be able to determine which situation exists. A standard approach to make this determination is to express each equation in slope-intercept form: y = mx + b, where “m” is the slope and “b” is the y-intercept. In other words, solve each equation for y.

      You will also recognize that “y = mx + b” is the formula for a line, and thus there is a line for each of the two equations in a system of linear equations. The slopes and y-intercepts of the two lines determine whether there is one solution, no solution, or an infinite number of solutions to the system of linear equations, as follows:

      One solution.

      If the two lines have different slopes, there will be one solution to the system of equations. Geometrically, the two lines will cross, and the point of intersection (x,y) will be the solution to the system of equations.

      No solution.

      If the two lines have the same slope but different y-intercepts, there will be no solution to the system of equations. Geometrically, the two lines are parallel and will never cross; thus there is no point of intersection, and there is no solution to the system of equations.

      An Infinite Number of Solutions.

      If the two lines have the same slope and the same y-intercept, there will be an infinite number of solutions to the system of equations. Geometrically, the two lines are really the same line, and one is a multiple of the other. Here is an example:

      y = 2x + 7
      2y = 4x + 14          The second equation is a multiple of the first one.

      The coefficients of the variables and y-intercept of one equation are multiples of the coefficients in the other equation. In this case:
      Variable y: 2 in the second equation is a multiple of the 1in the first equation
      Variable x: 4 in the second equation is a multiple of the 2 in the first equation
      y-intercept: 14 is a multiple of 7             The common multiple is 2.

      Keeping in mind the information above, answer the following system of equations questions.

      1. 6x + 2y = 3
        3x + y = 2
        How many solutions (x,y) are there to the system of equations above?

        1. Zero
        2. One
        3. Two
        4. More than two

      2. 8x – 2y = 12
        6y = kx – 42
        In the system of equations above, k represents a constant.  If the system of equations has no solution, what is the value of 2k?
        1. 6
        2. 12
        3. 24
        4. 48
      3.     3x – 2y = a
        -15x + by = 10

        In the system of equations above, a and b are constants. If the system of equations has infinitely many solutions, what is the value of a?

      SAT Math - Answers to Questions from Spring 2022, Week 15 (May 21, 2022)

      1. 6x + 2y = 3
        3x + y = 2
        How many solutions (x,y) are there to the system of equations above?

        1. Zero
        2. One
        3. Two
        4. More than two

        To determine whether there is one solution, no solution, or an infinite number of solutions, express
        each equation in slope-intercept form (solve for y) and compare the slopes and y-intercepts.

        6x + 2y = 3                                           3x +y = 2
        y = -3x + 1.5                                               y = -3x + 2

        The two equations have the same slope (-3) but different y-intercepts (1.5 and 2). Thus there is no solution to the system of equations; the lines are parallel. The answer is A.

        Let’s see what happens if we solve the equations by substitution. Solve the second equation for y:
        y = -3x + 2. Now substitute this value of y in the first equation and solve for x:
        6x + 2(-3x + 2) = 3
        6x – 6x + 4 = 3
        4 = 3                           Of course this is not true; thus there is no solution.

      2. 8x – 2y = 12
        6y = kx – 42
        In the system of equations above, k represents a constant.  If the system of equations has no solution, what is the value of 2k?
        1. 6
        2. 12
        3. 24
        4. 48

        There is no solution if the slopes are equal and the y-intercepts are unequal.  Let’s express each equation in slope-intercept form.There is no solution if the slopes are equal and the y-intercepts are unequal.  Let’s express each equation in slope-intercept form.

        8x – 2y = 12     
              -2y = -8x + 12                                             
                  y = 4x - 6 
                
           The y-intercepts are unequal; now we need to determine whether the slopes are equal.  They are equal if 4 = k/6. 
        Thus we need to solve for k.
        k/6 = 4                  
           k = 24         If the system of equations has no solution, then k must = 24.   The question asks us for 2k.  Thus the answer is 2(24) = 48  -------- D

      3.     3x – 2y = a
        -15x + by = 10

        In the system of equations above, a and b are constants. If the system of equations has infinitely many solutions, what is the value of a?

        If a system of equations has infinitely many solutions, the two lines have the same slope and the same y-intercept. The two lines are really the same line, and one is a multiple of the other.

        There is a common multiple that is applied to each of the two variables (x and y) and to the y-intercept.

        In this case the second equation is a multiple of the first equation. What is the common multiple?

        Look at the coefficients of x, the coefficients of y, and the y-intercepts. Is there a multiple in either of these?
        Yes, there is a multiple in the x coefficients; -15 is a multiple of 3.

        The multiple is -5 since 3(-5) = -15. -5 is the common multiple, and we will use it to determine b and a.

        – 2(-5) = b                                        a(– 5) = 10
        b = 10                                                      a = 10/(– 5) = – 2


        We can check our work by calculating the slope and y-intercept for each equation using the values of a and b that we calculated. The slopes and y-intercepts should be the same for both equations.

        First equation: 3x – 2y = a
        3x – 2y = – 2
             – 2y = – 3x – 2
                   y = 1.5x + 1

        Second equation:             – 15x + by = 10
                                                – 15x + 10y = 10
                                                             10y = 15x + 10
                                                                 y = 1.5x + 1

        The slopes and y-intercepts are the same.

      SAT Verbal - Questions from Spring 2022, Week 15 (May 21, 2022)

      SAT QUICK CHALLENGE EXERCISE O21
      The Relative Pronoun

      Using “When.” Use when to refer to times and events. Note the following: Do you know the dates where homecoming will be celebrated this year? Explanation. That sentence is incorrect. Since it talks about time, we need a word that refers to time – when, not where. Correction. Do you know the dates when homecoming will be celebrated this year?

      Using a Preposition (in, during, to, etc.) and “Which” Instead of “When.” Besides using “when,” you can refer to time by using a preposition that relates to time, followed by the word “which.” Note the following: Do you know the dates during which homecoming will be celebrated this year?

      Using “When.” Use when to refer to times and events. Note the following: Do you know the dates where homecoming will be celebrated this year? Explanation. That sentence is incorrect. Since it talks about time, we need a word that refers to time – when, not where. Correction. Do you know the dates when homecoming will be celebrated this year?

      Using a Preposition (in, during, to, etc.) and “Which” Instead of “When.” Besides using “when,” you can refer to time by using a preposition that relates to time, followed by the word “which.” Note the following: Do you know the dates during which homecoming will be celebrated this year?

      Using Where. Use where to refer to places. Do not use it to refer to times, books, films, etc. Note the following: Disneyland is a place when lots of people like to go during the summer. Explanation. Disneyland is a place, and we need a word that refers to placeswhere, not when. Correction. Disneyland is a place where lots of people like to go during the summer.

      Using a Preposition (in, during, to, etc.) and “Which” Instead of “Where.” Besides using the word“ where,” you can refer to places by using a preposition that relates to places, followed by the word “which.” Note the following: New York is a city in which many people travel by subway every day.

      . Use where to refer to places. Do not use it to refer to times, books, films, etc. Note the following: Disneyland is a place when lots of people like to go during the summer. Explanation. Disneyland is a place, and we need a word that refers to places – where, not when. Correction. Disneyland is a place where lots of people like to go during the summer.

      Using a Preposition (in, during, to, etc.) and “Which” Instead of “Where.” Besides using the word“ where,” you can refer to places by using a preposition that relates to places, followed by the word “which.” Note the following: New York is a city in which many people travel by subway every day.

      Now, keeping in mind the information above, complete the exercise below.

       

      SAT Quick Challenge O21
      The Relative Pronoun

      Directions.  Identify the letter of the answer choice that corrects each sentence below. If you think a statement is already correct, select choice A –                      NO CHANGE. Use the answer key to check your work.

      1.  Angie Jackson’s house is the place when the whole family gathers for dinner after church on Sundays.

      1. NO CHANGE
      2. where
      3. which
      4. during which

      2.  It was four days before Christmas where Matt Lucas’s three-year-old son unwrapped all the presents that were under the Christmas tree.

      1. NO CHANGE
      2. when
      3. which
      4. at which

      3.  2011- 2015 was the period of time during which Carl Joiner was the principal at the local high school.

      1. NO CHANGE
      2. where
      3. at which
      4. to which

      SAT Verbal - Answers to Questions from Spring 2022, Week 15 (May 21, 2022)

      1. B
      2. B
      3. A

      SAT Math - Questions fom Spring 2022, Week 14 (April 30, 2022)

      Systems of Linear Equations, A Short Cut - Part 4

      The traditional method of determining the answer in a system of linear equations problem involves calculating the values of each of the two variables by substitution, by elimination, or by using the answer choices. In some cases it is possible to find the answer without determining the values of each of the two variables. Consider the following example:
      3x + 2y = 8
      x - 4y = 12

      Based on the system of equations above, what is the value of 4x - 2y?
      Let’s solve this problem by substitution.

      Step 1: Solve for x in the second equation.
      x - 4y = 12
      x = 12 + 4y

      Step 2. Substitute this expression of x in the first equation, and solve for y.
      3x + 2y = 8
      3(12 + 4y) + 2y = 8
      36 + 12y + 2y = 8
      14y = 8 - 36
      14y = -28
      y = -2

      Step 3. Substitute this value of y in the second equation, and solve for x.
      x - 4y = 12
      x - 4(-2) = 12
      x + 8 = 12
      x = 4

      Now that we have values for x and y, we can find the value of 4x – 2y:
      4x – 2y = 4(4) – 2(-2) = 16 + 4 = 20

      In this case an alternative for finding our answer is simply to add the two equations:

      3x + 2y = 8
      x - 4y = 12
      ------------
      4x – 2y = 20

      We get 4x – 2y = 20, the same answer we got using the longer traditional approach. Whenever you have a problem similar to this one (where you are asked questions such as find the value of 3x + 5y or 8x + 10y or 3x + 7y), try adding the two equations or subtracting one equation from the other one.

      Keeping in mind the information above, answer the following systems of equations questions.

      1. 2x + 3y = 2
        x – 4y = 12
        Based on the system of equations above, what is the value of  3x – y?

      2. 4x + y = 14
        3x + 2y = 13
        Based on the system of equations above, what is the value of x - y?

      3. 2x + y = 6
        7x + 2y = 27
        Based on the system of equations above, what is the value of 3x + y?

      SAT Math - Answers to Questions from Spring 2022, Week 14 (April 30, 2022)

      1. 2x + 3y = 2
        x – 4y = 12
        Based on the system of equations above, what is the value of  3x – y?

        Solving this system of equations be substitution or by elimination, we find that x = 4 and y = -2; thus 3x – y = 3(4) – (-2) = 12 + 2 = 14

        The short cut, which will work in this case, is to add the two equations together:

        2x + 3y = 2
        x – 4y = 12
        ------------
        3x –y = 14             By adding, we get 3x – y = 14, which is our answer; we did not need to determine
                                      the individual values of x and y.

      2. 4x + y = 14
        3x + 2y = 13
        Based on the system of equations above, what is the value of x - y?

        Try adding the equations:
        4x + y = 14
        3x + 2y = 13
        -------------
        7x + 3y = 27

        Adding did not work. Try subtracting one equation from the other one. Subtract the second equation from the first equation.

        4x + y = 14
        -(3x + 2y = 13)
        ---------------

        4x + y = 14
        -3x - 2y = -13
        x – y = 1                  This is our answer, and again we did not need to determine the individual values
                                        of x and y. (For your information, if you choose to calculate the values of x
                                        and y, x = 3, y = 2.)

      3. 2x + y = 6
        7x + 2y = 27
        Based on the system of equations above, what is the value of 3x + y?

        Try adding the equations:
        2x + y =     6
        7x + 2y = 27
        ___________
        9x + 3y = 33

        This is not our answer, but we observe that 9x + 3y is a multiple of what we are trying to find, 3x + y.

        We can divide the 9x + 3y = 33 by 3, and we get 3x + y = 11. Thus 11 is our answer.
        (For your information, if you choose to calculate the values of x and y, x = 5, y = -4.)

      SAT Verbal - Questions from Spring 2022, Week 14 (April 30, 2022)

      SAT QUICK CHALLENGE EXERCISE N21
      The Relative Pronoun

      You will recall that a clause is a group of words that contains a subject and a verb. An independent clause expresses a complete thought and can stand alone as a complete sentence, but a dependent clause is simply a part of a sentence. It does not express a complete thought and cannot stand alone. A dependent clause that begins with a relative pronoun and modifies a noun in the independent clause of a sentence is called a relative clause. The tips below can help you remember how to use selected relative pronouns correctly (which, that, who, and whom). You might see questions about them when you take the SAT.

      Using That and Which. The word that can modify both things and people. It introduces essential clauses, so no comma should come before it because essential clauses must not be separated from the rest of the sentence. Note Examples A and B, which follow. Example A. Leon is very excited about the letter, that he received from his dad today. Explanation A. This sentence is incorrect because a comma separates the relative pronoun “that” from the rest of the sentence. Note the correction in Example B, which follows. Example B. Leo is excited about the letter that he received from his dad today.

      The word which modifies things only. When which introduces a nonessential clause, that clause must be enclosed in commas, but when it introduces an essential clause, commas must not separate the clause from the rest of the sentence. Note Examples C and D, which follow. Example C. Pecans which are very nutritious are one of my favorite snacks. Explanation C. Example C is incorrect because commas must separate the nonessential clause from the rest of the sentence. Note the correction in Example D. Example D. Pecans, which are very nutritious, are one of my favorite snacks. *CAUTION: Do not use commas for which when that word means which ones, as in the following sentence: Mrs. Lee said to give some of these canned goods to the soup kitchen, but she didn’t say which (ones) to give away. In this situation, the relative clause is quite essential and must not be separated from the rest of the sentence.

      Using Who and Whom. Who and whom both refer to people. Who must be used before a verb, and whom must be used after a preposition (such as about, for, from, on, to, or with). Note Examples E and F, which follow. Example E. Calvin Wilkins, whom coached at Rynolds College for 12 years, is our new coach. Explanation E. Example E is incorrect because who must be used before a verb. Note the correction in Example F, which follows. Example F. Calvin Wilkins, who coached at Reynolds College for 12 years, is our new coach. Finally, note Example G, which follows. Example G. His wife Helena is the person for who Marvin wrote this poem. Explanation G. The word who is incorrect. We need a relative pronoun that (1) refers to people and (2) follows a preposition. Note that Example H fulfills both requirements by using the word whom. Example H. Marvin’s wife Helena is the person for whom he wrote this poem.
      Now, keeping in mind the information above, complete the exercise below.

       

      Practice Exercise O21 - The Relative Pronoun

      Directions.  For each question, write the letter of the answer choice which corrects the underlined part of the sentence. If the underlined part is                            already correct, select choice A (NO CHANGE). Then, use the answer key to check your work.

      1.  Nora Carter is the student which will represent our school in the National Spelling Bee.

      1. NO CHANGE
      2. student, whom will
      3. student that will
      4. student, which will

      2.  The lady, whom lives in the big house on the corner is my math teacher.

      1. NO CHANGE
      2. whose lives
      3. which lives
      4. who lives

      3.  The person to who’s the trophy will be awarded is Marcus Baker.

      1. NO CHANGE
      2. to whom the trophy
      3. to who the trophy
      4. with which the trophy

      SAT Verbal - Answers to Questions from Spring 2022, Week 14 (April 30, 2022)

      1. C
      2. D
      3. B

      SAT Math - Questions fom Spring 2022, Week 13 (April 23, 2022)

      System of Linear Equations - Part 3


      A system of equations can be solved in three ways: by substitution, by elimination, and by using the answer choices. When elimination is used, we find a way to eliminate one of the variables from the calculation. The first step is to multiply one of the equations by a number so that when the two equations are added together, one of the variables falls out. (For example, we might get 6x – 6x = 0 or 8y – 8y = 0.) We then solve for the remaining variable, then use the value of that variable to find the value of the variable that was eliminated.

      Consider the following example:

      3x + 4y = -23
      -x + 2y = -19

      Which ordered pair (x,y) satisfies the system of equations above?

      1. (-5, -2)
      2. (3, -8)
      3. (4, -6)
      4. (9, -6)

      Step 1. To eliminate one of the variables, we could multiply the second equation by 3 so that the x variable would be eliminated (we would get 3x – 3x = 0). Or we could multiply the second equation by -2 so that the y variable could be eliminated (we would get 4y – 4y = 0).

      In this case, multiply the second equation by 3.

      -x + 2y = -19 -----------  -3x + 6y = -57

      Step 2. Add the two equations together and solve for y.

       3x + 4y = -23
      -3x + 6y = -57
      ---------------
                10y = -80 (The x variable is eliminated.)
                     y = -8

      Step 3. Substitute this value of y into either of the two equations and solve for x.
      First equation: 3x + 4y = -23
      3x + 4(-8) = -23
      3x - 32 = -23
      3x = 9
      x = 3

      Second equation: -x + 2y = -19
      -x + 2(-8) = -19
      -x - 16 = -19
      -x = -3
      x = 3

      Thus our answer choice is B: x = 3 and y = -8.


      Keeping in mind the information above, solve the following systems of equations by elimination.

      1.   x + y  =  0                    
        3x - 2y = 10
        Which of the following ordered pairs (x,y) satisfies the system of equations above?

        1. (3, -2)
        2. (2, -2)
        3. (-2, 2)
        4. (-2, -2)
      2.   x + 2y = 7
        2x + 3y = 11
        Which of the following ordered pairs (x,y) satisfies the system of equations above?

        1. (-1, -3)
        2. (-1, 3)
        3. (1, 3)
        4. (1, -3)
      3.  -2x = 4y + 6
        2(2y + 3) = 3x - 5
        Which of the following ordered pairs (x,y) satisfies the system of equations above?

        1. (1, 2)
        2. (1, -2)
        3. (-1, -1)
        4. (-1, 1)

      SAT Math - Answers to Questions from Spring 2022, Week 13 (April 23, 2022)

      1.   x + y  =  0                    
        3x - 2y = 10
        Which of the following ordered pairs (x,y) satisfies the system of equations above?

        1. (3, -2)
        2. (2, -2)
        3. (-2, 2)
        4. (-2, -2)

          Step 1. Multiply the first equation by 2 so that the y variable can be eliminated.
          x + y = 0 ----------- 2x + 2y = 0

          Step 2. Add the two equations together and solve for x.
          2x + 2y = 0
          3x - 2y = 10
          ------------
          5x          = 10
                      x = 2

          Step 3. Substitute this value of x into either of the two equations and solve for y.
          First equation:
          x + y = 0
          2 + y = 0
          y = -2

          Second equation:
          3x - 2y = 10
          3(2) -2y = 10
          6 – 2y = 10
          -2y = 4
          y = -2

          Thus our answer choice is B: x = 2 and y = -2.



      2.   x + 2y = 7
        2x + 3y = 11
        Which of the following ordered pairs (x,y) satisfies the system of equations above?

        1. (-1, -3)
        2. (-1, 3)
        3. (1, 3)
        4. (1, -3)

          Step 1. Multiply the first equation by -2 so that the x variable can be eliminated.
          x + 2y = 7 -----------     -2x - 4y = -14                        

          Step 2. Add the two equations together and solve for x.
          -2x - 4y = -14
          2x + 3y = 11
          ------------------
          -y = -3
          y = 3

          Step 3. Substitute this value of y into either of the two equations and solve for x.
          First equation: x + 2y = 7
          x +2(3) = 7
          x + 6 = 7
          x = 1

          Second equation: 2x +3y = 11
          2x + 3(3) = 11
          2x+ 9 = 11
          2x = 2
          x = 1

          Thus our answer choice is C: x = 1 and y = 3.



      3.  -2x = 4y + 6
        2(2y + 3) = 3x - 5
        Which of the following ordered pairs (x,y) satisfies the system of equations above?

        1. (1, 2)
        2. (1, -2)
        3. (-1, -1)
        4. (-1, 1)

          We first should rewrite both equations in standard form: ax + by = c
          -2x = 4y + 6 2(2y + 3) = 3x - 5
          -2x -4y = 6 4y + 6 = 3x – 5
          -3x + 4y = -11
          Now we have this system of equations:
          -2x -4y = 6
          -3x + 4y = -11

          Step 1. In this case we do not need to multiply either equation by a number since the two equations can be added and the y variable can be eliminated.

          Step 2. Add the two equations together and solve for x.
          -2x - 4y = 6
          -3x + 4y = -11
          ---------------
          -5x         = -5
                      x = 1

          Step 3. Substitute this value of x into either of the two equations and solve for y.
          First equation: -2x -4y = 6
          -2(1) - 4y = 6
          -2 - 4y = 6
          - 4y = 8
          y = -2

          Second equation: -3x +4y = -11
          -3(1) + 4y = -11
          -3 + 4y = -11
          4y = -8
          y = -2

          Thus our answer choice is B: x = 1 and y = -2.

      SAT Verbal - Questions from Spring 2022, Week 13 (April 23, 2022)

      SAT QUICK CHALLENGE EXERCISE M21
      Using the Dash Correctly

      Using the Dash To Set Off Non-Essential Information. You may recall that like the colon, the dash can be used to indicate that a list or an explanation will follow. Similarly, as in the case of the comma, the dash can be used to separate a non-essential phrase or clause from the rest of a sentence. (Re-minder: Non-essential information provides additional insight, but is not needed for expressing the main idea.) Note Example A, which follows. Example A. Jerry’s chorus teacher -- who has a powerful sing-ing voice -- will be a soloist at the State Symphony Orchestra’s annual “Spring Singsation.” Note that since the non-essential information is entirely within the sentence, rather than at the beginning or end, you need two dashes to separate that information from the rest of the sentence, just as you would use two commas if you had chosen to use commas in this situation. However, remember that you must not mix and match commas and dashes by using one of each.

      Using the Dash To Create a Deliberate Pause. Another function of the dash is to indicate a deliberate pause in order to highlight or emphasize a strong point. The pause may sometimes lead to a sense of drama, suspense, or humor. Note Example B, which follows. Example B. Baby Lori likes most vegetables, but not spinach – not one bit!  

      Use only one pair of dashes per sentence if the non-essential addition is completely within the sentence, but a single dash otherwise. Too many dashes can cause your message to become rather choppy, and understanding it could become quite challenging, as in Example C, which follows. Example C. Aunt Alisa – my favorite aunt of all – perhaps except for Aunt Loretta – or maybe Aunt Cindy – will have a birthday next week, and I plan to give her a surprise party on Friday – because she has never had one – and I think she deserves one. Note the improvement in Example D, which follows. Example D. We’d like for you to come to Aunt Alisa’s birthday party next Friday – her very first surprise party, so shhh!

      Now, keeping in mind the information above, complete the exercise below.

      SAT Quick Challenge M21
      using the Dash Correctly

      Directions.   Identify the letter of the answer choice in which the dash is used correctly in each sentence below. If you think a statement is already                       correct, select choice A – NO CHANGE. Use the answer key to check your work.

      1.  With six generations of preachers already in her family, Ada says that she wants to be, anything except a preacher!

      1. NO CHANGE
      2. be anything
      3. be anything --
      4. be -- anything --

      2.  My cousin Joy, a really great dancer – owns several dance studios in New York City.

      1. NO CHANGE
      2. -- a really great dancer --
      3. -- a really great dancer,
      4. -- a really great dancer

      3.  Mr. Dan Carter an amazing volunteer – received the Gold Leaf trophy for his service to our school.

      1. NO CHANGE
      2. -- an amazing volunteer,
      3. an amazing volunteer --
      4. -- an amazing volunteer --

      SAT Verbal - Answers to Questions from Spring 2022, Week 13 (April 23, 2022)

      1. C
      2. B
      3. D

      SAT Math - Questions fom Spring 2022, Week 12 (April 9, 2022)

      Systems of Linear Equations - Part 2

      As noted last week, a system of equations can be solved in three ways: by substitution, by elimination, and by using the answer choices. The alternative of using the answer choices can be used only when the answer choices include  values for both variables. Remember answer choices are not given for 5 of the problems in Section 3 of the SAT and answer choices are not given for 8 of the problems in Section 4. Also in some Problems the question might be “What is the value of x?” or “What is the value of x + y?” In these cases we would have to solve the problem by substitution or by elimination.
      Let’s look at the substitution process.

      Consider the following example:
      3x + 4y = -23
      -x + 2y = -19

      Which ordered pair (x,y) satisfies the system of equations above?

      1. (-5, -2)
      2. (3, -8)
      3. (4, -6)
      4. (9, -6)

      These are the steps for solving by substitution.

      Step 1: In one of the equations solve for one of the variables in terms of the other variable. Let’s solve for x in the second equation.
      3x + 4y = -23
               -x = -19 - 2y
                 x = 19 + 2y

      Step 2. Substitute this expression of x in the first equation, and solve for y.
      3x + 4y = -23
      3(19 + 2y) + 4y = -23
          57 + 6y + 4y = -23
                         10y = -23 - 57
                         10y = -80
                             y = -8

      Step 3. Substitute this value of y in the second equation, and solve for x.
      -x + 2y = -19
      -x + 2(-8) = -19
      -x -16 = -19
              -x = -3
                x = 3                                                  Thus the answer choice is B, x = 3 and y = -8.

      Keeping in mind the information above, solve the following systems of equations by substitution.

      1. x + y = 0
        3x - 2y = 10

        Which of the following ordered pairs (x,y) satisfies the system of equations above?
        1. (3, -2)
        2. (2, -2)
        3. (-2, 2)
        4. (-2, -2)

      2.  x + 2y  =  7 
        2x + 3y =  11

        Based on the system of equations above, what is the value of x?  (No answer choices are given here.)                          
      3. -2x = 4y + 6                                               
        2(2y + 3) = 3x - 5

        If the ordered pair (x, y) that satisfies the system of equations is (x, y) what is the value of (x + y)?

        1. -1
        2. 0
        3. 1
        4. 2

      SAT Math - Answers to Questions from Spring 2022, Week 12 (April 9, 2022)

      1. x + y = 0
        3x - 2y = 10
        Which of the following ordered pairs (x,y) satisfies the system of equations above?
        1. (3, -2)
        2. (2, -2)
        3. (-2, 2)
        4. (-2, -2)

        These are the steps for solving by substitution.

        Step 1: In one of the equations solve for one of the variables in terms of the other variable. Let’s solve for x in the first equation.
        x = -y

        Step 2. Substitute this expression of x in the second equation, and solve for y.
        3x - 2y = 10
        3(-y) - 2y = 10
        -3y -2y = 10
               -5y = 10
                   y = -2

        Step 3. Substitute this value of y in the first equation, and solve for x.
        x + y = 0
         x -2 = 0
               x = 2

        Thus our answer choice is B, x = 2 and y = -2.



      2.  x + 2y  =  7 
        2x + 3y =  11

        Based on the system of equations above, what is the value of x?  (No answer choices are given here.)

        These are the steps for solving by substitution.

        Step 1:  In one of the equations solve for one of the variables in terms of the other variable.  Let’s solve for x in the first equation.
                      x = -2y + 7 

        Step 2. Substitute this expression of x in the second equation, and solve for y.
        2x + 3y = 11
        2(-2y + 7) + 3y = 11
        -4y + 3y = 11 -14
                  -y = -3
                    y = 3

        Step 3. Substitute this value of y in the first equation, and solve for x.

          x + 2y = 7
        x + 2(3) = 7
            x + 6 = 7
                  x = 1                                                                     Thus, our answer is: x = 1

        1. -1
        2. 0
        3. 1
        4. 2
      3. We first should rewrite both equations in standard form: ax + by = c
        -2x = 4y + 6                                                2(2y + 3) = 3x - 5
        -2x -4y = 6                                                 4y + 6 = 3x – 5
                                                                        -3x + 4y = -11
        Now we have this system of equations:
         -2x -4y = 6
        -3x + 4y = -11

        These are the steps for solving by substitution.

        Step 1:
        In one of the equations solve for one of the variables in terms of the other variable. Let’s solve for x in the first equation.
        -2x -4y = 6
               -2x = 4y + 6
                   x = -2y -3

        Step 2. Substitute this expression of x in the second equation, and solve for y.
        -3x + 4y = 6
        -3(-2y -3) + 4y = -11
                6y +9 + 4y = -11
                           10y = -20
                                y = -2

        Step 3. Substitute this value of y in the first equation, and solve for x.
        -2x -4y = 6
        -2x -4(-2) = 6
        -2x +8 = 6
             -2x = -2
                 x = 1

        Now we have values for x and y: x = 1 and y = -2.
        Thus x + y = 1 + (-2) = -1, and our answer choice is A.

      SAT Verbal - Questions from Spring 2022, Week 12 (April 9, 2022)

      SAT QUICK CHALLENGE EXERCISE L21
      The Colon and the Dash

      The Colon. One role of the colon is to introduce a list, but what comes before the colon must be a complete sentence. What comes after the colon can be either a sentence or a fragment. Do not capitalize the first word after a colon unless there is a grammatical reason to capitalize it (if it is a proper noun or the first word of a complete sentence).

      Avoid answers with “including” or “such as” before a colon; such wording will not be a complete sentence. Note examples A1 and A2, which follow. Example A1. Ella makes yummy desserts, including: cakes, pies, cookies, and even candy. (Incorrect) Example A2. Ella makes yummy desserts: cakes, pies, cookies, and even candy. (Correct)

      The colon can also be used just before an explanation, and a complete sentence typically follows that colon. Since a colon used this way does the same job as a semicolon, a period, or a dash, the SAT generally will not ask you to select from among those identical choices. However, if you are faced with that situation, the colon will be correct only if the sentence following the colon explains what was stated before the colon. Note Examples B1 and B2, which follow. Example B1: We will watch the NCAA championship game on my friend’s big-screen TV: that game is always exciting. (Incorrect because the sentence after the colon does not explain the decision about watching the game. Example B2: We will watch the NCAA championship game on my friend’s big-screen TV: we’ll save lots of time and money, and we’ll still have a great time. (Correct because the sentence after the colon does explain the decision to watch the game on the friend’s big-screen TV.)

      The Dash. Like the colon, the dash can be used before a list or an explanation. Examples C1 and C2, below, show how to use the dash to write correctly the same sentences written correctly with colons in Examples A2 and B2 above. Example C1. Ella made some delicious desserts -- cakes, pies, cookies, and candy. Example C2. We will watch the NCAA championship game on my friend’s big-screen TV – We’ll save lots of time and money, and we’ll still have a great time.

      Now, keeping in mind the information above, complete the exercise below.

       

      SAT Quick Challenge L21
      The Colon and the Dash

      Directions.  Identify the letter of the answer choice that corrects each sentence below. If you think a statement is already correct, select choice A  - NO CHANGE. Use the answer key to check your work.

      1.  Mom said that Kevin can’t keep the little kitten who followed him home: it is really adorable.

      1. NO CHANGE
      2. home: he loves that little animal.
      3. home: she has cat allergies
      4. home, he is very angry with her.

      2.  I liked the food at the restaurant, but I couldn’t drink the coffee, it tasted very bitter.

      1. NO CHANGE
      2. coffee - it tasted
      3. coffee. it tasted
      4. coffee it tasted

      3.  My neighbor Rita often volunteers to help senior citizens in various ways such as: mowing grass, shoveling snow, delivering meals on wheels, and helping to build wheelchair ramps.

      1. NO CHANGE
      2. ways such as,: mowing
      3. ways, such as: mowing
      4. ways: mowing

      SAT Verbal - Answers to Questions from Spring 2022, Week 12 (April 9, 2022)

      1. C
      2. B
      3. D

      SAT Math - Questions fom Spring 2022, Week 11 (April 2, 2022)

      Systems of Linear Equations - Part 1


      One of the types of questions you will see on the SAT is solving systems of linear equations, sometimes called simultaneous equations or two equations in two unknowns.  A system of equations can be solved in three ways: by substitution, by elimination, and by using the answer choices (when answer choices are given).  When the alternative of using the answer choices is used, we substitute an answer choice in each of the two equations and determine which one works.  The correct answer choice must work in both equations.  Consider the following example:

      3x + 4y = -23
      -x + 2y  =  -19

      Which ordered pair (x,y) satisfies the system of equations above?

      1. (-5, -2)
      2. (3, -8)
      3. (4, -6)
      4. (9, -6)  

      Try choice A):  x = -5 and y = -2
      First equation: 
       
      3x +  4y  =  -23                                                                                                      3(-5) + 4(-2)  =  -23                           
      -15  -8  =  -23           YES            

      Second equation:   
      -x +  2y  =  -19                                                                                                     
      -(-5) + 2(-2)  =  -19                             
      5  -4  = 1, not -19         NO; choice A does not work; it needs to work in both equations.

      Now try choice B):  x = 3 and y = -8
      First equation:   
      3x +  4y  =  -23                                                                                                                                                                                                    
      3(3) + 4(-8)  =  -23                             
      9  -32  =  -23           YES            

      Second equation:   
      -x +  2y  =  -19                                                                                                                                                                                                      
      -(3) + 2(-8)  =  -19                               
      -3  -16  =  -19        YES           
      Thus our answer is B; these numbers work for both equations. Answer choices C and D do not work..

      Keeping in mind the information above, solve the following systems of equations by using the answer choices.

      1. x + y = 0
        3x - 2y = 10

        Which of the following ordered pairs (x,y) satisfies the system of equations above?

        1. (3, -2)
        2. (2, -2)
        3. (-2, 2)
        4. (-2, -2)


      2. x + 2y  =  7
        2x + 3y =  11
        Which of the following ordered pairs (x,y) satisfies the system of equations above?
        1. (-1, -3)
        2. (-1, 3)
        3. (1, 3)
        4. (1, -3)
      3. -2x = 4y + 6
        -2x = 4y + 6                           
        2(2y + 3) = 3x - 5

        Which of the following ordered pairs (x,y) satisfies the system of equations above?
        1. (1, 2)
        2. (1, -2)
        3. (-1, -1)
        4. (-1, 1)

      SAT Math - Answers to Questions from Spring 2022, Week 11 (April 2, 2022)

      1. x + y = 0
        3x - 2y = 10

        Which of the following ordered pairs (x,y) satisfies the system of equations above?

        1. (3, -2)
        2. (2, -2)
        3. (-2, 2)
        4. (-2, -2)

        We will substitute an answer choice in each of the two equations.

        Try choice A):  x = 3 and y = -2

        First equation:   
        x +  y  =  0             3 -2 = 1, not 0         NO; choice A does not work.                                                                                                                             
        Now try choice B):  x = 2 and y = -2
        First equation:    x +  y  =  0                          2  -2  =  0          YES             
        Second equation:    3x -  2y  =  10        3(2) -2(-2)  =  6 + 4 = 10   YES   
        Thus our answer is B; answer choices C and D do not work.

      2. x + 2y  =  7
        2x + 3y =  11
        Which of the following ordered pairs (x,y) satisfies the system of equations above?
        1. (-1, -3)
        2. (-1, 3)
        3. (1, 3)
        4. (1, -3)

        Try choice A):  x = -1 and y = -3
        First equation:
           
        x +  2y  =  7             
        -1 +2(-3) = -1 -6  =  -7, not 7         
        NO; choice A does not work                                                                                                                                                                                             

        Try choice B):   x = -1 and y = 3
        First equation:
           
        x +  2y  =  7             
        -1 +2(3) = -1 +6  =  5, not 7          NO; choice B does not work

        Try choice C):   x = 1 and y = 3
        First equation:
           
        x +  2y  =  7             
        1 +2(3) = 1 +6  =  7    YES
        Second equation:   
        2x +  3y  =  11      2(1)  +  3(3)  =  2 + 9 = 11   YES                                      Thus our answer is C; answer choice D does not work.

      3. -2x = 4y + 6
        -2x = 4y + 6                           
        2(2y + 3) = 3x - 5

        Which of the following ordered pairs (x,y) satisfies the system of equations above?
        1. (1, 2)
        2. (1, -2)
        3. (-1, -1)
        4. (-1, 1)

        Try choice A): x = 1 and y = 2
        First equation:
        -2x = 4y + 6
        -2x = -2(1) = -2 and 4y + 6 = 4(2) +6 = 8 + 6 = 14, not -2 NO; choice A does not work

        Now try choice B): x = 1 and y = -2
        First equation:
        -2x = 4y + 6 -2x = -2(1) = -2 and 4y + 6 = 4(-2) + 6 = -8 + 6 = -2 YES
        Second equation: 2(2y + 3) = 3x -5 2(2y + 3) = 2(2(-2) + 3) = 2(-4 + 3) = 2(-1) = -2 and 3x – 5 = 3(1) – 5 = 3 – 5 = -2 YES
        Thus our answer is B; answer choices C and D do not work.

      SAT Verbal - Questions from Spring 2022, Week 11 (April 2, 2022)

      SAT QUICK CHALLENGE EXERCISE K21
      Transitions Within a Sentence Rather Than at the Beginning

      The Relationship Between Two Sentences. An SAT question about an underlined transition enclosed in commas and located within a sentence tests your ability to identify the relationship between the underlined sentence and the sentence it follows. (For a quick refresher about selecting the right kind of transition, review the information on the TRANSITIONS MINI REVIEW CHART beneath the practice exercise below. Then, note Example A below.

      Example A.  (1) Dr. Hall has enjoyed snacking on peanuts since she was a child.  (2) As an adult, for example, she gets a severe itch when she eats them.  In Sentence 2, “for example” says that the “severe itch” is an example of the enjoyment mentioned in Sentence 1.  However, we need a contrast transition that shows the contrast between “enjoyment” and a “severe itch.”  The word “however” in Example B, which follows, provides that contrast.  Example B.  Dr. Hall has enjoyed snacking on peanuts since she was a child.  As an adult, however, she gets a severe itch when she eats them. 

      Combining Two Sentences. Sometimes an SAT transition question will ask you to identify the best way to combine two complete sentences. Note Example C, which follows. Example C. Directions. Select the best way to create a single sentence from the two sentences which follow. Combine the sentences at the underlined section where the two sentences meet.

      During the COVID pandemic, many workers with strong computer skills have been able to work from home. Workers who can not use the computer to do their jobs have had to keep going to the office.

      A. home, for, workers   B. home; and workers     C. home, whereas workers*   D. home moreover workers

      Explanation of Example C: Since the two sentences express opposite ideas, a contrast transition is required. Therefore, choice C is correct. *Merging two sentences into one usually means placing a semicolon between the two (and starting the second one with a lower case letter), but whenwhereasbegins the second sentence, a comma must be placed beforewhereas,” and no punctuation should follow it. Now, keeping in mind the information in this lesson, complete the exercise below.

       

      SAT Quick Challenge K21
      Transitions Within a Sentence Rather Than at the Beginning

      Directions.   Identify the letter of the answer choice that corrects each sentence below. If you think a statement is already correct,
                           select Choice A -- NO CHANGE. After completing this activity, use the answer key to check your work.

      1.  Forecasters said that stormy weather is heading this way. Consequently, we have to go to the pharmacy and get Grandma’s medicine.

      1. NO CHANGE
      2. Nevertheless
      3. Furthermore
      4. Since

      2.  Forecasters said that stormy weather is heading this way. Consequently, we have to go to the pharmacy and get Grandma’s medicine.

      1. NO CHANGE
      2. Rather
      3. Accordingly,
      4. However,

      3.  Select the best way to create a single sentence from the two sentences which follow. Combine the sentences at the underlined section where
           they meet.

            We must get our income tax forms filed on time. Otherwise, we will have to pay a late fee.

      1. on time, otherwise
      2. on time, otherwise;
      3. on time; otherwise,
      4. on time, otherwise,

      SAT Verbal - Answers to Questions from Spring 2022, Week 11 (April 2, 2022)

      1. B
      2. C
      3. C

      SAT Math - Questions from Spring 2022, Week 10 (March 26, 2022)

      Equations of a Line Word Problems

      SAT math often tests equation of a line concepts in the form of word problems. The questions will include phrases such as “at a constant rate” or “a flat fee.” In most cases you will create an equation in the form of y= mx + b, or interpret the components of an equation. The problems will usually start with a certain amount (a flat fee – meaning the y intercept) and increasing or decreasing at a constant rate (meaning the slope).

      Now examine examples 1, 2, and 3.

        1. The average number of students per classroom at Eastern High School from 2008 to 2021 can be modeled by the equation y = 0.38x + 24.5, where x represents the number of years since 2008, and y represents the average number of students per classroom. Which of the following best describes the meaning of the number 0.38 in the equation?

            1. The total number of students at the school in 2008
            2. The average number of students per classroom in 2008
            3. The estimated increase in the average number of students per classroom each year
            4. The estimated difference between the average number of students per classroom in 2021 and 2008

              The slope is 0.38, which means that for an increase of 1 in x, there is an increase of 0.38 in y, which is the average number of students per classroom. Thus the answer is C.

        2. The equation y = 21.50 + 1.75x models the total cost y, in dollars, that a company charges a customer to rent a truck for one day and drive the truck x miles. The total cost consists of a flat fee plus a charge per mile driven. When the equation is graphed in the xy-plane, what does the y-intercept of the graph represent in terms of the model?
          1. A flat fee of $21.50
          2. A charge per mile of $1.75
          3. A charge per mile of $21.50
          4. Total daily charge of $23.25

            The y-intercept is the value of y when x is zero. In this problem it is the flat fee to which a charge per mile driven is added. Thus the answer is A.

        3. To strengthen his abdominal muscles, Jerry has a 12 day plan to increase the number of sit-ups he can do each day. His plan calls for him to do 65 sit-ups by day 12 by increasing the number of sit-ups each day by a constant amount. If Jerry can do 20 sit-ups by day 3 and successfully complete his plan by day 12, how many sit-ups could he do before his plan was put into effect?
          1. 0
          2. 2
          3. 5
          4. 7


      In terms of y = mx + b, we are asked to find the y-intercept, b. We can find b if we have values for y, m, and x; we need the equation of a line.

      We have what can be considered to be coordinate points, (x1,y1) and (x2,y2): day 12, 65 sit-ups and day 3, 20 sit-ups, (12,65) and (3,20). We can use these 3 steps to solve the problem:

        1. Identify the coordinate points: (12,65) and (3,20).
        2. Calculate the slope. The slope of a line can be determined from any two points on the line by using the slope formula: slope = (y2 - y1)/(x2 - x1) Slope = (65 – 20)/(12 – 3) = 45/9 = 5
        3. Set up the equation of a line: y = mx + b. The slope, m, = 5. We can use either point for x and y, to solve for b. Using the point (12,65), we have 65 = 5(12) + b
          65 – 60 = b = 5

          Using the other point (3, 20), we have 20 = 5(3) + b
          20 – 15 = b = 5

          Thus, Jerry could do 5 sit-ups before he put his plan into effect. The answer is C.

          Keeping in mind the information above, solve the following problems.

        1. A start-up company opened with 8 employees. The company’s growth plan assumes that 2 new employees will be hired each quarter (every three months) for the first 5 years. If an equation is written in the form of y = ax + b to represent the number of employees, y, employed by the company, what is the value of b?

          1. The approximate femur length, in inches, of a man with a height of 32.01 inches
          2. The approximate increase in a man’s femur length, in inches, for each increase of 32.01 inches in his height
          3. The approximate increase in a man’s femur length, in inches, for each one-inch increase in his height
          4. The approximate increase in a man’s height, in inches, for each one-inch increase in his femur length.The formula H = 1.88L + 32.01 can be used to approximate the height, H, in inches, of an adult male based on the length L, in inches, of his femur. What is the meaning of 1.88 in this context? 
        2. Mr. Rollins is purchasing materials and renting tools needed for the concrete patio he is building in his back yard. The total cost, y, for buying the materials and renting the tools in terms of the number of days, x, is given by the equation y = M + (W + C)x where M is the materials cost in dollars, W is the rental cost of a wheelbarrow in dollars per day, and C is the rental cost of a concrete mixer in dollars per day. If this equation is graphed in the xy-plane, what does the slope of the line represent? 

          1. The total cost of the project
          2. The total cost of the materials
          3. The total daily cost of the project
          4. The total daily rental costs of the tools



        3. The equation C = 75h + 125 gives the amount C, in dollars, an electrician charges for a job that takes h hours. Mrs. Cox and Mr. Lee each hired this electrician. The electrician worked two hours longer on Mr. Lee’s job than on Mrs. Cox’s job. How much more did the electrician charge Mr. Lee than Mrs. Cox?

          1. $75
          2. $125
          3. $150
          4. $200


        4. A company purchased a machine valued at $120,000. The value of the machine depreciates by the same amount each year so that after 10 years the value will be $30,000. Which of the following equations gives the value, v, of the machine, in dollars, t years after it was purchased, where t is between 0 and 10 years?

          1. v = 30,000 – 9,000t
          2. v = 120,000 – 9,000t
          3. v = 120,000 + 9,000t
          4. v = 120,000 – 30,000t


        5. A copy machine is initially loaded with 5,000 sheets of paper. The machine starts a large job and copies at a constant rate. After 20 minutes it has used 30% of the paper. Which of the following equations models the number of sheets of paper, p, remaining in the machine m minutes after the machine started printing?

          1. p = 5,000 – 20m
          2. p = 5,000 – 75m
          3. p = 5,000 – 0.3m
          4. p = 5,000 – 30m


        6. The quantity of a product supplied in an economic market is a function of the price of the product. The estimated supply function is S = (1/2)P + 40 where S is the quantity of the product supplied to the market when the price is P dollars. How will the quantity supplied change if the price of the product is increased by $10?

          1. The quantity supplied will increase by 5 units
          2. The quantity supplied will decrease by 5 units
          3. The quantity supplied will increase by 10 units
          4. The quantity supplied will increase by 50 units


        SAT Math - Answers to Questions from Spring 2022, Week 10 (March 26, 2022)

        1. A start-up company opened with 8 employees. The company’s growth plan assumes that 2 new employees will be hired each quarter (every three months) for the first 5 years. If an equation is written in the form of y = ax + b to represent the number of employees, y, employed by the company, what is the value of b?

          b is the y-intercept, the amount we start with. Thus b is 8.

        2. The formula H = 1.88L + 32.01 can be used to approximate the height, H, in inches, of an adult male based on the length L, in inches, of his femur. What is the meaning of 1.88 in this context? 
          1. The approximate femur length, in inches, of a man with a height of 32.01 inches
          2. The approximate increase in a man’s femur length, in inches, for each increase of 32.01 inches in his height
          3. The approximate increase in a man’s femur length, in inches, for each one-inch increase in his height
          4. The approximate increase in a man’s height, in inches, for each one-inch increase in his femur length.

          1.88 is the slope; it tells us what happens to H for a one unit increase in L. The answer is D.
        3. Mr. Rollins is purchasing materials and renting tools needed for the concrete patio he is building in his back yard. The total cost, y, for buying the materials and renting the tools in terms of the number of days, x, is given by the equation y = M + (W + C)x where M is the materials cost in dollars, W is the rental cost of a wheelbarrow in dollars per day, and C is the rental cost of a concrete mixer in dollars per day. If this equation is graphed in the xy-plane, what does the slope of the line represent? 

          1. The total cost of the project
          2. The total cost of the materials
          3. The total daily cost of the project
          4. The total daily rental costs of the tools

          The slope is the total daily rental costs, not the total daily costs. The answer is D.

        4. The equation C = 75h + 125 gives the amount C, in dollars, an electrician charges for a job that takes h hours. Mrs. Cox and Mr. Lee each hired this electrician. The electrician worked two hours longer on Mr. Lee’s job than on Mrs. Cox’s job. How much more did the electrician charge Mr. Lee than Mrs. Cox?
          1. $75
          2. $125
          3. $150
          4. $200

          The slope is $75; if h increases by 1, C increases by $75. If h increases by 2, C increases by 2($75) or $150. The answer is C.

        5. A company purchased a machine valued at $120,000. The value of the machine depreciates by the same amount each year so that after 10 years the value will be $30,000. Which of the following equations gives the value, v, of the machine, in dollars, t years after it was purchased, where t is between 0 and 10 years?

          1. v = 30,000 – 9,000t
          2. v = 120,000 – 9,000t
          3. v = 120,000 + 9,000t
          4. v = 120,000 – 30,000t

          We need to determine the amount the machine depreciates each year.

          The amount of the decrease each year = ($120,000 - $30,000)/10 = $90,000/10 = $9,000. The slope is -$9,000; thus the answer is B since the value decreases each year.

        6. A copy machine is initially loaded with 5,000 sheets of paper. The machine starts a large job and copies at a constant rate. After 20 minutes it has used 30% of the paper. Which of the following equations models the number of sheets of paper, p, remaining in the machine m minutes after the machine started printing?

          1. p = 5,000 – 20m
          2. p = 5,000 – 75m
          3. p = 5,000 – 0.3m
          4. p = 5,000 – 30m

          We need to determine how much is used each minute. 30% of the total amount is used in 20 minutes. 30% of 5,000 = 0.3(5,000) = 1,500 Thus 1,500 sheets are used in 20 minutes. The number of sheets used per minute = 1,500/20 = 75. Thus the number of sheets remaining equals the 5,000 we started with minus the number of sheets used in m minutes after the machine started printing : p = 5,000 – 75m. the answer is B.

        7. The quantity of a product supplied in an economic market is a function of the price of the product. The estimated supply function is S = (1/2)P + 40 where S is the quantity of the product supplied to the market when the price is P dollars. How will the quantity supplied change if the price of the product is increased by $10?

          1. The quantity supplied will increase by 5 units
          2. The quantity supplied will decrease by 5 units
          3. The quantity supplied will increase by 10 units
          4. The quantity supplied will increase by 50 units

          The slope is ½. If the price increases by $1, the quantity supplied increases by ½ unit.
          If the price increases by $2, the quantity supplied increases by 1 unit.
          If the price increases by $10, the quantity supplied increases by 5 units.
          Thus the answer is A.

        SAT Verbal - Questions from Spring 2022, Week 10 (March 26, 2022)

        SAT QUICK CHALLENGE EXERCISE J21
        Parallel Structures with Multiple Sentences

        SAT parallelism questions often ask you to identify or create parallel structure within a single sentence, but some questions ask you to do that job when more than one sentence is involved. For those questions, the correct answer is always the choice that has the same pattern as the sentences which the underlined sentence comes after. Note Example A, which follows. Example A. Tell which choice best maintains the sentence pattern already established in the following paragraph.
        Freezing rain made driving hazardous. Ice storms caused schools and businesses to close. It was heavy, wet snowflakes that turned the grass white.  

        1. NO CHANGE
        2. The grass was turned white by heavy, wet snowflakes.
        3. Heavy, wet snowflakes turned the grass white.
        4. The white grass was caused by heavy, wet snowflakes. 

        Explanation of Example A. The first two sentences in the paragraph both begin with the adjective, noun, active (not passive) verb pattern. In fact, each of the first two sentences begins with an adjective phrase that is followed by an active verb, and then the rest of the sentence. The correct answer must follow that same pattern, and the correct answer is choice C.

        Now, keeping in mind the information above, complete Exercise J21 below. Then use the answer key to check your work.

         

        SAT Quick Challenge Exercise J21
        Parallel Structures with Multiple Sentences

        Directions.   For each question below, indicate which answer choice follows the pattern of the first two statements. If the underlined statement
                             already follows the pattern of the first two sentences, mark choice “A” as your answer.

        1.  Displaying a Bunsen burner, Lorie symbolizes scientists. Carrying a stethoscope, Rachel stands for doctors. The chalk and eraser that Ricky
             holds are the things that represent teachers

        1. NO CHANGE
        2. Teachers are represented by Ricky’s chalk and eraser.
        3. Holding chalk and an eraser, Ricky represents teachers.
        4. Ricky’s chalk and eraser are what represents teachers.

        2.  Prolonged drought destroyed our crops. Heavy flooding caused roads to be closed. What woke us up is booming thunder.  

        1. NO CHANGE
        2. Booming thunder woke us up.
        3. Booming thunder is what woke us up.
        4. Hearing booming thunder is what woke us up.

        3.   Teresa will do her homework after she finishes her piano lesson. Ron will wash the dishes after we have lunch. After we put the groceries away
              is when Mom will cook dinner.

        1. NO CHANGE
        2. When Mom will cook dinner is after we put away the groceries.
        3. After the groceries have been put away is when Mom will cook dinner.
        4. Mom will cook dinner after we put the groceries away.

        SAT Verbal - Answers to Questions from Spring 2022, Week 10 (March 26, 2022)

        1. C
        2. B
        3. D

        SAT Math - Questions fom Spring 2022, Week 9 (March 19, 2022)

        Equation of a Line


        These are some points to remember about the equation of a line..:

        1. The equation of a line is y = mx + b, where m is the slope of the line and b is the y-intercept.

        2. The slope of a line shows the direction and steepness of the line. The slope can be positive or negative. If the slope is positive, the line is upward sloping. If the slope is negative, the line is downward sloping.

          The line is more vertical if the absolute value of the slope is large, and the line is more horizontal if
          the absolute value of the slope is small; at the limit, the slope of a horizontal line is zero.

        3. The slope tells us what change takes place in y for a one unit increase is x. Consider the equation:
          y = 4x + 3 (the slope = 4)
          If x = 2, then y = 4(2) + 3 = 8 + 3 = 11
          If x increases by 1, then y increases by 4: y = 4(3) + 3 = 12 + 3 = 15 (an increase of 4)
          (The line is upward sloping; when x increases, y also increases.)

          Consider this equation: y = -3x + 6 (the slope = -3)
          If x = 4, then y = -3(4) + 6 = -12 + 6 = -6
          If x increases by 1, then y decreases by 3: y = -3(5) + 6 = -15 + 6 = - 9 (a decrease of 3)
          (The line is downward sloping; when x increases, y decreases.)

        4. The slope of a line can be calculated from any two points on the line. The slope formula is:
          m = (y2 – y1)/(x2 – x1) where the two points are (x1,y1) and (x2,y2).
          If the two points are (3,6) and (5,10), the slope = (10 – 6)/(5 – 3) = 4/2 = 2
        5. .The y-intercept is the value of y when the line crosses the y axis; it is the value of y when x = 0. In SAT problems, the y-intercept represents a base or starting amount before another amount is added. For example, you may be renting a car and the charge is a base amount plus an additional amount based on the number of miles driven.

        6. There is also an x-intercept. The x-intercept is the value of x when the line crosses the x axis; it is the value of x when y = 0.

        7. If a particular point is on a line, the equation of the line works out when that point is plugged into the equation for x and y. For example, to determine whether the point (2,10) is on the line y = 3x + 4, we substitute the x value, 2, into the equation, and we should get the y value, 10. In this case, the point (2,10) is on the line y = 3x + 4 since 3(2) + 4 = 6 + 4 = 10.
          But the point (4,5) is not on the line y = 3x + 4 since 3(4) + 4 = 12 + 4 = 16, not 5.
        8. Parallel lines: two lines are parallel if their slopes are the same.

        9. Perpendicular lines: two lines are perpendicular if the slope of one line is the negative reciprocal of the slope of the other line. For example, if the slope of line a is 2, the slope of a line perpendicular to line a is -1/2. If the slope of line j -4/5, the slope of a line perpendicular to line j is 5/4.

        10. If we are given the slope of a line and a point on that line, we can determine the equation of the line. For example, if the slope of a line = 2 and a point on that line is (4,11), we can proceed as follows to determine the equation of the line:

          Write the equation of a line:                    y = mx + b
          From the point (4,11), we have values for x and y; we need to find a value for b; lets call it h,
          and then solve for h:                                y = mx + b
                                                                          11 = 2(4) + h now solve for h
                                                                          11 = 8 + h
                                                                            3 = h
          Thus the equation of the line is y = 2x + 3
          We can check our results by plugging in the x and y values from the point (4,11) into our equation:
          y = mx + b
          y = 2(4) + 3 = 8 + 3 = 11

          We can now find other points on the line. For example, if x = 5, y = mx + b = 2(5) + 3 = 10 + 3 = 13. Thus the point (5, 13) is on the line.

        Keeping in mind the information above, solve the following problems.

        1. Which of the following equations represents a line that is parallel to the line with equation y = -3x + 4?

          1. 6x + 2y = 15
          2. 3x – y = 7
          3. 2x – 3y = 6
          4. x + 3y = 1

        2. The line with the equation (4/5)x + (1/3)y = 1 is graphed in the x,y-plane. What is the x coordinate of the x-intercept of the line?  

        3. Which of the following is the equation of a line that is perpendicular to y + 3 = 3x -8?

          1. y + 3x = 9
          2. y – 3x = 10
          3. 9y -6x = 18
          4. 3y + x = 27
        4. If a line has a slope of -2 and it passes through the point (-3,2), what is its y-intercept?

          1. 6
          2. 0
          3. -4
          4. -6
        5. The line y = mx + 7, where m is a constant, is graphed in the xy-plane. If the line contains the point (2,13), what is the slope of the line?

          1. 2
          2. 3
          3. 5
          4. 7

        6. The line y = kx + 4, where k is a constant, is graphed in the xy-plane. If the line contains the point (c,d), where c ≠ 0 and d ≠ 0, what is the slope of the line in terms of c and d?

          1. (d – 4)/c
          2. (c – 4)/d
          3. (4 – d)/c
          4. (4 – c)/d

        7. A line has intercepts at (a,0) and (0,b) in the x,y-plane. If a + b = 0 and a ≠ b, which of the following is true about the slope of the line?

          1. It is positive.
          2. It is negative.
          3. It equals 0.
          4. It is undefined.

        SAT Math - Answers to Questions from Spring 2022, Week 9 (March 19, 2022)

        1. The l

            1. Which of the following equations represents a line that is parallel to the line with equation y = -3x + 4?

                1. 6x + 2y = 15
                2. 3x – y = 7
                3. 2x – 3y = 6
                4. x + 3y = 1

              The slope of y = -3x + 4 is -3. The line with a slope of -3 is parallel to this line. Check each
              answer choice; solve for y and determine which one has a slope of -3.

              1. 6x + 2y = 15
                2y = -6x + 15
                y = -3x + 15/2                 Yes
              2. 3x – y = 7
                – y = -3x + 7
                y = 3x – 7                         No

              3. 2x – 3y = 6
                – 3y = -2x + 6
                y = (2/3)x – 2                    No

              4. x + 3y = 1
                3y = -x + 1
                y = -(1/3)x + 1/3               No The answer is A.
            2. The line with the equation (4/5)x + (1/3)y = 1 is graphed in the x,y-plane. What is the x coordinate of the x-intercept of the line?  

              The x-intercept is the value of x when y = 0. Let y = 0 in the equation and solve for x:

              (4/5)x + (1/3)y = 1
              (4/5)x + (1/3)(0) = 1
              (4/5)x = 1
              x = 5/4

            3. Which of the following is the equation of a line that is perpendicular to y + 3 = 3x -8?

              1. y + 3x = 9
              2. y – 3x = 10
              3. 9y -6x = 18
              4. 3y + x = 27

              Solve the equation, y + 3 = 3x -8, for y: y = 3x -11

              The slope of y = 3x -11 is 3. The slope of a line that is perpendicular to this line is the negative reciprocal of 3, which is -1/3. Check each answer choice; solve for y and determine which one has a slope of -1/3.

              1. y + 3x = 9
                y = - 3x + 9                           No

              2. y – 3x = 10
                y = 3x + 10                            No

              3. 9y -6x = 18
                9y = 6x + 18
                y = (6/9)x + 2
                y = (2/3)x + 2                           No

              4. 3y + x = 27
                3y = - x + 27
                y = -(1/3)x + 9                          Yes The answer is D
            4. If a line has a slope of -2 and it passes through the point (-3,2), what is its y-intercept?

              1. 6
              2. 0
              3. -4
              4. -6

              The y-intercept is the value of y when x = 0. Let x = 0 in the equation and solve for y. But we must first determine the equation.

              Write the equation of a line: y = mx + b
              From the point (-3,2), we have values for x and y; we need to find a value for b; let's call it k,
              and then solve for k:                          y = mx + b
                                                                           2 = -2(-3) + h                    Now solve for h
                                                                           2 = 6 + h
                                                                         -4 = h
              Thus the equation of the line is y = -2x -4
              Now let x = 0 in the equation and solve for y: y = -2(0) -4
              y = -4                       The answer is C.

            5. The line y = mx + 7, where m is a constant, is graphed in the xy-plane. If the line contains the point (2,13), what is the slope of the line?

              1. 2
              2. 3
              3. 5
              4. 7

              Substitute the values of the point, x = 2 and y = 13, into the equation and solve for the slope, m.

              y = mx + 7
              13 = 2m + 7
              6 = 2m
              m = 3                        The answer is B.                        

            6. The line y = kx + 4, where k is a constant, is graphed in the xy-plane. If the line contains the point (c,d), where c ≠ 0 and d ≠ 0, what is the slope of the line in terms of c and d?

              1. (d – 4)/c
              2. (c – 4)/d
              3. (4 – d)/c
              4. (4 – c)/d

              We solve this problem the same way we solved number 5:
              Substitute the values of the point, c = x and d = y, into the equation and solve for the slope, k.

              y = kx + 4
              d = kc + 4
              d – 4 = kc
              (d – 4)/c = k              The answer is A.

            7. A line has intercepts at (a,0) and (0,b) in the x,y-plane. If a + b = 0 and a ≠ b, which of the following is true about the slope of the line?

              1. It is positive.
              2. It is negative.
              3. It equals 0.
              4. It is undefined.

                The slope = m = (y2 – y1)/(x2 – x1) where the two points are (a,0) and (0,b).
                m = (b – 0)/(0 – a) = b/-a = -b/a
                Since the slope = -b/a, many students would choose B as the answer. However, we need to know the sign of b/a. We have a negative times b/a. If b is positive and a is positive, then b/a is positive; if b is negative and a is negative, then b/a is positive; but if one if them is positive and the other one is negative, then b/a is negative. We are told that a + b = 0. Therefore one of them must be positive and the other one is negative. Thus a/b is negative, and we have m = - (b/a);
                and m equals a negative times a negative. Thus the slope is positive. The answer is A.

          SAT Verbal - Questions from Spring 2022, Week 9 (March 19, 2022)

          SAT QUICK CHALLENGE EXERCISE I21
          The Pronoun and the Apostrophe 

          What is the difference between a noun and a pronoun? Well, there are many, but this lesson focuses on one -- the use of the apostrophe. We add an apostrophe and the letter "s" to a noun to show ownership. For example, we write Rick's book or Ava's dentist, or even the people's choice.

          The Pronoun and the Apostrophe. We never add an apostrophe and the letter "s" to a pronoun to show ownership. Instead, we use special pronoun ownership words. For Rick, we would write "his" book; for Ava, we would write "her" dentist; and for the people, we would write "their" choice.

          Sometimes, we do use the apostrophe and the letter "s" for pronouns -- but only to make contractions.
          You will recall that a contraction is a ''short cut;" it combines two words to make one word. Examples of contractions include "I'm" (I am), "You're" (You are), etc. Remembering the difference between how apostrophes are used for nouns and how they are used for pronouns will help you earn points each time an SAT question requires you to show that you know the difference.

          Study the possessive pronouns, subject pronouns, and pronoun contractions in the chart below. Then, complete Exercise I21 (beneath the chart), and use the Answer Key to check your work.

           

          SUBJECT PRONOUNS   POSSESSIVE PRONOUNS   PRONOUN CONTRACTIONS
          (with the verb "to be")
          Singular Plural   Singular Plural   Singular Plural
          I We My Our I'm (I am) We're (We are)
          You You Your Your You're (You are) You're (You are)
          He, She, It They His, Her, Its Their He's, She's, It's
          (He is, She is, It is)
          They're (They are)

          SAT Quick Challenge I21
          The Apostrophe

          Directions.   On the line that follows each statement below, place the letter of the answer choice that corrects any error in that statement. If there is
                               no error, mark choice "A" as your answer. When you have finished the exercise, use the answer key in the dropdown below to check
                                your work.

          1.  Its so much fun to ride on the roller coaster. ________

          1. NO CHANGE
          2. Its’ so much fun
          3. It’s so much fun
          4.  Its’s so much fun

          2.  The teacher said that you did such a great job on the project that your sure to win the contest. ________

          1. NO CHANGE
          2. you’re sure to win
          3. you’ve sure to win
          4. you’ll sure to win

          3.  These shoes feel much better because theyre the right size. ________

          1. NO CHANGE
          2. their the right size
          3. they're the right size
          4. there the right size

          SAT Verbal - Answers to Questions from Spring 2022, Week 9 (March 19, 2022)

          1. C
          2. B
          3. C

          SAT Math - Questions fom Spring 2022, Week 8 (March 12, 2022)

          Linear Equations: One Variable in Terms of Another

          When solving linear equations, sometimes there is a need to solve for one variable in terms of another variable. For example, an SAT problem may ask you to solve for “a” in terms of “b”, or solve for “x” in terms of “y and z”. Some students who can readily solve linear equations where there is only one variable are not sure what to do when they are asked to solve a problem involving one variable in terms of another variable.

          Actually, the process for determining one variable in terms of another variable is similar to solving an equation where there is only one variable. We isolate the variable for which we are solving on one side of the equal sign and put everything else on the other side of the equal sign, remembering that whatever we do to one side of the equation, we must do the same for the other side of the equation. We can perform several operations to both sides of the equation: add, subtract, multiply, divide, raise to a power, or take a root.
          Now examine examples 1, 2, and 3.

          Example 1. Solve for x in terms of m: 2x – 3 = m + x

                                       Solve for x in terms of m                                       Similar problem with one variable
                                                          2x – 3 = m + x                                      2x – 3 = 4 + x (m = 4)
          Subtract x from both sides:     x – 3 = m                                              x – 3 = 4
          Add 3 to both sides:                       x = m + 3                                  x = 4 + 3 = 7

          Example 2. Solve for x in terms of y: (5 + 6x)/2 = y + x

          Solve for x in terms of y                                                              Similar problem with one variable
          (5 + 6x)/2 = y + x                                                                             (5 + 6x)/2 = 3 + x (y = 3)

          Multiply both sides by 2:                     5 + 6x = 2(y + x)                                                            5 + 6x = 2(3 + x)
                                                                         5 + 6x = 2y + 2x                                                           5 + 6x = 2(3) + 2x
          Subtract 2x from both sides:                5 + 4x = 2y                                                                   5 + 4x = 2(3)
          Subtract 5 from both sides:                        4x = 2y – 5                                                                   4x = 2(3) – 5
          Divide both sides by 4:                                  x = (2y – 5)/4                                                                x = (2(3) – 5)/4 = (6 – 5)/4 = 1/4

          Example 3. Solve for b in terms of a, c, and x: a(b - 2)/(c – 3) = x

                                                        Solve for b in terms of a, c, and x                          Similar problem with one variable
                                                                   a(b - 2)/(c – 3) = x                                           4(b - 2)/(7 – 3) = 9 (a = 4, c = 7, x = 9)
          Multiply both sides by (c – 3)  :                     a(b-2) = x(c-3)                                              4(b-2) = 9(7 – 3)
          Divide both sides by a:                                  b – 2 = x(c-3)/a                                               b – 2 = 9(7-3)/4
          Add 2 to both sides:                                             b = x(c-3)/a + 2                                               b = 9(7-3)/4 + 2 = 9(4)/4 + 2 = 9+2=11

          Keeping in mind the information above, solve the following problems.

          1. Solve for x in terms of y: (x + 2y)/3 = 6 + 3x

          2. Solve for a in terms of x and b: x = – b/2a

          3. Solve for x in terms of a and n: (3ax – n)/5 = – 4

          4. Solve for y in terms of b and c: (by + 2)/3 = c

          5. Solve for x in terms of e, y, and z: ex – 2y = 3z

          6. Solve for b in terms of k, m, and x: k = (x/2)(b + m)

          SAT Math - Answers to Questions from Spring 2022, Week 8 (March 12, 2022)

          1. Solve for x in terms of y: (x + 2y)/3 = 6 + 3x
            Multiply both sides by 3: x + 2y = 3(6 + 3x)
            x + 2y = 18 + 9x
            Subtract x from both sides: 2y = 18 + 8x
            Subtract 18 from both sides: 2y - 18 = 8x
            Divide both sides by 8: (2y – 18)/8 = x
            x = (2y – 18)/8 = (y – 9)/4

          2. Solve for a in terms of x and b: x = – b/2a
            Multiply both sides by 2a: 2ax = – b
            Divide both sides by 2x: a = – b/2x

          3. Solve for x in terms of a and n: (3ax – n)/5 = – 4
            Multiply both sides by 5: 3ax – n = – 20
            Add n to both sides: 3ax = n – 20
            Divide both sides by 3a: x = (n – 20)/3a

          4. Solve for y in terms of b and c: (by + 2)/3 = c
            Multiply both sides by 3: by + 2 = 3c
            Subtract 2 from both sides: by = 3c – 2
            Divide both sides by b: y = (3c – 2)/b

          5. Solve for x in terms of e, y, and z: ex – 2y = 3z
            Add 2y to both sides: ex = 3z + 2y
            Divide both sides by e: x = (3z + 2y)/e

          6. Solve for b in terms of k, m, and x: k = (x/2)(b + m)
            Multiply both sides by 2: 2k = x(b + m)
            Divide both sides by x: 2k/x = b + m
            Subtract m from both sides: 2k/x – m = b

          SAT Verbal - Questions from Spring 2022, Week 8 (March 12, 2022)

          SAT QUICK CHALLENGE EXERCISE H21
          Parallelism and Faulty Word Pairs Review

          What Is Parallelism? When a sentence has two or more words, phrases, or clauses that have the same level of importance, maintain parallelism by using the same format for each of them. If you use a noun to identify one of the items, use a noun to identify each of the others. For instance, if Anna said, “I love to make ice cream and baking cakes,” the sentence would not be parallel because her sentence would have an infinitive (to make) and a gerund (baking). She could correct the sentence by saying “I love making ice cream and baking cakes.” (two gerunds) She could also say “I love to make ice cream and to bake cakes.” (two infinitives) However, since the infinitive “to” applies to both verbs, “make” and “bake,” the second “to” is optional and can be omitted for the sake of conciseness.


          Word Pairs and Comparisons. Some SAT questions ask about word pairs noted for pointing out similarities and differences such as “either…or, neither…nor,” or not only….but also. Always use the pairs together; do not mix or match them with any other words. Also, be sure that the words or phrases that follow the parts of a word pair are parallel. Note the examples in the chart below.

          WORD PAIRS AND COMPARISONS

          Word Pair Purpose Sample Sentence
          Either...or Tells that only one of two people or things will be involved in something. Either Missy or Marla will host the party.
          Neither...nor Excludes two or more people/things. Neither Missy nor Marla will host the party.
          Not only...but also Points out different qualities of a thing or person. Beyonce is famous not only for her singing, but also for her acting and dancing.* OR Beyonce is famous not only for her singing, but also her acting and dancing.*

          * Note:    Although the preposition “for” applies to Beyonce’s singing, acting, and dancing, repeating “for” with the “not only…but also” word pair s optional; the sentence is parallel with or without that repetition. Note that the situation is similar when the same linking verb is needed for a series of adjectives, as in the following: Watching Mike’s electrifying act, judges were captivated and spectators fascinated. Since the linking verb (“were”) applies to both adjectives, repeating that verb is optional.

          Practice Exercise

          Directions. For each question below, write the letter of the answer choice which corrects the underlined part of the sentence. If the underlined part is already correct, select choice A (NO CHANGE). Use the answer key beneath the questions to check your work.

           

          1.  Neither the original oil painting or George’s reproduction will be included in today’s art show.

          1. NO CHANGE
          2. and not
          3. nor
          4. nor subsequently

          2.  Sampling Grandma’s homemade apple pie, critics were amazed and judges delighted.

          1. NO CHANGE
          2. tasters will be
          3. tasters would be
          4. tasters could feel

          3.  Alvin Ailey ticket stubs, to buy miniature Statues of Liberty, and taking our pictures at a fancy restaurant are some of the souvenirs my class wants to bring back from New York City.

          1. NO CHANGE
          2. miniature Statues of Liberty, and
          3. miniature Statues of Liberty, and to take
          4. miniature Statues of Liberty, and taking

          SAT Verbal - Answers to Questions from Spring 2022, Week 8 (March 12, 2022)

          1. C
          2. A
          3. B

          SAT Math - Questions fom Spring 2022, Week 7 (March 5, 2022)

          Linear Equations: Additional Word Problems


          The lesson last week dealt with word problems that could be solved by setting up an equation. Because of the large number of word problems on the SAT, additional word problems will be reviewed this week. It was noted that the following steps can help solve word problems:

          1. Read the problem carefully and avoid misreading anything important.
          2. Identify the key values and determine how they are related mathematically by converting the words to math.
            1. It is necessary to define the variables and create equations to represent relationships. State in words what the unknown is.
            2. Carefully determine the equality: identify what goes on the left side of the equal sign and what goes on the right side of the equal sign.
          3. Solve the problem and interpret the solution.
          4. Make sure that the answer is reasonable. Ask yourself: Does it make sense? Does it answer the question that was asked?
          5. Warning: beware of a possible change in units in the problem. Information in the problem may be in one unit while the question to be answered may be in another unit. For example, information in the problem may be in feet, but the question to be answered may be in inches.

          Solve the following problems

          1. A website hosting service charges businesses a one time setup fee of $350 plus a monthly amount. If a business paid $1,010 for the first 12 months, including the setup fee, what was the monthly amount?
            1. $25
            2. $35
            3. $45
            4. $55

          2. The Andersons, a retired couple, made two trips to Europe last year. The first trip lasted 15 days longer than the second trip, and the two trips combined lasted a total of 287 days. How many days did the second trip last? 
            1. 122
            2. 136
            3. 146
            4. 155

          3. Marsha made a deposit into her investment account on July 1, 2011. The amount of money in the account doubled each year until Marsha had $2,400 in her investment account on July 1, 2015. What was the amount of the deposit made on July 1, 2011?
            1. 150
            2. 800
            3. 1,000
            4. 1,200

          4. Six years ago, I was three-fourths as old as I will be 7 years from now. How old am I now??
            1. 32
            2. 38
            3. 45
            4. 53

          5. The sum of the ages of three children is 29. The age of the oldest is three times the age of the youngest. The two oldest children differ in age by 6 years. What is the age of the youngest child?
            1. 5
            2. 7
            3. 10
            4. 16

          6. The Zylox Corporation produces several saline solutions that have a variety of medical and industrial uses. How many gallons of a 25% saline solution must be added to 40 gallons of a 10% saline solution to obtain a 15% saline solution?
            1. 9
            2. 14
            3. 20
            4. 25

          7. Charles drives an average of 100 miles each week. His car can travel an average of 25 miles per gallon of gasoline. Charles would like to reduce his weekly expenditure on gasoline by $5. If gasoline costs $4 per gallon, how many fewer average miles should he drive each week?
            1. 23.25
            2. 31.25
            3. 52.75
            4. 68.75

          SAT Math - Answers to Questions from Spring 2022, Week 7 (March 5, 2022)

            1. The variable is the monthly amount.
            2. The amount paid for the first 12 months, 1010, goes on the right side of the equal sign.
            3. The one time setup fee, 350, plus the monthly payments, 12x, goes on the left side of the equal sign.

              350 + 12x = 1010.
              12x = 660
              x = 55 -------- D      The monthly amount is $55.
              A website hosting service charges businesses a one time setup fee of $350 plus a monthly amount. If a business paid $1,010 for the first 12 months, including the setup fee, what was the monthly amount?
              1. $25
              2. $35
              3. $45
              4. $55

              To answer this question, we need to determine the elements of the equation: what the variable is, what goes on the right side of the equal sign, and what goes on the left side of the equation.

          1. The Andersons, a retired couple, made two trips to Europe last year. The first trip lasted 15 days longer than the second trip, and the two trips combined lasted a total of 287 days. How many days did the second trip last?
            1. 122
            2. 136
            3. 146
            4. 155

            The variable is the length of the second trip.
            The combined length of the two trips, 287 days, goes on the right side of the equal sign.
            The length of the second trip, x, plus the length of the first trip, 15 + x, goes on the left side of the equal sign.

            x + 15 + x = 287
                        2x = 272
                          x = 136 -------- B         The second trip lasted 136 days.

          2. Marsha made a deposit into her investment account on July 1, 2011. The amount of money in the account doubled each year until Marsha had $2,400 in her investment account on July 1, 2015. What was the amount of the deposit made on July 1, 2011?
            1. 150
            2. 800
            3. 1,000
            4. 1,200

            The variable is the amount deposited on July 1, 2011.
            The amount in the account on July 1, 2015, $2,400, goes on the right side of the equal sign.
            24x, goes on the left side of the equal sign. The “2” indicates that the amount doubled each year; the
            “4” indicates that there are 4 years between July 1, 2011 and July 1, 2015.

            24x = 2400
            16x = 2400
            x = 150 ----------- A           The amount deposited on July 1, 2011, was $150.

            If “24x = 2400” is not obvious, we could also solve this problem as follows, noting that the amount doubled each year:

            Amount on July 1, 2011 = x
            Amount on July 1, 2012 = 2x
            Amount on July 1, 2013 = 4x
            Amount on July 1, 2014 = 8x
            Amount on July 1, 2015 = 16x

            16x = 2,400
                x = 150

          3. Six years ago, I was three-fourths as old as I will be 7 years from now. How old am I now?
            1. 32
            2. 38
            3. 45
            4. 53

            The variable is my age now.
            ¾ of my age 7 years from now, ¾(x + 7), goes on the right side of the equal sign.
            My age 6 years ago, (x – 6), goes on the left side of the equal sign.

              (x – 6) = ¾(x + 7)
            4x – 24 = 3x + 21
                       x = 45 ---------- C                 My current age is 45. 

          4. The sum of the ages of three children is 29. The age of the oldest is three times the age of the youngest. The two oldest children differ in age by 6 years. What is the age of the youngest child?
            1. 5
            2. 7
            3. 10
            4. 16

            The variable is the age of the youngest child.
            The sum of the ages, 29, goes on the right side of the equal sign.
            The age of the youngest child + the age of the middle child + the age of the oldest child goes on the left side of the equal sign.

            The age of the youngest child = x
            The age of the oldest child = three times the youngest = 3x
            The age of the middle child is 6 less than the age of the oldest child = 3x – 6

            x + 3x + 3x – 6 = 29
                          7x – 6 = 29
                                7x = 35
                                  x = 5 ---------- A               The youngest child is 5, the oldest child is 15, and the
                                                                              middle child is 9.

          5. The Zylox Corporation produces several saline solutions that have a variety of medical and industrial uses. How many gallons of a 25% saline solution must be added to 40 gallons of a 10% saline solution to obtain a 15% saline solution?
            1. 9
            2. 14
            3. 20
            4. 25

            The variable is the number of gallons of the 25% saline solution.
            The 15% solution, 0.15(40 + x), goes on the right side of the equal sign.
            The 10% solution + the 25% solution, 0.10(40) + 0.25x, goes on the left side of the equal sign.

            0.10(40) + 0.25x = 0.15(40 + x)
                       4 + 0.25x = 6 + 0.15x
                             0.10x = 2
                                    x = 20 ----------- C 20 gallons of the 25% saline solution must be added.

          6. Charles drives an average of 100 miles each week. His car can travel an average of 25 miles per gallon of gasoline. Charles would like to reduce his weekly expenditure on gasoline by $5. If gasoline costs $4 per gallon, how many fewer average miles should he drive each week?
            1. 23.25
            2. 31.25
            3. 52.75
            4. 68.75

            The variable is the number of miles driven to result in a $5 reduction in weekly expenditure.

            The amount of weekly saving, $5, goes on the right side of the equal sign.

            The weekly amount spent now, (100/25)(4), minus the weekly amount spent to result in a $5 reduction in expenditure (x/25)(4), goes on the left side of the equal sign.

            (100/25)(4) - (x/25)(4) = 5
                          (4)(4) – 4x/25 = 5
                              16 – 4x/25 = 5
                                 400 – 4x = 125
                                         – 4x = – 275
                                 x = 68.75 = weekly miles needed to drive to reduce weekly expenditure by $5

                100 – 68.75 = 31.25 miles = fewer average miles he should drive each week -------- B

          SAT Verbal - Questions from Spring 2022, Week 7 (March 5, 2022)

          SAT QUICK CHALLENGE EXERCISE G1
          Correcting Modifier Errors

          Putting the Modifier in the Right Place. A modifier describes or provides additional information about someone or something. A modifier that is not placed next to (or as close as possible to) the word/word group it should describe is misplaced, and the sentence it is in does not convey the meaning intended. Note the underlined, misplaced modifier in Example A, which follows. Example A: Thoughtless telemarketers really annoy homeowners making calls early in the morning or late at night. Explanation. Since the modifier is next to the word “homeowners,” the sentence says that homeowners are making calls early in the morning or late at night. However, the modifier is intended to describe the telemarketers. Note the correction in Example B, which follows. Example B: Thoughtless telemarketers making calls early in the morning or late at night really annoy homeowners. Explanation. Now, the modifier is next to the word “telemarketers,” and the sentence says that “telemarketers” are making calls early in the morning or late at night. Therefore, the sentence is written correctly.

          The Introductory Modifying Phrase Error. “Misplaced modifier” is a general phrase that applies to any modifier that is not in the right place. However, sometimes more specific terminology is used to indicate exactly what kind of modifier error has been made. For instance, note the introductory modifying phrase error in Example C, which follows: Example C: Completely exhausted after working overtime for six hours, the bed was a very welcome sight to Lori. Explanation. Because the modifier at the beginning of the sentence is closest to the noun “bed,” the sentence says that the bed was completely exhausted after it had worked overtime for six hours. Of course, that sentence does not make sense. Lori,-- not the bed -- had worked overtime. Now, note Example D, which follows: Example D: Completely exhausted after working overtime for six hours, Lori found the bed to be a very welcome sight. Explanation. Since the modifier is right beside the word it should modify, “Lori,” the modifier is in the right place, and the sentence makes sense. Now, keeping in mind the information above, complete the exercise below.
           

          SAT Quick Challenge F1
          The Misplaced Modifier

          Directions.  Select the letter of the answer choice which corrects the underlined part of each statement below. If you believe that the underlined                          part is already correct, select Choice A -- NO CHANGE. After completing this activity, use the answer key to check your work.

          1.   Very conscientious about food safety, Kevin’s habit is to wash carefully surfaces touched by raw meat right away.

          1. NO CHANGE
          2. Kevin’s careful washing is for
          3. careful washing by Kevin is done on
          4. Kevin carefully washes

          2.  Watching from huge airport windows, it is exciting to see airplanes taking off and landing.

          1. NO CHANGE
          2. excitement is felt by watching
          3. passengers and visitors enjoy seeing
          4. sights enjoyed by passengers include

          3.  Peering at the delicious desserts in the display case, deciding what to order can be very difficult.

          1. NO CHANGE
          2. customers find that deciding
          3.  making decisions about
          4.  the decision about

          SAT Verbal - Answers to Questions from Spring 2022, Week 7 (March 5, 2022)

          1. D
          2. C
          3. B

          SAT Math - Questions fom Spring 2022, Week 6 (February 26, 2022)

          Linear Equations: Word Problems


          An increasing number of problems on the SAT are word problems. The following steps can help you solve word problems:

          1. Read the problem carefully and avoid misreading anything important.
          2. Identify the key values and determine how they are related mathematically by converting the words to math.  
            1. It is necessary to define the variables and create equations to represent relationships.  State in words what the unknown is.  
            2. Carefully determine the equality:  identify what goes on the left side of the equal sign and what goes on the right side of the equal sign.
          3. Solve the problem and interpret the solution.
          4. Make sure that the answer is reasonable.  Ask yourself:  Does it make sense?  Does it answer the question that was asked?
          5. Warning:  beware of a possible change in units in the problem.  Information in the problem may be in one unit while the question to be answered may be in another unit.  For example, information in the problem may be in feet, but the question to be answered may be in inches.

          Now examine examples 1, 2, and 3.

          Example 1:  Betty and her younger sister Lynn work together to peel potatoes.  Betty peels 15 more than twice the number of potatoes that Lynn peels.  Together they peel 75 potatoes.  How many potatoes does Lynn peel?   
          To answer this question, we need to determine the elements of the equation:  what the variable is, what goes on the right side of the equal sign, and what goes on the left side of the equation.  

          1. The variable is the number of potatoes that Lynn peels.
          2. The number of potatoes peeled, 75, goes on the right side of the equal sign.
          3. The number of potatoes Betty peels, 2x + 15,  plus the number of potatoes Lynn peels, x,  goes on the left side of the equal sign.     
            2x + 15 + x  =  75.
            3x  =  60
              x  =  20            Lynn peels 20 potatoes.    


          Example 2: Martin bought a laptop computer at a store that gave a 10% discount off the original price. The total amount he paid to the cashier was $715.50, which included a 6% sales tax that was added to the discounted price. What was the original price of the computer?

          1. The variable is the original price of the computer.
          2. The amount paid to the cashier, $715.50, goes on the right side of the equal sign.
          3. The discounted price of the computer, (1 – 0.1)x times the sales tax adjustment, (1-0.1)x(1.06), goes on the left side of the equal sign.
                     (1 – 0.1)x(1.06) = 715.50
                              (.9)x(1.06) = 715.50
                                     .954x = 715.5
                                             x = 750                    The original cost of the computer was $750.00.

          Example 3:  Oliver is selling photographs as part of a project for his entrepreneurship class.  He sells the first 20 photographs for $10 each.  Because the first 20 photographs sold so quickly, he raised the price of the photographs to $15 each for the rest of the project.  After his expenses, Oliver earns a profit of 80% of the revenues from his sales.  How many photographs must he sell for the rest of the project to earn a profit of $400?

          1. 18
          2. 20
          3. 24
          4. 32

          The variable is the number of additional photographs he must sell.  The profit of $400 goes on the right side of the equal sign.  The total revenues times 80% goes on the left side of the equal sign.
          The total revenues  =  $10 times 20  + ( $15 times x).  Thus we have:

          0.8(200 + 15x)  =  400
                 160 + 12x  =  400
                            12x  =  240
                                  x  =  20  ------------ B


          Keeping in mind the information above, answer the following questions.

          1. A customer paid $81.00 for a coat after an 8% sales tax was added. What was the price of the coat before the sales tax was added?

            1. $72.00
            2. $74.52
            3. $75.00
            4. $87.48

          2.  The weight limit for the contents in a paper bag is 25 pounds. A customer is purchasing a three pound cantaloupe and several boxes of spaghetti each weighing 24 ounces. What is the maximum number of boxes of spaghetti that can be put in the paper bag so that the weight of the contents is below the maximum weight limit?

          3. The cost of using a telephone in a hotel meeting room is $0.15 per minute. Which of the following equations represents the total cost C, in dollars, for x hours of phone use? 

            1. 0.15x  =  C
            2. 0.15(60x)  =  C
            3. 60x/0.15  =  C
            4. 0.15x/60  =  C

          4. A school district is forming a committee to discuss plans for the construction of a new high school. Of those invited to join the committee, 15% are parents of students, 45% are teachers from the current high school, 25% are school and district administrators, and 6 individuals are students. How many more teachers were invited to join the committee than school and district administrators?

          5. The mean score of 8 players in a basketball game was 14.5 points. If the highest individual score is removed, the mean score of the remaining 7 players becomes 12 points. What was the highest score?

            1. 20
            2. 24
            3. 32
            4. 36

          6. Max is working this summer as part of a crew on a farm. He earned $8 per hour for the first 10 hours he worked this week. Because of his performance, his crew leader raised his salary to $10 per hour for the rest of the week. Max saves 90% of his earnings from each week. How many hours must he work the rest of the week to save $270 for the week?

            1. 16
            2. 22
            3. 33
            4. 38

          SAT Math - Answers to Questions from Spring 2022, Week 6 (February 26, 2022)

          1. A customer paid $81.00 for a coat after an 8% sales tax was added. What was the price of the coat before the sales tax was added?
            1. $72.00
            2. $74.52
            3. $75.00
            4. $87.48

            The variable is the price of the coat before the sales tax is added. The amount the customer paid of $81.00 goes on the right side of the equal sign. The initial price plus the sales tax goes on the left side of the equal sign.
                x + .08x = 81
                    1.08x = 81
                            x = 75 ----------- C

          2. The weight limit for the contents in a paper bag is 25 pounds. A customer is purchasing a three pound cantaloupe and several boxes of spaghetti each weighing 24 ounces. What is the maximum number of boxes of spaghetti that can be put in the paper bag so that the weight of the contents is below the maximum weight limit?

            The variable is the number of boxes of spaghetti. The maximum weight of the contents that can be put in the paper bag, 25 pounds, goes on the right side of the equal sign. There is an issue of units in this problem; every item should be expressed in pounds or ounces. In this case we will express everything in pounds. The weight of the contents in the paper bag goes on the left side of the equal sign.

            Since we are expressing each item in pounds, change the weight of the spaghetti boxes from 24 ounces to 1.5 pounds.

                      3 + 1.5x = 25
                            1.5x = 22
                                 x = 14.67           The maximum number of boxes of spaghetti is 14.

            If we express the weight of each item in ounces, we have 3 pounds = 48 ounces and 25 pounds = 25 times 16 = 400 ounces. Thus our equation is:

            48 + 24x = 400
                    24x = 352
                        x = 14.67 The maximum number of boxes of spaghetti is 14.

          3. The cost of using a telephone in a hotel meeting room is $0.15 per minute. Which of the following equations represents the total cost C, in dollars, for x hours of phone use?
            1. 0.15x = C
            2. 0.15(60x) = C
            3. 60x/0.15 = C
            4. 0.15x/60 = C

            Since the hotel charges $0.15 per minute, we convert the per minute rate to an hourly rate: $0.15 per minute times 60 minutes per hour = 0.15(60) per hour. Then we multiply by the number of hours of phone use. Thus the equation is: 0.15(60x) = C Choice B is correct.

          4. A school district is forming a committee to discuss plans for the construction of a new high school. Of those invited to join the committee, 15% are parents of students, 45% are teachers from the current high school, 25% are school and district administrators, and 6 individuals are students. How many more teachers were invited to join the committee than school and district administrators?

            The variable is the number of people invited to join the committee. The total number of people invited to join the committee goes on the right side of the equal sign. The total of the individual groups goes on the left side of the equal sign.

            Thus our equation is:                    0.15x + 0.45x + 0.25x + 6 = x
                                                                                              0.85x + 6 = x
                                                                                                           6 = 0.15x
                                                                                                            x = 6/0.15 = 40

            40 people were invited to join the committee.

            Teachers:                                                                                 0.45(40) = 18
            School and district administrators:                                      0.25(40)  = 10
                                                                                                                               --
                                                                                                                                 8 Our answer is 8.

          5. The mean score of 8 players in a basketball game was 14.5 points. If the highest individual score is removed, the mean score of the remaining 7 players becomes 12 points. What was the highest individual score?
            1. 20
            2. 24
            3. 32
            4. 36

            The variable is the highest individual score. The total number of points scored by the
            remaining 7 players goes on the right side of the equal sign. The total number of
            points sored by all 8 players minus the highest individual score goes on the left side
            of the equal sign.

            Remember that in a mean (average) problem the first thing to do is to find the total. There are two ways to find the total: 1) add all of the items; 2) multiply the mean by the number of items. On the SAT the second method is used more often.

            The total of the seven players = 7(12) = 84
            The total of the eight players = 8(14.5) = 116

            The equation is: 116 – x = 84
                                                 -x = -32
                                                   x = 32 --------- C

          6. Max is working this summer as part of a crew on a farm. He earned $8 per hour for the first 10 hours he worked this week. Because of his performance, his crew leader raised his salary to $10 per hour for the rest of the week. Max saves 90% of his earnings from each week. How many hours must he work the rest of the week to save $270 for the week?
            1. 16
            2. 22
            3. 33
            4. 38

            The variable is the number of additional hours he must work. The savings of $270 goes on the right side of the equal sign. The total earnings times 90% goes on the left side of the equal sign. The total earnings = $8 times 10 + $10 times x. Thus we have: 0.9(80 + 10x) = 270
                                                                                      72 + 9x = 270
                                                                                              9x = 198
                                                                                                x = 22 -------------  B

          SAT Verbal - Questions from Spring 2022, Week 6 (February 26, 2022)

          SAT QUICK CHALLENGE EXERCISE F1
          The Misplaced Modifier

          A modifier is a word or word group that describes someone or something. If the modifier is not placed next to (or as close as possible to) the word/word group it should describe, it will be misplaced, and the sentence will not convey the meaning intended, as in the sentence which follows. Extremely hungry, the "snack” waiting for Dave was eaten right away. Because the modifier “extremely hungry” is closest to the noun “snack,” the sentence says that the snack was extremely hungry. However, Dave – not the snack – was extremely hungry. Note the correction which follows: “Extremely hungry,” Dave ate the snack that was waiting for him right away. Since the modifier is right beside the word it should modify --“Dave,” it is in the right place, and the sentence makes sense.

          "Possessive noun" and "descriptive aside" modifier errors, explained below, are often tested on the SAT., 

          The Possessive Noun Error. A possessive noun error occurs when a possessive noun is placed where a noun that is not possessive should be. Note the following sentence: With endless patience, Mrs. White's clear explanations helped me learn to do advanced math. EXPLANATION: The closest noun phrase to the introductory modifier ("With endless patience”) is "Mrs. White's clear explanations." Hence, the sentence says that Mrs. White's clear explanations had endless patience, but explanations do not have patience of any kind; teachers do. Hence, the sentence has a possessive noun error. Note the following correction: "With endless patience, Mrs. White used clear explanations that helped me learn to do advanced math." Now, the modifier correctly modify "Mrs. White," and the sentence makes sense.

          The Misplaced "Descriptive Aside" Modifier.   Sometimes, a modifier that is not part of a sentence's
          main idea is added to a sentence, as follows: "Diana Carter does breathtaking stunts, the drum major, at the games." Enclosed in commas because it is extra, nonessential information, "the drum major" describes the noun it follows -- "stunts." Accordingly, the sentence says that "stunts" are the drum major. Since the description is not in the right place, the idea that it gets across does not make sense because "stunts" can not be a drum major. Placed after "Diana Carter," the proper noun that it should modify, the modifier would say correctly that Diana Carter is the drum major, as indicated in the sentence which
          follows: "Diana Carter, the drum major, does breathtaking stunts at the games." Now, keep in mind the information above, and complete Exercise F1 below.

           

          SAT Quick Challenge F1
          The Misplaced Modifier

          Directions.   Select the letter of the answer choice which corrects the underlined part of each statement below. If you believe that the underlined                         part is already correct, select Choice A -- NO CHANGE. After completing this activity, use the answer key to check your work.

          1.  Arriving in the mail every day, gifts, family, and friends filled the nursery before the baby was born..

          1. NO CHANGE
          2. family and friends bringing gifts
          3. family bringing gifts and friends
          4. gifts from family and friends

          2.  Ron Hall won the rocket design contest, a really sharp eight-year-old, and got a free trip to New York.
               To make this sentence grammatically correct, the underlined phrase should be placed

          1. NO CHANGE
          2. after the word "rocket"
          3. before the word "won"
          4. after the word "trip"

          3.  Excited about homecoming, parties and other activities lasted for weeks. 

          1. NO CHANGE
          2. students partied and celebrated
          3. festivities and celebrations took place
            1. the ballroom was filled with music and dancing

          SAT Verbal - Answers to Questions from Spring 2022, Week 6 (February 26, 2022)

          1. D
          2. C
          3. B

          SAT Math - Questions fom Spring 2022, Week 5 (February 19, 2022)

          Linear Equations: one solution, no solution, and an infinite number of solutions

          Linear equations may have one solution, no solution, or an infinite number of solutions. On the SAT most of the linear equation problems will have one solution. It is important to recognize, however, that there will occasionally be problems where there is no solution or an infinite number of solutions.

          One Solution
          There will be one solution when we can isolate the variable for which we are solving on one side of the equal sign and the other parts of the equation on the other side. By adding, subtracting, multiply, dividing, raising to a power, or taking a root, we arrive at a situation such as “x = 4” or “x = 17”.

          No Solution
          There will be no solution when we make our algebraic manipulations and arrive at a situation such as “2x + 7 = 2x + 5” or “ + 7 = + 5”. The coefficient of the x term on the left side of the equation is the same as the coefficient of the x term on right side of the equation, and the constant on the left side of the equation is different from the constant on the right side of the equation. There is no value of x that will make both sides of the equation equal, and + 7 can never equal + 5.

          Infinite Number of Solutions
          There will be an infinite number of solutions when we make our algebraic manipulations and arrive at a situation such as “5x + 3 = 5x + 3”. The coefficient of the x term on the left side of the equation is the same as the coefficient of the x term on right side of the equation, and the constant on the left side of the equation is the same as the constant on the right side of the equation. All possible values of x will make both sides of the equation equal.

          Consider the following example (example 1):

          Solve the following equation: 7x – 6 –x = 12 + 2x + 10
          Add or subtract to combine like terms: 6x – 6 = 22 + 2x

          Subtract 2x from both sides and add 6 to both sides: 4x = 28

          Divide both sides by 4: x = 7. There is one solution for example 1.

          Consider another example (example 2):

          Solve the following equation: x + 5 + 3x = 6x – 2 – 6

          Add or subtract to combine like terms: 4x + 5 = 4x – 8

          There is no value of x that will make both sides of the equation equal, and + 5 can never equal – 8.
          There is no solution for example 2.

          Consider another example (example 3):

          Solve the following equation: 3x – 9 + 5x + 4 = 11x – 5 – 3x

          Add or subtract to combine like terms: 8x – 5 = 8x – 5      In example 3 there is an infinite number of solutions. All possible values of x will make both sides of the equation equal.

          Keeping in mind the information above, solve the following problems.

          1.  Solve the following equation:       3(x + 2) + 5x – 1 = 7x – 3 + x

          2. If 2x + 8 = 16, what is the value of x + 4?

          3. 2(9x – 6) - 12 = 3(9x – 6)
            Based on the equation above, what is the value of 3x – 2?

          4. Solve the following equation: 3(2x + 1) – 2x = 8x + 5 – 4x – 2

          5. 9ax + 9b – 6 = 21
            Based on the equation above, what is the value of ax + b?

            1. 3
            2. 6
            3. 8
            4. 12
          6.   2ax – 15 = 3(x + 5) + 5(x – 1)
            In the equation above, a is a constant. If no value of x satisfies the equation, what is the value of a? 

            1. 1
            2. 2
            3. 4
            4. 8
          7. a(x + b) = 4x + 10
            In the equation above, a and b are constants. If the equation has infinitely many solutions for x, what is the value of b? 

          SAT Math - Answers to Questions from Spring 2022, Week 5 (February 19, 2022)

          1.  Solve the following equation:       3(x + 2) + 5x – 1 = 7x – 3 + x
            3x + 6 + 5x – 1 = 6x – 3
            8x + 5 = 8x – 3
            There is no solution

          2. If 2x + 8 = 16, what is the value of x + 4?

            Solve for x; then solve for x + 4.

            2x + 8 = 16
            2x = 8
            x = 4
            x + 4 = 4 + 4 = 8 This is our answer.

            Recognizing that (2x + 8) is a multiple of (x + 4), there is an alternative method for finding our answer. We can divide both sides by 2.
            2x + 8 = 16
            Dividing both sides by 2, we get x + 4 = 8 This is our answer: x + 4 = 8, which is the same as the answer above.

          3. 2(9x – 6) - 12 = 3(9x – 6)
            Based on the equation above, what is the value of 3x – 2?

            Solve for x; then solve for 3x – 2.
            18x – 12 - 12 = 27x – 18
            18x – 24 = 27x - 18
            -9x = 6
            -x = 6/9 = 2/3
            x = -2/3

            3x – 2 = 3(-2/3) – 2 = - 6/3 – 2 = - 2 – 2 = -4 This is our answer.

            Recognizing that (9x – 6) is a multiple of (3x – 2), there is an alternative method for finding our answer. We can subtract 2(9x – 6) from both sides.

            2(9x – 6) - 12 = 3(9x – 6)

            Subtracting 2(9x – 6) from both sides, we get -12 = 9x – 6
            Divide both sides by 3: -4 = 3x – 2. This is our answer: 3x – 2 = -4, which is the same as the
            answer above.

          4. Solve the following equation:               3(2x + 1) – 2x = 8x + 5 – 4x – 2
            6x + 3 – 2x = 4x + 3
            4x + 3 = 4x + 3
            There is an infinite number of solutions.

          5. 9ax + 9b – 6 = 21
            Based on the equation above, what is the value of ax + b?

            1. 3
            2. 6
            3. 8
            4. 12

              Some students give up on problems of this type. They want to get the value of a, multiply it by the value of x, and add the value of b. Since they cannot find the values of these variables, they do not know what to do. Actually, we cannot determine the value of a, we cannot determine the value of x, and we cannot determine the value of b, but we can find the value of ax + b.

              9ax + 9b – 6 = 21
              Add 6 to both sides:                        9ax + 9b = 27
              Divide both sides by 9:                      ax + b = 3 ------- A     
                                                                          This is our answer.

          6.   2ax – 15 = 3(x + 5) + 5(x – 1)
            In the equation above, a is a constant. If no value of x satisfies the equation, what is the value of a? 

            1. 1
            2. 2
            3. 4
            4. 8

              For no solution, we want to arrive at a situation where the coefficient of the x term on the left side of the equation is the same as the coefficient of the x term on right side of the equation, and the constant on the left side of the equation is different from the constant on the right side of the equation. What is our approach here? We want to find a value of “a” which when multiplied by “2” will equal the coefficient of “x” on the right side of the equation.

              2ax – 15 = 3(x + 5) + 5(x – 1)
              2ax – 15 = 3x + 15 + 5x – 5
              2ax – 15 = 8x + 10

              The constant on the left side of the equation ( -15) is different from the constant on the right side of the equation (+10). Now we need the coefficients of the x terms to be the same. Therefore 2a must
              equal 8. 2a = 8
              a = 4 -------------- C




          7. a(x + b) = 4x + 10
            In the equation above, a and b are constants. If the equation has infinitely many solutions for x, what is the value of b? 

            For there to be infinitely many solutions, we want to arrive at a situation where the coefficient of the x term on the left side of the equation is the same as the coefficient of the x term on right side of the equation, and the constant on the left side of the equation is the same as the constant on the right side of the equation. What is our approach here? We want to find a value of “b” which when
            multiplied by “a” will equal the constant on the right side of the equation, 10. We first must find the value of “a”.

            a(x + b) = 4x + 10
            ax + ab = 4x + 10

            ax = 4x and ab = 10
            Thus a must equal 4 and ab must equal 10.
            a = 4
            ab = 10
            Since a = 4, 4b = 10
            b = 10/4 = 5/2 = 2.5

          SAT Verbal - Questions from Spring 2022, Week 5 (February 19, 2022)

          SAT QUICK CHALLENGE EXERCISE E1
          Parallelism

          Parallelism – Repeating Grammatical Constructions. Parallelism is a writing strategy that repeats the manner in which related ideas in the same sentence or paragraph are expressed (as in all nouns, all verbs, etc.). This repetition is important because it (1) indicates that the ideas involved are equally important and (2) makes the writing clearer and easier to understand. Note Example A, which follows.
          Example A: Eating, talking, or to send text messages while driving can cause accidents. Note that the three items at the beginning of the sentence are not parallel because the underlined phrase does not repeat the pattern used for the two words the phrase follows. Gerunds (which look like “ing” verbs, but function as nouns) begin the sentence, but the underlined item is an infinitive phrase (which begins with the infinitive “to,” followed by a verb and a direct object). Note the corrections in Examples B and C.

          Example B.    Eating, talking, or sending text messages while driving can cause accidents. Note that each item in the subject is a gerund or a gerund phrase. Hence, the sentence is now parallel.

          Example C.    To eat, talk, or send text messages while driving can cause accidents. Note that each item in the subject of this example is an infinitive – that is, the word “to” followed by a verb or verb phrase. Since all three items in the subject begin with the infinitive “to,” that word does not need to be repeated for each verb. Hence, this sentence is also parallel.

          Parallelism and the Word “and.” Another kind of parallelism error you might see on the SAT involves two words, phrases, or clauses that are connected by the word and.” The two words/word groups connected by that conjunction must match. Note Examples D and E, which follow.

          Example D
          .
          The number of deaths from the illness has decreased because of an increase in the number of people being tested and the number of people getting vaccinations has also increased. Note that the word group before the conjunction and” is a noun phrase, but the word group after the conjunction is an independent clause (or complete sentence). Since the two sides do not match, the sentence is not parallel. Note the correction in Example E.

          Example E The number of deaths from the illness has decreased because of an increase in the number of people being tested and the number of people getting vaccinations. Since there is a noun phrase both before and after the conjunction “and,” the two sides match, and the sentence is parallel.

          Keeping in mind the information above, complete the exercise below. 

           

          QUICK CHALLENGE Exercise E1

          Directions.   Replace the underlined words in each question below with the answer choice that
                               corrects the error in the sentence. If there is no error, select choice A -- NO CHANGE.

          1.  Kevin’s dog likes to grab ham bones from the kitchen, bury them in the back yard, and then digging
               them up later to gnaw on them.

          1. NO CHANGE
          2. they dug
          3. might can dig them
          4. dig them

          2.  The team that wins the Quiz Bowl can choose between having a pizza party or they can get free
               tickets to the Fall Fair Frolic next month.

          1. NO CHANGE
          2. or getting
          3. or they might can get
          4. but they could also win

          3.  Mrs. Ford agreed to let her teenager get an after-school job because of the significant improvement in
               his grades, and he now manages his time much better

          1. NO CHANGE
          2. his grades and how he stopped being so late
          3. his grades, yet he’s on time now
          4. his grades and his time management skills

          SAT Verbal - Answers to Questions from Spring 2022, Week 5 (February 19, 2022)

          1. D
          2. B
          3. D

          SAT Math - Questions from Spring 2022, Week 4 (February 12, 2022)

          Solving Equations by Cross Multiplying

          1. We cross multiply when both sides of an equation are fractions.
          2. Multiply the denominator of the left side of the equation by the numerator of the right side of the equation, and multiply the numerator of the left side of the equation by the denominator of the right side of the equation.
          3. Set the two items equal to each other, and solve the resulting equation.

                                                 6            3
          Here is an example:      -     =    ---
                                                 x            10

                                                 3x = 60
                                                   x = 20

          Now solve the following equations. The answer key is in the drop-down below.

          1.   5        2x + 7
            ---   =  ------
              2            3

            1. 0.25
            2. 1.00
            3. 2.5
            4. 3
          2.            x + 5
               If     ------   =   10, what is the value of x?
                       x - 5

            1. 45/11
            2. 5
            3. 11/2
            4. 55/9
          3.            3x + 2y      17                 x
               If       ------   =  --, what is   --- ?
                            y            4                   y

            1. 4/17
            2. 5/9
            3. 3/4
            4. 5/6
          4.           3x + 4         2x - 8
               If     ------   =   -------, what is x?
                           5                  7

            1. -68/11
            2. -16/7
            3. 13/5
            4. 8/17
          5.            4 - 7x            x
               If      ------   =     --- + 2, what is x?
                           6                8

            1. -35/27
            2. -32/31
            3. -19/24
            4. -7/15
          6.            p + q + r             p + q
               If     ---------     =     -----, then r =
                              3                    2

            1. q + p
            2. 2p + 2q
            3. 1/2 (p + q)
            4. 1

          SAT Math - Answers to Questions from Spring 2022, Week 4 (February 12, 2022)

          1.   5        2x + 7
            ---   =  ------
              2            3

            1. 0.25
            2. 1.00
            3. 2.5
            4. 3

              2(2x + 7) = 15
              4x + 14 = 15
              4x = 1
              x = 1/4 = 0.25 -------- A

          2.            x + 5
               If     ------   =   10, what is the value of x?
                       x - 5

            1. 45/11
            2. 5
            3. 11/2
            4. 55/9                 

                                                                  x + 5      10
              We can write the equation as   -----  =  ---
                                                                  x - 5        1
              Then cross multiply: 10(x - 5) = x + 5
                                                  10x - 50 = x + 5
                                                     9x = 55
                                                       x = 55 / 9 ----- D
          3.            3x + 2y      17                 x
               If       ------   =  --, what is   --- ?
                            y            4                   y

            1. 4/17
            2. 5/9
            3. 3/4
            4. 5/6

              3x + 2y         17
              -------    =   --
                   y               4
              17y = 4(3x + 2y)
              17y = 12x + 8y
              9y = 12x
               x         9         3
              --- =  ---   =  --- ----- C 
               y        12        4

          4.           3x + 4         2x - 8
               If     ------   =   -------, what is x?
                           5               7

            1. -68/11
            2. -16/7
            3. 13/5
            4. 8/17

              7(3x + 4) = 5(2x - 8)
              21x + 28 = 10x - 40
              11x = -68
                  x = -68
                       -----    ----- A
                         11
          5.            4 - 7x            x
               If      ------   =     --- + 2, what is x?
                           6                8

            1. -35/27
            2. -32/31
            3. -19/24
            4. -7/15

              Combine the values on the right side of the equation so that we get one fraction. We change the "2" so that there is a denominator of 8: 
                          16
              2    =    ---
                            8

                                           4 - 7x           x           16
              Now we have       ------    =   ---     +  ---
                                               6               8            8
                                          4 - 7x         x + 16
                                          ------    =   ------
                                                6                8
                                         6(x + 16) = 8(4 - 7x)
                                         6x + 96 = 32 - 56x
                                                  62x = -64
                                                       x = -64/62 = -32/31 ----- B
          6.            p + q + r             p + q
               If     ---------     =     -----, then r =
                              3                    2

            1. q + p
            2. 2p + 2q
            3. 1/2 (p + q)
            4. 1

              2(p + q + r) = 3(p + q)
              2p + 2q + 2r = 3p + 3q
              2r = p + q
                 r = p + q         1
                      -----   =   ---   (p + q) ----- C
                         2             2

          SAT Verbal - Questions from Spring 2022, Week 4 (February 12, 2022)

          SAT QUICK CHALLENGE D1
          Using Transitions Correctly

          Transitions at the Beginning of a Sentence. You may recall that when an underlined transition is at the beginning of an SAT question, the correct transition is the one required based on the relationship between the underlined sentence and the preceding sentence. Note Examples A and B below.
          Example A. Diana loves to go to the beach. Additionally, her sister Deena prefers to go to the mountains. Note that the two sentences express contrasting ideas. Therefore, a contrast transition is needed. However, "additionally" is a continuer. Therefore, the sentence is incorrect. Note the correction in Example B, which follows. Example B. Diana loves to go to the beach. On the other hand, her sister Deena prefers to go to the mountains. Since "On the other hand" is a contrast transition, Example B is correct.

          Transitions Within a Sentence. When two related sentences are combined to form a single sentence, the sentences appear as two independent clauses separated by the appropriate transition and the correct punctuation. Note Example C, which follows.

          Example C. Diana loves to go to the beach; on the other hand, her sister Deena prefers to go to the mountains. In this situation, the first independent clause must end with a semicolon, and the transition must (1) follow the semicolon, (2) begin with a lower case letter, and (3) be followed by a comma. Moreover, the transition must indicate the relationship between the part of the sentence that comes before the underlined words and the part that comes after them.

          When No Transition Is Needed. Sometimes, a sentence will explain, in a clear, obvious manner, information that is in the sentence it follows. In that situation, no transition is needed. Note Example D, which follows.

          Example D. We had really crazy weather this weekend. We got rain, sleet, snow, and ice all within the same day. A question (1) without a transition as an answer choice or (2) with a "DELETE the underlined portion..." option suggests that a transition may not be necessary. Such answer choices are rare, but if you see one, consider the possibility that no transition is needed for the question which offers such answer choices.

          Keeping in mind the information above, complete the exercise which follows. 

           

          QUICK CHALLENGE Exercise D1

          Directions.   Replace the underlined words in each question below with the answer choice that
                               corrects the error in the sentence. If there is no error, select choice A -- NO CHANGE.

          1. Mr. Jones is very thankful for his job, He does not make enough money to pay his monthly expenses.  

          1. NO CHANGE
          2. job. additionally, he
          3. job; however, he
          4. job, for example, he

          2.  Tameka got full scholarships to several colleges that she really would like to attend. Nevertheless; her twin sister Carlotta got full scholarships to several of her favorite colleges, also.

          1. NO CHANGE
          2. to attend; likewise
          3. Otherwise
          4. Consequently

          3.  There was a tremendous fire at a local nursing home in the middle of the night. Sleepy residents
          had to be awakened quickly and evacuated right away.

          1. NO CHANGE
          2. However; sleepy
          3. Additionally,
          4. Nevertheless, sleepy

          SAT Verbal - Answers to Questions from Spring 2022, Week 4 (February 12, 2022)

          1. C
          2. B
          3. A

          SAT Math - Questions from Spring 2022, Week 3 (February 5, 2022)

          Solving Linear Equations Continued
          Simple equations

          1. Solve for x (or for any variable) by isolating x on one side of the equal sign and everything else on
            the other side.
          2. Always remember to do the same thing to both sides of the equation: add, subtract, multiply,
            divide, raise to a power, or take the square root.


          Solve these problems, and then check your answers. The answer key is in the drop-down below.

          1. If c = (1/x) + (1/y) and x > y > 0, then which of the following is equal to 1/c?

            1. x + y
            2. x - y
            3. (x + y) / xy
            4. xy / (x + y)
          2. If c = (1/x) + (1/y), solve for x in terms of c and y.

          3. b = 2.35 + 0.25x
            c = 1.75 + 0.40x

            In the equations above, b and c represent the price per pound, in dollars, of beef and
            chicken, respectively, x weeks after July 1 during last summer. What was the price
            per pound of beef when it was equal to the price per pound of chicken?

            1. $2.60
            2. $2.85
            3. $2.95
            4. $3.35

          4. 3(1/2 - y) =  (3/5) + 15y
            What is the solution to the equation above?

          5. If (1/2)(x) + (1/3)(y) = 4, what is the value of 3x + 2y?

          SAT Math - Answers to Questions from Spring 2022, Week 3 (February 5, 2022)

          1. If c = (1/x) + (1/y) and x > y > 0, then which of the following is equal to 1/c?

            1. x + y
            2. x - y
            3. (x + y) / xy
              1. x + y = 6 + 3 = 9                                           NO
              2. x - y  = 6 - 3 = 3                                           NO
              3. (x + y) / (xy) 
                = (6 + 3) / (6) (3)
                = 9 / 18
                = 1 / 2                                                           NO
              4. (xy) / (x + y)
                = (6)(3) / (6 + 3)
                = 18 / 9 = 2                                                  YESxy / (x + y)

                This problem can be solved in one of two ways.

                Alternative One:
                Find the least common denominator and combine the two terms. The least common denominator is xy.

                c = (1 / x) + (1 / y)
                   = [(1 / x)] * [(xy) / (xy)]
                   + [(1 / y)] * [(xy) / (xy)]
                   =   (y / xy) + (x / xy)   
                   = (y + x) / (xy)

                Then (1 / c) = (xy) / (x + y)   -------- D

                Alternative Two:
                Pick numbers and substitute. This method can be used when there are variables in the answer choices. Let x = 6 and let y = 3.

                Then c = (1 / x) + (1 / y)
                            = (1 / 6) + (1 / 3)
                            = (1 / 6) + (2 / 6)
                            = (3 / 6)
                            = (1 / 2)

                Then (1 / c) = (2 / 1) = 2
                Now, using our numbers, which answer choice = 2?

          2. If c = (1 / x) + (1 / y), solve for x in terms of c and y.
            c - (1 / y) = (1 / x)
            (1 / x) = c - (1 / y)
                       = (cy / y) - (1 / y)
                       = (cy - 1) / y
            Then x = (y) / (cy - 1)

          3. b = 2.35 + 0.25x
            c = 1.75 + 0.40x

            In the equations above, b and c represent the price per pound, in dollars, of beef and
            chicken, respectively, x weeks after July 1 during last summer. What was the price
            per pound of beef when it was equal to the price per pound of chicken?

            1. $2.60
            2. $2.85
            3. $2.95
            4. $3.35

              Set the two expressions equal to each other, solve for x, then substitute the value of x into either expression.

              1.75 + 0.40x = 2.35 + 0.25x                                Now solve for x.
                         0.15x = 0.6
                                x = (0.6) / (0.15) = 4                            Now substitute into either expression.
                       b = 2.35 + 0.25(4) = 2.35 + 1.00 = 3.35   ------ D
               Also c = 1.75 + 0.4(4) = 1.75 + 1.60 = 3.35    ------ D

          4. 3(1/2 - y) =  (3/5) + 15y
            What is the solution to the equation above?

            3(1 / 2 - y) = (3 / 5) + 15y                           Solve for y. First, remove the parentheses.
            (3 / 2) - 3y = (3 / 5) + 15 y                          Get rid of the denominators. Multiply both sides by 2.
            3 - 6y = (6 / 5) + 30y                                  Now multiply both sides by 5.
            15 - 30y = 6 + 150y                                  (Note: instead of multiplying both sides by 2 and then by
                                                                                5, we could have initially multiplied both sides by 10.)

            Now solve for y:
            15 - 30y = 6 + 150y
                        9 = 180y
                        y = 9 / 180 = 1 / 20 = .05

          5. If (1 / 2)(x) + (1 / 3)(y) = 4, what is the value of 3x + 2y?
            This type of problem frustrates many students taking the SAT. They would like to get a value for x and multiply it by 3, get a value for y and multiply it by 2, and then add the two terms together. From the information given, however, we cannot get a value for x and we cannot get a value for y.
            We can try multiplying both sides by a constant or dividing both sides by a constant. In this case, multiply both sides by 6.

            [ (6)  (1 / 2) (x) ] + [ (6)  (1 / 3) (y) ] = (6) (4)
            3x + 2y = 24
            This is our answer. We got it without getting the individual values of x and y.


          SAT Verbal - Questions from Spring 2022, Week 3 (February 5, 2022)

          SAT QUICK CHALLENGE C1
          Combining Sentences Correctly

          The Comma Splice. You may recall that connecting two sentences or independent clauses with a comma creates an error called a comma splice.
          You probably also recall that various correction strategies that you might see on the SAT include the following;

          1. turning the two clauses into a single sentence by placing a semicolon between them
          2. connecting the clauses with a semicolon followed by a strong transition and a comma
          3. connecting the clauses with a comma followed by a participle. Note the comma splice in Example A, which follows, as well as the corrections in the chart below.

            Example A.  Leon was rewarded well for his invention, he got a free Hawaiian vacation.

          COMMA SPLICE CORRECTIONS

          Correction Type

          Correction

          Semicolon

          Leon was rewarded well for his invention; he got a free Hawaiian vacation.

          Semicolon, strong transition, and comma

          Leon was rewarded well for his invention; as a result, he got a free Hawaiian vacation.

          Comma and participle

          Leon was rewarded well for his invention, getting a free Hawaiian vacation.

           

          The With...ING Strategy.
          On the SAT, you might also see ", With...ING" used to correct comma splices. Note the ERRORS in Examples B and D below and the CORRECTIONS in Examples C and E.

                                     
              Example B    The new Math Team will start soon, members will practice after school on Mondays.
              Example C    The new Math Team will start soon, with members practicing after school on Mondays.
              Example D    Drama Club parents will meet in the Media Center at 10 am on Friday, Pep Club parents will meet there
                                    at 2:00 pm on Friday.
              Example E    Drama Club parents will meet in the Media Center at 10 am on Friday, with Pep
                                   Club parents meeting there at 2:00 pm on Friday.

                                                                  Now, keeping in mind the information above, complete the exercise below.

           

          QUICK CHALLENGE Exercise C1

          Directions.   Replace the underlined words in each question below with the answer choice that
                               corrects the error in the sentence. If there is no error, select choice A -- NO CHANGE.

          1.  The school's rally was great, the band put on an amazing show.

          1. NO CHANGE
          2. with the band putting on
          3. the band would have
          4. the band puts on

          2.  School buses ran on schedule despite the school bus driver strike because National Guard members driving the buses.

          1. NO CHANGE
          2. strike, because National Guard
          3. strike after National Guard
          4. strike, with National Guard

          3. Little Anna loved playing with the frisky little puppy, would laugh with delight as Midnight swatted at the bubbles she was blowing into the air. 

          1. NO CHANGE
          2. puppy, laughing with delight
          3. puppy, she laughed with delight
          4. puppy laughs with delight

          SAT Verbal - Answers to Questions from Spring 2022, Week 3 (February 5, 2022)

          1. B
          2. D
          3. B

          SAT Math - Questions from Spring 2022, Week 2 (January 29, 2022)

          Solving Linear Equations
          Simple equations

          1. Solve for x (or for any variable) by isolating x on one side of the equal sign and everything else on the other side.
          2. Always remember to do the same thing to both sides of the equation: add, subtract, multiply, divide, raise to a power, or take the square root


          Solve the following problems, then check your answers. The answer key is in the drop-down below.

          1. If x - 6 = 11, what is the value of x?

          2. If a + 1/2= 4 + 2/3 , what is the value of a?

          3. If 2(x – 40) = 3(x – 30), what is the value of x?

          4. If [(x-1) / 3] = k and k = 3, what is the value of x?

            1. 2
            2. 4
            3. 9
            4. 10

          5. If y = kx, where k is a constant, and y = 24 when x = 6, what is the value of y when x = 5?
            1. 6
            2. 15
            3. 20
            4. 23
          6. 1/4 + 1/5 = ?

          7. 2/7 + 3/y = ?

          SAT Math - Answers to Questions from Spring 2022, Week 2 (January 29, 2022)

          1. If x - 6 = 11, what is the value of x?
            x = 17

          2. If a + 1/2= 4 + 2/3 , what is the value of a?
            6a + 3 = 24 + 4
            (To simplify the equation, we multiplied both sides by 6 to get rid of the denominators; 6 is the least common denominator for 2 and 3.)
            6a = 24 + 4 – 3 = 25
            a = 25/6 = 4 1/6

          3. If 2(x – 40) = 3(x – 30), what is the value of x?
            2x - 80 = 3x – 90
            10 = x

          4. If [(x-1) / 3] = k and k = 3, what is the value of x?

            1. 2
            2. 4
            3. 9
            4. 10

              x – 1 = 9
              x = 10

          5. If y = kx, where k is a constant, and y = 24 when x = 6, what is the value of y when x = 5?
            1. 6
            2. 15
            3. 20
            4. 23

              24 = k(6)
              k = 4

              y = kx = 4(5) = 20

          We need the least common denominator for numbers 6 and 7.

          1. 1/4 + 1/5 = ?
            LCD = 20
            1/4 + 1/5 = 5/20 + 4/20 = 9/20

          2. 2/7 + 3/y = ?

            LCD = 7y
            2/7 + 3/y =
            (2y)/(7y) + (7)(3) / (7)(y) =
            (2y + 21) / 7y

          SAT Verbal - Questions from Spring 2022, Week 2 (January 29, 2022)

          SAT QUICK CHALLENGE B1
          Combining Sentences – The Strong Transition and the Participle

          The Strong Transition. A strong transition is a word or phrase that is considered clear enough to get across the meaning intended when it is used to create one single sentence from two separate ones.
          Strong transitions that are likely to be used for SAT questions include the following: however, moreover, therefore, thus, consequently, and nevertheless.

          Using a Strong Transition To Merge Two Separate Sentences into a Single One. The steps below tell how to turn two separate sentences into a single sentence by using a strong transition.
          Step 1. Place a semicolon after the first sentence.
          Step 2. Place the strong transition after that sentence and put a comma behind it.
          Step 3. Add the second sentence.

          Note that Example A below consists of two related sentences that are to be merged into a single sentence. Example B is that merged sentence, created with the help of the strong transition “therefore.”

          Example A. Mom was concerned about weather forecasts for a bad storm. She made sure that we had emergency kits in the car and in the house.
          Example B. Mom was concerned about weather forecasts for a bad storm; therefore, she made sure that we had emergency kits in the car and in the house.

          Using a Participle to Create a Single Sentence from Two Separate Ones. A participle is a word made from a verb, but that describes – just as an adjective does. Present participles end in ing. Past participle endings commonly used include ed, en, d, t, n, and ne, as in the words asked, eaten, saved, dealt, seen, and gone. See Example C1, which follows.

          Example C1
          . The burned beans filled the house with smoke. They made everybody cough. (“Burned” is a participle that describes beans.”) Note that the steps for forming a single sentence from the two related ones in Example C1 above are identified and illustrated below.

          Steps for Using a Participle to Merge the Related Sentences in Example C1 Above

          Step 1. Place a comma after the first sentence.
          Step 2. Add from the second sentence the participle and the related words that complete the thought.
          Step 3. End the sentence with a period.

               Step (1) – The burned beans filled the house with smoke.
               Step (2) – The burned beans filled the house with smoke, making everybody cough
               Step (3) – The burned beans filled the house with smoke, making everybody cough.

          Now, keeping in mind the information above, complete the exercise below.

          QUICK CHALLENGE Exercise B1

          Directions.   Replace the underlined words in each question below with the answer choice that
                               corrects the error in the sentence. If there is no error, select choice A -- NO CHANGE.

          1.  Baby Joey enjoys playing with the babysitter; He will laugh with excitement when she would peek out from behind the box and say playfully, “I see you!”

          1. NO CHANGE
          2. babysitter, he will laugh,
          3. babysitter, laughing
          4. babysitter and laugh

          2.   Several players on the team are ill. Nevertheless, the game had to be re-scheduled.

          1. NO CHANGE
          2. ill, consequently, the game
          3. ill; consequently, the game
          4. ill; however, the game

          3.  This tea kettle is a great reminder for me; whistling, to let me know when the water is hot. 

          1. NO CHANGE
          2. for me, whistling
          3. for me whistling
          4. for me, whistling;

          SAT Verbal - Answers to Questions from Spring 2022, Week 2 (January 29, 2022)

          1. C
          2. C
          3. B

          SAT Math - Questions from Spring 2022, Week 1 (January 22, 2022)

          Substitution

          Substitutions are the easiest type of SAT algebra problems. In each case, substitute a number for a letter, and perform the indicated operation.

          Examine examples 1, 2, and 3.

          Example 1.
          If x = 4 and y = – 4, then x – y =

          1. -8
          2. 0
          3. 4
          4. 8

          x – y = 4 – ( – 4) = 4 + 4 = 8 ------ D

          Example 2.
          If 2x + 7 = 13, then 3x + 1 =

          1. 3
          2. 6
          3. 9
          4. 10

          2x + 7 = 13
          2x = 6
          x = 3
          3(3) + 1 = 9 + 1 = 10 ------- D

          A common mistake here is to solve for x, which is 3, and mistakenly choose (a) for the answer.
          Always make sure that you are answering the question that was asked.

          Example 3.
          If a + b = 3, then 6 – a + 2 – b =

          1. 3
          2. 4
          3. 5
          4. 6

          This might not look like a substitution problem, but it really is. We can rewrite the expression in a form where we can substitute.
          6 – a + 2 – b
          6 + 2 – a – b
          6 +2 – (a + b)
          Now substitute 3 for (a + b)
          8 – 3 = 5 --------------- C

          Substitution can also be used as an alternative technique for solving some algebra problems. A problem may be difficult and a student may not know what to do to solve it. Other problems may involve substantial manipulations to arrive at a solution. This technique works well when the answer choices have variables in them; the technique involves picking a number (or numbers) and substituting it (or them) for the variable(s) in the problem. The following steps are used in applying the “pick numbers and substitute” method:

          1. Pick numbers for the variables in the problem.

          2. Using the numbers you picked, get an answer for the problem.

          3. Substitute the numbers you picked for the variables into each of the answer choices to see which answer is the same as the answer you calculated in step 2.

          Here is an example.
          If r and s are positive integers, how many integers are greater than rs and less than r(s + 1)?

          1. r + 1
          2. r – 1
          3. s + 1
          4. s – 1

          Step 1:
          Our first step is to pick numbers for the variables. Let r = 3 and s = 4

          Step 2:
          Next we answer the question “how many integers are greater than rs and less than r(s + 1)?”
          using the numbers we picked. rs = 3(4) = 12 and r(s + 1) = 3(4 + 1) = 3(5) = 15
          There are 2 integers, 13 and 14, that are greater than 12 and less than 15. Thus the answer to
          our question is 2.

          Step 3:
          Now we determine which answer choice equals 2 when we substitute the numbers we
          picked, r = 3 and s = 4, into each of the answer choices.

          1. r + 1 = 3 + 1 = 4 NO
          2. r – 1 = 3 – 1 = 2 YES
          3. s + 1 = 4 + 1 = 5 NO
          4. s – 1 = 4 – 1 = 3 NO              thus our answer is b.

          If we had picked different numbers, we would still get the same result – choice b is the correct answer.
          Let’s check it: Let r = 5 and s = 2
          Using the numbers we picked, rs = 5(2) = 10 and r(s + 1) = 5(2 + 1) = 5(3) = 15
          There are 4 integers, 11, 12, 13 and 14, that are greater than 10 and less than 15. Thus the answer to
          our question is 4.

          Now we determine which answer choice equals 4 when we substitute the numbers we picked,
          r = 5 and s = 2, into each of the answer choices.

          1. r + 1 = 5 + 1 = 6 NO
          2. r – 1 = 5 – 1 = 4 YES
          3. s + 1 = 2 + 1 = 3 NO
          4. s – 1 = 2 – 1 = 1 NO         thus our answer choice is still b.

          Note these suggestions when picking numbers:

          1. Select numbers that are easy to work with, such 2, 3, 4, 5, 10, or 100.
          2. Do not use 1, or 0 because using them may make more than one answer choice seem correct.
          3. If more than one answer choice matches our computed answer, pick a different number.

          Here is another example:

          If a + b = c, which of the following is equal to a2 + b2?

          1. c + 2ab
          2. c2
          3. c2 – ab
          4. c2 – 2ab

          Pick numbers: Let a = 2 and b = 3. Then c = a + b = 2 + 3 = 5
          a2 + b2 = 22 + 32 = 4 + 9 = 13

          Now see which answer choice = 13 when we substitute our values for a, b, and c.

          1. c + 2ab = 5 + 2(2)(3) = 5 + 12 = 17            NO
          2. c2 = 52 = 25                                                NO
          3. c2 – ab = 52 – 2ab 2(3) = 25 – 6 = 19        NO
          4. c2 – 2ab = 52 – 2(2)(3) = 25 – 12 = 13       YES          Thus our answer is D.


          Keeping in mind the information above, solve the following problems.

          1. If y = kx, where k is a constant, and y = 24 when x = 6, what is the value of y when x = 5?
            1. 6
            2. 15
            3. 20
            4. 23

          2. If a2 + b2 = x and ab = y, which of the following is equivalent to 4x + 8y?
            1. (a + 2b)2
            2. (2a + 2b)2
            3. (4a + 4b)2
            4. (4a + 8b)2

          3. The expression (5x – 2)/(x + 3) is equivalent to which of the following?
            1. (5 – 3)/2
            2. 5 – 2/3
            3. 5 – 2/(x + 3)
            4. 5 – 17/(x + 3)

          4. If a, b, and c are three positive integers and a > b > c, then which of the following is closest to the product abc?
            1. (a – 1)bc
            2. a(b – 1)c
            3. ab(c – 1)
            4. a(b + 1)c

          5. If y = 3x and x and y are both positive integers, which of the following is equivalent to 9x + 3x+1?
            1. y3
            2. 3y + 3
            3. y(y + 3)
            4. y2 + 3

          6. If m/n = 2/3, what is the value of 2n/m + 5? (There are no answer choices.)

          SAT Math - Answers to Questions from Spring 2022, Week 1 (January 22, 2022)

          1. If y = kx, where k is a constant, and y = 24 when x = 6, what is the value of y when x = 5?
            1. 6
            2. 15
            3. 20
            4. 23

              There are two steps in arriving at a solution here.
              First, substitute 24 for y and 6 for x in the equation y = kx, and solve for k:
              y = kx
              24 = 6k
              k = 4

              Then substitute 5 for x and 4 for k in the equation y = kx, and solve for y:
              y = kx
              y = 4(5) = 20 --------- C

          2. If a2 + b2 = x and ab = y, which of the following is equivalent to 4x + 8y?
            1. (a + 2b)2
            2. (2a + 2b)2
            3. (4a + 4b)2
            4. (4a + 8b)2
              Use the “pick numbers and substitute” method. Let a = 2 and b = 3;
              then x = a2 + b2 = 22 + 32 = 4 + 9 = 13; y = ab = 2(3) = 6
              Thus 4x + 8y = 4(13) + 8(6) = 52 + 48 = 100. Which answer choice = 100?
              1. (a + 2b)2 = (2 + 2(3))2 = (8)2 = 64 NO
              2. (2a + 2b)2 = (2(2) + 2(3))2 = (4 + 6)2 = 102 = 100 YES
              3. (4a + 4b)2 = (4(2) + 2(3))2 = (8 + 6)2 = 142 = 196 NO
              4. (4a + 8b)2 = (4(2) + 8(3))2 = (8 + 24)2 = 322 = 1024 NO

          3. The expression (5x – 2)/(x + 3) is equivalent to which of the following?
            1. (5 – 3)/2
            2. 5 – 2/3
            3. 5 – 2/(x + 3)
            4. 5 – 17/(x + 3)
              Use the “pick numbers and substitute” method.
              Let x = 2; then (5x – 2)/(x + 3) = (5(2) – 2)/(2 + 3) = 8/5. Which answer choice = 8/5?

              1. (5 – 3)/2 = 2/2 = 1 NO
              2. 5 – 2/3 = 4 1/3 NO
              3. 5 – 2/(x + 3) = 5 – 2/(2+3) = 5 – 2/5 = 4 3/5 NO
              4. 5 – 17/(x + 3) = 5 – 17/(2+3) = 5 – 17/5 = 5 – 3 2/5 = 1 3/5 = 8/5 YES

          4. If a, b, and c are three positive integers and a > b > c, then which of the following is closest to the product abc?
            1. (a – 1)bc
            2. a(b – 1)c
            3. ab(c – 1)
            4. a(b + 1)c

              Use the “pick numbers and substitute” method.
              Let a = 4, b = 3, and c = 2. Then abc = 4(3)(2) = 24. Which answer choice is closest to 24?

            1. (a – 1)bc = (4 – 1)(3)(2) = 3(3)(2) = 18 Maybe
            2. a(b – 1)c = 4(3 – 1)(2) = 4(2)(2) = 16 NO; 18 is closer to 24
            3. ab(c – 1) = 4(3)(2 – 1) = 4(3)(1) = 12 NO; 18 is closer to 24
            4. a(b + 1)c = 4(3 + 1)(2) = 4(4)(2) = 32 NO; 18 is closer to 24; the answer is A.

          5. If y = 3X and x and y are both positive integers, which of the following is equivalent to 9X + 3X+1?

            1. y3
            2. 3y + 3
            3. y(y + 3)
            4. yX + 3

              Use the “pick numbers and substitute” method.
              Let x = 2. Then y = 32 = 9; 9X + 3X+1= 92 + 32+1 = 81 + 27 = 108. Now which choice = 108?

            1. y3 = 93 = 729 NO
            2. 3y + 3 = 3(9) + 3 = 27 + 3 = 30 NO
            3. y(y + 3) = 9(9 + 3) = 9(12) = 108 YES
            4. y2 + 3 = 92 + 3 = 81 + 3 = 84 NO

          6. If m/n = 2/3, what is the value of 2n/m + 5? (There are no answer choices; a problem like this
            one could appear in the grid in section of the test.)
            Let m = 2 and n = 3. Then 2n/m = 2(3)/2 + 5 = 6/2 + 5 = 3 + 5 = 8.

          SAT Verbal - Questions from Spring 2022, Week 1 (January 22, 2022)

          SAT QUICK CHALLENGE A1
          Combining Closely Related Sentences

          Often, two closely related sentences are combined to make a single one, but the combined sentence must be punctuated properly. The correct punctuation depends on the strategy used to combine the original ones. Read the two sentences in Example A below, and note the punctuation in Examples B and C -- single sentences formed by combining the two original sentences in Example A.

          Example A -- Two Separate Sentences -- Consumers have become quite frustrated by rising grocery prices. Many products cost more today than they did a year ago.

          Example B--Combining Sentences with a Semicolon. One way to combine the two sentences is to , connect them with a semicolon. REMINDER: Remember to begin the word after the semicolon with a lower case letter -- not a capital letter (unless that letter routinely requires capitalization, such as the word "I" or someone's name), as follows. Example B. Consumers have become quite frustrated by rising grocery prices; many products cost more today than they did a year ago.

          Example C -- Combining Sentences with a Comma Followed by a FANBOYS Conjunction. You may recall that the words for, and, nor, but, or, yet, and so are known as FANBOYS conjunctions because all the words in that listing are conjunctions, and the word FANBOYS is created when you take in sequence the first letter of each of the listed conjunctions.

          Another way to form a single sentence from two separate but closely related ones is to place a comma and the appropriate FANBOYS conjunction, based on the meaning of your sentence, right after the first sentence. Then add the second one. CAUTION: Do not connect the two sentences with just a comma (but no FANBOYS conjunction) because connecting sentences with just a comma creates a highly criticized error called a comma splice. Now, see Example C, which follows.

          Example C. Consumers have become quite frustrated by rising grocery prices, as many products cost more today than they did a year ago.

          Now, keeping in mind the information above, complete SAT QUICK CHALLENGE EXERCISE A1 below.

          SAT QUICK CHALLENGE EXERCISE A1 -- Combining Closely Related Sentences

          Directions. On the blank line to the left of each sentence below, write the letter of the answer choice that corrects the underlined error in the sentence. If there is no error, select choice A -- NO CHANGE

          1.  Meals for children 6-12 are half price at Mama's Place meals for children under 6 are free.

          1. NO CHANGE
          2. Place; meals for children
          3. Place. meals for children
          4. Place, Meals for children

          2.   Della is very excited about her birthday on Saturday she will be 16 years old.

          1. NO CHANGE
          2. on Saturday. she will be
          3. on Saturday; she will be
          4. on Saturday, she will

          3.  Our teacher's son is very sick so she will not be at school today.

          1. NO CHANGE
          2. .sick, so she will
          3. sick so, she will
          4. sick so; she will

          SAT Verbal - Answers to Questions from Spring 2022, Week 1 (January 22, 2022)

          1. B
          2. C
          3. B

          SAT Math - Questions from Fall 2021, Week 10 (December 4, 2021)

          Equation of a Line

          These are some points to remember about the equation of a line.

          1. The equation of a line is y = mx + b, where m is the slope of the line and b is the y-intercept.

          2. The slope of a line shows the direction and steepness of the line. The slope can be positive or negative. If the slope is positive, the line is upward sloping. If the slope is negative, the line is downward sloping.

            The line is more vertical if the absolute value of the slope is large, and the line is more horizontal if the absolute value of the slope is small; at the limit, the slope of a horizontal line is zero.

          3. The slope of a line can be calculated from any two points on the line. The slope formula is:
            m = (y2 – y1)/(x2 – x1) where the two points are (x1,y1) and (x2,y2).
            If the two points are (3,6) and (5,10), the slope = (10 – 6)/(5 – 3) = 4/2 = 2

          4. The y-intercept is the value of y when the line crosses the y axis; it is the value of y when x = 0. In SAT problems, the y-intercept represents a base or starting amount before another amount is added. For example, you may be renting a car and the charge is a base amount plus an additional amount based on the number of miles driven.

          5. There is also an x-intercept. The x-intercept is the value of x when the line crosses the x axis; it is the value of x when y = 0.

          6. If a particular point is on a line, the equation of the line works out when that point is plugged into the equation for x and y. For example, to determine whether the point (2,10) is on the line y = 3x + 4, we substitute the x value, 2, into the equation, and we should get the y value, 10. In this case, the point (2,10) is on the line y = 3x + 4 since 3(2) + 4 = 6 + 4 = 10.

            But the point (4,5) is not on the line y = 3x + 4 since 3(4) + 4 = 12 + 4 = 16, not 5.

          7. Parallel lines: two lines are parallel if their slopes are the same.

          8. Perpendicular lines: two lines are perpendicular if the slope of one line is the negative reciprocal of the slope of the other line. For example, if the slope of line a is 2, the slope of a line perpendicular to line a is -1/2. If the slope of line j -4/5, the slope of a line perpendicular to line j is 5/4.

          9. If we are given the slope of a line and a point on that line, we can determine the equation of the line. For example, if the slope of a line = 2 and a point on that line is (4,11), we can proceed as follows to determine the equation of the line:

            Write the equation of a line: y = mx + b
            From the point (4,11), we have values for x and y; we need to find a value for b; lets call it h, and then solve for h:
            y = mx + b
            11 = 2(4) + h now solve for h
            11 = 8 + h
            3 = h

            Thus the equation of the line is y = 2x + 3

            We can check our results by plugging in the x and y values from the point (4,11) into our equation:

            y = mx + b
            y = 2(4) + 3 = 8 + 3 = 11


          Keeping in mind the information above, solve the following problems.

          1. Which of the following equations represents a line that is parallel to the line with equation y = -3x + 4?
            1. 6x + 2y = 15
            2. 3x – y = 7
            3. 2x – 3y = 6
            4. x + 3y = 1

          2. The line y = kx + 4, where k is a constant, is graphed in the xy-plane. If the line contains the point (c,d), where c ≠ 0 and d ≠ 0, what is the slope of the line in terms of c and d?
            1. (d – 4)/c
            2. (c – 4)/d
            3. (4 – d)/c
            4. (4 – c)/d

          3. The line with the equation (4/5)x + (1/3)y = 1 is graphed in the x,y-plane. What is the x coordinate of the x-intercept of the line?

          4. Which of the following is the equation of a line that is perpendicular to y + 3 = 3x -8?
            1. y + 3x = 9
            2. y – 3x = 10
            3. 9y -6x = 18
            4. 3y + x = 27

          5. If a line has a slope of -2 and it passes through the point (-3,2), what is its y-intercept?
            1. 6
            2. 0
            3. -4
            4. -6

          6. A line has intercepts at (a,0) and (0,b) in the x,y-plane. If a + b = 0 and a ≠ b, which of the following is true about the slope of the line?
            1. It is positive.
            2. It is negative.
            3. It equals 0.
            4. It is undefined.

          SAT Math - Answers to Questions from Fall 2021, Week 10 (December 4, 2021)

          1. Which of the following equations represents a line that is parallel to the line with equation y = -3x + 4?
            1. 6x + 2y = 15
            2. 3x - y = 7
            3. 2x - 3y = 6
            4. x + 3y = 1
            The slope of y = -3x + 4 is -3. The line with a slope of -3 is parallel to this line. Check each
            answer choice; solve for y and determine which one has a slope of -3.
            1. 6x + 2y = 15
                      2y = -6x + 15
                        y = -3x + 15/2    Yes
            2. 3x - y = 7
                   -y = -3x + 7
                     y = 3x - 7     No
            3. 2x - 3y = 6
                   -3y = -2x + 6
                       y = (2/3)x - 2    No
            4. x + 3y = 1
                    3y = -x + 1
                      y = -(1/3)x + 1/3    No    The answer is A.

          2. The line y = kx + 4, where k is a constant, is graphed in the xy-plane. If the line contains the point (c,d), where c ≠ 0 and d ≠ 0, what is the slope of the line in terms of c and d?
            1. (d - 4)/c
            2. (c - 4)/d
            3. (4 - d)/c
            4. (4 - c)/d
            Substitute the values of the point, c = x and d = y, into the equation and solve for the slope, k.
            y = kx + 4
            d = kc + 4
            d – 4 = kc
            (d – 4)/c = k     The answer is A.
          3. The line with the equation (4/5)x + (1/3)y = 1 is graphed in the x,y-plane. What is the x coordinate of the x-intercept of the line?

            The x-intercept is the value of x when y = 0. Let y = 0 in the equation and solve for x:

            (4/5)x + (1/3)y = 1
            (4/5)x + (1/3)(0) = 1
            (4/5)x  =  1
            x = 5/4
          4. Which of the following is the equation of a line that is perpendicular to y + 3 = 3x -8?
            1. y + 3x = 9
            2. y - 3x = 10
            3. 9y - 6x = 18
            4. 3y + x = 27
            Solve the equation, y + 3 = 3x -8, for y:     y = 3x -11

            The slope of y = 3x -11 is 3. The slope of a line that is perpendicular to this line is the negative reciprocal of 3, which is -1/3. Check each answer choice; solve for y and determine which one has a slope of -1/3.
            1. y + 3x = 9
                      y = -3x + 9    No
            2. y - 3x = 10
              y = 3x + 10           No
            3. 9y - 6x = 18
                      9y = 6x + 18
                        y = (6/9)x + 2
                        y = (2/3)x + 2    No
            4. 3y + x = 27
                    3y = -x + 27
                      y = -(1/3)x = 9     Yes
          5. If a line has a slope of -2 and it passes through the point (-3,2), what is its y-intercept?
            1. 6
            2. 0
            3. -4
            4. 6
            The y-intercept is the value of y when x = 0. Let x = 0 in the equation and solve for y. But we must first determine the equation.

            Write the equation of a line: y = mx + b
            From the point (-3,2), we have values for x and y; we need to find a value for b; let's call it k, and then solve for k:

            y = mx + b
            2 = -2(-3) + h     now solve for h
            2 = 6 + h
            -4 = h

            Thus the equation of the line is y = -2x -4
            Now let x = 0 in the equation and solve for y:  y = -2(0) -4

            y = -4    The answer is C.


          6. A line has intercepts at (a,0) and (0,b) in the x,y-plane. If a + b = 0 and a ≠ b, which of the following is true about the slope of the line?
            1. It is positive.
            2. It is negative.
            3. It equals 0.
            4. It is undefined.
            The slope = m = (y2 – y1)/(x2 – x1) where the two points are (a,0) and (0,b).
            m = (b – 0)/(0 – a) = b/-a = -b/a
            Since the slope = -b/a, many students would choose B as the answer. However, we need to know the sign of b/a. We have a negative times b/a. If b is positive and a is positive, then b/a is positive; if b is negative and a is negative, then b/a is positive; but if one if them is positive and the other one is negative, then b/a is negative. We are told that a + b = 0. Therefore one of them must be positive and the other one is negative. Thus a/b is negative, and we have m = - (b/a); and m equals a negative times a negative. Thus the slope is positive. The answer is A.
          -->

          SAT Verbal - Questions from Fall 2021, Week 10 (December 4, 2021)

          SAT QUICK CHALLENGE S21
          Parallelism and Sentence Patterns

          Parallelism Review. You may recall that when you list two or more of the same types of things in a sentence, (whether single words, phrases, clauses, or complete sentences), you must keep the sentence parallel by listing all the items in the same way. Note Sentences 1-3, which follow. (1) My brother likes playing basketball, but my sister prefers to play soccer. (2) My brother likes playing basketball, but my sister prefers playing soccer. (3) My brother likes to play basketball, but my sister prefers to play soccer.

          Sentence 1 is not parallel because the things the brother and sister like are not listed the same way, as in playing vs. to play. Sentences 2 and 3 are parallel because each one lists the activities the same way -- playing in Sentence 2 and to play in Sentence 3.

          Parallel Sentence Patterns. Some SAT questions show three complete sentences and ask you to correct the one (if any) that is not parallel. To make that decision, compare the sentence patterns, as in Sentences 4-6, which follow. (4) While the crowd gazed in amazement, Crissy recited a beautiful, dramatic poem that she had written. (5) Earl juggled balls in the air while the band played popular songs. (6) Crystal did an intricate ballet while the audience watched in tremendous admiration. Note that sentence 4 begins with a dependent clause, followed by an independent clause, but Sentences 5 and 6 each begins with an independent clause, followed by a dependent clause. To be parallel with the others, Sentence 4 must also start with an independent clause, which is then followed by a dependent clause, as follows: (4 revised): Crissy recited a beautiful, dramatic poem that she had written while the crowd gazed in amazement.

          Now with the information above in mind, complete SAT QUICK CHALLENGE EXERCISE S21 below.

          SAT Quick Challenge S21 - Parallelism

          Directions. For each question below, write the letter of the answer choice which completes the underlined part of the question correctly. If the underlined part is already correct, place the letter A (NO CHANGE) on that line. Then use the Answer Key to check your work.

          1.  Each of Grandma's finicky little cats has its own favorite sleeping spot in her room. Midnight curls up on Grandma's fluffy house shoes. Beside the heating vent near the desk is where Bo Peep stretches out. Snowball gets comfortable on the soft, fuzzy floor mat near the bed. 

          1. NO CHANGE
          2. Delete the underlined words.
          3. Place the underlined words after "out."
          4. Place the underlined words after "out," and delete "is where."

          2.   To surprise my grandparents on their thirtieth wedding anniversary, we left little gifts for them around their house. Dad arranged some of Granny's favorite flowers in a vase on the coffee table.
          Mom placed new golfing videos beside Grandpa's golf bag in the den. In a beautiful gift box on the dining room table that's the place where my siblings and I put theater tickets for the couple.

          1. NO CHANGE
          2. Delete the underlined words.
          3. Put the underlined words after "couple."
          4. Place the underlined words after the word "couple,"and end the sentence with the word "table."

          3.  The storm last week was so ferocious that we wondered if we would survive it. In some instances, strong winds blew trees onto houses. In other cases, flooding forced people to go upstairs to wait out the storm. Additionally, power failures left some families in terrible distress without lights or heat.

          1. NO CHANGE
          2. Place "is how" after heat, and put the underlined words in front of "power failures."
          3. Place a comma after "heat," and place the underlined words in front of "Additionally."
          4. Place a comma after "heat," and place the underlined words after "Additionally."

          SAT Verbal - Answers to Questions from Fall 2021, Week 10 (December 4, 2021)

          1. D
          2. D
          3. A

          SAT Math - Questions from Fall 2021, Week 9 (November 20, 2021)

          Absolute Value Equations

          The absolute value of a number is the distance of that number from 0 on the number line. The symbol |x| is used to designate the absolute value of x. Since distance is nonnegative, the absolute value of a number is always positive or zero. Thus, |7| = 7 because 7 is seven units away from 0 on the number line. Also, |– 6| = 6 because 6 is six units away from 0 on the number line.

          Some equations contain an absolute value expression, which take the form |ax + b| = c. To solve an absolute value equation, it is necessary to consider the possibility that “ax + b” may be positive and “ax + b” may be negative. To solve for x in the equation, we solve both for “ax + b = c” and “ax + b = – c”. Thus there may be two solutions for the equation. It is always necessary to check the solutions in the original equation to verify the solutions.

          Examine examples 1, 2, 3, and 4 below; solve each of the equations.

          Example 1.                |2x + 3| = 15
                                               2x + 3 = 15                2x + 3 = – 15
                                                     2x = 12                       2x = – 18
                                                        x = 6                           x = – 9

          We have two answers, 6 and – 9. Now we can check our answers for each value of x in the original equation.
           
                     x = 6                                                        x = – 9

                    |2x + 3| = 15                                        |2x + 3| = 15
                  |2(6) + 3| = 15                                  |2(– 9) + 3| = 15
                    |12 + 3| = 15                                    |– 18 + 3| = 15
                          |15| = 15                                          |– 15| = 15
                            15 = 15                                                15 = 15                 Our answers check successfully.

          Example 2.                 4|6 – 2x| – 5 = 27
          To solve the absolute value equation, the absolute value term must be alone on one side of the equation. Thus we need to add 5 to both sides of the equation and then divide both sides by 4.
                    4|6 – 2x| = 32
                      |6 – 2x| = 8
                        6 – 2x = 8                                           6 – 2x = – 8
                           – 2x = 2                                              – 2x = – 14
                                x = – 1                                                 x = 7

          We have two answers, – 1 and 7. We can check our answers for each value of x in the original equation.
                                 x = – 1                                                x = 7

               4|6 – 2x| – 5 = 27                             4|6 – 2x| – 5 = 27
           4|6 – 2(–1)| – 5 = 27                          4|6 – 2(7)| – 5 = 27
                  4|6 +2| – 5 = 27                            4|6 – 14| – 5 = 27
                       4|8| – 5 = 27                                  4|– 8| – 5 = 27
                       4(8) – 5 = 27                                      4(8) – 5 = 27
                         32 – 5 = 27                                        32 – 5 = 27
                                27 = 27                                             27 = 27                Our answers check successfully.

          Example 3.                        |3x + 9| + 4 = 0
                         |3x + 9| = – 4

          This equation has no solution. Since the absolute value of a number is always positive or zero, there is no value of x that would make this sentence true.

          Example 4.                             |x – 2| = 2x – 10

                         |x – 2| = 2x – 10                                   |x – 2| = 2x – 10
                           x – 2 = 2x – 10                                      x – 2 = – (2x – 10)
                              – 2 = x                                        – 10 x – 2 = – 2x + 10
                                 8 = x                                                     3x = 12
                                                                                                x = 4
          We have two answers, 8 and 4. We can check our answers for each value of x in the original equation.
                              x = 8                                                          x = 4

                            |x – 2| = 2x – 10                                  |x – 2| = 2x – 10
                            |8 – 2| = 2(8) – 10                               |4 – 2| = 2(4) – 10
                                  |6| = 16 – 10                                        |2| = 8 – 10
                                    6 = 6                                                     2 = – 2          NO

          In this case, only one value of x works; x = 8 works, but x = 4 does not. The only solution is x = 8.

          Keeping in mind the information above, solve the following problems.

          1. |x – 3| = 17

          2. |x + 11| = 42

          3. 11|x – 9| = 121

          4. |2x + 7| = x – 4

          5. 8|x – 3| = 88

          6. 3|x + 6| = 9x – 6

          SAT Math - Answers to Questions from Fall 2021, Week 9 (November 20, 2021)

            1. |x – 3| = 17

                                x – 3 = 17                                                x – 3 = – 17
                                       x = 20                                                      x = – 14
              We have two answers, 20 and – 14. We can check our answers for each value of x in the original equation.

                                        x = 20                                                 x = – 14
                                |x – 3| = 17                                        |x – 3| = 17
                         
                               |20 – 3| = 17                                | – 14 – 3| = 17
                                     |17| = 17                                      | – 17| = 17
                                       17 = 17                                             17 = 17         Our answers check successfully.

            2. |x + 11| = 42

                                   x + 11 = 42                                       x + 11 = – 42
                                           x = 31                                                x = – 53
              We have two answers, 31 and – 53. We can check our answers for each value of x in the original equation.

                                             x = 31                                               x = – 53

                                   |x + 11| = 42                                    |x + 11| = 42
                                |31 + 11| = 42                              | – 53 + 11| = 42
                                        |42| = 42                                       | – 42| = 42
                                          42 = 42                                              42 = 42      Our answers check successfully.

            3. 11|x – 9| = 121

                                   |x – 9| = 11
                                     x – 9 = 11                                           x – 9 = – 11
                                           x = 20                                                 x = – 2

              We have two answers, 20 and – 2. We can check our answers for each value of x in the original equation.

                                            x = 20                                                 x = – 2

                                11|x – 9| = 121                                  11|x – 9| = 121
                             11|20 – 9| = 121                              11| – 2 – 9| = 121
                                   11|11| = 121                                  11| – 11| = 121
                                  11(11)| = 121                                      11(11) = 121
                                        121 = 121                                         121 = 121    Our answers check successfully.

            4. |2x + 7| = x – 4

              There is no solution for this equation. There is no value of x that will make the two sides of the
              equation equal. The absolute value of 2x + 7 will always be greater than x – 4.

            5. 8|x – 3| = 88

                                     |x – 3| = 11
                                       x – 3 = 11                                           x – 3 = – 11
                                             x = 14                                                 x = – 8

              We have two answers, 14 and – 8. We can check our answers for each value of x in the original equation.


                                                x = 14                                              x = – 8

                                      8|x – 3| = 88                                      8|x – 3| = 88
                                   8|14 – 3| = 88                                  8| – 8 – 3| = 88
                                         8|11| = 88                                      8| – 11| = 88
                                         8(11) = 88                                           8(11) = 88
                                             88 = 88                                               88 = 88  Our answers check successfully.

            6. 3|x + 6| = 9x – 6

                                         |x + 6| = 3x – 2
                                           x + 6 = 3x – 2                                       x + 6 = – (3x – 2)
                                                 8 = 2x                                             x + 6 = – 3x + 2
                                                 4 = x                                                   4x = – 4
                                                                                                              x = – 1

              We have two answers, 4 and – 1. We can check our answers for each value of x in the original equation.

                                                     x = 4                                                  x = – 1
                                           3|x + 6| = 9x – 6                               3|x + 6| = 9x – 6
                                           3|4 + 6| = 9(4) – 6                        3| – 1 + 6| = 9(– 1) – 6
                                               3|10| = 36 – 6                                    3|5| = – 9 – 6
                                               3(10) = 30                                           3(5) = – 15
                                                   30 = 30                                             15 = – 15            NO

              In this case, only one value of x works; x = 4 works, but x = – 1 does not. The only solution is x = 4.

          SAT Verbal - Questions from Fall 2021, Week 9 (November 20, 2021)

          SAT QUICK CHALLENGE R21
          Parallelism

          What Is Parallelism? Parallelism is defined as the repetition of a specific grammatical construction within a sentence or paragraph. This writing strategy makes writing clear and easy for the reader to follow. Its use indicates that the ideas involved are equally important. For example, in a listing of two or more items, each item must appear in the same format, as in all verb phrases, all infinitives, all phrases, etc. Note the error and corrections in Sentences A-C, which follow.

          Sentence A. Flo likes flying better than to drive. Incorrect: mixes a gerund (flying) with an infinitive  (to drive)

          Sentence B
          .
          Flo likes flying better than driving. Correct: contains two gerunds (flying and driving)

          Sentence C. Flo likes to fly better than to drive. Correct: contains two infinitives (to fly and to drive)

          Do Not Mix Phrases and Clauses in a Series in the Same Sentence. Parallelism requires that similar items in a sentence be expressed in the same way-- whether those items are clauses, phrases, or even individual words. Note Sentence D, which follows. Sentence D. Fred says that he would enjoy playing football, running track, and he wants to sing in the choir. This sentence is not parallel because the two phrases in it ("playing football" and "running track") are followed by a clause ("he wants to sing in the choir"). However, all three of the elements in the series must follow the same format, as illustrated in Sentences E and F, which follow. Sentence E. Fred says that he would enjoy playing football, running track, and singing in the choir. This sentence is correct because all three elements are expressed as gerund phrases. Sentence F. Willie says that he would like to play football, run track, and sing in the choir. (Note that the infinitive "to" does not have to be repeated for each infinitive phrase; the sentence means the same even when the word "to" is not repeated.) This sentence is also correct because all three items in the series are expressed as infinitive phrases.

          Now, keeping in mind the information above, complete the Quick Challenge exercise below.

          SAT Quick Challenge R21 - Parallelism

          Directions. Replace the underlined word(s) in each question below with the answer choice that corrects the error in the sentence. If there is no error, select choice A -- NO CHANGE.

          1.  Whether you coordinate the kitchen activities, oversee the cleaning crew, or if you are managing the financial responsibilities, we truly appreciate your help with our annual "Fall Frolic."

          1. NO CHANGE
          2. manage
          3. operating
          4. if you operate

          2.   Our family will go to New York to visit popular landmarks, dine in famous restaurants, and we would love to watch live plays in the theaters there.

          1. NO CHANGE
          2. shows in the theaters would be great
          3. enjoy watching live theater productions
          4. going to see theater shows

          3.  Little Leon says that he would love playing basketball, running track, and to race sports cars when he grows up.

          1. NO CHANGE
          2. racing sports cars
          3. he'd enjoy racing sports cars
          4. he loves sports car races

          4.  People form opinions of you based on what you say and your actions.

          1. NO CHANGE
          2. promises made by you
          3. your promises
          4. what you do

          SAT Verbal - Answers to Questions from Fall 2021, Week 9 (November 20, 2021)

          1. B
          2. C
          3. B
          4. D

          SAT Math - Questions from Fall 2021, Week 8 (November 13, 2021)

          Linear Equations Review

          When solving linear equations, sometimes there is a need to solve for one variable in terms of another variable. For example, an SAT problem may ask you to solve for “a” in terms of “b”, or solve for “x” in terms of “y and z”. Some students who can readily solve linear equations where there is only one variable are not sure what to do when they are asked to solve a problem involving one variable in terms of another variable.

          Actually, the process for determining one variable in terms of another variable is similar to solving an equation where there is only one variable. We isolate the variable for which we are solving on one side of the equal sign and put everything else on the other side of the equal sign, remembering that whatever we do to one side of the equation, we must do the same for the other side of the equation. We can perform several operations to both sides of the equation: add, subtract, multiply, divide, raise to a power, or take a root.


          Now examine examples 1, 2, and 3.

          Example 1.
          Solve for x in terms of m:     2x – 3 = m + x

          Solve for x in terms of m                                                                        Similar problem with one variable
          2x – 3 = m + x                                                                                              2x – 3 = 4 + x
             Subtract x from both sides:          x – 3 = m                                              x – 3 = 4
             Add 3 to both sides: x = m + 3                                                                   x = 4 + 3 = 7


          Example 2.     
          Solve for x in terms of y: (5 + 6x)/2 = y + x
                                                   
           Solve for x in terms of y                                                                            Similar problem with one variable
               (5 + 6x)/2 = y + x                                                                                        (5 + 6x)/2 = 3 + x
               Multiply both sides by 2:  5 + 6x = 2(y + x)                                              5 + 6x = 2(3 + x)
                                                      5 + 6x = 2y + 2x                                                  5 + 6x = 2(3) + 2x
           Subtract 2x from both sides: 5 + 4x = 2y                                                     5 + 4x = 2(3)
           Subtract 5 from both sides:          4x = 2y – 5                                                    4x = 2(3) – 5
           Divide both sides by 4:                    x = (2y – 5)/4                                                 x = (2(3) – 5)/4 = (6 – 5)/4 = 1/4

          Example 3. Solve for b in terms of a, c, and x: a(b - 2)/(c – 3) = x
                     Solve for b in terms of a, c, and x                                                     Similar problem with one variable
                                   a(b - 2)/(c – 3) = x                                                                   4(b - 2)/(7 – 3) = 9
          Multiply both sides by (c – 3):   a(b-2) = x(c-3)                                             4(b-2) = 9(7 – 3)
          Divide both sides by a:           b – 2 = x(c-3)/a                                                  b – 2 = 9(7-3)/4
          Add 2 to both sides:                    b = x(c-3)/a + 2                                             b = 9(7-3)/4 + 2 = 9(4)/4 + 2 = 9+2=11


          Keeping in mind the information above, solve the following problems.

          1. Solve for x in terms of y: (x + 2y)/3 = 6 + 3x
          2. Solve for a in terms of x and b: x = – b/2a
          3. Solve for x in terms of a and n: (3ax – n)/5 = – 4
          4. Solve for y in terms of b and c: (by + 2)/3 = c
          5. Solve for x in terms of e, y, and z: ex – 2y = 3z
          6. Solve for b in terms of k, m, and x: k = (x/2)(b + m)

          SAT Math - Answers to Questions from Fall 2021, Week 8 (November 13, 2021)

            1. Solve for x in terms of y: (x + 2y)/3 = 6 + 3x
              Multiply both sides by 3: x + 2y = 3(6 + 3x)
                                                        x + 2y = 18 + 9x
              Subtract x from both sides:   2y = 18 + 8x
              Subtract 18 from both sides: 2y - 18 = 8x
              Divide both sides by 8: (2y – 18)/8 = x
                                      x = (2y – 18)/8 = (y – 9)/4
            2. Solve for a in terms of x and b: x = – b/2a 

              Multiply both sides by 2a:    2ax = – b
              Divide both sides by 2x:            a = – b//2x

            3. Solve for x in terms of a and n: (3ax – n)/5 = – 4

              Multiply both sides by 5:    3ax – n = – 20
              Add n to both sides: 3ax = n – 20
              Divide both sides by 3a: x = (n – 20)/3a

            4. Solve for y in terms of b and c: (by + 2)/3 = c

              Multiply both sides by 3: by + 2 = 3c
              Subtract 2 from both sides: by = 3c – 2
              Divide both sides by b: y = (3c – 2)/b

            5. Solve for x in terms of e, y, and z: ex – 2y = 3z

              Add 2y to both sides: ex = 3z + 2y
              Divide both sides by e: x = (3z + 2y)/e

            6. Solve for b in terms of k, m, and x: k = (x/2)(b + m)

              Multiply both sides by 2: 2k = x(b + m)
              Divide both sides by x: 2k/x = b + m
              Subtract m from both sides: 2k/x – m = b

          SAT Verbal - Questions from Fall 2021, Week 8 (November 13, 2021)

          SAT QUICK CHALLENGE Q21
          The Comma and Nonessential vs. Essential Clauses

          Essential and Nonessential Information. Information that is needed so that the reader can understand clearly what the writer is saying is called essential information. However, nonessential information gives the reader "extra" insight about the writer's point, but the sentence will still make sense without that extra information. Its removal will not affect a sentence's grammatical structure or its essential meaning. Since the added information is not a part of the writer's main point, that information must be separated from the rest of the sentence with commas. Note Sentences A - D, which follow.

          The Words "Who" and "Which": Essential or Nonessential Information? To address this question, let's examine Sentences A - D below, beginning with Sentences A and B. Sentence A. My sister who lives in Florida likes to spend her vacations skiing in the mountains. Sentence B. My sister, who lives in Florida, likes to spend her vacations skiing in the mountains. Note that both sentences talk about a sister who likes skiing, but in Sentence A, the clause "who lives in Florida" is not cut off from the rest of that sentence because the clause modifies "sister" by telling which sister likes skiing in the mountains. However, in Sentence B, commas do separate "who lives in Florida" from the rest of the sentence because (1) the writer has only one sister, and (2) the clause simply provides extra information about her; it does not affect the sentence's main point, and if the quote is removed, the sentence will still make sense.

          Note that the logic just discussed also applies to the word "which," as used in Sentences C and D, which follow. Sentence C. Because of allergies, Ava can't drink cocoa which is made from cow's milk. Sentence D. Many traditional foods, which are now hard to get because of supply shortages, may be missing from holiday meals this year.

          Essential Clauses and the Word "That." Generally speaking, clauses beginning with the word "that" are always essential, and the word "that" should not have commas or any other form of punctuation in front of or behind it. However, an interrupter must always be separated from a sentence because an interrupter is not critical to the meaning of a sentence -- even when that interrupter follows the word "that," as it does in Sentence E, which follows. Sentence E: We discovered that, as we had expected, our star player couldn't play because of a serious injury.

          While a comma does follow the word "that" in Sentence E, the purpose of the comma is just to help separate the interrupter (as we had expected) from the rest of the sentence. The interrupter simply happens to follow the word "that."

          Now, using the information above, complete the exercise below.

          SAT Quick Challenge Q21 - The Comma and Nonessential vs. Essential Clauses

          Directions. . For each question below, select the answer choice which shows where the comma(s) must be placed in order for the sentence to be punctuated correctly. Select choice A -- no CHANGE -- if you think that the sentence is already punctuated correctly.

          1.   Mrs. Evans who owns several day care centers provides reading classes for all her students.

          1. NO CHANGE
          2. after "Evans" and after "centers"
          3. after "who" and after "centers"
          4. before "who" and after "classes"

          2.   Chocolate is the ice cream flavor that my mom likes best.

          1. NO CHANGE
          2. after "Chocolate" and after "flavor"
          3. after "that"
          4. after "flavor"

          3.  Several kinds of apple pie which is Little Al's favorite dessert were served at his birthday party.

          1. NO CHANGE
          2. after "apple pie"
          3. after "served"
          4. after "apple pie" and after "dessert"

          SAT Verbal - Answers to Questions from Fall 2021, Week 8 (November 13, 2021)

          1. B
          2. A
          3. D

          SAT Math - Questions from Fall 2021, Week 7 (November 6, 2021)

          INEQUALITIES: Word Problems Continued

          The following symbols are used in inequalities:

          ≠    is not equal to
          >    is greater than
          <    is less than
          ≥     is greater than or equal to
          ≤    is less than or equal to

          Some word problems involving inequalities require that you only write the inequality; it is not necessary to solve the inequality. These problems are usually in a multiple choice context, where you must select the appropriate choice.

          It is important to read the problem carefully, identify the relevant information, and use the appropriate direction of the inequality signs.

          Examine examples 1, 2, and 3.

          1. An architect in an arid region determines that a building’s current landscaping uses $1700 worth of water monthly. The architect plans to replace the current landscaping with arid zone landscaping at a cost of $18,000, which will reduce the monthly watering cost to $820. Which of the following inequalities can be used to find x, the number of months after replacement that the savings in water costs will be at least as much as the cost of replacing the landscaping?

            1. 820x ≥ 18,000
            2. (1,700 – 820)x ≤ 18,000
            3. 820x < 18,000
            4. (1,700 – 820)x ≥ 18,000

            We want to know how long it will take for our total savings to be greater than our cost of replacement. Our monthly savings will be (1700 – 820). Eliminate choices A and C since they do not contain our savings. Our total savings will be (1700 – 820)x. We want this value to be greater than or equal to the replacement cost of 18,000. Thus our answer is D.

          2. Mr. Cox is an insurance agent who sells two types of life insurance policies: a $40,000 policy and a $200,000 policy. Last month his goal was to sell at least 70 insurance policies. While he did not meet his goal, the total value of the policies he sold was over $4,000,000. Which of the following inequalities describes x, the possible number of $40,000 policies, and y, the possible number of $200,000 policies, that Mr. Cox sold last month?

            1. x + y < 70
              40,000x + 200,000y < 4,000,000

            2. x + y > 70
              40,000x + 200,000y > 4,000,000

            3. x + y < 70
              40,000x + 200,000y > 4,000,000

            4. x + y > 70
              40,000x + 200,000y < 4,000,000

            He did not meet his goal of selling at least 70 policies; thus eliminate choices B and D. The total value of the policies he sold was over $4,000,000. Eliminate choice A. Thus our answer is C.

          3. Delores needs to hire at least 10 staff members for an upcoming project. The staff members will be made up of junior directors, who will be paid $640 per week, and senior directors, who will be paid $880 per week. Her budget for hiring the staff members is no more than $9,700 per week. She must hire at least 3 junior directors and at least 1 senior director. Which of the following inequalities represent the conditions described if x is the number of junior directors and y is the number of senior directors?
            1. 640x + 880y ≥ 9,700
              x + y ≤ 10
              x ≥ 3
              y ≥ 1

            2. 640x + 880y ≤ 9,700
              x + y ≥ 10
              x ≥ 3
              y ≥ 1

            3. 640x + 880y ≥ 9,700
              x + y ≥ 10
              x ≤ 3
              y ≤ 1

            4. 640x + 880y ≤ 9,700
              x + y ≤ 10
              x ≤ 3
              y ≤ 1

            The amount to be paid to the staff members, 640x + 880y, can be no more than 9,700. Eliminate A and C. Delores will hire at least 10 staff members. Eliminate D. The answer is B, which meets all of the inequality conditions.

          Keeping in mind the information above, solve the following problems.

          1. The average annual energy cost for a certain home is $4,800. The homeowner plans to spend $30,000 to install a geothermal heating system. The homeowner estimates that the average annual cost will be reduced to $3,100. Which of the following inequalities can be used to find x, the number of years after installation at which the total amount of energy cost savings will exceed the installation cost?

            1. 3,100x > 30,000
            2. 3,100x < 30,000 – 4,800
            3. 4,800 – 3,100)x > 30,000
            4. 4,800 – 3,100)x < 30,000

          2. A psychologist set up an experiment to study the tendency of a person to select the first item when presented with a series of items. In the experiment, 300 people were presented with a set of five pictures arranged in random order. Each person was asked to choose the most appealing picture. Of the first 150 participants, 36 choose the first picture in the set. Among the remaining 150 participants, x people chose the first picture in the set. If more than 20% of all participants chose the first picture in the set, which of the following inequalities best describes the possible values of x?

            1. x > 0.20(300 – 36), where x ≤ 150
            2. x > 0.20(300 + 36), where x ≤ 150
            3. x -36 > 0.20(300), where x ≤ 150
            4. x + 36 > 0.20(300), where x ≤ 150

          3. Jackie has two summer jobs. She works as a tutor, which pays $12 per hour and she works as a lifeguard, which pays $8 per hour. She can work no more than 20 hours per week, but she wants to earn at least $220 per week. Which of the following inequalities represent this situation in terms of x and y, where x is the number of hours she tutors and y is the number of hours she works as a lifeguard?

            1. 12x + 8y ≤ 220
              x + y ≥ 20

            2. 12x + 8y ≤ 220
              x + y ≤ 20

            3. 12x + 8y ≥ 220
              x+ y ≤ 20

            4. 12x + 8y ≥ 220
              x + y ≥ 20

          4. Dogs need 8.5 to 17 ounces of water each day for every 10 pounds of their weight. Ronald has two dogs – Rover is a 35 pound black lab mix, and Rex is a 55 pound beagle. Which of the following ranges represents the approximate total number of ounces of water , x, that Rover and Rex need in a week?

            1. 77 ≤ w ≤ 153
            2. 109 ≤ w ≤ 218
            3. 536 ≤ w ≤ 1,071
            4. 765 ≤ w ≤ 1,530

          5. A laundry service is buying detergent and fabric softener from its supplier. The supplier will deliver no more than 300 pounds in a shipment. Each container of detergent weighs 7.35 pounds and each container of fabric softener weighs 6.2 pounds. The service wants to buy at least twice as many containers of detergent as containers of fabric softener. Let d represent the number of containers of detergent, and let s represent the number of containers of fabric softener, where d and s are nonnegative integers. Which of the following inequalities best represents this situation?

            1. 7.35d + 6.2s ≤ 300
              d ≥ 2s

            2. 7.35d + 6.2s ≤ 300
              2d ≥ s

            3. 4,800 – 3,100)x > 30,000
              d ≥ 2s

            4. 4,800 – 3,100)x < 30,000
              2d ≥ s
          6. David must buy at least 100 shares of stock for his portfolio. The shares he buys will be from Stock X, which costs $22 per share and Stock Y, which costs $35 per share. His budget for buying stock is no more than $4,500. He must buy at least 20 shares of Stock X and 15 shares of Stock Y. Which of the following represents the condition described if x is the number of shares of Stock X purchased and y is the number of shares of Stock Y purchased?

            1. 22x + 35y ≤ 4,500
              x + y ≥ 100
              x ≤ 20
              y ≤ 15

            2. 22x + 35y ≤ 4,500
              x + y ≤ 100
              x ≤ 20
              y ≤ 15

            3. 22x + 35y ≤ 4,500
              x + y ≤ 100
              x ≥ 20
              y ≥ 15

            4. 22x + 35y ≤ 4,500
              x + y ≥ 100
              x ≥ 20
              y ≥ 15


          SAT Math - Answers to Questions from Fall 2021, Week 7 (November 6, 2021)

            1. The average annual energy cost for a certain home is $4,800. The homeowner plans to spend $30,000 to install a geothermal heating system. The homeowner estimates that the average annual cost will be reduced to $3,100. Which of the following inequalities can be used to find x, the number of years after installation at which the total amount of energy cost savings will exceed the installation cost?

              1. 3,100x > 30,000
              2. 3,100x < 30,000 – 4,800
              3. 4,800 – 3,100)x > 30,000
              4. 4,800 – 3,100)x < 30,000

              We want to know how long it will take for our total savings to be greater than our installation cost. Our annual savings will be (4,800 – 3,100). Eliminate choices A and B since they do not contain our savings. Our total savings will be (4,800 – 3,100)x. We want this value to be equal to or greater than the installation cost of 30,000. Thus our answer is C.

            2. A psychologist set up an experiment to study the tendency of a person to select the first item when presented with a series of items. In the experiment, 300 people were presented with a set of five pictures arranged in random order. Each person was asked to choose the most appealing picture. Of the first 150 participants, 36 choose the first picture in the set. Among the remaining 150 participants, x people chose the first picture in the set. If more than 20% of all participants chose the first picture in the set, which of the following inequalities best describes the possible values of x?

              1. x > 0.20(300 – 36), where x ≤ 150
              2. x > 0.20(300 + 36), where x ≤ 150
              3. x -36 > 0.20(300), where x ≤ 150
              4. x + 36 > 0.20(300), where x ≤ 150

              More than 20% of all participants chose the first picture in the set. 20% of 300 = 0.20(300)= 60. 36 of the first 150 participants chosen the first picture in the set. From the remaining 150, some amount, x, chose the first picture, such that x + 36 is greater than 60. Thus x + 6 > 60, or x + 36 > 0.20(300). The answer is D.

            3. Jackie has two summer jobs. She works as a tutor, which pays $12 per hour and she works as a life guard, which pays $8 per hour. She can work no more than 20 hours per week, but she wants to earn at least $220 per week. Which of the following inequalities represent this situation in terms of x and y, where x is the number of hours she tutors and y is the number of hours she works as a lifeguard?

              1. 12x + 8y ≤ 220
                x + y ≥ 20

              2. 3,100x < 30,000 – 4,800
                x + y ≤ 20

              3. 12x + 8y ≥ 220
                x+ y ≤ 20

              4. 12x + 8y ≥ 220
                x + y ≥ 20

              Jackie can work no more than 20 hours per week; eliminate A and D. She wants to earn at least $220 per week; eliminate B. Thus the answer is C.

            4. Dogs need 8.5 to 17 ounces of water each day for every 10 pounds of their weight. Ronald has two dogs – Rover is a 35 pound black lab mix, and Rex is a 55 pound beagle. Which of the following ranges represents the approximate total number of ounces of water , x, that Rover and Rex need in a week?

              1. 77 ≤ w ≤ 153
              2. 109 ≤ w ≤ 218
              3. 536 ≤ w ≤ 1,071
              4. 765 ≤ w ≤ 1,530

              The dogs need 8.5 to 17 ounces of water for every 10 pounds of weight. Together the dogs weigh 90 pounds (35 + 55 = 90). There are 9 ten pounds in 90. Thus the range would be 8.5(9) = 76.5 and 17(9) = 153. This is A: 77 ≤ w ≤ 153. But this is a trap answer that many students would choose if they did not read the question carefully. What is the trap? The range in A is for a day; the question asks for the range in a week. Thus the correct range is 76.5(7) = 535.5 and 153(7) = 1,071 (since there are 7 days in a week). The answer is C.

            5. A laundry service is buying detergent and fabric softener from its supplier. The supplier will deliver no more than 300 pounds in a shipment. Each container of detergent weighs 7.35 pounds and each container of fabric softener weighs 6.2 pounds. The service wants to buy at least twice as many containers of detergent as containers of fabric softener. Let d represent the number of containers of detergent, and let s represent the number of containers of fabric softener, where d and s are nonnegative integers. Which of the following inequalities best represents this situation?

              1. 7.35d + 6.2s ≤ 300
                d ≥ 2s
              2. 7.35d + 6.2s ≤ 300
                2d ≥ s
              3. 4,800 – 3,100)x > 30,000
                d ≥ 2s 4,800 – 3,100)x < 30,000
                2d ≥ s

              The service wants to buy at least twice as many containers of detergent as containers of fabric softener: d ≥ 2s. Eliminate B and D. 7.35 is multiplied by d and 6.2 is multiplied by s. Eliminate C. the answer is A.

            6. David must buy at least 100 shares of stock for his portfolio. The shares he buys will be from Stock X, which costs $22 per share and Stock Y, which costs $35 per share. His budget for buying stock is no more than $4,500. He must buy at least 20 shares of Stock X and 15 shares of Stock Y. Which of the following represents the condition described if x is the number of shares of Stock X purchased and y is the number of shares of Stock Y purchased?

              1. 22x + 35y ≤ 4,500
                x + y ≥ 100
                x ≤ 20
                y ≤ 15

              2. 22x + 35y ≤ 4,500
                x + y ≤ 100
                x ≤ 20
                y ≤ 15

              3. 22x + 35y ≤ 4,500
                x + y ≤ 100
                x ≥ 20
                y ≥ 15

              4. 22x + 35y ≤ 4,500
                x + y ≥ 100
                x ≥ 20
                y ≥ 15


              The total number of shares is equal to or greater than 100. Eliminate B and C. He must buy at least 20 shares of Stock X and 15 shares of Stock Y. Eliminate A. The answer is D.

          SAT Verbal - Questions from Fall 2021, Week 7 (November 6, 2021)

          SAT QUICK CHALLENGE P21
          Review -- Using Commas Correctly

          Some SAT questions test to see how well students remember how to use commas correctly, This exercise will help you maintain mastery of some of the comma skills tested frequently on the SAT.

          Separating Items in a List or Series. A very common use for commas is to separate items in a list or series, as in Sentences A1 and A2, which follow.

          Sentence A1: Missy wants shrimp, green beans, candied yams, and rolls for her birthday dinner.
          Sentence A2: Missy wants shrimp, green beans, candied yams and rolls for her birthday dinner.

          Introductory Words and Phrases. An introductory word/phrase (1) comes at the beginning of a sentence, (2) sets the tone for what the sentence will say, and (3) needs a comma after it. Examples of those words/phrases include the following: in fact, by the way, as you might expect, as you know, yes, no, initially, before, and however. Note the bold, underlined introductory phrase at the beginning of Sentence B.
          Sentence B: As we had anticipated,
          Lee took a long nap after he ran in the marathon.

          Interrupters. An interrupter (1) comes within a sentence, (2) shows emotion, tone, or emphasis by temporarily breaking the flow of the thought the sentence is expressing, and (3) must be enclosed in commas. Many expressions routinely used as introductory words/phrases (in fact, indeed, initially, however, as you might expect, for example, etc.) are also used as interrupters. Even a person's name can be used as an interrupter. Note the interrupters in Sentences C and D, which follow.
          Sentence C
          : What did you do, Roger, when you realized that you were locked out of your house?
          Sentence D: After Lee ran in the marathon, as we had anticipated, he took a very long nap.

          Now, keeping in mind the information above, complete the exercise below.

          SAT Quick Challenge P21
          Using Commas Correctly

          Directions. . For each statement below, select the answer choice which shows where the comma9s0 must be placed in order for the sentence to be punctuated correctly. Select choice A -- no CHANGE -- if you think that the sentence is already punctuated correctly. Then use the answer key in the dropdown below to check your work.

          1.   We found the keys Lynn exactly where you said they would be.

          1. NO CHANGE
          2. before and after "Lynn"
          3. after "Lynn"
          4. after "said"

          2.   As you know people who go to see the play must wear masks during the entire event.

          1. NO CHANGE
          2. after "know"
          3. before "must" and after "masks"
          4. after "play"

          3.  After the art lesson, paint brushes, paint smears, markers, glue, and scraps of paper were scattered all over the floor.

          1. NO CHANGE
          2. after "scattered"
          3. after "and"
          4. after "paper"

          SAT Verbal - Answers to Questions from Fall 2021, Week 7 (November 6, 2021)

          1. B
          2. B
          3. A


          SAT Math - Questions from Fall 2021, Week 6 (October 23, 2021)

          INEQUALITIES: Word Problems

          The following symbols are used in inequalities:

          ≠    is not equal to
          >    is greater than
          <    is less than
          ≥     is greater than or equal to
          ≤    is less than or equal to

          Just as some word problems can be approached by solving equations (see Linear Equations: Word Problems, Week 3, Fall 2021), some other word problems are solved by inequalities.

          The following steps can help you solve inequality word problems:

          1. Read the problem carefully and be sure that you understand key words and concepts.

          2. Determine how the key words and concepts are related mathematically by
            converting the words to math; write the inequality in mathematical terms.

            1. It is important to determine what the variable is; express the unknown as a
              mathematical symbol (x or any other latter).

            2. It is also important to determine accurately the direction of the inequality sign:
              is it greater than or is it less than?

            3. Determine what goes on either side of the inequality sign.
          3. Solve the problem and interpret the solution.

          4. Be sure that the answer is reasonable and answers the question that was asked.

          Examples;
          Now examine examples 1, 2, and 3.

          1. Six more than twice a number is at least 60. What is the minimum value of the number?

            Our unknown, our variable, is a number. Let’s call it x. Twice x plus 6 is greater than 60; it could be 80, or 300, or 20,000, or any number greater than 60. But we want the smallest possible value that meets our conditions. Thus the direction of the inequality sign is greater than or equal to. We want the smallest number that is greater than or equal to 27.

            2x + 6 ≥ 60
            2x ≥ 54
            x ≥ 27 Our answer is 27.

          2. The Petty family is considering renting a boat for a fun ride on the Dan River. Booker Boats charges $260 per week, plus $2 per hour, for use of its boats. A competitor, Saulter Ships, charges $40 per week, plus $8 per hour, for use of its boats. Boats of both companies are similar. How much would the Petty family have to use the boat in a week in order for Booker Boats to be the better deal?

            Because of the small weekly cost, $40 vs $260, Saulter Ships is the better deal if we use the boat just a few hours per week. But if we use the boat for many hours during the week, Booker Boats is the better choice because of its lower per hour charge. Our question, then, is when does Booker Boats become the better deal?

            Our variable is the number of hours per week; let’s call it x. We want to find x when the total cost for Booker Boats is less than the total cost for Saulter Ships. Thus the direction for the inequality is less than.

            260 + 2h < 40 + 8h
            220 < 6h
            36.667 < h

            Therefore the Petty family would have to use the boat for 37 or more hours per week for Booker Boats to be the better deal.

          3. Tickets for A&T’s winter concert cost $4.00 for students and $10 for adults. If Mr. Harrison spends at least $42 but no more than $70 on x student tickets and 3 adult tickets, what is one possible value of x?

            We have a compound inequality: an amount is between two values. For example, if x is an integer and we have

            5 < x ≤ 9 then x could be 6, 7, 8, or 9.

            In our ticket example, we have a lower amount, $42, and a higher amount, $70, and the variable is the number of student tickets.

            42 ≤ 4x + 3(10) ≤ 70
            42 ≤ 4x + 30 ≤ 70
            12 ≤ 4x ≤ 40
            3 ≤ x ≤ 10

            Thus the number of student tickets could be any number between 3 and 10, including 3 and 10.

            Keeping in mind the information above, answer the following questions.

          Keeping in mind the information above, solve the following problems.

          1. Dave and Ron are the top scorers on their high school basketball team. If the team is to have any chance of winning its next game, Dave and Ron together must score at least 50 points. What is the minimum number of points that Dave must score if Ron scores 4 points less than twice the number of Dave’s points?

          2. I have 200 shares of IBM stock. My uncle, who recently retired, has 2000 shares of IBM stock. He has substantial investments in other companies, and wants to give the IBM stock to his five nephews. Starting tomorrow, he will give each of his five nephews, including me, one share of his IMB stock every day. How many shares of IBM stock will I have on the first day that I have more shares of the stock than my uncle has?

          3. Marcus rented a motor cycle. The rental cost $15 per hour, and he also had to pay for a helmet that costs $20. In total, he spent more than $90 for the rental and helmet. If the motor cycle was available for only a whole number of hours, what was the minimum number of hours that Marcus could have rented the motor cycle?

          4.  C = 25x + 8y

            The formula above gives the monthly cost C, in dollars, of operating a printer when a technician works a total of x hours and when y reams of paper are used. If, in July it costs no more than $5,800 to operate the printer and 125 reams of paper were used, what is the maximum number of hours the technician could have worked?

            1. 144
            2. 160
            3. 192
            4. 240
          5.  A restaurant is making its weekly purchase of beef and chicken from its supplier. The supplier will deliver no more than 250 pounds in a shipment. Each package of beef
            weighs 20 pounds and each package of chicken weighs 15 pounds. The restaurant wants to purchase 2 times as many packages of chicken as packages of beef, and the restaurant wants to purchase at least 100 pounds of meat. What is the minimum number of packages of beef that should be purchased?

          6. The average monthly cable television cost for the Adams family is $120. The family plans to spend $3,000 to install a satellite television system. The family estimates that the average annual satellite television cost will be $1,100. How many years after installation of the satellite system will the total amount of television cost savings exceed
            the installation cost?

          SAT Math - Answers to Questions from Fall 2021, Week 6 (October 23, 2021)

            1. Dave and Ron are the top scorers on their high school basketball team. If the team is to have any chance of winning its next game, Dave and Ron together must score at least 50 points. What is the minimum number of points that Dave must score if Ron scores 4 points less than twice the number of Dave’s points?

              The variable is the number of points that Dave must score. The direction of the inequality sign is greater than. Dave’s score is x and Ron’s score is 2x – 4. Thus we have
              x + 2x – 4 ≥ 50
              3x ≥ 54
              x ≥ 18

              Thus Dave must score at least 18 points.

            2. I have 200 shares of IBM stock. My uncle, who recently retired, has 2000 shares of IBM stock. He has substantial investments in other companies, and wants to give the IBM stock to his five nephews. Starting tomorrow, he will give each of his five nephews, including me, one share of his IMB stock every day. How many shares of IBM stock will I have on the first day that I have more shares of the stock than my uncle has?

              Each day my uncle gives away share of stock, I gain one share. After he has given away stock on x days, I have 200 + x shares of stock. Each day my uncle gives away stock, he loses 5 shares; thus he has given away 5x shares after x days. This leaves him with 2000 – 5x shares of IBM stock. We want to know when I will have more shares than he has, so we want to know when

              200 + x > 2000 – 5x

              We now solve for x.
              200 + x > 2000 – 5x
              6x > 1800
              x > 300

              The first time I will have more shares of IBM stock than my uncle is just after he has given away shares of stock 301 times (301 is the smallest number that is greater than 300). At that point, I will have 200 + 301 = 501 shares, and he will have 2000 – 5(301) = 2000 – 1505 = 495 shares.

              Therefore I will have 501 shares of stock the first time I have more shares than my uncle has.

            3. Marcus rented a motor cycle. The rental cost $15 per hour, and he also had to pay for a helmet that costs $20. In total, he spent more than $90 for the rental and helmet. If the motor cycle was available for only a whole number of hours, what was the minimum number of hours that Marcus could have rented the motorcycle?

              The variable, x, is the number of hours Marcus rents the motorcycle.

              There is a cost of $12 per hour; if he rents the motorcycle of x hours, he will spend 12x dollars plus an initial one time payment of $20 for the helmet. Thus the total cost of the rental and helmet is 12x + 20.

              The amount he will spend is more than 90; thus we have this inequality:
              12x + 20 > 90

              Now solve for x:
              12x + 20 > 90
              12x > 70
              x > 70/12 = 5 10/12 = 5 5/6

              Since he must rent for a whole number of hours, the minimum number of hours is 6.

            4.  C = 25x + 8y

              The formula above gives the monthly cost C, in dollars, of operating a printer when a technician works a total of x hours and when y reams of paper are used. If, in July it costs no more than $5,800 to operate the printer and 125 reams of paper were used, what is the maximum number of hours the technician could have worked?

              1. 144
              2. 160
              3. 192
              4. 240

                The variable is the number of hours the technician could have worked; the direction of the inequality is less than. Thus we have the following inequality:

                25x + 8y ≤ 5,800
                25x + 8(125) ≤ 5,800
                25x + 1,000 ≤ 5,800
                25x ≤ 4,800
                x ≤ 192 -------> C

            5.  A restaurant is making its weekly purchase of beef and chicken from its supplier. The supplier will deliver no more than 250 pounds in a shipment. Each package of beef weighs 20 pounds and each package of chicken weighs 15 pounds. The restaurant wants to purchase 2 times as many packages of chicken as packages of beef, and the restaurant wants to purchase at least 100 pounds of meat. What is the minimum number of packages of beef that should be purchased?

              We have a compound inequality: an amount is between two values, 100 and 250. Our variable, x, is the number of packages of beef; the number of packages of chicken is 2x.

              100 ≤ 20x + 15(2x) ≤ 250
              100 ≤ 20x + 30x ≤ 250
              100 ≤ 50x ≤ 250
              2 ≤ x ≤ 5

              The minimum number of packages of beef is 2.

            6. The average monthly cable television cost for the Adams family is $120. The family plans to spend $3,000 to install a satellite television system. The family estimates that the average annual satellite television cost will be $1,100. How many years after installation of the satellite system will the total amount of television cost savings exceed
              the installation cost?

              Our variable, x, is the number of years it takes for the savings to exceed the installation cost. There will be annual savings since the annual cost after installation, $1,100, is less than the family’s current annual cost of $1,440 ($120(12) = $1,440). The direction of the inequality sign is greater than since the total savings will exceed the installation
              cost. Thus the inequality is:
              (1440 – 1100)x > 3,000
              340x > 3,000
              x > 8.82

              It will take 9 years of savings for the family to recoup the installation cost.

          SAT Verbal - Questions from Fall 2021, Week 6 (October 23, 2021)

          SAT QUICK CHALLENGE
          Exercise O21 -- Placing Modifiers Correctly

          The Modifier and the Introductory Modifying Phrase (IMP). You may recall that a modifier is a word/word group that provides information about another word/word group in a sentence and must be placed as close as possible to the word/word group it modifies. For example, the introductory modifying phrase (IMP) comes at the beginning of a sentence, is followed by a comma, and describes the subject of the sentence -- which comes right after the IMP, as in Sentence A, which follows.

          Sentence A. Excited that the fair starts today, crowds began lining up at the gates early this morning.
          Note the bold, underlined IMP at the beginning of the sentence, the comma that follows it, and the bold, italicized subject right after the comma. As stated in the paragraph above, the IMP modifies that subject.

          The Modifier Followed by Parentheses Instead of a Comma. Some people ignore information in parentheses, including modifiers, because they erroneously assume that anything in parentheses is unimportant. However, some students taking the SAT have lost points because they have made that same mistake. Do not let that happen to you! Remember that whether enclosed in commas or parentheses, a modifier that is placed as close as possible to the word it modifies is always placed correctly, and that kind of answer choice could be the correct answer. Note Sentences B and C, which follow.

          Sentence B. Mrs. Hall must take her baby to the pediatrician, a doctor who specializes in treating children.
          Sentence C. Mrs. Hall must take her baby to the pediatrician (a doctor who specializes in treating children).

          Although the modifier in Sentence B follows a comma, and the modifier in Sentence C is in parentheses, each one modifies the noun that it follows ("pediatrician"). Accordingly, each modifier is in the right place. Now, keeping in mind the information above, complete the exercise below.

          SAT Quick Challenge O21
          Placing Modifiers Correctly

          Directions. . For each statement below, select the letter of the answer choice which places the underlined modifier in the right place. If you think that the modifier is already in the right place, select choice A -- NO CHANGE. Then use the answer key in the dropdown below to check your work.

          1.  At First Community Church, Freedom School is held during the summer (a local church that promotes better education for low income children,.

          1. NO CHANGE
          2. after "during the summer"
          3. after "Freedom School is held"
          4. after "At First Community Church"

          2.  So that interested people can find good jobs, job fairs are held on a quarterly basis at Albany's Mt. Zion (a local church that helps workers get
               good jobs).

          1. NO CHANGE
          2. after "So that interested people"
          3. after "job fairs are held on a quarterly basis"
          4. after "can find good jobs,"

          3. For Halloween, children and adults love to dress up in costumes (a time when people wear disguises and have fun with their families and friends). .

          1. NO CHANGE
          2. after "love to dress up"
          3. after "Halloween"
          4. after "children and adults"

          SAT Verbal - Answers to Questions from Fall 2021, Week 6 (October 23, 2021)

          1. D
          2. A
          3. C

          SAT Math - Questions from Fall 2021, Week 5 (October 16, 2021)

          INEQUALITIES: No Solution; All Real Numbers Solution
          As with linear equations, it is possible for inequalities to have no solution. And it is also possible that the solution to an inequality can include all real numbers.

          The following symbols are used in inequalities:

          ≠    is not equal to
          >    is greater than
          <    is less than
          ≥     is greater than or equal to
          ≤    is less than or equal to

          An inequality will have no solution if the “solution” states something false or gives a contradiction. For example, if we work through an inequality and arrive either at the following as our solution, there is no solution.

          4 < -4
          6 < x < -6

          Statements similar to these are false and each inequality has no solution.

          We may have other inequalities where all real numbers are solutions to an inequality. This is the case when it is true for any value of the variable, whether it is 0, less than 0, or greater than 0. For example, if we work through an inequality and arrive something similar to either of the following as our solution, the inequality is true for all real numbers:

          -3x + 7 ≥ 7 -3x
          -1 ≤ x2 -1

          Examples;
          Now examine examples 1, 2, and 3.

          1. What values of x satisfy 5x + 2 < 6x < 5x - 2
            Subtract 5x from each section: 5x + 2 < 6x < 5x – 2 becomes 2 < x < - 2

            We cannot pick a value of x that is greater than 2 and less than -2. Thus this inequality has no solution.

          2. What values of x satisfy 7x -11x + 3 ≥ 3 -4x

            Combine like terms:
            7x -11x + 3 ≥ 3 -4x becomes -4x + 3 ≥ 3 -4x

            Add 4x to both sides:
            -4x + 3 ≥ 3 -4x becomes 3 ≥ 3

            This statement is true for all real numbers.

          3. What values of x satisfy 2(3x – 4) < 4 + 6x - 15
            Remove parentheses:
            2(3x – 4) < 4 + 6x – 15 becomes 6x – 8 < 4 + 6x - 15
            Combine like terms:
            6x – 8 < 4 + 6x - 15 becomes 6x – 8 < 6x - 11

            Subtract 6x from both sides: 6x – 8 < 6x - 11 becomes – 8 < - 11

            This statement is false because – 8 is not less than –11. Therefore the inequality has no solution.

          Keeping in mind the information above, solve the following problems.

          1. What values of x satisfy 9 + 2x – 5x ≥ –x + 12 – 2x ?

          2. What values of x satisfy 10(x -2) < 6x?

          3. What values of c satisfy 2 – 3c ≥ 6c – 3 – 9c?

          4. What values of m satisfy 9m + 5 – 12m ≥ 7 + 3m +10?

          5. What values of y satisfy 9 + 6(y + 1) ≥ 18 + 3(3y – 1) – 3y?

          6. What values of x satisfy 5(x – 1) +7 ≤ 2(x – 4) +3x + 1?

          SAT Math - Answers to Questions from Fall 2021, Week 5 (October 16, 2021)

          1. What values of x satisfy 9 + 2x – 5x ≥ –x + 12 – 2x ?
            Combine like terms:
            9 + 2x – 5x ≥ –x + 12 – 2x becomes 9 – 3x ≥ 12 – 3x

            Add 3x to both sides:
            9 – 3x ≥ 12 – 3x becomes 9 ≥ 12

            This statement is false because 9 is not greater than or equal to 12. Therefore the inequality has no solution.

          2. What values of x satisfy 10(x -2) < 6x?
            Remove parentheses:
            10(x -2) < 6x becomes 10x - 20 < 6x

            Subtract 6x from both sides and add 20 to both sides:
            10x -20 < 6x becomes 4x < 20

            Solve for x: x < 5

            This is a true statement. Here x could be any value less than 5, not including 5.

          3. What values of c satisfy 2 – 3c ≥ 6c – 3 – 9c?
            Combine like terms:
            2 – 3c ≥ 6c – 3 – 9c becomes 2 – 3c ≥ – 3 – 3c

            Add 3c to both sides:
            2 – 3c ≥ – 3 – 3c becomes 2 ≥ – 3

            This statement is true for all real numbers of c. Here c could be any real number to solve the inequality.

          4. What values of m satisfy 9m + 5 – 12m ≥ 7 + 3m +10?
            Combine like terms:
            9m + 5 – 12m ≥ 7 + 3m +10 becomes 5 – 3m ≥ 17 + 3m

            Subtract 17 from both sides and add 3m to both sides:
            5 – 3m ≥ 17 + 3m becomes – 12 ≥ 6m

            Solve for m: – 2 ≥ m
            This is a true statement. Here m could be any value less than or equal to 2, including 2.

          5. What values of y satisfy 9 + 6(y + 1) ≥ 18 + 3(3y – 1) – 3y?
            Remove parentheses:
            9 + 6(y + 1) ≥ 18 + 3(3y – 1) – 3y becomes 9 + 6y +6 ≥ 18 + 9y –3 – 3y

            Combine like terms:
            9 + 6y +6 ≥ 18 + 9y –3 – 3y becomes 15 + 6y ≥ 15 + 6y

            Subtract 6y from both sides: 15 + 6y ≥ 15 + 6y becomes 15 ≥ 15

            This statement is true for all real numbers of y. Here y could be any real number to solve the inequality.

          6. What values of x satisfy 5(x – 1) +7 ≤ 2(x – 4) +3x + 1?
            Remove parentheses: 5(x – 1) +7 ≤ 2(x – 4) +3x + 1 becomes 5x – 5 +7 ≤ 2x – 8+3x +1

            Combine like terms:
            5x – 5 +7 ≤ 2x – 8+3x +1 becomes 5x +2 ≤ 5x –7

            Subtract 5x from both sides:
            5x +2 ≤ 5x –7 becomes 2 ≤ –7

            This statement is false because 2 is not less than or equal to -7. Therefore the inequality has no solution.

          SAT Verbal - Questions from Fall 2021, Week 5 (October 16, 2021)

          SAT QUICK CHALLENGE
          Exercise N21 -- Dangling Modifiers

          The Modifier. A modifier is a word/word group that adds extra information about another word/word group in a sentence. The modifier must be placed as close as possible to the word/word group being modified. An introductory modifying phrase (IMP) comes at the beginning of a sentence; is followed by a comma; and describes the subject of the sentence, which must come right after the IMP.

          The Dangling Modifier. When a sentence has a modifier in it, but does not have the word that is being modified, that modifier is said to be "dangling," and the sentence's message can be illogical, unclear, or absurd, as you will note in Sentence A, which follows.

          Sentence A: After undergoing extensive physical therapy, Grandma's balance improved greatly. Note that the IMP at the beginning of this sentence tells us that Grandma's balance (the subject of the sentence) has had extensive physical therapy. Of course, that idea does not make sense. However, we can tell (1) that it was actually Grandma -- not her balance -- that had physical therapy, and (2) that her balance improved after the physical therapy. Hence, the IMP is a dangling modifier.

          Correcting Dangling Modifiers. Sentences B and C below show two ways of correcting Sentence A's dangling modifier. One correction strategy is to ask yourself who/what the IMP logically refers to. (In this case, the answer is "Grandma.") Then, place that word right into the IMP. Finally, following the comma after the IMP, complete the point being made about that word -- as in Sentence B, which follows. Sentence (B). After Grandma had extensive physical therapy, her balance improved tremendously. Now, there is no dangling modifier, and the sentence makes sense.

          Another correction strategy is to leave intact the IMP, place the subject right after the comma that follows the IMP (in this case, "Grandma"), and end the sentence with an appropriately revised main clause that tells about the results of the physical therapy, as in Sentence C, which follows: 
          Sentence (C). After undergoing extensive physical therapy, Grandma found that her balance had improved tremendously. Again, the dangling modifier is gone, and the sentence now makes sense.


          Keeping in mind the information above, complete the exercise below.




          SAT Quick Challenge N21
          Misplaced Modifiers

          Directions. . For each statement below, select the letter of the answer choice which corrects the underlined part of that statement. If you believe that the underlined part is already correct, select Choice A -- NO CHANGE. After completing this activity, use the answer key in the dropdown below to check your work.

          1. Watching the evening news, the smoke alarm went off, and smoke spread throughout the house.

          1. NO CHANGE
          2. With the evening news
          3. While Mr. Carter watched the evening news
          4. Coughing as smoke filled the air

          2.  To become such a great singer, regular, faithful practice daily was for Ava

          1. NO CHANGE
          2. regular, faithful practice was every day for Ava.
          3. Ava's dedicated practice was on a daily basis.
          4. Ava practiced faithfully each day.

          3. Flying along the road in the new sports car, the world seemed to whiz by rapidly.

          1. NO CHANGE
          2. As Olivia raced
          3. Olivia speeding
          4. Dashing and zipping

          SAT Verbal - Answers to Questions from Fall 2021, Week 5 (October 16, 2021)

          1. C
          2. D
          3. B


          SAT Math - Questions from Fall 2021, Week 4 (October 9, 2021)

          INEQUALITIES

          In an equation, one side equals the other. In an inequality, the two sides are not equal.

          The following symbols are used in inequalities:

          ≠    is not equal to
          >    is greater than
          <    is less than
          ≤    is less than or equal to
          ≥     is greater than or equal to

          You can usually work with inequalities in exactly the same way you work with equations. You can collect similar terms, and you can simplify by doing the same thing to both sides: adding, subtracting, multiplying, dividing, raising to a power, or taking a root. An important caution: multiplying both sides of an inequality by a negative number reverses the direction of the inequality.

          Examine examples 1, 2, and 3.

          1. x > y    
              a. Add 2 to both sides
              b. Multiply both sides by 10
              c. Multiply both sides by -2

              a. x > y becomes x + 2 > y + 2
              b. x > y becomes 10x > 10y
              c. x > y becomes -2x < -2y

          2. If 2x < 3 and 3x > 4, what is one possible value of x?
              The best approach here is to solve for x and then express each fraction as a decimal.
              2x < 3                                       3x > 4
                x < 3/2                                      x > 4/3
                x < 1.5                                       x > 1.33333

                 Thus x is between 1.33333 and 1.5, not including 1.33333 and 1.5. Any value in this range would be acceptable.

          3.   What values of x satisfy 7 + 2x – 5x ≥ – x + 13 – 4x ?

                Solve for x by adding, subtracting, multiplying, dividing, raising to a power, or taking a root.
                -3x + 7 ≥ –5x + 13
                  2x ≥ 6
                    x ≥ 3

                    Thus x is any value greater than or equal to 3, including 3.

             Keeping in mind the information above, solve the following problems.

          1. 6x - 9y > 12
            Which of the following inequalities is equivalent to the inequality above?

            1. x - y > 2
            2. 2x - 3y > 4
            3. 3x - 2y > 4
            4. 3y - 2x > 2

          2. If -4 < x < -2, which of the following could be the value of 3x?

            1. -2.5
            2. -3.5
            3. -4.5
            4. -7.5
            5. -14.5

          3. What values of x satisfy 7x + 3 – 10x ≥ 4 + 2x + 14?

          4. If x + 6 > 0 and 1 – 2x > - 1, then x could equal each of the following EXCEPT
            1. -6
            2. -4
            3. 0
            4. 1/2

          5. If -6 < -4x + 10 ≤ 2, what is the least possible value of 4x + 3?
            1. 2
            2. 5
            3. 8
            4. 11

          6. If -8 < -(3/5)r + 1 ≤ -(16/5) , what is one possible value of r?
            1. 3
            2. 5
            3. 8
            4. 16

          SAT Math - Answers to Questions from Fall 2021, Week 4 (October 9, 2021)

              1. 6x - 9y > 12
                Which of the following inequalities is equivalent to the inequality above?

                1. x - y > 2
                2. 2x - 3y > 4
                3. 3x - 2y > 4
                4. 3y - 2x > 2

                Divide each term by 3: 6x – 9x > 12 becomes 2x – 3y > 4                  --------> B
              2. If -4 < x < -2, which of the following could be the value of 3x?

                1. -2.5
                2. -3.5
                3. -4.5
                4. -7.5
                5. -14.5

                Multiply each term by 3: -4 < x < -2 becomes -12 < 3x < -6 It could be -7.5.             --------> D
              3. What values of x satisfy 7x + 3 – 10x ≥ 4 + 2x + 14?

                Solve for x: 7x + 3 – 10x ≥ 4 + 2x + 14
                                             3 – 3x ≥ 2x + 18
                                                – 5x ≥ 15
                                                  – x ≥ 3
                                                    x ≤ -3

              4. If x + 6 > 0 and 1 – 2x > - 1, then x could equal each of the following EXCEPT
                  1. -6
                  2. -4
                  3. 0
                  4. 1/2

                Solve each inequality and determine the range of values for x.
                x + 6 > 0                                  1 – 2x > - 1
                x > -6                                           –2x > - 2
                                                                         x < 1

                Here x could be any value between -6 and 1, not including -6 and 1. Thus x could be -4, 0, or ½. It could not be -6. The answer is A.

              5. If -6 < -4x + 10 ≤ 2, what is the least possible value of 4x + 3?
                  1. 2
                  2. 5
                  3. 8
                  4. 11

                What do we do here? We want to add, subtract, multiply, or divide so that -4x + 10 becomes 4x + 3.
                The first step is to change the -4x to 4x.
                We can make this change by multiplying each section by -1:
                -6 < -4x + 10 ≤ 2 becomes
                 6 > 4x - 10 ≥ - 2

                Next we need to change the -10 to +3. We can make this change by adding 13 to each section:
                6 > 4x - 10 ≥ - 2 becomes
                19 > 4x +3 ≥ 11

                Thus 4x + 3 is between 11 and 19, not including 19. The lowest value is 11.     --------> D

              6. If -6 < -4x + 10 ≤ 2, what is the least possible value of 4x + 3?
                1. 3
                2. 5
                3. 8
                4. 16

                                      We will solve for r. We will begin by multiplying each section by -1:
                                      -8 < -(3/5)r + 1 ≤ -(16/5) becomes
                                        8 > (3/5)r - 1 ≥ (16/5)

                                        Now we multiply each section by 5 (to get rid of the 5 in the denominator).
                                        8 > (3/5)r - 1 ≥ (16/5) becomes
                                        40 > 3r - 5 ≥ 16

                                        Now we add 5 to each section.
                                        40 > 3r - 5 ≥ 16 becomes
                                        45 > 3r ≥ 21

                                        Finally we divide each section by 3. 45 > 3r ≥ 21
                                        becomes 15 > r ≥ 7

                                        Thus r is any value between 7 and 15, not including 15.

          SAT Verbal - Questions from Fall 2021, Week 4 (October 9, 2021)

          SAT QUICK CHALLENGE
          Exercise M21 -- The Misplaced Modifier

          A modifier is a word or word group that describes someone or something. If the modifier is not placed next to (or as close as possible to) the word/word group it should describe, it will be misplaced, and the sentence will not convey the meaning intended. The "possessive noun" and "descriptive aside" modifier errors are often tested on the SAT. Both are explained below.

          The Possessive Noun Error. A possessive noun error is a situation in which a possessive noun is placed where a noun that is not possessive should be. Note the following sentence: Using endless patience and encouragement, Mrs. White's clear, understandable explanations helped me learn to solve math problems I had never been able to solve before.

          EXPLANATION
          : The closest noun phrase to the introductory modifier ("Using endIess patience and encouragement") is "Mrs. White's clear, understandable explanations." Accordingly, the sentence says that Mrs. White's explanations were using endless patience and encouragement to help me learn to do math, but explanations do not use endless patience and encouragement, and they do not teach; teachers do those things. Hence, the sentence has a possessive noun error.  
          Note the following correction: "Using endless patience and encouragement, as well as clear, understandable explanations, Mrs. White helped me learn to solve math problems I had never been able to solve before." Now, the modifying phrases correctly modify "Mrs. White," and the sentence makes sense.

          The Misplaced "Descriptive Aside" Modifier. Sometimes, a modifier that is not part of a sentence's main idea is added to a sentence, as follows: "Diana Carter does breathtaking stunts, the drum major, at the games." Enclosed in commas because it is extra, nonessential information, "the drum major" is interpreted as describing the noun it follows -- "stunts." Accordingly, the sentence says that "stunts" are the drum major. Obviously, the description is not in the right place, and the point that it actually makes where it is does not make sense because it identifies the word "stunts" as the drum major. Placed after "Diana Carter," the proper noun that it should modify, the modifier would say correctly that Diana is the drum major, as indicated in the sentence which follows: "Diana Carter, the drum major, does breathtaking stunts at the games." Now, keep in mind the information above, and complete Exercise M21 below. 


          SAT Quick Challenge M21
          The Misplaced Modifier

          Directions. Select the letter of the answer choice which corrects the underlined part of each statement below. If you believe that the underlined part is already correct, select Choice A -- NO CHANGE. After completing this activity, use the answer key to check your work.

          1. Practicing two hours every weekday and four hours every Saturday and Sunday, Ava's saxophone skills improved a great deal in just
              a few months.. _______

          1. NO CHANGE
          2. Ava improved her saxophone skills
          3. the saxophone skills of Ava improved
          4. improvement was noticed in Ava's playing

          2. Granny Hawkins mixes her own spices, an energetic little lady in her late eighties, for her delicious, award-winning apple pies. To make this sentence grammatically correct, the underlined phrase should be placed

          1. NO CHANGE
          2. after the word "mixes"
          3. after the word "pies"
          4. after the word "Hawkins"

          3. Overjoyed about finally beating their longtime competitor, the crowd's cheers and celebrations continued long after the game had ended.  ______

          1. NO CHANGE
          2. cheering and celebrating continued
          3. fans continued to cheer and celebrate
          4. the stadium was filled with cheering crowds

          SAT Verbal - Answers to Questions from Fall 2021, Week 4 (October 9, 2021)

          1. B
          2. D
          3. C


          SAT Math - Questions from Fall 2021, Week 3 (October 2, 2021)

          Linear Equations: Word Problems

          In recent administrations of the SAT there has been an increase in the number of word problems included. The following steps can help you solve word problems:

          1. Read the problem carefully and avoid misreading anything important.
            1. It is necessary to define the variables and create equations to represent relationships. State in words what the unknown is.
            2. Carefully determine the equality: identify what goes on the left side of the equal sign and what goes on the right side of the equal sign. Identify the key values and determine how they are related mathematically by converting the words to math.
          2. Solve the problem and interpret the solution.
          3. Make sure that the answer is reasonable. Ask yourself: Does it make sense? Does it answer the question that was asked?

          Now examine examples 1, 2, and 3.

          Example 1:
          A computer that costs $800 is to be purchased with a down payment of $80 and weekly payments of $40. How many weekly payments will be necessary to complete the purchase, assuming that there are no taxes or fees?

          To answer this question, we need to determine the elements of the equation: what the variable is, what goes on the right side of the equal sign, and what goes on the left side of the equation.
          1. The variable is the number of weekly payments necessary to complete the purchase.
          2. The cost of the computer, $800, goes on the right side of the equal sign.
          3. The amount needed to purchase the computer, the down payment plus the weekly payments, goes on the left side of the equal sign.
            80 + 40x = 800
            40x = 720
            x = 18                                    18 weekly payments are needed.

          Example 2:
          Harold and his roommate are buying a printer for the computer that they share. Since Harold will use the printer much more often than his roommate, Harold will pay $70 more than his roommate. If the printer costs $260, how much will the roommate pay?
          1. The variable is the amount the roommate will have to pay.
          2. The cost of the printer, $260, goes on the right side of the equal sign.
          3. The amount the roommate will pay, x, plus the amount Harold will pay, $70 + x, goes on the left side of the equal sign.
            x + 70 + x = 260
            70 + 2x = 260
            2x = 190
            x = 95                         The roommate will pay $95 and Harold will pay $165.

          Example 3:
          Oliver is selling photographs as part of a project for his entrepreneurship class. He sells the first 20 photographs for $10 each. Because the first 20 photographs sold so quickly, he raised the price of the photographs to $15 each for the rest of the project. After his expenses, Oliver earns a profit of 80% of the revenues from his sales. How many photographs must he sell for the rest of the project to earn a profit of $400?
          1. 18
          2. 20
          3. 24
          4. 32
          The variable is the number of additional photographs he must sell. The profit of $400 goes on the right side of the equal sign. The total revenues times 80% goes on the left side of the equal sign. The total revenues = $10 times 20 + $15 times x. Thus we have:

          0.8(200 + 15x) = 400
          160 + 12x = 400
          12x = 240
          x = 20 ----------------------> B



          Keeping in mind the information above, answer the following questions.

          1. A customer paid $53.00 for a jacket after a 6% sales tax was added. What was the price of the jacket before the sales tax was added?

            1. $47.60
            2. $50.00 
            3. $52.60
            4. $52.84
          2. Lynn has $8.00 to spend on apples and oranges. Apples cost $0.65 each, and oranges cost $0.75 each. If there is no sales tax on this purchase and she buys 5 apples, what is the maximum number of whole oranges she can buy?

          3. Sam plans to rent a boat. The boat rental costs $60 per hour plus a water safety course that costs $10. Sam has budgeted $280 for the rental and the course. If the boat rental is available only for a whole number of hours, what is the maximum number of hours for which Sam can rent the boat?

          4. A school district is forming a committee to discuss plans for the construction of a new high school. Of those invited to join the committee, 15% are parents of students, 45% are teachers from the current high school, 25% are school and district administrators, and 6 individuals are students. How many more teachers were invited to join the committee than school and district administrators?

          5. The mean score of 8 players in a basketball game was 14.5 points. If the highest individual score is removed, the mean score of the remaining 7 players becomes 12 points. What was the highest score?
            1. 20
            2. 24
            3. 32
            4. 36

          6. Max is working this summer as part of a crew on a farm. He earned $8 per hour for the first 10 hours he worked this week. Because of his performance, his crew leader raised his salary to $10 per hour for the rest of the week. Max saves 90% of his earnings from each week. How many hours must he work the rest of the week to save $270 for the week?
            1. 16
            2. 22
            3. 33
            4. 38


          SAT Math - Answers to Questions from Fall 2021, Week 3 (October 2, 2021)

          1. A customer paid $53.00 for a jacket after a 6% sales tax was added. What was the price of the jacket before the sales tax was added?

            1. $47.60
            2. $50.00 
            3. $52.60
            4. $52.84

            The variable is the price of the jacket before the sales tax is added. The amount the customer paid of $53.00 goes on the right side of the equal sign. The initial price plus the sales tax goes on the left side of the equal sign.

            x + .06x = 53
            1.06x = 53
            x = 50 ----------------> B
          2. Lynn has $8.00 to spend on apples and oranges. Apples cost $0.65 each, and oranges cost $0.75 each. If there is no sales tax on this purchase and she buys 5 apples, what is the maximum number of whole oranges she can buy?

            The variable is the number of oranges she can buy. The amount she can spend, $8.00, goes on the right side of the equal sign. The amount spent on apples and oranges goes on the left side of the equal sign.

            0.65 times 5 + 0.75 times x = 8.00
            3.25 + 0.75x = 8
            0.75x = 4.75
            x = 6.33                          She can buy 6 whole oranges.

          3. Sam plans to rent a boat. The boat rental costs $60 per hour plus a water safety course that costs $10. Sam has budgeted $280 for the rental and the course. If the boat rental is available only for a whole number of hours, what is the maximum number of hours for which Sam can rent the boat?

            The variable is the number of hours he can rent the boat. The amount he has budgeted, $280, goes on the right side of the equal sign. The rental cost plus the cost of the water safety course goes on the left side of the equal sign.

            Thus our equation is:
            60x + 10 = 280
            60x = 270
            x = 4.5
            He can rent a boat for a maximum of 4 hours since the boat can be rented only for a whole number of hours.

          4. A school district is forming a committee to discuss plans for the construction of a new high school. Of those invited to join the committee, 15% are parents of students, 45% are teachers from the current high school, 25% are school and district administrators, and 6 individuals are students. How many more teachers were invited to join the committee than school and district administrators?

            The variable is the number of people invited to join the committee. The total number of people invited to join the committee goes on the right side of the equal sign. The total of the individual groups goes on the left side of the equal sign.

            Thus our equation is:
            0.15x + 0.45x + 0.25x + 6 = x
            0.85x + 6 = x
            6 = 0.15x
            x = 6 / 0.15 = 40
            40 people were invited to join the committee.

            Teachers:                                                  0.45(40) = 18
            School and district administrators:        0.25(40) = 10
                                                                                                ---
                                                                                                   8    Our answer is 8.

          5. The mean score of 8 players in a basketball game was 14.5 points. If the highest individual score is removed, the mean score of the remaining 7 players becomes 12 points. What was the highest score?
            1. 20
            2. 24
            3. 32
            4. 36

            The variable is the highest individual score. The total number of points scored by the remaining 7 players goes on the right side of the equal sign. The total number of points scored by all 8 players minus the highest individual score goes on the left side of the equal sign.

            Remember that in a mean (average) problem the first thing to do is to find the total. There are two ways to find the total: 1) add all of the items; 2) multiply the mean by the number of items. On the SAT the second method is used more often.

            The total of the seven players = 7(12) = 84
            The total of the eight players = 8(14.5) = 116

            The equation is:
            116 - x = 84
            -x = -32
            x = 32 --------------------> C

          6. Max is working this summer as part of a crew on a farm. He earned $8 per hour for the first 10 hours he worked this week. Because of his performance, his crew leader raised his salary to $10 per hour for the rest of the week. Max saves 90% of his earnings from each week. How many hours must he work the rest of the week to save $270 for the week?
              1. 16
              2. 22
              3. 33
              4. 38

            The variable is the number of additional hours he must work. The savings of $270 goes on the right side of the equal sign. The total earnings times 90% goes on the left side of the equal sign. The total earnings = $8 times 10 + $10 times x. Thus we have: 0.9(80 + 10x) = 270

            72 + 9x = 270
            9x = 198
            x = 22 --------------------> B

          SAT Verbal - Questions from Fall 2021, Week 3 (October 2, 2021)

          SAT QUICK CHALLENGE
          Exercise L21 -- Noun Agreement Errors and Faulty Comparison

          Then or Than for Comparisons? (1) Use "then" to indicate "when," as in the following: "First, do your homework. Then, you can get on social media." Getting on social media must wait until after the homework has been done. Did you note (a) that "then" and "when" rhyme and (b) that both words end with "en"? (2) Use "than" to make a comparison, as in the following: "Cindy runs faster than Diana." Did you note that both "comparison" and "than" have the letter "a" in them? Use than to compare!

          Consistency -- Noun and Pronoun Synonyms. When a writer uses a noun repeatedly, a synonymous noun or pronoun should replace some of those repetitions. Moreover, to maintain consistency, the writer must replace singular nouns with singular counterparts and plural nouns with plural counterparts.

          Using "That of" and "Those of." Both "that of" and "those of" (1) show ownership and (2) help you avoid being repetitive. For example, to compare the success of two singers, you could say, "Singer A's annual income is higher than the annual income of Singer B." However, to be concise, you could say, "Singer A's annual income is higher than that of Singer B." For plural nouns, use the pronoun "those" in place of "that." Hence, you could say, "Singer A's annual sales are higher than those of Singer B."

          Comparing Similar Things. Because there is no basis for comparison without recognized similarities or differences, writers must compare the same kinds of things: living things with living things, cars with cars, sports with sports, etc. In fact, a common kind of comparison error on the SAT is a statement which does not compare the same kinds of things. Therefore, you must be able to recognize and correct such errors, often referred to as "faulty comparisons." Note the faulty comparison error in the next sentence, as well as various corrections beneath the sentence: Tamela Mann's music is less popular than Kirk Franklin.

          EXPLANATION AND CORRECTIONS: That sentence does not compare the same kinds of things. It compares music (Tamela Mann's music) with a person (singer Kirk Franklin). Note corrections 1-3 below.

          1. Tamela Mann's music is less popular than the music of Kirk Franklin. (compares music with music)
          2. Tamela Mann's music is less popular than Kirk Franklin. (compares music with music)
          3. Tamela Mann's music is less popular than that of Kirk Franklin. (compares music with music; the singular pronoun "that" takes the place of the singular noun phrase "the music")

            Now, using the information above, complete the exercise below. Then use the answer key in the dropdown below to check your work.

          SAT QUICK CHALLENGE Exercise L21
          Noun Agreement Errors and Faulty Comparisons

          Directions. For each sentence below, select the answer choice that corrects the error in the underlined part. If you think the underlined part is already correct, select choice A -- NO CHANGE. When you have finished the exercise, use the answer key to check your work.

          1. Cindy Davis said that her cousin's jokes are funnier than Jeff Allen. _______

          1. NO CHANGE
          2. than those of Jeff Allen's
          3. than those of Jeff Allen
          4. as those of Jeff Allen's

          2. Mr. Stuart's trained service dogs have instincts of a mature adult. ____

          1. NO CHANGE
          2. similar to those of a mature adult
          3. of a young adult
          4. similar to a young adult

          3. After weeks of private coaching sessions, Olivia now sings better than anyone else in her choir. ______

          1. NO CHANGE
          2. now sings better then anybody else
          3. now sings better then anyone
          4. now sings better than anyone else

          SAT Verbal - Answers to Questions from Fall 2021, Week 3 (October 2, 2021)

          1. C
          2. B
          3. D

          SAT Math - Questions from Fall 2021, Week 2 (September 25, 2021)

          Linear Equations: one solution, no solution, and an infinite number of solutions

          Linear equations may have one solution, no solution, or an infinite number of solutions. On the SAT most of the linear equation problems will have one solution. It is important to recognize, however, that there will occasionally be problems where there is no solution or an infinite number of solutions.

          There will be one solution when we can isolate the variable for which we are solving on one side of the equal sign and the other parts of the equation on the other side. By adding, subtracting, multiplying, dividing, raising to a power, or taking a root, we arrive at a situation such as “x = 4” or “x = 17”.

          There will be no solution when we make our algebraic manipulations and arrive at a situation such as “2x + 7 = 2x + 5” or “ + 7 = + 5”. The coefficient of the x term on the left side of the equation is the same as the coefficient of the x term on right side of the equation, and the constant on the left side of the equation is different from the constant on the right side of the equation. There is no value of x that will make both sides of the equation equal, and + 7 can never equal + 5.

          There will be an infinite number of solutions when we make our algebraic manipulations and arrive at a situation such as “5x + 3 = 5x + 3”. The coefficient of the x term on the left side of the equation is the same as the coefficient of the x term on the right side of the equation, and the constant on the left side of the equation is the same as the constant on the right side of the equation. All possible values of x will make both sides of the equation equal.

          Consider the following example (example 1):

          Solve the following equation: 7x – 6 –x = 12 + 2x + 10
          Add or subtract to combine like terms: 6x – 6 = 22 + 2x

          Subtract 2x from both sides and add 6 to both sides: 4x = 28
          Divide both sides by 4: x = 7. There is one solution for example 1.


          Consider another example (example 2):

          Solve the following equation: x + 5 + 3x = 6x – 2 – 6
          Add or subtract to combine like terms: 4x + 5 = 4x – 8
          There is no value of x that will make both sides of the equation equal, and + 5 can never equal – 8.
          There is no solution for example 2.



          Consider another example (example 3):


          Solve the following equation: 3x – 9 + 5x + 4 = 11x – 5 – 3x
          Add or subtract to combine like terms: 8x – 5 = 8x – 5
          In example 3 there is an infinite number of solutions. All possible values of x will make both sides of the equation equal.


          Keeping in mind the information above, solve the following problems.
          1. Solve the following equation: 3(x + 2) + 5x – 1 = 7x – 3 + x

          2. If 2x + 8 = 16, what is the value of x + 4?

          3. 2(9x – 6) - 12 = 3(9x – 6)
            Based on the equation above, what is the value of 3x – 2?

          4. Solve the following equation: 3(2x + 1) – 2x = 8x + 5 – 4x – 2

          5. 9ax + 9b – 6 = 21
            Based on the equation above, what is the value of ax + b?

            1. 3
            2. 6
            3. 8
            4. 12
          6. 2ax – 15 = 3(x + 5) + 5(x – 1)
            In the equation above, a is a constant. If no value of x satisfies the equation, what is the value of a?

            1. 1
            2. 2
            3. 4
            4. 8

          7. a(x + b) = 4x + 10
            In the equation above, a and b are constants. If the equation has infinitely many solutions for x, what is the value of b?

          SAT Math - Answers to Questions from Fall 2021, Week 2 (September 25, 2021)

          1. Solve the following equation: 3(x + 2) + 5x – 1 = 7x – 3 + x
            3x + 6 + 5x – 1 = 6x – 3
            8x + 5 = 8x – 3
            There is no solution.

          2. If 2x + 8 = 16, what is the value of x + 4?

            Solve for x; then solve for x + 4.   
            2x + 8   =  16
            2x  =  8
             x  =  4

            x + 4  =  4 + 4  =  8      This is our answer.


            Recognizing that (2x + 8) is a multiple of (x + 4), there is an alternative method for finding our answer. We can divide both sides by 2.
            2x + 8 = 16

            Dividing both sides by 2, we get x + 4 = 8.
            This is our answer: x + 4 = 8, which is the same as the answer above.

          3. 2(9x – 6) - 12 = 3(9x – 6)
            Based on the equation above, what is the value of 3x – 2?

            18x – 12 - 12 = 27x – 18
            18x – 24 = 27x - 18
            -9x = 6
            -x = 6/9 = 2/3
            x = -2/3
            3x – 2 = 3(-2/3) – 2 = - 6/3 – 2 = - 2 – 2 = -4 This is our answer.


            Recognizing that (9x – 6) is a multiple of (3x – 2), there is an alternative method for finding our answer. We can subtract 2(9x – 6) from both sides.
            2(9x – 6) - 12 = 3(9x – 6)

            Subtracting 2(9x – 6) from both sides, we get -12 = 9x – 6.
            Divide both sides by 3: -4 = 3x – 2. This is our answer: 3x – 2 = -4, which is the same as the answer above.

          4. Solve the following equation: 3(2x + 1) – 2x = 8x + 5 – 4x – 2

            6x + 3 – 2x = 4x + 3
            4x + 3 = 4x + 3
            There is an infinite number of solutions.

          5. 9ax + 9b – 6 = 21
            Based on the equation above, what is the value of ax + b?

            1. 3
            2. 6
            3. 8
            4. 12


            Some students give up on problems of this type. They want to get the value of a, multiply it by the value of x, and add the value of b. Since they cannot find the values of these variables, they do not know what to do. Actually, we cannot determine the value of a, we cannot determine the value of x, and we cannot determine the value of b, but we can find the value of ax + b.

            9ax + 9b - 6 - 21
            Add 6 to both sides:  9ax + 9b = 27
            Divide both sides by 9:  ax + b = 3 ----------> A     This is our answer.

          6. 2ax – 15 = 3(x + 5) + 5(x – 1)
            In the equation above, a is a constant. If no value of x satisfies the equation, what is the value of a?
            1. 1
            2. 2
            3. 4
            4. 8

            For no solution, we want to arrive at a situation where the coefficient of the x term on the left side of the equation is the same as the coefficient of the x term on right side of the equation, and the constant on the left side of the equation is different from the constant on the right side of the equation.

            2ax – 15 = 3(x + 5) + 5(x – 1)
            2ax – 15 = 3x + 15 + 5x – 5
            2ax – 15 = 8x + 10

            The constant on the left side of the equation ( -15) is different from the constant on the right side of the equation (+10). Now we need the coefficients of the x terms to be the same. Therefore 2a must equal 8. 2a = 8.
            a = 4 ----------> C

          7. a(x + b) = 4x + 10
            In the equation above, a and b are constants. If the equation has infinitely many solutions for x, what is the value of b?

            For there to be infinitely many solutions, we want to arrive at a situation where the coefficient of the x term on the left side of the equation is the same as the coefficient of the x term on right side of the equation, and the constant on the left side of the equation is the same as the constant on the right side of the equation.

            a(x + b) = 4x + 10
            ax + ab = 4x + 10

            ax = 4x and ab = 10
            Thus a must equal 4 and ab must equal 10.

            a = 4
            ab = 10
            Since a = 4, 4b = 10
            b = 10/4 = 5/2 = 2.5

          SAT Verbal - Questions from Fall 2021, Week 2 (September 25, 2021)

          SAT Quick Challenge K21 - Faulty Comparisons

          Some SAT questions test to see if you know how to use correctly word pairs routinely used to point out similarities and differences. Note the commonly used word pairs in the chart below. Remember that you must always use the pairs together; do not mix or match them with one another, or with any other words. After studying the information, complete the exercise that follows.


          WORD PAIRS ROUTINELY USED FOR MAKING COMPARISONS

          Comparison Word Pair What the Comparison Does Sample Sentence(s) Explanation of What Job Word Pair Does
          1. As...as Indicates that two people or things are equal Ann runs as fast as Doris does. Indicates that Ann and Doris run equally fast.
          2. Not only...but also  Points out two different qualities of a person on thing Beyonce is not only a great singer, but also a fantastic dancer. Names two of Beyonce's great characteristics
          3. More/...er than Shows that one of two people or things has a larger or (smaller) portion or amount of a characteristic or quality (1) than that person or thing had at another time or (2) than another person or thing has or had This movie was more frightening than the other one.

          This chair is stronger than that one.

          Jeff Bezos is much wealthier than Richard Branson.

          Tells which of two movies frightened people more.

          Tells which of two chairs has more strength.

          Tells which of two billionaires has more wealth.

          4. Neither...nor Indicates that none of two or more possible participants will actually participate. Neither Victoria nor her mother will go to see the play. Indicates that not even one of two possible participants -- Victoria and her mother -- will go to see the play.

          SAT QUICK CHALLENGE EXERCISE K21 -- Faulty Comparisons

          Directions. Keeping in mind the information above, select the answer choice that corrects the underlined part in each question. If you think that part is already correct, select choice A (NO CHANGE) as your answer.

          1. The two little boys were arguing about which of them has the most strongest muscles.
            1. NO CHANGE
            2. most strong
            3. stronger
            4. strongest

          2. It's only spring, but last week's temperatures were as hottest than the hottest days last summer.
            1. NO CHANGE
            2. more hot than
            3. as hotter than
            4. as hot as

          3. Neither the schools or the churches will be open for regular activities for the next three months
            because of the pandemic that is causing so much severe illness and death.
            1. NO CHANGE
            2. the schools nor
            3. the schools and
            4. the schools yet

          SAT Verbal - Answers to Questions from Fall 2021, Week 2 (September 25, 2021)

          1. C
          2. D
          3. B

          SAT Math - Questions from Fall 2021, Week 1 (September 18, 2021)

          Linear Equations

          Understanding how to solve linear equations is essential for solving higher level math problems. Performing well on many of the problems on the SAT will require a mastery of equation manipulation.

          An equation will show that two expressions are equal. To solve the equation, we isolate the variable for which we are solving (usually x, but it can be any letter) on one side of the equal sign and the other parts of the equation on the other side.

          Important note: whatever we do to one side of the equation, we must do the same for the other side of the equation. We can perform several operations to both sides of the equation: add, subtract, multiply, divide, raise to a power, or take a root. For example, if we add 3x to one side of the equation, we must add 3x to the other side of the equation; if we divide one side of the equation by 2, we must also divide the other side by 2.

          Keeping in mind the information above, solve the following problems.


          Solve the following equations:

          1. 4x + 8 = 36

          2. 2x – 7 = 3x + 4

          3. 6x + 3 = 4x – 9

          4. 4(b – 5) = 3(4b +4)

          5. (8x +5)/4 = (3x - 9)/2

          6. Solve for x in terms of a, b, and c.
            5a = (3x + b)/4c

          7. For what value of m do the equations 3x – 4 = 11 and mx – 3 = 27 have the same value of x?


          SAT Math - Answers to Questions from Fall 2021, Week 1 (September 18, 2021)

          1. 4x + 8  =  36 
            4x  =  28 
            x  =   7 

          2. 2x – 7  =  3x + 4   
            – x   =  11         
            x =  –11  

          3. 6x + 3  =  4x – 9
            2x  =  – 12 
            x   =  – 6

          4. 4(x – 5)  =  3(4x +4)
            4x – 20  =  12x + 12   
            – 8x  =  32         
            x  = – 4

          5. (6x +5)/4  =  (3x - 9)/2
            Cross multiply:  2(8x + 5)  =  4(3x – 9)                             
                                       16x + 10  =  12x – 36
                                       4x  =  – 46    
                                         x  =  – 46/4  =  – 23/2

          6. Solve for x in terms of a, b, and c.  
            5a  =  (3x + b)/4c
            20ac  =  3x + b
            20ac – b  =  3x           
                        x  =  (20ac – b)/3  

          7. For what value of m do the equations 3x – 4  =  11   and   mx – 3  =  27   have the same value of x?

            Solve for x in the first equation; then substitute this value of x in the second equation and solve for m.           
            3x – 4  =  11                 
                  3x  =  15                   
                    x  =  5

             mx – 3  =  27              
             5m – 3  =  27                    
             5m  =  30                      
               m  =  6

          SAT Verbal - Questions from Fall 2021, Week 1 (September 18, 2021)

          SAT Quick Challenge J21 - Using the Correct Word

          Deciding which of two commonly confused words should be used may not be as difficult as one might think
          initially. The tips below explain how to use some words in that category correctly.

          Words in Question A: "less" vs. "fewer"
          RULE: If the word in question is followed by a singular noun and refers to something that cannot
          be counted, use the word "less." If the word in question is followed by a plural noun and refers to
          something that can be counted, use the word fewer.

          Sentence 1
          : During meals, Mrs. Parker usually has _____ food on her plate than her son has on
          his. Explanation 1: A singular noun (food) follows the word in question and names something
          that cannot be counted. (NOTE: You can count individual food items, but not the concept of
          "food.") Therefore, the missing word must be "less." Correct Sentence: Mrs. Parks usually
          has less food on her plate than her son has on his.

          Sentence 2
          :
          When I take my time, I make _____ mistakes than I do when I rush. Explanation 2: A
          plural noun (mistakes) follows the word in question and refers to something that can be
          counted. Therefore, the missing word is "fewer." Correct Sentence: When I take my time, I
          make fewer mistakes than I do when I rush.

          Words in Question B: "much" vs. "many"
          RULE: If the word in question is followed by a singular noun and refers to something that cannot
          be counted, use the word "much." If the word in question is followed by a plural noun and refers
          to something that can be counted, use the word many.
          Sentence 3: During meals, Mrs. Parker's son usually has _____ more food on his plate than she
          has on hers. Explanation 3: A singular noun (food) follows the word in question and names
          something that cannot be counted. (NOTE: You can count individual food items, but not the
          concept of "food.") Therefore, the missing word is "much." Correct Sentence: Mrs. Hall's
          son usually has much more food on his plate than she has on hers.
          Sentence 4: When I rush, I make _____ more mistakes than I do when I take my time. Explanation
          4: A plural noun ("mistakes") follows the word in question and refers to something that can be
          counted. Therefore, the missing word is "many." Correct Sentence: When I rush, I make many
          more mistakes than I do when I take my time.


          Keeping in mind the information above, complete Exercise J21 below.

          Exercise J21

          Directions. On the line that follows each statement below, place the letter of the answer choice that corrects any error in the statement. If there is no error, mark choice "A" as your answer. When you have finished the exercise, use the answer key to check your work.

          1. When Ella works carefully, she makes far less mistakes than she makes when she rushes carelessly. _______

          1. NO CHANGE
          2. lesser
          3. lots of fewer
          4. fewer

          2. We scored much more points during the game this week than we scored last week. ____

          1. NO CHANGE
          2. lots of more
          3. many more
          4. more better

          3. Our Booster Club sold less boxes of cookies at the school fair than they sold at the mall. ______

          1. NO CHANGE
          2. fewer boxes
          3. lesser boxes
          4. fewest boxes

          SAT Verbal - Answers to Questions from Fall 2021, Week 1 (September 18, 2021)

          1. A
          2. C
          3. B

          SAT Math - Questions from Summer 2021, Week 7 (August 28, 2021)

          Percent Change

          Percent change problems can be somewhat confusing because there may be a question in the minds of some students as to which value to use as a denominator when trying to determine the percent increase or decrease of an amount. We should note that for finding percent changes, the denominator is always the original value or beginning value.

          Actually percentage change problems can involve finding:
          1. the percent change,
          2. the amount of change,
          3. the final amount after a percent increase, and
          4. the final amount after a percent decrease.
          1. The percent change = the amount of change/the beginning amount
            Example: A department store was selling a shirt for $20. To cover increased costs, the price was increased by $3. What was the percent change? 3/20 = 0.15 = 15%

          2. There are two ways to determine the amount of change:
            1. The amount of change = The final amount - the beginning amount
            2. The amount of change = The beginning amount x the percent change

              Example:
              • The price of KYZ stock closed at $75 per share today.  The closing price yesterday was $70.  What was the amount of change?     $75  -  $70  =  $5a) The price of KYZ stock closed at $75 per share today.  The closing price yesterday was $70.  What was the amount of change?     $75  -  $70  =  $5

              • Yesterday the closing price of MYQ stock was $35 per share.  The price today increased by 12%.  What was the amount of change?    $35  x  0.12  =  $4.20

          3. There are two ways to determine the final amount after a percent increase.
            1. The final amount   =  The beginning amount + the amount of increase

            2. The final amount  =  The beginning amount(1 + percent increase)  where the percent                                    increase is expressed as a decimal
              Example:
              1. There were 40 members in the Honor Society in 2020, and the number of members increased by 10% in 2021.  How many members were there in 2021?a) There were 40 members in the Honor Society in 2020, and the number of members increased by 10% in 2021.  How many members were there in 2021?
                Amount of increase = 40 x 0.10  =  4
                Final amount  =  40  +  4  =  44

              2. There were 40 members in the Honor Society in 2020, and the number of members increased by 10% in 2021.  How many members were there in 2021?
                Final amount  =  40(1 + 0.10)  =  40(1.10)  =  44

          4. There are two ways to determine the final amount after a percent decrease.
            1. The final amount   =  The beginning amount - the amount of decrease

            2. The final amount  =  The beginning amount(1 - percent decrease)  where the percent                              decrease is expressed as a decimal
                Example:
                1. The budget for the recreation department was $120,000 in 2019. Because of the pandemic, it was decreased by 30% in 2020. What was the budget amount for 2020?
                  Amount of decrease = $120,000 x 0.30 = $36,000
                  Final amount = $120,000 - $36,000 = $84,000

                2. The budget for the recreation department was $120,000 in 2019. Because of the pandemic, it was decreased by 30% in 2020. What was the budget amount for 2020?
                  Final amount = $120,000 (1 - 0.3) = $120,000(0.7) = $84,000

          Successive percent increases and percent decreases can be multiplied together.

          Note: they are not added together; they are multiplied together.

          Example: A clothing store has a suit that has attracted Martin’s attention. The price of the suit has been $250 but the store just announced a discount of 20%. Martin has a coupon for an additional 15% discount to be applied to the already discounted price of the suit. If Martin buys the suit, how much will be pay after both discounts are applied and a 7% sales tax is added to the final price?
          - Price after 20% discount = $250(1 – 0.2) = $250(0.8) = $200
          - Price after 15% coupon = $200(1 – 0.15) = $200(0.85) = $170
          - Final price after 7% tax is added = $170(1 + 0.07) = $170(1.07) = $181.90

          We could also write: final price = $250(0.8)(0.85)(1.07) = $181.90.

          When successive percent increases or decreases are the same for each successive period, we can use an exponent.

          Example: David deposits $1,000 into a bank account that earns an annual interest rate of 3%. How much would he have in the account at the end of one year?
          - Final amount = $1,000(1.04) = $1,040

          If he does not make any additional deposits and makes no withdrawals, how much would he have in the account at the end of two years?
          - Final amount = $1,040(1.04) = $1,081.60 or
          - Final amount = $1,000(1.04)(1.04) = $1,081.60 or
          - Final amount = $1,000(1.04)2 = $1,000(1.0816) = $1,081.60

          If he does not make any additional deposits and makes no withdrawals, how much would he have in the account at the end of three years?
          - Final amount = $1,081.60(1.04) = $1,124.86 or
          - Final amount = $1,000(1.04)(1.04)(1.04) = $1,124.86 or
          - Final amount = $1,000(1.04)3 = $1,000(1.124864) = $1,124.86

          This pattern can be used for any number of periods, as long as the interest rate (or percent increase) remains constant and there are no more deposits and no withdrawals.

          How much would he have in the account at the end of 12 years?
          - Final amount = $1,000(1.04)12 = $1,000(1.60103) = $1,601.03


          Keeping in mind the information above, answer the following questions.

          1. The Mason Company increased the salary of its treasurer from $380,000 to $400,000. What was the
            percent increase?

          2. Lloyd received a hot tip about the Zylon Company, a new company in the computer industry with
            great prospects for the future. He was persuaded to invest $10,000 in the stock of the company. Shortly after making the investment the value of the stock dropped by 30%. Several months later the value of the stock increased by 30%. Was this increase sufficient to bring Lloyd’s value back to its original value?

          3. Jackie is a sales clerk at Dade Clothing Store where a 25% clearance sale is currently in effect. Jackie is planning to buy a pair of shoes that originally cost $40, and she will apply her 10% employee discount in addition to the 25% clearance sale discount. How much will she pay for the shoes when an 8% sales tax is added to the final price?

          4. In the current school year 400 students are enrolled in a new major in cyber security at Goldsboro College. The college administration expects an increase of 5% enrollment in the major each year for the next three academic years. If no students drop out of the major, approximately how many students will be enrolled in the major at the end of 3 academic years?

          SAT Math - Answers to Questions from Summer 2021, Week 7 (August 28, 2021)

          1. The Mason Company increased the salary of its treasurer from $380,000 to $400,000. What was the
            percent increase?
            Percent increase = ($400,000 - $380,000)/$380,000 = $20,000/$380,000 = 0.0526 = 5.26%

          2. Lloyd received a hot tip about the Zylon Company, a new company in the computer industry with
            great prospects for the future. He was persuaded to invest $10,000 in the stock of the company. Shortly after making the investment the value of the stock dropped by 30%. Several months later the value of the stock increased by 30%. Was this increase sufficient to bring Lloyd’s value back to its original value?

            - Value after 30% drop = $10,000(0.70) = $7,000
            - Value after 30% increase = $7,000 = $7,000(1.30) = $9,100

            The answer is no; a 30% increase is not sufficient to bring the value of the investment back to
            $10,000. This problem illustrates an important fact: percent decrease and percent increase are
            not mirror images of each other.

          3. Jackie is a sales clerk at Dade Clothing Store where a 25% clearance sale is currently in effect. Jackie is planning to buy a pair of shoes that originally cost $40, and she will apply her 10% employee discount in addition to the 25% clearance sale discount. How much will she pay for the shoes when an 8% sales tax is added to the final price?

            Price after 25% discount = $40(1 – 0.25) = $40(0.75) = $30
            Price after 10% employee discount = $30(1 – 0.10) = $30(0.9) = $27
            Final price after 8% tax is added = $27(1 + 0.08) = $27(1.08) = $29.16

            We could also write: final price = $40(0.75)(0.9)(1.08) = $29.16

          4. In the current school year 400 students are enrolled in a new major in cyber security at Goldsboro College. The college administration expects an increase of 5% enrollment in the major each year for the next three academic years. If no students drop out of the major, approximately how many students will be enrolled in the major at the end of 3 academic years?

            Number projected to be enrolled = 400(1.05)3 = 400(1.157625) = 463.05
            Therefore, approximately 463 students will be enrolled in the major.

          SAT Verbal - Questions from Summer 2021, Week 7 (August 28, 2021)

          SAT Quick Challenge U21C - Parallelism


          Infinitives and Gerunds. An infinitive is often described as a word (1) that looks like a verb with the word "to" in front of it, but (2) does not function as a verb. Although an infinitive can function as different parts of speech, it often functions as a noun, as in Sentence A, which follows.
          (A) To play the drums in the band has always been Lynn's dream. Since the subject of Sentence A is the infinitive "To play," it is functioning as a noun. A gerund looks like a verb that ends in "ing," but it functions as a noun, Note Sentence B, which follows.
          (B) Singing in the choir is Maggie's favorite activity at church. In this sentence, the subject is the gerund "Singing."

          Although both infinitives and gerunds can function as nouns, they must not be mixed within the same sentence. Note Sentences C, D, and E, which follow. (C) Ella loves hiking in the woods and to swim at the beach. This sentence is not parallel because it contains both a gerund (hiking) and an infinitive (to swim). Now, note Sentences D and E. (D) Ella loves hiking in the woods and swimming at the beach. Note that the infinitive "to swim" (from Sentence C) has been replaced with the gerund "swimming." Hence, the sentence is now parallel. (E) Ella loves to hike in the woods and to swim at the beach. The gerund hiking (from Sentence C) has been replaced with the infinitive to hike, and this sentence is now parallel.

          Do Not Mix Phrases and Clauses in the Same Sentence. Parallelism requires that similar items in a sentence (a list, a series, etc.) be expressed in the same way-- whether those items are individual words, phrases, or clauses. Note Sentence F, which follows. (F) Ernie says that he would enjoy playing football, running track, and he wants to play in the band. This sentence is not parallel because the two phrases in it ("playing football" and "running track") are followed by a clause ("he wants to play in the band"). All three elements in the series must follow the same format, as illustrated in Sentences G and H. Sentence G. Ernie says that he would enjoy playing football, running track, and playing in the band. This sentence is correct because all three elements are expressed as gerund phrases. Sentence H. Ernie says that he would like to play football, run track, and play in the band. This sentence is correct because all three items in the series are expressed as infinitive phrases.

          Now, complete the Quick Challenge Exercise below.

          QUICK CHALLENGE Practice Exercise U21C - Parallelism

          DirectionsReplace the underlined word(s) in each question below with the answer choice that corrects the error in the sentence. If there is no error, select choice A -- NO CHANGE.

          1. Whether you will be serving food, cleaning up, or when you operate the cash register, we thank you for helping with our annual spaghetti dinner.

          1. NO CHANGE
          2. near the porch, the mailbox, and along
          3. near the porch, the mailbox, and
          4. near the porch, mailbox, and

          2. The Browns want to visit New York to see interesting places, to sample the food, and they would love going to the theaters there.  

          1. NO CHANGE
          2. shows in the theater will be great
          3. they want to see the theater productions
          4. to enjoy the shows

          3. Ernie says that he would like playing basketball, running track, and to race sports cars.

          1. NO CHANGE
          2. racing sports cars
          3. he'd enjoy racing sports cars
          4. he loves sports car races

          4. People form opinions of you based on what you say and your words.

          1. NO CHANGE
          2. promises made by you
          3. your promises
          4. what you do

          SAT Verbal - Answers to Questions from Summer 2021, Week 7 (August 28, 2021)

          1. C
          2. D
          3. B
          4. D

          SAT Math - Questions from Summer 2021, Week 6 (August 21, 2021)

          Percent Word Problems

          In recent administrations of the SAT, there has been an increase in the number of word problems included. The following steps can help you solve word problems:
          1. Read the problem carefully an avoid misreading anything important.
          2. Identify the key values and determine how they are related mathematically.
          3. Solve the problem.
          4. Make sure that the answer is reasonable. Ask yourself: does it make sense?

          Now examine examples 1 and 2.

          Example 1:
          All of the students in Mr. Taylor’s math took Exam 1, and each student either passed or failed the exam. 80% of the students passed and 5 students failed. How many students passed the exam?

          To answer this question, we need to determine the total number of students in the class. We know that 5 students failed the test and these 5 students represent 20% of the total number of students in the class (100% total - 80% that passed = 20% that failed).
          - Thus 20% of the total = 5, or 0.2(Total) = 5;
          - Thus the total = 5/0.2 = 25
          - There are 25 students in the class. We then multiply 80% times 25 to get the number that passed: 25 x 0.8 = 20.

          Example 2:
          The Girl Scouts take 500 palm leaves to pin on church members on Palm Sunday. They pinned 80% of the palm leaves on the adult members. They then pinned 40% of the remaining leaves on teen age members. Finally, they pinned 30% on the remaining leaves on the younger members. How many leaves did they have left?

          First, they pinned 80% of the leaves on the adult members.
          We compute 80% of 500: 0.8 x 500 = 400.
          400 are pinned on the adult members.
          Thus, they had 500 – 400 = 100 remaining.

          Next, they pinned 40% of their remaining leaves on the teenagers.
          We compute 40% of 100: 0.4 x 100 = 40.
          40 were pinned on the teenagers.
          Thus, they had 100 – 40 = 60 remaining.

          Finally, they pinned 30% of their remaining leaves on the younger members.
          We compute 30% of 60: 0.3 x 60 = 18.
          18 are pinned on the younger children.
          Thus, they had 60 – 18 = 42 left.

          Here is an alternative calculation:
          500 x 0.2 = 100 remaining after pinning on the adult members (100% - 80% = 20%) 100 x 0.6 = 60 remaining after pinning on the teenagers.
          Finally, 60 x 0.7 = 42 were left after pinning on the younger children. 

          Keeping in mind the information above, answer the following questions.
          1. John and Sam played chess against each other many times during the pandemic. Sam won 15% of the time and John won the other 51 games. There were no draws or ties. How many games did Sam win?

          2. In a tryout for the football team, 20 students trying out were seniors and 80 were underclassmen. What percent of the students trying out were seniors?

          3. A large 200 quart container of juice consisted of 60% orange juice and 40% lemonade. 30% of the juice was used, and then 20 quarts of lemonade were added to the container. What percentage of the mixture was lemonade after the new addition?

          4. 800 students applying to Atlantic University took the entrance exam. 80% of the students passed the exam and 20% failed it. Suppose instead only 5% of the students had failed the exam. How many more students would have passed the exam?

          SAT Math - Answers to Questions from Summer 2021, Week 6 (August 21, 2021)

          1. John and Sam played chess against each other many times during the pandemic. Sam won 15% of the time and John won the other 51 games. There were no draws or ties. How many games did Sam win?
            - Sam won 15% of the total number of games played, and John won the other 85%.
            - John’s 51 games represent 85% of the total number of games, or 85% of the total = 51, or 0.85 (total) = 51.
            - Thus, the total = 51/0.85 = 60 games played. Sam won 0.15 x 60 = 9 games.

          2. In a tryout for the football team, 20 students trying out were seniors and 80 were underclassmen. What percent of the students trying out were seniors?
            - Some students will incorrectly divide 20 by 80 and get 20/80 = 0.25 = 25%.
            - Actually, we should divide 20 by the total, which is 100:
               20/100 = 0.2 = 20%.

          3. A large 200 quart container of juice consisted of 60% orange juice and 40% lemonade. 30% of the juice was used, and then 20 quarts of lemonade were added to the container. What percentage of the mixture was lemonade after the new addition?

            Amount of orange juice = 0.6 x 200 = 120 quarts
            Amount of lemonade = 0.4 x 200 = 80 quarts

            Amount of each remaining after 30% was used:
            - Amount of orange juice = 0.7 x 120 = 84 quarts
            - Amount of lemonade = 0.7 x 80 = 56 quarts

            Now add 20 quarts of lemonade:
            - Amount of orange juice = 84 quarts
            - Amount of lemonade = 56 quarts + 20 quarts = 76 quarts

            Total juice = 84 + 76 = 160
            Percentage of mixture that is lemonade = 76/160 = 0.475 = 47.5%

          4. 800 students applying to Atlantic University took the entrance exam. 80% of the students passed the exam and 20% failed it. Suppose instead only 5% of the students had failed the exam. How many more students would have passed the exam?

            800 x 0.8 = 640 passed initially.
            If only 5% failed, 800 x .05 = 40 would have failed.

            Total number that would have passed = 800 – 40 = 760
            Total number that initially passed =                           640
                                                                                                  ----
                                                                                                  120
            Thus, 120 more students would have passed the exam if only 5% of the students had failed the exam.


            Alternative solution:
            If only 5% of the students had failed the exam, 95% would have passed:
            800 x 0.95 =                                                                 760
            Total number that initially passed =                          640
                                                                                                  ----
                                                                                                 120
            Thus, 120 more students would have passed the exam if only 5% of the students had failed the exam.

          SAT Verbal - Questions from Summer 2021, Week 6 (August 21, 2021)

          SAT Quick Challenge T21B - Parallelism



          Which Word(s) Can You Change To Correct the Error in the Question?  SAT grammar and writing questions ask you to replace the error in the underlined part of a sentence with the answer choice with corrects that error. However, parallelism questions like Question A lbelow can be confusing if you are not familiar with that type of question. Take a look at Question A.

          Question A: Compared to a personal vehicle, public transportation is cheaper, safer, and environmentally friendly.
          Discussion: The underlined words in this statement (cheaper and safer) are parallel to each other because they are both comparative adjectives (which are used specifically to compare two things), and they compare public transportation to personal vehicles.
          Environmentally friendly is not parallel to the underlined words because it is not a comparative adjective; it is an adjective phrase with no limit to the number of things that it can describe. Since the correct answer to Question A requires all the descriptions  of public transportation to be parallel with another, something must be changed. Because only the underlined words in SAT writing and grammar questions can be changed, we must leave environmentally friendly as it is and select an answer choice that is parallel to it.  Now, keeping the information about in mind, answer Question A.

          Question A. Compared to a personal vehicle, public transportation is cheaper, safer, and environmentally friendly.   
          A. NO CHANGE
          B. cheaper, safe
          C. cheap, safe
          D. cheap, safer

          Answer: The correct answer is Choice C because the two words in that choice are adjectives that do the same thing that "environmentally friendly" does. They all describe public transportation, and there is no limit to the number of thing each adjective/adjective phrase can describe.

          Prepositions Repeated in a Series of Prepositional Phrases.
          When each phrase in a series of prepositional phrases begins with the same preposition, place that preposition in front of the first prepositional phrase only, as in the following: The children left paint smudges on the walls, the sink, and the windows. However, if the same preposition is needed at the beginning of one or more of the phrases, but not all of them, place the preposition needed in front of each phrase, as follows: The children left paint smudges on the walls, on the sink, and under the windows.

          Now, keeping in mind the information above, complete the exercise below. Then, use the dropdown in the next section to check your work.

          QUICK CHALLENGE Practice Exercise T21B - Parallelism

          DirectionsReplace the underlined word(s) in each question below with the answer choice that corrects the error in the sentence. If there is no error, select choice A -- NO CHANGE.

          1. Grandma planted flowers near the porch, near the mailbox, and along the driveway.

          1. NO CHANGE
          2. near the porch, the mailbox, and along
          3. near the porch, the mailbox, and
          4. near the porch, mailbox, and

          2. Unlike the Afghan army, the Taliban army is toughest, strongest, and extremely vicious.

          1. NO CHANGE
          2. violence, harsher
          3. tough, strong
          4. crueler, meaner

          3. My neighbor received a scam call from someone extending her a(n) clever, new offer to make $500,000 in one year.

          1. NO CHANGE
          2. clever, safe
          3. unwise, untrue
          4. sensible, profitable

          4. For lunch, we will have vegetable gumbo made with tomatoes, with okra, with corn, and with rice.

          1. NO CHANGE
          2. with tomatoes, okra, corn, and rice
          3. with tomatoes, okra with corn, and rice
          4. with tomatoes, okra, and corn with rice

          SAT Verbal - Answers to Questions from Summer 2021, Week 6 (August 21, 2021)

          1. A
          2. C
          3. C
          4. B

          SAT Math - Questions from Summer 2021, Week 5 (August 14, 2021)

          Decimals and Percent

          A percent is a fraction whose denominator is 100:  25%  =  25/100
          Percent means “per 100”
          If there are 100 questions on your math test and you answer 80 of them correctly, you have answered 80 of 100 correctly, or 80/100, or 80%.

          Think of it this way:  part/whole  =  percent
          80/100 = 80%

          Converting percentages to fractions: sometimes it will be useful to express a percentage as a fraction: put the percentage over a denominator of 100 and reduce.
          60% = 60/100 = 6/10 = 3/5
          150% = 150/100 = 15/10 = 3/2
          25% = 25/100 = 1/4

          Converting fractions to percentages:
          Divide the numerator by the denominator
          Move the decimal point in the result two places to the right
          ¾ = 0.75 = 75%
          1/5 = 0.20 = 20%

          Converting percentages to decimals: move the decimal point two places to the left.
          16% = .16
          2% = .02
          .5% = .005

          Converting decimals to percentages: move the decimal point two places to the right.
          0.85 = 85%
          0.4 = 40%
          1.7 = 170%
          .003 = .3%

          Remember that a percent relates part to a whole: 20 is 50% of 40 Thus, there are three problem types involving percent:
          1. Find the percent: 20 is what percent of 40? 20/40 = 0.5 = 50%
          2. Find the part: What number is 50% of 40? 40 x 0.5 = 20 (change the percent to a decimal and multiply)
          3. Find the whole: 20 is 50% of what amount? 20/0.5 = 200/5 = 40 (change the percent to a decimal and divide)

          Keeping in mind the information above, answer the following questions without using your calculator.

          1. Find the percent:
            1. 4 is what percent of 8?
            2. 2 is what percent of 5?
            3. 6 is what percent of 4?

          2. Find the percentage amount (or part.)
            1. What number is 50% of 8?
            2. What is 30% of 50?
            3. What is 15% of 60?
            4. What number is 250% of 2?

          3. Find the base (or whole.)
            1. 4 is 25% of what number?
            2. 70 is 40% of what amount?
            3. 12 is 30% of what amount?
            4. 3 is 0.2% of what number?

          4. A summer beach volleyball league has 750 players in it. At the start of the season, 150 of the players are randomly chosen and polled on whether games should be played while it is raining, of if the games should be cancelled. The results of the poll show that 42 of the polled players would prefer to play in the rain. The margin of error is ±4%. What is the range of players in the entire league that would be expected to prefer to play volleyball in the rain rather than cancel the game? (You may use your calculator for this one.)
            1. 24-32
            2. 38-46
            3. 146-154
            4. 180-240

          SAT Math - Answers to Questions from Summer 2021, Week 5 (August 14, 2021)

          1. Find the percent:
            1. 4 is what percent of 8?
              4/8 = 0.5 = 50%
            2. 2 is what percent of 5?
              2/5 = 0.4 = 40%
            3. 6 is what percent of 4?
              6/4 = 1.5 = 150%

          2. Find the percentage amount (or part.)
            1. What number is 50% of 8?
              8 x 0.5 = 4
            2. What is 30% of 50?
              50 x 0.3 = 15
            3. What is 15% of 60?
              60 x 0.15 = 9
            4. What number is 250% of 2?
              2 x 2.5 = 5

          3. Find the base (or whole.)
            1. 4 is 25% of what number?
              4/0.25  =  400/25  =  80/5  =  16
            2. 70 is 40% of what amount?
              70/0.4 = 700/4 = 175
            3. 12 is 30% of what amount?
              12/0.3 = 120/3 = 40
            4. 3 is 0.2% of what number?
              3/0.002 = 3000/2 = 1500

          4. A summer beach volleyball league has 750 players in it. At the start of the season, 150 of the players are randomly chosen and polled on whether games should be played while it is raining, of if the games should be cancelled. The results of the poll show that 42 of the polled players would prefer to play in the rain. The margin of error is ±4%. What is the range of players in the entire league that would be expected to prefer to play volleyball in the rain rather than cancel the game? (You may use your calculator for this one.)
              1. 24-32
              2. 38-46
              3. 146-154
              4. 180-240

                First, determine the percent of polled players who wanted to play in the rain: 42/150 = 0.28 = 28%

                Next, apply this percent to the entire population of the league: 750 x 0.28 = 210. The only range that contains this value is D; thus, this is the answer.


                To calculate the actual range, we add and subtract 4% to the 28% to get a range of 24% - 32% of the total:

                750 x 0.24 = 180

                750 x 0.32 = 240

              Thus, the actual range is 180 to 240.

          SAT Verbal - Questions from Summer 2021, Week 5 (August 14, 2021)

          SAT Quick Challenge T21 - Parallelism



          What is Parallelism?  Parallelism involves using the same structure or format to link two or more related words, phrases, or clauses in a sentence. This strategy makes the writing clearer and easier to understand. Note Sentences A and B, which follow. (A) Pam loves cooking food, to clean, and sewing. (B) Pam loves cooking, cleaning, and sewing. The two sentences say similar things, but not in the same way. The verbs in Sentence A are not listed the same way (cooking food, to clean, and sewing), so that sentence is not parallel. The verbs in Sentence B are listed the same way (cooking, cleaning, sewing), so that sentence is parallel.

          Phrases or Independent Clauses. You will recall that parallelism requires using the same structure or format to list two or more related words or word groups in a sentence. They can be phrases only or independent clauses only, but not a mixture of both. Note Sentences C and D, which follow. (C) Poor eating habits can cause serious health problems, from high blood pressure to you could get an increased risk of diabetes. Explanation: Sentence C is not parallel because it combines a prepositional phrase (from high blood pressure) with an independent clause (you could get an increased risk of diabetes). Sentence D is parallel because it contains two prepositional phrases (from high blood pressure) and (to an increased risk of diabetes).

          Avoiding Unnecessary Prepositions. When prepositional phrases which all begin with the same preposition appear one after another in the same sentence, place that preposition in front of the first prepositional phrase only, and the preposition will apply to all the linked prepositional phrases in the sentence. This strategy avoids the wordiness and redundancy that would result from repeating the preposition for each prepositional phrase. Note Sentences E and F, which follow. (E) On her vacation, Rev. Jones went to London, to Paris, and to Washington, DC. (F) On her vacation, Rev. Jones went to London, Paris, and Washington, DC. Explanation: Sentence E is not correct because it repeats the preposition "to" in front of each prepositional phrase. Sentence F is correct because the common preposition "to" is placed only in front of the first prepositional phrase.



          Now, complete QUICK CHALLENGE Practice Exercise T21 below. Then, use the dropdown in the next section to check your work.

          QUICK CHALLENGE Exercise T21
          Parallelism

          DirectionsReplace the underlined word(s) in each question below with the answer choice that corrects the error in the sentence. If there is no error, select choice A -- NO CHANGE.

          1. We searched carefully in the house, the garage, and in the car, but we couldn't find the keys.

          1. NO CHANGE
          2. in the house, the garage, and the car
          3. in the house, in the garage, and in the car
          4. in the house, in the garage, and the car

          2. Earl's game show prizes are fantastic, especially his new car and he got a trip to Paris.

          1. NO CHANGE
          2. Consequently
          3. Otherwise
          4. Nevertheless

          3. My brother likes waffles with butter, syrup, and with cinnamon.

          1. NO CHANGE
          2. with butter, syrup, and cinnamon
          3. with butter, with syrup, and with cinnamon
          4. with butter, with syrup and cinnamon

          4. Good study habits can lead to excellent grades and you could get a college scholarship.

          1. NO CHANGE
          2. you might get money for college
          3. college scholarships
          4. colleges could give you scholarships

          SAT Verbal - Answers to Questions from Summer 2021, Week 5 (August 14, 2021)

          1. B
          2. D
          3. B
          4. C

          SAT Math - Questions from Summer 2021, Week 4 (August 7, 2021)

          Decimals

          A lesson on decimals is basic and may seem unnecessary. However, many individual problems on the SAT contain more than one math concept, and understanding decimals will be necessary for solving some of them. It is also important to understand the relation between decimals and fractions, and between decimals and percents.

          A decimal is just another way of expressing a fraction.
          1/5 = 1 ÷ 5 = 0.2 3/4 = 3 ÷ 4 = 0.75

          Not all decimals have visible decimal points.
          2 is a decimal: 2 = 2.0 (or 2.00 or 2.000); 3 = 3.0 (or 3.00 or 3.000)

          Adding and subtracting decimals:
          Line up the decimal points and add or subtract (or use your calculator when appropriate - if you are in the calculator section of the test)
          Add: 3.7, 14.23, and 9
            3.7
          14.23
             9.00
          ------
           26.93

          Multiplying Decimals:
          • Multiply exactly as you would integers
          • Count the total number of digits located to the right of the decimal points in the numbers you are multiplying
          • Place the decimal point in your answer so that there are the same number of digits to the right of it
          (or use your calculator when appropriate)
                 0.4              2.3              1.23     
               x0.4               x 3              x0.4                  
               ----               ---               ----
               0.16              6.9            0.492

          Multiply a decimal by 10: Move the decimal point 1 place to the right
          1.06 x 10 = 10.6 0.0076 x 10 = 0.076

          Multiply a decimal by 100: Move the decimal point 2 places to the right
          62.4 x 100 = 6240 0.0017 x 100 = 0.17

          Dividing decimals:

          1. Convert the denominator to a whole number by moving the decimal point in the denominator a sufficient number of places to the right to make it a whole number
          2. The decimal point in the numerator must be moved the same number of places
          3. Then divide

          (or use your calculator when appropriate)

          Divide 6 by 0.48 6/0.48 = 600/48 reduce (when possible) then divide
                                                     600/48 = 100/8 = 25/2 = 12.5

          Divide a decimal by 10:  Move the decimal point 1 place to the left
          12.635 ÷ 10  =  1.2635                   
          0.0076 ÷ 10  =  0.00076

          Divide a decimal by 100:  Move the decimal point 2 places to the left   
          62.4 ÷ 100  =  .624                     
          397 ÷ 100  =  3.97

          Convert a fraction to a decimal:  just divide       
          2/5  =  0.4                   
          7/4  =  1.75

          Convert a decimal to a fraction:

          1. Count the number of digits to the right of the decimal point and call this “n”
          2. numerator will be the decimal number without the decimal point
          3. The denominator will be a “1” followed by “n” zeros

          Reduce to lowest terms
          0.75 = ?
          n = 2
          75/100 = ¾

          0.2 = ?
          n = 1
          2/10 = 1/5

          Stay out of trouble by converting decimals to fractions or by using a calculator when appropriate.

          1. Confusion about decimal points causes more errors on the SAT than confusion about fractions
          2. Therefore, whenever you can conveniently convert a decimal to a fraction, you should do so, especially when the answer choices are fractions.


          Keeping in mind the information above, answer the following questions without using your calculator.

          1. Multiply the following decimals.
            1. 0.6 x 84.2
            2. 75 x 0.3
            3. 0.14 x 0.009

          2. Divide the following decimals.
            1. 8/0.32
            2. 48/2.5
            3. 0.9/0.04
            4. 14.4/0.12

          3. Convert the following decimals to fractions.
            1. 0.8
            2. 2.62
            3. .004

          4. Convert the following fractions to decimals.
            1. 3/15
            2. 15/24
            3. 24/32

          5. A coastal geologist estimates that a certain country’s beaches are eroding at a rate of 1.5 feet per year. According to the geologist’s estimate, how long will it take, in years, for the country’s beaches to erode by 21 feet?
            1. 7
            2. 14
            3. 22.5
            4. 31.5

          6. If 0.6w = 4/3, what is the value of w?
            1. 9/20
            2. 4/5
            3. 5/4
            4. 20/9

          SAT Math - Answers to Questions from Summer 2021, Week 4 (August 7, 2021)

          1. Multiply the following decimals.
            1. 0.6 x 84.2 = 50.52
            2. 75 x 0.3 = 22.5
            3. 0.14 x 0.009 = 0.00126

          2. Divide the following decimals.
            1. 8/0.32 = 800/32 = 100/4 = 25
            2. 48/2.5 = 480/25 = 96/5 = 19.2
            3. 0.9/0.04 = 90/4 = 45/2 = 22.5
            4. 14.4/0.12 = 1440/12 = 360/3 = 120

          3. Convert the following decimals to fractions.
            1. 0.8                  n = 1             8/10 = 4/5
            2. 2.62                n = 2             262/100 = 131/50
            3. .004                n = 3             4/1000 = 1/250

          4. Convert the following fractions to decimals.
            1. 3/15 = 1/5 = 0.2
            2. 15/24 = 5/8 = 0.625
            3. 24/32 = 3/4 = 0.75

          5. A coastal geologist estimates that a certain country’s beaches are eroding at a rate of 1.5 feet per year. According to the geologist’s estimate, how long will it take, in years, for the country’s beaches to erode by 21 feet?
            1. 7
            2. 14
            3. 22.5
            4. 31.5

            21/1.5 = 210/15 = 70/5 = 14 --------> B

          6. If 0.6w = 4/3, what is the value of w?
            1. 9/20
            2. 4/5
            3. 5/4
            4. 20/9

            Multiply both sides by 3:
            1.8w = 4
            Then w = 4/1.8 = 40/18 = 20/9 ----->D

            An alternative approach: since the answer choices are in fractions, convert the decimal (0.6) to a fraction and then solve.
            0.6 = 3/5
            (3/5)w = 4/3
            3w /5= 4/3
            Cross multiply: 9w = 20
            w = 20/9 ----->D

          SAT Verbal - Questions from Summer 2021, Week 4 (August 7, 2021)

          SAT Quick Challenge S21
          Transition Words and Phrases



          What is a Transition?  Sometimes, two sentences that are written one right after the other in a paragraph have a very specific relationship. For instance, the two might contrast with each other, or the second sentence might be an explanation or example of what is stated in the first sentence. In such situations, a word or phrase may be placed in the second sentence to show how the two sentences are related. That word or phrase is called a transition. In some cases, a transition may come between two independent clauses that are in the same sentence. Even so, the transition still shows how the two clauses are related to each other.

          How To Determine Which Transition To Use. To decide which transition to use for an SAT question, draw a line through the underlined transition in the question so that it will not distract you. Then read the two sentences and determine their relationship. Finally, select the kind of transition which is needed for that relationship. Note the chart below.

          TRANSITION RELATIONSHIPS FREQUENTLY TESTED ON THE SAT

          Type of Transition and Job it Does Example Sample Sentences (and How to Select Their Transitions)
          Reason/cause effect:
          Shows why something did or did not happen
          consequently, as
          a result, hence, since, therefore,
          accordingly
          The children were exhausted. Therefore, they went to sleep at the dinner table. (Because the first sentence gives the reason for what happened in the second sentence, a reason or cause effect transition is needed.)
          Addition/example:
          Points out added information that explains further a point already made
          moreover, for example, furthermore, indeed, for instance Dr. Taylor always has huge science classes because the students say that she is an excellent professor. Moreover, her math classes are also large because students also love the way she teaches math. (Since the second sentence further explains how well Dr. Taylor teaches, an addition/example transition is needed.)
          Contrast:
          Points out conflicting ideas
          although, even so, however Frank loves cakes, pies, and other tasty desserts. Nevertheless, he avoids those sweet treats because he has diabetes. (Since the two sentences are in conflict, a contrast transition is needed. )



          Keeping in mind the information above, complete QUICK CHALLENGE Exercise S21 below. Then use the dropdown in the next section to check your work.

          QUICK CHALLENGE Exercise S21

          DirectionsReplace the underlined words in each question below with the answer choice that corrects the error in the sentence. If there is no error, select choice A -- NO CHANGE.

          1. Some people disagree with the doctor's thoughts about the cause of this disease. Even so, his theory has never been proven wrong.

          1. NO CHANGE
          2. Therefore
          3. Accordingly
          4. As a result

          2. There was a terrible accident on the highway this morning. On the other hand, I did not get to work today until lunch time. 

          1. NO CHANGE
          2. Consequently
          3. Otherwise
          4. Nevertheless

          3. Dina is a highly talented writer who does excellent work. However, she has written award-winning commercials for numerous companies.

          1. NO CHANGE
          2. Nevertheless
          3. Since
          4. For instance

          SAT Verbal - Answers to Questions from Summer 2021, Week 4 (August 7, 2021)

          1. A
          2. B
          3. D

          SAT Math - Questions from Summer 2021, Week 3 (July 24, 2021)

          The Average (Mean) and Standard Deviation

          Questions about the standard deviation are also asked on the SAT. The standard deviation is a measure of how closely clustered a data set is about the mean of the data set (how spread out the values are).

          The standard deviation is low if most of the values are near the mean and close together (narrow spread).

          The standard deviation is high if most of the values are spread out over the range of values (wide spread).

          The SAT will not require you to calculate the standard deviation, but you need to understand the concept.

          Consider the following example:

          Arnold, Ronald, and David are the top three scorers on the Eastern High School basketball team. During the first five games of the season, they scored the following points:

          Game 1 Game 2 Game 3 Game 4 Game 5
          Arnold 18 21 15 24 22
          Ronald 14 10 26 32 18
          David 20 20 20 20 20

          In each case the mean is 20:
          Arnold: (18 + 21 + 15 + 24 + 22)/ 5 = 100/5 = 20
          Ronald: (14 + 10 + 26 + 32 + 18)/5 = 100/5 = 20
          David: (20+ 20 + 20 + 20 + 20)/5 = 100/5 = 20

          Let’s compare Arnold and Ronald: The points scored by Arnold are close to the mean, while the points scored by Ronald are more spread out. The standard deviation of Arnold is lower than the standard deviation of Ronald. At the extreme, the points scored by David are all the same; there is no variation in the value of the points. Thus the standard deviation for David is zero.

          Keeping in mind the information above, answer the following questions.

            1. Set A 37 42 58 19 66 97 22
              Set B 48 63 55 59 42 51 65
              The table above shows two sets of data. Which of the following statements is true about the standard deviation of the two sets?

              1. The standard deviation of Set A is lower than the standard deviation of Set B.
              2. The standard deviation of Set A is greater than the standard deviation of Set B.
              3. The standard deviation of Set A is equal to the standard deviation of Set B.
              4. The standard deviation of Set A and the standard deviation Set B cannot be compared.

            2. The tables below give the distribution of speeding tickets given by patrol officers in Greensboro and Winston-Salem for the 30 days in June.

              Greensboro   Winston-Salem
              Number of Speeding Tickets Given Frequency   Number of Speeding Tickets Given Frequency
              13 1   13 6
              14 5   14 7
              15 19   15 7
              16 3   16 6
              17 2   17 4
              Which of the following is true about the data shown for these 30 days?

              1. The standard deviation of the number of speeding tickets given in Greensboro is smaller.
              2. The standard deviation of the number of speeding tickets given in Winston-Salem is smaller.
              3. The standard deviation of the number of speeding tickets given in Greensboro is the same as that in Winston-Salem.
              4. The standard deviation shows that the number of speeding tickets given in Winston-Salem is too large.

            3. 13, 22, 29, 17, 9, 15, 27, 21
              If the number 43 is added to the above set of numbers, how will the standard deviation of the set change?

              1. The standard deviation of the set will be higher.
              2. The standard deviation of the set will be lower.
              3. The standard deviation of the set will be unchanged.
              4. The standard deviation of the set cannot be compared.

            4. The weights, in pounds, for 45 players on the football team were reported, and the mean, median, range, and standard deviation were calculated for the data. The player for the lowest reported weight was found to actually weigh 15 pounds less than his reported weight. What value remains unchanged if the four values are reported using the corrected weight?

              1. Mean
              2. Median
              3. Range
              4. Standard deviation

          SAT Math - Answers to Questions from Summer 2021, Week 3 (July 24, 2021)

              1. Set A 37 42 58 19 66 97 22
                Set B 48 63 55 59 42 51 65
                The table above shows two sets of data. Which of the following statements is true about the standard deviation of the two sets?

                1. The standard deviation of Set A is lower than the standard deviation of Set B.
                2. The standard deviation of Set A is greater than the standard deviation of Set B.
                3. The standard deviation of Set A is equal to the standard deviation of Set B.
                4. The standard deviation of Set A and the standard deviation Set B cannot be compared.

                The numbers in Set A are spread out farther than the numbers in Set B. Thus Set A has a greater standard deviation than Set B. The answer is B.

              2. The tables below give the distribution of speeding tickets given by patrol officers in Greensboro and Winston-Salem for the 30 days in June.

                Greensboro   Winston-Salem
                Number of Speeding Tickets Given Frequency   Number of Speeding Tickets Given Frequency
                13 1   13 6
                14 5   14 7
                15 19   15 7
                16 3   16 6
                17 2   17 4
                Which of the following is true about the data shown for these 30 days?

                1. The standard deviation of the number of speeding tickets given in Greensboro is smaller.
                2. The standard deviation of the number of speeding tickets given in Winston-Salem is smaller.
                3. The standard deviation of the number of speeding tickets given in Greensboro is the same as that in Winston-Salem.
                4. The standard deviation shows that the number of speeding tickets given in Winston-Salem is too large.

                The number of tickets given out in Greensboro are much more tightly concentrated than those in Winston-Salem. Since most of the values in Greensboro are very closely clustered around 15 tickets, the number of tickets given in Greensboro has a lower standard deviation than that of Winston-Salem. The answer is A.

              3. 13, 22, 29, 17, 9, 15, 27, 21
                If the number 43 is added to the above set of numbers, how will the standard deviation of the set change?

                1. The standard deviation of the set will be higher.
                2. The standard deviation of the set will be lower.
                3. The standard deviation of the set will be unchanged.
                4. The standard deviation of the set cannot be compared.

                Adding the number 43 to the set of numbers will increase the spread of the numbers. Thus the standard deviation will be higher. The answer is A.

              4. The weights, in pounds, for 45 players on the football team were reported, and the mean, median, range, and standard deviation were calculated for the data. The player for the lowest reported weight was found to actually weigh 15 pounds less than his reported weight. What value remains unchanged if the four values are reported using the corrected weight?

                1. Mean
                2. Median
                3. Range
                4. Standard deviation


                The answer is B.

            SAT Verbal - Questions from Summer 2021, Week 3 (July 24, 2021)

            SAT Quick Challenge R21C
            The Semicolon



            Using the Semicolon Correctly. You will recall that an independent clause consists of a subject and verb that work together to express a complete thought that stands alone as a complete sentence. SAT questions about the semicolon often focus on using that punctuation mark to create a new sentence by connecting two closely related independent clauses, as shown in sentences A, B, and C, which follow. (A) Elaine enjoys cooking. (B) She especially loves to create tasty desserts. (C) Elaine enjoys cooking; she especially loves to create tasty desserts. Since sentence B provides further insight into what was said in sentence A, the two clauses are closely related, and the semicolon in sentence C creates a new sentence by connecting A and B properly.

            The Comma Splice Error. A common mistake tested on the SAT is the comma splice, an error created when a writer connects two independent clauses with just a comma. You may recall that a comma can help create a single sentence from two independent clauses if the comma comes just before an appropriate FANBOYS conjunction (for, and, nor, but, or, yet, and so). However, the comma cannot do the job without the FANBOYS conjunction. Note Sentences D-H, which follow. (D) Elaine enjoys cooking. (E) She especially loves to create tasty desserts. (F) Elaine enjoys cooking, she especially loves to create tasty desserts. (G) Elaine enjoys cooking, and she especially loves to create tasty desserts. (H) Elaine enjoys cooking; she especially loves to create tasty desserts. Note the comma splice in sentence F and the corrections in sentences G and H.

            List of Paired Items Connected with Commas. When the items listed in a sentence consist of paired words that must be connected to each other with a comma (such as a city and state), a semicolon instead of a comma must be placed between the listed pairs to make the writing easier to read, as in Sentence J. (J) During Lisa's vacation, she plans to visit Cape Town, South Africa; Atlanta, Georgia; Albany, New York; and Las Vegas, Nevada. As shown in Sentence K, which follows (and which is almost completely identical to sentence J), the semicolon just before the word "and" at the end of the listing in sentence J is optional. Therefore, Sentences J and K are both correct. (K) During Lisa's vacation, she plans to visit Cape Town, South Africa; Atlanta, Georgia; Albany, New York and Las Vegas, Nevada. Now, keeping in mind the information above, complete exercise R21C below.

            QUICK CHALLENGE R21C: The Semicolon

            DirectionsReplace the underlined words in each question below with the answer choice that corrects the error in the sentence. If there is no error, select choice A -- NO CHANGE.

            1. Next summer, students from our College International Fellowship will have summer internships in New Delhi, India, London, England, and Paris, France.

            1. NO CHANGE
            2. India, London; England,
            3. India; London, England;
            4. India, London England

            2. Donnie Gray does not plan to get a COVID 19 shot, he is extremely afraid of needles. 

            1. NO CHANGE
            2. COVID, 19 shot he is;
            3. COVID 19 shot; he is
            4. COVID 19 shot, he is;

            3. We slept too late this morning; we need Mom to take us to school so we won't be late.

            1. NO CHANGE
            2. late this morning; we need Mom,
            3. late this morning, we need Mom
            4. late this morning; we need Mom;

            SAT Verbal - Answers to Questions from Summer 2021, Week 3 (July 24, 2021)

            1. C
            2. C
            3. A

            SAT Math - Questions from Summer 2021, Week 2 (July 17, 2021)

            The Average (Mean) Revisited

            It was noted in the last lesson that on the SAT the use of the word average usually refers to the mean and is indicated by “average (arithmetic mean).”

            It was also noted that the key to solving any problem involving the average (mean) is to find the total of the items before you do anything else. There are two ways to find the total: (1) add the numbers and (2) multiply the average (mean) by the total number of items. The second method is frequently used on the SAT.

            In some problems it will be necessary to calculate two or three totals. Some problems require maximizing or minimizing values. In order to minimize one value, you will need to maximize some other values. For example, if you have an upcoming test in a class and you are trying to determine the minimum score you can get on that test to give you a specific average for all of the tests in that class, you will need to assume that you will get the maximum score of 100 on the remaining tests.

            Consider the following example:

            An online store receives customer satisfaction ratings between 0 and 100, inclusive. In the first 10 ratings the store received, the average (arithmetic mean) of the ratings was 75. What is the least value the store can receive for the 11th rating and still be able to have an average of at least 85 for the first 20 ratings?

            We begin by getting the total for the first 10 ratings. Total = 10 x 75 = 750.

            Since we want an average of 85 for 20 ratings, we get the total for 20 ratings: Total = 20 x 85 = 1700.

            We can now determine the total for the last 10 ratings: 1700 - 750 = 950.

            Now if one rating, the 11th, is as small as possible, the other 9 ratings must be as large as possible. The highest possible rating is 100. Thus the maximum total for the other 9 ratings = 9 x 100 = 900.

            Thus the least possible rating for the 11th rating = 950 - 900 = 50.


            Keeping in mind the information above, answer the following questions.

            1. In a set of 15 integers, three of the integers are 12, 19, and 23. The mean of the 15 integers is 44. If 12, 19, and 23 are removed from the set, what is the mean of the remaining 12 numbers in the set?

            2. A new computer game receives critical reviews between 1 and 50, inclusive. In the first 6 ratings, the average (mean) of the ratings was 40. What is the least value the game can receive for the 8th rating and still be able to have an average of at least 42 for the first 10 ratings?

            3. Jackie took 6 tests in the fall semester of school. The mean score of the 6 tests was 84. If the mean score of the first four tests is 80, what is the mean score of the last 2 tests?

            SAT Math - Answers to Questions from Summer 2021, Week 2 (July 17, 2021)

            1. In a set of 15 integers, three of the integers are 12, 19, and 23. The mean of the 15 integers is 44. If 12, 19, and 23 are removed from the set, what is the mean of the remaining 12 numbers in the set?

              We begin by getting the total for the 15 integers. Total = 15 x 44 = 660
              We then subtract 12, 19, and 23 from 660 to get the new total: 660 – 12 – 19 – 23 = 606

              This new total divided by 12 gives us the mean of the remaining 12 numbers:
              606/12 = 50.5


            2. A new computer game receives critical reviews between 1 and 50, inclusive. In the first 6 ratings, the average (mean) of the ratings was 40. What is the least value the game can receive for the 8th rating and still be able to have an average of at least 42 for the first 10 ratings?

              We begin by getting the total for the first 10 ratings. Total = 6 x 40 = 240.

              Since we want an average of 42 for 10 ratings, we get the total for 10 ratings:
              Total = 10 x 42 = 420

              We can now determine the total for the last 4 ratings: 420 – 240 = 180.

              Now if one rating, the 8th, is as small as possible, the other 3 ratings must be as large as possible. The highest possible rating is 50. Thus the maximum total for the other 3 ratings = 3 x 50 = 150.

              Thus the least possible rating for the 8th rating = 180 - 150 = 30.


            3. Jackie took 6 tests in the fall semester of school. The mean score of the 6 tests was 84. If the mean score of the first four tests is 80, what is the mean score of the last 2 tests?

              The total for the 6 tests: total = 6 x 84 = 504 The total for the first 4 tests: total = 4 x 80 = 320
              Total for the last 2 tests = 504 - 320 = 184
              Thus the mean of the last 2 tests = 184/2 = 92

            SAT Verbal - Questions from Summer 2021, Week 2 (July 17, 2021)

            SAT Quick Challenge R21B
            Routine Uses of the Comma, Part II



            The Versatile Comma.  As noted in previous lessons, the comma does different kinds of jobs. For instance, when two separate independent clauses (also sentences by definition) are combined to make a single sentence, a FANBOYS conjunction (for, and, nor, but, or, yet, and so) can be used to create that new sentence. Note Sentences A, B, and C, which follow. (A) We expected our cousins to arrive this morning. (B) They missed their flight and won't get here until tonight. (C) We expected our cousins to arrive this morning, but they missed their flight and won't get here until tonight. As Sentence C shows, a comma must be placed in front of "but," the FANBOYS conjunction that turns the two independent clauses into a single sentence. As you will note below, today's lesson identifies additional ways in which commas are used in writing.

            Separating Items in a List. When a list of items is written as part of a sentence, a comma is often placed just before the word "and" to show that the last item is about to be named. However, that comma is optional. Note Sentences D and E, which follow. (D) People are advised to keep on hand masks, gloves, hand sanitizer, first aid kits, non-perishable foods, water, and other essential emergency supplies. (E) People are advised to keep on hand masks, gloves, hand sanitizer, first aid kits, non-perishable foods, water and other essential emergency supplies. Since the comma is optional, sentences D and E are both correct, so the SAT will not ask you to choose between the two; you will just need to remember that both are correct.

            Introductory Words and Phrases. An introductory word/phrase comes at the beginning of a sentence and sets the tone for what will be said. As illustrated in Sentence F, which follows, a comma must be used after an introductory word/phrase (in fact, initially, nevertheless, however, etc.) in a sentence. (F) Initially, we were going to the beach, but we decided to cook out at home, instead.

            The "Self" Words Exception. "Self" pronouns are used to emphasize the fact that a particular person or thing is being referred to. Every "object" pronoun (both direct and indirect objects) has a "self" counterpart, as in the following: me/myself, you/yourself, he/himself, she/herself, it/itself, etc. As illustrated in Sentence G, which follows, generally, a writer must not place a comma in front of or behind a self word. (G) Britney Spears herself will sing "America, the Beautiful" during the halftime show at the game on Friday. However, if a grammar rule requires a comma where a self word is used, you must follow that rule. (That is why commas follow the bold, italicized self words listed above in this paragraph.) Another example of the exception is the required comma for combining two independent clauses with a FANBOYS conjunction, as in Sentence H, which follows: (H) I will take Eric to school myself, but Mom will pick him up after school.

            Keeping in mind the information above, complete QUICK CHALLENGE R21B below. Then use the dropdown in the next section to check your work.

            QUICK CHALLENGE R21B: Routine Uses of the Comma, Part II

            DirectionsReplace the underlined words in each question below with the answer choice that corrects the error in the sentence. If there is no error, select choice A -- NO CHANGE.

            1. Mom said that you, yourself, must was the dishes after lunch today.

            1. NO CHANGE
            2. you, yourself must,
            3. you yourself, must
            4. you yourself must

            2. Little Andy wants to have cupcakes, pie ice cream: and candy for his birthday dinner. 

            1. NO CHANGE
            2. cupcakes, pie. ice cream: and
            3. cupcakes, pie, ice cream and
            4. cupcakes, pie ice cream, and

            3. The hotel overbooked its rooms; so our guests will be staying at our home during their visit.

            1. NO CHANGE
            2. its rooms; so our guests,
            3. its rooms, so our guests
            4. its rooms so our guests;

            4. Finally, Mom said, that we could go to the beach, but bad weather spoiled our plans. 

            1. NO CHANGE
            2. Finally, Mom said
            3. Finally Mom said
            4. Finally, Mom, said

            SAT Verbal - Answers to Questions from Summer 2021, Week 2 (July 17, 2021)

            1. D
            2. C
            3. C
            4. B

            SAT Math - Questions from Summer 2021, Week 1 (July 10, 2021)

            Averages and Range

            There are three averages that are tested on the SAT: mean, median, and mode.
            • Mean: the total of the items divided by the number of items
            • Median: the number that is exactly in the middle of a group of numbers when the numbers are arranged from smallest to largest; the median is always the middle value in a data set
            • Mode: the number that appears most often
            Range: the largest number – the smallest number

            Find the averages and range of these numbers: 6, 18, 12, 6, 8
            • Mean = (6 + 18 + 12 + 6 + 8)/5 = 50/5 = 10
            • Median; arrange the items in order: 6, 6, 8, 12, 18
            • Median = 8
            • Mode = 6
            • Range = 18 – 6 = 12
            Find the median of these numbers: 7, 4, 15, 20, 8, 15
            • Arrange in order: 4, 7, 8, 15, 15, 20
            • In this case, the median is midway between the two middle numbers:
              Median = (8 + 15)/2 = 23/2 = 11.5
            On the SAT the use of the word average usually refers to the mean and is indicated by “average (arithmetic mean).” Questions involving the median and mode will have those terms stated as part of the question’s text.

            The key to solving any problem involving the average (mean) is to find the total of the items before you do anything else. There are two ways to find the total: (1) add the numbers and (2) multiply the average (mean) by the total number of items. The second method is frequently used on the SAT.

            The average of 4 numbers is 5. If three of the four numbers are 3, 4, and 5, what is the fourth number? The first thing to do is to get the total. The total = 4 x 5 = 20.
            Thus the sum of the four numbers must total 20.
            3 + 4 + 5 = 12; to make the total = 20, the fourth number must be 8.

            Keeping in mind the information above, answer the following questions.

            1. 1, 6, 4, 10, 16, 4, 10, 25, 4, 20
              Calculate the mean, median, mode, and range for the above set of numbers.

            2. Weight (Pounds) Number of Dumbbells
              5 6
              10 10
              20 4

              David bought dumbbells in three different weights, in pounds. The table above shows the weight of the dumbbells, in pounds, and the number of dumbbells for each weight David bought. What is the mean weight of the dumbbells, in pounds?
              1. 11.67
              2. 10.50
              3. 8.50
              4. 6.67


            3. Player Height Player Height
              Alice 77 Florence 73
              Barbara 69 Geraldine 76
              Carolyn 71 Helen 68
              Denise 72 Ivey 70
              Edith 67 Jane 74

              The table above shows the heights of 10 players on the Greensboro High School women’s basketball team. If the coach takes Alice out of the game and substitutes Geraldine in her place, and makes no other substitutions, which of the following must be true? (In basketball, exactly five players from a team are allowed on the court at a time.)

              1. The median height of players on the court from Greensboro High School will not change.
              2. The median height of players on the court from Greensboro High School will increase.
              3. The median height of players on the court from Greensboro High School will decrease.
              4. A change in the median height of players on the court from Greensboro High School cannot be determined from the information given.

            SAT Math - Answers to Questions from Summer 2021, Week 1 (July 10, 2021)

              1. 1, 6, 4, 10, 16, 4, 10, 25, 4, 20
                Calculate the mean, median, mode, and range for the above set of numbers.

                • Mean = (1+6+4+10+16+4+10+25+4+20)10 = 100/10 = 10
                • Median: arrange the numbers in order from lowest to highest:
                  1, 4, 4, 4, 6, 10, 10, 16, 20, 25 The two middle numbers are 6 and 10.
                  Thus the median = (6 + 10)/2 = 16/2 = 8
                • Mode = 4 (the number that occurs most often)
                • Range = highest - lowest = 25 – 1 = 24

              2. Weight (Pounds) Number of Dumbbells
                5 6
                10 10
                20 4

                David bought dumbbells in three different weights, in pounds. The table above shows the weight of the dumbbells, in pounds, and the number of dumbbells for each weight David bought. What is the mean weight of the dumbbells, in pounds?
                1. 11.67
                2. 10.50
                3. 8.50
                4. 6.67

                Remember in an average problem, the first thing to do is to find the total.

                • The total weight of 6 dumbbells that weigh 5 pounds each is 6 x 5 = 30.
                • The total weight of 10 dumbbells that weigh 10 pounds each is 10 x 10 = 100.
                • The total weight of 4 dumbbells that weigh 20 pounds each is 4 x 20 = 80.
                • There are 20 dumbbells (6 + 10 + 4 = 20).
                • The total weight of the 20 dumbbells = 30 + 100 + 80 = 210.
                • Thus, the mean = 210/20 = 10.50 ------------------> B



              3. Player Height Player Height
                Alice 77 Florence 73
                Barbara 69 Geraldine 76
                Carolyn 71 Helen 68
                Denise 72 Ivey 70
                Edith 67 Jane 74

                The table above shows the heights of 10 players on the Greensboro High School women’s basketball team. If the coach takes Alice out of the game and substitutes Geraldine in her place, and makes no other substitutions, which of the following must be true? (In basketball, exactly five players from a team are allowed on the court at a time.)

                1. The median height of players on the court from Greensboro High School will not change.
                2. The median height of players on the court from Greensboro High School will increase.
                3. The median height of players on the court from Greensboro High School will decrease.
                4. A change in the median height of players on the court from Greensboro High School cannot be determined from the information given.


                1. This problem does not require any calculation or rearranging of data; it requires only an understanding of the concept of the median.
                2. If Alice is playing and Geraldine is not, it does not matter who the other four players on the team are playing with her; Alice is the tallest player.
                3. If Alice is taken out of the game and is replaced by Geraldine, then Geraldine is the tallest player on the court. All that has happened is that the tallest player on the court from Greensboro High School has changed.
                4. Since the median is the middle value in a data set, changing only the greatest value has no impact on the median. Thus the answer is A.

            SAT Verbal - Questions from Summer 2021, Week 1 (July 10, 2021)

            SAT Quick Challenge R21-A
            Routine Uses of the Comma, Part I



            Punctuating Nonessential Information (NESI).  A nonessential word or word group provides extra information (NOT the main idea) about another word or word group in a sentence -- generally the word that the nesi follows. If the nesi is removed, we will still have the main idea of the sentence, and the sentence will still make sense. Therefore, the nesi should be enclosed in commas. However, if the word or word group is needed to express the main idea clearly, do not use commas.

            When deciding whether to place commas around a word group, and where to place them if they are needed, draw a line under the entire possible nesi. The underlining can help you remember (1) to omit the nesi when you are reading to see if the sentence makes sense without it and (2) to read the entire nesi as you decide whether or not to use commas when an SAT question underlines only a small part of a long nesi.

            A relative pronoun (such as who, which, or that) is sometimes used to introduce a relative clause that modifies the word the clause follows. If the clause provides nonessential information (nesi) that does not help the reader understand clearly the main idea the writer is trying to state, use commas to separate the nesi from the rest of the sentence. If the clause is needed to help convey the main idea, do not separate it.

            Comparing NESI Punctuation. The three sentences which follow show how relative pronouns are sometimes used to introduce nesi. (A) Nan Epps, who just graduated from Northwest High School, has accepted a scholarship to Salem College. (B) Nan Epps has accepted a scholarship to Salem College. (C) The student who has accepted a scholarship to Salem College is Nan Epps.

            Commas separate the underlined relative clause in sentence A from the main idea because the reader does not need the information in the clause in order to understand the main idea. Sentence B simply states its main idea; it has no relative clause. In sentence C, commas do not separate the underlined relative clause from the rest of the sentence because the reader needs that clause in order to understand who Nan Epps is.

            The Interrupter. An interrupter comes within a sentence and creates emphasis or shows emotion by temporarily breaking the flow of thought in that sentence. Expressions routinely used as introductory words/phrases (in fact, however, initially, etc.) are also used as interrupters. Other common interrupters include expressions such as as you know and for example. Even a person's name can be used as an interrupter.

            You can use two commas, two dashes, or two parentheses to separate an interrupter from the rest of the sentence, as in sentences C and D, which follow. (C) What, Kenny, did you think would happen after you broke that window? (D) The heroic teenagers -- as we had expected -- received awards for their bravery.

            Keeping in mind the information above, complete QUICK CHALLENGE R21-A below. Then use the dropdown in the next section to check your work.

            QUICK CHALLENGE R21-A: Routine Comma Uses, Part I

            DirectionsReplace the underlined words in each question below with the answer choice that corrects the error in the sentence. If there is no error, select choice A -- NO CHANGE.

            1. Our Business Honor Society speaker is a lady who, we were told --  ran a business when she was 11 years old.

            1. NO CHANGE
            2. who we were told,
            3. who -- we were told
            4. who, we were told,

            2. A musical instrument, that, has been mishandled frequently, may never work properly again. 

            1. NO CHANGE
            2. instrument -- that has been mishandled frequently,
            3. instrument, that has been mishandled frequently
            4. instrument that has been mishandled frequently

            3. Crystal Beach, which attracts huge crowds to its annual Summer Fest events have great restaurants.

            1. NO CHANGE
            2. events; have
            3. events, has
            4. events, having

            SAT Verbal - Answers to Questions from Summer 2021, Week 1 (July 10, 2021)

            1. D
            2. D
            3. C

            SAT Math Formulas

            1. At the beginning of each math section, these formulas are given in the test booklet. If you haven’t memorized them, you should be familiar with what they mean.
              1. the length of the hypotenuse = twice the length of the side opposite the 30° angle.
              2. the length of the side opposite the 60° angle = the length of the side opposite the 30° angle times √3
              3. the length of the side opposite the 30° angle = ½ the length of the hypotenuse
              1. The two legs are equal
              2. the length of the hypotenuse = the length of the either leg times √2
              1. Area of a circle: A = π r2
              2. Circumference of a circle: c = 2 π r
              3. Area of a rectangle: A = lw
              4. Area of a triangle: A = ½ bh
              5. Pythagorean theorem: c2 = a2 + b2
              6. 30° – 60° right triangle:
              7. 45° – 45° right triangle:
              8. The volume of a rectangular solid: V = lwh
              9. The volume of a cylinder: V = π r2h
              10. The volume of a sphere: V = (4/3) π r3
              11. The volume of a cone: V = (1/3) π r2h
              12. The volume of a pyramid: V = (1/3)lwh
              13. The number of degrees in a circle = 360
              14. The number of degrees in a triangle = 180
              15. The number of radians in a circle = 2π

                You are given these 12 formulas and three geometry laws on the test itself. It can be helpful and save you time and effort to memorize the given formulas, but it is ultimately unnecessary, as they are given on every SAT math section.

            2. The following formulas are not printed on the test booklet; you will have to memorize them.
              1. Slope = rise (vertical change)/run (horizontal change) 
              2. Given two points on a line, (x1, y1) and (x2, y2), the slope = (y2 – y1)/(x2 – x1).
              3. If the equation of the line is in the slope/intercept form, y = mx + b, the slope = m
              4. If the equation of the line is in standard form, Ax + By = C, the slope = -A/B
              1. Total = sum of the items
              2. Total = the average times the number of items. This method is usually required on SAT problems. 
              1. sine of an angle = side opposite the angle over the hypotenuse (SOH)
              2. cosine of an angle = side adjacent to the angle over the hypotenuse (CAH)
              3. tangent of an angle = side opposite the angle over the side adjacent to the angle (TOA)
              1. i = √-1
              2. i2 = -1
              3. i= -i
              4. i= i
              5. i= i
              6. i= -i
              7. i= -i
              8. i= i
                etc.
              1. The volume of a square: A = s2
              2. The perimeter of figure = the sum of all of the sides
              3. Area of a parallelogram: A = lw
              4. Area of a trapezoid: A = ½ h(b1 + b2)
              5. Given a radius and a degree measure of an arc from the center of a circle, find the area of the sector that is defined by the angle and the arc:
                Area of a sector of a circle: A = (t/360) π r2 when t = the number of degrees in the central angle
              6. Given a radius and a degree measure of an arc from the center, find the length of the arc:
                Length of an arc: L = (t/360) (2 π r) when t = the number of degrees in the central angle
              7. When the angles of triangle A are equal to the angles of triangle B, the sides of triangle A are proportional to the sides of triangle B.
              8. x2 – y2 = (x + y)(x – y)
              9. (x + y)2 = x2 + 2xy + y2
              10. (x - y)2 = x2 - 2xy + y2
              11. A function in the form of f(x) = 3x + 12 is the same as y = 3x + 12.
              12. The equation of the line in the slope/intercept form: y = mx + b, where the slope = m, and the y-intercept = b.
              13. The equation of the line in standard form: Ax + By = C, where the slope = -A/B and the
                y-intercept = C/B
              14. Slope – four ways to determine the slope:
              15. The standard form of a parabola equation: y = ax2 + bx + c
              16. Vertex form of the parabola equation: y = a(x – h)2 + k, where the vertex is the point (h,k).
              17. Equation of a circle?    (x – h)2 + (y – k)2 = r2 where the center of the circle is the point (h,k)
                and the radius of the circle is r.
              18. The quadratic formula:
                For ax2 + bx + c = 0, the value of x is given by:

                 Quadratic Equation

              19. The key to solving average problems is to find the total of the items before doing anything else. There are two ways to find the total:
              20. Average speed = total distance / total time; Distance = (speed) x (time)
              21. SOHCAHTOA (applies to a right triangle)
              22. 180 degrees = π radians
              23. Imaginary Numbers
              24. A present amount P increases at an annual rate r for t years. The future amount F in t years is:
                F = P(1 + r) 
              25. A present amount P decreases at an annual rate r for t years. The future amount F in t years is:
                F = P(1 - r) 
              26. Item sold at discount:  discount amount = original price x discount percent  
              27. Item sold at discount:  reduced price = original price x (1-discount percent)
              28. Given two points, A(x1,y1) , B(x2,y2), find the midpoint of the line that connects them:
                Midpoint = the average of the x coordinates and the y coordinates:  (x1 + x2) / 2 , (y1 + y2) / 2 
              29. Given two points, A(x1,y1) , B(x2,y2), find the distance between them:
                Distance= √[ (x2 - x1)2 + (y2 - y1)
              30. Probability of x = (number of outcomes that are x) / (total number of possible outcomes)

            SAT Math Operations

            Operations You Need to be Able to Perform
            1. Substitute values for a variable and simplify.
            2. Add fractions with different denominators, where the denominators are numbers.
            3. Add fractions with different denominators, where the denominators are variables.
            4. Know how to simplify complex fractions.
            5. In a fraction, the denominator cannot equal zero. If an equation is solved and the value of the variable makes the denominator = zero, then that value cannot be a solution to the problem.
            6. When picking numbers, consider positive numbers, negative numbers, zero, decimals, and extreme numbers.
            7. Understand the definitions of the terms digit, integer, number, prime number, factor, multiple, divisible, reciprocal of a number, absolute value of a number.
            8. Know the absolute value sign.
            9. Know the common fraction-decimal-percent equivalents.
            10. Know how to change a fraction to a decimal or to a percent.
            11. Know how to change a decimal to a percent.
            12. Know how to change a percent to a decimal.
            13. Understand the factorial concept.
            14. Know how to compute permutations and combinations: n items taken x at a time.
            15. Know when to use Venn diagrams.
            16. When angles are formed when a line crosses parallel lines, several equal angles are created.
            17. When a diagram is given in a geometry problem, consider adding one or more lines to create another figure.
            18. Geometric figures are not necessarily drawn to scale; lines that look equal may not be equal; angles that look equal may not be equal.
            19. In a triangle, the length of sides opposite equal angles are equal.
            20. In a triangle, the length of a side opposite a larger angle is greater than the side opposite a smaller angle.
            21. Know the third side rule for triangles: the length of any one side of a triangle must be less than the sum of the other two sides, and greater than the difference between the other two sides.
            22. Two triangles are congruent if the sides of one triangle are equal to the corresponding sides of the other triangle and the angles of one triangle are equal to the corresponding angles of the other triangle.
            23. Two triangles are similar if the angles of one triangle are equal to the corresponding angles of the other triangle and the sides of one triangle are not equal to the corresponding sides of the other triangle.
            24. If two triangles are similar, their corresponding sides are proportional.
            25. The measure of an angle inscribed in a circle is half the measure of the central angle that intercepts the same arc.
            26. The length of an arc is a fraction of the circumference of a circle.
            27. A line tangent to a circle produces a right angle at the point of tangency between the line and another line that connects the point of tangency to the center of the circle.
            28. Know the exponent rules.
            29. Know how to express a number with alternative bases using appropriate exponents; the most common problems involve changing a number to a base of 2 or a base of 3.
            30. Know how to determine the three averages: mean, median, and mode.
            31. Know how the normal curve, mean, and standard deviation interact.
            32. Read ratio problems carefully,
              1. A ratio can express a part to part relationship.
                For example, a ratio of 1 to 2 = 1:2 = ½.
              2. A ratio can express a part to whole relationship.
                For example, a ratio of 1 to 2 has two parts and a whole (1 + 2 = 3). One part is ⅓, the other part is ⅔.
            33. Solve linear equations when the answer is a number (one equation and one unknown.)
            34. Solve linear equations when one variable is in terms of other variables (one equation with all variables.)
            35. Solve simultaneous equations (two equations in 2 unknowns.)
            36. Solve quadratic equations by factoring, by using the quadratic equation, and by completing the square.
            37. Find the radius of a circle from the formula of a circle.
            38. Know how to write the formula of a circle in standard form.
            39. Factor an expression.
              1. Type 1: 3xy + 7x = x(3y + 7)
              2. Type 2: 2x2 + 13x + 15 = (2x + 3)(x + 5)
              3. Type 3: 213 – 211 = 211(22 – 1) = 211(4-1) = 211(3)
            40. Solve inequalities.
            41. Find the price of an item after a sales tax is added.
            42. Find the price of an item after a percent increase.
            43. Find the price of an item after a percent decrease.
            44. Find the percent of a number.
            45. When one number is greater than another, find the percent greater.
            46. When an amount changes, find the percent change.
            47. Know the three averages: mean, median, and mode.
            48. In an average problem, the first thing to do is to find the total; there are two ways to find the total.
            49. Find the average of a set of numbers.
            50. Find the missing number in a set of numbers when the mean is known.
            51. From the equation of a line, determine the y intercept, x intercept, and slope.
            52. Understand positive slopes, negative slopes, and slope = zero.
            53. Equation of a parabola.
            54. Coordinate geometry: locate points in the xy plane.
            55. Know the I, II, III, and IV quadrants.
            56. Evaluate information in a chart.
            57. Word problems: write down each detail; proceed in a step by step fashion.
            58. Trigonometry: find the sine of an angle; find the cosine of an angle; find the tangent of an angle (remember SOH-CAH-TOA.)
              The sine of an angle = the side opposite the angle divided by the hypotenuse (SOH)
              The cosine of an angle = the side adjacent to the angle divided by the hypotenuse (CAH)
              The tangent of an angle = the side opposite the angle divided by the side adjacent to the angle (TOA)
            59. In right triangle ABC, if angle B is the right angle, then the sine of angle A = the cosine of angle C.


            2021 Lessons