SAT Preparation Classes

SAT Quick Challenge - Week 5
Fall 2021

Saturday, October 16, 2021

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. Each Saturday except those occurring during holiday weekends, this site will be updated with a set of activities that will help students acquire and maintain mastery of important skills that help lead to high SAT scores. Be sure to review the SAT Math Formulas and the SAT Math Operations each week. You can find them in the dropdowns at the bottom of this page.              

Please stay healthy and safe. Remember the three Ws: Wear, Wait, Wash. May God bless you and yours.

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.

    February 2021 Lessons
    March 2021 Lessons
    April 2021 Lessons
    May 2021 Lessons
    June 2021 Lessons