Math Word Problems for CBEST Exam Study Guide

Writing Word Problems in Algebraic Form

The following are exercises that will help you translate words into equations.

Sample Word Conversion Questions

The following are simple problems to rewrite in algebraic form. Using N for a number, try writing out the problems below. Remember to add parentheses as needed to avoid order of operation problems.

1. Three added to a number gives us 6.
2. Six subtracted from a number is 50.

Answers

Use the four Success Steps to find the answer to question 1.

1. "Gives us" is the verb. Put in an equal sign: =
2. 3, 6, and N are the numbers: 3 N = 6
3. Added means +: 3 + N = 6
4. No parentheses are needed.

Follow the Success Steps for question 2.

1. "Is" is the verb. Put in an equal sign: =
2. 6, N, and 50 are the numbers: 6 N = 50.
3. Subtracted from means –, but it is a backwardsword: N – 6 = 50.
4. No parentheses are needed.

Practice

Underline the backwardswords, then write the equations.

3. A number subtracted from 19 is 7.
4. Three less a number is 5.
5. Three less than a number is 5.
6. Nine less a number is –8.
7. A number taken from 6 is –10.
8. Thirty deducted from a number is 99.
9. The quotient of 4 and a number equals 2.
10. The reciprocal of 5 over a number is 10.
11. Six divided into a number is 3.

Change the following sentences into algebraic equations.

12. The sum of 60 and a number all multiplied by 2 amounts to 128.
13. Forty combined with twice a number is 46.
14. Nine dollars fewer than a number costs $29.
15. Seven feet lengthened by a number of feet all divided by 5 is equivalent to 4 feet.
16. Ninety subtracted from the sum of a number and one gives us 10.
17. Half a number plus 12 is the same as 36.

Answers

3. subtracted from, 19 – N = 7
4. 3 – N = 5
5. less than, N – 3 = 5
6. 9 – N = –8
7. taken from, 6 – N = –10
8. deducted from, N – 30 = 99
9. = 2
10. reciprocal of, = 10
11. divided into, = 3
12. (60 + N)2 = 128 or 2(60 + N) = 128
13. 40 + 2N = 46
14. N – $9 = $29
15. = 4
16. (N + 1) – 90 = 10
17. +12 = 36 or + 12 = 36

Words or Numbers?

Try these two problems and determine which is easier for you.

1. Three more than five times a number equals 23.
2. Jack had three more than five times the number of golf balls that Ralph had. If Jack had 23 golf balls, how many did Ralph have?

Answers

1. 23 = 3 + 5N
2. 23 = 3 + 5N

Did you notice that the two problems were the same, but the second one was more wordy? If question 1 was easier, you can work word problems more easily by eliminating nonessential words. If question 2 was easier, you can work out problems more easily by picturing actual situations. If they were both equally easy, then you have mastered this section. Go on to the section on two-variable problems, which is a little more difficult.

Practice

If you found wordy word problems difficult, here are some more to try:

18. Sally bought 6 less than twice the number of boxes of CDs that Raphael (R) bought. If Sally bought 4 boxes, how many did Raphael buy?
19. A 1-inch by 13-inch rectangle is cut off a piece of linoleum that was made up of three squares in a row; each had N inches on a side. This left 62 square inches in the original piece of linoleum. How long was each of the sides of the square?
20. Six was added to the number of sugar cubes in a jar. After that, the number was divided by 5. The result was 6.How many sugar cubes were in the jar?

Answers

18. Sally = 2R – 6. Substitute 4 for Sally: 4 = 2R – 6
19. The first step is to find the dimensions of the original piece of linoleum by adding the area of the piece that was cut (1 × 13 = 13 square inches) to the area of the remaining piece: 13 + 62 = 75 square inches. If we use S to represent the length of a side of one square, then the area of the square is S², and the area of all three squares is 3S². The equation, then, will be 3S²= 75. Dividing both sides by 3, we get S²= 25. Finally, take the square root of each side to get S = 5. The length of each side is 5 inches.
20. = 6.Multiply both sides of the equation by 5, then subtract 6 from both sides of the equation: 5 () = 6 (5), 6 + N = 30, N = 24. There are 24 sugar cubes in the jar.

Three Success Steps for Problems with Two Variables

When turning "as many as" sentences into equations, consider the following steps.

1. Read the problem to decide which variable is least.
2. Combine the number given with the least variable.
3. Make the combined number equal to the larger amount.

Problems with Two Variables

In solving problems with two variables, you have to watch out for another backwards phrase: as many as.

Sample Two-Variable Questions

The following equations require the use of two variables. Choose the answers from the following:

1. 2x = y
2. 2y = x
3. 2 + x = y
4. 2 + y = x
5. none of the above

21. Twice the number of letters Joey has equals the number of letters Tina has. Joey = x, Tina = y.
22. Tuli corrected twice as many homework assignments as tests. Homework = x, tests = y.

Answers

21. a. Equals is the verb. Joey, or x, is on one side of the verb; Tina, or y, is on the other. A straight rendering will give you choice a, or 2x = y, because Tina has twice as many letters as Joey has. To check, plug in 6 for y. If Tina has 6 letters, Joey will have 6 ÷ 2, or 3 letters. The answer makes sense.
22. b. Corrected is the verb. Which did Tuli correct fewer of? Tests. You need to multiply 2 times the number of tests to reach the number of homework assignments. Check: If there are 6 tests, then there are 12 homework assignments: 2 × 6 = 12. This answer makes sense.

Practice

Now that you are clued in, try the following using these same answer choices.

1. 2x = y
2. 2y = x
3. 2 + x = y
4. 2 + y = x
5. none of the above

23. Sandra found two times as many conch shells as mussel shells. Conch = x, mussel = y.
24. Sharon walked two more miles today than she walked yesterday. Today = x, yesterday = y.
25. Martin won two more chess games than his brother won. His brother = x, Martin = y.

Answers

23. b.
24. d.
25. c.