Preparing for the Pre-Algebra Math test? Try these free Pre-Algebra Math Practice questions. Reviewing practice questions is the best way to brush up on your Math skills. Here, we walk you through solving 10 common Pre-Algebra Math practice problems covering the most important math concepts on the Pre-Algebra Math test.

These Pre-Algebra Math practice questions are designed to be similar to those found on the real Pre-Algebra Math test. They will assess your level of preparation and will give you a better idea of what to study for your exam.

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## 10 Sample **Pre-Algebra** Math Practice Questions

1- Jason needs a score of 75 average in his writing class to pass. On his first 4 exams, he earned scores of 68, 72, 85, and 90. What is the minimum score Jason can earn on his fifth and final test to pass?_________

2- Anita’s trick–or–treat bag contains 12 pieces of chocolate, 18 suckers, 18 pieces of gum, 24 pieces of licorice. If she randomly pulls a piece of candy from her bag, what is the probability of her pulling out a piece of sucker?

A. \(\frac{1}{3} \)

B. \(\frac{1}{4} \)

C. \(\frac{1}{6} \)

D. \(\frac{1}{12} \)

3- What is the perimeter of a square in centimeters that has an area of 595.36 cm\(^2\)?

A. 97.2

B. 97.6

C. 97.7

D. 97.9

4- The perimeter of a rectangular yard is 60 meters. What is its length if its width is twice its length?

A. 10 meters

B. 18 meters

C. 20 meters

D. 24 meters

5- The average of 6 numbers is 12. The average of 4 of those numbers is 10. What is the average of the other two numbers?

A. 10

B. 12

C. 14

D. 16

6- If \(40\%\) of a number is 4, what is the number?

A. 4

B. 8

C. 10

D. 12

7- The average of five numbers is 24. If a sixth number 42 is added, then, which of the following is the new average?

A. 25

B. 26

C. 27

D. 42

8- The ratio of boys and girls in a class is 4:7. If there are 44 students in the class, how many more boys should be enrolled to make the ratio 1:1?

A. 8

B. 10

C. 12

D. 14

9-What is the slope of the line: \(4x-2y=6\):_______

10- A football team had $20,000 to spend on supplies. The team spent $14,000 on new balls. New sports shoes cost $120 each. Which of the following inequalities represent the number of new shoes the team can purchase.

A. 120\(x\)+14,000 \(\leq \) 20,000

B. 20\(x\)+14,000 \(\geq \) 20,000

C. 14,000\(x\)+120 \(\leq \) 20,000

D. 14,000\(x\)+12,0 \(\geq \) 20,000

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## Answers:

1- **60**

Jason needs a score 75 average to pass for five exams. Therefore, the sum of 5 exams must be at lease 5 \(\times \) 75 = 375

The sum of 4 exams is:

68 + 72 + 85 + 90 = 315

The minimum score Jason can earn on his fifth and final test to pass is:

\(375 – 315 = 60\)

2- **B**

\(Probability = \frac{number \space of \space desired \space outcomes}{number \space of \space total \space outcomes} = \frac{18}{12+18+18+24} = \frac{18}{72} = \frac{1}{4}\)

3-** B**

The area of the square is 595.36. Therefore, the side of the square is the square root of the area.

\(\sqrt{595.36}=24.4\)

Four times the side of the square is the perimeter:

\(4 {\times} 24.4 = 97.6\)

4- **A**The width of the rectangle is twice its length. Let \(x\) be the length. Then, width=\(2x\)

Perimeter of the rectangle is \(2 (width + length) = 2(2x+x)=60 {\Rightarrow} 6x=60 {\Rightarrow} x=10 \)

The length of the rectangle is 10 meters.

5- **D**

average \(= \frac{sum \space of \space terms}{number \space of \space terms} {\Rightarrow} (average \space of \space 6 \space numbers) \space 12 = \frac{sum \space of \space terms}{6} ⇒sum \space of \space 6 \space numbers\space is \)

\(12 {\times} 6 = 72\)

\((average \space of \space 4 \space numbers) \space 10 = \frac{sum \space of \space terms}{4}{\Rightarrow} sum \space of \space 4 \space numbers \space is \space 10 {\times} 4 = 40\)

sum of 6 numbers – sum of 4 numbers = sum of 2 numbers

\(72 – 40 = 32\)

average of 2 numbers = \(\frac{32}{2} = 16 \)

6- **C**

Let \(x\) be the number. Write the equation and solve for \(x\).

\(40\% \space of \space x=4{\Rightarrow} 0.40 \space x=4 {\Rightarrow} x=4 {\div}0.40=10\)

7- **C**

First, find the sum of five numbers.

\(average =\frac{ sum \space of \space terms }{ number \space of \space terms } ⇒ 24 = \frac{ sum \space of \space 5 \space numbers }{5}\)

\( ⇒ sum \space of \space 5 \space numbers = 24 × 5 = 120\)

The sum of 5 numbers is 120. If a sixth number that is 42 is added to these numbers, then the sum of 6 numbers is 162.

120 + 42 = 162

average \(==\frac{ sum \space of \space terms }{ number \space of \space terms } = \frac{162}{6}=27\)

8- **C**

The ratio of boys to girls is 4:7.

Therefore, there are 4 boys out of 11 students.

To find the answer, first, divide the total number of students by 11, then multiply the result by 4.

\(44 {\div} 11 = 4 {\Rightarrow} 4 {\times} 4 = 16\)

There are 16 boys and 28 \((44 – 16)\) girls. So, 12 more boys should be enrolled to make the ratio 1:1

9- **2**

Solve for y.

\(4x-2y=6 {\Rightarrow} -2y=6-4x {\Rightarrow} y=2x-3\)

The slope of the line is 2.

10- **A**

Let \(x\) be the number of new shoes the team can purchase. Therefore, the team can purchase 120 \(x\).

The team had $20,000 and spent $14000. Now the team can spend on new shoes $6000 at most.

Now, write the inequality:

\(120x+14,000 {\leq}20,000\)

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