# TExES Core Math Practice Test Questions

Preparing for the TExES Core Math test? Try these free TExES Core 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 TExES Core Math practice problems covering the most important math concepts on the TExES Core Math test.

These TExES Core Math practice questions are designed to be similar to those found on the real TExES Core 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 TExES Core 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. \(120x+14,000 \leq 20,000\)

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

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

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

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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 a square root of the area.

\(\sqrt{595.36}=24.4\)

Four times the size 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=2\(x\)

Perimeter of the rectangle is (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 \(2\)8 \((44 – 16)\) 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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