# Top 10 TExES Core Subjects EC-6 Math Practice Questions

Preparing for the TExES Core Subjects Math test? The best way to prepare for your TExES Core Subjects EC-6 Math test is to work through as many TExES Math practice questions as possible. Here are the top 10 TExES Core Subjects EC-6 Math practice questions to help you review the most important TExES Math concepts. These TExES Math practice questions are designed to cover mathematics concepts and topics that are found on the actual test. The questions have been fully updated to reflect the latest 2022 TExES Core Subjects guidelines. Answers and full explanations are provided at the end of the post.

Start your TExES Math test prep journey right now with these sample TExES Core Subjects EC-6 Math questions.

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**TExES Core Subjects EC-6 Math Practice Questions**

1- What is the value of the expression \((5(x-2y)+(2-x))^2 \) when \(x=3\) and \(y=-2 \)?

A. \(20\)

B. \(50\)

C. \(1156\)

D. \(36\)

2- What is the slope of a line that is perpendicular to the line

\(4x-2y=12\)?

A. \(-2\)

B. \(4\)

C. \(2\)

D. \(-\frac{1}{2}\)

3- Which of the following is equal to the expression below?

\((2x+2y)(2x-y)\)

A. \(4x^2- 2y^2\)

B. \(2x^2+6xy-2y^2\)

C. \(24x^2+2xy-2y^2\)

D. \(4x^2+2xy-2y^2\)

4- What is the product of all possible values of \(x\) in the following equation?

\(|x-10| = 3\)

A. \(3\)

B. \(7\)

C. \(13\)

D. \(91\)

5- \( [6 \times (- 24) + 8] – (-4) + [4 \times 5] \div 2 = \)?

A. \(111\)

B. \(-112\)

C. \(-122\)

D. \(-144\)

6- What is the value of \(x\) in the following system of equations?

\(2x+5y=11\)

\(4x-2y=-14\)

A. \(-1\)

B. \(1\)

C. \(4\)

D. \(-2\)

7- What is the median of these numbers? \(2, 27, 28, 19, 67, 44, 35\)

A. \(19\)

B. \(28\)

C. \(44\)

D. \(35\)

8- A swimming pool holds \(2,000\) cubic feet of water. The swimming pool is \(25\) feet long and \(10\) feet wide. How deep is the swimming pool? ____________

9- The area of a circle is \(64 π\). Which of the following is the circumference of the circle?

A. \(8 π\)

B. \(12 π\)

C. \(16 π\)

D. \(64 π\)

10- What is the value of \(3^6 \)? ____________

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## Answers:

1- **D**

Plug in the value of \(x\) and \(y\).

\(x=3\) and \(y=-2\)

\((5(x-2y)+(2-x))^2=(5(3-2(-2))+(2-3))^2=(5(3+4)+(-1))^2 = (34)^2=36\)

2- **D**

The equation of a line in slope intercept form is: \(y=mx+b\)

Solve for \(y\).

\(4x-2y=12 {\Rightarrow} -2y=12-4x {\Rightarrow} y=(12-4x){\div}(-2) {\Rightarrow} y=2x-6\)

The slope of this line is \(2\).

The product of the slopes of two perpendicular lines is \(-1\).

Therefore, the slope of a line that is perpendicular to this line is:

\(m_{1} {\times} m_{2} = -1 {\Rightarrow} 2 {\times} m_{2} = -1 {\Rightarrow} m_{2} = \frac{-1}{2}\)

3- **D**

Use FOIL method.

\((2x+2y)(2x-y) = 4x^2-2xy+4xy-2y^2=4x^2+2xy-2y^2\)

4- **D**

To solve absolute values equations, write two equations.

\(x-10\) could be positive \(3\), or negative \(3\). Therefore,

\(x-10=3 \Rightarrow x=13\)

\(x-10=-3 \Rightarrow x=7\)

Find the product of solutions: \((7 \times\) \(13 = 91\))

5- **C**

Use PEMDAS (order of operation):

\([6 {\times} (- 24) + 8] – (- 4) + [4 × 5] {\div} 2\)

\([- 144 + 8] – (- 4) + [20] {\div} 2 =

[- 144 + 8] – (- 4) + 10 \)

\([- 136] – (- 4) + 10 = [- 136] + 4 + 10 = – 122\)

6- **D**

Solving Systems of Equations by Elimination

Multiply the first equation by \((-2)\), then add it to the second equation.

\({-2(2x+5y= 11) \ 4x-2y=-14} {\Rightarrow} {-4x-10y= -22 \ 4x-2y=-14} {\Rightarrow} {-12y= -36} {\Rightarrow} {y= 3}\)

Plug in the value of \(y\) into one of the equations and solve for \(x\).

\(2x+5(3)= 11 {\Rightarrow} 2x+15= 11 {\Rightarrow} 2x= -4 {\Rightarrow} x= -2\)

7- **B**

Write the numbers in order:

\(2, 19, 27, 28, 35, 44, 67\)

Median is the number in the middle. So, the median is \(28\).

8- **8**

Use the formula of rectangle prism volume.

V = (length) (width) (height) \({\Rightarrow}\) \(2000 = (25) (10) (height)\) \({\Rightarrow}\) \(height = 2000 {\div}\) \(250 = 8\)

9- **C**

The area of the circle is \(16 π\). Use the formula of areas of circles.

Area \(= πr^2 ⇒ 64 π> πr^2⇒ 64 > r^2⇒ r < 8\)

The radius of the circle is 8. Let’s put 8 for the radius. Now, use the circumference formula:

Circumference \(=2πr=2π (8)=16π\)

10- **729**

\(3^6 = 3 × 3 × 3 × 3 × 3 × 3 = 729\)

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