What do you think is the best way to prepare your student for the 6th Grade NYSE Math exam? Participating in pre-tests and solving sample practice test questions can help your student prepare for the 6th Grade NYSE Math test. The more your students becomes familiar with the different types of test’s questions, the less anxious they will be on the day of the test. This type of preparation can lead to Better answers to the various challenges and questions that your students might pose. The test takers can never predict what kind of questions await them in the exam. So it is better to be 100% ready before the test.

Therefore, in this blog post, we provide you with a collection of 10 commonly used practice questions and their step-by-step solutions to help your student prepare for the 6th Grade NYSE Math test as much as possible.

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## 10 Sample **6th Grade NYSE** Math Practice Questions

1- What is the missing prime factor of number 420?

\(420=2^2×3^1×…\)

A. \(2^2×3^1×5^1×7^1\)

B. \(2^2×3^1×7^1×9^1\)

C. \(1^2×2^3×2^1×3^1\)

D. \(3^2×5^1×7^1×9^1\)

2- If the area of the following trapezoid is equal to \(A\), which equation represent \(x\)?

A. \( x = \frac{13}{A}\)

B. \( x = \frac{A}{13}\)

C. \( x=A+13\)

D. \( x=A-13\)

3- By what factor did the number below change from first to fourth number?

\(8, 104, 1352, 17576\)

A. 13

B. 96

C. 1456

D. 17568

4- 170 is equal to …

A. \( -20-(3×10)+(6×40)\)

B. \(((\frac{15}{8})×72 )+ (\frac{125}{5}) \)

C. \(((\frac{30}{4} + \frac{15}{2})×8) – \frac{11}{2} + \frac{222}{4}\)

D. \(\frac{481}{6} + \frac{121}{3}+50\)

5- The distance between two cities is 3,768 feet. What is the distance of the two cities in yards?

A. 1,256 yd

B. 11,304 yd

C. 45,216 yd

D. 3,768 yd

6- Mr. Jones saves $3,400 out of his monthly family income of $74,800. What fractional part of his income does Mr. Jones save?

A. \(\frac{1}{22}\)

B. \(\frac{1}{11}\)

C. \(\frac{3}{25}\)

D. \(\frac{2}{15}\)

7- What is the lowest common multiple of 12 and 20?

A. 60

B. 40

C. 20

D. 12

8- Based on the table below, which expression represents any value of f in term of its corresponding value of \(x\)?

A. \(f=2x-\frac{3}{10}\)

B. \(f=x+\frac{3}{10}\)

C. \(f=2x+2 \frac{2}{5}\)

D. \(2x+\frac{3}{10}\)

9- 96 kg \(=\)… ?

A. 96 mg

B. 9,600 mg

C. 960,000 mg

D. 96,000,000 mg

10- Calculate the approximate area of the following circle? (the diameter is 25)

A. 78

B. 491

C. 157

D. 1963

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## Answers:

1- **A**

\(420=2^2×3^1×5^1×7^1\)

2- **B**

The area of the trapezoid is: area= \(\frac{(base 1+base 2)}{2})×height= ((\frac{10 + 16}{2})x = A\)

\( →13x = A→x = \frac{A}{13}\)

3-** A**

\(\frac{104}{8}=13, \frac{1352}{104}=13, \frac{17576}{1352}=13\)

Therefore, the factor is 13

4- **C**Simplify each option provided.

\( A. -20-(3×10)+(6×40)=-20-30+240=190\)

\( B. (\frac{15}{8})×72 + (\frac{125}{5}) =135+25=160\)

\(C. ((\frac{30}{4} + \frac{15}{2})×8) – \frac{11}{2} + \frac{222}{4} = ((\frac{30 + 30}{4})×8)- \frac{11}{2}+ \frac{111}{2}=(\frac{60}{4})×8) + \frac{100}{2}= 120 + 50 = 170\)this is the answer

\(D. \frac{481}{6} + \frac{121}{3}+50= \frac{481+242}{6}+50=120.5+50=170.5\)

5- **A**

1 yard \(= \)3 feet

Therefore, \(3,768 ft × \frac{1 \space yd }{3 \space ft}=1,256 \space yd\)

6- **A**

3,400 out of 74,800 equals to \(\frac{3,400}{74,800}=\frac{17}{374}=\frac{1}{22}\)

7- **A**

Prime factorizing of \(20=2×2×5\)

Prime factorizing of \(12=2×2×3\)

LCM\(=2×2×3×5=60\)

8- **C**

Plug in the value of \(x\) into the function f. First, plug in 3.1 for \(x\).

\(A. f=2x-\frac{3}{10}=2(3.1)-\frac{3}{10}=5.9≠8.6\)

\(B. f=x+\frac{3}{10}=3.1+\frac{3}{10}=3.4≠10.8\)

\(C. f=2x+2 \frac{2}{5}=2(3.1)+2 \frac{2}{5}=6.2+2.4=8.6 \)

This is correct!

Plug in other values of \(x. x=4.2\)

\(f=2x+2\frac{2}{5} =2(4.2)+2.4=10.8 \)

This one is also correct.

\(x=5.9\)

\(f=2x+2 \frac{2}{5}=2(5.9)+2.4=14.2 \)

This one works too!

\(D. 2x+\frac{3}{10}=2(3.1)+\frac{3}{10}=6.5≠8.6\)

9- **D**

1 kg\(=\) 1000 g and 1 g \(=\) 1000 mg

96 kg\(=\) 96 \(×\) 1000 g \(=\)96 \(×\) 1000 \(×\) 1000 \(=\)96,000,000 mg

10- **B**

The diameter of a circle is twice the radius. Radius of the circle is \(\frac{25}{2}\).

Area of a circle = \(πr^2=π(\frac{25}{2})^2=156.25π=156.25×3.14=490.625≅491\)

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