6th Grade FSA Math Practice Test Questions

6th Grade FSA Math Practice Test Questions

For your 6th Grade students to perform well on the 6th Grade FSA Math test, they need to practice enough of what they have learned. The use of practice test questions strengthens 6th Grade students’ testing skills. You can use the top 10 6th Grade FSA Math practice questions collected in this article to help strengthen your students’ math skills. The answers to the questions are explained in detail so that students are well acquainted with how each question works.

Make sure to follow some of the related links at the bottom of this post to get a better idea of what kind of mathematics questions students need to practice.

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10 Sample 6th Grade FSA Math Practice Questions

1- What is the missing prime factor of the 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 represents \(x\)?

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 the first to the 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 the two cities is 3,768 feet. What is the distance between 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

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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