# 6th Grade Georgia Milestones Assessment System math Practice Test Questions

As you know, the proper study is only one part of the learning process and preparation for the 6th Grade Georgia Milestones Assessment System. Better learning is possible with a lot of practice. One way to practice the lessons is to work on the 6th Grade Georgia Milestones Assessment System practice test questions. In this article, our experts have prepared 10 common 6th Grade math questions so that you can use them to improve your student’s skills. You also have a step-by-step guide to solving the questions.

We suggest that you also check the relevant links at the bottom of the article to find other related practice questions

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**10 Sample 6th Grade Georgia Milestones Assessment System 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 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 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

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

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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