FREE 6th Grade FSA Math Practice Test

B. 162 cm\(^2\)

C. 99 cm\(^2\)

D. 81cm\(^2\)

6- Which of the following expressions has the greatest value?

A. \( 3^1+12\)

B. \( 3^3-3^2\)

C. \( 3^4-60\)

D. \( 3^5-218\)

7- Alfred has \(x\) apples. Alvin has 40 apples, which is 15 apples less than number of apples Alfred owns. If Baron has \(\frac{1}{5}\) times as many apples as Alfred has. How many apples does Baron have?

A. 5

B. 11

C. 55

D. 275

8- In the following triangle find \(α\).

A. \(100^\circ\)

B. \(90^\circ\)

C. \(60^\circ\)

D. \(30^\circ\)

9- The price of a laptop is decreased by \(15\%\) to $425. What is its original price?

A. $283

B. $430

C. $500

D. $550

10- Find the perimeter of the shape in the following figure. (all angles are right angles)

A. 21

B. 22

C. 24

D. 20

11- What are the values of mode and median in the following set of numbers?
\(1,3,3,6,6,5,4,3,1,1,2\)

A. Mode: 1, 2, Median: 2

B. Mode: 1, 3, Median: 3

C. Mode: 2, 3, Median: 2

D. Mode: 1, 3, Median: 2.5

12- Which expression equivalent to \(x × 92\)?

A. \((x×90)+2\)

B. \(x×9×2\)

C. \((x×90)+(x×2)\)

D. \((x×90)+2\)

13- The ratio of pens to pencils in a box is 3 to 5. If there are 96 pens and pencils in the box altogether, how many more pens should be put in the box to make the ratio of pens to pencils 1: 1?

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A. 22

B. 23

C. 24

D. 25

14- If point A placed at \(-\frac{24}{3}\) on a number line, which of the following points has a distance equal to 5 from point A?

A. \(-13\)

B. \(-3\)

C. \(-2\)

D. A and B

15- Which of the following shows the numbers in increasing order?

A. \(\frac{3}{13}, \frac{4}{11}, \frac{5}{14}, \frac{2}{5}\)

B. \(\frac{3}{13}, \frac{5}{14}, \frac{4}{11}, \frac{2}{5}\)

C. \(\frac{3}{13}, \frac{5}{14}, \frac{2}{5}, \frac{4}{11}\)

D. \(\frac{5}{14}, \frac{3}{13}, \frac{2}{5}, \frac{4}{11}\)

16- If \(x=- 4\), which of the following equations is true?

A. \(x(3x-1)=50\)

B. \(5(11-x^2 )=-25\)

C. \(3(-2x+5)=49\)

D. \(x(-5x-19)=-3\)

17- What is the missing prime factor of number 450?
\(450=2^1×3^2×…\) _________

18- What is the perimeter of the following shape? (it’s a right triangle)

A. 14 cm

B. 18 cm

C. 24 cm

D. 32 cm

19- 65 is what percent of 50?

A. \(50 \%\)

B. \(77 \%\)

C. \(130 \%\)

D. \(140 \%\)

20- Which of the following expressions has a value of \(-23\)?

A. \(-10+(-8)+ \frac{5}{2}×(-2)\)

B. \(5×3+(-2)×18\)

C. \(-10+6×8÷(-4)\)

D. \((-3) × (-7) + 2\)

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Answers:

2- C
\(2205÷315=\frac{2205}{315}=\frac{441}{63}=\frac{147}{21}= 7\)

3- B
\(112=22+x \)
Subtract 22 from both sides of the equation. Then:
\(x=112-22=90\)

4- D
Distance that car B travels \(=1.2 ×\) distance that car A travels
=\(1.2×221.5=265.8 \) km

5- D
The perimeter of the trapezoid is 38.
Therefore, the missing side (height) is \(= 38 – 8 – 10 – 11 = 9\)
Area of the trapezoid: \(A = \frac{1}{2} h (b_1 + b_2) = \frac{1}{2}1 (9) (8 + 10) = 81\)

6- D
A. \(3^1+12=3+12=15\)
B. \(3^3-3^2=27-9=18\)
C. \(3^4-60=81-60=21\)
D. \(3^5-218=243-218=25\)

7- B
Alfred has \(x\) apple which is 15 apples more than a number of apples Alvin owns. Therefore:
\(x-15=40→x=40+15=55\)
Alfred has 55 apples.
Let \(y\) be the number of apples that Baron has. Then: \(y=\frac{1}{5}×55=11\)

8- A
Complementary angles add up to 180 degrees.
\( β+150^\circ=180^\circ→β=180^\circ-150^\circ=30^\circ\)
The sum of all angles in a triangle is 180 degrees. Then:
\(α+β+50^\circ=180^\circ→α+30^\circ+50^\circ=180^\circ\)
\(→α+80^\circ=180^\circ→α=180^\circ-80^\circ=100^\circ\)

9- C
Let \(x\) be the original price.
If the price of a laptop is decreased by \(15\%\) to $425, then:
\(85 \% \space of \space x=425⇒ 0.85x=425 ⇒ x=425÷0.85=500\)

10- C
Let \(x\) and \(y\) be two sides of the shape. Then:
\(x+1=1+1+1→x=2\)
\(y+6+2=5+4→y+8=9→y=1\)
Then, the perimeter is:
\(1+5+1+4+1+2+1+6+2+1=24\)

11- B
First, put the numbers in order from least to greatest: \(1, 1, 1, 2, 3, 3, 3, 4, 5, 6, 6\)
The Mode of the set of numbers is: 1 and 3 (the most frequent numbers)
The median is: 3 (the number in the middle)

12- C
\(x×92=x×(90+2)=(x×90)+(x×2)\)

13- C
The ratio of pens to pencils is \(3: 5\). Therefore there are 3 pens out of all 8 pens and pencils. To find the answer, first dived 96 by 8 then multiply the result by 3.
\(96÷8=12→12×3=36\)
There are 36 pens and 60 pencils \((96-36)\). Therefore, 24 more pens should be put in the box to make the ratio \(1: 1\)

14- D
If the value of point A is greater than the value of point B, then the distance of two points on the number line is: value of A- value of B
A. \(-\frac{24}{3}-(-13)=-8+13=5=5\)
B. \(-3-(-\frac{24}{3})=-3+8=5=5\)
C. \(-2-(-\frac{24}{3})=-2+8=6≠5\)

15- B
\(\frac{3}{13}≅0.23, \frac{5}{14}≅0.357, \frac{4}{11}≅0.36, \frac{2}{5}=0.4\)

16- B
Plugin the value of \(x\) in the equations. \(x = -4\), then:
A.\(x(3x-1)=50→-4(3(-4)-1)=-4(-12-1)=-4(-13)=52≠50\)
B. \(5(11-x^2 )=-25→5(11-(-4)^2 )= 5(11-16)=5(-5)=-25\)
C. \(3(-2x+5)=49→3(-2(-4)+5)=3(8+5)=39≠49\)
D. \(x(-5x-19)=-3→-4(-5(-4)-19=-4(20-19)=-4≠-3\)

17- 5
Let \(x\) be the missing prime factor of 450.
\(450= 2 × 3 × 3 × x ⇒ x =\frac{450}{18} ⇒ x = 25=5×5\)

18- C
Use the Pythagorean theorem to find the hypotenuse of the triangle.
\(a^2+b^2=c^2→6^2+8^2=c^2→36+64=c^2→100=c^2→c=10\)
The perimeter of the triangle is: \(6+8+10=24\)

19- C
Use the percent formula:
\(Part = \frac{percent}{100} × whole\)
\(65= \frac{percent}{100} × 50⇒ 65 = \frac{percent ×50}{100}⇒ 65=\frac{percent ×5}{10}\)
multiply both sides by 10.
\(650 =percent ×5, \space divide \space both \space sides \space by \space 5.\)
130 = percent
The answer is \(130\%\)

20- A
Let’s check the options provided.
A. \(-10+(-8)+ (\frac{5}{2})×(-2)=-10+(-8)+(-5)=-10-13=-23\)
B. \(5×3+(-2)×18=15+(-38)=-21\)
C. \(-10+6×8÷(-4)=-10+48÷(-4)=-10-12=-22\)
D. \((-3)× (-7)+ 2=21+2=23\)

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