ASVAB Arithmetic Reasoning Practice Test 921567 Results

Your Results Global Average
Questions 5 5
Correct 0 3.46
Score 0% 69%

Review

1

What is \( \frac{1}{9} \) x \( \frac{2}{8} \)?

72% Answer Correctly
\(\frac{1}{4}\)
\(\frac{1}{36}\)
\(\frac{4}{25}\)
\(\frac{1}{7}\)

Solution

To multiply fractions, multiply the numerators together and then multiply the denominators together:

\( \frac{1}{9} \) x \( \frac{2}{8} \) = \( \frac{1 x 2}{9 x 8} \) = \( \frac{2}{72} \) = \(\frac{1}{36}\)


2

Find the average of the following numbers: 14, 6, 12, 8.

74% Answer Correctly
6
10
15
11

Solution

To find the average of these 4 numbers add them together then divide by 4:

\( \frac{14 + 6 + 12 + 8}{4} \) = \( \frac{40}{4} \) = 10


3

What is \( \sqrt{\frac{64}{16}} \)?

70% Answer Correctly
\(\frac{2}{5}\)
2
\(\frac{5}{8}\)
\(\frac{2}{3}\)

Solution

To take the square root of a fraction, break the fraction into two separate roots then calculate the square root of the numerator and denominator separately:

\( \sqrt{\frac{64}{16}} \)
\( \frac{\sqrt{64}}{\sqrt{16}} \)
\( \frac{\sqrt{8^2}}{\sqrt{4^2}} \)
\( \frac{8}{4} \)
2


4

What is \( \frac{-8y^8}{1y^3} \)?

60% Answer Correctly
-8y-5
-8y11
-\(\frac{1}{8}\)y11
-8y5

Solution

To divide terms with exponents, the base of both exponents must be the same. In this case they are so divide the coefficients and subtract the exponents:

\( \frac{-8y^8}{y^3} \)
\( \frac{-8}{1} \) y(8 - 3)
-8y5


5

Which of the following is an improper fraction?

70% Answer Correctly

\({a \over 5} \)

\(1 {2 \over 5} \)

\({2 \over 5} \)

\({7 \over 5} \)


Solution

A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form \({a \over b}\) where a is the numerator and b is the denominator. An improper fraction (\({5 \over 3} \)) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number (\(1 {2 \over 3} \)) which has a whole number part and a fractional part.