ASVAB Arithmetic Reasoning Practice Test 20733 Results

Your Results Global Average
Questions 5 5
Correct 0 3.50
Score 0% 70%

Review

1

If \( \left|y - 9\right| \) + 3 = -5, which of these is a possible value for y?

62% Answer Correctly
-6
5
-3
1

Solution

First, solve for \( \left|y - 9\right| \):

\( \left|y - 9\right| \) + 3 = -5
\( \left|y - 9\right| \) = -5 - 3
\( \left|y - 9\right| \) = -8

The value inside the absolute value brackets can be either positive or negative so (y - 9) must equal - 8 or --8 for \( \left|y - 9\right| \) to equal -8:

y - 9 = -8
y = -8 + 9
y = 1
y - 9 = 8
y = 8 + 9
y = 17

So, y = 17 or y = 1.


2

If \(\left|a\right| = 7\), which of the following best describes a?

67% Answer Correctly

a = 7

a = 7 or a = -7

none of these is correct

a = -7


Solution

The absolute value is the positive magnitude of a particular number or variable and is indicated by two vertical lines: \(\left|-5\right| = 5\). In the case of a variable absolute value (\(\left|a\right| = 5\)) the value of a can be either positive or negative (a = -5 or a = 5).


3

Roger loaned Alex $1,200 at an annual interest rate of 2%. If no payments are made, what is the interest owed on this loan at the end of the first year?

74% Answer Correctly
$84
$20
$99
$24

Solution

The yearly interest charged on this loan is the annual interest rate multiplied by the amount borrowed:

interest = annual interest rate x loan amount

i = (\( \frac{6}{100} \)) x $1,200
i = 0.02 x $1,200
i = $24


4

What is \( \frac{1}{6} \) ÷ \( \frac{4}{5} \)?

68% Answer Correctly
\(\frac{5}{6}\)
\(\frac{1}{8}\)
\(\frac{5}{24}\)
\(\frac{1}{5}\)

Solution

To divide fractions, invert the second fraction and then multiply:

\( \frac{1}{6} \) ÷ \( \frac{4}{5} \) = \( \frac{1}{6} \) x \( \frac{5}{4} \)

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

\( \frac{1}{6} \) x \( \frac{5}{4} \) = \( \frac{1 x 5}{6 x 4} \) = \( \frac{5}{24} \) = \(\frac{5}{24}\)


5

What is (a2)3?

80% Answer Correctly
3a2
2a3
a6
a-1

Solution

To raise a term with an exponent to another exponent, retain the base and multiply the exponents:

(a2)3
a(2 * 3)
a6