ASVAB Arithmetic Reasoning Practice Test 877135 Results

Your Results Global Average
Questions 5 5
Correct 0 3.74
Score 0% 75%

Review

1

What is the distance in miles of a trip that takes 2 hours at an average speed of 75 miles per hour?

87% Answer Correctly
225 miles
150 miles
260 miles
525 miles

Solution

Average speed in miles per hour is the number of miles traveled divided by the number of hours:

speed = \( \frac{\text{distance}}{\text{time}} \)

Solving for distance:

distance = \( \text{speed} \times \text{time} \)
distance = \( 75mph \times 2h \)
150 miles


2

Monty loaned Diane $900 at an annual interest rate of 2%. If no payments are made, what is the total amount owed at the end of the first year?

71% Answer Correctly
$981
$936
$918
$945

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 $900
i = 0.02 x $900

No payments were made so the total amount due is the original amount + the accumulated interest:

total = $900 + $18
total = $918


3

What is \( \frac{2}{9} \) x \( \frac{4}{7} \)?

72% Answer Correctly
\(\frac{8}{63}\)
\(\frac{8}{9}\)
\(\frac{1}{8}\)
\(\frac{1}{24}\)

Solution

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

\( \frac{2}{9} \) x \( \frac{4}{7} \) = \( \frac{2 x 4}{9 x 7} \) = \( \frac{8}{63} \) = \(\frac{8}{63}\)


4

Simplify \( \frac{28}{56} \).

77% Answer Correctly
\( \frac{3}{8} \)
\( \frac{1}{2} \)
\( \frac{5}{6} \)
\( \frac{1}{3} \)

Solution

To simplify this fraction, first find the greatest common factor between them. The factors of 28 are [1, 2, 4, 7, 14, 28] and the factors of 56 are [1, 2, 4, 7, 8, 14, 28, 56]. They share 6 factors [1, 2, 4, 7, 14, 28] making 28 their greatest common factor (GCF).

Next, divide both numerator and denominator by the GCF:

\( \frac{28}{56} \) = \( \frac{\frac{28}{28}}{\frac{56}{28}} \) = \( \frac{1}{2} \)


5

What is 9b7 + 7b7?

66% Answer Correctly
16b7
16b14
2b-7
-2b7

Solution

To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so add the coefficients and retain the base and exponent:

9b7 + 7b7
(9 + 7)b7
16b7