ASVAB Arithmetic Reasoning Practice Test 944800 Results

Your Results Global Average
Questions 5 5
Correct 0 3.21
Score 0% 64%

Review

1

What is \( \frac{1}{5} \) x \( \frac{2}{5} \)?

72% Answer Correctly
\(\frac{8}{35}\)
\(\frac{2}{27}\)
\(\frac{8}{21}\)
\(\frac{2}{25}\)

Solution

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

\( \frac{1}{5} \) x \( \frac{2}{5} \) = \( \frac{1 x 2}{5 x 5} \) = \( \frac{2}{25} \) = \(\frac{2}{25}\)


2

Alex loaned Monica $1,300 at an annual interest rate of 8%. If no payments are made, what is the total amount owed at the end of the first year?

71% Answer Correctly
$1,339
$1,365
$1,391
$1,404

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 $1,300
i = 0.08 x $1,300

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

total = $1,300 + $104
total = $1,404


3

What is \( \sqrt{\frac{81}{4}} \)?

70% Answer Correctly
\(\frac{5}{8}\)
1\(\frac{2}{3}\)
1
4\(\frac{1}{2}\)

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:

\( \sqrt{\frac{81}{4}} \)
\( \frac{\sqrt{81}}{\sqrt{4}} \)
\( \frac{\sqrt{9^2}}{\sqrt{2^2}} \)
\( \frac{9}{2} \)
4\(\frac{1}{2}\)


4

How many 1\(\frac{1}{2}\) gallon cans worth of fuel would you need to pour into an empty 9 gallon tank to fill it exactly halfway?

52% Answer Correctly
2
6
3
5

Solution

To fill a 9 gallon tank exactly halfway you'll need 4\(\frac{1}{2}\) gallons of fuel. Each fuel can holds 1\(\frac{1}{2}\) gallons so:

cans = \( \frac{4\frac{1}{2} \text{ gallons}}{1\frac{1}{2} \text{ gallons}} \) = 3


5

\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?

56% Answer Correctly

commutative property for multiplication

commutative property for division

distributive property for multiplication

distributive property for division


Solution

The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).