ASVAB Arithmetic Reasoning Practice Test 193109 Results

Your Results Global Average
Questions 5 5
Correct 0 3.13
Score 0% 63%

Review

1

Simplify \( \sqrt{12} \)

62% Answer Correctly
2\( \sqrt{3} \)
3\( \sqrt{6} \)
8\( \sqrt{3} \)
6\( \sqrt{6} \)

Solution

To simplify a radical, factor out the perfect squares:

\( \sqrt{12} \)
\( \sqrt{4 \times 3} \)
\( \sqrt{2^2 \times 3} \)
2\( \sqrt{3} \)


2

Simplify \( \frac{32}{56} \).

77% Answer Correctly
\( \frac{3}{8} \)
\( \frac{5}{14} \)
\( \frac{4}{7} \)
\( \frac{1}{3} \)

Solution

To simplify this fraction, first find the greatest common factor between them. The factors of 32 are [1, 2, 4, 8, 16, 32] and the factors of 56 are [1, 2, 4, 7, 8, 14, 28, 56]. They share 4 factors [1, 2, 4, 8] making 8 their greatest common factor (GCF).

Next, divide both numerator and denominator by the GCF:

\( \frac{32}{56} \) = \( \frac{\frac{32}{8}}{\frac{56}{8}} \) = \( \frac{4}{7} \)


3

Alex loaned Monica $900 at an annual interest rate of 7%. If no payments are made, what is the total amount owed at the end of the first year?

71% Answer Correctly
$963
$927
$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 = 0.07 x $900

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

total = $900 + $63
total = $963


4

Solve for \( \frac{2!}{4!} \)

66% Answer Correctly
\( \frac{1}{42} \)
\( \frac{1}{12} \)
120
336

Solution

A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:

\( \frac{2!}{4!} \)
\( \frac{2 \times 1}{4 \times 3 \times 2 \times 1} \)
\( \frac{1}{4 \times 3} \)
\( \frac{1}{12} \)


5

What is \( 6 \)\( \sqrt{45} \) + \( 6 \)\( \sqrt{5} \)

35% Answer Correctly
12\( \sqrt{45} \)
36\( \sqrt{45} \)
12\( \sqrt{225} \)
24\( \sqrt{5} \)

Solution

To add these radicals together their radicands must be the same:

6\( \sqrt{45} \) + 6\( \sqrt{5} \)
6\( \sqrt{9 \times 5} \) + 6\( \sqrt{5} \)
6\( \sqrt{3^2 \times 5} \) + 6\( \sqrt{5} \)
(6)(3)\( \sqrt{5} \) + 6\( \sqrt{5} \)
18\( \sqrt{5} \) + 6\( \sqrt{5} \)

Now that the radicands are identical, you can add them together:

18\( \sqrt{5} \) + 6\( \sqrt{5} \)
(18 + 6)\( \sqrt{5} \)
24\( \sqrt{5} \)