ASVAB Arithmetic Reasoning Practice Test 652227 Results

Your Results Global Average
Questions 5 5
Correct 0 3.39
Score 0% 68%

Review

1

What is (z4)4?

79% Answer Correctly
4z4
z16
z0
z8

Solution

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

(z4)4
z(4 * 4)
z16


2

Simplify \( \frac{36}{72} \).

77% Answer Correctly
\( \frac{9}{19} \)
\( \frac{1}{2} \)
\( \frac{6}{17} \)
\( \frac{5}{16} \)

Solution

To simplify this fraction, first find the greatest common factor between them. The factors of 36 are [1, 2, 3, 4, 6, 9, 12, 18, 36] and the factors of 72 are [1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72]. They share 9 factors [1, 2, 3, 4, 6, 9, 12, 18, 36] making 36 their greatest common factor (GCF).

Next, divide both numerator and denominator by the GCF:

\( \frac{36}{72} \) = \( \frac{\frac{36}{36}}{\frac{72}{36}} \) = \( \frac{1}{2} \)


3

If all of a roofing company's 12 workers are required to staff 4 roofing crews, how many workers need to be added during the busy season in order to send 8 complete crews out on jobs?

55% Answer Correctly
17
10
12
11

Solution

In order to find how many additional workers are needed to staff the extra crews you first need to calculate how many workers are on a crew. There are 12 workers at the company now and that's enough to staff 4 crews so there are \( \frac{12}{4} \) = 3 workers on a crew. 8 crews are needed for the busy season which, at 3 workers per crew, means that the roofing company will need 8 x 3 = 24 total workers to staff the crews during the busy season. The company already employs 12 workers so they need to add 24 - 12 = 12 new staff for the busy season.


4

The __________ is the greatest factor that divides two integers.

67% Answer Correctly

greatest common factor

absolute value

greatest common multiple

least common multiple


Solution

The greatest common factor (GCF) is the greatest factor that divides two integers.


5

What is \( \frac{4}{5} \) - \( \frac{2}{7} \)?

61% Answer Correctly
\(\frac{18}{35}\)
1 \( \frac{2}{9} \)
\( \frac{5}{35} \)
1 \( \frac{5}{10} \)

Solution

To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 5 are [5, 10, 15, 20, 25, 30, 35, 40, 45, 50] and the first few multiples of 7 are [7, 14, 21, 28, 35, 42, 49, 56, 63, 70]. The first few multiples they share are [35, 70] making 35 the smallest multiple 5 and 7 share.

Next, convert the fractions so each denominator equals the lowest common multiple:

\( \frac{4 x 7}{5 x 7} \) - \( \frac{2 x 5}{7 x 5} \)

\( \frac{28}{35} \) - \( \frac{10}{35} \)

Now, because the fractions share a common denominator, you can subtract them:

\( \frac{28 - 10}{35} \) = \( \frac{18}{35} \) = \(\frac{18}{35}\)