ASVAB Arithmetic Reasoning Practice Test 651238 Results

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

Review

1

Convert y-5 to remove the negative exponent.

67% Answer Correctly
\( \frac{-5}{y} \)
\( \frac{1}{y^5} \)
\( \frac{5}{y} \)
\( \frac{-1}{-5y} \)

Solution

To convert a negative exponent to a positive exponent, calculate the positive exponent then take the reciprocal.


2

52% Answer Correctly
1
1.2
6.3
0.8

Solution


1


3

Simplify \( \frac{28}{48} \).

77% Answer Correctly
\( \frac{9}{20} \)
\( \frac{7}{15} \)
\( \frac{6}{11} \)
\( \frac{7}{12} \)

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 48 are [1, 2, 3, 4, 6, 8, 12, 16, 24, 48]. They share 3 factors [1, 2, 4] making 4 their greatest common factor (GCF).

Next, divide both numerator and denominator by the GCF:

\( \frac{28}{48} \) = \( \frac{\frac{28}{4}}{\frac{48}{4}} \) = \( \frac{7}{12} \)


4

If all of a roofing company's 6 workers are required to staff 2 roofing crews, how many workers need to be added during the busy season in order to send 4 complete crews out on jobs?

55% Answer Correctly
11
16
6
17

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 6 workers at the company now and that's enough to staff 2 crews so there are \( \frac{6}{2} \) = 3 workers on a crew. 4 crews are needed for the busy season which, at 3 workers per crew, means that the roofing company will need 4 x 3 = 12 total workers to staff the crews during the busy season. The company already employs 6 workers so they need to add 12 - 6 = 6 new staff for the busy season.


5

What is \( \frac{5}{2} \) + \( \frac{3}{6} \)?

59% Answer Correctly
1 \( \frac{4}{6} \)
1 \( \frac{3}{6} \)
2 \( \frac{1}{6} \)
3

Solution

To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 2 are [2, 4, 6, 8, 10, 12, 14, 16, 18, 20] and the first few multiples of 6 are [6, 12, 18, 24, 30, 36, 42, 48, 54, 60]. The first few multiples they share are [6, 12, 18, 24, 30] making 6 the smallest multiple 2 and 6 share.

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

\( \frac{5 x 3}{2 x 3} \) + \( \frac{3 x 1}{6 x 1} \)

\( \frac{15}{6} \) + \( \frac{3}{6} \)

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

\( \frac{15 + 3}{6} \) = \( \frac{18}{6} \) = 3