ASVAB Arithmetic Reasoning Practice Test 920915 Results

Your Results Global Average
Questions 5 5
Correct 0 2.89
Score 0% 58%

Review

1

If all of a roofing company's 12 workers are required to staff 3 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
16
13
20
4

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 3 crews so there are \( \frac{12}{3} \) = 4 workers on a crew. 8 crews are needed for the busy season which, at 4 workers per crew, means that the roofing company will need 8 x 4 = 32 total workers to staff the crews during the busy season. The company already employs 12 workers so they need to add 32 - 12 = 20 new staff for the busy season.


2

What is 4b7 - 2b7?

71% Answer Correctly
2b7
6b49
-2b7
2b-7

Solution

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

4b7 - 2b7
(4 - 2)b7
2b7


3

What is \( 5 \)\( \sqrt{18} \) - \( 2 \)\( \sqrt{2} \)

38% Answer Correctly
3\( \sqrt{9} \)
10\( \sqrt{9} \)
13\( \sqrt{2} \)
3\( \sqrt{-5} \)

Solution

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

5\( \sqrt{18} \) - 2\( \sqrt{2} \)
5\( \sqrt{9 \times 2} \) - 2\( \sqrt{2} \)
5\( \sqrt{3^2 \times 2} \) - 2\( \sqrt{2} \)
(5)(3)\( \sqrt{2} \) - 2\( \sqrt{2} \)
15\( \sqrt{2} \) - 2\( \sqrt{2} \)

Now that the radicands are identical, you can subtract them:

15\( \sqrt{2} \) - 2\( \sqrt{2} \)
(15 - 2)\( \sqrt{2} \)
13\( \sqrt{2} \)


4

What is \( \frac{4}{8} \) x \( \frac{4}{6} \)?

72% Answer Correctly
\(\frac{4}{63}\)
\(\frac{1}{3}\)
\(\frac{3}{32}\)
\(\frac{3}{20}\)

Solution

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

\( \frac{4}{8} \) x \( \frac{4}{6} \) = \( \frac{4 x 4}{8 x 6} \) = \( \frac{16}{48} \) = \(\frac{1}{3}\)


5

A circular logo is enlarged to fit the lid of a jar. The new diameter is 30% larger than the original. By what percentage has the area of the logo increased?

50% Answer Correctly
15%
32\(\frac{1}{2}\)%
17\(\frac{1}{2}\)%
37\(\frac{1}{2}\)%

Solution

The area of a circle is given by the formula A = πr2 where r is the radius of the circle. The radius of a circle is its diameter divided by two so A = π(\( \frac{d}{2} \))2. If the diameter of the logo increases by 30% the radius (and, consequently, the total area) increases by \( \frac{30\text{%}}{2} \) = 15%