ASVAB Arithmetic Reasoning Practice Test 600130 Results

Your Results Global Average
Questions 5 5
Correct 0 3.04
Score 0% 61%

Review

1

If \(\left|a\right| = 7\), which of the following best describes a?

67% Answer Correctly

a = -7

none of these is correct

a = 7 or a = -7

a = 7


Solution

The absolute value is the positive magnitude of a particular number or variable and is indicated by two vertical lines: \(\left|-5\right| = 5\). In the case of a variable absolute value (\(\left|a\right| = 5\)) the value of a can be either positive or negative (a = -5 or a = 5).


2

A circular logo is enlarged to fit the lid of a jar. The new diameter is 35% larger than the original. By what percentage has the area of the logo increased?

50% Answer Correctly
22\(\frac{1}{2}\)%
20%
17\(\frac{1}{2}\)%
30%

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 35% the radius (and, consequently, the total area) increases by \( \frac{35\text{%}}{2} \) = 17\(\frac{1}{2}\)%


3

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

66% Answer Correctly
\( \frac{1}{12} \)
8
\( \frac{1}{210} \)
\( \frac{1}{15120} \)

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:

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


4

Convert x-5 to remove the negative exponent.

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

Solution

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


5

Solve 2 + (2 + 5) ÷ 5 x 3 - 22

52% Answer Correctly
1\(\frac{1}{3}\)
1
2\(\frac{1}{5}\)
1\(\frac{1}{6}\)

Solution

Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):

2 + (2 + 5) ÷ 5 x 3 - 22
P: 2 + (7) ÷ 5 x 3 - 22
E: 2 + 7 ÷ 5 x 3 - 4
MD: 2 + \( \frac{7}{5} \) x 3 - 4
MD: 2 + \( \frac{21}{5} \) - 4
AS: \( \frac{10}{5} \) + \( \frac{21}{5} \) - 4
AS: \( \frac{31}{5} \) - 4
AS: \( \frac{31 - 20}{5} \)
\( \frac{11}{5} \)
2\(\frac{1}{5}\)