ASVAB Arithmetic Reasoning Practice Test 650418 Results

Your Results Global Average
Questions 5 5
Correct 0 2.94
Score 0% 59%

Review

1

Solve 2 + (5 + 3) ÷ 4 x 2 - 42

52% Answer Correctly
2
2\(\frac{1}{4}\)
-10
1

Solution

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

2 + (5 + 3) ÷ 4 x 2 - 42
P: 2 + (8) ÷ 4 x 2 - 42
E: 2 + 8 ÷ 4 x 2 - 16
MD: 2 + \( \frac{8}{4} \) x 2 - 16
MD: 2 + \( \frac{16}{4} \) - 16
AS: \( \frac{8}{4} \) + \( \frac{16}{4} \) - 16
AS: \( \frac{24}{4} \) - 16
AS: \( \frac{24 - 64}{4} \)
\( \frac{-40}{4} \)
-10


2

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

66% Answer Correctly
4
\( \frac{1}{120} \)
\( \frac{1}{12} \)
120

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:

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


3

If \( \left|a - 1\right| \) - 6 = -1, which of these is a possible value for a?

62% Answer Correctly
7
-11
-4
-5

Solution

First, solve for \( \left|a - 1\right| \):

\( \left|a - 1\right| \) - 6 = -1
\( \left|a - 1\right| \) = -1 + 6
\( \left|a - 1\right| \) = 5

The value inside the absolute value brackets can be either positive or negative so (a - 1) must equal + 5 or -5 for \( \left|a - 1\right| \) to equal 5:

a - 1 = 5
a = 5 + 1
a = 6
a - 1 = -5
a = -5 + 1
a = -4

So, a = -4 or a = 6.


4

Simplify \( \sqrt{8} \)

62% Answer Correctly
8\( \sqrt{4} \)
8\( \sqrt{2} \)
2\( \sqrt{2} \)
5\( \sqrt{4} \)

Solution

To simplify a radical, factor out the perfect squares:

\( \sqrt{8} \)
\( \sqrt{4 \times 2} \)
\( \sqrt{2^2 \times 2} \)
2\( \sqrt{2} \)


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
30%
15%
25%
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%