ASVAB Arithmetic Reasoning Practice Test 748944 Results

Your Results Global Average
Questions 5 5
Correct 0 3.18
Score 0% 64%

Review

1

Which of the following is not an integer?

77% Answer Correctly

\({1 \over 2}\)

-1

1

0


Solution

An integer is any whole number, including zero. An integer can be either positive or negative. Examples include -77, -1, 0, 55, 119.


2

If \( \left|b + 8\right| \) + 9 = -5, which of these is a possible value for b?

62% Answer Correctly
-15
-17
-22
-1

Solution

First, solve for \( \left|b + 8\right| \):

\( \left|b + 8\right| \) + 9 = -5
\( \left|b + 8\right| \) = -5 - 9
\( \left|b + 8\right| \) = -14

The value inside the absolute value brackets can be either positive or negative so (b + 8) must equal - 14 or --14 for \( \left|b + 8\right| \) to equal -14:

b + 8 = -14
b = -14 - 8
b = -22
b + 8 = 14
b = 14 - 8
b = 6

So, b = 6 or b = -22.


3

Find the average of the following numbers: 10, 6, 9, 7.

75% Answer Correctly
10
8
7
11

Solution

To find the average of these 4 numbers add them together then divide by 4:

\( \frac{10 + 6 + 9 + 7}{4} \) = \( \frac{32}{4} \) = 8


4

A circular logo is enlarged to fit the lid of a jar. The new diameter is 55% larger than the original. By what percentage has the area of the logo increased?

51% Answer Correctly
27\(\frac{1}{2}\)%
17\(\frac{1}{2}\)%
15%
32\(\frac{1}{2}\)%

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 55% the radius (and, consequently, the total area) increases by \( \frac{55\text{%}}{2} \) = 27\(\frac{1}{2}\)%


5

A machine in a factory has an error rate of 9 parts per 100. The machine normally runs 24 hours a day and produces 10 parts per hour. Yesterday the machine was shut down for 4 hours for maintenance.

How many error-free parts did the machine produce yesterday?

49% Answer Correctly
192.1
182
193.2
123.7

Solution

The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:

\( \frac{9}{100} \) x 10 = \( \frac{9 \times 10}{100} \) = \( \frac{90}{100} \) = 0.9 errors per hour

So, in an average hour, the machine will produce 10 - 0.9 = 9.1 error free parts.

The machine ran for 24 - 4 = 20 hours yesterday so you would expect that 20 x 9.1 = 182 error free parts were produced yesterday.