ASVAB Arithmetic Reasoning Practice Test 893259 Results

Your Results Global Average
Questions 5 5
Correct 0 3.08
Score 0% 62%

Review

1

A circular logo is enlarged to fit the lid of a jar. The new diameter is 75% larger than the original. By what percentage has the area of the logo increased?

51% Answer Correctly
37\(\frac{1}{2}\)%
17\(\frac{1}{2}\)%
15%
22\(\frac{1}{2}\)%

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 75% the radius (and, consequently, the total area) increases by \( \frac{75\text{%}}{2} \) = 37\(\frac{1}{2}\)%


2

Find the average of the following numbers: 14, 8, 13, 9.

75% Answer Correctly
9
11
16
6

Solution

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

\( \frac{14 + 8 + 13 + 9}{4} \) = \( \frac{44}{4} \) = 11


3

A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 15% off." If Charlie buys two shirts, each with a regular price of $40, how much money will he save?

70% Answer Correctly
$12.00
$8.00
$6.00
$14.00

Solution

By buying two shirts, Charlie will save $40 x \( \frac{15}{100} \) = \( \frac{$40 x 15}{100} \) = \( \frac{$600}{100} \) = $6.00 on the second shirt.


4

What is the least common multiple of 8 and 16?

73% Answer Correctly
16
31
83
29

Solution

The first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80] and the first few multiples of 16 are [16, 32, 48, 64, 80, 96]. The first few multiples they share are [16, 32, 48, 64, 80] making 16 the smallest multiple 8 and 16 have in common.


5

What is \( 6 \)\( \sqrt{28} \) - \( 2 \)\( \sqrt{7} \)

39% Answer Correctly
12\( \sqrt{4} \)
4\( \sqrt{196} \)
10\( \sqrt{7} \)
4\( \sqrt{45} \)

Solution

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

6\( \sqrt{28} \) - 2\( \sqrt{7} \)
6\( \sqrt{4 \times 7} \) - 2\( \sqrt{7} \)
6\( \sqrt{2^2 \times 7} \) - 2\( \sqrt{7} \)
(6)(2)\( \sqrt{7} \) - 2\( \sqrt{7} \)
12\( \sqrt{7} \) - 2\( \sqrt{7} \)

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

12\( \sqrt{7} \) - 2\( \sqrt{7} \)
(12 - 2)\( \sqrt{7} \)
10\( \sqrt{7} \)