ASVAB Arithmetic Reasoning Practice Test 846411 Results

Your Results Global Average
Questions 5 5
Correct 0 3.02
Score 0% 60%

Review

1

A bread recipe calls for 2\(\frac{1}{2}\) cups of flour. If you only have \(\frac{1}{8}\) cup, how much more flour is needed?

62% Answer Correctly
1\(\frac{1}{2}\) cups
2\(\frac{7}{8}\) cups
2\(\frac{3}{8}\) cups
2\(\frac{3}{4}\) cups

Solution

The amount of flour you need is (2\(\frac{1}{2}\) - \(\frac{1}{8}\)) cups. Rewrite the quantities so they share a common denominator and subtract:

(\( \frac{20}{8} \) - \( \frac{1}{8} \)) cups
\( \frac{19}{8} \) cups
2\(\frac{3}{8}\) cups


2

What is \( \sqrt{\frac{25}{4}} \)?

70% Answer Correctly
2\(\frac{1}{2}\)
\(\frac{6}{7}\)
4
2\(\frac{2}{3}\)

Solution

To take the square root of a fraction, break the fraction into two separate roots then calculate the square root of the numerator and denominator separately:

\( \sqrt{\frac{25}{4}} \)
\( \frac{\sqrt{25}}{\sqrt{4}} \)
\( \frac{\sqrt{5^2}}{\sqrt{2^2}} \)
\( \frac{5}{2} \)
2\(\frac{1}{2}\)


3

A circular logo is enlarged to fit the lid of a jar. The new diameter is 70% larger than the original. By what percentage has the area of the logo increased?

50% Answer Correctly
35%
22\(\frac{1}{2}\)%
30%
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 70% the radius (and, consequently, the total area) increases by \( \frac{70\text{%}}{2} \) = 35%


4

A factor is a positive __________ that divides evenly into a given number.

78% Answer Correctly

mixed number

improper fraction

integer

fraction


Solution

A factor is a positive integer that divides evenly into a given number. For example, the factors of 8 are 1, 2, 4, and 8.


5

What is \( 4 \)\( \sqrt{175} \) - \( 5 \)\( \sqrt{7} \)

38% Answer Correctly
20\( \sqrt{175} \)
-1\( \sqrt{1225} \)
15\( \sqrt{7} \)
20\( \sqrt{7} \)

Solution

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

4\( \sqrt{175} \) - 5\( \sqrt{7} \)
4\( \sqrt{25 \times 7} \) - 5\( \sqrt{7} \)
4\( \sqrt{5^2 \times 7} \) - 5\( \sqrt{7} \)
(4)(5)\( \sqrt{7} \) - 5\( \sqrt{7} \)
20\( \sqrt{7} \) - 5\( \sqrt{7} \)

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

20\( \sqrt{7} \) - 5\( \sqrt{7} \)
(20 - 5)\( \sqrt{7} \)
15\( \sqrt{7} \)