ASVAB Arithmetic Reasoning Practice Test 594458 Results

Your Results Global Average
Questions 5 5
Correct 0 2.97
Score 0% 59%

Review

1

This property states taht the order of addition or multiplication does not mater. For example, 2 + 5 and 5 + 2 are equivalent.

59% Answer Correctly

PEDMAS

commutative

associative

distributive


Solution

The commutative property states that, when adding or multiplying numbers, the order in which they're added or multiplied does not matter. For example, 3 + 4 and 4 + 3 give the same result, as do 3 x 4 and 4 x 3.


2

What is 9y6 + 7y6?

66% Answer Correctly
16y-12
16y36
2y6
16y6

Solution

To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so add the coefficients and retain the base and exponent:

9y6 + 7y6
(9 + 7)y6
16y6


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
15%
35%
17\(\frac{1}{2}\)%
25%

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

Which of the following is a mixed number?

82% Answer Correctly

\({5 \over 7} \)

\({7 \over 5} \)

\({a \over 5} \)

\(1 {2 \over 5} \)


Solution

A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form \({a \over b}\) where a is the numerator and b is the denominator. An improper fraction (\({5 \over 3} \)) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number (\(1 {2 \over 3} \)) which has a whole number part and a fractional part.


5

What is \( 6 \)\( \sqrt{18} \) + \( 7 \)\( \sqrt{2} \)

35% Answer Correctly
13\( \sqrt{9} \)
25\( \sqrt{2} \)
13\( \sqrt{2} \)
42\( \sqrt{9} \)

Solution

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

6\( \sqrt{18} \) + 7\( \sqrt{2} \)
6\( \sqrt{9 \times 2} \) + 7\( \sqrt{2} \)
6\( \sqrt{3^2 \times 2} \) + 7\( \sqrt{2} \)
(6)(3)\( \sqrt{2} \) + 7\( \sqrt{2} \)
18\( \sqrt{2} \) + 7\( \sqrt{2} \)

Now that the radicands are identical, you can add them together:

18\( \sqrt{2} \) + 7\( \sqrt{2} \)
(18 + 7)\( \sqrt{2} \)
25\( \sqrt{2} \)