ASVAB Arithmetic Reasoning Practice Test 691705 Results

Your Results Global Average
Questions 5 5
Correct 0 3.16
Score 0% 63%

Review

1

Simplify \( \frac{20}{68} \).

77% Answer Correctly
\( \frac{6}{17} \)
\( \frac{4}{7} \)
\( \frac{5}{17} \)
\( \frac{8}{19} \)

Solution

To simplify this fraction, first find the greatest common factor between them. The factors of 20 are [1, 2, 4, 5, 10, 20] and the factors of 68 are [1, 2, 4, 17, 34, 68]. They share 3 factors [1, 2, 4] making 4 their greatest common factor (GCF).

Next, divide both numerator and denominator by the GCF:

\( \frac{20}{68} \) = \( \frac{\frac{20}{4}}{\frac{68}{4}} \) = \( \frac{5}{17} \)


2

The __________ is the smallest positive integer that is a multiple of two or more integers.

56% Answer Correctly

least common factor

least common multiple

greatest common factor

absolute value


Solution

The least common multiple (LCM) is the smallest positive integer that is a multiple of two or more integers.


3

Christine scored 75% on her final exam. If each question was worth 4 points and there were 240 possible points on the exam, how many questions did Christine answer correctly?

57% Answer Correctly
45
35
47
44

Solution

Christine scored 75% on the test meaning she earned 75% of the possible points on the test. There were 240 possible points on the test so she earned 240 x 0.75 = 180 points. Each question is worth 4 points so she got \( \frac{180}{4} \) = 45 questions right.


4

What is \( \sqrt{\frac{9}{4}} \)?

70% Answer Correctly
1\(\frac{1}{2}\)
1\(\frac{1}{8}\)
1
2

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:

\( \sqrt{\frac{9}{4}} \)
\( \frac{\sqrt{9}}{\sqrt{4}} \)
\( \frac{\sqrt{3^2}}{\sqrt{2^2}} \)
\( \frac{3}{2} \)
1\(\frac{1}{2}\)


5

\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?

55% Answer Correctly

commutative property for multiplication

distributive property for division

distributive property for multiplication

commutative property for division


Solution

The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).