ASVAB Arithmetic Reasoning Practice Test 336864 Results

Your Results Global Average
Questions 5 5
Correct 0 3.55
Score 0% 71%

Review

1

Simplify \( \frac{40}{44} \).

77% Answer Correctly
\( \frac{5}{9} \)
\( \frac{10}{11} \)
\( \frac{5}{11} \)
\( \frac{9}{16} \)

Solution

To simplify this fraction, first find the greatest common factor between them. The factors of 40 are [1, 2, 4, 5, 8, 10, 20, 40] and the factors of 44 are [1, 2, 4, 11, 22, 44]. They share 3 factors [1, 2, 4] making 4 their greatest common factor (GCF).

Next, divide both numerator and denominator by the GCF:

\( \frac{40}{44} \) = \( \frac{\frac{40}{4}}{\frac{44}{4}} \) = \( \frac{10}{11} \)


2

4! = ?

84% Answer Correctly

4 x 3 x 2 x 1

4 x 3

5 x 4 x 3 x 2 x 1

3 x 2 x 1


Solution

A factorial has the form n! and is the product of the integer (n) and all the positive integers below it. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.


3

What is \( \frac{5}{3} \) + \( \frac{6}{5} \)?

60% Answer Correctly
2\(\frac{13}{15}\)
1 \( \frac{5}{14} \)
2 \( \frac{5}{11} \)
1 \( \frac{9}{15} \)

Solution

To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 3 are [3, 6, 9, 12, 15, 18, 21, 24, 27, 30] and the first few multiples of 5 are [5, 10, 15, 20, 25, 30, 35, 40, 45, 50]. The first few multiples they share are [15, 30, 45, 60, 75] making 15 the smallest multiple 3 and 5 share.

Next, convert the fractions so each denominator equals the lowest common multiple:

\( \frac{5 x 5}{3 x 5} \) + \( \frac{6 x 3}{5 x 3} \)

\( \frac{25}{15} \) + \( \frac{18}{15} \)

Now, because the fractions share a common denominator, you can add them:

\( \frac{25 + 18}{15} \) = \( \frac{43}{15} \) = 2\(\frac{13}{15}\)


4

The __________ is the greatest factor that divides two integers.

67% Answer Correctly

greatest common multiple

absolute value

least common multiple

greatest common factor


Solution

The greatest common factor (GCF) is the greatest factor that divides two integers.


5

What is the next number in this sequence: 1, 4, 10, 19, 31, __________ ?

69% Answer Correctly
43
49
54
46

Solution

The equation for this sequence is:

an = an-1 + 3(n - 1)

where n is the term's order in the sequence, an is the value of the term, and an-1 is the value of the term before an. This makes the next number:

a6 = a5 + 3(6 - 1)
a6 = 31 + 3(5)
a6 = 46