ASVAB Arithmetic Reasoning Practice Test 959799 Results

Your Results Global Average
Questions 5 5
Correct 0 3.32
Score 0% 66%

Review

1

Solve for \( \frac{5!}{3!} \)

67% Answer Correctly
20
\( \frac{1}{6720} \)
\( \frac{1}{4} \)
60480

Solution

A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:

\( \frac{5!}{3!} \)
\( \frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} \)
\( \frac{5 \times 4}{1} \)
\( 5 \times 4 \)
20


2

Which of the following is an improper fraction?

71% Answer Correctly

\(1 {2 \over 5} \)

\({2 \over 5} \)

\({7 \over 5} \)

\({a \over 5} \)


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.


3

If there were a total of 150 raffle tickets sold and you bought 3 tickets, what's the probability that you'll win the raffle?

60% Answer Correctly
10%
8%
15%
2%

Solution

You have 3 out of the total of 150 raffle tickets sold so you have a (\( \frac{3}{150} \)) x 100 = \( \frac{3 \times 100}{150} \) = \( \frac{300}{150} \) = 2% chance to win the raffle.


4

How many 6-passenger vans will it take to drive all 66 members of the football team to an away game?

81% Answer Correctly
12 vans
5 vans
8 vans
11 vans

Solution

Calculate the number of vans needed by dividing the number of people that need transported by the capacity of one van:

vans = \( \frac{66}{6} \) = 11


5

Solve 5 + (5 + 2) ÷ 5 x 3 - 42

53% Answer Correctly
3\(\frac{1}{2}\)
\(\frac{3}{4}\)
-6\(\frac{4}{5}\)
\(\frac{2}{9}\)

Solution

Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):

5 + (5 + 2) ÷ 5 x 3 - 42
P: 5 + (7) ÷ 5 x 3 - 42
E: 5 + 7 ÷ 5 x 3 - 16
MD: 5 + \( \frac{7}{5} \) x 3 - 16
MD: 5 + \( \frac{21}{5} \) - 16
AS: \( \frac{25}{5} \) + \( \frac{21}{5} \) - 16
AS: \( \frac{46}{5} \) - 16
AS: \( \frac{46 - 80}{5} \)
\( \frac{-34}{5} \)
-6\(\frac{4}{5}\)