ASVAB Arithmetic Reasoning Practice Test 497454 Results

Your Results Global Average
Questions 5 5
Correct 0 3.38
Score 0% 68%

Review

1

What is 2c7 + 8c7?

66% Answer Correctly
-6c-7
-6c7
10c-14
10c7

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:

2c7 + 8c7
(2 + 8)c7
10c7


2

What is the least common multiple of 8 and 16?

72% Answer Correctly
16
106
24
77

Solution

The first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80] and the first few multiples of 16 are [16, 32, 48, 64, 80, 96]. The first few multiples they share are [16, 32, 48, 64, 80] making 16 the smallest multiple 8 and 16 have in common.


3

In a class of 25 students, 7 are taking German and 7 are taking Spanish. Of the students studying German or Spanish, 3 are taking both courses. How many students are not enrolled in either course?

63% Answer Correctly
11
20
23
14

Solution

The number of students taking German or Spanish is 7 + 7 = 14. Of that group of 14, 3 are taking both languages so they've been counted twice (once in the German group and once in the Spanish group). Subtracting them out leaves 14 - 3 = 11 who are taking at least one language. 25 - 11 = 14 students who are not taking either language.


4

Which of the following is an improper fraction?

70% Answer Correctly

\({7 \over 5} \)

\({2 \over 5} \)

\(1 {2 \over 5} \)

\({a \over 5} \)


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.


5

Solve for \( \frac{2!}{6!} \)

66% Answer Correctly
\( \frac{1}{360} \)
15120
6720
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:

\( \frac{2!}{6!} \)
\( \frac{2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \)
\( \frac{1}{6 \times 5 \times 4 \times 3} \)
\( \frac{1}{360} \)