ASVAB Arithmetic Reasoning Practice Test 641231 Results

Your Results Global Average
Questions 5 5
Correct 0 3.06
Score 0% 61%

Review

1

What is \( \frac{5}{4} \) + \( \frac{9}{12} \)?

60% Answer Correctly
2
\( \frac{7}{12} \)
1 \( \frac{5}{12} \)
2 \( \frac{8}{12} \)

Solution

To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 4 are [4, 8, 12, 16, 20, 24, 28, 32, 36, 40] and the first few multiples of 12 are [12, 24, 36, 48, 60, 72, 84, 96]. The first few multiples they share are [12, 24, 36, 48, 60] making 12 the smallest multiple 4 and 12 share.

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

\( \frac{5 x 3}{4 x 3} \) + \( \frac{9 x 1}{12 x 1} \)

\( \frac{15}{12} \) + \( \frac{9}{12} \)

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

\( \frac{15 + 9}{12} \) = \( \frac{24}{12} \) = 2


2

If \(\left|a\right| = 7\), which of the following best describes a?

67% Answer Correctly

a = 7 or a = -7

none of these is correct

a = -7

a = 7


Solution

The absolute value is the positive magnitude of a particular number or variable and is indicated by two vertical lines: \(\left|-5\right| = 5\). In the case of a variable absolute value (\(\left|a\right| = 5\)) the value of a can be either positive or negative (a = -5 or a = 5).


3

How many 1 gallon cans worth of fuel would you need to pour into an empty 8 gallon tank to fill it exactly halfway?

51% Answer Correctly
8
4
4
6

Solution

To fill a 8 gallon tank exactly halfway you'll need 4 gallons of fuel. Each fuel can holds 1 gallons so:

cans = \( \frac{4 \text{ gallons}}{1 \text{ gallons}} \) = 4


4

Which of these numbers is a factor of 32?

68% Answer Correctly
4
34
6
12

Solution

The factors of a number are all positive integers that divide evenly into the number. The factors of 32 are 1, 2, 4, 8, 16, 32.


5

What is \( \frac{-8x^7}{6x^2} \)?

60% Answer Correctly
-1\(\frac{1}{3}\)x14
-1\(\frac{1}{3}\)x\(\frac{2}{7}\)
-1\(\frac{1}{3}\)x9
-1\(\frac{1}{3}\)x5

Solution

To divide terms with exponents, the base of both exponents must be the same. In this case they are so divide the coefficients and subtract the exponents:

\( \frac{-8x^7}{6x^2} \)
\( \frac{-8}{6} \) x(7 - 2)
-1\(\frac{1}{3}\)x5