ASVAB Arithmetic Reasoning Practice Test 18340 Results

Your Results Global Average
Questions 5 5
Correct 0 3.08
Score 0% 62%

Review

1

4! = ?

84% Answer Correctly

4 x 3 x 2 x 1

4 x 3

3 x 2 x 1

5 x 4 x 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.


2

A factor is a positive __________ that divides evenly into a given number.

78% Answer Correctly

improper fraction

mixed number

integer

fraction


Solution

A factor is a positive integer that divides evenly into a given number. For example, the factors of 8 are 1, 2, 4, and 8.


3

On average, the center for a basketball team hits 50% of his shots while a guard on the same team hits 60% of his shots. If the guard takes 20 shots during a game, how many shots will the center have to take to score as many points as the guard assuming each shot is worth the same number of points?

42% Answer Correctly
24
48
41
28

Solution
If the guard hits 60% of his shots and takes 20 shots he'll make:

guard shots made = shots taken x \( \frac{\text{% made}}{100} \) = 20 x \( \frac{60}{100} \) = \( \frac{60 x 20}{100} \) = \( \frac{1200}{100} \) = 12 shots

The center makes 50% of his shots so he'll have to take:

shots made = shots taken x \( \frac{\text{% made}}{100} \)
shots taken = \( \frac{\text{shots taken}}{\frac{\text{% made}}{100}} \)

to make as many shots as the guard. Plugging in values for the center gives us:

center shots taken = \( \frac{12}{\frac{50}{100}} \) = 12 x \( \frac{100}{50} \) = \( \frac{12 x 100}{50} \) = \( \frac{1200}{50} \) = 24 shots

to make the same number of shots as the guard and thus score the same number of points.


4

What is \( 4 \)\( \sqrt{50} \) + \( 4 \)\( \sqrt{2} \)

35% Answer Correctly
24\( \sqrt{2} \)
16\( \sqrt{25} \)
8\( \sqrt{100} \)
8\( \sqrt{25} \)

Solution

To add these radicals together their radicands must be the same:

4\( \sqrt{50} \) + 4\( \sqrt{2} \)
4\( \sqrt{25 \times 2} \) + 4\( \sqrt{2} \)
4\( \sqrt{5^2 \times 2} \) + 4\( \sqrt{2} \)
(4)(5)\( \sqrt{2} \) + 4\( \sqrt{2} \)
20\( \sqrt{2} \) + 4\( \sqrt{2} \)

Now that the radicands are identical, you can add them together:

20\( \sqrt{2} \) + 4\( \sqrt{2} \)
(20 + 4)\( \sqrt{2} \)
24\( \sqrt{2} \)


5

In a class of 28 students, 7 are taking German and 10 are taking Spanish. Of the students studying German or Spanish, 4 are taking both courses. How many students are not enrolled in either course?

63% Answer Correctly
24
12
21
15

Solution

The number of students taking German or Spanish is 7 + 10 = 17. Of that group of 17, 4 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 17 - 4 = 13 who are taking at least one language. 28 - 13 = 15 students who are not taking either language.