ASVAB Arithmetic Reasoning Practice Test 187260 Results

Your Results Global Average
Questions 5 5
Correct 0 2.89
Score 0% 58%

Review

1

A sports card collection contains football, baseball, and basketball cards. If the ratio of football to baseball cards is 7 to 2 and the ratio of baseball to basketball cards is 7 to 1, what is the ratio of football to basketball cards?

53% Answer Correctly
49:2
5:4
5:1
7:8

Solution

The ratio of football cards to baseball cards is 7:2 and the ratio of baseball cards to basketball cards is 7:1. To solve this problem, we need the baseball card side of each ratio to be equal so we need to rewrite the ratios in terms of a common number of baseball cards. (Think of this like finding the common denominator when adding fractions.) The ratio of football to baseball cards can also be written as 49:14 and the ratio of baseball cards to basketball cards as 14:2. So, the ratio of football cards to basketball cards is football:baseball, baseball:basketball or 49:14, 14:2 which reduces to 49:2.


2

a(b + c) = ab + ac defines which of the following?

75% Answer Correctly

distributive property for multiplication

commutative property for division

distributive property for division

commutative property for multiplication


Solution

The distributive property for multiplication helps in solving expressions like a(b + c). It specifies that the result of multiplying one number by the sum or difference of two numbers can be obtained by multiplying each number individually and then totaling the results: a(b + c) = ab + ac. For example, 4(10-5) = (4 x 10) - (4 x 5) = 40 - 20 = 20.


3

This property states taht the order of addition or multiplication does not mater. For example, 2 + 5 and 5 + 2 are equivalent.

60% Answer Correctly

associative

PEDMAS

distributive

commutative


Solution

The commutative property states that, when adding or multiplying numbers, the order in which they're added or multiplied does not matter. For example, 3 + 4 and 4 + 3 give the same result, as do 3 x 4 and 4 x 3.


4

If all of a roofing company's 10 workers are required to staff 5 roofing crews, how many workers need to be added during the busy season in order to send 9 complete crews out on jobs?

55% Answer Correctly
15
8
6
2

Solution

In order to find how many additional workers are needed to staff the extra crews you first need to calculate how many workers are on a crew. There are 10 workers at the company now and that's enough to staff 5 crews so there are \( \frac{10}{5} \) = 2 workers on a crew. 9 crews are needed for the busy season which, at 2 workers per crew, means that the roofing company will need 9 x 2 = 18 total workers to staff the crews during the busy season. The company already employs 10 workers so they need to add 18 - 10 = 8 new staff for the busy season.


5

If a rectangle is twice as long as it is wide and has a perimeter of 54 meters, what is the area of the rectangle?

47% Answer Correctly
162 m2
98 m2
50 m2
8 m2

Solution

The area of a rectangle is width (w) x height (h). In this problem we know that the rectangle is twice as long as it is wide so h = 2w. The perimeter of a rectangle is 2w + 2h and we know that the perimeter of this rectangle is 54 meters so the equation becomes: 2w + 2h = 54.

Putting these two equations together and solving for width (w):

2w + 2h = 54
w + h = \( \frac{54}{2} \)
w + h = 27
w = 27 - h

From the question we know that h = 2w so substituting 2w for h gives us:

w = 27 - 2w
3w = 27
w = \( \frac{27}{3} \)
w = 9

Since h = 2w that makes h = (2 x 9) = 18 and the area = h x w = 9 x 18 = 162 m2