ASVAB Math Knowledge Practice Test 618159 Results

Your Results Global Average
Questions 5 5
Correct 0 3.03
Score 0% 61%

Review

1

If a = c = 2, b = d = 6, what is the area of this rectangle?

80% Answer Correctly
42
56
18
12

Solution

The area of a rectangle is equal to its length x width:

a = l x w
a = a x b
a = 2 x 6
a = 12


2

Solve 2c + 5c = 8c + 8y + 4 for c in terms of y.

34% Answer Correctly
\(\frac{7}{9}\)y + \(\frac{8}{9}\)
2\(\frac{2}{5}\)y - \(\frac{3}{5}\)
-\(\frac{1}{2}\)y - \(\frac{2}{3}\)
-\(\frac{5}{17}\)y + \(\frac{1}{17}\)

Solution

To solve this equation, isolate the variable for which you are solving (c) on one side of the equation and put everything else on the other side.

2c + 5y = 8c + 8y + 4
2c = 8c + 8y + 4 - 5y
2c - 8c = 8y + 4 - 5y
-6c = 3y + 4
c = \( \frac{3y + 4}{-6} \)
c = \( \frac{3y}{-6} \) + \( \frac{4}{-6} \)
c = -\(\frac{1}{2}\)y - \(\frac{2}{3}\)


3

On this circle, line segment AB is the:

70% Answer Correctly

radius

chord

circumference

diameter


Solution

A circle is a figure in which each point around its perimeter is an equal distance from the center. The radius of a circle is the distance between the center and any point along its perimeter. A chord is a line segment that connects any two points along its perimeter. The diameter of a circle is the length of a chord that passes through the center of the circle and equals twice the circle's radius (2r).


4

Breaking apart a quadratic expression into a pair of binomials is called:

74% Answer Correctly

squaring

normalizing

deconstructing

factoring


Solution

To factor a quadratic expression, apply the FOIL (First, Outside, Inside, Last) method in reverse.


5

Solve for b:
5b - 5 > \( \frac{b}{4} \)

44% Answer Correctly
b > \(\frac{5}{18}\)
b > 1\(\frac{1}{19}\)
b > 1\(\frac{1}{71}\)
b > -1\(\frac{15}{34}\)

Solution

To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the > sign and the answer on the other.

5b - 5 > \( \frac{b}{4} \)
4 x (5b - 5) > b
(4 x 5b) + (4 x -5) > b
20b - 20 > b
20b - 20 - b > 0
20b - b > 20
19b > 20
b > \( \frac{20}{19} \)
b > 1\(\frac{1}{19}\)