ASVAB Math Knowledge Practice Test 980760 Results

Your Results Global Average
Questions 5 5
Correct 0 2.84
Score 0% 57%

Review

1

If the base of this triangle is 1 and the height is 5, what is the area?

58% Answer Correctly
2\(\frac{1}{2}\)
105
48
30

Solution

The area of a triangle is equal to ½ base x height:

a = ½bh
a = ½ x 1 x 5 = \( \frac{5}{2} \) = 2\(\frac{1}{2}\)


2

To multiply binomials, use the FOIL method. Which of the following is not a part of the FOIL method?

84% Answer Correctly

First

Odd

Last

Inside


Solution

To multiply binomials, use the FOIL method. FOIL stands for First, Outside, Inside, Last and refers to the position of each term in the parentheses.


3

Which of the following statements about a triangle is not true?

57% Answer Correctly

exterior angle = sum of two adjacent interior angles

perimeter = sum of side lengths

sum of interior angles = 180°

area = ½bh


Solution

A triangle is a three-sided polygon. It has three interior angles that add up to 180° (a + b + c = 180°). An exterior angle of a triangle is equal to the sum of the two interior angles that are opposite (d = b + c). The perimeter of a triangle is equal to the sum of the lengths of its three sides, the height of a triangle is equal to the length from the base to the opposite vertex (angle) and the area equals one-half triangle base x height: a = ½ base x height.


4

On this circle, a line segment connecting point A to point D is called:

46% Answer Correctly

radius

circumference

diameter

chord


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).


5

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

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

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.

-4c + 2y = -c - 8y - 5
-4c = -c - 8y - 5 - 2y
-4c + c = -8y - 5 - 2y
-3c = -10y - 5
c = \( \frac{-10y - 5}{-3} \)
c = \( \frac{-10y}{-3} \) + \( \frac{-5}{-3} \)
c = 3\(\frac{1}{3}\)y + 1\(\frac{2}{3}\)