ASVAB Math Knowledge Practice Test 863628 Results

Your Results Global Average
Questions 5 5
Correct 0 3.07
Score 0% 61%

Review

1

Which of the following is not a part of PEMDAS, the acronym for math order of operations?

88% Answer Correctly

addition

division

pairs

exponents


Solution

When solving an equation with two variables, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.)


2

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

79% Answer Correctly
32
12
4
6

Solution

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

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


3

Solve for y:
9y + 5 = 8 - 2y

58% Answer Correctly
-7
\(\frac{1}{9}\)
\(\frac{3}{11}\)
-1

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 equal sign and the answer on the other.

9y + 5 = 8 - 2y
9y = 8 - 2y - 5
9y + 2y = 8 - 5
11y = 3
y = \( \frac{3}{11} \)
y = \(\frac{3}{11}\)


4

If the length of AB equals the length of BD, point B __________ this line segment.

45% Answer Correctly

trisects

midpoints

bisects

intersects


Solution

A line segment is a portion of a line with a measurable length. The midpoint of a line segment is the point exactly halfway between the endpoints. The midpoint bisects (cuts in half) the line segment.


5

Solve -3a - 4a = 9a - 5z - 4 for a in terms of z.

34% Answer Correctly
\(\frac{1}{15}\)z - \(\frac{1}{5}\)
\(\frac{1}{2}\)z + 1\(\frac{1}{6}\)
-\(\frac{1}{4}\)z - \(\frac{1}{2}\)
\(\frac{1}{12}\)z + \(\frac{1}{3}\)

Solution

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

-3a - 4z = 9a - 5z - 4
-3a = 9a - 5z - 4 + 4z
-3a - 9a = -5z - 4 + 4z
-12a = -z - 4
a = \( \frac{-z - 4}{-12} \)
a = \( \frac{-z}{-12} \) + \( \frac{-4}{-12} \)
a = \(\frac{1}{12}\)z + \(\frac{1}{3}\)