ASVAB Math Knowledge Practice Test 216591 Results

Your Results Global Average
Questions 5 5
Correct 0 2.70
Score 0% 54%

Review

1

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

88% Answer Correctly

division

exponents

pairs

addition


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

45% Answer Correctly

midpoints

bisects

trisects

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.


3

A(n) __________ is to a parallelogram as a square is to a rectangle.

51% Answer Correctly

quadrilateral

rhombus

triangle

trapezoid


Solution

A rhombus is a parallelogram with four equal-length sides. A square is a rectangle with four equal-length sides.


4

Solve -4c + 3c = 4c - 4y + 6 for c in terms of y.

34% Answer Correctly
\(\frac{7}{8}\)y - \(\frac{3}{4}\)
\(\frac{1}{5}\)y + 1\(\frac{1}{5}\)
-\(\frac{2}{3}\)y - 1\(\frac{1}{3}\)
-2\(\frac{1}{2}\)y - \(\frac{1}{4}\)

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 + 3y = 4c - 4y + 6
-4c = 4c - 4y + 6 - 3y
-4c - 4c = -4y + 6 - 3y
-8c = -7y + 6
c = \( \frac{-7y + 6}{-8} \)
c = \( \frac{-7y}{-8} \) + \( \frac{6}{-8} \)
c = \(\frac{7}{8}\)y - \(\frac{3}{4}\)


5

For this diagram, the Pythagorean theorem states that b2 = ?

47% Answer Correctly

a2 - c2

c - a

c2 - a2

c2 + a2


Solution

The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c2) is equal to the sum of the two perpendicular sides squared (a2 + b2): c2 = a2 + b2 or, solved for c, \(c = \sqrt{a + b}\)