ASVAB Math Knowledge Practice Test 53219 Results

Your Results Global Average
Questions 5 5
Correct 0 3.92
Score 0% 78%

Review

1

A(n) __________ is two expressions separated by an equal sign.

77% Answer Correctly

formula

equation

expression

problem


Solution

An equation is two expressions separated by an equal sign. The key to solving equations is to 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.


2

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

91% Answer Correctly

addition

pairs

division

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.)


3

A coordinate grid is composed of which of the following?

91% Answer Correctly

x-axis

origin

all of these

y-axis


Solution

The coordinate grid is composed of a horizontal x-axis and a vertical y-axis. The center of the grid, where the x-axis and y-axis meet, is called the origin.


4

Simplify (y - 2)(y - 7)

64% Answer Correctly
y2 - 9y + 14
y2 + 5y - 14
y2 - 5y - 14
y2 + 9y + 14

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:

(y - 2)(y - 7)
(y x y) + (y x -7) + (-2 x y) + (-2 x -7)
y2 - 7y - 2y + 14
y2 - 9y + 14


5

If the area of this square is 64, what is the length of one of the diagonals?

68% Answer Correctly
9\( \sqrt{2} \)
8\( \sqrt{2} \)
7\( \sqrt{2} \)
5\( \sqrt{2} \)

Solution

To find the diagonal we need to know the length of one of the square's sides. We know the area and the area of a square is the length of one side squared:

a = s2

so the length of one side of the square is:

s = \( \sqrt{a} \) = \( \sqrt{64} \) = 8

The Pythagorean theorem defines the square of the hypotenuse (diagonal) of a triangle with a right angle as the sum of the squares of the other two sides:

c2 = a2 + b2
c2 = 82 + 82
c2 = 128
c = \( \sqrt{128} \) = \( \sqrt{64 x 2} \) = \( \sqrt{64} \) \( \sqrt{2} \)
c = 8\( \sqrt{2} \)