ASVAB Math Knowledge Practice Test 793615 Results

Your Results Global Average
Questions 5 5
Correct 0 3.24
Score 0% 65%

Review

1

The formula for volume of a cube in terms of height (h), length (l), and width (w) is which of the following?

68% Answer Correctly

2lw x 2wh + 2lh

h2 x l2 x w2

lw x wh + lh

h x l x w


Solution

A cube is a rectangular solid box with a height (h), length (l), and width (w). The volume is h x l x w and the surface area is 2lw x 2wh + 2lh.


2

Solve for b:
5b - 6 > 5 - 9b

55% Answer Correctly
b > 2\(\frac{1}{2}\)
b > -2\(\frac{1}{4}\)
b > -1\(\frac{1}{2}\)
b > \(\frac{11}{14}\)

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 - 6 > 5 - 9b
5b > 5 - 9b + 6
5b + 9b > 5 + 6
14b > 11
b > \( \frac{11}{14} \)
b > \(\frac{11}{14}\)


3

If the area of this square is 16, what is the length of one of the diagonals?

69% Answer Correctly
5\( \sqrt{2} \)
3\( \sqrt{2} \)
4\( \sqrt{2} \)
9\( \sqrt{2} \)

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 = \( \sqrt{a} \) = \( \sqrt{16} \) = 4

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 = 42 + 42
c2 = 32
c = \( \sqrt{32} \) = \( \sqrt{16 x 2} \) = \( \sqrt{16} \) \( \sqrt{2} \)
c = 4\( \sqrt{2} \)


4

Simplify 7a x 5b.

86% Answer Correctly
12ab
35ab
35\( \frac{a}{b} \)
35a2b2

Solution

To multiply monomials, multiply the coefficients (the numbers that come before the variables) of each term, add the exponents of like variables, and multiply the different variables together.

7a x 5b = (7 x 5) (a x b) = 35ab


5

The endpoints of this line segment are at (-2, -7) and (2, 1). What is the slope of this line?

46% Answer Correctly
\(\frac{1}{2}\)
2\(\frac{1}{2}\)
2
1\(\frac{1}{2}\)

Solution

The slope of this line is the change in y divided by the change in x. The endpoints of this line segment are at (-2, -7) and (2, 1) so the slope becomes:

m = \( \frac{\Delta y}{\Delta x} \) = \( \frac{(1.0) - (-7.0)}{(2) - (-2)} \) = \( \frac{8}{4} \)
m = 2