ASVAB Math Knowledge Practice Test 533016 Results

Your Results Global Average
Questions 5 5
Correct 0 3.15
Score 0% 63%

Review

1

If c = 6 and x = -2, what is the value of 5c(c - x)?

69% Answer Correctly
-64
14
312
240

Solution

To solve this equation, 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.)

5c(c - x)
5(6)(6 + 2)
5(6)(8)
(30)(8)
240


2

If side a = 4, side b = 1, what is the length of the hypotenuse of this right triangle?

64% Answer Correctly
\( \sqrt{17} \)
\( \sqrt{145} \)
\( \sqrt{85} \)
\( \sqrt{26} \)

Solution

According to the Pythagorean theorem, the hypotenuse squared is equal to the sum of the two perpendicular sides squared:

c2 = a2 + b2
c2 = 42 + 12
c2 = 16 + 1
c2 = 17
c = \( \sqrt{17} \)


3

Which of the following statements about math operations is incorrect?

71% Answer Correctly

you can add monomials that have the same variable and the same exponent

all of these statements are correct

you can multiply monomials that have different variables and different exponents

you can subtract monomials that have the same variable and the same exponent


Solution

You can only add or subtract monomials that have the same variable and the same exponent. For example, 2a + 4a = 6a and 4a2 - a2 = 3a2 but 2a + 4b and 7a - 3b cannot be combined. However, you can multiply and divide monomials with unlike terms. For example, 2a x 6b = 12ab.


4

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

h x l x w

lw x wh + lh


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.


5

Solve for x:
-4x + 5 < \( \frac{x}{-1} \)

44% Answer Correctly
x < -\(\frac{21}{41}\)
x < 1\(\frac{10}{71}\)
x < \(\frac{36}{71}\)
x < 1\(\frac{2}{3}\)

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.

-4x + 5 < \( \frac{x}{-1} \)
-1 x (-4x + 5) < x
(-1 x -4x) + (-1 x 5) < x
4x - 5 < x
4x - 5 - x < 0
4x - x < 5
3x < 5
x < \( \frac{5}{3} \)
x < 1\(\frac{2}{3}\)