ASVAB Math Knowledge Practice Test 525030 Results

Your Results Global Average
Questions 5 5
Correct 0 3.47
Score 0% 69%

Review

1

Solve for x:
7x - 9 < 7 - 8x

55% Answer Correctly
x < -\(\frac{4}{5}\)
x < -\(\frac{5}{7}\)
x < 1\(\frac{1}{15}\)
x < -1\(\frac{2}{5}\)

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.

7x - 9 < 7 - 8x
7x < 7 - 8x + 9
7x + 8x < 7 + 9
15x < 16
x < \( \frac{16}{15} \)
x < 1\(\frac{1}{15}\)


2

A right angle measures:

91% Answer Correctly

360°

45°

90°

180°


Solution

A right angle measures 90 degrees and is the intersection of two perpendicular lines. In diagrams, a right angle is indicated by a small box completing a square with the perpendicular lines.


3

A trapezoid is a quadrilateral with one set of __________ sides.

70% Answer Correctly

parallel

right angle

equal angle

equal length


Solution

A trapezoid is a quadrilateral with one set of parallel sides.


4

Simplify (5a)(7ab) - (4a2)(8b).

62% Answer Correctly
144a2b
144ab2
67a2b
3a2b

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.

(5a)(7ab) - (4a2)(8b)
(5 x 7)(a x a x b) - (4 x 8)(a2 x b)
(35)(a1+1 x b) - (32)(a2b)
35a2b - 32a2b
3a2b


5

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

68% Answer Correctly
\( \sqrt{2} \)
4\( \sqrt{2} \)
2\( \sqrt{2} \)
3\( \sqrt{2} \)

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\( \sqrt{2} \)