ASVAB Math Knowledge Practice Test 196548 Results

Your Results Global Average
Questions 5 5
Correct 0 3.28
Score 0% 66%

Review

1

If a = 1, b = 8, c = 9, and d = 3, what is the perimeter of this quadrilateral?

88% Answer Correctly
23
20
21
24

Solution

Perimeter is equal to the sum of the four sides:

p = a + b + c + d
p = 1 + 8 + 9 + 3
p = 21


2

Breaking apart a quadratic expression into a pair of binomials is called:

74% Answer Correctly

deconstructing

factoring

normalizing

squaring


Solution

To factor a quadratic expression, apply the FOIL (First, Outside, Inside, Last) method in reverse.


3

The dimensions of this cylinder are height (h) = 8 and radius (r) = 9. What is the volume?

62% Answer Correctly
288π
648π
16π
36π

Solution

The volume of a cylinder is πr2h:

v = πr2h
v = π(92 x 8)
v = 648π


4

If the area of this square is 9, what is the length of one of the diagonals?

68% Answer Correctly
8\( \sqrt{2} \)
3\( \sqrt{2} \)
7\( \sqrt{2} \)
6\( \sqrt{2} \)

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 = \( \sqrt{a} \) = \( \sqrt{9} \) = 3

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 = 32 + 32
c2 = 18
c = \( \sqrt{18} \) = \( \sqrt{9 x 2} \) = \( \sqrt{9} \) \( \sqrt{2} \)
c = 3\( \sqrt{2} \)


5

Solve 8b + 9b = 2b - 8x + 1 for b in terms of x.

34% Answer Correctly
\(\frac{11}{14}\)x + \(\frac{4}{7}\)
-2\(\frac{5}{6}\)x + \(\frac{1}{6}\)
-\(\frac{5}{12}\)x - \(\frac{1}{4}\)
-x + 6

Solution

To solve this equation, isolate the variable for which you are solving (b) on one side of the equation and put everything else on the other side.

8b + 9x = 2b - 8x + 1
8b = 2b - 8x + 1 - 9x
8b - 2b = -8x + 1 - 9x
6b = -17x + 1
b = \( \frac{-17x + 1}{6} \)
b = \( \frac{-17x}{6} \) + \( \frac{1}{6} \)
b = -2\(\frac{5}{6}\)x + \(\frac{1}{6}\)