ASVAB Math Knowledge Practice Test 923087 Results

Your Results Global Average
Questions 5 5
Correct 0 3.33
Score 0% 67%

Review

1

A coordinate grid is composed of which of the following?

88% Answer Correctly

y-axis

x-axis

all of these

origin


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.


2

What is the area of a circle with a radius of 2?

69% Answer Correctly
81π

Solution

The formula for area is πr2:

a = πr2
a = π(22)
a = 4π


3

The dimensions of this cube are height (h) = 2, length (l) = 7, and width (w) = 9. What is the surface area?

51% Answer Correctly
102
190
322
258

Solution

The surface area of a cube is (2 x length x width) + (2 x width x height) + (2 x length x height):

sa = 2lw + 2wh + 2lh
sa = (2 x 7 x 9) + (2 x 9 x 2) + (2 x 7 x 2)
sa = (126) + (36) + (28)
sa = 190


4

Factor y2 - 6y + 8

54% Answer Correctly
(y + 4)(y + 2)
(y + 4)(y - 2)
(y - 4)(y + 2)
(y - 4)(y - 2)

Solution

To factor a quadratic expression, apply the FOIL method (First, Outside, Inside, Last) in reverse. First, find the two Last terms that will multiply to produce 8 as well and sum (Inside, Outside) to equal -6. For this problem, those two numbers are -4 and -2. Then, plug these into a set of binomials using the square root of the First variable (y2):

y2 - 6y + 8
y2 + (-4 - 2)y + (-4 x -2)
(y - 4)(y - 2)


5

If the area of this square is 1, what is the length of one of the diagonals?

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

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 = \( \sqrt{a} \) = \( \sqrt{1} \) = 1

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 = 12 + 12
c2 = 2
c = \( \sqrt{2} \)