ASVAB Math Knowledge Practice Test 692529 Results

Your Results Global Average
Questions 5 5
Correct 0 2.96
Score 0% 59%

Review

1

If the base of this triangle is 7 and the height is 2, what is the area?

58% Answer Correctly
36
17\(\frac{1}{2}\)
20
7

Solution

The area of a triangle is equal to ½ base x height:

a = ½bh
a = ½ x 7 x 2 = \( \frac{14}{2} \) = 7


2

A coordinate grid is composed of which of the following?

91% Answer Correctly

origin

y-axis

all of these

x-axis


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.


3

Solve for x:
-7x + 4 = \( \frac{x}{-7} \)

46% Answer Correctly
\(\frac{7}{12}\)
1\(\frac{2}{7}\)
1
\(\frac{9}{14}\)

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 equal sign and the answer on the other.

-7x + 4 = \( \frac{x}{-7} \)
-7 x (-7x + 4) = x
(-7 x -7x) + (-7 x 4) = x
49x - 28 = x
49x - 28 - x = 0
49x - x = 28
48x = 28
x = \( \frac{28}{48} \)
x = \(\frac{7}{12}\)


4

If side a = 1, side b = 5, what is the length of the hypotenuse of this right triangle?

64% Answer Correctly
\( \sqrt{8} \)
\( \sqrt{26} \)
\( \sqrt{85} \)
\( \sqrt{145} \)

Solution

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

c2 = a2 + b2
c2 = 12 + 52
c2 = 1 + 25
c2 = 26
c = \( \sqrt{26} \)


5

Solve 5a + 9a = -4a + 3y + 3 for a in terms of y.

34% Answer Correctly
-\(\frac{2}{3}\)y + \(\frac{1}{3}\)
-\(\frac{1}{7}\)y + 1
\(\frac{4}{11}\)y + \(\frac{5}{11}\)
-1\(\frac{3}{5}\)y + 1\(\frac{3}{5}\)

Solution

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

5a + 9y = -4a + 3y + 3
5a = -4a + 3y + 3 - 9y
5a + 4a = 3y + 3 - 9y
9a = -6y + 3
a = \( \frac{-6y + 3}{9} \)
a = \( \frac{-6y}{9} \) + \( \frac{3}{9} \)
a = -\(\frac{2}{3}\)y + \(\frac{1}{3}\)