ASVAB Math Knowledge Practice Test 423192 Results

Your Results Global Average
Questions 5 5
Correct 0 2.75
Score 0% 55%

Review

1

Solve for z:
z2 + 17z + 31 = 5z - 1

48% Answer Correctly
-6 or -9
3 or -5
8 or -9
-4 or -8

Solution

The first step to solve a quadratic expression that's not set to zero is to solve the equation so that it is set to zero:

z2 + 17z + 31 = 5z - 1
z2 + 17z + 31 + 1 = 5z
z2 + 17z - 5z + 32 = 0
z2 + 12z + 32 = 0

Next, factor the quadratic equation:

z2 + 12z + 32 = 0
(z + 4)(z + 8) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (z + 4) or (z + 8) must equal zero:

If (z + 4) = 0, z must equal -4
If (z + 8) = 0, z must equal -8

So the solution is that z = -4 or -8


2

Solve -2a - 4a = 4a - 7z - 7 for a in terms of z.

34% Answer Correctly
-2\(\frac{3}{4}\)z + \(\frac{1}{4}\)
\(\frac{1}{2}\)z + 1\(\frac{1}{6}\)
-5\(\frac{2}{3}\)z + 3
-\(\frac{5}{7}\)z + \(\frac{3}{7}\)

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.

-2a - 4z = 4a - 7z - 7
-2a = 4a - 7z - 7 + 4z
-2a - 4a = -7z - 7 + 4z
-6a = -3z - 7
a = \( \frac{-3z - 7}{-6} \)
a = \( \frac{-3z}{-6} \) + \( \frac{-7}{-6} \)
a = \(\frac{1}{2}\)z + 1\(\frac{1}{6}\)


3

Solve for a:
4a + 6 = 4 + 6a

58% Answer Correctly
1\(\frac{3}{4}\)
1\(\frac{4}{5}\)
1
-\(\frac{4}{9}\)

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.

4a + 6 = 4 + 6a
4a = 4 + 6a - 6
4a - 6a = 4 - 6
-2a = -2
a = \( \frac{-2}{-2} \)
a = 1


4

If AD = 14 and BD = 8, AB = ?

75% Answer Correctly
6
11
12
13

Solution

The entire length of this line is represented by AD which is AB + BD:

AD = AB + BD

Solving for AB:

AB = AD - BD
AB = 14 - 8
AB = 6


5

Solve for y:
y2 - y - 20 = 0

58% Answer Correctly
-4 or 5
4 or -6
3 or -4
6 or -2

Solution

The first step to solve a quadratic equation that's set to zero is to factor the quadratic equation:

y2 - y - 20 = 0
(y + 4)(y - 5) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (y + 4) or (y - 5) must equal zero:

If (y + 4) = 0, y must equal -4
If (y - 5) = 0, y must equal 5

So the solution is that y = -4 or 5