ASVAB Math Knowledge Practice Test 542797 Results

Your Results Global Average
Questions 5 5
Correct 0 2.60
Score 0% 52%

Review

1

On this circle, a line segment connecting point A to point D is called:

46% Answer Correctly

circumference

radius

diameter

chord


Solution

A circle is a figure in which each point around its perimeter is an equal distance from the center. The radius of a circle is the distance between the center and any point along its perimeter. A chord is a line segment that connects any two points along its perimeter. The diameter of a circle is the length of a chord that passes through the center of the circle and equals twice the circle's radius (2r).


2

Solve for z:
z2 + 4z - 27 = 2z - 3

48% Answer Correctly
-7 or -9
4 or -6
-3 or -6
1 or -5

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 + 4z - 27 = 2z - 3
z2 + 4z - 27 + 3 = 2z
z2 + 4z - 2z - 24 = 0
z2 + 2z - 24 = 0

Next, factor the quadratic equation:

z2 + 2z - 24 = 0
(z - 4)(z + 6) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (z - 4) or (z + 6) must equal zero:

If (z - 4) = 0, z must equal 4
If (z + 6) = 0, z must equal -6

So the solution is that z = 4 or -6


3

If BD = 9 and AD = 14, AB = ?

75% Answer Correctly
10
18
5
8

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 - 9
AB = 5


4

For this diagram, the Pythagorean theorem states that b2 = ?

47% Answer Correctly

a2 - c2

c2 - a2

c2 + a2

c - a


Solution

The Pythagorean theorem defines the relationship between the side lengths of a right triangle. The length of the hypotenuse squared (c2) is equal to the sum of the two perpendicular sides squared (a2 + b2): c2 = a2 + b2 or, solved for c, \(c = \sqrt{a + b}\)


5

Solve for c:
6c + 3 > \( \frac{c}{1} \)

44% Answer Correctly
c > -\(\frac{48}{53}\)
c > \(\frac{4}{27}\)
c > -\(\frac{4}{17}\)
c > -\(\frac{3}{5}\)

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.

6c + 3 > \( \frac{c}{1} \)
1 x (6c + 3) > c
(1 x 6c) + (1 x 3) > c
6c + 3 > c
6c + 3 - c > 0
6c - c > -3
5c > -3
c > \( \frac{-3}{5} \)
c > -\(\frac{3}{5}\)