ASVAB Math Knowledge Practice Test 925244 Results

Your Results Global Average
Questions 5 5
Correct 0 2.72
Score 0% 54%

Review

1

Solve -b - 6b = -6b + 5z - 4 for b in terms of z.

34% Answer Correctly
2\(\frac{1}{5}\)z - \(\frac{4}{5}\)
-1\(\frac{2}{3}\)z - 3
12z - 9
z - 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.

-b - 6z = -6b + 5z - 4
-b = -6b + 5z - 4 + 6z
-b + 6b = 5z - 4 + 6z
5b = 11z - 4
b = \( \frac{11z - 4}{5} \)
b = \( \frac{11z}{5} \) + \( \frac{-4}{5} \)
b = 2\(\frac{1}{5}\)z - \(\frac{4}{5}\)


2

If angle a = 25° and angle b = 59° what is the length of angle d?

56% Answer Correctly
155°
160°
114°
127°

Solution

An exterior angle of a triangle is equal to the sum of the two interior angles that are opposite:

d° = b° + c°

To find angle c, remember that the sum of the interior angles of a triangle is 180°:

180° = a° + b° + c°
c° = 180° - a° - b°
c° = 180° - 25° - 59° = 96°

So, d° = 59° + 96° = 155°

A shortcut to get this answer is to remember that angles around a line add up to 180°:

a° + d° = 180°
d° = 180° - a°
d° = 180° - 25° = 155°


3

The dimensions of this cylinder are height (h) = 5 and radius (r) = 8. What is the surface area?

48% Answer Correctly
208π
20π
252π
224π

Solution

The surface area of a cylinder is 2πr2 + 2πrh:

sa = 2πr2 + 2πrh
sa = 2π(82) + 2π(8 x 5)
sa = 2π(64) + 2π(40)
sa = (2 x 64)π + (2 x 40)π
sa = 128π + 80π
sa = 208π


4

If side a = 4, side b = 6, what is the length of the hypotenuse of this right triangle?

63% Answer Correctly
\( \sqrt{10} \)
\( \sqrt{52} \)
\( \sqrt{34} \)
\( \sqrt{89} \)

Solution

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

c2 = a2 + b2
c2 = 42 + 62
c2 = 16 + 36
c2 = 52
c = \( \sqrt{52} \)


5

What is the circumference of a circle with a diameter of 5?

70% Answer Correctly
5π
4π
7π
30π

Solution

The formula for circumference is circle diameter x π:

c = πd
c = 5π