ASVAB Math Knowledge Practice Test 433812 Results

Your Results Global Average
Questions 5 5
Correct 0 3.44
Score 0% 69%

Review

1

The dimensions of this cylinder are height (h) = 8 and radius (r) = 2. What is the surface area?

48% Answer Correctly
40π
14π
120π
156π
Solution

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

sa = 2πr2 + 2πrh
sa = 2π(22) + 2π(2 x 8)
sa = 2π(4) + 2π(16)
sa = (2 x 4)π + (2 x 16)π
sa = 8π + 32π
sa = 40π

2

If a = 9 and x = 2, what is the value of -6a(a - x)?

68% Answer Correctly
-378
-288
-24
3
Solution

To solve this equation, replace the variables with the values given and then solve the now variable-free equation. (Remember order of operations, PEMDAS, Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.)

-6a(a - x)
-6(9)(9 - 2)
-6(9)(7)
(-54)(7)
-378

3

If AD = 25 and BD = 20, AB = ?

76% Answer Correctly
5
7
8
1
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 = 25 - 20
AB = 5

4

The dimensions of this cube are height (h) = 4, length (l) = 7, and width (w) = 8. What is the volume?

83% Answer Correctly
21
48
18
224
Solution

The volume of a cube is height x length x width:

v = h x l x w
v = 4 x 7 x 8
v = 224

5

If the area of this square is 36, what is the length of one of the diagonals?

68% Answer Correctly
9\( \sqrt{2} \)
8\( \sqrt{2} \)
6\( \sqrt{2} \)
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 = \( \sqrt{a} \) = \( \sqrt{36} \) = 6

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 = 62 + 62
c2 = 72
c = \( \sqrt{72} \) = \( \sqrt{36 x 2} \) = \( \sqrt{36} \) \( \sqrt{2} \)
c = 6\( \sqrt{2} \)