ASVAB Math Knowledge Practice Test 853372 Results

Your Results Global Average
Questions 5 5
Correct 0 3.36
Score 0% 67%

Review

1

The dimensions of this cube are height (h) = 4, length (l) = 7, and width (w) = 5. What is the surface area?

51% Answer Correctly
192
112
166
56

Solution

The surface area of a cube is (2 x length x width) + (2 x width x height) + (2 x length x height):

sa = 2lw + 2wh + 2lh
sa = (2 x 7 x 5) + (2 x 5 x 4) + (2 x 7 x 4)
sa = (70) + (40) + (56)
sa = 166


2

What is 5a5 + 9a5?

75% Answer Correctly
14
-4a10
14a10
14a5

Solution

To combine like terms, add or subtract the coefficients (the numbers that come before the variables) of terms that have the same variable raised to the same exponent.

5a5 + 9a5 = 14a5


3

What is the circumference of a circle with a radius of 10?

71% Answer Correctly
20π
17π

Solution

The formula for circumference is circle diameter x π. Circle diameter is 2 x radius:

c = πd
c = π(2 * r)
c = π(2 * 10)
c = 20π


4

Which of the following statements about math operations is incorrect?

70% Answer Correctly

you can multiply monomials that have different variables and different exponents

you can subtract monomials that have the same variable and the same exponent

all of these statements are correct

you can add monomials that have the same variable and the same exponent


Solution

You can only add or subtract monomials that have the same variable and the same exponent. For example, 2a + 4a = 6a and 4a2 - a2 = 3a2 but 2a + 4b and 7a - 3b cannot be combined. However, you can multiply and divide monomials with unlike terms. For example, 2a x 6b = 12ab.


5

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

68% Answer Correctly
2\( \sqrt{2} \)
5\( \sqrt{2} \)
4\( \sqrt{2} \)
6\( \sqrt{2} \)

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\( \sqrt{2} \)