ASVAB Math Knowledge Practice Test 783257 Results

Your Results Global Average
Questions 5 5
Correct 0 3.30
Score 0% 66%

Review

1

If side x = 9cm, side y = 5cm, and side z = 9cm what is the perimeter of this triangle?

84% Answer Correctly
39cm
25cm
23cm
30cm

Solution

The perimeter of a triangle is the sum of the lengths of its sides:

p = x + y + z
p = 9cm + 5cm + 9cm = 23cm


2

Factor y2 + 13y + 42

53% Answer Correctly
(y - 6)(y - 7)
(y + 6)(y + 7)
(y + 6)(y - 7)
(y - 6)(y + 7)

Solution

To factor a quadratic expression, apply the FOIL method (First, Outside, Inside, Last) in reverse. First, find the two Last terms that will multiply to produce 42 as well and sum (Inside, Outside) to equal 13. For this problem, those two numbers are 6 and 7. Then, plug these into a set of binomials using the square root of the First variable (y2):

y2 + 13y + 42
y2 + (6 + 7)y + (6 x 7)
(y + 6)(y + 7)


3

What is 3a8 + 6a8?

74% Answer Correctly
-3
18a16
9a16
9a8

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.

3a8 + 6a8 = 9a8


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

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

all of these statements are correct


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

The dimensions of this cylinder are height (h) = 1 and radius (r) = 2. What is the surface area?

48% Answer Correctly
20π
12π
40π
180π

Solution

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

sa = 2πr2 + 2πrh
sa = 2π(22) + 2π(2 x 1)
sa = 2π(4) + 2π(2)
sa = (2 x 4)π + (2 x 2)π
sa = 8π + 4π
sa = 12π