ASVAB Math Knowledge Practice Test 772388 Results

Your Results Global Average
Questions 5 5
Correct 0 2.69
Score 0% 54%

Review

1

What is the area of a circle with a diameter of 10?

69% Answer Correctly
64π
25π
81π

Solution

The formula for area is πr2. Radius is circle \( \frac{diameter}{2} \):

r = \( \frac{d}{2} \)
r = \( \frac{10}{2} \)
r = 5
a = πr2
a = π(52)
a = 25π


2

Solve for b:
b2 - 4b + 3 = 0

58% Answer Correctly
9 or 6
8 or 5
1 or 3
9 or 3

Solution

The first step to solve a quadratic equation that's set to zero is to factor the quadratic equation:

b2 - 4b + 3 = 0
(b - 1)(b - 3) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (b - 1) or (b - 3) must equal zero:

If (b - 1) = 0, b must equal 1
If (b - 3) = 0, b must equal 3

So the solution is that b = 1 or 3


3

Simplify (2a)(4ab) + (9a2)(6b).

65% Answer Correctly
-46ab2
46ab2
90ab2
62a2b

Solution

To multiply monomials, multiply the coefficients (the numbers that come before the variables) of each term, add the exponents of like variables, and multiply the different variables together.

(2a)(4ab) + (9a2)(6b)
(2 x 4)(a x a x b) + (9 x 6)(a2 x b)
(8)(a1+1 x b) + (54)(a2b)
8a2b + 54a2b
62a2b


4

Factor y2 + 6y + 9

53% Answer Correctly
(y + 3)(y - 3)
(y - 3)(y + 3)
(y + 3)(y + 3)
(y - 3)(y - 3)

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 9 as well and sum (Inside, Outside) to equal 6. For this problem, those two numbers are 3 and 3. Then, plug these into a set of binomials using the square root of the First variable (y2):

y2 + 6y + 9
y2 + (3 + 3)y + (3 x 3)
(y + 3)(y + 3)


5

The formula for the area of a circle is which of the following?

24% Answer Correctly

c = π d

c = π r2

c = π r

c = π d2


Solution

The circumference of a circle is the distance around its perimeter and equals π (approx. 3.14159) x diameter: c = π d. The area of a circle is π x (radius)2 : a = π r2.