ASVAB Math Knowledge Practice Test 901591 Results

Your Results Global Average
Questions 5 5
Correct 0 3.20
Score 0% 64%

Review

1

If side x = 10cm, side y = 15cm, and side z = 6cm what is the perimeter of this triangle?

84% Answer Correctly
33cm
21cm
40cm
31cm

Solution

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

p = x + y + z
p = 10cm + 15cm + 6cm = 31cm


2

To multiply binomials, use the FOIL method. Which of the following is not a part of the FOIL method?

83% Answer Correctly

Last

Odd

Inside

First


Solution

To multiply binomials, use the FOIL method. FOIL stands for First, Outside, Inside, Last and refers to the position of each term in the parentheses.


3

Solve 7b - 4b = -7b - 2x - 7 for b in terms of x.

34% Answer Correctly
2\(\frac{3}{4}\)x - 1\(\frac{1}{4}\)
2\(\frac{1}{2}\)x + 1
\(\frac{1}{7}\)x - \(\frac{1}{2}\)
-\(\frac{7}{13}\)x + \(\frac{8}{13}\)

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.

7b - 4x = -7b - 2x - 7
7b = -7b - 2x - 7 + 4x
7b + 7b = -2x - 7 + 4x
14b = 2x - 7
b = \( \frac{2x - 7}{14} \)
b = \( \frac{2x}{14} \) + \( \frac{-7}{14} \)
b = \(\frac{1}{7}\)x - \(\frac{1}{2}\)


4

If the area of this square is 4, what is the length of one of the diagonals?

68% Answer Correctly
7\( \sqrt{2} \)
4\( \sqrt{2} \)
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 = \( \sqrt{a} \) = \( \sqrt{4} \) = 2

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 = 22 + 22
c2 = 8
c = \( \sqrt{8} \) = \( \sqrt{4 x 2} \) = \( \sqrt{4} \) \( \sqrt{2} \)
c = 2\( \sqrt{2} \)


5

Solve for a:
a2 + 10a - 6 = 4a + 1

48% Answer Correctly
6 or -1
-1 or -7
3 or -4
1 or -7

Solution

The first step to solve a quadratic expression that's not set to zero is to solve the equation so that it is set to zero:

a2 + 10a - 6 = 4a + 1
a2 + 10a - 6 - 1 = 4a
a2 + 10a - 4a - 7 = 0
a2 + 6a - 7 = 0

Next, factor the quadratic equation:

a2 + 6a - 7 = 0
(a - 1)(a + 7) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (a - 1) or (a + 7) must equal zero:

If (a - 1) = 0, a must equal 1
If (a + 7) = 0, a must equal -7

So the solution is that a = 1 or -7