ASVAB Math Knowledge Practice Test 140644 Results

Your Results Global Average
Questions 5 5
Correct 0 3.09
Score 0% 62%

Review

1

Which of the following is not required to define the slope-intercept equation for a line?

42% Answer Correctly

slope

x-intercept

y-intercept

\({\Delta y \over \Delta x}\)


Solution

A line on the coordinate grid can be defined by a slope-intercept equation: y = mx + b. For a given value of x, the value of y can be determined given the slope (m) and y-intercept (b) of the line. The slope of a line is change in y over change in x, \({\Delta y \over \Delta x}\), and the y-intercept is the y-coordinate where the line crosses the vertical y-axis.


2

What is the area of a circle with a radius of 4?

70% Answer Correctly
16π
36π

Solution

The formula for area is πr2:

a = πr2
a = π(42)
a = 16π


3

Which of the following statements about math operations is incorrect?

71% Answer Correctly

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

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

all of these statements are correct

you can multiply monomials that have different variables and different exponents


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.


4

Solve for x:
5x + 1 = -9 + x

59% Answer Correctly
-2\(\frac{1}{2}\)
-\(\frac{3}{4}\)
-1
1\(\frac{1}{6}\)

Solution

To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the equal sign and the answer on the other.

5x + 1 = -9 + x
5x = -9 + x - 1
5x - x = -9 - 1
4x = -10
x = \( \frac{-10}{4} \)
x = -2\(\frac{1}{2}\)


5

If the area of this square is 25, what is the length of one of the diagonals?

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

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 = \( \sqrt{a} \) = \( \sqrt{25} \) = 5

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 = 52 + 52
c2 = 50
c = \( \sqrt{50} \) = \( \sqrt{25 x 2} \) = \( \sqrt{25} \) \( \sqrt{2} \)
c = 5\( \sqrt{2} \)