ASVAB Math Knowledge Practice Test 227987 Results

Your Results Global Average
Questions 5 5
Correct 0 2.48
Score 0% 50%

Review

1

If the base of this triangle is 5 and the height is 3, what is the area?

58% Answer Correctly
7\(\frac{1}{2}\)
67\(\frac{1}{2}\)
55
39

Solution

The area of a triangle is equal to ½ base x height:

a = ½bh
a = ½ x 5 x 3 = \( \frac{15}{2} \) = 7\(\frac{1}{2}\)


2

The endpoints of this line segment are at (-2, -4) and (2, -2). What is the slope-intercept equation for this line?

41% Answer Correctly
y = \(\frac{1}{2}\)x - 1
y = 2\(\frac{1}{2}\)x - 2
y = \(\frac{1}{2}\)x - 3
y = 2\(\frac{1}{2}\)x + 2

Solution

The slope-intercept equation for a line is y = mx + b where m is the slope and b is the y-intercept of the line. From the graph, you can see that the y-intercept (the y-value from the point where the line crosses the y-axis) is -3. The slope of this line is the change in y divided by the change in x. The endpoints of this line segment are at (-2, -4) and (2, -2) so the slope becomes:

m = \( \frac{\Delta y}{\Delta x} \) = \( \frac{(-2.0) - (-4.0)}{(2) - (-2)} \) = \( \frac{2}{4} \)
m = \(\frac{1}{2}\)

Plugging these values into the slope-intercept equation:

y = \(\frac{1}{2}\)x - 3


3

The dimensions of this cylinder are height (h) = 9 and radius (r) = 1. What is the volume?

63% Answer Correctly
18π
405π
150π

Solution

The volume of a cylinder is πr2h:

v = πr2h
v = π(12 x 9)
v = 9π


4

The dimensions of this cube are height (h) = 4, length (l) = 3, and width (w) = 3. What is the surface area?

51% Answer Correctly
76
66
118
96

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 3 x 3) + (2 x 3 x 4) + (2 x 3 x 4)
sa = (18) + (24) + (24)
sa = 66


5

Solve b - 3b = -2b + 2y - 2 for b in terms of y.

35% Answer Correctly
1\(\frac{1}{8}\)y + \(\frac{1}{4}\)
-\(\frac{9}{13}\)y + \(\frac{9}{13}\)
-2\(\frac{1}{3}\)y + 1\(\frac{1}{6}\)
1\(\frac{2}{3}\)y - \(\frac{2}{3}\)

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.

b - 3y = -2b + 2y - 2
b = -2b + 2y - 2 + 3y
b + 2b = 2y - 2 + 3y
3b = 5y - 2
b = \( \frac{5y - 2}{3} \)
b = \( \frac{5y}{3} \) + \( \frac{-2}{3} \)
b = 1\(\frac{2}{3}\)y - \(\frac{2}{3}\)