ASVAB Math Knowledge Practice Test 301766 Results

Your Results Global Average
Questions 5 5
Correct 0 2.84
Score 0% 57%

Review

1

On this circle, a line segment connecting point A to point D is called:

46% Answer Correctly

diameter

chord

radius

circumference


Solution

A circle is a figure in which each point around its perimeter is an equal distance from the center. The radius of a circle is the distance between the center and any point along its perimeter. A chord is a line segment that connects any two points along its perimeter. The diameter of a circle is the length of a chord that passes through the center of the circle and equals twice the circle's radius (2r).


2

Which of the following is not true about both rectangles and squares?

63% Answer Correctly

all interior angles are right angles

the lengths of all sides are equal

the area is length x width

the perimeter is the sum of the lengths of all four sides


Solution

A rectangle is a parallelogram containing four right angles. Opposite sides (a = c, b = d) are equal and the perimeter is the sum of the lengths of all sides (a + b + c + d) or, comonly, 2 x length x width. The area of a rectangle is length x width. A square is a rectangle with four equal length sides. The perimeter of a square is 4 x length of one side (4s) and the area is the length of one side squared (s2).


3

If side a = 2, side b = 3, what is the length of the hypotenuse of this right triangle?

63% Answer Correctly
\( \sqrt{82} \)
\( \sqrt{10} \)
\( \sqrt{52} \)
\( \sqrt{13} \)

Solution

According to the Pythagorean theorem, the hypotenuse squared is equal to the sum of the two perpendicular sides squared:

c2 = a2 + b2
c2 = 22 + 32
c2 = 4 + 9
c2 = 13
c = \( \sqrt{13} \)


4

If a = c = 1, b = d = 7, what is the area of this rectangle?

79% Answer Correctly
54
42
7
1

Solution

The area of a rectangle is equal to its length x width:

a = l x w
a = a x b
a = 1 x 7
a = 7


5

Solve 2a - 3a = 4a - 2z - 5 for a in terms of z.

34% Answer Correctly
-\(\frac{1}{2}\)z + 2\(\frac{1}{2}\)
-\(\frac{2}{7}\)z + 1
-3\(\frac{1}{2}\)z + \(\frac{1}{2}\)
z - 3

Solution

To solve this equation, isolate the variable for which you are solving (a) on one side of the equation and put everything else on the other side.

2a - 3z = 4a - 2z - 5
2a = 4a - 2z - 5 + 3z
2a - 4a = -2z - 5 + 3z
-2a = z - 5
a = \( \frac{z - 5}{-2} \)
a = \( \frac{z}{-2} \) + \( \frac{-5}{-2} \)
a = -\(\frac{1}{2}\)z + 2\(\frac{1}{2}\)