ASVAB Math Knowledge Practice Test 26763 Results

Your Results Global Average
Questions 5 5
Correct 0 2.76
Score 0% 55%

Review

1

Solve for c:
-8c + 7 = 5 + 3c

59% Answer Correctly
3
-7
\(\frac{2}{11}\)
1\(\frac{1}{7}\)

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.

-8c + 7 = 5 + 3c
-8c = 5 + 3c - 7
-8c - 3c = 5 - 7
-11c = -2
c = \( \frac{-2}{-11} \)
c = \(\frac{2}{11}\)


2

A(n) __________ is to a parallelogram as a square is to a rectangle.

51% Answer Correctly

rhombus

triangle

quadrilateral

trapezoid


Solution

A rhombus is a parallelogram with four equal-length sides. A square is a rectangle with four equal-length sides.


3

If angle a = 34° and angle b = 22° what is the length of angle d?

56% Answer Correctly
135°
155°
146°
130°

Solution

An exterior angle of a triangle is equal to the sum of the two interior angles that are opposite:

d° = b° + c°

To find angle c, remember that the sum of the interior angles of a triangle is 180°:

180° = a° + b° + c°
c° = 180° - a° - b°
c° = 180° - 34° - 22° = 124°

So, d° = 22° + 124° = 146°

A shortcut to get this answer is to remember that angles around a line add up to 180°:

a° + d° = 180°
d° = 180° - a°
d° = 180° - 34° = 146°


4

Find the value of b:
3b + x = -6
-b + x = -8

42% Answer Correctly
\(\frac{15}{46}\)
\(\frac{1}{2}\)
-\(\frac{26}{29}\)

Solution

You need to find the value of b so solve the first equation in terms of x:

3b + x = -6
x = -6 - 3b

then substitute the result (-6 - 3b) into the second equation:

-b + 1(-6 - 3b) = -8
-b + (1 x -6) + (1 x -3b) = -8
-b - 6 - 3b = -8
-b - 3b = -8 + 6
-4b = -2
b = \( \frac{-2}{-4} \)
b = \(\frac{1}{2}\)


5

If the area of this square is 9, what is the length of one of the diagonals?

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

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 = \( \sqrt{a} \) = \( \sqrt{9} \) = 3

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 = 32 + 32
c2 = 18
c = \( \sqrt{18} \) = \( \sqrt{9 x 2} \) = \( \sqrt{9} \) \( \sqrt{2} \)
c = 3\( \sqrt{2} \)