ASVAB Math Knowledge Practice Test 521301 Results

Your Results Global Average
Questions 5 5
Correct 0 2.89
Score 0% 58%

Review

1

This diagram represents two parallel lines with a transversal. If y° = 158, what is the value of z°?

73% Answer Correctly
22
154
20
163

Solution

For parallel lines with a transversal, the following relationships apply:

  • angles in the same position on different parallel lines equal each other (a° = w°, b° = x°, c° = z°, d° = y°)
  • alternate interior angles are equal (a° = z°, b° = y°, c° = w°, d° = x°)
  • all acute angles (a° = c° = w° = z°) and all obtuse angles (b° = d° = x° = y°) equal each other
  • same-side interior angles are supplementary and add up to 180° (e.g. a° + d° = 180°, d° + c° = 180°)

Applying these relationships starting with y° = 158, the value of z° is 22.


2

If angle a = 43° and angle b = 35° what is the length of angle c?

71% Answer Correctly
102°
70°
49°
97°

Solution

The sum of the interior angles of a triangle is 180°:
180° = a° + b° + c°
c° = 180° - a° - b°
c° = 180° - 43° - 35° = 102°


3

Solve for y:
-4y + 2 = \( \frac{y}{7} \)

46% Answer Correctly
\(\frac{14}{29}\)
\(\frac{3}{7}\)
\(\frac{8}{23}\)
-10\(\frac{1}{2}\)

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.

-4y + 2 = \( \frac{y}{7} \)
7 x (-4y + 2) = y
(7 x -4y) + (7 x 2) = y
-28y + 14 = y
-28y + 14 - y = 0
-28y - y = -14
-29y = -14
y = \( \frac{-14}{-29} \)
y = \(\frac{14}{29}\)


4

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

51% Answer Correctly

triangle

quadrilateral

rhombus

trapezoid


Solution

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


5

Solve for y:
y2 + 4y - 25 = 3y + 5

48% Answer Correctly
5 or -6
5 or -9
-3 or -3
7 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:

y2 + 4y - 25 = 3y + 5
y2 + 4y - 25 - 5 = 3y
y2 + 4y - 3y - 30 = 0
y2 + y - 30 = 0

Next, factor the quadratic equation:

y2 + y - 30 = 0
(y - 5)(y + 6) = 0

For this expression to be true, the left side of the expression must equal zero. Therefore, either (y - 5) or (y + 6) must equal zero:

If (y - 5) = 0, y must equal 5
If (y + 6) = 0, y must equal -6

So the solution is that y = 5 or -6