ASVAB Math Knowledge Practice Test 346956 Results

Your Results Global Average
Questions 5 5
Correct 0 3.29
Score 0% 66%

Review

1

A(n) __________ is two expressions separated by an equal sign.

76% Answer Correctly

formula

problem

expression

equation


Solution

An equation is two expressions separated by an equal sign. The key to solving equations is to 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.


2

Simplify (3a)(6ab) + (7a2)(3b).

65% Answer Correctly
39ab2
39a2b
-3a2b
90ab2

Solution

To multiply monomials, multiply the coefficients (the numbers that come before the variables) of each term, add the exponents of like variables, and multiply the different variables together.

(3a)(6ab) + (7a2)(3b)
(3 x 6)(a x a x b) + (7 x 3)(a2 x b)
(18)(a1+1 x b) + (21)(a2b)
18a2b + 21a2b
39a2b


3

This diagram represents two parallel lines with a transversal. If b° = 141, what is the value of y°?

72% Answer Correctly
163
21
141
147

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 b° = 141, the value of y° is 141.


4

If the area of this square is 1, what is the length of one of the diagonals?

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

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 = \( \sqrt{a} \) = \( \sqrt{1} \) = 1

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 = 12 + 12
c2 = 2
c = \( \sqrt{2} \)


5

Solve for y:
-2y + 1 = \( \frac{y}{2} \)

46% Answer Correctly
-7\(\frac{1}{5}\)
\(\frac{2}{5}\)
10\(\frac{4}{5}\)
-3\(\frac{1}{2}\)

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.

-2y + 1 = \( \frac{y}{2} \)
2 x (-2y + 1) = y
(2 x -2y) + (2 x 1) = y
-4y + 2 = y
-4y + 2 - y = 0
-4y - y = -2
-5y = -2
y = \( \frac{-2}{-5} \)
y = \(\frac{2}{5}\)