ASVAB Arithmetic Reasoning Practice Test 674054 Results

Your Results Global Average
Questions 5 5
Correct 0 3.22
Score 0% 64%

Review

1

If \( \left|y + 0\right| \) + 7 = 5, which of these is a possible value for y?

62% Answer Correctly
1
4
2
-11

Solution

First, solve for \( \left|y + 0\right| \):

\( \left|y + 0\right| \) + 7 = 5
\( \left|y + 0\right| \) = 5 - 7
\( \left|y + 0\right| \) = -2

The value inside the absolute value brackets can be either positive or negative so (y + 0) must equal - 2 or --2 for \( \left|y + 0\right| \) to equal -2:

y + 0 = -2
y = -2 + 0
y = -2
y + 0 = 2
y = 2 + 0
y = 2

So, y = 2 or y = -2.


2

What is -6b2 + 3b2?

66% Answer Correctly
-9b2
9b-2
-3b-4
-3b2

Solution

To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so add the coefficients and retain the base and exponent:

-6b2 + 3b2
(-6 + 3)b2
-3b2


3

Bob loaned Monty $200 at an annual interest rate of 6%. If no payments are made, what is the interest owed on this loan at the end of the first year?

74% Answer Correctly
$36
$30
$12
$18

Solution

The yearly interest charged on this loan is the annual interest rate multiplied by the amount borrowed:

interest = annual interest rate x loan amount

i = (\( \frac{6}{100} \)) x $200
i = 0.06 x $200
i = $12


4

Which of the following is a mixed number?

82% Answer Correctly

\({a \over 5} \)

\({5 \over 7} \)

\(1 {2 \over 5} \)

\({7 \over 5} \)


Solution

A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form \({a \over b}\) where a is the numerator and b is the denominator. An improper fraction (\({5 \over 3} \)) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number (\(1 {2 \over 3} \)) which has a whole number part and a fractional part.


5

What is \( 9 \)\( \sqrt{50} \) + \( 4 \)\( \sqrt{2} \)

35% Answer Correctly
36\( \sqrt{25} \)
49\( \sqrt{2} \)
13\( \sqrt{2} \)
13\( \sqrt{100} \)

Solution

To add these radicals together their radicands must be the same:

9\( \sqrt{50} \) + 4\( \sqrt{2} \)
9\( \sqrt{25 \times 2} \) + 4\( \sqrt{2} \)
9\( \sqrt{5^2 \times 2} \) + 4\( \sqrt{2} \)
(9)(5)\( \sqrt{2} \) + 4\( \sqrt{2} \)
45\( \sqrt{2} \) + 4\( \sqrt{2} \)

Now that the radicands are identical, you can add them together:

45\( \sqrt{2} \) + 4\( \sqrt{2} \)
(45 + 4)\( \sqrt{2} \)
49\( \sqrt{2} \)