ASVAB Arithmetic Reasoning Practice Test 679937 Results

	Your Results	Global Average
Questions	5	5
Correct	0	3.12
Score	0%	62%

Review

Roger loaned April $1,000 at an annual interest rate of 8%. If no payments are made, what is the total amount owed at the end of the first year?

71% Answer Correctly

	$1,060
	$1,080
	$1,090
	$1,010

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 $1,000
i = 0.08 x $1,000

No payments were made so the total amount due is the original amount + the accumulated interest:

total = $1,000 + $80
total = $1,080

Find the average of the following numbers: 19, 11, 16, 14.

75% Answer Correctly

	13
	16
	15
	12

Solution

To find the average of these 4 numbers add them together then divide by 4:

$ \frac{19 + 11 + 16 + 14}{4} $ = $ \frac{60}{4} $ = 15

A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 20% off." If Alex buys two shirts, each with a regular price of $48, how much will he pay for both shirts?

57% Answer Correctly

	$57.60
	$67.20
	$9.60
	$86.40

Solution

By buying two shirts, Alex will save $48 x $ \frac{20}{100} $ = $ \frac{$48 x 20}{100} $ = $ \frac{$960}{100} $ = $9.60 on the second shirt.

So, his total cost will be
$48.00 + ($48.00 - $9.60)
$48.00 + $38.40
$86.40

What is 7c⁵ x 9c³?

75% Answer Correctly

	16c³
	63c³
	63c⁸
	16c⁵

Solution

To multiply terms with exponents, the base of both exponents must be the same. In this case they are so multiply the coefficients and add the exponents:

7c⁵ x 9c³
(7 x 9)c^{(5 + 3)}
63c⁸

What is $ 3 $$ \sqrt{8} $ + $ 7 $$ \sqrt{2} $

35% Answer Correctly

	21$ \sqrt{16} $
	13$ \sqrt{2} $
	10$ \sqrt{16} $
	10$ \sqrt{2} $

Solution

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

3$ \sqrt{8} $ + 7$ \sqrt{2} $
3$ \sqrt{4 \times 2} $ + 7$ \sqrt{2} $
3$ \sqrt{2^2 \times 2} $ + 7$ \sqrt{2} $
(3)(2)$ \sqrt{2} $ + 7$ \sqrt{2} $
6$ \sqrt{2} $ + 7$ \sqrt{2} $

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

6$ \sqrt{2} $ + 7$ \sqrt{2} $
(6 + 7)$ \sqrt{2} $
13$ \sqrt{2} $

	21\( \sqrt{16} \)
	13\( \sqrt{2} \)
	10\( \sqrt{16} \)
	10\( \sqrt{2} \)