ASVAB Arithmetic Reasoning Practice Test 591518 Results

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

Review

The total water usage for a city is 25,000 gallons each day. Of that total, 19% is for personal use and 34% is for industrial use. How many more gallons of water each day is consumed for industrial use over personal use?

58% Answer Correctly

	3,750
	11,200
	1,050
	3,850

Solution

34% of the water consumption is industrial use and 19% is personal use so (34% - 19%) = 15% more water is used for industrial purposes. 25,000 gallons are consumed daily so industry consumes $ \frac{15}{100} $ x 25,000 gallons = 3,750 gallons.

What is $ \frac{4}{5} $ ÷ $ \frac{2}{6} $?

68% Answer Correctly

	$\frac{1}{54}$
	$\frac{1}{21}$
	$\frac{2}{81}$
	2$\frac{2}{5}$

Solution

To divide fractions, invert the second fraction and then multiply:

$ \frac{4}{5} $ ÷ $ \frac{2}{6} $ = $ \frac{4}{5} $ x $ \frac{6}{2} $

To multiply fractions, multiply the numerators together and then multiply the denominators together:

$ \frac{4}{5} $ x $ \frac{6}{2} $ = $ \frac{4 x 6}{5 x 2} $ = $ \frac{24}{10} $ = 2$\frac{2}{5}$

Bob loaned Damon $600 at an annual interest rate of 5%. If no payments are made, what is the interest owed on this loan at the end of the first year?

74% Answer Correctly

	$4
	$30
	$120
	$24

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 $600
i = 0.05 x $600
i = $30

What is $ \frac{3b^5}{7b^2} $?

60% Answer Correctly

	$\frac{3}{7}$b³
	2$\frac{1}{3}$b⁷
	$\frac{3}{7}$b⁷
	2$\frac{1}{3}$b^-3

Solution

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

$ \frac{3b^5}{7b^2} $
$ \frac{3}{7} $ b^{(5 - 2)}
$\frac{3}{7}$b³

How many 1 gallon cans worth of fuel would you need to pour into an empty 6 gallon tank to fill it exactly halfway?

52% Answer Correctly

	2
	6
	3
	8

Solution

To fill a 6 gallon tank exactly halfway you'll need 3 gallons of fuel. Each fuel can holds 1 gallons so:

cans = $ \frac{3 \text{ gallons}}{1 \text{ gallons}} $ = 3

	\(\frac{1}{54}\)
	\(\frac{1}{21}\)
	\(\frac{2}{81}\)
	2\(\frac{2}{5}\)

	\(\frac{3}{7}\)b³
	2\(\frac{1}{3}\)b⁷
	\(\frac{3}{7}\)b⁷
	2\(\frac{1}{3}\)b^-3