ASVAB Arithmetic Reasoning Practice Test 336839 Results

Your Results Global Average
Questions 5 5
Correct 0 3.52
Score 0% 70%

Review

1

What is (y2)5?

79% Answer Correctly
y-3
y10
2y5
y7

Solution

To raise a term with an exponent to another exponent, retain the base and multiply the exponents:

(y2)5
y(2 * 5)
y10


2

Bob loaned Monica $700 at an annual interest rate of 1%. If no payments are made, what is the total amount owed at the end of the first year?

71% Answer Correctly
$763
$735
$707
$756

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 $700
i = 0.01 x $700

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

total = $700 + $7
total = $707


3

What is the next number in this sequence: 1, 9, 17, 25, 33, __________ ?

92% Answer Correctly
41
32
36
34

Solution

The equation for this sequence is:

an = an-1 + 8

where n is the term's order in the sequence, an is the value of the term, and an-1 is the value of the term before an. This makes the next number:

a6 = a5 + 8
a6 = 33 + 8
a6 = 41


4

A machine in a factory has an error rate of 3 parts per 100. The machine normally runs 24 hours a day and produces 10 parts per hour. Yesterday the machine was shut down for 5 hours for maintenance.

How many error-free parts did the machine produce yesterday?

49% Answer Correctly
184.3
117.2
89.3
117.6

Solution

The hourly error rate for this machine is the error rate in parts per 100 multiplied by the number of parts produced per hour:

\( \frac{3}{100} \) x 10 = \( \frac{3 \times 10}{100} \) = \( \frac{30}{100} \) = 0.3 errors per hour

So, in an average hour, the machine will produce 10 - 0.3 = 9.7 error free parts.

The machine ran for 24 - 5 = 19 hours yesterday so you would expect that 19 x 9.7 = 184.3 error free parts were produced yesterday.


5

What is \( \frac{5}{2} \) - \( \frac{8}{4} \)?

61% Answer Correctly
1 \( \frac{6}{14} \)
\(\frac{1}{2}\)
\( \frac{3}{7} \)
\( \frac{7}{15} \)

Solution

To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 2 are [2, 4, 6, 8, 10, 12, 14, 16, 18, 20] and the first few multiples of 4 are [4, 8, 12, 16, 20, 24, 28, 32, 36, 40]. The first few multiples they share are [4, 8, 12, 16, 20] making 4 the smallest multiple 2 and 4 share.

Next, convert the fractions so each denominator equals the lowest common multiple:

\( \frac{5 x 2}{2 x 2} \) - \( \frac{8 x 1}{4 x 1} \)

\( \frac{10}{4} \) - \( \frac{8}{4} \)

Now, because the fractions share a common denominator, you can subtract them:

\( \frac{10 - 8}{4} \) = \( \frac{2}{4} \) = \(\frac{1}{2}\)