ASVAB Arithmetic Reasoning Practice Test 124574 Results

Your Results Global Average
Questions 5 5
Correct 0 2.76
Score 0% 55%

Review

1

A circular logo is enlarged to fit the lid of a jar. The new diameter is 70% larger than the original. By what percentage has the area of the logo increased?

51% Answer Correctly
15%
35%
17\(\frac{1}{2}\)%
32\(\frac{1}{2}\)%

Solution

The area of a circle is given by the formula A = πr2 where r is the radius of the circle. The radius of a circle is its diameter divided by two so A = π(\( \frac{d}{2} \))2. If the diameter of the logo increases by 70% the radius (and, consequently, the total area) increases by \( \frac{70\text{%}}{2} \) = 35%


2

Solve 3 + (4 + 3) ÷ 4 x 5 - 22

52% Answer Correctly
7\(\frac{3}{4}\)
\(\frac{5}{7}\)
1\(\frac{3}{5}\)
\(\frac{3}{8}\)

Solution

Use PEMDAS (Parentheses, Exponents, Multipy/Divide, Add/Subtract):

3 + (4 + 3) ÷ 4 x 5 - 22
P: 3 + (7) ÷ 4 x 5 - 22
E: 3 + 7 ÷ 4 x 5 - 4
MD: 3 + \( \frac{7}{4} \) x 5 - 4
MD: 3 + \( \frac{35}{4} \) - 4
AS: \( \frac{12}{4} \) + \( \frac{35}{4} \) - 4
AS: \( \frac{47}{4} \) - 4
AS: \( \frac{47 - 16}{4} \)
\( \frac{31}{4} \)
7\(\frac{3}{4}\)


3

What is 8\( \sqrt{9} \) x 8\( \sqrt{6} \)?

41% Answer Correctly
192\( \sqrt{6} \)
16\( \sqrt{6} \)
64\( \sqrt{6} \)
16\( \sqrt{9} \)

Solution

To multiply terms with radicals, multiply the coefficients and radicands separately:

8\( \sqrt{9} \) x 8\( \sqrt{6} \)
(8 x 8)\( \sqrt{9 \times 6} \)
64\( \sqrt{54} \)

Now we need to simplify the radical:

64\( \sqrt{54} \)
64\( \sqrt{6 \times 9} \)
64\( \sqrt{6 \times 3^2} \)
(64)(3)\( \sqrt{6} \)
192\( \sqrt{6} \)


4

Which of the following is not an integer?

77% Answer Correctly

1

-1

0

\({1 \over 2}\)


Solution

An integer is any whole number, including zero. An integer can be either positive or negative. Examples include -77, -1, 0, 55, 119.


5

How many 2 gallon cans worth of fuel would you need to pour into an empty 20 gallon tank to fill it exactly halfway?

52% Answer Correctly
9
10
2
5

Solution

To fill a 20 gallon tank exactly halfway you'll need 10 gallons of fuel. Each fuel can holds 2 gallons so:

cans = \( \frac{10 \text{ gallons}}{2 \text{ gallons}} \) = 5