ASVAB Arithmetic Reasoning Practice Test 217752 Results

Your Results Global Average
Questions 5 5
Correct 0 2.78
Score 0% 56%

Review

1

What is the least common multiple of 8 and 16?

72% Answer Correctly
16
45
32
79

Solution

The first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80] and the first few multiples of 16 are [16, 32, 48, 64, 80, 96]. The first few multiples they share are [16, 32, 48, 64, 80] making 16 the smallest multiple 8 and 16 have in common.


2

What is 8\( \sqrt{7} \) x 8\( \sqrt{3} \)?

41% Answer Correctly
64\( \sqrt{3} \)
64\( \sqrt{21} \)
16\( \sqrt{3} \)
16\( \sqrt{21} \)

Solution

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

8\( \sqrt{7} \) x 8\( \sqrt{3} \)
(8 x 8)\( \sqrt{7 \times 3} \)
64\( \sqrt{21} \)


3

What is \( \frac{1y^7}{7y^4} \)?

60% Answer Correctly
7y3
7y-3
\(\frac{1}{7}\)y1\(\frac{3}{4}\)
\(\frac{1}{7}\)y3

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:

\( \frac{y^7}{7y^4} \)
\( \frac{1}{7} \) y(7 - 4)
\(\frac{1}{7}\)y3


4

A menswear store is having a sale: "Buy one shirt at full price and get another shirt for 35% off." If Charlie buys two shirts, each with a regular price of $22, how much will he pay for both shirts?

57% Answer Correctly
$7.70
$25.30
$36.30
$27.50

Solution

By buying two shirts, Charlie will save $22 x \( \frac{35}{100} \) = \( \frac{$22 x 35}{100} \) = \( \frac{$770}{100} \) = $7.70 on the second shirt.

So, his total cost will be
$22.00 + ($22.00 - $7.70)
$22.00 + $14.30
$36.30


5

If a rectangle is twice as long as it is wide and has a perimeter of 30 meters, what is the area of the rectangle?

47% Answer Correctly
98 m2
50 m2
2 m2
8 m2

Solution

The area of a rectangle is width (w) x height (h). In this problem we know that the rectangle is twice as long as it is wide so h = 2w. The perimeter of a rectangle is 2w + 2h and we know that the perimeter of this rectangle is 30 meters so the equation becomes: 2w + 2h = 30.

Putting these two equations together and solving for width (w):

2w + 2h = 30
w + h = \( \frac{30}{2} \)
w + h = 15
w = 15 - h

From the question we know that h = 2w so substituting 2w for h gives us:

w = 15 - 2w
3w = 15
w = \( \frac{15}{3} \)
w = 5

Since h = 2w that makes h = (2 x 5) = 10 and the area = h x w = 5 x 10 = 50 m2