ASVAB Arithmetic Reasoning Practice Test 853845 Results

Your Results Global Average
Questions 5 5
Correct 0 3.23
Score 0% 65%

Review

1

What is \( \frac{10\sqrt{6}}{2\sqrt{2}} \)?

71% Answer Correctly
5 \( \sqrt{3} \)
5 \( \sqrt{\frac{1}{3}} \)
3 \( \sqrt{\frac{1}{5}} \)
\(\frac{1}{5}\) \( \sqrt{\frac{1}{3}} \)

Solution

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

\( \frac{10\sqrt{6}}{2\sqrt{2}} \)
\( \frac{10}{2} \) \( \sqrt{\frac{6}{2}} \)
5 \( \sqrt{3} \)


2

What is -3z3 - 6z3?

71% Answer Correctly
9z-3
3z3
3z-6
-9z3

Solution

To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so subtract the coefficients and retain the base and exponent:

-3z3 - 6z3
(-3 - 6)z3
-9z3


3

Solve for \( \frac{4!}{6!} \)

67% Answer Correctly
504
\( \frac{1}{30} \)
4
7

Solution

A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:

\( \frac{4!}{6!} \)
\( \frac{4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \)
\( \frac{1}{6 \times 5} \)
\( \frac{1}{30} \)


4

This property states taht the order of addition or multiplication does not mater. For example, 2 + 5 and 5 + 2 are equivalent.

60% Answer Correctly

distributive

associative

PEDMAS

commutative


Solution

The commutative property states that, when adding or multiplying numbers, the order in which they're added or multiplied does not matter. For example, 3 + 4 and 4 + 3 give the same result, as do 3 x 4 and 4 x 3.


5

\({b + c \over a} = {b \over a} + {c \over a}\) defines which of the following?

55% Answer Correctly

commutative property for division

distributive property for multiplication

distributive property for division

commutative property for multiplication


Solution

The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).