GRE Math : GRE Quantitative Reasoning

Study concepts, example questions & explanations for GRE Math

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Example Questions

Example Question #2 : Basic Squaring / Square Roots

Simplify.

\(\displaystyle \sqrt{624}\) 

Possible Answers:

\(\displaystyle 8\sqrt{39}\)

\(\displaystyle \sqrt{39}\)

\(\displaystyle 16\sqrt{39}\)

\(\displaystyle 4\sqrt{39}\)

\(\displaystyle 2\sqrt{39}\)

Correct answer:

\(\displaystyle 4\sqrt{39}\)

Explanation:

To simplify, we must try to find factors which are perfect squares. In this case 16 is a factor of 624 and is also a perfect square.

Therefore we can rewrite the square root of 624 as:

\(\displaystyle \sqrt{624}=\sqrt{16\times 39}=\sqrt{16}\sqrt{39}=4\sqrt{39}\)

Example Question #1 : Basic Squaring / Square Roots

Reduce \(\displaystyle \sqrt{400}\) to its simplest form. 

Possible Answers:

\(\displaystyle \sqrt{20}\)

\(\displaystyle 20\)

\(\displaystyle 2\sqrt{20}\)

\(\displaystyle \sqrt{200}\)

\(\displaystyle 4\sqrt2\)

Correct answer:

\(\displaystyle 20\)

Explanation:

To simplify, we must try to find factors which are perfect squares. In this case 20 is a factor of 400 and is also a perfect square.

Thus we can rewrite the problem as:

\(\displaystyle \sqrt{400}=\sqrt{20\times 20}=20\)

Note: \(\displaystyle 20^{2}=20\times20=400\)

Example Question #1 : How To Find The Common Factors Of Squares

Simplify.

\(\displaystyle \sqrt{720}\)

Possible Answers:

\(\displaystyle \sqrt{720}\)

\(\displaystyle 144\sqrt{5}\)

\(\displaystyle 12\sqrt{5}\)

\(\displaystyle 5\sqrt{144}\)

\(\displaystyle \sqrt{12}\)

Correct answer:

\(\displaystyle 12\sqrt{5}\)

Explanation:

Use the following steps to reduce this square root.

To simplify, we must try to find factors which are perfect squares. In this case 144 is a factor of 720 and is also a perfect square.

Thus we can rewrite the problem as follows.

\(\displaystyle \sqrt{720}=\sqrt{144\times 5}=\sqrt{144}\sqrt{5}=12\sqrt{5}\)

Example Question #4 : Arithmetic

Find the square root of \(\displaystyle 1,800\).

Possible Answers:

\(\displaystyle 60\)

\(\displaystyle 30\sqrt{2}\)

\(\displaystyle 900\)

\(\displaystyle \sqrt{32}\)

\(\displaystyle 2\sqrt{60}\)

Correct answer:

\(\displaystyle 30\sqrt{2}\)

Explanation:

Use the following steps to find the square root of \(\displaystyle 1,800:\)

To simplify, we must try to find factors which are perfect squares. In this case 900 is a factor of 1800 and is also a perfect square.

Thus we can rewrite the problem as follows.

\(\displaystyle \sqrt{1,800}=\sqrt{900\times 2}=\sqrt{900}\sqrt{2}=30\sqrt{2}\)

Example Question #1 : Arithmetic

Simplify. 

\(\displaystyle \sqrt{54}\)

Possible Answers:

\(\displaystyle 3\sqrt{6}\)

\(\displaystyle 2\sqrt{6}\)

\(\displaystyle \sqrt{3}\)

\(\displaystyle 6\sqrt{6}\)

\(\displaystyle 9\sqrt{6}\)

Correct answer:

\(\displaystyle 3\sqrt{6}\)

Explanation:

To simplify, we must try to find factors which are perfect squares. In this case 9 is a factor of 54 and is also a perfect square.

To reduce this expression, use the following steps: 

\(\displaystyle \sqrt{54}=\sqrt{9\times 6}=\sqrt{9}\sqrt{6}=3\sqrt{6}\)

Example Question #1 : How To Find The Common Factors Of Squares

Reduce.

\(\displaystyle \sqrt{72}\)

Possible Answers:

\(\displaystyle 2\sqrt{6}\)

\(\displaystyle 36\sqrt{2}\)

\(\displaystyle 4\sqrt{36}\)

\(\displaystyle 6\sqrt{2}\)

\(\displaystyle 2\sqrt{21}\)

Correct answer:

\(\displaystyle 6\sqrt{2}\)

Explanation:

To simplify, we must try to find factors which are perfect squares. In this case 36 is a factor of 72 and is also a perfect square.

To reduce this expression, use the following arithmetic steps: 

\(\displaystyle \sqrt{72}=\sqrt{36\times 2}=\sqrt{36}\sqrt{2}=6\sqrt{2}\)

Example Question #1 : How To Find The Common Factors Of Squares

Which quantity is greater: \(\displaystyle \sqrt{900}\) or \(\displaystyle 30\)?

Possible Answers:

Not enough information to determine the relationship between these two quantities. 

\(\displaystyle 30>\sqrt{900}\)

\(\displaystyle 30< \sqrt{900}\)

\(\displaystyle \sqrt{900}=30\)

\(\displaystyle \sqrt{900}=\pm30\) 

Correct answer:

\(\displaystyle \sqrt{900}=\pm30\) 

Explanation:

To simplify, we must try to find factors which are perfect squares. In this case 30 is a factor of 900 and is also a perfect square.

The square root of \(\displaystyle 900\) is equal to: 

\(\displaystyle \sqrt{900}=\sqrt{30\times 30}=30\)

However,

\(\displaystyle \sqrt{900}=\sqrt{-30\times -30}=-30\)

Thus, \(\displaystyle \pm30\) \(\displaystyle =\sqrt{900}\)

Example Question #11 : Arithmetic

Reduce. 

\(\displaystyle \sqrt{32}\)

Possible Answers:

\(\displaystyle 8\sqrt{2}\)

\(\displaystyle 4\sqrt{2}\)

\(\displaystyle 16\sqrt{2}\)

\(\displaystyle \sqrt{4}\)

\(\displaystyle 2\sqrt{2}\)

Correct answer:

\(\displaystyle 4\sqrt{2}\)

Explanation:

To simplify, we must try to find factors which are perfect squares. In this case 16 is a factor of 32 and is also a perfect square.

To reduce this expression, use the following steps:

\(\displaystyle \sqrt{32}=\sqrt{16\times 2}=\sqrt{16}\sqrt{2}=4\sqrt{2}\)

Example Question #11 : How To Find The Common Factors Of Squares

Find the square root of \(\displaystyle 164\).

Possible Answers:

\(\displaystyle 4\sqrt{41}\)

\(\displaystyle \sqrt{43}\)

\(\displaystyle 2\sqrt{41}\)

\(\displaystyle 8\sqrt{2}\)

\(\displaystyle 2\sqrt{43}\)

Correct answer:

\(\displaystyle 2\sqrt{41}\)

Explanation:

To simplify, we must try to find factors which are perfect squares. In this case 4 is a factor of 164 and is also a perfect square.

To find the square root of \(\displaystyle 164\), use the following steps:

\(\displaystyle \sqrt{164}=\sqrt{4\times 41}=\sqrt{41}\sqrt{4}=2\sqrt{41}\)

Example Question #11 : Arithmetic

Reduce. 

\(\displaystyle \sqrt{192}\)

Possible Answers:

\(\displaystyle 8\sqrt{3}\)

\(\displaystyle 3\sqrt{4}\)

\(\displaystyle 4\sqrt{3}\)

\(\displaystyle 64\sqrt{3}\)

\(\displaystyle 64\sqrt{4}\)

Correct answer:

\(\displaystyle 8\sqrt{3}\)

Explanation:

Use the following arithmetic steps to reduce \(\displaystyle \sqrt{192}\).

To simplify, we must try to find factors which are perfect squares. In this case 64 is a factor of 192 and is also a perfect square.

Note \(\displaystyle 64\) and \(\displaystyle 3\) are both factors of \(\displaystyle 192\), however only \(\displaystyle \sqrt{64}\) can be reduced.


\(\displaystyle \sqrt{192}=\sqrt{64\times 3}=\sqrt{64}\sqrt{3}=8\sqrt{3}\)

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