GRE Math : Basic Squaring / Square Roots

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Example Questions

Example Question #1 : Basic Squaring / Square Roots

Which quantity is greater? When 0<x<1

Quantity A

x^2

Quantity B

Possible Answers:

Quantity A is greater.

The relationship cannot be determined from the information given.

Quantity B is greater.

The two quantities are equal.

Correct answer:

Quantity B is greater.

Explanation:

X must equal a number between 0 and 1.

If you square any decimal number in this range you will get an answer less than 1. Try (0.999)^2 = 0.998.

Therefore Quantity B is always larger.

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Example Question #2 : Basic Squaring / Square Roots

Which is greater, when $\small \small \small 0<x<1?$

Quantity A

$\small x^2$

Quantity B

$\small \frac{x}{2}$

Possible Answers:

The two quantities are equal

Quantity A is greater

The relationship cannot be determined from the information given

Quantity B is greater

Correct answer:

The relationship cannot be determined from the information given

Explanation:

When $\small 0<x<1$ , $\small x$ must be a decimal. Therefore it will decrease when it is squared. To find this answer we can substitute two values for $\small x$ .

$\small x = 0.9$ and $\small x = 0.1$

When $\small x = 0.9$

$\small \small 0.9^2 = 0.81 > \frac{0.9}{2} = 0.45$

But when $\small x = 0.1$

$\small 0.1^2 = 0.01 < \frac{0.1}{2} = 0.05$

Therefore we cannot tell which is greater based on the information given.

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Example Question #3 : Basic Squaring / Square Roots

Which is greater, when $\small \small 0<x<1?$

Quantity A

$\small \small x$

Quantity B

$\small 2x^2$

Possible Answers:

Quantity B is greater

The two quantities are equal

Quantity A is greater

The relationship cannot be determined from the information given

Correct answer:

The relationship cannot be determined from the information given

Explanation:

When $\small 0<x<1$ , $\small x$ must be a decimal. Therefore it will decrease when it is squared. To find this answer we can substitute two values for $\small x$ .

$\small x = 0.9$ and $\small x = 0.1$

When $\small x = 0.9$

$\small \small \small 2\cdot0.9^2 = 1.62 > 0.9$

But when $\small x = 0.1$

$\small \small 2\cdot0.1^2 = 0.02 < 0.1$

Therefore we cannot tell which is greater based on the information given.

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Example Question #4 : Basic Squaring / Square Roots

Which is greater, when $\small -1<x<0?$

Quantity A

$\small \left | x\right |$

Quantity B

$\small x^2$

Possible Answers:

Quantity A is greater

Quantity B is greater

The two quantities are equal

The relationship cannot be determined from the information given

Correct answer:

Quantity A is greater

Explanation:

When $\small \small -1<x<0$ , $\small x$ must be a negative decimal. Therefore it will decrease when it is squared and become positive. The first quantity $\small \left | x\right |$ is the distance to 0 from $\small x$ , in other words the positive value of $\small x$ . The second quantity $\small x^2$ will always be positive, but because it is a decimal it will always be less than $\small x$ , if $\small x$ was a positive quantity. Since $\small \left | x\right |\equiv x$ , then $\small \left | x\right |$ wiill always be greater than $\small x^2$ .

Quantity A is greater.

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Example Question #5 : Basic Squaring / Square Roots

Which is greater, when $\small \small -1<x<0?$

Quantity A

$\small x^2$

Quantity B

$\small 1$

Possible Answers:

Quantity B is greater

Quantity A is greater

The relationship cannot be determined from the information given

The two quantities are equal

Correct answer:

Quantity B is greater

Explanation:

When $\small \small -1<x<0$ , $\small x$ must be a negative decimal. Therefore it will decrease when it is squared and become positive. Because $\small x$ must be a decimal, $\small x^2$ will also be a decimal. Therefore $\small x^2$ will never be greater than 1.

Quantity B is greater.

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Example Question #2 : Basic Squaring / Square Roots

Find the square root of the following decimal:

$\sqrt{.00081}=$

Possible Answers:

0.028

0.09

0.9

0.009

Correct answer:

0.028

Explanation:

The easiest way to find the square root of a fraction is to convert it into scientific notation.

$\dpi{100} \small .00081 = 8.1 \times 10^{-4}$

The key is that the exponent in scientific notation has to be even for a square root because the square root of an exponent is diving it by two. The square root of 9 is 3, so the square root of 8.1 is a little bit less than 3, around 2.8

$\dpi{100} \small \sqrt{8.1 \times 10^{-4}} \approx 2.8 \times 10^{-2} \approx 0.028$

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Example Question #7 : Basic Squaring / Square Roots

Find the square root of the following decimal:

$\small \sqrt{0.0049}$

Possible Answers:

$\small 0.07$

$\small 0.022$

$\small 0.7$

$\small 0.007$

Correct answer:

$\small 0.07$

Explanation:

To find the square root of this decimal we convert it into scientific notation.

$\small 0.0049 = 49 \cdot10^{-4}$

Because $\small 10^{-4}$ has an even exponent, we can divide the exponenet by 2 to get its square root.

$\small \sqrt{0.0049} = \sqrt{49}\cdot\sqrt{10^{-4}} = 7\cdot10^{-2} = 0.07$

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Example Question #8 : Basic Squaring / Square Roots

Find the square root of the following decimal:

$\small \sqrt{0.025}$

Possible Answers:

$\small 0.005$

$\small 0.05$

$\small 0.158$

$\small 0.625$

Correct answer:

$\small 0.158$

Explanation:

This problem can be solve more easily by rewriting the decimal into scientific notation.

$\small 0.025 = 2.5 \times 10^{-2}$

Because $\small 10^{-2}$ has an even exponent, we can take the square root of it by dividing it by 2. The square root of 4 is 2, and the square root of 1 is 1, so the square root of 2.5 is less than 2 and greater than 1.

$\small \sqrt{0.025} = \sqrt{2.5}\times \sqrt{10^{-2}} = 1.58\times 10^{-1} = 0.158$

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Example Question #9 : Basic Squaring / Square Roots

Find the square root of the following decimal:

$\small \sqrt{0.00036}$

Possible Answers:

$\small 0.006$

$\small 0.019$

$\small 0.013$

$\small 0.06$

Correct answer:

$\small 0.019$

Explanation:

This problem becomes much simpler if we rewrite the decimal in scientific notation

$\small 0.00036 = 3.6\times 10^{-4}$

Because $\small 10^{-4}$ has an even exponent, we can take its square root by dividing it by two. The square root of 4 is 2, and because 3.6 is a little smaller than 4, its square root is a little smaller than 2, around 1.9

$\small \sqrt{0.025} = \sqrt{3.6}\times \sqrt{10^{-4}} \approx 1.9\times 10^{-2} = 0.019$

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Example Question #21 : Decimals

Find the square root of the following decimal:

$\small \sqrt{0.00064}$

Possible Answers:

$\small 0.08$

$\small 0.0253$

$\small 0.008$

$\small 0.8$

Correct answer:

$\small 0.0253$

Explanation:

To find the square root of this decimal we convert it into scientific notation.

$\small \small 0.00064 = 6.4 \times10^{-4}$

Because $\small 10^{-4}$ has an even exponent, we can divide the exponenet by 2 to get its square root. The square root of 9 is 3, and the square root of 4 is two, so the square root of 6.4 is between 3 and 2, around 2.53

$\small \small \sqrt{0.00064} = \sqrt{6.4}\times\sqrt{10^{-4}} \approx 2.53 \times10^{-2} = 0.0253$

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GRE Math : Basic Squaring / Square Roots

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #1 : Basic Squaring / Square Roots

Example Question #2 : Basic Squaring / Square Roots

Example Question #3 : Basic Squaring / Square Roots

Example Question #4 : Basic Squaring / Square Roots

Example Question #5 : Basic Squaring / Square Roots

Example Question #2 : Basic Squaring / Square Roots

Example Question #7 : Basic Squaring / Square Roots

Example Question #8 : Basic Squaring / Square Roots

Example Question #9 : Basic Squaring / Square Roots

Example Question #21 : Decimals

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