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Expert replies
by VJesus12 » Thu Nov 16, 2017 7:19 am
If $$x=\sqrt{y},$$ which of the following could be equal to $$\frac{1}{\left(x^{-2}\right)^{-2}}\ ?$$

$$A.\ y\ ^{-8}$$
$$B.\ y\ ^{-4}$$
$$C.\ y\ ^{-2}$$
$$D.\ \ y\ $$
$$E.\ \ y\ ^2$$

The OA is A .

How can I show that A is the correct answer? Can any Expert help me?
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Source: — Problem Solving |

by EconomistGMATTutor » Thu Nov 16, 2017 8:35 am
Hello Vjesus12.

Let's take a look at your question. $$If\ \ \ \sqrt{x}=y,\ then\ x=y^2\ and\ then\ x^4=\left(y^2\right)^4=y^8.$$ By the other hand, $$\left(x^{-2}\right)^{-2}=x^4.$$ So, $$\frac{1}{\left(x^{-2}\right)^{-2}}=\frac{1}{x^4}=\frac{1}{y^8}=y^{-8}.$$

So, the correct answer should be A .

I hope this explanation may help you.

Feel free to ask me again if you have a doubt.
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by regor60 » Thu Nov 16, 2017 9:12 am
EconomistGMATTutor wrote:Hello Vjesus12.

Let's take a look at your question. $$If\ \ \ \sqrt{x}=y,\ then\ x=y^2\ and\ then\ x^4=\left(y^2\right)^4=y^8.$$ By the other hand, $$\left(x^{-2}\right)^{-2}=x^4.$$ So, $$\frac{1}{\left(x^{-2}\right)^{-2}}=\frac{1}{x^4}=\frac{1}{y^8}=y^{-8}.$$

So, the correct answer should be A .

I hope this explanation may help you.

Feel free to ask me again if you have a doubt.
Some correction in order. The question states that X = Y^1/2.

So 1/X^4 = X^-4 = Y^-2 C
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by Brent@GMATPrepNow » Thu Nov 16, 2017 9:29 am
VJesus12 wrote:If $$x=\sqrt{y},$$ which of the following could be equal to $$\frac{1}{\left(x^{-2}\right)^{-2}}\ ?$$

$$A.\ y\ ^{-8}$$
$$B.\ y\ ^{-4}$$
$$C.\ y\ ^{-2}$$
$$D.\ \ y\ $$
$$E.\ \ y\ ^2$$

The OA is A .

How can I show that A is the correct answer? Can any Expert help me?
I'm not a big fan of this question.
The key word is COULD
Notice that the values x = 1 and y = 1 satisfy the condition that x = √y
Also notice that, when x = 1, 1/(x^-2)^-2 = 1
Also notice that, when y = 1, all 5 answer choices evaluate to be 1

So, all 5 answer choices COULD equal 1/(x^-2)^-2

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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