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# If √x=y, which of the following co

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Joined: 04 Jan 2015
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If √x=y, which of the following co  [#permalink]

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24 May 2017, 12:06
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Question Stats:

69% (01:10) correct 31% (01:21) wrong based on 127 sessions

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Q.

If $$\sqrt{x}=y,$$ which of the following could be equal to $$\frac{1}{(x^{−2})^{−2}}$$

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

Thanks,
Saquib
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Re: If √x=y, which of the following co  [#permalink]

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24 May 2017, 12:07
Reserving this space to post the official solution.
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Re: If √x=y, which of the following co  [#permalink]

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24 May 2017, 13:09
$$\sqrt{x}$$ = y --- we can now square both side

$$x = y^2$$ --- square two more times on both sides

$$x^4 = y^8$$

$$\frac{1}{(x^{−2})^{−2}}$$ = $$\frac{1}{x^{4}}$$ = $$\frac{1}{y^{8}}$$ = $$y^{-8}$$

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If √x=y, which of the following co  [#permalink]

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24 May 2017, 22:28
1
EgmatQuantExpert wrote:
Q.

If $$\sqrt{x}=y,$$ which of the following could be equal to $$\frac{1}{(x^{−2})^{−2}}$$

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

Thanks,
Saquib
Quant Expert
e-GMAT

$$\sqrt{x}=y,$$
Squaring both sides; x = $$y^2$$

$$\frac{1}{(x^{−2})^{−2}}$$ = $$\frac{1}{(x^{4})}$$

Putting value of x in the fraction; $$\frac{1}{(x^{4})}$$

$$\frac{1}{(y^2)^{4}}$$ = $$\frac{1}{y^{8}}$$ = $$y^{-8}$$
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Re: If √x=y, which of the following co  [#permalink]

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31 May 2017, 06:52

Official Solution

Given:

• $$\sqrt{x}=y$$

Working Out:

• $$\frac{1}{(x^{−2})^{−2}}$$

• $$\frac{1}{x^4}$$

It is given that

• $$\sqrt{x}=y$$

Taking 8th power on both sides we get:

• $$x^4 = y^8$$

Therefore,

• $$\frac{1}{x^4} =\frac{1}{y^8} =y^{-8}$$

Thanks,
Saquib
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If √x=y, which of the following co  [#permalink]

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04 Jun 2017, 09:01
EgmatQuantExpert wrote:
Q.
If $$\sqrt{x}=y,$$ which of the following could be equal to $$\frac{1}{(x^{−2})^{−2}}$$

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

Thanks,
Saquib

Pick nice numbers, and y is positive. When the principal square root symbol is written in the problem, we need to consider the positive square root only.

$$\sqrt{x}=y$$, and $$\sqrt{4}= 2$$, so x = 4 and y = 2

$$\frac{1}{(x^{−2})^{−2}}$$ = $$\frac{1}{x^4}$$ ---> substitute 4 for x

= $$\frac{1}{4^4}$$

= $$\frac{1}{(2^{2})^{4}}$$

= $$\frac{1}{2^8}$$

= $$2^{-8}$$

Substitute y for 2 ---> $$y^{-8}$$

Shorten the math: $$\frac{1}{4^4}$$ = $$\frac{1}{256}$$, and $$\frac{1}{256}$$ = $$\frac{1}{2^8}$$, which is $$2^{-8}$$. Substitute y for 2 ---> $$y^{-8}$$
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If √x=y, which of the following co  [#permalink]

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04 Jun 2017, 09:44
EgmatQuantExpert wrote:
Q.

If $$\sqrt{x}=y,$$ which of the following could be equal to $$\frac{1}{(x^{−2})^{−2}}$$

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

$$\sqrt{x}=y,$$

Or, $$x = y^2$$

Quote:
which of the following could be equal to $$\frac{1}{(x^{−2})^{−2}}$$

$$\frac{1}{(x^{−2})^{−2}}$$

= $$\frac{1}{(x^{−4})}$$

= $$x^4$$

= $$y^{2*4}$$

= $$y^{8}$$

Hence, answer will be (A) $$y^{8}$$
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Re: If √x=y, which of the following co  [#permalink]

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10 Jun 2017, 06:52
1
I consider it very easy EGMAT is rarely kind for students
Time consumed:0:50(I cross check since was in doubt of EGMAT kindness)
√x=y
x^1/2=y
we have to make it x^4
So x^1/2*8=y^8
Multiply both sides with 8
Sp,
X^4=y^8
Since
1/(x^-2)^-2
1/x^4
put equivalent of x^4
it is
1/y^8
which is
y^-8
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Re: If √x=y, which of the following co  [#permalink]

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14 Feb 2019, 20:26

Kind regards!
Re: If √x=y, which of the following co   [#permalink] 14 Feb 2019, 20:26
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# If √x=y, which of the following co

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