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If x > 0 and y > 0

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Intern
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If x > 0 and y > 0 [#permalink] New post 06 Jan 2017, 08:12
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Question Stats:

85% (00:36) correct 14% (01:41) wrong based on 21 sessions
If \(x >\) 0 and \(y > 0\), which of the following os equivalent to \(\frac{x}{y}\sqrt{\frac{y}{x^2}}\)


A. \(1\)

B. \(\frac{\sqrt{x}}{\sqrt{y}}\)

C. \(\sqrt{x}\)

D. \(\frac{1}{\sqrt{x}}\)

E. \(\frac{1}{\sqrt{y}}\)
[Reveal] Spoiler: OA
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Joined: 07 Jun 2014
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GRE 1: Q167 V156
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Re: If x > 0 and [#permalink] New post 06 Jan 2017, 16:14
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Expert's post
Hey,

This question is pretty straight forward.
We have

\(\frac{x}{y}\sqrt{\frac{y}{x^2}}\)

rewritten as

\(\sqrt{\frac{x}{y}} \times \sqrt{\frac{x}{y}} \times \sqrt{\frac{y}{x}}\times\sqrt{\frac{1}{x}}\)

cancelling out \(\sqrt{\frac{x}{y}}\) and \(\sqrt{\frac{y}{x}}\)

As x > 0 and y > 0 \(\sqrt{x}\) and \(\sqrt{y}\) are defined and \(\frac{x}{y}\) and \(\frac{y}{x}\) are also defined.

\(\sqrt{\frac{x}{y}} \times \sqrt{\frac{1}{x}}\)

cancelling out \(\sqrt{x}\)

\(\sqrt{\frac{1}{y}}\)

Hence E is the correct answer.
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Re: If x > 0 and [#permalink] New post 29 Mar 2019, 06:33
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Expert's post
HarveyKlaus wrote:
I got the answer right but Im not sure regarding the method I used. Any ideas how to solve this?

Thanks


Another approach is to choose some x- and y-vales to evaluate the original expression, and then see which answer choice evaluates to the same value.

For example, if x = 2 and y = 4, then \(\frac{x}{y}\sqrt{\frac{y}{x^2}}=\frac{2}{4}\sqrt{\frac{4}{2^2}}=\frac{1}{2}\sqrt{1}=\frac{1}{2}\)

So, when x = 2 and y = 4, the original expression evaluates to be 1/2
So, the CORRECT answer must also evaluate to be 1/2, when x = 2 and y = 4

Let's take each answer choice and plug in x = 2 and y = 4

A) 1
Doesn't equal 1/2
ELIMINATE

B) (√x)/(√y) = (√2)/(√4) ≈ 1.4/2
Doesn't equal 1/2
ELIMINATE

C) √x = √2 ≈ 1.4
Doesn't equal 1/2
ELIMINATE

D) 1/√x = 1/√2 ≈ 1/1.4
Doesn't equal 1/2
ELIMINATE

E) 1/√y = 1/√4 = 1/2
VOILA!!

Answer: E

Cheers,
Brent
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Brent Hanneson – Creator of greenlighttestprep.com
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Re: If x > 0 and   [#permalink] 29 Mar 2019, 06:33
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