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# x > y xy 0

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Retired Moderator
Joined: 07 Jun 2014
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GRE 1: Q167 V156
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Kudos [?]: 3036 [1] , given: 394

x > y xy 0 [#permalink]  20 May 2018, 16:56
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Question Stats:

61% (01:01) correct 38% (01:04) wrong based on 36 sessions
$$x > y$$
$$xy \neq 0$$

 Quantity A Quantity B $$\frac{x^2}{y+\frac{1}{y}}$$ $$\frac{y^2}{x+\frac{1}{x}}$$

A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Intern
Joined: 20 Mar 2018
Posts: 39
GRE 1: Q163 V149

GRE 2: Q168 V162
GPA: 3.5
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Kudos [?]: 40 [0], given: 8

Re: x > y xy 0 [#permalink]  20 May 2018, 22:57
We can spend a lot of time looking for an algebraic solution. However, the best strategy is to just plug in numbers.

Given: x>y
1) x = 3 and y = 2 gives quantity A greater than quantity B

2) x = 3 and y = -2 gives quantity B greater than quantity A

Hence, D
Re: x > y xy 0   [#permalink] 20 May 2018, 22:57
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# x > y xy 0

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