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# Is x > y?

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Manager
Joined: 16 Feb 2012
Posts: 193
Concentration: Finance, Economics

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01 Jul 2012, 02:12
1
1
00:00

Difficulty:

35% (medium)

Question Stats:

66% (00:41) correct 34% (00:47) wrong based on 175 sessions

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Is x > y?

(1) ax > ay
(2) a^2x > a^2y

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Math Expert
Joined: 02 Sep 2009
Posts: 47946
Re: Is x > y?  [#permalink]

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01 Jul 2012, 02:34
2
Stiv wrote:
Is x > y?

(1) ax > ay
(2) a^2x > a^2y

Is x>y?

Question: is $$x-y> 0$$

(1) $$ax>ay$$ --> $$a(x-y)>0$$, two cases:
A. $$a>0$$ and $$x-y>0$$;
OR
B. $$a<0$$ and $$x-y<0$$;
Depending on $$a$$, $$x-y$$ may or may not be greater than zero. Not sufficient.

(2) $$a^2x > a^2y$$. Since $$a^2>0$$ ($$a^2$$ can not be zero, because if $$a=0$$, then the inequality won't hold true) we can safely reduce by it: $$x>y$$. Sufficient.

Hope it's clear.
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Re: Is x > y?  [#permalink]

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20 Sep 2016, 08:04
2
1
Stiv wrote:
Is x > y?

(1) ax > ay
(2) a^2x > a^2y

Responding to a pm:

Is x > y?

(1) ax > ay

We don't know whether a is positive or negative.

Say a is positive.
Divide both sides by a. You get x > y

Say a is negative.
Divide both sides by a. You get x < y (the inequality sign will flip)

Is x > y? We cannot say. Not sufficient.

(2) a^2x > a^2y
a^2 cannot be negative since its the square of a real number. a^2 must be positive only.
So when you divide both sides by a^2, you get x > y
Sufficient alone.

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Re: Is x > y?  [#permalink]

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26 Jan 2018, 13:29
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Re: Is x > y? &nbs [#permalink] 26 Jan 2018, 13:29
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