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Is x > y?

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Joined: 16 Feb 2012
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Is x > y?  [#permalink]

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New post 01 Jul 2012, 02:12
2
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A
B
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D
E

Difficulty:

  25% (medium)

Question Stats:

67% (01:01) correct 33% (01:22) wrong based on 175 sessions

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Is x > y?

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

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Re: Is x > y?  [#permalink]

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New post 01 Jul 2012, 02:34
2
1
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.

Answer: B.

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

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New post 20 Sep 2016, 08:04
2
2
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.

Answer (B)
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Re: Is x > y?  [#permalink]

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New post 10 Feb 2019, 00:21
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Re: Is x > y?   [#permalink] 10 Feb 2019, 00:21
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