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# Is x > y ? (1) ax > ay (2) a^2 * x > a^2 * y

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Is x > y ? (1) ax > ay (2) a^2 * x > a^2 * y [#permalink]

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10 Aug 2009, 17:21
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Question Stats:

67% (00:38) correct 33% (00:31) wrong based on 219 sessions

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Is x > y ?

(1) ax > ay
(2) a^2 * x > a^2 * y
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Apr 2012, 00:52, edited 1 time in total.
Edited the question and added the OA
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Re: Is x > y ? [#permalink]

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10 Aug 2009, 17:57
2
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1. ax > ay. The inequality between x and y depends on a, which can be either positive or negative.

2. a^2.x > a^2.y
Though a is positive or negative a^2 will always be positive.
So we can conclude x > y.
2 alone is sufficient, where as 1 alone is not sufficient.
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Re: Is x > y ? [#permalink]

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10 Aug 2009, 22:09
Is x > y ?

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

from 1

ax-ay>0

a(x-y)>0 we dont know whether a is -ve or +ve...insuff

from 2

a^2(x-y)>0 a^2 is +ve thus x-y is +ve thus x>y...suff

B
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Re: Is x > y ? [#permalink]

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10 Aug 2009, 22:14
B it is ...

by squaring 'a' you are making it +ve ....
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Joined: 30 Jan 2011
Posts: 18
Re: Is x > y ? [#permalink]

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11 Jul 2011, 16:02
Statement 1: If a is negative and we divide both side of the inequality by a, then the order of the inequality reverses: x < y
If a is positive --> x > y
==> INSUFF.

Statement 2: SUFF.
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Re: Is x > y ? [#permalink]

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13 Apr 2012, 19:52
AshForGMAT wrote:
B it is ...

by squaring 'a' you are making it +ve ....

What if a=sqrt(-1)?
a^2 will then be negative. I've seen another gmat question where it's tricky like that.

Posted from GMAT ToolKit
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Joined: 02 Sep 2009
Posts: 43853
Re: Is x > y ? [#permalink]

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14 Apr 2012, 01:02
Expert's post
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Avh86 wrote:
AshForGMAT wrote:
B it is ...

by squaring 'a' you are making it +ve ....

What if a=sqrt(-1)?
a^2 will then be negative. I've seen another gmat question where it's tricky like that.

Posted from GMAT ToolKit

First of all, the GMAT is dealing only with Real Numbers, so even roots from negative numbers are not defined. So, $$a^{even}\geq{0}$$.

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) $$a^2x > a^2y$$ --> $$a^2(x-y)>0$$ --> since $$a^2>0$$ ($$a^2$$ can not be zero, since the product is more than zero) we can safely reduce by it: $$x-y>0$$. Sufficient.

Hope it's clear.
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Re: Is x > y ? (1) ax > ay (2) a^2 * x > a^2 * y [#permalink]

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14 Apr 2012, 20:50
Jivana wrote:
Is x > y ?

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

Explanation:

1: AX > AY
no mention of whether A, X & Y are +ve or -ve....
so A = -ve
X = -ve
Y = -ve
two conditions arise
X< Y, AX> AY
X>Y, AX> Ay
hence Not sufficient

2. A2 * X > A2 * Y
so X> Y irrespective of a2 values...
hence SUFFICIENT
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Posts: 10
Re: Is x > y ? [#permalink]

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24 Oct 2013, 23:29
Bunuel wrote:
Avh86 wrote:
AshForGMAT wrote:
B it is ...

by squaring 'a' you are making it +ve ....

What if a=sqrt(-1)?
a^2 will then be negative. I've seen another gmat question where it's tricky like that.

Posted from GMAT ToolKit

First of all, the GMAT is dealing only with Real Numbers, so even roots from negative numbers are not defined. So, $$a^{even}\geq{0}$$.

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) $$a^2x > a^2y$$ --> $$a^2(x-y)>0$$ --> since $$a^2>0$$ ($$a^2$$ can not be zero, since the product is more than zero) we can safely reduce by it: $$x-y>0$$. Sufficient.

Hope it's clear.

Hi Bunuel

What if question is

Is x = y

A) ax = bx
B ) a^2x = a^2y

Would then the answer be D

Regards
Math Expert
Joined: 02 Sep 2009
Posts: 43853
Re: Is x > y ? [#permalink]

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25 Oct 2013, 00:55
DivyanshuRohatgi wrote:
Bunuel wrote:
Avh86 wrote:

What if a=sqrt(-1)?
a^2 will then be negative. I've seen another gmat question where it's tricky like that.

Posted from GMAT ToolKit

First of all, the GMAT is dealing only with Real Numbers, so even roots from negative numbers are not defined. So, $$a^{even}\geq{0}$$.

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) $$a^2x > a^2y$$ --> $$a^2(x-y)>0$$ --> since $$a^2>0$$ ($$a^2$$ can not be zero, since the product is more than zero) we can safely reduce by it: $$x-y>0$$. Sufficient.

Hope it's clear.

Hi Bunuel

What if question is

Is x = y

A) ax = bx
B ) a^2x = a^2y

Would then the answer be D

Regards

No, in this case the answer would be E. Consider: x=y=0 or a=b=0.
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Re: Is x > y ? (1) ax > ay (2) a^2 * x > a^2 * y [#permalink]

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08 Jan 2018, 01:01
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Re: Is x > y ? (1) ax > ay (2) a^2 * x > a^2 * y   [#permalink] 08 Jan 2018, 01:01
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