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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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New post 10 Aug 2009, 17:21
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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
2 KUDOS received
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Re: Is x > y ? [#permalink]

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New post 10 Aug 2009, 17:57
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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.
So answer is B
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Re: Is x > y ? [#permalink]

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

by squaring 'a' you are making it +ve ....
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Re: Is x > y ? [#permalink]

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New post 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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New post 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.

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Re: Is x > y ? [#permalink]

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New post 14 Apr 2012, 01:02
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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.

Image Posted from GMAT ToolKit


Welcome to GMAT Club. Below is an answer to your question.

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.

Answer: B.

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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New post 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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Re: Is x > y ? [#permalink]

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New post 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.

Image Posted from GMAT ToolKit


Welcome to GMAT Club. Below is an answer to your question.


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.

Answer: B.

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
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Re: Is x > y ? [#permalink]

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New post 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.

Image Posted from GMAT ToolKit


Welcome to GMAT Club. Below is an answer to your question.


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.

Answer: B.

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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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

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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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