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

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Manager
Joined: 02 Sep 2008
Posts: 102
Is x > y ? (1) ax > ay (2) a^2x > a^2y  [#permalink]

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31 Mar 2009, 21:33
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Question Stats:

67% (00:13) correct 33% (00:00) wrong based on 11 sessions

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18. Is x > y ?

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

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Director
Joined: 04 Jan 2008
Posts: 821

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31 Mar 2009, 22:11
Its B

(2) says (a^2)(x-y)>0
=>(+ve)(x-y)>0

hence x-y=+ve
=>x>y--SUFF

(1) alone is NOT SUFF

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Current Student
Joined: 13 Jan 2009
Posts: 341
Location: India

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02 Apr 2009, 15:55
Agree B. Same reasoning.

St 1 a could be positive or negative.
VP
Joined: 28 Dec 2005
Posts: 1482

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02 Apr 2009, 17:35
B as well. From stat 1, you dont know sign of a, so you cant conclude if x>y. From stat 2, you have a^2, which is always positive, therefore, x must be greater than y
Director
Joined: 01 Apr 2008
Posts: 827
Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014

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02 Apr 2009, 20:52
Agree with B.
1) we cant divide by a on both sides because we don't know the sign of a. ..insuff
2) a^2 is always positive, so we can divide both sides by a^2 without reversing the inequality...suff
Manager
Joined: 25 Mar 2009
Posts: 52

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07 Apr 2009, 05:08
E

if a=0, then (1) and (2) are insufficient
Manager
Joined: 19 Aug 2006
Posts: 226

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07 Apr 2009, 10:19
Agree with B.
Same reasoning as the guys above with the same answer.
Intern
Joined: 29 Dec 2006
Posts: 31

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15 Apr 2009, 23:19
Mikko wrote:
E

if a=0, then (1) and (2) are insufficient

it can't be e, then conditions 1 and 2 wouldn't be true. they are always true
Manager
Joined: 22 Feb 2009
Posts: 129
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)

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17 Apr 2009, 10:30
Mikko wrote:
E

if a=0, then (1) and (2) are insufficient

a cannot be 0. B

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Re: S7-DS &nbs [#permalink] 17 Apr 2009, 10:30
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