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

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

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31 Mar 2009, 20:33
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18. Is x > y ?

(1) ax > ay
(2)$$a^2x > a^2y$$
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31 Mar 2009, 21:11
Its B

in this kind of Q, start with the easier condition
(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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02 Apr 2009, 14:55
Agree B. Same reasoning.

St 1 a could be positive or negative.
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02 Apr 2009, 16:35
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
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02 Apr 2009, 19:52
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
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07 Apr 2009, 04:08
E

if a=0, then (1) and (2) are insufficient
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07 Apr 2009, 09:19
Agree with B.
Same reasoning as the guys above with the same answer.
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15 Apr 2009, 22:19
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
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17 Apr 2009, 09:30
Mikko wrote:
E

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

a cannot be 0. B

B is the answer
Re: S7-DS   [#permalink] 17 Apr 2009, 09:30
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# Is x > y ? (1) ax > ay (2) a^2x > a^2y

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