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Manager
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Is x > y ? (1) ax > ay (2) a^2x > a^2y [#permalink]
 31 Mar 2009, 21:33
Question Stats:
50% (01:27) correct
50% (00:00) wrong based on 1 sessions
18. Is x > y ?
(1) ax > ay
(2)a^2x > a^2y
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Director
Joined: 04 Jan 2008
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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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Current Student
Joined: 13 Jan 2009
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Agree B. Same reasoning.
St 1 a could be positive or negative.
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SVP
Joined: 28 Dec 2005
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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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Director
Joined: 01 Apr 2008
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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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Manager
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E
if a=0, then (1) and (2) are insufficient
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Senior Manager
Joined: 19 Aug 2006
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Agree with B.
Same reasoning as the guys above with the same answer.
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Intern
Joined: 29 Dec 2006
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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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Manager
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Mikko wrote: E
if a=0, then (1) and (2) are insufficient a cannot be 0. B B is the answer
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