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Is xy + xy < xy ?

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Intern
Joined: 28 Mar 2014
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Is xy + xy < xy ? [#permalink]  17 Jun 2014, 18:45
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Difficulty:

75% (hard)

Question Stats:

38% (02:15) correct 63% (01:10) wrong based on 32 sessions
Is xy + xy < xy ?

(1) x^2/y < 0

(2) x^9*(y^3)^3 < (x^2)^4*y^8
[Reveal] Spoiler: OA

Last edited by Bunuel on 18 Jun 2014, 00:04, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.
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Re: Is xy + xy < xy ? [#permalink]  18 Jun 2014, 00:17
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nitin1negi wrote:
Is xy + xy < xy ?

(1) x^2/y < 0

(2) x^9*(y^3)^3 < (x^2)^4*y^8

Is xy+xy<xy?

Is xy+xy<xy? --> cancel xy in both sides: is xy<0? So, the question basically asks whether x and y have the opposite signs.

(1) x^2/y<0. This statement implies that y is negative and x can be anything but zero. Not sufficient.

(2) x^9*(y^3)^3 < (x^2)^4*y^8 --> (xy)^9<(xy)^8. Since (xy)^8>0 (from the inequality we can deduce that neither of the unknowns is zero, thus even power of xy will ensure that it's positive), then we can safely reduce by it: xy<1 (xy\neq{0}). So, we cannot say whether xy<0. Not sufficient.

(1)+(2) y<0 and xy<1. If x=-\frac{1}{2} and y=-1, then the answer is NO but if x=1 and y=-1, then the answer is YES. Not sufficient.

Hope it's clear.

_________________
Re: Is xy + xy < xy ?   [#permalink] 18 Jun 2014, 00:17
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