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# Is xy + xy < xy ?

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Intern
Joined: 28 Mar 2014
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Location: India
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WE: Business Development (Retail Banking)
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Is xy + xy < xy ? [#permalink]

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17 Jun 2014, 19:45
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44% (02:45) correct 56% (01:16) wrong based on 61 sessions

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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, 01:04, 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]

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18 Jun 2014, 01:17
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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.

Answer: E.

Hope it's clear.

P.S. Please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html pay attention to rules, 2, 3, 5 and 7. Thank you.

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Re: Is xy + xy < xy ?   [#permalink] 18 Jun 2014, 01:17
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# Is xy + xy < xy ?

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