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# Is xy + xy < xy ?

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Is xy + xy < xy ? [#permalink]  02 Mar 2013, 12:03
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Question Stats:

45% (02:43) correct 55% (01:22) wrong based on 138 sessions
Is xy + xy < xy ?

(1) x^2/y < 0
(2) x^3y^3 < (xy)^2
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jul 2014, 08:55, edited 2 times in total.
Edited the question
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Re: Is xy + xy < xy ? [#permalink]  02 Mar 2013, 13:54
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Hello Mun23,

Let me try answering this question for you. This question asks us whether 2xy<xy. Now the question we need to ask ourselves is when would this turn out to be true?

If x and y are both positive, then 2xy>xy.
If x or y =0, then 2xy=xy.
If either x or y is negative and the other is positive, then 2xy<xy. In short xy needs to be <0 or negative.

So, we need to find whether exactly one of those variables is positive and the other negative.

Now, 1 gives us x^2/y<0. Since, x^2 is always positive, this tells us that y is negative. However, we have no idea about the actual value of x.
x^2=some positive number
x could be a positive number or negative number. However, it cannot be 0. Hence, INSUFFICIENT.
Statement 2 gives us x^3*y^3<(xy)^2
Since (xy)^2 is always positive, we can divide the inequality by (xy)^2 without changing the sign.

Hence, (xy)<1. However, we still don't know their exact value. If xy<0, then one of the variables is positive and the other negative and this would be sufficient. However, if x or y=0, even then x*y<1. This does not give us conclusive proof that one of the variables is positive and the other negative(xy is negative). Hence, INSUFFICIENT.

From statement 1 we know that y<0. However, x can be negative or positive. However, x cannot be 0 as then x^2/y=0.
From statement 2 we know that x*y<1. We also know from 1) that y is negative. However, x could still be negative or positive.

Now, what if x=1/3 and y=-1, x^2/y=-1/9<0 and xy=-1/3<1. Is xy negative? Yes
What if x=-1/3 and y=-1. x^2/y=-1/9. However, xy=1/3<1. iS xy negative? No

Hence, INSUFFICIENT and the answer is E.

Hope this helps! Let me know if I can help you any further.

mun23 wrote:
Is xy+xy<xy?
(1)x^2/y<0
(2)X^3y^3<(xy)^2

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Re: Is xy + xy < xy ? [#permalink]  02 Mar 2013, 20:07
mun23 wrote:
Is xy+xy<xy?
(1)x^2/y<0
(2)X^3y^3<(xy)^2

it's an easy question provided basics of inequalities are clear.

question is asking for xy +xy <xy i.e. 2xy <xy i.e. xy < 0
this is possible if a) x is negative and y is positive or b) x is positive and y is negative.

from 1) x^2/y < 0 , as square of something is always positive hence x^2 is +ve , so y has to be -ve
....nothing can be concluded about xy ....hence insufficient
from 2) x^3y^3 < x^2y^2 i.e. x^3y^3 - x^2y^2 < 0
i.e. x^2y^2(xy - 1) < 0 as square of something is always +ve
hence xy - 1 < 0 i.e. xy < 1..............not sufficient as xy can be +ve or -ve
now two situation possible
a) if y +ve then xy < 1
=> x < 1/y ............since y is +ve this means x < +ve no. e.g 0.2 < 1 and -2 < 1
hence x can be +ve or -ve

b) if y -ve then xy <1
=> x >1/y (inequality will change)
since y is -ve this means x > some -ve no. e.g. 0.2 > -1 and -0.2 > -1
hence x can be +ve or -ve.

using both Conditions 1 and 2
i.e. y is -ve....from 1
and case b holds from 2
still x can be +ve or -ve
hence xy can be +ve or -ve

hence Option (E)
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Re: Is xy + xy < xy ? [#permalink]  04 Mar 2013, 02:58
Expert's post
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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^3y^3<(xy)^2 --> $$x^2y^2(xy-1)<0$$ --> this statement implies that $$xy\neq{0}$$ and $$xy<1$$. 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 asnwer is YES. Not sufficient.

Hope it's clear.
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Re: Is xy + xy < xy ? [#permalink]  04 Mar 2013, 03:07
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Expert's post
Kris01 wrote:

Now, 1 gives us x^2/y<0. Since, x^2 is always positive...

Statement 2 gives us x^3*y^3<(xy)^2
Since (xy)^2 is always positive, we can divide the inequality by (xy)^2

jbisht wrote:
from 1) x^2/y < 0 , as square of something is always positive hence x^2 is +ve , so y has to be -ve

from 2) x^3y^3 < x^2y^2 i.e. x^3y^3 - x^2y^2 < 0
i.e. x^2y^2(xy - 1) < 0 as square of something is always +ve
hence xy - 1 < 0 i.e. xy < 1..............not sufficient as xy can be +ve or -ve

The red parts above are not correct. Square of a number is always non-negative: $$x^2\geq{0}$$ and $$(xy)^2\geq{0}$$.

For (2) it's not correct to write that xy<1, without mentioning that $$xy\neq{0}$$, because if $$xy=0<1$$, then $$x^3y^3 =0= x^2y^2$$.

Hope it's clear.
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Re: Is xy + xy < xy ? [#permalink]  17 Jul 2014, 08:35
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Is xy + xy < xy? (1) x/(y^2)< 0 (2) x^3y^3 < (xy)^2 [#permalink]  19 Dec 2014, 08:11
Is xy + xy < xy?
(1) x/(y^2)< 0
(2) x^3y^3 < (xy)^2
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Re: Is xy + xy < xy ? [#permalink]  19 Dec 2014, 08:15
Expert's post
JusTLucK04 wrote:
Is xy + xy < xy?
(1) x/(y^2)< 0
(2) x^3y^3 < (xy)^2

Merging topics. Please refer to the discussion above.
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Re: Is xy + xy < xy ?   [#permalink] 19 Dec 2014, 08:15
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