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Manager  Status: struggling with GMAT
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Is xy + xy < xy ?  [#permalink]

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Question Stats: 52% (02:04) correct 48% (02:02) wrong based on 336 sessions

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Is xy + xy < xy ?

(1) x^2/y < 0
(2) x^3y^3 < (xy)^2

Originally posted by mun23 on 02 Mar 2013, 13:03.
Last edited by Bunuel on 17 Jul 2014, 09:55, edited 2 times in total.
Edited the question
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Re: Is xy + xy < xy ?  [#permalink]

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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]

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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]

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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]

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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]

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Bunuel wrote:
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.

Statement 2 states that x^3*y^3<x^2*y^2; this implies that xy is negative (xy<0) which is sufficient. So I chose answer B.

Can you please explain what am I doing wrong? Thank you for your help.
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Re: Is xy + xy < xy ?  [#permalink]

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MeghaP wrote:
Bunuel wrote:
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.

Statement 2 states that x^3*y^3<x^2*y^2; this implies that xy is negative (xy<0) which is sufficient. So I chose answer B.

Can you please explain what am I doing wrong? Thank you for your help.

How did you deduce that xy < 0? Please check the highlighted part: (2) gives that xy<1.
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Re: Is xy + xy < xy ?  [#permalink]

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mun23 wrote:
Is xy + xy < xy ?

(1) x^2/y < 0
(2) x^3y^3 < (xy)^2

Is xy + xy < xy ?
is 2xy < xy
is xy < 0 ? x and y have diff sign ??

stmt-1:
x^2/y < 0

x^2 is positive, so y has to be negative. now that we do not know sign of x, we cant be certain if xy <0 or >0. insuff

stmt-2:
x^3y^3 < (xy)^2
(xy)^2(xy-1)<0
as (xy)^2 will be positive xy-1 has to be negative. so we can write xy<1

so xy can be netween 0 and 1 or less than zero. insuff

combining:

again we dont know sign of x. hence xy could be between 0 and 1 or negative. insuff.
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Re: Is xy + xy < xy ?  [#permalink]

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Bunuel wrote:
MeghaP wrote:
Bunuel wrote:
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.

Statement 2 states that x^3*y^3<x^2*y^2; this implies that xy is negative (xy<0) which is sufficient. So I chose answer B.

Can you please explain what am I doing wrong? Thank you for your help.

How did you deduce that xy < 0? Please check the highlighted part: (2) gives that xy<1.

I assumed that x^3*y^3 < x^2*y^2 is only possible if x*y is negative. I failed to take into account that certain positive fractions also fulfill the criteria. I apologize for my mistake. Thank you for your help.
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Re: Is xy + xy < xy ?  [#permalink]

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mun23 wrote:
Is xy + xy < xy ?

(1) x^2/y < 0
(2) x^3y^3 < (xy)^2

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

The question $$xy + xy < xy$$ is equivalent to $$xy < 0$$.
Since we have 2 variables and 0 equation, C is most likely to be the answer.

1) & 2)
$$\frac{x^2}{y} < 0 ⇔ x^2 y < 0$$.
This is equivalet to$$x≠0$$ and $$y < 0$$.

$$x^3 y^3 < (xy)^2 ⇔ x^3 y^3 - x^2 y^2 < 0 ⇔ x^2 y^2 ( xy - 1 ) < 0$$
This is equivalent to $$x≠0$$, $$y≠0$$ and $$xy < 1$$

$$x = \frac{1}{2}$$ and $$y = - \frac{1}{2} ⇒ xy = - \frac{1}{4} < 0$$ : Yes
$$x = - \frac{1}{2}$$ and $$y = - \frac{1}{2} ⇒ xy = \frac{1}{4} > 0$$ : No

The answer is E.

Normally for cases where we need 2 more equations, such as original conditions with 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, we have 1 equation each in both 1) and 2). Therefore C has a high chance of being the answer, which is why we attempt to solve the question using 1) and 2) together. Here, there is 70% chance that C is the answer, while E has 25% chance. These two are the key questions. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer according to DS definition, we solve the question assuming C would be our answer hence using 1) and 2) together. (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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