Is xy + xy

Question

Is  xy + xy < xy  ?

(1)  \frac{x^2}{y} < 0

(2)  x^3y^3 < (xy)^2

Bunuel · Accepted Answer

Is 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 eq{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. Answer: E. Hope it's clear.

Kris01 · Answer

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.

jbisht · Answer

it's an easy question provided basics of inequalities are clear. question is asking for xy +xy 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)

Bunuel · Answer

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
eq{0} , because if  xy=0<1 , then  x^3y^3 =0= x^2y^2 .

Hope it's clear.

bethebest · Answer

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.

Bunuel · Answer

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

HKD1710 · Answer

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.

bethebest · Answer

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.

MathRevolution · Answer

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.

jabhatta2 · Answer

Hi Experts --  chetan2u ,  VeritasKarishma ,  Bunuel ,  nick1816 ,  GMATBusters

I have done quite a few "MUST BE TRUE" questions.

If xy < 1, then WHAT MUST BE TRUE

a) XY < 0

Isn't this accurate ?

I thought S2 was saying something like this and thus i chose B for this question

KarishmaB · Answer

Yes, this is accurate. If some variable is less than 0, it is certainly less than 1. 
But stmnt 2 does not give you that xy < 0. It says 
(xy)^3 < (xy)^2
(xy)^3 - (xy)^2 < 0
(xy)^2 * (xy - 1) < 0
So one of these factors must be negative. Since (xy)^2 cannot be negative, (xy - 1) < 0
So xy < 1

What I KNOW is that xy < 1.

Can I say now whether xy is less than 0? No, I don't know. I know that it is less than 1 but it could be 1/2 or 0 or -20 etc. I do not know whether it is less than 0. It may or may not be.

bv8562 · Answer

Bunuel

Can I deduce from this part of the expression:

(xy)^2 (xy-1)<0

that either 0<xy<1 or xy<0 and xy≠0?

Bunuel · Answer

(xy)^2 (xy-1)<0

(xy)^2 is the square of a number so it must be positive or 0, so for  (xy)^2 (xy-1)<0   to be true, we must have  (xy)^2 
eq 0  and  xy-1<0 . Thus,  (xy)^2 (xy-1)<0   means  xy<1  and  (xy)^2 
eq 0  .

bv8562 · Answer

Sorry asked the wrong question. This is what I want to know:

Can I deduce from this part of the expression:

(xy)^2 (xy-1)<0

since (xy)^2≠0, therefore xy≠0, so, either 0<xy<1 or xy<0 ?

Bunuel · Answer

_________________________
Yes.

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards