Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Hard inequality: Is xy < 0 [#permalink]
12 Nov 2009, 18:44

10

This post received KUDOS

1

This post was BOOKMARKED

Statement 1: \(\frac{x^3*y^5}{x*y^2} < 0\)

\(x^2*y^3 < 0\)

We know that \(x^2\) must be positive, so therefore y < 0. Since we do not know whether x is positive or negative, however, this is insufficient.

Statement 2: \(|x|-|y| < |x-y|\)

I'll approach this the way I would as if I was writing it on the GMAT with time constraints, by picking numbers:

(1) Is it possible for xy > 0 AND the above criteria to be satisfied? Yes. (i.e. x = 2, y = 3) (2) Is it possible for xy < 0 AND the above criteria to be satisfied? Yes. (i.e. x = 2, y = -3)

Therefore, not sufficient.

Evaluating Both Statements

We know from Statement 1 that y < 0. Therefore we can revise the above approach for Statement 2 to specifics:

(1) Is it possible for y < 0, x < 0 (i.e. xy > 0) AND the above criteria to be satisfied? Yes. (i.e. x = -2, y = -3) (2) Is it possible for y < 0, x > 0 (i.e. xy < 0) AND the above criteria to be satisfied? Yes. (i.e. x = 2, y = -3)

Re: Hard inequality: Is xy < 0 [#permalink]
12 Nov 2009, 18:50

2

This post received KUDOS

Bunuel wrote:

Is \(xy<0\)?

(1) \(\frac{x^3*y^5}{x*y^2}<0\)

(2) \(|x|-|y|<|x-y|\)

I'm going with e on this one

(1) \(\frac{x^3*y^5}{x*y^2}<0\)

x^3-1 * y*5-2 < 0 x^2*y^3<0 y < 0 but x<0 or x > 0 so insufficient because it doesn't matter what x is the answer will be negative however plugging back into the original equation gives two different answers

(2) \(|x|-|y|<|x-y|\)[/quote]

if x = 1 and y = -2 |1| - |-2| < |1 + 2|

if x>0 and y > 0 then the equations will be equal if x> 0 and y < 0 yes if x<0 and Y < 0 yes and both would make \(xy<0\) a different answer

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
07 Mar 2012, 09:19

2

This post received KUDOS

Expert's post

pavanpuneet wrote:

For the second case, I tried to solve by squaring on both the sides. So you get, -2|x||y|<-2xy, which is ....|x||y|>xy...only case possible is when one of them is negative... will it then not prove that xy<0. Please let me know where am I wrong?

Welcome to GMAT Club. Below is an answer to your question.

You can square both sides of an inequality if and only both sides are non-negative.

For |x|-|y|<|x-y| we know that |x-y| is non-negative, but |x|-|y| can be negative, as well as positive (non-negative), hence you cannot apply squaring.

GENERAL RULE: A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: \(2<4\) --> we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work. For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: \(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\); \(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
07 Mar 2012, 09:05

For the second case, I tried to solve by squaring on both the sides. So you get, -2|x||y|<-2xy, which is ....|x||y|>xy...only case possible is when one of them is negative... will it then not prove that xy<0. Please let me know where am I wrong?

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
19 Nov 2012, 18:42

I really like one of the ideas forwarded above.

data point 1 was fairly obvious. with data point 2, we need only to stare at it and the question xy<0 for a few seconds. the more you think about it the more it becomes obvious that you would need to pick numbers such that xy<0 can be proven and xy<0 can be debunked. you could do this by using x>0 and y<0 (to prove xy<0) and using x>0,y>0 (to debunk).

Re: Hard inequality: Is xy < 0 [#permalink]
13 May 2013, 11:13

Bunuel wrote:

Yes, OA is E.

Number plugging as AKProdigy87 did is probably the best way here. But still if needed below is short overview of algebra of this question:

(1) y<0. Not sufficient.

(2) Holds true when: A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x

Not sufficient.

(1)+(2) C. and D. options are left from (2) but still insufficient, as xy may or may not be negative.

Answer: E.

Hi Bunuel, I want to double check the following with you ... the 4 cases u ve mentioned is detailed version of the following or not for the inequality to hold true

1) xy<0 2) /y/>/x/

and this is because the equality (=) in /y-x/ > = /x/ - /y/ holds true only in 2 cases /x/ = /y/ or /x/>/y/ and xy>0

am i right ?? plz let me know.. thanking you in advance.

Re: Hard inequality: Is xy < 0 [#permalink]
17 May 2013, 15:45

Sergiy wrote:

Bunuel wrote:

Yes, OA is E.

Number plugging as AKProdigy87 did is probably the best way here. But still if needed below is short overview of algebra of this question:

(1) y<0. Not sufficient.

(2) Holds true when: A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x

Not sufficient.

(1)+(2) C. and D. options are left from (2) but still insufficient, as xy may or may not be negative.

Answer: E.

Hi Bunuel,

Can you explain how you got the following set of inequality? :

A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x

Thanks

/x-y/ >= /x/-/y/ is always true ( try it ) and the equality (equal sign) holds in 2 cases 1) when xy> 0 ( same sign) and /x/>/y/ ( both conditions together) 2)

x=y=0

Now , the inequality holds true in any other case (/x-y/ > /x/-/y/) i.e. xy<0 ( different signs) or /x/</y/ if u translate this last bit into inequalities it gives the above 4 scenarios that i reformatted below

A. xy>0 but /y/> /x/ b. xy<0 , y>x c. xy<0 and /y/ > /x/ d. xy<0 , x>y

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
30 Jun 2013, 12:18

Is xy<0?

Is either x or y negative?

(1)(x^3*y^5)/(x*y^2)<0

(x^3 - x^1)*(y^5 - y^2) < 0 (x^2)*(y^3) < 0

(x^2) is positive regardless of whether x is positive or negative. If (x^2)*(y^3) < 0 then (y^3) must be negative and because a number raised to an odd power holds its sign, y must be negative. However, we do not know whether x is positive or negative meaning ab could be (a)*(-b) = -ab OR (-a)*(-b) = ab. INSUFFICIENT

There are two valid statements where x and y are negative and where x is positive and y is negative meaning we don't know whether ab is negative or positive. INSUFFICIENT

Re: Hard inequality: Is xy < 0 [#permalink]
06 Nov 2013, 10:41

Bunuel wrote:

Yes, OA is E.

Number plugging as AKProdigy87 did is probably the best way here. But still if needed below is short overview of algebra of this question:

(1) y<0. Not sufficient.

(2) Holds true when: A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x

Not sufficient.

(1)+(2) C. and D. options are left from (2) but still insufficient, as xy may or may not be negative.

Answer: E.

Bunuel, I'm finding this confusing. Can you help me out. Your post on this thread is-x-y-x-y-123108.html says that for statement 2 to be true both x and y need to have the same sign.

Now, this question if you look at statement 2, it says that x and y have opposite signs. Hence, it is never possible for |x|-|y|<|x-y| to be true. So statement 2 seems sufficient. Where is my error in reasoning? Thank you.

Re: Hard inequality: Is xy < 0 [#permalink]
06 Nov 2013, 23:13

emailmkarthik wrote:

Bunuel wrote:

Yes, OA is E.

Number plugging as AKProdigy87 did is probably the best way here. But still if needed below is short overview of algebra of this question:

(1) y<0. Not sufficient.

(2) Holds true when: A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x

Not sufficient.

(1)+(2) C. and D. options are left from (2) but still insufficient, as xy may or may not be negative.

Answer: E.

Bunuel, I'm finding this confusing. Can you help me out. Your post on this thread is-x-y-x-y-123108.html says that for statement 2 to be true both x and y need to have the same sign.

Now, this question if you look at statement 2, it says that x and y have opposite signs. Hence, it is never possible for |x|-|y|<|x-y| to be true. So statement 2 seems sufficient. Where is my error in reasoning? Thank you.

I agree. Inequality rule saying that |x|-|y|<|x-y| is true only when x and y are having different signs exists. Does not it? If so, B is correct

gmatclubot

Re: Hard inequality: Is xy < 0
[#permalink]
06 Nov 2013, 23:13

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...