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 500,000 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:

Is \(\frac{x-y}{x+y}>1\)? --> Is \(0>1-\frac{x-y}{x+y}\)? --> Is \(0>\frac{x+y-x+y}{x+y}\)? --> Is \(0>\frac{2y}{x+y}\)?

(1) \(x>0\) --> Not sufficient.

(2) \(y<0\) --> Not sufficient.

(1)+(2) \(x>0\) and \(y<0\) --> numerator (y) is negative, but we cannot say whether the denominator {positive (x)+negative (y)} is positive or negative. Not sufficient.

Answer: E.

The problem with your solution is that when you are then writing \(x-y>x+y\), you are actually multiplying both sides of inequality by \(x+y\): never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable). Because if \(x+y>0\) you should write \(x-y>x+y\) BUT if \(x+y<0\), you should write \(x-y<x+y\) (flip the sign when multiplying by negative expression).

So again: given inequality can be simplified as follows: \(\frac{x-y}{x+y}>1\) --> \(0>1-\frac{x-y}{x+y}<0\) --> \(0>\frac{x+y-x+y}{x+y}\) --> \(0>\frac{2y}{x+y}\) --> we can drop 2 and finally we'll get: \(0>\frac{y}{x+y}\).

Now, numerator is negative (\(y<0\)), but we don't know about the denominator, as \(x>0\) and \(y<0\) can not help us to determine the sign of \(x+y\). So the answer is E.

twi ways to approach: 1) Check it out by plugging in numbers: a) x < y: makes the numerator negative, so doesn't work b) x = y: creates a 0, so doesn't work c) x > y: makes the denominator greater than the numerator, so doesn't work d) y < 0: makes numerator greater than denominator, so WORKS

2) solve the equation: (x-y)/(x+y)>1 => x-y > x + y => x-x > y + y => 0 > 2 y => y must be negative

\(\frac{x-y}{x+y}>1\) Let us make cross multiplication with respect to the sign of x+y. - Case 1: x-y>x+y>0 or x+y>0 and y<0 - Case 2: x-y<x+y<0 or x+y<0 and y>0

Both statement 1 and 2 together cannot say about the sign of x+y, thereofore correct answer is E.
_________________

Rephrasing the question \(\frac{(x-y)}{(x+y)} > 1\) \((x-y) > (x+y)\) , Since x <> -y, we can multiply (x+y) both sides Adding -x both sides -y > y Adding y both sides 0 > y , which is y < 0, given by statement 2, so why Answer is E(both together not sufficient)? Am I missing something?

What you did is only true is (x+y)>0. If you multiply both sides of an inequality with a number, the sign remains same if the number or expression is positive, whereas it flips if it is negative.

A counter example to show answer is indeed (e) is taking x=5 and y=-20 ... you'll get 25/-15 = -5/3 which is less than 1 ... on the other hand x=5 & y=-1 would get you 6/4 which is greater than 1.
_________________

if x is different from -y, in (x-y)/(x+y) greater than 1 ? 1. x > 0 2. y< 0

Even though fluke has already provided the detailed solution, there are a couple of points I would like to reinforce here.

People often get confused when they read 'if x is different from -y'. Why do they give this information? They do that because you have (x+y) in the denominator and hence it cannot be 0. That is, x + y = 0 or x = -y is not valid. Hence, they are just clarifying that the fraction is indeed defined.

Also, we are used to having 0 on the right of an inequality. What do we do when we have a 1? We take the 1 to the left hand side and get a 0 on the right. Is \(\frac{(x-y)}{(x+y)} > 1\)? (Don't forget it is a question, not given information) Is \(\frac{(x-y)}{(x+y)} - 1 > 0\)? which simplifies to: Is y/(x+y) < 0 ?

We know how to deal with this inequality! Note here that we have no information about x and y as yet (except that x is not equal to -y which is more of a technical issue rather than actual information)

1. x > 0 No information about y so not sufficient. 2. y< 0 No information about x so not sufficient.

Both together, we know that y is negative. We need the sign of (x+y) now. But (x+y) may be positive or negative depending on whether x or y has greater absolute value. Hence not sufficient. Answer (E)
_________________

Re: Did Sal had this question wrong ? [#permalink]

Show Tags

14 Apr 2011, 22:31

1) + 2) When x > 0 and y < 0, x- y will always be +ve but nothing can be concluded about (x + y). y can be very small -Infinity or y can be close to zero. Hence sign of x - y is unknown. Hence the division i.e x-y / (x + y) may or may not be greater than 1. I don't think any calculation is required in this problem.

Case I: If y<0 or -y>0 -----A Then, x+y>0 x>-y-----B Combining A and B: x>-y>0

Case II: If y>0 or -y<0 -----C Then, x+y<0 x<-y-----D Combining C and D: x<-y<0

1. x > 0 CaseI: x>0 Is -y between 0 and x

If x=5 y=-3 -y=-(-3)=3 will be between 0 and x(5) and the expression will be true.

If x=5 y=-7 -y=-(-7)=7 will NOT be between 0 and x(5) and the expression will be false. Not Sufficient.

2. y < 0 CaseI: x>0 Is -y between 0 and x

If x=5 y=-3; Here y<0 -y=-(-3)=3 will be between 0 and x(5) and the expression will be true.

If x=5 y=-7; Here y<0 -y=-(-7)=7 will NOT be between 0 and x(5) and the expression will be false. Not Sufficient.

Combing both; we use the same sample set: x=5 y=-3 AND x=5 y=-7 To prove its insufficiency.

Ans: "E"

Dear Fluke i had below approach please correct me if i am wrong

the term simplifies to \(-\frac{y}{x+y}>0\)

Case I Numerator and denominator both are positive -y > 0 y < 0 and x + y > 0 x > -y but since -y>0 so x also also be >0

Case 2 Numerator and denominator both are negative

-y < 0 y > 0 and x + y < 0 x < -y but since -y<0 so x will also be < 0

now Clearly none of the statement alone is sufficient and taking them together gives us only Case I but not takes care about the case 2 so answer is E
_________________

WarLocK _____________________________________________________________________________ The War is oNNNNNNNNNNNNN for 720+ see my Test exp here http://gmatclub.com/forum/my-test-experience-111610.html do not hesitate me giving kudos if you like my post.

If x is not equal to -y,is (x -y)/(x+y) >1? [#permalink]

Show Tags

15 Nov 2012, 18:55

Hi,

Request help with the following question.Thanks..

If x is not equal to -y,is (x -y)/(x+y) >1?

1.x > 0 2.y < 0

P.S. - While i can solve the question by testting numbers,i would be most grateful if somebody could help me understand the fraction theory being tested on this question.

Re: If x is not equal to -y,is (x -y)/(x+y) >1? [#permalink]

Show Tags

15 Nov 2012, 19:14

shivanigs wrote:

Hi,

Request help with the following question.Thanks..

If x is not equal to -y,is (x -y)/(x+y) >1?

1.x > 0 2.y < 0

P.S. - While i can solve the question by testting numbers,i would be most grateful if somebody could help me understand the fraction theory being tested on this question.

Here you go: First, lets rephrase the question- \((x -y)/(x+y) >1\) \((x -y)/(x+y) -1 >0\) \(-2y/(x+y) >0\) or is \(y/(x+y) < 0\) ?

Statement 1: x >0 This doesnt help us understanding the sign of numerator or denominator. Not sufficient. Statement 2: y<0 now we know, that numerator is postive, but we still dont know what is the sign for x+y. Not sufficient.

Combining 1 and 2. x>0, y<0 We want to know if y/(x+y) < 0 Numerator is negative here, and denominator can be positive or negative depending upon absolute values of x and y. Not sufficien

P.S. - While i can solve the question by testting numbers,i would be most grateful if somebody could help me understand the fraction theory being tested on this question.

Responding to a pm:

Question: Is (x -y)/(x+y) >1?

What do you do when you are given >1 ? It is hard to find implications of '>1'. It is much easier to handle '>0'

Is \(\frac{(x -y)}{(x+y)} -1 > 0\) ? Is \(\frac{-2y}{(x+y)} > 0\) ?

When will this be positive? In 2 cases: 1. When y is negative and (x+y) is positive i.e. x is positive and x has greater absolute value than y's absolute value. 2. When y is positive and (x+y) is negative i.e. x is negative and x has greater absolute value than y's absolute value.

Both statements together tell us that y is negative and x is positive but we don't know whether 'x's absolute value is greater than y's absolute value'. Hence (E)
_________________

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...