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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. _________________

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

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14 Apr 2011, 18:18

Expert's post

whichscore wrote:

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]

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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 _________________

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

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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

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

Show Tags

15 Nov 2012, 19:21

Expert's post

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.

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) _________________

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