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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.

Re: GMATPrep DS question [#permalink]
01 Aug 2010, 14:03

Hi,

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

Re: GMATPrep DS question [#permalink]
01 Aug 2010, 14:37

\(\frac{x-y}{x+y}>1\) 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 not equal to -y,is (x -y)/(x+y) >1? [#permalink]
15 Nov 2012, 17:55

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]
15 Nov 2012, 18:14

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]
15 Nov 2012, 18:21

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) _________________

Re: x ≠ -y, is (x-y)/(x+y) > 1 [#permalink]
16 May 2013, 12:48

gmacforjyoab wrote:

if x ≠ -y, is (x-y)/(x+y) > 1?

1) x> 0 2) y<0

The answer is [E]. I marked it [B], but as I began to write the solution, the obviousness of the solution hit me!

The expression, \((x-y)/(x+y) > 1\) when simplified gives us y < 0. Clearly the above will drive us directly to [B]. But when we consider this, in the case y < 0 and x > 0, and no relation is given between x and y, what happens if |x| < |y|. In such a case x+ y < 0. But clearly, x-y is greater than 0! Hence the solution doesn't stick! This is the reason why [B] can not be the answer! Similarly if x < 0 and y < 0, we cannot for sure determine the sign of the fraction.

Also, clearly the individual choice x > 0 does not help! so, neither of the choices lead to our answer. To actually solve the above we need a relation determined between x and y so as to determine the sign of (x-y)/(x+y). Wonderful question gmacforjyoab!

Hope my procedure is what you expected as an answer!

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