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If x#-y is (x-y)/(x+y)>1? (1) x>0 (2) y<0 [#permalink]
 19 Jun 2010, 03:45
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If x#-y is (x-y)/(x+y)>1? (1) x>0 (2) y<0
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gmatcracker2010 wrote: Please explain the attached problem.
My take on it:
is (X+y)/ (x-y) > 1
ie. x + y > x -y
ie. 0 > y
which is B but OA is E. If x\neq{-y} is \frac{x-y}{x+y}>1?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. Hope it helps.
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Re: GMATPrep DS question [#permalink]
01 Aug 2010, 15:03
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
Cheers
Tim
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Re: GMATPrep DS question [#permalink]
01 Aug 2010, 15:28
The OA is E. I thought it would be B. I think the catch is that x cannot be cancelled from both sides... If x - y > x + y Cancel x y < 0 Should be B. OA says E.
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Re: GMATPrep DS question [#permalink]
01 Aug 2010, 15:37
\frac{x-y}{x+y}>1Let 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.
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Re: GMATPrep DS question [#permalink]
01 Aug 2010, 15:56
But Bogos, what is wrong in just cancelling x from both sides of x-y > x+y? You mentioned about w.r.t the sign of x+y, why is that important?
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Re: GMATPrep DS question [#permalink]
03 Aug 2010, 10:22
bogos wrote: You can refer to this review (3rd point): inequalities-review-69016.htmlBogos I read about inequalities and it says that one can subtract the same amout from both sides so if x-y/x+y > 1 => x-y > x+y => y<0 where is the flaw?
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Re: GMATPrep - DS - Inequalities [#permalink]
27 Oct 2010, 14:20
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thirst4edu wrote: 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.
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Thanks Bunuel, that explains it perfectly
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(x-y/x+y)>1 What is the problem if i try numbers (1) x>0 no information about y clearly insuf. (2) y <0 no information about x. insuf. For C x=2, y=-1 2+1/2-1>1 again x=2, y=-3 2+3/2-3<1 so ans E. Help if i wrong.
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If x is not equal to -y,is (x -y)/(x+y) >1? [#permalink]
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.
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Re: If x is not equal to -y,is (x -y)/(x+y) >1? [#permalink]
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) >0or 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 Hence, Ans E it is!
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Re: If x is not equal to -y,is (x -y)/(x+y) >1? [#permalink]
15 Nov 2012, 19:21
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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x ≠ -y, is (x-y)/(x+y) > 1 [#permalink]
16 May 2013, 13:38
if x ≠ -y, is (x-y)/(x+y) > 1?
1) x> 0
2) y<0
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Re: x ≠ -y, is (x-y)/(x+y) > 1 [#permalink]
16 May 2013, 13:48
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! Regards, Arpan
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Re: x ≠ -y, is (x-y)/(x+y) > 1 [#permalink]
16 May 2013, 13:58
if x ≠ -y, is (x-y)/(x+y) > 1?
1) x> 0
2) y<0
is [(x-y)-(x+y)]/(x+y) > 0 , i.e. is -2y / x+y > 0 i.e. is 2y / x+y < 0? true if 1) y <0 , x>0 and /y/ > /x/ 2) Y>0 AND X<0 AND /X/>/Y/
from 1
x is +ve .... insuff
from 2
y -ve insuff
both
this looks like the 1st case described above but we dont know whether /y/ > /x/ ..... insuff
E
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Re: x ≠ -y, is (x-y)/(x+y) > 1 [#permalink]
16 May 2013, 14:03
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Re: x ≠ -y, is (x-y)/(x+y) > 1
[#permalink]
16 May 2013, 14:03
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