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If x#y is (xy)/(x+y)>1? (1) x>0 (2) y<0 [#permalink]
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19 Jun 2010, 03:45
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If x#y is (xy)/(x+y)>1? (1) x>0 (2) y<0
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Re: gmat prep DS [#permalink]
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19 Jun 2010, 04:34
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gmatcracker2010 wrote: Please explain the attached problem.
My take on it:
is (X+y)/ (xy) > 1
ie. x + y > x y ie. 0 > y
which is B but OA is E. If \(x\neq{y}\) is \(\frac{xy}{x+y}>1\)?Is \(\frac{xy}{x+y}>1\)? > Is \(0>1\frac{xy}{x+y}\)? > Is \(0>\frac{x+yx+y}{x+y}\)? > 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 \(xy>x+y\), 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 \(xy>x+y\) BUT if \(x+y<0\), you should write \(xy<x+y\) (flip the sign when multiplying by negative expression). So again: given inequality can be simplified as follows: \(\frac{xy}{x+y}>1\) > \(0>1\frac{xy}{x+y}<0\) > \(0>\frac{x+yx+y}{x+y}\) > \(0>\frac{2y}{x+y}\) > 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]
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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: (xy)/(x+y)>1 => xy > x + y => xx > y + y => 0 > 2 y => y must be negative
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Re: GMATPrep DS question [#permalink]
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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]
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01 Aug 2010, 15:37
\(\frac{xy}{x+y}>1\) Let us make cross multiplication with respect to the sign of x+y.  Case 1: xy>x+y>0 or x+y>0 and y<0  Case 2: xy<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]
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01 Aug 2010, 15:56
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But Bogos, what is wrong in just cancelling x from both sides of xy > 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]
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03 Aug 2010, 10:22
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bogos wrote: You can refer to this review (3rd point): inequalitiesreview69016.htmlBogos I read about inequalities and it says that one can subtract the same amout from both sides so if xy/x+y > 1 => xy > x+y => y<0 where is the flaw?
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Re: GMATPrep  DS  Inequalities [#permalink]
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27 Oct 2010, 14:20
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thirst4edu wrote: Rephrasing the question \(\frac{(xy)}{(x+y)} > 1\) \((xy) > (x+y)\) , 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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Re: gmat prep DS [#permalink]
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22 Dec 2010, 07:59
Thanks Bunuel, that explains it perfectly



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Re: gmat prep DS [#permalink]
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13 Mar 2011, 08:17
(xy/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+1/21>1 again x=2, y=3 2+3/23<1 so ans E. Help if i wrong.
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if x is different from y, in (xy)/(x+y) greater than 1 ? [#permalink]
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14 Apr 2011, 02:55
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if x is different from y, in (xy)/(x+y) greater than 1 ? 1. x > 0 2. y< 0



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Re: Did Sal had this question wrong ? [#permalink]
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14 Apr 2011, 04:10
whichscore wrote: if x is different from y, in (xy)/(x+y) greater than 1 ? 1. x > 0 2. y< 0 \(\frac{xy}{x+y}>1\) \(\frac{xy}{x+y}1>0\) \(\frac{xyxy}{x+y}>0\) \(\frac{2y}{x+y}>0\) \(\frac{y}{x+y}>0\) \(\frac{y}{x+y}<0\) Case I: If y<0 or y>0 A Then, x+y>0 x>yB Combining A and B: x>y>0 Case II: If y>0 or y<0 C Then, x+y<0 x<yD 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"
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Re: Did Sal had this question wrong ? [#permalink]
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14 Apr 2011, 05:11
I plugged numbers to get the answer : A per (1) x = 5, y = 5 then 0/10 < 1 x = 5, y = 1 6/4 > 1 So insufficient As per (2) x = 6, y = 2 8/4 > 1 x = 2 , y = 3 5/1 < 1 So insufficient As evident above, (1) and (2) together are also insufficient Answer  E
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Re: Did Sal had this question wrong ? [#permalink]
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14 Apr 2011, 18:18
whichscore wrote: if x is different from y, in (xy)/(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 \(\frac{(xy)}{(x+y)} > 1\)? (Don't forget it is a question, not given information) Is \(\frac{(xy)}{(x+y)}  1 > 0\)? 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)
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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 xy / (x + y) may or may not be greater than 1. I don't think any calculation is required in this problem.



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Re: Did Sal had this question wrong ? [#permalink]
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17 Apr 2011, 05:55
fluke wrote: whichscore wrote: if x is different from y, in (xy)/(x+y) greater than 1 ? 1. x > 0 2. y< 0 \(\frac{xy}{x+y}>1\) \(\frac{xy}{x+y}1>0\) \(\frac{xyxy}{x+y}>0\) \(\frac{2y}{x+y}>0\) \(\frac{y}{x+y}>0\) \(\frac{y}{x+y}<0\) Case I: If y<0 or y>0 A Then, x+y>0 x>yB Combining A and B: x>y>0 Case II: If y>0 or y<0 C Then, x+y<0 x<yD 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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Re: Did Sal had this question wrong ? [#permalink]
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19 May 2011, 23:45
by resolving  x/ (x+y) > 1 a + b, x>0 y<0 x=2,y= 1 LHS > RHS x=1,y =2 LHS < RHS 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.



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Re: If x is not equal to y,is (x y)/(x+y) >1? [#permalink]
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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 Hence, Ans E it is!
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Re: If x is not equal to y,is (x y)/(x+y) >1? [#permalink]
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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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Re: If x is not equal to y,is (x y)/(x+y) >1?
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