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If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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19 Jun 2010, 02:45
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If x ≠ y is \(\frac{xy}{x+y}>1\)? (1) x > 0 (2) y < 0
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Re: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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19 Jun 2010, 03:34
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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: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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01 Aug 2010, 14:37
\(\frac{xy}{x+y}>1\) 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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If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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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?
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Re: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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27 Oct 2010, 13:20
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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: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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22 Dec 2010, 06:59
Thanks Bunuel, that explains it perfectly



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Re: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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13 Mar 2011, 07:17
(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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Re: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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14 Apr 2011, 03:10
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: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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14 Apr 2011, 04:11
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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: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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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: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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14 Apr 2011, 21:31
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: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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17 Apr 2011, 04:55
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: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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19 May 2011, 22:45
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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Re: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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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 Hence, Ans E it is!
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If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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15 Nov 2012, 18:21
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 ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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16 May 2013, 12:48
gmacforjyoab wrote: if x ≠ y, is (xy)/(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, \((xy)/(x+y) > 1\) 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, xy 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 (xy)/(x+y). Wonderful question gmacforjyoab! Hope my procedure is what you expected as an answer! Regards, Arpan
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Re: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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16 May 2013, 12:58
if x ≠ y, is (xy)/(x+y) > 1?
1) x> 0 2) y<0
is [(xy)(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: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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03 Oct 2014, 14:13
in your explanation i just dont see the progression how the "1" goes away and we change the numerator to "x+yx+y" to arrive at 2y



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Re: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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Re: If x ≠ y is (x  y)/(x + y) > 1? (1) x > 0 (2) y < 0 [#permalink]
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19 Oct 2015, 22:50
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. If x#y is (xy)/(x+y)>1? (1) x>0 (2) y<0 If we modify the original condition, as squared numbers are always positive, the inequality sign does not change even if we multiply certain positive integers. So if we multiply (x+y)^2, (xy)(x+y)>(x+y)^2, > (xy)(x+y)(x+y)^2>0, (x+y)(xyxy)>0, (x+y)(2y)>0, and if we divide both sides by 2, the inequality sign changes, and ultimately we want to know whether (x+y)y<0. There are 2 variables, so we need 2 equations, which are provided by the 2 conditions. If we look at the conditions together, the answer to what we want to know becomes 'yes' for x=2, y=1, but 'no' for x=2,y=3. Hence, the conditions are insufficient and the answer becomes (E). For cases where we need 2 more equation, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.
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