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Is 1 / (x-y) < y - x [#permalink]
 07 Jan 2011, 09:09
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Is 1 / (x-y) < y - x (1) 1 / x < 1 / y (2) 2x = 3y
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Last edited by Bunuel on 27 Aug 2012, 08:12, edited 1 time in total.
Edited the OA.
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Re: Tough DS Algebra [#permalink]
07 Jan 2011, 09:37
Proposition 1:
1/X < 1/Y
This is equal to X > Y | x, y both > 0 or both < 0
Case 1: Both > 0. X > Y, therefore X-Y >0 , Y-X <0, so 1/(X-Y) > 0 and Y-X<0 so FALSE
Case 2: X < 0, Y>0. Cross multiple and switch the signs, you get Y>X. There for Y-X>0, X-Y<0.
1/(X-Y)<0, Y-X>0, so case 2, 1/(X-Y) < (Y-X) is TRUE
Therefore statement 1 is insufficient by contradiction.
Proposition 2:
2X=3Y
Case 1: X=3, Y=2 - therefore X-Y=1, Y-X = -1
1/(X-Y) < Y-X ----> 1/1 < -1 = FALSE
Case 2: X=-3, Y=-2 - therefore X-Y=-1, Y-X = 1
1/(X-Y) < Y-X ----> 1/-1 < 1 = TRUE
Insufficient by contradicition
Both Statments together - Intutitively, neither statement removes the ambiguity of the sign of X or Y. I can't think of a proof.
Therefore answer is E.
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Re: Tough DS Algebra [#permalink]
07 Jan 2011, 10:15
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rxs0005 wrote: Is 1 / (x-y) < y - x
1 / x < 1 / y
2x = 3y First of all if it were realistic GMAT question it would most likely state that xy\neq{0} and x\neq{y} (as x, y and x-y are in denominators). Is \frac{1}{x-y}<y-x? --> is \frac{1}{x-y}+x-y<0 --> is \frac{1+(x-y)^2}{x-y}<0? as the nominator ( 1+(x-y)^2) is always positive then the question basically becomes whether denominator ( x-y) is negative --> is x-y<0? or is x<y? (1) \frac{1}{x}<\frac{1}{y} --> if both unknowns are positive or both unknowns are negative then y<x (if both are positive cross multiply to get y<x and if both are negative cross multiply and flip the sigh twice to get y<x again) and the answer will be NO but if x<0<y given inequality also holds true and in this case the answer will be YES (if x is any negative number and y is any positive number then \frac{1}{x}=negative<positive=\frac{1}{y}). Not sufficient. (2) 2x=3y --> x and y have the same sign, next: \frac{x}{y}=\frac{3}{2}: if both x and y are positive (for example 3 and 2 respectively) then 0<y<x and the answer will be NO but if both x and y are negative (for example -3 and -2 respectively) then x<y<0 and the answer will be NO. Not sufficient. (1)+(2) As from (2) x and y have the same sign then from (1) y<x and the answer to the question is NO. Sufficient. Answer: C.
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Re: Tough DS Algebra [#permalink]
07 Sep 2011, 04:24
** Quote: Is 1 / (x-y) < y - x?
1. 1 / x < 1 / y
2. 2x = 3y equation: \frac{1}{(x-y)} < y-xStatement 1Given \frac{1}{x}<\frac{1}{y}, we have 2 possibilities: One: if both x and y share the same signs (i.e. both are positive or both are negative) --> y<xleft side of equation will be a positive fraction; right side will be negative integer --> equation = true Two: if x is negative and y is positive (e.g. x=-2 and y=3), left side of equation will be negative fraction; right side will be positive integer --> equation = false Insufficient. Statement 2Given 2x=3y --> x/y = 3/2 -->i.e. x and y share the same sign (i.e. both positive or both negative). 2 possibilities: One: if both x and y are positive (i.e. x=3, y=2), results in 1/1 < -1 --> equation = falseTwo: if both x and y are negative (i.e. x=-3, y=-2), results in 1/-1 < 1 --> equation = true Insufficient. Statement 1 + 2Applying statement 2 and 1 together, we know that x and y: i. share the same sign, and ii. that y<x --> which means, equation = false i.e. No Sufficient. Answer: C
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Is 1/(x-y) < y - x? [#permalink]
27 Aug 2012, 07:45
Is \frac{1}{(x-y)} < y - x ? (1) \frac{1}{x} < \frac{1}{y}(2) 2x = 3yI don't agree with the OA. It must be C. If 2x = 3y, then x and yare both positive or both negative. So, if we know that: \frac{1}{x} < \frac{1}{y}, then x>y --> x -y > 0With that information we can conclude that the answer is No. C is correct. Please, your comments.
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Re: Is 1/(x-y) < y - x? [#permalink]
27 Aug 2012, 08:14
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Re: Is 1 / (x-y) < y - x [#permalink]
03 Sep 2012, 05:16
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rxs0005 wrote: Is 1 / (x-y) < y - x
(1) 1 / x < 1 / y
(2) 2x = 3y If we denote by A=x-y, the question is "Is \frac{1}{A}<-A?" If A>0, the above inequality cannot hold (a positive number is not smaller than a negative number). So, the question can be reworded as is A<0, or is x-y<0 which is the same as is x<y?(1) The given inequality is equivalent to \frac{x-y}{xy}>0.If xy>0, or in other words if x and y have the same sign, then necessarily x must be greater than y.If xy<0, or in other words if x and y have opposite signs, then necessarily x must be smaller than y.Not sufficient. (2) x=\frac{3}{2}y. If y<0, then x<y. But if y>0, then x>y.Not sufficient. (1) and (2) together: Since from (2) we have that x and y have the same sign, using (1) we deduce that necessarily x-y>0.So, the answer to the question "Is x<y" is a definite NO. Sufficient. Answer C.
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Re: Is 1 / (x-y) < y - x [#permalink]
25 Nov 2012, 19:54
Hi Buneul,
Can you please explain this part?
if both are negative cross multiply and flip the sigh twice to get y<x again)
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Re: Is 1 / (x-y) < y - x [#permalink]
26 Nov 2012, 02:28
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Re: Is 1 / (x-y) < y - x [#permalink]
14 Jan 2013, 06:16
stem reduces to Is x<y ? [-1/(y-x) = (y-x) => (y-x) >0 ]
A. Depends on sign of x,y. (both positive or both negative for eg. Vs positive-negative give both YES and NO)
B. x= 3k, y= 2k. K>=0, Answer NO. K<0 Answer is YES.
C.
Using 2. x =3k when y =2k.
using the above in 1. 1/x < 1/ y or 1/3k < 1/2k holds only for k>0. For k>0, Using 2, we definitely get the single answer as NO.
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Re: Tough DS Algebra [#permalink]
14 Jan 2013, 23:43
Bunuel wrote: rxs0005 wrote: Is 1 / (x-y) < y - x
1 / x < 1 / y
2x = 3y First of all if it were realistic GMAT question it would most likely state that xy\neq{0} and x\neq{y} (as x, y and x-y are in denominators). Is \frac{1}{x-y}<y-x? --> is \frac{1}{x-y}+x-y<0 --> is \frac{1+(x-y)^2}{x-y}<0? as the nominator ( 1+(x-y)^2) is always positive then the question basically becomes whether denominator ( x-y) is negative --> is x-y<0? or is x<y? (1) \frac{1}{x}<\frac{1}{y} --> if both unknowns are positive or both unknowns are negative then y<x (if both are positive cross multiply to get y<x and if both are negative cross multiply and flip the sigh twice to get y<x again) and the answer will be NO but if x<0<y given inequality also holds true and in this case the answer will be YES (if x is any negative number and y is any positive number then \frac{1}{x}=negative<positive=\frac{1}{y}). Not sufficient. (2) 2x=3y --> x and y have the same sign, next: \frac{x}{y}=\frac{3}{2}: if both x and y are positive (for example 3 and 2 respectively) then 0<y<x and the answer will be NO but if both x and y are negative (for example -3 and -2 respectively) then x<y<0 and the answer will be NO. Not sufficient. (1)+(2) As from (2) x and y have the same sign then from (1) y<x and the answer to the question is NO. Sufficient. Answer: C. Hello Bunuel, Agree with your approach but I wanted to plug in nos and check so here it goes. we get the expression ( 1+ (x-y)^2 )/(x-y) <0----> Q becomes Is x<y ? From St 1, we get 1/x<1/y ----> 1/x-1/y <0 and therefore expression becomes (y-x)/xy < 0 which means is y<x Now lets take values y=2 , x=3, the expression is true i.e <0 y=-2 and x=-3, the expression is false ie >0 So not sufficient from St 2, we get x= 3/2y Now y=4, x=6, Expression is False ie >0 y=-4, x=-6, expression is true i.e <0 So alone not sufficient Now Combining we get y<x and x=3/2y ----> y< 3/2y which means Y > 0 and x>y. So we have x>y>0. For this condition, the expression is always false ie. ( 1+ (x-y)^2 )/(x-y) >0 Can't thank you enough in solving inequalities the way you just did. Thanks Mridul
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Re: Tough DS Algebra
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14 Jan 2013, 23:43
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