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# Is 1 / (x-y) < y - x

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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
[Reveal] Spoiler: OA

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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

Both Statments together - Intutitively, neither statement removes the ambiguity of the sign of X or Y. I can't think of a proof.

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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.

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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-x

Statement 1
Given \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<x
left 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 2
Given 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 = false
Two: if both x and y are negative (i.e. x=-3, y=-2), results in 1/-1 < 1 --> equation = true
Insufficient.

Statement 1 + 2
Applying 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.

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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 = 3y

I 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 > 0

With that information we can conclude that the answer is No. C is correct.

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Re: Is 1/(x-y) < y - x? [#permalink]  27 Aug 2012, 08:14
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metallicafan wrote:
Is \frac{1}{(x-y)} < y - x ?

(1) \frac{1}{x} < \frac{1}{y}
(2) 2x = 3y

I 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 > 0

With that information we can conclude that the answer is No. C is correct.

Merging similar topics. You are right, answer should be C, not E.
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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.

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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
shankar245 wrote:
Hi Buneul,

Can you please explain this part?
if both are negative cross multiply and flip the sigh twice to get y<x again)

Given: 1/x<1/y. Now, if both x and y are negative, then when we multiply both parts by negative x we should flip the sign and write 1>x/y. Now, multiply both sides by negative y and flip the sign again to get y<x.

Hope it's clear.
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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.

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
Re: Tough DS Algebra   [#permalink] 14 Jan 2013, 23:43
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