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

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Is 1/(x-y) < y - x [#permalink]  07 Jan 2011, 08:09
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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 10 Jul 2013, 01:31, edited 2 times in total.
Edited the OA.
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Re: Tough DS Algebra [#permalink]  07 Jan 2011, 08:37
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.

Therefore answer is E.
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Re: Tough DS Algebra [#permalink]  07 Jan 2011, 09:15
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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, 03:24
**
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, 06:45
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 $$y$$are 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, 07:14
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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 $$y$$are 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, 04:16
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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, 18:54
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, 01:28
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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, 05:16
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, 22:43
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
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Re: Is 1/(x-y) < y - x [#permalink]  14 Jul 2013, 23:44
Expert's post
From 100 hardest questions
Bumping for review and further discussion.
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Re: Is 1/(x-y) < y - x [#permalink]  07 Aug 2014, 21:46
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Re: Is 1/(x-y) < y - x [#permalink]  15 Sep 2014, 12:54
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.

Hi Bunuel.
When taking both statements together, I substituted x=3/2 and y=2/3, or x=-3/2 and y=-2/3
When taking into account statement 1, I noticed that we can't have the negative value for the numbers, as then statement 1 won't hold true.
Is this true? Am I missing something?
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Re: Is 1/(x-y) < y - x [#permalink]  15 Sep 2014, 20:00
Expert's post
ronr34 wrote:
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.

Hi Bunuel.
When taking both statements together, I substituted x=3/2 and y=2/3, or x=-3/2 and y=-2/3
When taking into account statement 1, I noticed that we can't have the negative value for the numbers, as then statement 1 won't hold true.
Is this true? Am I missing something?

Yes, when we combine the statements we get that x > y and 2x = 3y, so x > y > 0.
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Re: Is 1/(x-y) < y - x [#permalink]  07 Oct 2015, 03:27
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Re: Is 1/(x-y) < y - x [#permalink]  07 Oct 2015, 04:33
Hi All,
I attended this question quite late from the posting date but i am getting the different answer D.

1) 1/x<1/y --- which means x>y..so for the positive values the given expression holds NO, and for the negaive values X=-1 and Y=-2 (X>Y) the expression says NO.

2) 2x=3y --- which means x>y..so the same solution like above

So both gives the answer.

Can you help.
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Re: Is 1/(x-y) < y - x [#permalink]  07 Oct 2015, 04:44
Expert's post
asethi wrote:
Hi All,
I attended this question quite late from the posting date but i am getting the different answer D.

1) 1/x<1/y --- which means x>y..so for the positive values the given expression holds NO, and for the negaive values X=-1 and Y=-2 (X>Y) the expression says NO.

2) 2x=3y --- which means x>y..so the same solution like above

So both gives the answer.

Can you help.

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Re: Is 1/(x-y) < y - x [#permalink]  07 Oct 2015, 04:46
asethi wrote:
Hi All,
I attended this question quite late from the posting date but i am getting the different answer D.

1) 1/x<1/y --- which means x>y..so for the positive values the given expression holds NO, and for the negaive values X=-1 and Y=-2 (X>Y) the expression says NO.

2) 2x=3y --- which means x>y..so the same solution like above

So both gives the answer.

Can you help.

Be very careful with multiplying or diving inequalities with variables for which you do not know the signs of.

In this case as well, you do not know whether x and y are both positive or both negative or one negative one positive etc.

Without this information you can not say x>y when you are given 1/x<1/y

Case in point, consider the cases (3,2) and (-2,-3) you get different answers and because of this the first statement is not sufficient.

You are making the same mistake when you are analyzing statement 2 alone.

Make sure to be extra careful when dealing with unknown variables in inequalities.

Hope this helps.
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Re: Is 1/(x-y) < y - x   [#permalink] 07 Oct 2015, 04:46
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