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

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Is 1/(x-y) < (y - x) ? [#permalink]

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15 Jan 2012, 06:26
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Is 1/(x-y) < (y - x) ?

(1) y is positive.
(2) x is negative.
[Reveal] Spoiler: OA
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15 Jan 2012, 06:35
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Expert's post
Apoorva81 wrote:
Is 1/x-y < y - x ?

(1) y is positive.
(2) x is negative.

can you please provide detailed explanation..??

Is $$\frac{1}{x-y}<y - x$$?

(1) $$y$$ is positive, clearly insufficient, as no info about $$x$$;
(2) $$x$$ is negative, also insufficient, as no info about $$y$$;

(1)+(2) We have $$y=positive$$ and $$x=negative$$, thus $$y>x$$ (this can be rewritten as $$y-x>0$$ or $$0>x-y$$). Now: $$LHS=\frac{1}{x-y}=\frac{1}{negative}=negative$$, and $$RHS=y-x=positive$$ thus $$\frac{1}{x-y}=negative<y-x=positive$$. Sufficient.

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Re: Is 1/x-y<y-x? (1) y is positive (2) x is negative [#permalink]

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15 Jan 2012, 10:07
Is 1/(x-y) < y - x ?

1
______ - (y-x) < 0
(x-y)

(1) y is positive.
(2) x is negative.

From 1 and 2 -

(x-y) is always negative.
(y-x) will be positive
-(y-x) will be negative

If it were 1/(x-y) < x - y then answer would have been E
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Re: Is 1/x-y<y-x? (1) y is positive (2) x is negative [#permalink]

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09 May 2012, 17:18
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Slightly different approach, I hope everyone can see the attached images with solutions.

Dabral

Apoorva81 wrote:
Is 1/x-y < y - x ?

(1) y is positive.
(2) x is negative.

can you please provide detailed explanation..??

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02 Apr 2013, 20:52
In order to separate your x's and y's from each other, you need to know which term is larger. This is because you need to know whether x-y or y-x is negative (one of them will be, unless they are equal). Taking both statements together, you know that the two variables are not equal, and you can manipulate the inequality, keeping in mind that you need to flip the sign when multiplying or dividing by a negative number.
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02 Apr 2013, 20:53
I explained it in kind of a backwards way. The first thing you need to know in that problem is whether x and y are equal. No single statement tells you that.

Sorry for the double post, but it seems I can't edit my previous post while viewing the forum with Tapatalk.

Sent from my HTC Glacier using Tapatalk 2
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Re: Is 1/(x-y) < (y - x) ? [#permalink]

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22 Feb 2014, 14:15
Bumping for review and further discussion.
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09 May 2014, 08:38
Bunuel wrote:
Apoorva81 wrote:
Is 1/x-y < y - x ?

(1) y is positive.
(2) x is negative.

can you please provide detailed explanation..??

Is $$\frac{1}{x-y}<y - x$$?

(1) $$y$$ is positive, clearly insufficient, as no info about $$x$$;
(2) $$x$$ is negative, also insufficient, as no info about $$y$$;

(1)+(2) We have $$y=positive$$ and $$x=negative$$, thus $$y>x$$ (this can be rewritten as $$y-x>0$$ or $$0>x-y$$). Now: $$LHS=\frac{1}{x-y}=\frac{1}{negative}=negative$$, and $$RHS=y-x=positive$$ thus $$\frac{1}{x-y}=negative<y-x=positive$$. Sufficient.

Why can't we do the following rephrase?

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

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

-1 < (y-x)^2

since the RHS is a square, irrespective of x and y values the equation is satisfied.
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09 May 2014, 09:25
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Expert's post
Bunuel wrote:
Apoorva81 wrote:
Is 1/x-y < y - x ?

(1) y is positive.
(2) x is negative.

can you please provide detailed explanation..??

Is $$\frac{1}{x-y}<y - x$$?

(1) $$y$$ is positive, clearly insufficient, as no info about $$x$$;
(2) $$x$$ is negative, also insufficient, as no info about $$y$$;

(1)+(2) We have $$y=positive$$ and $$x=negative$$, thus $$y>x$$ (this can be rewritten as $$y-x>0$$ or $$0>x-y$$). Now: $$LHS=\frac{1}{x-y}=\frac{1}{negative}=negative$$, and $$RHS=y-x=positive$$ thus $$\frac{1}{x-y}=negative<y-x=positive$$. Sufficient.

Why can't we do the following rephrase?

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

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

-1 < (y-x)^2

since the RHS is a square, irrespective of x and y values the equation is satisfied.

The point is that we don't know whether y-x is positive or negative. If it's positive, then yes we'd have -1 < (y-x)^2 but if it's negative, then we'd have -1 > (y-x)^2 (flip the sign when multiplying by negative value).

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero.
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Is 1/(x-y) < (y-x) ? [#permalink]

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09 Aug 2014, 01:53
Is 1/(x-y) < (y-x) ?

1) x is +ve
2) y is -ve

I took a very long time to solve this

Does anyone know a shortcut? or a simpler method?

Source: 4gmat
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Re: Is 1/(x-y) < (y-x) ? [#permalink]

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09 Aug 2014, 02:15
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alphonsa wrote:
Is 1/(x-y) < (y-x) ?

1) x is +ve
2) y is -ve

I took a very long time to solve this

Does anyone know a shortcut? or a simpler method?

Source: 4gmat

The question can be re written as

$$\frac{1}{(x-y)} +( x-y )<0$$

Or
$$\frac{1+ (x-y)^2}{(x-y)} <0$$
For the expression to be less than zero,numerator and denominator should be of opp sign
Now$$(x-y)^2\geq{0}$$

So numerator is positive ,therefore if denominator is negative then answer to the question is yes or else no

st 1 x is positive let x=1 and y=-10 then x-y=9answer is no
But if x=1 and y=3 then answer to the question is yes

St 1 not sufficient .. Option A and D ruled out

St 2 y is negative
If y= -3 and x =-8 then ans to the question is yes
If y=-3 and x =8 then answer to the q. Is no

St 2 alone not sufficient

Combining we see that x-y=+ve

So ans is C
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Re: Is 1/(x-y) < (y-x) ? [#permalink]

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09 Aug 2014, 02:54
With the way you put it, this question seemed like a below 500 question :\

I don't know why I took so long to decipher it was option C .
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Re: Is 1/(x-y) < (y-x) ? [#permalink]

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09 Aug 2014, 03:05
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WoundedTiger wrote:

The question can be re written as

$$\frac{1}{(x-y)} +( x-y )<0$$

Or
$$\frac{1+ (x-y)^2}{(x-y)} <0$$
For the expression to be less than zero,numerator and denominator should be of opp sign
Now$$(x-y)^2\geq{0}$$

So numerator is positive ,therefore if denominator is negative then answer to the question is yes or else no

st 1 x is positive let x=1 and y=-10 then x-y=9answer is no
But if x=1 and y=3 then answer to the question is yes

St 1 not sufficient .. Option A and D ruled out

St 2 y is negative
If y= -3 and x =-8 then ans to the question is yes
If y=-3 and x =8 then answer to the q. Is no

St 2 alone not sufficient

Combining we see that x-y=+ve

So ans is C

WoundedTiger, it's better to do one step more. In data sufficiency if you have very simple statements you need to understand what you are looking for.

So, a little bit different solution:

The question can be re written as

$$\frac{1}{(x-y)} +( x-y )<0$$

Or
$$\frac{1+ (x-y)^2}{(x-y)} <0$$

Since $$1+(x-y)^2>0$$, denominator must be negative: $$x-y<0$$ or $$x<y$$. Hence, we need to find: is $$x$$ less than $$y$$?

(1) Don't know anything about $$y$$. Insufficient
(2) Don't know anything about $$x$$. Insufficient

(1)+(2) Since $$x>0$$ and $$y<0$$, we have $$x>y$$. Sufficient

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Re: Is 1/(x-y) < (y-x) ? [#permalink]

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09 Aug 2014, 03:31
alphonsa wrote:
With the way you put it, this question seemed like a below 500 question :\

I don't know why I took so long to decipher it was option C .

It happens to the best of people especially if the question is done during exam time....
One more thing..I think you need to look at these links for books for your preparation

best-gmat-math-prep-books-reviews-recommendations-77291-180.html
gmat-prep-software-analysis-and-what-if-scenarios-146146.html#p1171760
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Re: Is 1/(x-y) < (y-x) ? [#permalink]

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09 Aug 2014, 04:43
smyarga wrote:
WoundedTiger wrote:

The question can be re written as

$$\frac{1}{(x-y)} +( x-y )<0$$

Or
$$\frac{1+ (x-y)^2}{(x-y)} <0$$
For the expression to be less than zero,numerator and denominator should be of opp sign
Now$$(x-y)^2\geq{0}$$

So numerator is positive ,therefore if denominator is negative then answer to the question is yes or else no

st 1 x is positive let x=1 and y=-10 then x-y=9answer is no
But if x=1 and y=3 then answer to the question is yes

St 1 not sufficient .. Option A and D ruled out

St 2 y is negative
If y= -3 and x =-8 then ans to the question is yes
If y=-3 and x =8 then answer to the q. Is no

St 2 alone not sufficient

Combining we see that x-y=+ve

So ans is C

WoundedTiger, it's better to do one step more. In data sufficiency if you have very simple statements you need to understand what you are looking for.

So, a little bit different solution:

The question can be re written as

$$\frac{1}{(x-y)} +( x-y )<0$$

Or
$$\frac{1+ (x-y)^2}{(x-y)} <0$$

Since $$1+(x-y)^2>0$$, denominator must be negative: $$x-y<0$$ or $$x<y$$. Hence, we need to find: is $$x$$ less than $$y$$?

(1) Don't know anything about $$y$$. Insufficient
(2) Don't know anything about $$x$$. Insufficient

(1)+(2) Since $$x>0$$ and $$y<0$$, we have $$x>y$$. Sufficient

I tried this way.

we have 1/(x-y) < y-x

1) says x is positive, we have no value of Y, not sufficient

2) says y is negative, no value of x defined, not sufficient.

1+2 , x is +ve and y is -ve => 1/+ve < -ve => +ve < -ve, not possible.

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Re: Is 1/(x-y) < (y - x) ? [#permalink]

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12 Aug 2014, 07:17
alphonsa wrote:
Is 1/(x-y) < (y-x) ?

1) x is +ve
2) y is -ve

I took a very long time to solve this

Does anyone know a shortcut? or a simpler method?

Source: 4gmat

Merging topics. Please refer to the discussion above.

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Re: Is 1/(x-y) < (y - x) ? [#permalink]

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25 Aug 2015, 22:48
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Re: Is 1/(x-y) < (y - x) ? [#permalink]

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13 Mar 2016, 07:16
Excellent Explanation dabral
Also i did it a bit differently..
I made two cases => one for Y>x and the other for Y<x
hence i marked C
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Re: Is 1/(x-y) < (y - x) ? [#permalink]

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02 Dec 2016, 02:27
In the question it does not say it has to be an integer. The solution does not work for rational numbers. Could you explain?

Posted from GMAT ToolKit
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Re: Is 1/(x-y) < (y - x) ? [#permalink]

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02 Dec 2016, 03:49
In the question it does not say it has to be an integer. The solution does not work for rational numbers. Could you explain?

Posted from GMAT ToolKit

It's not clear what you mean at all. The solutions above does not distinguish between rational and irrational numbers. Why/how should they? The solutions are fine and are correct generally for all real numbers.
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Re: Is 1/(x-y) < (y - x) ?   [#permalink] 02 Dec 2016, 03:49

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