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

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VP
Joined: 23 Feb 2015
Posts: 1257

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07 Mar 2019, 09:50
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76% (01:22) correct 24% (01:50) wrong based on 29 sessions

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Is $$\frac{1}{x-y}$$< $$(y-x)$$?
1) $$x>0$$
2) $$y<0$$

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Joined: 02 Aug 2009
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07 Mar 2019, 19:29
Is $$\frac{1}{x-y}$$< $$(y-x)$$?
1) $$x>0$$...x=2, y=1, Ans is NO, but X=2 and y=3, the answer will be yes..
2) $$y<0$$...x=2, y=-1, answer is NO, but x=-3 and y=-1, the answer will be yes..

Combined..
X>0 and y<0, so x-y>0 and $$\frac{1}{x-y}$$ will always be positive..
On the other hand y-x will be negative.. so $$\frac{1}{(x-y)}>y-x$$
Sufficient

C

SECOND way..
We can modify the original condition...
$$\frac{1}{x-y}$$< $$(y-x)$$.....$$\frac{1}{x-y}$$- $$(y-x)<0$$....$$\frac{1+(x-y)^2}{x-y}<0$$
The numerator is always positive, so basically question becomes .." Is x-y<0?" or Is x<y?
1) $$x>0$$...x=2, y=1, Ans is NO, but X=2 and y=3, the answer will be yes..
2) $$y<0$$...x=2, y=-1, answer is NO, but x=-3 and y=-1, the answer will be yes..
Combined x>0 and y<0.....so y<0<x, that is x>y
sufficient
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VP
Joined: 23 Feb 2015
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07 Mar 2019, 22:23
chetan2u wrote:
Is $$\frac{1}{x-y}$$< $$(y-x)$$?
1) $$x>0$$...x=2, y=1, Ans is NO, but X=2 and y=-1, the answer will be yes..

Hi chetan2u,
If x=2, and y=-1
$$\frac{1}{2-(-1)}$$<$$(-1-2)$$?
$$\frac{1}{2+1}$$<$$-3$$?
$$\frac{1}{3}$$<$$-3$$?
Is it possible in real life?
Did you miss anything or am I?
Thanks__
_________________
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Math Expert
Joined: 02 Aug 2009
Posts: 7971

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07 Mar 2019, 23:09
chetan2u wrote:
Is $$\frac{1}{x-y}$$< $$(y-x)$$?
1) $$x>0$$...x=2, y=1, Ans is NO, but X=2 and y=-1, the answer will be yes..

Hi chetan2u,
If x=2, and y=-1
$$\frac{1}{2-(-1)}$$<$$(-1-2)$$?
$$\frac{1}{2+1}$$<$$-3$$?
$$\frac{1}{3}$$<$$-3$$?
Is it possible in real life?
Did you miss anything or am I?
Thanks__

It was a typo .. was meant to be 3 and not -1.. Thanks
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Re: Is 1/x-y<y-x?   [#permalink] 07 Mar 2019, 23:09
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