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

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Is 1/(x-y) < (x-y)?  [#permalink]

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Updated on: 01 Jul 2018, 23:45
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65% (hard)

Question Stats:

60% (01:16) correct 40% (01:37) wrong based on 72 sessions

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

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

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/is-1-x-y-y-x-126110.html

Originally posted by Anasthaesium on 03 Jan 2012, 05:58.
Last edited by Bunuel on 01 Jul 2018, 23:45, edited 3 times in total.
Renamed the topic and edited the question.
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Re: Is 1/(x-y) < (x-y)?  [#permalink]

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03 Jan 2012, 06:59
Just look at the question and you will realise that it is only possible if (x-y) is greater than 1. So the question is asking if (x-y) is greater than 1. Now the answer you give "C" is only possible if x and y are integers, which i don't find in the question. For example:

Let x = 0.5
Let y = -0.1

Then (x-y) = 0.6

so 1/(x-y) is greater than than (x-y)

Lets assume, x = 1 and y = -1

than x-y = 2

So 1/(x-y) = 0.5 which satisfies but we don't know if x and y are integers... Did you miss any information in the question while posting. Because the answer to the question you posed is E. If x and y are integers, than the answer is C.
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Re: Is 1/(x-y) < (x-y)?  [#permalink]

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03 Jan 2012, 09:53
Nope. I am in the same confusion. Got E too. So was hoping someone could help me out with that.
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Re: Is 1/(x-y) < (x-y)?  [#permalink]

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03 Jan 2012, 10:33
Anasthaesium wrote:
Nope. I am in the same confusion. Got E too. So was hoping someone could help me out with that.

I second that. E is the answer.
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Re: Is 1/(x-y) < (x-y)?  [#permalink]

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03 Jan 2012, 13:25
1
Anasthaesium wrote:
is 1/(x-y) < (x-y)?
1) x is positive
2) y is negative

+1 E

Combining (1) and (2), let's pick numbers:

a) $$x=2 , y= -1$$, then $$\frac{1}{(2+1)} < (2+1)$$

b) $$x=0.25 , y= -0.25$$, then $$\frac{1}{(0.50)} > (0.50)$$. Simplifying $$2 > 0.5$$

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

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03 Jan 2012, 14:19
1
Absolutely, E. Took 2.5 minutes to solve though. Number plugging is the key.

1. x +ve -> integers [4-2]>1/[4-2], but (3-2) = 1/(3-2) Insufficient
2. y -ve -> integers [2-(-1)]>1/[2-(-1)], but [0-(-1)] = 1/[0-(-1)]

Now, it gets tricky. Together, use both integers and fractions
[2-(-1)]>1/[2-(-1)]. However, [0.3-(-0.2)]<1/[0.3-(-0.2)]

TAKEAWAY: Start by proving insufficiency with integers only first, then move to fractions/Zero plugging if need be.

The OA here is incorrect.
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Re: Is 1/(x-y) < (x-y)?  [#permalink]

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03 Apr 2014, 05:42
Anasthaesium wrote:
Is 1/(x-y) < (x-y)?

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

Getting E too
Yes OA seems incorrect.Can Moderator Please change the OA or provide explanation for C. Thank you.
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Re: Is 1/(x-y) < (x-y)?  [#permalink]

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09 Aug 2016, 10:10
It is clear that both statements are insufficient since we don't know anything about either x or y

Now from 1+2 we know that (x-y) is positive. And hence we can multiply both sides by (x-y) Hence we have (x-y)^2>1 If x and y are integeres then the answer is Yes. But if x and y are fractions then answer will be No. Hence E
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Re: Is 1/(x-y) < (x-y)?  [#permalink]

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01 Jul 2018, 23:45
Anasthaesium wrote:
Is 1/(x-y) < (x-y)?

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

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.

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/is-1-x-y-y-x-126110.html
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Re: Is 1/(x-y) < (x-y)? &nbs [#permalink] 01 Jul 2018, 23:45
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