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# If x ≠ y, is 1/(x - y) < y - x?

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Director
Joined: 07 Jun 2004
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If x ≠ y, is 1/(x - y) < y - x? [#permalink]

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24 Jan 2011, 05:34
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If x ≠ y, is 1/(x - y) < y - x?

(1) x^2 + y^2 < 4xy

(2) y < x
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 43810
If x ≠ y, is 1/(x - y) < y - x? [#permalink]

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24 Jan 2011, 06:54
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If $$x\neq{y}$$ is $$\frac{1}{x-y}<y-x$$?

(1) x^2 + y^2 < 4xy --> if $$x=1>y=\frac{1}{2}$$ then the answer will be NO as in this case $$LHS=\frac{1}{x-y}>0$$ and $$RHS=y-x<0$$, so $$LHS>RHS$$ but if $$x=\frac{1}{2}<y=1$$ then the answer will be YES, as in this case $$LHS=\frac{1}{x-y}<0$$ and $$RHS=y-x>0$$, so $$LHS<RHS$$. Not sufficient.

(2) y < x --> we can rewrite it as $$y-x<0$$ or $$x-y>0$$ --> in this case $$LHS=\frac{1}{x-y}>0$$ and $$RHS=y-x<0$$, so $$LHS>RHS$$ directly gives the answer NO to he question. Sufficient.

Or you can simplify the question: is $$\frac{1}{x-y}<y-x$$? --> is $$y-x+\frac{1}{y-x}>0$$? is $$\frac{(y-x)^2+1}{y-x}>0$$? As the nominator ($$(y-x)^2+1$$) is always positive then the question basically becomes whether denominator ($$y-x$$) is positive too --> is $$y-x>0$$? or is $$y>x$$?

(1) Not sufficient.
(2) Answers NO to the question. Sufficient.

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Re: If x ≠ y, is 1/(x - y) < y - x? [#permalink]

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24 Jan 2011, 07:00
The question is $$\frac{1}{(x - y)} < -(x - y)$$
or Is $$1 < -(x - y)^2$$? or Is $$(x - y)^2 < -1$$?

(1) $$x^2 + y^2 < 4xy$$ implies $$x^2 + y^2 -2xy < 2xy$$ or $$(x - y)^2 < 2xy$$
Since we don't know any relation between x and y, so statement 1 is NOT SUFFICIENT.

(2) y < x implies (x - y) > 0 or $$(x - y)^2 > 0$$
So, answer to the main question is "No"
Statement 2 is SUFFICIENT.

[Reveal] Spoiler:

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Anurag Mairal, Ph.D., MBA
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Gurome, Inc.
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If x ≠ y, is 1/(x - y) < y - x? [#permalink]

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24 Jan 2011, 07:35
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Anurag@Gurome wrote:
The question is $$\frac{1}{(x - y)} < -(x - y)$$
or Is $$1 < -(x - y)^2$$? or Is $$(x - y)^2 < -1$$?

(1) $$x^2 + y^2 < 4xy$$ implies $$x^2 + y^2 -2xy < 2xy$$ or $$(x - y)^2 < 2xy$$
Since we don't know any relation between x and y, so statement 1 is NOT SUFFICIENT.

(2) y < x implies (x - y) > 0 or $$(x - y)^2 > 0$$
So, answer to the main question is "No"
Statement 2 is SUFFICIENT.

[Reveal] Spoiler:

The red part is not correct. You cannot multiply both sides of the inequality by $$x-y$$ as you don't know whether this expression is positive or negative. Also answer to the question "is $$(x - y)^2 < -1$$?" is always NO as the square of a number can not be negative. Basically if $$x-y>0$$ the answer is always NO and if $$x-y<0$$ the answer is always YES.

You can simplify the question as follows: is $$\frac{1}{x-y}<y-x$$? --> is $$y-x+\frac{1}{y-x}>0$$? is $$\frac{(y-x)^2+1}{y-x}>0$$? As the nominator ($$(y-x)^2+1$$) is always positive then the question basically becomes whether denominator ($$y-x$$) is positive too --> is $$y-x>0$$? or is $$y>x$$?
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Re: If x ≠ y, is 1/(x - y) < y - x? [#permalink]

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24 Jan 2011, 20:03
You are right, thanks for pointing that.
_________________

Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)
+91-99201 32411 (India)

Senior Manager
Joined: 02 Apr 2014
Posts: 395
If x ≠ y, is 1/(x - y) < y - x? [#permalink]

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10 Dec 2017, 13:36
$$1/(x-y) < (y-x)$$?
rearranging,
$$(1/(x-y)) + (x-y) < 0$$ ?

if $$x > y$$, then $$(1/(x-y)) + (x-y)$$ is positive
if $$x < y$$, then $$(1/(x-y)) + (x-y)$$ is negative,
so question becomes $$x < y$$ ?

Statement 1:
$$x^2 + y^2 < 4xy$$
=>$$x^2 + y^2 - 2xy < 2xy$$
=> $$(x - y)^2 < 2xy$$
since $$(x - y)^2$$ is greater than zero,
so $$2xy > 0$$, but this does not tell whether $$x < y$$ => Insuff

Statement 2: Directly answers the question => suff

If x ≠ y, is 1/(x - y) < y - x?   [#permalink] 10 Dec 2017, 13:36
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