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# If x ≠ y, is 1/(x - y) < y - x?

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

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24 Jan 2011, 06:34
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Question Stats:

59% (01:56) correct 41% (01:16) wrong based on 46 sessions

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

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24 Jan 2011, 07:54
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rxs0005 wrote:
if x <> y

is 1 / x - y < y - x

S1 x^2 + y^2 < 4xy

S2 y < x

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, 08:00
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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If x ≠ y, is 1/(x - y) < y - x? [#permalink]

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24 Jan 2011, 08:35
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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, 21:03
You are right, thanks for pointing that.
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Anurag Mairal, Ph.D., MBA
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Re: If x ≠ y, is 1/(x - y) < y - x? [#permalink]

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10 Oct 2017, 00:22
rxs0005 wrote:
If x ≠ y, is 1/(x - y) < y - x?

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

(2) y < x

Isn't the condition always true?

1/(x - y) < y - x => 1/(-1)(y - x) < y - x => -1 < (y - x)^2

RHS is a square and will always be positive, there equation is true no matter what. Therefore, D should be the answer.

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

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10 Oct 2017, 00:25
praneetkakani wrote:
rxs0005 wrote:
If x ≠ y, is 1/(x - y) < y - x?

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

(2) y < x

Isn't the condition always true?

1/(x - y) < y - x => 1/(-1)(y - x) < y - x => -1 < (y - x)^2

RHS is a square and will always be positive, there equation is true no matter what. Therefore, D should be the answer.

Hope it helps.
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Re: If x ≠ y, is 1/(x - y) < y - x?   [#permalink] 10 Oct 2017, 00:25
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# If x ≠ y, is 1/(x - y) < y - x?

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