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

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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.

Answer: B.

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.

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

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New post 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:
The correct answer is B.

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

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New post 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 correct answer is B.


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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New post 24 Jan 2011, 20:03
You are right, thanks for pointing that.
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If x ≠ y, is 1/(x - y) < y - x? [#permalink]

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New post 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

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