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24 Jan 2011, 06:34
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
60% (02:32) correct
40%
(02:27)
wrong
based on 144
sessions
History
Date
Time
Result
Not Attempted Yet
24 Jan 2011, 07:54
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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.
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.
(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.
24 Jan 2011, 08:00
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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.
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.
The correct answer is B.
