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12 Nov 2009, 09:45
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
22% (01:52) correct
78%
(01:54)
wrong
based on 893
sessions
History
Date
Time
Result
Not Attempted Yet
12 Nov 2009, 10:14
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Economist
This is a good one. +1 Economist for it.
If xy ≠ 0, is x > y?
Question asks whether \(x>y\)?
(1) 4x = 3y --> \(x=\frac{3}{4}y\), well this one is relatively easy. This statement only tells that x and y have the same sign. When they are both positive then \(x<y\), BUT when they are both negative \(y<x\). Not sufficient.
(2) \(|y - x| = x - y\) --> \(|y - x| = -(y-x)\). This means that \(y - x\leq{0}\) --> \(x\geq{y}\). Thus, this statement says that x can be more than or equal to y. Not sufficient.
(1)+(2) From (1) (\(x=\frac{3}{4}y\)) it follows that \(x\) is not equal to \(y\) (bearing in mind that xy ≠ 0), hence from (2): \(x>y\). Sufficient.
Answer: C.
12 Nov 2009, 10:32
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lagomez
One more thing: the red part is not correct.
\(|y - x| = x - y\). Look at the LHS it's absolute value, it's never negative, hence RHS or \(x-y\) is never negative, \(x-y>=0\) or \(y-x<=0\). If so then only one possibility for \(|y - x|\), it must be \(-y+x\).
So we'll have:
Condition: \(y-x<=0\), which is the same as \(x>=y\)
And: \(|y - x| = x - y\)--> \(-y+x=x-y\), --> \(0=0\).
The above means that \(|y - x| = x - y\) is always true for \(x>=y\). But as we concluded earlier it's not enough. We need to be sure that \(x>y\). And \(x>=y\) leaves the possibility that they are equal, which is removed with the statement (1) when considering together. Hence C.
> 
silly Mistake and end up with 
