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03 May 2009, 16:16
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03 May 2009, 16:52
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S2 is clearly insufficient alone.
S1 will always be true if x is negative and y is positive, in which case xy < 0. S1 can also be true when both x and y are positive, or when both are negative (e.g. take x = 2 and y = 1), in which case xy > 0. So S1 is insufficient.
Taken together, if x > 0, then 1/x > 0. Since 1/y > 1/x > 0, y must also be positive, and xy must be positive. So we can answer the question, and the answer is 'no'. C.
S1 will always be true if x is negative and y is positive, in which case xy < 0. S1 can also be true when both x and y are positive, or when both are negative (e.g. take x = 2 and y = 1), in which case xy > 0. So S1 is insufficient.
Taken together, if x > 0, then 1/x > 0. Since 1/y > 1/x > 0, y must also be positive, and xy must be positive. So we can answer the question, and the answer is 'no'. C.
03 May 2009, 17:08
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IanStewart
Thanks for replying Ian,
I appreciate your help.
The question is similar to the one in Quotes.
Statement 1 can also be simplified like this
(1/x) < (1/y)
1 < (x/y) (multiplying both sides by x)
(x/y) > 1
If the above is correct, then it is the same question as in Quotes.
Then, How can we have Two different answers ???
I have tried and could not get over this.
