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Re: 700+ INIQUALITIES ( HELP NEEDED) [#permalink]
09 Aug 2006, 13:16

yezz wrote:

rephrasing the question is x-y/x+y > 1 ie : is x-y > x+y ps: ( x+ y can not be zero)

This is wrong. We don't know the sign of (x+y) so you can not multiply. If x+y is -ve then the sign of equality should be reversed.

Answer should be E.

Here is the explanation:

Question:Is (x-y)/(x+y) >1 ?

Lets draw a matrix
x-------y-------------answer to above question
pos---pos------------------always less than 1.
pos---neg------------------may be less or greater depending on values of x and y
neg---pos------------------may be less or greater depending on values of x and y
neg---neg----------------- always less than 1.

St1: x>0: Look at choices 1 and 2: INSUFF

St2: y<0: Look at choices 2 and 4: INSUFF

Combined: x> 0 and y <0: Look at choice 2.: INSUFF _________________

Re: 700+ INIQUALITIES ( HELP NEEDED) [#permalink]
09 Aug 2006, 13:53

yezz wrote:

If x not equal to y , is x-y/x+y > 1 ? 1. x > 0 2. y < 0

rephrasing the question is x-y/x+y > 1 ie : is x-y > x+y ps: (x+y cannot be zero) x-y-x-y > 0 ie: is y > zero Clearly the answer is b. am i right ????

first, the red part above is not correct. we cannot do that without knowing the valaues for (x-y) and (x+y). if they are +ves, we cannot do that. if (x-y) = 5 and (x+y) =7 and if we do so, 5>7 and this is not true.

using 1 and 2, we cannot say that (x-y)/(x+y) >1.

from i, x-y is +ve.
from ii, x+y can be -ve or +ve. if it is +ve, it is <(x-y) and the answer is yes. if it is -ve, it is again <(x-y) but (x-y)/(x+y) < 1 and the answer is no.