Bunuel wrote:
GMAT98 wrote:
I want to know where is the error in following method for statement 2
1 < la-bl
la-bl > 1
then
a-b > 1 or a-b< -1
a-b > 1 or b-a> 1
since both sides are positive for first possibility
1/(a-b) < 1 or 1 < b-a ( i think problem is here)![:x]( "Mad")
combining
1/(a-b) < 1 < b-a
1/(a-b)< b-a
therefore yes... i know i am wrong somewhere.. but where i dontknow
1 < la-bl
la-bl > 1
then
a-b > 1 or a-b< -1
a-b > 1 or b-a> 1
since both sides are positive for first possibility
1/(a-b) < 1 or 1 < b-a ( i think problem is here)
combining
1/(a-b) < 1 < b-a
1/(a-b)< b-a
therefore yes... i know i am wrong somewhere.. but where i dontknow
Please elaborate your question.
Hi Bunuel
Pl help me with following problem
Is 1/ a-b < b-a
(1) a<b
(2) 1<|a-b|
My thought process is as follows
(1) a <b
a-b < 0 or 0 < b-a
1/(a-b) < 0 or 0 < b-a
negative will always be less than positive, therefore, statement one is sufficient
(2) 1 < la-bl
la-bl > 1
then opening absolute bracket
a-b > 1 or a-b< -1
a-b > 1 or b-a> 1
since a-b > 1 it implies a-b is positive and 1 is also positive therefore taking reciprocal
1/(a-b) < 1 or b-a > 1
combining
1/(a-b) < 1 < b-a
1/(a-b)< b-a
therefore statement two is also sufficient Therefore answer D but answer is A
... i know i am wrong somewhere.. but where i dont know
can you please go through my solution for second statement and pin-point the error