Meetup wrote:
Hi,
I tried solving this question as below:
(m-n)/(m+n)>1,
Square both sides, now since the denominator is squared - even if the value is -ve it becomes +ve. So, (m+n)^2 is positive and you can remove the same by multiplying it to both sides without changing the sign. So,
(m-n)^2 > 1(m+n)^2,
m^2 - 2mn + n^2 > m^2 + 2mn + n^2,
-2mn > 2mn,
0 > 2mn + 2mn
0 > 4mn.
The question then becomes: Do m and n have opposite signs?
st 1: Insufficient as we do not know what is m,
st 2: Insufficient as we do not know what is n,
Combining both,
we know m is +ve and n is -ve. Hence (c).
I know i goofed up somewhere, can someone please help is locating my error?
Thanks.
You are making the cardinal sin of inequalities, squaring variables in an inequality. You can not do that unless you know what sign are the variables.
You must proceed like this:
Is (m-n)/(m+n)>1, ---> Is (m-n)/(m+n) - 1>0 ---> \(\frac{m-n-m-n}{m+n}>0\) --->\(\frac{-2n}{m+n}>0\) ---> \(\frac{2n}{m+n}<0\) (you change the sign when multiplying by -1 or any negative number)
---> Is \(\frac{n}{m+n}<0\)
Per statement 1, n<-2 ----> 2 Cases, n=-3, m=5 ---> \(\frac{-3}{-3+5} = <0\) but with m=-1 and n=-3 --->\(\frac{-3}{-3-1} = >0\). Thus not sufficient.
Per statement 2, Proceeding as above, not sufficient. 2 cases , n =-6, m =5 or n=-3, m =10
Combining the 2 Statements you get, n<-2 and m>1 again clearly you can get 2 cases giving contradictory answers, making E the correct answer.
Hope this helps.
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