If m ≠ -n, is (m – n)/(m + n) > 1? (1) n < -2 (2) m > 1

You can not multiply or divide by variables when you do not know whether m+n >0 or <0. Refer to my post for the explanation.

Rearranging question we get (m-n)>(m+n) therefore 2n<0 or n<0. so question becomes is n<0?
1. n<-2 hence n will be<0 Sufficient.
2. m>1, not sufficient to answer the question
Hence answer is A

E should be the answer.
Per statement 1, lets assume m = -4 and n = -3 then output = -1/-7 which is less than 1
second possibility m = 4 n = -3 then output 7/1 which is greater than 1. hence 1 out.
Per statement 2, lets assume m = 4 and n = 8 , answer 12/-4 which is less than 1,
even combining them will not give correct answer

IMO : E

Question Stem: If m ≠ -n, is (m – n)/(m + n) > 1?

\(\frac{(m – n)}{(m + n)}\) - 1 >0
\(\frac{(m – n)}{(m + n)}\) - \(\frac{(m + n)}{(m + n)}\) >0
\(\frac{(–2n)}{(m + n)}\) >0

St 1: n < -2

\(\frac{(–2n)}{(m + n)}\) >0
Here Numerator will be positive.
Denominator?
Positive when m+n >0
Negative when m+n < 0
Hence nt suff

St 2: m > 1

We can't determine as nature of n is not known. As it is required to determine the nature of fraction.

Combined: n < -2 & m > 1

\(\frac{(–2n)}{(m + n)}\) >0
Here Numerator will be positive.

Positive when m+n >0 --[If n = -3 & m = 5 ]
Negative when m+n < 0 --[If n = -3 & m = 2]
Hence not suff

If n is negative, then irrespective of the value or sign of m, m - n will be greater than m + n.
Take 3 cases, with m as +, - or 0. You would find the answer.

Hence I think, (a) is the answer.

How about 2 cases:

n = -3, m = -4, ---> m-n / m+n = -4+3/-4-3 = -1/-7 = 1/7 < 1 but

n = -3, m = 7, ---> m-n / m+n = 7+3/7-3 = 10/4 = 2.5 > 1

So for statement 1, you get 2 different answers for n<-2. Hence this statement is NOT sufficient and hence A can not be the answer.

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.

it will be >1 only if |m|>|n|, and only when n is negative.

1. n<-2.
suppose m=4, n=-3
4+3/4-3 = 7>1 yes
suppose m=3, n=-4
3+4/3-4 = -7 <1 - no. so 1 out

2 alone is insufficient, as we don't know anything about n.

1+2
same example given above applies here as well. so insufficient.

E

Here Our best approach is to take the test cases and Prove the inequality both correct and incorrect.
hence we get E (choose M=100 and M=10 and N=-3 and N=-100 for the combination statement)

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