OA C.
1)
Let m=1/3. n = 1/2, x =2
m/n = 2/3
(m+x)/(n+x) = 7/3 / 5/2 = 14/15 .
(m+x)/(n+x) > m/n
But m=5. n = 6 if x = -1
m/n = 5/6 = 0.83...
(m+x)/(n+x) = 4/5 = 0.8
(m+x)/(n+x) < m/n
Stmt 2:
x>0.
NS
But m=2. n = 3, x = 1
m/n = 2/3
(m+x)/(n+x) = 3/4
(m+x)/(n+x) > m/n
But m=3. n = 2, x = 1
m/n = 3/2
(m+x)/(n+x) = 4/3
(m+x)/(n+x) < m/n
Combining the two.
m<n
m/n <1.
So the conditions set in each other are cancelled out.
Hence sufficient.
C
I know you can take similar values in all cases during the exam. I just tried to illustrate with as many values as possible.
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