m and n are both integers. Is m>n?

Hello happyapple123

try m = -2 and n = -1,here n >m

For (2) to get a NO answer, try m < 0 and n > 0.

Those values do not satisfy (m-n)/n < (m-n)/m.

Correct sir. thanks.

Bunuel

Thanks. Does it mean the answer is C not E then?

The question is flawed. m/n > 1 and (m-n)/n < (m-n)/m cannot simultaneously be true. Is the question really from egmat?

Statement 1: \(m/n>1\) implies both m and n are either positive or negative.
Case1: m,n>0 => m>n
Case2: m,n<0 => m<n

Not sufficient.

Statement 2: \((m-n)/n\) < \((m-n)/m\)
Solving the above equation we'll obtain \(m/n+n/m<2\).
Case1: m<n, m=-2, n=5 => -0.4-2.5 < 0
Case2: m>n, m=5, n =-2 => -2.5-0.4 < 0

Not sufficient.

S1 and S2 together:
From S1 we know m,n>0 => m>n----(1) and m,n<0 => m<n-----(2)
m,n>0 => m-n>0. Cancelling the numerator on both sides we obtain \(1/n<1/m\) => \(m<n\). In contradiction with equation (1).
m,n<0 => m-n<0. The negative signs will cancel out in numerator and denominator and we obtain LHS > RHS. This also fails.

S1 contradicts S2. Even for option E, we need S1 and S2 complementing each other but resulting in more than 1 solution in which case we can't determine the exact solution.

Not sure, where I'm going wrong.

Given:m and n are both integers.
Asked: Is m>n?

(1) m/n > 1
m/n - 1 > 0
(m-n)/n>0
If n>0; m>n
But if n<0; m<n
NOT SUFFICIENT

(2) (m-n)/n < (m-n)/m
(m-n)/n - (m-n)/m < 0
(m-n)(1/n- 1/m) < 0
(m-n)(m-n)/mn < 0
1/mn < 0; Since (m-n)^2>=0
mn < 0
NOT SUFFICIENT

(1) + (2)
(1) m/n > 1
m/n - 1 > 0
(m-n)/n>0
(2) (m-n)/n < (m-n)/m
(m-n)/n - (m-n)/m < 0
(m-n)(1/n- 1/m) < 0
(m-n)(m-n)/mn < 0
1/mn < 0; mn<0; Since (m-n)^2>=0
If n<0; m>0; m>n
But if n>0; m<0; m<n
NOT SUFFICIENT

IMO E

Excatly.
From 1
m/n>1
and From 2
mn<0

From 1
m/n>1
From 2
mn<0
How it is possible that Both are simultaneously true?