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happyapple123
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m and n are both integers. Is m>n? 30 May 2020, 09:24
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47% (02:11) correct 53% (01:40) wrong based on 55 sessions

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

(1) m/n > 1
(2) (m-n)/n < (m-n)/m

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E
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NitishJain
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m and n are both integers. Is m>n? 30 May 2020, 09:36
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E.

Statement1: m/n >1

it is possible if both are +ve or both are -ve

let's say m=6, n =2. m/n >1. so is m>n. YES
let's say m=-6, n =-2. m/n >1. so is m>n. NO

NOT SUFFICIENT

Statement 2: (m-n)/n < (m-n)/m. or m/n - 1 < 1- n/m. or m/n + n/m < 2

Here also combination of +ve and -ve values for m,n fails to uniquely suggest m>n NOT SUFFICIENT

Combining both:

m/n> 1
m/n + n/m < 2

=> n/m < 1

Here also the combination of +ve and -ve values for m,n fails to uniquely suggest m>n NOT SUFFICIENT
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m and n are both integers. Is m>n? 30 May 2020, 09:53
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NitishJain

I think B is sufficient?

B is sufficient if you plug in m=2 and n=1. Thus m>n. Can't find any examples to prove m<n.
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m and n are both integers. Is m>n? 30 May 2020, 09:57
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happyapple123
NitishJain

I think B is sufficient?

B is sufficient if you plug in m=2 and n=1. Thus m>n. Can't find any examples to prove m<n.
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Hello happyapple123

try m = -2 and n = -1,here n >m
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m and n are both integers. Is m>n? 30 May 2020, 09:59
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happyapple123
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I think B is sufficient?

B is sufficient if you plug in m=2 and n=1. Thus m>n. Can't find any examples to prove m<n.
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For (2) to get a NO answer, try m < 0 and n > 0.
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m and n are both integers. Is m>n? 30 May 2020, 10:00
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NitishJain

I think B is sufficient?

B is sufficient if you plug in m=2 and n=1. Thus m>n. Can't find any examples to prove m<n.

Hello happyapple123

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

Those values do not satisfy (m-n)/n < (m-n)/m.
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m and n are both integers. Is m>n? 30 May 2020, 10:05
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Bunuel
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NitishJain

I think B is sufficient?

B is sufficient if you plug in m=2 and n=1. Thus m>n. Can't find any examples to prove m<n.

Hello happyapple123

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

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

Correct sir. thanks.
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m and n are both integers. Is m>n? 30 May 2020, 10:22
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Bunuel

Thanks. Does it mean the answer is C not E then?
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m and n are both integers. Is m>n? 30 May 2020, 10:28
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Bunuel

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

The question is flawed. m/n > 1 and (m-n)/n < (m-n)/m cannot simultaneously be true. Is the question really from egmat?
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m and n are both integers. Is m>n? 30 May 2020, 11:17
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happyapple123
m and n are both integers. Is m>n?

(1) m/n > 1
(2) (m-n)/n < (m-n)/m
Show more

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.
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m and n are both integers. Is m>n? 30 May 2020, 23:24
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happyapple123
m and n are both integers. Is m>n?

(1) m/n > 1
(2) (m-n)/n < (m-n)/m
Show more

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
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m and n are both integers. Is m>n? 31 May 2020, 04:26
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Bunuel
happyapple123
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?
Show more
Excatly.
From 1
m/n>1
and From 2
mn<0
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m and n are both integers. Is m>n? 31 May 2020, 04:28
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Kinshook
happyapple123
m and n are both integers. Is m>n?

(1) m/n > 1
(2) (m-n)/n < (m-n)/m

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

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