Is m>n?

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Source: — Data Sufficiency |

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by Neo Anderson » Wed Mar 14, 2012 11:02 am
from statement 2 you can get:
(m-n)^2/mn > 0
=> mn > 0
thus if m>0; n>0 or if m<0 ; n<0
no specific answer thus insufficient

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by sam2304 » Wed Mar 14, 2012 11:53 pm
Neo Anderson wrote:from statement 2 you can get:
(m-n)^2/mn > 0
=> mn > 0
thus if m>0; n>0 or if m<0 ; n<0
no specific answer thus insufficient
How did you get this, "(m-n)^2/mn > 0" ?

Is m > n ?

1.m/n > 1 => m > n if both are +ve
m < n if both are -ve

2. (m-n)/n > (m-n)/m
divide by m-n we get 1/n > 1/m
multiply by mn we get m > n
again we don't know the sign of both m and n. So m > n and m < n both are possible.

Combining both we don't get any information.

Hence E.
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by GMATGuruNY » Thu Mar 15, 2012 3:59 am
pappueshwar wrote:Is m>n?

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

OA IS E
Algebra often is the best way to prove that a statement is SUFFICIENT.
To prove that a statement is INSUFFICIENT, plugging in numbers often is faster.

Statement 1: m/n > 1.
It's possible that m=2 and n=1, in which case m>n.
It's possible that m=-2 and n=-1, in which case m<n.
INSUFFICIENT.

Statement 2: (m-n)/n > (m-n)/m

If m=2 and n=1, then m-n=1.
1/1 > 1/2
1 > 1/2.
In this case, m>n.

If m=-2 and n=-1, then m-n = -2-(-1) = -1.
-1/-1 > -1/-2
1 > 1/2.
In this case, m<n.

Since both statements are satisfied by m=2,n=1 and by m=-2,n=-1, and in the first case m>n and in the second case m<n, INSUFFICIENT.

The correct answer is E.
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by [email protected] » Sat Mar 17, 2012 6:08 am
Whenever I take this question, I get the answer as D and hence get it wrong... According to me the solution is as follows:

Statement 1: m/n > 1
= m/n - 1 > 0

= (m - n)/n > 0

= (m - n) > 0

when you say that m - n > 0, then the equation m>n is proved to be correct...


Statement 2:

Dividing both sides by (m-n), all we are left with is:

1/n > 1/m

I tried putting all the values and got the answer as m > n...

Hence D... Could you please suggest as to where am I going wrong....
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by sam2304 » Sat Mar 17, 2012 6:17 am
[email protected] wrote: Statement 1: m/n > 1
= m/n - 1 > 0

= (m - n)/n > 0

= (m - n) > 0
You are multiplying by n in both sides. You assume n to be +ve. There is no information to support this. We should have two cases if the sign is not known

m - n > 0 if n is +ve and m - n < 0 if n is -ve
Statement 2:

Dividing both sides by (m-n), all we are left with is:

1/n > 1/m
Same mistake again. We don't know the sign of both m and n. So you have two cases m > n and m < n as well.
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