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Re: m < 0 < n [#permalink]
Hi Brent GreenlightTestPrep, I got D here. Not sure where did it go wrong?
Thanks Brent

m^(-1) - n^(-1)
1/m - 1/n
n-m/mn
multiply with mn and left with n-m
n=1, m=-1 > n-m = 0 = with B :0
n=2, m=-3 > n-m = 5 > B :0
So is D.
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m < 0 < n [#permalink]
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I am not sure how you did manipulate

we do have \(\frac{1}{m} -\frac{1}{n}\)

\(\frac{n-m}{mn}\)

Now, iff you pick numbers such as m=-2 and n =1

we have

\(\frac{1+2}{-2*1}=\frac{2}{-2}=-1\) A will be always negative

B is the answer
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Re: m < 0 < n [#permalink]
Carcass wrote:
I am not sure how you did manipulate

we do have \(\frac{1}{m} -\frac{1}{n}\)

\(\frac{n-m}{mn}\)

Now, iff you pick numbers such as m=-2 and n =1

we have

\(\frac{1+2}{-2*1}=\frac{2}{-2}=-1\) A will be always negative

B is the answer


Thanks Carcass for your reply.
m < 0 < n , think I probably made the mistake here by taking m as positive here too of doing multiplication of mn to both side to elimiate it. That's why it has gone wrong.
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Re: m < 0 < n [#permalink]
Expert Reply
You are doing careless mistakes that a student cannot afford when he/she takes the GRE test.

Be careful
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Re: m < 0 < n [#permalink]
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