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# If a*b = a/b - b/a and m > n > 0, then which of following must be true

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Re: If a*b = a/b - b/a and m > n > 0, then which of following must be true [#permalink]
What does a and b have to do with m and n? The question is unclear.
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Re: If a*b = a/b - b/a and m > n > 0, then which of following must be true [#permalink]
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GittinGud wrote:
What does a and b have to do with m and n? The question is unclear.

It basically tells you the operations you have to perform. You can think of a and b as placeholders that are just there to demonstrate how the equation works and m, n as you actual inputs.

Hope this helps.
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Re: If a*b = a/b - b/a and m > n > 0, then which of following must be true [#permalink]
stonecold wrote:
If $$a*b=\frac{a}{b}-\frac{b}{a}$$ and $$m>n>0$$, then which of following must be true?

(I) $$\frac{1}{m}*\frac{1}{n} > \frac{1}{n} *\frac{1}{m}$$

(II) $$\frac{1}{m} *\frac{1}{n} < \frac{1}{n} *\frac{1}{m}$$

(III) $$m*n<0$$

(IV) $$m*n>0$$

A) I and III
B) I and IV
C) II and III
D) II and IV
E) only IV

The format of the question is incorrect. It is not given as a function or user defined operator. You cannot use a standard operator as a user defined operator that too without explaining explicitly. We can guess that the relation given for a and b is supposed to hold for m and n too to solve it but an actual GMAT question would need to be formatted differently.

Given the question as is, I would just say that m > n > 0 so m and n are positive and hence m*n > 0. This is the ONLY statement that will be true.

If you define a new operator such as
a#b = a/b - b/a

now, you can worry about (1/m) # (1/n) and (1/n)#(1/m)

(1/m) # (1/n) = n/m - m/n = (n - m)/mn (since n < m, this is negative)
(1/n)#(1/m) = m/n - n/m = (m - n)/mn (since m > n, this is positive)

So II is correct.

m#n = m/n - n/m = (m - n)/mn (since m > n, this is positive)
So IV is correct.

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Re: If a*b = a/b - b/a and m > n > 0, then which of following must be true [#permalink]
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Re: If a*b = a/b - b/a and m > n > 0, then which of following must be true [#permalink]
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