Alternative resolution to find a ratio from averages. 
Problem: A club sold an average(arithmetic mean) of 92 raffle tickets per number. Among the female members, average number sold was 84, and among the male members, the average number sold was 96. What was the ratio of the number of male members to the number of female members in the club?

I solved it with an equation $92(m+n)=96m+84n\implies \frac{m}{n}=\frac{2}{1}$. So the answer is 2:1. However the solution presented an alterneaitve method that I have never seen but it is very practical.Since I intend to submit to the GRE I need to be as fast as possible.

Alternative Resolution:
$92-84=8\::\:96-92=4\implies2:1$.

I tried to prove this out, in order to understand the mechanism. But I have no idea on how to prove this alternative method is right. And it weird to me why the differences invert.
Question:
How to prove the alternative method(abstractly)?
Thanks in advance!

Assume the actual average is $k$, the average of men is $m$, and the average of women is $n$, with $m>k>n$.
Assume there are $a$ men and $b$ women.
Then $(a+b)k=am+bn$.
This gives $ak+bk=am+bn$.
Since $m>k$ and $k>n$,
$a(m-k)=b(k-n)$
This gives $$\frac ab={k-n\over m-k}$$
$\text W^5.$
