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There is a question for which I can't find an equation to solve it:

After $m$ couples leave a party, the total number of people drops by 20%, and the number of men drops by 12.5%.

Which case is greater:

  • One-third of the men before the couples leave
  • The number of women before the couples leave

I tried to fix a number, like 100 members. Then 10 couples leave (10 men, 10 women). The 10 men constitute 12.5% of whole men. So, the number of men must be 80 men. So the number of women must be 20, while a third of the men is 80/3 = 26. So, the first choice should be greater. Am I correct? What would be a formula to solve this problem, without fixing a number for people?

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    $\begingroup$ Does a couple have to be one man and one woman? $\endgroup$ – Max Punck-Institut Feb 14 '18 at 8:41
  • $\begingroup$ @MaxPunck-Institut yes $\endgroup$ – Ahmad Feb 14 '18 at 8:45
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You do fix the number of people, but you don't use a number, you use a letter. That way you know that your argument works no matter what number you could've chosen. Let $n$ signify the number of people before couples leave.

So, $m$ couples leaving means $2m$ people leaving, and this is $20\% = \frac15$ of the total number of people. Thus $$ 2m =\frac15n\\ $$ which is to say $n = 10m$. This second version is what we'll use later. (I didn't know from the start that I needed this second version. I added this line here after I realized way further down that the second version was what I actually needed.)

Also, we need a letter to signify the number of men before couples leaving. Let's use $k$, because $m$ is taken already. $m$ couples leaving means $m$ men leaving (assuming each couple is one man and one woman), and this is $12.5\% = \frac18$ of the total number of men: $$ m = \frac18k $$ Finally, the number of women before couples leave is $n-k$, since it is a standard assumption in math problems that everyone who is not a man is a woman.

So, we are asked to compare $n-k$ with $\frac13k$. We do this using the two equations we found earlier. First, $$ n-k = 10m - k $$ from the first equation. Then the second equation tells us that $$ 10m-k = 10\left(\frac18k\right) - k\\ = \frac{10}8k - k = \frac28k = \frac14k $$ So the number of women before the couples leave is equal to $\frac14k$, while a third of the number of men before the couples leaving is $\frac13k$. Now it is easy to see which is larger.

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Let $M$ be a number of men (I'd like to use $m$ here, but that has a meaning assigned already) and $W$ a number of women in the party. Then $m$ couples, meaning $m$ men and $m$ women is $20\% = \frac 15$ of $(W+M)$: $$2m=\frac 15(M+W)$$ and at the same time $m$ men from those couples is $12.5\% = \frac 18$ of $M$: $$m=\frac 18M$$ Substituting $m$ from the latter to the former we get $$\frac 14M = \frac 15(M+W)$$ hence, by multiplying both sides by $20$ $$5M=4M+4W$$ $$M=4W$$ Then one-third of the initial number of men $$\frac 13M = \frac 43W > W$$ is greater than the initial number of women.

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After $m$ couples leave: $$\begin{cases} N-2m=0.8N \\ M-m=\frac78M \end{cases} \Rightarrow \begin{cases} m=0.1N \\ M=8m=0.8N \ \ (\text{so} \ W=0.2N \end{cases}$$ Now compare: $$\frac13M > W \iff \frac{0.8N}{3}>0.2N \iff \frac{4}{15}N>\frac{3}{15}N.$$

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