2000 AMC 10 Problem 22

1. Set up names. Let $p$ be the number of people, $m$ the total milk and $c$ the total coffee, in ounces. Both $m$ and $c$ are more than $0$.
2. Two facts. Everyone drinks $8$ ounces, so all together $m + c = 8p$. Angela's cup is
$$\frac{m}{4} + \frac{c}{6} = 8.$$
3. Could there be $4$ people? Then Angela's $8$ ounces would be $\frac14$ of all the drink, a quarter of the milk plus a quarter of the coffee. But she drank only a sixth of the coffee, which is less than a quarter. So her cup would be less than $8$. Four people is impossible, and fewer people is even worse.
4. Could there be $6$ people? Then her $8$ ounces would be $\frac16$ of everything, a sixth of the milk plus a sixth of the coffee. But she drank a quarter of the milk, which is more than a sixth. So her cup would be more than $8$. Six people is impossible, and more people is even worse.
5. So $p$ is more than $4$ and less than $6$, which leaves $p = 5$.
6. Check that $5$ really works. Try $16$ ounces of milk and $24$ of coffee, $40$ ounces in all, which is $5 \cdot 8$. Angela drinks $\frac{16}{4} = 4$ milk and $\frac{24}{6} = 4$ coffee, $8$ ounces. The other four people share $12$ milk and $20$ coffee: $3$ milk and $5$ coffee each, $8$ ounces. Every cup has some of both.

The answer is $\boxed{\textbf{(C)}\ 5}$.

Angela drank exactly one person's share, $\frac1p$ of all the drink. She took a bigger fraction ($\frac14$) of the milk and a smaller fraction ($\frac16$) of the coffee, so her fraction of the total lands somewhere in between. The only fraction $\frac1p$ strictly between $\frac16$ and $\frac14$ is $\frac15$. The words "never zero" are what rule out the edge cases $4$ and $6$.

Use algebra. Multiply Angela's equation by $12$: $3m + 2c = 96$. Multiply the total by $2$: $2m + 2c = 16p$. Subtract the second from the first: $m = 96 - 16p$. Since $m > 0$, $p < 6$. Then $c = 8p - m = 24p - 96$, and since $c > 0$, $p > 4$. So $p = 5$.