2000 AMC 10 Problem 22

Let there be $p$ people, and let $m$ and $c$ be the total ounces of milk and coffee, both positive. Every cup is $8$ ounces, so $m + c = 8p$, which means one cup is exactly $\frac{1}{p}$ of everything poured. Angela's cup is
$$\frac{m}{4} + \frac{c}{6} = 8 = \frac{m + c}{p}.$$

Now compare. If $p \le 4$, then $\frac{m+c}{p} \ge \frac{m}{4} + \frac{c}{4} > \frac{m}{4} + \frac{c}{6}$, since $c > 0$. Contradiction. If $p \ge 6$, then $\frac{m+c}{p} \le \frac{m}{6} + \frac{c}{6} < \frac{m}{4} + \frac{c}{6}$, since $m > 0$. Contradiction again.

So $4 < p < 6$, and $p = 5$. This is consistent: the equation $\frac{m}{4} + \frac{c}{6} = \frac{m+c}{5}$ reduces to $3m = 2c$, e.g. $m = 16$ and $c = 24$, a total of $40 = 8 \cdot 5$ ounces, with Angela drinking $4$ and $4$ and each of the other four drinking $3$ milk and $5$ coffee.

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

Angela's fraction of the total, $\frac{1}{p}$, is a weighted average of $\frac14$ and $\frac16$ with weights $m$ and $c$: $\frac{1}{p} = \frac{m \cdot \frac14 + c \cdot \frac16}{m + c}$. A weighted average with positive weights sits strictly between the two values, and the only reciprocal of an integer strictly between $\frac16$ and $\frac14$ is $\frac15$. The phrase "never zero" is what makes the inequalities strict and the answer unique.

Algebra: multiply Angela's equation by $12$ to get $3m + 2c = 96$, and double the total to get $2m + 2c = 16p$. Subtracting gives $m = 96 - 16p$, so $m > 0$ forces $p < 6$; then $c = 8p - m = 24p - 96 > 0$ forces $p > 4$. Hence $p = 5$.

• Writing c + m = 8p and m/4 + c/6 = 8, then stalling with two equations in three unknowns instead of using that c and m are positive.
• Assuming Angela's cup must be larger than average because she drank "a quarter of the milk"; every cup is the same 8 ounces.
• Picking $p = 4$ or $p = 6$ by allowing equality, which would require zero coffee or zero milk in total.