Problem 8.2. The arithmetic mean of three two-digit natural numbers $x, y, z$ is 60. What is the maximum value that the expression $\frac{x+y}{z}$ can take?
Answer: 17.

Solution. From the condition, it follows that the sum of the numbers $x, y, z$ is $60 \cdot 3=180$. Then

$$
\frac{x+y}{z}=\frac{180-z}{z}=\frac{180}{z}-1 \leqslant \frac{180}{10}-1=17
$$

since $z \geqslant 10$. Note also that when $x=90, y=80, z=10$, the value 17 is achieved.