$A$ walks $30 \mathrm{~km}$ per day from $P$ to $Q$. After walking $72 \mathrm{~km}$, $B$ starts from $Q$ towards $P$ on the same road, and walks $\frac{1}{10}$ of the entire distance each day. After $B$ has walked for as many days as the $\frac{1}{8}$ of the distance he walks per day, he meets $A$. What is the distance between $P$ and $Q$?
I. Solution: Let the distance $P Q$ be $x \text{ km}$. While traveling from $B Q$, it takes $\frac{x}{10} \cdot \frac{1}{8} = \frac{x}{80}$ days to meet $A$. Therefore, by the time they meet, $A$ has traveled $72 + 30 \cdot \frac{x}{80} \text{ km}$, while $B$ has traveled $\frac{x}{10} \cdot \frac{x}{80} \text{ km}$. The sum of the distances traveled is $P Q$, so

$$
72 + 30 \frac{x}{80} + \frac{x^2}{800} = x
$$

Removing the fractions and rearranging,

$$
x^2 - 500 x + 57600 = 0
$$

from which

$$
x_1 = 300, \quad x_2 = 180
$$

Thus, the problem has two solutions:

$$
\begin{aligned}
& P_1 Q_1 = 320 \text{ km} \\
& P_2 Q_2 = 180 \text{ km}
\end{aligned}
$$

Bartha László (Balassagyarmat, Balassa g. I. o. t.)

II. Solution: Let $x$ be the number of days that elapse from $B$'s departure until the meeting with $A$. According to the problem, $x = \frac{P Q}{10} \cdot \frac{1}{8}$, so $P Q = 80 x \text{ km}$.

By the time they meet, $A$ has traveled $72 + 30 x \text{ km}$, and $B$ has traveled $x \cdot \frac{P Q}{10} = x \cdot 8 x = 8 x^2 \text{ km}$.

Thus,

$$
72 + 30 x + 8 x^2 = 80 x
$$

Rearranging and simplifying by 2,

$$
4 x^2 - 25 x + 36 = 0
$$

from which

$$
x_1 = 4 \text{ days}, \quad x_2 = \frac{9}{4} \text{ days.}
$$

Therefore, the sought distance $P Q$ is either $80 \cdot 4 = 320 \text{ km}$ or $80 \cdot \frac{9}{4} = 180 \text{ km}$.

Endrödy Tamás (Bp., III., Árpád g. I. o. t.)