# 7.2. (7 points)

Two pedestrians set out at dawn. Each walked at a constant speed. One walked from $A$ to $B$, the other from $B$ to $A$. They met at noon and, without stopping, arrived: one - in $B$ at 4 PM, and the other - in $A$ at 9 PM. At what hour was dawn that day?
Answer: at 6 o'clock.

Solution: Let $x$ be the number of hours from dawn to noon. The first pedestrian walked $x$ hours before noon and 4 after, the second - $x$ before noon and 9 after. Note that the ratio of times is equal to the ratio of the lengths of the paths before and after the meeting point, so $\frac{x}{4}=\frac{9}{x}$. From this proportion, we find that $x=6$.

Indeed, let the speed of the first pedestrian (walking from $A$ to $B$) be $V_{1}$, and the speed of the second pedestrian (walking from $B$ to $A$) be $V_{2}$. Then before the meeting point, the first pedestrian walked $V_{1} \cdot x$ hours, and the second $-V_{2} \cdot x$. After noon, the first pedestrian walked $V_{1} \cdot 4$, and the second $-V_{2} \cdot 9$. But after noon, the first pedestrian walked as much as the second did before noon. And vice versa, after noon, the second pedestrian walked as much as the first did before noon. We have:

$$
\begin{aligned}
& V_{1} \cdot x=V_{2} \cdot 9, \frac{V_{1}}{V_{2}}=\frac{9}{x} \\
& V_{1} \cdot 4=V_{2} \cdot x, \frac{V_{1}}{V_{2}}=\frac{x}{4}
\end{aligned}
$$

From this, $\frac{x}{4}=\frac{9}{x}$.