478. Two pedestrians set out simultaneously from point A in the same direction. The first one met a tourist walking towards A, 20 minutes after leaving A, while the second one met the tourist 5 minutes later than the first. Ten minutes after the second meeting, the tourist arrived at A. Find the ratio of the speeds of the pedestrians.
Let $B$ and $M$ be the points of meeting of the first and second pedestrian with the tourist (Fig. 3).

![](https://cdn.mathpix.com/cropped/2024_05_21_88152b3867f80e3be9a3g-166.jpg?height=57&width=896&top_left_y=734&top_left_x=480)

Fig. 3

Let the path $AB$ be taken as a unit. Denote the speeds of the first and second pedestrian by $v_{1}$ and $v_{2}$, respectively, and the speed of the tourist by $v$ (in fractions of the path $AB=1$ per hour). Then the distance $AB$ is, on the one hand, $\frac{1}{3} v_{1}$, and on the other, $\frac{1}{4} v$, since the tourist arrived at $A$ 15 minutes $=\frac{1}{4}$ hour after the first meeting. This gives us the first equation:

$$
\frac{1}{3} v_{1}=\frac{1}{4} v
$$

Similarly, we express the distance $AM$ in two ways and equate these expressions:

$$
\frac{5}{12} v_{2}=\frac{1}{6} v
$$

We now have a system of two equations with three unknowns:

$$
\left\{\begin{array}{l}
\frac{1}{3} v_{1}=\frac{1}{4} v \\
\frac{5}{12} v_{2}=\frac{1}{6} v
\end{array}\right.
$$

We need to find the ratio $\frac{v_{1}}{v_{2}}$. To do this, we divide the first equation by the second:

$$
\frac{4}{5} \cdot \frac{v_{1}}{v_{2}}=\frac{3}{2}, \quad \frac{v_{1}}{v_{2}}=\frac{3}{2} : \frac{4}{5}=\frac{15}{8}
$$

Answer: $15 / 8$.