## Task 23/62

A society of 12 people wanted to reach a place 20 km away. However, they only had one taxi available, which could transport three people in addition to the driver. They devised a "transport plan" that guaranteed that if all people set off simultaneously, they would also arrive at the destination simultaneously. An average speed of the taxi of $65 \frac{\mathrm{~km}}{\mathrm{h}}$ and of the pedestrians of $5 \frac{\mathrm{km}}{\mathrm{h}}$ was assumed.

What did the transport plan look like?
If all persons are to cover the distance in the same time, they must have the same average speed. This is achieved by each of them walking the same distance and driving the same distance. Therefore, the taxi, to transport each person a part of the way, must drive four times in the direction of the destination and three times back.

Let $x$ be the distance each person travels in the taxi, and $y$ the return distance. Then $4 x - 3 y = 20$. (1)

During the time the taxi makes one round trip, a pedestrian covers the distance $x - y$. However, the distances covered in the same time behave like the speeds. Thus, $\frac{x + y}{x - y} = \frac{65}{5} = \frac{13}{1} = 13$. (2)

This results in a system of equations with two unknowns; the solutions are $x = 14$ and $y = 12$. This means that the taxi drives $14 \text{ km}$ in the direction of the destination and $12 \text{ km}$ back. Therefore, $6 \text{ km}$ must be walked. It is also easily calculated that the group will take about $1 \text{ hour} 25 \text{ minutes}$ to reach the destination.