Tokarev S.I.

Doughnut was snacking at a roadside cafe when a bus drove past him. Three doughnuts after the bus, a motorcycle passed Doughnut, and another three doughnuts later, a car passed. Past Syrup, who was snacking at another cafe on the same road, they drove by in a different order: first the bus, three doughnuts later the car, and another three doughnuts later the motorcycle. It is known that Doughnut and Syrup always eat doughnuts at the same constant speed. Find the speed of the bus if the speed of the car is 60 km/h and the speed of the motorcycle is 30 km/h.
Let at the moment when the bus passed by Donut, the motorcycle had $x$ km left to the first cafe. This means that in the time of three donuts, the car travels $x$ km. The car moves twice as fast as the motorcycle and in the same time travels $2x$ km. It travels the same distance in the next time of three donuts. Therefore, at the moment when the bus passed by Donut, the car was four times farther from him than the motorcycle.

When the bus passed by Syrup, both the car and the motorcycle were behind him at a distance of $2x$ km. Thus, while the bus was traveling from Donut to Syrup, the car caught up with the motorcycle, eliminating the lag of $4x - x = 3x$ km. In the same time, the car reduced the lag from the bus by $4x - 2x = 2x$ km. This means that the speed at which the car catches up with the motorcycle (30 km/h) is $3/2$ of the speed at which it catches up with the bus. That is, the car catches up with the bus at a speed of 20 km/h, so the bus speed is 60 - 20 = 40 km/h.

## Answer

40 km/h.

## Problem