The insidious education system now and then throws up "impossible" problems to solve, which most children are often unable to cope with. Even more interesting is that most adults are not able to cope with such tasks. About one such will now be discussed.

This task was included in the 5th grade textbook for Belarusian institutions of general secondary education. The same problem was used in Magnitogorsk in the tournament of young mathematicians among grades 6-8. The problem appeared in Barnaul at the 9th grade competition, as well as at the school Olympiad in Nizhny Novgorod for 10th grade.

## Condition

On the way, a bus, a motorcycle and a car drove past the observer at regular intervals. Past another observer, vehicles passed at the same intervals in time, but in a different order: bus, car, motorcycle. What was the speed of the bus if the speed of the car was 60 km / h, and the speed of the motorcycle was 30 km / h.

## Solution

There are several solutions to the problem. The Novate.ru edition will cite one of them as an example.

Let's say Vx is the bus speed to find. Let t be the time that the car spent on the road between the observers, and the time interval with which a bus, car and motorcycle passed by the observers.

Then, the time that the bus spent on the road between the two observers will be t + a, and the time of the motorcycle will be t + 2a. You can now express the distance for each vehicle.

Car: S = 60 ⋅ t

Motorcycle: S = 30 ⋅ (t + 2a)

Bus: S = Vx ⋅ (t + a)

Since the distance for all vehicles was the same, we compose the following equation.

For car and motorcycle distance:

60t = 30 (t + 2a)

60t = 30t + 60a

30t = 60a

a = 0.5t

For car and bus distance:

60t = Vx ⋅ (t + a)

60t = Vx ⋅ (t + 0.5t)

60t = Vx ⋅1, 5t

Vx = 60t / 1.5t

Vx = 40