Thank **Justin from Skyview High School, Billings, MT,
USA** for this solution and well
done!

Rate, time and distance are connected by the equation r =d/t .

Call the rate (or speed) of the car r _{c} and the rate
of every bus r _{b} . Each bus is a constant distance from
the bus preceding it and the bus following it; call this distance
d.

For a bus approaching on the other highway and coming towards
the car, the rate of the bus relative to the car, considering the
bus still, is (r _{b} + r _{c} ). This rate
multiplied by the time it takes (three minutes) for the car to
close the gap between it and the bus is equal to d, hence:

3(r _{b} + r _{c} ) = d.

The rate of the bus which is travelling in the same direction as
the car, relative to the rate of the car (considering the car
still) is (r _{b} - r _{c} ). This rate multiplied
by the time it takes (six minutes) for the bus to close the gap
between it and the car is also equal to d _{1} , hence

6(r _{b} - r _{c} ) = d.

Multiplying the first equation by 2 and add the two equations, one obtains

12r _{b} = 3d.

But the distance d between the buses divided by the rate of the
bus is equal to the time interval between the buses therefore the
time interval = d/r _{b} = 4 minutes.

So the buses leave the depot at intervals of 4 minutes.