Two towns `A` and `B` are connected by a regular bus service with a bus leaving in either direction every `T` min. `A` man cycling with a speed of `20 km h^(-1)` in the direction `A` to `B` notices that a bus goes past him every `18 min` in the direction of his motion, and every `6 min` in the opposite direction. What is the period `T` of the bus service and with what speed (assumed constant )do the buses ply on the road?

Explanation

Let `V_(b)=` speed of bus
`V_(c)`=speed of cyclist.
The relative velocity of bus in the direction of motion of cyclist `=V_(b)-V_(c)=V_(b)-20`
The relative velocity of bus in opposite direction of motion of cyclist `=V_(b)+V_(c)`
`=V_(b)+20`
`therefore` The distance travelled by bus in time `T=V_(b)T`
`(V_(b)T)/(V_(b)-20)=18`
`(V_(b)T)/(V_(b)+20)=6`
Solving we get `V_(b)=40kmh^(-1)`
T=9min