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 be the speed of the bus running between towns A and B.
Speed of the cyclist, v = 20 km/h
Relative speed of the bus moving in the direction of the cyclist = V – v = (V – 20) km/h
The bus went past the cyclist every 18 min i.e., `18/60`h (when he moves in the direction of the bus).
Distance covered by the bus = `(V-20)18/60 Km`.... (i)
Since one bus leaves after every T minutes, the distance travelled by the bus will be equal to `VxxT/60`...(i)
Both equations (i) and (ii) are equal.
`(V-20)xx18/60=(VT)/60`...(iii)
Relative speed of the bus moving in the opposite direction of the cyclist =(V+20) km/h
Time taken by the bus to go past the cyclist =6 min =`6/60`h
`therefore (V+20)6/60=(VT)/60`...(iv)
From equations (iii) and (iv) , we get
`(V+20)xx6/60=(V-20)xx18/60`
V+20=3V-60
2V=80
V=40 km/h
Substituting the value of V in equation (iv), we get 40+20×660 = 40T60T = 36040T = 9 min