Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. 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?

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., (when he moves in the direction of the bus).

Distance covered by the bus =  … (i)

Since one bus leaves after every T minutes, the distance travelled by the bus will be equal to 

Both equations (i) and (ii) are equal.

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 

From equations (iii) and (iv), we get

Substituting the value of V in equation (iv), we get

Let V be speed of each bus and T be time interval Then distance between any two bus in same direction= VTIn first case this distance is travelled with relative velocity V-20 and take 18/60 hr .In second case this distance is travelled with relative velocity V+20 and take 6/60 hr.(V-20)*18/60=(V+20)*3/60=VTSolve for V and T