Two towns A and B are connected by regular bus service . . .

Question : Two towns A and B are connected by regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km/h 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. 

Doubt by Manasvini

Solution : 

Let the constant speed of each bus = v km/s

and the distance between the two town A and B = x km

Time period of each bus service in going from either Town A to Town B or Town B to Town A
= T minutes 
= T/60 hours

Speed of the cyclist = 20 km/h

Relative Velocity of the Bus in the direction of motion of cyclist = (v-20) km/h

We know, 
Speed = Distance /Time 
(v-20) = x/(18/60)
(v-20)(18/60) = x
x=(v-20)(18/60) — (1) 

Relative Velocity of the Bus in the direction opposite to the motion of cyclist = (v+20) km/h

We know, 
Speed = Distance /Time 
(v+20) = x/(6/60)
(v+20)(6/60) = x
x=(v+20)(6/60) — (2) 

Equating equation (1) and (2)

(v-20)(18/60)=(v+20)(6/60)
(v-20)(3)=(v+20)
3v-60=v+20
3v-v=20+60
2v=80
v=80/2
v=40 km/h

substituting in equation (1)

x=(40-20)(18/60)
x=(20)(3/10)
x=6 km

We know, speed = Distance / Time

Time = Distance/speed

T = 6/40 
T = (3/20) Hours
T = (3/20)×60 Minutes
T= 3×3 Minutes
T= 9 minutes

Hence, the period T of the bus service is 9 minutes and 40 km/h is the speed with which the buses are plying on the road.