Problem On Relative Motion

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 in the direction A to B notices that a bus goes past him every 18 minute in the direction of his motion, and every 6 minute 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 ?
Solution

Let speed of each bus = v km/h
The distance between the nearest buses plying on either side = vT km.. (i)
For buses going from town A to B :
Relative speed of bus in the direction of motion of man = (v – 20)
Buses plying in this direction go past the cyclist after every 18 min. Therefore separation
between the buses = (v – 20) × 18
From equation (i),
(v – 20) ×(18/60) = vT ......(ii)
For buses coming from B to A :
The relative velocity of bust with respect to man = (v + 20)
Buses coming from town B past the cyclist after every 6 min therefore
(v + 20) ×(6/60) = vT ......(iii)
Solving equations (ii) and (iii), we get
v = 40 km/h and T =(3/20) h