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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?

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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)18/60km...............(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 =6 min=6/60 h

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

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

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