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. The period T of the bus service and speed of buses on the road are

Let speed of the bus be u km/h
Speed of cyclist = 20km/h
Rel. vel. of cyclist w.r.t. bus along
AB = u – 20
Rel. vel. of cyclist w.r.t. bus along BA = u + 20 

As time of encountering the bus is inversely proportional to relative velocity, therefore
$\frac{18}{6}=\frac{\mathrm{u}+20}{\mathrm{u}-20}$
$\Rightarrow$ 3u – 60 = u + 20; 2u = 80, u = 40km/h
If the first bus from A crosses the cyclist at a distance of 40 km from A after one hour of starting from A, the 2nd bus crosses him after 18 minutes
When cyclist has travelled a distance 20/60 x 18
= 6 km. Thus 2nd bus has travelled a distance
= 40 + 6 = 46 km in a time = 1hr 18 min - T
= (1.3 – T) hr, where T is time internal between two successive buses.
∴ Velocity of bus $=\frac{46}{1.3-\mathrm{T}}=40$
$\Rightarrow$ 46 = 52 – 40 T
$\Rightarrow$ 40 T = 52 – 46 = 6
$\Rightarrow$ $\mathrm{T}=\frac{6}{40}=\frac{6}{40}\times 60\min =9\min$