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^(-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?

Explanation

Let v be the speed bus in km/h. Separation between two consective buses can be written as vT.
For the buses moving in the same direction as that of cyclist , relative speed is v-20. Hence we can write the following :
`(vT)/(v-20)=(18)/(60)`
`rArr (vT)/(v-20)=(3)/(10)`
For the buses moving in the opposite as that of cyclist , relative speed is v+20 . Hence we can write the following:
`(vT)/(v+20)=(6)/(60)`
`(vT)/(v+20)=(1)/(10)`
Dividing equations (i) and (ii) , we get the following :
`(v+20)/(v-20)=3`
`rArrv+20=3v-60rArrv=40km//h`.
Substituting value of equation (ii) , we get the following
`(40T0)/(40+20)=(1)/(10)rArrT=(3)/(20)hr=9min`
Hence buses are moving with speed 40 km/h and time interval between buses is 9 min .