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=speed of buses playing between A and B
The relative speed of bus going from A to B w.r.t. cyclist `=(v-20)` km `hr^(-1)`
and relative speed of bus goint from B to A w.r.t. cyclist `=(v+20)km` `hr^(-1)`
when bus and cyclist are in same directin then `(vT)/(v-20)=18` min ....(1)
when bus and cyclist are in opposite direction then `(VT)/(v-20)=6` min.....(2)
dividing eq (1) by eq(2) `(v+20)/(v-20)=(18)/(6)=3rArrv=40km hr^(-1)`
putting value v in eq `(1) (40xxT)/(40-20)=18` min `rArr(40)/(20)T=18` min `rArr=T=9` min