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?

Options

A. 4.5 min
B. 9 min
C. 12 min
D. 24 min

Answer

B

Explanation

Let v km `h^(-1)` be the constant speed with which the buses ply between the towns A and B.
Relative velocity of the bus from A to B with respect to the cyclist Relative velocity of the bus from A to B with respect to the cyclist `=(v-20)km h^(-1)`
Distance travelled by the bus in time T (minutes) = vT
As per question `vT=18 (v-20)`
or `vT = 18 v - 18 xx 20" "...(i) `
and `vT = 6 (v+20) " or " vT = 6v + 20 xx 6 " "...(ii)`
Equating (i) and (ii) ,we get
`18 v - 18 xx 20 = 6v + 20 xx 6 " or " v = 40 km h^(-1)`
Putting this value of in (i) , we get
`40 T = 18 xx 40 - 18 xx 20 = 18 xx 20 `
or `T=(18xx20)/40 = 9 min`