Two town A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man leaving in either direction every 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 is

Options

A. 4.5 min
B. 9 min
C. 1.2 min
D. 24 min

Answer

B

Explanation

Let `v "km"h^(-1)` be the constant speed with which the bus travel ply between the towns A and B.
Relative velocity of the bus from A to B with respect to the cyclist = `(v - 20) km h^(-1)`
Relative velocity of the bus from B to A with respect to the cyclist `= (v + 20) km h^(-1)`
Distance travelled by the bus in time T (minutes) = vT
As per question
`(vT)/v - 20 = 18` or `vT = 18v - 18 xx 20` and `(vT)/v + 20 = 6` ...(i)
or `vT = 6v + 20 xx 6` ....(ii)
Equaitons (i) and (ii), we get
`18v - 18 xx 20 = 6v + 20 xx 6`
or `12 v = 20 xx 6 + 18 xx 20 = 480` or v = `40 km h^(-1)`
Putting this value of v in (i) , we get
`40T = 18 xx 40 - 18 xx 20 = 18 xx 20`
or T = `18 xx 20/40` = 9 min