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

Explanation

Let speed of buses ` v km h^(-1) ,` Given speed of cyclist =`10 km h^(-1)`
Relative speed of a bus (going from Chandigarh to Ambala) w.r.t. the cyclist ` km h^(-1) `
Relative speed of a bus (going from Ambala to Chandigarh) w.r.t. the cyclist ` = (v+10 ) km h^(-1)`
Time interval of a bus leaving the bus stop =T
` therefore " " `Distance travelled by the bus in either direction =vT ., since ,time ` =("Distance ")/("Velocity ") `
`therefore " " 15` minute `=(vT)/( v-10 ) or (vT)/((v-10)) =(15)/( 60) =(1)/(4) " "[because 1h=60 "min"]`
or ` 4vt =v-10 `
Similarly , 5min =` (vT)/( (v+10 )) or (vT)/((V+10 )) = ( 5)/( 60) =(1)/(12) `
` therefore " "1 2vT=v+10 `
Dividing equation (1) by equation (2), we get
` (1)/(3) =(v-10 )/( v+10 ) " "or v+ 10 =3v -30 or , 2v = 40 " " or v= 20 km h^(-1) `
Substituting v= 20 ` km h^(-1) ` in equation (1), we get
` 4 xx 20 T = 20 - 10 " " or T=(10)/( 80) =(1)/(8) h=(1)/(8) xx 60 "min" .=7.5` minutes