Question

Two towns A and B are connected by a regulaar 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 nus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the perild T of the bus service and with what speed (assumed constant )do the buses ply on the road?

Answer 1

Answer:

Hello!

Let speed of bus = V

relative speed when cylist and bus both move same direction = (V - 20) km/h

We know ,

Distance = speed x time

A/C to question ,

Bus went past cyclist every 18min when in the motion of his direction .

e.g distance covered by bus = ( V-20) x 18/60 km

Every T time bus travels distance = VT

e.g. ( V - 20) x 18/60 = VT ----------(1)

Similarly ,

Relative speed of bus when bus and cyclist move in opposite direction = ( v+ 20) km/h

Bus went past every 6 min in opppsite direction of his motion .

then,

( V + 20) x 6/60 = VT ----------(2)

Solve both equation ,

( V -20)x 18/60 =( V +20) x 6/60

=> 3V - 60 = V + 20

=> 2V = 80

=> V = 40 Km/h put equation (1)

(40 - 20)x18/60 = 40T

T = 6/40 hour = 9 min

Hence,

V= 40 km/h and T = 9 min

hope it helps..