Question

Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h^-1 in the direction A to B notices that a bus goes post 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

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

Answer 1

Answer: (B)

9 min

Answer 2

Let v $kmh^{- 1}$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

= (v – 20) km $h^{- 1}$

Relative velocity of the bust form B to A with respect to the cyclist

= (v + 20) km $h^{- 1}$

Distance travel by the bus in time T (minutes) = Vt

As per questions

$\frac{vT}{v - 20} = 18$

Or vT = 18v - 18 × 20 …….(i)

And $\frac{vT}{v + 20} = 6$

Or vT = 6v+ 20 × 6 ……..(ii)

Equating (i) and (ii), we get

18v – 18 × 20 = 6v + 20 × 6

Or 12v = 20 × 6 + 18 × 20 = 480 or v = 40 km $h^{- 1}$

Putting this value of v in (i), we get

40T = 18 × 40 – 18 × 20 = 18 × 20

Or $T = \frac{18 \times 20}{40} = 9\min$