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 20 $km{h}^{-1}$ in the direction A to B notices that a bus goes past him every 10 min in the direction of his motion, and every 2 min in the opposite direction. The speed of the bus on the road is:

• A. 10 kmph
• B. 20 kmph
• C. 40 kmph
• D. 30 kmph

Answer 1

Answer: (C)

Let V be the speed of the bus running between A and B

Speed of the cyclist, $v=20kmph$

Relative speed of the bus moving in the direction of the cyclist$=V-v=(V-20)kmph$

The bus went past the cyclist every 18 min i.e. $\frac{18}{60}h$(when he moves in the direction of the bus)

Distance covered by the bus$=(V-20)\times \frac{18}{60}...........\left(i\right)$

Since one bus leaves after every T minutes, the distance traveled by the bus will be$=V\times \frac{T}{60}.................\left(ii\right)$

Both equations (i) and (ii) are equal,

$(V-20)\times \frac{18}{60}=V\times \frac{T}{60}...........\left(iii\right)$

Relative speed of the bus moving in the opposite direction of the cyclist$=(V+20)kmph$

Time taken by the bus to go past the cyclist$=6min=\frac{6}{60}h$

$\therefore (V+20)\times \frac{6}{60}=V\times \frac{T}{60}......\left(iv\right)$

From equation (iii) and (iv) , we get

$(V+20)\times \frac{6}{60}=(V-20)\times \frac{18}{60}$

$V+20=3V-60$

$2V=80$

$V=40kmph$