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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/hr in the direction A to B notices that a bus goes past him every 18 minutes in the direction of his motion, and every 6 minutes in the opposite direction. What is the period T of the bus service and with what speed do the buses ply on the road. Assume the buses ply with constant speed on the road.

A
T=6 min
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B
T=9 min
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C
V=40 km/hr
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D
V=60 km/hr
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Solution

The correct option is C V=40 km/hr
Let V be the speed of the bus running between towns A and B.

Given, speed of the cyclist =20 Km/h

Relative speed of bus moving in the direction of cyclist with respect to cyclist =(V20) km/h

Every 18 mins, the bus went past the cyclist.

So, the distance covered by the bus d=(V20) 1860 km .......(1)

As, one bus leaves every T mins

So, d=V(T60) ......(2)

Using equations (1) and (2) we have,

(V20) 1860=VT60 ......(3)

Relative speed of the bus moving in the opposite direction of the cyclist =(V+20) km/h

Time taken by the bus to go past the cyclist =660 h

(V+20)(660)=VT60 .....(4)

Using equations (3) and (4) we have:

(V+20)660=(V20)1860

2V=80

V=40 km/h

Putting the value of V in equation (4) we have,

T=36040=9 min

Hence, option (B) and (C) are the correct answers.

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