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  • Two town 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 <img alt="20km{{h}^{-1}}" src="https://entran

Two town 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 20km{{h}^{-1}} 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 time period T of the bus service is

Option: 1

4.5 min


Option: 2

9 min


Option: 3

12 min

 


Option: 4

24 min


Answers (1)

best_answer

Let v km{{h}^{-1}} be the constant speed with which the bus travel 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 bus from B to A with respect to the cyclist =(v+20)km{{h}^{-1}}

Distance travelled by the bus in time T (minutes) =vT

As per question, 

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

vT=18(v-20)

vT=18\times v-18\times 20\; \; \; \; \; ....(i)

and \frac{vT}{v+20}=6

vT=6v+20\times 6 ......(ii)

Equating (i) and (ii), we get

\\18v-18\times 20=6v+20\times 6\\ \\12v=20\times 6+18\times 20=480\\ \\ v=40km{{h}^{-1}}

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

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

Posted by

Shailly goel

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