In search of the elusive answer.

Recently I came across a question which I am unable to solve. Can you guys help me?

Two tourists who are at a distance of 40 km from their camp must reach the camp together in the shortest possible time. They have one bicycle which they decide to ride in turn. One of them started walking at a speed ${ v }_{ 1 }=\quad5\quad km{ h }^{ -1 }$ and the other rode off on the bicycle at a a speed ${ v }_{ 2 }=\quad15\quad km{ h }^{ -1 }$. The tourists agreed to leave the bicycle at intermediate points between which one walks and the other rides. What is the mean speed of the tourists? How long will the bicycle remain unused?

I figured that if one tourist rides for half the distance and then walks the remaining half then both will reach the camp together. But I believe it can't be that simple. However if it is then do tell me! All solutions are welcome. Note by Siddharth Singh
5 years, 3 months ago

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Considering they ride and walk for equal distances (ie 20kms) they both reach the camp at the same time. Time taken by them to reach the camp would be the total time or riding the bike for 20kms at 15kmph and walking for 20kms at 5kmph. Time = distance/speed. Hence the time taken would be 4/3 hrs and 4hrs respectively. Hence the time the bicycle remains unused = 4 - 4/3 = 8/3 hrs Total time taken to reach the camp would be 4 + 4/3= 16/3 hrs Hence mean speed = distance/time =40 ÷ 16/3 = 15/2 = 7.5kmph

- 5 years, 3 months ago

So basically my reasoning was correct?

- 5 years, 3 months ago