Hike and Hitch
The only completely correct solution to this question came, with a very nice diagram, from Vassil Vassilev, age 14, Lawnswood High School, Leeds.
If nobody hiked, the car would need two return journeys plus one single journey, making a total distance of $5\times 60$ miles and hence a total time of $(5\times 60)/40 = 7\,{\rm hrs}\ 30\,{\rm mins}$. You have to find out how much time is saved by the students hiking whenever they are not in the car.
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- The distance 10 children moved when the car was going towards the destination point. | |
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| - The distance the car moved towards the destination point. | ||
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- The distance 15 children moved when the car was going back. | |
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| - The distance the car moved back to pick up the rest of the children. | ||
You could work out all the distances but that involves a lot of equations. It is best to focus on the time taken. Let $T$ hours be the total time in which 5 people (plus the driver) are in the car going towards the destination point, and$t$ hours be the total time in which the driver only is in the car going back. We assume that the optimal solution is such that all of the people reach the destination at the same time , that is after$T+t$ hours.
In total, the fifteen people cover $15\times 60$ ($=900$ miles, and this is made up from the miles hitched in the car, and miles hiked on foot. The number of person-miles hitched in the car is$5\times 40T$ , and the number of person-miles hiked is
The car spends $T$ hours travelling forwards, and$t$ hours travelling back towards the start. It therefore travels forwards a distance of$40T$, and backwards a distance of $40t$. Hence$40(T-t)=60$ , or
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