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Topic: Arrive on time (Read 553 times) 

Altamira_64
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Posts: 116


Arrive on time
« on: Feb 21^{st}, 2016, 1:26pm » 
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David, Jason and Adam are stuck in traffic on a bridge in NY. They must be arrive at work the soonest possible but they are currently 30 miles away! At the edge of the road they find a motorbike and a racing bicycle, which they decide to “borrow”, as a last desperate effort to finally get to the office on time! The motorbike is quite old and can only run at a maximum speed of 30 miles per hour, while the bicycle can run at 20 miles per hour. Fortunately, all three are very fit and can also run at 10 miles per hour. If they want, they can leave the motorbike or the bicycle at the edge of the street, for someone else to take it. What is the minimum time they will require to reach the office (all three of them), assuming that the streets are always open for motorbikes / bicycles and joggers?


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rmsgrey
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Re: Arrive on time
« Reply #1 on: Feb 22^{nd}, 2016, 8:49am » 
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I make it 1h50 Possible schedule: hidden:  David sets off on the motorbike, goes 6.25 miles in 12m30, then runs another 8.75 miles in 52m30 (15 miles in 1h5) finding the bicycle at the halfway mark (where Adam leaves it) and cycles the remaining 15 miles in 45m  total time 1h50 Jason sets off running, runs the 6.25 miles in 37m30, gets on the motorbike David left him, rides for 17.5 miles, taking 35m, (23.75 miles in 1h12m30) and runs the remaining 6.25 miles in 37m30  total time 1h50 Adam starts off on the bike, cycles 15 miles in 45m, runs 8.75 miles in 52m30 (23.75 miles in 1h37m30) and rides the motorbike Jason left him for the remaining 6.25 miles in 12m30  total time 1h50.  Reasoning: hidden:  My starting point was establishing the lower bound  if you look at the distances covered by each form of transport, if they all move forward the whole time, each vehicle covers 30 miles, leaving 30 miles to cover running (obviously any backtracking adds to the distance a particular form of transport covers without reducing any other form of transport's distance)  so the 90 person miles take a total of 5h30  if everyone keeps moving the whole time and all three finish at once, then that time is divided evenly  1h50 each. My first thought was dividing the transport evenly between the 3 men  so each travel 10 miles by each method  it sounds good, but you always end up with someone waiting at some point. It's fairly easy to get a 2h trip that way, but impossible to reach the lower bound. So, having convinced myself that you couldn't do it "fairly", I realised that you could make the trip in the right time with just a combination of motorbike and running, leaving the other two sharing the bicycle and using the remaining distance on the motorbike and on foot. Doing things symmetrically from there gave the solution above. 


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Altamira_64
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Posts: 116


Re: Arrive on time
« Reply #2 on: Feb 26^{th}, 2016, 3:40am » 
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Very cool!!


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rmsgrey
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Re: Arrive on time
« Reply #3 on: Feb 26^{th}, 2016, 8:21am » 
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Maybe a little cooler: hidden:  David rides the motorbike 17.5 miles in 35 minutes, then runs the remaining 12.5 miles in 75 minutes. Jason starts on the bicycle for 6⅔ miles in 20 minutes, leaves it and runs 10⅚ miles in 65 minutes, then gets on the motorbike for the remaining 12.5 miles, taking 25 minutes. Adam runs 6⅔ miles in 40 minutes, then gets on the bike for the remaining 23⅓ miles, taking another 70 minutes. There's also a solution family where one person only runs and cycles, while the other two use all three modes of transport.  The fact there are several schedules that get the last of the three to arrive at the same time means only the time taken is wellspecified  though demonstrating a suitable schedule does require at least one meaningful insight...


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