## Teaser 3140: Enjoy every minute

**From The Sunday Times, 27th November 2022** [link] [link]

On rugby international morning, I found myself, along with eight friends, in a pub 5.8 miles from the match ground. We were enjoying ourselves, and so wished to delay our departure for the ground until the last possible minute. The publican, wishing to keep our custom for as long as possible, offered to help us get there by carrying us, one at a time, as pillion passengers on his motorbike.

We could walk at 2.5mph and the bike would travel at 30mph. We all left the pub together, and arrived at the ground in time for kick-off.

Ignoring the time taken getting on and off the bike, what was our minimum travelling time in minutes?

[teaser3140]

## Jim Randell 4:58 pm

on25 November 2022 Permalink |(Curiously this week’s

New Scientistback page puzzle is a simpler variation on the problem [link]).If the friends just walked from pub to the stadium it would take 2.32 hours.

If the barman ferried each of them individually between the pub and the stadium it would take 3.29 hours.

But we can do better.

If the barman takes the first friend on his motorbike, and the rest of the friends start walking towards the stadium. Then the barman drops the first friend off to walk the remainder of the distance to the stadium and returns to the group (now a short distance from the pub) and ferries the next friend to join the first friend, and so on until he collects the final friend and drops him off at the stadium at the same time that all the other friends arrive, then we can achieve a minimum overall time.

The only thing to work out is how far before the stadium to drop the first friend, then it is just a matter of ferrying friends from the trailing to the leading group.

If each of the

kfriends walks a distancexat velocityw, and travels by motorbike a distance(d − x)at a velocityv, then each journey takes:And the total time taken for the barman to ferry them all is:

Everyone arrives at the stadium at the same time, so:

or:

The answer can then be calculated directly.

Run:[ @replit ]Solution:The minimum travelling time is 82 minutes.We calculate:

So the first friend is dropped off 3.2 miles from the stadium (and walks the remainder of the way).

Each friend walks for 76.8 minutes and is on the motorbike for 5.2 minutes. Giving each a total journey time of 82 minutes.

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## Jim Randell 10:56 pm

on30 November 2022 Permalink |(See also:

Enigma 977).Here is a numerical approach, based on the barman dropping the first friend off after a distance

x(from the pub) while the remaining friends set off on foot.The barman then returns to the trailing group and ferries a single friend from the trailing to the leading group until everyone has been transferred to the leading group. And we record the maximal journey time for the friends to give us a total time to get all the friends to the stadium.

We then use the [[

`find_min()`

]] function from theenigma.pylibrary to determine at what distancexthe shortest total journey time occurs.Unsurprisingly this gives us the same answer as the analytical approach above (but with considerably more effort). But it does show that the logic used in the analysis does indeed produce the minimum time.

Run:[ @replit ]Here is a graph of the total journey time against the distance

xthat the first friend is taken from the pub. We see that the minimum time is achieved whenx= 2.6 miles.LikeLike

## NigelR 4:29 pm

on1 December 2022 Permalink |Hi Jim. I very much enjoyed this puzzle (more for the logic than the Python content!) but I’m intrigued to know how you got to the term 2kx in your second equation. I got to it indirectly by looking at the vector sum of all the motorbike journeys out: [(k)(d-x)] +back: [(k)(d-x)-d] given that he finishes up a distance d from where he started. That then reduces to:

[(2k-1)d -2kx] but is there a more intuitive way of getting to the second term?

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## Jim Randell 10:50 pm

on1 December 2022 Permalink |@Nigel: The way I thought of it the barman makes a journey towards the stadium for each of the

kfriends. And in between friends he has to make(k − 1)return journeys towards the pub.If he made complete journeys between the pub and the stadium this would be

(2k − 1)journeys of distancedfor a total of(2k − 1)d.But if he made complete journeys he would be travelling over the ground that each friend walks,

in both directions, so this amount is overcounting the barmans distance by2xfor each of thekfriends.So the actual overall distance travelled by the barman is:

And this actual distance is travelled at velocity

v, so we can determine the total time taken for the barman.LikeLike

## NigelR 9:31 am

on3 December 2022 Permalink |Thanks Jim. It was the ‘in both directions’ that I was struggling to get my head around. This week’s puzzle is trivial by comparison!

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## Hugh Casement 9:45 am

on4 December 2022 Permalink |There were some early Brain-Teasers of a similar kind by Brigadier A.G.H. Brousson (who died in 1976), involving the Smart brothers who took part in races with one bicycle between them.

Numbers 640, 663, 686, 722, 741, and 792 would presumably yield to a variant of Jim’s analysis as given here. 792 (at least) is flawed because it doesn’t specify that they all arrived together — nor, for that matter, whether riding pillion on the rear carrier is permitted (I know from experience that it seriously upsets the balance and steering!).

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## Jim Randell 10:53 am

on8 December 2022 Permalink |@Hugh: Thanks. I’ll have a look at those. None of them are in the published books I’m working from, so it would have been a while before I got round to looking at them.

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