by Howard Williams
Published Sunday November 27 2022
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?
Sunday Times Teasers Discussion Group 
This is fairly gentle. The condition for the minimum time leads to an integer solution.
My fist attempt on this one neglected that leaving people inactive during the transport operation was a waste of time (and gave an ugly answer as well).
My second attempt uses the optimal time to let the first passenger off, as variable, but reaches the answer only by adjustments, not algebraic (yet).
Hi Erling,
it seems I made initially the same futile attempt of bringing all persons per moter byke right away to the destination. This gave an about 28% longer travel time than the simple procedure of enforcing that all persons walk the same distance and arrive simultaneously at the destination.
sorry, about 29%.
I can’t make this one work.
It seems to me that there is a group of people who set off walking at the same speed – that gives me a time for them to get to the rugby ground. The size of this group will be reduced as people are taken away as pillion passengers, but the remaining group will still travel at the same speed. The first pillion passenger will reach the rugby ground at a time set by the motorbike speed, but the bike must return to collect the next passenger – it will of course meet the walking group, so return time is less. I can calculate all these times, but get a (non integer) result which is larger than Brian’s hidden answer. Not sure if it’s my analysis or my maths that’s wrong!
Hi Robert, Its your analysis.
If you deliver one of the friends to the ground before another, you should have dropped them off earlier and let them walk the rest of the way because this allows the barman to return earlier to collect the next friend.
In short all the friends walking power and the barman’s riding power must be used, which means no-one can be idle until they all arrive at the ground at the same time.
Thanks Brian
I noticed the anomaly that while the “barman” is on the bike, he can’t be serving at the pub !! (Joke)
Maybe he leaves 8 beers at one pick up point, 7 at the next, etc.😆
There is a shortcut on this one… Consider the final walker, who has to get to the game (on the bike) after walking a certain distance in a certain time.
I used this route (of the final pillion rider) but used equalities given from the first and second pillion riders to get the overall least time.
I used graphs and deduced equations from it.
My initial futile attempt required more calculations than the simple case where the nine friends arrived simultaneously at the match ground.
Hi Peter,
We have for sure followed the same path step by step with this one. My last is now available as ‘solution’, above.
The two scenarios discussed have analytic solutions for “transit times” whereby TM Is the minimum time (as required) with all group members arriving at the ground simultaneously and TS is the time taken for all group members to be dropped off at the ground in a staggered way.
Defining S = 5.8 as the pub-ground distance, V = 1/2 as the bike speed, v = 1/24 as the walking speed and a parameter R=(V-v)/(V+v) = 11/13 we find the time in minutes as:
TM = (S/V) (17V+v)/(V+17v) and
TS = (S/V) { [ (1-R^9)(V+v)/v] – 1}
An interesting teaser – especially the unwanted scenario.
I came to this one late on account of being away treking in Nepal, followed by a short break walking in Wales.
I then made the same error as some others, assuming the first person didn’t walk at all and each one walked a bit more, with the bike taking them the whole rest of the way. When I reached an answer I came here to check it, without reading comments. Then I went back and tried to find my mistake and couldn’t.
Then I came back to read the comments which immediately showed me my mistake! It’s a clever and enjoyable teaser. It was a much easier problem to solve doing it the correct way.
Now on to teaser 3141 which I am also late in starting…
This teaser took on a life of its own…….
If the fans walk x miles between bike pick-ups, the bike travels 12x miles, ie 6.5x (= y) miles out and 5.5x miles return.
The total distance travelled by each fan = 8x + y = 14.5x = 5.8 miles
x, y and the time taken by each fan (205x minutes) follow in very short order.