A Scooter Riddle

Puzzle. Alice, Bob, and Charlie are at Alice’s house. They are going to Bob’s house which is 33 miles away. They have a 2-seat scooter which rides at 25 miles per hour with 1 rider on it; or, at 20 miles per hour with 2 riders. Each of the 3 friends walks at 5 miles per hour. How fast can all three of them make it to Bob’s house?



  1. Jonathan Kariv:

    I can do 3 hours and suspect this is optimal (no proof as yet). Steps:
    1. A+B take the scooter for 72 minutes (6/5 hours). C walks. At this point C has travelled 6 miles and A and B have travelled 24.
    2. B takes the scooted back to meet C while A and C continue walking. B and C meet after 36 minutes. At this point B and C are at 9 miles while A is at 27 miles.
    3. Spend the final 72 minutes with B and C on the scooter and A walking all in the direction of B’s house. They all arrive after another 72 minutes.

    In coming up with this I thought about “go that way for 2 hours and that way for 1” and then rescaled at the end.

  2. Chris Tunstall:

    They can make it in 3 hours.

  3. lvps1000vm:

    It is optimal for any stretch that two people ride 8/11 of the length and then one rides back to pick up the remaining, who should’ve walked 3/11 by the time.

    The interesting thing is, one can partition the route arbitrarily in segments and as long as this strategy is followed in every segment the solution is optimal, so there are infinite valid and different solutions.

    (That includes the case “partition in a single segment” of Jonathan above)

  4. TR:

    Follow up: using variables
    w = walking speed
    s1 = speed of scooter with one passenger
    s2 = speed of scooter with two passengers
    express the fraction of the total distance at which the scooter should turn back.

  5. Fabio:

    The time that matters is the time the last person gets to Bob’s house. The best way the last person can improve his/her time without letting others behind is getting to the destination together with the others, since everybody’s velocity has impacts on each other’s velocity. The scooter gets nothing by advancing with only one rider or coming back with two riders, it must always advance with two and come back with one. So, if two of them must walk and get to Bob’s house together, they should walk the same as each other and ride the same as each other. The way the scooter changes its velocity less(one of the possible solutions) is by leaving one at a distance him/her can get to Bob’s house walking and coming back to pick up the other who started walking the moment the scooter left Alice’s house. I’m not sure if I missed anything, but it was as close as a verbal proof I got till now. But, to make it more fun for Alice, Bob and Charlie, they should change the pilot at both stops, so everybody can get the fun of driving the scooter a little bit and the problem stays the same 🙂

  6. TR:

    Fabio: well put. That’s why I thought the formula is interesting. It lets one fiddle with questions like, what if w=0? w=s2? s1=s2? s1<s2? that test the robustness of the model. No takers, alas.

  7. DKT:

    3:30:22.5 is what I came up with.

    Alice and Bob take off on the scooter together. Charlie starts walking.

    1.65hrs later:
    Alice and Bob arrive at Bob’s house at 1.65hrs.
    Charlie has traveled 8.25m, leaving 24.75m between them.
    Alice takes off alone back for Charlie.

    0.825hrs later:
    Charlie walks 1/6 the remaining distance, 4.125m. (5/(5+25)) = 1/6
    Alice meets up with Charlie and heads back to Bob’s
    Alice and Charlie are 20.625m away from Bob’s house.

    1.03125hrs later:
    Alice and Charlie arrive back at Bob’s.

    Total time = 3.50625hrs or 3hrs 30 minutes and 22.5 seconds.

    …I think?

  8. dontuchmysolderingun:

    They all ride on the scooter so it goes at 15 mph, and they get there in 2hrs and 12 minutes.

  9. Max:

    I also get 3 hours with 2 on the scooter for x miles (which takes t1 = x/20) while one walks, one goes back with the scooter and fetches the one behind, then they head to the destination where all arrive at the same time after an additional t2 = (33-x)/5 = (x-y)/25 + (33-y)/20, y=5*(t1+(x-y)/25). Solving this for y(x), then t2 for x, I get x = 24. By symmetry, both walkers have probably walked the same distance (y = 33-x = 9), but I don’t see immediately how this could help to make the calculation much easier.

  10. pepe:

    Under the same logic it is easy to show that n people can ‘travel’ at a speed of 5 + 30/(3n-4) per time unit. (n=3 gives the 11, so 3 hours for 33).

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