We have three riders which need to travel some distance (say, 13.5km) through a rough terrain (no running, no skiing etc., only walking or horse trot), but they have only two horses. The speed of a walking rider is, say, 4km/h, and the speed of a rider riding a horse is, say, 6km/h. The horses can't be left unattended, and only a single person can travel on one horse. What's the fastest way for them to travel the given distance, and what's the time it takes to travel using it?

Note: no calling taxi cabs with cell phones, no blue portals, no jet engines, no timespace warps etc. this time, please? This is not a nor puzzle.


5 Answers 5


A 2 hour 45 minutes (165 minutes) solution:

Two riders ride until they reach 5/9ths of the way there (7.5km), while the third walks. One of the riders gets off the horse and starts to walk, and the other leads both horses back to fetch the original walker at 6km and reach the destination at the same time as the one who got off the horses.

As for the time taken, the horse travels the 13.5 km plus an extra 1.5km roundtrip to go back and return to where he was, or 16.5km overall. This takes 2.75 hours.

I did most of the math in my head, so it's a bit messy. How I got to that number:

Let the relative speed between a horse and a walker in the same direction be $v$, and the relative speed between them in different directions be $V$, and the total distance be $D$.

Assuming they all arrive at the same time, both walkers walk for the same amount of time. The horse rider that doesn't get off rides for $x$ distance, then moves $x \frac{v}{V}$ to go back, then rides for $x$ again.

The horse must have ridden exactly the total distance in the forward direction, i.e., $x + x - x \frac{v}{V} = D$. Substituting $\frac{v}{V} = \frac{6-4}{6+4} = \frac{1}{5}$, we have that $\frac{9x}{5} = D$, or $x = \frac{5}{9}D$.

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    $\begingroup$ I thought the time would be fractionary so I got scared, but the numbers are actually nice. $\endgroup$
    – ffao
    Apr 4, 2016 at 13:07
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    $\begingroup$ Does the rider that leads both horses backwards ride or walk? Seems to me that I've seen riders towing a second horse in many westerns. $\endgroup$ Apr 4, 2016 at 13:57
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    $\begingroup$ I assumed he could ride; if he had to walk, the optimal distance to turn back would have been 4/7ths of the whole way $\endgroup$
    – ffao
    Apr 4, 2016 at 14:05
  • $\begingroup$ If he had to walk the horses back, optimal distance would have been to turn back at 6/11 of the whole way, not 4/7. $\endgroup$ Apr 4, 2016 at 22:29
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    $\begingroup$ @aslum In these kinds of puzzles it's common to assume turning around, mounting and accelerating all happen instantly (or at least fast enough that it doesn't matter) $\endgroup$
    – ffao
    Apr 5, 2016 at 16:40

Person 1 and 2 ride for 1.25 hours (7.5km, 6km to go) whilst person 3 walks (5km, 8.5km to go)

Person 1 rides back, leading the other horse until they meet person 3 (15 minutes, 1.5km by horse, 1km walking, 7.5km to go) whilst person 2 walks to the destination (5km to go)

Persons 1 and 3 ride to the end (1.25 hours), person 2 walks the rest of the way (1.25 hours), everyone's there within 2.75 hours (2 hours, 45 minutes).

Apparently my maths was terrible when I first typed this up and I somehow confused myself into thinking it was a 15km journey

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    $\begingroup$ I think it's on the right path. It's the kind of solution I think about, but I can't do maths until tonight. The optimisation will be the solution, I think. $\endgroup$
    – Shkeil
    Apr 4, 2016 at 12:57
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    $\begingroup$ There will be only 7.5km to go, rather than 9km to go. ​ ​ $\endgroup$
    – user1579
    Apr 4, 2016 at 12:59
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    $\begingroup$ Did you just guess at the 1.25 hour number? Your intuition is good! $\endgroup$
    – ffao
    Apr 4, 2016 at 13:55
  • $\begingroup$ @ffao Sort of. I'd like to say it was mathematical, but I can't remember how I came up with that number. Slightly surprised to find that it's the best case! $\endgroup$ Apr 4, 2016 at 14:42




they start as this:
2 ride, 1 walks.
after 1h15m (7.5km) one of the riders starts walking and the other goes back with both horses.
The guy that walks will have 6 more km so 1.5h (1h30m).
So total 2h45m.
The first one will walk 5km till one of the riders decide for one to go back.
At this point they will be at 2.5km apart walking on opposite directions each. It will take 15m to meet.
total so far 1h30m.
at this point they will be 6km away from the start and 7.5km from the end.
They ride for 1h15 minutes more. Total 2h45m.
They all arrive in the same time.

The math way.
Speed of the horse is $v_h$
Speed of the person is $v_p$
Distance to cover is $d$

2 ride one walks.
After some time $t_1$ one of the riders start walking till the end.
it will take him the time $t_2$
So his time in total will be $t_1 + t_2$. But...
$t_1*v_h + t_2 * v_p = d$ (let's call this eq 1).
now let's handle the guy that goes back with both horses.
The time passed since he went back till he meets the guy that started on foot is $t_3$.
The time it takes after they meet till the finish is $t_4$.
Now we get this equation:
$t_1 * v_h - t_3*v_h + t_4 * v_h = d$ (2)
Now let's look from the perspective of the guy that starts the journey walking.
$t_1 * v_p + t_3 * v_p + t_4 * v_h = d$ (3).
but the time it takes the guy that is always on a horse is $t_1 + t_3 + t_4$.
the time it takes the guy that starts on a horse and finishes on foot is $t_1 + t_2$.
To minimize the time needed, the 2 times above should be equal.
So we get: $t_1 + t_3 + t_4 = t_1 + t_2$.
Which translates to $t_2 = t_3 + t_4$ (4).
Now we have 4 equations with 4 unknown values.
$t_1 * v_h + t_2 * v_p = d$
$t_1 * v_h - t_3 * v_h + t_4 * v_h = d$
$t_1 * v_p + t_3 * v_p + t_4 * v_h = d$
$t_2 = t_3 + t_4$
Filling in the numbers for $v_p$, $v_h$ and $d$ should make the problem trivial.


Our three travelers shall be dubbed Adam, Bill and Carl.

! Adam and Bill set off on the horses for a time, then Adam dismounts and proceeds to walk the remainder of the distance. Bill rides his horse, accompanied by the empty horse, and returns to Carl, who has been walking the entire time thus far. Then Carl mounts up and he and Bill proceed to the end, where they arrive at the exact same time as Adam.

! Let us examine the distance that Bill will travel. Bill’s journey can be described as (a) to the point where he will eventually return for Carl; (b) to the point where he drops off Adam; (c) back to the point where he picks up Carl; (d) back to the point where he originally dropped off Adam; and (e) to the end. We can see that b = c = d, since these three legs are all from point 1 to point 2, back to point 1, and back to point 2. Bill then traveled a distance of a + b + c + d + e, but c was backtracking, so his total progress was a + b – c + d + e.

!We now proceed to examine this problem from another angle, leaving Bill of the equations for now. Adam’s journey can be broken into three distances of R1T1, R1T2 and R2T3. Carl, on the other hand, has R2T1, R1T2 and R1T3. [R1 = horse rate, R2 = foot rate] Thus they are walking the same distance for the middle segment, which is intuitive, since they are both on foot. Since they are traveling the same total distance, we have R1T1 + R2T3 = R2T1 + R1T3, or T1(R1-R2) = T3(R1-R2), and T1 = T3. But this means that Carl must now take the same amount of time to get to the end as Adam took to accomplish the first segment – which means that Adam’s first segment is the same distance as Carl’s last segment, since they are both on horseback for that respective segment.

!But we also know that Adam’s first journey of R1T1 = Bill’s a + b, and Carl’s last journey of R1T3 = Bill’s d + e. So the total distance accomplished by Bill is R1T1 – c + R1T3, but R1T1 = R1T3, so we can simplify this to 2xR1T1 – c.

!For calculation purposes, we will state that D1 = R1T1. During this time, Carl has traveled D1 x (R2 / R1), where R2 is the rate of walking. Thus, in order to return to Carl, Bill must travel D1 – D1 x (R2 / R1), or D1 x (1 – R2 / R1), or (D1 x R1 – D1 x R2) / R1, or (D1 / R1) x (R1 – R2). The closing rate of the two is R1 + R2, and he will travel R1 / (R1 + R2) of that distance, or D1 x (R1 – R2) / (R1 + R2). This distance is the same as b = c = d. So the total distance traveled by Adam or Carl is 2D1 – D1 (R1 – R2) / (R1 + R2), or D1(2 - (R1 – R2) / (R1 + R2)). From here it is elementary to determine the distances traveled and their respective times.

!For the given numbers, for instance, we have D1(2 – (6 – 4) / (6 + 4)) = 13.5, so D1 (1.8) = 13.5, and D1 = 7.5. c = 7.5 (0.2) = 1.5. Adam and Bill should ride for 7.5km / 6 kmh = 1.25 hours, and then Bill should go back and get Carl. They will all meet at the end at the same time.


The one rider take Two Hours and Fifteen minutes the two people who have to walk take Three hours and Fifteen minutes. The other horse must be sold, or shot and buried.

You say only single person can ride a horse, so the second horse is entirely extraneous.


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