Two guys and a horse This problem is from my general maths textbook.
Two guys will travel to a place $20$ miles away from their house. They both will start at the same time, and will walk at a constant velocity of $4$ miles/hour.
However, they own a horse that can carry one person at a time and can run at $10$ miles/hour, regardless of whether someone is riding or not. Find the minimum time required for both of them to reach their destination.  
This is what I got till now:
The horse will travel x miles with person 1, drop him off, and turn back to catch person 2 who had been walking that whole time.
We will choose x in such a way that person 1 reaches the destination by the time the horse reaches person 2, and then the horse will carry person 2 the rest of the way, solving it gives $3 \frac{19}{41}$ hours. The answer provided is $3 \frac{1}{11}$ hours. What did I miss?
Update 1:
Apparently we can improve our choice of x. We'll be choosing it in such a way that both persons reach the destination at the same time. This resulted in overall $3 \frac{1}{3}$ hours.
 A: Let $t$ be the minimum time for the trip in hours.  The optimal solution is for the horse to take one person $x$ miles, drop him off, go back to pick up the other person, and ride $x$ miles with the second person in such a way that they all reach the destination at the same time.  In this way, each person rode for $x$ miles and walked for $20-x$ miles.  Since the points where the people switch from walking to riding are each $20-x$ miles from an endpoint, the total number of miles the horse traveled is $$2x+(20-2(20-x))=2x+(2x-20)=4x-20$$
Since time is distance divided by speed, we have the following equations (the first since the horse travelled for a total of $t$ hours and the second since the people rode and walked respectively for a total of $t$ hours):
$$\frac{4x-20}{10}=t\\\frac x{10}+\frac{20-x}{4}=t$$
The solution to this system of equations is $x=\frac{140}{11}$ miles and $t=\frac{34}{11}$ hours, as the book suggested.
A: To get an answer of $3\frac1{11}$ hours, we have to assume that (1) the horse is sufficiently trained to run back down the road on its own to pick up the walking person, (2) the initial rider knows the exact moment to dismount to achieve the optimal arrival time, and (3) the time needed to mount or dismount the horse is negligible. It's probably easier to see what's going on if we look at the general case rather than jumping straight into the arithmetic. Let the distance to travel be $d$, the walking speed be $u$, and the trotting speed $v$ (respectively 20 miles, 4 mph, and 10 mph in the question). Person A sets out on foot, with person B on the horse. At an intermediate time $t$, B dismounts, sends the horse back to pick up A, and continues the journey on foot. Person A meets the horse later at time $t'$, mounts it, and arrives at the destination at the same time $T$ as B does.
Since the total distance travelled by A is $d$, we have $ut'+v(T-t')=d,$ or$$(v-u)t'=vT-d\qquad(1)$$and similarly for B, we get $vt+u(T-t)=d$, or:$$(v-u)t=d-uT.\qquad(2)$$The net distance of the horse from the start, after its backward leg, equals the distance covered by A when they meet at time $t'$: namely $vt-v(t'-t)=ut'$, or$$(u+v)t'=2vt.\qquad\qquad(3)$$Considering the ratio of $t'$ to $t$ from eqns 1 and 2, and then from eqn 3, we have$$\frac{t'}t=\frac{vT-d}{d-uT}=\frac{2v}{u+v},$$from which we obtain$$T=\frac{u+3v}{3u+v}\cdot\frac dv.$$Substituting the given values for the distance and speeds then gives the required result.
