Time, Speed and Distance : Returning back to pick up problem $A$,$B$ and $C$ started out in a journey to a place which was $120$ km away. $A$ and $B$ went by car at the speed of $50$ kmph while $C$ travelled by tonga at $10$ kmph. After a certain distance $B$ got off and travelled the rest distance by another tonga at $10$ kmph, while $A$ went back for $C$ and reached the destination at the same time as $B$ arrived. The number of hours required for the trip was ____ $?$
I tried to approach the problem by equating the time taken by $A$ and $B$ as both of them has reached the destination at the same time after starting together. That means they must have taken the same amount of time.
Time taken by $B=\frac{d}{50} + \frac{120-d}{10}$ ; where $d$ is the distance travelled by $A$ and $B$ together on a car together.
Time take by $A=\frac{d}{50}$ + Time taken to return back an pick up $C$ + Time taken to reach the destination after picking up $C$
Time taken to return back an pick up C $\Rightarrow$ this I am not able to figure out as when A will return back to pick up C and until them meet, the C will be moving forward for some distance while A would be coming towards him and here the concept of relative distance will come into the picture. Getting confused here!!!
Please help me out.
Answer Provided : 4.8 hours
 A: B and C cover the same distance in the same time using a mix of the two modes of transport. Therefore the amount of time they go by car must be the same, as must the amount of time in the tonga.
Let $t_c$ be the time B (or C) spends in the car, and $t_t$ the time spent in a tonga. We want to know the total time $t_c+t_t$.
Person A drops off person B after $t_c$ hours, and starts travelling back until he meets person C who has now been travelling by tonga for $t_t$ hours. Person A therefore moves backwards for $t_t-t_c$ hours. The net distance he has travelled is $50t_c - 50(t_t-t_c)$, which is the same as the distance C has travelled, $10t_t$. So:
$$50t_c - 50(t_t-t_c) = 10t_t$$
which simplifies to:
$$5t_c = 3t_t$$
We also know that B (or C) traveled 120km in total by car and tonga, so
$$50t_c+10t_t = 120$$
Substituting the first equation into the second, we get
$$30t_t+10t_t = 120\\
t_t = 3$$
So
$$t_c = \frac{3t_t}5 = \frac95$$
And the total time is
$$t_t+t_c = 3 + \frac95 = 4 + \frac45 = 4.8$$
A: This will be equivalent to the other methods, but is another way of dividing up the trip.  A and B travel together at 50 kph for a distance $ \ x \ $ ; at that point, $ \ B \ $ continues for the remaining distance at 10 kph.  The total travel time for B is then
$$ \frac{x}{50} \ + \ \frac{120-x}{10} \ \ = \ \ T \ \ \Rightarrow \ \ x \ + \ (600 - 5x)  \ \ = \ \ 600 - 4x \ \ = \ \ 50T \ \ . $$
A is the only person who makes the entire trip at 50 kph, but has to "backtrack" a distance $ \ y \ $ in order to intercept C .  So A will have traveled $ \ 120 + 2y \ $ kilometers, for a total travel time
$$ \frac{120 + 2y}{50} \ \ = \ \ T \ \ \Rightarrow \ \ 120 + 2y \ \ = \ \ 50T \ \ . $$
When C is met by A, C will have covered a distance of $ \ x - y \ $ at 10 kph, after which A and C complete the trip at 50 kph.  Thus,
$$ \frac{x-y}{10} \ + \ \frac{120-(x-y)}{50} \ \ = \ \ T \ \ \Rightarrow \ \ 5·(x-y) \ + \ 120 \ - \ (x-y) \ \ = \ \ 4·(x-y) \ + \ 120 \ \ = \ \ 50T \ \ . $$
We can put the second and third of these equations together to write $ \ 2y \ = \ 4·(x-y) \ \Rightarrow \ 6y \ = \ 4x \  $ or
$ \ y \ = \ \frac{2}{3} x \ \ . $  [Now the relation to saulspatz's solution becomes clear.]
The third and first equations together become
$$ 4·(\frac13 x) \ + \ 120 \ \ = \ \ 600 \ - \ 4x \ \ \Rightarrow \ \ \frac{16}{3}x \ \ = \ \ 480 \ \ \Rightarrow \ \ x = 90 \ , \ y = 60 \ \ , $$
and now any of the equations can be used to find $ \ T \ = 4.8 \ $ hours.
[We also have a chronology of events:  B was "dropped off" at 1.8 hours into the journey, and A met C at 3 hours.]
$$ $$
ADDENDUM: This suggests a "time-symmetric" approach somewhat related to what Jaap Scherphuis has done.  B is taken some distance along the route at 50 kph and has to complete the remainder at 10 kph, while C does the reverse, traveling at 10 kph for a portion of the trip and being driven the rest of the way at 50 kph.
As arnold points out, when C has traveled a distance $ \ d \ $ , A and B have covered $ \ 5d \ $ ; B is then left to travel $ \ 120 - 5d \ $ on their own.  C goes on for a distance $ \ s \ $ until being met by A .  So A and C come together at "mile-marker" $ \ d + s \ $ at a certain time, by which time A has gone a total distance of $ \ 5d \ + \ [ \ 5d - (d+s) \ ] \ = \ 9d - s \ . $
This amount of time is
$$ \frac{d+s}{10} \ = \ \frac{9d-s}{50} \ \ \Rightarrow \ \ 5d  +  5s \ = \ 9d - s \ \ \Rightarrow \ \ 4d \ = \ 6s \ \ , $$
as we've seen elsewhere.  Since B takes as long to travel $ \ 120 - 5d \ $ as C takes to cover $ \ d + s \ $ at the same speed, we have
$$ 120 - 5d \ = \ d + s \ \ \Rightarrow \ \ s \ = \ \frac{2}{3} d \ = \ 120 - 6d \ \ \Rightarrow \ \  \frac{20}{3}·d \ = \ 120  $$
$$ \Rightarrow \ \ d \ = \ 18 \ \ , \ \ s \ = \ 12 \ \ . $$
So B and C have each traveled alone for $ \ \frac{18+12}{10} \ = \ \frac{120-5·18}{10} \ = \ 3 \ $ hours, and each has traveled with A in the car for $ \ \frac{120 - (18+12)}{50} \ = \ \frac{5·18}{50} \ = \ 1.8 \ $ hours, making a total trip time of $ \ 4.8 \ $ hours.
A: I think it's easier to do this if we make times the unknows, rather than distances, but of course, you can do it either way.  Let us say that A and B drive for $t_1$ hours before they stop, and that it takes A $t_2$ hours to go back and pick up C.
It takes B $$t_1+\frac{120-50t_1}{10}=12-4t_1$$ hours to complete the trip.
When A and C meet, C has traveled a distance of $10(t_1+t_2)$ and A's distance from the staring point is $50t_1-50t_2$.  Equating these distances gives $2t_1=3t_2$.
Now the time it take until the car arrives at the destination is $$t_1+t_2+\frac{120-10(t_1+t_2)}{50}$$
Equating the two arrival times, and using $2t_1=3t_2$ should give the right answer.  (I haven't checked this.)
A: At the moment A turns around, C will have traveled a distance of $10\frac{d}{50}$, one-fifth the distance as A and B given that it is moving at one-fifth the speed. A and C will meet when they have together covered the distance between them, $4d/5$. We can expect that A will be doing most of the legwork, and that it will cover five times the separating distance as C.
We're interested in both the time ($t_{A\rightarrow C}$) and place ($x_{A\rightarrow C}$) of A and C's reunion. We'll measure time from the moment of the turn-around and distance from the starting point. We can set up a simple system of equations to express the unknowns in terms of d.
$$50t_{A\rightarrow C}+10t_{A\rightarrow C}=\frac{4d}{5}$$
$$x_{A\rightarrow C}=\frac{d}{5}+10t_{A\rightarrow C}$$
The solution is $t_{A\rightarrow C}=2d/150$ and $x_{A\rightarrow C}=d/3$. The total time for A to arrive at the end is
$$T=\frac{d}{50}+\frac{2d}{150}+\frac{120-d/3}{50}=\frac{360+4d}{150}$$
while B gets to the end in
$$T=\frac{d}{50}+\frac{120-d}{10}=\frac{600-4d}{50}.$$
As we are told they arrive at the same instant, then we find
$$4T=960/50$$
$$T=4.8 hrs$$
A: I have another solution that I must add because it shows the power of reframing the problem geometrically.
Let's construct a cartesian coordinate system containing order pairs of time and position, which we'll refer to as "points". There are four points of interest: O, the starting point; U, the u-turn point; R, the rendezvous point; and F, the finishing point. Note that all of these points refer to both a time and position. As the travelers are moving from point to point with constant velocity, their paths through the cartesian coordinate system can be represented by line segments. In the diagram, the line segments are labeled to refer to who takes the path; elsewhere, I will refer to the line stretching from some point P to some point Q as $\overline{PQ}$.

The big revelation of this diagram is that $\overline{OR}$ parallels $\overline{UF}$ and $\overline{OU}$ parallels $\overline{RF}$, as  the traveler(s) on each pair of paths are moving with the same velocity. That means quadrilateral $OUFR$ is a parallelogram. It is not hard to show by a similarity argument that the immediate consequence is $\overline{OR}$ is congruent to $\overline{UF}$ and $\overline{OU}$ is congruent to $\overline{RF}$.
Translating this geometric result back into the original language, we find some results that may have not been immediately obvious:

*

*Traveler B takes the same amount of time and covers the same amount of distance to arrive at the finish line after boarding the tonga as does C to get from the starting point to the rendezvous with A.

*Travelers A and B take the same amount of time and cover the same amount of distance to arrive at the U-turn point after leaving the starting point as do A and C to drive to the finish line after their rendezvous.

This is the same observation explicitly stated by Jaap Scherphuis, now made obvious if it were not initially. With this insight in mind, let's solve the problem of finding the finish time. We can immediately construct a system of equations in terms of $(t_P,x_P)$, the time and position at a point P with reference to O.
\begin{array}{lcl} x_U+x_R & = & 120 \\ x_U & = & 50t_U
\\ x_R & = & 10t_R \\ x_R-x_U & = & -50(t_R-t_U)\end{array}
The first equation says the total distance is 120km and the remaining equations are the speeds (slopes) along each path. The solution is $(t_U,x_U)=(9/5,90)$ and $(t_R,x_R)=(3,30)$. The total time is the sum of $t_U$ and $t_R$, which is 4.8 hours.
A: Let fraction of distance travelled by tonga
by $C$ (at start) and $B$ (towards end)  be  $f$ each,
Then the middle gap $|\; f\;|\; gap\;|\; f\;| = 1-2f$  is the backtrack for the car,  thus the car travels $2(1-f)+(1-2f) =3-4f$
Equating times for car and mixed conveyance,
$\dfrac{3-4f}{5} = f + \dfrac{1-f}{5} \Rightarrow f = \dfrac14$
Finally, time taken $={\dfrac14}\dfrac{120}{10} +\dfrac34\dfrac{120}{50} = 4.8 \;hrs$
