$A$,$B$ 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
Best Answer
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
$$t_t+t_c = 3 + \frac95 = 4 + \frac45 = 4.8$$