mathematics - The bird and the train

This question is one I found in a competitive exam paper, and has two solutions: one simple and one complex.

A train starts from the east end of a 200 km long east-west track at 50 km/h. At the same time, a bird starts flying in a straight line from the west end along the same track at 60 km/h. When the train and bird meet, the bird immediately turns around and goes to its starting point, then again turns around and meets the train. This goes on till the train reaches the bird's origin point. How much total distance has the bird covered?

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  Both the bird and the train neither speed up nor slow down during this entire time.

  
  


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  Bonus for posting more than one solution.

  
  

Answer

The bird travels

$240$ km in total.

Take the general case where the train starts $x$ km from the west end, and the bird starts at the west end. They travel towards each other at a relative speed of $110$ km/h, so they will meet each other after $\frac{x}{110}$ hours. The bird will then take the same amount of time to travel back, at which point the train will have travelled $2\times\frac{50x}{110}=\frac{10x}{11}$ km and the bird will have travelled $2\times\frac{60x}{110}=\frac{12x}{11}$ km. At this point we are back in the initial position, except that the train is now only $\frac{x}{11}$ km from the west end.

Now if we start the train at $200$ km, we can see that the bird travels $\frac{12}{11}\times 200$ km on its first round trip, $\frac{12}{11}\times \frac{200}{11}$ km on its second, and so on. Thus we can construct the total distance travelled as the infinite sum

$\begin{align}\frac{2400}{11}\sum\limits_{n=0}^{\infty}\frac{1}{11^n} &=\frac{2400}{11}\times\frac{1}{1-\frac{1}{11}} \\ &=240\end{align}$

So, as stated, the bird travels

$240$ km in total.