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?

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

  • Bonus for posting more than one solution.

  • $\begingroup$ i am not sure what makes this question interesting... simple middle school math question... if u have asked how many turn arounds happened, it would be much more interesting... $\endgroup$ – Oray Apr 2 '16 at 6:22
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    $\begingroup$ simple middle school math question Yup, if you only consider the simple solution. Turn arounds are infinite(mathematically). $\endgroup$ – cst1992 Apr 2 '16 at 6:23
  • $\begingroup$ @Oray No, the answer is not infinite. That's a common trap ;) $\endgroup$ – cst1992 Apr 2 '16 at 6:25
  • $\begingroup$ well yes the answer is 240 okay, but what is the interesting part of this question? $\endgroup$ – Oray Apr 2 '16 at 6:26
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    $\begingroup$ von Neumann calculated this infinite sum in his head, can't be that difficult. $\endgroup$ – ffao Apr 2 '16 at 6:30

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.

  • $\begingroup$ Bonus: you can instead compute the sum for the time the bird travels and combine the two answers! $\endgroup$ – Zandar Apr 2 '16 at 8:10

The bird flies:

240 km.


The train travels at 50km/h for 200km so that takes it 4 hours.
The bird is constantly flying at 60km/h for these 4 hours so it covers a distance of 240 km.


The bird travels 109.09km.

Bonus because:

It then splats into the front of the train since neither slow down!

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    $\begingroup$ how do you know the bird doesn't smash the train up? $\endgroup$ – JMP Apr 2 '16 at 6:46
  • $\begingroup$ then the bird flies back which takes x hours, in which the train travels 50*x km, leaving a gap of y km, and B+T meet again when (50+60)z=y, etc... $\endgroup$ – JMP Apr 2 '16 at 6:49
  • $\begingroup$ @Zandar; yeah thanks... i going to delete it $\endgroup$ – JMP Apr 2 '16 at 7:06

Simple solution is by logic:

Time taken for both the bird and the train to reach the bird's origin point is the same in the end, i.e. 4 hours. If the speeds don't change, we can simply do $distance$ = $speed$ x $time$ and get the answer for the bird as 240 km.

Complex solution is by math:

We trace the path of the bird in this whole event, and calculate the distance travelled in each round trip. Since the bird has to travel less distance each time but not zero, we'll have infinite round trips, but in a way that converge (so infinite distance is the wrong answer). Adding distances covered by the bird, we again get 240 km.

i.e. basically the answer by @Zandar.


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