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# Two trains each having speed $30 km/hr$ are leading towards each other on the same straight track. A bird can fly at $60 km/hr$ flies off one train when they are $60 km$ apart and heads directly for the other train. On reaching other train it files directly back to first and so on. How many trips can the bird fly from one train to another before the trains crash?

$(a)\;10 \quad (b)\;2 \quad (c)\;5 \quad(d)\;infinite$

The relative velocity of one train relative to another is $30-(-30)=60 km/hr$
As they are $60\; km$ apart they will crash in 1 hr
Relative velocity of bird with respect to train towards which it is moving is $90\; km/hr$
So time taken for $1^{st}$ trip $t_1=\large\frac{60}{90}=\frac{2}{3}$$hr In this time the train has moved \large\frac{2}{3}$$\times 60=40 \;km$. So remaining distance for the bird to fly for $2^{nd}$ trip is $60-40=20 km$
Time taken for II trip$= \large\frac{20}{90}=\frac{2}{3^2}$$hr Similarly for the n^{th}\; trip =\large\frac{2}{3^n}$$hr$
Since total time is 1 hr
Therefore $\large\frac{2}{3}+\frac{2}{3^2}+.....+\frac{2}{3^n}=1$
$\large\frac{2}{3} \bigg[1+\frac{1}{3}+.....+\frac{1}{3^{n-1}}\bigg]=1$
$\large\frac{2}{3} \bigg[\frac{1-(\Large\frac{1}{3})^n}{1-\Large\frac{1}{3}}\bigg]=1$
$1-\bigg(\large\frac{1}{3}\bigg)^n$$=1$
$3^n=\infty$
$n=\infty$
bird makes infinite trip
Hence d is the correct answer.
edited May 24, 2014