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A train and a bus arrive at random between 9 am and 10 am. The train stops for 10 minutes whereas the bus stops for 'x' minutes. Find 'x' such that the probability that the bus and train meet is 0.5.

How to approach such type of questions ?

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There are two cases to consider. One where the bus arrives before the train and the other way around. Representing the arrival times of the bus and the train as x and y coordinate respectively. We have the following equations representing the situation. 

$y < x + t$ (i.e. train should arrive within t minutes after the bus.)

$x < y + 10$ (i.e. bus should arrive within 10 minutes after the train.)

We thus obtain the following plot:

The area of the required region represents time duration during which the train and the bus can meet. The sample space is represented by 60x60 plot (60 mins from 9 to 10). And we know that there is a 0.5 probability of the conditions being met. Thus, by dividing the area of the required region by total area of the sample space we get 0.5. 

Area of the two triangles = $\frac{1}{2} *[(60-t)^2 + 50*50]$.

The sum of areas of the two triangles would equate to half the total area ($\because probability = 0.5$).

Thus we have:
=> $\frac{1}{2} *[(60-t)^2 + 50*50] = 3600/2 \\
 = (60-t)^2 = 1100\\
 = 60 - t = 10*\sqrt{11}\\
 => t = 26.83 mins$

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