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