Two trains travel from A to B and B to A in 6 and 10 hours respectively. They start simultaneously at 10 a.m.. If the first train meets wit…

Let x km be the distance between the two points A and B and let us assume that the accident occurred after t hours.

Speed of train going from A to B = $(\frac{x}{6})$ kmph

Speed of train going from B to A = $(\frac{x}{{10}})$ kmph

Now, distances travelled by the trains in t hours are $(\frac{x}{6} \times t = \frac{{xt}}{6}km)$ and $(\frac{x}{{10}} \times t = \frac{{xt}}{{10}}km)$ respectively.

Distance between the trains at the time of accident = $(x - \frac{{xt}}{6} - \frac{{xt}}{{10}} = x\left( {1 - \frac{t}{6} - \frac{t}{{10}}} \right))$

Speed of train going from A to B after accident = $(\frac{1}{2} \times \frac{x}{6} = \frac{x}{{12}})$ kmph

The two trains meet at 2 p.m. which is 4 hours from 10 a.m.

Time taken by the two trains to meet after accident = 4 - t

$(Relative\;speed = \frac{{Distance\;between\;the\;trains}}{{Time\;taken\;to\;meet}})$

$(\Rightarrow \frac{x}{{12}} + \frac{x}{{10}} = \frac{{x\left( {1 - \frac{t}{6} - \frac{t}{{10}}} \right)}}{{4 - t}})$

On solving, we get t = $(3\frac{1}{3})$ hours = 3 hours 20 minutes

∴ The first train meets with an accident after 3 hours 20 minutes of the start i.e. at 1 : 20 p.m..