A train passes two persons walking in the same direction at a speed of 3 kmph and 5 kmph respectively in 10 seconds and 11 seconds respectively. The speed of the train is

Explanation:-

1^st method:
Let the speed of the train be S. And length of the train be x.
When a train crosses a man, its travels its own distance.
$$\eqalign{ & {\text{According to question}}; \cr & \frac{x}{{\left[ {\left( {s - 3} \right) \times \left( {\frac{5}{{18}}} \right)} \right]}} = 10 \cr & {\text{or}},\,18x = 50 \times s - 150.....({\text{i}}) \cr & {\text{and}} \cr & \frac{x}{{\left[ {\left( {x - 5} \right) \times \left( {\frac{5}{{18}}} \right)} \right]}} = 11 \cr & 18x = 55 \times s - 275......({\text{ii}}) \cr & {\text{Equating equation }}\left( {\text{i}} \right){\text{ and }}\left( {{\text{ii}}} \right) \cr & 50 \times s - 150 = 55 \times s - 275 \cr & {\text{or}},\,5 \times s = 125 \cr & {\text{or}},\,s = 25\,{\text{kmph}} \cr}$$
2^nd method:
In the both cases train has to travel its own length that means, distance is constant.
So, we have,
When distance is constant, speed is inversely proportional to time.
i.e. speed α 1/time.
In such case, the following ratios will be valid