Let the distance between the man's home and the railway station be $d$ km.

Step 1: Set up equations based on the two scenarios.

1. First Scenario:

   • The man walks at $12$ km/hr and reaches the station $10$ minutes (or $\frac{1}{6}$ hours) after the train has departed.
   • Time taken by the man at $12$ km/hr to reach the station: $\frac{d}{12}$.
   • Since he is $\frac{1}{6}$ hours late, we can write: $T + \frac{1}{6} = \frac{d}{12}$ where $T$ is the time of the train's departure from his home.

2. Second Scenario:

   • The man walks at $15$ km/hr and reaches the station $10$ minutes (or $\frac{1}{6}$ hours) before the train departs.
   • Time taken by the man at $15$ km/hr to reach the station: $\frac{d}{15}$.
   • Since he is $\frac{1}{6}$ hours early, we can write: $T - \frac{1}{6} = \frac{d}{15}$

Step 2: Solve the equations.

We have two equations:

1. $T + \frac{1}{6} = \frac{d}{12}$
2. $T - \frac{1}{6} = \frac{d}{15}$

Subtract the second equation from the first: $\left(T + \frac{1}{6}\right) - \left(T - \frac{1}{6}\right) = \frac{d}{12} - \frac{d}{15}$ $\frac{2}{6} = \frac{d}{12} - \frac{d}{15}$ $\frac{1}{3} = \frac{5d - 4d}{60}$ $\frac{1}{3} = \frac{d}{60}$ $d = 20 \, \text{km}$

Conclusion

The distance from his home to the railway station is 20 km.

Would you like further details on any part of this solution?

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Here are five related questions for further exploration:

1. What if the man’s walking speeds were both slower—how would that affect the distance?
2. How would the problem change if he was late by 15 minutes instead of 10 in both scenarios?
3. What if the train's timing had been delayed—how would we adjust the equations?
4. Can we solve the problem graphically by plotting the two scenarios?
5. How would adding rest stops along the way affect the calculations?

Tip: For problems involving different speeds and times, setting up equations based on known times and distances simplifies the solution process.