With an average speed of 60 km/h a train reaches to its destination on time.…

Concept: For two journeys covering the SAME distance, time and speed are inversely proportional (Time = Distance ÷ Speed). If travelling at a lower speed makes a traveller late by a known amount, that lateness equals the difference between the two travel times for that same distance — captured by x/s2 − x/s1 = Δt (s1 > s2), which rearranges to the shortcut Distance = (s1 × s2)/(s1 − s2) × Δt.

Application: Applying this to the given journey:

1. Let the distance travelled be x km.

2. Time taken at 60 km/h = x/60 hours; time taken at 45 km/h = x/45 hours.

3. Travelling at the lower speed (45 km/h) causes a 20-minute delay, i.e. 20/60 hour, so x/45 − x/60 = 20/60.

4. Combine over a common denominator of 180: (4x − 3x)/180 = 20/60, which simplifies to x/180 = 1/3.

5. Solving gives x = 60.

Cross-check: Applying the shortcut identity directly: Distance = (60 × 45)/(60 − 45) × (20/60) = (2700/15) × (1/3) = 180 × (1/3) = 60 km — the same value obtained above, confirming the result independently.

∴ Result: The distance travelled by the train is 60 km.