A train reaches its destination on time if its speed is 60 km/hr from its…

Concept

When the same distance is covered at two different speeds, the times taken differ. If a distance d is covered at speed v1 in time t1 = d/v1, and at a slower speed v2 in time t2 = d/v2, the extra time taken at the slower speed is the difference d/v2 minus d/v1. This relation connects the distance directly to the two speeds and the time difference between them, without needing either travel time individually.

Application

Let the distance between the two stations be d km.

1. At 60 km/hr, the time taken is d/60 hours.

2. At 50 km/hr, the time taken is d/50 hours.

3. The train is late by 15 minutes, which is 15/60 hour, that is 1/4 hour, when it runs at 50 km/hr, so d/50 minus d/60 equals 1/4.

4. Taking a common denominator of 300, this becomes (6d minus 5d) divided by 300 equals 1/4, that is, d/300 equals 1/4.

5. Solving, d equals 300 divided by 4, that is, d equals 75.

So the distance between the two stations is 75 km.

Cross-check

At 60 km/hr, the time is 75/60, that is 1 hour 15 minutes. At 50 km/hr, the time is 75/50, that is 1 hour 30 minutes. The difference is exactly 15 minutes, matching the given delay, confirming the distance is 75 km.