A train travelling from Jaipur to Agra meets with an accident after 1 hour. It…

Concept Used:

When a train's speed drops to a fraction of its usual speed for the remaining part of a journey, the extra time taken over any fixed stretch equals (that distance) × (1/reduced speed − 1/usual speed). Comparing two scenarios where the point of speed-reduction differs by a known extra distance isolates the delay caused purely by that extra distance, which lets us solve for the usual speed directly — without needing the total distance or the exact location of the accident.

Calculation:

1. Let the usual speed of the train be v km/hr. After the accident, the train proceeds at (5/8)v km/hr (given).

2. The difference between the two delays is 2 hr 24 min minus 1 hr 48 min = 36 min = 3/5 hr. This difference is caused entirely by the last 80 km, which was covered at the reduced speed in the first scenario but at the usual speed in the second scenario (there, the accident happens 80 km further along, after that stretch).

3. So the extra time taken to cover this 80 km at the reduced speed instead of the usual speed equals 3/5 hr: 80/((5/8)v) − 80/v = 3/5.

4. Simplify the left side: (80 × 8)/(5v) − 80/v = 128/v − 80/v = 48/v.

5. So 48/v = 3/5, which gives v = 48 × 5/3 = 80.

Cross-check:

With v = 80 km/hr, the reduced speed is (5/8) × 80 = 50 km/hr. Time for 80 km at 50 km/hr = 80/50 = 1.6 hr = 96 min; time for the same 80 km at 80 km/hr = 60 min. The difference is 96 − 60 = 36 min, exactly matching the gap between the two given delay figures — confirming the usual speed is 80 km/hr.