A train has to complete a journey of 800 km. If it meets a minor accident, its speed becomes half of the existing speed. If there is a mechanical defect, the speed becomes one-fourth of the existing speed. On its way, the train meets with a minor accident after 200 km; and 400 km thereafter, it develops a mechanical defect. Had the train developed the mechanical defect after 200 km and met the minor accident 400 km thereafter, it would have taken 4 more hours to reach its destination. What was the original speed of the train in km per hour?
Answer & explanation
Answer: (a) 200
With original speed v, the actual trip takes 200/v + 400/(v/2) + 200/(v/8) = 2600/v hours, and the alternative takes 200/v + 400/(v/4) + 200/(v/8) = 3400/v hours. The gap 800/v = 4 hours gives v = 200 km/h.
- Actual: first 200 km at v; next 400 km at v/2; last 200 km at (v/2)/4 = v/8.
- Time = 200/v + 800/v + 1600/v = 2600/v.
- Alternative: 200 km at v; 400 km at v/4; last 200 km at (v/4)/2 = v/8.
- Time = 200/v + 1600/v + 1600/v = 3400/v.
- 3400/v − 2600/v = 800/v = 4, so v = 200 km/h.
- Check: 13 h and 17 h — a 4-hour difference.
Remember · Write each scenario's time as distance ÷ speed with the speed in terms of v, then equate the given difference.
Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·