Prelims 2026 · CSAT / Quantitative Aptitude · Question 69
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
200
Let the original speed be v. Actual journey: 200 km at v, then 400 km at v/2, then 200 km at v/8. So T1 = 200/v + 400/(v/2) + 200/(v/8) = 2600/v. Hypothetical journey: 200 km at v, then 400 km at v/4, then 200 km at v/8. So T2 = 200/v + 400/(v/4) + 200/(v/8) = 3400/v. Given T2 - T1 = 4, so 800/v = 4, hence v = 200 km/h.