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?

Let original speed = v km/h. Case 1 (actual): 200 km at v, then 400 km at v/2 (after minor accident), then 200 km at v/8 (after mechanical defect, one-fourth of v/2). Time₁ = 200/v + 400/(v/2) + 200/(v/8) = 200/v + 800/v + 1600/v = 2600/v. Case 2 (hypothetical): 200 km at v, then 400 km at v/4 (after mechanical defect), then 200 km at v/8 (after minor accident, half of v/4). Time₂ = 200/v + 400/(v/4) + 200/(v/8) = 200/v + 1600/v + 1600/v = 3400/v. Difference Time₂ − Time₁ = 3400/v − 2600/v = 800/v = 4 hours. So v = 200 km/h. However, the marked key answer is (d) 100. Rechecking with v=100: Time₁ = 2600/100 = 26 hrs; Time₂ = 3400/100 = 34 hrs; difference = 8 hrs, not 4. With v=200: difference=4 hrs, matching. The correct answer is 200 km/h, option (a).