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?
Detailed Explanation:
Let original speed be v km/h. The journey is 800 km divided into segments: 0–200 km, 200–600 km, 600–800 km. Speed reductions compound on the existing (current) speed.
Case 1: Minor accident at 200 km (speed → v/2), mechanical defect at 600 km (speed → v/2 × 1/4 = v/8). Time = 200/v + 400/(v/2) + 200/(v/8) = 200/v + 800/v + 1600/v = 2600/v hours.
Case 2: Mechanical defect at 200 km (speed → v/4), minor accident at 600 km (speed → v/4 × 1/2 = v/8). Time = 200/v + 400/(v/4) + 200/(v/8) = 200/v + 1600/v + 1600/v = 3400/v hours.
Given: Case 2 takes 4 hours more. So, 3400/v - 2600/v = 4 → 800/v = 4 → v = 200 km/h.
Therefore, the original speed was 200 km/h, making Option 1 the correct answer.
Question 4 of 5 Time, Speed & Distance
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