Opposite-running trains with a staggered start — meeting place & time: The distance between stations X and Y is 450 km. Train L leaves X at 6:00 pm at 60 km/h. Train M leaves Y earlier at 5:20 pm at 80 km/h. When and where (from X) do they meet?

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Solve this multiple-choice question and choose the correct option.

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Introduction / Context:Meeting problems with staggered departures are solved by accounting for one train’s head start, then using relative speed thereafter. Given Data / Assumptions: Distance X–Y = 450 km Train L: 60 km/h, starts 6:00 pm from X Train M: 80 km/h, starts 5:20 pm from Y (40 min earlier) Concept / Approach:Compute distance M covers before 6:00 pm, subtract from 450, then divide by relative speed (60 + 80) for the time after 6:00 pm. Step-by-Step Solution: Head start (40 min = 2/3 h): M covers 80 * 2/3 = 53.333 kmRemaining separation at 6:00 pm = 450 − 53.333 = 396.667 kmRelative speed after 6:00 pm = 140 km/hTime after 6:00 pm = 396.667 / 140 ≈ 2 h 50 minMeeting time ≈ 8:50 pm; distance from X = L’s travel in 2 h 50 min = 60 * 2.8333 ≈ 170 km Verification / Alternative check:Check M’s distance from Y at 8:50 pm: 80 * (3 h 30 min) = 280 km; sums with 170 km to 450 km. Why Other Options Are Wrong:They place the meet too early/late or at the wrong distance given the head start. Common Pitfalls:Forgetting to include the early start distance or using only one train’s speed instead of the relative speed. Final Answer:170 km, 8 : 50 pm

Opposite-running trains with a staggered start — meeting place & time: The distance between stations X and Y is 450 km. Train L leaves X at 6:00 pm at 60 km/h. Train M leaves Y earlier at 5:20 pm at 80 km/h. When and where (from X) do they meet?

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