Problems on Trains — Advanced Exam Problems

🎯 Learning Objective

Combine lengths, relative speed and equal-time logic to crack multi-equation train problems.

💡 Concept

• System questions: pole/man time and platform time give two equations — subtracting them isolates the speed
• Two trains crossing each other: distance = both lengths, speed = relative; with equal lengths, L cancels
• Meeting problems: both trains share the same travel time — the gap uses the speed DIFFERENCE, the total uses the SUM
• Harmonic shortcut: equal-length trains with pole times t₁ and t₂ cross each other (opposite) in 2t₁t₂/(t₁ + t₂)
• Hard train questions are 90% setup — name L and v, then translate each sentence into one equation

🧮 Key Formulas

Equal lengths, pole times t₁, t₂ → opposite-crossing time = 2t₁t₂/(t₁ + t₂)

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Man: L = v·t₁; platform: L + P = v·t₂ → v = P/(t₂ − t₁)

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Meet with d km extra: t = d/(v₁ − v₂); total distance = (v₁ + v₂) × t

✏️ Easy Example

Q. A 240 m long train crosses a platform twice its own length in 40 seconds. Find its speed in km/h.

1. Platform length = 2 × 240 = 480 m — 'twice its length' is measured against the train
2. Crossing distance = train + platform = 240 + 480 = 720 m
3. Speed = 720/40 = 18 m/s
4. In km/h: 18 × 18/5 = 64.8

Answer: 64.8 km/h

🇮🇳 Real-Life Example

Control rooms compute crossing windows for single-track sections with exactly these equations — two lengths, relative speed, one safe time slot.

📝 Exam-Level Example

Q. Two trains of equal length cross a pole in 4 seconds and 5 seconds respectively. Running in opposite directions, how long will they take to cross each other?

1. Let each train be L metres — pole time gives the speed: v₁ = L/4, v₂ = L/5
2. Crossing each other needs distance L + L = 2L — both lengths always add
3. Opposite directions → relative speed = L/4 + L/5 = (5L + 4L)/20 = 9L/20
4. Time = distance ÷ relative speed = 2L ÷ (9L/20) — the unknown L cancels, so no length was needed
5. = 2 × 20/9 = 40/9 seconds ≈ 4.44 s

Answer: 40/9 s (≈ 4.44 seconds)

📝 Exam-Level Example

Q. A train crosses a man standing on a platform in 8 seconds and the 180 m platform itself in 20 seconds. Find the length and the speed of the train.

1. Crossing the man: he is a point, so distance = the train's own length → L = 8v
2. Crossing the platform: the platform's length joins in → L + 180 = 20v
3. Substitute L = 8v into the second equation: 8v + 180 = 20v — two equations collapse into one
4. 12v = 180 → v = 15 m/s
5. L = 8 × 15 = 120 m; speed in km/h = 15 × 18/5 = 54

Answer: Length 120 m, speed 54 km/h

📝 Exam-Level Example

Q. Two trains start at the same time from stations P and Q towards each other at 50 km/h and 40 km/h. When they meet, the faster train has travelled 60 km more than the slower one. Find the distance between P and Q.

1. Both trains travel for the SAME time t until they meet — the meeting freezes the clock for both
2. The gap between them grows at 50 − 40 = 10 km/h and must reach 60 km
3. t = 60/10 = 6 hours
4. Together they close the whole P-to-Q distance at 50 + 40 = 90 km/h
5. Distance = 90 × 6 = 540 km; check: 300 − 240 = 60 ✓

Answer: 540 km

🪄 Memory Trick

Equal-length pole-time pattern: opposite-direction crossing time = 2t₁t₂/(t₁ + t₂) — the harmonic shortcut. For 4 s and 5 s it is 40/9 s.

⚠️ Common Mistakes

• ❌ Taking only one train's length when they cross each other (both lengths always add)
• ❌ Giving the standing man a length — he is a point, only the train's length counts
• ❌ Using the speed SUM for the 60-km gap (the gap grows at the DIFFERENCE)

🏆 Exam Tips

• ✅ Man/pole time and platform time → two equations; subtract them to isolate the speed
• ✅ Meeting problems: equal travel time is the hidden equation — write it first
• ✅ Stay in m/s throughout; convert to km/h only at the final line

📌 Summary

• Two-train crossing: distance = both lengths, speed = relative
• Man + platform pair → subtract equations for speed
• Meeting: equal times; gap uses difference, total uses sum
• L cancels in ratio questions — assume it freely