Problems on Trains

Let train length = L meters, speed = S m/s. Crossing signal: L/S = 21.6, so L = 21.6S. Crossing platform: (L + 160)/S = 28. Substituting L: (21.6S + 160)/S = 28. So 21.6 + 160/S = 28, giving 160/S = 6.4, thus S = 25 m/s. Then L = 21.6 × 25 = 540 meters.

Total distance to cross the bridge = train length + bridge length = 280 m + 320 m = 600 m. Speed = 60 km/h = 60 * (5/18) = 50/3 m/s. Time = distance / speed = 600 / (50/3) = 600 * 3 / 50 = 36 seconds.

Train length = speed × 18. Platform length = 220m. Time difference = 29 - 18 = 11 seconds. Speed = 220/11 = 20 m/s. Convert to kmph: 20 × 18/5 = 72 kmph.

Speed = 90 km/h = 90 × 5/18 = 25 m/s. Distance covered in 18 seconds = 25 × 18 = 450 m. This distance equals train length + platform length. Platform length = 450 - 150 = 300 m.

Relative speed = (200 + 200) / 16 = 400/16 = 25 m/s. This is the SUM of both trains' speeds. Without additional information about either train's individual speed or the ratio between them, we cannot determine train A's speed specifically. The answer 'Cannot be determined' is correct.

First train speed = 150/12 m/s = 12.5 m/s = 45 km/hr. Let second train speed = v km/hr. Relative speed (opposite direction) = (45 + v) km/hr = (45 + v) × 5/18 m/s. Time = 150/[(45+v)×5/18] = 15 seconds. Solving: 45 + v = (150 × 18)/(15 × 5) = 36, so v = 27 km/hr. Option B matches.

Train length = 300 m, Speed = 54 kmph. Convert speed to m/s: 54 × 5/18 = 15 m/s. Time to cross a pole = Length/Speed = 300/15 = 20 seconds. The train needs to travel its own length to completely cross the pole.

Let speeds be v1 and v2 (v1 > v2). Total distance = 300 + 500 = 800m. Same direction: relative speed = v1 - v2 = 800/40 = 20 m/s. Opposite direction: relative speed = v1 + v2 = 800/10 = 80 m/s. Adding: 2v1 = 100, v1 = 50 m/s. Converting to km/hr: 50 × (3600/1000) = 180 km/hr.

Train length = platform length = L. Total distance to cross = L + L = 2L. Speed = 25 m/s, time = 60 seconds. Distance = speed × time = 25 × 60 = 1500 m. So 2L = 1500, L = 750 m. The train must travel its own length plus the platform length to completely cross.

Relative speed = 36 + 54 = 90 km/hr = 90 × 5/18 = 25 m/s. Total distance to cover = 132 + 118 = 250 m. Time = Distance/Speed = 250/25 = 10 seconds. Option B is correct.

Let the train's length be L meters and speed be s m/s. From crossing the pole: L = 10s, so s = L/10. From crossing the platform: L + 200 = 20s = 20(L/10) = 2L. Solving: L + 200 = 2L gives L = 200 meters. The train takes 10 seconds to cross a pole at its own speed, and 20 seconds to cover both its length and the 200m platform.

Speed ratio 5:8, time ratio 4:3. Distance = Speed × Time. Length1 = 5k×4t = 20kt. Length2 = 8k×3t = 24kt. Sum of lengths: 20kt+24kt=44kt=660, so kt=15. Difference in lengths: 24kt-20kt=4kt=4×15=60. The answer is 60.

Statement II: Train length = speed × time = 45 × (5/18) × 12 = 150m. Statement I and III together give length from (train speed - walking speed) × 60 = train speed × 30. Both approaches yield same answer, so II alone OR I+III are sufficient.

When trains move in the same direction, their relative speed equals the difference of individual speeds. Train 1's length = 54 kmph × 30 sec = 450 m. Combined length = (54-36) kmph × 200 sec = 1000 m. Therefore train 2's length = 1000 - 450 = 550 m.

First 4 primes: 2, 3, 5, 7. First train length = (2 × 7)^2 - 1 = 196 - 1 = 195m. Second train length = (3 × 5)^2 = 225m. Total length = 420m. Crossing time = 12s, so relative speed = 420/12 = 35 m/s = 126 kmph. Since speed of first train is 54 kmph, speed of second = 126 - 54 = 72 kmph.