Problems on Trains

Distance covered = train length + platform length = L + L = 2L. Speed = 90 kmph = 90×(5/18) = 25 m/s. Time = 1 minute = 60 seconds. Distance = 25×60 = 1500 m. So 2L = 1500, L = 750 m. Options A (500), B (600), D (900) don't match.

Let train length be L meters and speed be v m/s. When passing a man, distance = L, time = 20s, so L = 20v. When passing the platform, distance = L + 100, time = 28s, so L + 100 = 28v. Substituting: 20v + 100 = 28v, giving 8v = 100 or v = 12.5 m/s. Converting to km/hr: 12.5 × 18/5 = 45 km/hr.

Speed = 30 m/sec, bridge length = 600 m, time = 30 sec. Distance covered = speed × time = 30 × 30 = 900 m. This distance equals bridge length + train length, so train length = 900 - 600 = 300 m. Option D (300 m) is correct.

When a train crosses a platform, total distance covered = train length + platform length = 160 + 146 = 306 m. Speed = distance / time = 306 / 20.4 = 15 m/s. Converting to km/hr: 15 × 18/5 = 54 km/hr.

When trains move in the same direction, subtract their speeds to get relative speed. Relative speed = 35 - 25 = 10 km/hr, which converts to 25/9 m/s. Total crossing distance is sum of train lengths: 80 + 120 = 200 m. Time = 200 ÷ (25/9) = 72 seconds. Option A (48) would be the answer if trains were moving in opposite directions.

The total distance covered is the sum of the train length and bridge length: 130 m + 245 m = 375 m. Speed = distance/time = 375/30 = 12.5 m/s. Converting to km/h: 12.5 × 3.6 = 45 km/h. The key insight is that the train must completely clear the bridge, so we add both lengths.

III gives train length = 200 m and time to cross pole = 10 s, so speed = 20 m/s = 72 km/h. II gives departure time (7:15 am) and distance (560 km). Time = distance/speed = 560/72 ≈ 7.78 hours ≈ 7 hours 47 minutes. Arrival ≈ 3:02 pm. Statement I about another train is unnecessary. A is correct.

From I: train length L = 13×speed. From II: L + 250 = 27×speed. Combining: 13s + 250 = 27s, so 14s = 250, speed = 250/14 m/s. Statements I and II together are sufficient. Statement III involves another train with unknown length/speed, so it doesn't help without additional information.

Train length = 250 m. Time to cross pole = 15 seconds (train covers its length). Speed = 250/15 m/s. Time to cross platform = 39 seconds (train covers platform length + train length). Distance = (250/15) × 39 = 650 m. Platform length = 650 - 250 = 400 m. Option A (450 m) is incorrect as it doesn't account for the train's length properly.

When trains move in opposite directions, their relative speed is the sum of individual speeds. Convert 77 + 67 = 144 km/hr to m/s by multiplying by 5/18, giving 40 m/s. Total distance to cross is sum of train lengths: 160 + 140 = 300 m. Time = distance/speed = 300/40 = 7.5 seconds.

Total distance to cover=sum of train lengths=108+112=220m. Relative speed when moving towards each other=45+54=99 km/hr=99×(5/18)=27.5 m/s. Time=220/27.5=8 seconds.

Total distance to cross = length of train + length of wall = 260 + 120 = 380 meters. Time taken = 19 seconds. Speed = distance/time = 380/19 = 20 m/s. Convert to km/hr: 20 × (3600/1000) = 20 × 3.6 = 72 km/hr. Option A (27) is incorrectly calculated. Option B (49) might result from using only train length. Option C (70) is close but incorrect. Remember to add both lengths when crossing a platform, bridge, or wall.

Total distance to cross = train length + bridge length = 110 + 132 = 242 meters. Speed = 72 km/hr = 72 × (1000/3600) = 20 m/s. Time = Distance/Speed = 242/20 = 12.1 seconds. Option A (9.8 sec) would require a higher speed, while options C and D don't match the calculation.

Total distance to cover = train length + platform length = 100 + 125 = 225 m. Speed = 54 km/hr = 54 × 1000/3600 = 15 m/s. Time = Distance/Speed = 225/15 = 15 seconds. Option B (12 seconds) would only account for 180m, not the full 225m needed.

The second train is stationary, so only the first train's speed matters. Total distance to cover = 110 + 170 = 280 m. Speed = 60 km/hr = 60 × 5/18 = 50/3 m/s. Time = 280/(50/3) = 280 × 3/50 = 16.8 seconds.