A train travelling at a speed of 60 km/hr crosses a platform in 20 seconds. The same train crosses a person who is walking at a speed of 6 km/hr in the same direction as that of the train in 12 seconds. What is the length of the train and that of the platform, respectively?
- (a)160 m and 153·33 m
- (b)170 m and 166·66 m
- (c)180 m and 153·33 m
- (d)180 m and 170 m
Correct — C, 180 m and 153·33 m. Crossing a person means covering the train's own length, and the person is walking the same way, so the closing speed is 60 − 6 = 54 km/hr, which is 54 × 5/18 = 15 m/s. In 12 seconds the train covers 15 × 12 = 180 m, so the train is 180 m long. Crossing a platform means covering the train's length plus the platform's, at the train's own speed of 60 km/hr = 50/3 m/s. In 20 seconds that is 1000/3 = 333.33 m. Subtracting the train gives 333.33 − 180 = 153.33 m of platform.
- (a)160 m and 153·33 m — The platform figure is right but the train is not. A 160 m train would need a closing speed of 13.33 m/s against the walker, that is 48 km/hr rather than 54.
- (b)170 m and 166·66 m — Neither figure survives. The two lengths must add to 333.33 m, and 170 + 166.66 is 336.66.
- (d)180 m and 170 m — The train length is right, but 180 + 170 = 350 m, which at 50/3 m/s would take 21 seconds rather than the 20 given.
Two rules cover every train question. First, the distance covered in crossing an object equals the train's length plus the object's length, so a person or a pole adds nothing while a platform or a bridge adds its own. Second, the speed to use is the relative speed: the difference for objects moving the same way, and the sum for objects approaching each other. A stationary platform leaves the train's own speed unchanged.
The item hands you the platform crossing first, but it cannot be used first — it has two unknowns in it. The person crossing has only one, so start there, and the walker's own speed must be subtracted because he is moving in the train's direction. Conversion is where marks leak: multiply km/hr by 5/18 for m/s, so 54 becomes a clean 15 while 60 becomes the recurring 16.67. Keeping 60 km/hr as 50/3 m/s and multiplying by 20 gives the exact 1000/3 rather than a rounded 333.
- Crossing a pole or a person covers only the train's length; crossing a platform covers train plus platform.
- Relative speed is the difference when both move the same way, the sum when they move towards each other.
- To convert km/hr to m/s, multiply by 5/18; 54 km/hr = 15 m/s and 60 km/hr = 50/3 m/s.
- Train length = 15 m/s × 12 s = 180 m.
- Train plus platform = 50/3 m/s × 20 s = 333.33 m, so the platform is 153.33 m.
The person carries one unknown and the platform two, which fixes the order in which the two events must be used.
- Using 60 km/hr against the walker and forgetting to subtract his 6 km/hr.
- Treating the person as having a length, or the platform as having none.
- Rounding 1000/3 to 333 and losing the 0.33 that the option list preserves.
A two-event train item where one event fixes the train's length and the other then yields the platform, so the order of working is itself the test.
A thief running at 8 km/hr is chased by a policeman whose speed is 10 km/hr. If the thief is 100 metres ahead of the policeman, then the time required for the policeman to catch the thief will be
- (a) 2 minutes
- (b) 6 minutes
- (c) 10 minutes
- (d) 3 minutes
Answer(d) 3 minutes
Same-direction relative speed in its purest form. The gap of 100 m closes at the difference of the two speeds, exactly as the train closes its own length on a walker moving the same way.
- practice — not a real PYQ
A 200 m long train crosses a pole in 10 seconds. Its speed is
- (a)36 km/hr
- (b)54 km/hr
- (c)72 km/hr
- (d)90 km/hr
Answer(c) 72 km/hr — 200 ÷ 10 = 20 m/s, and 20 × 18/5 = 72 km/hr.
- practice — not a real PYQ
A train 150 m long travelling at 45 km/hr crosses a platform 200 m long in
- (a)24 seconds
- (b)28 seconds
- (c)30 seconds
- (d)35 seconds
Answer(b) 28 seconds — the distance is 350 m at 12.5 m/s, which is 28 seconds.