Problems on Trains Questions and Answers

Q: Two trains running in opposite direction cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speed is

The ratio of their speeds is 3:2.Let the speeds be \(x\) and \(y\). Passing a standing man, each train covers its own length, so the lengths are \(27x\) and \(17y\). Crossing each other in opposite directions, they cover the sum of their lengths at the sum of their speeds:\[ \begin{aligned} \frac{27x + 17y}{x + y} &= 23 \\ 27x + 17y &= 23x + 23y \\ 4x = 6y &\Rightarrow x : y = 3 : 2 \end{aligned} \]2:3 is the same ratio reversed; the working shows the train that takes 27 seconds to pass the man is the faster one, so it must also be much the longer. 1:3 and 3:4 do not satisfy the equation.

Q: A train 140m long running at 60kmph. In how much time will it pass a platform 260m long?

The train passes the platform in 24 sec.To pass the platform completely, the train covers its own length plus the platform's: \(140 + 260 = 400\) m. Convert the speed to metres per second:\[ \begin{aligned} 60\ \text{km/h} &= 60 \times \frac{5}{18} = \frac{50}{3}\ \text{m/s} \\ \text{time} &= 400 \div \frac{50}{3} = 24\ \text{s} \end{aligned} \]Using only the platform's 260 m would give 15.6 s, and only the train's 140 m gives 8.4 s; neither is offered, and 36, 44 and 58 s do not follow from the 400 m distance.

Q: A train running at the speed of 72 km/ hr passed a man in 14 sec and a platform in 32 sec respectively. Find the length of the platform.

The platform is 360 meter long.Convert the speed: \(72 \times \frac{5}{18} = 20\) m/s. Passing a man, the train covers its own length; passing a platform, its length plus the platform's.\[ \begin{aligned} \text{train} &= 20 \times 14 = 280\ \text{m} \\ \text{train + platform} &= 20 \times 32 = 640\ \text{m} \\ \text{platform} &= 640 - 280 = 360\ \text{m} \end{aligned} \]280 m is the train's own length, the usual trap.

Q: A train can cross a pole in 15 seconds and travelling at the same speed it can cross a bridge of 500 m in 45 seconds, then find the length of the train (in metre).

The train is 250 m long.Crossing a pole, the train covers its own length \(L\) in 15 s. Crossing the bridge, it covers \(L + 500\) in 45 s at the same speed.\[ \begin{aligned} \frac{L}{15} &= \frac{L + 500}{45} \\ 3L &= L + 500 \\ L &= 250\ \text{m} \end{aligned} \]Check: speed is \(250 / 15 = 16.67\) m/s, and \(750 / 16.67 = 45\) s for the bridge. The bridge takes three times as long, so it adds twice the train's length.

Q: A 200m long train crosses a platform of certain length in 15 seconds. Speed of the train is 90 km/hr. Then the length of the platform is

The platform is 175m long.To cross a platform, the train must cover its own length plus the platform's. Convert the speed to metres per second first:\[ \begin{aligned} 90\ \text{km/h} &= 90 \times \frac{5}{18} = 25\ \text{m/s} \\ \text{distance} &= 25 \times 15 = 375\ \text{m} \\ \text{platform} &= 375 - 200 = 175\ \text{m} \end{aligned} \]375 m is the total distance, which forgets to subtract the train's length; 225 m and 150 m do not follow from the data.

Q: If a train passes an electrical pole in 30 seconds and passes a platform of length 200m in 50 seconds, the length of the train is

The train is 300 m long.Passing a pole, the train covers its own length \(L\) in 30 s. Passing the platform, it covers \(L + 200\) in 50 s, at the same speed.\[ \begin{aligned} \frac{L}{30} &= \frac{L + 200}{50} \\ 50L &= 30L + 6000 \\ L &= 300\ \text{m} \end{aligned} \]Shortcut: the extra 20 seconds are spent covering the 200 m platform, so the speed is 10 m/s, and in 30 s the train covers 300 m. 320, 280 and 250 m do not give the same speed in both cases.

Q: Two trains are moving in the same direction at 50 km/h and 30 km/h. The faster train crosses a man in the slower train in 18 seconds. Find the length of the faster train.

The faster train is 100 m long.The trains move in the same direction, so the faster train passes the man at the difference of their speeds. In crossing him, it covers its own length.\[ \begin{aligned} v &= 50 - 30 = 20 \text{ km/h} \\ &= 20 \times \frac{5}{18} = \frac{50}{9} \text{ m/s} \\ L &= \frac{50}{9} \times 18 = 100 \text{ m} \end{aligned} \]Using the sum of the speeds (80 km/h), as for trains moving in opposite directions, would give 400 m, which is not offered. 120 m, 90 m and 150 m do not fit a relative speed of 20 km/h over 18 seconds.

Q: If a train 150 meter long crosses a pole in 12 seconds, then what is the speed of the train in km/h ?

The speed is 45 km/h.To cross a pole, the train covers its own length.\[ \begin{aligned} v &= \frac{150}{12} = 12.5\ \text{m/s} \\ &= 12.5 \times \frac{18}{5} = 45\ \text{km/h} \end{aligned} \]Multiplying by \(\frac{5}{18}\) instead of \(\frac{18}{5}\) gives about 3.5, and 50 or 60 come from rounding 12.5 m/s up before converting.

Q: A 300 meter long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform?

A signal pole is a point, so that crossing gives the speed.\[ v = \frac{300}{18} = \frac{50}{3} \text{ m/s} \]\[ \text{total distance over the platform} = \frac{50}{3} \times 39 = 650 \text{ m} \]\[ \text{platform} = 650 - 300 = 350 \text{ m} \]The data are complete, so "data inadequate" is a trap; 650 m is the combined distance, not the platform.

Q: A train having length 150 m passes a platform of 550 m length. The time taken for it is 56 seconds. In how much time will this train take to pass the platform of 250 m length?

Find the speed from the first crossing.\[ v = \frac{150 + 550}{56} = \frac{700}{56} = 12.5 \text{ m/s} \]\[ t = \frac{150 + 250}{12.5} = \frac{400}{12.5} = 32 \text{ seconds} \]The train's own length must be added to both platform lengths; leaving it out of the second would give 20 seconds.

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11.

Two trains running in opposite direction cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speed is

A.1:3
B.2:3
C.3:2
D.3:4

Exam Year(s)

Northern Coalfields Limited - 2020, NCL Technician - 2020,

Answer: C

The ratio of their speeds is 3:2.

Let the speeds be \(x\) and \(y\). Passing a standing man, each train covers its own length, so the lengths are \(27x\) and \(17y\). Crossing each other in opposite directions, they cover the sum of their lengths at the sum of their speeds:

\[ \begin{aligned} \frac{27x + 17y}{x + y} &= 23 \\ 27x + 17y &= 23x + 23y \\ 4x = 6y &\Rightarrow x : y = 3 : 2 \end{aligned} \]

2:3 is the same ratio reversed; the working shows the train that takes 27 seconds to pass the man is the faster one, so it must also be much the longer. 1:3 and 3:4 do not satisfy the equation.

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12.

A train 140m long running at 60kmph. In how much time will it pass a platform 260m long?

A.44 sec
B.24 sec
C.36 sec
D.58 sec

Exam Year(s)

Northern Coalfields Limited - 2020, NCL Technician - 2020,

Answer: B

The train passes the platform in 24 sec.

To pass the platform completely, the train covers its own length plus the platform's: \(140 + 260 = 400\) m. Convert the speed to metres per second:

\[ \begin{aligned} 60\ \text{km/h} &= 60 \times \frac{5}{18} = \frac{50}{3}\ \text{m/s} \\ \text{time} &= 400 \div \frac{50}{3} = 24\ \text{s} \end{aligned} \]

Using only the platform's 260 m would give 15.6 s, and only the train's 140 m gives 8.4 s; neither is offered, and 36, 44 and 58 s do not follow from the 400 m distance.

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13.

A train running at the speed of 72 km/ hr passed a man in 14 sec and a platform in 32 sec respectively. Find the length of the platform.

A.360 meter
B.280 meter
C.210 meter
D.240 meter

Exam Year(s)

Tripura Public Service Commission - 2025,

Answer: A

The platform is 360 meter long.

Convert the speed: \(72 \times \frac{5}{18} = 20\) m/s. Passing a man, the train covers its own length; passing a platform, its length plus the platform's.

\[ \begin{aligned} \text{train} &= 20 \times 14 = 280\ \text{m} \\ \text{train + platform} &= 20 \times 32 = 640\ \text{m} \\ \text{platform} &= 640 - 280 = 360\ \text{m} \end{aligned} \]

280 m is the train's own length, the usual trap.

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14.

A train can cross a pole in 15 seconds and travelling at the same speed it can cross a bridge of 500 m in 45 seconds, then find the length of the train (in metre).

A.250
B.300
C.200
D.240

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: A

The train is 250 m long.

Crossing a pole, the train covers its own length \(L\) in 15 s. Crossing the bridge, it covers \(L + 500\) in 45 s at the same speed.

\[ \begin{aligned} \frac{L}{15} &= \frac{L + 500}{45} \\ 3L &= L + 500 \\ L &= 250\ \text{m} \end{aligned} \]

Check: speed is \(250 / 15 = 16.67\) m/s, and \(750 / 16.67 = 45\) s for the bridge. The bridge takes three times as long, so it adds twice the train's length.

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15.

A 200m long train crosses a platform of certain length in 15 seconds. Speed of the train is 90 km/hr. Then the length of the platform is

A.300m
B.225m
C.150m
D.175m

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: D

The platform is 175m long.

To cross a platform, the train must cover its own length plus the platform's. Convert the speed to metres per second first:

\[ \begin{aligned} 90\ \text{km/h} &= 90 \times \frac{5}{18} = 25\ \text{m/s} \\ \text{distance} &= 25 \times 15 = 375\ \text{m} \\ \text{platform} &= 375 - 200 = 175\ \text{m} \end{aligned} \]

375 m is the total distance, which forgets to subtract the train's length; 225 m and 150 m do not follow from the data.

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16.

If a train passes an electrical pole in 30 seconds and passes a platform of length 200m in 50 seconds, the length of the train is

A.320m
B.300m
C.280m
D.250m

Exam Year(s)

Tripura Public Service Commission - 2019,

Answer: B

The train is 300 m long.

Passing a pole, the train covers its own length \(L\) in 30 s. Passing the platform, it covers \(L + 200\) in 50 s, at the same speed.

\[ \begin{aligned} \frac{L}{30} &= \frac{L + 200}{50} \\ 50L &= 30L + 6000 \\ L &= 300\ \text{m} \end{aligned} \]

Shortcut: the extra 20 seconds are spent covering the 200 m platform, so the speed is 10 m/s, and in 30 s the train covers 300 m. 320, 280 and 250 m do not give the same speed in both cases.

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17.

Two trains are moving in the same direction at 50 km/h and 30 km/h. The faster train crosses a man in the slower train in 18 seconds. Find the length of the faster train.

A.100m
B.120m
C.90m
D.150m

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: A

The faster train is 100 m long.

The trains move in the same direction, so the faster train passes the man at the difference of their speeds. In crossing him, it covers its own length.

\[ \begin{aligned} v &= 50 - 30 = 20 \text{ km/h} \\ &= 20 \times \frac{5}{18} = \frac{50}{9} \text{ m/s} \\ L &= \frac{50}{9} \times 18 = 100 \text{ m} \end{aligned} \]

Using the sum of the speeds (80 km/h), as for trains moving in opposite directions, would give 400 m, which is not offered. 120 m, 90 m and 150 m do not fit a relative speed of 20 km/h over 18 seconds.

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18.

If a train 150 meter long crosses a pole in 12 seconds, then what is the speed of the train in km/h ?

A.45
B.60
C.40
D.50

Exam Year(s)

UKPSC Lower PCS - 2025,

Answer: A

The speed is 45 km/h.

To cross a pole, the train covers its own length.

\[ \begin{aligned} v &= \frac{150}{12} = 12.5\ \text{m/s} \\ &= 12.5 \times \frac{18}{5} = 45\ \text{km/h} \end{aligned} \]

Multiplying by \(\frac{5}{18}\) instead of \(\frac{18}{5}\) gives about 3.5, and 50 or 60 come from rounding 12.5 m/s up before converting.

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19.

A 300 meter long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. What is the length of the platform?

A.150 m
B.200 m
C.350 m
D.400 m

Answer: C

A signal pole is a point, so that crossing gives the speed.

\[ v = \frac{300}{18} = \frac{50}{3} \text{ m/s} \]

\[ \text{total distance over the platform} = \frac{50}{3} \times 39 = 650 \text{ m} \]

\[ \text{platform} = 650 - 300 = 350 \text{ m} \]

The data are complete, so "data inadequate" is a trap; 650 m is the combined distance, not the platform.

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20.

A train having length 150 m passes a platform of 550 m length. The time taken for it is 56 seconds. In how much time will this train take to pass the platform of 250 m length?

A.32 seconds
B.24 seconds
C.28 seconds
D.30 seconds

Answer: A

Find the speed from the first crossing.

\[ v = \frac{150 + 550}{56} = \frac{700}{56} = 12.5 \text{ m/s} \]

\[ t = \frac{150 + 250}{12.5} = \frac{400}{12.5} = 32 \text{ seconds} \]

The train's own length must be added to both platform lengths; leaving it out of the second would give 20 seconds.

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