Problems on Trains Questions and Answers

Q: A train takes 20 seconds to pass completely through a station 160 m long and 15 seconds through another station 110 m long. Find the length of the train.

Each crossing covers the train's length plus the station's.\[ L + 160 = 20v, \qquad L + 110 = 15v \]Subtracting:\[ 50 = 5v \quad\Longrightarrow\quad v = 10 \text{ m/s} \]\[ L = 20 \times 10 - 160 = 40 \text{ m} \]Subtracting the two equations eliminates the unknown length in one step.

Q: How many seconds will a 500 m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr?

The man walks the same way, so the speeds subtract.\[ \text{relative speed} = 63 - 3 = 60 \text{ km/hr} = 60 \times \frac{5}{18} = \frac{50}{3} \text{ m/s} \]\[ t = \frac{500 \times 3}{50} = 30 \text{ seconds} \]Only the train's own length counts, since the man occupies a single point.

Q: A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?

\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]\[ 45 \text{ km/hr} = 12.5 \text{ m/s} \]\[ t = \frac{360 + 140}{12.5} = \frac{500}{12.5} = 40 \text{ seconds} \]

Q: A train 110 meters long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?

The man walks toward the train, so the speeds add.\[ \text{relative speed} = 60 + 6 = 66 \text{ kmph} = 66 \times \frac{5}{18} = \frac{55}{3} \text{ m/s} \]A man is a point, so only the train's own length is covered:\[ t = \frac{110 \times 3}{55} = 6 \text{ seconds} \]Had he walked the same way as the train, the relative speed would be 54 kmph and the answer about 7.3 seconds.

Q: A train 280 m long, running with a speed of 63 km/hr will pass a tree in?

\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]\[ 63 \text{ km/hr} = 63 \times \frac{5}{18} = 17.5 \text{ m/s} \]A tree is a point, so the train covers only its own length:\[ t = \frac{280}{17.5} = 16 \text{ seconds} \]The conversion is where the answer is decided: at 17.5 m/s the 280 m takes exactly 16 seconds, not 15.

Q: A train passes a stationary pole in 8 seconds. The train also passes a 200 m long bridge in 28 seconds. What is the length and the speed of the train?

The pole crossing covers only the train:\[ L = 8v \]The bridge crossing covers the train plus 200 m:\[ L + 200 = 28v \quad\Longrightarrow\quad 8v + 200 = 28v \]\[ 20v = 200 \quad\Longrightarrow\quad v = 10 \text{ m/s} \]\[ L = 80 \text{ m} \]Only one option gives a pair consistent with both crossings.

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31.

A train takes 20 seconds to pass completely through a station 160 m long and 15 seconds through another station 110 m long. Find the length of the train.

A.100 m
B.70 m
C.60 m
D.40 m

Answer: D

Each crossing covers the train's length plus the station's.

\[ L + 160 = 20v, \qquad L + 110 = 15v \]

Subtracting:

\[ 50 = 5v \quad\Longrightarrow\quad v = 10 \text{ m/s} \]

\[ L = 20 \times 10 - 160 = 40 \text{ m} \]

Subtracting the two equations eliminates the unknown length in one step.

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32.

How many seconds will a 500 meter long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr?

A.25 sec
B.28 sec
C.30 sec
D.35 sec

Answer: C

Let length of tunnel is x meter

Distance = 800+x meter

Time = 1 minute = 60 seconds

Speed = 78 km/hr = 78*5/18 m/s = 65/3 m/s

Distance = Speed*Time

800+x = (65/3) * 60

800+x = 20 * 65 = 1300

x = 1300 - 800 = 500 meters

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33.

How many seconds will a 500 m long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr?

A.25
B.30
C.40
D.45

Answer: B

The man walks the same way, so the speeds subtract.

\[ \text{relative speed} = 63 - 3 = 60 \text{ km/hr} = 60 \times \frac{5}{18} = \frac{50}{3} \text{ m/s} \]

\[ t = \frac{500 \times 3}{50} = 30 \text{ seconds} \]

Only the train's own length counts, since the man occupies a single point.

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34.

A train is 31m long. It passes a platform which is 175 m long. What is the distance covered by the train in passing the platform?

A.206 m
B.144 m
C.175/31 m
D.Cannot be determined

Answer: A

Passing a platform means the train travels its own length plus the platform's, from the moment its front enters to the moment its rear leaves.

\[ 31 + 175 = 206 \text{ m} \]

No speed or time is needed, so "cannot be determined" is a trap - the distance follows from the two lengths alone.

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35.

A train 240 m long passed a pole in 24 sec. How long will it take to pass a platform 650 m long?

A.65 sec
B.89 sec
C.100 sec
D.150 sec

Answer: B

A pole has no length, so the first crossing gives the speed.

\[ v = \frac{240}{24} = 10 \text{ m/s} \]

\[ t = \frac{240 + 650}{10} = \frac{890}{10} = 89 \text{ seconds} \]

Using only the platform's 650 m would give 65 seconds - the first distractor.

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36.

A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?

A.40 sec
B.42 sec
C.45 sec
D.48 sec

Answer: A

\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]

\[ 45 \text{ km/hr} = 12.5 \text{ m/s} \]

\[ t = \frac{360 + 140}{12.5} = \frac{500}{12.5} = 40 \text{ seconds} \]

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37.

A train 110 meters long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?

Answer: B

The man walks toward the train, so the speeds add.

\[ \text{relative speed} = 60 + 6 = 66 \text{ kmph} = 66 \times \frac{5}{18} = \frac{55}{3} \text{ m/s} \]

A man is a point, so only the train's own length is covered:

\[ t = \frac{110 \times 3}{55} = 6 \text{ seconds} \]

Had he walked the same way as the train, the relative speed would be 54 kmph and the answer about 7.3 seconds.

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38.

Two, trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:

A.2 : 3
B.4 : 3
C.6 : 7
D.9 : 16

Answer: B

Two trains that meet and then take \( t_{1} \) and \( t_{2} \) to finish have speeds in the ratio

\[ \frac{v_{1}}{v_{2}} = \sqrt{\frac{t_{2}}{t_{1}}} \]

\[ = \sqrt{\frac{16}{9}} = \frac{4}{3} \quad\Longrightarrow\quad 4 : 3 \]

The square root is essential; taking the times directly would give 9 : 16, which is offered as the last option and is also the wrong way round.

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39.

A train 280 m long, running with a speed of 63 km/hr will pass a tree in?

A.15 sec
B.16 sec
C.18 sec
D.20 sec

Answer: B

\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]

\[ 63 \text{ km/hr} = 63 \times \frac{5}{18} = 17.5 \text{ m/s} \]

A tree is a point, so the train covers only its own length:

\[ t = \frac{280}{17.5} = 16 \text{ seconds} \]

The conversion is where the answer is decided: at 17.5 m/s the 280 m takes exactly 16 seconds, not 15.

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40.

A train passes a stationary pole in 8 seconds. The train also passes a 200 m long bridge in 28 seconds. What is the length and the speed of the train?

A.Length = 200m; Speed = 36 m/s
B.Length = 160m; Speed = 20 m/s
C.Length = 80m; Speed = 10 m/s
D.Length = 100m; Speed = 20 m/s

Answer: C

The pole crossing covers only the train:

\[ L = 8v \]

The bridge crossing covers the train plus 200 m:

\[ L + 200 = 28v \quad\Longrightarrow\quad 8v + 200 = 28v \]

\[ 20v = 200 \quad\Longrightarrow\quad v = 10 \text{ m/s} \]

\[ L = 80 \text{ m} \]

Only one option gives a pair consistent with both crossings.

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