A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?
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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A train 150 m long is running at 72 km/hr. How long will it take to cross a pole ?
Answer: C
Correct answer: 7.5 seconds
Convert the speed to metres per second by multiplying by \(\tfrac{5}{18}\).
\[72 \times \frac{5}{18} = 20 \text{ m/s}\]
Crossing a pole means covering just the train’s own length, since a pole has no length of its own.
\[t = \frac{150}{20} = 7.5 \text{ seconds}\]
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A train 120 m long passes a telegraph post in 6 seconds. Find the speed of the train.
Answer: D
A telegraph post has no length, so the train covers only its own 120 m.
\[ v = \frac{120}{6} = 20 \text{ m/s} \]
\[ 20 \times \frac{18}{5} = 72 \text{ km/hr} \]
The factor \( \tfrac{18}{5} \) converts m/s to km/hr.
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Two goods train each 500 m long, are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one.
Answer: B
Trains moving in the same direction close at the difference of their speeds; in opposite directions, at the sum.
Passing the driver of the other train means covering only the slower train's own length.
\[ \text{relative speed} = 45 + 30 = 75 \text{ km/hr} = 75 \times \frac{5}{18} = \frac{125}{6} \text{ m/s} \]
\[ t = \frac{500 \times 6}{125} = 24 \text{ seconds} \]
Passing the whole train would need 1000 m and take 48 seconds - which is why that value is on the list.
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Two trains A and B are moving in the same direction. A has speed of 8 km/h and B has speed of 13 km/h. What is relative speed of B with respect to A?
Answer: B
Trains moving in the same direction close at the difference of their speeds; in opposite directions, at the sum.
\[ 13 - 8 = 5 \text{ km/hr} \]
Because they travel the same way, only the excess of B's speed over A's brings them together. In opposite directions the answer would be their sum, 21 km/hr - which is why that value is offered.
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A train moves fast a telegraph post and a bridge 264 m long in 8 sec and 20 sec respectively. What is the speed of the train?
Answer: D
The post crossing covers the train alone; the bridge crossing adds 264 m.
\[ L = 8v, \qquad L + 264 = 20v \]
\[ 12v = 264 \quad\Longrightarrow\quad v = 22 \text{ m/s} \]
\[ 22 \times \frac{18}{5} = 79.2 \text{ km/hr} \]
The train works out at 176 m long, though the question asks only for the speed.
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A 1200 m long train crosses a tree in 120 sec, how much time will I take to pass a platform 700 m long?
Answer: B
A tree is a point, so the first crossing gives the speed.
\[ v = \frac{1200}{120} = 10 \text{ m/s} \]
\[ t = \frac{1200 + 700}{10} = 190 \text{ seconds} \]
Timing only the platform's 700 m would give 70 seconds; the train's own length is part of the crossing.
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A train passes a station platform in 36 sec and a man standing on the platform in 20 sec. If the speed of the train is 54 km/hr. What is the length of the platform?
Answer: B
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
\[ 54 \text{ km/hr} = 15 \text{ m/s} \]
The man is a point, so that crossing gives the train's length:
\[ L = 15 \times 20 = 300 \text{ m} \]
The platform crossing covers train plus platform:
\[ 15 \times 36 = 540 \quad\Longrightarrow\quad P = 540 - 300 = 240 \text{ m} \]
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Two trains are moving in the same direction at 50 km/hr and 30 km/hr. The faster train crosses a man in the slower train in 18 seconds. Find the length of faster train.
Answer: B
Crossing a man in the other train covers only the faster train's length.
\[ \text{relative speed} = 50 - 30 = 20 \text{ km/hr} = 20 \times \frac{5}{18} = \frac{50}{9} \text{ m/s} \]
\[ L = \frac{50}{9} \times 18 = 100 \text{ m} \]
The slower train's length is irrelevant, because the reference point is a single person, not the whole train.
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A train travels a certain distance by taking 3 stops of 20 minutes each. Considering the period of stoppage, the overall speed of the train comes to 40 kmph; while without consideration of stoppage, it is 60 kmph. How much distance must the train have travelled ?
Answer: B
Three stops of 20 minutes cost exactly one hour, so the journey with stops takes an hour longer.
\[ \frac{D}{40} - \frac{D}{60} = 1 \]
\[ D\left(\frac{3 - 2}{120}\right) = 1 \quad\Longrightarrow\quad D = 120 \text{ km} \]
Checking: 3 hours with stops against 2 hours of pure running.
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