A goods train runs at the speed of 72 km/hr and crosses a 250 m long platform in 26 sec. What is the length of the goods train?
Answer: D
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
\[ 72 \text{ kmph} = 20 \text{ m/s} \]
\[ \text{total distance} = 20 \times 26 = 520 \text{ m} \]
\[ \text{train} = 520 - 250 = 270 \text{ m} \]
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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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A 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec. What is the length of the platform?
Answer: B
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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A train speeds past a pole in 15 sec and a platform 100 m long in 25 sec, its length is?
Answer: B
The pole crossing covers the train alone, the platform crossing the train plus 100 m.
\[ L = 15v, \qquad L + 100 = 25v \]
Subtracting:
\[ 100 = 10v \quad\Longrightarrow\quad v = 10 \text{ m/s} \]
\[ L = 15 \times 10 = 150 \text{ m} \]
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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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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?
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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Two trains are running at 40 km/hr and 20 km/hr respectively in the same direction. Fast train completely passes a man sitting in the slower train in 5 sec. What is the length of the fast train?
Answer: D
Passing a man sitting in the slower train means covering only the fast train's own length, at the difference of the speeds.
\[ \text{relative speed} = 40 - 20 = 20 \text{ km/hr} = 20 \times \frac{5}{18} = \frac{50}{9} \text{ m/s} \]
\[ L = \frac{50}{9} \times 5 = \frac{250}{9} = 27\tfrac{7}{9} \text{ m} \]
The slower train's length never enters, because the reference point is one seated person.
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Two trains, each 100 m long, moving in opposite directions, cross other in 8 sec. If one is moving twice as fast the other, then the speed of the faster train is?
Answer: C
Let the speed of the slower train be x m/sec.
Then, speed of the train = 2x m/sec.
Relative speed = ( x + 2x) = 3x m/sec.
(100 + 100)/8 = 3x => x = 25/3.
So, speed of the faster train = 50/3 = 50/3 * 18/5 = 60 km/hr.
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