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 300 metre long train crosses a platform in 39 seconds while it crosses a signal pole in 18 seconds. 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 travelling at 48 kmph completely crosses another train having half its length and travelling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. The length of the platform is
Answer: A
Trains moving in the same direction close at the difference of their speeds; in opposite directions, at the sum.
Let the first train be \( L \) long; the second is \( \tfrac{L}{2} \).
\[ \text{relative speed} = 48 + 42 = 90 \text{ kmph} = 25 \text{ m/s} \]
\[ L + \frac{L}{2} = 25 \times 12 = 300 \quad\Longrightarrow\quad 1.5L = 300 \]
\[ L = 200 \text{ m} \]
Crossing the platform, the train travels alone at 48 kmph \( = \tfrac{40}{3} \) m/s:
\[ 200 + P = \frac{40}{3} \times 45 = 600 \quad\Longrightarrow\quad P = 400 \text{ m} \]
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A train overtakes two boys who are running at the rate of 8 km/hr and 16 km/hr in the same direction as the train. The train completely passes them in 36 seconds and 40 seconds respectively. What is the length of the train?
Answer: C
Let the train's speed be \( v \) km/hr. Passing each boy covers the train's own length at the relative speed.
\[ L = (v - 8) \times \frac{5}{18} \times 36 = (v - 16) \times \frac{5}{18} \times 40 \]
The factor \( \tfrac{5}{18} \) is common:
\[ 36(v - 8) = 40(v - 16) \quad\Longrightarrow\quad 36v - 288 = 40v - 640 \]
\[ 4v = 352 \quad\Longrightarrow\quad v = 88 \text{ km/hr} \]
\[ L = (88 - 8) \times \frac{5}{18} \times 36 = 80 \times 10 = 800 \text{ m} \]
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A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. Its length is:
Answer: D
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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Two trains of equal length run on parallel tracks in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower one in 36 seconds. What is the length of each train ?
Answer: B
Correct answer: 50 m
The relative speed is the difference, since they travel the same way.
\[46 - 36 = 10 \text{ km/hr} = 10 \times \frac{5}{18} = \frac{25}{9} \text{ m/s}\]
\[\text{Distance covered} = \frac{25}{9} \times 36 = 100 \text{ m}\]
That 100 m is the combined length of both trains, so each is 50 m long.
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A train 800 m long is running at a speed of 78 km/hr. If it crosses a tunnel in 1 min, then the length of the tunnel is?
Answer: C
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
\[ 78 \text{ km/hr} = 78 \times \frac{5}{18} = \frac{65}{3} \text{ m/s} \]
\[ \text{total distance in 60 s} = \frac{65}{3} \times 60 = 1300 \text{ m} \]
\[ \text{tunnel} = 1300 - 800 = 500 \text{ metres} \]
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A train overtakes two persons walking along a railway track. The first one walks at 4.5 km/hr. The other one walks at 5.4 km/hr. The train needs 8.4 and 8.5 seconds respectively to overtake them. What is the speed of the train if both the persons are walking in the same direction as the train?
Answer: D
Both walkers move the same way as the train, so the relative speeds are differences.
\[ L = (v - 4.5) \times \frac{5}{18} \times 8.4 = (v - 5.4) \times \frac{5}{18} \times 8.5 \]
\[ 8.4(v - 4.5) = 8.5(v - 5.4) \]
\[ 8.4v - 37.8 = 8.5v - 45.9 \quad\Longrightarrow\quad 0.1v = 8.1 \]
\[ v = 81 \text{ km/hr} \]
The train's length works out at 75.6 m, though the question asks only for the speed.
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A train speeds past a pole in 15 seconds and a platform 100 m long in 25 seconds. 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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In what time will a train 100 m long cross an electric pole, it its speed be 144 km/hr?
Answer: A
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
\[ 144 \text{ km/hr} = 144 \times \frac{5}{18} = 40 \text{ m/s} \]
An electric pole has no length, so only the train's own 100 m is covered:
\[ t = \frac{100}{40} = 2.5 \text{ seconds} \]
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