A train 240 m long crosses a bridge in 30 seconds while running at 72 km/hr. What is the length of the bridge ?
Answer: D
Correct answer: 360 m
\[\text{Speed} = 72 \times \frac{5}{18} = 20 \text{ m/s}\]
\[\text{Total distance covered} = 20 \times 30 = 600 \text{ m}\]
That total is the train plus the bridge, so the bridge is \(600 - 240 = 360\) m.
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How long does a train 110 m long running at the speed of 72 km/hr takes to cross a bridge 132 m length?
Answer: B
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
\[ 72 \text{ km/hr} = 20 \text{ m/s} \]
\[ t = \frac{110 + 132}{20} = \frac{242}{20} = 12.1 \text{ seconds} \]
Both the train and the bridge must be cleared, so their lengths add.
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A train running at the speed of 60 km/hr crosses a pole in 9 sec. What is the length of the train?
Answer: B
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
\[ 60 \text{ km/hr} = \frac{50}{3} \text{ m/s} \]
\[ L = \frac{50}{3} \times 9 = 150 \text{ m} \]
150 m is not among the listed values, so the answer is none of these. The data are complete, which rules out "cannot be determined".
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How long does a train 90 metres long running at the speed of 71 km/hr take to cross a bridge 132 metres in length?
Answer: B
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
\[ 71 \text{ km/hr} = 71 \times \frac{5}{18} = 19.72 \text{ m/s} \]
\[ t = \frac{90 + 132}{19.72} = \frac{222}{19.72} = 11.26 \text{ seconds} \]
Crossing a bridge covers the train plus the bridge; the train's own 90 m must not be dropped.
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Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. At what time will they meet?
Answer: B
Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x - 1) hours = 25(x - 1) km.
20x + 25(x - 1) = 110
" src="https://www.indiabix.com/_files/images/aptitude/1-sym-imp.gif" /> 45x = 135
" src="https://www.indiabix.com/_files/images/aptitude/1-sym-imp.gif" /> x = 3.
So, they meet at 10 a.m.
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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.
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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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?
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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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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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?
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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A train 240 m long passed a pole in 24 sec. 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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