Problems on Trains

1.A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?

1. 120 metres
2. 180 metres
3. 324 metres
4. 150 metres

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Speed = (60 \times \frac{5}{18}) m/sec = (\frac{50}{3}) m/sec \\ Length of the train = (Speed \times Time) \\ ∴ Length of the train \\ = (\frac{50}{3} \times 9) m = 150 m \end{aligned}

$\eqalign{ & {\text{Speed}} = \left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} = \left( {\frac{{50}}{3}} \right){\text{m/sec}} \cr & {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} = \left( {{\text{Speed}} \times {\text{Time}}} \right) \cr & \therefore {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} \cr & = \left( {\frac{{50}}{3} \times 9} \right)m = 150m \cr}$

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2.A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:

1. 45 km/hr
2. 50 km/hr
3. 54 km/hr
4. 55 km/hr

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Speed of the train relative to man \\ = (\frac{125}{10}) m/sec \\ = (\frac{25}{2}) m/sec \\ = (\frac{25}{2} \times \frac{18}{5}) km/hr \\ = 45 km/hr \\ Let the speed of the train be x km/hr . \\ T h e n, r e l a t i v e s p e e d = (x - 5) km/hr \\ ∴ x - 5 = 45 \Rightarrow x = 50 km/hr \end{aligned}

$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{man}} \cr & = \left( {\frac{{125}}{{10}}} \right){\text{m/sec}} \cr & = \left( {\frac{{25}}{2}} \right){\text{m/sec}} \cr & = \left( {\frac{{25}}{2} \times \frac{{18}}{5}} \right){\text{km/hr}} \cr & = 45\,{\text{km/hr}} \cr & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{km/hr}}. \cr & Then,\,relative\,speed = \left( {x - 5} \right)\,{\text{km/hr}} \cr & \therefore x - 5 = 45 \Rightarrow x = 50\,{\text{km/hr}} \cr}$

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3.The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:

1. 200 m
2. 225 m
3. 245 m
4. 250 m

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Speed = (45 \times \frac{5}{18}) m/sec \\ = (\frac{25}{2}) m/sec \\ Time = 30 sec \\ Let the length of bridge be x metres \\ Then, \frac{130 + x}{30} = \frac{25}{2} \\ \Rightarrow 2 (130 + x) = 750 \\ \Rightarrow x = 245 m \end{aligned}

$\eqalign{ & {\text{Speed}} = \left( {45 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{25}}{2}} \right)\,{\text{m/sec}} \cr & {\text{Time}} = 30\,{\text{sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{bridge}}\,{\text{be}}\,x\,{\text{metres}} \cr & {\text{Then}},\,\frac{{130 + x}}{{30}} = \frac{{25}}{2} \cr & \Rightarrow 2\left( {130 + x} \right) = 750 \cr & \Rightarrow x = 245\,m \cr}$

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4.A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?

1. 120 m
2. 240 m
3. 300 m
4. None of these

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Speed = (54 \times \frac{5}{18}) m/sec = 15 m/sec \\ Length of the train = (15 \times 20) m = 300 m \\ Let the length of the platform be x metres \\ Then, \frac{x + 300}{36} = 15 \\ \Rightarrow x + 300 = 540 \\ \Rightarrow x = 240 m \end{aligned}

$\eqalign{ & {\text{Speed}} = \left( {54 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} = 15\,{\text{m/sec}} \cr & {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} = \left( {15 \times 20} \right){\text{m}} = 300\,{\text{m}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{platform}}\,{\text{be}}\,x\,{\text{metres}} \cr & {\text{Then}},\,\frac{{x + 300}}{{36}} = 15 \cr & \Rightarrow x + 300 = 540 \cr & \Rightarrow x = 240\,{\text{m}} \cr}$

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5.A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?

1. 65 sec
2. 89 sec
3. 100 sec
4. 150 sec

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Speed = (\frac{240}{24}) m/sec = 10 m/sec \\ ∴ Required time \\ = (\frac{240 + 650}{10}) sec . \\ = 89 s e c . \end{aligned}

$\eqalign{ & {\text{Speed}} = \left( {\frac{{240}}{{24}}} \right)\,{\text{m/sec}} = 10\,{\text{m/sec}} \cr & \therefore {\text{Required}}\,{\text{time}} \cr & {\text{ = }}\,\left( {\frac{{240 + 650}}{{10}}} \right)\,{\text{sec}}. \cr & = 89\,sec. \cr}$

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6.Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:

1. 50 m
2. 72 m
3. 80 m
4. 82 m

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Let the length of each train be x metres . \\ Then, distance covered = 2 x metres . \\ Relative speed \\ = (46 - 36) km/hr \\ = (10 \times \frac{5}{18}) m/sec \\ = (\frac{25}{9}) m/sec \\ ∴ \frac{2 x}{36} = \frac{25}{9} \\ \Rightarrow 2 x = 100 \\ \Rightarrow x = 50 \end{aligned}

$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{each}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}. \cr & {\text{Then,}}\,{\text{distance}}\,{\text{covered}} = 2x\,{\text{metres}}. \cr & {\text{Relative}}\,{\text{speed}} \cr & = \left( {46 - 36} \right)\,{\text{km/hr}} \cr & = \left( {10 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & = \left( {\frac{{25}}{9}} \right)\,{\text{m/sec}} \cr & \therefore \frac{{2x}}{{36}} = \frac{{25}}{9} \cr & \Rightarrow 2x = 100 \cr & \Rightarrow x = 50 \cr}$

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7.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. 40 sec
2. 42 sec
3. 45 sec
4. 48 sec

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Formula for converting from km/hr to m/s: \\ X km/hr = (X \times \frac{5}{18}) m/s \\ Therefore, Speed \\ = (45 \times \frac{5}{18}) m/sec = \frac{25}{2} m/sec \\ Total distance to be covered \\ = (360 + 140) m = 500 m \\ Formula for finding Time \\ = (\frac{Distance}{Speed}) \\ ∴ Required time \\ = (\frac{500 \times 2}{25}) \sec \\ = 40 \sec . \end{aligned}

$\eqalign{ & {\text{Formula}}\,{\text{for}}\,{\text{converting}}\,{\text{from}}\,{\text{km/hr}}\,{\text{to}}\,{\text{m/s:}} \cr & X\,{\text{km/hr}} = \left( {X \times \frac{5}{{18}}} \right)\,{\text{m/s}} \cr & {\text{Therefore,}}\,{\text{Speed}} \cr & = \left( {45 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} = \frac{{25}}{2}{\text{m/sec}} \cr & {\text{Total}}\,{\text{distance}}\,{\text{to}}\,{\text{be}}\,{\text{covered}} \cr & = \left( {360 + 140} \right)m = 500\,m \cr & {\text{Formula}}\,{\text{for}}\,{\text{finding}}\,{\text{Time}} \cr & = \left( {\frac{{{\text{Distance}}}}{{{\text{Speed}}}}} \right) \cr & \therefore {\text{Required}}\,{\text{time}} \cr & = \left( {\frac{{500 \times 2}}{{25}}} \right)\,\sec \cr & = 40\,\sec . \cr}$

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8.A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?

1. 3.6 sec
2. 18 sec
3. 36 sec
4. 72 sec

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Speed of train relative to jogger \\ = (45 - 9) km/hr \\ = 36 km/hr \\ (36 \times \frac{5}{18}) m/sec \\ = 10 m/sec \\ Distance to be covered \\ = (240 + 120) m \\ = 360 m \\ ∴ Time taken \\ = (\frac{360}{10}) sec \\ = 36 sec \end{aligned}

$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{jogger}} \cr & = \left( {45 - 9} \right)\,{\text{km/hr}} \cr & = 36\,{\text{km/hr}} \cr & \left( {36 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & = 10\,{\text{m/sec}} \cr & {\text{Distance}}\,{\text{to}}\,{\text{be}}\,{\text{covered}} \cr & = \left( {240 + 120} \right)\,m \cr & = 360\,m \cr & \therefore {\text{Time}}\,{\text{taken}} \cr & = \left( {\frac{{360}}{{10}}} \right)\,{\text{sec}} \cr & = 36\,{\text{sec}} \cr}$

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9.A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?

1. 230 m
2. 240 m
3. 260 m
4. 320 m

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Relative speed \\ = (120 + 80) km/hr \\ = (200 \times \frac{5}{18}) m/sec \\ = (\frac{500}{9}) m/sec \\ Let the length of the other train be x metres . \\ Then, \frac{x + 270}{9} = \frac{500}{9} \\ \Rightarrow x + 270 = 500 \\ \Rightarrow x = 230 \end{aligned}

$\eqalign{ & {\text{Relative}}\,{\text{speed}} \cr & = \left( {120 + 80} \right)\,{\text{km/hr}} \cr & = \left( {200 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & = \left( {\frac{{500}}{9}} \right)\,{\text{m/sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{other}}\,{\text{train}}\,{\text{be}}\,{\text{x}}\,{\text{metres}}{\text{.}} \cr & {\text{Then,}}\,\frac{{x + 270}}{9} = \frac{{500}}{9} \cr & \Rightarrow x + 270 = 500 \cr & \Rightarrow x = 230 \cr}$

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10.A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?

1. 230 m
2. 240 m
3. 260 m
4. 270 m

Answer:Option 1

Explanation:

Solution:

\begin{aligned} Speed = (72 \times \frac{5}{18}) m/sec \\ = 20 m/sec \\ Time = 26 sec \\ Let the length of the train be x metres . \\ Then, \frac{x + 250}{26} = 20 \\ \Rightarrow x + 250 = 520 \\ \Rightarrow x = 270 \end{aligned}

$\eqalign{ & {\text{Speed}} = \left( {72 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20\,{\text{m/sec}} \cr & {\text{Time}} = 26\,{\text{sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}{\text{.}} \cr & {\text{Then}},\,\frac{{x + 250}}{{26}} = 20 \cr & \Rightarrow x + 250 = 520 \cr & \Rightarrow x = 270 \cr}$

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