Home Mathematics Mcqs Problems on Train

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

A. 100 m
B. 150 m
C. 200 m
D. 250 m

Solution:

\begin{array} { l } { \text { Speed } = ( 60 \times \frac { 5 } { 18 } ) m / sec = ( \frac { 50 } { 3 } ) m / sec } \\ { \text { Length of the train } = ( \text { Speed } x \text { Time) } } \\ { \therefore \text { Length of the train } = ( \frac { 50 } { 3 } \times 9 ) m}\\{ = 150 m } \end{array}

# 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:

A. 50 km/hr
B. 60 km/hr
C. 40 km/hr
D. 35 km/hr

Solution:

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

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

A. 245 m
B. 240 m
C. 248 m
D. 242 m

Solution:

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

# Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:

A. 3:9
B. 3:5
C. 2:1
D. 3:2

Solution:

$\left. \begin{array} { l } { \text { Let the speeds of the two trains be } xand}\\ { \text { m/sec } y \text { m/sec respectively. } } \\ { \text { Then, length of the first train } = 27 x \text { metres, } } \\ { \text { and length of the second train } = 17 y \text { metres } } \\ { \therefore \frac { 27 x + 17 y } { x + y } = 23 } \\ { \Rightarrow 27 x + 17 y = 23 x + 23 y } \\ { \Rightarrow 4 x = 6 y } \\ { \Rightarrow \frac { x } { y } = \frac { 3 } { 2 } } \end{array} \right.$

# 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?

A. 254 m
B. 250 m
C. 240 m
D. 280 m

Solution:

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

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

A. 89 sec
B. 80 sec
C. 81 sec
D. 87 sec

Solution:

$\left. \begin{array} { l } { \text { Speed } = ( \frac { 240 } { 24 } ) m / sec = 10 m / sec } \\ { \therefore \text { Required time } = ( \frac { 240 + 650 } { 10 } ) sec }\\{= 89 sec } \end{array} \right.$

# 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:

A. 60 m
B. 40 m
C. 50 m
D. 30 m

Solution:

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

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