Practice MCQ Questions and Answer on Problems on Trains

Train A traveling at 63 kmph can cross a platform 199.5 m long in 21 seconds. How much would train A take to completely cross (from the moment they meet ) train B, 157 m long and traveling at 54 kmph in opposite direction which train A is traveling? (in seconds)

(A) 16
(B) 18
(C) 12
(D) 10

Solution: $\begin{aligned} Speed of train A \\ = 63 kmph \\ = (\frac{63 \times 5}{18}) m/sec \\ = 17 .5 m/sec \\ Speed of train B \\ = 54 kmph \\ = (\frac{54 \times 5}{18}) m/sec = 15 m/sec \\ If the length of train A be x metre, \\ then \\ Speed of train A \\ = \frac{Length of train + length of platform}{Time taken in crossing} \\ \Rightarrow 17.5 = \frac{x + 199.5}{21} \\ \Rightarrow 17.5 \times 21 = x + 199.5 \\ \Rightarrow 367.5 = x + 199.5 \\ \Rightarrow x = 367.5 - 199.5 \\ \Rightarrow 168 metres \\ Relative speed \\ = ( Speed train A + Speed train B) \\ = (17 .5 + 15) m/sec \\ = 32 .5 m/sec \\ Required time \\ = \frac{Length of train A + Length of train B}{Relative speed} \\ = (\frac{168 + 157}{32.5}) seconds \\ = 10 seconds \end{aligned}$

A train 132 m long passes a telegraph pole in 6 seconds. Find the speed of the train?

(A) 70 km/hr
(B) 72 km/hr
(C) 79.2 km/hr
(D) 80 km/hr

Solution: $\begin{aligned} Speed \\ = (\frac{132}{6}) m/sec \\ = (22 \times \frac{18}{5}) km/sec \\ = 79 .2 km/hr \end{aligned}$

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) 45 km/hr
(B) 50 km/hr
(C) 54 km/hr
(D) 55 km/hr

Solution: $\begin{aligned} Speed of the train relative to man \\ = \frac{125}{10} m/sec \\ = \frac{25}{2} m/sec \\ = \frac{25}{2} \times \frac{18}{5} km/hr \\ = 45 km/hr \\ Let the speed of the train be x km/hr . \\ Then, relative speed = (x - 5) km/hr \\ ∴ x - 5 = 45 \\ \Rightarrow x = 50 km/hr \end{aligned}$

Two trains of equal lengths take 10 seconds and 15 seconds respectively to cross a telegraph post. If the length of each train be 120 metres, in what time (in seconds) will they cross each other travelling in opposite direction?

(A) 10
(B) 12
(C) 15
(D) 20

Solution: $\begin{aligned} Speed of the first train \\ = \frac{120}{10} m/sec \\ = 12 m/sec \\ Speed of the second train \\ \frac{120}{15} m/sec \\ = 8 m/sec \\ Relative speed = 12 + 8 = 20 m/sec \\ ∴ Required time \\ = \frac{120 + 120}{20} sec \\ = 12 sec \end{aligned}$

A train 108 m long moving at a speed of 50 km/hr crosses a train 112 m long coming from opposite direction in 6 seconds. The speed of the second train is:

(A) 48 km/hr
(B) 54 km/hr
(C) 66 km/hr
(D) 82 km/hr

Solution: $\begin{aligned} Let the speed of the second train be x km/hr . \\ Relative speed \\ = (x + 50) km/hr \\ = [(x + 50) \times \frac{5}{18}] m/sec \\ = \frac{250 + 5 x}{18} m/sec \\ Distance covered \\ = (108 + 112) = 220 m \\ ∴ \frac{220}{\frac{250 + 5 x}{18}} = 6 \\ \Rightarrow 250 + 5 x = 660 \\ \Rightarrow x = 82 km/hr \end{aligned}$

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?

(A) 66 km/hr
(B) 72 km/hr
(C) 78 km/hr
(D) 81 km/hr

Solution: $\begin{aligned} 4.5 km/hr = (4.5 \times \frac{5}{18}) m/sec \\ = \frac{5}{4} m/sec \\ = 1.25 m/sec, and \\ 5.4 k m / h r = (5.4 \times \frac{5}{18}) m/sec \\ = \frac{3}{2} m/sec \\ = 1.5 m/sec \\ Let the speed of the train be x m/sec \\ Then, (x - 1.25) \times 8.4 = (x - 1.5) \times 8.5 \\ \Rightarrow 8.4 x - 10.5 = 8.5 x - 12.75 \\ \Rightarrow 0.1 x = 2.25 \\ \Rightarrow x = 22.5 \\ ∴ Speed of the train \\ = (22.5 \times \frac{18}{5}) km/hr \\ = 81 km/hr \end{aligned}$

The time taken by a train 180 m long, travelling at 42 kmph, in passing a person walking in the same direction at 6 kmph, will be

(A) 18 sec
(B) 21 sec
(C) 24 sec
(D) 25 sec

Solution: Speed of train relative to man = (42 - 6) kmph = 36 kmph = $(36 \times \frac{5}{18})$ m/sec = 10 m/sec ∴ Time taken to pass the man = $\frac{180}{10}$ sec = 18 sec

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?

(A) 230 m
(B) 240 m
(C) 260 m
(D) 270 m

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

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 seconds. What is the length of the fast train?

(A) 23 m
(B) $23 \frac{2}{9}$ m
(C) $27 \frac{7}{9}$ m
(D) 29 m

Solution: $\begin{aligned} Relative speed = (40 - 20) km/hr \\ = (20 \times \frac{5}{18}) m/sec \\ = \frac{50}{9} m/sec \\ ∴ Length of faster train \\ = (\frac{50}{9} \times 5) m \\ = \frac{250}{9} m \\ = 27 \frac{7}{9} m \end{aligned}$

10.

Two trains start simultaneously (with uniform speeds) from two stations 270 km apart, each to the opposite station; they reach their destinations in $6 \frac{1}{4}$ hours and 4 hours after they meet. The rate at which the slower train travels is :

(A) 16 km/hr
(B) 24 km/hr
(C) 25 km/hr
(D) 30 km/hr

Solution: $\begin{aligned} Ratio of speeds \\ = \sqrt{4} : \sqrt{6 \frac{1}{4}} \\ = \sqrt{4} : \sqrt{\frac{25}{4}} \\ = 2 : \frac{5}{2} \\ = 4 : 5 \end{aligned}$ Let the speeds of the two trains be 4x and 5x km/hr respectively Then time taken by trains to meet each other $\begin{aligned} = (\frac{270}{4 x + 5 x}) hr \\ = (\frac{270}{9 x}) hr = (\frac{30}{x}) hr \\ Time taken by slower train to travel \\ 270 km = (\frac{270}{4 x}) hr \\ ∴ \frac{270}{4 x} = \frac{30}{x} + 6 \frac{1}{4} \\ \Rightarrow \frac{270}{4 x} - \frac{30}{x} = \frac{25}{4} \\ \Rightarrow \frac{150}{4 x} = \frac{25}{4} \\ \Rightarrow 100 x = 600 \\ \Rightarrow x = 6 \\ Hence speed of slower train \\ = 4 x \\ = 24 km/hr \end{aligned}$

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Practice MCQ Questions and Answer on Problems on Trains

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