Practice MCQ Questions and Answer on Problems on Trains

31.

Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is:

(A) 30 km/hr
(B) 45 km/hr
(C) 60 km/hr
(D) 75 km/hr

Solution: $\begin{aligned} Let the speed of the slower train be x m/sec \\ Then, speed of the faster train = 2 x m/sec \\ Relative speed = (x + 2 x) m/sec = 3 x m/sec \\ ∴ \frac{100 + 100}{8} = 3 x \\ \Rightarrow 24 x = 200 \\ \Rightarrow x = \frac{25}{3} \\ So, speed of the faster train = \frac{50}{3} m/sec \\ = \frac{50}{3} \times \frac{18}{5} km/hr \\ = 60 km/hr \end{aligned}$

32.

A train moves with a speed of 30 kmph for 12 minutes and for next 8 minutes at a speed of 45 kmph. Find the average speed of the train?

(A) 37.50 kmph
(B) 36 kmph
(C) 48 kmph
(D) 30 kmph

Solution: $\begin{aligned} Distance = Speed \times Time \\ Distance covered by train with the \\ speed of 30 kmph in 12 minutes is \\ = 30 \times \frac{12}{60} = 6 km \\ Distance covered by the same train \\ with the speed of 45 kmph in 8 minutes is \\ = 45 \times \frac{8}{60} = 6 km \\ Average speed \\ = \frac{total distance}{total time} . \\ \Rightarrow \frac{(6 + 6) km}{(12 + 8) min} = \frac{12}{20} \times 60 \\ = 36 kmph \end{aligned}$

33.

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

34.

A train takes 5 minutes to cross a telegraphic post. Then the time taken by another train whose length is just double of the first train and moving with same speed to cross a platform of its own length is :

(A) 10 minutes
(B) 15 minutes
(C) 20 minutes
(D) Data inadequate

Solution: Let the length of the train be x metres. Time taken to cover x meters = 5 min = (5 × 60) sec = 300 sec Speed of the train = $\frac{x}{300}$ m/sec Length of the second train = 2x meters Length of the platform = 2x meters ∴ Required time $\begin{aligned} = [\frac{2 x + 2 x}{(\frac{x}{300})}] sec \\ = (\frac{4 x \times 300}{x}) sec \\ = 1200 sec \\ = \frac{1200}{60} min \\ = 20 minutes \end{aligned}$

35.

A train B speeding with 120 kmph crosses another train C running in the same direction, in 2 minutes. If the lengths of the trains B and C be 100m and 200m respectively, what is the speed (in kmph) of the train C?

(A) 111 km
(B) 123 km
(C) 127 km
(D) 129 km

Solution: $\begin{aligned} Relative speed of the trains \\ = (\frac{100 + 200}{2 \times 60}) m/sec \\ = (\frac{5}{2}) m/sec \\ Speed of train B \\ = 120 kmph \\ = (120 \times \frac{5}{18}) m/sec \\ = (\frac{100}{3}) m/sec \\ Let the speed of second train be x m/sec \\ Then, \frac{100}{3} - x = \frac{5}{2} \\ \Rightarrow x = (\frac{100}{3} - \frac{5}{2}) \\ = (\frac{185}{6}) m/sec \\ ∴ Speed of second train \\ = (\frac{185}{6} \times \frac{18}{5}) kmph \\ = 111 kmph \end{aligned}$

36.

Two, trains, one from Howrah to Patna and the other from Patna to Howrah, start simultaneously. After they meet, the trains reach their destinations after 9 hours and 16 hours respectively. The ratio of their speeds is:

(A) 2 : 3
(B) 4 : 3
(C) 6 : 7
(D) 9 : 16

Solution: $\begin{aligned} Let us name the trains as A and B . Then, \\ (A's speed) : (B's speed) \\ = \sqrt{b} : \sqrt{a} \\ = \sqrt{16} : \sqrt{9} \\ = 4 : 3 \end{aligned}$

37.

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

38.

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

39.

A train passes two bridges of length 500 m and 250 m in 100 seconds and 60 seconds respectively. The length of the train is?

(A) 152 m
(B) 125 m
(C) 250 m
(D) 120 m

Solution: $\begin{aligned} Let the length of train x m \\ Speed of train \\ = \frac{(Length of train + length of bridge)}{Time taken in crossing} \\ According to information we get \\ \Rightarrow \frac{x + 500}{100} = \frac{x + 250}{60} \\ \Rightarrow 60 (x + 500) = 100 (x + 250) \\ \Rightarrow 3 (x + 500) = 5 (x + 250) \\ \Rightarrow 5 x + 1250 = 3 x + 1500 \\ \Rightarrow 5 x - 3 x = 1500 - 1250 \\ \Rightarrow 2 x = 250 \\ \Rightarrow x = \frac{250}{2} = 125 m \end{aligned}$

40.

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

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Practice MCQ Questions and Answer on Problems on Trains

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