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Problems On Trains - Aptitude MCQ Questions and Solutions with Explanations

Problems On Trains MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Problems On Trains with easy and logical explanations. Problems On Trains MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc.

1
A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
Discuss
Answer & Solution
Answer: Option D
Solution:
$$\eqalign{ & {\text{Speed}} = \left( {60 \times \frac{5}{{18}}} \right){\text{m/sec}} = {\frac{{50}}{3}} {\text{m/sec}} \cr & {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} = \left( {{\text{Speed}} \times {\text{Time}}} \right) \cr & \therefore {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} \cr & = \left( {\frac{{50}}{3} \times 9} \right)m = 150m \cr} $$
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:
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{man}} \cr & = {\frac{{125}}{{10}}} {\text{ m/sec}} \cr & = {\frac{{25}}{2}} {\text{ m/sec}} \cr & = {\frac{{25}}{2} \times \frac{{18}}{5}} {\text{ km/hr}} \cr & = 45\,{\text{km/hr}} \cr & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{km/hr}}. \cr & \text{Then, relative speed} = \left( {x - 5} \right)\,{\text{km/hr}} \cr & \therefore x - 5 = 45 \cr & \Rightarrow x = 50\,{\text{km/hr}} \cr} $$
3
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Speed}} = {45 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\frac{{25}}{2}} \,{\text{m/sec}} \cr & {\text{Time}} = 30\,{\text{sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{bridge}}\,{\text{be}}\,x\,{\text{metres}} \cr & {\text{Then}},\,\frac{{130 + x}}{{30}} = \frac{{25}}{2} \cr & \Rightarrow 2\left( {130 + x} \right) = 750 \cr & \Rightarrow x = 245\,m \cr} $$
4
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:
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speeds}}\,{\text{of}}\,{\text{the}}\,{\text{two}}\,{\text{trains}}\,{\text{be}}\,x\,{\text{m/sec}} \cr & {\text{and}}\,y\,{\text{m/sec}}\,{\text{respectively}}. \cr & {\text{Then,}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} = 27x\,{\text{metres}}, \cr & {\text{and}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}} = 17y\,{\text{metres}}. \cr & \therefore \frac{{27x + 17y}}{{x + y}} = 23 \cr & \Rightarrow 27x + 17y = 23x + 23y \cr & \Rightarrow 4x = 6y \cr & \Rightarrow \frac{x}{y} = \frac{3}{2} \cr} $$
5
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?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Speed}} = {54 \times \frac{5}{{18}}} \,{\text{m/sec}} = 15\,{\text{m/sec}} \cr & {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}} = \left( {15 \times 20} \right){\text{m}} = 300\,{\text{m}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{platform}}\,{\text{be}}\,x\,{\text{metres}} \cr & {\text{Then}},\,\frac{{x + 300}}{{36}} = 15 \cr & \Rightarrow x + 300 = 540 \cr & \Rightarrow x = 240\,{\text{m}} \cr} $$
6
A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{Speed}} = {\frac{{240}}{{24}}} \,{\text{m/sec}} = 10\,{\text{m/sec}} \cr & \therefore {\text{Required}}\,{\text{time}} \cr & {\text{ = }}\, {\frac{{240 + 650}}{{10}}} \,{\text{sec}}. \cr & = 89\,sec. \cr} $$
7
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:
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{each}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}. \cr & {\text{Then,}}\,{\text{distance}}\,{\text{covered}} = 2x\,{\text{metres}}. \cr & {\text{Relative}}\,{\text{speed}} \cr & = \left( {46 - 36} \right)\,{\text{km/hr}} \cr & = {10 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = {\frac{{25}}{9}} \,{\text{m/sec}} \cr & \therefore \frac{{2x}}{{36}} = \frac{{25}}{9} \cr & \Rightarrow 2x = 100 \cr & \Rightarrow x = 50 \cr} $$
8
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?
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & {\text{Formula}}\,{\text{for}}\,{\text{converting}}\,{\text{from}}\,{\text{km/hr}}\,{\text{to}}\,{\text{m/s:}} \cr & X\,{\text{km/hr}} = {X \times \frac{5}{{18}}} \,{\text{m/s}} \cr & {\text{Therefore,}}\,{\text{Speed}} \cr & = {45 \times \frac{5}{{18}}} \,{\text{m/sec}} = \frac{{25}}{2}{\text{m/sec}} \cr & {\text{Total}}\,{\text{distance}}\,{\text{to}}\,{\text{be}}\,{\text{covered}} \cr & = \left( {360 + 140} \right)m = 500\,m \cr & {\text{Formula}}\,{\text{for}}\,{\text{finding}}\,{\text{Time}} \cr & = {\frac{{{\text{Distance}}}}{{{\text{Speed}}}}} \cr & \therefore {\text{Required}}\,{\text{time}} \cr & = \left( {\frac{{500 \times 2}}{{25}}} \right)\,\sec \cr & = 40\,\sec . \cr} $$
9
Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is:
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Relative}}\,{\text{speed}} = \left( {60 + 90} \right)\,{\text{km/hr}} \cr & = {150 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = {\frac{{125}}{3}} \,{\text{m/sec}} \cr & {\text{Distance}}\,{\text{covered}} \cr & = \left( {1.10 + 0.9} \right)\,km \cr & = 2\,km \cr & = \,2000\,m \cr & {\text{Required}}\,{\text{time}} \cr & = {2000 \times \frac{3}{{125}}} \,{\text{sec}} \cr & = 48\,sec \cr} $$
10
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?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{jogger}} \cr & = \left( {45 - 9} \right)\,{\text{km/hr}} \cr & = 36\,{\text{km/hr}} \cr & {36 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = 10\,{\text{m/sec}} \cr & {\text{Distance}}\,{\text{to}}\,{\text{be}}\,{\text{covered}} \cr & = \left( {240 + 120} \right)\,m \cr & = 360\,m \cr & \therefore {\text{Time}}\,{\text{taken}} \cr & = {\frac{{360}}{{10}}} \,{\text{sec}} \cr & = 36\,{\text{sec}} \cr} $$