## Problems on Trains Questions and Answers Part-1

1. A train of length 150 meters takes 40.5 seconds to cross a tunnel of length 300 meters. What is the speed of the train in km/hr?
a) 13.33
b) 26.67
c) 40
d) 66.67

Explanation:
\eqalign{ & {\text{Speed = }}\left( {\frac{{150 + 300}}{{40.5}}} \right)m/\sec \cr & = \left( {\frac{{450}}{{40.5}} \times \frac{{18}}{5}} \right)km/hr \cr & = 40km/hr \cr}

2. A 280 meter long train crosses a platform thrice its length in 50 seconds. What is the speed of the train in km/hr?
a) 60.48
b) 64.86
c) 80.64
d) 82.38

Explanation:
\eqalign{ & {\text{Length of train}} = 280 m \cr & {\text{Length of platform}} \cr & {\text{ = (3}} \times {\text{280) m = 840m}} \cr & {\text{Speed of train}} \cr & {\text{ = }}\left( {\frac{{280 + 840}}{{50}}} \right)m/\sec \cr & = \frac{{1120}}{{50}}m/\sec \cr & = \left( {\frac{{1120}}{{50}} \times \frac{{18}}{5}} \right)km/hr \cr & = 80.64\,km/hr \cr}

3. A train 110 meters long is running with a speed of 60 kmph. In what time will it pass a man who is running at 6 kmph in the direction opposite to that in which the train is going?
a) 5 sec
b) 6 sec
c) 7 sec
d) 10 sec

Explanation:
\eqalign{ & {\text{Speed of train relative to man}} \cr & {\text{ = }}\left( {60 + 6} \right){\text{km/hr}} \cr & = 66\,{\text{km/hr}} \cr & = \left( {66 \times \frac{5}{{18}}} \right)m/\sec \cr & = \left( {\frac{{55}}{3}} \right)m/\sec \cr & {\text{Time taken to pass the man}} = \left( {110 \times \frac{3}{{55}}} \right)\sec \cr & = 6\,\sec \cr}

4. Two trains A and B start running together from the same point in the same direction, at the speed of 60 kmph and 72 kmph respectively. If the length of each of the trains is 240 meters, how long will it take for B to cross train A?
a) 1 min 12 sec
b) 1 min 24 sec
c) 2 min 12 sec
d) 2 min 24 sec

Explanation:
\eqalign{ & {\text{Relative speed}} = (72 - {\text{60) km/hr}} \cr & {\text{ = 12 km/hr}} \cr & = \left( {12 \times \frac{5}{{18}}} \right)m/\sec \cr & = \left( {\frac{{10}}{3}} \right)m/\sec \cr & {\text{Total distance covered}} \cr & {\text{ = Sum of lengths of trains}} \cr & {\text{ = (240 + 240) m}} \cr & {\text{ = 480 m}} \cr & {\text{Time taken}} \cr & {\text{ = }}\left( {480 \times \frac{3}{{10}}} \right)\sec \cr & = 144\sec \cr & = 2\min \,24sec \cr}

5. 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 second is?
a) 36
b) 45
c) 48
d) 49

Explanation:
\eqalign{ & {\text{Relative speed}} \cr & {\text{ = (60 + 90) km/hr}} \cr & {\text{ = }}\left( {150 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & {\text{ = }}\left( {\frac{{125}}{3}} \right){\text{m/sec}} \cr & {\text{Distance coverd}} \cr & {\text{ = (1}}{\text{.10 + 0}}{\text{.9)km}} \cr & {\text{ = 2 km}} \cr & {\text{ = 2000 m}}{\text{}} \cr & {\text{Required time}} = \left( {2000 \times \frac{3}{{125}}} \right)\sec \cr & = 48{\text{ sec}}\cr}

6. 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 kmph
b) 123 kmph
c) 127 kmph
d) 129 kmph

Explanation:
\eqalign{ & {\text{Relative speed of the trains }} \cr & {\text{ = }}\left( {\frac{{100 + 200}}{{2 \times 60}}} \right){\text{m/sec}} \cr & {\text{ = }}\left( {\frac{5}{2}} \right){\text{m/sec}} \cr & {\text{Speed of train B}} \cr & {\text{ = 120 kmph}} \cr & = \left( {120 \times \frac{5}{{18}}} \right){\text{m/sec}} \cr & {\text{ = }}\left( {\frac{{100}}{3}} \right){\text{m/sec}} \cr & {\text{Let the speed of second train be }}x{\text{ m/sec}} \cr & {\text{Then, }} \frac{{100}}{3} - x = \frac{5}{2} \cr & \Rightarrow x = \left( {\frac{{100}}{3} - \frac{5}{2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{185}}{6}} \right){\text{m/sec}} \cr & {\text{Speed of second train}} \cr & {\text{ = }}\left( {\frac{{185}}{6} \times \frac{{18}}{5}} \right){\text{ kmph}} \cr & {\text{ = 111 kmph}} \cr}

7. What is the speed of a train if it overtakes two persons who are walking in the same direction at the rate of a m/s and (a + 1) m/s and passes them completely in b seconds and (b + 1) seconds respectively?
a) (a + b) m/s
b) (a + b + 1) m/s
c) (2a + 1) m/s
d) $$\frac{{2{\text{a}} + 1}}{2}$$ m/s

Explanation:
\eqalign{ & {\text{Let the length of the train be }}x{\text{ metres}} \cr & {\text{and its speed be }}y{\text{ m/s}} \cr & {\text{ }}\frac{x}{{y - a}}{\text{ = b}}\,\,{\text{and}}\, \cr & \,\frac{x}{{y - \left( {a + 1} \right)}} = \left( {b + 1} \right) \cr & \Leftrightarrow {\text{ }}x{\text{ = }}b\left( {y - a} \right){\text{ and}} \cr & \,\,\,\,\,\,\,\,\,\,{\text{ }}x = \left( {b + 1} \right)\left( {y - a - 1} \right) \cr & \Leftrightarrow b\left( {y - a} \right) = \left( {b + 1} \right)\left( {y - a - 1} \right) \cr & \Leftrightarrow by - ba = by - ba - b + y - a - 1 \cr & y = \left( {a + b + 1} \right) \cr}.

8. A train passes a 50 meter long platform in 14 seconds and a man standing on platform 10 seconds.The speed of the train is?
a) 24 km/hr
b) 36 km/hr
c) 40 km/hr
d) 45 km/hr

Explanation:
\eqalign{ & {\text{Distance travelled in 14 sec}} \cr & {\text{ = 50 + }}l \cr & {\text{Distance travelled in 10 sec}} = l \cr & {\text{Speed of train}} \cr & {\text{ = }}\frac{{50}}{{14 - 10}}{\text{m/sec}} \cr & {\text{ = }}\frac{{50}}{4} \times \frac{{18}}{5}{\text{km/hr}} \cr & {\text{ = 45 km/hr}} \cr}

9. A train is moving at a speed of 132 km/hr. If the length of the train is 110 meters, how long it will take to cross a railway platform 165 meter long?
a) 5 second
b) 7.5 second
c) 10 second
d) 15 second

\eqalign{ & {\text{Speed = 132 km/hr }} \cr & {\text{ = 132}} \times \frac{5}{{18}}{\text{m/sec}} \cr & {\text{ = }}\frac{{110}}{3}m/\sec \cr & T = \frac{D}{S} \cr & \,\,\,\,\,\, = \frac{{110 + 165}}{{\frac{{100}}{3}}} \cr & \,\,\,\,\,\, = \frac{{3\left( {275} \right)}}{{110}} \cr & \,\,\,\,\,\, = 7.5\sec \cr}
\eqalign{ & {\text{Speed of the train}} \cr & {\text{ = }}\frac{{700 + 500}}{{10}} \cr & {\text{ = 120 ft/second}} \cr}