Problems on Trains Questions and Answers Part-5

1. How many seconds will a 500 metre long train take to cross a man walking with a speed of 3 km/hr in the direction of the moving train if the speed of the train is 63 km/hr?
a) 25
b) 30
c) 40
d) 45

Answer: b
Explanation:
$$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{man}} \cr & = \left( {63 - 3} \right)\,{\text{km/hr}} \cr & = 60\,{\text{km/hr}} \cr & = \left( {60 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & = \frac{50}{3}\, \text{m/sec} \cr & {\text{Time}}\,{\text{taken}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{man}} \cr & = \left( {500 \times \frac{3}{{50}}} \right)\,\sec \cr & = 30\,\sec \cr} $$

2. Two goods train each 500 m long, are running in opposite directions on parallel tracks. Their speeds are 45 km/hr and 30 km/hr respectively. Find the time taken by the slower train to pass the driver of the faster one.
a) 12 sec
b) 24 sec
c) 48 sec
d) 60 sec

Answer: b
Explanation:
$$\eqalign{ & {\text{Relative}}\,{\text{speed}} \cr & = \left( {45 + 30} \right)\,{\text{km/hr}} \cr & = \left(75 \times \frac{5}{18} \right)\, \text{m/sec} \cr & = {\frac{{125}}{6}} \,{\text{m/sec}} \cr & {\text{We}}\,{\text{have}}\,{\text{to}}\,{\text{find}}\,{\text{the}}\,{\text{time}}\,{\text{taken}}\,{\text{by}}\,{\text{the}} \cr & {\text{slower}}\,{\text{train}}\,{\text{to}}\,{\text{pass}}\,{\text{the}}\,{\text{DRIVER}}\,{\text{of}}\, \cr & {\text{The}}\,{\text{faster}}\,{\text{train}}\,{\text{and}}\,{\text{not}}\,{\text{the}}\,{\text{complete}}\,{\text{train}}{\text{.}} \cr & {\text{So,}}\,{\text{distance}}\,{\text{covered = Length}}\,{\text{of}}\,{\text{the}}\,{\text{slower}}\,{\text{train}}. \cr & {\text{Therefore,}}\,{\text{Distance}}\,{\text{covered = 500}}\,{\text{m}}. \cr & {\text{Required}}\,{\text{time}} = {500 \times \frac{6}{{125}}} \cr & = 24\,\sec \cr} $$

3. Two trains are running in opposite directions with the same speed. If the length of each train is 120 metres and they cross each other in 12 seconds, then the speed of each train (in km/hr) is:
a) 10
b) 18
c) 36
d) 72

Answer: c
Explanation:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{each}}\,{\text{train}}\,{\text{be}}\,x\,{\text{m/sec}}. \cr & {\text{Then,}}\,{\text{relative}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{two}}\,{\text{trains}} = 2x\,{\text{m/sec}} \cr & {\text{So}},\,2x = \frac{{ {120 + 120} }}{{12}} \cr & 2x = 20 \cr & x = 10 \cr & {\text{Speed}}\,{\text{of}}\,{\text{each}}\,{\text{train}} = 10\,{\text{m/sec}} \cr & = {10 \times \frac{{18}}{5}} \,{\text{km/hr}} \cr & = 36\,{\text{km/hr}} \cr} $$

4. 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

Answer: b
Explanation:
$$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} \cr & = {\frac{{120}}{{10}}} \,{\text{m/sec}} \cr & = 12\,{\text{m/sec}} \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}} \cr & {\frac{{120}}{{15}}} \,{\text{m/sec}} \cr & = 8\,{\text{m/sec}} \cr & {\text{Relative}}\,{\text{speed}} = {12 + 8} = 20\,{\text{m/sec}} \cr & {\text{Required}}\,{\text{time}} \cr & = {\frac{{ {120 + 120} }}{{20}}} \,{\text{ sec}} \cr & = 12\,{\text{sec}} \cr} $$

5. 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

Answer: d
Explanation:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{be}}\,x\,{\text{km/hr}}. \cr & {\text{Relative}}\,{\text{speed}}\, \cr & = \,\left( {x + 50} \right)\,{\text{km/hr}} \cr & = \left[ {\left( {x + 50} \right) \times \frac{5}{{18}}} \right]\,{\text{m/sec}} \cr & = {\frac{{250 + 5x}}{{18}}} \,{\text{m/sec}} \cr & {\text{Distance}}\,{\text{covered}} \cr & = \left( {108 + 112} \right) = 220\,m \cr & \frac{{220}}{{ {\frac{{250 + 5x}}{{18}}} }} = 6 \cr & 250 + 5x = 660 \cr & x = 82\,{\text{km/hr}} \cr} $$.

6. A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?
a) 65 sec
b) 89 sec
c) 100 sec
d) 150 sec

Answer: b
Explanation:
$$\eqalign{ & {\text{Speed}} = {\frac{{240}}{{24}}} \,{\text{m/sec}} = 10\,{\text{m/sec}} \cr & {\text{Required}}\,{\text{time}} \cr & {\text{ = }}\, {\frac{{240 + 650}}{{10}}} \,{\text{sec}}. \cr & = 89\,sec. \cr} $$

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:
a) 50 m
b) 72 m
c) 80 m
d) 82 m

Answer: a
Explanation:
$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{each}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}. \cr & {\text{Then,}}\,{\text{distance}}\,{\text{covered}} = 2x\,{\text{metres}}. \cr & {\text{Relative}}\,{\text{speed}} \cr & = \left( {46 - 36} \right)\,{\text{km/hr}} \cr & = {10 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = {\frac{{25}}{9}} \,{\text{m/sec}} \cr & \therefore \frac{{2x}}{{36}} = \frac{{25}}{9} \cr & 2x = 100 \cr & x = 50 \cr} $$

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?
a) 40 sec
b) 42 sec
c) 45 sec
d) 48 sec

Answer: a
Explanation:
$$\eqalign{ & {\text{Formula}}\,{\text{for}}\,{\text{converting}}\,{\text{from}}\,{\text{km/hr}}\,{\text{to}}\,{\text{m/s:}} \cr & X\,{\text{km/hr}} = {X \times \frac{5}{{18}}} \,{\text{m/s}} \cr & {\text{Therefore,}}\,{\text{Speed}} \cr & = {45 \times \frac{5}{{18}}} \,{\text{m/sec}} = \frac{{25}}{2}{\text{m/sec}} \cr & {\text{Total}}\,{\text{distance}}\,{\text{to}}\,{\text{be}}\,{\text{covered}} \cr & = \left( {360 + 140} \right)m = 500\,m \cr & {\text{Formula}}\,{\text{for}}\,{\text{finding}}\,{\text{Time}} \cr & = {\frac{{{\text{Distance}}}}{{{\text{Speed}}}}} \cr & {\text{Required}}\,{\text{time}} \cr & = \left( {\frac{{500 \times 2}}{{25}}} \right)\,\sec \cr & = 40\,\sec . \cr} $$

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:
a) 36
b) 45
c) 48
d) 49

Answer: c
Explanation:
$$\eqalign{ & {\text{Relative}}\,{\text{speed}} = \left( {60 + 90} \right)\,{\text{km/hr}} \cr & = {150 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = {\frac{{125}}{3}} \,{\text{m/sec}} \cr & {\text{Distance}}\,{\text{covered}} \cr & = \left( {1.10 + 0.9} \right)\,km \cr & = 2\,km \cr & = \,2000\,m \cr & {\text{Required}}\,{\text{time}} = {2000 \times \frac{3}{{125}}} \,{\text{sec}} \cr & = 48\,sec \cr} $$

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?
a) 3.6 sec
b) 18 sec
c) 36 sec
d) 72 sec

Answer: c
Explanation:
$$\eqalign{ & {\text{Speed}}\,{\text{of}}\,{\text{train}}\,{\text{relative}}\,{\text{to}}\,{\text{jogger}} \cr & = \left( {45 - 9} \right)\,{\text{km/hr}} \cr & = 36\,{\text{km/hr}} \cr & {36 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = 10\,{\text{m/sec}} \cr & {\text{Distance}}\,{\text{to}}\,{\text{be}}\,{\text{covered}} \cr & = \left( {240 + 120} \right)\,m \cr & = 360\,m \cr & {\text{Time}}\,{\text{taken}} = {\frac{{360}}{{10}}} \,{\text{sec}} \cr & = 36\,{\text{sec}} \cr} $$