## Problems on Trains Questions and Answers Part-3

1. A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
a) 230 m
b) 240 m
c) 260 m
d) 320 m

Explanation:
\eqalign{ & {\text{Relative}}\,{\text{speed}} = \left( {120 + 80} \right)\,{\text{km/hr}} \cr & = {200 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = {\frac{{500}}{9}} \,{\text{m/sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{other}}\,{\text{train}}\,{\text{be}}\,{\text{x}}\,{\text{metres}}{\text{.}} \cr & {\text{Then,}}\,\frac{{x + 270}}{9} = \frac{{500}}{9} \cr & x + 270 = 500 \cr & x = 230 \cr}

2. 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

Explanation:
\eqalign{ & {\text{Speed}} = {72 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20\,{\text{m/sec}} \cr & {\text{Time}} = 26\,{\text{sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}{\text{.}} \cr & {\text{Then}},\,\frac{{x + 250}}{{26}} = 20 \cr & x + 250 = 520 \cr & x = 270 \cr}

3. 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

Explanation:
\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{slower}}\,{\text{train}}\,{\text{be}}\,x\,{\text{m/sec}} \cr & {\text{Then,}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{faster}}\,{\text{train}} = 2x\,{\text{m/sec}} \cr & {\text{Relative}}\,{\text{speed}} = \,\left( {x + 2x} \right)\,{\text{m/sec}} = 3x\,{\text{m/sec}} \cr & \frac{{ {100 + 100} }}{8} = 3x \cr & 24x = 200 \cr & x = \frac{{25}}{3} \cr & {\text{So,}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{faster}}\,{\text{train}}\, = \frac{{50}}{3}\,{\text{m/sec}} \cr & = {\frac{{50}}{3} \times \frac{{18}}{5}} \,{\text{km/hr}} \cr & = 60\,{\text{km/hr}} \cr}

4.Two trains 140 m and 160 m long run at the speed of 60 km/hr and 40 km/hr respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other, is:
a) 9
b) 9.6
c) 10
d) 10.8

Explanation:
\eqalign{ & {\text{Relative}}\,{\text{speed}} = \left( {60 + 40} \right)\,{\text{km/hr}} \cr & = {100 \times \frac{5}{{18}}} \,{\text{m/sec}} \cr & = {\frac{{250}}{9}} \,{\text{m/sec}}. \cr & {\text{Distance}}\,{\text{covered}}\,{\text{in}}\,{\text{crossing}}\,{\text{each}}\,{\text{other}} \cr & = \left( {140 + 160} \right)m = 300\,m \cr & {\text{Required}}\,{\text{time}} = {300 \times \frac{9}{{250}}} \,{\text{sec}} \cr & = \frac{{54}}{5}\,{\text{sec}} \cr & = 10.8\,{\text{sec}} \cr}

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

6.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

Explanation:
\eqalign{ & {\text{Relative}}\,{\text{speed}} = \left( {40 - 20} \right)\,{\text{km/hr}} \cr & = \left( {20 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & = {\frac{{50}}{9}} \,{\text{m/sec}} \cr & {\text{Length}}\,{\text{of}}\,{\text{faster}}\,{\text{train}} = \left( {\frac{{50}}{9} \times 5} \right)\,m \cr & = \frac{{250}}{9}\,m \cr & = 27\frac{7}{9}\,m \cr}

7. A train overtakes two persons who are walking in the same direction in which the train is going, at the rate of 2 kmph and 4 kmph and passes them completely in 9 and 10 seconds respectively. The length of the train is:
a) 45 m
b) 50 m
c) 54 m
d) 72 m

Explanation:
\eqalign{ & 2\,kmph = \left( {2 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{9}\,{\text{m/sec}} \cr & 4\,kmph = \left( {4 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{10}}{9}\,{\text{m/sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}}\, \cr & {\text{and}}\,{\text{its}}\,{\text{speed}}\,{\text{be}}\,y\,{\text{m/sec}} \cr & {\text{Then}},\, {\frac{x}{{y - \frac{5}{9}}}} = 9\,{\text{and}}\, {\frac{x}{{y - \frac{{10}}{9}}}} = 10 \cr & 9y - 5 = x\,{\text{and}}\,10\left( {9y - 10} \right) = 9x \cr & \Rightarrow 9y - x = 5\,{\text{and}}\,90y - 9x = 100 \cr & {\text{On}}\,{\text{solving,}}\,{\text{we}}\,{\text{get}}:\,x = 50 \cr & {\text{Length}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{is}}\,50\,m \cr}.

8.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) 76 km/hr
d) 81 km/hr

Explanation:
\eqalign{ & 4.5\,{\text{km/hr}} = \left( {4.5 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{5}{4}\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.25\,{\text{m/sec,}}\,{\text{and}} \cr & 5.4\,km/hr = \left( {5.4 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{2}\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1.5\,{\text{m/sec}} \cr & {\text{Let}}\,{\text{the}}\,{\text{speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}}\,{\text{be}}\,x\,{\text{m/sec}} \cr & {\text{Then}},\,\left( {x - 1.25} \right) \times 8.4 = \left( {x - 1.5} \right) \times 8.5 \cr & 8.4x - 10.5 = 8.5x - 12.75 \cr & 0.1x = 2.25 \cr & x = 22.5 \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{train}} \cr & = \left( {22.5 \times \frac{{18}}{5}} \right)\,{\text{km/hr}} \cr & = 81\,{\text{km/hr}} \cr}

9. A train travelling at 48 kmph completely crosses another train having half its length and travelling in opposite direction at 42 kmph, in 12 seconds. It also passes a railway platform in 45 seconds. The length of the platform is
a) 400 m
b) 450 m
c) 560 m
d) 600 m

\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}}\,{\text{be}}\,x\,{\text{metres}} \cr & {\text{Then,}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{the}}\,{\text{second}}\,{\text{train}}\,{\text{is}}\, {\frac{x}{2}} \,{\text{metres}} \cr & {\text{Relative}}\,{\text{speed}} = \left( {48 + 42} \right)\,{\text{kmph}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {90 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 25\,{\text{m/sec}} \cr & \frac{{ {x + \left( {x/2} \right)} }}{{25}} = 12 \cr & \frac{{3x}}{2} = 300 \cr & x = 200 \cr & {\text{Length}}\,{\text{of}}\,{\text{first}}\,{\text{train}} = 200\,{\text{m}} \cr & {\text{Let}}\,{\text{the}}\,{\text{length}}\,{\text{of}}\,{\text{platform}}\,{\text{be}}\,y\,{\text{metres}} \cr & {\text{Speed}}\,{\text{of}}\,{\text{the}}\,{\text{first}}\,{\text{train}} \cr & = \left( {48 \times \frac{5}{{18}}} \right)\,{\text{m/sec}} \cr & = \frac{{40}}{3}\,{\text{m/sec}} \cr & \therefore \left( {200 + y} \right) \times \frac{3}{{40}} = 45 \cr & \Rightarrow 600 + 3y = 1800 \cr & \Rightarrow y = 400\,{\text{m}} \cr}.
\eqalign{ & {\text{Suppose}}\,{\text{they}}\,{\text{meet}}\,x\,{\text{hours}}\,{\text{after}}\,{\text{7}}\,{\text{a}}{\text{.m}}. \cr & {\text{Distance}}\,{\text{covered}}\,{\text{by}}\,{\text{A}}\, \cr & {\text{in}}\,x\,{\text{hours = 20x}}\,{\text{km}}{\text{.}} \cr & {\text{Distance}}\,{\text{covered}}\,{\text{by}}\,{\text{B}} \cr & \,{\text{in}}\,\left( {x - 1} \right)\,{\text{hours}} = 25\left( {x - 1} \right)\,km \cr & 20x + 25\left( {x - 1} \right) = 110 \cr & 45x = 135 \cr & x = 3 \cr & {\text{So,}}\,{\text{they}}\,{\text{meet}}\,{\text{at}}\,{\text{10}}\,{\text{a}}{\text{.m}}{\text{.}}\, \cr}