1. A man sitting in a train is counting the pillars of electricity. The distance between two pillars is 60 meters, and the speed of the train is 42 km/hr. In 5 hours, how many pillars will he count?
a) 3501
b) 3600
c) 3800
d) None of these
Explanation: Distance covered by the train in 5 hours = (42 × 5) km
= 210 km
= 210000 m
Number of pillars counted by the man
= $$\left( {\frac{{210000}}{{60}} + 1} \right)$$
= 3500 + 1
= 3501
2. A 120 meter long train is running at a speed of 90 km/hr. It will cross a railway platform 230 m long in :
a) 4 seconds
b) 7 seconds
c) 12 seconds
d) 14 seconds
Explanation: Speed = $$\left( {90 \times \frac{5}{{18}}} \right)$$ m/sec = 25 m/sec
Total distance covered = (120 + 230) m
= 350 m
Required time
= $$\frac{{350}}{{25}}$$ seconds
= 14 seconds
3. A 50 meter long train passes over a bridge at the speed of 30 km per hour. If it takes 36 seconds to cross the bridge, what is the length of the bridge?
a) 200 meters
b) 250 meters
c) 300 meters
d) 350 meters
Explanation: Speed = $$\left( {30 \times \frac{5}{{18}}} \right)$$ m/sec = $$\frac{{25}}{3}$$ m/sec
Time = 36 second
Let the length of the bridge be x meters.
Then, $$\frac{{50 + {\text{x}}}}{{36}}$$ = $$\frac{{25}}{3}$$
3(50 + x) = 900
50 + x = 300
x = 250 meters
4. 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
Explanation: 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 = $$\frac{{\text{x}}}{{300}}$$ m/sec
Length of the second train = 2x meters
Length of the platform = 2x meters
Required time
$$\eqalign{ & = \left[ {\frac{{2{\text{x}} + 2{\text{x}}}}{{\left( {\frac{{\text{x}}}{{300}}} \right)}}} \right]{\text{sec}} \cr & = \left( {\frac{{4{\text{x}} \times 300}}{{\text{x}}}} \right){\text{sec}} \cr & = 1200\,{\text{sec}} \cr & = \frac{{1200}}{{60}}\,{\text{min}} \cr & = 20\,{\text{minutes}} \cr} $$
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?
a) 225 meters
b) 240 meters
c) 230 meters
d) 235 meters
Explanation: Speed = $$\left( {54 \times \frac{5}{{18}}} \right)$$ m/sec = 15 m/sec
Length of the train = (15 × 20) m = 300 m
Let the length of the platform be x meters
Then, $$\frac{{{\text{x}} + 300}}{{36}}$$ = 15
x + 300 = 540
x = 240 meters
6. A train 75 m long overtook a person who was walking at the rate of 6 km/hr in the same direction and passed him in $$7\frac{1}{2}$$ seconds. Subsequently, it overtook a second person and passed him in $$6\frac{3}{4}$$ seconds. At what rate was the second person travelling?
a) 1 km/hr
b) 2 km/hr
c) 4 km/hr
d) 5 km/hr
Explanation: Speed of the train relative to first man
$$\eqalign{ & = \frac{{75}}{{7.5}}{\text{m/sec}} = 10\,{\text{m/sec}} \cr & = \left( {10 \times \frac{{18}}{5}} \right){\text{km/hr}} = 36\,{\text{km/hr}} \cr} $$
Let the speed of the train be x km/hr.
Then, relative speed = (x - 6) km/hr
x - 6 = 36
x = 42 km/hr
Speed of the train relative to second man
$$\eqalign{ & {\text{ = }}\frac{{75}}{{6\frac{3}{4}}}\,{\text{m/sec}} \cr & = \left( {75 \times \frac{4}{{27}}} \right){\text{m/sec}} \cr & = \frac{{100}}{9}{\text{m/sec}} \cr & = \left( {\frac{{100}}{9} \times \frac{{18}}{5}} \right){\text{km}} \cr & = 40\,{\text{km/hr}} \cr} $$
Let the speed of the second man be y kmph.
Then, relative speed = (42 - y) kmph
42 - y = 40
y = 2 km/hr
7.Two trains are running in opposite directions with the same speed. If the length of each train is 120 meters and they cross each other in 12 seconds, then the speed of each train (in km/hr) is?
a) 10 km/hr
b) 18 km/hrs
c) 72 km/hr
d) 36 km/hr
Explanation: Let the speed of each train be x m/sec.
Then, relative speed of the two trains = 2x m/sec
So, 2x = $$\frac{{120 + 120}}{{12}}$$
2x = 20
x = 10
Speed of each train = 10 m/sec
= $$\left( {10 \times \frac{{18}}{5}} \right)$$ km/hr
= 36 km/hr
8. A 150 m long train crosses a milestone in 15 seconds and a train of same length coming from the opposite direction in 12 seconds. The speed of the other train is?
a) 36 kmph
b) 45 kmph
c) 50 kmph
d) 54 kmph
Explanation: Speed of first train = $$\frac{{150}}{{15}}$$ m/sec = 10 m/sec
Let the speed of second train be x m/sec
Relative speed = (10 + x) m/sec
$$\frac{{300}}{{10 + {\text{x}}}}$$ = 12
300 = 120 + 12x
12x = 180
x = $$\frac{{180}}{{12}}$$ = 15 m/sec
Hence, speed of other train = $$\left( {15 \times \frac{{18}}{5}} \right)$$ kmph
= 54 kmph
9.A man standing on a platform finds that a train takes 3 seconds to pass him and another train of the same length moving in the opposite direction takes 4 seconds. The time taken by the trains to pass each other will be :
a) $$2\frac{3}{7}$$ seconds
b) $$3\frac{3}{7}$$ seconds
c) $$4\frac{3}{7}$$ seconds
d) $$5\frac{3}{7}$$ seconds
Explanation: Let the length of each train be x meters
Then, speed of first train = $$\frac{{\text{x}}}{3}$$ m/sec
Speed of second train = $$\frac{{\text{x}}}{4}$$ m/sec
Required time
$$\eqalign{ & = \left[ {\frac{{{\text{x}} + {\text{x}}}}{{\left( {\frac{{\text{x}}}{3} + \frac{{\text{x}}}{4}} \right)}}} \right]{\text{sec}} \cr & = \left[ {\frac{{2{\text{x}}}}{{\left( {\frac{{7{\text{x}}}}{{12}}} \right)}}} \right]{\text{sec}} \cr & = \left( {2 \times \frac{{12}}{7}} \right){\text{sec}} \cr & = \frac{{24}}{7}{\text{sec}} \cr & = 3\frac{3}{7}{\text{sec}} \cr} $$.
10. Two trains, 130 and 110 meters long, are going in the same direction. The faster train takes one minute to pass the other completely. If they are moving in opposite directions, they pass each other completely in 3 seconds. Find the speed of the faster train.
a) 38 m/sec
b) 42 m/sec
c) 46 m/sec
d) 50 m/sec
Explanation: Let the speeds of the faster and slower trains be x m/sec and y m/sec respectively.
Then, $$\frac{{240}}{{{\text{x}} - {\text{y}}}}$$ = 60
x - y = 4 . . . . . . . . (i)
And, $$\frac{{240}}{{{\text{x}} + {\text{y}}}}$$ = 3
x + y = 80 . . . . . . . . (ii)
Adding (i) and (ii)
2x = 84
x = 42
Putting x = 42 in (i), we get: y = 38
Hence, speed of faster train = 42 m/sec