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