1. Two trains start at the same time for two station A and B toward B and A respectively. If the distance between A and B is 220 km and their speeds are 50 km/hr and 60 km/hr respectively then after how much time will they meet each other?

a) 2 hr

b) $$2\frac{1}{2}$$ hr

c) 3 hr

d) 1 hr

Explanation:

$$\eqalign{ & {\text{Relative speed}} = 60 + 50 \cr & {\text{ = 110 km/h}} \cr & {\text{Time taken}} \cr & {\text{ = }}\frac{{220}}{{110}} \cr & {\text{ = 2 hr}} \cr} $$

2. A train 100 meter long meets a man going in opposite direction at 5 km/h and passes him in 7^{1}/_{5} seconds. What is the speed of the train (in km/hr)?

a) 45 km/h

b) 60 km/h

c) 55 km/hr

d) 50 km/hr

Explanation:

$$\eqalign{ & {\text{Relative speed of man & train}} \cr & {\text{ = }}\frac{{100 \times 5}}{{36}} \times \frac{{18}}{5} \cr & {\text{ = 50km/hr}} \cr & {\text{speed of train}} \cr & {\text{ = 50}} - {\text{5}} \cr & {\text{ = 45 km/hr}} \cr} $$

3. A train takes 9 sec to cross a pole. If the speed of the train is 48 kmph, then length of the train is?

a) 150 m

b) 120 m

c) 90 m

d) 80 m

Explanation:

$$\eqalign{ & {\text{Time taken by train to cross a pole}} = 9 sec \cr & {\text{Distance covered in crossing a pole}} \cr & {\text{ = length of train}} \cr & {\text{Speed of the train}} = 48 km/h \cr & = \left( {\frac{{48 \times 5}}{{18}}} \right)m/\sec \cr & = \frac{{40}}{3}m/\sec \cr & {\text{Length of the train}} \cr & {\text{ = Speed }} \times {\text{Time}} \cr & {\text{ = }}\frac{{40}}{3} \times 9 \cr & {\text{ = 120 m}} \cr} $$

4. Two trains start at the same time from A and B and proceed toward each other at the sped of 75 km/hr and 50 km/hr respectively. When both meet at a point in between, one train was found to have traveled 175 km more then the other. Find the distance between A and B?

a) 875 km

b) 785 km

c) 758 km

d) 857 km

Explanation:

$$\eqalign{ & {\text{Let the trains meet after t hours}} \cr & {\text{Speed of train A}} = 75 km/hr \cr & {\text{Speed of train B}} = 50 km/hr \cr & {\text{Distance covered by train A}} \cr & {\text{ = 75}} \times {\text{t = 75t}} \cr & {\text{Distance covered by train B}} \cr & {\text{ = 50}} \times {\text{t = 50t}} \cr & {\text{Distance}}\,{\text{ = Speed }} \times {\text{Time}} \cr & {\text{According to question}} \cr & 75{\text{t}} - 50{\text{t}} = 175 \cr & \Rightarrow 25{\text{t}} = 175 \cr & \Rightarrow {\text{t}} = \frac{{175}}{{25}} = 7\,{\text{hour}} \cr & {\text{Distance between A and B }} \cr & {\text{ = 75t}} + 50{\text{t}} = 125{\text{t}} \cr & = 125 \times 7 = 875\,{\text{km}} \cr} $$

5. Two trains 180 meters and 120 meters in length are running towards each other on parallel tracks, one at the rate 65 km/hr and another at 55 km/hr. In how many seconds will they be cross each other from the moment they meet?

a) 6 seconds

b) 9 seconds

c) 12 seconds

d) 15 seconds

Explanation:

$$\eqalign{ & {\text{Time taken by trains to cross each }} \cr & {\text{other in opposite direction}} \cr & {\text{ = }}\frac{{{l_1} + {l_2}}}{{{\text{relative speed in opposite direction}}}} \cr & {\text{ = }}\frac{{\left( {180 + 120} \right)}}{{\left( {65 + 55} \right)}} \cr & {\text{ = }}\frac{{300}}{{120 \times \frac{5}{{18}}}} \cr & {\text{ = 9 seconds}} \cr} $$

6. A train starts from A at 7 a.m. towards B with speed 50 km/h. Another train starts from B at 8 a.m. with speed of 60 km/h towards A. Both of them meet at 10 a.m. at C. The ratio of the distance AC to BC is?

a) 5 : 6

b) 5 : 4

c) 6 : 5

d) 4 : 5

Explanation: The speed of train A is 50km/hr and A starts its journey at 7 AM and reaches C at 10 AM. Total Travel time = 3hr

Distance cover by A in 3hr = 50 × 3 = 150KM

Similarly, the speed of train B is 60km/hr and B starts its journey at 8 AM and reaches C at 10 AM. Total Travel time = 2hr

Distance cover by B in 2hr = 60 × 2 = 120KM

The ratio of the distance between AC : BC

= 150 : 120

= 5 : 4

7. Train A passes a lamp post in 9 seconds and 700 meter long platform in 30 seconds. How much time will the same train take to cross a platform which is 800 meters long? (in seconds)

a) 32 seconds

b) 31 seconds

c) 33 seconds

d) 30 seconds

Explanation:

$$\eqalign{ & {\text{Let the length of train be x m}} \cr & {\text{When a train crosses a light }} \cr & {\text{post in 9 second the distance covered}} \cr & {\text{ = length of train }} \cr & \Rightarrow {\text{speed of train = }}\frac{x}{9} \cr & {\text{Distance covered in crossing a}} \cr & {\text{700 meter platfrom in 30 seconds}} \cr & {\text{ = Length of platfrom + length of train}} \cr & {\text{Speed of train = }}\frac{{x + 700}}{30} \cr & \Rightarrow \frac{x}{9} = \frac{{x + 700}}{{30}}\left[ { {\text{Speed = }}\frac{{{\text{Distance}}}}{{{\text{Time}}}}} \right] \cr & \frac{x}{3} = \frac{{x + 700}}{{10}} \cr & 10x = 3x + 2100 \cr & 10x - 3x = 2100 \cr & 7x = 2100 \cr & x = \frac{{2100}}{7} = 300{\text{m}} \cr & {\text{When the length of the platform be 800m,}} \cr & {\text{then time T be taken by train to cross 800m}} \cr & {\text{long platform}} \cr & \frac{x}{9} = \frac{{x + 800}}{T} \cr & Tx = 9x + 7200 \cr & 300T = 2700 + 7200 \cr & 300T = 9900 \cr & T = \frac{{9900}}{{300}} = 33{\text{ seconds}} \cr} $$

8. Train A traveling at 63 kmph can cross a platform 199.5 m long in 21 seconds. How much would train A take to completely cross (from the moment they meet ) train B, 157 m long and traveling at 54 kmph in opposite direction which train A is traveling? (in seconds)

a) 16

b) 18

c) 12

d) 10

Explanation:

$$\eqalign{ & {\text{Speed of train A}} \cr & {\text{ = 63 kmph}} \cr & {\text{ = }}\left( {\frac{{63 \times 5}}{{18}}} \right){\text{m/sec}} \cr & {\text{ = 17}}{\text{.5 m/sec}} \cr & {\text{Speed of train B}} \cr & {\text{ = 54 kmph}} \cr & {\text{ = }}\left( {\frac{{54 \times 5}}{{18}}} \right){\text{m/sec = 15 m/sec}} \cr & {\text{If the length of train A be }}x{\text{ metre,}} \cr & {\text{then}} \cr & {\text{Speed of train A}} \cr & {\text{ = }}\frac{{{\text{Length of train + length of platform}}}}{{{\text{Time taken in crossing}}}}{\text{ }} \cr & \Rightarrow 17.5 = \frac{{x + 199.5}}{{21}} \cr & 17.5 \times 21 = x + 199.5 \cr & 367.5 = x + 199.5 \cr & x = 367.5 - 199.5 \cr & 168\,{\text{metres}} \cr & {\text{Relative speed}} \cr & {\text{ = ( Speed train A + Speed train B)}} \cr & {\text{ = (17}}{\text{.5 + 15) m/sec}} \cr & {\text{ = 32}}{\text{.5 m/sec}} \cr & {\text{Required time}} \cr & {\text{ = }}\frac{{{\text{ Length of train A + Length of train B}}}}{{{\text{Relative speed }}}} \cr & = \left( {\frac{{168 + 157}}{{32.5}}} \right){\text{seconds}} \cr & = 10\,{\text{seconds}} \cr} $$

9. A train which is moving at an average speed of 40 km/h reaches its destination on time. When its average speed reduces to 35 km/h, then it reaches its destination 15 minutes late. The distance traveled by the train is?

a) 70 km

b) 80 km

c) 40 km

d) 30 km

Explanation:

$$\eqalign{ & {\text{Average speed of train}} = 40 km/hr \cr & {\text{Reach at its destination at on time }} \cr & {\text{New average speed of train}} = 35 km/h \cr & {\text{Time = 15 minutes}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{ = }}\frac{{15}}{{60}}{\text{hours }} \cr & {\text{Then distance travelled}} \cr & {\text{ = }}\frac{{40 \times 35}}{{40 - 35}}{\text{ }} \times \frac{{15}}{{60}} \cr & {\text{ = }}\frac{{40 \times 35}}{5}{\text{ }} \times \frac{{15}}{{60}} \cr & {\text{ = 70}}\,{\text{km}} \cr} $$

10. A train moves with a speed of 30 kmph for 12 minutes and for next 8 minutes at a speed of 45 kmph. Find the average speed of the train?

a) 37.50 kmph

b) 36 kmph

c) 48 kmph

d) 30 kmph

Explanation:

$$\eqalign{ & {\text{Distance = Speed }} \times {\text{Time}} \cr & {\text{Distance covered by train with the}} \cr & {\text{speed of 30 kmph in 12 minutes is }} \cr & {\text{ = 30}} \times \frac{{12}}{{60}} = 6{\text{km}} \cr & {\text{Distance covered by the same train}} \cr & {\text{with the speed of 45 kmph in 8 minutes is }} \cr & {\text{ = 45}} \times \frac{8}{{60}} = 6{\text{km}} \cr & {\text{Average speed}} \cr & {\text{ = }}\frac{{{\text{total distance}}}}{{{\text{total time}}}}. \cr & \Rightarrow \frac{{\left( {6 + 6} \right){\text{km}}}}{{\left( {12 + 8} \right)\min }} = \frac{{12}}{{20}} \times 60 \cr & {\text{ = 36 kmph}} \cr} $$