Problems on trains are very common in competitive exams. Various types of questions are asked on trains. Questions on trains are solved using the concept of time, speed and distance i.e. we use the formulas of time, speed and distance to solve questions on trains.

Given below is a list of some important points that need to be kept in mind while solving questions on trains.

- When two trains are going in the same direction, then their relative speed is the difference between the two speeds.
- When two trains are moving in the opposite direction, then their relative speed is the sum of the two speeds.
- When a train crosses a stationary man/ pole/ lamp post/ sign post- in all these cases, the object which the train crosses is stationary and the distance travelled is the length of the train.
- When it crosses a platform/ bridge- in these cases, the object which the train crosses is stationary and the distance travelled is the length of the train and the length of the object.
- When two trains are moving in same direction, then their speed will be subtracted.
- When two trains are moving in opposite directions, then their speed will be added.
- In both the above cases, the total distance is the sum of the length of both the trains.
- When a train crosses a car/ bicycle/ a mobile man- in these cases, the relative speed between the train and the object is taken depending upon the direction of the movement of the other object relative to the train- and the distance travelled is the length of the train.

Now, let us try doing some questions and understand the concept of solving train related problems

Time taken to cross the man = 6 secs

Therefore, Distance = (50/3)* 6 = 100 metres (i.e. the length of the train)

Speed of train B = = 50 kmph = 50 *(5/18) = 125/9 m/sec

The relative speed =(50/3)-(125/9)=25/9 m/s (we have subtracted the two values because both the trains are going in the same direction)

Time taken by train A to cross train B = 30 secs

Distance = Speed * Time

Distance =25/9 * 30 = 250/3 metres (i.e. the combined length of both trains)

Total Speed = 64 + 80 = 144 kmph (added because they are travelling in opposite directions)

In m/sec, speed = 144 *(5/18) = 40 m/sec

Distance = Speed * Time

600 = 40 * Time

Therefore, Time = 15 seconds

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Distance_{1} = Speed * Time = 80 * t = 80t

Distance_{2} = Speed * Time = 95 * t = 95t

As the distance gap between both trains is 180 kms

Therefore, we can say that:

95t - 80t = 180

15t = 180

t = 12 seconds

Total Distance, (95+80) t = 175 * 12 (t = 12)

Distance = 2100 kms

Time required by the train starting from A to cover x is x/50 hr

Time taken by the other train starting from B to cover (570 - x) km = (570-x)/80

But the first train has started 1 hr early. So, it has travelled 50 km in this 1 hr.

Therefore, x/50 - 1= (570-x)/80

On Solving, x = 250

So, they will meet at a distance of 250 km from place A

So the time at which they will meet will be (250/50) = 5 hrs (after 6 am)

Hence, they will meet at 11 am.

With stoppages, the average speed reduces by (48-40) = 8 km

Therefore, (8/48)* 60 minutes = 10 minutes

Hence, 10 minutes would be the time per hour the train stops on an average.

Indrayani’s Speed =(d/4) kmph and that of Shatabdi = (d/3.5)kmph

Let t be the time in hrs after Shatabdi has left for Pune, when the two trains meet

Therefore, distance travelled by Shatabdi = (d/3.5)* t

And that of Indrayani =(d/4) * (t-30/60)

The sum of the distances travelled by the two trains = distance between Bombay and Pune = d km

Therefore, (d/3.5)* t +(d/4) * (t-30/60)=d

Solving for t, we get t = 2.1 hrs or 2 hrs and 6 mins

Hence, the two trains meet at 19:06 hrs

Speed = 60 kmph = 60 *(5/18)=(50/3) m/sec

Distance = Speed * Time

450 =(50/3) * Time

Time = 27 seconds