Aptitude - Problems on Trains - Discussion

Hi, well in this question. We need the location where the man is sitting the slower train. So that we can find the distance which the faster train has to travel to cross the man sitting in slower train.

The solution to this assumes that the man is sitting at the end of the train. So in this case, length of slower train does not matter.

But assume if the man is sitting in middle of slower train, then faster train has to travel, it's own length plus half the length of slower train.

@Dwivedi.

Here what they asked is, the length of the faster train.

If you want the length generally we multiply speed and time.

In this question, you think that you are in a slower train it's speed is 20kmph in which ever the place either in front middle or in last your speed is 20kmph. The faster train has to cross you so the speed of train relative to you is 20kmph or 50/9mps.

Length of the faster train = relative speed in mps*time in sec = 50/9 * 5 = 250/9 = 27 * 7/9.
.

The question focus on the fast train and the man sitting somewhere in the train.

Here we can consider the slow train as a man running with 20km/hr.

So now the question can be understood in this way as well:.

There is a train running at speed of 40km/hr and a man running with 20km/hr in the same direction of the train. Calculate the length of the train?

Now solve it. Hope now it will become easy for everyone.

@Siddarth.

You don't need to divide 250/9 completely because we don't need the answer in decimals according to the options.

You should divide 250 by 9 until you cannot divide it further without the need of a decimal i.e. when the quotient is 27 and the remainder is 7.

And then you can write it in the form of a mixed fraction i.e. quotient remainder/divisor or 27 7/9 as given in option C.

1)First calculate the relative speed of two trains ( as they are moving in same direction it will be difference of two train speeds).

Relative speed = 40-20 = 20 km/hr = 20*5/18 =50/9 m/s

2)Now faster crosses the man sitting in the slower one (irrespective of the position of man it may be starting,middle or last) now use the formula

length of train (faster) = speed X time = 50/9*5 = 27 7/9 m

This question definitely depends upon where the person is sitting in the second train - middle, end or int engine of the train depending on which the total time taken by the fast train to pass him completely will change. And the solution given is correct only if the person is sitting at the end of the slower train. Can someone explain this, please?

The relative speed b/w the train is 20kmph. It is easier to understand if you look in this way-- The faster train is moving at a speed of 20kmph and the slower train is 0kmph i.e. static. So now since the person sitting inside the train is also static we can simply treat this problem like "the time taken by a train to cross a pole" type.

Simple method.

Speed =distance/time.
Relative speed (if same direction)=s1-s2.
Consider usefull formula s1-s2=(l1+l2)/time.
Here l1= trian 1 length= x meters.
L2= 0 meters ( person who is sitting in train),
Time taken to cross person is t=5 sec,
40-20 km/h = x/5,
20km/h =x/5,
20*5/18 =x/5,
(50/9)*5 =x,
250/9=x,
27.77 => answer.

Use the concept of relative speed i.e. slower train is fixed and the faster train is moving with the speed of (40-20)km/hr=20km/hr. Then the men sitting in the slower train is also a stationary one wrt. faster train .then simply use the formula of a train moving with a velocity of 20 km/hr to pass a stationary object in 5 seconds.

The thing is that when we find the relative speed, it means how their respective speeds change their positions and how their distance is varying, so with the help of the relative speed and the 5 second time which is the crossing time, for the tenure of 5 seconds we can find the length of the fastest train.