Aptitude - Problems on Trains - Discussion

125/10 = 12.5,
12.5 * 3.6 = 45,
45 + 5 = 50.

3600sec then, 1sec = 1/3600hr.
1km =1000m then, 1m = 1/1000km.
So,
km/hr to m/s is : 1/1000km bigger divided by 1/3600hr.
So, they put 18/5.

Given man's speed = 5km per hour.
The train length is 125m.
The train crosses man in 10 seconds.

It means both the train and the man are moving in the same direction.

The speed related to man is 125m/10sec, which is 12.5m/s.

{convert this into km/hr:- [basic idea:-km/hr to m/s or viceversa;
we know that:1km=1000m, 1hr = 3600sec,
1km/hr to m/s:-1000/3600 = 5/18m/s.
For n*km/hr to m/s = n * (5/18)m/s.
And for viceversa(m/s to km/hr) :n*(18/5)km/hr}.

Now, convert 12.5m/s into km/hr:-12.5*18/5 = 45km/hr.
This is nothing but relative speed, according to man, is 45km/hr.

In basic;

Relative speed = |speed of object1 - speed of object2|(if in the same direction).
Relative speed = |speed of object1 + speed of object2|(if in opposite direction).

The problem is in the same direction,
So;
45 = speed of train - speed of man.
Therefore, the speed of the train = 45 + 5 = 50km/hr.

@Hemanth sairam.

Why 60 mins and 60 seconds taken there?

Please explain me.

How - 5 came? Please explain this step.

How this 18/5 come? please explain to me.

How 18/5 came after 25/2? Please answer.

How it 18÷5 will become? Please explain.

18/5 is the direct formula to convert the m/seconds to km/hr.

In this given problem, we can convert the m/sec to km/hr so we use 18/5.

This is the concept of reference of frame.

Train speed is (x)
Let's say A person in a train says that a man(outside of the train(or) beside of train running at a speed of 5 km/ hr) running slow as (x-5)km/hr.

This means if the train speed is 20 km/hr. Then the man's relative speed is as slow as 15 km/hr when compared to train speed.
This relation is mentioned in the mathematical formula as( x-5)km/hr or ( example:20-5 km/hr).
So, the relative speed is;
X - 5 = (120/10)*(18/5)
So x = 50 km/ hr.
*18/5 is used to convert m/s into km/ hr.