Aptitude - Problems on Trains - Discussion

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

I am not getting this. Anyone, help me to understand this.

How it is 50? please explain to me.

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.

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.

How answer is 50 km/hr? Please explain to me.

Sp of man = 5 * 5/18
= 25/18 m/sec.

Let Sp of train be x, then;
x - 25/18 = 125/10
So x = 125/10 + 25/18 = 125/9 m/sec.
So x = 125/9 * 18/5 =50 Km/h.

@All.

Why do we minus the speed of the man moving in the same direction? Please explain to me.

Consider we need to convert m/sec to km/hr..
If we need to convert something from m to km, 1km=1000m. Thus 1m= (1/1000)km
So we multiply the numerator by 1/1000.

Similarly, (denominator part) for a sec to hr, 1hr=3600sec. Thus 1sec=(1/3600)hr.
So we multiply the denominator by 1/3600.

This conversion part becomes 3600/1000 thus simplifying this gives an std form of 18/5.. which is always directly used in converting m/sec conversion to km/hr.. helps in just multiplying this 18/5, instead of separate conversion of numerator and denominator to km and hr separately which is time-consuming..

Similarly, the reciprocal of this is just the vice versa case from km/hr to m/sec just multiply with 5/18.

According to me, it should be 25/2 × 18/5 = 45.