Aptitude - Problems on Trains - Discussion

Relative speed of any two object say x km/h and why km/h moving in the same direction would always be (x-y) km/h.

Here the speed of the train is related to the man.

Hence relative speed = 125/10 m/s = 25/2 m/s.

The speed of the man is 5 km/h. When you convert it for m/s the speed will be (5x5/18) m/s = 25/18 m/s.

We know that relative speed,

R.S = (x-y) m/s or km/h.

Here x (unknown) , y= 25/18 m/s and R.S = 25/2 m/s.

Substituting we get,

X = (25/2)+(25/18).
X = 250/18 m/s (or) (250/18) x (18/5) km/h.

So X = 50 km/h.

Therefore the speed of the train will be 50 km/h. Hope you got it!.

Why is the distance covered by man in 10 sec not considered? I believe the total distance would be the length of train + distance covered by man in 10 sec. Considering this the relative speed of train comes out to 55 km/hr. Please suggest.

Let the speed of train= x km/hr.

So relativee speed = distance/time.

Relative speed of train = (x- 5) km/hr.

x-5 = 0.125/10/3600.

So x- 5 = 0.125x3600/10 = 45.

x = 45+5 = 50 km/hr answer.

In 10 sec man will cover 125/9 m. In the relative motion distance should be added in either cases (opposite direction or same direction). So relative speed should be (125+(125/9))/10. Please explain.

Nice trick.

I made a relation like this (125m/v1-5*5/18) =10 sec. But the answer is not coming ? I converted 5 km/hr to m/s. Then I put it in above equation.

I put all the terms in m and sec (not in km and hour). My answer is coming to be 30. How is it? Help me out

This concept is relative to distance not to time so you cannot take a relation like you have mentioned.

Correct your equation to 125/v1-(10*5)/(5*18) = 10 you will get.

v1 = 13.88 m/s.
==> v1 = 50 kmph.

In first problem you mentioned that meters per sec is equal to 5/18 but then you are taking like 18/5 why so please tell me.

Well,

10 sec ----125 mt.

36 sec----- = (36/10)*125 = 450 mt = 450/10 = 45 + 5 = 50 km/hr.