Difficult Rate Problem

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by Night reader » Fri Mar 04, 2011 11:18 am
attaching the less attractive pic :)
if train with the length l passes a pole (point) in t seconds, the train's length must be l*t;
the same train passing a platform in 3t seconds defines the length of the platform as l*3t-l*t=2t*l OR 2l with the account of second unit t.

IOM D, 2l
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by vineeshp » Fri Mar 04, 2011 11:20 am
First case.
A train of length L units passes a pole in t seconds. Imagine a pole stationed with the train's front most point at it. When the train's rear end crosses this pole, the front end of the train will have moved a distance equal to the length of this train.
So the speed of the train is total distance / total time = L/t

Second case.
Assume the platform is of length P.
The front end of the train starts at one end of the station. By the time the rear end of the train passes the other end of the station, the front end would have travelled L plus P distance. They say that this took 3t seconds.
So speed of train is total distance / total time = (L+P)/3t

According to them, the train moves at the same speed in both cases.

L/t = (L+P)/3t
Cancelling t and multiplying both sides by 3.
3L = L+P
P=2L

Answer D
Vineesh,
Just telling you what I know and think. I am not the expert. :)