Aptitude - Problems on Trains - Discussion

Please explain sum one in simple method.

Thanks @Nagarjun.

Hi, everyone.

If a train pass any object whose length is negligible (ex:- man, post) we don't need to add d length of d train and d object, so we use speed= (distance/time) formula. Time is specified so (x/y) =8 ->1. If a train passes any entity which can b calculated (ex: bridge, tunnel) we have to add length of d train+length of entity,

Hence (x+264)/y = 20 ->2.

By solving 1 and 2 you will get answer.

Please any one explain why we divide 176/8?

@Priya,

Speed=Distance/Time.
here Distance=length of train + length of bridge.

in this question we have no idea about the length of train and speed of the train.
so let the length of the train be x metres and speed be y m/sec.
then speed of train y = (x+264)/20.

we can't solve because vale of x and y is unknown.
now speed of the train relative to telegraph post
i.e. y = x/8.
comparing the two above equation
x/8 = (x+264)/20
3x = 528
x = 176 which is length of train.
now put the value of x in any of the above two equation.
therefor speed y = 176/8
y = 176/8 *18/5 (because to convert in kmph)
y = 79.2 km/hr

Hi, I didn't understand the solution. Can someone explain step by step ?