well i came up with a solution for this which is feel is maybe right but i dont get the ans as 72.

the process is as follows:

assuming both the boys walk with same speed i.e. Sb and speed of train as St and lenght of the train be 'l'

1st boy : l/(St-Sb) 20 secs and 2nd boy : l/(St+Sb) 18 secs

solving above eqtns we get Sb 1/19 St

now time taken by two boys to meet distance travelled by train in 10 mins / Sb+Sb

10 * 60 secs * St / 2 Sb

substituting Sb 1/19 St we get answer as 95 secs

Now i m not sure if i made any calculation mistake or some error in considering values but thts wat i get as answer.

thanks buddy..

theres another question coming ur way...

A Boy has taken a short cut on which there is a railway track that goes through a narrow bridge. When the boy covered 7/16^th of the track in the bridge he heard the whistle of the train. Then he ran out of that bridge when the train came to that bridge. If he were taken actual root he would have reached the destination on time rather taking the shortcut and running out of that bridge. If the boy covers 4km/hr what was the speed of the train?

95 mins is right.

He actually got it right.

distance covered by the train in 10 mins

10*60*St

and time for the boys (this distance / 2Sb) sec

i.e. 600St/2Sb sec (60 * 5St/Sb) sec (60*95) sec 95 min

Thanks...

Priyankar

(Ghutuda)

dear mr friends .

u all forgot one thing........by the time the train reaches the other boy in 10 mins...........the first person l also ve moved some distance which 10*60*x..

the solution s

x be speed of train and a be the boys........

so l/(x-a) 20 l/(x+a) 18 therefore <x+a>/<x-a> 10/9;

x 19a

now the distance is 60*!0*x 600*19a.

but the actual distance between the boys l be 600*19a-600a {as the 1st boy would ve moved some distance }

so 600*18a...........(600*18a)/(a+a) 600*9 sec 90 mins..........

note if the second boy s moving also n same direction then we ve to take the second boy relative distance also nto account......which is important.thank u enjoy....