nilesh

• Feb 20th, 2006

 

 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.

infytak

 Profile Answers by infytak Questions by infytak

• Feb 20th, 2006

 

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?

Manas

• Feb 28th, 2006

 

if the train took 10 mins to reach the 2nd boy, and the boys walk 1/19 the speed of the train, then the boys will take somewhere around half of 10*19 mins????therefore around 95 mins?how the does the answer come in seconds?

Priyankar Ghosh

• Mar 7th, 2006

 

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)

arvind

• Apr 3rd, 2006

 

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....

bijoy joseph

• Apr 11th, 2006

 

Hai freinds, i have got another soln d/(Vtrain-Vman)=20 sec therefor d=20*(vtrain-vman) |||y d=18*(vtrain+vman) so vtrain=19vman now in 10min train travels 10*19vman distance since both men r moving to each other there relative speed is 2vman Hence time taken to cover the distance=10*19v/(2v)=19*5=95

NARENDER SINGH

• Jun 30th, 2006

 

check this:letspeed of boy=bspeed of train=tx be length of traincase 1: time=20=x/(t-b)..............................(I)case 2: time=18=x/(t+b)..............................(II)div (I) by (II),and solve we get t=19b................................................(III) DISTANCE BETWEEN 2 BOYS WHEN TRAIN MEETS 2 BOY = 10*60*t - b*10*60 =10*60*19*b - b*10*60 =10*60*18*bBUT WHEN TRAIN PASSES 2 BOY THEN DISTANCE BETWEEN 2 BOYS: =10*60*18*b - 2*b*18 =18*b*(600-2)=18*b*598TIME TO MEET: =18*b*598/(2*b) =5382 sec or 5382/60 =89.7 min