Aptitude - Problems on Trains - Discussion

@Randhir

In the problem what you have solved above,the distance between point A and B is 6Y. I am not getting how you got 6Y. Can you please explain me that?

@Faisel,

If they meet 8am then

Distance covered by b in xhrs is 25x
Distance covered by a in(X+1)HRS=20(X+1)
(x+1) coz v r considering 1 hr aft train A has started..

So,
25X+20(X+1)=110
45X=90
X=2

Hence 8am + 2 = 10am

what if we say they meet after 8AM
Distance covered by B in x Hrs =25X
Distance covered by A in (x-1)Hrs = 20(x-1)
25x+20(x-1)=110
25x+20x-20=110
25x+20x=130
45x=130
x= 2.8 ( Which in incorrect how come )

Hi Erererer,
It was assumed that trains r meeting after x hours,
so 5 pm(starting time) + 3.20 = 8.20pm


Thanks Randhir.....

X=3 how it convet to 10 am

hi Easwar,
The speed of one train is twice the other so let
speed of 1st train is Y.
speed of 2nd train is 2Y.
then distance between point A and B is 6Y.

Suppose they meet x hours after 5 a.m.

Distance covered by 1st train in X hours = YX km.

Distance covered by B in (X - 2) hours = 2Y(X - 2) km.

Therefore YX + 2Y(X - 2) = 6Y

=> YX + 2YX - 4Y = 6Y

=> 3YX = 10Y

=> X =10 / 3 (in hours)
=> X =10/3*60 (because converting into minutes)
=> X =200 minutes
200 minutes means 3 hour 20 minutes

So, they meet at 8 : 20 p.m.

A train leaves a point A at 5 p.m. and reaches another point B a 11 p.m. Another train leaves point B at 7 p.m. and reaches point A at 10 p.m. At what time will the two trains meet?
8 : 15 p.m.
8 : 30 p.m.
8 : 20 p.m.
8 : 00 p.m.
can anyone solve this problem

Though the train B starts 1 hour late than train A, The speed of the train varies.

Hi Shri,
We can write here
(X-1) b'coz train 'B' starts travelling 1 hr later agains train 'A'.
So, At the end, train B will travel 1 hr short, out of full running time to meet each other. i.e 'X'
So, we can write total time for train 'B' = (X-1)
Thanks

Why (x-1) term is used ?