Aptitude - Problems on Trains - Discussion

Let's assume the speed of train A = x
Speed of train B = y
Trains A& B meet after 't' hours.

(Distance traveled by train A in 't' hours(distance travelled=speed*time=x*t) will be the distance train B going to travel in 16 hours after they meet(speed*time=y*16), in the same way, the distance traveled by train B in t hours(y*t) will be the distance train A going to travel in 9 hours after the meet(9*x).)
from the above statement.

xt=16y ---> equation1
yt=9x ---> equation2

From eq 2,
t=9x/y ----> put this in eq 1
x*9x/y = 16y.
9x^2 = 16y^2.
x^2 / y^2 = 16/9.
Taking the root in both sides of the equation.
x/y = 4/3.
x:y = 4:3.

Let v be the speed of train 1 (H to P) and u be the speed of train 2(P to H).
Time taken by both trains to meet is equal; so,
x/v = y/u.

Also,
9 = y/v and 16 = x/u.
=> y=9v and x=16u.
put in first eqn;
16u/v=9v/u.
(v/u)^2 = 16/9.
v/u = 4/3.

That is nice. Thanks all for the explanation.

@All.

We know that speed =distance/time.
If the ratio of the speeds of A and B is a : b, then the ratio of
The times taken by them to cover the same distance is 1/a : 1/b or b: a. Because the distance travelled is the same.

Very helpful. Thanks all.

Very helpful, Thanks for all your explanations.

Thanks for very simple answer @Ajinkya.

Consider first train speed v' & other v"

After meeting point v'=(meeting point to patna)/9 = b/9.
v"=(meeting point to howrah)/16 =a/16
v'/v"=[b/a][16/9] ------> (1)
when train start and meet each other
time taken is same ie. T.

v'= (howrah to meeting point)/T = a/T.
v"=(patna to meeting point)/T = b/T.
then from above eq. v'/v"=a/b -----> (2)put in eq. 1.
From eq. 1;
v'/v" = (v"/v')[16/9].
v'^2/v"^2 = 16/9.
v'/v" = 4/3.

Let us take both trains to meet after time t.

And the speed of train A and B as x and y.

So now whatever distance covered by train A in time t is covered by train B in 16 hours.
So x*t = y*16 --> eq(1)

Similarly whatever distance covered by train B in time t is covered by train A in 9 hours.
so y*t = x*4 --> eq(2).

From eq(1) and eq(2),
We get (x^2 / y^2) =( 16/4).

Let say the distance between them is d, and it takes time t when both the trains meet each other.

So relative speed = v1+v2.

So t = d/ (v1+v2) -----> (1).

Now for the train 1 distance remains = d - t*v1.
For the train 2 distance remains = d-t*v2.

According to the question, it takes 9 hrs for the first train and 16 hrs for the second train.
So, 9 = (d-t*v1) /v1 and 16 = (d-t*v2) /v2.

Now from the equation 1, d = t*v1+t*v2 substitue above.
So 9 = t*v2/v1 and 16 = t*v1/v2.

Equating t, 9v1/v2 = 16v2/v1 so v1/v2 = 4/3.