Aptitude - Problems on Trains - Discussion

Thank you @Ishita Narula.

Thanks a lot @Sourav Sinha.

Best answer, thanks a lot @Chunky Kumar.

Let speed of train from Howrah to Patna = u.
And speed of train from Patna to Howrah= v.

Now let they meet at a distance x from Howrah and y from Patna after time t.

So t= (x/u ) = (y/ v)
=> x/y = u/v ----------------------------(1)

Now we are given that y/u = 9 --------(A)
and x/v =16 ----------------(B)

Divide B by A. We will get, (x/y) * (u/v ) = 16/9.

From equation (1), replace x/y by u/v,
= > (u/v)2 = 16/9
= > u/v = 4/3.

Lets assume that:

1. They meet at time t.
2. total distance between trains before they start is d.
3. when they meet, one of the trains having speed "v1" has travelled "d1" distance. Hence other train having speed "v2" must have "d-d1" distance traveled.

Explanation:

Using the formula, distance=speed * time.
d1=v1*t -----> (1) for train 1.
d-d1=v2*t -----> (2) for train 2.

By using (1)and (2).
v1/v2=d1/(d-d1) -----> (3).

Now, the second train takes t+16 hours to cover "d" distance and other train take t+9 hours for the same distance.

So,

d=v1*(t+9)-----> (4).
d=v2*(t+16)-----> (5).

For result (4),

d=t*v1+9*v1.
Replacing t*v1 by the result (1).

Then we get,
d=d1+9v1-----> (6).

For result (5),
d=t*v2+16*v2.

Replacing t*v2 by the result (2).
Then we get,
d=d-d1+16*v2-----> (7).

From (6) and (7) we can write,
9v1/16v2 = (d-d1)/(d1)-----> (8).

From (8) and (3) we write that,
9v1/16v2=v2/v1.

Hence:
v1*v1/v2*v2=16/9.
v1/v2=sqrt(16/9).
v1/v2=4/3.

Let a be the speed of train A.
b be the speed of train B.

Time taken by train to meet each other = t.
at is distance covered by A in meeting with train B,
bt is distance covered by B in meeting with train A.
A/Q
bt/a = 9 ----> (1)
at/b = 16 ----> (2).

Solving equation (1) and (2) we get,
a/b = 4/3.

Time to meet the train t=(t1 * t2)^1/2.

Where t1=time take by 1 train to reach at the other ends just after using two train and similarly t2.

Thank you

Ratio of time of A and B =9:16.

Then the ratio of distance is the reverse of the ratio of time because when the time of A is less than B to cover a distance than the speed of A is greater than B so the ratio of the speed of A:B=16:9.

Simply A:B=4:3.

Let the train meet after time t at point O and train from Howrah to Patna travels X1 distance with speed v1 and train from Patna travels X2 distance with speed v2 to reach O.
X2=9 *v1 ; X1= 16*v2
and X1 = t*v1 ; X2= t*v2.
From above equations
16* V2^2=9*V1^2,
V1/V2=4/3.

Howrah(A) ------------------------- Patna(B).

Let, distance between Howrah and Patna be x; S1 = speed of 1st train from A to B; S2 = speed of 2nd train from B to A.

Let, after t hours both trains meet each other.

So, distance covered by 1st train in t hours = S1 * t = distance covered by 2nd train after he meets the 1st train and goes towards station A.

So, S1 * t = S2 * 16 Or, S1/S2 = 16/t--> (I)

Again, distance covered by 2nd train in t hours = S2 * t = distance covered by 1st train after he meets the 2nd train and goes towards station B.

So, S2 * t = S1 * 9 Or, S1/S2 = t/9--> (II)
From equations (I) and (II), we get,
16/t = t/9
=> t^2 = 9*16 = 144
=>t = 12 hours ( as t is time hence -12 is neglected).

Hence, ratio of speeds,
S1/S2 = 12/9 = 4/3 (putting t=12 in equation(II)).
Or, S1 : S2 = 4 : 3.