Aptitude - Problems on Trains - Discussion

Here in this question, one train had reached in 9 hrs and another one in 16 hrs.The first train reached faster than the second one ,clearly it should have more speed than the second one, as we know that time and speed are inversely proportional and comparing all answers, 4:3 is most optimal.

Let's say.

Train A travels 'x' is the distance from Patna with speed 'V1' till the point of meet and
train B travels 'y' is the distance from Patna with speed 'V2' till the point of the meet.

the time taken for them to meet 't' = x/V1= y/V2 ---- (1)

Now Train A will travel 'y' distance that was previously travelled by train 'B' and vice-versa.

As per given info;
y/V1 = 9 and x/V2 = 16.

Taking ratio:
y/V1*V2/x = 9/16.

but from (1) -----> y/x = V1/V2,

=> ( V1/V2)^2 = 9/16.
=> V1:V2 = 3:4.

Thanks all for explaining this.

Why are getting the square root. Can anyone explain to me?

Solving the eqns..
D=(v1+v2)t ---------> 1
D=(t+9)v1---------> 2
D=(t+16)v2---------> 3.

Putting the value of t from eq 1 in eq 2 & 3 then taking all the term having D on one side and then taking ratio of eq.2 & 3.

And we will get v1/v2=√16/√9= 4/3.

The correct answer is 4/3 because in the question both trains meets with each other before the now covered their remaining distance in 9hr and 16hrs respectively.

Speed = Distance/Time; Then how will you write root of time?

How can you calculate speed as the square-root of time?

It is formula, see below for important formulas.

Problems on Trains - Important Formulas

Can someone prove that?