Two trains problem I have been doing these train problems lately and I am stuck at a problem that I can't figure out.


*

*Two trains start moving at the same time towards each other from point $\mathbf{A}$ and $\mathbf{B}$, respectively.

*They meet at a point and after that one reaches the dest in $1$ hour and the other takes $4$ hrs.

*How fast is one train compared to the other ?.


Note that one is leaving from $\mathbf{A}$ towards $\mathbf{B}$ and the other from $\mathbf{B}$ towards $\mathbf{A}$.
 A: The problem can be solved to find the ratio between the speeds of the trains.
Let the first train have speed $v_1$ and let it take $1$ hour to reach the other terminus after the meeting. Let the second train have speed $v_2$ and let it take $4$ hours to reach the other terminus after the meeting.
The distance travelled by train $1$ to the meeting point is therefore $4v_2$, and that travelled by train $2$ to the same point is $v_1$.
The time taken to reach that meeting point from their respective starting points is the same.
Hence, $\displaystyle \frac{4v_2}{v_1} = \frac{v_1}{v_2}$, which gives us $v_1 = 2v_2$.
So one train is twice as fast as the other.
A: Here is how I would solve the question. Let the speeds of the trains be $v_1$, $v_2$, and suppose they met t hours after starting. Let train 2 reach 1 hour later, and train 1 reach 4 hours later.
Then, distance covered by train 1 in time $t$ hrs=distance covered by train $2$ in time 1 hr
And,
distance covered by train 2 in time $t$ hrs=distance covered by train $1$ in time 4 hrs
Hence, $v_1t=v_2*1 $
and $v_2t=v_1*4 $
Hence, dividing LHS by LHS and RHS by RHS, $\frac{v_1}{v_2}=\frac{v_2}{4v_1}$
$\frac{v_1}{v_2}=\frac{1}{2}$
A: If one train takes one hour to reach a destination and the other takes four, the relative speed is simply 4:1; one train is four times as fast as the other.
