Two trains $A$ and $B$ start from station $X$ and $Y$ towards each other. $B$ leaves station $Y$ half an hour after train $A$ leaves station $X$. Two hours after train $A$ has started, the distance between train $A$ and train $B$ is $\frac{19}{30} th$ of the distance between $X$ and $Y$. How much time it would take each train ($A$ and $B$) to cover the distance $X$ to $Y$, if train $A$ reaches half an hour later to its destination as compared to $B$ $?$
My solution approach :-
Let the distance between $X$ and $Y$ be $x$.
Let the speed of train $A$ be $a$ kmph and of train $B$ be $b$ kmph.
As per question $2a + 1.5b = \frac{11x}{30}$ –Eq.(i) (Distance travelled by them i.e. Total distance $-$ Distance left between them $= x-\frac{19x}{30}$
Now we know that train $A$ reaches half an hour later to its destination as compared to $B$, so:-
$x/b + 0.5 = x/a$ –Eq.(ii)
I am stuck here as you can see that I have got three variables and just two equations I can form from the question. What am I missing here? Please help!
Best Answer
You can simplify the working. Say, time taken by $B$ to cover distance $d$ between stations $X$ and $Y$ is $t$ hours. Then time taken by $A$ is $(t+1)$ hours (as $A$ starts $30$ mins earlier and reaches $30$ mins later) and speed of train $A$ is $\displaystyle \frac{d}{t+1}$ and of train $B$ is $\displaystyle \frac{d}{t}$.
So, $\displaystyle \frac{2d}{t+1} + \frac{1.5 d}{t} = \frac{11d}{30}$
Take out $d$ from both sides and solve for $t$ which comes to $9$ hours. That is time taken by train $B$. So time taken by $A$ is $10$ hours.
Note: While the question most likely meant that they have not crossed each other but it should have been more explicit. They can be at a distance of $\frac{19d}{30}$ even after having crossed each other, which is represented by the equation $\displaystyle \frac{2d}{t+1} + \frac{1.5 d}{t} = \frac{49d}{30}$ and it does have a valid solution.