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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!

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    $\begingroup$ The question asks for the values of x/a and x/b. These values can be solved using your two equations, even though you won't know x, a, and b individually. So make a change of variables to r=x/a and s=x/b and try to solve r and s. $\endgroup$ Mar 12 '21 at 14:49
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    $\begingroup$ Hint: B started a half hour late, and finished a half hour early. Therefore, B took exactly 1 hour less than A to cover the same distance. $\endgroup$ Mar 12 '21 at 15:01
  • $\begingroup$ ohhhk....such a silly mistake i did with the 2nd equation..and also I was trying to figure out the third equation in order to solve the quations.....i got it now...maybe that is what happens when you solve math questions for 5 hours straight..i should take a break now...thanks for all the help from everyone... $\endgroup$
    – Ganit
    Mar 12 '21 at 15:43
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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.

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  • $\begingroup$ yeah...that could be a scenario too...that thought never came to my mind... if you don't mind..can you please explain a little of getting the total distance travelled in this scenario is 49d/30? $\endgroup$
    – Ganit
    Mar 13 '21 at 3:07
  • $\begingroup$ Also when i solved the equation the solution came out to be t = 1.6872 hours. $\endgroup$
    – Ganit
    Mar 13 '21 at 3:17
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    $\begingroup$ i got it... 1 + 19d/30 = 49d/30... $\endgroup$
    – Ganit
    Mar 13 '21 at 3:25
  • $\begingroup$ Yes, B takes 1.6872 hours and A takes 2.6872 hours is another solution. $\endgroup$
    – Math Lover
    Mar 13 '21 at 4:17
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    $\begingroup$ As you said it is d + 19d /30. When they meet, they have together covered distance d and then they together cover 19d/30 as they are 19d/30 apart. $\endgroup$
    – Math Lover
    Mar 13 '21 at 4:23
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As noted in a comment, your second equation is incorrect. B started half an hour earlier than A, and arrived half an hour sooner, so took one hour less to cover the distance. We have two equations:

$$ \begin{align}\frac{11}{30}x &= 2a+1.5b\\ \frac xa &= 1+\frac xb \end{align}$$ The point you have missed is that we are not asked to find $a,b,$ and $x$ but $\frac xa$ and $\frac xb$. If we write $y=\frac xa,\ z=\frac xb$ then the equations become $$\begin{align} \frac{11}{30}&=\frac2y+\frac{1.5}{z}\\ y&=1+z \end{align}$$ Substituting the second in the first, clearing denominators and simplifying gives $$11z^2-94z-45=0$$ whose only positive root is $z=9$.

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    $\begingroup$ Based on how the question reads, can we confidently eliminate the case where they have crossed each other and are at a distance of $\frac{19x}{31}$ from each other? That gives a valid solution as well. $\endgroup$
    – Math Lover
    Mar 12 '21 at 16:14
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    $\begingroup$ @MathLover That's a good point that never occurred to me. $\endgroup$
    – saulspatz
    Mar 12 '21 at 16:25

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