# Time, Speed and Distance : Trains Partial Distance

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!

• 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. Mar 12 '21 at 14:49
• 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. Mar 12 '21 at 15:01
• 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... Mar 12 '21 at 15:43

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.

• 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? Mar 13 '21 at 3:07
• Also when i solved the equation the solution came out to be t = 1.6872 hours. Mar 13 '21 at 3:17
• i got it... 1 + 19d/30 = 49d/30... Mar 13 '21 at 3:25
• Yes, B takes 1.6872 hours and A takes 2.6872 hours is another solution. Mar 13 '21 at 4:17
• 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. Mar 13 '21 at 4:23

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 $$z=9$$.

• 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. Mar 12 '21 at 16:14
• @MathLover That's a good point that never occurred to me. Mar 12 '21 at 16:25