Two trains run in opposite directions on a circular track.

The trains will travel the perimeter of the circle which is 12π miles. The relative speed of the trains is 2π miles per hour. Time taken to meet = 12π/2π = 6 hours

In cases like this I find it best to just calculate at what times
each train will reach point S and take the first common time.

a: 3 6 9
b: 2 4 6

so we can see that the first time they will meet back at point S is 6

Just wanted to clarify .I think some of the pals are confused about the relative speed concepts.

When bodies move in the same direction ,there relative speeds must be subtracted.
When bodies move in the opposite direction their relative speeds must be added.

Some of them have applied it wrongly .It can lead to errors in the exam .

Sorry but move in the same direction is not the same to say: move toward each other and eventually crash ?? and in this case the relative speed is not the sum of the respective rates ??

here we do not have the situation move toward but meet at some point after rouded a circle.

Please some expert can clarify this situation of relative speed toward and relative speed in the problem at end ???

Thanks

Bunuel,

Shouldn't we directly add the relative speeds of train A and B? Is circular approach different from linear one? Please explain

I had the same doubt as above.
Shouldn't we add the relative speed?

But the first post mentions that the relative speed is subtracted ?

Yes, when two objects run in opposite directions, their relative speed is given by adding the two speeds. But think what will \(12\pi/(4\pi + 6\pi)\) give us. It will give us the time taken by the two of them to complete one circle together. They will meet somewhere on the circle but not at the starting point in this much time.

For the two objects to meet again at the starting point, one object must complete one full circle more than the other object. In every hour, the train B - that runs at a speed of \(6\pi\) - covers \(2\pi\) extra miles compared with train A which runs at a speed of \(4\pi\). We want train B to complete one full circle more than train A. In how much time will train B cover \(12\pi\) (a full circle) more than train A? In \(12\pi/ 2\pi\) hrs = 6 hrs.

Or another way to think about it is this:

Time taken by train A to complete one full circle \(= 12\pi/4\pi = 3\) hrs
Time taken by train B to complete one full circle \(= 12\pi/6\pi = 2\) hrs

So every 3 hrs train A is at S and every 2 hrs train B is at S. When will they both be together at S?
train A at S -> 3 hrs, 6 hrs, 9 hrs
train B at S -> 2 hrs, 4 hrs, 6 hrs, 8 hrs

In 6 hrs - the first common time (the LCM of 3 and 2)

Answer (B)­

Another thing - this method of dividing the distance by difference of the speed will not work in all cases.

Try the same question with speed of the trains as \(3\pi\) and \(5\pi\) and radius of the circle as 7.5 miles. The two methods give different answers - figure out why!­

Karishma,

The right method should be finding the LCM only for the trains have to meet at the same point. In the 3 Pi & 5 Pi problem, the trains will meet at the same point in 15 hrs.

Adding speeds can give the time at which they meet but not at the same meet e.g. in the 4 Pi & 6 Pi problem, the trains will first meet 1.2 hrs from the starting time i.e the 4 Pi train would have traveled 4.8Pi km & the 6Pi tarin would have traveled 7.2Pi kms [the total distance is 12Pi]


[Only if the trains travel in the same direction, the speeds should be subtracted...and again only the catch up time can be calculated from this. To get the time at which they will meet at the same point, LCM is the route]

??

Try the same question with speed of the trains as 3\pi and 5\pi and radius of the circle as 7.5 miles. The two methods give different answers - figure out why!

Method 1

Ratio of speed of Train A and Train B is 3:5.
Considering the circle as 8 units, the 1st meeting point is 3 units away from starting point.
For A & B to meet first time at starting point, we find lowest common multiple of 3 and 8 i.e. 24 units; thus, they must meet (24 units/3 units) 8 times.
Time for each meet = \((\frac{15}{8})hrs\)
Or time to meet 8th time=\((\frac{15}{8})*8=15 hrs.\)

Method 2

For the two objects to meet again at the starting point, one object must complete one full circle more than the other object .
However, in order to satisfy the meeting point to be same as starting point, it is necessary that the difference of distance covered is a exact multiple of full circular distance.
Relative speed per hr =\(2\pi\)
Total circular distance =\(15\pi\)
LCM of relative speed and circular distance =\(30\pi\)
Or, Time to meet at starting point = \(\frac{30\pi}{2\pi per hr} =15 hrs\)

The method of "dividing the distance by difference of the speed" is not necessarily wrong. In the \(3\pi\) and \(5\pi\) problem, the circumference of the circle is \(15\pi\). Train B gets ahead of train A by \(2\pi\) every hour. After 7.5 hours, it is \(15\pi\) ahead but it is not at S at that time because it reaches S in only integral hours. So train B needs to complete 2 full circles more than train A which it will do in 15 hours. In 15 hours, both trains will be at S.

When two trains are running in opposite direction, the relative speeds are added . Am I correct here ?

kshitij89 , yes, you are correct.

VeritasPrepKarishma

here:

https://gmatclub.com/forum/two-trains-run-in-opposite-directions-on-a-circular-track-132630.html#p1417394

See this post, too:

https://gmatclub.com/forum/two-trains-run-in-opposite-directions-on-a-circular-track-132630.html#p1418111

Maybe re-read the whole thread? Rates aren't the best method to solve this problem.

Hope that helps.

Sorry for dragging up an old post, but is this solution right? The objects are moving in opposite direction so relative speed should be sum of individual speeds right? Can relative speed be applied to solve such a problem?

EMPOWERgmatRichC VeritasKarishma ScottTargetTestPrep

This solution is correct and you are right too.

Yes, since the trains are moving in opposite directions, their "relative speed" will be sum of their speeds. But the concept of relative speed does not help us here because the trains need to meet back at point S. We can find the the trains will cover 1 full round together in 12π/10π hrs but they will not meet at S in this case.

The best way is to figure at what time each train comes to S and the common point when they both will be at S.

What is done in this solution is this:

Assume the trains are moving in same direction. Then in 1 hr, they will create a gap of 2π miles between them. In 2 hrs, they will create a gap of 4π miles between them and so on till in 6 hrs, they create a gap of 12π miles between them. So they would create a gap of one full circle and would both be back at S. In this time, one train would have made 2 full rounds and the other would have made 3 full rounds. Now think - does it matter in which direction the trains were moving?

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