# Are your pupils creative enough to solve this one? How fast are the trains?

I’ve yet to find a pupil that can solve the problem below but what is lovely about this problem is that most pupils can access it and have some ideas about how to solve it. It hooks them but provides a real challenge to solve it. There are many methods you can use to solve the problem and it is lovely to see some pupils using algebra, some using speed = distance / time, some using trial and error, some using speed-time graphs, some using inequalities etc.

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The solution is that one train is twice as fast as the other. Here’s the algebraic proof that I came up with:

*Definitions*

Two trains A and B.

Speed of train A = A

Speed of train B = B

Speed = distance/ time therefore distance = speed X time

T = time from start of train journey until the passing point.

*Distance from London to passing point*

We don’t know the exact distance from London to the passing point but we do know that this is the same for both trains. Using distance = speed X time for each train and the fact that the distances are the same we can say:

1 X A = B X T (i.e. train A is travelling at speed A and has 1 hour to go, train B is travelling at speed B and has been travelling for T hours)

A = BT

*Distance from Liverpool to passing point*

Using the same approach that the distances are the same for both trains (so speed X time for each train is equal):

A X T = 4 X B (i.e. train A is travelling at speed A for T hours and train B is travelling at speed B for 4 hours)

AT = 4B

*Simultaneous equations*

Using:

A = BT

AT = 4B

Rearrange the first formula in terms of T (T=A/B) and substitute into the second formula to give:

A^2 / B = 4B

So:

A^2 / B^2 = 4

Therefore:

A/B = 2

**Train A is travelling at twice the speed of B**

I heard this problem on the TES podcast, and having spent a number of hours trying to solve it, I was a little disappointed with the eventual answer. Am I right in saying that if the total distance is x, the problem only works if the faster train is travelling at exactly (1/3)x miles per hour? E.g. if the distance is 300 miles, the faster train must be travelling at 100mph. At any other speed, the remaining time will not allow the trains to meet at the point where one is 1 hour away and the other is 4 hours away – e.g. taking 300 miles as the distance again, if train A travels at 20mph and train B at 10mph, when train A is 1 hour away (after 14 hours) it will have travelled 280 miles, whereas train B will only have travelled 140 miles. Indeed, by the time train B is 4 hours away, train A will long since have arrived!