BLY shared this Question and a solution using Algebra. Even though the question asked for an algebraic approach I wanted to look at other methods for solving.

I liked BLYs algebraic method

but I wanted to visualise the problem so used the knowledge that if the lorry is 12mph slower than the car and took an hour longer then you could plot two speed-time graphs. One T=240/x where x is the speed of the car and 240/(x-12) then look for the value on the graph where the T values are 1 hour apart

So I can see the red graph (240/x) is at 60, 4 and the blue is at 60,5 which is an hour apart. So the cars speed was 60mph.

The second method used a bit of trial and errors. I made x represent the speed of the lorry and so 5x was the distance over 5 hours and that gave me 240

If the lorry travels at a 12mph slower speed but takes an hour longer then the speed of the car would need to equal 12mph x how many hours the lorry travelled for.

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