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Rocco's Race

Age 7 to 11 Short Challenge Level:

Hmmm, let's see if we can figure out the solution by seeing how you might solve the problem. There seemed to be two ways that this problem was tackled. One way was to use manipulatives - tools - to help discover the number of people who took part and Rocco's position in the race.

Joanne and Holly from Moorfield Primary School used manipulatives. Here they explain what they used and how:

We started of by using multi-link cubes. We arranged the cubes so there was an odd number and Rocco was one before the middle. We tried it first with $5$ cubes but that didn't work. We kept adding on $2$ so that the number stayed as an odd number. We kept it as an odd because if you put one runner from behind in front, Rocco will be in the middle. Seven didn't work so we tried $9$. This did work so everything fitted in to place. This showed that Rocco came $4$th.

Someone else said:

We started off with three people in the race but it didn't work because there can't be three times as many people behind Rocco as in front of him. Then we tried ten but that wouldn't work because ten is an even number and Rocco can't fit in the middle of them. Then we said $9$ and it was right because it is an odd number and he can fit in the middle and there were $6$ behind him and $2$ in front of him so he finished $4$th out of $9$ people.

The other strategy used was 'guess and check'. This can be combined with using manipulatives. You have to have a number to start with that seems reasonable. Let's see how some people used 'guess and check'.

Matthew and Steven of Moorfield Junior School explain:

We started off with $5$ people but it didn't work so we then tried $15$. That didn't work because there were $6$ people in front of him and $8$ behind him. Then we tried $9$ and it worked. When you move $1$ in front of him it made him in the middle and when you move one behind him it makes $2$ in front and $6$ behind, $2 \times3 = 6$. Rocco came $4$th.

Robin, also of Moorfield Junior School, did a great job in showing how you can combine 'guess and check' with drawing a diagram to help you thinking through the problem. Robin writes:

I knew that it had to be an odd number if Rocco was to finish in the middle of the race to make the same number of runners in front as behind. Starting at five we placed Rocco second in the race but this didn't work because when a runner finished after Rocco the amount of runners behind was not $3$ times the runners in front.

Then we tried $9$ runners, it would look like this: - - - - R - - - - -

So based on these diagrams, Rocco starts off in fourth place

If one more runner finishes before him it would look like this: - - - - R- - - -
and in this case there is an equal number of runners in front and behind.

If one runner finished behind him, however, it looks like this: - - R - -- - - -

Rocco has $2$ people in front of him and six people behind. $6$ is three times $2$.

At the finishing line, Rocco came fourth in the race.

Thomas here did so well, despite being frustrated by the computer - perhaps it was the font you chose Thomas!

Thomas's solution.