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Multilink Cubes

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This Pied Piper of Hamelin

Age 7 to 11 Challenge Level:
We had a good number of solutions sent in, thank you.  Here we will feature those of you who've looked at many possibilities. From George at Linton Heights Junior School
 we had the following:

a) $0$ rats and $300$ children
b) $1$ rat and $298$ children
c) $5$ rats and $290$ children
d) $10$ rats and $280$ children
e) $25$ rats and $250$ children
f) $50$ rats and $200$ children
g) $100$ rats and $100$ children
h) $125$ rats and $50$ children
i) $150$ rats and $0$ children 

From Patrick at Manorcroft Primary School we had this good explanation, well done:

There are $148$ different combinations of child and rat because I figured out that you could replace $1$ rat with $2$ children (because one rat has twice as many legs as a child) so the maximum possible children to rats are $298$ children to $1$ rat and the maximum possible rats to children is $149$ rats to $2$ children. If you take $1$ from $149$ to get $148$ possiblities.

Year $5
$ pupils from St Ambrose Catholic Primary School said;

There are many possible answers to this question, to find out how many children, you can start with the number of rats. You can go from $1$ rat to $149$ rats to work out how many children.  The rule is: multiply by $4$, take away from $600$, divide by $2$. 
For example, $1$ rat= $298$ children because $1$x$4=4 600-4=596, 596$ divided by $2=298$.
If we start with the number of children the rule is: multiply by $2$, take away from $600$, divide by $4$.
For example, $2$ children $ = 149$ rats because $2$x$2=4$, $600-4= 596, 596$ divided by $4= 149$.
There are some patterns that we noticed, such as: If you take $1$ away from the amount of rats, it adds $2$ to the amount of children. For example: $86$ children, $107$ rats; $106$ rats, $88$ children.

, who calls his school  "BG"
 sent in the following good explanation:

When working out the pattern for children/rats, I thought it would be a good idea to start with either everything as children or everything as rats. I decided to start with everything as children, which would be $300$ children and $0$ rats, since $600$ halved is $300$.

Then, I knew if I wanted to get all possible solutions, I'd need to come up with a pattern. My pattern was adding some of the children's legs, to the rats each time. But of course, since rats have twice as many legs as children, that wouldn't work, so I took away two children for every rat I added, as shown in the pattern below:

Children Rats Legs
$300$ $0$ $600 + 0$
$298$ $1$ $596+ 4$
$296$ $2$ $592 + 8$
$294$ $3$ $588 + 12$

I did that on and on, until I was certain, that every time, the legs would total up to $600$. By looking at the number of children, and noticing that it started at $300$, and got $2$ fewer each time, I divided $300$ by $2$, and then added on the $1$ possibilty, with $0$ children, to work out there would be $151$ possible solutions for this.

These solutions are really good. Well done, keep submitting solutions to other activities.