You may also like

problem icon


There are three tables in a room with blocks of chocolate on each. Where would be the best place for each child in the class to sit if they came in one at a time?

problem icon

Four Triangles Puzzle

Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?

problem icon

Cut it Out

Can you dissect an equilateral triangle into 6 smaller ones? What number of smaller equilateral triangles is it NOT possible to dissect a larger equilateral triangle into?

Fractions in a Box

Age 7 to 11 Challenge Level:

This was a tricky problem. Well done to those of you who had a go. We had some very clearly explained answers. The key was to work out the size of the booklet first.

Rachel, Ol, Jack and Alex from Moretonhampstead Primary said:

First we worked out how many squares the booklet is. The number has to be a square number and has to be even and $\frac{1}{4}$ of that number has to be even again. The only number possible for that is $1$6 (four squared).
Then we took $16$ (that was how big the booklet was) from a $100$ which is $84$ (there are $84$ squares to play the game with). With $84$ we can answer the first question - how many discs are there altogether? ($84$).
After that we worked out how many discs there would be for the colours. We worked out there would be $42$ discs ($\frac{1}{2}$ of $84$), $21$ black discs ($\frac{1}{4}$ of $84$) and $7$ blue ($\frac{1}{12}$ of $84$).
We put that on the grid as it says on the sheet. Next we used the last full column for blue and green. We know that there are $7$ blues (because of what we worked out earlier) which means there are $3$ green discs ($7 +3 =10$ (which is how many in a column)).
There were five squares left. It says that there is $1$ white square then the leftovers are yellow so we had $4$ yellows and $1$ white disc.
Now we hade completed the grid we could answer question $2$ and $3$. For question $2$ we counted up all the orange squares ($6$) and the fraction is $\frac{6}{84}$ but we had to simplify it to $\frac{1}{14}$ (the answer).
Lastly we did question $3$. The fraction of green is $\frac{3}{84}$ and simplified for the actual answer is $\frac{1}{28}$.
The fraction of yellow is $\frac{4}{84}$ and simplified for the answer is $\frac{1}{21}$.
There is only one white square so the answer is $\frac{1}{84}$, for white.

Hamish, Rory, Sarah, Jesse and Samuel from Rutherglen Primary also reasoned very clearly and they sent us a picture of the full box which they modelled using cubes:

Sophie and Claire from The Downes School wrote:

$1\times 1$ didn't work because it said that two shortened rows have red discs.
$2\times 2$ didn't work because you need two shortened rows of red and one of orange.
$3\times 3$ didn't work because the total number of discs would be odd and you couldn't halve it. This means all odd numbers didn't work.
$4\times 4$ did work because you had the right amount of shortened rows.
$6 \times 6$ didn't work because you can't divide $64$ by $12$.
$8 \times 8$ didn't work because you need six whole rows.

Emma, Abi, Matthew B and Yuji from Moorfield Junior School; Keshinie and Sharon at Kilvington GGS Victoria, Australia; Gideon from Newberries Primary School and Hannah, Georgia, Patrick; Hana from Bali International School and Matthew from Brighton College Prep School realised that the number left after taking away the booklet must be a multiple of $12$. Keshinie and Sharon describe how they continued from there:

So that made it $84$.
Half of the disks are red so that made the amount of red $42$.
Then it said that a quarter is black so that made it $21$.
Then it said that one twelfth is blue so that made it $7$.
Then it said that one complete row was filled with all of blue and green and the remainder of $10$ if you take away $7$ made it $3$ green.
Then it said that one of the shortened rows is exactly filled with all the orange disks so that makes it $6$.
Then it said that there was only one white disk.
Then we added all the numbers together making $80$ disks so there was a remainder of $4$ which had to be yellow.

We divided the $84$ disks by the $6$ orange ones that made it $14$. So the fraction of orange had to be $1$ out of $14$ ($\frac{1}{14}$).
We divided the $84$ disks by the $3$ green disks making the answer $28$. So the fraction of green had to be $\frac{1}{28}$.
We already knew that the fraction of white disk was $\frac{1}{84}$.
We divided the $84$ disks by the $4$ yellow ones making it $21$ so the fraction of yellow had to be $\frac{1}{21}$.

James from the Charter School explained very well how he went about the problem:

Each of the sides is $10$ units and I called each of the sides of the booklet $x$. this means that the equation for finding the number of discs was $N=100-x^2$. ($N$ being the number of remaining discs).
The amount of Blue discs was $\frac{N}{12}$ meaning that $N$ was a multiple of $12$. So I then collected all of the multiples of $12$: $12$, $24$, $36$, $48$, $60$, $72$, $84$, $96$. I then eliminated those that did not fit the earlier equation because there was not a square number that fitted. This left: $36$, $84$, $96$.
I eliminated $96$ because $x$ had to be more that three for there was one complete row of orange discs and two of red discs.
This left: $36$, $84$. From this I deduced that $x$ had to be $4$ or $8$. This means that the amount of red disks had to end in a $2$ or a $4$, because there are two incomplete rows of red these either have to be a length of $6$ or $2$. Since the amount of red discs is half of $N$ I halved both my possible $N$'s which came up with $18$ and $42$. This means that the amount of reds was $42$ and $N$ was $84$.
This means that $x$ is $4$ and that the amount of orange discs was $6$ meaning the fraction is $\frac{1}{14}$.
The amount of blue was $\frac{N}{12}$ which was $7$.
This means that the amount of green discs was $3$ because blues + greens = $10$. That means the fraction was $\frac{1}{28}$.
The amount of white was $1$ meaning the fraction was $\frac{1}{84}$.
Finally the rest were yellow.
Red was $42$. Blacks was $\frac{N}{4}$ which was $21$. Blue was $7$. Orange was $6$. Green was $3$. White was $1$.
If you take all those away from $84$ you end up with $4$. That is the amount of yellows. This means the fraction is $\frac{1}{21}$.

Well done too to Harriet and Harah from Greenacre School for Girls, Ruairidh from St Mary's High School, Anne-Marie, Emma, Katherine, Laura from Gorseland Primary and Eulalie and Holly who go to Lympstone Primary School.