### Chocolate

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?

### Tweedle Dum and Tweedle Dee

Two brothers were left some money, amounting to an exact number of pounds, to divide between them. DEE undertook the division. "But your heap is larger than mine!" cried DUM...

### Matching Fractions, Decimals and Percentages

An activity based on the game 'Pelmanism'. Set your own level of challenge and beat your own previous best score.

# Between a Sixth and a Twelfth

##### Age 11 to 14 Short Challenge Level:

Finding the distance
The distance between $\frac16$ and $\frac1{12}$ is $\frac16-\frac1{12}=\frac{2}{12}-\frac1{12}=\frac1{12}$

So the length of each of the 3 smaller sections is equal to $\frac13$ of $\frac1{12}$, which is $\frac13$ of $\frac3{36}$, which is $\frac1{36}$. This is labelled below:

So the number indicated is $\frac1{12}+\frac1{36}=\frac3{36}+\frac1{36}=\frac4{36}=\frac19$

Using a weighted average
To find the point half way between $\frac16$ and $\frac1{12}$, we would add $\frac16$ and $\frac1{12}$ and divide by $2.$

We want the point that is twice as close to $\frac1{12}$ as it is to $\frac16$ - so instead of adding $\frac16$ and $\frac1{12}$, we can add $\frac16$, $\frac1{12}$ and $\frac1{12}$ - and then divide the result by $3$. That way, we give twice as much importance to $\frac1{12}$ as to $\frac16.$ This is called a weighted average.
$$\begin{split}\left(\tfrac16+2\times\tfrac1{12}\right)\div3&=\left(\tfrac16+\tfrac16\right)\div3\\ &=\tfrac26\div3\\&=\tfrac13\div3\\&=\tfrac19\end{split}$$
You can find more short problems, arranged by curriculum topic, in our short problems collection.