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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.