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# Changing Averages

Using each piece of information, we can construct the list, with the numbers ordered from smallest to largest.

*'The smallest number in the list is $10$'* allows us to begin constructing the list: $$ 10, \underline{ }, \underline{ }, ... $$

*'The median is $m$. $m$ is one of the numbers on the list'*, means $m$ must be the middle number on the list.

The*'list has a mode of $32$'*, so $32$ must appear at least twice on the list. We can't be sure, but it is likely that $32$ is greater than $m$: $$10, ..., m, ..., 32, 32, ...$$

The mean is $22$, but*'If $m$ were replaced with $m + 10$, the mean of the new list would be $24$.'* So increasing one of the numbers by $10$ increases the mean by $2$, which is $10\div5$. So there must be $5$ numbers on the list: $$10, \underline{ }, m, 32, 32$$

*'If $m$ were instead replaced with $m - 8$, the median of the new list would be $m - 4$.'* So $m-4$ must also be a number on the list: $$10, m-4, m, 32, 32$$

The mean is $22$, so we can set up an equation to find $m$:

$$\begin{align} \frac{10+(m-4)+m+32+32}5&=22\\

\Rightarrow \frac{70+2m}{5}&=22\\

\Rightarrow \frac{35}{5}+\frac{m}{5}&=11\\

\Rightarrow 7+\frac{m}5&=11\\

\Rightarrow \frac{m}5&=11-7\\

\Rightarrow m&=4\times5\\

\Rightarrow m&=20\end{align}$$

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Age 14 to 16

ShortChallenge Level

- Problem
- Solutions

Using each piece of information, we can construct the list, with the numbers ordered from smallest to largest.

The

The mean is $22$, but

The mean is $22$, so we can set up an equation to find $m$:

$$\begin{align} \frac{10+(m-4)+m+32+32}5&=22\\

\Rightarrow \frac{70+2m}{5}&=22\\

\Rightarrow \frac{35}{5}+\frac{m}{5}&=11\\

\Rightarrow 7+\frac{m}5&=11\\

\Rightarrow \frac{m}5&=11-7\\

\Rightarrow m&=4\times5\\

\Rightarrow m&=20\end{align}$$

You can find more short problems, arranged by curriculum topic, in our short problems collection.