# Advanced (for me) algebra and mean

I have been struggling over questions like these which my maths teacher has been throwing into our weekly papers for about a week now, and it has stumped all of us. Can you help? Question: There are n people in a room. Their mean IQ is 111. A person with an IQ of 93 walks into the room. The mean IQ is now 110. Find the value of n. Thanks! Freddie, Age 12

• The sum of the $n$ peoples' IQ is $n\cdot 111$. Write the mean for the $n+1$ people in terms of $n$. You know this is $110$. Solve the equation for $n$. – David Mitra Apr 25 '15 at 13:30

Call $s$ the sum of the individual IQs. What we know:
• $\dfrac sn=111$;
• $\dfrac{s+93}{n+1}=110.$
These equalities can be rewritten as: $$\begin{cases} s=111n,\\ s+93=110(n+1). \end{cases}$$ From these we deduce: $$111n+93=110n+110,\enspace\text{whence}\quad n=\color{red}{17}.$$
Bernard’s answer is excellent, but I’m going to suggest a slightly different way to think about essentially the same calculation that may also be helpful. Instead of thinking directly in terms of the mean IQ, think in terms of the total IQ in the room. Initially that must be $111n$, just as if each of the $n$ people in the room had an IQ of $111$. After the $(n+1)$-st person enters the room, the total IQ is $110(n+1)$. That person raised the total IQ by $93$, so $111n+93=110(n+1)$.
Yet another way, one that is good for mental calculation because the numbers are smaller, is to change the base point for measuring IQs to something different from $0$. Suppose that we measure from a base point of $110$, the final mean IQ. Then the original mean adjusted IQ in the room is $1$, the newcomer has an adjusted IQ of $-17$, and the final mean adjusted IQ in the room is $0$. This is exactly as if each of the $n$ people originally in the room had an adjusted IQ of $1$, the newcomer had an adjusted IQ of $-17$, and the new mean was $0$. The only way to get a mean of $0$ is to get a total of $0$, and that happens if and only if $n=17$.