
\documentclass[12pt]{article}
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{lscape}
\usepackage{verbatim}
%
\font\elevenss=cmss11
\font\tenss=cmss10
\font\niness=cmss9
\font\eightss=cmss8
\font\sevenss=cmss8 at 7pt
\font\sixss=cmss8 at 6pt
\font\fivess=cmss8 at 5pt
\newfam\ssfam
\textfont\ssfam=\elevenss \scriptfont\ssfam=\eightss
 \scriptscriptfont\ssfam=\sixss
\def\ss{\fam\ssfam \elevenss}
%
\setlength{\evensidemargin}{-.05in}
\setlength{\oddsidemargin}{-.05in}
\setlength{\textwidth}{6.4in}
\renewcommand{\baselinestretch}{1.0}
\setlength{\parskip}{15pt plus 2pt minus 1pt}

\def\ray{\overrightarrow}
\def\pb{\pagebreak}
\def\blank{\underline{\;\;\;\;\;\;}}
\def\W{{\mathbb W}}
\def\br{\hfill \break}
\def\ssc{\noindent} 

\begin{document}

\begin{center}
{\Large \bf Banned Book Survey -- model solution via guess and check}
\end{center}

Not knowing how to solve this algebraically I decided to try a guess.
I guessed that 2\% of respondents had read all three books.  From this
I was able to work backward.  First I entered 2\% in the center part
of a Venn diagram.  Then I computed more entries in the Venn diagram:
\begin{quote}
Since 25\% read both HF and BT, my assumption that 2\% read all three meant
that 23\% read exactly HF and BT; now I entered 23\% in the Venn diagram
as shown below.

Since 14\% read BT and LCL, that meant 12\% read exactly these two books.

Since 10\% read HF and LCL, 8\% read exactly these two books.  
\end{quote}
So far, my Venn diagram looks like this.

\begin{figure}[h!]
\hspace{1in} \includegraphics[scale=0.5]{books1} 
\end{figure}

Now, knowing that 53\% read HF, I was able to subtract the 23\%, the 8\%
and the 2\% from 53\% to find that 20\% read only HF; I entered that
in the Venn diagram.  Similarly, I found that $42\% - 23\% - 12% - 2\% = 
5\%$ read only BT and that $36\% - 12\% - 8\% - 2\% = 14\%$ read
only LCL.  My Venn diagram now looks like this.

\begin{figure}[h!]
\hspace{1in} \includegraphics[scale=0.5]{books2} 
\end{figure}

Adding up all the entries in the Venn diagram, we find that 84\% of
the respondents are accounted for.  We have not yet entered the percentage
of respondents who didn't read any of the three books, but we  are 
told that theis is 12\%.  Writing that in somewhere outside the three 
circles, we add it to the 84\% and get 96\%.  Oh no, it doesn't add up to
100\%, so our guess was wrong!

Next, we don't panic, instead we adjust our guess.  I am tempted to
adjust it by 4\% one way or the other, but I'm not sure that will work
so I adjust it by 1\% instead, to 3\%, just to see what happens.
The numbers come out similarly, but all different by 1\% in one
direction or the other.

\begin{figure}[h!]
\hspace{1in} \includegraphics[scale=0.5]{books3} 
\end{figure}

The total is now 97\%.  From the pattern, I am now confident that I can 
do what I wanted to do in the first place and change it to 6\% (an increase
of four percentage points over the original guess of 2\%) and that this
will result in something adding to 100\%.  It works!

\begin{figure}[h!]
\hspace{1in} \includegraphics[scale=0.5]{books4} 
\end{figure}

This meets all the specifications of the problem, so the answer is 6\%.
I observe also that the given information of 12\% non-readers did matter.
If that were 16\%, for example, my very first guess would have been right
and the number of readers of all three books would have been just 2\%.

{\bf Up to here the solution would merit perhaps 18 out of 21.  The
correct problem has been answered, the correct answers given, a clear
explanation of the process is provided, and part of the justification
is present.  In particular, the answer to the question of whether the 
given figure of 12\% matters is quite complete, including an example
of what would happen were that figure to be different.  All that is 
lacking is some generality and a discussion of whether any other answers
are possible.  The model solution continues...}

Now I see how to make this into algebra.  I can let my initial guess be $x$,
and then see algebraically what $x$ has to be.  Now the Venn diagram
after the first step looks like this.

\begin{figure}[h!]
\hspace{1in} \includegraphics[scale=0.5]{books5} 
\end{figure}

To see what it looks like after the next step, I will go through the
computation for HF (the other two are similar).  53\% read HF, but
I have already said that $x\%$ read all three, and reasoned that
$(25-x)\%$ read exactly HF and BT, while $(10-x)\%$ read HF and LCL.
Thus $53 - x - (25-x) - (10-x) \, \% = (18+x)\%$ read only HF.  After
working out the other two, my Venn diagram is as follows.

\begin{figure}[h!]
\hspace{1in} \includegraphics[scale=0.5]{books6} 
\end{figure}

Adding up the total, including the 12\% of non-readers, gives $(94 + x)\%$.
I know the total number of respondents is 100\%, so I set 
$$(94 + x)\% = 100\%$$
and solve to find that $x = 6$.  This is the only solution to
$94 + x = 100$, so it is the only possible solution to the problem.
I believe this method generalizes to any numbers you could have put in
for the survey data.  For example if the 53\% is replaced by 54\% and
all other numbers remain the same, we'll wind up with $(95 + x)\% = 100\%$,
and the percentage $x$ of readers of all three books will be 5.


\end{document}

