Initial commitHEADmaster

8 files changed, 894 insertions, 0 deletions

diff --git a/.gitignore b/.gitignore
new file mode 100644
index 0000000..bd31fd2
--- /dev/null
+++ b/.gitignore
@@ -0,0 +1,26 @@
+*.aux
+*.glo
+*.idx
+*.log
+*.toc
+*.ist
+*.acn
+*.acr
+*.alg
+*.bbl
+*.blg
+*.dvi
+*.glg
+*.gls
+*.ilg
+*.ind
+*.lof
+*.lot
+*.maf
+*.mtc
+*.mtc1
+*.out
+*.synctex.gz
+*.pdf
+*.swp
+*.fdb_latexmk
diff --git a/Makefile b/Makefile
new file mode 100644
index 0000000..292b057
--- /dev/null
+++ b/Makefile
@@ -0,0 +1,41 @@
+#SOURCES=linear.tex ratios.tex dataprocessing.tex
+NORMAL=-pdf -silent
+SHOW_ANSWERS=-silent -pdf -pdflatex="pdflatex --synctex=1 %O '\def\showanswers{1} \input { %S }'"
+
+all: linear.pdf linear-answers.pdf ratios.pdf ratios-answers.pdf dataprocessing.pdf dataprocessing-answers.pdf
+
+linear.pdf: linear.tex
+	latexmk $(NORMAL) linear.tex
+
+linear-answers.pdf: linear.tex
+	latexmk -jobname=linear-answers $(SHOW_ANSWERS) linear.tex
+
+linear-pvc: linear.tex
+	latexmk -pvc $(NORMAL) linear.tex
+
+ratios.pdf: ratios.tex
+	latexmk $(NORMAL) ratios.tex
+
+ratios-answers.pdf: ratios.tex
+	latexmk -jobname=ratios-answers $(SHOW_ANSWERS) ratios.tex
+
+ratios-pvc: ratios.tex
+	latexmk -pvc $(NORMAL) ratios.tex
+
+dataprocessing.pdf: dataprocessing.tex
+	latexmk $(NORMAL) dataprocessing.tex
+
+dataprocessing-answers.pdf: dataprocessing.tex
+	latexmk -jobname=dataprocessing-answers $(SHOW_ANSWERS) dataprocessing.tex
+
+dataprocessing-pvc: dataprocessing.tex
+	latexmk -pvc $(NORMAL) dataprocessing.tex
+
+clean:
+	latexmk -C
+	latexmk -jobname=linear-answers -C linear.tex
+	latexmk -jobname=ratios-answers -C ratios.tex
+	latexmk -jobname=dataprocessing-answers -C dataprocessing.tex
+	rm dataprocessing-answers.synctex.gz
+	rm linear-answers.synctex.gz
+	rm ratios-answers.synctex.gz
diff --git a/README.md b/README.md
new file mode 100644
index 0000000..89aa5f8
--- /dev/null
+++ b/README.md
@@ -0,0 +1,15 @@
+Mathematical Worksheets
+=======================
+
+These worksheets probably best serve as examples of how to create, hopefully
+passable worksheets (in terms of layout and methodology), the content has not
+been checked, and is not complete.
+
+Building the Worksheets
+-----------------------
+
+The source files (e.g. linear.tex) serve as the source for both the worksheet,
+and the completed worksheet (with answers). To build all the worksheets and
+answer sheets, just run `make` in the root of the repository. If this works,
+there should be pdf's created with names such as linear.pdf and
+linear-answers.pdf.
diff --git a/common.sty b/common.sty
new file mode 100644
index 0000000..115e8b5
--- /dev/null
+++ b/common.sty
@@ -0,0 +1,25 @@
+\NeedsTeXFormat{LaTeX2e}[1994/06/01]
+\ProvidesPackage{common}[2013/01/13 Custom Package]
+
+\renewcommand{\familydefault}{\sfdefault}
+
+\ifdefined\showanswers
+  \printanswers
+\else
+  \noprintanswers
+\fi
+
+\setlength{\droptitle}{-8em}
+
+% Create a custom float for the key points section
+\floatstyle{boxed}
+\newfloat{keypoints}{tbhp}{lop}
+\floatname{keypoints}{Key Points}
+
+\setlength{\columnsep}{0.6in}
+
+% use ' marks' instead of points
+\pointname{\ marks}
+\pointsdroppedatright
+
+\endinput
\ No newline at end of file
diff --git a/dataprocessing.tex b/dataprocessing.tex
new file mode 100644
index 0000000..9b946cc
--- /dev/null
+++ b/dataprocessing.tex
@@ -0,0 +1,144 @@
+\documentclass[12pt,a4paper,twocolumn,landscape]{exam}
+
+\usepackage{xparse}
+\usepackage{tikz}
+   \usetikzlibrary{arrows,calc,fit}
+\usepackage{mathtools}
+\usepackage{fullpage}
+\usepackage{todonotes}
+\usepackage{float}
+\usepackage[compact,explicit]{titlesec}% http://ctan.org/pkg/titlesec
+\usepackage[utf8]{inputenc}
+\usepackage{titling}
+\usepackage{pgfplots}
+\usepackage{wrapfig}
+\usepackage{boxedminipage}
+\usepackage{parskip} % Don't indent paragraphs
+
+\usepackage{fib}
+\usepackage{common}
+
+\newcommand{\slice}[4]{
+  \pgfmathparse{0.5*#1+0.5*#2}
+  \let\midangle\pgfmathresult
+
+  % slice
+  \draw[thick,fill=black!10] (0,0) -- (#1:1) arc (#1:#2:1) -- cycle;
+
+  % outer label
+  \node[label=\midangle:#4] at (\midangle:1) {};
+
+  % inner label
+  \pgfmathparse{min((#2-#1-10)/110*(-0.3),0)}
+  \let\temp\pgfmathresult
+  \pgfmathparse{max(\temp,-0.5) + 0.8}
+  \let\innerpos\pgfmathresult
+  \node at (\midangle:\innerpos) {#3};
+}
+
+\pagestyle{headandfoot}
+
+\firstpageheadrule
+\firstpageheader{Ratios}
+                {Level 5-6}
+                {\today}
+
+\runningheadrule
+\runningheader{Ratios}
+              {Ratios, Page \thepage\ of \numpages}
+              {\today}
+
+\firstpagefooter{}{}{}
+\runningfooter{}{}{}
+
+\begin{document}
+\section*{Data Processing - Level 5-6}
+
+\begin{wrapfigure}{R}{0.25\linewidth}
+    \centering
+    \begin{boxedminipage}{\linewidth}
+        Keywords:
+        \begin{itemize}
+            \item Chart
+            \item Graph
+            \item Data
+            \item Stem
+        \end{itemize}
+    \end{boxedminipage}
+\end{wrapfigure}
+
+Data can be represented in many ways, one useful way is to represent it as a
+graph or chart.
+
+\subsection*{Stem and Leaf Diagrams}
+
+Stem and leaf diagrams are useful for displaying small to medium amounts of
+numerical data.
+
+For example, in the stem and leaf diagram below, you can see some test
+results. By looking at the length of the row of numbers, you can tell how many
+results fall in to each range.
+
+A very important part of the stem and leaf diagram is the key, this tells you
+how to interpret the data in the diagram.
+
+\begin{table}[htbp]
+    \centering
+    \caption{Stem Plot of test results Key: $1 | 1= 11$}
+    \begin{tabular}{r|*{9}{l}}
+        Stem & \multicolumn{9}{l}{Leaf} \\
+        \hline
+        1 & 1 &   &   &   &   &   &   &   &   \\
+        2 & 0 & 3 &   &   &   &   &   &   &   \\
+        3 & 5 & 6 & 6 & 6 & 7 & 7 & 7 & 8 & 9 \\
+        4 & 5 & 7 & 7 & 8 &   &   &   &   &   \\
+    \end{tabular}
+    \label{tab:addlabel}
+\end{table}
+
+\subsection*{Bar Charts}
+
+Bar charts are another way of representing data, here the chart below
+represents the number of different colour cars that passed a point on a road
+over a hour.
+
+\begin{figure}[H]
+    \centering
+    \begin{tikzpicture}
+        \begin{axis}[
+            symbolic x coords={Red, Green, Blue},
+            xtick=data
+          ]
+            \addplot[ybar,fill=black!10] coordinates {
+                (Red,   42)
+                (Green,  50)
+                (Blue,   80)
+            };
+        \end{axis}
+    \end{tikzpicture}
+\end{figure}
+
+\subsection*{Pie Charts}
+
+\begin{figure}[H]
+    \centering
+    \begin{tikzpicture}[scale=3]
+        \newcounter{a}
+        \newcounter{b}
+        \foreach \p/\t in {20/type A, 4/type B, 11/type C,
+        49/type D, 16/other}
+        {
+            \setcounter{a}{\value{b}}
+            \addtocounter{b}{\p}
+            \slice{\thea/100*360}
+            {\theb/100*360}
+            {\p\%}{\t}
+        }
+    \end{tikzpicture}
+\end{figure}
+
+\subsection*{Questions}
+
+TODO
+
+\end{document}
diff --git a/fib.sty b/fib.sty
new file mode 100644
index 0000000..e5465a5
--- /dev/null
+++ b/fib.sty
@@ -0,0 +1,68 @@
+\NeedsTeXFormat{LaTeX2e}[1994/06/01]
+\ProvidesPackage{fib}[2013/01/13 Custom Package]
+
+% Source:
+% https://tex.stackexchange.com/questions/147701/graphical-placeholder-for-variables
+% adjusted slightly to fit better with the exam class (printanswers)
+
+\tikzset{box/.style={draw, rectangle, rounded corners, thick, node distance=7em, text width=6em, text centered, minimum height=3.5em}}
+\tikzset{container/.style={draw, rectangle, dashed, inner sep=2em}}
+\tikzset{line/.style={draw, thick, -latex'}}
+
+\makeatletter
+\newlength\fib@width
+\def\fib@widthfactor{1.75}
+
+\tikzset{
+   every fill in box/.style={
+     inner xsep=0pt,
+     minimum height=3ex,
+     align=center,
+     font={\sffamily\slshape},
+   },
+   colored box/.style={
+      every fill in box,
+      fill=yellow!50!white,
+   },
+   framed box/.style={
+      every fill in box,
+      draw,
+   },
+   fill in/.style={
+      framed box,
+   }
+}
+\NewDocumentCommand { \fib@hide } { m } {%
+   \ifprintanswers
+        #1%
+   \else
+        \phantom{#1}%
+   \fi
+}
+\NewDocumentCommand { \fib@makebox }{ m }{%
+   \settowidth{\fib@width}{\tikz\node[fill in]{#1};}%
+   \begin{tikzpicture}[baseline=(fill in node.base)]
+      \node (fill in node) [text width=\fib@widthfactor*\fib@width,fill in] {%
+         \fib@hide{#1}%
+      };
+   \end{tikzpicture}%
+}
+\NewDocumentCommand { \fib } { s d{<}{>} o m }{{%
+   \IfBooleanT{#1}{\noprintanswers}%
+   \IfValueT{#2}{\only<#2>{\noprintanswers}}%
+   \IfValueT{#3}{\tikzset{fill in/.style={#3}}}%
+   \ifmmode
+      \mathchoice
+         {\fib@makebox{$\displaystyle#4$}}
+         {\fib@makebox{$\textstyle#4$}}
+         {\fib@makebox{$\scriptstyle#4$}}
+         {\fib@makebox{$\scriptscriptstyle#4$}}
+   \else
+      \fib@makebox{#4}%
+   \fi
+   \IfValueT{#2}{}%
+}}
+\makeatother
+% end fib definition
+
+ \endinput
\ No newline at end of file
diff --git a/linear.tex b/linear.tex
new file mode 100644
index 0000000..7710933
--- /dev/null
+++ b/linear.tex
@@ -0,0 +1,360 @@
+\documentclass[12pt,a4paper,twocolumn,landscape]{exam}
+
+\usepackage{xparse}
+\usepackage{tikz}
+    \usetikzlibrary{arrows,calc,fit}
+\usepackage{mathtools}
+\usepackage{fullpage}
+\usepackage{todonotes}
+\usepackage{float}
+\usepackage[compact,explicit]{titlesec} % http://ctan.org/pkg/titlesec
+\usepackage[utf8]{inputenc}
+\usepackage{titling}
+\usepackage{wrapfig}
+\usepackage{boxedminipage}
+\usepackage{parskip} % Don't indent paragraphs
+
+\usepackage{fib}
+\usepackage{common}
+
+\pagestyle{headandfoot}
+
+\firstpageheadrule
+\firstpageheader{Solving Linear Equations}
+                {Level 7-8}
+                {\today}
+
+\runningheadrule
+\runningheader{Solving Linear Equations}
+              {Solving Linear Equations, Page \thepage\ of \numpages}
+              {\today}
+
+\firstpagefooter{}{}{}
+\runningfooter{}{}{}
+
+\begin{document}
+
+\section*{Solving Linear Equations - Level 7-8}
+
+\begin{minipage}{\linewidth}
+
+\begin{wrapfigure}{R}{0.3\linewidth}
+    \centering
+    \begin{boxedminipage}{\linewidth}
+        Keywords:
+        \begin{itemize}
+            \item Linear
+            \item Solve
+            \item Unknown
+        \end{itemize}
+    \end{boxedminipage}
+\end{wrapfigure}
+
+Linear equations are written with one unknown variable which is shown by a
+letter.
+
+It is necessary to \textbf{rearrange} the equation in order to find the value
+of x.
+
+\end{minipage}
+
+\begin{questions}
+
+\question Solve 2x+3 = 7 for the unknown.
+
+\begin{minipage}{\linewidth}
+
+    \begin{wrapfigure}{R}{0.3\linewidth}
+        \vspace{-25pt}
+        \centering
+        \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm,
+            thick,main node/.style={draw=none,fill=none}]
+
+            \node[main node] (1) {$2x + 3 = 7$};
+            \node[main node] (2) [below of=1] {$2x = 4$};
+
+            \path[every node/.style={font=\sffamily\small}]
+            (1.west) edge [bend right] node[left]{$-3$} (2.west)
+            (1.east) edge [bend left] node[right] {$-3$} (2.east);
+        \end{tikzpicture}
+    \end{wrapfigure}
+
+    Start by looking at the section of the equation that contains the unknown. In
+    this case the left hand side.
+
+\end{minipage}
+
+The next step is to collect all constants (numbers) on the side of the equation
+that does not contain the unknown. We want to move the 3 to the right hand
+side. Because it is +3, to move it to the other side we must -3.
+
+\begin{minipage}{\linewidth}
+
+    \begin{wrapfigure}{R}{0.3\linewidth}
+        \centering
+        \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm,
+            thick,main node/.style={draw=none,fill=none}]
+
+            \node[main node] (1) {$2x = 4$};
+            \node[main node] (2) [below of=1] {$x = 2$};
+
+            \path[every node/.style={font=\sffamily\small}]
+            (1.west) edge [bend right] node[left]{$\div 2$} (2.west)
+            (1.east) edge [bend left] node[right] {$\div 2$} (2.east);
+        \end{tikzpicture}
+    \end{wrapfigure}
+
+    We can see that there are still numbers on the right hand side with the
+    unknown. The coefficient (multiplier) of x is 2. Since the 2 is times x we
+    need to divide by 2 to get the x by itself.
+
+\end{minipage}
+
+We can see that there are no more numbers on the side of x so the solution is:
+$x = 2$
+
+To check the solution, put the unknown back into the original equation. $2
+\times 2 + 3 = 7$
+
+\begin{keypoints}[t]
+    Key things to remember:
+    \begin{itemize}
+        \item Collect all \textbf{terms} involving the \textbf{unknown} to one
+              side of the equation.
+        \item The opposite of addition is subtraction.
+        \item The opposite of multiplication is division.
+        \item If you do something to one side of the equation you MUST do the
+              same thing to the other side.
+    \end{itemize}
+\end{keypoints}
+
+\subsection*{Higher Level Questions}
+
+A harder question is shown below, the steps taken are shown on the arrows.  The
+first step is to bring all \textbf{unknowns} to the same side of the equation.
+
+\question Find the value of x: $6x - 12 = x + 8$
+
+\begin{figure}[H]
+    \centering
+    \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm,
+        thick,main node/.style={draw=none,fill=none}]
+
+        \node[main node] (1) {$6x - 12 = x + 8$};
+        \node[main node] (2) [below of=1] {$5x - 12 = 8$};
+        \node[main node] (3) [below of=2] {$5x = 20$};
+        \node[main node] (4) [below of=3] {$x = 4$};
+
+        \path[every node/.style={font=\sffamily\small}]
+            (1.west) edge [bend right] node[left]{$-x$} (2.west)
+            (1.east) edge [bend left] node[right] {$-x$} (2.east)
+            (2.west) edge [bend right] node[left]{$+12$} (3.west)
+            (2.east) edge [bend left] node[right] {$+12$} (3.east)
+            (3.west) edge [bend right] node[left]{$\div 5$} (4.west)
+            (3.east) edge [bend left] node[right] {$\div 5$} (4.east);
+    \end{tikzpicture}
+\end{figure}
+
+Check:
+
+On the left hand side of the \textbf{original} equation:\\
+$6(4) - 12 = 24 - 12 = 12$
+
+On the right hand side of the \textbf{original} equation:\\
+$4 + 8 = 12$
+
+The left hand side and right hand side of the equation are equal so you know
+that your solution is correct: $x = 4$
+
+\end{questions}
+
+\newpage
+
+\subsection*{Practice Questions}
+
+\begin{questions}
+
+\question Find the value of x: $6+2x = x-6$
+
+\begin{figure}[H]
+    \centering
+    \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm,
+        thick,main node/.style={draw=none,fill=none}]
+
+        \node[main node] (1) {$6 + 2x = x - 6$};
+        \node[main node] (2) [below of=1] {$\fib{6 + x} = -6$};
+        \node[main node] (3) [below of=2] {$\fib{x} = \fib{-12}$};
+
+        \path[every node/.style={font=\sffamily\small}]
+            (1.west) edge [bend right] node[left]{$-x$} (2.west)
+            (1.east) edge [bend left] node[right] {$-x$} (2.east)
+            (2.west) edge [bend right] node[left]{$-6$} (3.west)
+            (2.east) edge [bend left] node[right] {$\fib{-6}$} (3.east);
+    \end{tikzpicture}
+\end{figure}
+
+Check:
+
+$6 + \fib{2}x = 6 + 2 \times -12 = \fib{-18}$\\
+$\fib{x} - 6 = \fib{-12} - 6 = \fib{-18}$
+
+\end{questions}
+
+\begin{questions}
+
+\newpage
+
+\subsection*{Solving Linear Equations}
+
+\question[3] Find the value of x: $6x + 13 = 4 + 9x$ \droppoints
+
+\begin{minipage}{0.69\linewidth}
+\begin{figure}[H]
+    \centering
+    \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm,
+        thick,main node/.style={draw=none,fill=none}]
+
+        \node[main node] (1) {$6x + 13 = 4 + 9x$};
+        \node[main node] (2) [below of=1] {$\fib{13} = 4 + 3x$};
+        \node[main node] (3) [below of=2] {$\fib{9} = \fib{3x}$};
+        \node[main node] (4) [below of=3] {$\fib{x} = \fib{3}$};
+
+        \path[every node/.style={font=\sffamily\small}]
+        (1.west) edge [bend right] node[left]{$-6x$} (2.west)
+        (1.east) edge [bend left] node[right] {$-6x$} (2.east)
+        (2.west) edge [bend right] node[left]{$-4$} (3.west)
+        (2.east) edge [bend left] node[right] {$-4$} (3.east)
+        (3.west) edge [bend right] node[left]{$\fib{\div 3}$} (4.west)
+        (3.east) edge [bend left] node[right] {$\fib{\div 3}$} (4.east);
+    \end{tikzpicture}
+\end{figure}
+\end{minipage}
+\begin{boxedminipage}{0.3\linewidth}
+    TIP:
+
+    When you have unknowns on both sides of the equations you can choose which
+    one to move.
+
+    Try and move the lower number of unknowns to avoid negatives.
+
+\end{boxedminipage}
+
+Check:
+
+$6\fib{x} + 13 = 6\fib{3} + 13 = 31$\\
+$4 + 9\fib{x} = 4 + \fib{27} = 31$
+
+\newpage
+
+\question[3] Find the value of x: $7x + 8 = 2x - 2$ \droppoints
+
+\begin{figure}[H]
+    \centering
+    \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm,
+        thick,main node/.style={draw=none,fill=none}]
+
+        \node[main node] (1) {$7x + 8 = 2x - 2$};
+        \node[main node] (2) [below of=1] {$5x + 8 = \fib{-2}$};
+        \node[main node] (3) [below of=2] {$\fib{5x} = \fib{-10}$};
+        \node[main node] (4) [below of=3] {$\fib{x} = \fib{-2}$};
+
+        \path[every node/.style={font=\sffamily\small}]
+            (1.west) edge [bend right] node[left]{$-2x$} (2.west)
+            (1.east) edge [bend left] node[right] {$-2x$} (2.east)
+            (2.west) edge [bend right] node[left]{$\fib{-8}$} (3.west)
+            (2.east) edge [bend left] node[right] {$\fib{-8}$} (3.east)
+            (3.west) edge [bend right] node[left]{$\div \fib{5}$} (4.west)
+            (3.east) edge [bend left] node[right] {$\div \fib{5}$} (4.east);
+    \end{tikzpicture}
+\end{figure}
+
+Check:
+
+$7\fib{x} + 8 = \fib{-14} + 8 = \fib{-4}$\\
+$2\fib{x} - 2 = \fib{-6} - 2 = \fib{-4}$
+
+\newpage
+
+\question[3]
+
+Susie clears out boxes on the weekend. She gets paid £10 for the weekend and an
+extra £2 for each box she clears. If Susie is paid £40, how many boxes did she
+clear? \droppoints
+
+\begin{minipage}{0.69\linewidth}
+\begin{figure}[H]
+    \centering
+    \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm,
+        thick,main node/.style={draw=none,fill=none}]
+
+        \node[main node] (1) {$\fib{10 + 2x} = 40$};
+        \node[main node] (2) [below of=1] {$\fib{2x} = \fib{30}$};
+        \node[main node] (3) [below of=2] {$\fib{x} = \fib{15}$};
+
+        \path[every node/.style={font=\sffamily\small}]
+            (1.west) edge [bend right] node[left]{$-\fib{10}$} (2.west)
+            (1.east) edge [bend left] node[right] {$-\fib{10}$} (2.east)
+            (2.west) edge [bend right] node[left]{$\div\fib{2}$} (3.west)
+            (2.east) edge [bend left] node[right] {$\div\fib{2}$} (3.east);
+    \end{tikzpicture}
+\end{figure}
+\end{minipage}
+\begin{boxedminipage}{0.3\linewidth}
+    LOOK!:
+
+    This question is asking you to make the linear equation yourself.
+
+    HINT:
+
+    What is the unknown?
+\end{boxedminipage}
+
+Check:
+
+$\fib{2}\fib{x} + \fib{10} = \fib{30} + \fib{10} = \fib{40}$
+
+\newpage
+
+\question[3]
+
+Rearrange to find the unknown:\\
+$3z - 18 = 7z - 10$ \droppoints
+
+\begin{minipage}{0.69\linewidth}
+\begin{figure}[H]
+    \centering
+    \begin{tikzpicture}[->,>=stealth',shorten >=1pt,auto,node distance=1.5cm,
+        thick,main node/.style={draw=none,fill=none}]
+
+        \node[main node] (1) {$3z - 18 = 7z - 10$};
+        \node[main node] (2) [below of=1] {$\fib{3z} = \fib{7z + 8}$};
+        \node[main node] (3) [below of=2] {$\fib{-4z} = \fib{8}$};
+        \node[main node] (4) [below of=3] {$\fib{z} = \fib{-2}$};
+
+
+        \path[every node/.style={font=\sffamily\small}]
+            (1.west) edge [bend right] node[left]{$\fib{+ 18}$} (2.west)
+            (1.east) edge [bend left] node[right] {$\fib{+ 18}$} (2.east)
+            (2.west) edge [bend right] node[left]{$\fib{- 7z}$} (3.west)
+            (2.east) edge [bend left] node[right] {$\fib{- 7z}$} (3.east)
+            (3.west) edge [bend right] node[left]{$\fib{\div -2}$} (4.west)
+            (3.east) edge [bend left] node[right] {$\fib{\div -2}$} (4.east);
+    \end{tikzpicture}
+\end{figure}
+\end{minipage}
+\begin{boxedminipage}{0.3\linewidth}
+    NOTE:
+
+    This question uses z as the unknown. Any letter may be used as the
+    unknown.
+
+\end{boxedminipage}
+
+Check:
+
+$3\fib{z} - 18 = \fib{-6} - 18 = \fib{-24}$\\
+$7\fib{z} - 10 = \fib{-14} - 10 = \fib{-24}$
+
+\end{questions}
+
+\end{document}
diff --git a/ratios.tex b/ratios.tex
new file mode 100644
index 0000000..a9a9ec0
--- /dev/null
+++ b/ratios.tex
@@ -0,0 +1,215 @@
+\documentclass[12pt,a4paper,twocolumn,landscape]{exam}
+
+\usepackage{xparse}
+\usepackage{tikz}
+   \usetikzlibrary{arrows,calc,fit}
+\usepackage{mathtools}
+\usepackage{fullpage}
+\usepackage{todonotes}
+\usepackage{float}
+\usepackage[compact,explicit]{titlesec}% http://ctan.org/pkg/titlesec
+\usepackage[utf8]{inputenc}
+\usepackage{titling}
+\usepackage{boxedminipage}
+\usepackage{wrapfig}
+\usepackage{parskip} % Don't indent paragraphs
+
+\usepackage{fib}
+\usepackage{common}
+
+\pagestyle{headandfoot}
+
+\firstpageheadrule
+\firstpageheader{Ratios}
+                {Level 5-6}
+                {\today}
+
+\runningheadrule
+\runningheader{Ratios}
+              {Ratios, Page \thepage\ of \numpages}
+              {\today}
+
+\firstpagefooter{}{}{}
+\runningfooter{}{}{}
+
+\begin{document}
+\section*{Ratios - Level 5-6}
+
+\begin{wrapfigure}{R}{0.3\linewidth}
+    \centering
+    \begin{boxedminipage}{\linewidth}
+        Keywords:
+        \begin{itemize}
+            \item Ratio
+            \item Comparison
+        \end{itemize}
+    \end{boxedminipage}
+\end{wrapfigure}
+
+\textbf{Ratios} are used to show how much of one thing there is compared to
+another thing.
+
+Ratios will be written in two ways, either in words "2 to 1" or as "2:1". Both
+are read the same way.
+
+A question involving \textbf{ratios} will often ask you to find out how many of
+one thing there is.
+
+\subsection*{Worked Example}
+
+\begin{questions}
+
+\question
+
+20 sweets are shared between James and Sasha in the ration 1:3, how many sweets
+do James and Sasha each receive?
+
+The first step is to look at what the ratio means for each sweet that James
+gets, Sasha gets three so we can expect Sasha to have more sweets than James in
+our answer.
+
+The ratio 1:3 means that 4 parts are shared between the boys.
+
+To find out how many sweets are in 1 part, we divide 20 by 4.  $20 \div 4 = 5$
+sweets are in 1 part.
+
+Since James gets one part, we now know that he receives 5 sweets.
+
+Sasha gets 3 parts, so we times the number of sweets in 1 part by 3. $3 \times
+5 = 15$ sweets in 3 parts.
+
+Sasha gets 15 sweets.
+
+\textbf{Check}
+
+We can write this back in ration form as the number of sweets James has to
+Sasha is. 5:15 We know there has to be 20 sweets in total so we can check that
+we have shared all 20 sweets out by adding the numbers of sweets that James and
+Sasha have. $5 + 15 = 20$, so this supports our answer.
+
+\end{questions}
+
+\subsection*{Questions}
+
+\begin{questions}
+
+\question
+What is the simplest form of the following ratios?
+
+\begin{parts}
+\part
+6:15
+
+\begin{solution}[0.2in]
+2:5
+\end{solution}
+
+\part
+2:8
+
+\begin{solution}[0.2in]
+1:4
+\end{solution}
+
+\part
+4:12
+
+\begin{solution}[0.2in]
+1:3
+\end{solution}
+
+\part
+9:3
+
+\begin{solution}[0.2in]
+3:1
+\end{solution}
+
+\part
+4:2
+
+\begin{solution}[0.2in]
+2:1
+\end{solution}
+
+\part
+10:14
+
+\begin{solution}[0.2in]
+5:7
+\end{solution}
+
+\end{parts}
+
+\newpage
+
+\question
+
+John and Sam stand on a pair of scales, together the weigh 200kg, if the ratio
+of John and Sam's weights  is 2:3, how heavy is Sam?
+
+\begin{solution}[2in]
+
+Sum the sides of the ratio, to convert the ratio in to two fractions.
+
+$2 + 3 = 5$
+
+Therefore, Sam weighs $3 \div 5$'s of 200kg.
+
+Sam weighs 120kg.
+
+\end{solution}
+
+\question
+
+A recipe for a cake requires 200g of flour, 100g of butter, 50g of sugar and 2
+eggs.  The cake recipe indicates that the cake will serve 4. How much of each
+ingredient is necessary for the cake to serve 8?
+
+\begin{boxedminipage}{\linewidth}
+    Hint: What is the ratio of cake servings?
+\end{boxedminipage}
+
+\begin{solution}[2in]
+
+    Create a ratio between the people that the recipe serves (4) and the people to
+    serve (8).
+
+    4:8
+
+    Simplify the ratio.
+
+    1:2
+
+    So it takes 2 lots of each ingredient, for every 1 in the recipe.
+
+    This means:
+    \begin{itemize}
+        \item 400g of flour
+        \item 200g of butter
+        \item 100g of sugar
+        \item 4 eggs
+    \end{itemize}
+
+\end{solution}
+
+\question
+
+John and Sam run a race, in total the race takes them 1 hour, however John is
+twice as fast as Sam. How long does Sam take to finish the race?
+
+\begin{solution}[2in]
+
+    Create a ratio between John's time, and Sam's.
+
+    1:2
+
+    Therefore, Sam took twice (2 times) as long as John.
+
+    $2 \times 1hour = 2hours$
+
+\end{solution}
+
+\end{questions}
+
+\end{document}