From 723b6ded0066651ac8521ce1320371c8430b1434 Mon Sep 17 00:00:00 2001 From: Christopher Baines Date: Tue, 1 Apr 2014 19:10:41 +0100 Subject: Initial commit --- .gitignore | 26 ++++ Makefile | 41 ++++++ README.md | 15 +++ common.sty | 25 ++++ dataprocessing.tex | 144 +++++++++++++++++++++ fib.sty | 68 ++++++++++ linear.tex | 360 +++++++++++++++++++++++++++++++++++++++++++++++++++++ ratios.tex | 215 ++++++++++++++++++++++++++++++++ 8 files changed, 894 insertions(+) create mode 100644 .gitignore create mode 100644 Makefile create mode 100644 README.md create mode 100644 common.sty create mode 100644 dataprocessing.tex create mode 100644 fib.sty create mode 100644 linear.tex create mode 100644 ratios.tex 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} -- cgit v1.2.3