\begin{onehalfspacing}
There are 2 types of interest, \emph{simple} and \emph{compound}. In this section we will look at \textbf{simple} interest and understand what makes it so simple...\\

One thing that you must realise is that simple interest always applies only to the original amount. For example, if I borrow \$1000 from you and you charge me 5\% interest on that every year, in one year I will owe you \$1050, then the year after that \$1100, etc, etc with the amount owed increasing by 5\% or \$50 a year.\\

On the other hand, we have compound interest which will apply the interest to however much money is \emph{owed}. Take the above example again. I borrow \$1000, you give a rate of 5\% interest - but this time you tell me it's \textbf{compounded}!\\

The difference is after the first year I'll owe you \$1050, but the year after that I will owe you 5\% ontop of \$1050! Which comes out to be \$1102.5. See how that is more than \$1100?\\

Now whilst it might not seem like much at the start, \textsc{compound} interest is \textit{\textbf{very}} powerful.\\

\end{onehalfspacing}

\fbox{
\begin{minipage}{\textwidth}
    \begin{flushleft}
        \emph{Compound interest is the eighth wonder of the world. S/he who understands it, earns it, s/he who doesn't, pays it.}
    \end{flushleft}
    \begin{flushright}--- Albert Einstein\end{flushright}
\end{minipage}
}

\begin{examplebox}
    \subsection{Examples:}
    \begin{questions}
    \question Find the simple interest on \(\$ 8000\) for eight years at 9.5\% p.a.
    \begin{solutionordottedlines}[1in]
        \[
        \begin{aligned}
        I & =P R T \\
        & =8000 \times 9.5 \% \times 8 \\
        & =8000 \times 0.095 \times 8 \\
        & =\$ 6080
        \end{aligned}
        \]
    \end{solutionordottedlines}
    \question Jessie borrows \(\$ 3000\) from her parents to help buy a car. They agree that she should only pay simple interest. Five years later she pays them back \(\$ 3600\), which includes simple interest on the loan. What was the interest rate?
    \begin{solutionordottedlines}[1in]
        The total interest paid was \(\$ 600\), the principal was \(\$ 3000\) and the time was 5 years.

        \[
        \begin{aligned}
        I & =P R T \\
        600 & =3000 \times R \times 5 \\
        600 & =15000 \times R \\
        R & =\frac{600}{15000} \times 100 \% \\
        & =4 \%
        \end{aligned}
        \]
    \end{solutionordottedlines}
    \end{questions}
\end{examplebox}

\begin{theorembox}
    \section*{Simple interest formula}
    \begin{itemize}
      \item Suppose that a principal \(P\) is invested for \(T\) years at an interest rate \(R\) p.a. Then the total interest \(I\) is given by:
    \end{itemize}

    \[
    I=P R T
    \]

    Remember that \(R\) is a percentage. If the interest rate is \(5 \%\), then \(R=0.05\).

    \begin{itemize}
      \item If the interest rate \(R\) is given per year, the time \(T\) must be given in years.

      \item The formula has four pronumerals. If any three are known, the fourth can be found by substitution.

    \end{itemize}

\end{theorembox}

\begin{exercisebox}
    \subsection{Exercises:}
    \begin{questions}
        \question[2] \(\$ 12000\) is invested at \(7 \%\) p.a. simple interest for five years.
        \begin{parts}
            \part How much interest will be earned each year?
                \begin{solutionordottedlines}[0.5in]
                    Annual interest = \(\$ 12000 \times 7\% = \$ 840\)
                \end{solutionordottedlines}
            \part Find how much interest will be earned over the five-year period.
                \begin{solutionordottedlines}[0.5in]
                    Total interest = \(\$ 840 \times 5 = \$ 4200\)
                \end{solutionordottedlines}
        \end{parts}
        \question \(\$ 2000\) is invested at \(6.75 \%\) p.a. simple interest for three years.
        \begin{parts}
            \part How much interest will be earned each year?
                \begin{solutionordottedlines}[0.5in]
                    Annual interest = \(\$ 2000 \times 6.75\% = \$ 135\)
                \end{solutionordottedlines}
            \part Find how much interest will be earned over the three-year period.
                \begin{solutionordottedlines}[0.5in]
                    Total interest = \(\$ 135 \times 3 = \$ 405\)
                \end{solutionordottedlines}
        \end{parts}
        \question[3] Find the total simple interest earned in each of these investments.
        \begin{parts}
            \part \(\$ 400\) for three years at \(6 \%\) p.a.
                \begin{solutionordottedlines}[0.5in]
                    Total interest = \(\$ 400 \times 6\% \times 3 = \$ 72\)
                \end{solutionordottedlines}
            \part \(\$ 850\) for six years at \(4.5 \%\) p.a.
                \begin{solutionordottedlines}[0.5in]
                    Total interest = \(\$ 850 \times 4.5\% \times 6 = \$ 229.50\)
                \end{solutionordottedlines}
            \part \(\$ 15000\) for 12 years at \(8.4 \%\) p.a.
                \begin{solutionordottedlines}[0.5in]
                    Total interest = \(\$ 15000 \times 8.4\% \times 12 = \$ 15120\)
                \end{solutionordottedlines}
        \end{parts}
        \question[1] Find the time \(T\) for \(\$ 2000\) of simple interest on a principal of \(\$ 8000\) at a rate of \(5 \%\) p.a.
        \begin{solutionordottedlines}[1in]
            \(2000 = 8000 \times 5\% \times T\)
            \(T = \frac{2000}{8000 \times 5\%} = 5\) years
        \end{solutionordottedlines}
        \question[1] Find the rate \(R\) p.a. for \(\$ 7200\) of simple interest on a principal of \(\$ 8000\) over 12 years.
        \begin{solutionordottedlines}[1in]
            \(7200 = 8000 \times R \times 12\)
            \(R = \frac{7200}{8000 \times 12} = 0.075\) or \(7.5\% \) p.a.
        \end{solutionordottedlines}
        \question Find the principal \(P\) for \(\$ 3500\) of simple interest at a rate of \(7 \%\) p.a. over 10 years.
        \begin{solutionordottedlines}[1in]
            \(3500 = P \times 7\% \times 10\)
            \(P = \frac{3500}{7\% \times 10} = \$ 5000\)
        \end{solutionordottedlines}
        \question[2] Regan has arranged to borrow \(\$ 10000\) at \(9.5 \%\) p.a. for four years. She will pay simple interest to the bank every year for the loan, with the principal remaining unchanged. How much interest will Regan pay over the four years of the loan?
        \begin{solutionordottedlines}[1in]
            Annual interest = \(\$ 10000 \times 9.5\% = \$ 950\) \\
            Total interest over 4 years = \(\$ 950 \times 4 = \$ 3800\)
        \end{solutionordottedlines}
        \question[2] Tyler intends to live on the interest on an investment with the bank at \(8.6 \%\) p.a. simple interest. She will receive \(\$ 68000\) simple interest every year from the investment. How much money has she invested?
        \begin{solutionordottedlines}[1in]
            Principal = \(\$ 68000\) / \(8.6\%\) \\
            Principal = \(\$ 68000\) / \(0.086 \approx \$ 790697.67\)
        \end{solutionordottedlines}
    \end{questions}
\end{exercisebox}