Remember we introduced compound interest back in section 4? Here are some practise questions now.

\fbox{\parbox{\textwidth}{\textbf{Recall that:} Compound interest applies interest to the \emph{outstanding amount}, unlike \textsc{simple interest} which applies it to the \emph{principal} amount.}}

\begin{examplebox}
    \subsection{Examples:}
    \begin{questions}
        \question[6] Gail has invested \(\$ 100000\) for six years with the Mountain Bank. The bank pays her interest at the rate of \(7.5 \%\) p.a., compounded annually.
        \begin{parts}
            \part How much will the investment be worth at the end of one year?
            \begin{solutionordottedlines}[1in]
                Each year the investment is worth \(107.5 \%\) of its value the previous year.

                Amount at the end of one year \(=100000 \times 1.075\)

                \[
                =\$ 107500
                \]
            \end{solutionordottedlines}
            \part How much will the investment be worth at the end of two years?
            \begin{solutionordottedlines}[1in]
                Amount at the end of two years \(=100000 \times 1.075 \times 1.075\)

                \[
                \begin{aligned}
                & =100000 \times(1.075)^{2} \\
                & =\$ 115562.50
                \end{aligned}
                \]

            \end{solutionordottedlines}
            \part How much will the investment be worth at the end of six years?
            \begin{solutionordottedlines}[1in]
                Amount at the end of six years \(=100000 \times(1.075)^{6}\)

                \[
                \approx \$ 154330.15
                \]
            \end{solutionordottedlines}
            \part What is the percentage increase on her original investment at the end of six years?
            \begin{solutionordottedlines}[1in]
                Final amount \(=\) original amount \(\times(1.075)^{6}\)

                \[
                \approx \text { original amount } \times 1.5433
                \]

                So the total increase over six years is about \(54.33 \%\).
            \end{solutionordottedlines}
            \part What is the total interest earned over the six years?
            \begin{solutionordottedlines}[1in]
                Total interest \(\approx 154330.15-100000\)

                \[
                =\$ 54330.15
                \]
            \end{solutionordottedlines}
            \part What would the simple interest on the investment have been, assuming the same interest rate of \(7.5 \%\) p.a.?
            \begin{solutionordottedlines}[1in]
                Simple interest \(=P R T\)

                \[
                \begin{aligned}
                & =100000 \times 0.075 \times 6 \\
                & =\$ 45000
                \end{aligned}
                \]

                Note: Compound interest for two or more years is always greater than simple interest for two or more years.
            \end{solutionordottedlines}
        \end{parts}
        \question[4] \textbf{Compound Interest on a loan:} Hussain is setting up a plumbing business and needs to borrow \(\$ 150000\) from a bank to buy a truck and other equipment. The bank will charge him interest of \(11 \%\) p.a., compounded annually. Hussain will pay the whole loan off all at once four years later.
        \begin{parts}
            \part How much will Hussain owe the bank at the end of four years?
            \begin{solutionordottedlines}[1in]
                Each year Hussain owes \(111 \%\) of what he owed the previous year.

                Amount at the end of four years \(=150000 \times(1.11)^{4}\)

                \[
                \approx \$ 227710.56
                \]

            \end{solutionordottedlines}
            \part What is the percentage increase in the money owed at the end of four years?
            \begin{solutionordottedlines}[1in]
                Final amount \(=\) original amount \(\times(1.11)^{4}\)

                \[
                \approx \text { original amount } \times 1.5181
                \]

                So the total increase over four years is about \(51.81 \%\).

            \end{solutionordottedlines}
            \part What is the total interest that Hussain will pay on the loan?
            \begin{solutionordottedlines}[1in]
                \(\quad\) Total interest \(\approx 227710.56-150000\)

                \[
                =\$ 77710.56
                \]
            \end{solutionordottedlines}
            \part What would the simple interest on the loan have been, assuming the same interest rate of \(11 \%\) p.a.?
            \begin{solutionordottedlines}[1in]
                Simple interest \(=P R T\)

                \[
                \begin{aligned}
                & =150000 \times 0.11 \times 4 \\
                & =66000
                \end{aligned}
                \]

                Note: Making no repayments on a loan that is accruing compound interest is a risky business practice because, as this example makes clear, the amount owing grows with increasing rapidity as time goes on. This is particularly relevant to credit card debt.
            \end{solutionordottedlines}
        \end{parts}
        \question[2] Eleni wants to borrow money for three years to start a business, and then pay all the money back, with interest, at the end of that time. The bank will not allow her final debt, including interest, to exceed \(\$ 300000\). Interest is \(9 \%\) p.a., compounded annually. What is the maximum amount that Eleni can borrow?
        \begin{solutionordottedlines}[1in]
            Each year Eleni will owe \(109 \%\) of what she owed the previous year.

            Hence \(\quad\) final debt \(=\) original debt \(\times 1.09 \times 1.09 \times 1.09\)

            \[
            \text { final debt }=\text { original debt } \times(1.09)^{3}
            \]

            Reversing this, original debt \(=\) final debt \(\div(1.09)^{3}\)

            \[
            \begin{aligned}
            & =300000 \div(1.09)^{3} \\
            & \approx \$ 231655
            \end{aligned}
            \]
        \end{solutionordottedlines}
    \end{questions}


\end{examplebox}

\begin{exercisebox}
    \subsection{Exercises:}
\begin{questions}
    \question[5] Christine invested \(\$ 100000\) for six years at 5\% p.a. interest, compounded annually.
    \begin{parts}
        \part By multiplying by 1.05, find the value of the investment after one year.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 100000 \times 1.05 = \$ 105000\)
        \end{solutionordottedlines}
        \part By multiplying by \((1.05)^{2}\), find the value of the investment after two years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 100000 \times (1.05)^{2} = \$ 110250\)
        \end{solutionordottedlines}
        \part By multiplying by \((1.05)^{6}\), find the value of the investment after six years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 100000 \times (1.05)^{6} \approx \$ 134009.56\)
        \end{solutionordottedlines}
        \part Find the percentage increase in the investment over the six years.
        \begin{solutionordottedlines}[0.5in]
            Percentage increase = \(\frac{\$ 134009.56 - \$ 100000}{\$ 100000} \times 100 \approx 34.01\%\)
        \end{solutionordottedlines}
        \part Find the total interest earned over the six years.
        \begin{solutionordottedlines}[0.5in]
            Total interest = \(\$ 134009.56 - \$ 100000 = \$ 34009.56\)
        \end{solutionordottedlines}
    \end{parts}
    \question[1] Find the simple interest on the principal of \(\$ 100000\) over the six years at the same rate of \(5 \%\) p.a.
    \begin{solutionordottedlines}[1in]
        Simple interest = \(\$ 100000 \times 5\% \times 6 = \$ 30000\)
    \end{solutionordottedlines}
    \question[5] Gary has borrowed \(\$ 200000\) for six years at \(8 \%\) p.a. interest, compounded annually, in order to start his indoor decorating business. He intends to pay the whole amount back, plus interest, at the end of the six years.
    \begin{parts}
        \part Find the amount owing after one year.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 200000 \times 1.08 = \$ 216000\)
        \end{solutionordottedlines}
        \part Find the amount owing after two years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 200000 \times (1.08)^{2} \approx \$ 233280\)
        \end{solutionordottedlines}
        \part Find the amount owing after six years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 200000 \times (1.08)^{6} \approx \$ 318111.47\)
        \end{solutionordottedlines}
        \part Find the percentage increase in the loan over the six years.
        \begin{solutionordottedlines}[0.5in]
            Percentage increase = \(\frac{\$ 318111.47 - \$ 200000}{\$ 200000} \times 100 \approx 59.06\%\)
        \end{solutionordottedlines}
        \part Find the total interest charged over the six years.
        \begin{solutionordottedlines}[0.5in]
            Total interest = \(\$ 318111.47 - \$ 200000 = \$ 118111.47\)
        \end{solutionordottedlines}
    \end{parts}
    \question[1] Find the simple interest on the principal of \(\$ 200000\) over the six years at the same rate of \(8 \%\) p.a.
    \begin{solutionordottedlines}[1in]
        Simple interest = \(\$ 200000 \times 8\% \times 6 = \$ 96000\)
    \end{solutionordottedlines}
    \question[1] A couple take out a housing loan of \(\$ 320000\) over a period of 20 years. They make no repayments over the 20-year period of the loan. Compound interest is payable at \(6 \frac{1}{2} \%\) p.a., compounded annually. How much would they owe at the end of the 20-year period, and what is the total percentage increase?
    \begin{solutionordottedlines}[0.5in]
        Owing amount = \(\$ 320000 \times (1 + 6.5\%)^{20} \approx \$ 1126970.22\) \\
        Percentage increase = \(\frac{\$ 1126970.22 - \$ 320000}{\$ 320000} \times 100 \approx 252.18\%\)
    \end{solutionordottedlines}
    \question[1] The population of a city increases annually at a compound rate of \(3.2 \%\) for five years. If the population is 21000 initially, what is it at the end of the five-year period, and what is the total percentage increase?
    \begin{solutionordottedlines}[0.5in]
        Final population = \(21000 \times (1 + 3.2\%)^{5} \approx 24423.60\) \\
        Percentage increase = \(\frac{24423.60 - 21000}{21000} \times 100 \approx 16.3\%\)
    \end{solutionordottedlines}
    \question[1] Suzette wants to invest a sum of money now so that it will grow to \(\$ 180000\) in 10 years' time. How much should she invest now, given that the interest rate is \(6 \%\) compounded annually?
    \begin{solutionordottedlines}[0.5in]
        Present value = \(\$ 180000 / (1.06)^{10} \approx \$ 100927.08\)
    \end{solutionordottedlines}
    \question[4] A bank offers \(8 \%\) p.a. compound interest. How much needs to be invested if the investment is to be worth \(\$ 100000\) in:
    \begin{parts}\begin{multicols}{2}
        \part 10 years?
        \begin{solutionordottedlines}[0.5in]
            Present value = \(\$ 100000 / (1.08)^{10} \approx \$ 46319.38\)
        \end{solutionordottedlines}
        \part 20 years?
        \begin{solutionordottedlines}[0.5in]
            Present value = \(\$ 100000 / (1.08)^{20} \approx \$ 21454.70\)
        \end{solutionordottedlines}
        \part 25 years?
        \begin{solutionordottedlines}[0.5in]
            Present value = \(\$ 100000 / (1.08)^{25} \approx \$ 14693.57\)
        \end{solutionordottedlines}
        \part 100 years?
        \begin{solutionordottedlines}[0.5in]
            Present value = \(\$ 100000 / (1.08)^{100} \approx \$ 214.55\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
\end{questions}
\end{exercisebox}