This final section is about \emph{things losing value}. Here our percentages reduce the original value of the items:

\begin{examplebox}
    \subsection{Examples:}
    \begin{questions}
        \question[4] The Medicine Home Delivery Company bought a car four years ago for \(\$ 40000\), and assumed that the value of the car would depreciate at \(20 \%\) p.a.
        \begin{parts}
            \part What value did the car have at the end of two years?
            \begin{solutionordottedlines}[0.5in]
                The value each year is taken to be \(100 \%-20 \%=80 \%\) of the value in the previous year.

                Value at the end of two years \(=40000 \times 0.80 \times 0.80\)

                \[
                \begin{aligned}
                & =40000 \times(0.80)^{2} \\
                & =\$ 25600
                \end{aligned}
                \]

            \end{solutionordottedlines}
            \part What value does the car have now, after four years?
            \begin{solutionordottedlines}[0.5in]
                Value at the end of four years \(=40000 \times 0.80 \times 0.80 \times 0.80 \times 0.80\)

                \[
                \begin{aligned}
                & =40000 \times(0.80)^{4} \\
                & =\$ 16384
                \end{aligned}
                \]


            \end{solutionordottedlines}
            \part What is the percentage decrease in value over the four years?
            \begin{solutionordottedlines}[0.5in]
                \(\quad\) Final value \(=\) original value \(\times(0.80)^{4}\)

                \[
                =\text { original value } \times 0.4096
                \]

                Hence the percentage decrease over four years is \(100 \%-40.96 \%=59.04 \%\)

            \end{solutionordottedlines}
            \part What is the average depreciation on the car over the four years? (Express your answer in dollars per year)
            \begin{solutionordottedlines}[0.5in]
                Depreciation over four years \(=40000-16384\)

                \[
                =\$ 23616
                \]

                Average depreciation per year \(=23616 \div 4\)

                \[
                =\$ 5904 \text { per year }
                \]
            \end{solutionordottedlines}
        \end{parts}
        \question[2] A school buys new computers every four years. At the end of the four years, it offers them for sale to the students on the assumption that they have depreciated at \(35 \%\) p.a. (per annum). The school is presently advertising some computers at \(\$ 400\) each.
        \begin{parts}
            \part What did each computer cost the school originally?
            \begin{solutionordottedlines}[1in]
                Each year a computer is worth \(100 \%-35 \%=65 \%\) of its value the previous year.

                Hence final value \(=\) original value \(\times 0.65 \times 0.65 \times 0.65 \times 0.65\)

                \[
                \text { final value }=\text { original value } \times(0.65)^{4}
                \]

                Reversing this, original value \(=\) final value \(\div(0.65)^{4}\)

                \[
                \begin{aligned}
                & =400 \div(0.65)^{4} \\
                & \approx \$ 2241
                \end{aligned}
                \]


            \end{solutionordottedlines}
            \part What is the average depreciation on each computer, in dollars per year?
            \begin{solutionordottedlines}[1in]
                Depreciation over four years \(\approx 2241-400\)

                \[
                =\$ 1841
                \]

                Average depreciation per year \(\approx 1841 \div 4\)

                \[
                \approx \$ 460
                \]
            \end{solutionordottedlines}
        \end{parts}

    \end{questions}


\end{examplebox}

\begin{exercisebox}
    \subsection{Exercises:}
\begin{questions}
    \question[6] The landlord of a large block of home units purchased washing machines for its units four years ago for \(\$ 400000\), and is assuming a depreciation rate of \(30 \%\).
    \begin{parts}
        \part By multiplying by 0.70, find the value after one year.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 400000 \times 0.70 = \$ 280000\)
        \end{solutionordottedlines}
        \part By multiplying by \((0.70)^{2}\), find the value after two years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 400000 \times (0.70)^{2} = \$ 196000\)
        \end{solutionordottedlines}
        \part By multiplying by \((0.70)^{3}\), find the value after three years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 400000 \times (0.70)^{3} = \$ 137200\)
        \end{solutionordottedlines}
        \part By multiplying by \((0.70)^{4}\), find the value after four years.
        \begin{solutionordottedlines}[0.5in]
            \(\$ 400000 \times (0.70)^{4} = \$ 96040\)
        \end{solutionordottedlines}
        \part What is the percentage decrease in value over the four years?
        \begin{solutionordottedlines}[0.5in]
            Percentage decrease = \((1 - \frac{\$ 96040}{\$ 400000}) \times 100 \approx 76\%\)
        \end{solutionordottedlines}
        \part What is the average depreciation on the washing machines, in dollars p.a. over the four years?
        \begin{solutionordottedlines}[0.5in]
            Average depreciation = \(\frac{\$ 400000 - \$ 96040}{4} \approx \$ 76049\) per year
        \end{solutionordottedlines}
    \end{parts}
    \question[1] A business spent \(\$ 560000\) installing alarms at its premises and then depreciated them at \(20 \%\) p.a. Find the value after five years, and the percentage depreciation of their value.
        \begin{solutionordottedlines}[0.5in]
            Value after five years = \(\$ 560000 \times (0.80)^{5} \approx \$ 180224\) \\
            Percentage depreciation = \((1 - \frac{\$ 180224}{\$ 560000}) \times 100 \approx 67.82\%\)
        \end{solutionordottedlines}
    \question[1] The population of a sea lion colony decreases at a compound rate of \(2 \%\) p.a. for 10 years. If the population is 8000 initially, what is it at the end of the 10-year period?
        \begin{solutionordottedlines}[0.5in]
            Final population = \(8000 \times (0.98)^{10} \approx 6612\) sea lions
        \end{solutionordottedlines}
    \question[1] The Northern Start Bus Company bought a bus for \(\$ 480000\), depreciated it at \(30 \%\) p.a., and sold it again seven years later for \(\$ 60000\). Was the price that they obtained better or worse than the depreciated value, and by how much?
        \begin{solutionordottedlines}[0.5in]
            Depreciated value = \(\$ 480000 \times (0.70)^{7} \approx \$ 33684\) \\
            The sale price was better by \(\$ 60000 - \$ 33684 = \$ 26316\)
        \end{solutionordottedlines}
    \question[5] Mr Wong's 10-year-old used car is worth \(\$ 4000\), and has been depreciating at \(22.5 \%\) p.a. (Calculate amounts of money in whole dollars.)
    \begin{parts}
        \part Use division by 0.775 to find how much it was worth a year ago.
        \begin{solutionordottedlines}[0.5in]
            Value a year ago = \(\$ 4000 / 0.775 \approx \$ 5161\)
        \end{solutionordottedlines}
        \part Find how much it was worth two years ago.
        \begin{solutionordottedlines}[0.5in]
            Value two years ago = \(\$ 5161 / 0.775 \approx \$ 6660\)
        \end{solutionordottedlines}
        \part Find how much it was worth 10 years ago.
        \begin{solutionordottedlines}[0.5in]
            Value 10 years ago = \(\$ 4000 / (0.775)^{10} \approx \$ 20860\)
        \end{solutionordottedlines}
        \part What is the total percentage depreciation on the car over the 10-year period?
        \begin{solutionordottedlines}[0.5in]
            Percentage depreciation = \((1 - \frac{\$ 4000}{\$ 20860}) \times 100 \approx 80.8\%\)
        \end{solutionordottedlines}
        \part What was the average depreciation in dollars per year over the 10-year period?
        \begin{solutionordottedlines}[0.5in]
            Average depreciation = \(\frac{\$ 20860 - \$ 4000}{10} \approx \$ 1686\) per year
        \end{solutionordottedlines}
    \end{parts}
\end{questions}
\end{exercisebox}
