Now we take the methods from the last section and apply them more than once to the same problem:

\begin{examplebox}
    \subsection{Examples:}
    \begin{questions}
        \question[4] \textbf{Repeated Increase:}\\The population of Abelsburg in the census three years ago was 46430 . In the three years after the census, however, its population has risen by \(6.2 \%, 8.5 \%\) and \(13.1 \%\), respectively.
        \begin{parts}
            \part What was its population one year after the census?
            \begin{solutionordottedlines}[1in]
                One year after the census, the population was \(106.2 \%\) of its original value.

                Hence population after one year \(=46430 \times 1.062\)

                \[
                \approx 49309 \text {, correct to the nearest whole number }
                \]
            \end{solutionordottedlines}
            \part What was its population two years after the census?
            \begin{solutionordottedlines}[1in]
                Two years afterwards, the population was \(108.5 \%\) of its value one year afterwards.

                Hence population after two years \(=(46430 \times 1.062) \times 1.085\)

                \[
                \approx 53500 \text {, correct to the nearest whole number }
                \]

                Note: Do not use the approximation from part \(\mathbf{a}\) to calculate part \(\mathbf{b}\). Either start the calculation again, or use the unrounded value from part \(\mathbf{a}\).
            \end{solutionordottedlines}
            \part What is its population now, three years after the census?
            \begin{solutionordottedlines}[1in]
                Three years afterwards, the population was \(113.1 \%\) of its value two years afterwards.

                Hence population after three years \(=(46430 \times 1.062 \times 1.085) \times 1.131\)

                \[
                \approx 60508 \text {, correct to the nearest whole number }
                \]
            \end{solutionordottedlines}
            \part What was the percentage increase in population over the three years, correct to the nearest \(0.1 \%\) ?
            \begin{solutionordottedlines}[1in]
                \[
                \begin{aligned}
                & =\text { original population } \times 1.062 \times 1.085 \times 1.131 \\
                & \approx \text { original population } \times 1.303 \\
                & \approx \text { original population } \times 130.3 \%
                \end{aligned}
                \]

                Hence the population has increased over the three years by about \(30.3 \%\).

                Note: The percentage increase of \(30.3 \%\) is significantly larger than the sum of the three percentage increases,

                \[
                6.2 \%+8.5 \%+13.1 \%=27.8 \% .
                \]

                Note: The answer to part \(\mathbf{d}\) does not depend on what the original population was.
            \end{solutionordottedlines}
        \end{parts}
        \question[3] \textbf{Repeated Decrease:}\\ Teresa invested \(\$ 75000\) from her inheritance in a mining company that has not been very successful. In the first year, she lost \(55 \%\) of the money, and in the second year, she lost \(72 \%\) of what remained.
        \begin{parts}
            \part How much does she have left after one year?
            \begin{solutionordottedlines}[1in]
                One year later, the percentage remaining was \(100 \%-55 \%=45 \%\).

                Hence amount left after one year \(=75000 \times 0.45\)

                \[
                =\$ 33750
                \]
            \end{solutionordottedlines}
            \part How much does she have left after two years?
            \begin{solutionordottedlines}[1in]
                Two years later, she had \(100 \%-72 \%=28 \%\) of what she had after one year.

                Hence amount left after two years \(=75000 \times 0.45 \times 0.28\)

                \[
                =\$ 9450
                \]

            \end{solutionordottedlines}
            \part What percentage of the original inheritance has she lost over the two years?
            \begin{solutionordottedlines}[1in]
                Amount left after two years \(=\) original investment \(\times 0.45 \times 0.28\)

                \[
                \begin{aligned}
                & =\text { original investment } \times 0.126 \\
                & =\text { original investment } \times 12.6 \%
                \end{aligned}
                \]
                So she has lost \(100 \%-12.6 \%=87.4 \%\) of her investment over the two years.
            \end{solutionordottedlines}
        \end{parts}
        \question[2] \textbf{Combinations of increases and decreases:}\\ The volume of water in the Welcome Dam has varied considerably over the last three years. During the first year the volume rose by \(27 \%\), then it fell \(43 \%\) during the second year, and it rose \(16 \%\) in the third year.
        \begin{parts}
            \part What is the percentage increase or decrease over the three years, correct to the nearest \(1 \%\) ?
            \begin{solutionordottedlines}[1in]
                Final volume \(=\) original volume \(\times 1.27 \times 0.57 \times 1.16\)

                \[
                \approx \text { original volume } \times 0.84
                \]

                Since \(0.84<1\), the volume has decreased. The percentage decrease is about \(100 \%-84 \%=16 \%\) over the three years.
            \end{solutionordottedlines}
            \part If there were 366500 megalitres of water in the dam three years ago, how much water is in the dam now, correct to the nearest 500 megalitres?
            \begin{solutionordottedlines}[1in]
                Final volume \(=\) original volume \(\times 1.27 \times 0.57 \times 1.16\)

                \[
                \begin{aligned}
                & =366500 \times 1.27 \times 0.57 \times 1.16 \\
                & \approx 308000 \text { megalitres, correct to the nearest } 500 \text { megalitres. }
                \end{aligned}
                \]

                This time the sum of the percentages is \(27 \%-43 \%+16 \%=0 \%\), but the volume of water has changed.
            \end{solutionordottedlines}
        \end{parts}
    \end{questions}

\end{examplebox}

\begin{exercisebox}
    \subsection{Exercises:}
\begin{questions}
    \question[1] Oranges used to cost \(\$ 2.80\) per kg, but the price has increased by \(5 \%, 10 \%\) and \(12 \%\) in three successive years. Multiply by \(1.05 \times 1.1 \times 1.12\) to find their price now.
    \begin{solutionordottedlines}[1in]
        New price = \(\$ 2.80 \times 1.05 \times 1.1 \times 1.12 \approx \$ 3.50\) per kg
    \end{solutionordottedlines}
    \question[4] The dividend per share in the Electron Computer Software Company has risen over the last four years by 10\%, 15\%, 5\% and 12\%, respectively. Find the latest dividend received by a shareholder whose dividend four years ago was:
    \begin{parts}\begin{multicols}{2}
        \part \(\$ 1000\)
        \begin{solutionordottedlines}[0.5in]
            Latest dividend = \(\$ 1000 \times 1.1 \times 1.15 \times 1.05 \times 1.12 \approx \$ 1408.62\)
        \end{solutionordottedlines}
        \part \(\$ 1678\)
        \begin{solutionordottedlines}[0.5in]
            Latest dividend = \(\$ 1678 \times 1.1 \times 1.15 \times 1.05 \times 1.12 \approx \$ 2363.62\)
        \end{solutionordottedlines}
        \part \(\$ 28.46\)
        \begin{solutionordottedlines}[0.5in]
            Latest dividend = \(\$ 28.46 \times 1.1 \times 1.15 \times 1.05 \times 1.12 \approx \$ 40.15\)
        \end{solutionordottedlines}
        \part \(\$ 512.21\)
        \begin{solutionordottedlines}[0.5in]
            Latest dividend = \(\$ 512.21 \times 1.1 \times 1.15 \times 1.05 \times 1.12 \approx \$ 722.15\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[4] Rates in Bullimbamba Shire have risen 7\% every year for the last seven years. Find the rates now payable by a landowner whose rates seven years ago were:
    \begin{parts}\begin{multicols}{2}
        \part \(\$ 1000\)
        \begin{solutionordottedlines}[0.5in]
            Current rates = \(\$ 1000 \times (1.07)^7 \approx \$ 1605.78\)
        \end{solutionordottedlines}
        \part \(\$ 346\)
        \begin{solutionordottedlines}[0.5in]
            Current rates = \(\$ 346 \times (1.07)^7 \approx \$ 556.14\)
        \end{solutionordottedlines}
        \part \(\$ 2566.86\)
        \begin{solutionordottedlines}[0.5in]
            Current rates = \(\$ 2566.86 \times (1.07)^7 \approx \$ 4121.98\)
        \end{solutionordottedlines}
        \part \(\$ 788.27\)
        \begin{solutionordottedlines}[0.5in]
            Current rates = \(\$ 788.27 \times (1.07)^7 \approx \$ 1265.76\)
        \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[2] A tree, whose original foliage was estimated to have a mass of \(1500 \mathrm{~kg}\), lost \(20 \%\) of its foliage in a storm, then lost \(15 \%\) of what was left in a storm the next day, then lost \(40 \%\) of what was left in a third storm. Estimate the remaining mass of foliage.
    \begin{solutionordottedlines}[1in]
        Remaining mass = \(1500 \times 0.8 \times 0.85 \times 0.6 \approx 612 \mathrm{~kg}\)
    \end{solutionordottedlines}
\end{questions}
\end{exercisebox}
