This could either be really fun or really chaotic. Either way, here are 8 challenging consumer arithmetic questions for you all to chew on.

\begin{problembox}
  \begin{questions}
    \question[2] The population of a town decreases by \(15 \%\) during 2010. What percentage increase, correct to 2 decimal places, is necessary during 2011 for the population to be restored to the level it was at immediately before the decrease in 2010?
          \begin{solutionordottedlines}[2in]
            Let the original population be \( P \). After a decrease of \( 15\% \), the new population is \( P \times (1 - 0.15) = P \times 0.85 \). To restore the population to \( P \), we need to solve for the percentage increase \( x \) such that:
            \[ P \times 0.85 \times (1 + \frac{x}{100}) = P \]
            Simplifying, we get:
            \[ 0.85 \times (1 + \frac{x}{100}) = 1 \]
            \[ 1 + \frac{x}{100} = \frac{1}{0.85} \]
            \[ \frac{x}{100} = \frac{1}{0.85} - 1 \]
            \[ x = ( \frac{1}{0.85} - 1 ) \times 100 \]
            \[ x = (1.17647 - 1) \times 100 \]
            \[ x = 0.17647 \times 100 \]
            \[ x = 17.647 \]
            The necessary percentage increase is \( 17.65\% \) (rounded to two decimal places).
          \end{solutionordottedlines}
    \question[2] The length of a rectangle is increased by \(15 \%\) and the width is decreased by \(11 \%\). What is the exact percentage change in the area?
          \begin{solutionordottedlines}[2in]
            Let the original length be \( L \) and the original width be \( W \). The original area is \( A = L \times W \).
            The new length is \( L \times 1.15 \) and the new width is \( W \times 0.89 \). The new area is:
            \[ A' = (L \times 1.15) \times (W \times 0.89) \]
            \[ A' = L \times W \times 1.15 \times 0.89 \]
            \[ A' = A \times 1.15 \times 0.89 \]
            The percentage change in the area is:
            \[ \frac{A' - A}{A} \times 100\% \]
            \[ = (1.15 \times 0.89 - 1) \times 100\% \]
            \[ = (1.0235 - 1) \times 100\% \]
            \[ = 0.0235 \times 100\% \]
            \[ = 2.35\% \]
            The exact percentage change in the area is \( 2.35\% \).
          \end{solutionordottedlines}
    \question[2] The radius of a circular pool of water increases by \(8 \%\). What is the exact percentage change in the area of the pool of water?
          \begin{solutionordottedlines}[2in]
            Let the original radius be \( r \). The area of the circle is \( A = \pi r^2 \).
            The new radius is \( r' = r \times 1.08 \). The new area is:
            \[ A' = \pi (r')^2 \]
            \[ A' = \pi (r \times 1.08)^2 \]
            \[ A' = \pi r^2 \times 1.08^2 \]
            \[ A' = A \times 1.08^2 \]
            The percentage change in the area is:
            \[ \frac{A' - A}{A} \times 100\% \]
            \[ = (1.08^2 - 1) \times 100\% \]
            \[ = (1.1664 - 1) \times 100\% \]
            \[ = 0.1664 \times 100\% \]
            \[ = 16.64\% \]
            The exact percentage change in the area of the pool of water is \( 16.64\% \).
          \end{solutionordottedlines}
    \question[2] The area of a circular pool of oil is increased by \(8 \%\). What is the percentage increase of the radius?
          \begin{solutionordottedlines}[2in]
            Let the original area be \( A \) and the original radius be \( r \). The new area is \( A' = A \times 1.08 \).
            Since the area of a circle is \( A = \pi r^2 \), we have:
            \[ A' = \pi (r')^2 \]
            \[ A \times 1.08 = \pi (r')^2 \]
            \[ \frac{A \times 1.08}{\pi} = (r')^2 \]
            \[ r' = \sqrt{\frac{A \times 1.08}{\pi}} \]
            \[ r' = \sqrt{\frac{\pi r^2 \times 1.08}{\pi}} \]
            \[ r' = r \times \sqrt{1.08} \]
            The percentage increase of the radius is:
            \[ \frac{r' - r}{r} \times 100\% \]
            \[ = (\sqrt{1.08} - 1) \times 100\% \]
            \[ \approx (1.03923 - 1) \times 100\% \]
            \[ \approx 0.03923 \times 100\% \]
            \[ \approx 3.923\% \]
            The percentage increase of the radius is approximately \( 3.92\% \) (rounded to two decimal places).
          \end{solutionordottedlines}
    \question[2] A man earns a salary of \(\$ 1440\) for a 44-hour week. His weekly salary is increased by \(10 \%\) and his hours are reduced by \(10 \%\). Find his new hourly salary.
          \begin{solutionordottedlines}[2in]
            The original hourly salary is \( \frac{\$1440}{44 \text{ hours}} = \$32.7272 \) per hour.
            The new weekly salary is \( \$1440 \times 1.10 = \$1584 \).
            The new number of hours is \( 44 \times 0.90 = 39.6 \) hours.
            The new hourly salary is:
            \[ \frac{\$1584}{39.6 \text{ hours}} \approx \$40 \] per hour.
          \end{solutionordottedlines}
    \question[2] In a particular country in \(2010, 15 \%\) of the population is unemployed and \(85 \%\) is employed. In 2011, 10\% of the people unemployed became employed and \(10 \%\) of those employed became unemployed. What percentage of the population is employed now?
          \begin{solutionordottedlines}[2in]
            Let the total population be \( P \). Initially, \( 0.85P \) is employed and \( 0.15P \) is unemployed.
            In 2011, \( 0.10 \times 0.15P = 0.015P \) of the unemployed become employed, and \( 0.10 \times 0.85P = 0.085P \) of the employed become unemployed.
            The new number of employed people is:
            \[ 0.85P - 0.085P + 0.015P \]
            \[ = 0.85P - 0.07P \]
            \[ = 0.78P \]
            The percentage of the population that is employed now is \( 78\% \).
          \end{solutionordottedlines}
    \question[2] The number of trees on Green Plateau fell by \(5 \%\) every year for 10 years. Then the numbers rose by \(5 \%\) every year for 20 years. What was the total percentage gain or loss of trees over the 30-year period?
          \begin{solutionordottedlines}[2in]
            Let the original number of trees be \( T \).
            After 10 years of decreasing by \(5\%\), the number of trees is:
            \[ T \times (1 - 0.05)^{10} \]
            After 20 more years of increasing by \(5\%\), the number of trees is:
            \[ T \times (1 - 0.05)^{10} \times (1 + 0.05)^{20} \]
            The total percentage gain or loss is:
            \[ \frac{T \times (1 - 0.05)^{10} \times (1 + 0.05)^{20} - T}{T} \times 100\% \]
            \[ = ((1 - 0.05)^{10} \times (1 + 0.05)^{20} - 1) \times 100\% \]
            \[ = ((0.95)^{10} \times (1.05)^{20} - 1) \times 100\% \]
            \[ \approx (0.59874 \times 2.65330 - 1) \times 100\% \]
            \[ \approx (1.58846 - 1) \times 100\% \]
            \[ \approx 0.58846 \times 100\% \]
            \[ \approx 58.846\% \]
            The total percentage gain of trees over the 30-year period is approximately \( 58.85\% \) (rounded to two decimal places).
          \end{solutionordottedlines}
    \question[2] Particular shares were released in the stock market and lost, per day, an average of \(2.23 \%\) of their original value over the first four days. Over the first day, the shares increased in value by \(15 \%\), and over the second day, a further \(10 \%\) increase was recorded. However, a \(20 \%\) decrease in the share value occurred on the third day. What percentage decrease was recorded over the fourth day?
          \begin{solutionordottedlines}[2in]
            Let the original value of the shares be \( S \). After the first day, the value is \( S \times 1.15 \). After the second day, the value is \( S \times 1.15 \times 1.10 \). After the third day, the value is \( S \times 1.15 \times 1.10 \times 0.80 \).
            The total loss over four days is \( 2.23\% \times 4 = 8.92\% \) of the original value.
            The value after four days is \( S \times (1 - 0.0892) \).
            We have:
            \[ S \times 1.15 \times 1.10 \times 0.80 \times (1 - x) = S \times (1 - 0.0892) \]
            \[ 1.15 \times 1.10 \times 0.80 \times (1 - x) = 1 - 0.0892 \]
            \[ 0.92 \times (1 - x) = 0.9108 \]
            \[ 1 - x = \frac{0.9108}{0.92} \]
            \[ 1 - x = 0.99043 \]
            \[ x = 0.00957 \]
            \[ x = 0.957\% \]
            The percentage decrease recorded over the fourth day is \( 0.957\% \).
          \end{solutionordottedlines}
    \end{questions}
\end{problembox}
