Here are some (mostly) unseen problems. If you have been attempting \textit{ALL} of your homework carefully, then you may be able to recognise a question or two from there...

\begin{problembox}
    \subsection*{Problems:}
    \begin{questions}
        \question[2] There are 740 students at a primary school, \(5 \%\) of whom have red hair. Calculate the number of students in the school who have red hair.
        \begin{solutionordottedlines}[1in]
        \(740 \times 0.05 = 37\) students
        \end{solutionordottedlines}
        \question[2] A soccer match lasted 92 minutes (including injury time). If team A was in possession for \(55 \%\) of the match, for how many minutes and seconds was team A in possession?
        \begin{solutionordottedlines}[1in]
        \(92 \times 0.55 = 50.6\) minutes, which is 50 minutes and \(0.6 \times 60 = 36\) seconds.
        \end{solutionordottedlines}
      \question[4]
      The label on a Sunnyvale tomato paste bottle says that in every \(25 \mathrm{~g}\) serving, there are \(3.6 \mathrm{~g}\) of carbohydrate, \(0.1 \mathrm{~g}\) of fat, and \(105 \mathrm{mg}\) of sodium.
      \begin{parts}
        \part Express as a percentage of the \(25 \mathrm{~g}\) serving:
        \begin{parts}
          \part the mass of carbohydrate
          \begin{solutionordottedlines}[1in]
          \(\frac{3.6}{25} \times 100 = 14.4\%\)
          \end{solutionordottedlines}
          \part the mass of fat
          \begin{solutionordottedlines}[1in]
          \(\frac{0.1}{25} \times 100 = 0.4\%\)
          \end{solutionordottedlines}
          \part the mass of sodium
          \begin{solutionordottedlines}[1in]
          First convert sodium to grams: \(105 \mathrm{mg} = 0.105 \mathrm{g}\)\\
          \(\frac{0.105}{25} \times 100 = 0.42\%\)
          \end{solutionordottedlines}
        \end{parts}
        \part The Sunnyvale website claims that the percentage of protein is \(3.2\mathrm{~\%}\). What mass of protein is that per \(25 \mathrm{~g}\) serving?
          \begin{solutionordottedlines}[2in]
          \(25 \times \frac{3.2}{100} = 0.8 \mathrm{~g}\)
          \end{solutionordottedlines}
      \end{parts}
      \question[2]
      Mt Kosciusko has a height of \(2228 \mathrm{~m}\), while the height of Mt Everest is \(8848 \mathrm{~m}\). Calculate your answers to this question correct to 3 decimal places.
      \begin{parts}
        \part What percentage is the height of Mt Everest of the height of Mt Kosciusko?
          \begin{solutionordottedlines}[1in]
          \(\frac{8848}{2228} \times 100 \approx 397.129\%\)
          \end{solutionordottedlines}
        \part The Earth's radius is about \(6400 \mathrm{~km}\). What percentage of the radius of the Earth is the height of Mt Everest?
          \begin{solutionordottedlines}[1in]
          \(\frac{8848}{6400000} \times 100 \approx 0.138\%\)
          \end{solutionordottedlines}
      \end{parts}
      \question[2] In the Federal Parliament, there are 150 members in the House of Representatives, of whom 37 are from Victoria.
      \begin{parts}
        \part Correct to 1 decimal place, what percentage of members are from Victoria?
          \begin{solutionordottedlines}[1in]
          \(\frac{37}{150} \times 100 \approx 24.7\%\)
          \end{solutionordottedlines}
        \part The population of Australia is about 22.6 million. What percentage of Australians are members of the House of Representatives?
          \begin{solutionordottedlines}[1in]
          \(\frac{150}{22600000} \times 100 \approx 0.000664\%\)
          \end{solutionordottedlines}
      \end{parts}
      \question[2] The distance by air from Melbourne to Darwin is \(3346 \mathrm{~km}\), and from Melbourne to Singapore it is \(6021 \mathrm{~km}\). What percentage, correct to the nearest percent, is:
      \begin{parts}
        \part the Melbourne-Darwin distance of the Melbourne-Singapore distance?
          \begin{solutionordottedlines}[1in]
          \(\frac{3346}{6021} \times 100 \approx 55.6\%\) (rounded to 56\%)
          \end{solutionordottedlines}
        \part the Melbourne-Singapore distance of the Melbourne-Darwin distance?
          \begin{solutionordottedlines}[1in]
          \(\frac{6021}{3346} \times 100 \approx 179.9\%\) (rounded to 180\%)
          \end{solutionordottedlines}
      \end{parts}
    \question[4] A book dealer sells rare books and charges a commission of \(8 \%\) on the selling price. Find, correct to the nearest cent, the commission charged on a book that sells for these prices, and the amount that the seller eventually receives.
    \begin{parts}\begin{multicols}{2}
      \part \(\$ 400\)
          \begin{solutionordottedlines}[1cm]
          Commission: \(400 \times 0.08 = \$32\)\\
          Seller receives: \(400 - 32 = \$368\)
          \end{solutionordottedlines}
      \part \(\$ 1300\)
          \begin{solutionordottedlines}[1cm]
          Commission: \(1300 \times 0.08 = \$104\)\\
          Seller receives: \(1300 - 104 = \$1196\)
          \end{solutionordottedlines}
      \part \(\$ 575\)
          \begin{solutionordottedlines}[1cm]
          Commission: \(575 \times 0.08 = \$46\)\\
          Seller receives: \(575 - 46 = \$529\)
          \end{solutionordottedlines}
      \part \(\$ 142.50\)
          \begin{solutionordottedlines}[1cm]
          Commission: \(142.50 \times 0.08 = \$11.40\)\\
          Seller receives: \(142.50 - 11.40 = \$131.10\)
          \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[4] Shara the stockbroker charges \(0.15 \%\) commission on all shares that she sells for clients. In each case, find the price at which a parcel of shares was sold if her commission was:
    \begin{parts}\begin{multicols}{2}
      \part \(\$ 30.00\)
          \begin{solutionordottedlines}[1cm]
          Price of shares: \(\frac{30}{0.0015} = \$20000\)
          \end{solutionordottedlines}
      \part \(\$ 67.35\)
          \begin{solutionordottedlines}[1cm]
          Price of shares: \(\frac{67.35}{0.0015} = \$44900\)
          \end{solutionordottedlines}
      \part \(\$ 384.75\)
          \begin{solutionordottedlines}[1cm]
          Price of shares: \(\frac{384.75}{0.0015} = \$256500\)
          \end{solutionordottedlines}
      \part \(\$ 36.51\)
          \begin{solutionordottedlines}[1cm]
          Price of shares: \(\frac{36.51}{0.0015} = \$24340\)
          \end{solutionordottedlines}
    \end{multicols}\end{parts}
    \question[2] Madeline has received \(\$ 168000\) in total simple interest payments on an investment of \(\$ 400000\) that she made six years ago. What rate of interest has the bank been paying?
          \begin{solutionordottedlines}[1in]
          Interest rate: \(\frac{168000}{400000 \times 6} \times 100 = 7\%\)
          \end{solutionordottedlines}
    \question[2] An investor wishes to earn \(\$ 240000\) interest over a five-year period from an account that earns \(12.5 \%\) p.a. simple interest. How much does the investor have to deposit into the account?
          \begin{solutionordottedlines}[1in]
          Principal: \(\frac{240000}{5 \times 0.125} = \$384000\)
          \end{solutionordottedlines}
    \question[5]
    \begin{parts}
      \part Find, correct to 2 decimal places, the percentage decrease necessary to restore a quantity to its original value if it has been increased by:
      \begin{parts}\begin{multicols}{2}
        \part \(10 \%\)
          \begin{solutionordottedlines}[1cm]
          Decrease: \(\frac{10}{110} \times 100 \approx 9.09\%\)
          \end{solutionordottedlines}
        \part \(22 \%\)
          \begin{solutionordottedlines}[1cm]
          Decrease: \(\frac{22}{122} \times 100 \approx 18.03\%\)
          \end{solutionordottedlines}
        \part \(240 \%\)
          \begin{solutionordottedlines}[1cm]
          Decrease: \(\frac{240}{340} \times 100 \approx 70.59\%\)
          \end{solutionordottedlines}
        \part \(2.3 \%\)
          \begin{solutionordottedlines}[1cm]
          Decrease: \(\frac{2.3}{102.3} \times 100 \approx 2.25\%\)
          \end{solutionordottedlines}
      \end{multicols}\end{parts}
      \part Find, correct to 2 decimal places, the percentage increase necessary to restore a quantity to its original value if it has been decreased by:
      \begin{parts}
        \part \(10 \%\)
          \begin{solutionordottedlines}[1cm]
          Increase: \(\frac{10}{90} \times 100 \approx 11.11\%\)
          \end{solutionordottedlines}
        \part \(22 \%\)
          \begin{solutionordottedlines}[1cm]
          Increase: \(\frac{22}{78} \times 100 \approx 28.21\%\)
          \end{solutionordottedlines}
        \part \(75 \%\)
          \begin{solutionordottedlines}[1cm]
          Increase: \(\frac{75}{25} \times 100 = 300\%\)
          \end{solutionordottedlines}
        \part \(2.3 \%\)
          \begin{solutionordottedlines}[1cm]
          Increase: \(\frac{2.3}{97.7} \times 100 \approx 2.35\%\)
          \end{solutionordottedlines}
      \end{parts}
    \end{parts}
    \question[3] The radioactivity of any sample of the element strontium- 90 decreases by \(90.75 \%\) every century. Find the percentage reduction in radioactivity over each of the periods given below. (Calculate percentages correct to 3 decimal places.)
    \begin{parts}
      \part Two centuries
          \begin{solutionordottedlines}[1cm]
          Reduction: \(100 - (100 - 90.75)^2 \approx 99.437\%\)
          \end{solutionordottedlines}
      \part Three centuries
          \begin{solutionordottedlines}[1cm]
          Reduction: \(100 - (100 - 90.75)^3 \approx 99.969\%\)
          \end{solutionordottedlines}
      \part Five centuries
          \begin{solutionordottedlines}[1cm]
          Reduction: \(100 - (100 - 90.75)^5 \approx 99.999\%\)
          \end{solutionordottedlines}
    \end{parts}
    \question Here is a table of the annual inflation rate in Australia in the years ending 30 June 2001 to 30 June 2006 (from the Reserve Bank of Australia website).

\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|}
\hline
\begin{tabular}{l}
Year \\
\begin{tabular}{l}
Inflation \\
rate \\
\end{tabular} \\
\end{tabular} & 2001 & 2002 & 2003 & 2004 & 2005 & 2006 \\
\hline
\end{tabular}
\end{center}

Suppose the salary for certain jobs at Company \(\mathrm{X}\) rises on the 1 July every year, in line with Australia's inflation rate for the financial year just past (ending 30 June).
\begin{parts}
  \part A junior secretary earned \(\$ 40000\) from 1 July 2000 to 30 June 2001. Determine how much someone in that position would earn from:
  \begin{parts}
    \part 1 July 2001 to 30 June 2002
          \begin{solutionordottedlines}[1cm]
          Assuming an inflation rate of \(x\%\), the salary would be: \(40000 \times (1 + \frac{x}{100})\)
          \end{solutionordottedlines}
    \part 1 July 2006 to 30 June 2007
          \begin{solutionordottedlines}[1cm]
          Assuming an inflation rate of \(y\%\), the salary would be: \(40000 \times (1 + \frac{y}{100})\)
          \end{solutionordottedlines}
  \end{parts}
  \part A team manager was on an annual salary of \(\$ 100000\) from 1 July 2006 to 30 June 2007.
  Determine how much someone in that position would earn:
  \begin{parts}
    \part in the previous financial year
          \begin{solutionordottedlines}[1cm]
          Assuming an inflation rate of \(z\%\), the salary would be: \(100000 \div (1 + \frac{z}{100})\)
          \end{solutionordottedlines}
    \part from 1 July 2003 to 30 June 2004
          \begin{solutionordottedlines}[1cm]
          Assuming an inflation rate of \(w\%\), the salary would be: \(100000 \times (1 + \frac{w}{100})\)
          \end{solutionordottedlines}
  \end{parts}
  \end{parts}
    \question[2] One six-year loan attracts compound interest calculated at 2\%, 4\%, 6\%, 8\%, 10\% and \(12 \%\) in successive years. Another six-year loan attracts compound interest calculated at \(12 \%, 10 \%, 8 \%, 6 \%, 4 \%\) and \(2 \%\) in successive years. Find the total percentage increase in money owing in both cases, compare the two results, and explain what has happened.
    \begin{solutionordottedlines}[2in]
    First loan: \(P(1.02)(1.04)(1.06)(1.08)(1.10)(1.12)\)\\
    Second loan: \(P(1.12)(1.10)(1.08)(1.06)(1.04)(1.02)\)\\
    The total percentage increase for both loans will be the same due to the commutative property of multiplication.
    \end{solutionordottedlines}
    \question[2] An investment at an interest rate of \(10 \%\) p.a., compounded annually, returned interest of \(\$ 40000\) after five years. Calculate the original amount invested.
    \begin{solutionordottedlines}[2in]
    Let \(P\) be the original amount.\\
    \(P(1.10)^5 - P = 40000\)\\
    \(P = \frac{40000}{(1.10)^5 - 1}\)\\
    \(P \approx \$24457.88\)
    \end{solutionordottedlines}
    \question[4] Ms Wu's seven-year-old car is worth \(\$ 5600\), and has been depreciating at \(22.5 \%\) p.a. Calculate your answers to the nearest dollar.
    \begin{parts}
      \part How much was it worth four years ago?
          \begin{solutionordottedlines}[1in]
          \(P(1 - 0.225)^3 = 5600\)\\
          \(P \approx \$10177\)
          \end{solutionordottedlines}
      \part How much was it worth seven years ago?
          \begin{solutionordottedlines}[1in]
          \(P(1 - 0.225)^7 = 5600\)\\
          \(P \approx \$20000\)
          \end{solutionordottedlines}
      \part \(\mathrm{Ms} \mathrm{Wu}\), however, only bought the car four years ago, at its depreciated value at that time. What has been Ms Wu's average depreciation in dollars over the four years she has owned the car?
          \begin{solutionordottedlines}[1in]
          Ms Wu bought the car for \( \$ 15542 \) and now it's worth \( \$ 5600 \).
          Depreciation over four years is \( 15542 - 5600 = 9942 \).
          Average annual depreciation is \( \frac{9942}{4} \approx 2486 \).
          Ms Wu's average depreciation in dollars over the four years is approximately \( \$ 2486 \) per annum.
          \end{solutionordottedlines}
      \part What was the average depreciation in dollars over the first three years of the car's life?
          \begin{solutionordottedlines}[1in]
          The value of the car after three years is \( P(0.775)^3 \).
          \( P(0.775)^3 = \frac{24834}{(0.775)^3} \).
          \( P(0.775)^3 \approx 24834 \times 0.4420 \).
          \( P(0.775)^3 \approx 10976 \).
          Depreciation over the first three years is \( 24834 - 10976 = 13858 \).
          Average annual depreciation is \( \frac{13858}{3} \approx 4619 \).
          The average depreciation in dollars over the first three years is approximately \( \$ 4619 \) per annum.
          \end{solutionordottedlines}
    \end{parts}
    \question[4] I take \(900 \mathrm{~mL}\) of a liquid and dilute it with \(100 \mathrm{~mL}\) of water. Then I take \(900 \mathrm{~mL}\) of the mixture and again dilute it with \(100 \mathrm{~mL}\) of water. I repeat this process 20 times.
    \begin{parts}
      \part What proportion of the original liquid remains in the mixture at the end?
          \begin{solutionordottedlines}[1in]
          Each dilution is a reduction to \( \frac{900}{1000} = 0.9 \) of the previous amount.
          After 20 dilutions, the proportion remaining is \( 0.9^{20} \).
          \( 0.9^{20} \approx 0.1216 \).
          The proportion of the original liquid remaining in the mixture is approximately \( 0.1216 \).
          \end{solutionordottedlines}
      \part How much mixture should I take if I want it to contain \(20 \mathrm{~mL}\) of the original liquid?
          \begin{solutionordottedlines}[1in]
          Let \( x \) be the amount of mixture to take.
          \( x \times 0.1216 = 20 \).
          \( x = \frac{20}{0.1216} \).
          \( x \approx 164.47 \).
          Approximately \( 164.47 \mathrm{~mL} \) of the mixture should be taken to contain \( 20 \mathrm{~mL} \) of the original liquid.
          \end{solutionordottedlines}
    \end{parts}
    \question[2] I take a sealed glass container and remove \(80 \%\) of the air. Then I remove \(80 \%\) of the remaining air. I do this process six times altogether. What percentage of the original air is left in the container?
          \begin{solutionordottedlines}[1in]
          Each removal is keeping \( 20 \% \) of the previous amount.
          After six removals, the percentage remaining is \( (0.2)^6 \).
          \( (0.2)^6 = 0.000064 \).
          \( 0.000064 \times 100 = 0.0064 \% \).
          The percentage of the original air left in the container is approximately \( 0.0064 \% \).
          \end{solutionordottedlines}
  \end{questions}
\end{problembox}
