A car park operator charges permit holders an annual fee that rises by the same percentage, \(r\%\), at the start of each year. In Year 1 the fee is \(£300\...
Question 1 Report
A car park operator charges permit holders an annual fee that rises by the same percentage, \(r\%\), at the start of each year. In Year 1 the fee is \(£300\).
- Write down an expression, in terms of \(r\), for the fee in Year 3. (2)
- The fee in Year 3 is \(£349.92\). Show that \(r = 8\). (2)
- Using this rate, work out the fee in Year 5, giving your answer to the nearest penny. (2)
- The operator wants the Year 6 fee to be no more than \(£450\). Determine, with a reason, whether continuing to raise the fee by 8% each year keeps the Year 6 fee within this limit. (2)
Answer Details
This question builds a general compound-growth expression, uses given data to confirm the growth rate, then projects forward and checks the projection against a stated limit.
With the fee rising by \(r\%\) at the start of each year, the fee in Year 3 has grown for \(2\) years from the Year 1 fee of \(£300\):
\[300\left(1+\dfrac{r}{100}\right)^2\][2 marks]
Setting this equal to \(£349.92\):
\[300\left(1+\dfrac{r}{100}\right)^2 = 349.92\] \[\left(1+\dfrac{r}{100}\right)^2 = 1.1664\] \[1+\dfrac{r}{100} = \sqrt{1.1664} = 1.08\]so \(r = 8\), as required [2 marks].
The Year 5 fee has grown for \(4\) years from Year 1:
\[£300 \times 1.08^4 = £300 \times 1.36048896 = £408.15 \text{ (nearest penny)}\][2 marks]
The Year 6 fee has grown for \(5\) years:
\[£300 \times 1.08^5 = £440.80 \text{ (nearest penny)}\]Since \(£440.80 \lt £450\), continuing to raise the fee by \(8\%\) each year does keep the Year 6 fee within the operator's limit [2 marks].
Counting years of growth carefully is essential: the Year 1 fee has had no rises yet, so the Year \(n\) fee has grown for \((n-1)\) years, not \(n\) years.
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