| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2008 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Exponential growth/decay model setup |
| Difficulty | Moderate -0.3 This is a straightforward exponential model question requiring substitution to find A, solving k^60 = 100 using logarithms, evaluating P at t=123, and solving an exponential equation by equating and using logs. All steps are routine C4 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context |
| Answer | Marks |
|---|---|
| \(A = 20\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{2000}{A} = k^{60}\) | M1 | |
| \(k = (100)^{\frac{1}{2}} = 1.079775\) | A1 | AG; or \(k = 10^{\frac{\log 100}{60}} = 10^{0.0333}\) or \(\sqrt[60]{100}\) or \(e^{\frac{\ln 100}{60}} = e^{0.0776}\) or \(e^{0.077}\) or 1.079775(6) seen |
| Answer | Marks | Guidance |
|---|---|---|
| \(P = 20 \times k^{2008-1883} = 251780 = 252000\) | M1 A1 | CAO nearest 1000 |
| Answer | Marks | Guidance |
|---|---|---|
| \(15 \times 1.082709^t = 20 \times 1.079775^t\) | M1 | equate prices |
| \(\frac{15}{20} = \left(\frac{1.079775}{1.082709}\right)^t\) | M1 | \(t\) as a single index, or correct log expression at this stage |
| \(t = \frac{\log 0.75}{\log 0.997290}\) | m1 | expression for \(t\) |
| \(t = 106.017 \Rightarrow 1991\) | A1 | SC Answer only/Trial and error 106 seen (2 out of 4); 1991 (4 out of 4) |
**4(a)(i)**
| $A = 20$ | B1 | |
**4(a)(ii)**
| $\frac{2000}{A} = k^{60}$ | M1 | |
| $k = (100)^{\frac{1}{2}} = 1.079775$ | A1 | AG; or $k = 10^{\frac{\log 100}{60}} = 10^{0.0333}$ or $\sqrt[60]{100}$ or $e^{\frac{\ln 100}{60}} = e^{0.0776}$ or $e^{0.077}$ or 1.079775(6) seen |
**4(a)(iii)**
| $P = 20 \times k^{2008-1883} = 251780 = 252000$ | M1 A1 | CAO nearest 1000 |
**4(b)**
| $15 \times 1.082709^t = 20 \times 1.079775^t$ | M1 | equate prices |
| $\frac{15}{20} = \left(\frac{1.079775}{1.082709}\right)^t$ | M1 | $t$ as a single index, or correct log expression at this stage |
| $t = \frac{\log 0.75}{\log 0.997290}$ | m1 | expression for $t$ |
| $t = 106.017 \Rightarrow 1991$ | A1 | SC Answer only/Trial and error 106 seen (2 out of 4); 1991 (4 out of 4) |
**Total: 9 marks**
---4 David is researching changes in the selling price of houses. One particular house was sold on 1 January 1885 for $\pounds 20$. Sixty years later, on 1 January 1945, it was sold for $\pounds 2000$. David proposes a model
$$P = A k ^ { t }$$
for the selling price, $\pounds P$, of this house, where $t$ is the time in years after 1 January 1885 and $A$ and $k$ are constants.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the value of $A$.
\item Show that, to six decimal places, $k = 1.079775$.
\item Use the model, with this value of $k$, to estimate the selling price of this house on 1 January 2008. Give your answer to the nearest $\pounds 1000$.
\end{enumerate}\item For another house, which was sold for $\pounds 15$ on 1 January 1885, David proposes the model
$$Q = 15 \times 1.082709 ^ { t }$$
for the selling price, $\pounds Q$, of this house $t$ years after 1 January 1885. Calculate the year in which, according to these models, these two houses would have had the same selling price.
\end{enumerate}
\hfill \mbox{\textit{AQA C4 2008 Q4 [9]}}