| Exam Board | AQA |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Finding x from given y value |
| Difficulty | Moderate -0.3 This is a standard exponential modelling question with routine algebraic manipulation and logarithm application. Parts (a) and (b)(i) are straightforward substitution, (b)(ii) is a textbook logarithm exercise, and part (c) requires equating two exponentials and rearranging—all standard C4 techniques with no novel insight required. Slightly easier than average due to heavy scaffolding and step-by-step guidance. |
| Spec | 1.06f Laws of logarithms: addition, subtraction, power rules1.06g Equations with exponentials: solve a^x = b |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = 5000\) | B1 | Accept £5000 |
| Answer | Marks | Guidance |
|---|---|---|
| On 1 April 2011, \(t = 10\), so \(V = 5000p^{10} = 25000\) | M1 | Substituting \(t=10\) and \(V=25000\) |
| \(p^{10} = \frac{25000}{5000} = 5\) | A1 | Shown convincingly |
| Answer | Marks | Guidance |
|---|---|---|
| \(5000p^t = 75000\) | M1 | Setting up equation |
| \(p^t = 15\) | ||
| \(t\ln p = \ln 15\) | M1 | Taking logarithms |
| \(p = 5^{\frac{1}{10}}\), so \(\ln p = \frac{\ln 5}{10}\) | ||
| \(t = \frac{\ln 15}{\frac{\ln 5}{10}} = \frac{10\ln 15}{\ln 5}\) | A1 | Correct expression |
| \(t \approx 17.1\) years after 2001, so year 2018 | A1 | 2018 |
| Answer | Marks | Guidance |
|---|---|---|
| First painting at time \(T\) years after 1991: \(t = T - 10\), so \(V = 5000p^{T-10}\) | M1 | Expressing \(V\) in terms of \(T\) from 1991 |
| Second painting: \(W = 2500q^T\) | ||
| Setting equal: \(5000p^{T-10} = 2500q^T\) | M1 | Equating the two expressions |
| \(2p^{T-10} = q^T\) | ||
| \(2 \cdot \frac{p^T}{p^{10}} = q^T\) | ||
| \(\frac{2}{5}p^T = q^T\) (since \(p^{10}=5\)) | A1 | Correct simplification |
| \(T\ln p - T\ln q = \ln\frac{5}{2}\) | M1 | Taking logs |
| \(T = \frac{\ln\!\left(\frac{5}{2}\right)}{\ln\!\left(\frac{p}{q}\right)}\) | A1 | Fully shown |
| Answer | Marks | Guidance |
|---|---|---|
| If \(p = 1.029q\), then \(\frac{p}{q} = 1.029\) | M1 | Substituting ratio |
| \(T = \frac{\ln(2.5)}{\ln(1.029)} \approx \frac{0.9163}{0.02859} \approx 32.05\) | ||
| \(T \approx 32\) years after 1991 = year 2023 | A1 | 2023 |
# Question 4:
## Part (a)
| $A = 5000$ | B1 | Accept £5000 |
## Part (b)(i)
| On 1 April 2011, $t = 10$, so $V = 5000p^{10} = 25000$ | M1 | Substituting $t=10$ and $V=25000$ |
| $p^{10} = \frac{25000}{5000} = 5$ | A1 | Shown convincingly |
## Part (b)(ii)
| $5000p^t = 75000$ | M1 | Setting up equation |
| $p^t = 15$ | | |
| $t\ln p = \ln 15$ | M1 | Taking logarithms |
| $p = 5^{\frac{1}{10}}$, so $\ln p = \frac{\ln 5}{10}$ | | |
| $t = \frac{\ln 15}{\frac{\ln 5}{10}} = \frac{10\ln 15}{\ln 5}$ | A1 | Correct expression |
| $t \approx 17.1$ years after 2001, so year **2018** | A1 | 2018 |
## Part (c)(i)
| First painting at time $T$ years after 1991: $t = T - 10$, so $V = 5000p^{T-10}$ | M1 | Expressing $V$ in terms of $T$ from 1991 |
| Second painting: $W = 2500q^T$ | | |
| Setting equal: $5000p^{T-10} = 2500q^T$ | M1 | Equating the two expressions |
| $2p^{T-10} = q^T$ | | |
| $2 \cdot \frac{p^T}{p^{10}} = q^T$ | | |
| $\frac{2}{5}p^T = q^T$ (since $p^{10}=5$) | A1 | Correct simplification |
| $T\ln p - T\ln q = \ln\frac{5}{2}$ | M1 | Taking logs |
| $T = \frac{\ln\!\left(\frac{5}{2}\right)}{\ln\!\left(\frac{p}{q}\right)}$ | A1 | Fully shown |
## Part (c)(ii)
| If $p = 1.029q$, then $\frac{p}{q} = 1.029$ | M1 | Substituting ratio |
| $T = \frac{\ln(2.5)}{\ln(1.029)} \approx \frac{0.9163}{0.02859} \approx 32.05$ | | |
| $T \approx 32$ years after 1991 = year **2023** | A1 | 2023 |4 A painting was valued on 1 April 2001 at $\pounds 5000$.\\
The value of this painting is modelled by
$$V = A p ^ { t }$$
where $\pounds V$ is the value $t$ years after 1 April 2001, and $A$ and $p$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Write down the value of $A$.
\item According to the model, the value of this painting on 1 April 2011 was $\pounds 25000$.
Using this model:
\begin{enumerate}[label=(\roman*)]
\item show that $p ^ { 10 } = 5$;
\item use logarithms to find the year in which the painting will be valued at $\pounds 75000$.
\end{enumerate}\item A painting by another artist was valued at $\pounds 2500$ on 1 April 1991. The value of this painting is modelled by
$$W = 2500 q ^ { t }$$
where $\pounds W$ is the value $t$ years after 1 April 1991, and $q$ is a constant.
\begin{enumerate}[label=(\roman*)]
\item Show that, according to the two models, the value of the two paintings will be the same $T$ years after 1 April 1991,
$$\text { where } T = \frac { \ln \left( \frac { 5 } { 2 } \right) } { \ln \left( \frac { p } { q } \right) }$$
\item Given that $p = 1.029 q$, find the year in which the two paintings will have the same value.\\[0pt]
[1 mark]
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C4 2014 Q4 [11]}}