| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Linear transformation to find constants |
| Difficulty | Moderate -0.8 This is a standard A-level question on exponential models using logarithmic linearization. It requires routine application of log laws (log₁₀(ab^N) = log₁₀a + Nlog₁₀b) and reading values from a given line equation. The multi-part structure guides students through each step methodically, making it easier than average despite covering several marks. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context |
| Year | 2009 | 2010 | 2011 | 2012 | 2013 | 2014 | 2015 | 2016 | ||
| 56.4 | 74.5 | 86.9 | 97.7 | 109.3 | 141.9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\log_{10} a = 1.76\), \(a = 10^{1.76}\), \(a = 57.5\) | B1 | Writes at least one of: \(\log_{10} a = 1.76\) or \(\log a = 1.76\) or \(a = 10^{1.76}\); AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(b = 1.14\) | B1 | AWRT 1.14 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(100(b-1)\) = 14% | B1F | FT their \(b\) where \(b > 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(V = 57.5 \times 1.14^{16} = 467.9\); £467 900 000 | M1 | Substitutes \(N=16\) into \(\log_{10} V = 0.057N + 1.76\) or into \(V = a \times b^N\) using their \(b\) and \(a = 57.5\) or AWRT 57.5; PI AWRT 467.9 or 469.9 |
| Value in interval [£467 800 000, £470 000 000] | A1 | Must include £ or pounds; accept use of millions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| e.g. "Sales may suddenly fall due to unforeseen circumstances such as a pandemic." | E1 | Gives a reason in context why extrapolation from the model may not be valid; must include reference to sales or shopping |
## Question 6:
**Part 6(a)(i):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\log_{10} a = 1.76$, $a = 10^{1.76}$, $a = 57.5$ | B1 | Writes at least one of: $\log_{10} a = 1.76$ or $\log a = 1.76$ or $a = 10^{1.76}$; AG |
**Part 6(a)(ii):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $b = 1.14$ | B1 | AWRT 1.14 |
**Part 6(b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $100(b-1)$ = 14% | B1F | FT their $b$ where $b > 1$ |
**Part 6(c)(i):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $V = 57.5 \times 1.14^{16} = 467.9$; £467 900 000 | M1 | Substitutes $N=16$ into $\log_{10} V = 0.057N + 1.76$ or into $V = a \times b^N$ using their $b$ and $a = 57.5$ or AWRT 57.5; PI AWRT 467.9 or 469.9 |
| Value in interval [£467 800 000, £470 000 000] | A1 | Must include £ or pounds; accept use of millions |
**Part 6(c)(ii):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. "Sales may suddenly fall due to unforeseen circumstances such as a pandemic." | E1 | Gives a reason in context why extrapolation from the model may not be valid; must include reference to sales or shopping |6 Victoria, a market researcher, believes the average weekly value, $\pounds V$ million, of online grocery sales in the UK has grown exponentially since 2009.
Victoria models the incomplete data, shown in the table, using the formula
$$V = a \times b ^ { N }$$
where $N$ is the number of years since 2009 and $a$ and $b$ are constants.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Year & 2009 & 2010 & 2011 & 2012 & 2013 & 2014 & 2015 & 2016 \\
\hline
\begin{tabular}{ l }
Average Weekly Sales \\
$\pounds V$ million \\
\end{tabular} & 56.4 & & 74.5 & 86.9 & 97.7 & 109.3 & & 141.9 \\
\hline
\end{tabular}
\end{center}
6
\begin{enumerate}[label=(\alph*)]
\item Victoria wishes to determine the values of $a$ and $b$ in her formula.\\
To do this she plots a graph of $\log _ { 10 } V$ against $N$ and then draws a line of best fit as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{de8a7d38-a665-4feb-854e-ac83f413d133-08_757_1040_1169_589}
The equation of Victoria's line of best fit is
$$\log _ { 10 } V = 0.057 N + 1.76$$
6 (a) (i) Use the equation of Victoria's line of best fit to show that, correct to three significant figures, $a = 57.5$\\[0pt]
[1 mark]\\
6 (a) (ii) Use the equation of Victoria's line of best fit to find the value of $b$\\
Give your answer to three significant figures.
6
\item According to Victoria's model, state the yearly percentage increase in the average weekly value of online grocery sales.
6
\item (i) Use Victoria's model to predict the average weekly value of online grocery sales in 2025.\\
6 (c) (ii) Explain why the prediction made in part (c)(i) may be unreliable.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2023 Q6 [6]}}