| Exam Board | AQA |
|---|---|
| Module | AS Paper 2 (AS Paper 2) |
| Session | Specimen |
| Marks | 8 |
| 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 straightforward exponential modeling question requiring basic substitution (parts a,b), a simple logarithm equation (part c), and solving an inequality with logarithms (part d). While it has multiple parts and covers several techniques, each step is routine AS-level work with no novel insight required—slightly easier than average due to the scaffolded structure and standard procedures. |
| Spec | 1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| 10(a)(i) | States correct value CAO | AO3.4 |
| (a)(ii) | States correct integer value | |
| CAO | AO3.4 | B1 |
| (b) | Forms correct equation and | |
| rearranges to obtain e0.5t = … | AO3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Must give answer to 3 sf | AO1.1b | A1 |
| (c) | 1 mark for any clear valid |
| Answer | Marks | Guidance |
|---|---|---|
| of the question | AO3.5b | E1 |
| Answer | Marks |
|---|---|
| (d) | Forms an equation with |
| Answer | Marks | Guidance |
|---|---|---|
| PI | AO3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| correctly | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| NMS scores full marks for 2023 | AO3.2a | A1 |
| Total | 8 | |
| Q | Marking Instructions | AO |
Question 10:
--- 10(a)(i) ---
10(a)(i) | States correct value CAO | AO3.4 | B1 | 50
(a)(ii) | States correct integer value
CAO | AO3.4 | B1 | 609
(b) | Forms correct equation and
rearranges to obtain e0.5t = … | AO3.4 | M1 | 150 = 50e0.5t
so e0.5t = 3
Obtains the correct solution.
Must give answer to 3 sf | AO1.1b | A1 | t = 2ln 3 = 2.20
(c) | 1 mark for any clear valid
reason, must be set in context
of the question | AO3.5b | E1 | No constraint on the number of
rabbits (ie could go off to infinity)
OR
Model is only based on the 3 years
of the study. Things may change
OR
Continuous model but number of
rabbits is discrete
OR
Ignores extraneous factors such as
disease, predation, limited food
supply
(d) | Forms an equation with
exponentials by letting R = C
PI | AO3.4 | M1 | 1000e0.1t = 50e0.5t
20 = e0.4t
t = ln20 ÷ 0.4
= 7.49
2023
Solves ‘their’ equation
correctly | AO1.1a | M1
States correct answer as the
year 2023 CAO
NMS scores full marks for 2023 | AO3.2a | A1
Total | 8
Q | Marking Instructions | AO | Marks | Typical SolutionDavid has been investigating the population of rabbits on an island during a three-year period.
Based on data that he has collected, David decides to model the population of rabbits, $R$, by the formula
$$R = 50e^{0.5t}$$
where $t$ is the time in years after 1 January 2016.
\begin{enumerate}[label=(\alph*)]
\item Using David's model:
\begin{enumerate}[label=(\roman*)]
\item state the population of rabbits on the island on 1 January 2016; [1 mark]
\item predict the population of rabbits on 1 January 2021. [1 mark]
\end{enumerate}
\item Use David's model to find the value of $t$ when $R = 150$, giving your answer to three significant figures. [2 marks]
\item Give one reason why David's model may not be appropriate. [1 mark]
\item On the same island, the population of crickets, $C$, can be modelled by the formula
$$C = 1000e^{-0.1t}$$
where $t$ is the time in years after 1 January 2016.
Using the two models, find the year during which the population of rabbits first exceeds the population of crickets. [3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 2 Q10 [8]}}