| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | Finding x from given y value |
| Difficulty | Moderate -0.8 This is a straightforward exponential decay question with standard parts: (a) tests conceptual understanding of exponential vs linear decay with simple arithmetic, (b) requires routine application of exponential model with logarithms (finding constants then solving), and (c) asks for a basic contextual limitation. All techniques are standard AS-level with no novel problem-solving required, making it easier than average. |
| Spec | 1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| 10(a) | States that Kaya is correct | |
| and/or Charlie is wrong | 2.3 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| sign. | 3.3 | B1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 10(b) | Uses 18 000 for value of A | 3.1b |
| Answer | Marks | Guidance |
|---|---|---|
| model | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| exact or AWRT 0.203 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| their value of k | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone t = 3 | 1.1b | A1 |
| Subtotal | 5 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 10(c) | Gives a reason in context why |
| Answer | Marks | Guidance |
|---|---|---|
| suggests. | 3.5b | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Question 10 Total | 8 | |
| Q | Marking instructions | AO |
Question 10:
--- 10(a) ---
10(a) | States that Kaya is correct
and/or Charlie is wrong | 2.3 | E1 | Charlie is wrong.
Over the same time, the value
goes down by the same proportion.
Two thirds of £18 000 is £12 000
so two thirds of £12 000 is £8 000
Shows where the value £8000
has come from.
Ignore missing or incorrect £
sign. | 3.3 | B1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 10(b) ---
10(b) | Uses 18 000 for value of A | 3.1b | B1 | 12 000 = 18 000 e–2k
1
k = ln 1.5 = 0.203
2
10 000 = 18 000 e–kt
t = 2.9
Substitutes 12 000 and 2 into
model | 3.4 | M1
Solves to find correct value of k,
exact or AWRT 0.203 | 1.1b | A1
Uses model with V = 10 000 and
their value of k | 3.4 | M1
Obtains the correct value of t
AWRT 2.9
Condone t = 3 | 1.1b | A1
Subtotal | 5
Q | Marking instructions | AO | Marks | Typical solution
--- 10(c) ---
10(c) | Gives a reason in context why
the model will not be suitable.
For example:
• Car will be worthless by
then.
• Car will have been
scrapped after 30 years.
• Model gives an
unrealistic value of £41.
• Scrap value will be worth
more than model
suggests. | 3.5b | E1 | The car will probably have been
scrapped by then.
Subtotal | 1
Question 10 Total | 8
Q | Marking instructions | AO | Marks | Typical solutionCharlie buys a car for £18000 on 1 January 2016.
The value of the car decreases exponentially.
The car has a value of £12000 on 1 January 2018.
\begin{enumerate}[label=(\alph*)]
\item Charlie says:
• because the car has lost £6000 after two years, after another two years it will be worth £6000.
Charlie's friend Kaya says:
• because the car has lost one third of its value after two years, after another two years it will be worth £8000.
Explain whose statement is correct, justifying the value they have stated.
[2 marks]
\item The value of Charlie's car, £$V$, $t$ years after 1 January 2016 may be modelled by the equation
$$V = Ae^{-kt}$$
where $A$ and $k$ are positive constants.
Find the value of $t$ when the car has a value of £10000, giving your answer to two significant figures.
[5 marks]
\item Give a reason why the model, in this context, will not be suitable to calculate the value of the car when $t = 30$
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2023 Q10 [8]}}