| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Moderate -0.8 This is a straightforward AS-level question on exponential models and logarithms. Part (a) tests basic interpretation, (b) is routine log manipulation, (c) involves standard linear regression using logs with clear scaffolding, and (d) asks for standard modelling assumptions. All techniques are direct applications with no problem-solving insight required, making it easier than average. |
| Spec | 1.02z Models in context: use functions in modelling1.06f Laws of logarithms: addition, subtraction, power rules1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Year | 2013 | 2015 |
| \(t\) | 3 | |
| \(P\) | 10200 | 12800 |
| \(\log_{10} P\) | 4.0086 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a) | Explains that a represents the | |
| initial population. OE | 2.4 | E1 |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 7(b) | Takes logarithms to base 10 of | |
| both sides | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| convincingly AG | 2.1 | R1 |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 7(c)(i) | Completes table correctly, |
| Answer | Marks | Guidance |
|---|---|---|
| used in part (c)(ii) | 3.3 | B1 |
| Year | 2013 | 2015 |
| t | 3 | 5 |
| P | 10 200 | 12 800 |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | 4.0086 | 4.1072 |
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 7(c)(ii) | Uses data to set up a pair of | |
| simultaneous equations | 3.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 10 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| calculate a | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| to 7300 | 1.1b | A1 |
| Subtotal | 4 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 7(c)(iii) | Substitutes their values of a and | |
| b into model and t = 14 | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| FT provided >12800 | 1.1b | A1F |
| Subtotal | 2 | |
| Q | Marking instructions | AO |
| Answer | Marks | Guidance |
|---|---|---|
| 7(d)(i) | States an appropriate | |
| assumption about the model. | 3.5b | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 7(d)(ii) | Makes appropriate comment |
| Answer | Marks | Guidance |
|---|---|---|
| specific factor. | 3.5a | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Question Total | 12 | |
| Q | Marking instructions | AO |
Question 7:
--- 7(a) ---
7(a) | Explains that a represents the
initial population. OE | 2.4 | E1 | a is the population in 2010
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 7(b) ---
7(b) | Takes logarithms to base 10 of
both sides | 1.1a | M1 | P = a(10bt)
log P = log (a10bt)
10 10
= log a + log (10bt)
10 10
= log a + bt
10
Completes derivation
convincingly AG | 2.1 | R1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 7(c)(i) ---
7(c)(i) | Completes table correctly,
figures seen in table or text or
used in part (c)(ii) | 3.3 | B1
Year | 2013 | 2015
t | 3 | 5
P | 10 200 | 12 800
log P
10 | 4.0086 | 4.1072
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 7(c)(ii) ---
7(c)(ii) | Uses data to set up a pair of
simultaneous equations | 3.1a | M1 | 4.0086 = 3b + log a
10
4.1072 = 5b + log a
10
b = 0.0493
log a = 3.8607
10
a = 7256
Solves equations for either b or
log a correct
10 | 1.1b | A1
Converts log a to obtain a
10
value of a
or uses their b and data to
calculate a | 1.1a | M1
Obtains both b and a correct.
AWRT 0.049 and AWFW 7200
to 7300 | 1.1b | A1
Subtotal | 4
Q | Marking instructions | AO | Marks | Typical solution
--- 7(c)(iii) ---
7(c)(iii) | Substitutes their values of a and
b into model and t = 14 | 3.4 | M1 | 7256 × 10(14 × 0.0493)
= 35555
Calculates correct value of
population AWFW 35500 to
35600
FT provided >12800 | 1.1b | A1F
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 7(d)(i) ---
7(d)(i) | States an appropriate
assumption about the model. | 3.5b | E1 | The value of constant b does not
change after 2020
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 7(d)(ii) ---
7(d)(ii) | Makes appropriate comment
about limited data, or length of
extrapolation, changing food
supply, disease or equivalent
specific factor. | 3.5a | E1 | Not very reliable, because it is only
based on data from two years
Subtotal | 1
Question Total | 12
Q | Marking instructions | AO | Marks | Typical solution
Scientists observed a colony of seabirds over a period of 10 years starting in 2010.
They concluded that the number of birds in the colony, its population $P$, could be modelled by a formula of the form
$$P = a(10^{bt})$$
where $t$ is the time in years after 2010, and $a$ and $b$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Explain what the value of $a$ represents.
[1 mark]
\item Show that $\log_{10} P = bt + \log_{10} a$
[2 marks]
\item The table below contains some data collected by the scientists.
\begin{center}
\begin{tabular}{|c|c|c|}
\hline
Year & 2013 & 2015 \\
\hline
$t$ & 3 & \\
\hline
$P$ & 10200 & 12800 \\
\hline
$\log_{10} P$ & 4.0086 & \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Complete the table, giving the $\log_{10} P$ value to 5 significant figures.
[1 mark]
\item Use the data to calculate the value of $a$ and the value of $b$.
[4 marks]
\item Use the model to estimate the population of the colony in 2024.
[2 marks]
\end{enumerate}
\item \begin{enumerate}[label=(\roman*)]
\item State an assumption that must be made in using the model to estimate the population of the colony in 2024.
[1 mark]
\item Hence comment, with a reason, on the reliability of your estimate made in part (c)(iii).
[1 mark]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2021 Q7 [12]}}