| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Moderate -0.3 This is a standard A-level question on exponential models and logarithmic transformations. Part (a) is routine algebra, part (b) requires reading a graph to find gradient and intercept, part (c) applies the model, and part (d) asks for basic interpretation. While it spans multiple techniques, each step follows textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form1.06i Exponential growth/decay: in modelling context |
| Year | 1970 | 1980 | 1990 | 2000 | 2010 |
| Valuation price | £8000 | £19000 | £36000 | £82000 | £205000 |
| Answer | Marks |
|---|---|
| 8(a) | Takes logs of both sides of the |
| Answer | Marks | Guidance |
|---|---|---|
| rule | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone missing base | 2.1 | R1 |
| Answer | Marks |
|---|---|
| 8(b) | Equates log p to 3.90 |
| Answer | Marks | Guidance |
|---|---|---|
| line of best fit only | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| equations to obtain p and q | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| CSO | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| CSO | 1.1b | A1 |
| Answer | Marks |
|---|---|
| 8(c) | Substitutes V= 500000 into their |
| Answer | Marks | Guidance |
|---|---|---|
| PI by correct t value | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Must have t >40 | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Must be later than 2010 | 3.2a | A1F |
| Answer | Marks |
|---|---|
| 8(d) | Explains that their 2023 |
| Answer | Marks | Guidance |
|---|---|---|
| the range of data collected | 3.5b | E1F |
| Answer | Marks | Guidance |
|---|---|---|
| crash | 3.2b | E1 |
| Total | 11 | |
| Q | Marking instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Takes logs of both sides of the
equation and applies addition
rule | 1.1a | M1 | log V = log p qt
10 10
log V = log p + log qt
10 10 10
log V = log p + t log q
10 10 10
Completes rigorous argument to
show required result
Condone missing base | 2.1 | R1
--- 8(b) ---
8(b) | Equates log p to 3.90
10
or
Forms two simultaneous
equations using points from the
line of best fit only | 3.4 | M1 | log p = 3.90
10
p = 7940
5.28−3.90
log q = =0.0345
40−0
q = 1.08
Calculates gradient and
equates to log q
10
or
Solves their pair of simultaneous
equations to obtain p and q | 3.4 | M1
Obtains correct
AWRT 8000
CSO | 1.1b | A1
Obtains correct q
AWRT 1.1
CSO | 1.1b | A1
--- 8(c) ---
8(c) | Substitutes V= 500000 into their
V = 7940 × 1.08 t
or into their
log V = log 7940 + t log 1.08
10 10 10
to form an equation for t
PI by correct t value | 3.4 | M1 | 500000=7940×1.08t
t =53.82
The house will first be worth half a
million pounds during 2023
Solves their equation for t
Must have t >40 | 1.1a | M1
States their correct year using
1970+ their integer part of t
Must be later than 2010 | 3.2a | A1F
--- 8(d) ---
8(d) | Explains that their 2023
(FT later than 2010) is outside
the range of data collected | 3.5b | E1F | The model is only based on data
between 1970 and 2010
House prices may not continue to
grow in the same way indefinitely
Explains that house prices may
not continue to grow in the same
way
Must refer to context not just to
extrapolation/pattern
Can be implied by comments
such as:
Theresa may have made
improvements by adding a new
room
Prices could fall in a market
crash | 3.2b | E1
Total | 11
Q | Marking instructions | AO | Mark | Typical solutionTheresa bought a house on 2 January 1970 for £8000.
The house was valued by a local estate agent on the same date every 10 years up to 2010.
The valuations are shown in the following table.
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Year & 1970 & 1980 & 1990 & 2000 & 2010 \\
\hline
Valuation price & £8000 & £19000 & £36000 & £82000 & £205000 \\
\hline
\end{tabular}
The valuation price of the house can be modelled by the equation
$$V = pq^t$$
where $V$ pounds is the valuation price $t$ years after 2 January 1970 and $p$ and $q$ are constants.
\begin{enumerate}[label=(\alph*)]
\item Show that $V = pq^t$ can be written as $\log_{10} V = \log_{10} p + t \log_{10} q$ [2 marks]
\item The values in the table of $\log_{10} V$ against $t$ have been plotted and a line of best fit has been drawn on the graph below.
\includegraphics{figure_8b}
Using the given line of best fit, find estimates for the values of $p$ and $q$.
Give your answers correct to three significant figures. [4 marks]
\item Determine the year in which Theresa's house will first be worth half a million pounds. [3 marks]
\item Explain whether your answer to part (c) is likely to be reliable. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2019 Q8 [11]}}