| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| 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 log-linear modelling question requiring routine application of logarithms to linearize exponential data, plotting points, reading gradient/intercept from a graph, and basic interpretation. While multi-part with 5 sections, each step follows a predictable textbook pattern with no novel problem-solving required. Slightly easier than average due to the scaffolded structure and straightforward calculations. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form2.02c Scatter diagrams and regression lines |
| Number of months after start-up \(( x )\) | 1 | 2 | 3 | 4 | 5 | 6 |
| Profit for this month \(( \pounds y )\) | 500 | 800 | 1200 | 1900 | 3000 | 4800 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = k \times 10^{ax}\) | ||
| \(\log_{10} y = \log_{10} k + ax\) | M1 | Taking \(\log_{10}\) of both sides |
| Straight line with gradient \(a\) and intercept \(\log_{10} k\) | A1 | Both identified |
| Answer | Marks | Guidance |
|---|---|---|
| Table values: \(\log_{10} y\): \(2.70,\ 2.90,\ 3.08,\ 3.28,\ 3.48,\ 3.68\) | B2 | Plotted correctly with line of best fit |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient \(a \approx 0.196\) (from graph) | M1 A1 | Reading from graph |
| Intercept \(\log_{10} k \approx 2.50\), so \(k \approx 316\) | A1 | |
| Equation: \(y = 316 \times 10^{0.196x}\) | A1 | (values may vary slightly) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log_{10}(75000) = 4.875\) | M1 | |
| Reading from graph: \(x \approx 12\) months | A1 M1 A1 | Correct use of equation or graph |
| Answer | Marks | Guidance |
|---|---|---|
| E.g. profits cannot keep increasing indefinitely / model predicts unrealistically large profits | B1 | Any valid comment |
## Question 13(i):
$y = k \times 10^{ax}$ |
$\log_{10} y = \log_{10} k + ax$ | M1 | Taking $\log_{10}$ of both sides
Straight line with gradient $a$ and intercept $\log_{10} k$ | A1 | Both identified
## Question 13(ii):
Table values: $\log_{10} y$: $2.70,\ 2.90,\ 3.08,\ 3.28,\ 3.48,\ 3.68$ | B2 | Plotted correctly with line of best fit
## Question 13(iii):
Gradient $a \approx 0.196$ (from graph) | M1 A1 | Reading from graph
Intercept $\log_{10} k \approx 2.50$, so $k \approx 316$ | A1 |
Equation: $y = 316 \times 10^{0.196x}$ | A1 | (values may vary slightly)
## Question 13(iv):
$\log_{10}(75000) = 4.875$ | M1 |
Reading from graph: $x \approx 12$ months | A1 M1 A1 | Correct use of equation or graph
## Question 13(v):
E.g. profits cannot keep increasing indefinitely / model predicts unrealistically large profits | B1 | Any valid comment13 Answer part (ii) of this question on the insert provided.
The table gives a firm's monthly profits for the first few months after the start of its business, rounded to the nearest $\pounds 100$.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | }
\hline
Number of months after start-up $( x )$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
Profit for this month $( \pounds y )$ & 500 & 800 & 1200 & 1900 & 3000 & 4800 \\
\hline
\end{tabular}
\end{center}
The firm's profits, $\pounds y$, for the $x$ th month after start-up are modelled by
$$y = k \times 10 ^ { a x }$$
where $a$ and $k$ are constants.\\
(i) Show that, according to this model, a graph of $\log _ { 10 } y$ against $x$ gives a straight line of gradient $a$ and intercept $\log _ { 10 } k$.\\
(ii) On the insert, complete the table and plot $\log _ { 10 } y$ against $x$, drawing by eye a line of best fit.\\
(iii) Use your graph to find an equation for $y$ in terms of $x$ for this model.\\
(iv) For which month after start-up does this model predict profits of about $\pounds 75000$ ?\\
(v) State one way in which this model is unrealistic.
\hfill \mbox{\textit{OCR MEI C2 2007 Q13 [12]}}