| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | ln(y) vs ln(x) linear graph |
| Difficulty | Moderate -0.3 This is a standard logarithmic linearization question requiring students to take logs of both sides, plot given data, draw a line of best fit, and use it for prediction. While it involves multiple steps (12 marks total), each step follows a routine procedure taught in C2: algebraic manipulation of logs, plotting points, reading gradient/intercept from a graph, and substitution. No novel problem-solving or insight is required—it's a textbook application of the log-linear model technique, making it slightly easier than average. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form2.02c Scatter diagrams and regression lines |
| Year | 1986/87 | 1991/92 | 1996/97 | 1999/00 | 2000/01 | 2001/02 |
| Percentage of the adult population visiting the cinema | 31 | 44 | 54 | 56 | 55 | 57 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log P = \log a + b \log t\) www | 1 | must be with correct equation |
| comparison with \(y = mx + c\) | 1 | condone omission of base |
| intercept \(= \log a\) | 1 | |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\log t\) | 0 | 0.78 |
| 1.20 | ||
| \(\log P\) | 1.49 | 1.64 |
| 1.76 | ||
| plots f.t. ruled line of best fit | 1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| gradient rounding to 0.22 or 0.23 | 2 | M1 for y step / x-step / accept 1.47 – 1.50 for intercept accept answers that round to 30 – 32 , their positive \(m\) |
| \(a = 10^{1.49}\) s.o.i. | 1 | |
| \(P = 31t^m\) allow the form \(P = 10^{0.22\log t + 1.49}\) | 1 | |
| 4 |
| Answer | Marks |
|---|---|
| answer rounds in range 60 to 63 | 1 |
| 1 |
### Part i
$\log P = \log a + b \log t$ www | 1 | must be with correct equation
comparison with $y = mx + c$ | 1 | condone omission of base
intercept $= \log a$ | 1 |
| 3 |
### Part ii
$\log t$ | 0 | 0.78 | 1.15 | 1.18 | 1 | accept to 2 or more dp
| 1.20 |
$\log P$ | 1.49 | 1.64 | 1.75 | 1.74 | 1 |
| 1.76 |
plots f.t. ruled line of best fit | 1 |
| 4 |
### Part iii
gradient rounding to 0.22 or 0.23 | 2 | M1 for y step / x-step / accept 1.47 – 1.50 for intercept accept answers that round to 30 – 32 , their positive $m$
$a = 10^{1.49}$ s.o.i. | 1 |
$P = 31t^m$ allow the form $P = 10^{0.22\log t + 1.49}$ | 1 |
| 4 |
### Part iv
answer rounds in range 60 to 63 | 1 |
| 1 |The percentage of the adult population visiting the cinema in Great Britain has tended to increase since the 1980s. The table shows the results of surveys in various years.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline
Year & 1986/87 & 1991/92 & 1996/97 & 1999/00 & 2000/01 & 2001/02 \\
\hline
Percentage of the adult population visiting the cinema & 31 & 44 & 54 & 56 & 55 & 57 \\
\hline
\end{tabular}
\end{center}
Source: Department of National Statistics, www.statistics.gov.uk
This growth may be modelled by an equation of the form
$$P = at^b,$$
where $P$ is the percentage of the adult population visiting the cinema, $t$ is the number of years after the year 1985/86 and $a$ and $b$ are constants to be determined.
\begin{enumerate}[label=(\roman*)]
\item Show that, according to this model, the graph of $\log_{10} P$ against $\log_{10} t$ should be a straight line of gradient $b$. State, in terms of $a$, the intercept on the vertical axis. [3]
\item Complete the table of values on the insert, and plot $\log_{10} P$ against $\log_{10} t$. Draw by eye a line of best fit for the data. [4]
\item Use your graph to find the equation for $P$ in terms of $t$. [4]
\item Predict the percentage of the adult population visiting the cinema in the year 2007/2008 (i.e. when $t = 22$), according to this model. [1]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2008 Q13 [12]}}