| Exam Board | OCR MEI |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2013 |
| Session | January |
| Marks | 13 |
| 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 exponential modeling question requiring logarithmic transformation to linearize data. The steps are routine: take logs of both sides, plot log p vs t, find gradient/intercept from the graph, and use the model for predictions. While it involves multiple parts (13 marks total), each step follows a well-established procedure taught explicitly in C2 with no novel problem-solving required. Slightly easier than average due to the highly structured, step-by-step guidance through a textbook application. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form2.02c Scatter diagrams and regression lines |
| Year | 1969 | 1979 | 1989 | 1999 | 2009 |
| Population in millions (\(p\)) | 58.81 | 80.35 | 105.27 | 134.79 | 169.71 |
The table shows population data for a country.
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
\hline
Year & 1969 & 1979 & 1989 & 1999 & 2009 \\
\hline
Population in millions ($p$) & 58.81 & 80.35 & 105.27 & 134.79 & 169.71 \\
\hline
\end{tabular}
\end{center}
The data may be represented by an exponential model of growth. Using $t$ as the number of years after 1960, a suitable model is $p = a \times 10^{kt}$.
\begin{enumerate}[label=(\roman*)]
\item Derive an equation for $\log_{10} p$ in terms of $a$, $k$ and $t$. [2]
\item Complete the table and draw the graph of $\log_{10} p$ against $t$, drawing a line of best fit by eye. [3]
\item Use your line of best fit to express $\log_{10} p$ in terms of $t$ and hence find $p$ in terms of $t$. [4]
\item According to the model, what was the population in 1960? [1]
\item According to the model, when will the population reach 200 million? [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C2 2013 Q12 [13]}}