| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Time to reach target in exponential model |
| Difficulty | Moderate -0.8 This is a straightforward application of an exponential model requiring simple substitution (part a) and solving a basic exponential equation using logarithms (part b). Both parts are routine C2-level exercises with no conceptual challenges beyond standard technique recall. |
| Spec | 1.06i Exponential growth/decay: in modelling context |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(11\) a.m. \(\therefore t = 3\) | M1 A1 | |
| \(N = 20\,000 \times (1.06)^3 = 23\,820\) (nearest unit) | M1 A1 | |
| (b) \(40\,000 = 20\,000 \times (1.06)^t\) | M1 | |
| \((1.06)^t = 2\) | M1 | |
| \(t = \frac{\lg 2}{\lg 1.06} = 11.8957\) | M1 A1 | |
| \(11.8957\) hours \(= 11\) hours \(54\) mins \(\therefore 7.54\) p.m. | A1 | (6 marks) |
(a) $11$ a.m. $\therefore t = 3$ | M1 A1 |
$N = 20\,000 \times (1.06)^3 = 23\,820$ (nearest unit) | M1 A1 |
(b) $40\,000 = 20\,000 \times (1.06)^t$ | M1 |
$(1.06)^t = 2$ | M1 |
$t = \frac{\lg 2}{\lg 1.06} = 11.8957$ | M1 A1 |
$11.8957$ hours $= 11$ hours $54$ mins $\therefore 7.54$ p.m. | A1 | (6 marks)\begin{enumerate}
\item During one day, a biological culure is allowed to grow under controlled conditions.
\end{enumerate}
At 8 a.m. the culture is estimated to contain 20000 bacteria. A model of the growth of the culture assumes that $t$ hours after 8 a.m., the number of bacteria present, $N$, is given by
$$N = 20000 \times ( 1.06 ) ^ { t } .$$
Using this model,\\
(a) find the number of bacteria present at 11 a.m.,\\
(b) find, to the nearest minute, the time when the initial number of bacteria will have doubled.\\
\hfill \mbox{\textit{Edexcel C2 Q1 [6]}}