| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2017 |
| Session | Specimen |
| Marks | 12 |
| Topic | Differential equations |
| Type | Exponential growth/decay - direct proportionality (dN/dt = kN) |
| Difficulty | Standard +0.3 This is a standard differential equations question from Stats 1 covering exponential growth. Part (a) requires writing dy/dt = ky (basic recall), part (b) is routine separation of variables with given initial condition, part (c) is straightforward substitution, and part (d) requires setting up equations for two different periods but follows a predictable structure. While it has multiple parts and 12 marks total, each step uses well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.06i Exponential growth/decay: in modelling context1.08k Separable differential equations: dy/dx = f(x)g(y) |
Helga invests £4000 in a savings account.
After $t$ days, her investment is worth $£y$.
The rate of increase of $y$ is $ky$, where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Write down a differential equation in terms of $t$, $y$ and $k$. [1]
\item Solve your differential equation to find the value of Helga's investment after $t$ days. Give your answer in terms of $k$ and $t$. [4]
\end{enumerate}
It is given that $k = \frac{r}{365}\ln\left(1 + \frac{r}{100}\right)$ where $r\%$ is the rate of interest per annum.
During the first year the rate of interest is 6% per annum.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the value of Helga's investment after 90 days. [2]
\end{enumerate}
After one year (365 days), the rate of interest drops to 5% per annum.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Find the total time that it will take for Helga's investment to double in value. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/02 2017 Q6 [12]}}