| Exam Board | OCR |
|---|---|
| Module | Further Pure Core AS (Further Pure Core AS) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Applied matrix modeling problems |
| Difficulty | Standard +0.3 This is a straightforward application of matrices to solve a system of three linear equations. The setup requires translating a word problem into equations (a standard skill), then solving using matrix methods. The calculations are routine with no conceptual surprises, making it slightly easier than average for A-level Further Maths. |
| Spec | 4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let \(x\), \(y\) and \(z\) be the amount invested in each account: \(\begin{pmatrix}1&1&1\\0.025&0.037&0.049\\1&-2&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2000\\92\\0\end{pmatrix}\) | B2 | B1 for writing down matrix alone; OR B2 let \(s\) be amount invested in savings account: \((0.025\ \ 0.037\ \ 0.049)\begin{pmatrix}2s\\s\\2000-3s\end{pmatrix}=92\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1&1&1\\0.025&0.037&0.049\\1&-2&0\end{pmatrix}^{-1}\begin{pmatrix}2000\\92\\0\end{pmatrix}\) | M1 | Attempt to find \(x,y,z\); OR M1 by multiplying out and solving for \(s\) |
| \(x=200\), \(y=100\), \(z=1700\); so he invests £200 in the current account, £100 in the savings account, £1700 in the supersaver account | A1 | Interpret in context; A1 \(s=100\), same conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The \(92\) in part (ii) should be \(92\pm0.5\) | E1 | |
| therefore giving a range of answers for each account | E1 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let $x$, $y$ and $z$ be the amount invested in each account: $\begin{pmatrix}1&1&1\\0.025&0.037&0.049\\1&-2&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2000\\92\\0\end{pmatrix}$ | B2 | **B1** for writing down matrix alone; **OR B2** let $s$ be amount invested in savings account: $(0.025\ \ 0.037\ \ 0.049)\begin{pmatrix}2s\\s\\2000-3s\end{pmatrix}=92$ |
---
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1&1&1\\0.025&0.037&0.049\\1&-2&0\end{pmatrix}^{-1}\begin{pmatrix}2000\\92\\0\end{pmatrix}$ | M1 | Attempt to find $x,y,z$; **OR M1** by multiplying out and solving for $s$ |
| $x=200$, $y=100$, $z=1700$; so he invests £200 in the current account, £100 in the savings account, £1700 in the supersaver account | A1 | Interpret in context; **A1** $s=100$, same conclusion |
---
## Question 6(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| The $92$ in part (ii) should be $92\pm0.5$ | E1 | |
| therefore giving a range of answers for each account | E1 | |
---6 At the beginning of the year John had a total of $\pounds 2000$ in three different accounts. He has twice as much money in the current account as in the savings account.
\begin{itemize}
\item The current account has an interest rate of $2.5 \%$ per annum.
\item The savings account has an interest rate of $3.7 \%$ per annum.
\item The supersaver account has an interest rate of $4.9 \%$ per annum.
\end{itemize}
John has predicted that he will earn a total interest of $\pounds 92$ by the end of the year.\\
(i) Model this situation as a matrix equation.\\
(ii) Find the amount that John had in each account at the beginning of the year.\\
(iii) In fact, the interest John will receive is $\pounds 92$ to the nearest pound. Explain how this affects the calculations.
\hfill \mbox{\textit{OCR Further Pure Core AS Q6 [6]}}