| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Matrices |
| Type | Applied matrix modeling problems |
| Difficulty | Standard +0.3 This is a straightforward applied matrices problem requiring students to set up a system of three linear equations from a word problem and solve using matrix methods. While it involves multiple steps (defining variables, forming equations, writing in matrix form, and solving), each step is routine and the problem follows a standard textbook pattern. The context is accessible and the algebraic manipulation required is basic for Further Maths students. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution4.03r Solve simultaneous equations: using inverse matrix |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x/C\) = number of Construction students, \(y/D\) = number of Design students, \(z/H\) = number of Hospitality students | B1 | Defines 3 variables, minimum e.g. construction = \(C\), Design = \(D\), Hospitality = \(H\). Abbreviations may be used |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Increase in 2020: \(1110 \times 0.0027 = 2.997 \approx 3\), OR number in 2020: \(1110 \times 1.0027 = 1112.997 \approx 1113\) | M1 | Finds either the increase or number of students in 2020. Implied by any equation equalling 1113 or 1112.997. If 1100 used instead of 1110, condone as slip |
| \(x + y + z = 1110\), \(x - z = 370\), \(0.0125x + 0.025y - 0.02z = 3\) or equivalent forms | M1 | Attempts to use model to set up at least 2 equations |
| All 3 simplified equations correct | A1 | Award even if B0 scored; ignore additional incorrect equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & -1 \\ 1.0125 & 1.025 & 0.98 \end{pmatrix}\begin{pmatrix} C \\ D \\ H \end{pmatrix} = \begin{pmatrix} 1110 \\ 370 \\ 1113 \end{pmatrix}\) | M1 | Uses equations from (a) to set up matrix equation of the form \(\begin{pmatrix} \cdots \end{pmatrix}\begin{pmatrix} C \\ D \\ H \end{pmatrix} = \begin{pmatrix} \cdots \end{pmatrix}\) |
| Correct matrix equation for their equations | A1ft | Follow through on their equations |
| \(\begin{pmatrix} C \\ D \\ H \end{pmatrix} = \begin{pmatrix} \cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots \end{pmatrix}^{-1} \begin{pmatrix} 1110 \\ 370 \\ 1113 \end{pmatrix} = \begin{pmatrix} \cdots \\ \cdots \\ \cdots \end{pmatrix}\) | dM1 | Dependent on previous M1. Writes \(A()^{-1}\begin{pmatrix} 1110 \\ \text{their "370"} \\ \text{their "3"} \end{pmatrix}\) and obtains at least one value of \(C\), \(D\) or \(H\) |
| \(C = 720\), \(D = 40\), \(H = 350\) in 2019 | A1 | Interprets answer in context; minimum \(C=720\), \(D=40\), \(H=350\) with their variables |
## Question 2:
### Part (a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x/C$ = number of Construction students, $y/D$ = number of Design students, $z/H$ = number of Hospitality students | B1 | Defines 3 variables, minimum e.g. construction = $C$, Design = $D$, Hospitality = $H$. Abbreviations may be used |
### Part (a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Increase in 2020: $1110 \times 0.0027 = 2.997 \approx 3$, OR number in 2020: $1110 \times 1.0027 = 1112.997 \approx 1113$ | M1 | Finds either the increase or number of students in 2020. Implied by any equation equalling 1113 or 1112.997. If 1100 used instead of 1110, condone as slip |
| $x + y + z = 1110$, $x - z = 370$, $0.0125x + 0.025y - 0.02z = 3$ or equivalent forms | M1 | Attempts to use model to set up at least 2 equations |
| All 3 simplified equations correct | A1 | Award even if B0 scored; ignore additional incorrect equations |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\begin{pmatrix} 1 & 1 & 1 \\ 1 & 0 & -1 \\ 1.0125 & 1.025 & 0.98 \end{pmatrix}\begin{pmatrix} C \\ D \\ H \end{pmatrix} = \begin{pmatrix} 1110 \\ 370 \\ 1113 \end{pmatrix}$ | M1 | Uses equations from (a) to set up matrix equation of the form $\begin{pmatrix} \cdots \end{pmatrix}\begin{pmatrix} C \\ D \\ H \end{pmatrix} = \begin{pmatrix} \cdots \end{pmatrix}$ |
| Correct matrix equation for their equations | A1ft | Follow through on their equations |
| $\begin{pmatrix} C \\ D \\ H \end{pmatrix} = \begin{pmatrix} \cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots \\ \cdots & \cdots & \cdots \end{pmatrix}^{-1} \begin{pmatrix} 1110 \\ 370 \\ 1113 \end{pmatrix} = \begin{pmatrix} \cdots \\ \cdots \\ \cdots \end{pmatrix}$ | dM1 | Dependent on previous M1. Writes $A()^{-1}\begin{pmatrix} 1110 \\ \text{their "370"} \\ \text{their "3"} \end{pmatrix}$ and obtains at least one value of $C$, $D$ or $H$ |
| $C = 720$, $D = 40$, $H = 350$ in 2019 | A1 | Interprets answer in context; minimum $C=720$, $D=40$, $H=350$ with their variables |
---\begin{enumerate}
\item In this question you must show all stages of your working.
\end{enumerate}
A college offers only three courses: Construction, Design and Hospitality. Each student enrols on just one of these courses.
In 2019, there was a total of 1110 students at this college.\\
There were 370 more students enrolled on Construction than Hospitality.\\
In 2020 the number of students enrolled on
\begin{itemize}
\item Construction increased by $1.25 \%$
\item Design increased by $2.5 \%$
\item Hospitality decreased by $2 \%$
\end{itemize}
In 2020, the total number of students at the college increased by $0.27 \%$ to 2 significant figures.\\
(a) (i) Define, for each course, a variable for the number of students enrolled on that course in 2019.\\
(ii) Using your variables from part (a)(i), write down three equations that model this situation.\\
(b) By forming and solving a matrix equation, determine how many students were enrolled on each of the three courses in 2019.
\hfill \mbox{\textit{Edexcel CP2 2022 Q2 [8]}}