| Exam Board | Edexcel |
|---|---|
| Module | D1 (Decision Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear Programming |
| Type | Three-variable constraint reduction |
| Difficulty | Standard +0.8 This question requires translating verbal constraints into mathematical inequalities, then using the ratio constraint (x = 5z/2) to eliminate a variable and reduce a 3D problem to 2D. Part (b) involves substitution and solving a constrained optimization. While systematic, it demands careful algebraic manipulation, understanding of LP formulation, and multi-step reasoning beyond routine D1 exercises. |
| Spec | 7.06a LP formulation: variables, constraints, objective function7.06b Slack variables: converting inequalities to equations |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| 5(a) | \(\frac{1}{2}(x + y + z) \leq y\) (\(\Rightarrow x - y + z \leq 0\)) | M1 |
| \(\frac{3}{20}(x + y + z) \geq z\) (\(\Rightarrow 3x + 3y - 17z \geq 0\)) | M1 | |
| \(2x - 5z\) | B1 | |
| \(5x + 12y + 15z \leq 834\) | B1 | |
| Eliminating \(z\) from the objective \(x + y + z\) and at least one correct constraint or states the objective and eliminates \(z\) from at least two correct constraints | M1 | |
| Maximise \((P =) 1.4x + y\) subject to \(11x + 12y \leq 834\); \(7x - 5y \leq 0\); \(19x - 15y \leq 0\); (\(x \geq 0, y \geq 0\)) | A1, A1 (7) | |
| 5(b)(i) | Substitute \(y = 42\) into LP gives \(x \leq 30\), \(x \leq 30\), \(x \leq \frac{630}{19}\) which implies that \(x = 30\) | M1 |
| 5(b)(ii) | Total number of reams ordered is \(1.4(30) + 42 = 84\); 12 reams of graph paper ordered | A1, A1 (3) |
| Total: 10 marks |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| 5(a) | $\frac{1}{2}(x + y + z) \leq y$ ($\Rightarrow x - y + z \leq 0$) | M1 | |
| | $\frac{3}{20}(x + y + z) \geq z$ ($\Rightarrow 3x + 3y - 17z \geq 0$) | M1 | |
| | $2x - 5z$ | B1 | |
| | $5x + 12y + 15z \leq 834$ | B1 | |
| | Eliminating $z$ from the objective $x + y + z$ and at least one correct constraint or states the objective and eliminates $z$ from at least two correct constraints | M1 | |
| | Maximise $(P =) 1.4x + y$ subject to $11x + 12y \leq 834$; $7x - 5y \leq 0$; $19x - 15y \leq 0$; ($x \geq 0, y \geq 0$) | A1, A1 (7) | |
| 5(b)(i) | Substitute $y = 42$ into LP gives $x \leq 30$, $x \leq 30$, $x \leq \frac{630}{19}$ which implies that $x = 30$ | M1 | |
| 5(b)(ii) | Total number of reams ordered is $1.4(30) + 42 = 84$; 12 reams of graph paper ordered | A1, A1 (3) | |
| | **Total: 10 marks** | | |
---5. The head of a Mathematics department needs to order three types of paper. The three types of paper are plain, lined and graph.
All three types of paper are sold in reams. (A ream is 500 sheets of paper.)\\
Based on the last academic year the head of department formed the following constraints.
\begin{itemize}
\item At least half the paper must be lined
\item No more than $15 \%$ of the paper must be graph paper
\item The ratio of plain paper to graph paper must be $5 : 2$
\end{itemize}
The cost of each ream of plain, lined and graph paper is $\pounds 5 , \pounds 12$ and $\pounds 15$ respectively. The head of department has at most $\pounds 834$ to spend on paper.
The head of department wants to maximise the total number of reams of paper ordered.\\
Let $x , y$ and $z$ represent the number of reams of plain paper, lined paper and graph paper ordered respectively.
\begin{enumerate}[label=(\alph*)]
\item Formulate this information as a linear programming problem in $x$ and $y$ only, stating the objective and listing the constraints as simplified inequalities with integer coefficients.
The head of department decides to order exactly 42 reams of lined paper and still wishes to maximise the total number of reams of paper ordered.
\item Determine
\begin{enumerate}[label=(\roman*)]
\item the total number of reams of paper to be ordered,
\item the number of reams of graph paper to be ordered.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel D1 2024 Q5 [10]}}