| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Optimisation via quadratic model |
| Difficulty | Moderate -0.8 This is a straightforward applied linear (not quadratic despite the topic label) problem requiring students to form equations from word problems, solve simultaneous equations, and interpret parameters. All steps are routine: setting up y = mx + c, using profit = revenue - cost to create two equations, solving for m and c, then finding break-even point. No novel insight or complex problem-solving required, making it easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02z Models in context: use functions in modelling |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=kx+c\) where \(k\) and \(c\) are constants | B1 | Correct form; ignore how constants are labelled (e.g. \(k\), \(c\), \(m\) etc.); must be seen in (a); work cannot be recovered in (b) or (c) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x=800 \Rightarrow y=2(800)-500=1100\), so \((x,y)=(800,1100)\); \(x=300 \Rightarrow y=2(300)+80=680\), so \((x,y)=(300,680)\) | M1 | Translates problem into model; allow 1st M1 for \(1600-y=500\) or \(600-y=-80\) |
| Applies \((800, 1100)\) and \((300, 680)\) to give \(1100=800k+c\) and \(680=300k+c\) | dM1 | Dependent on previous M mark |
| \(k=0.84\), \(c=428\), states \(y=0.84x+428\) * | A1* | No errors in working; answer \(y=0.84x+428\) must be stated |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Same two points as Way 1 | M1 | |
| \(k=\frac{1100-680}{800-300} = 0.84\); \((800,1100) \Rightarrow 1100=800(0.84)+c \Rightarrow c=\ldots\) | dM1 | Dependent on previous M mark |
| \(k=0.84\), \(c=428\), states \(y=0.84x+428\) * | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Same two points as Way 1 | M1 | |
| \(\{y=0.84x+428 \Rightarrow\}\) \(x=800 \Rightarrow y=(0.84)(800)+428=1100\); \(x=300 \Rightarrow y=(0.84)(300)+428=680\) | dM1 | Dependent on previous M mark; confirms true for both points |
| Hence \(y=0.84x+428\) * | A1* | Conclusion could be "\(y=0.84x+428\)" or "QED" or "proved" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.84\) (in £s) represents the cost of making each extra bar of soap / direct cost / marginal cost / cost of making a bar (condone this) | B1 | Do not allow: profit per bar or selling price per bar; give B0 for incorrect units e.g. "cost is £84"; condone use of £0.84p; also condone "rate of change of cost", "cost of making a bar", "constant of proportionality for cost per bar", "rate of increase in cost" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2n=0.84n+428 \Rightarrow n=\ldots\) (condone \(2x=0.84x+428 \Rightarrow x=\ldots\)) | M1 | Using model to construct argument leading to critical value; ignore inequality symbols |
| Answer of \(369\) bars | A1 | 369 only; do not accept decimal answers; give M1A1 for no working leading to 369; condone final A1 in words "at least 369 bars must be made/sold" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Trial 1: \(n=368 \Rightarrow y=(0.84)(368)+428=737.12\); revenue \(=2(368)=736\), loss \(=1.12\) | M1 | Uses 368 or 369 to find cost \(y=\ldots\) |
| Trial 2: \(n=369 \Rightarrow y=(0.84)(369)+428=737.96\); revenue \(=2(369)=738\), profit \(=0.04\); leading to answer of \(369\) bars | A1 | Attempts both trials to find cost \(y=\ldots\) and arrives at 369 only; give final A0 for \(x>369\) or \(x>368\) or \(x\geq 369\) without \(x=369\) or 369 as final answer |
## Question 7:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=kx+c$ where $k$ and $c$ are constants | B1 | Correct form; ignore how constants are labelled (e.g. $k$, $c$, $m$ etc.); must be seen in (a); work cannot be recovered in (b) or (c) |
### Part (b) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x=800 \Rightarrow y=2(800)-500=1100$, so $(x,y)=(800,1100)$; $x=300 \Rightarrow y=2(300)+80=680$, so $(x,y)=(300,680)$ | M1 | Translates problem into model; allow 1st M1 for $1600-y=500$ or $600-y=-80$ |
| Applies $(800, 1100)$ and $(300, 680)$ to give $1100=800k+c$ and $680=300k+c$ | dM1 | Dependent on previous M mark |
| $k=0.84$, $c=428$, states $y=0.84x+428$ * | A1* | No errors in working; answer $y=0.84x+428$ must be stated |
### Part (b) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Same two points as Way 1 | M1 | |
| $k=\frac{1100-680}{800-300} = 0.84$; $(800,1100) \Rightarrow 1100=800(0.84)+c \Rightarrow c=\ldots$ | dM1 | Dependent on previous M mark |
| $k=0.84$, $c=428$, states $y=0.84x+428$ * | A1* | |
### Part (b) Way 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Same two points as Way 1 | M1 | |
| $\{y=0.84x+428 \Rightarrow\}$ $x=800 \Rightarrow y=(0.84)(800)+428=1100$; $x=300 \Rightarrow y=(0.84)(300)+428=680$ | dM1 | Dependent on previous M mark; confirms true for both points |
| Hence $y=0.84x+428$ * | A1* | Conclusion could be "$y=0.84x+428$" or "QED" or "proved" |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.84$ (in £s) represents the cost of making each extra bar of soap / direct cost / marginal cost / cost of making a bar (condone this) | B1 | Do not allow: profit per bar or selling price per bar; give B0 for incorrect units e.g. "cost is £84"; condone use of £0.84p; also condone "rate of change of cost", "cost of making a bar", "constant of proportionality for cost per bar", "rate of increase in cost" |
### Part (d) Way 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2n=0.84n+428 \Rightarrow n=\ldots$ (condone $2x=0.84x+428 \Rightarrow x=\ldots$) | M1 | Using model to construct argument leading to critical value; ignore inequality symbols |
| Answer of $369$ bars | A1 | 369 only; do not accept decimal answers; give M1A1 for no working leading to 369; condone final A1 in words "at least 369 bars must be made/sold" |
### Part (d) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Trial 1: $n=368 \Rightarrow y=(0.84)(368)+428=737.12$; revenue $=2(368)=736$, loss $=1.12$ | M1 | Uses 368 or 369 to find cost $y=\ldots$ |
| Trial 2: $n=369 \Rightarrow y=(0.84)(369)+428=737.96$; revenue $=2(369)=738$, profit $=0.04$; leading to answer of $369$ bars | A1 | Attempts both trials to find cost $y=\ldots$ and arrives at 369 only; give final A0 for $x>369$ or $x>368$ or $x\geq 369$ without $x=369$ or 369 as final answer |\begin{enumerate}
\item A small factory makes bars of soap.
\end{enumerate}
On any day, the total cost to the factory, $\pounds y$, of making $x$ bars of soap is modelled to be the sum of two separate elements:
\begin{itemize}
\item a fixed cost
\item a cost that is proportional to the number of bars of soap that are made that day\\
(a) Write down a general equation linking $y$ with $x$, for this model.
\end{itemize}
The bars of soap are sold for $\pounds 2$ each.\\
On a day when 800 bars of soap are made and sold, the factory makes a profit of £500 On a day when 300 bars of soap are made and sold, the factory makes a loss of $\pounds 80$ Using the above information,\\
(b) show that $y = 0.84 x + 428$\\
(c) With reference to the model, interpret the significance of the value 0.84 in the equation.
Assuming that each bar of soap is sold on the day it is made,\\
(d) find the least number of bars of soap that must be made on any given day for the factory to make a profit that day.
\hfill \mbox{\textit{Edexcel Paper 2 2019 Q7 [7]}}