OCR MEI AS Paper 1 2022 June — Question 8 7 marks

Exam BoardOCR MEI
ModuleAS Paper 1 (AS Paper 1)
Year2022
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicInequalities
TypeCombined linear and quadratic inequalities
DifficultyStandard +0.3 This is a straightforward multi-part question involving basic linear modeling and solving simultaneous linear inequalities. Parts (a)-(c) require simple substitution and interpretation. Part (d) involves setting up and solving two linear inequalities (supply ≥ demand), which is a standard AS-level technique requiring no novel insight—slightly easier than average due to the scaffolded structure and routine algebraic manipulation.
Spec1.02c Simultaneous equations: two variables by elimination and substitution1.02z Models in context: use functions in modelling

8 A team of volunteers donates cakes for sale at a charity stall. The number of cakes that can be sold depends on the price. A model for this is \(\mathrm { y } = 190 - 70 \mathrm { x }\), where \(y\) cakes can be sold when the price of a cake is \(\pounds\) x.
  1. Find how many cakes could be given away for free according to this model. The number of volunteers who are willing to donate cakes goes up as the price goes up. If the cakes sell for \(\pounds 1.20\) they will donate 50 cakes, but if they sell for \(\pounds 2.40\) they will donate 140 cakes. They use the linear model \(\mathrm { y } = \mathrm { mx } + \mathrm { c }\) to relate the number of cakes donated, \(y\), to the price of a cake, \(\pounds x\).
  2. Find the values of the constants \(m\) and \(c\) for which this linear model fits the two data points.
  3. Explain why the model is not suitable for very low prices.
  4. The team would like to sell all the cakes that they donate. Find the set of possible prices that the cakes could have to achieve this.

Question 8:
Part (a):
AnswerMarks Guidance
When \(x = 0\), number of cakes is 190B1 Allow www
Part (b):
AnswerMarks Guidance
\((1.20, 50)\) gives \(50 = 1.2m + c\)M1 Setting up simultaneous equations for \(m\) and \(c\)
\((2.40, 140)\) gives \(140 = 2.4m + c\)
AnswerMarks Guidance
\(m = 75,\ c = -40\)A1 Allow for values given or \(y = 75x - 40\) seen
Alternative method:
AnswerMarks Guidance
\(m = \frac{140-50}{2.4-1.2}\)M1 Using data to calculate \(m\)
\(m = 75,\ c = -40\)A1 Allow for values given or \(y = 75x - 40\) seen
Part (c):
AnswerMarks Guidance
[When \(x\) is small,] \(y\) is negative and number of cakes donated cannot be negativeE1 Negative \(y\)-values and \(y\) cannot be negative must both be stated or implied
Part (d):
AnswerMarks Guidance
Upper bound for demand to exceed supplyM1 Attempt to find one of the bounds for \(x\) using \(y = 75x - 40\)
Lower bound for \(y\) positive
\(190 - 70x = 75x - 40\)
AnswerMarks Guidance
So \(x < \frac{46}{29}\)A1 Accept \(x < 1.586\) or \(x < 1.59\) or \(x < 1.58\) or \(x \leq 1.56\); Allow use of \(<\) or \(\leq\); Note prices £1.58 and £1.57 do not lead to integer values, fully correct answer is \(x \leq 1.56\)
\(y = 75x - 40 > 0\)
AnswerMarks Guidance
\(x > \frac{40}{75}\)A1 Accept \(0.533 < x\) or \(0.53 < x\) or \(0.54 < x\) or \(0.56 \leq x\); Allow use of \(<\) or \(\leq\); Note prices £0.54 and £0.55 do not lead to integer values, fully correct answer is \(0.56 \leq x\)
# Question 8:

## Part (a):
When $x = 0$, number of cakes is 190 | **B1** | Allow www

## Part (b):
$(1.20, 50)$ gives $50 = 1.2m + c$ | **M1** | Setting up simultaneous equations for $m$ and $c$

$(2.40, 140)$ gives $140 = 2.4m + c$

$m = 75,\ c = -40$ | **A1** | Allow for values given or $y = 75x - 40$ seen

**Alternative method:**

$m = \frac{140-50}{2.4-1.2}$ | **M1** | Using data to calculate $m$

$m = 75,\ c = -40$ | **A1** | Allow for values given or $y = 75x - 40$ seen

## Part (c):
[When $x$ is small,] $y$ is negative and number of cakes donated cannot be negative | **E1** | Negative $y$-values and $y$ cannot be negative must both be stated or implied

## Part (d):
Upper bound for demand to exceed supply | **M1** | Attempt to find one of the bounds for $x$ using $y = 75x - 40$

Lower bound for $y$ positive

$190 - 70x = 75x - 40$

So $x < \frac{46}{29}$ | **A1** | Accept $x < 1.586$ or $x < 1.59$ or $x < 1.58$ or $x \leq 1.56$; Allow use of $<$ or $\leq$; Note prices £1.58 and £1.57 do not lead to integer values, fully correct answer is $x \leq 1.56$

$y = 75x - 40 > 0$

$x > \frac{40}{75}$ | **A1** | Accept $0.533 < x$ or $0.53 < x$ or $0.54 < x$ or $0.56 \leq x$; Allow use of $<$ or $\leq$; Note prices £0.54 and £0.55 do not lead to integer values, fully correct answer is $0.56 \leq x$

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8 A team of volunteers donates cakes for sale at a charity stall. The number of cakes that can be sold depends on the price. A model for this is $\mathrm { y } = 190 - 70 \mathrm { x }$, where $y$ cakes can be sold when the price of a cake is $\pounds$ x.
\begin{enumerate}[label=(\alph*)]
\item Find how many cakes could be given away for free according to this model.

The number of volunteers who are willing to donate cakes goes up as the price goes up. If the cakes sell for $\pounds 1.20$ they will donate 50 cakes, but if they sell for $\pounds 2.40$ they will donate 140 cakes. They use the linear model $\mathrm { y } = \mathrm { mx } + \mathrm { c }$ to relate the number of cakes donated, $y$, to the price of a cake, $\pounds x$.
\item Find the values of the constants $m$ and $c$ for which this linear model fits the two data points.
\item Explain why the model is not suitable for very low prices.
\item The team would like to sell all the cakes that they donate.

Find the set of possible prices that the cakes could have to achieve this.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI AS Paper 1 2022 Q8 [7]}}