| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2020 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Linear modelling problems |
| Difficulty | Easy -1.2 This is a straightforward linear modelling question requiring only basic coordinate geometry (finding equation of a line through two points) and simple interpretation. Part (a) involves calculating gradient and using y = mx + c with given data points. Part (b) requires only a basic comparison of predicted vs actual value. No complex problem-solving or novel insight needed—this is routine GCSE-level algebra dressed as an A-level context question. |
| Spec | 1.02z Models in context: use functions in modelling1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03c Straight line models: in variety of contexts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(A = mn+c\) with either \((0,190)\) or \((8,169)\), or gradient \(m = \pm\frac{190-169}{8}\ (=-2.625)\) | M1 | Accept sight of \(190=0n+c\) or \(169=8m+c\) or \(A-169=m(n-8)\) or \(\pm2.625\) or \(\pm\frac{21}{8}\) |
| Full method to find both constants e.g. solves \(190=0n+c\) and \(169=8n+c\) simultaneously | dM1 | Method 1: simultaneous equations. Method 2: gradient and point. Condone different variable names |
| \(A = -2.625n+190\) | A1 | Or \(A = -\frac{21}{8}n+190\) oe |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Attempts \(A = -2.625\times19+190 = \ldots\) | M1 | Substitutes \(n=19\) into linear model. Alt: substitutes \(A=120\) to find \(n\) |
| \(A = 140.125 \text{ g km}^{-1}\) | A1 | Allow \(A=140\). Or \(n=26/27\) following \(A=120\) |
| It is predicting a much higher value and so is not suitable | B1ft | Requires correct calculation, correct statement and conclusion. Allow anything in \([114,126]\) as suitable. Values less than 108 or more than 132 should be justified as unsuitable. B0 for "recorded value does not equal predicted value" |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $A = mn+c$ with either $(0,190)$ or $(8,169)$, or gradient $m = \pm\frac{190-169}{8}\ (=-2.625)$ | M1 | Accept sight of $190=0n+c$ or $169=8m+c$ or $A-169=m(n-8)$ or $\pm2.625$ or $\pm\frac{21}{8}$ |
| Full method to find both constants e.g. solves $190=0n+c$ and $169=8n+c$ simultaneously | dM1 | Method 1: simultaneous equations. Method 2: gradient and point. Condone different variable names |
| $A = -2.625n+190$ | A1 | Or $A = -\frac{21}{8}n+190$ oe |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $A = -2.625\times19+190 = \ldots$ | M1 | Substitutes $n=19$ into linear model. Alt: substitutes $A=120$ to find $n$ |
| $A = 140.125 \text{ g km}^{-1}$ | A1 | Allow $A=140$. Or $n=26/27$ following $A=120$ |
| It is predicting a much higher value and so is not suitable | B1ft | Requires correct calculation, correct statement and conclusion. Allow anything in $[114,126]$ as suitable. Values less than 108 or more than 132 should be justified as unsuitable. B0 for "recorded value does not equal predicted value" |\begin{enumerate}
\item In 1997 the average $\mathrm { CO } _ { 2 }$ emissions of new cars in the UK was $190 \mathrm {~g} / \mathrm { km }$.
\end{enumerate}
In 2005 the average $\mathrm { CO } _ { 2 }$ emissions of new cars in the UK had fallen to $169 \mathrm {~g} / \mathrm { km }$.\\
Given $\mathrm { Ag } / \mathrm { km }$ is the average $\mathrm { CO } _ { 2 }$ emissions of new cars in the UK $n$ years after 1997 and using a linear model,\\
(a) form an equation linking $A$ with $n$.
In 2016 the average $\mathrm { CO } _ { 2 }$ emissions of new cars in the UK was $120 \mathrm {~g} / \mathrm { km }$.\\
(b) Comment on the suitability of your model in light of this information.
\hfill \mbox{\textit{Edexcel AS Paper 1 2020 Q4 [6]}}