October 2020 Paper 1 Q4
4. In 1997 the average CO2 emissions of new cars in the UK was 190 g/km.
In 2005 the average CO2 emissions of new cars in the UK had fallen to 169 g/km.
Given \(A\) g/km is the average CO2 emissions of new cars in the UK \(n\) years after 1997 and using a linear model,
In 2016 the average CO2 emissions of new cars in the UK was 120 g/km.
| Scheme | Marks | AO |
|---|---|---|
| Attempts \(A = mn + c\) with either \((0, 190)\) or \((8, 169)\) Or attempts gradient eg \(m = \pm\dfrac{190 - 169}{8}\ (= -2.625)\) | M1 | 3.3 |
| Full method to find a linear equation linking \(A\) with \(n\) E.g. Solves \(190 = 0n + c\) and \(169 = 8n + c\) simultaneously | dM1 | 3.1b |
| \(A = -2.625n + 190\) | A1 | 1.1b |
| (3) |
Notes
M1: Attempts \(A = mn + c\) with either \((0, 190)\) or \((8, 169)\) considered.
Eg Accept sight of \(190 = 0n + c\) or \(169 = 8m + c\) or \(A - 169 = m(n - 8)\) or \(A = 190 + mn\) where \(m\) could be a value.
Also accept an attempt to find the gradient \(\pm\dfrac{190 - 169}{8}\) or sight of \(\pm 2.625\) or \(\pm\dfrac{21}{8}\) oe
dM1: A full method to find both constants of a linear equation
Method 1: Solves \(190 = 0n + c\) and \(169 = 8n + c\) simultaneously
Method 2: Uses gradient and a point Eg \(m = \pm\dfrac{190 - 169}{8}\ (= -2.625)\) and \(c = 190\)
Condone different variables for this mark. Eg. \(y\) in terms of \(x\).
A1: \(A = -2.625n + 190\) or \(A = -\dfrac{21}{8}n + 190\) oe
| Scheme | Marks | AO |
|---|---|---|
| Attempts \(A = -2.625 \times 19 + 190 = \ldots\) | M1 | 3.4 |
| \(A = 140.125\ \mathrm{g\ km^{-1}}\) | A1 | 1.1b |
| It is predicting a much higher value and so is not suitable | B1ft | 3.5a |
| (3) | ||
| (6 marks) |
Notes
M1: Attempts to substitute "\(n\)" = 19 into their linear model to find \(A\). They may call it \(x = 19\)
Alternatively substitutes \(A = 120\) into their linear model to find \(n\).
A1: \(A = 140.125\) from \(n = 19\) Allow \(A = 140\)
or \(n = 26/27\) following \(A = 120\)
B1ft: Requires a correct calculation for their model, a correct statement and a conclusion
E.g For correct (a) \(A = 140\) is (much) higher than 120 so the model is not suitable/appropriate.
Follow through on a correct statement for their equation. As a guide allow anything within [114, 126] to be regarded as suitable. Anything less than 108 or more than 132 should be justified as unsuitable.
Note B0 Recorded value is not the same as/does not equal/does not match the value predicted