June 2024 Paper 2 Q2
2. Jamie takes out an interest-free loan of £8100
Jamie makes a payment every month to pay back the loan.
Jamie repays £400 in month 1, £390 in month 2, £380 in month 3, and so on, so that the amounts repaid each month form an arithmetic sequence.
After Jamie’s \(N\)th payment, the loan is completely paid back.
| Scheme | Marks | AO |
|---|---|---|
| \(\left(u_{12} =\right) 400 + 11 \times -10 = 290\ *\) or e.g. \(\left(u_{12} =\right) 400 - 110 = 290\ *\) or e.g. \(\left(u_{12} =\right) 400 + (12-1) \times -10 = 290\ *\) or e.g. \(\left(u_{12} =\right) 410 + 12 \times -10 = 290\ *\) | B1* | 1.1b |
| (1) |
Notes
B1*: Correct working to obtain 290. Must be a correct calculation so do not condone missing brackets unless they are recovered. E.g. \(\left(u_{12} =\right) 400 + 12 - 1 \times -10 = 290\) scores B0 unless followed by \(= 400 + 11 \times -10 = 290\). Condone \(\left(u_{12} =\right) 400 + (12-1) - 10 = 290\)
The “£” symbol is not required but the “290” must appear.
Alternative 1:
| Scheme | Marks |
|---|---|
| \(400 + (n-1) \times -10 = 290\) \(\Rightarrow 400 - 10n + 10 = 290 \Rightarrow 10n = 120 \Rightarrow n = 12\ *\) | B1* |
B1*: Correct working using the 290 to obtain \(n = 12\).
There must be at least one intermediate line after setting up the equation and must be correct work so do not condone missing brackets unless they are recovered (as above).
A conclusion is not required with this approach as long as 12 is correctly obtained.
Alternative 2:
| Scheme | Marks |
|---|---|
| \(290 = 400 + (12-1)d \Rightarrow 11d = -110 \Rightarrow d = -10\ *\) | B1* |
B1*: Correct working using the 290 and 400 to obtain \(d = -10\).
There must be at least one intermediate line after setting up the equation and must be correct work so do not condone missing brackets unless they are recovered (as above).
A conclusion is not required with this approach as long as \(-10\) is correctly obtained.
Allow candidates to list terms and show the 12th term is 290 e.g.
400, 390, 380, 370, 360, 350, 340, 330, 320, 310, 300, 290
Must list all 12 terms which must be correct and end with 290
Condone if missing 400, 390, 380 as these are given in the question.
| Scheme | Marks | AO |
|---|---|---|
| \(8100 = \dfrac{1}{2}N\left(2 \times 400 + (N-1) \times -10\right)\) or e.g. \(8100 = \dfrac{1}{2}N\left(400 + 400 + (N-1) \times -10\right)\) | M1 | 1.1b |
| \(8100 = \dfrac{1}{2}N\left(2 \times 400 + (N-1) \times -10\right)\) \(\Rightarrow 16200 = 800N - 10N^2 + 10N\) or e.g. \(\Rightarrow 8100 = 400N - 5N^2 + 5N\) \(\Rightarrow N^2 - 81N + 1620 = 0\ *\) | A1* | 2.1 |
| (2) |
Notes
Mark (b) and (c) together
M1: Uses a correct sum formula in terms of \(N\) or \(n\) with \(a = 400\) and \(d = -10\) or \(+10\) and sets = 8100. Condone e.g. > 8100 and allow A1 if this is recovered to become “=” before the final line.
Condone \(8100 = \dfrac{1}{2}N\left(2 \times 400 + (N-1) - 10\right)\) if recovered or not.
A1*: Fully correct proof with sufficient working shown and no unrecovered errors.
Do not condone e.g. missing brackets or e.g. a missing \(N/n\) unless recovered before the final given answer.
Condone the use of \(n\) instead of \(N\) for both marks.
Condone terms in a different order as long as they are correct.
Condone \(0 = N^2 - 81N + 1620\ *\)
Sufficient working requires all brackets to be removed to obtain an unsimplified expanded quadratic before proceeding to the given answer including the “=0”.
Alternative (further maths method): Series summation approach:
\[\begin{aligned}&\sum_{r=1}^{N}(410 - 10r) = 8100 \Rightarrow 410N - 10 \times \frac{1}{2}N(N+1)\\&\Rightarrow 410N - 5N^2 - 5N = 8100 \Rightarrow N^2 - 81N + 1620 = 0\ *\end{aligned}\]M1: Attempt to sum an appropriate series with first term 400. Condone use of \(+10\) as in the main scheme.
A1*: As main scheme.
| Scheme | Marks | AO |
|---|---|---|
| \(N^2 - 81N + 1620 = 0 \Rightarrow (N-45)(N-36) = 0 \Rightarrow N = 45, 36\) | M1 | 1.1b |
| \((N =)\ 36\) | A1 | 2.3 |
| (2) | ||
| (5 marks) |
Notes
M1: Solves the given quadratic equation by any correct method including a calculator to obtain at least one value for \(N\). See general guidance for solving a 3-term quadratic.
If values are just written down and only one value is given it must be 45 or 36.
If both values are just written down they must both be correct.
A1: Realises that the smaller value is required and so selects \((N =)\ 36\).
Ignore any units if given.
The “\(N =\)” is not required, just look for the correct value.
It must be clear that this value has been selected. This may be indicated by e.g. underlining the 36 or the omission of the 45. If the 45 is not rejected score A0.
\(N = 36\) with no working scores M1A1