C1 January 2012 Q9
9. A company offers two salary schemes for a 10-year period, Year 1 to Year 10 inclusive.
| Scheme 1: | Salary in Year 1 is £\(P\). Salary increases by £\((2T)\) each year, forming an arithmetic sequence. |
| Scheme 2: | Salary in Year 1 is £\((P + 1800)\). Salary increases by £\(T\) each year, forming an arithmetic sequence. |
For the 10-year period, the total earned is the same for both salary schemes.
For this value of \(T\), the salary in Year 10 under Salary Scheme 2 is £29 850
| Scheme | Marks |
|---|---|
| \(S_{10} = \dfrac{10}{2}\left[2P + 9\times 2T\right]\) or \(\dfrac{10}{2}\left(P + [P + 18T]\right)\) e.g. \(5[2P + 18T]\) | M1 |
| \(= (\text{£})\,(10P + 90T)\) or \((\text{£})\ 10P + 90T \qquad (*)\) | A1cso |
| (2) |
Notes
M1: for identifying \(a = P\) or \(d = 2T\) and attempt at \(S_{10}\). Using \(n = 10\) and one of \(a\) or \(d\) correct. Must see evidence for M mark, at least one line before the answer.
A1cso: for simplifying to given answer. No incorrect working seen.
Do not penalise missing end bracket in working eg \(5(2P + 18T\)
List: M1A1 for a full list seen (with + signs or written in columns) and no incorrect working seen. Any missing terms is M0A0
| Scheme | Marks |
|---|---|
| Scheme 2: \(S_{10} = \dfrac{10}{2}\left[2(P + 1800) + 9T\right] = \left\{10P + 18000 + 45T\right\}\) | M1A1 |
| \(10P + 90T = 10P + 18000 + 45T\) | M1 |
| \(90T = 18000 + 45T\) \(T = 400\) (only) | A1 |
| (4) |
Notes
1st M1: for attempting \(S_{10}\) for scheme 2 (allow missing (…) brackets e.g. \(2P + 1800 + 9T\))
Using \(n = 10\) and at least one of \(a\) or \(d\) correct.
1st A1: for a correct expression for \(S_{10}\) using scheme 2 (needn’t be multiplied out)
List: Allow M1A1 if they reach \(10P + 18000 + 45T\) with no incorrect working seen
\(10P + 18000 + 45T\) with no working is M1A1
2nd M1: for forming an equation using the two sums that would enable \(P\) to be eliminated.
Follow through their expressions provided \(P\) would disappear.
2nd A1: for \(T = 400\) Answer only (4/4)
| Scheme | Marks |
|---|---|
| Scheme 2, Year 10 salary: \([a + (n - 1)d =](P + 1800) + 9T\) | B1ft |
| \(P + 1800 + \text{“}3600\text{”} = 29850\) | M1 |
| \(P = (\text{£})\ \underline{24450}\) | A1 |
| (3) | |
| (9 marks) |
Notes
B1: for using \(u_{10}\) for scheme 2. Can be \(9T\) or follow through their value of \(T\)
M1: for forming an equation based on \(u_{10}\) for scheme 2 and using 29850 and their value of \(T\)
A1: for 24450 seen Answer only (3/3)
MR: If they misread scheme 2 as scheme 1 in part (c) apply MR rule and award B0M1A0 max for an equation based on \(u_{10}\) for scheme 1 and using 29850 and their value of \(T\)