C2 June 2005 Q9
9.
Mr. King will be paid a salary of £35 000 in the year 2005. Mr. King’s contract promises a 4% increase in salary every year, the first increase being given in 2006, so that his annual salaries form a geometric sequence.
Mr. King will receive a salary each year from 2005 until he retires at the end of 2024.
| Scheme | Marks |
|---|---|
| \((S =)\ a + ar + \ldots + ar^{n-1}\) “\(S =\)” not required. Addition required. | B1 |
| \((rS =)\ ar + ar^2 + \ldots + ar^n\) “\(rS =\)” not required (M: Multiply by \(r\)) | M1 |
| \(S(1 - r) = a(1 - r^n) \qquad S = \dfrac{a(1 - r^n)}{1 - r}\) (M: Subtract and factorise) (*) | M1 A1cso |
| (4) |
Notes
B1: At least the 3 terms shown above, and no extra terms.
A1: Requires a completely correct solution.
Alternative for the 2 M marks:
M1: Multiply numerator and denominator by \(1 - r\).
M1: Multiply out numerator convincingly, and factorise.
| Scheme | Marks |
|---|---|
| \(ar^{n-1} = 35000 \times 1.04^3 = 39\,400\) (M: Correct \(a\) and \(r\), with \(n = 3, 4\) or 5). | M1 A1 |
| (2) |
Notes
M1 can also be scored by a “year by year” method.
Answer only: 39 400 scores full marks, 39 370 scores M1 A0.
Failure to round correctly in (b) and (c):
Penalise once only (first occurrence).
| Scheme | Marks |
|---|---|
| \(n = 20\) (Seen or implied) | B1 |
| \(S_{20} = \dfrac{35000(1 - 1.04^{20})}{(1 - 1.04)}\) (M1: Needs any \(r\) value, \(a = 35000\), \(n = 19, 20\) or 21). (A1ft: ft from \(n = 19\) or \(n = 21\), but \(r\) must be 1.04). | M1 A1ft |
| \(= 1\,042\,000\) | A1 |
| (4) | |
| (10 marks) |
Notes
M1 can also be scored by a “year by year” method, with terms added. In this case the B1 will be scored if the correct number of years is considered.
Answer only: Special case: 1 042 000 scores 2 B marks, scored as 1, 0, 0, 1 (Other answers score no marks).
Failure to round correctly in (b) and (c):
Penalise once only (first occurrence).