C1 January 2014 (IAL) Q7
7. Shelim starts his new job on a salary of £14 000. He will receive a rise of £1500 a year for each full year that he works, so that he will have a salary of £15 500 in year 2, a salary of £17 000 in year 3 and so on. When Shelim’s salary reaches £26 000, he will receive no more rises. His salary will remain at £26 000.
Anna starts her new job at the same time as Shelim on a salary of £\(A\). She receives a rise of £1000 a year for each full year that she works, so that she has a salary of £\((A + 1000)\) in year 2, £\((A + 2000)\) in year 3 and so on. The maximum salary for her job, which is reached in year 10, is also £26 000.
| Scheme | Marks |
|---|---|
| \(14000 + 8\times 1500 = 14000 + 12000\) | M1 |
| \(= \text{£}26000\) | A1* |
| (2) |
Notes
M1: Uses \(S = a + (n - 1)d\) with \(a\)=14000, \(d\)=1500 and \(n\)=8, 9 or 10 in an attempt to find salary in year 9
Accept a sequence written out only if all terms up to year 9 are included-Allow no errors.
A1*: csa 26000. It is acceptable to write a sequence for both the 2 marks
FYI the terms are 14000, 15500, 17000, 18500, 20000, 21500, 23000, 24500, 26000
Alt (a) Alternative working backwards
M1: Uses \(S = a + (n - 1)d\) with \(a\)=14000, \(d\)=1500 and \(S\) =26000 in attempt to find \(n\). It must reach \(n\)=..
A1: \(n\)=9
| Scheme | Marks |
|---|---|
| \(S_n = \dfrac{n}{2}(a + l) = \dfrac{9}{2}\times(14000 + 26000)\) OR \(S_9 = \dfrac{n}{2}(2a + (n - 1)d) = \dfrac{9}{2}\times(28000 + 8\times 1500)\) | M1 |
| \(= \text{£}180000\) | A1 |
| (2) |
Notes
M1: Uses \(S_n = \dfrac{n}{2}(a + l)\) with \(a\)=14000, \(l\)=26000 and \(n\)=8, 9 or 10. Do not allow ft’s on incorrect \(l\)’s .
Alternatively uses \(S_n = \dfrac{n}{2}(2a + (n - 1)d)\) with \(a\)=14000, \(d\)=1500 and \(n\)=8, 9 or 10.
Weaker candidates may list the individual salaries. This is acceptable as long as all terms are included.
For example
\(14000 + 15500 + 17000 + 18500 + 20000 + 21500 + 23000 + 24500 + 26000\)
A1: Cao (£) 180000.
| Scheme | Marks |
|---|---|
| Use \(a + (n - 1)d\) to find \(A\) | |
| \(A + (10 - 1)\times 1000 = 26000\) | M1 |
| \(A = 17000\) | A1 |
| Use \(S_n = \dfrac{n}{2}(a + l)\) or \(S_n = \dfrac{n}{2}(2a + (n - 1)d)\) to find \(S\) for Anna | |
| \(S_{10} = \dfrac{10}{2}(17000 + 26000)\ (= \text{£}215000)\) or \(S_{10} = \dfrac{10}{2}(2\times 17000 + 9\times 1000)\ (= \text{£}215000)\) | M1A1 |
| Shelim earns 180000+26000 in 10 years =(£206000) | B1ft |
| Difference= £9000 | A1 |
| (6) | |
| (10 marks) |
Notes
M1: Use \(l = a + (n - 1)d\) to find \(A\).
It must be a full method with \(d\)=1000, \(l\)=26000, \(a\)=\(A\) and \(n\)=9, 10 or 11 leading to a value for \(A\)
A1: \(A\)=17000.
Accept \(A\)=17000 written down for 2 marks as long as no incorrect work seen in its calculation.
M1: Use \(S_n = \dfrac{n}{2}(a + l)\) to find \(S\) for Anna. Follow through on their \(A\), but \(l\)=26000 and \(n\)=9, 10 or 11
Alternatively uses \(S_n = \dfrac{n}{2}(2a + (n - 1)d)\) with their numerical value of \(A\), \(d\)=1000 and \(n\)=9, 10 or 11
Accept a series of terms with their value of A, rising in £1000’s up to a maximum of £26000.
A1: Anna earns \(S_{10} = \dfrac{10}{2}(17000 + 26000)\) OR \(S_{10} = \dfrac{10}{2}(2\times 17000 + 9\times 1000)\) in 10 years
This is an intermediate answer. There is no requirement to state the value £215 000
B1ft: Shelim earns (b)+26000 in 10 years. This may be scored at the start of part c.
A1: CAO and CSO Difference =£9000