C1 June 2018 Q4
4. Each year, Andy pays into a savings scheme. In year one he pays in £600. His payments increase by £120 each year so that he pays £720 in year two, £840 in year three and so on, so that his payments form an arithmetic sequence.
Kim starts paying money into a different savings scheme at the same time as Andy. In year one she pays in £130. Her payments increase each year so that she pays £210 in year two, £290 in year three and so on, so that her payments form a different arithmetic sequence.
At the end of year \(N\), Andy has paid, in total, twice as much money into his savings scheme as Kim has paid, in total, into her savings scheme.
| Scheme | Marks |
|---|---|
| \(a + (n - 1)d = 600 + 9\times 120\) | M1 |
| \(= (\text{£})1680\) | A1 |
| Answer only scores both marks | |
| Listing M1: Lists ten terms starting £600, £720, £840, £960, … A1: Identifies the 10th term as (£)1680 | |
| (2) |
Notes
M1: This mark is for: \(600 + 9\times 120\) or \(600 + 8\times 120\)
A1: 1680 with or without the “£”
| Scheme | Marks |
|---|---|
| Allow the use of \(n\) instead of \(N\) throughout in (b) | |
| \(d = 80\) for Kim | B1 |
| \(\dfrac{N}{2}\left\{2\times 600 + \left(N - 1\right)\times 120\right\}\) OR \(\dfrac{N}{2}\left\{2\times 130 + \left(N - 1\right)\times 80\right\}\) | M1 |
| \(\dfrac{N}{2}\left\{2\times 600 + \left(N - 1\right)\times 120\right\} = 2\times\dfrac{N}{2}\left\{2\times 130 + \left(N - 1\right)\times 80\right\}\) A correct equation in any form | A1 |
| \(20N = 360 \Rightarrow N = \ldots\) | dM1 |
| \(\left(N =\right)18\) | A1 |
| See below for listing approach If you see \(N = 18\) with no working send to Review | |
| (5) | |
| (7 marks) |
Notes
B1: Identifies or uses \(d = 80\) for Kim
M1: Attempts a sum formula for Andy or Kim. A correct formula must be seen or implied with: \(a = 600\), \(d = 120\) for Andy or \(a = 130\), \(d = 80\) for Kim. If B0 was scored, allow M1 here if Kim’s incorrect “\(d\)” is used.
dM1: Proceeds to find a value for \(N\). (Allow if it leads to \(N < 0\)) Dependent on the first method mark and must be an equation that uses Andy’s and Kim’s sum.
A1: Ignore \(N/n = 0\) and if a correct value of \(N\) is seen, isw any further reference to years etc.
Listing approach
| Year | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Andy | 600 | 1320 | 2160 | 3120 | 4200 | 5400 | 6720 | 8160 | 9720 | 11400 | 13200 | 15120 | 17160 | 19320 | 21600 | 24000 | 26520 | 29160 |
| Kim | 130 | 340 | 630 | 1000 | 1450 | 1980 | 2590 | 3280 | 4050 | 4900 | 5830 | 6840 | 7930 | 9100 | 10350 | 11680 | 13090 | 14580 |
| Kimx2 | 260 | 680 | 1260 | 2000 | 2900 | 3960 | 5180 | 6560 | 8100 | 9800 | 11660 | 13680 | 15860 | 18200 | 20700 | 23360 | 26180 | 29160 |
B1: States or uses \(d = 80\) for Kim
M1: Attempts to find the total savings for Andy or Kim – must see the correct pattern for Andy (600, 1320, 2160,…) or Kim (130, 340, 630,…) (or Kimx2)
A1: Correct totals for Andy and Kim (or Kimx2) at least as far as \(n = 18\)
M1: Identifies when Andy’s total = 2xKim’s total
A1: \(N = 18\)