C1 June 2013 (R) Q7
7. Each year, Abbie pays into a savings scheme. In the first year she pays in £500. Her payments then increase by £200 each year so that she pays £700 in the second year, £900 in the third year and so on.
Abbie pays into the scheme for \(n\) years until she has paid in a total of £67 200.
| Scheme | Marks |
|---|---|
| \(U_{10} = 500 + (10 - 1)\times 200\) | M1 |
| \(= (\text{£})2300\) | A1 |
| If the term formula is not quoted and the numerical expression is incorrect score M0. A correct answer with no working scores full marks. | |
| (2) |
Notes
M1: Uses \(a + (n - 1)d\) with \(a\)=500, \(d\)=200 and \(n\) = 9,10 or 11
Mark parts (b) and (c) together
| Scheme | Marks |
|---|---|
| \(\dfrac{n}{2}\left\{2\times 500 + (n - 1)\times 200\right\} = 67200\) | M1A1 |
| If the sum formula is not quoted and the equation is incorrect score M0. | |
| \(n^2 + 4n - 672 = 0\) | dM1A1 |
| E.g. allow \(n^2 + 4n = 672\), \(n^2 = 672 - 4n\), \(672 - 4n - n^2 = 0\), \(200n^2 + 800n = 134400\) etc. | |
| \(n^2 + 4n - 24\times 28 = 0\,*\) | A1 |
| (5) |
Notes
M1: Attempt to use \(S = \dfrac{n}{2}\left\{2a + (n - 1)d\right\}\) with , \(S_n = 67200\), \(a = 500\) and \(d = 200\)
A1: Correct equation
dM1: An attempt to remove brackets and collect terms. Dependent on the previous M1
A1: A correct three term equation in any form
A1: Replaces 672 with 24×28 with the equation as printed (including = 0) with no errors. (= 0 may not appear on the last line but must be seen at some point)
| Scheme | Marks |
|---|---|
| \((n - 24)(n + 28) = 0 \Rightarrow n = ..\) or \(n(n + 4) = 24\times 28 \Rightarrow n = ..\) | M1 |
| 24 | A1 |
| Allow correct answer only in (c) | |
| (2) | |
| (9 marks) |
Notes
M1: Solves the given quadratic in an attempt to find \(n\). They may use the quadratic formula.
A1: States that \(n = 24\), or the number of years is 24