C2 January 2010 Q6
6. A car was purchased for £18 000 on 1st January.
On 1st January each following year, the value of the car is 80% of its value on 1st January in the previous year.
The value of the car falls below £1000 for the first time \(n\) years after it was purchased.
An insurance company has a scheme to cover the maintenance of the car.
The cost is £200 for the first year, and for every following year the cost increases by 12% so that for the 3rd year the cost of the scheme is £250.88
| Scheme | Marks |
|---|---|
| \(18000 \times (0.8)^3\quad = \text{£}9216\ *\) [may see \(\dfrac{4}{5}\) or 80% or equivalent]. | B1cso |
| (1) |
Notes
B1 NB Answer is printed so need working. May see as above or \(\times 0.8\) in three steps giving 14400, 11520, 9216. Do not need to see £ sign but should see 9216 .
| Scheme | Marks |
|---|---|
| \(18000 \times (0.8)^n < 1000\) | M1 |
| \(n\log(0.8) < \log\left(\tfrac{1}{18}\right)\) | M1 |
| \(n > \dfrac{\log\left(\frac{1}{18}\right)}{\log(0.8)} = 12.952\ldots.\qquad\) so \(n = 13\). | A1 cso |
| (3) |
Notes
1st M1 for an attempt to use \(n\)th term and 1000. Allow \(n\) or \(n - 1\) and allow > or =
2nd M1 for use of logs to find \(n\) Allow \(n\) or \(n - 1\) and allow > or =
A1 Need \(n = 13\) This is an accuracy mark and must follow award of both M marks but should not follow incorrect work using \(n - 1\) for example.
Condone slips in inequality signs here.
Alternative Methods
Trial and Improvement
See 989.56 ( or 989 or 990) identified with 12, 13 or 14 years for first M1
See 1236.95 ( or 1236 or 1237) identified with 11, 12 or 13 years for second M1
Then \(n = 13\) is A1 (needs both Ms)
Special case \(18000 \times (0.8)^n < 1000\) so \(n = 13\) as 989.56<1000 is M1M0A0 (not discounted \(n = 12\))
| Scheme | Marks |
|---|---|
| \(u_5 = 200 \times (1.12)^4,\qquad = \text{£}314.70\) or £314.71 | M1, A1 |
| (2) |
Notes
M1 for use of their \(a\) and \(r\) in formula for 5th term of GP
A1 cao need one of these answers – answer can imply method here
NB 314.7 – A0
Alternative
May see the terms 224, 250.88, 280.99, 314.71 with a small slip for M1 A0, or done accurately for M1A1
| Scheme | Marks |
|---|---|
| \(S_{15} = \dfrac{200\left(1.12^{15} - 1\right)}{1.12 - 1}\) or \(\dfrac{200\left(1 - 1.12^{15}\right)}{1 - 1.12}\), \(=\ 7455.94\ldots..\) awrt £7460 | M1A1, A1 |
| (3) | |
| [9] |
Notes
M1 for use of sum to 15 terms of GP using their \(a\) and their \(r\) ( allow if formula stated correctly and one error in substitution, but must use \(n\) not \(n\) - 1)
1st A1 for a fully correct expression ( not evaluated)
Alternative
Adds 15 terms 200 + 224 + 250.88+… + (977.42) M1
Seeing 977… is A1
Obtains answer 7455.94 A1 or awrt £7460 NOT 7450