| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Geometric Sequences and Series |
| Type | Compound growth applications |
| Difficulty | Moderate -0.3 This is a straightforward application of geometric sequences to depreciation and compound interest. Part (a) is simple verification requiring two multiplications by 0.8. Part (b) requires setting up an inequality and solving using logarithms (standard C2 technique). Part (c) applies the geometric series formula to calculate savings with compound interest. While multi-part and requiring several steps, all techniques are routine C2 content with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.04i Geometric sequences: nth term and finite series sum1.04k Modelling with sequences: compound interest, growth/decay |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(15000 \times (0.8)^2 = 9600\) | M1 | for \(\times\) by 0.8 |
| \(= 9600\) | A1 cso | |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(15000 \times (0.8)^n < 500\) | M1 | Suitable equation or inequality |
| \(n\log(0.8) < \log(\frac{1}{30})\) | M1 | Take logs |
| \(n > 15.24...\) | A1 | \(n\) is OK |
| So machine is replaced in 2015 | A1 | |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a = 1000,\ r = 1.05,\ n = 16\) | M1 | \(\geq 2\) correct |
| \(S_{16} = \frac{1000(1.05^{16}-1)}{1.05-1}\) | M1 A1 | |
| \(= £23\,657.49 = £23\,700\) or \(£23\,660\) or \(£23657\) | A1 | |
| (4) |
# Question 6:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $15000 \times (0.8)^2 = 9600$ | M1 | for $\times$ by 0.8 |
| $= 9600$ | A1 cso | |
| | **(2)** | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $15000 \times (0.8)^n < 500$ | M1 | Suitable equation or inequality |
| $n\log(0.8) < \log(\frac{1}{30})$ | M1 | Take logs |
| $n > 15.24...$ | A1 | $n$ is OK |
| So machine is replaced in 2015 | A1 | |
| | **(4)** | |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a = 1000,\ r = 1.05,\ n = 16$ | M1 | $\geq 2$ correct |
| $S_{16} = \frac{1000(1.05^{16}-1)}{1.05-1}$ | M1 A1 | |
| $= £23\,657.49 = £23\,700$ or $£23\,660$ or $£23657$ | A1 | |
| | **(4)** | |
---6. At the beginning of the year 2000 a company bought a new machine for $\pounds 15000$. Each year the value of the machine decreases by $20 \%$ of its value at the start of the year.
\begin{enumerate}[label=(\alph*)]
\item Show that at the start of the year 2002, the value of the machine was $\pounds 9600$.
When the value of the machine falls below $\pounds 500$, the company will replace it.
\item Find the year in which the machine will be replaced.
To plan for a replacement machine, the company pays $\pounds 1000$ at the start of each year into a savings account. The account pays interest at a fixed rate of $5 \%$ per annum. The first payment was made when the machine was first bought and the last payment will be made at the start of the year in which the machine is replaced.
\item Using your answer to part (b), find how much the savings account will be worth immediately after the payment at the start of the year in which the machine is replaced.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 Q6 [10]}}