C2 January 2013 Q3
3. A company predicts a yearly profit of £120 000 in the year 2013. The company predicts that the yearly profit will rise each year by 5%. The predicted yearly profit forms a geometric sequence with common ratio 1.05
| Scheme | Marks |
|---|---|
| \(120000 \times (1.05)^3 = 138915\ *\) | B1 |
| (1) |
Notes
Or \(120000 \times 1.05 \times 1.05 \times 1.05 = 138915\)
Or 120000, 126000, 132300, 138915
Or \(a = 120000\) and \(a \times (1.05)^3 = 138915\)
| Scheme | Marks |
|---|---|
| \(120000 \times (1.05)^{n-1} > 200000\) | M1 |
| \(\log 1.05^{n-1} > \log\left(\dfrac{5}{3}\right)\) | M1 |
| \((n - 1 >)\ \dfrac{\log\left(\frac{5}{3}\right)}{\log 1.05}\) or equivalent e.g \((n >)\ \dfrac{\log\left(\frac{7}{4}\right)}{\log 1.05}\) | A1 |
| 2024 | M1A1 |
| (5) |
Notes
M1: Allow \(n\) or \(n - 1\) and “\(>\)”, “\(<\)”, or “\(=\)” etc.
M1: Takes logs correctly. Allow \(n\) or \(n - 1\) and “\(>\)”, “\(<\)”, or “\(=\)” etc.
A1: Allow \(n\) or \(n - 1\) and “\(>\)”, “\(<\)”, or “\(=\)” etc. Allow \(1.\dot{6}\) or awrt 1.67 for 5/3.
M1: Identifies a calendar year using their value of \(n\) or \(n - 1\)
A1: 2024
Listing or trial/improvement in (b)
| Scheme | Marks |
|---|---|
| \(U_{10} = 186\,159.39,\ U_{11} = 195\,467.36,\ U_{12} = 205\,240.72\) | |
| Attempt to find at least the 10th or 11th or 12th terms correctly using a common ratio of 1.05 (all the terms need not be listed) | M1 |
| Forms the geometric progression correctly to reach a term > 200 000 | M1 |
| Obtains an “11th” term of awrt 195 500 and a “12th” term of awrt 205 200 | A1 |
| Uses their number of terms to identify a calendar year | M1 |
| 2024 | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\dfrac{a(1 - r^n)}{1 - r} = \dfrac{120000\left(1 - 1.05^{11}\right)}{1 - 1.05}\) | M1 A1 |
| 1704814 | A1 |
| (3) | |
| [9] |
Notes
M1: Correct sum formula with \(n = 10\), 11 or 12
A1: Correct numerical expression with \(n = 11\)
A1: Cao (Allow 1704814.00)