June 2024 Paper 2 Q7
7 On the first day of each month, Kate pays £50 into a savings account.
Interest is paid on the total amount in the account on the last day of each month.
The interest rate is 0.2%
At the end of the \(n\)th month, the total amount of money in Kate’s savings account is £\(T_n\)
Kate correctly calculates \(T_1\) and \(T_2\) as shown below:
\[T_1 = 50 \times 1.002 = 50.10\]\[\begin{aligned} T_2 &= (T_1 + 50) \times 1.002 \\ &= \big((50 \times 1.002) + 50\big) \times 1.002 \\ &= 50 \times 1.002^2 + 50 \times 1.002 \\ &\approx 100.30 \end{aligned}\](a) Show that \(T_3\) is given by\[T_3 = 50 \times 1.002^3 + 50 \times 1.002^2 + 50 \times 1.002\] [1 mark]
(b) Kate uses her method to correctly calculate how much money she can expect to have in her savings account at the end of 10 years.
(i) Find the amount of money Kate expects to have in her savings account at the end of 10 years. [3 marks]
(ii) The amount of money in Kate’s savings account at the end of 10 years may not be the amount she has correctly calculated.
Explain why. [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| Obtains \(50 \times 1.002^3 + 50 \times 1.002^2 + 50 \times 1.002\) with \(\left(50 \times 1.002^2 + 50 \times 1.002 + 50\right) \times 1.002\) or better seen | R1 | 2.1 |
| (1) |
Typical solution
\[T_3 = \left(50 \times 1.002^2 + 50 \times 1.002 + 50\right) \times 1.002\]\[= 50 \times 1.002^3 + 50 \times 1.002^2 + 50 \times 1.002\]| Scheme | Marks | AO |
|---|---|---|
| (i) Models the total as the sum to n terms of a geometric sequence Evidence for this could include at least two of \(a\), \(r\) or \(n\) substituted into sum formula \(a\) = 50.1, \(r\) = 1.002, \(n\) = 120 Condone \(a\) = 50 or \(n\) = 10 or 119 PI by AWRT 6724, 6737, 6801 or 505.53 | M1 | 3.3 |
| Forms the correct expression for the correct total \(\left(T_{120} =\right)\dfrac{50.1\left(1 - 1.002^{120}\right)}{1 - 1.002}\) or \(\displaystyle\sum_{x=1}^{120} 50 \times 1.002^x\) | A1 | 3.3 |
| Obtains £6 787, £6 787.15 or £6 787.16 | A1 | 3.2a |
| (3) | ||
| (ii) Makes a reasonable comment in context. For example: The interest rate is unlikely to remain fixed. Or May have needed to withdraw some amount. Or May change the monthly payments. | E1 | 3.5b |
| (1) | ||
| (5 marks) |
Typical solution
(i)
\[T_{120} = \frac{50.1\left(1 - 1.002^{120}\right)}{1 - 1.002}\]\[= 6787.1595\ldots\]Total in account = £ 6 787
(ii)
The interest rate is unlikely to remain fixed for the whole 10 years