Numeracy Higher November 2017 Unit 2 Q7
7. Iestyn opened a savings account on 1 August 2017, investing £2800.
On 1 October 2017, he viewed his savings account online.
The table below shows all the transactions that had taken place since he opened the account.
| Date | Details | Paid in (£) | Paid out (£) | Balance (£) |
|---|---|---|---|---|
| 01/08/17 | Account opened | 2800.00 | 2800.00 | |
| 31/08/17 | Interest | 14.00 | 2814.00 | |
| 30/09/17 | Interest | 14.07 | 2828.07 |
(a) Calculate the nominal interest rate per annum. [3]
(b) Calculate the AER the account was paying.
Give your answer as a percentage, correct to 2 decimal places. [3]
From the formula list: Annual Equivalent Rate (AER): AER, as a decimal, is calculated using the formula \(\left(1+\dfrac{i}{n}\right)^{n}-1\), where \(i\) is the nominal interest rate per annum as a decimal and \(n\) is the number of compounding periods per annum.
Give your answer as a percentage, correct to 2 decimal places. [3]
| Scheme | Marks | Comments |
|---|---|---|
| \(\dfrac{14}{2800}\) (× 100) ×12 OR \(\dfrac{14.07}{2814}\) (× 100) ×12 | M2 | M1 for \(\dfrac{14}{2800}\) (× 100) OR \(\dfrac{14.07}{2814}\) (× 100) Or M1 for 0∙005 or 0∙5% |
| = 6(%) or 0∙06 | A1 | A0 for 0∙06% |
| Scheme | Marks | Comments |
|---|---|---|
| \(\left(1 + \dfrac{0{\cdot}06}{12}\right)^{12} - 1\) or equivalent | M1 | FT their final answer from (a) |
| = 6·17 (%) | A2 | A1 for 0·061(67…) or 0·062, OR A1 for 6·1(67…%) or 6·2(%) Alternative methods: M1 for 2800 × 1·00512 M1 for \(\dfrac{2972.69(7873) - 2800}{2800}\) (× 100) or equivalent FT ‘their 2972.69(7873) provided previous M1 awarded. A1 for 6·17 (%), OR M1 for \(\left(1 + \dfrac{0{\cdot}06015}{6}\right)^{6} - 1\) A2 for 6·17 (%) A1 for 0·06(167…), or for a correct percentage but not correct to 2 dp. |