Numeracy Higher June 2023 Unit 2 Q9
9.
On 1st April 2023, Caitlin invested £10 000 into a savings account.
The account had a nominal annual interest rate of 3%, with interest paid on the last day of every month.
Caitlin did not plan to make any further payments into the account or make any withdrawals.
Calculate the date when Caitlin will first have over £10 500 in her account. [4]
Date when Caitlin will first have over £10 500 in her account is ……………………………
Calculate the AER for Caitlin’s savings account.
Give your answer as a percentage, correct to 2 decimal places. [2]
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.
| Scheme | Marks | Comments |
|---|---|---|
| (Monthly rate =) 0.25% or 0.0025 | B1 | B0 for 0.0025%. May be implied in further work |
| 10 000 × 1.0025\(^{n}\) | B1 | For use of ‘n’ or for use of any value of n |
| 10 000 × 1.0025\(^{20}\) (= (£)10 512(.055…)) | M1 | Allow (£)10 000 × 1.0025\(^{19}\) (= (£)10 485(.84…) or (£)10 486) with convincing work that over (£)10 500 will be reached in a months’ time An answer of (n=) 20 (months) implies B1B1M1 provided no incorrect work seen |
| (Date =) 30th November 2024 | A1 | CAO Allow 31st November 2024 or 1st December 2024 If only the first B1 awarded and using the calculation (10000×1.03) × 1.0025\(^{8}\) or 10300 × 1.0025\(^{8}\), award a further SC1 for (£)10507(.81…) or (£)10508 AND 30th November (allowing 31st) or 1st December 2024 If no marks awarded and from using a multiplier of 1.03\(^{2}\), award SC1 for (£)10 609 AND 31st May (allowing 30th) or 1st June (2023), OR If no marks awarded and from using a multiplier of 1.025\(^{2}\), award SC1 for (£)10 506(.25) AND 31st May (allowing 30th) or 1st June (2023) |
Alternative
9(a) Alternative method:
| Scheme | Marks | Comments |
|---|---|---|
| (Monthly rate =) 0.25% or 0.0025 | B1 | B0 for 0.0025%. May be implied in further work |
| 10 000 × 1.0025\(^{n}\) = 10500 | B1 | Or 1.0025\(^{n}\) = \(\dfrac{10500}{10000}\) or equivalent |
| (Number of months =) \(\dfrac{\log(10500/10000)}{\log 1.0025}\) or \(\dfrac{\log 1.05}{\log 1.0025}\) or \(\log_{1.0025}(1.05)\) (= 19.5(404…) or 20 months) | M1 | Implies previous B1B1 Note: The logs in the first 2 possibilities for M1 can be to any base An answer of (n=) 19.5(404…) or 20 (months) implies B1B1M1 provided no incorrect work seen |
| (Date =) 30th November 2024 | A1 | CAO Allow 31st November 2024 or 1st December 2024 If no marks awarded and from using 1.03 instead of 1.0025, award SC1 for 1.6(506…) or 2 (months) AND 31st May (allowing 30th) or 1st June (2023), OR If no marks awarded and from using 1.025 instead of 1.0025, award SC1 for 1.9(759…) or 2 (months) AND 31st May (allowing 30th) or 1st June (2023) |
| Scheme | Marks | Comments |
|---|---|---|
| (AER =) \(\left(1 + \dfrac{0.03}{12}\right)^{12} - 1\) | M1 | M0A0 if n ≠ 12 substituted into the AER formula |
| = 3.04 (%) | A1 | Allow an answer of 3.04(…) % An unsupported 3.04% is awarded M1A1 |