Numeracy Higher November 2024 Unit 2 Q7
7. Isaac and his sister Mari are both going to open savings accounts.
Isaac decides to invest some of his savings in the Hereford Saver account.
Details of the account are shown below.
| Hereford Saver account |
|---|
| Monthly interest rate 0·26% |
Calculate the AER that the Hereford Saver account offers.
Give your answer as a percentage. [2]
Mari invests £3000 in the Silver Plus savings account.
Details of the account are shown below.
| Account name | Nominal annual rate | Interest paid |
|---|---|---|
| Silver Plus | 2·48% | Quarterly |
Mari does not withdraw any money or make any further payments into the account.
Mari closes the account after 10 years.
Calculate the percentage increase in the value of her investment.
You must show all your working. [5]
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 |
|---|---|---|
| (AER =) \((1 + 0.0026)^{12} - 1\) or equivalent | M1 | e.g. \(\left(1 + \dfrac{12 \times 0.0026}{12}\right)^{12} - 1\) or \(\left(1 + \dfrac{0.0312}{12}\right)^{12} - 1\) |
| = 3.16(500…) or 3.17 or 3.2 (%) | A1 |
| Scheme | Marks | Comments |
|---|---|---|
| \(AER = \left(1 + \dfrac{2.48 \div 100}{4}\right)^4 - 1 =\) (= 0.025(03…) or 2.5(03…) (%)) | M1 | The −1 may be implied in further working |
| (Amount in account after 10 years =) 3000 × \((1 + 0.025(03\ldots))^{10}\) | M1 | FT ‘their derived 0.025(03…)’ provided it comes from the AER formula with 1 < \(n\) \(\leqslant\) 12 |
| = (£)3841.43(752…) or (£)3841.44 | A1 | CAO. Must come from M1M1 Accept (£)3840.25 from the use of the multiplier 1.025 provided M1M1 previously awarded |
| (Percentage increase =) \(\dfrac{3841.43(752\ldots) - 3000}{3000}\) (×100) or \(\dfrac{3841.43(752\ldots)}{3000}\) – 1 (×100) | M1 | FT ‘their 3841.43(752…)’ provided at least one M1 previously awarded |
| = 28(.04) to 28.05 (%) | A1 | An amount in the account after 10 years of (£)3840.25 leads to 28(.008) (%) If no marks awarded, SC1 for an answer of 27.7(58…) or 27.8% from use of \(\dfrac{3000 \times (1.0248)^{10} - 3000}{3000}\) × 100 or \(((1.0248)^{10} - 1)\) × 100 or equivalent |
Alternative
Alternative method 1:
| Scheme | Marks | Comments |
|---|---|---|
| (Quarterly rate =) \(\dfrac{2.48}{4}\) (%) or \(\dfrac{2.48 \div 100}{4}\) (= 0.62(%) or 0.0062) | M1 | May be implied in further working |
| (Amount in account after 10 years =) \(3000 \times \left(1 + \dfrac{2.48 \div 100}{4}\right)^{10\times 4}\) or \(3000 \times 1.0062^{40}\) | M1 | FT ‘their 0.0062’ provided it comes from 2.48(÷100) /\(n\) provided 1 < \(n\) \(\leqslant\) 12 Allow \(3000 \times \left(1 + \dfrac{2.48 \div 100}{n}\right)^{10\times n}\) provided their value of \(n\) has been used consistently and 1 < \(n\) \(\leqslant\) 12 |
| = (£)3841.43(752…) or (£)3841.44 | A1 | CAO. Must come from M1M1 |
| (Percentage increase =) \(\dfrac{3841.43(752\ldots) - 3000}{3000}\) (×100) or \(\dfrac{3841.43(752\ldots)}{3000}\) – 1 (×100) | M1 | FT ‘their 3841.43(752…)’ provided at least one M1 previously awarded |
| = 28(.04) to 28.05 (%) | A1 |
Alternative
Alternative method 2:
| Scheme | Marks | Comments |
|---|---|---|
| \(AER = \left(1 + \dfrac{2.48 \div 100}{4}\right)^4 - 1 =\) (= 0.025(03…) or 2.5(03…) (%)) | M1 | The −1 may be implied in further working |
| (Percentage increase =) \((1 + 0.025(03\ldots))^{10} - 1\) | M3 | FT ‘their 0.025(03…)’ provided it comes from \(\left(1 + \dfrac{2.48 \div 100}{n}\right)^{n} - 1\) where \(n\) has been used consistently and 1 < \(n\) \(\leqslant\) 12 M2 for \((1 + 0.025(03\ldots))^{10}\) |
| = 28(.04) to 28.05 (%) | A1 | CAO. Must come from M1M3 |
Alternative
Alternative method 3:
| Scheme | Marks | Comments |
|---|---|---|
| (Quarterly rate =) \(\dfrac{2.48}{4}\) (%) or \(\dfrac{2.48 \div 100}{4}\) (= 0.62(%) or 0.0062) | M1 | |
| (Percentage increase =) \((1 + 0.0062)^{10\times 4} - 1\) | M3 | FT ‘their 0.0062’ provided it comes from 2.48(÷100) /\(n\) provided 1 < \(n\) \(\leqslant\) 12 Allow \(\left(1 + \dfrac{2.48 \div 100}{n}\right)^{10\times n} - 1\) provided their value of \(n\) has been used consistently and 1 < \(n\) \(\leqslant\) 12 M2 for \((1 + 0.0062)^{10\times 4}\) |
| = 28(.04) to 28.05 (%) | A1 | CAO |