Numeracy Higher November 2016 Unit 2 Q10
10. Huw wants to open a savings account.
Here are the details of savings accounts advertised by two local Welsh banks.
| Banc Padarn Nominal interest rate of 1·98% per annum Interest paid monthly | Banc Teilo AER 1·99% |
Circle your answer. [1]
0·0198 0·198 1·098 1·98 98·0
You must show your working. [4]
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.
| Tax rates for savings | |
|---|---|
| Basic rate taxpayer | 20% on annual interest earned above £1000 |
| Higher rate taxpayer | 40% on annual interest earned above £500 |
Matthew is a higher rate taxpayer.
Therefore, any savings interest he earns over £500 within a year is taxed at 40%.
On 1st May 2016, he invested £150 000 in a savings account that pays interest at a rate of 1·98% per annum.
Circle your answer. [1]
0·0033 0·00495 0·00165 0·0099 0·0066
Savings interest is added at the end of every month.
Date ……………………… Tax Matthew would have to pay ………………………
| Scheme | Marks | Comments |
|---|---|---|
| (i) 0·0198 | B1 | |
| (ii) Banc Padarn For use of n = 12 (AER=) (1 + 0·0198/12)12 – 1 = 0·01998(0…) OR 1·998(0…) % | B1 M1 A1 | FT 'their 0·0198' for M1 only Allow 0·02(00..) OR 2(·00) % Do not accept 0·01998 % or 0·02(00..) % |
| Banc Padarn’s savings account would give most interest per annum. | E1 | FT ‘their AER for Banc Padarn’ provided at least M1 awarded AND they are comparing like percentages or decimals Alternative method: B1 for monthly interest rate of 0·00165 M1 for the method of calculating the value of an investment for Banc Padarn after a year (initial investment × 1·0016512) and Banc Teilo (initial investment × 1·0199) A1 for accurate values for both banks (with appropriate rounding) E1 for Banc Padarn chosen |
| Scheme | Marks | Comments |
|---|---|---|
| (i) 0·00165 | B1 | |
| (ii) (31st May OR 1st June) (Interest =) (0·00165 × 150 000) (£)247.5(0) | B1 | |
| (30th June OR 1st July) (Interest =) (0·00165 × 150 247.5(0)) (£)247.90(84) OR Total interest of (£)495.40(84) | B1 | FT ‘their 247.5(0)’ |
| (31st July OR 1st August) (Interest =) (0·00165 × 150 495.40(84)) (£)248.31(74) OR Total interest of (£)743.72(58) | B1 | FT ‘their 247.90(84)’ Alternative method: B1 for (1·00165×150 000) (£)150247.5(0) B1 for (1·001652×150 000) (£)150495.40(84) This B1 implies the 1st B1 B1 for (1·001653×150 000) (£)150743.72(58) The last B1 implies the previous two B1 marks |
| (Tax to pay =) (‘their 743.72(58)’ – 500) × 0·4 | M1 | FT ‘their 743.72(58)’ provided compound interest attempted |
| 31st July OR 1st August AND (£)97.49(032) | A1 | Allow ‘End of July’ OR ‘30th July’ for the date Allow (£)97.48 from rounding down of the monthly interest payments |