Percentages: WJEC GCSE Numeracy Higher November 2022 Unit 2 Q6

Numeracy Higher November 2022 Unit 2 Q6

WJECHigherCurrent spec7 marksPercentages Substitution

6. Rebecca needs to save £10 000 for the deposit to buy a new house.
She has opened a savings account with Dragon Building Society. The account has a nominal annual rate of 5·4%, with interest paid on the last day of every month.

The formula for calculating the amount of money in the account at the end of every month is:

\[A = M\left(\dfrac{\big((1+r)^{n}-1\big)(1+r)}{r}\right)\]

where:

\(A\) is the amount of money in the account at the end of every month,
\(M\) is the amount deposited into the account on the 1st day of each month,
\(r\) is the monthly interest rate written as a decimal,
\(n\) is the number of months the account has been open.

Rebecca opened the account on 1st August 2022, depositing £335 into the account.
She will deposit £335 into the account on the 1st day of every month.

(a)

At the end of which month, and in which year, will Rebecca have the £10 000 she needs?
You must show all your working. [4]

Rebecca will have the £10 000 at the end of …………… in the year ……………

(b)

Calculate the AER for Rebecca’s savings account.
Give your answer as a percentage, correct to 2 decimal places. [2]

(c)

Rebecca’s friend, Seren, opened a similar savings account with Dragon Building Society on 1st June 2020, depositing £300.
Seren then deposited £300 into the account on the 1st day of every month.
By 30th November 2021, Seren had £5636.84 in the account.

How much interest had Seren received?
Circle your answer. [1]

£236.84 £636.84 £836.84 £3836.84 £4136.84

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.

Mark scheme (a)

Scheme	Marks	Comments
		For all the amounts of money shown below, accept other reasonable approximations e.g. nearest 10p, rounded or truncated Only allow a misread of the amount deposited, not of the nominal annual rate
\(335\left(\dfrac{\left(\left(1+\frac{5.4/100}{12}\right)^{n}-1\right)\left(1+\frac{5.4/100}{12}\right)}{\frac{5.4/100}{12}}\right)\) or \(335\left(\dfrac{((1+0.0045)^{n}-1)(1+0.0045)}{0.0045}\right)\)	B1	For any value of n
\(335\left(\dfrac{\left(\left(1+\frac{5.4/100}{12}\right)^{28}-1\right)\left(1+\frac{5.4/100}{12}\right)}{\frac{5.4}{12\times 100}}\right)\) or \(335\left(\dfrac{((1+0.0045)^{28}-1)(1+0.0045)}{0.0045}\right)\)	M1	OR \(335\left(\dfrac{\left(\left(1+\frac{5.4/100}{12}\right)^{n}-1\right)\left(1+\frac{5.4/100}{12}\right)}{\frac{5.4/100}{12}}\right)\) = 10000 (or ≥ 10000) May be implied by amounts of (£)9259.52(1991) for n=26 AND (£)9637.69(734) for n=27
= (£) 10 017(.57…) or (£) 10 018	A1	Or evidence that n = 28 An answer of 28 months with no incorrect work seen can be awarded B1M1A1 and possible final A1
(Date when Rebecca has £10 000) November 2024	A1	CAO If first B1 only awarded, SC1 for a correct evaluation of the formula for any value of n from 20 to 30 n=20 leads to (£)7025(.78…) or (£)7026 n=21 leads to (£)7393(.906…) or (£)7394 n=22 leads to (£)7763(.686..) or (£)7764 n=23 leads to (£)8135(.13…) n=24 leads to (£)8508(.246…) n=25 leads to (£)8883(.04…) n=26 leads to (£)9259(.52…) or (£)9260 n=27 leads to (£)9637(.697…) or (£)9638 n=29 leads to (£)10399(.16…) n=30 leads to (£)10782(.46…) If no marks awarded and from using a rate of ^5.4/₁₂ = 0.45, SC2 for (£)13467(.65…) or (£)13468 AND Feb 2023 from \(335\left(\dfrac{((1+0.45)^{7}-1)(1+0.45)}{0.45}\right)\), or SC1 for (£)13467(.65) or (£)13468 If no marks awarded and from using a rate of 0.045, SC2 for (£)10174(.426…) AND February 2024 from \(335\left(\dfrac{((1+0.045)^{19}-1)(1+0.045)}{0.045}\right)\), or SC1 for (£)10174(.426...) If no marks awarded and from using a rate of 0.054, SC2 for (£)10312(.178…) AND January 2024 from \(335\left(\dfrac{((1+0.054)^{18}-1)(1+0.054)}{0.054}\right)\), or SC1 for (£)10312(.178…)