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Date	November 2016	Marks available	5	Reference code	16N.2.AHL.TZ0.H_12
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Determine	Question number	H_12	Adapted from	N/A

Question

On the day of her birth, 1st January 1998, Mary’s grandparents invested $\$ x$ in a savings account. They continued to deposit $\$ x$ on the first day of each month thereafter.

The account paid a fixed rate of 0.4% interest per month. The interest was calculated on the last day of each month and added to the account.

Let $\$ {A_n}$ be the amount in Mary’s account on the last day of the $n{\text{th}}$ month, immediately after the interest had been added.

Find an expression for ${A_1}$ and show that ${A_2} = {1.004^2}x + 1.004x$ .

[2]

(i) Write down a similar expression for ${A_3}$ and ${A_4}$ .

(ii) Hence show that the amount in Mary’s account the day before she turned 10 years old is given by $251({1.004^{120}} - 1)x$ .

[6]

Write down an expression for ${A_n}$ in terms of $x$ on the day before Mary turned 18 years old showing clearly the value of $n$ .

[1]

Mary’s grandparents wished for the amount in her account to be at least $\$ 20\,000$ the day before she was 18. Determine the minimum value of the monthly deposit $\$ x$ required to achieve this. Give your answer correct to the nearest dollar.

[4]

As soon as Mary was 18 she decided to invest $\$ 15\,000$ of this money in an account of the same type earning 0.4% interest per month. She withdraws $\$ 1000$ every year on her birthday to buy herself a present. Determine how long it will take until there is no money in the account.

[5]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

${A_1} = 1.004x$ A1

${A_2} = 1.004(1.004x + x)$ A1

$= {1.004^2}x + 1.004x$ AG

Note: Accept an argument in words for example, first deposit has been in for two months and second deposit has been in for one month.

[2 marks]

(i) ${A_3} = 1.004({1.004^2}x + 1.004x + x) = {1.004^3}x + {1.004^2}x + 1.004x$ (M1)A1

${A_4} = {1.004^4}x + {1.004^3}x + {1.004^2}x + 1.004x$ A1

(ii) ${A_{120}} = ({1.004^{120}} + {1.004^{119}} + \ldots + 1.004)x$ (A1)

$= \frac{{{{1.004}^{120}} - 1}}{{1.004 - 1}} \times 1.004x$ M1A1

$= 251({1.004^{120}} - 1)x$ AG

[6 marks]

${A_{216}} = 251({1.004^{216}} - 1)x{\text{ }}\left( { = x\sum\limits_{t = 1}^{216} {{{1.004}^t}} } \right)$ A1

[1 mark]

$251({1.004^{216}} - 1)x = 20\,000 \Rightarrow x = 58.22 \ldots$ (A1)(M1)(A1)

Note: Award (A1) for $251({1.004^{216}} - 1)x > 20\,000$ , (M1) for attempting to solve and (A1) for $x > 58.22 \ldots$ .

$x = 59$ A1

Note: Accept $x = 58$ . Accept $x \geqslant 59$ .

[4 marks]

$r = {1.004^{12}}{\text{ }}( = 1.049 \ldots )$ (M1)

$15\,000{r^n} - 1000\frac{{{r^n} - 1}}{{r - 1}} = 0 \Rightarrow n = 27.8 \ldots$ (A1)(M1)(A1)

Note: Award (A1) for the equation (with their value of $r$ ), (M1) for attempting to solve for $n$ and (A1) for $n = 27.8 \ldots$

$n = 28$ A1

Note: Accept $n = 27$ .

[5 marks]

Examiners report

[N/A]

Syllabus sections

Topic 1—Number and algebra » SL 1.2—Arithmetic sequences and series

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21N.2.SL.TZ0.2e:
Find $k$ .

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