| Date | November 2016 | Marks available | 6 | Reference code | 16N.2.hl.TZ0.12 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Hence and Write down | Question number | 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 \({A_2} = {1.004^2}x + 1.004x\).
(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 \(251({1.004^{120}} - 1)x\).
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 \(n\).
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.
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.
Markscheme
\({A_1} = 1.004x\) 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]
