| Date | November 2017 | Marks available | 3 | Reference code | 17N.2.hl.TZ0.12 |
| Level | HL only | Paper | 2 | Time zone | TZ0 |
| Command term | Show that | Question number | 12 | Adapted from | N/A |
Question
Phil takes out a bank loan of $150 000 to buy a house, at an annual interest rate of 3.5%. The interest is calculated at the end of each year and added to the amount outstanding.
To pay off the loan, Phil makes annual deposits of $P at the end of every year in a savings account, paying an annual interest rate of 2% . He makes his first deposit at the end of the first year after taking out the loan.
David visits a different bank and makes a single deposit of $Q , the annual interest rate being 2.8%.
Find the amount Phil would owe the bank after 20 years. Give your answer to the nearest dollar.
Show that the total value of Phil’s savings after 20 years is \(\frac{{({{1.02}^{20}} - 1)P}}{{(1.02 - 1)}}\).
Given that Phil’s aim is to own the house after 20 years, find the value for \(P\) to the nearest dollar.
David wishes to withdraw $5000 at the end of each year for a period of \(n\) years. Show that an expression for the minimum value of \(Q\) is
\(\frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}} + \ldots + \frac{{5000}}{{{{1.028}^n}}}\).
Hence or otherwise, find the minimum value of \(Q\) that would permit David to withdraw annual amounts of $5000 indefinitely. Give your answer to the nearest dollar.
Markscheme
\(150000 \times {1.035^{20}}\) (M1)(A1)
\( = \$ 298468\) A1
Note: Only accept answers to the nearest dollar. Accept $298469.
[3 marks]
attempt to look for a pattern by considering 1 year, 2 years etc (M1)
recognising a geometric series with first term \(P\) and common ratio 1.02 (M1)
EITHER
\(P + 1.02P + \ldots + {1.02^{19}}P{\text{ }}\left( { = P(1 + 1.02 + \ldots + {{1.02}^{19}})} \right)\) A1
OR
explicitly identify \({u_1} = P,{\text{ }}r = 1.02\) and \(n = 20\) (may be seen as \({S_{20}}\)). A1
THEN
\({s_{20}} = \frac{{({{1.02}^{20}} - 1)P}}{{(1.02 - 1)}}\) AG
[3 marks]
\(24.297 \ldots P = 298468\) (M1)(A1)
\(P = 12284\) A1
Note: Accept answers which round to 12284.
[3 marks]
METHOD 1
\(Q({1.028^n}) = 5000(1 + 1.028 + {1.028^2} + {1.028^3} + \ldots + {1.028^{n - 1}})\) M1A1
\(Q = \frac{{5000\left( {1 + 1.028 + {{1.028}^2} + {{1.028}^3} + ... + {{1.028}^{n - 1}}} \right)}}{{{{1.028}^n}}}\) A1
\( = \frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}} + \ldots + \frac{{5000}}{{{{1.028}^n}}}\) AG
METHOD 2
the initial value of the first withdrawal is \(\frac{{5000}}{{1.028}}\) A1
the initial value of the second withdrawal is \(\frac{{5000}}{{{{1.028}^2}}}\) R1
the investment required for these two withdrawals is \(\frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}}\) R1
\(Q = \frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}} + \ldots + \frac{{5000}}{{{{1.028}^n}}}\) AG
[3 Marks]
sum to infinity is \(\frac{{\frac{{5000}}{{1.028}}}}{{1 - \frac{1}{{1.028}}}}\) (M1)(A1)
\( = 178571.428 \ldots \)
so minimum amount is $178572 A1
Note: Accept answers which round to $178571 or $178572.
[3 Marks]
