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Date	November 2017	Marks available	3	Reference code	17N.2.AHL.TZ0.H_12
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 0
Command term	Find	Question number	H_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.

[3]

Show that the total value of Phil’s savings after 20 years is $\frac{{({{1.02}^{20}} - 1)P}}{{(1.02 - 1)}}$ .

[3]

Given that Phil’s aim is to own the house after 20 years, find the value for $P$ to the nearest dollar.

[3]

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

$\frac{{5000}}{{1.028}} + \frac{{5000}}{{{{1.028}^2}}} + \ldots + \frac{{5000}}{{{{1.028}^n}}}$ .

[3]

d.i.

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.

[3]

d.ii.

Markscheme

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

$150000 \times {1.035^{20}}$ (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

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]

d.i.

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]

d.ii.

Examiners report

[N/A]

d.i.

[N/A]

d.ii.

Syllabus sections

Topic 1—Number and algebra » SL 1.4—Financial apps – compound interest, annual depreciation

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