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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]
a.

Show that the total value of Phil’s savings after 20 years is ( 1.02 20 1 ) P ( 1.02 1 ) .

[3]
b.

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

[3]
c.

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

5000 1.028 + 5000 1.028 2 + + 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 × 1.035 20     (M1)(A1)

= $ 298468     A1

 

Note:     Only accept answers to the nearest dollar. Accept $298469.

 

[3 marks]

a.

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 + 1.02 P + + 1.02 19 P   ( = P ( 1 + 1.02 + + 1.02 19 ) )     A1

OR

explicitly identify u 1 = P ,   r = 1.02 and n = 20 (may be seen as S 20 ).     A1

THEN

s 20 = ( 1.02 20 1 ) P ( 1.02 1 )     AG

[3 marks]

b.

24.297 P = 298468     (M1)(A1)

P = 12284     A1

 

Note:     Accept answers which round to 12284.

 

[3 marks]

c.

METHOD 1

Q ( 1.028 n ) = 5000 ( 1 + 1.028 + 1.028 2 + 1.028 3 + + 1.028 n 1 )     M1A1

Q = 5000 ( 1 + 1.028 + 1.028 2 + 1.028 3 + . . . + 1.028 n 1 ) 1.028 n     A1

= 5000 1.028 + 5000 1.028 2 + + 5000 1.028 n     AG

 

METHOD 2

the initial value of the first withdrawal is 5000 1.028     A1

the initial value of the second withdrawal is 5000 1.028 2     R1

the investment required for these two withdrawals is 5000 1.028 + 5000 1.028 2     R1

Q = 5000 1.028 + 5000 1.028 2 + + 5000 1.028 n     AG

 

[3 Marks]

d.i.

sum to infinity is 5000 1.028 1 1 1.028     (M1)(A1)

= 178571.428

so minimum amount is $178572     A1

 

Note:     Accept answers which round to $178571 or $178572.

 

[3 Marks]

d.ii.

Examiners report

[N/A]
a.
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b.
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c.
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d.i.
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d.ii.

Syllabus sections

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

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