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Date November 2016 Marks available 4 Reference code 16N.2.hl.TZ0.12
Level HL only Paper 2 Time zone TZ0
Command term Determine 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 $An be the amount in Mary’s account on the last day of the nth month, immediately after the interest had been added.

Find an expression for A1 and show that A2=1.0042x+1.004x.

[2]
a.

(i)     Write down a similar expression for A3 and A4.

(ii)     Hence show that the amount in Mary’s account the day before she turned 10 years old is given by 251(1.0041201)x.

[6]
b.

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

[1]
c.

Mary’s grandparents wished for the amount in her account to be at least $20000 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]
d.

As soon as Mary was 18 she decided to invest $15000 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]
e.

Markscheme

A1=1.004x    A1

A2=1.004(1.004x+x)    A1

=1.0042x+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]

a.

(i)     A3=1.004(1.0042x+1.004x+x)=1.0043x+1.0042x+1.004x     (M1)A1

A4=1.0044x+1.0043x+1.0042x+1.004x    A1

(ii)     A120=(1.004120+1.004119++1.004)x     (A1)

=1.00412011.0041×1.004x    M1A1

=251(1.0041201)x    AG

[6 marks]

b.

A216=251(1.0042161)x (=x216t=11.004t)    A1

[1 mark]

c.

251(1.0042161)x=20000x=58.22    (A1)(M1)(A1)

 

Note: Award (A1) for 251(1.0042161)x>20000, (M1) for attempting to solve and (A1) for x>58.22.

 

x=59    A1

 

Note: Accept x=58. Accept x59.

 

[4 marks]

d.

r=1.00412 (=1.049)    (M1)

15000rn1000rn1r1=0n=27.8    (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

 

n=28    A1

 

Note: Accept n=27.

 

[5 marks]

e.

Examiners report

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Syllabus sections

Topic 1 - Core: Algebra » 1.1 » Arithmetic sequences and series; sum of finite arithmetic series; geometric sequences and series; sum of finite and infinite geometric series.
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