User interface language: English | Español

Date November 2016 Marks available 1 Reference code 16N.2.AHL.TZ0.H_12
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Write down Question number H_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 th month, immediately after the interest had been added.

Find an expression for A 1  and show that A 2 = 1.004 2 x + 1.004 x .

[2]
a.

(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 .

[6]
b.

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 .

[1]
c.

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.

[4]
d.

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.

[5]
e.

Markscheme

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

A 1 = 1.004 x    A1

A 2 = 1.004 ( 1.004 x + x )    A1

= 1.004 2 x + 1.004 x    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)     A 3 = 1.004 ( 1.004 2 x + 1.004 x + x ) = 1.004 3 x + 1.004 2 x + 1.004 x      (M1)A1

A 4 = 1.004 4 x + 1.004 3 x + 1.004 2 x + 1.004 x    A1

(ii)     A 120 = ( 1.004 120 + 1.004 119 + + 1.004 ) x      (A1)

= 1.004 120 1 1.004 1 × 1.004 x    M1A1

= 251 ( 1.004 120 1 ) x    AG

[6 marks]

b.

A 216 = 251 ( 1.004 216 1 ) x   ( = x t = 1 216 1.004 t )    A1

[1 mark]

c.

251 ( 1.004 216 1 ) x = 20 000 x = 58.22    (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 .

 

x = 59    A1

 

Note: Accept x = 58 . Accept x 59 .

 

[4 marks]

d.

r = 1.004 12   ( = 1.049 )    (M1)

15 000 r n 1000 r n 1 r 1 = 0 n = 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

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.

Syllabus sections

Topic 1—Number and algebra » SL 1.2—Arithmetic sequences and series
Show 111 related questions
Topic 1—Number and algebra

View options