IBDP Maths AA: Topic : SL 1.4: Financial applications of geometric sequences and series: IB style Questions SL Paper 2

Question:

In this question, give all answers correct to two decimal places.
Sam invests \($1700 \) in a savings account that pays a nominal annual rate of interest of 2.74%, compounded half-yearly. Sam makes no further payments to, or withdrawals from, this account.
(a) Find the amount that Sam will have in his account after 10 years.
David also invests $1700 in a savings account that pays an annual rate of interest of r%,  compounded yearly. David makes no further payments or withdrawals from this account.
(b) Find the value of r required so that the amount in David’s account after 10 years will be equal to the amount in Sam’s account.
(c) Find the interest David will earn over the 10 years

Answer/Explanation

Ans:

Note: The first time an answer is not given to two decimal places, the final A1 in that part is not awarded.
(a) EITHER

Note: Award (M1) for an attempt to use a financial app in their technology with at least two entries seen, and award (A1) for all entries correct. Accept a positive or negative value for PV.

Note: Award (M1) for substitution into compound interest formula. Award (A1) for correct substitution.

THEN
$2231.71

(b) EITHER
N =10
PV = ± 1700
FV = ± 2231.71…
P/Y = 1
C/Y = 1

Note: Award (M1) for an attempt to use a financial app in their technology with at least two entries seen.

OR

Note: Ignore omission of opposite signs for PV and FV if r = 2.76 is obtained.
(c) $531.71

Question:

Gemma and Kaia started working for different companies on January 1st 2011.
Gemma’s starting annual salary was $45000, and her annual salary increases 2% on January 1st each year after 2011.
(a) Find Gemma’s annual salary for the year 2021, to the nearest dollar.
Kaia’s annual salary is based on a yearly performance review. Her salary for the years 2011, 2013, 2014, 2018, and 2022 is shown in the following table.

year (x)20112013201420182022
annual salary ($S )4500047200485005300057000

(b) Assuming Kaia’s annual salary can be approximately modelled by the equation S = ax + b, show that Kaia had a higher salary than Gemma in the year 2021, according to the model.

Answer/Explanation

Ans:

(a) METHOD 1
using geometric sequence with r =1.02
correct expression or listing terms correctly
45000 × 1.0210 OR  45000 × 1.0211-1 OR listing terms
Gemma’s salary is $54855 (must be to the nearest dollar)

METHOD 2
N =10
PV = ± 45000 
I% = 2
P/Y = 1
C/Y = 1 
FV = ±54854.7489…
Gemma’s salary is $54855 (must be to the nearest dollar)
(b) finds a =1096.89… and b = −2160753.8… (accept 6 b = − 2.16 × 106 )
Note: Award (A1)(A1) for S = 1096.89…x + 33028.49… , or S = 1096.89…x + 43997.4… , or S = 1096.89…x + 45094.3…
Kaia’s salary in 2021 is $56063.21 (accept $56817.09 from b = -2.16 × 106)
Kaia had a higher salary than Gemma in 2021

Question

Two friends Falicia and Dominic, each set themselves a target of saving $20 000. They each have $9000 to invest.

    1. Falicia invests her $9000 in an account that offers an interest rate of 7 % per annum compounded annually.

      1. Find the value of Falicia’s investment after 5 years to the nearest hundred dollars.

      2. Determine the number of years required for Falicia’s investment to reach the target. [5]

    2. Dominic invests his $9000 in an account that offers an interest rate of r % per annum compounded monthly, where r is set to two decimal places.

      Find the minimum value of r needed for Dominic to reach the target after 10 years. [3]

    3. A third friend Aayush also wants to reach the $20 000 target. He puts his money in a safe where he does not earn any interest. His system is to add more money to this safe each year. Each year he will add half the amount added in the previous year.

      1. Show that Aayush will never reach the target if his initial deposit is $9000.

      2. Find the amount Aayush needs to deposit initially in order to reach the target after 5 years. Give your answer to the nearest dollar. [8]

Answer/Explanation

Ans: 

(a)

(i)

EITHER

\(9000\times (1+\frac{7}{100})^{5}\)

12622.965..

OR

n=5

I%=7

\(PV = \pm 9000\)

P/Y =1

C/Y = 1

\(\pm 12622.965..\)

THEN 

($) 12600

(ii)

EITHER 

\(9000(1+ \frac{7}{100})^{x}\)= 20000

OR

I%=7

\(PV = \mp 9000\)

\(FV = \pm 20000\)

P/Y = 1

C/Y = 1

THEN 

= 12 (years)

(b)

METHOD 1

attempt to substitute into compound interest formula (condone absence of compounding periods)

\(9000(1+\frac{r}{100\times 12})^{12\times 10}= 20000\)

8.01170..

r = 8.02(%)

METHOD2

n=10

PV = \(\pm 9000\)

\(FV = \pm 20000\)

P/Y = 1

C/Y = 12

r = 8.01170..

r = 8.02 %

(c)

(i)

recognising geometric series (seen anywhere)

\(r= \frac{4500}{900}=\frac{1}{2}\)

EITHER

considering \(S_\infty \)

\(\frac{9000}{1-0.5}=18000\)

correct reasoning that 18000< 20000

  OR

considering \(S_\infty \) for a large value of \(n,n\geq 80\)

correct value of  \(S_\infty \) for their \(n\)
valid reason why Aayush will not reach the target, which involves their choice of \(n\), their value of \(S_\infty \)  and Aayush’ age OR using two large
values of \(n\) to recognize asymptotic behaviour of  \(S_\infty \)  as \(n \rightarrow \infty \)

THEN

Therefore, Aayush will never reach the target.

(ii)

recognising geometric sum 

\(\frac{u_{1(1-0.5^{5})}}{0.5}\)= 20000

10322.58..

($) 10323

Question

Give your answers in this question correct to the nearest whole number.

Shahid invested 25 000 Singapore dollars (SGD) in a fixed deposit account with a nominal

annual interest rate of 3.6 %, compounded monthly.

    1. Calculate the value of Shahid’s investment after 5 years. [3]

      At the end of the 5 years, Shahid withdrew x SGD from the fixed deposit account and

      reinvested this into a super-savings account with a nominal annual interest rate of

      5.7 %, compounded half-yearly.

      The value of the super-savings account increased to 20 000 SGD after 18 months.

    2. Find the value of x . [3]

Answer/Explanation

Ans:

(a)

\(FV=25000\times (1+ \frac{3.5}{100\times 12})^{12\times 5}\)

OR

N= 5

I%=3.6

PV= \(\mp 25000\)

\(P/Y\) = 1

C/Y = 12

OR

N= 60

I%=3.6

PV = \(\mp 25000\)

P/V = 12

C/Y = 12

FV=29922(SGD)

(b)

2000= PV\times \((1+\frac{5.7}{100\times 2})^{2\times 1.5}\)

OR

N= 1.5

I%=5.7

FV=\(\pm 20000\)

\(P/Y\) = 1

C/Y= 2

OR

N=3

I%= 5.7

FV = \(\pm 20000\)

P/V=2

C/Y = 2

X= 18383 (SGD)

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