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

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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