Home / IB Mathematics SL 1.4 Financial applications of geometric sequences and series AI HL Paper 1- Exam Style Questions

IB Mathematics SL 1.4 Financial applications of geometric sequences and series AI HL Paper 1- Exam Style Questions

IB Mathematics SL 1.4 Financial applications of geometric sequences and series AI HL Paper 1- Exam Style Questions- New Syllabus

Question

Angel has \( \$520 \) in his savings account. Angel considers investing the money for 5 years with a bank. The bank offers an annual interest rate of 1.2% compounded quarterly.

a) Calculate the amount of money Angel would have at the end of 5 years with the bank. Give your answer correct to two decimal places.

b) Instead of investing the money, Angel decides to buy a phone that costs \( \$520 \). At the end of 5 years, the phone will have a value of \( \$30 \). It may be assumed that the depreciation rate per year is constant. Calculate the annual depreciation rate of the phone.

▶️ Answer/Explanation
Markscheme

a) To calculate the amount after 5 years:
Method 1 (Financial Calculator):
Use financial calculator inputs:
\( N = 20 \)
\( I\% = 1.2 \)
\( PV = \pm 520 \)
\( P/Y = 4 \)
\( C/Y = 4 \)

Compute:
\( S = 552.11 \)

Method 2 (Formula):
Use the compound interest formula:
\( A = P \times \left(1 + \frac{r}{n}\right)^{n \times t} \)
where \( P = 520 \), \( r = 0.012 \), \( n = 4 \), \( t = 5 \)

Substitute:
\( A = 520 \times \left(1 + \frac{0.012}{4}\right)^{4 \times 5} \)
\( A = 520 \times (1.003)^{20} \)

Calculate:
\( 1.003^{20} \approx 1.061834 \)
\( A = 520 \times 1.061834 \approx 552.1537 \)

Round to two decimal places:
\( A \approx 552.11 \)

Thus:
The amount is \( \$552.11 \) [3]

b) To calculate the annual depreciation rate:
Method 1 (Financial Calculator):
Use financial calculator inputs:
\( N = 5 \)
\( PV = \pm 520 \)
\( FV = \mp 30 \)

Compute:
\( I\% = 43.5 \)

Method 2 (Formula):
Use the compound depreciation formula:
\( A = P \times \left(1 – \frac{r}{100}\right)^n \)
where \( P = 520 \), \( A = 30 \), \( n = 5 \)

Substitute:
\( 30 = 520 \times \left(1 – \frac{r}{100}\right)^5 \)

Simplify:
\( \left(1 – \frac{r}{100}\right)^5 = \frac{30}{520} \approx 0.057692 \)

Take the 5th root:
\( 1 – \frac{r}{100} = (0.057692)^{1/5} \approx 0.565 \)

Solve:
\( \frac{r}{100} \approx 0.435 \)
\( r \approx 43.5 \)

Thus:
The annual depreciation rate is \( 43.5\% \) [3]

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