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

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

IB DP Maths AI SL Paper 1 - All Topics

Question

Sophie invests $3000 in a bank that pays a nominal annual interest rate of 1.25 % compounded monthly.
(a) Determine the amount of money Sophie will have in the bank at the end of 6 years. Give your answer correct to two decimal places. [3]
(b) Determine the number of months it takes until Sophie has at least $3550 in the bank. [2]
Sophie uses the \($\)3550 as a partial payment for a used car costing $22 000. For the remainder they take out a loan from a bank.
(c) Write down the amount of money that Sophie takes out as a loan. [1]
The loan is for 8 years and the nominal annual interest rate is 12.6 % compounded monthly. Sophie will pay the loan in fixed monthly instalments at the end of each month.
(d) Determine the amount, correct to the nearest dollar, that Sophie will have to pay the bank each month. [3]
▶️ Answer/Explanation
Markscheme

Given/Setup

Nominal annual rate \(i_{\text{nom}}=1.25\%\Rightarrow\) monthly rate \(r=\dfrac{0.0125}{12}\).
Compounding monthly. Use \(A=P(1+r)^n\). For annuity repayments use \( \text{PMT}=\dfrac{Pi}{1-(1+i)^{-N}} \).

(a) Amount after 6 years, compounded monthly

\(P=3000,\; n=6\times 12=72,\; r=\dfrac{0.0125}{12}\).
\(A=3000\left(1+\dfrac{0.0125}{12}\right)^{72}=3000(1.001041\ldots)^{72}\).
\(A= \boxed{\$\,3233.53}\) (to 2 d.p.). M1 A1 A1

(b) Months to reach at least \$3550

Solve \(3000(1+r)^{n}\ge 3550\).
\(n=\dfrac{\ln(3550/3000)}{\ln(1+0.0125/12)} = 161.686\ldots\ \text{months}\).
Minimum whole months: \(\boxed{162}\). A1 A1

(c) Loan principal

\(22\,000-3550=\boxed{\$\,18\,450}\). A1

(d) Monthly instalment over 8 years at 12.6 % (compounded monthly)

Monthly rate \(i=\dfrac{0.126}{12}=0.0105\), number of payments \(N=8\times 12=96\), present value \(P=18450\).
\(\displaystyle \text{PMT}=\frac{18450(0.0105)}{1-(1+0.0105)^{-96}}= \$\,305.98\ldots\).
Nearest dollar: \(\boxed{\$\,306}\). M1 A1 A1
Total Marks: 9
Question
In this question, give all answers correct to 2 decimal places.
Raul and Rosy want to buy a new house and need a loan of 170,000 Australian dollars (AUD) from a bank. The loan is for 30 years with an annual interest rate of 3.8%, compounded monthly. They will pay the loan in fixed monthly instalments at the end of each month.
(a) Determine the monthly payment to the bank. [3]
(b)(i) Determine the amount Raul and Rosy will still owe the bank at the end of the first 10 years. [3]
(b)(ii) Using your answers to parts (a) and (b)(i), calculate the total interest paid during the first 10 years. [3]
▶️ Answer/Explanation
Markscheme
(a)
Use annuity formula:
\[ \begin{aligned} PMT &= \frac{PV \times r}{1 – (1 + r)^{-N}} \\ &\text{where } PV = 170,000, \, r = \frac{3.8}{100 \times 12} \approx 0.00316667, \, N = 30 \times 12 = 360 \end{aligned} \] M1
Substitute:
\[ \begin{aligned} PMT &= \frac{170,000 \times 0.00316667}{1 – (1.00316667)^{-360}} \end{aligned} \] A1
Compute: \( PMT \approx 792.13 \) AUD. A1
Monthly payment: 792.13 AUD.
Note: Accept -792.13. Answer must be to 2 decimal places.
[3 marks]
(b)(i)
Use future value formula:
\[ \begin{aligned} FV &= PV \times (1 + r)^N – PMT \times \frac{(1 + r)^N – 1}{r} \\ &\text{where } PV = 170,000, \, r = 0.00316667, \, PMT = -792.13, \, N = 10 \times 12 = 120 \end{aligned} \] M1
Substitute:
\[ \begin{aligned} FV &= 170,000 \times (1.00316667)^{120} \\ &\quad – 792.13 \times \frac{(1.00316667)^{120} – 1}{0.00316667} \end{aligned} \] A1
Compute: \( FV \approx 133,019.94 \) AUD. A1
Remaining balance: 133,019.94 AUD.
Note: Accept 133,020.30 from exact \( PMT \). Answer must be to 2 decimal places.
[3 marks]
(b)(ii)
Total paid: \( 120 \times 792.13 = 95,055.60 \) AUD. M1
Principal repaid: \( 170,000 – 133,019.94 = 36,980.06 \) AUD. M1
Interest paid: \( 95,055.60 – 36,980.06 = 58,075.54 \) AUD. A1
Note: Accept 58,075.60 or 58,075.90 from exact values. Follow-through from parts (a) and (b)(i).
[3 marks]
Total Marks: 9
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