IB Mathematics SL 1.4 Financial applications of geometric sequences and series AI SL Paper 1- Exam Style Questions- New Syllabus
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]
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]
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 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
Question
Thomas and Peter have each been given 1500 euro to save for college.
Peter invests his money in an account that pays a nominal annual interest rate of 2.75%, compounded half-yearly.
(a) Determine the amount Peter will have in his account after 5 years. Give your answer correct to 2 decimal places. [3]
Thomas wants to invest his money in an account such that his investment will increase to 1.5 times the initial amount in 5 years. Assume the account pays a nominal annual interest of r% compounded quarterly.
(b) Determine the value of r%. [3]
▶️ Answer/Explanation
Markscheme
(a)
Method 1 (Financial Calculator):
Set: \( N = 10 \), \( I\% = \frac{2.75}{2} = 1.375 \), \( PV = -1500 \), \( PMT = 0 \), \( P/Y = 1 \), \( C/Y = 2 \).
Compute future value: \( FV \approx 1719.49 \) euro.
Note: \( PV \) and \( FV \) must have opposite signs.
Method 2 (Formula):
Use the compound interest formula \( A = P \left(1 + \frac{r}{n}\right)^{nt} \),
where P = 1500 euro, r = 2.75% = 0.0275, n = 2 (half-yearly), t = 5 years.
Number of compounding periods = \( n \times t = 2 \times 5 = 10 \).
Interest rate per period = \( \frac{r}{n} = \frac{0.0275}{2} = 0.01375 \).
Substitute: \(A = 1500 \left(1 + \frac{2.75}{2 \times 100}\right)^{2 \times 5} = 1500 \left(1.01375\right)^{10} \).
Compute: \( 1.01375^{10} \approx 1.146326 \), so \( A = 1500 \times 1.146326 \approx 1719.49 \) euro. M1 A1 A1
[3 marks]
Method 1 (Financial Calculator):
Set: \( N = 10 \), \( I\% = \frac{2.75}{2} = 1.375 \), \( PV = -1500 \), \( PMT = 0 \), \( P/Y = 1 \), \( C/Y = 2 \).
Compute future value: \( FV \approx 1719.49 \) euro.
Note: \( PV \) and \( FV \) must have opposite signs.
Method 2 (Formula):
Use the compound interest formula \( A = P \left(1 + \frac{r}{n}\right)^{nt} \),
where P = 1500 euro, r = 2.75% = 0.0275, n = 2 (half-yearly), t = 5 years.
Number of compounding periods = \( n \times t = 2 \times 5 = 10 \).
Interest rate per period = \( \frac{r}{n} = \frac{0.0275}{2} = 0.01375 \).
Substitute: \(A = 1500 \left(1 + \frac{2.75}{2 \times 100}\right)^{2 \times 5} = 1500 \left(1.01375\right)^{10} \).
Compute: \( 1.01375^{10} \approx 1.146326 \), so \( A = 1500 \times 1.146326 \approx 1719.49 \) euro. M1 A1 A1
[3 marks]
(b)
Method 1 (Financial Calculator):
Set: \( N = 20 \), \( PV = \pm 1500 \), \( FV = \mp 2250 \), \( PMT = 0 \), \( P/Y = 1 \), \( C/Y = 4 \).
Compute interest rate: \( I\% \approx 8.19206 \), rounded to 8.19%.
Note: \( PV \) and \( FV \) must have opposite signs.
Method 2 (Formula):
Use \( A = P \left(1 + \frac{r}{n}\right)^{nt} \), where P = 1500, A = \( 1.5 \times 1500 = 2250 \), n = 4 (quarterly), t = 5, and r is the annual rate as a decimal.
Number of periods = \( 4 \times 5 = 20 \).
Set up: \( 1500 \left(1 + \frac{r}{4 \times 100}\right)^{4 \times 5} = 2250 \).
Simplify: \( \left(1 + \frac{r}{4 \times 100}\right)^{20} = \frac{2250}{1500} = 1.5 \).
Solve: \( 1 + \frac{r}{400} = 1.5^{1/20} \).
Compute: \( 1.5^{1/20} \approx 1.020480 \), so \( \frac{r}{400} = 0.020480 \),
\( r = 0.020480 \times 400 \approx 8.19206 \), rounded to 8.19%. M1 A1 A1
[3 marks]
Method 1 (Financial Calculator):
Set: \( N = 20 \), \( PV = \pm 1500 \), \( FV = \mp 2250 \), \( PMT = 0 \), \( P/Y = 1 \), \( C/Y = 4 \).
Compute interest rate: \( I\% \approx 8.19206 \), rounded to 8.19%.
Note: \( PV \) and \( FV \) must have opposite signs.
Method 2 (Formula):
Use \( A = P \left(1 + \frac{r}{n}\right)^{nt} \), where P = 1500, A = \( 1.5 \times 1500 = 2250 \), n = 4 (quarterly), t = 5, and r is the annual rate as a decimal.
Number of periods = \( 4 \times 5 = 20 \).
Set up: \( 1500 \left(1 + \frac{r}{4 \times 100}\right)^{4 \times 5} = 2250 \).
Simplify: \( \left(1 + \frac{r}{4 \times 100}\right)^{20} = \frac{2250}{1500} = 1.5 \).
Solve: \( 1 + \frac{r}{400} = 1.5^{1/20} \).
Compute: \( 1.5^{1/20} \approx 1.020480 \), so \( \frac{r}{400} = 0.020480 \),
\( r = 0.020480 \times 400 \approx 8.19206 \), rounded to 8.19%. M1 A1 A1
[3 marks]
Total Marks: 6