IB Mathematics SL 1.7 Amortization and annuities AI HL Paper 2 - Exam Style Questions- New Syllabus
Claire intends to purchase a home valued at 285,000 US Dollars (USD). She visits a financial institution to secure a loan for the purchase. To qualify for the loan, Claire must provide an initial deposit equivalent to 15% of the home’s value.
The financial institution offers a 30-year loan for the outstanding amount, featuring a 4% annual interest rate, compounded monthly. Claire will settle the loan with consistent payments due at the end of each month.
(a) (i) Determine the initial loan amount after the deposit is made. Provide the precise answer.
(ii) Calculate Claire’s monthly payment for this loan, rounded to two decimal places. [5]
(b) Using your result from part (a)(ii), compute the total amount Claire will pay over the loan term, rounded to the nearest dollar. Exclude the initial deposit. [2]
Claire wishes to clear the loan more quickly and opts to increase her payments to 1300 USD per month.
(c) Ascertain the total number of monthly payments required to fully repay the loan. [2]
This approach will lead to Claire’s last payment being less than 1300 USD.
(d) Calculate the amount of Claire’s final payment, rounded to two decimal places. [4]
(e) Thus, determine the total savings Claire will achieve, rounded to the nearest dollar, by opting for the higher monthly payments. [3]
▶️ Answer/Explanation
(a)
(i) Calculating 15% of the house price to determine the down payment, then subtracting it from the total price to find the loan amount.
\[ 285000 \times 0.15 = 42750 \; (\text{USD}) \]
\[ 285000 – 42750 = 242250 \; (\text{USD}) \]
242250 (USD)
(ii) Using the loan amortization formula with \( N = 360 \) months, \( I\% = \frac{4}{12} = 0.333\ldots\% \) per month, \( PV = 242250 \), \( FV = 0 \), \( P/Y = 12 \), \( C/Y = 12 \).
\[ PMT = \frac{PV \cdot \frac{I\%}{100}}{\left(1 – \left(1 + \frac{I\%}{100}\right)^{-N}\right)} \]
Substituting values:
\[ PMT = \frac{242250 \cdot \frac{0.333\ldots}{100}}{\left(1 – \left(1 + \frac{0.333\ldots}{100}\right)^{-360}\right)} \]
Using a financial calculator or approximation, \( PMT = 1156.54 \; (\text{USD}) \)
[5 marks]
(b)
Multiplying the monthly payment by the total number of payments over 30 years (360 months).
\[ 1156.54 \times 360 = 416354.4 \]
Rounding to the nearest dollar, 416354 (USD)
[2 marks]
(c)
Applying the loan repayment formula with \( I\% = \frac{4}{12} = 0.333\ldots\% \) per month, \( PV = 242250 \), \( PMT = 1300 \), \( FV = 0 \), \( P/Y = 12 \), \( C/Y = 12 \).
Using the formula to solve for \( N \):
\[ N = -\frac{\log\left(1 – \frac{PV \cdot \frac{I\%}{100}}{PMT}\right)}{\log\left(1 + \frac{I\%}{100}\right)} \]
\[ N \approx 292 \]
[2 marks]
(d)
Method 1
Calculating the remaining balance after 291 payments with \( N = 291 \), \( I\% = 0.333\ldots\% \), \( PV = 242250 \), \( PMT = 1300 \), \( P/Y = 12 \), \( C/Y = 12 \).
\[ FV = 871.91 \; (871.908\ldots) \]
Determining the interest for the final month and subtracting from the next payment:
\[ N = 1, \; I\% = 0.333\ldots\%, \; PV = 871.91, \; FV = 0, \; P/Y = 12, \; C/Y = 12 \]
\[ PMT = 874.82 \; (\text{USD}) \]
[4 marks]
Method 2
Calculating the remaining balance after 292 payments with \( N = 292 \), \( I\% = 0.333\ldots\% \), \( PV = 242250 \), \( PMT = 1300 \), \( P/Y = 12 \), \( C/Y = 12 \).
\[ FV = 425.185\ldots \]
\[ 1300 – 425.185\ldots = 874.814\ldots \]
Rounding to two decimal places, \( PMT = 874.82 \; (\text{USD}) \)
[4 marks]
(e)
Calculating the total paid with higher payments: 291 months at 1300 USD plus the final payment.
\[ 291 \times 1300 + 874.82 = 379174.82 \; (\text{USD}) \]
Finding the difference between the original total and the new total:
\[ 416354 – 379174.82 = 37179.18 \]
Rounding to the nearest dollar, 37179 (USD)
[3 marks]
[Total: 16 marks]