Year 12 Maths Advanced Module 7 ~35 min Checkpoint 3 of 4

Checkpoint 3, Loans and Recurrence

Lessons 11–15 cover recurrence relations, superannuation, loan mathematics, repayment calculations, and extra repayment strategies. This checkpoint tests your ability to model financial situations iteratively, calculate repayments, and compare debt-reduction strategies. Aim for 80%+ before moving to Inquiry Question 4.

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Comprehensive Answers

Multiple Choice Explanations

Q1. The investment recurrence is $A_{n+1} = (1+r)A_n + a$. Answer: A

Q2. $A_1 = 1.005(5{,}000) + 200 = 5{,}225$. Answer: C

Q3. Net return = gross return - fees = 7% - 1.2% = 5.8%. Answer: B

Q4. The loan recurrence is $A_{n+1} = (1+r)A_n - M$. Answer: D

Q5. $M = 400{,}000 x 0.004167 / [1 - (1.004167)^{-360}] = \\$2{,}147.29$. Answer: B

Q6. $P = 2{,}000 x [1 - (1.004)^{-240}] / 0.004 = \\$303{,}000$ (approx). Answer: C

Q7. Offset reduces the balance on which interest is calculated. Answer: A

Q8. $n = -\ln(1 - 250{,}000 x 0.005/1{,}800) / \ln(1.005) = 196$ months ≈ 16.3 years. Answer: B

Short Answer Model Answers

Q9 (3 marks): (a) $A_{n+1} = 1.006A_n + 150$ [1]. (b) $A_1 = 1.006(4{,}000) + 150 = \\$4{,}174$ [1]. $A_2 = 1.006(4{,}174) + 150 = \\$4{,}349.04$ [1].

Q10 (3 marks): (a) $C = 80{,}000 x 0.115 = \\$9{,}200$ [1]. (b) $r_{net} = 7% - 1% = 6%$ [1]. (c) $A_{30} = 20{,}000(1.06)^{30} + 9{,}200 x [(1.06)^{30} - 1]/0.06 = 114{,}870 + 726{,}340 = \\$841{,}210$ [1].

Q11 (4 marks): (a) $M = 300{,}000 x 0.005 / [1 - (1.005)^{-240}] = \\$2{,}149.29$ [1]. (b) Total repaid = $2{,}149.29 x 240 = \\$515{,}830$ [1]. Total interest = $515{,}830 - 300{,}000 = \\$215{,}830$ [1]. (c) With $3{,}000/month: n = -\ln(1 - 300{,}000 x 0.005/3{,}000) / \ln(1.005) = 126$ months = 10.5 years. Saves 9.5 years [1].

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