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.
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
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].
Tick when you've finished all questions and reviewed your answers.