This checkpoint assesses your understanding of compound interest, the compound interest formula, comparisons with simple interest, depreciation, and problem-solving with all four methods. It covers Lessons 11–15.
❓ Multiple Choice (10 questions)
1. What is the key difference between simple and compound interest?
2. In the formula A = P(1 + r)^n, what does n represent when interest is compounded monthly for 3 years?
3. $6,000 is invested at 4% p.a. compounded quarterly for 5 years. What is the final amount?
4. $10,000 at 5% p.a. for 8 years. Approximately how much more does compound interest earn than simple interest?
5. A term deposit pays 5.8% p.a. simple interest for 2 years. A savings account pays 5.0% p.a. compounded monthly. For $5,000, which is better after 2 years?
6. A $15,000 car depreciates by $2,500 per year using straight-line. What is its value after 4 years?
7. Equipment worth $8,000 depreciates at 15% p.a. using reducing balance. What is its value after 3 years?
8. Which statement about depreciation is true?
9. A $20,000 car is bought with a flat-rate loan at 8% p.a. simple interest over 4 years. The car depreciates at 20% p.a. reducing balance. After 2 years, what is the owner's approximate equity position?
10. You need $12,000 in 3 years. You have $7,000 to invest at 5.5% p.a. compounded annually. Approximately how much extra must you save in total?
✍ Short Answer (4 questions)
11. Calculate the final amount and total interest for each investment:
(a) $9,000 at 4.5% p.a. compounded annually for 6 years. (2 marks)
(b) $5,000 at 3.8% p.a. compounded monthly for 4 years. (2 marks)
4 MARKS
12. A student is comparing two 3-year investment options for $6,000:
Option A: 5.0% p.a. simple interest.
Option B: 4.8% p.a. compounded quarterly.
(a) Calculate the total for Option A. (2 marks)
(b) Calculate the total for Option B. (2 marks)
(c) Which is better? By how much? (1 mark)
5 MARKS
13. A business buys a delivery van for $32,000. It depreciates at 14% p.a. using reducing balance.
(a) Calculate its value after 4 years. (2 marks)
(b) Calculate the total depreciation over 4 years. (1 mark)
(c) If the business had used straight-line depreciation of $5,000 per year, what would the value be after 4 years? (1 mark)
(d) Which method gives the higher value after 4 years? By how much? (1 mark)
5 MARKS
14. Real-World Problem: Sarah wants to buy a car for $18,000 in 4 years. She currently has $8,000 saved. She is deciding between two options:
Option A: Invest the $8,000 at 5.0% p.a. compounded monthly and save $200 per month.
Option B: Buy a $8,000 used car now. It depreciates at 12% p.a. reducing balance. She saves $200 per month separately.
(a) Calculate the value of the investment in Option A after 4 years. (2 marks)
(b) Calculate the total savings from $200/month over 4 years. (1 mark)
(c) Calculate the total available in Option A after 4 years. (1 mark)
(d) Calculate the value of the used car in Option B after 4 years. (2 marks)
(e) Calculate the total available in Option B after 4 years. (1 mark)
(f) Which option gets Sarah closer to $18,000? By how much? (1 mark)
(g) Explain one advantage of each option beyond the dollar amount. (1 mark)
9 MARKS
1. C — Compound interest earns interest on interest, causing exponential growth. Simple interest only earns on the principal.
2. B — Monthly compounding for 3 years: n = 3 × 12 = 36 periods.
3. D — r = 0.04/4 = 0.01. n = 5 × 4 = 20. A = $6,000 × (1.01)^20 = $6,000 × 1.22019 = $7,321.14 ≈ $7,320.
4. B — Simple: $10,000 + $10,000 × 0.05 × 8 = $14,000. Compound: $10,000 × (1.05)^8 = $10,000 × 1.47746 = $14,774.55. Difference = $14,774.55 − $14,000 = $774.55 ≈ $775.
5. C — Term deposit: $5,000 + $5,000 × 0.058 × 2 = $5,580. Savings: r = 0.05/12 = 0.004167, n = 24. A = $5,000 × (1.004167)^24 = $5,000 × 1.10494 = $5,524.70. Term deposit wins by $55.30 ≈ $55.
6. B — V = $15,000 − $2,500 × 4 = $15,000 − $10,000 = $5,000.
7. B — V = $8,000 × (0.85)^3 = $8,000 × 0.614125 = $4,913.
8. A — Reducing balance uses V = P(1 − r)^n, which has the same structure as A = P(1 + r)^n.
9. C — Loan: I = $20,000 × 0.08 × 4 = $6,400. Total = $26,400. Monthly = $550. After 2 years: repaid = $13,200. Owed = $13,200. Car: $20,000 × (0.80)^2 = $20,000 × 0.64 = $12,800. Negative equity = $13,200 − $12,800 = $400.
10. B — $7,000 × (1.055)^3 = $7,000 × 1.17424 = $8,219.68. Extra needed = $12,000 − $8,219.68 = $1,780.32 ≈ $1,780.
Q11 (4 marks): (a) A = $9,000 × (1.045)^6 = $9,000 × 1.30226 = $11,720.34 [1]. I = $11,720.34 − $9,000 = $2,720.34 [1]. (b) r = 0.038/12 = 0.003167. n = 48. A = $5,000 × (1.003167)^48 = $5,000 × 1.16285 = $5,814.25 [1]. I = $814.25 [1].
Q12 (5 marks): (a) A = $6,000 + $6,000 × 0.05 × 3 = $6,900 [2]. (b) r = 0.048/4 = 0.012. n = 12. A = $6,000 × (1.012)^12 = $6,000 × 1.15385 = $6,923.10 [2]. (c) Option B is better by $23.10 [1].
Q13 (5 marks): (a) V = $32,000 × (0.86)^4 = $32,000 × 0.547008 = $17,504.26 [2]. (b) Total depreciation = $32,000 − $17,504.26 = $14,495.74 [1]. (c) V = $32,000 − $5,000 × 4 = $12,000 [1]. (d) Reducing balance is higher by $5,504.26 [1].
Q14 (9 marks): (a) r = 0.05/12 = 0.004167. n = 48. A = $8,000 × (1.004167)^48 = $8,000 × 1.22090 = $9,767.20 [2]. (b) $200 × 48 = $9,600 [1]. (c) Total = $9,767.20 + $9,600 = $19,367.20 [1]. (d) V = $8,000 × (0.88)^4 = $8,000 × 0.599695 = $4,797.56 [2]. (e) Total = $4,797.56 + $9,600 = $14,397.56 [1]. (f) Distance from $18,000: Option A = $1,367.20 over. Option B = $3,602.44 under. Option A is closer by $2,235.24 [1]. (g) Option A: no car maintenance costs for 4 years. Option B: has a car to use immediately [1].
Tick when you have finished all questions and checked your answers.