Review Solutions — Financial Mathematics

1. Simple interest. Fluency

   I = Prn = 3500×0.06×4 = $840. A = P+I = 3500+840 = $4340.

2. Compound interest — annual. Fluency

   A = 4000×(1.07)⁸ = 4000×1.71819 ≈ $6873.

3. Compound interest — monthly. Fluency

   A = 2000×(1.0075)³⁶ ≈ 2000×1.30865 ≈ $2617. Period rate = 9%÷12 = 0.75%; periods = 12×3 = 36.

4. Straight-line depreciation. Fluency

   V = 12000(1−0.15×5) = 12000×0.25 = $3000. Annual depreciation = $1800; after 5 years only $3000 of the original $12000 remains.

5. Reducing balance depreciation. Fluency

   V = 18000×(0.80)⁴ = 18000×0.4096 ≈ $7373.

6. Appreciation. Fluency

   V = 250000×(1.04)⁶ = 250000×1.26532 ≈ $316330. The house gained over $66000 in value.

7. Find the principal. Understanding

   13310 = P×(1.10)³ = P×1.331 ⇒ P = 13310÷1.331 = $10000. Key step: recognise 1.1³=1.331.

8. Inflation. Understanding

   C = 4.50×(1.03)⁵ = 4.50×1.15927 ≈ $5.22. A modest 3% inflation still adds 72 cents to a cup of coffee over 5 years.

9. Present value. Understanding

   P = 30000 ÷ (1.05)⁸ = 30000 ÷ 1.47746 ≈ $20298. Investing $20298 today at 5% p.a. compound will grow to exactly $30000 in 8 years.

10. Simple vs compound interest. Understanding

    • (a) Final values: Option A (SI 8%): A = 6000(1+0.08×6) = 6000×1.48 = $8880. Option B (CI 7%): A = 6000×(1.07)⁶ = 6000×1.50073 ≈ $9004.
    • (b) Which is better?: Option B ($9004 > $8880). Despite a 1% lower annual rate, compound interest overtakes simple interest by year 6. This illustrates why compound interest always wins over a long enough time period.

11. Loan growth. Problem Solving

    • (a) Balance after 5 years: A = 15000×(1.06)⁵ = 15000×1.33823 ≈ $20073.
    • (b) Interest accumulated (compound): 20073 − 15000 = $5073.
    • (c) Interest under simple: I = 15000×0.06×5 = $4500.
    • (d) Extra from compounding: 5073 − 4500 = $573 extra. The compounding effect costs Ben an extra $573 compared with simple interest over 5 years.

12. Multi-step depreciation and appreciation. Problem Solving

    • (a) After 3 years at 15% RB: V = 40000×(0.85)³ = 40000×0.61413 ≈ $24565.
    • (b) After 2 more years at 5% appreciation: V = 24565×(1.05)² = 24565×1.1025 ≈ $27093.
    • (c) Net change over 5 years: 27093 − 40000 = −$12907. Despite 2 years of appreciation, the car is still worth $12907 less than when purchased.
    • (d) Compare with 5 years all at 15% RB: V = 40000×(0.85)⁵ = 40000×0.44371 ≈ $17748. Yes — the 2 years of appreciation brought the car value to $27093 vs only $17748 under continuous depreciation. The owner is $9345 better off because of the appreciation period.