Compound Interest — Solutions

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1. Total amount after compound interest

   1. P=$2000, r=5%, n=2: A = 2000 × (1.05)² = 2000 × 1.1025 = $2205.00
   2. P=$3000, r=4%, n=3: A = 3000 × (1.04)³ = 3000 × 1.124864 = $3374.59
   3. P=$10 000, r=3%, n=5: A = 10 000 × (1.03)&sup5; = 10 000 × 1.159274 = $11 592.74
   4. P=$1500, r=8%, n=4: A = 1500 × (1.08)&sup4; = 1500 × 1.360489 = $2040.73
   5. P=$7500, r=6%, n=3: A = 7500 × (1.06)³ = 7500 × 1.191016 = $8932.62
   6. P=$500, r=2%, n=10: A = 500 × (1.02)¹⁰ = 500 × 1.218994 = $609.50
   7. P=$4000, r=5.5%, n=6: A = 4000 × (1.055)&sup6; = 4000 × 1.379463 = $5517.85
   8. P=$20 000, r=3.5%, n=4: A = 20 000 × (1.035)&sup4; = 20 000 × 1.147523 = $22 950.46

2. Compound interest earned (CI = A − P)

   1. P=$6000, r=4%, n=3: A = 6000 × (1.04)³ = $6749.18. CI = $6749.18 − $6000 = $749.18
   2. P=$8000, r=5%, n=2: A = 8000 × (1.05)² = $8820.00. CI = $8820.00 − $8000 = $820.00
   3. P=$12 000, r=3%, n=4: A = 12 000 × (1.03)&sup4; = $13 502.04. CI = $13 502.04 − $12 000 = $1502.04
   4. P=$9500, r=6%, n=5: A = 9500 × (1.06)&sup5; = $12 713.64. CI = $12 713.64 − $9500 = $3213.64

3. Depreciation A = P(1 − r)ⁿ

   1. Laptop $2400, 20%, 3 yr: A = 2400 × (0.80)³ = 2400 × 0.512 = $1228.80
   2. Machinery $50 000, 15%, 5 yr: A = 50 000 × (0.85)&sup5; = 50 000 × 0.443705 = $22 185.25
   3. Phone $900, 30%, 2 yr: A = 900 × (0.70)² = 900 × 0.49 = $441.00
   4. Boat $40 000, 10%, 6 yr: A = 40 000 × (0.90)&sup6; = 40 000 × 0.531441 = $21 257.64

4. Compare CI and SI

   1. P=$5000, r=6%, n=4 yr:
      1. SI: I = 5000×6×4÷100 = $1200. Total = $6200
      2. CI: A = 5000×(1.06)&sup4; = 5000×1.26248 = $6312.38. CI earned = $1312.38. Total = $6312.38
      3. Compound earns $6312.38 − $6200 = $112.38 more
   2. Why CI grows faster: With compound interest, each year’s interest is added to the principal. So the next year’s interest is calculated on a larger amount, causing accelerating growth. With simple interest, only the original principal earns interest each period.
   3. When are SI and CI equal?: At n = 1 (the end of the first compounding period), both give the same result because there has been exactly one interest calculation on the original principal.
   4. $10 000, Option A (7% SI, 5 yr) vs B (6% CI, 5 yr): A: I = 10 000×7×5÷100 = $3500. Total = $13 500. B: A = 10 000×(1.06)&sup5; = 10 000×1.338226 = $13 382.26. Option A gives more: $13 500 > $13 382.26

5. Real-world CI and depreciation problems

   1. Mia: $15 000 at 4.5% CI, target $18 000: n=4: 15000×(1.045)&sup4; = $17 801.35 (not enough). n=5: 15000×(1.045)&sup5; = $18 602.41 > $18 000. Minimum: 5 years
   2. Car $45 000, depreciation 18% p.a.:
      1. A = 45 000×(0.82)³ = 45 000×0.551368 = $24 811.56
      2. n=4: 45000×(0.82)&sup4; = $20 345.48 (above $20k). n=5: 45000×(0.82)&sup5; = $16 683.29 (below $20k). Value first falls below $20 000 after 5 complete years.
   3. James: P×(1.05)² = $11 025: P = 11 025 ÷ (1.05)² = 11 025 ÷ 1.1025 = $10 000
   4. Town populations after 10 years:
      1. Town A (grows 3%): 12 000×(1.03)¹⁰ = 12 000×1.343916 ≈ 16 127. Town B (declines 2%): 15 000×(0.98)¹⁰ = 15 000×0.817073 ≈ 12 256
      2. Town A has the larger population after 10 years (16 127 > 12 256).

6. Comparing simple and compound interest over time

   1. Year 1: SI = $5300; CI = $5300; Difference = $0
   2. Year 2: SI = $5600; CI = $5618.00; Difference = $18.00
   3. Year 3: SI = $5900; CI = $5955.08; Difference = $55.08
   4. Year 4: SI = $6200; CI = $6312.38; Difference = $112.38
   5. The CI − SI difference grows larger each year (the gap accelerates). Compound interest earns interest on previous interest, so the advantage compounds over time.

7. Choosing an investment

   1. Option A: $10 000 + $10 000 × 0.07 × 5 = $13 500
   2. Option B: $10 000 × (1.06)&sup5; = $10 000 × 1.338226 = $13 382.26
   3. Option A gives more by $13 500 − $13 382.26 = $117.74
   4. At 10 years: A = $10 000 + $7000 = $17 000. B = $10 000 × (1.06)¹⁰ = $17 908.47. Option B is better by $908.47 — compound interest wins over the long run.

8. Finding the principal

   1. 11 025 = P × (1.05)²
   2. P = 11 025 ÷ 1.1025 = $10 000
   3. CI = $11 025 − $10 000 = $1 025

9. Reaching a savings goal

   1. n=4: $15 000 × (1.045)&sup4; ≈ $17 888 (not enough). n=5: $15 000 × (1.045)&sup5; ≈ $18 693 > $18 000. Minimum = 5 years.
   2. At 6%: n=3: $15 000 × (1.06)³ ≈ $17 865 (not enough). n=4: $15 000 × (1.06)&sup4; ≈ $18 937 ✓. Yes, she reaches $18 000 in 4 years at 6%.
   3. A higher rate means each year’s balance grows by a greater amount, so the target is reached in fewer compounding periods.

10. Car depreciation investigation

    1. $45 000 × (0.82)³ = $45 000 × 0.551368 = $24 811.56
    2. n=4: $45 000 × (0.82)&sup4; = $20 345.48 (above $20 000). n=5: $45 000 × (0.82)&sup5; = $16 683.29 (below). First falls below $20 000 after 5 complete years.
    3. Second car: $60 000 × (0.85)³ = $60 000 × 0.614125 = $36 847.50. First car after 3 years = $24 811.56. The $60 000 car is worth more.
    4. Half of $45 000 = $22 500. n=3: $24 811.56 (above); n=4: $20 345.48 (below). Falls below half after 4 complete years.