Solutions — Simple and Compound Interest

1. Calculate simple interest. Fluency

   • (a) P=$500, r=4%, n=3: I = 500×0.04×3 = $60. A = 500+60 = $560.
   • (b) P=$2000, r=6%, n=5: I = 2000×0.06×5 = $600. A = 2000+600 = $2600.
   • (c) P=$800, r=2.5%, n=4: I = 800×0.025×4 = $80. A = 800+80 = $880.
   • (d) P=$1500, r=8%, n=2.5: I = 1500×0.08×2.5 = $300. A = 1500+300 = $1800.

2. Calculate compound interest. Fluency

   • (a) P=$1000, r=5%, n=4: A = 1000×(1.05)⁴ = 1000×1.21551 ≈ $1215.51.
   • (b) P=$3000, r=3%, n=10: A = 3000×(1.03)¹⁰ = 3000×1.34392 ≈ $4031.75.
   • (c) P=$500, r=8%, n=6: A = 500×(1.08)⁶ = 500×1.58687 ≈ $793.44.
   • (d) P=$2000, r=4%, n=3: A = 2000×(1.04)³ = 2000×1.12486 ≈ $2249.73.

3. Find the unknown (simple interest). Fluency

   • (a) Find n: I = Prn ⇒ 200 = 1000×0.05×n = 50n ⇒ n = 4 years.
   • (b) Find r: 240 = 1600×r×3 = 4800r ⇒ r = 240/4800 = 0.05 = 5% p.a.
   • (c) Find n: I = 1400−1000 = 400. 400 = 1000×0.05×n ⇒ n = 400/50 = 8 years.
   • (d) Find P: A = P(1+rn) ⇒ 1500 = P(1+0.10×5) = 1.5P ⇒ P = 1500/1.5 = $1000.

4. Compound interest with multiple compounding periods. Fluency

   • (a) Monthly, 6%, 2 years: A = 1000×(1+0.06/12)²⁴ = 1000×(1.005)²⁴ ≈ 1000×1.12716 ≈ $1127.16.
   • (b) Quarterly, 4%, 3 years: A = 5000×(1+0.04/4)¹² = 5000×(1.01)¹² ≈ 5000×1.12683 ≈ $5634.13.
   • (c) Semi-annual, 8%, 5 years: A = 2000×(1+0.08/2)¹⁰ = 2000×(1.04)¹⁰ ≈ 2000×1.48024 ≈ $2960.49.
   • (d) Monthly, 12%, 1 year: A = 3000×(1+0.12/12)¹² = 3000×(1.01)¹² ≈ 3000×1.12683 ≈ $3380.50.

5. Read from the graph: simple vs compound interest. Understanding

   • (a) At year 3: Account A (simple 12%): A = 1000+120×3 = $1360. Account B (compound 10%): A = 1000×1.1³ = $1331. Account A has more by about $29.
   • (b) Crossover: At n=4: A=$1480, B=$1464 — A leads. At n=5: A=$1600, B=$1611 — B leads. Compound overtakes simple between year 4 and year 5.
   • (c) At year 7: Account A: $1000+120×7 = ≈$1840. Account B: 1000×1.1⁷ ≈ ≈$1949.
   • (d) Why compound wins: Compound interest earns interest on interest — each year's interest is added to the principal, so the base grows and the interest amount increases every year. Simple interest always adds the same fixed amount. Over time, exponential growth always outpaces linear growth.

6. Which account earns more? Understanding

   • (a) Account A (simple 7%, 5 yr): A = 4000(1+0.07×5) = 4000×1.35 = $5400.
   • (b) Account B (compound 5%, 5 yr): A = 4000×(1.05)⁵ = 4000×1.27628 ≈ $5105.12.
   • (c) Which is better after 5 years?: Account A ($5400 > $5105). Simple interest at a higher rate wins in the short term.
   • (d) When does B overtake A?: n=10: A=$6800, B=4000×1.05¹⁰≈$6516 — A leads. n=14: A=$7920, B=4000×1.05¹⁴≈$7920 — nearly equal! n=15: A=$8200, B=4000×1.05¹⁵≈$8316 — B leads. B overtakes A in year 15 (approximately between year 14 and 15).

7. Interpret the compound interest formula. Understanding

   • (a) Monthly rate: r/k = 6%/12 = 0.5% per month (= 0.005 as a decimal).
   • (b) Compounding periods: kn = 12×2 = 24 periods.
   • (c) Formula: A = 4000×(1 + 0.06/12)^12×2 = 4000×(1.005)²⁴
   • (d) Final amount: A = 4000×(1.005)²⁴ ≈ 4000×1.12716 ≈ $4508.64.

8. Find the principal. Understanding

   • (a) I=$600, r=5%, n=4 (SI): I = Prn ⇒ 600 = P×0.05×4 = 0.2P ⇒ P = 600/0.2 = $3000.
   • (b) A=$2662, r=10%, n=3 (CI): 2662 = P×(1.1)³ = P×1.331 ⇒ P = 2662/1.331 = $2000.
   • (c) A=$1300, r=5%, n=6 (SI): A = P(1+rn) ⇒ 1300 = P(1+0.05×6) = P×1.3 ⇒ P = 1300/1.3 = $1000.
   • (d) A=$1276.28, r=5%, n=5 (CI): 1276.28 = P×(1.05)⁵ = P×1.27628 ⇒ P = 1276.28/1.27628 ≈ $1000.

9. Compare over multiple time periods. Problem Solving

   • (a) After 3 years: X = 5000(1+0.08×3) = 5000×1.24 = $6200. Y = 5000×(1.06)³ = 5000×1.19102 ≈ $5955. Option X is better.
   • (b) After 8 years: X = 5000(1+0.08×8) = 5000×1.64 = $8200. Y = 5000×(1.06)⁸ = 5000×1.59385 ≈ $7969. Option X is still better.
   • (c) After 15 years: X = 5000(1+0.08×15) = 5000×2.2 = $11 000. Y = 5000×(1.06)¹⁵ = 5000×2.39656 ≈ $11 983. Option Y is now better.
   • (d) When does Y overtake X?: n=10: X = $9000, Y = 5000×1.06¹⁰≈$8954 — X barely leads. n=11: X = $9400, Y = 5000×1.06¹¹≈$9491 — Y leads! Y overtakes X between year 10 and year 11.

10. Doubling time. Problem Solving

    • (a) Double at 6%: Need (1.06)ⁿ ≥ 2. Trial: (1.06)¹¹≈1.898 (not yet); (1.06)¹²≈2.012 (doubled!). Answer: 12 complete years.
    • (b) Double at 8%: Need (1.08)ⁿ ≥ 2. Trial: (1.08)⁹≈1.999 (just under); (1.08)¹⁰≈2.159. Answer: 9 complete years (3 fewer than at 6%).
    • (c) Rule of 72: 6%: 72÷6 = 12 years (exact answer: 12). 8%: 72÷8 = 9 years (exact answer: 9). The rule is very accurate at these rates.
    • (d) Simple to double in 10 years: A = 2P ⇒ P(1+10r) = 2P ⇒ 1+10r = 2 ⇒ r = 0.1 = 10% p.a.