Practice Maths

Solutions — Simple Interest

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  1. Q1 — Calculate simple interest

    (a) I = 2000 × 0.05 × 3 = $300

    (b) I = 5500 × 0.04 × 2 = $440

    (c) t = 18/12 = 1.5 years. I = 800 × 0.075 × 1.5 = $90

    (d) t = 6/12 = 0.5 years. I = 12000 × 0.032 × 0.5 = $192

  2. Q2 — Total amount A

    (a) I = 4000×0.06×5 = $1200. A = 4000+1200 = $5 200

    (b) I = 750×0.08×3 = $180. A = 750+180 = $930

    (c) I = 1600×0.045×2.5 = $180. A = 1600+180 = $1 780

  3. Q3 — Find the missing variable

    (a) P = I ÷ (rt) = 360 ÷ (0.06 × 3) = 360 ÷ 0.18 = $2 000

    (b) r = I ÷ (Pt) = 875 ÷ (2500 × 7) = 875 ÷ 17500 = 0.05 = 5% p.a.

    (c) t = I ÷ (Pr) = 480 ÷ (3200 × 0.05) = 480 ÷ 160 = 3 years

  4. Q4 — Time conversions

    (a) t = 8/12 years. I = 900 × 0.06 × (8/12) = 900 × 0.06 × 0.667 = $36

    (b) t = 146/365 = 0.4 years. I = 2400 × 0.05 × 0.4 = $48

    (c) t = 2.5 years. I = 6000 × 0.03 × 2.5 = $450

  5. Q5 — Working backwards from total amount

    (a) A = P(1+rt) → 3920 = P(1 + 0.05×4) = P(1.2) → P = 3920÷1.2 = $3 267 (to nearest dollar)

    (b) I = 6250−5000 = $1250. r = I÷(Pt) = 1250÷(5000×5) = 1250÷25000 = 0.05 = 5% p.a.

    (c) I = 2300−2000 = $300. t = I÷(Pr) = 300÷(2000×0.06) = 300÷120 = 2.5 years

  6. Q6 — Comparing investments

    (a) Bank A: I = 4000×0.055×3 = $660. Bank B: I = 4000×0.05×3 = $600. Difference = $60 more with Bank A.

    (b) Bank B earns: I = 1000×0.06×2 = $120. Set equal: 1000×0.04×t = 120 → 40t = 120 → t = 3 years

  7. Q7 — Reading the graph

    (a) At t=3: A = 1000(1 + 0.08×3) = 1000×1.24 = $1 240

    (b) Doubled means A = $2000. 2000 = 1000(1+0.08t) → 1+0.08t = 2 → 0.08t = 1 → t = 12.5 years

    (c) Gradient = dA/dt = Pr = 1000×0.08 = $80 per year. It represents the interest earned per year (constant under simple interest).

  8. Q8 — Simple interest on a loan

    (a) I = 8500×0.09×3 = $2 295. Total repaid = 8500+2295 = $10 795

    (b) Monthly repayments = $10 795 ÷ 36 = $299.86/month

    (c) At 6%: I = 8500×0.06×3 = $1530. Saving = 2295−1530 = $765

  9. Q9 — Finding rate given repayments

    (a) Total interest = 7440 − 6000 = $1 440

    (b) r = I ÷ (Pt) = 1440 ÷ (6000×2) = 1440÷12000 = 0.12 = 12% p.a.

  10. Q10 — Further problem solving

    Answers will depend on the specific Q10 in your questions file. Use I = Prt, A = P(1+rt), and rearrange as needed. Show all substitution steps clearly.