Mathematics Standard • Year 11 • Module 3 • Lesson 14
Interest and Depreciation — Exam Practice
Build fluency in identifying the correct formula from question wording — simple interest, compound interest, straight-line and declining balance depreciation — and applying each cleanly.
1. Quick recall — formula chooser
Answer each part. 1 mark each
Q1.1 Write each formula:
Simple interest: I = ____________; A = ____________
Compound interest: A = ____________; I = ____________
Straight-line dep.: S = ____________; D = ____________
Declining balance dep.: S = ____________
Q1.2 Match each cue to a formula (SI / CI / SL / DB):
"flat rate of 6% per annum" → ____ "compounds monthly at 5% p.a." → ____
"depreciates by $4,000/yr" → ____ "depreciates at 15% per annum" → ____
Q1.3 For a loan of $10,000 at 6% per annum compounding monthly for 2 years: state the adjusted values: r = ____________, n = ____________
2. Worked example — simple vs compound interest loan
Follow each line. Every step has a reason on the right.
Problem. Aiko needs to borrow $18,000 for 3 years. Lender A offers simple interest at 7.5% p.a. Lender B offers compound interest at 6.8% p.a. compounding annually. (a) Total repaid under each lender, (b) which is cheaper and by how much.
Step 1 — Lender A (simple interest).
P = $18,000, r = 0.075, n = 3
I = Prn = 18,000 × 0.075 × 3 = $4,050.00
Total repaid (A) = P + I = $18,000 + $4,050 = $22,050.00
Reason: simple interest is a flat rate applied to the original principal for n years.
Step 2 — Lender B (compound interest).
P = $18,000, r = 0.068, n = 3
(1.068)3 = 1.21819 (calculator)
A = 18,000 × 1.21819 = $21,927.36
Reason: compound interest formula gives the total amount repaid directly. Show (1 + r)n on its own line.
Step 3 — Compare totals and conclude.
Saving = $22,050.00 − $21,927.36 = $122.64
Reason: for a BORROWER, the lower total repayment is cheaper. Don't assume the lower rate always wins — always calculate both totals first.
Conclusion. Aiko should choose Lender B — it is cheaper by $122.64 over 3 years.
3. Faded example — investment and depreciation in one scenario
A business spends $28,000 on a machine depreciating at 15% per annum declining balance, and simultaneously invests $28,000 at 4.2% p.a. compounding semi-annually. After 4 years, find (a) the machine's value, (b) the investment's value, (c) the difference. Fill in each blank. 5 marks
Part (a) — Machine (declining balance).
(1 − r) = 1 − ____ = ____________
(____)4 = ____________
S = $28,000 × ____________ = $ ____________
Part (b) — Investment (compound interest, semi-annual).
r = 0.042 ÷ ____ = ____________ per half-year; n = 4 × ____ = ____________
(1 + ____)____ = ____________
A = $28,000 × ____________ = $ ____________
Part (c) — Difference.
Difference = $ __________ − $ __________ = $ ____________
Conclusion. The investment is worth $ ____________ more than the machine after 4 years.
4. Graduated practice — mixed formula recognition
For each question, identify which formula applies, then solve. Show one line of substitution.
Foundation — formula identification + single-step (4 questions)
| Q | Problem | Answer |
|---|---|---|
| 4.1 1 | $5,000 invested at 4% p.a. flat rate for 3 years. Find the interest. | |
| 4.2 1 | $5,000 invested at 4% p.a. compounding annually for 3 years. Find the amount. | |
| 4.3 1 | An asset of $20,000 depreciates by $1,800 per year SL. Value after 5 years? | |
| 4.4 1 | An asset of $20,000 depreciates at 18% per annum DB. Value after 2 years? |
Standard — typical HSC difficulty (6 questions)
Identify the formula first, then show one line of substitution and the value of (1 ± r)n where relevant.
4.5 $7,200 is invested at 3.8% p.a. simple interest for 6 years. Find the total amount after 6 years. 2 marks
4.6 $15,000 is invested at 5.2% p.a. compounding quarterly for 4 years. Find the final amount. 2 marks
4.7 A machine costs $60,000 and depreciates by $5,500 per year SL. After how many complete years does its value first fall below $20,000? 2 marks
4.8 A vehicle bought for $48,000 depreciates at 22% per annum DB. Find its value after 4 years and the total depreciation. 2 marks
4.9 A loan of $25,000 at 6% p.a. compounding monthly for 3 years (no repayments). Find the amount owed. 2 marks
4.10 Find r if an asset of $54,000 has fallen to $29,160 after 4 years of declining balance depreciation. Give r as a percentage. 2 marks
Extension — combine two concepts (2 questions)
4.11 $25,000 is invested for 5 years. Compare: Option A = simple interest at 6% p.a., Option B = compound interest at 5.6% p.a. compounding annually. State which option earns more interest and by how much. 3 marks
4.12 A business buys equipment for $32,000 (depreciating at 12% p.a. DB) and at the same time deposits $32,000 in a savings account (5% p.a. compounding annually). After 5 years, find the difference between the savings balance and the equipment's book value. 3 marks
5. Self-check the easy 3
Tick the first three once you've checked your method works.
How did this worksheet feel?
What I'll revisit before next class:
Q1.1 — Formula list
SI: I = Prn, A = P(1 + rn). CI: A = P(1 + r)n, I = A − P. SL: S = V₀ − Dn, D = (V₀ − Sn) ÷ n. DB: S = V₀(1 − r)n.
Q1.2 — Cue match
"flat rate" → SI; "compounds monthly" → CI; "$4,000/yr" → SL; "15% p.a." (depreciation) → DB.
Q1.3 — 6% p.a. monthly for 2 years
r = 0.06 ÷ 12 = 0.005 per month; n = 2 × 12 = 24 months.
Q3 — Faded example (machine DB + investment CI semi-annual, 4 years)
(a) Machine: (1 − r) = 1 − 0.15 = 0.85; (0.85)4 = 0.52201…; S = $28,000 × 0.52201 = $14,616.18.
(b) Investment: r = 0.042 ÷ 2 = 0.021; n = 4 × 2 = 8. (1 + 0.021)8 = 1.18088…; A = $28,000 × 1.18088 = $33,064.65.
(c) Difference: $33,064.65 − $14,616.18 = $18,448.48.
Conclusion: investment worth $18,448.48 more than the machine after 4 years.
Q4.1 — SI on $5,000 at 4% for 3 years
I = 5,000 × 0.04 × 3 = $600.00.
Q4.2 — CI on $5,000 at 4% (annual) for 3 years
A = 5,000 × (1.04)3 = 5,000 × 1.124864 = $5,624.32.
Q4.3 — SL on $20,000, $1,800/yr, 5 years
S = 20,000 − (1,800 × 5) = 20,000 − 9,000 = $11,000.00.
Q4.4 — DB on $20,000 at 18% for 2 years
(0.82)2 = 0.6724. S = 20,000 × 0.6724 = $13,448.00.
Q4.5 — SI on $7,200 at 3.8% for 6 years, total amount
I = 7,200 × 0.038 × 6 = $1,641.60. A = 7,200 + 1,641.60 = $8,841.60.
Q4.6 — CI on $15,000 at 5.2% quarterly for 4 years
r = 0.052 ÷ 4 = 0.013; n = 4 × 4 = 16. (1.013)16 = 1.229822…; A = 15,000 × 1.229822 = $18,447.33.
Q4.7 — SL inequality: $60,000, $5,500/yr, S < $20,000?
60,000 − 5,500n < 20,000 ⇒ 5,500n > 40,000 ⇒ n > 7.272… ⇒ n = 8 complete years. Check: year 7 → $21,500 (above); year 8 → $16,000 (first below). ✓
Q4.8 — DB on $48,000 at 22% for 4 years
(0.78)4 = 0.370151…; S = 48,000 × 0.370151 = $17,767.26. Total depreciation = 48,000 − 17,767.26 = $30,232.74.
Q4.9 — Loan $25,000 at 6% p.a. monthly for 3 years
r = 0.06 ÷ 12 = 0.005; n = 36. (1.005)36 = 1.196680…; A = 25,000 × 1.196680 = $29,917.00.
Q4.10 — Find r from $54,000 → $29,160 over 4 years DB
(1 − r)4 = 29,160 ÷ 54,000 = 0.54. 1 − r = (0.54)0.25 ≈ 0.85730. r ≈ 1 − 0.85730 = 0.14270 = 14.27% per annum.
Q4.11 — SI vs CI on $25,000, 5 years
SI at 6%: I = 25,000 × 0.06 × 5 = $7,500.00.
CI at 5.6% annual: (1.056)5 = 1.313174…; A = 25,000 × 1.313174 = $32,829.36; I = $32,829.36 − $25,000 = $7,829.36.
Compound (Option B) earns more by $7,829.36 − $7,500.00 = $329.36 over 5 years, despite the lower nominal rate.
Q4.12 — Equipment DB vs savings CI, both $32,000, 5 years
Equipment (12% DB): (0.88)5 = 0.527732; S = 32,000 × 0.527732 = $16,887.42.
Savings (5% CI annual): (1.05)5 = 1.276282; A = 32,000 × 1.276282 = $40,841.01.
Difference = $40,841.01 − $16,887.42 = $23,953.59. (The savings balance is worth almost $24k more than the equipment's book value after 5 years.)