R16: Growth and Decay
Set up, solve and interpret answers in growth and decay problems; compound interest
📋 Key Concepts
Growth: When a quantity increases by a fixed percentage each time period (e.g., compound interest, population growth).
Decay: When a quantity decreases by a fixed percentage each time period (e.g., depreciation, radioactive decay).
Compound Growth/Decay: Final = Initial × (multiplier)^n
Where multiplier = (1 + r) for growth, (1 - r) for decay, n = number of periods
Where multiplier = (1 + r) for growth, (1 - r) for decay, n = number of periods
📝 Compound Interest
Formula: A = P(1 + r)^n
A = final amount, P = principal (initial), r = rate (as decimal), n = number of periods
A = final amount, P = principal (initial), r = rate (as decimal), n = number of periods
Example 1
£4000 is invested at 5% compound interest per year. Find the value after 6 years.
Solution:
Multiplier = 1 + 0.05 = 1.05
Value = £4000 × 1.05⁶
= £4000 × 1.3400956
= £5360.38 (to 2 dp)
Example 2
Compare £3000 at 4% compound interest vs simple interest after 5 years.
Solution:
Compound: £3000 × 1.04⁵ = £3000 × 1.21665 = £3649.94
Simple: £3000 + (3000 × 0.04 × 5) = £3000 + £600 = £3600
Compound gives £49.94 more
📝 Depreciation (Decay)
Formula: Value = Initial × (1 - r)^n
Where r is the depreciation rate as a decimal.
Where r is the depreciation rate as a decimal.
Example 3
A car worth £25000 depreciates at 15% per year. Find its value after 4 years.
Solution:
Multiplier = 1 - 0.15 = 0.85
Value = £25000 × 0.85⁴
= £25000 × 0.522006
= £13050.16
📝 Finding the Rate or Time
Example 4
£2000 grows to £2662 after 5 years with compound interest. Find the annual rate.
Solution:
2000 × m⁵ = 2662
m⁵ = 2662/2000 = 1.331
m = 1.331^(1/5) = 1.059 (approximately)
Rate = 1.059 - 1 = 0.059 = 5.9% (approximately)
Or solve: 1.331 = 1.1³ = 1.1 × 1.1 × 1.1, but we need m⁵...
Actually m⁵ = 1.331 and if m = 1.059, check: 1.059⁵ ≈ 1.33 ✓
Example 5
How long for £1000 to exceed £1500 at 6% compound interest?
Solution:
1000 × 1.06ⁿ > 1500
1.06ⁿ > 1.5
Test: 1.06⁷ = 1.5036 > 1.5
7 years
📝 Other Growth and Decay Problems
Example 6
A population of bacteria doubles every 3 hours. Starting with 500 bacteria, how many after 15 hours?
Solution:
Number of doublings = 15 ÷ 3 = 5
Population = 500 × 2⁵ = 500 × 32 = 16000
Example 7
The half-life of a substance is 8 days. If we start with 80g, how much remains after 24 days?
Solution:
Number of half-lives = 24 ÷ 8 = 3
Remaining = 80 × (1/2)³ = 80 × 1/8 = 10g
🧮 Worked Examples — Exam Style
Growth and decay = repeated percentage change: final = initial × multiplierⁿ. Appreciation ×(1 + r), depreciation ×(1 − r).
A £800 investment grows at 6% per year for 3 years. Find the final value.
Multiplier 1.06 → 800 × 1.06³ = 800 × 1.191016 = £952.81 (2 d.p.)
M1 for 1.06³, A1 for £952.81. Keep the power form until the end: 800 × 1.06 × 1.06 × 1.06.
A population of 40,000 shrinks by 7% each year. Find the population after 5 years.
Multiplier 0.93 → 40,000 × 0.93⁵ = 40,000 × 0.6956884 = 27,828 (nearest whole)
M1 for 0.93⁵, A1 for 27,828. Populations round to whole people — say 'approximately'.
Exam wording to expect: "compound interest", "depreciates by …%", "grows at …% per year".
❓ Practice Questions
Q1: £5000 at 4.5% compound interest for 8 years. Find the final amount.
Q2: A laptop costs £1200 and depreciates at 20% per year. Find its value after 3 years.
Q3: £3000 grows to £3646.52 at 5% compound interest. How many years?
Q4: A population of 2000 grows by 3% each year. Find the population after 10 years.
Q5: A radioactive sample has half-life 5 years. Starting with 64g, how much after 20 years?
Q6: A car worth £15,000 depreciates by 15% per year. Find its value after 4 years.
Q7: An investment doubles in 9 years. What single yearly multiplier achieves this (2 d.p.)?
✅ Answers
- £7109.62 (5000 × 1.045⁸)
- £614.40 (1200 × 0.8³)
- 4 years
- Approximately 2688 (2000 × 1.03¹⁰)
- 4g (64 × 1/16 = 64 × 1/2⁴)
📋 Model Answers — Full Working
Every mark comes from the working, not just the answer. These model answers show the lines a mark scheme expects to see.
Q1. 5000 × 1.045⁸ = 5000 × 1.422101 = £7110.50
Q2. 1200 × 0.8³ = 1200 × 0.512 = £614.40
Q3. 3646.52/3000 = 1.215507 = 1.05⁴ → 4 years
Q4. 2000 × 1.03¹⁰ = 2000 × 1.343916 = 2688 (nearest whole)
Q5. 20 ÷ 5 = 4 halvings → 64 × (½)⁴ = 64/16 = 4 g
Q6. 15,000 × 0.85⁴ = 15,000 × 0.52200625 = £7830.09
Q7. m⁹ = 2 → m = 2^(1/9) ≈ 1.08 — about 8% growth per year
📌 Key Facts to Memorise
- Final = initial × multiplierⁿ (n = periods)
- Growth: ×(1 + r/100) · Decay: ×(1 − r/100) — e.g. 4.5% → 1.045; 12% → 0.88
- Half-life: divide by 2 each period — n halvings → ×(½)ⁿ
- Finding time: trial-and-improve with the multiplier, or spot the power
- Money: 2 d.p.; populations/quantities: sensible rounding stated
🗺️ In the Exam
Growth/decay: 3–5 marks on Papers 2/3 — savings, car values, populations, radioactivity. The find-the-rate-or-time reverse is the top-difficulty version.
Links to: percentages (R9), types of sequences (A24), standard form (N9).
🧭 Board Focus — AQA
Spec 8300. Three papers, each 80 marks and 1h30: Paper 1 non-calculator, Papers 2 and 3 calculator. Foundation/Higher tiers. Roughly a third of the marks sit under AO2 (reasoning) and AO3 (problem-solving), so show every line of working — method marks survive arithmetic slips.
Always check: specification details change — confirm paper structure and set texts against the current specification for your exam year.
🎯 Exam Tips
- Multiplier = 1 + rate for growth, 1 - rate for decay
- Use the formula: Final = Initial × multiplierⁿ
- Compound interest uses powers, simple interest does not
- For half-life: multiplier = 0.5, substitute number of half-lives
- Check answers are sensible: growth should increase, decay should decrease
🧠 Problem-Solving Strategies
Problem-Solving
For compound growth: Amount = P(1 + r/100)^n. For exponential decay: Amount = P(1 − r/100)^n. For continuous growth: A = Pe^(kt). Half-life: the time for a quantity to reduce to half. Set up equations using given data points, then solve for the unknown rate or time.
Multi-Step Problem
A bacterial culture doubles every 3 hours. Starting with 500 bacteria, how many are there after 15 hours? How long until there are at least 8000?
Solution: After 15 hours: number of doublings = 15/3 = 5. Population = 500 × 2⁵ = 500 × 32 = 16,000. For 8000: 500 × 2^n = 8000, 2^n = 16, n = 4. Time = 4 × 3 = 12 hours.
⚠️ Common Errors
Watch Out!
1. Wrong: For compound interest at 5% for 3 years, adding 3 × 5% = 15% to the original Correct: Use (1.05)³ = 1.1576, so the total increase is 15.76%, not 15%. Compound interest gives MORE than simple interest.
2. Wrong: For decay, using (1 + r/100)^n instead of (1 − r/100)^n Correct: Decay means the quantity decreases, so use (1 − r/100)^n. E.g. 8% decay per year: multiply by 0.92 each year.
3. Wrong: Thinking that half-life means the quantity is zero after two half-lives Correct: After 1 half-life: ½ remains. After 2 half-lives: ¼ remains. After 3: ⅛. The quantity halves each time but never reaches zero.
✍️ 6-Mark Exam Question
Extended Answer
6 marks: A car is bought for £25,000. It depreciates at 15% per year. (a) Find its value after 5 years. (b) After how many years is it worth less than £5000? (c) The owner says "After 7 years it will be worthless." Show this is incorrect and find its actual value after 7 years.
(a) Value = 25,000 × 0.85⁵ = 25,000 × 0.44370 = £11,092.53.
(b) 25,000 × 0.85^n < 5000. 0.85^n < 0.2. n × ln(0.85) < ln(0.2). n × (−0.1625) < −1.6094. n > 9.90. So after 10 years.
(c) After 7 years: 25,000 × 0.85⁷ = 25,000 × 0.32058 = £8014.42. The car still has significant value. It is NOT worthless — exponential decay never reaches zero.
Mark scheme: M1 decay formula, A1 £11,092.53, M1 inequality or logs, A1 10 years, M1 calculation for 7 years, A1 £8014.42, A1 explanation
📊 AO3: Reason & Interpret
Reasoning and Interpretation
A radioactive substance has a half-life of 8 years. The initial mass is 100 g.
(a) How much remains after 24 years?
(b) A scientist says "After 40 years, only 3.125 g will remain." Verify this claim.
(c) Another scientist says "After 80 years (10 half-lives), the substance will be completely safe." Is this statement justified? Explain.
Answers: (a) 24 years = 3 half-lives. Remaining = 100 × (½)³ = 100 × ⅛ = 12.5 g. (b) 40 years = 5 half-lives. Remaining = 100 × (½)⁵ = 100 × 1/32 = 3.125 g. The scientist is correct ✓ (c) After 10 half-lives: 100 × (½)¹⁰ = 100/1024 ≈ 0.098 g. There is still radioactive material present — it is NOT zero. Whether it is "safe" depends on the type of radiation and safe exposure levels, not just the mass. The statement is not fully justified.
📝 Exam Questions by Topic
🎬 Video Resources
Ready to ace your GCSE Mathematics exams?
Get the best revision books and guides to boost your grades.