Every exponential growth and decay question from 6 Cambridge IGCSE Mathematics (0580) past papers (Oct/Nov 2025 to May/June 2026), syllabus section 1.17. 8 Extended-tier questions. Each question is labelled with its question paper code (for example 0580/42/M/J/26) and every final answer matches the official Cambridge mark scheme. Try each question first, then click “Show solution” to check your working. This page is part of our IGCSE Maths 0580 Number past paper questions.
0580 Exponential Growth and Decay past paper questions: Extended (Papers 2 and 4)
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q7
(a) Tom invests $2500 at a rate of 4.4% per year compound interest. Calculate the value of the investment at the end of 6 years.
(b) Behnaz invests $4500 at a rate of r% per year compound interest. At the end of 5 years, the value of the investment is $5420. Calculate the value of r.
Show solution
- A 4.4% increase each year means multiply by 1.044 each year.
- Value after 6 years = 2500 × 1.044⁶ = 2500 × 1.29480… = 3237.00…
- $3237 ($3240 to 3 s.f.)
- Let the yearly multiplier be m. Then 4500 × m⁵ = 5420.
- Divide by 4500: m⁵ = 5420⁄4500 = 1.2044…
- Take the fifth root: m = 1.2044…1/5 = 1.03790…
- The multiplier is 1 + r⁄100, so r = (1.03790… − 1) × 100 = 3.790…
- r = 3.79
May/June 2026 · Paper 4 · 0580/42/M/J/26 · Q12
A scientist records the mass of a radioactive substance at the same time each day. On 10 January, the scientist records a mass of 150 grams. Each day, the mass of the substance decreases by 5%. Find the date on which the scientist first records a mass of less than 96 grams.
Show solution
- A 5% decrease each day means multiply by 0.95 each day. After n days the mass is 150 × 0.95ⁿ.
- Try values of n until the mass drops below 96 g.
- n = 8: 150 × 0.95⁸ = 99.5 g (still above 96).
- n = 9: 150 × 0.95⁹ = 94.5 g (below 96).
- 9 days after 10 January is 19 January
May/June 2026 · Paper 4 · 0580/43/M/J/26 · Q19(b,c)
(b) Mia invests $3000 in an account that pays compound interest at a rate of 3.2% per year. Find the number of complete years it takes for the value of Mia’s investment to become greater than $4800.
(c) Ryan invests some money. The value of the investment increases exponentially at a rate of r% per year. At the end of 6 years, the value of the original investment has increased by 19.4%. Find the value of r.
Show solution
- Compound interest at 3.2%: multiply by 1.032 each year. After n years the value is 3000 × 1.032ⁿ.
- Try values of n until the value goes over $4800.
- n = 14: 3000 × 1.032¹⁴ = $4662.70 (not enough yet).
- n = 15: 3000 × 1.032¹⁵ = $4811.90 (greater than $4800).
- 15 years
- An increase of 19.4% means the value is multiplied by 1.194 over the 6 years.
- If the yearly multiplier is m, then m⁶ = 1.194.
- Take the sixth root: m = 1.1941/6 = 1.0299…
- m = 1 + r⁄100, so r = (1.0299… − 1) × 100 = 2.99… ≈ 3
Feb/Mar 2026 · Paper 4 · 0580/42/F/M/26 · Q16
The temperature, T °C, of a liquid is given by the formula T = abˣ, where x is the number of minutes after 2 pm. T = 80 at 2 pm and T = 20 at 2.30 pm. Find the value of a and the value of b.
Show solution
- At 2 pm, x = 0. Any number to the power 0 is 1, so T = a × 1 = a.
- T = 80 at 2 pm, so a = 80
- At 2.30 pm, x = 30: 20 = 80 × b³⁰.
- Divide by 80: b³⁰ = 0.25.
- Take the 30th root: b = 0.251/30 = 0.95484…
- b = 0.955
Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q19
The mass of a radioactive substance decays exponentially at a rate of 10% per day. The initial mass is 20 g. Find the number of whole days it takes for the mass to first be less than 1 g.
Show solution
- A 10% decrease each day leaves 90%, so multiply by 0.9 each day.
- After n days the mass is 20 × 0.9ⁿ grams.
- Try values of n until the mass first drops below 1 g.
- n = 28: 20 × 0.9²⁸ = 1.047 g (still more than 1 g).
- n = 29: 20 × 0.9²⁹ = 0.942 g (now less than 1 g).
- It takes 29 days
Oct/Nov 2025 · Paper 4 · 0580/41/O/N/25 · Q22
Alex invests $200 at a rate of r% per year compound interest. At the end of 25 years the value of this investment is $301.10. Find the value of r.
Show solution
- Compound interest: final value = 200 × (multiplier)²⁵, where the multiplier is 1 + r⁄100.
- 200 × m²⁵ = 301.10
- Divide by 200: m²⁵ = 1.5055
- Take the 25th root: m = ²⁵√1.5055 = 1.01649…
- r = (1.01649… − 1) × 100 = 1.65
Oct/Nov 2025 · Paper 4 · 0580/42/O/N/25 · Q14
(a) Carlos invests $24 000 at a rate of 3.2% per year compound interest. Calculate the value of his investment at the end of 4 years.
(b) Carlos buys a painting for $x. He sells it for $40 870 and makes a profit of 34%. Calculate the value of his profit.
(c) Carlos buys a car with a value of $32 500. The value decreases exponentially by 23% each year. Find a formula for the value, $V, of the car at the end of n years.
Show solution
- A 3.2% increase each year means multiply by 1.032 each year.
- Value after 4 years = 24 000 × 1.032⁴ = 24 000 × 1.13427… = 27 222.6…
- $27 223
- A profit of 34% means the selling price is 100% + 34% = 134% of the cost.
- Cost = 40 870 ÷ 1.34 = $30 500.
- Profit = 40 870 − 30 500 = $10 370
- A 23% decrease each year means multiply by 1 − 0.23 = 0.77 each year.
- After n years the value has been multiplied by 0.77 n times.
- V = 32 500 × 0.77ⁿ
Oct/Nov 2025 · Paper 4 · 0580/43/O/N/25 · Q15
The population of a town is 54 000. The population is decreasing exponentially at a rate of 2% per year.
(a) Calculate the decrease in the population at the end of 4 years.
(b) Find the number of complete years it takes for the population to first fall below 44 000.
Show solution
- A 2% decrease each year means multiply by 0.98 each year.
- Population after 4 years = 54 000 × 0.98⁴ = 54 000 × 0.92236… = 49 807.9.
- Decrease = 54 000 − 49 807.9 = 4192.1 ≈ 4192
- Try values of n in 54 000 × 0.98ⁿ until the result is below 44 000.
- n = 10: 54 000 × 0.98¹⁰ = 44 121.9 (still above 44 000).
- n = 11: 54 000 × 0.98¹¹ = 43 239.5 (below 44 000).
- 11 years
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