Launching your learning experience…
Cambridge IGCSE Mathematics 0580 · Extended · Syllabus 2025-2027
Exam-style worked examples on compound interest, depreciation, population growth and decay — including reverse problems and the trial-and-improvement question that asks for a number of years. Diagram questions cover reading a decay graph, finding a growth model from two points, and a bouncing ball.
This page has 11 exam-style IGCSE Maths 0580 exponential growth and decay worked examples — 0 Core, 8 Extended and 3 Challenge, 3 with diagrams. Each one shows the full method step by step, the final answer in the form a marker expects, and an examiner tip on where marks are commonly lost.
$2000 is invested at a compound interest rate of per year.
Calculate the value of the investment after years.
Use a single multiplier. Growing by means multiplying by each year.
Substitute the values.
Evaluate.
Answer
$2315.25
Examiner tip
Use the power directly rather than working year by year — it is faster and avoids rounding errors accumulating. Note the multiplier is , not .
$2000 is invested for years at per year.
Calculate how much more interest is earned under compound interest than under simple interest.
Simple interest is the same amount each year, calculated on the original sum only.
Simple interest total value.
Compound interest total value.
Find the difference.
Answer
$15.25 more
Examiner tip
Compound always beats simple over more than one period, because later interest is earned on earlier interest. If your compound figure comes out lower, you have used the wrong multiplier.
A car is bought for $18,000. Its value depreciates by each year.
Calculate its value after years. Give your answer correct to the nearest dollar.
Decreasing by means keeping , so the multiplier is .
Evaluate the power.
Multiply.
Round to the nearest dollar.
Answer
$9396
Examiner tip
For decay the multiplier is , so — not . Multiplying by gives the amount lost in a single year, which is a different quantity entirely.
An investment grows at compound interest per year. After years it is worth $2315.25.
Calculate the amount originally invested.
Write the growth relationship.
Substitute the known final value.
Divide to reverse the multiplier.
Answer
$2000
Examiner tip
Divide by the multiplier — do not subtract three times. Check by working forwards: . ✓
A population of increases by each year.
Find the smallest number of complete years after which the population first exceeds .
Write the growth relationship.
Try values of until the population passes .
is still below , so try the next year.
, and was below, so is the first year to exceed .
Answer
years
Examiner tip
Show both trials — the one that fails and the one that succeeds. Showing only does not prove it is the smallest number of years.
A radioactive sample has mass . Its mass halves every years.
(a) Find its mass after years. (b) Explain why the mass never reaches exactly zero.
(a) Find how many halvings occur in years.
Apply the multiplier three times.
(b) Each step multiplies by , which always leaves a positive amount. Halving a positive number can never produce zero.
Answer
(a) (b) each halving multiplies by , which always leaves a positive mass — it approaches zero but never reaches it
Examiner tip
Count the number of periods, not years — years is halvings, not . Part (b) tests understanding that exponential decay is asymptotic; a sentence naming the multiplier earns the mark.
The graph shows the amount of caffeine, in milligrams, in a person's body during the hours after drinking a strong coffee.
(a) Use the graph to find
(i) the amount of caffeine after hours,
(ii) the time it takes for the amount to fall to mg.
(b) Use the graph to find how long it takes for the amount to fall from mg to mg. Comment on your answer.
(c) The amount decreases by the same percentage each hour. Use the graph to find this percentage.
(a)(i) Go up from hours to the curve and across to the vertical axis. The curve is just above the line.
(a)(ii) Go across from mg to the curve and down to the time axis. It meets it a little after hours.
(b) Read the time for mg in the same way, then subtract the time for mg.
This is the same as the time taken to fall from mg to mg. For exponential decay the amount halves in equal time periods, whatever amount you start from.
(c) Read an accurate pair of values one hour apart: at hours the amount is mg and at hour it is mg. Find the percentage decrease.
Answer
(a)(i) about mg ( to ) (a)(ii) about hours (b) about hours — the same as the time to halve from mg, because the amount halves in equal intervals (c)
Examiner tip
Graph readings earn the mark within a tolerance, so draw your reading lines and read to the nearest small square. In (c), choose a point that is easy to read accurately; gives the multiplier, and the percentage decrease is , not .
The number of followers, , of a new online account weeks after it started is modelled by
The graph of this model passes through the points and , as shown.
(a) Find the value of .
(b) Find the value of , and explain what represents on the graph.
(c) Write down the percentage increase in followers each week.
(d) Use the model to find the number of followers after weeks.
(a) Substitute each point into the model to get two equations.
Divide the second equation by the first: cancels, and .
The number of followers is increasing, so take the positive root.
(b) Substitute into the equation for .
When , , so . So is where the curve crosses the -axis: the number of followers at the start. The graph confirms this — the curve starts at about .
(c) The multiplier means each week's total is of the week before.
(d) Substitute .
Answer
(a) (b) , the -intercept (the number of followers at the start) (c) (d)
Examiner tip
The efficient method divides one equation by the other so that disappears. A common error is to find from the difference , which treats the growth as linear. In (c), the percentage increase is , not or .
A ball is dropped from a height of m. After each bounce it rises to of the height it fell from, as shown in the diagram.
(a) Find , the height reached after the third bounce.
(b) Show that the height reached after the th bounce is metres.
(c) Find the number of the first bounce after which the ball rises to less than cm.
(d) Find the total distance the ball travels from when it is dropped until it hits the ground for the fourth time.
(a) Each height is times the one before. The diagram gives m and m, so multiply once more.
(b) Starting from m, every bounce multiplies the height by . After bounces the multiplier has been used times.
Check against the diagram: gives m and gives m. ✓
(c) Work in metres: cm m. Try values of until the height first drops below m.
Bounce is still above cm, so try bounce .
(d) The ball falls m before the first hit. Between the first and fourth hits it makes three bounces, and on each it travels up and back down.
Evaluate.
Answer
(a) m (b) the height is multiplied by at each of the bounces, giving (c) the th bounce (d) m
Examiner tip
In (c), the units must match — comparing in metres with is a common slip. Show the trial either side of the boundary ( and ). In (d), each bounce is travelled twice (up and down), but the first m drop only once; forgetting either of these is a common way to lose marks.
A bowl of soup is left to cool. Its temperature, C, after minutes is modelled by
When , . When , .
(a) Write down the value of .
(b) Find the value of , correct to significant figures.
(c) Use the model to find the temperature of the soup after minutes.
(a) Any number to the power is , so at the model gives .
(b) Substitute , and make the subject.
Undo the power by taking the sixth root, not by dividing by .
(c) Substitute , using the unrounded value of stored in the calculator.
Answer
(a) (b) (c) C
Examiner tip
The March 2026 examiner report noted that "Some candidates gave the 2 significant figure value 0.95 for b rather than 3 significant figures" — the in a model is often close to , so read the accuracy asked for. In (c), keep the full value of : using gives C. The paper-wide advice from the June 2025 report applies: "intermediate values within longer methods should not be rounded, with only the final answer rounded to the appropriate degree of accuracy."
An investment earns compound interest at per year. The value of the investment trebles in exactly years.
(a) Find the value of . Give your answer correct to significant figures.
(b) Maya invests $2500 at this rate of interest. Calculate the value of her investment after years. Give your answer correct to the nearest cent.
(a) Let the multiplier be . Trebling in years means multiplications give times the start value, whatever that value is.
Undo the power with the 14th root.
Convert the multiplier to a percentage rate.
(b) Use the unrounded multiplier for years.
Round to the nearest cent.
Answer
(a) (b) $12 009.97
Examiner tip
On finding a rate from a doubling time, the March 2025 examiner report found that "Having set up a correct starting equation some struggled to rearrange it correctly." and that "Some opted to use a trial-and-error approach to find the rate of interest, but this often led to an inaccurate answer." The June 2025 report noted that "Others divided by 17 instead of taking the 17th root." — here, is wrong; take the 14th root. Using the rounded rate in (b) gives $12 002.55, which is not accurate to the cent.
These worked examples are written by Tutopiya to cover the same syllabus content as the official Cambridge IGCSE 0580 specification (2025-2027). They are exam-style questions, not reproductions of past paper questions.