Worked example 1 · Easy
Worked method
An account has £500 and increases by 6% each year. Find its value after two years.
\(k=1.06\). So \(A=500(1.06)^2=500(1.1236)=\boxed{£561.80}\).
Chapter 17
Exponential growth and decay
Repeated changes by constant multipliers. Original explanations and examination-focused practice based on the supplied textbook's topic sequence.
By the end of this chapter, you should be able to
- Explain the key ideas and use correct mathematical terminology in each textbook subtopic.
- Use the appropriate methods and show enough working for examination marks.
- Check intermediate arithmetic, signs, units and the final answer.
1
Subtopic 1
Exponential growth
Exponential growth occurs when a value increases by the same percentage each period. The amount added grows as the starting amount grows; it is not a fixed increase.
For a percentage increase of \(r\%\), first find the growth multiplier \(k=1+\frac{r}{100}\). The value after \(n\) periods is \(A=P\times k^n\), where \(P\) is the original value.
Use the correct unit for the exponent: if the population doubles every hour, \(n\) must be hours; if it rises per day, \(n\) must be days. At \(n=0\), the starting value is \(P\).
Rules and important ideas
- \(A=P(1+\frac r{100})^n\)
- \(k>1\) means growth; doubling has \(k=2\).
Worked example 2 · Medium
Worked method
A culture starts with 240 cells and triples every two hours. How many after eight hours?
There are \(8\div2=4\) two-hour periods. \(A=240\times3^4=240\times81=\boxed{19\,440}\) cells.
Worked example 3 · Exam-style
Worked method
A town has 18 000 people and grows by 2.5% each year. Write a model and estimate the population after five years.
\(P(t)=18000(1.025)^t\). Then \(P(5)=18000(1.025)^5\approx\boxed{20\,365}\) people, to the nearest whole person.
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
2
Subtopic 2
Exponential decay
Exponential decay is a repeated percentage decrease. A decline of \(r\%\) leaves \((100-r)\%\) of the current value at the end of every period.
The decay multiplier is \(k=1-\frac r{100}\), with \(0<k<1\). Use \(A=P\times k^n\). Depreciation, declining populations and radioactive material can be modelled this way.
Watch the wording: a reduction of 15% each year is not the same as subtracting 15% of the original value every year. Keep calculator precision until the final rounding.
Rules and important ideas
- \(A=P(1-\frac r{100})^n\)
- A 20% decrease gives multiplier \(0.80\), not \(0.20\).
Worked example 4 · Easy
Worked method
A machine worth £5000 depreciates by 10% per year. Find its value after one year.
\(k=1-0.10=0.90\). So \(5000(0.9)=\boxed{£4500}\).
Worked example 5 · Medium
Worked method
A chemical sample of 80 g loses 12% of its mass each year. Find its mass after 4 years.
\(A=80(0.88)^4=80(0.59969536)=\boxed{47.98\text{ g}}\), to 2 decimal places.
Worked example 6 · Exam-style
Worked method
A car was bought for £14 500 and depreciates by 18% annually. Write a model and find its value after 5 years to the nearest pound.
\(V(t)=14500(0.82)^t\). At \(t=5\), \(V\approx5375.73\), so \(\boxed{£5376}\).
Exam technique
Identify what is given, write the correct rule, show the substitution or algebraic steps, and check the final result before moving on.
Chapter 17 practice
These are newly written practice questions, not past-paper reproductions. Questions progress from confidence-building to exam-style. Answers are hidden until you open each question’s Show Answer section.
Exponential growth
1Confidence practice[2 marks]
A price of £80 increases by 10% once. Find the new price.
Show Answer
Step-by-step solution: \(80\times1.10=\boxed{£88}\).
2Confidence practice[2 marks]
A quantity grows by 7% each month. State the multiplier.
Show Answer
Step-by-step solution: \(1+7/100=\boxed{1.07}\).
3Developing practice[3 marks]
A £1250 investment grows by 4% each year. Find the value after three years.
Show Answer
Step-by-step solution: \(1250(1.04)^3=1250(1.124864)=\boxed{£1406.08}\).
4Developing practice[3 marks]
There are 90 bacteria, doubling every 30 minutes. How many after 2 hours?
Show Answer
Step-by-step solution: \(120\div30=4\) periods; \(90\times2^4=\boxed{1440}\).
5Exam-style practice[4 marks]
A population is \(N=800(1.08)^t\), where \(t\) is years. State the initial population and annual percentage change, then calculate \(N\) at \(t=4\).
Show Answer
Step-by-step solution: At \(t=0\), \(N=\boxed{800}\). Multiplier 1.08 means \(\boxed{8\%\text{ growth}}\). \(N(4)=800(1.08)^4\approx\boxed{1088}\).
6Exam-style practice[4 marks]
A rare book is worth £1600 and rises 12% each year for 6 years. Find its final value to the nearest pound.
Show Answer
Step-by-step solution: \(A=1600(1.12)^6\approx3158.116\). To the nearest pound, \(\boxed{£3158}\).
Exponential decay
7Confidence practice[2 marks]
A toy costing £60 falls in value by 25%. Find its new value.
Show Answer
Step-by-step solution: \(60(1-0.25)=60(0.75)=\boxed{£45}\).
8Confidence practice[2 marks]
State the multiplier for a 6% yearly decrease.
Show Answer
Step-by-step solution: \(1-0.06=\boxed{0.94}\).
9Developing practice[3 marks]
A laptop worth £900 loses 15% of its value each year. Find its value after 2 years.
Show Answer
Step-by-step solution: \(900(0.85)^2=900(0.7225)=\boxed{£650.25}\).
10Developing practice[3 marks]
A mass starts at 200 g and halves every hour. Find the mass after 6 hours.
Show Answer
Step-by-step solution: \(200(0.5)^6=200/64=\boxed{3.125\text{ g}}\).
11Exam-style practice[4 marks]
The number of fish in a lake is modelled by \(F=2400(0.97)^t\), with \(t\) in years. Find the annual percentage decrease and estimate \(F\) after 8 years.
Show Answer
Step-by-step solution: \(1-0.97=0.03\), so decrease is \(\boxed{3\%}\). \(2400(0.97)^8\approx\boxed{1881}\) fish.
12Exam-style practice[4 marks]
A motorbike initially costs £7200 and depreciates at 9% a year for 4 years. Find the total decrease in value, to the nearest pound.
Show Answer
Step-by-step solution: Value after 4 years: \(7200(0.91)^4\approx4937.40\). Decrease \(=7200-4937.40\approx\boxed{£2263}\).
High-grade revision checklist
- I can explain each rule in the exact textbook topic order.
- I can select and apply the correct method, including signs, powers and brackets.
- I can show clear working in harder multi-step questions.
- I can check by reversing the operation, estimating, or substituting a value, as appropriate.