What is R16: Growth and decay?

Repeated percentage growth by r% uses multiplier 1+r/100 each period; repeated decay uses 1-r/100. After n equal periods, final=initial×(multiplier)^n.

How many marks are R16 questions worth?

In verified past-paper appearances, R16 questions on Edexcel GCSE Maths 1MA1 have been worth 2–6 marks.

Which past exam papers has R16 appeared in?

R16 has appeared in 24 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

R16

Set up, solve and interpret the answers in growth and decay problems, including compound interest and work with general iterative processes

Growth and decay

Worked answers, methods and verified real exam appearances for R16 on Edexcel GCSE Maths 1MA1.

Explanation

Repeated percentage growth by $r\%$ uses multiplier $1+\dfrac{r}{100}$ each period; repeated decay uses $1-\dfrac{r}{100}$ . After $n$ equal periods, $\text{final}=\text{initial}\times(\text{multiplier})^n$ .
Compound interest is repeated growth because each period acts on the latest balance.
Higher tier: an iterative process defines each new term from the previous term, so calculate successive values in order and identify the first one satisfying the condition.
Keep full calculator precision until the requested final rounding.
Examiners expect the multiplier, exponent or iteration trail, followed by an interpretation in context.

Worked example

Higher tier: A cooling model is $T_{n+1}=0.65T_n+12$ with $T_0=80$ . Find the first $n$ for which $T_n<40$ .

1. $T_1=64$ , $T_2=53.6$ and $T_3=46.84$ .
2. $T_4=42.446$ and $T_5=39.5899$ .
3. $T_4$ is not below $40$ but $T_5$ is, so the first value is $n=5$ .

Answer: $n=5$ , with $T_5=39.6$ to one decimal place.

Common mistakes

Don't calculate repeated percentage change as simple change from the original amount.
Don't use $1.15$ for a $15\%$ decay instead of $0.85$ .
Don't round each intermediate iteration and change the first term meeting the condition.

Exam tip

For “first time” questions, show the last value that fails and the next value that meets the condition.

Worked practice

Tier 1 · Easy

£600 is invested at

4\%

compound interest per year. Work out the value after

2

years.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	£648.96	2	Use multiplier $1.04$ twice: £ $600\times1.04^2=£648.96$ .

Tier 2 · Standard

A culture initially contains

960

cells and decreases by

15\%

each hour. Calculate the expected number after

3

hours.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	About $590$ cells Unrounded model value $=589.56$	3	The decay multiplier is $0.85$ . After $3$ hours the model gives $960\times0.85^3=589.56$ , which is about $590$ whole cells.

Tier 1 · Easy

Write down the multiplier for an increase of

6.5\%

(1)

(Total for Question 3 is 1 mark)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$1.065$	1	An increase of $6.5\%$ uses multiplier $1+0.065=1.065$ .

Tier 2 · Standard

An account pays

5\%

compound interest each year, and the interest earned in the second year is £31.50. Work out the amount originally invested.

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	£600	3	The balance before the second-year interest was added is £ $31.50\div0.05=£630$ . This is $105\%$ of the original amount, so the original investment was £ $630\div1.05=£600$ .

Tier 3 · Hard

A forestry model starts with

4096

young trees. Model S increases the number each year by

12.5\%

of the original number. Model C increases the previous year's number by

12.5\%

each year. Work out how many more young trees Model C predicts than Model S after

3

years.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$200$ more young trees	3	Model S adds $4096\times0.125=512$ trees each year, so after $3$ years it predicts $4096+3\times512=5632$ . Model C predicts $4096\times1.125^3=5832$ . The difference is $5832-5632=200$ young trees.

Tier 2 · Standard

A warehouse starts with

6400

filters. It dispatches

15\%

of the current stock in the first week and

10\%

of the remaining stock in the second week. Work out the number of filters left.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$4896$ filters	3	After the first week, $6400\times0.85=5440$ filters remain. After the second week, $5440\times0.90=4896$ filters remain.

Tier 3 · Hard

3000

is invested. The value grows by

4\%

in year one, then by

1.5\%

in each of the following two years. Work out the overall percentage increase across the three years, giving the percentage correct to

1

decimal place.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$7.1\%$	4	The overall multiplier is $1.04\times1.015^2=1.04\times1.030225=1.071434\ldots$ , so the value is multiplied by $1.0714\ldots$ overall. The overall increase is $7.1434\ldots\%$ , which is $7.1\%$ correct to $1$ decimal place. (Equivalently, £ $3000$ grows to £ $3214.30$ , an increase of £ $214.30$ , and $214.30\div3000=7.14\ldots\%$ .)

Tier 3 · Hard

A ball is dropped from a height of

10\,\text{m}

. After each impact, its greatest rebound height is

72\%

of the preceding greatest height. Work out the total vertical distance travelled from release until the ball hits the floor after its third rebound. Give the distance to the nearest centimetre.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$42.23\,\text{m}$	5	The first three rebound heights are $10\times0.72=7.2$ , $7.2\times0.72=5.184$ and $5.184\times0.72=3.73248$ metres. Each rebound height is travelled upwards and downwards, so the total distance is $10+2(7.2+5.184+3.73248)=42.23296\,\text{m}$ . This rounds to $42.23\,\text{m}$ .

Tier 3 · Hard

Higher tier: A tank initially contains

500

litres. At the start of each week,

80

litres are added. During that week,

20\%

of the resulting amount is used. Let

S_n

be the amount left after

n

weeks. Write an iterative formula for

S_{n+1}

in terms of

S_n

, and work out the amount left after

4

weeks.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$S_{n+1}=0.8(S_n+80)$ , with $S_0=500$ $393.728$ litres	5	Adding $80$ litres and then keeping $80\%$ gives $S_{n+1}=0.8(S_n+80)$ . Starting with $S_0=500$ , the values are $S_1=464$ , $S_2=435.2$ , $S_3=412.16$ and $S_4=393.728$ . Therefore $393.728$ litres remain after $4$ weeks.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2019-06	3F	Q25	3	Allowed	Foundation	QP MS
2022-11	2H	Q10	3	Allowed	Higher	QP MS
2022-06	2H	Q6	3	Allowed	Higher	QP MS
2022-11	2H	Q6	3	Allowed	Higher	QP MS
2022-11	2H	Q12	3	Allowed	Higher	QP MS
2024-06	3H	Q4	4	Allowed	Higher	QP MS
2023-06	2H	Q8	4	Allowed	Higher	QP MS
2024-06	3H	Q14	4	Allowed	Higher	QP MS
2021-11	3H	Q19	3	Allowed	Higher	QP MS
2024-11	3F	Q26	3	Allowed	Foundation	QP MS
2019-11	2H	Q13	3	Allowed	Higher	QP MS
2019-06	3H	Q2	3	Allowed	Higher	QP MS
2019-11	2H	Q22	4	Allowed	Higher	QP MS
2024-11	3H	Q12	6	Allowed	Higher	QP MS
2022-11	2F	Q23	3	Allowed	Foundation	QP MS
2022-06	3H	Q10	3	Allowed	Higher	QP MS
2023-06	2F	Q27	4	Allowed	Foundation	QP MS
2023-11	1H	Q16	4	Non-calculator	Higher	QP MS
2021-11	2H	Q10	4	Allowed	Higher	QP MS
2022-06	2F	Q26	3	Allowed	Foundation	QP MS
2023-11	2H	Q3	2	Allowed	Higher	QP MS
2022-06	3H	Q20	4	Allowed	Higher	QP MS
2024-06	3F	Q25	4	Allowed	Foundation	QP MS
2023-11	2F	Q24	2	Allowed	Foundation	QP MS

Other points in R Ratio, proportion and rates of change

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