What is R16: Growth and decay?

Growth or decay by a fixed percentage each period is repeated multiplication: the multiplier is 1/100, and after n periods the amount is 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.

Definition

Specification R16: set up, solve and interpret the answers in growth and decay problems, including compound interest, and work with general iterative processes.

Explanation

Growth or decay by a fixed percentage each period is repeated multiplication: the multiplier is $1\pm\dfrac{r}{100}$ , and after $n$ periods the amount is $\text{initial}\times(\text{multiplier})^n$ .
Compound interest earns interest on the interest, so the multipliers are applied in turn and the rates are never added; simple interest, by contrast, is the same amount every year and scores at most one mark when compound was asked for.
With different rates in different years, multiply by each year's multiplier, $7000\times1.03\times1.015$ .
To find the rate or the number of years, test powers of the multiplier or use the given data to solve.
Higher tier: an iterative formula such as $P_{n+1}=1.2P_n-50$ makes each term from the previous one, so evaluate the terms in order, stopping at the term asked for, and keep full precision until the end.

Worked example

Priya invests

\pounds5200

for

2

years in an account paying compound interest. In the first year the rate of interest is

2.5\%

. In the second year the rate of interest is

4\%

. Work out the value of Priya's investment at the end of

2

years.

(3)

(Total for Question 1 is 3 marks)

Working

1.The multiplier for the first year is $1.025$ : $5200\times1.025=5330$ .
2.The multiplier for the second year is $1.04$ , applied to the new balance: $5330\times1.04$ .
3. $5330\times1.04=5543.2$ , so the investment is worth $\pounds5543.20$ .

Answer: $\pounds5543.20$

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for a correct first step, eg $5200\times1.025$ (= 5330) or $5200\times2.5\div100$ (= 130), or for the multiplier for both years, eg $1.025\times1.04$ (= 1.066)	M1	oe
	for a compound method, eg $5200\times1.025\times1.04$ or “5330” $\times1.04$	M1	oe
	for $5543.20$	A1	cao; accept 5543.2

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.

Mark-scheme wording to use

“for a correct first step eg working out increase for one year … or find the multiplier for both years” — Pearson 1MA1, November 2022 Paper 2H, Q6
“M1 works out decrease for one year eg 679 × 4 ÷ 100 (= 27.16) oe or 679 × (100 − 4) ÷ 100 (= 651.84) oe … M1 for compound method, eg 679 × “0.96”^t, t ≥ 2 … A1 for 600.71 or 600.72 or 600.73 or 600.74” — Pearson 1MA1, June 2022 Paper 2H, Q6
“P1 uses formula eg 1.2 × 200 – 50 (= 190) P1 for complete process, eg May: 1.2 × “190” – 50 (= 178) and June: 1.2 × “178” – 50 (= 163.6) A1 for 163 or 164” — Pearson 1MA1, November 2019 Paper 2H, Q22(a)

What the examiner wants

Multipliers, not added percentages. The November 2024 Paper 3H report says the most successful solutions used percentage multipliers, while common errors were adding the percentages together and raising that multiplier to the power $3$ , working with simple interest for a maximum of one mark, and using $1.006$ instead of $1.06$ .

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

Want help turning this into marks?

Bring R16 or any tricky specification point, and we can work through the method and exam wording together.

Book a free intro call