What is R9: Percentages?

A percentage is a number of parts per hundred, so p%=p/100 and the multiplier for a change of p% is 1/100. Percentage change is always measured against the original value: percentage change=change/original100, so a rise from 245 to 883 is 638/245100260%, above 100% because the value more than doubled.

How many marks are R9 questions worth?

In verified past-paper appearances, R9 questions on Edexcel GCSE Maths 1MA1 have been worth 1–6 marks.

Which past exam papers has R9 appeared in?

R9 has appeared in 80 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Define percentage as 'number of parts per hundred'; interpret percentages and percentage changes as fractions/decimals, multiplicatively; percentages > 100%; percentage change and simple interest

Percentages

Worked answers, methods and verified real exam appearances for R9 on Edexcel GCSE Maths 1MA1.

Definition

Specification R9: define percentage as “number of parts per hundred”; interpret percentages and percentage changes as a fraction or a decimal, and interpret these multiplicatively; express one quantity as a percentage of another; compare two quantities using percentages; work with percentages greater than $100\%$ ; solve problems involving percentage change, including percentage increase or decrease and original value problems, and simple interest including in financial mathematics.

Explanation

A percentage is a number of parts per hundred, so $p\%=\dfrac{p}{100}$ and the multiplier for a change of $p\%$ is $1\pm\dfrac{p}{100}$ . Percentage change is always measured against the original value: $\text{percentage change}=\dfrac{\text{change}}{\text{original}}\times100$ , so a rise from $245$ to $883$ is $\dfrac{638}{245}\times100\approx260\%$ , above $100\%$ because the value more than doubled.
Percentage profit uses the cost price as the denominator. For an original value problem, the given amount is the multiplier times the original, so divide by the multiplier: a price of $\pounds46.75$ after a $15\%$ reduction was $46.75\div0.85$ .
Applying the percentage to the reduced price is the standard error and scores nothing.
Simple interest is the same amount each year, calculated from the original principal only.
Keep unrounded values until the comparison or final answer.

Worked example

In a sale the price of a machine is reduced by

12\%

. The sale price is

\pounds704

. Work out the price of the machine before the sale.

(2)

(Total for Question 1 is 2 marks)

Working

1.A reduction of $12\%$ leaves $100\%-12\%=88\%$ of the original price, so the multiplier is $0.88$ .
2.Form the equation $0.88\times\text{original}=704$ .
3.Divide by the multiplier: $\text{original}=704\div0.88=800$ .

Answer: $\pounds800$

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for $704\div0.88$ or $704\div88\times100$ or $0.88x=704$	M1	oe
	for $800$	A1	cao

Common mistakes

Don't add $12\%$ to the reduced value when reversing a $12\%$ decrease.
Don't use the new value rather than the original value as the denominator for percentage change.
Don't calculate simple interest from a growing balance as though it were compound interest.

Mark-scheme wording to use

“for a complete method to find the percentage increase … Accept answers in the range 260 to 260.5” — Pearson 1MA1, November 2018 Paper 3H, Q2
“(dep) for start to a process to find percentage profit, … oe with consistent units” — Pearson 1MA1, June 2020 Paper 1H, Q5
“M1 for a complete method eg 46.75 ÷ 0.85 oe or for a correct equation eg 0.85x = 46.75 oe … A1 accept trailing zeros eg 55.00” — Pearson 1MA1, November 2024 Paper 3H, Q10

What the examiner wants

The original value as the denominator and no early rounding. The November 2024 Paper 2H report lists the errors that cost marks: choosing the new rather than the original value as the denominator, forgetting to subtract $100$ from the percentage, and rounding $11.84$ to $12$ before comparing; the Paper 3H report adds increasing or decreasing $\pounds46.75$ by $15\%$ in a reverse percentage question.

Exam tip

For reverse percentage, write the multiplier equation and divide by the multiplier.

Worked practice

Tier 1 · Easy

Work out

35\%

240

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$84$	2	$35\%=0.35$ , so $0.35\times240=84$ .

Tier 2 · Standard

After a

12\%

decrease, a machine is valued at £704. Work out its value before the decrease.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	£800	3	A $12\%$ decrease leaves $88\%=0.88$ of the original value. Divide by the multiplier: £ $704/0.88=£800$ .

Tier 3 · Hard

A saver deposits £2500 in an account paying

3.6\%

simple interest each year. After

5

years, a fee equal to

2\%

of the final balance is charged. Calculate the amount left after the fee.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	£2891	5	The yearly simple interest is $0.036\times£2500=£90$ . Over $5$ years this is £450, giving £2950. The fee is $0.02\times£2950=£59$ , so £ $2950-£59=£2891$ remains.

Tier 1 · Easy

Work out

135\%

80

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$108$	2	$135\%=1.35$ , so $1.35\times80=108$ .

Tier 2 · Standard

The price of an item increases from £72 to £81. Work out the percentage increase.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$12.5\%$	3	The increase is £ $81-£72=£9$ . As a fraction of the original price this is $\dfrac{9}{72}=0.125$ , so the percentage increase is $12.5\%$ .

Tier 3 · Hard

An investment earns £378 in simple interest over

3

years at a rate of

4.5\%

per year. Work out the original amount invested.

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	£2800	3	Over $3$ years the simple interest is $3\times4.5\%=13.5\%$ of the original amount. Therefore the original amount is £ $378\div0.135=£2800$ .

Tier 2 · Standard

A workshop makes

840

parts in one month. This is

140\%

of the number made in the previous month. Work out the number of parts made in the previous month.

(2)

(Total for Question 7 is 2 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$600$ parts	2	$140\%=1.4$ , so $1.4\times\text{previous number}=840$ . The previous number was $840\div1.4=600$ .

Tier 3 · Hard

Nadia invests £1800 at

4\%

simple interest per year. She invests another amount at

2.5\%

simple interest per year. After

3

years, the total interest from the two investments is £396. Work out the second amount invested.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	£2400	4	The first investment earns $1800\times0.04\times3=£216$ . The second investment therefore earns £ $396-£216=£180$ . Over $3$ years its simple-interest rate is $3\times2.5\%=7.5\%$ , so the second amount is £ $180\div0.075=£2400$ .

Tier 3 · Hard

A quantity is increased by

20\%

and then decreased by

p\%

. Its final value is

108\%

of its original value. Work out

p

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$p=10$	4	The multipliers satisfy $1.20\left(1-\dfrac{p}{100}\right)=1.08$ . Hence $1-\dfrac{p}{100}=1.08\div1.20=0.9$ , so $p=10$ .

Q10

Tier 3 · Hard

A tank is initially

80\%

full. Then

15\%

of the water in it is used. After

102

litres are added, the tank is

92\%

full. Work out the capacity of the tank.

(4)

(Total for Question 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$425$ litres	4	Let the capacity be $C$ litres. After $15\%$ is used, $0.8C\times0.85=0.68C$ litres remain. Therefore $0.68C+102=0.92C$ , so $0.24C=102$ and $C=425$ litres.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2024-06	3F	Q15	3	Allowed	Foundation	QP MS
2019-11	3F	Q21	3	Allowed	Foundation	QP MS
2024-11	2F	Q17	6	Allowed	Foundation	QP MS
2023-11	3F	Q1	1	Allowed	Foundation	QP MS
2022-06	3F	Q1	1	Allowed	Foundation	QP MS
2021-11	3F	Q1	1	Allowed	Foundation	QP MS
2023-11	1H	Q8	2	Non-calculator	Higher	QP MS
2019-11	2F	Q20	4	Allowed	Foundation	QP MS
2023-11	2H	Q2	4	Allowed	Higher	QP MS
2024-11	2F	Q12	4	Allowed	Foundation	QP MS
2024-11	2F	Q25	4	Allowed	Foundation	QP MS
2019-11	2F	Q3	1	Allowed	Foundation	QP MS
2022-11	3H	Q25	5	Allowed	Higher	QP MS
2024-11	1F	Q1	1	Non-calculator	Foundation	QP MS
2023-06	3F	Q21	5	Allowed	Foundation	QP MS
2022-06	3F	Q30	2	Allowed	Foundation	QP MS
2022-06	3H	Q4	4	Allowed	Higher	QP MS
2019-06	2H	Q9	4	Allowed	Higher	QP MS
2019-06	1F	Q20	3	Non-calculator	Foundation	QP MS
2024-11	3F	Q21	5	Allowed	Foundation	QP MS
2024-11	3F	Q9	5	Allowed	Foundation	QP MS
2024-06	3F	Q1	1	Allowed	Foundation	QP MS
2021-11	1F	Q19	3	Non-calculator	Foundation	QP MS
2021-11	1F	Q1	1	Non-calculator	Foundation	QP MS
2024-06	1F	Q26	5	Non-calculator	Foundation	QP MS
2021-11	1H	Q4	4	Non-calculator	Higher	QP MS
2023-11	1F	Q15	4	Non-calculator	Foundation	QP MS
2019-11	3F	Q15	4	Allowed	Foundation	QP MS
2023-06	1F	Q14	5	Non-calculator	Foundation	QP MS
2023-06	1H	Q7	2	Non-calculator	Higher	QP MS
2024-11	2H	Q6	4	Allowed	Higher	QP MS
2024-06	1H	Q9	1	Non-calculator	Higher	QP MS
2023-06	1F	Q26	2	Non-calculator	Foundation	QP MS
2019-11	1F	Q5	1	Non-calculator	Foundation	QP MS
2024-11	1H	Q9	6	Non-calculator	Higher	QP MS
2022-11	1F	Q18	5	Non-calculator	Foundation	QP MS
2019-06	3H	Q2	3	Allowed	Higher	QP MS
2023-06	2F	Q19	5	Allowed	Foundation	QP MS
2024-06	2F	Q18	3	Allowed	Foundation	QP MS
2022-06	2F	Q5	1	Allowed	Foundation	QP MS
2024-11	3H	Q12	6	Allowed	Higher	QP MS
2019-06	3H	Q10	3	Allowed	Higher	QP MS
2019-06	1H	Q12	3	Non-calculator	Higher	QP MS
2022-11	2F	Q2	1	Allowed	Foundation	QP MS
2024-06	1F	Q27	2	Non-calculator	Foundation	QP MS
2024-06	1H	Q7	5	Non-calculator	Higher	QP MS
2022-06	2F	Q19	5	Allowed	Foundation	QP MS
2023-06	2F	Q25	3	Allowed	Foundation	QP MS
2022-06	3H	Q10	3	Allowed	Higher	QP MS
2023-06	2H	Q6	3	Allowed	Higher	QP MS
2024-11	3H	Q5	5	Allowed	Higher	QP MS
2021-11	2F	Q28	2	Allowed	Foundation	QP MS
2023-06	3F	Q2	1	Allowed	Foundation	QP MS
2024-11	3H	Q10	2	Allowed	Higher	QP MS
2023-06	2F	Q27	4	Allowed	Foundation	QP MS
2023-06	3H	Q2	5	Allowed	Higher	QP MS
2024-11	2F	Q1	1	Allowed	Foundation	QP MS
2019-06	3F	Q23	5	Allowed	Foundation	QP MS
2021-11	2F	Q1	1	Allowed	Foundation	QP MS
2021-11	2F	Q17	5	Allowed	Foundation	QP MS
2023-06	3F	Q19	3	Allowed	Foundation	QP MS
2022-06	1F	Q19	3	Non-calculator	Foundation	QP MS
2022-11	2F	Q26	2	Allowed	Foundation	QP MS
2021-11	1F	Q23	4	Non-calculator	Foundation	QP MS
2022-11	2H	Q8	3	Allowed	Higher	QP MS
2019-06	1F	Q2	1	Non-calculator	Foundation	QP MS
2024-06	1F	Q2	1	Non-calculator	Foundation	QP MS
2023-06	3H	Q9	2	Allowed	Higher	QP MS
2019-11	2F	Q16	2	Allowed	Foundation	QP MS
2023-06	1F	Q17	4	Non-calculator	Foundation	QP MS
2019-06	3F	Q5	1	Allowed	Foundation	QP MS
2023-11	3F	Q14	2	Allowed	Foundation	QP MS
2022-06	3F	Q19	4	Allowed	Foundation	QP MS
2021-11	3F	Q13	4	Allowed	Foundation	QP MS
2023-11	2F	Q23	4	Allowed	Foundation	QP MS
2023-11	1F	Q16	2	Non-calculator	Foundation	QP MS
2022-11	2F	Q11	4	Allowed	Foundation	QP MS
2023-06	1F	Q1	1	Non-calculator	Foundation	QP MS
2022-06	3F	Q25	4	Allowed	Foundation	QP MS
2024-06	2F	Q15	2	Allowed	Foundation	QP MS

Other points in R Ratio, proportion and rates of change

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