Percentages

Topic

Higher June 2025 Paper 1 Q13

EdexcelCurrent spec3 marksPercentages

13 Charlotte bought an apartment for $750 000

In the first year, the value of the apartment decreased by 4%
In the second year, the value of the apartment decreased by 6.5%

In the third year, the value of the apartment increased by $x$%

At the end of the third year, the value of Charlotte’s apartment was $698 445

Work out the value of $x$

(3)

Mark scheme

Scheme	Marks
750 000 × (1 – 0.04) (= 720 000) oe and “720 000” × (1 – 0.065) (= 673 200) or 750 000 × (1 – 0.04) × (1 – 0.065) (= 673 200) or 750 000 × 0.96 × 0.935 (= 673 200)	M1
eg $\dfrac{698\,445 - \text{``}{673\,200}\text{''}}{\text{``}{673\,200}\text{''}}(\times 100)$ (= 0.0375) or $\dfrac{698\,445}{\text{``}{673\,200}\text{''}} - 1$ (= 0.0375) or 1.0375 – 1 (= 0.0375) oe or $\dfrac{698\,445}{\text{``}{673\,200}\text{''}} \times 100 \;(= 103.75)$	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 3.75	A1
	(3)
	(3 marks)

Notes

M1: (NB: accept $\left(1 - \dfrac{4}{100}\right)$ for 0.96 but not (1 – 4%) and accept $\left(1 - \dfrac{6.5}{100}\right)$ for 0.935 but not (1 – 6.5%))

Calculations may be seen as part of an equation eg $750\,000 \times 0.96 \times 0.935 \times \left(1 + \dfrac{x}{100}\right) = 698\,445$

1.0375 or 25245 imply M1

M1: for a method to reach value one step away from $x$ ie a method leading to 0.0375 or 103.75

A1: oe eg $3\dfrac{3}{4}$ or $\dfrac{15}{4}$

Higher June 2025 Paper 2R Q12

EdexcelCurrent spec3 marksPercentages

12 Osman buys a car for $16 000

The car depreciates at a rate of 12% each year for the first 2 years.
In the third year, the car depreciates at a rate of $x$%

At the end of 3 years, the value of the car is $11 461.12

Work out the value of $x$

(3)

Mark scheme

Scheme	Marks
$16\,000 \times \left(1 - \dfrac{12}{100}\right)^2$ (= 12 390.4) or 16 000 × 0.7744 (= 12 390.4) oe or $16\,000 \times \left(1 - \dfrac{12}{100}\right)$ (= 14 080) and $\text{``}{14\,080}\text{''} \times \left(1 - \dfrac{12}{100}\right)$ (=12 390.4) oe or $\left(1 - \dfrac{12}{100}\right)^2$ (= 0.7744) and $\dfrac{11461.12}{16000}\left(= \dfrac{4477}{6250} = 0.71632\right)$	M1
eg $\dfrac{\text{``}{12390.4}\text{''} - 11461.12}{\text{``}{12390.4}\text{''}}\;(\times 100)\;(= 0.075)$ or $1 - \dfrac{11461.12}{\text{``}{12390.4}\text{''}}\;(\times 100)$ or 1 – 0.925 (= 0.075) or $\dfrac{11461.12}{\text{``}{12390.4}\text{''}}\;(\times 100)\;(= 0.925)$ or $\dfrac{\text{``}{0.71632}\text{''}}{\text{``}{0.7744}\text{''}}\;(\times 100)\;(= 0.925)$	M1
Correct answer only scores full marks (unless from obviously incorrect working) Answer: 7.5	A1
	(3)
	(3 marks)

Notes

M1: for a method to find the value of the car after two years

or
a method to find the overall percentage multiplier after two years and the overall percentage multiplier for the three years

May be seen embedded, eg in an equation

Do not allow (1 – 12%) unless processed correctly

M1: for a complete method to find the value of the decimal equivalent of $x$%
or
for a complete method to find the percentage multiplier for the third year

May be seen embedded, eg within a correct equation rearranged to one of these equivalent forms

A1: oe

SCB2 for answer –7.5

Higher June 2025 Paper 1R Q11

EdexcelCurrent spec5 marksPercentages Ratio & Basic Proportion

11 Zhou invests some money for 2 years.
He invests $2500 with Bank A and $3000 with Bank B.

Bank A

Invests $2500

amount of money invested : total amount of interest after 2 years = 20 : 3

Bank B

Invests $3000

4% per year compound interest for 2 years

Zhou receives more interest from Bank A than from Bank B.

How much more?

(5)

Mark scheme

Scheme	Marks
2500 ÷ 20 × 3 (= 375) oe or 125 × 3 (=375) or 7500 ÷ 20 (= 375) 3000 ÷ 20 × 3 (= 450) oe or 150 × 3 (= 450) or 9000 ÷ 20 (= 450)	M1
for 0.04 × 3000 (= 120) or 0.04 × 2500 oe (= 100) or 1.04 × 3000 oe (= 3120) or 1.04 × 2500 oe (= 2600) or 3000 × 1.04² (= 3244.8) or 2500 × 1.04² (=2704)	M1
1.04 × “3120” oe (= 3244.8) 1.04 × “2600” oe (= 2704)	M1
eg “3244.8” – 3000 (= 244.8) or “2704” – 2500 (= 204)	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 130.2(0)	A1
	(5)
	(5 marks)

Notes

M1: for a method to find the interest for Bank A
2875 or 3450 implies this method mark

M1: for finding 4% or 104% of 3000 or 2500

M1: for completing method to find the total amount for Bank B

M1: for a complete method to find the interest for Bank B

OR M2 for 3000 × 1.04² (= 3244.8) or 2500 × 1.04² (=2704) or 3000 × 1.04³ (= 3374.59) or 2500 × 1.04³ (= 2812.16)

SC: if none of the 2^nd or 3^rd M marks gained award SCM1 for
0.08 × 3000 oe or 240 or 1.08 × 3000 or 3240 or 0.08 × 2500 oe or 200 or 1.08 × 2500 or 2700 or 3000 × (1 – 0.04)² (= 2764.8(0)) or 2500 × (1 – 0.04)² (= 2304)

accept (1 + 0.04) or $\left(1 + \dfrac{4}{100}\right)$ as equivalent to 1.04 throughout

Higher June 2025 Paper 2R Q9

EdexcelCurrent spec3 marksPercentages

9 In a sale, normal prices are reduced by 15%

The sale price of a dishwasher is 612 Swiss francs.

Work out the normal price of the dishwasher.

(3)

Mark scheme

Scheme	Marks
1 – 0.15 (= 0.85) or 100(%) – 15(%) (= 85(%)) or $\dfrac{612}{85}$ (= 7.2) oe	M1
612 ÷ “0.85” oe or 612 ÷ “85” × 100 oe or “7.2” × 100	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 720	A1
	(3)
	(3 marks)

Notes

M1: may be seen embedded

Do not allow (1 – 15%) unless processed correctly

M1: for a complete method

Higher June 2025 Paper 2 Q7

EdexcelCurrent spec3 marksPercentages

7 Tim buys a bracelet for 4000 Swiss francs.
The value of the bracelet increases by 7% each year.

Work out the value of the bracelet at the end of 3 years.
Give your answer correct to the nearest Swiss franc.

(3)

Mark scheme

Scheme	Marks
0.07 × 4000 (= 280) or 1.07 × 4000 ( = 4280)	M1
4000 + “280” (= 4280) oe and 0.07 × “4280” ( = 299.6) and “4280” + “299.6” (= 4579.6) and 0.07 × “4579.6” (= 320.572) or “280” + “299.6” + “320.572” (= 900(.172))	M1
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 4900	A1
	(3)
	(3 marks)

Notes

M1: for finding 7% of 4000 or 107% of 4000

A1: allow answers in the range 4900 – 4901

If no other mark awarded, SCB1 for
4000 × 0.07 × 3 (= 840) or
4000 × 0.21 (= 840) or
4000 + 4000 × 0.07 × 3 (= 4840) or
4000 × 1.21 (= 4840) or
0.93 × 4000 (= 3720) or
0.79 × 4000 (= 3160) or
$0.93^3 \times 4000$ (= 3217…) or
$4000 \times 1.07^2$ (= 4579…)

M2 for $1.07^3 \times 4000$ or $1.07^4 \times 4000$ ( = 5243….)

Higher June 2025 Paper 1 Q5

EdexcelCurrent spec3 marksPercentages

5 In a sale, normal prices are reduced by 28%
The sale price of a watch is 198 euros.

Work out the normal price of the watch.

(3)

Mark scheme

Scheme	Marks
eg 1 − 0.28 (= 0.72) oe or 0.72$x$ = 198 or 100(%) − 28(%) (= 72(%)) or $\dfrac{198}{72}$ (= 2.75) oe	M1
eg ($x$ = ) 198 ÷ “0.72” oe or 198 ÷ “72” × 100 oe or “2.75” × 100	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 275	A1
	(3)
	(3 marks)

Notes

M1: for a correct first step

M1: for a complete method

A1: cao

Higher June 2025 Paper 2R Q4

EdexcelCurrent spec5 marksPercentages

4 Joshua is going to cover a floor with tiles for a customer.
The area of the floor is 45 m²

Joshua buys one box of tiles for each 1.5 m² of floor area.
Each box of tiles costs £64

Joshua also buys 5 bags of tile adhesive.
Each bag of tile adhesive costs £12

Joshua charges the customer £3000

Work out his percentage profit.
Give your answer correct to one decimal place.

(5)

Mark scheme

Scheme	Marks
45 ÷ 1.5 (= 30) or 5 × 12 (= 60) or $\dfrac{64}{1.5}\left(= \dfrac{128}{3} = 42.6(6\ldots)\right)$	M1
“30” × 64 (= 1920) or “42.6(6…)” × 45 (=1920)	M1
“1920” + “60” (= 1980) or 3000 – 1920 – 60 (=1020)	M1
eg $\dfrac{3000 - \text{``}{1980}\text{''}}{\text{``}{1980}\text{''}}$ (= 0.515...) or $\dfrac{3000 - \text{``}{1980}\text{''}}{\text{``}{1980}\text{''}} \times 100$ or $\dfrac{3000}{\text{``}{1980}\text{''}}$ (= 1.515…) or $\dfrac{3000}{\text{``}{1980}\text{''}} \times 100$ (= 151.5…) or $\dfrac{3000}{\text{``}{1980}\text{''}} \times 100 - 100$	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 51.5	A1
	(5)
	(5 marks)

Notes

M1: for a method to find the number of boxes needed or the cost of adhesive or cost of tiles per m²

M1: for a method to find the cost of the boxes of tiles

M1: for a method to find the total cost or the profit

M1: for a method to find the percentage profit or be one step away

A1: awrt 51.5

SCB3 for answer 56.3 or 56.25 (from use of 1920 instead of 1980 as total cost)

Higher June 2025 Paper 1 Q2

EdexcelCurrent spec6 marksPercentages Upper & Lower Bounds

2 Anna makes cups.
Each cup costs 6 Swiss francs to make.

Anna puts the cups into boxes to sell.
Each box contains 4 cups.

Anna sells 80 boxes of cups for a total of 2160 Swiss francs.

(a) Work out the percentage profit Anna makes.
Show your working clearly. (4)

The height of each cup is 9 cm, correct to the nearest cm

(b) Write down the lower bound of the height. (1)

The weight of each cup is 120 g, correct to the nearest 10 g

Mark scheme (a)

Scheme	Marks
4 × 6 × 80 (= 1920) or 4 × 6 (= 24) or 2160 ÷ 80 (= 27) or 2160 ÷ (80 × 4) (= 6.75)	M1
2160 – “1920” (= 240) or $\dfrac{2160}{\text{``}{1920}\text{''}}(= 1.125)$ or “27” – “24” (= 3) or $\dfrac{\text{``}{27}\text{''}}{\text{``}{24}\text{''}}(= 1.125)$ or “6.75” – 6 (= 0.75) or $\dfrac{\text{``}{6.75}\text{''}}{6}(= 1.125)$	M1
$\dfrac{\text{``}{240}\text{''}}{\text{``}{1920}\text{''}}(\times 100)$ or 0.125 (× 100) or $\left(\dfrac{2160}{\text{``}{1920}\text{''}} - 1\right)(\times 100)$ or (“1.125” – 1) (× 100) or “1.125” × 100 (= 112.5) or $\dfrac{\text{``}{3}\text{''}}{\text{``}{24}\text{''}}(\times 100)$ or 0.125 (× 100) or $\left(\dfrac{\text{``}{27}\text{''}}{\text{``}{24}\text{''}} - 1\right)(\times 100)$ or (“1.125” – 1) (× 100) or “1.125” × 100 (= 112.5) or $\dfrac{\text{``}{0.75}\text{''}}{6}(\times 100)$ or 0.125 (× 100) or $\left(\dfrac{\text{``}{6.75}\text{''}}{6} - 1\right)(\times 100)$ or (“1.125” – 1) (× 100) or “1.125” × 100 (= 112.5)	M1
Working required Answer: 12.5	A1
	(4)

Notes

M1: for method to work out total income or income from one box or expenditure for one box or income per cup

M1: for working out the profit or income ÷ expenditure

M1: for a method to reach one step from the answer ie getting to $\dfrac{1}{8}$ oe or 0.125 or 112.5

A1: dep on M1

Scheme	Marks
8.5	B1
	(1)

Scheme	Marks
125	B1
	(1)
	(6 marks)

Higher November 2024 Paper 2 Q9

EdexcelCurrent spec3 marksPercentages

9 Nisha invests 20 000 euros for 3 years in a savings account.
She gets 3.5% per year compound interest.

Work out how much money Nisha will have in her savings account at the end of the 3 years.
Give your answer correct to the nearest euro.

(3)

Mark scheme

Scheme	Marks
20 000 × 1.035 (= 20 700) oe or 20 000 × 0.035 (= 700) oe (NB: accept $\left(1 + \dfrac{3.5}{100}\right)$ for 1.035 but not (1 + 3.5%))	M1
“20 700” × 1.035 (= 21 424.5) “21 424.5” × 1.035 oe eg $20700 \times 0.035 = 724.5$ & $20700 + 724.5 = 21424.5$ $21424.5 \times 0.035 = 749.85\ldots$ & $21424.5 + 749.85\ldots$	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 22 174	A1
	(3)
	(3 marks)

Notes

M1: For finding 103.5% or 3.5% of 20 000

M1: dep for a complete method

A1: Allow 22 174 - 22175

(if you see the correct answer and then 20 000 is subtracted to give 2174 - 2175 then award full marks

2174 – 2175 with no working gains 2 marks)

SCB2 for $\left(2000 \times 1.035^3\right) = 2217\ldots$ [misread]

SCB2 for 22160 (20 000 × 1.108)

SCB2 for 22180 (20 000 × 1.109)

SCB1 if no marks awarded for any of these are seen (not necessarily the answer)

(20 000 × 1.035ⁿ)

20 000 × 0.965³ (= 17 972.……)

20 000 × 0.105 (= 2100)

20 000 × 1.105 ( (= 22 100)

20 000 × 1.105² ( = 21 424.5)

M2 for 20 000 × 1.035³
[NB: $1.035^3 = 1.108717\ldots$]
or
20 000 × 1.035⁴
(= 22 950…)

Higher November 2024 Paper 1 Q7

EdexcelCurrent spec4 marksPercentages

7 Shop A and Shop B have offers for buying the same type of laptop.

The normal price of the laptop in Shop A is different to the normal price of the laptop in Shop B

Shop A

Our normal price
£475

Get 16% off our
normal price

Shop B

Get 15% off our
normal price

Only pay £408

Which shop gives more money off their normal price?
Show your working clearly.

(4)

Mark scheme

Scheme	Marks
475 × 0.16 (= 76) oe or 475 × (1 – 0.16) (= 399) oe	M1
1 – 0.15 (= 0.85) or $x - 0.15x = 408$ or 100(%) – 15(%) (= 85(%)) or $\dfrac{408}{85}$ (= 4.8) oe	M1
408 ÷ “0.85” (= 480) or 408 ÷ “85” × 100 (= 480) or 408 × 100 ÷ “85” (= 480) oe or “4.8” × 100 (= 480) or $\dfrac{408}{85} \times 15\;(= 72)$	M1
Working required Answer: A and 72 and 76 seen	A1
	(4)
	(4 marks)

Notes

M1: (working for shop A)

M1: (working for shop B)

A1: dep on M2 for A with correct working (72 and 76 seen)

Higher November 2024 Paper 2 Q4

EdexcelCurrent spec5 marksPercentages

4 $N$ is a number.
17% of $N$ is 357

(a) Work out the value of $N$ (2)

In 2019, the population of a village was 650
In 2020, the population of the village was 806

(b) Work out the percentage increase in the population. (3)

Mark scheme (a)

Scheme	Marks
357 ÷ 0.17 oe or $0.17N = 357$ or $\dfrac{17}{100} \times N = 357$ oe or $\dfrac{357 \times 100}{17}$ oe eg 357 × 5.8(82…)	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 2100	A1
	(2)

Notes

M1: a correct calculation for $N$ or a correct equation in $N$
(not 17% × $N$ = 357)

A1: cao

Mark scheme (b)

Scheme	Marks
806 – 650 (= 156) or $\dfrac{806}{650}$ (=1.24) oe	M1
$\dfrac{806 - 650}{650}(\times 100)\;(= 0.24(\times 100))$ or “1.24” ×100 (= 124) or “1.24” – 1 (= 0.24)	M1
Correct answer scores full marks (unless from obvious incorrect working) Answer: 24	A1
	(3)
	(5 marks)

Notes

M1: a correct calculation for the percentage increase or seeing 124 or 0.24 as either the answer or in part of the working.

A1: cao

(SCB1 if no marks scored for an answer of 19.3 – 19.4)