Ratio & Basic Proportion

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Edexcel

AQA

Foundation June 2025 Paper 3 Q25

EdexcelCurrent spec4 marksRatio & Basic Proportion

25 In London, 2 kg of carrots cost £3.75
In New York, 5 lbs of carrots cost 4.90 US dollars.

£1 = 1.20 US dollars
1 kg = 2.2 lbs

In which city are carrots better value for money, London or New York?
You must show how you get your answer. (4)

Mark scheme

Answer	Mark	Mark scheme
New York (supported)	P1	for a conversion between kg and lbs, eg $2 \times 2.2\ (= 4.4)$ or $5 \div 2.2\ (= 2.27\ldots)$
	P1	for a conversion between pounds and US dollars, eg $3.75 \times 1.2(0)\ (= 4.50)$ or $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$
	P1	for a process to find figures to compare, eg $3.75 \div \text{``}4.4\text{''}\ (= 0.85\ldots)$ and $\text{``}4.08\ldots\text{''} \div 5\ (= 0.81(6\ldots))$ or $\text{``}4.50\text{''} \div 2\ (= 2.25)$ and $4.9(0) \div \text{``}2.27\ldots\text{''}\ (= 2.15(6\ldots))$ or $\text{``}4.4\text{''} \div 3.75\ (= 1.17\ldots)$ and $5 \div \text{``}4.08\text{''}\ (= 1.22\ldots)$ or $2 \div \text{``}4.5\text{''}\ (= 0.44\ldots)$ and $\text{``}2.27\text{''} \div 4.9(0)\ (= 0.46\ldots)$
	C1	for New York with correct comparable figures, eg 0.85 and 0.81(6…) OR 2.25 and 2.15(6…)

Additional guidance

See page at end of mark scheme

£ per lb
US dollars per kg
lbs per £
kg per US dollar

Condone the incorrect assignment of units for all 3 P marks

Final comparable values can be truncated or rounded, they just need to be accurate enough to allow a comparison

Question 25 – Additional information - NOT Exhaustive

Final comparable values can be truncated or rounded and just need to be accurate enough to allow a comparison

Cost per lb in $\$$: $2 \times 2.2\ (= 4.4)$ $3.75 \times 1.2(0)\ (= 4.5)$ $\text{``}4.5\text{''} \div \text{``}4.4\text{''}\ (= 1.02\ldots)$ and $4.9(0) \div 5\ (= 0.98)$ New York and 1.02… and 0.98	Cost per lb in £: $2 \times 2.2\ (= 4.4)$ $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$ $3.75 \div \text{``}4.4\text{''}\ (= 0.85\ldots)$ and $\text{``}4.08\ldots\text{''} \div 5\ (= 0.81(6\ldots))$ New York and 0.85… and 0.81(6…)
Cost per kg in $\$$: $5 \div 2.2\ (= 2.27\ldots)$ $3.75 \times 1.2(0)\ (= 4.5)$ $\text{``}4.5\text{''} \div 2\ (= 2.25)$ and $4.9(0) \div \text{``}2.27\ldots\text{''}\ (= 2.15(6))$ New York and 2.25 and 2.15(6)	Cost per kg in £: $5 \div 2.2\ (= 2.27\ldots)$ $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$ $3.75 \div 2\ (= 1.87(5))$ and $\text{``}4.08\ldots\text{''} \div \text{``}2.27\ldots\text{''}\ (= 1.79(6\ldots))$ New York and 1.87(5) and 1.79(6…)
Number of lbs per $\$$: $2 \times 2.2\ (= 4.4)$ $3.75 \times 1.2(0)\ (= 4.5)$ $\text{``}4.4\text{''} \div \text{``}4.5\text{''}\ (= 0.97(7\ldots))$ and $5 \div 4.9(0)\ (= 1.02\ldots)$ New York and 0.97(7…) and 1.02…	Number of lbs per £: $2 \times 2.2\ (= 4.4)$ $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$ $\text{``}4.4\text{''} \div 3.75\ (= 1.17\ldots)$ and $5 \div \text{``}4.08\ldots\text{''}\ (= 1.22\ldots)$ New York and 1.17… and 1.22…
Number of kg per $\$$: $5 \div 2.2\ (= 2.27\ldots)$ $3.75 \times 1.2(0)\ (= 4.5)$ $2 \div \text{``}4.5\text{''}\ (= 0.44\ldots)$ and $\text{``}2.27\ldots\text{''} \div 4.9(0)\ (= 0.46\ldots)$ New York and 0.44… and 0.46…	Number of kg per £: $5 \div 2.2\ (= 2.27\ldots)$ $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$ $2 \div 3.75\ (= 0.53\ldots)$ and $\text{``}2.27\ldots\text{''} \div \text{``}4.08\ldots\text{''}\ (= 0.55(6\ldots))$ New York and 0.53… and 0.55(6…)
Cost of 5 lb in $\$$: $2 \times 2.2\ (= 4.4)$ $3.75 \times 1.2(0)\ (= 4.5)$ $\text{``}4.5\text{''} \div \text{``}4.4\text{''} \times 5\ (= 5.11\ldots)$ New York and 5.11…	Cost of 5 lb in £: $2 \times 2.2\ (= 4.4)$ $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$ $3.75 \div \text{``}4.4\text{''} \times 5\ (= 4.26\ldots)$ New York and 4.08… and 4.26…
Cost of 2 kg in $\$$: $5 \div 2.2\ (= 2.27\ldots)$ $3.75 \times 1.2(0)\ (= 4.5)$ $4.9(0) \div \text{``}2.27\ldots\text{''} \times 2\ (= 4.31\ldots)$ New York and 4.5 and 4.31…	Cost of 2 kg in £: $5 \div 2.2\ (= 2.27\ldots)$ $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$ $\text{``}4.08\ldots\text{''} \div \text{``}2.27\ldots\text{''} \times 2\ (= 3.59\ldots)$ New York and 3.59…

Foundation June 2025 Paper 1 Q22

EdexcelCurrent spec6 marksFractions Ratio & Basic Proportion

22 There are 280 chocolates in a box.
There are only dark chocolates, milk chocolates and white chocolates.

$\dfrac{1}{7}$ of the 280 chocolates are dark chocolates.

The number of milk chocolates : the number of white chocolates = 1 : 3

The number of white chocolates : the number of dark chocolates = $n : 1$

(a) Work out the value of $n$.
You must show all your working. (5)

10 milk chocolates from the box are eaten.

(b) Does this affect your answer to part (a)?
Give a reason for your answer. (1)

Mark scheme (a)

Answer	Mark	Mark scheme
4.5	P1	for a process to find the number of dark chocolates, eg $280 \times \dfrac{1}{7}$ oe (= 40)
	P1	for a process to find one part of the ratio for milk chocolates to white chocolates eg $(280 - \text{``}40\text{''}) \div (1 + 3)\ (= 60)$
	P1	for process to find the number of white chocolates, eg $\text{``}60\text{''} \times 3\ (= 180)$ or $280 - \text{``}40\text{''} - \text{``}60\text{''}\ (= 180)$
	P1	for a ratio of white chocolates : dark chocolates eg $\text{``}180\text{''} : \text{``}40\text{''}$ oe or for a process to find the value of $n$ eg $\text{``}180\text{''} \div \text{``}40\text{''}$
	A1	(dep P3) for 4.5 oe

Additional guidance

Condone milk, white incorrectly labelled

Labels not required for this mark, but if seen must be correct
Must be a 2-part ratio.
Allow $\text{``}40\text{''} : \text{``}180\text{''}$ if labelled as dark chocolate : white chocolate oe

A correct answer with no supportive working gets 0 marks
Condone answer of 4.5:1

Mark scheme (b)

Answer	Mark	Mark scheme
No with reason	C1	for No with reason Acceptable examples No, the number of dark chocolates and white chocolates has not changed No, the ratio of $n$:1 only involved white and dark chocolates No, eating milk chocolates doesn’t change the rest No, the ratio of dark to white chocolates doesn’t change. Not acceptable examples Yes,… No, it doesn’t matter how many milk chocolates are eaten No, the number of white chocolates doesn’t change No, the number of dark chocolates doesn’t change

Additional guidance

Reason must refer to or imply both dark and white chocolates

Foundation June 2025 Paper 2 Q21

EdexcelCurrent spec3 marksRatio & Basic Proportion

21 Sid, Tam and Musa share £6900 in the ratio 2 : 3 : 5

Work out how much money each person receives. (3)

Mark scheme

Answer	Mark	Mark scheme
1380 2070 3450	M1	for beginning to work with ratio, eg $6900 \div (2 + 3 + 5)\ (= 690)$ or $6900 \div 2\ (= 3450)$ or $6900 \div 5\ (= 1380)$ or one accurate value correctly assigned
	M1	for full method to find the value for two people, eg two of: $\text{``}690\text{''} \times 2\ (= 1380)$ or $\text{``}690\text{''} \times 3\ (= 2070)$ or $\text{``}690\text{''} \times 5\ (= 3450)$ or $(6900 - \text{``}3450\text{''}) \div 5 \times 2\ (= 1380)$ or $(6900 - \text{``}3450\text{''}) \div 5 \times 3\ (= 2070)$ or $(6900 - \text{``}1380\text{''}) \div 8 \times 3\ (= 2070)$ or $(6900 - \text{``}1380\text{''}) \div 8 \times 5\ (= 3450)$ or $6900 \div 2\ (= 3450)$ and $6900 \div 5\ (= 1380)$
	A1	for Sid (£) 1380 Tam (£) 2070 Musa (£) 3450 If M1M0, award SCB2 for Sid (£) 3450 Tam (£) 2070 Musa (£) 1380

Additional guidance

Award M0 for a correct value assigned to an incorrect person

Correct values not assigned or assigned to an incorrect person, coming from an incorrect method,
eg (Sid =) $6900 \div 2 = 3450$ and (Tam =) $6900 \div 3 = 2300$ and (Musa =) $6900 \div 5 = 1380$
will score no marks

Foundation June 2025 Paper 1 Q17

EdexcelCurrent spec3 marksRatio & Basic Proportion

17 Sam wants to use this recipe to make 15 pancakes.

Ingredients for 10 pancakes
100 g flour 200 m$l$ milk 40 g butter 2 eggs

Sam has
200 g flour
250 m$l$ milk
70 g butter
5 eggs

Does Sam have enough flour, enough milk, enough butter and enough eggs to make 15 pancakes?
You must show all your working. (3)

Mark scheme

Answer	Mark	Mark scheme
No, not enough milk (supported)		Process to find amount needed (starting from recipe quantities)
	P1	for a correct process to find amount needed of one ingredient, eg $100 \div 10 \times 15$ oe (= 150) or $200 \div 10 \times 15$ oe (= 300) or $100 \div 2 + 100\ (= 150)$ or $200 \div 2 + 200\ (= 300)$ or $40 \div 2 \times 3$ oe (= 60) or $2 \div 2 \times 3$ oe (= 3) or scale factor of 1.5 seen
	P1	for a correct process to find amount needed of all the ingredients, eg $100 \div 10 \times 15$ oe (= 150) and $200 \div 10 \times 15$ oe (= 300) and $40 \div 10 \times 15$ oe (= 60) and $2 \div 10 \times 15$ oe (= 3)
	C1	for statement that there is not enough milk with supporting evidence
		OR Process to find quantity of pancakes that can be made (working from the quantities of ingredients available)
	P1	for a correct process to find the number of pancakes that can be made with one ingredient, eg $200 \div 100 \times 10$ oe (= 20) or $250 \div 200 \times 10$ oe (= 12.5) or $70 \div 40 \times 10$ oe (= 17.5) or $5 \div 2 \times 10$ oe (= 25)
	P1	for a correct process to find the number of pancakes all ingredients could produce eg $200 \div 100 \times 10$ oe (= 20) and $250 \div 200 \times 10$ oe (= 12.5) and $70 \div 40 \times 10$ oe (= 17.5) and $5 \div 2 \times 10$ oe (= 25)
	C1	for statement that there is not enough milk with supporting evidence

Additional guidance

C1 (both methods): Must have correct figures for all ingredients, but it is not necessary to comment on the other ingredients.
Any comment on the other ingredients must be correct.

Second method, first P1: Award this mark for a complete process to find the amount of ingredients for a quantity of pancakes that uses all of one of the ingredients eg process to find amount of ingredients for 20 pancakes (flour).

Foundation June 2025 Paper 1 Q14

EdexcelCurrent spec3 marksRatio & Basic Proportion Units & Time

14 A road has a length of 1.6 kilometres.

The road is shown on a map with a scale of 1 : 20 000

Work out the length, in centimetres, of this road on the map. (3)

Mark scheme

Answer	Mark	Mark scheme
8	M1	for unit conversion for $1.6 \times 100\,000\ (= 160\,000\text{ (cm)})$ or $1.6 \times 1000\ (= 1600\text{ (m)})$ or for $20\,000 \div 100\,000\ (= 0.2)$ or $20\,000 \div 100\ (= 200)$ or $20\,000 \div 1000\ (= 20)$ or $\text{``}0.00008\text{''} \times 1000\ (= 0.08)$ or $\text{``}0.00008\text{''} \times 100\,000\ (= 8)$
	M1	for use of scale [distance] $\div\, 20\,000$ or $1.6 \div \text{``}0.2\text{''}$ or $16 \div (\text{``}0.2\text{''} \times 10)$ or $1.6 \div \text{``}20\text{''}\ (= 0.08)$ or $\text{``}1600\text{''} \div \text{``}200\text{''}\ (= 8)$ or $1.6 \div 20\,000\ (= 0.00008)$
	A1	cao

Additional guidance

ie conversions to cm or m
Do not award if conversion linked to incorrect unit eg $1.6 \times 1000 = 1600$ cm
Answer of 160 000 centimetres scores M1M0A0

Where [distance] is an attempt at converting km to m or cm, ie figs 16

Foundation June 2025 Paper 2 Q10

EdexcelCurrent spec2 marksPercentages Ratio & Basic Proportion

10 20% of cars sold last year were electric cars.

Write down the ratio of

the number of electric cars sold last year to the number of non-electric cars sold last year.

Give your ratio in its simplest form. (2)

Mark scheme

Answer	Mark	Mark scheme
1 : 4	M1	for 20 : 80 oe ratio or 4 : 1
1 : 4	A1	1 : 4

Additional guidance

eg 10 : 40, 5 : 20, 2 : 8, 20% : 80%
or 0.2 : 0.8 oe

Condone 1% : 4%
but not 0.01 : 0.04
Accept 0.25 : 1 or $\dfrac{1}{4} : 1$

Higher June 2025 Paper 1 Q4

EdexcelCurrent spec6 marksFractions Ratio & Basic Proportion

4 There are 280 chocolates in a box.
There are only dark chocolates, milk chocolates and white chocolates.

$\dfrac{1}{7}$ of the 280 chocolates are dark chocolates.

The number of milk chocolates : the number of white chocolates = 1 : 3

The number of white chocolates : the number of dark chocolates = $n : 1$

(a) Work out the value of $n$.
You must show all your working. (5)

10 milk chocolates from the box are eaten.

(b) Does this affect your answer to part (a)?
Give a reason for your answer. (1)

Mark scheme (a)

Answer	Mark	Mark scheme
4.5	P1	for a process to find the number of dark chocolates, eg $280 \times \dfrac{1}{7}$ oe (= 40)
	P1	for a process to find one part of the ratio for milk chocolates to white chocolates eg $(280 - \text{``}40\text{''}) \div (1 + 3)\ (= 60)$
	P1	for a process to find the number of white chocolates, eg $\text{``}60\text{''} \times 3\ (= 180)$ or $280 - \text{``}40\text{''} - \text{``}60\text{''}\ (= 180)$
	P1	for a ratio of white chocolates : dark chocolates eg $\text{``}180\text{''} : \text{``}40\text{''}$ oe or for a process to find the value of $n$ eg $\text{``}180\text{''} \div \text{``}40\text{''}$
	A1	(dep P3) for 4.5 oe

Additional guidance

Condone milk, white incorrectly labelled

Labels not required for this mark, but if seen must be correct
Must be a 2-part ratio
Allow $\text{``}40\text{''} : \text{``}180\text{''}$ if labelled as dark chocolate : white chocolate oe

A correct answer with no supportive working gets 0 marks
Condone answer of 4.5 : 1

Mark scheme (b)

Answer	Mark	Mark scheme
No with reason	C1	for No with reason Acceptable examples No, the number of dark chocolates and white chocolates has not changed No, the ratio of $n$:1 only involved white and dark chocolates No, eating milk chocolates doesn’t change the rest No, the ratio of dark to white chocolates doesn’t change Not acceptable examples Yes,… No, it doesn’t matter how many milk chocolates are eaten No, the number of white chocolates doesn’t change No, the number of dark chocolates doesn’t change

Additional guidance

Reason must refer to or imply both dark and white chocolates

Higher June 2025 Paper 3 Q4

EdexcelCurrent spec4 marksRatio & Basic Proportion

4 In London, 2 kg of carrots cost £3.75
In New York, 5 lbs of carrots cost 4.90 US dollars.

£1 = 1.20 US dollars
1 kg = 2.2 lbs

In which city are carrots better value for money, London or New York?
You must show how you get your answer. (4)

Mark scheme

Answer	Mark	Mark scheme
New York (supported)	P1	for a conversion between kg and lbs, eg $2 \times 2.2\ (= 4.4)$ or $5 \div 2.2\ (= 2.27\ldots)$
	P1	for a conversion between pounds and US dollars, eg $3.75 \times 1.2(0)\ (= 4.50)$ or $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$
	P1	for a process to find figures to compare, eg $3.75 \div \text{``}4.4\text{''}\ (= 0.85\ldots)$ and $\text{``}4.08\ldots\text{''} \div 5\ (= 0.81(6\ldots))$ or $\text{``}4.50\text{''} \div 2\ (= 2.25)$ and $4.9(0) \div \text{``}2.27\ldots\text{''}\ (= 2.15(6\ldots))$ or $\text{``}4.4\text{''} \div 3.75\ (= 1.17\ldots)$ and $5 \div \text{``}4.08\text{''}\ (= 1.22\ldots)$ or $2 \div \text{``}4.5\text{''}\ (= 0.44\ldots)$ and $\text{``}2.27\text{''} \div 4.9(0)\ (= 0.46\ldots)$
	C1	for New York with correct comparable figures, eg 0.85… and 0.81(6…) OR 2.25 and 2.15(6…)

Additional guidance

See page at end of mark scheme

£ per lb
US dollars per kg
lbs per £
kg per US dollar

Condone the incorrect assignment of units for all 3 P marks

Final comparable values can be rounded or truncated, they just need to be accurate enough to allow a comparison.

Question 4 – Additional information NOT Exhaustive

Final comparable values, can be truncated or rounded, they just need to be accurate enough to allow a comparison

Cost per lb in $\$$: $2 \times 2.2\ (= 4.4)$ $3.75 \times 1.2(0)\ (= 4.5)$ $\text{``}4.5\text{''} \div \text{``}4.4\text{''}\ (= 1.02\ldots)$ and $4.9(0) \div 5\ (= 0.98)$ New York and 1.02… and 0.98	Cost per lb in £: $2 \times 2.2\ (= 4.4)$ $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$ $3.75 \div \text{``}4.4\text{''}\ (= 0.85\ldots)$ and $\text{``}4.08\ldots\text{''} \div 5\ (= 0.81(6\ldots))$ New York and 0.85… and 0.81(6…)
Cost per kg in $\$$: $5 \div 2.2\ (= 2.27\ldots)$ $3.75 \times 1.2(0)\ (= 4.5)$ $\text{``}4.5\text{''} \div 2\ (= 2.25)$ and $4.9(0) \div \text{``}2.27\ldots\text{''}\ (= 2.15(6))$ New York and 2.25 and 2.15(6)	Cost per kg in £: $5 \div 2.2\ (= 2.27\ldots)$ $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$ $3.75 \div 2\ (= 1.87(5))$ and $\text{``}4.08\ldots\text{''} \div \text{``}2.27\ldots\text{''}\ (= 1.79(6\ldots))$ New York and 1.87(5) and 1.79(6…)
Number of lbs per $\$$: $2 \times 2.2\ (= 4.4)$ $3.75 \times 1.2(0)\ (= 4.5)$ $\text{``}4.4\text{''} \div \text{``}4.5\text{''}\ (= 0.97(7\ldots))$ and $5 \div 4.9(0)\ (= 1.02\ldots)$ New York and 0.97(7…) and 1.02…	Number of lbs per £: $2 \times 2.2\ (= 4.4)$ $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$ $\text{``}4.4\text{''} \div 3.75\ (= 1.17\ldots)$ and $5 \div \text{``}4.08\ldots\text{''}\ (= 1.22\ldots)$ New York and 1.17… and 1.22…
Number of kg per $\$$: $5 \div 2.2\ (= 2.27\ldots)$ $3.75 \times 1.2(0)\ (= 4.5)$ $2 \div \text{``}4.5\text{''}\ (= 0.44\ldots)$ and $\text{``}2.27\ldots\text{''} \div 4.9(0)\ (= 0.46\ldots)$ New York and 0.44… and 0.46…	Number of kg per £: $5 \div 2.2\ (= 2.27\ldots)$ $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$ $2 \div 3.75\ (= 0.53\ldots)$ and $\text{``}2.27\ldots\text{''} \div \text{``}4.08\ldots\text{''}\ (= 0.55(6\ldots))$ New York and 0.53… and 0.55(6…)
Cost of 5 lb in $\$$: $2 \times 2.2\ (= 4.4)$ $3.75 \times 1.2(0)\ (= 4.5)$ $\text{``}4.5\text{''} \div \text{``}4.4\text{''} \times 5\ (= 5.11\ldots)$ New York and 5.11…	Cost of 5 lb in £: $2 \times 2.2\ (= 4.4)$ $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$ $3.75 \div \text{``}4.4\text{''} \times 5\ (= 4.26\ldots)$ New York and 4.08… and 4.26…
Cost of 2 kg in $\$$: $5 \div 2.2\ (= 2.27\ldots)$ $3.75 \times 1.2(0)\ (= 4.5)$ $4.9(0) \div \text{``}2.27\ldots\text{''} \times 2\ (= 4.31\ldots)$ New York and 4.5 and 4.31…	Cost of 2 kg in £: $5 \div 2.2\ (= 2.27\ldots)$ $4.9(0) \div 1.2(0)\ (= 4.08\ldots)$ $\text{``}4.08\ldots\text{''} \div \text{``}2.27\ldots\text{''} \times 2\ (= 3.59\ldots)$ New York and 3.59…

Higher June 2025 Paper 2 Q2

EdexcelCurrent spec3 marksRatio & Basic Proportion

2 Sid, Tam and Musa share £6900 in the ratio 2 : 3 : 5

Work out how much money each person receives. (3)

Mark scheme

Answer	Mark	Mark scheme
1380 2070 3450	M1	for beginning to work with ratio, eg $6900 \div (2 + 3 + 5)\ (= 690)$ or $6900 \div 2\ (= 3450)$ or $6900 \div 5\ (= 1380)$ or one accurate value correctly assigned
	M1	for full method to find the value for two people, eg two of: $\text{``}690\text{''} \times 2\ (= 1380)$ or $\text{``}690\text{''} \times 3\ (= 2070)$ or $\text{``}690\text{''} \times 5\ (= 3450)$ or $(6900 - \text{``}3450\text{''}) \div 5 \times 2\ (= 1380)$ or $(6900 - \text{``}3450\text{''}) \div 5 \times 3\ (= 2070)$ or $(6900 - \text{``}1380\text{''}) \div 8 \times 3\ (= 2070)$ or $(6900 - \text{``}1380\text{''}) \div 8 \times 5\ (= 3450)$ or $6900 \div 2\ (= 3450)$ and $6900 \div 5\ (= 1380)$
	A1	for Sid (£) 1380 Tam (£) 2070 Musa (£) 3450 If M1M0, award SCB2 for Sid (£) 3450 Tam (£) 2070 Musa (£) 1380

Additional guidance

Award M0 for a correct value assigned to an incorrect person

Foundation November 2024 Paper 2 Q22

EdexcelCurrent spec3 marksRatio & Basic Proportion

Two Choci bar labels: London, 200 g, £3.50; Zurich, 360 g, 7.20 Swiss francs

In London, a 200 g Choci bar costs £3.50
In Zurich, a 360 g Choci bar costs 7.20 Swiss francs.

The exchange rate is £1 = 1.25 Swiss francs.

In which city is the Choci bar the better value for money, in London or in Zurich?
You must show how you get your answer. (3)

Mark scheme

Answer	Mark	Mark scheme
Zurich (supported)	P1	for one process to compare, eg eg Currency conversion, $3.5 \times 1.25\ (= 4.375)$ or $7.20 \div 1.25\ (= 5.76)$ or finds 1g in one place £3.50 $\div 200\ (= 0.0175)$ or $7.20 \div 360\ (= 0.02)$ or finds 200g in Zurich, $7.2 \div 360 \times 200\ (= 4.0)$ or finds 360g in London, $3.5 \div 200 \times 360\ (= 6.30)$ or finds grams per unit cost, $200 \div 3.50\ (= 57.1\ldots)$ or $360 \div 7.20\ (= 50)$
	P1	for a complete process to find comparable figures in the same currency, eg comparing 200g in £ or francs $3.5 \times 1.25\ (= 4.375)$ and $7.2 \div 360 \times 200\ (= 4.0)$ or $\text{``}4.0\text{''} \div 1.25\ (= 3.20)$ OR comparing 360g in £ or francs $\text{``}6.30\text{''} \times 1.25\ (= 7.875)$ or $3.5 \div 200 \times 360\ (= 6.30)$ and $7.20 \div 1.25\ (= 5.76)$ OR comparing 1g in £ or francs $\text{``}0.0175\text{''} \times 1.25\ (= 0.0218\ldots)$ and $7.20 \div 360\ (= 0.02)$ or £3.50 $\div 200\ (= 0.0175)$ and $\text{``}0.02\text{''} \div 1.25\ (= 0.016)$ OR comparing quantity per unit cost in £ or francs $200 \div 3.50\ (= 57.1\ldots)$ and $360 \div \text{``}5.76\text{''}\ (= 62.5)$ or $200 \div \text{``}4.375\text{''}\ (= 45.7\ldots)$ and $360 \div 7.20\ (= 50)$
	C1	for Zurich supported by correct comparable values, eg 4.3(75 F) and 4(.0 F) or (£)3.2(0) or 7.8(75 F) or (£)6.3(0) and (£)5.76 or 0.021(8… F) and 0.02 (F) or (£)0.017(5) and (£)0.016 or 57(.1… g/£) and 62(.5 g/£) or 45(.7… g/F) and 50 (g/F)

Additional guidance

Accept figures rounded or truncated to 2sf throughout

Accept working in pence
Ignore incorrect units for P marks
Award of this mark implies the previous mark

Clear indication that bar is better value for money in Zurich supported by correct values for comparison
Units not needed but if stated must be correct.
Table with examples at end of mark scheme

Question 22 additional guidance

	London	Zurich
100g	$3.5 \div 2 =$ £1.75 $1.75 \times 1.25 =$ 2.1875 F	$7.2 \div 360 =$ 2.00 F $2.00 \div 1.25 =$ £1.60
200g	£3.50 $3.5 \times 1.25 =$ 4.375 F	$7.2 \div 360 \times 200 =$ 4.0 F $4.0 \div 1.25 =$ £3.20
360g	$3.5 \div 200 \times 360 =$ £6.30 $6.30 \times 1.25 =$ 7.875 F	7.20 F $7.20 \div 1.25 =$ £5.76
1g	£3.50 $\div 200 =$ £0.0175 $\times 1.25 =$ 0.021875 F	$7.20 \div 360 =$ 0.02 F $\div 1.25 =$ £0.016
40g	£3.50 $\div 5 =$ £0.70 $0.7 \times 1.25 =$ 0.875 F	$7.20 \div 9 =$ 0.8 F $0.8 \div 1.25 =$ £0.64
By weight	$350 \div 200 =$ 1.75 p/g $350 \times 1.25 = 4.375$ $4.375 \div 200 =$ 0.021875 F/g	$720 \div 360 =$ 0.02 F/g $720 \div 1.25 = 576$ $576 \div 360 =$ 1.6 p/g
By cost	$200 \div 350 =$ 0.571 g/p $350 \times 1.25 = 437.5$ $200 \div 437.5 =$ 45.7 g/F	$360 \div 720 =$ 50 g/F $720 \div 1.25 = 576$ $360 \div 576 =$ 0.625 g/P

Foundation November 2024 Paper 3 Q21

EdexcelCurrent spec5 marksPercentages Ratio & Basic Proportion

21 Robyn buys a total of 240 pens and pencils, where

number of pens : number of pencils = 3 : 5

Robyn pays 9p for each pen.
She sells each pen for 11p.

Robyn pays 6p for each pencil.
She sells each pencil for 10p.

Robyn sells all of the pens and pencils.

Work out Robyn’s percentage profit.
Give your answer correct to 1 decimal place.
You must show all your working. (5)

Mark scheme

Answer	Mark	Mark scheme
45.6	P1	for a process to start to work with the ratio, eg $240 \div (3 + 5)\ (= 30)$ or pens $= 3n$ and pencils $= 5n$ where $n$ is a positive integer
	P1	for a complete process to find the number of pens and pencils, eg $\text{``}30\text{''} \times 3\ (= 90)$ and $\text{``}30\text{''} \times 5\ (= 150)$ OR for process to find one cost or amount to sell for one item eg $[\text{pens}] \times 9\ (= 810)$ or $[\text{pens}] \times 11\ (= 990)$ or $[\text{pencils}] \times 6\ (= 900)$ or $[\text{pencils}] \times 10\ (= 1500)$ OR for process to find the profit for one pen or one pencil eg $11 - 9\ (= 2)$ or $10 - 6\ (= 4)$
	P1	for a process to find the total cost to buy or the total amount to sell for both, eg $[\text{pens}] \times 9 + [\text{pencils}] \times 6\ (= 1710)$ or $[\text{pens}] \times 11 + [\text{pencils}] \times 10\ (= 2490)$ OR process to find the profit for one item eg $[\text{pens}] \times 11 - [\text{pens}] \times 9\ (= 180)$ or $[\text{pens}] \times (11 - 9)\ (= 180)$ or $[\text{pencils}] \times 10 - [\text{pencils}] \times 6\ (= 600)$ or $[\text{pencils}] \times (10 - 6)\ (= 600)$
	P1	for a complete process to find the profit as a percentage or a decimal, eg $\dfrac{[2490] - [1710]}{[1710]} \times 100$ or $\dfrac{[2490] - [1710]}{[1710]}\ (= 0.456\ldots)$ or for a process to find the amount to sell as a percentage of the cost eg $\dfrac{[2490]}{[1710]} \times 100\ (= 145.6\ldots)$
	A1	answer in the range 45.6 to 45.62

Additional guidance

Can work in £ or pence but must be consistent, 90 or 150 imply P1
This mark can be awarded at any stage

[pens] could be $\text{``}30\text{''} \times 3$ or their number of pens
[pencils] could be $\text{``}30\text{''} \times 5$ or their number of pencils
[pens], [pencils] $\neq 1$

180 or 600 or 780 implies P3

[2490] is their amount to sell for both pens and pencils
[1710] is their cost of pens and pencils
[2490] – [1710] may be [180] + [600]

If an answer is given in the range in working and then rounded incorrectly award full marks.
A correct answer with no supportive working gets 0 marks

Higher November 2024 Paper 3 Q18

EdexcelCurrent spec4 marksRatio & Basic Proportion

18 Maria wants to find an estimate for the number of frogs in a lake.

On Saturday she catches 40 of the frogs.
She puts a tag on each frog and releases them.

On Monday she catches 55 of the frogs.
11 of the frogs have tags.

(a) Work out an estimate for the total number of frogs in the lake.
You must show all your working. (3)

(b) State one assumption you have made. (1)

Mark scheme (a)

Answer	Mark	Mark scheme
200	P1	for $\frac{55}{11}\ (= 5)$ or $\frac{11}{55}\ (= 0.2)$ or $\frac{40}{11}\ (= 3.63..)$ or $\frac{11}{40}\ (= 0.275)$ or for starting to use ratio eg tagged : untagged = 11 : 44 oe
	P1	for $\frac{11}{55} = \frac{40}{n}$ or $\frac{11}{40} = \frac{55}{n}$ or $\frac{55 \times 40}{11}$ oe or for process to find the number of untagged frogs eg $\dfrac{55 - 11}{11} \times 40\ (= 160)$
	A1	cao

Additional guidance

Allow decimal values truncated or rounded to at least 2dp

May be seen in ratio eg 40 : 160

A correct answer with no supportive working gets 0 marks.

Mark scheme (b)

Answer	Mark	Mark scheme
Assumption	C1	for a suitable assumption, Acceptable examples no frogs have left/arrived at the lake no tags have fallen off / no lost tags sample of frogs taken was a random sample / this is a representative sample tagged frogs are distributed around the lake no baby frogs / no frogs have died There are (still) 4 untagged for every tagged frog. The tagged frogs had time to blend into the population. It is a closed environment There aren’t any more frogs in the lake No other frogs in the lake Not acceptable examples She has caught all the frogs There are lots of frogs. This is the same lake

Foundation November 2024 Paper 1 Q17

EdexcelCurrent spec3 marksRatio & Basic Proportion

17 Gita wants to make 24 litres of green paint.
She is going to mix blue paint, yellow paint and white paint in the ratio 4 : 3 : 1

She has
12 litres of blue paint
7 litres of yellow paint
5 litres of white paint.

Does Gita have enough blue paint, yellow paint and white paint to make 24 litres of green paint?
You must show how you get your answer. (3)

Mark scheme

Answer	Mark	Mark scheme
No, with correct figures	P1	for start to process to find amount of each paint colour needed eg $24 \div (4 + 3 + 1)\ (= 3)$ or 8 : 6 : 2 or for start to process to simplify 12 : 7 : 5 to a ratio that can be compared to 4 : 3 : 1 eg $12 \div 4\ (= 3)$ or $7 \div 3\ (= 2.3\ldots)$ or $5 \div 1\ (= 5)$
	P1	for complete process to find amount of each paint colour needed eg $\text{``}3\text{''} \times 4\ (= 12)$ and $\text{``}3\text{''} \times 3\ (= 9)$ and $\text{``}3\text{''} \times 1\ (= 3)$ or 12 : 9 : 3 or for complete process to cancel 12 : 7 : 5 to a ratio that can be compared to 4 : 3 : 1 eg 12 : 7 : 5 as $4 : \text{``}\tfrac{7}{3}\text{''} : \text{``}\tfrac{5}{3}\text{''}$ or $4 : \text{``}2.3\ldots\text{''} : \text{``}1.6\ldots\text{''}$ or 12 : 7 : 5 as $\text{``}5\tfrac{1}{7}\text{''} : 3 : \text{``}2\tfrac{1}{7}\text{''}$ or $\text{``}5.14\ldots\text{''} : 3 : \text{``}2.14\ldots\text{''}$ or 12 : 7 : 5 as $\text{``}2\tfrac{2}{5}\text{''} : \text{``}1\tfrac{2}{5}\text{''} : 1$ or $\text{``}2.4\text{''} : \text{``}1.4\text{''} : 1$
	C1	No, with correct figure(s) for comparison. eg No with 12 : 9 : 3 No, 9 litres of yellow needed No with $4 : 2.3\ldots : 1.6\ldots$

Additional guidance

No may be indicated by eg ‘not enough yellow’.

Foundation November 2024 Paper 2 Q15

EdexcelCurrent spec3 marksRatio & Basic Proportion Units & Time

15 A map has a scale of 1 : $25\,000$
On the map, a road has a length of 14 cm.

Work out the real length of the road.
Give your answer in kilometres. (3)

Mark scheme

Answer	Mark	Mark scheme
3.5	M1	for a correct first step, eg $14 \times 25\,000\ (= 350\,000)$ or digits 14 $\times$ digits 25 or $25\,000 \div 100\,000\ (= 0.25)$ oe or $14 \div 100\,000\ (= 0.00014)$ or [distance] $\div\ 100\,000$
	M1	for a complete method, eg $\text{``}350\,000\text{''} \div 100\,000$ oe or $\text{``}0.25\text{''} \times 14$ or $\text{``}0.00014\text{''} \times 25\,000$
	A1	for 3.5 oe

Additional guidance

[distance] is any calculated value using digits 14 and digits 25

Foundation November 2024 Paper 2 Q9

EdexcelCurrent spec2 marksRatio & Basic Proportion

9 There are 90 red counters and 150 yellow counters in a bag.

Write down the ratio of the number of red counters to the number of yellow counters.
Give your ratio in its simplest form. (2)

Answer	Mark	Mark scheme
3 : 5	M1	for 90 : 150 oe ratio or 5 : 3
A1	cao

Higher November 2024 Paper 3 Q5

EdexcelCurrent spec5 marksPercentages Ratio & Basic Proportion

5 Robyn buys a total of 240 pens and pencils, where

number of pens : number of pencils = 3 : 5

Robyn pays 9p for each pen.
She sells each pen for 11p.

Robyn pays 6p for each pencil.
She sells each pencil for 10p.

Robyn sells all of the pens and pencils.

Work out Robyn’s percentage profit.
Give your answer correct to 1 decimal place.
You must show all your working. (5)

Mark scheme

Answer	Mark	Mark scheme
45.6	P1	for a process to start to work with the ratio, eg $240 \div (3 + 5)\ (= 30)$ or pens $= 3n$ and pencils $= 5n$ where $n$ is a positive integer
	P1	for a complete process to find the number of pens and pencils, eg $\text{``}30\text{''} \times 3\ (= 90)$ and $\text{``}30\text{''} \times 5\ (= 150)$ OR for process to find one cost or amount to sell for one item eg $[\text{pens}] \times 9\ (= 810)$ or $[\text{pens}] \times 11\ (= 990)$ or $[\text{pencils}] \times 6\ (= 900)$ or $[\text{pencils}] \times 10\ (= 1500)$ OR process to find the profit for one pen or one pencil eg $11 - 9\ (= 2)$ or $10 - 6\ (= 4)$
	P1	for a process to find the total cost to buy or the total amount to sell for both, eg $[\text{pens}] \times 9 + [\text{pencils}] \times 6\ (= 1710)$ or $[\text{pens}] \times 11 + [\text{pencils}] \times 10\ (= 2490)$ OR process to find the profit for one item eg $[\text{pens}] \times 11 - [\text{pens}] \times 9\ (= 180)$ or $[\text{pens}] \times (11 - 9)\ (= 180)$ or $[\text{pencils}] \times 10 - [\text{pencils}] \times 6\ (= 600)$ or $[\text{pencils}] \times (10 - 6)\ (= 600)$
	P1	for a complete process to find the profit as a percentage or a decimal, eg $\dfrac{[2490] - [1710]}{[1710]} \times 100$ or $\dfrac{[2490] - [1710]}{[1710]}\ (= 0.456\ldots)$ or for a process to find the amount to sell as a percentage of the cost eg $\dfrac{[2490]}{[1710]} \times 100\ (= 145.6\ldots)$
	A1	answer in the range 45.6 to 45.62

Additional guidance

Can work in £ or pence but must be consistent, 90 or 150 imply P1
This mark can be awarded at any stage

[pens] could be $\text{``}30\text{''} \times 3$ or their number of pens
[pencils] could be $\text{``}30\text{''} \times 5$ or their number of pencils
[pens], [pencils] $\neq 1$

180 or 600 or 780 implies P3

[2490] is their amount to sell for both pens and pencils
[1710] is their cost of pens and pencils
[2490] – [1710] may be [180] + [600]

If an answer is given in the range in working and then rounded incorrectly award full marks.
A correct answer with no supportive working gets 0 marks

Higher November 2024 Paper 2 Q3

EdexcelCurrent spec3 marksRatio & Basic Proportion

In London, a 200 g Choci bar costs £3.50
In Zurich, a 360 g Choci bar costs 7.20 Swiss francs.

The exchange rate is £1 = 1.25 Swiss francs.

In which city is the Choci bar the better value for money, in London or in Zurich?
You must show how you get your answer. (3)

Mark scheme

Answer	Mark	Mark scheme
Zurich (supported)	P1	for one process to compare, eg eg Currency conversion, $3.5 \times 1.25\ (= 4.375)$ or $7.20 \div 1.25\ (= 5.76)$ or finds 1g in one place £3.50 $\div 200\ (= 0.0175)$ or $7.20 \div 360\ (= 0.02)$ or finds 200g in Zurich, $7.2 \div 360 \times 200\ (= 4.0)$ or finds 360g in London, $3.5 \div 200 \times 360\ (= 6.30)$ or finds grams per unit cost, $200 \div 3.50\ (= 57.1\ldots)$ or $360 \div 7.20\ (= 50)$
	P1	for a complete process to find comparable figures in the same currency, eg comparing 200g in £ or francs $3.5 \times 1.25\ (= 4.375)$ and $7.2 \div 360 \times 200\ (= 4.0)$ or $\text{``}4.0\text{''} \div 1.25\ (= 3.20)$ OR comparing 360g in £ or francs $\text{``}6.30\text{''} \times 1.25\ (= 7.875)$ or $3.5 \div 200 \times 360\ (= 6.30)$ and $7.20 \div 1.25\ (= 5.76)$ OR comparing 1g in £ or francs $\text{``}0.0175\text{''} \times 1.25\ (= 0.0218\ldots)$ and $7.20 \div 360\ (= 0.02)$ or £3.50 $\div 200\ (= 0.0175)$ and $\text{``}0.02\text{''} \div 1.25\ (= 0.016)$ OR comparing quantity per unit cost in £ or francs $200 \div 3.50\ (= 57.1\ldots)$ and $360 \div \text{``}5.76\text{''}\ (= 62.5)$ or $200 \div \text{``}4.375\text{''}\ (= 45.7\ldots)$ and $360 \div 7.20\ (= 50)$
	C1	for Zurich supported by correct comparable values, eg 4.3(75 F) and 4(.0 F) or (£)3.2(0) or 7.8(75 F) or (£)6.3(0) and (£)5.76 or 0.021(8… F) and 0.02 (F) or (£)0.017(5) and (£)0.016 or 57(.1… g/£) and 62(.5 g/£) or 45(.7… g/F) and 50 (g/F)

Additional guidance

Accept figures rounded or truncated to 2sf throughout

Accept working in pence
Ignore incorrect units for P marks
Award of this mark implies the previous

Clear indication that bar is better value for money in Zurich supported by correct values for comparison
Units not needed but if given must be correct
Table with examples at end of mark scheme

Question 3 – Table with examples

	London	Zurich
100g	$3.5 \div 2 =$ £1.75 $1.75 \times 1.25 =$ 2.1875 F	$7.2 \div 360 \times 100 =$ 2.00 F $2.00 \div 1.25 =$ £1.60
200g	£3.50 $3.5 \times 1.25 =$ 4.375 F	$7.2 \div 360 \times 200 =$ 4.0 F $4.0 \div 1.25 =$ £3.20
360g	$3.5 \div 200 \times 360 =$ £6.30 $6.30 \times 1.25 =$ 7.875 F	7.20 F $7.20 \div 1.25 =$ £5.76
1g	£3.50 $\div 200 =$ £0.0175 $\times 1.25 =$ 0.021875 F	$7.20 \div 360 =$ 0.02 F $\div 1.25 =$ £0.016
40g	£3.50 $\div 5 =$ £0.70 $0.7 \times 1.25 =$ 0.875 F	$7.20 \div 9 =$ 0.8 F $0.8 \div 1.25 =$ £0.64
By weight	$350 \div 200 =$ 1.75 p/g $3.50 \times 1.25 = 4.375$ $4.375 \div 200 =$ 0.021875 F/g	$7.20 \div 360 =$ 0.02 F/g $720 \div 1.25 = 576$ $576 \div 360 =$ 1.6 p/g

Foundation June 2025 Paper 3 Q21

AQACurrent spec3 marksFractions Ratio & Basic Proportion

21 Ary, Bea and Cat each have an amount of money.

Cat has £280

Cat’s amount is $\dfrac{2}{3}$ of Bea’s amount.
Ary’s amount : Bea’s amount = 5 : 12

Work out how much money Ary has. [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1: works with the amount first
$280 \div 2 \times 3$ or 420	M1	oe
their $420 \div 12 \times 5$	M1dep	oe eg $35 \times 5$ their 420 cannot be 280
175	A1	SC2 77.77 or 77.78
Alternative method 2: works with the ratio first
$\dfrac{2}{3} \times 12$ or 8	M1	oe may be embedded in a ratio
$280 \div$ their $8 \times 5$	M1dep	oe eg $35 \times 5$
175	A1	SC2 77.77 or 77.78

Additional guidance

Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
SC2 is for using $\dfrac{2}{3}$ of Cat’s amount
5 : 12 : 8 in any order	M1

Foundation June 2025 Paper 2 Q18

AQACurrent spec4 marksRatio & Basic Proportion

(a) Sam uses either the bus or the train to get to work.\[\frac{\text{number of days using bus}}{\text{number of days using train}} = \frac{1}{8}\]

Write down the ratio $\quad$ number of days using bus : number of days using train [1 mark]

(b) Write the ratio $\quad 6a : 5a \quad$ in the form $\ n : 1\ $ where $n$ is a decimal. [2 marks]

Work out the value of $x$ [1 mark]

Answer	Mark	Comments
1 : 8 or 0.125 : 1	B1	oe eg 0.25 : 2

Answer	Mark	Comments
1.2 ( : 1)	B2	B1 \(1.2a\) or \(\dfrac{6}{5}\) or \(1\dfrac{1}{5}\) or \(6 \div 5\)

Answer	Mark	Comments
2	B1	accept \(-2\)

Foundation June 2025 Paper 2 Q17

AQACurrent spec3 marksRatio & Basic Proportion

Two packets of rice. Packet A: Rice, £1.10, 200 g. Packet B: Rice, £2.60, 500 g

Which packet of rice is better value for money?

Tick a box.

You must show your working. [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1: uses values from both A and B
Method to find the cost of any mass for either packet or a correct cost stated or Method to find the number of grams for any cost for either packet or a correct mass stated or Method to find the differences between masses and costs or £1.50 and 300g stated	M1	eg $1.1(0) \div 2$ eg $2.6(0) \times 2$ eg $200 \div 1.1(0)$
Fully correct method or values for two comparable/corresponding masses or costs	M1dep	eg $1.10 \times 5$ and $2.60 \times 2$ eg $1.10 \div 2$ and $2.60 \div 5$
Packet B with correct comparable values	A1	eg B and 0.55 and 0.52 eg B and 182 and 192
Alternative method 2: scales up from A or down from B
$\dfrac{2.6}{1.1}$ or 2.36(3…) or 2.364 or $\dfrac{1.1}{2.6}$ or 0.423(0…) or 0.4231 or $\dfrac{500}{200}$ or 2.5 or $\dfrac{200}{500}$ or 0.4	M1	oe
$\dfrac{2.6}{1.1} \times 200$ or [472, 473] or $\dfrac{1.1}{2.6} \times 500$ or [211, 212] or $\dfrac{500}{200} \times 1.1(0)$ or 2.75 or $\dfrac{200}{500} \times 2.6(0)$ or 1.04	M1dep	oe
Packet B and one correct comparable value stated	A1	eg B and 472 eg B and (£)2.75

Additional guidance

Up to M2 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
Working may be in pence
Two correct comparable values found for 100g followed by incorrect units being stated such as 55p $=$ 1kg and 52p $=$ 1kg and B	M1M1A1
Alt 1: If comparing, eg cost of 1g of each, there must be enough decimal places to enable a comparison to be made to award the A mark
Alt 1: Just repeating £1.10 and 200g or £2.60 and 500g	M0
Alt 1: Once they have a value from either A or B they may choose to use a value from the difference between the packets for comparison. eg obtains 55p for 100g for packet A £1.50 for 300g (the difference between the two packs) 50p for 100g Packet B	M1 M1 A1
Some examples of commonly seen answers awarded M1M1A1: B and 55 and 52 (cost in p per 100g) B and 0.0055 and 0.0052 (cost in £s per 1g) B and 5.2 and 5.5 (cost of 1kg of each in £) B and 181.8 and 192.3 (mass per £1 for each) B and 2.75 (cost of 500g of bag A)

Foundation June 2025 Paper 1 Q12

AQACurrent spec6 marksFractions Ratio & Basic Proportion

(a) One day,

a company runs 240 trains
$\dfrac{1}{8}$ of these trains are late.

The company is charged £350 for each late train.

How much is the company charged that day? [3 marks]

(b) Hot drinks are sold at a station in the ratio\[\text{tea} : \text{coffee} : \text{hot chocolate} = 3 : 5 : 2\]

2600 hot drinks are sold.

How many coffees are sold? [3 marks]

Answer	Mark	Comments
\(240 \div 8 \times 350\)	M2	oe M1 correct first step using one operator eg \(240 \times 350\) or 84 000
10 500	A1

Mark scheme (b)

Answer	Mark	Comments
$2600 \div (3 + 2 + 5)$ or 260	M1	oe
their $260 \times 5$	M1dep	oe $2600 \div 2$ implies M2
1300	A1	SC2 780 or 520 or 6500

Additional guidance

SC2 is for working out the number of teas / hot chocolates or for the misconception that 2600 was the number of hot chocolates
Answer 780 : 1300 : 520	M1M1A0

Foundation June 2025 Paper 2 Q12

AQACurrent spec2 marksRatio & Basic Proportion

12 4 apples cost £2.20

Work out the cost of 7 of these apples. [2 marks]

Mark scheme

Answer	Mark	Comments
$2.2(0) \div 4$ or 0.55 or $\dfrac{7}{4}$ or $\dfrac{4}{7}$ or 0.57(1…)	M1	oe eg $220 \div 4$ or 55 implied by 385
3.85	A1

Additional guidance

Allow working in pence
Answer 3.85p Answer 385p with £ sign crossed out Answer 385p without £ sign crossed out	M1A1 M1A1 M1A0

Foundation June 2025 Paper 1 Q9

AQACurrent spec4 marksRatio & Basic Proportion

9 Numbered discs are in two boxes.

Box A contains discs numbered 2, 4, 2, 3, 9. Box B contains discs numbered 1, 1, 2, 4, 4, 1, 3, 9

(a) Work out the ratio

total value of the numbers in Box A : total value of the numbers in Box B

Give your answer in its simplest form. [3 marks]

(b) One disc is picked at random from Box A.

Write down the probability that the number on the disc is greater than 6 [1 mark]

Mark scheme (a)

Answer	Mark	Comments
$2 + 2 + 3 + 4 + 9$ or 20 or $1 + 1 + 1 + 2 + 3 + 4 + 4 + 9$ or 25	M1	oe
20 : 25	A1	oe
4 : 5 or 0.8 : 1 or 1 : 1.25	A1ft	ft their ratio written in its simplest form with M1 awarded if it can be simplified SC2 5 : 4

Additional guidance

SC2 is for correct ratio reversed
Condone any ratio simplified to unitary form, with M1 awarded, for A1ft eg1 20 : 19 answer $1 : \dfrac{19}{20}$ or $\dfrac{20}{19} : 1$ eg2 20 : 19	M1A0A1ft M1A0A0ft
20 : 15 simplified to 4 : 3	M1A0A1ft

Mark scheme (b)

Answer	Mark	Comments
$\dfrac{1}{5}$	B1	oe fraction, decimal or percentage

Additional guidance

Ignore simplification or conversion attempts after correct probability seen
Ignore use of probability words
1 : 5 or 1 in 5 (chance)	B0

Foundation June 2025 Paper 1 Q8

AQACurrent spec3 marksRatio & Basic Proportion

8 Here is a list of ingredients for lentil soup.

Lentil soup for 4 people
Lentils	170 g
Water	600 ml
Onion	50 g

How many grams of lentils are needed to make this soup for 10 people? [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1: uses only multiplication and division
$170 \div 4 \times 10$ or $170 \div 2 \times 5$	M2	oe eg $170 \times 2.5$ M1 correct first step using one operator eg $170 \times 10$ or $5 \div 2$
425	A1	SC2 125 or 1500
Alternative method 2: uses addition or subtraction
$170 \div 2$ or 85	M1	oe
$170 \times 3 -$ their 85 or $170 \times 2 +$ their 85	M1dep	oe
425	A1	SC2 125 or 1500

Additional guidance

SC2 is from working with onions or water

Foundation June 2025 Paper 2 Q2

AQACurrent spec2 marksRatio & Basic Proportion

2 The only animals in a field are sheep and cows.

The ratio $\quad$ sheep : cows $= 31 : 47$

There are fewer than 100 animals in the field.

(a) Write down the number of cows in the field. [1 mark]

(b) In total, how many animals are there in the field? [1 mark]

Answer	Mark	Comments
47	B1

Answer	Mark	Comments
78	B1

Foundation November 2024 Paper 3 Q24

AQACurrent spec4 marksRatio & Basic Proportion Rounding

(a) At a school

there are 912 students
the ratio of students to teachers is $\quad$ 15.2 : 1

The number of students stays the same.

The number of teachers increases by 2

Work out the new ratio of students to teachers.

Give your answer in the form $\quad n : 1$ [3 marks]

(b) On a school trip, one teacher is needed for every group of 10 or fewer students.

72 students want to go on the trip.

Lexi tries to work out how many teachers are needed.

$72 \div 10 = 7.2$
7 teachers are needed.

What is wrong with her answer? [1 mark]

Mark scheme (a)

Answer	Mark	Comments
$912 \div 15.2$ or 60	M1	oe implied by 62
$912 \div (\text{their } 60 + 2)$ or $912 \div 62$ or [14.7, 14.71]	M1dep	oe
[14.7, 14.71] : 1 or 15 : 1 with M2 awarded	A1	oe eg $\dfrac{456}{31} : 1$

Additional guidance

M1 may be awarded for correct work, with no answer or incorrect answer, even if this is seen amongst multiple attempts

Do not allow misreads for 15.2

Mark scheme (b)

Answer	Mark	Comments
Valid explanation	B1	eg she needs to round up

Additional guidance

She rounded down	B1
Needs 8 (teachers)	B1
Need (one) more	B1
Only 70 students could go	B1
2 students do not have a teacher	B1
There are students without a teacher	B1
Have groups of 8 with 9 teachers	B0
8	B0
Cannot have 0.2 of a teacher	B0

Foundation November 2024 Paper 2 Q17

AQACurrent spec3 marksRatio & Basic Proportion

17 A map has a scale of $\quad$ 1 : 4000

On the map, the distance from a station to a museum is 7 cm

Is the actual distance from the station to the museum more than 300 m?

Tick a box.

Show working to support your answer. [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
$7 \times 4000 \div 100$ or 280	M2	oe M1 $7 \times 4000$ or 28000 or $7 \div 100$ or 0.07 or $4000 \div 100$ or 40
280 and No	A1	oe eg 20 less and No
Alternative method 2
$300 \times 100 \div 4000$ or 7.5	M2	oe M1 $300 \times 100$ or 30000 or $300 \div 4000$ or 0.075 or $100 \div 4000$ or 0.025
7.5 and No	A1
Alternative method 3
$300 \times 100 \div 7$ or 4285(.7…) or 4286	M2	oe M1 $300 \times 100$ or 30000 or $300 \div 7$ or [42.8, 42.9] or $100 \div 7$ or [14.2, 14.3]
[4200, 4300] and No with M2 seen	A1
Alternative method 4
$7 \times 4000$ or 28000	M1	oe
$300 \times 100$ or 30000	M1	oe
28000 and 30000 and No	A1
Alternative method 5
$300 \div 4000$ or 0.075	M1	oe
$7 \div 100$ or 0.07	M1	oe
0.075 and 0.07 and No	A1

Additional guidance

M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
No may be indicated by selecting the box or a statement in the working lines
No cannot be implied only by an inequality
A correct value is sufficient to show working eg 280 and No (except in alt 3)	M2A1
20 less in alt 1 implies M2

Foundation November 2024 Paper 3 Q15

AQACurrent spec3 marksFractions Ratio & Basic Proportion

15 To get to college, 120 students either walk, cycle or travel by car.

$\dfrac{3}{8}$ of the students walk.

students who cycle : students who travel by car = 1 : 2

How many students travel by car? [3 marks]

Mark scheme

Answer	Mark	Comments
$\dfrac{3}{8} \times 120$ or 45	M1	oe may be implied or embedded eg $\left(1 - \dfrac{3}{8}\right) \times 120$ or $\dfrac{5}{8} \times 120$ or 75
$(120 - \text{their } 45) \div (2 + 1)$ or $75 \div 3$ or 25	M1dep	oe may be implied or embedded eg $75 \times \dfrac{2}{3}$ or $\dfrac{5}{8} \times 120 \times \dfrac{2}{3}$
50	A1

Additional guidance

M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
25 : 50 on answer line	M1M1A0

Foundation November 2024 Paper 1 Q15

AQACurrent spec3 marksRatio & Basic Proportion

15 35 books are either for adults or for children.

number for adults : number for children = 6 : 1

How many more books are for adults than for children? [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 – Finding the value of one part first
$35 \div (6 + 1)$ or 5	M1	oe
$6 \times$ their 5 or 30 or $(6 - 1) \times$ their 5	M1dep	oe
25	A1
Alternative method 2 – Finding the difference in the number of parts first
$6 - 1$ or 5	M1	oe
$\dfrac{\text{their } 5}{6 + 1} \times 35$ or $\dfrac{5}{7} \times 35$	M1dep	oe
25	A1

Additional guidance

Allow a maximum of one error overall in build-up methods

Foundation November 2024 Paper 1 Q12

AQACurrent spec1 markRatio & Basic Proportion

12 Write the ratio 6 : 2 in the form $n : 1$ [1 mark]

Foundation November 2024 Paper 3 Q3

AQACurrent spec4 marksDecimals Ratio & Basic Proportion

(a) 3 apples cost 96p

Work out the cost, in pounds (£), of 12 of these apples. [2 marks]

(b) In total, the cost of 40 cartons of apple juice and 20 cartons of orange juice is £21.50

Work out the cost of

4 cartons of apple juice and 2 cartons of orange juice.

[2 marks]

Mark scheme (a)

Answer	Mark	Comments
$12 \div 3$ or 4 or $96 \div 3$ or 32 or $96 \times 12$ or 1152 or $96 \times 12 \div 3$ or 384	M1	oe accept working in pounds or pence
3.84	A1

Additional guidance

Condone £3.84p

Mark scheme (b)

Answer	Mark	Comments
$40 \div 4$ or $20 \div 2$ or 10 or $\dfrac{21.5(0)}{40 + 20}$ or 215	M1	oe accept working in pounds or pence
2.15	A1

Additional guidance

Condone £2.15p
Do not ignore further working eg 2.15 followed by $2.15 \times 4 + 2.15 \times 2 = 12.90$	M1A0
10 can be implied by $\dfrac{21.5}{40} \times 4$ or $\dfrac{21.5}{20} \times 2$	M1

Foundation June 2024 Paper 3 Q17

AQACurrent spec4 marksFractions Ratio & Basic Proportion

(a) There are 30 students in a class.

12 of the students have school lunch.

Work out the ratio

students having school lunch : students not having school lunch

Give your answer in its simplest form. [2 marks]

(b) In a different class

students wearing glasses : students not wearing glasses $= 3 : 8$

What fraction of students in this class wear glasses? [1 mark]

Work out the value of $n$. [1 mark]

Mark scheme (a)

Answer	Mark	Comments
$12 : 18$	B1	oe eg $6 : 9$ may be implied by correct answer
$2 : 3$ or $1 : 1.5$ or $1 : \dfrac{3}{2}$ or $\dfrac{2}{3} : 1$	B1ft	ft their ratio

Additional guidance

Accept [0.66, 0.67] for $\dfrac{2}{3}$
$2 : 3$	B1B1
Answer $1 : \dfrac{6}{4}$	B1B0
$12 : 30$ followed by $2 : 5$	B0B1ft

Foundation June 2024 Paper 1 Q13

AQACurrent spec3 marksFractions Ratio & Basic Proportion

(a) Two friends share £240 in the ratio 1 : 3

Work out the larger share. [2 marks]

(b) A tennis player wins or loses matches in the ratio win : lose = 5 : 9

What fraction of the matches do they win? [1 mark]

Mark scheme (a)

Answer	Mark	Comments
$240 \div (1 + 3)$ or $240 \div 4$ or 60	M1	oe
180	A1

Additional guidance

$240 \div 4$ and $240 \div 3$ is choice unless the answer comes from $240 \div 4$
60 : 180 or 180 : 60 with no answer chosen	M1A0

Foundation June 2024 Paper 2 Q9

AQACurrent spec3 marksRatio & Basic Proportion Units & Time

9 An electric car uses 1 unit of electricity to travel 3 miles.

1 unit of electricity costs 50 pence.

Work out the cost of electricity, in pounds, to travel 270 miles. [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
$270 \div 3$ or 90 or $270 \times 50$ or 13 500 or $270 \times 0.5(0)$ or 135	M1	oe
$270 \div 3 \times 50$ or $270 \div 3 \times 0.5(0)$	M1dep	oe
45	A1
Alternative method 2
Correctly finds units and cost for at least 30 miles	M1	eg 30 miles $=$ 10 units and 10 units cost (£)5
Correct method for cost of 270 miles	M1dep
45	A1

Additional guidance

45.00(p)	M1M1A1
45.0	M1M1A0
Accept 0.33(…) or 33.(…)% for $\dfrac{1}{3}$
Further work eg $45 \div 2$	M1M1A0
Embedded answer of 45 but not selected	M1M1A0

Foundation June 2024 Paper 2 Q4

AQACurrent spec2 marksDecimals Ratio & Basic Proportion

(a) One lettuce costs £1.29

How much do seven of these lettuces cost? [1 mark]

(b) Five cucumbers cost £6.40 in total.

How much do two of these cucumbers cost? [1 mark]

Foundation June 2024 Paper 1 Q2

AQACurrent spec4 marksRatio & Basic Proportion Units & Time

(a) Complete the statement. [1 mark]

2 metres = $\ldots\ldots\ldots$ centimetres

(b) Complete the statement. [1 mark]

8 kilograms = $\ldots\ldots\ldots$ grams

Use 8 kilometres = 5 miles [2 marks]

Mark scheme (c)

Answer	Mark	Comments
$24 \div 8$ or 3 or $5 \div 8$ or 0.625 or $8 \div 5$ or 1.6 or $24 \times 5$ or 120 or $24 \div 8 \times 5$ or 24 : 15	M1	oe eg $\dfrac{5}{8} \times 24$
15	A1

Additional guidance

$8 - 5 = 3$

Foundation November 2023 Paper 1 Q29

AQACurrent spec3 marksRatio & Basic Proportion Standard Form

(a) Write $\quad 2 \times 10^3 \quad$ as an ordinary number. [1 mark]

(b) Simplify $\quad (2 \times 10^3) : (5 \times 10^{-1})$

Give your answer in the form $\quad n : 1$ [2 marks]

Mark scheme (b)

Answer	Mark	Comments
0.5 or $\dfrac{2 \times 10^3}{5 \times 10^{-1}}$ or $\dfrac{\text{their } 2000}{5 \times 10^{-1}}$ or $0.4 \times 10^{3 - (-1)}$ or $0.4 \times 10^4$	M1	oe their 2000 from part (a)
4000 or $4 \times 10^3$	A1ft	ft $2 \times$ their 2000 in part (a)

Foundation November 2023 Paper 2 Q16

AQACurrent spec1 markRatio & Basic Proportion

16 $r : w = 1 : 4$

Write down the ratio $\quad r^2 : w^2$ [1 mark]

Foundation November 2023 Paper 3 Q14

AQACurrent spec4 marksDecimals Ratio & Basic Proportion

14 Des delivers takeaways to houses A, B and C.

Scale: 1 cm represents 3 miles

Not drawn accurately

Map of the routes. T (Takeaway) to A is 0.6 cm. T to B is 1.3 cm. B to C is 0.8 cm. C to T is 1.7 cm

Des drives

from T to A and back

and

from T to B, then B to C, then C to T.

Des is paid

40p for each mile he drives

and

£1.35 for each house he delivers to.

How much is Des paid in total for this work? [4 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
$0.6 + 0.6 + 1.3 + 0.8 + 1.7$ or 5	M1	oe condone one missing or incorrect distance
their $5 \times 3$ or 15	M1dep
their $15 \times 0.4(0) + 3 \times 1.35$ or $6(.00) + 4.05$	M1dep	oe working in pence their 15 can be a truncated or rounded up decimal
10.05	A1	SC2 6.05
Alternative method 2
$0.6 \times 3$ or 1.8 or $1.2 \times 3$ or 3.6 or $1.3 \times 3$ or 3.9 or $0.8 \times 3$ or 2.4 or $1.7 \times 3$ or 5.1	M1
$0.6 \times 3 + 0.6 \times 3 + 1.3 \times 3 + 0.8 \times 3 + 1.7 \times 3$ or $1.8 + 1.8 + 3.9 + 2.4 + 5.1$ or 15	M1dep	oe condone one missing or incorrect distance
their $15 \times 0.4(0) + 3 \times 1.35$ or $6(.00) + 4.05$	M1dep	oe working in pence their 15 can be a truncated or rounded up decimal
10.05	A1	SC2 6.05
Alternative method 3
$0.6 \times 3$ or 1.8 or $1.2 \times 3$ or 3.6 or $1.3 \times 3$ or 3.9 or $0.8 \times 3$ or 2.4 or $1.7 \times 3$ or 5.1	M1
$0.6 \times 3 \times 0.4$ or 0.72 or $1.2 \times 3 \times 0.4$ or 1.44 or $1.3 \times 3 \times 0.4$ or 1.56 or $0.8 \times 3 \times 0.4$ or 0.96 or $1.7 \times 3 \times 0.4$ or 2.04	M1dep	oe working in pence
$0.6 \times 3 \times 0.4 \times 2 + 1.3 \times 3 \times 0.4 + 0.8 \times 3 \times 0.4 + 1.7 \times 3 \times 0.4 + 3 \times 1.35$ or $0.72 \times 2 + 1.56 + 0.96 + 2.04 + 3 \times 1.35$ or $6(.00) + 4.05$	M1dep	oe working in pence condone one missing or incorrect distance
10.05	A1	SC2 6.05

Additional guidance

Units must be consistent unless recovered
All correct except for missing the return distance from A eg $0.6 + 1.3 + 0.8 + 1.7$ or 4.4 $4.4 \times 3 = 13.2$ $13.2 \times 0.4 + 3 \times 1.35 = 9.33$	M1M1M1A0
Example using their 15 truncated or rounded up for each mile $0.6 + 1.3 + 0.8 + 1.7$ or 4.4 $4.4 \times 3 = 13.2$ $13 \times 0.4 + 3 \times 1.35 = 9.25$ or $14 \times 0.4 + 3 \times 1.35 = 9.65$	M1M1M1A0
$3.6 + 3.9 + 5.1$ (BC missing)	M1M1
$3.9 + 2.4 + 5.1$ without further working (TA and AT missing)	M1M0

Foundation November 2023 Paper 2 Q12

AQACurrent spec3 marksFractions Ratio & Basic Proportion

(a) A sauce is made from cream and stock in the ratio

cream : stock $= 1 : 6$

How much sauce can be made using 80 millilitres of cream? [2 marks]

(b) A different sauce is made from olive oil and tomato juice in the ratio

olive oil : tomato juice $= 1 : 14$

What fraction of this sauce is olive oil? [1 mark]

Mark scheme (b)

Answer	Mark	Comments
$\dfrac{1}{15}$	B1	oe fraction

Additional guidance

Decimal, percentage or ratio answer	B0
Do not allow 1 in 15 or 1 out of 15 unless the correct fraction seen
$\dfrac{6}{100}$ or $\dfrac{7}{100}$ or $\dfrac{6.\dot{6}}{100}$ or $\dfrac{6.7}{100}$	B0

Foundation November 2023 Paper 1 Q11

AQACurrent spec5 marksDecimals Ratio & Basic Proportion

11 Last week, Erik worked for 10 hours.

He paid no tax and received £90

(a) This week, Erik works for 15 hours.

He assumes he will be paid at the same rate and pay no tax.

How much does he expect to receive this week? [3 marks]

(b) In fact, this week, Erik

is paid an extra 50p per hour
pays £8.90 tax.

What does this tell you about the amount he receives?

Tick one box.

It is less than he expected
It is the same as he expected
It is more than he expected

Show working to support your answer. [2 marks]

Mark scheme (a)

Answer	Mark	Comments
Alternative method 1
$90 \div 10$ or 9	M1	oe
their $9 \times 15$	M1dep	oe
135(.00)	A1
Alternative method 2
$15 \div 10$ or 1.5 or $10 \div 15$ or $\dfrac{2}{3}$	M1	oe
$90 \times$ their 1.5 or $90 \div$ their $\dfrac{2}{3}$	M1dep	oe
135(.00)	A1
Alternative method 3
$90 \div (10 \div (15 - 10))$ or 45	M1	oe
$90 +$ their 45 or their $45 \times 3$	M1dep	oe
135(.00)	A1

Additional guidance

Allow one error in a build-up method

Mark scheme (b)

Answer	Mark	Comments
Alternative method 1: works out extra from hourly increase or totals (and compares to tax)
$15 \times (0.)50$ or $15 \div 2$ or $15 \times ($their $9 + 0.5) - 15 \times$ their 9 or $142.5(0) -$ their 135 or 7.5(0)	M1	oe may be working in pence or pounds
It is less than he expected and 7.5(0)	A1ft	correct or ft their hourly rate and/or their answer in (a)
Alternative method 2: works out actual amount (and compares to expected amount)
$15 \times ($their $9 + 0.5) - 8.9(0)$ or $142.5(0) - 8.9(0)$ or 133.6(0)	M1	oe may be working in pence or pounds
Correct box ticked for comparison with their (a) and 133.6(0)	A1ft	correct or ft their hourly rate and/or their answer in (a)

Additional guidance

It’s 1.40 less with 135 in (a) (with no box ticked)	M1A1
Allow one error in a build-up method
Ignore further work after correct answer seen

Foundation November 2023 Paper 3 Q7

AQACurrent spec3 marksRatio & Basic Proportion Units & Time

7 The total cost of broadband for 5 months is £99.20

At the same monthly rate, work out the total cost of broadband for 2 years. [3 marks]

Mark scheme

Answer	Mark	Comments
$99.2(0) \div 5 \times 24$ or $19.84 \times 24$ or $4.8 \times 99.2(0)$ or $2380.8(0) \div 5$ or $\dfrac{11\,904}{25}$	M2	oe M1 $99.2(0) \div 5$ or 19.84 oe or $24 \div 5$ or 4.8 oe or $24 \times 99.2(0)$ or 2380.8(0) oe
476.16	A1

Additional guidance

(One year’s broadband cost) 238.08 oe implies M1

Foundation June 2017 Paper 1 Q24

AQACurrent spec1 markFractions Ratio & Basic Proportion

\[\dfrac{1}{2} : \dfrac{2}{3} = x : 1\]

Circle the value of $x$. [1 mark]

$\dfrac{1}{3}$
$\dfrac{3}{5}$
$\dfrac{3}{4}$
$\dfrac{4}{3}$

Foundation June 2017 Paper 3 Q21

AQACurrent spec3 marksRatio & Basic Proportion

21 Purple paint is made by mixing red paint and blue paint in the ratio 5 : 2

Yan has 30 litres of red paint and 9 litres of blue paint.

What is the maximum amount of purple paint he can make? [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
Any correct scaling of the ratio 5 : 2 eg 10 (:) 4 or 20 (:) 8 or 25 (:) 10	M1	oe
22.5 (:) 9 or 22.5 (red) or 30 (:) 12 or 12 (blue)	M1dep	oe
31.5 or $31\dfrac{1}{2}$ or $\dfrac{63}{2}$	A1
Alternative method 2
$9 \div 2$ or 4.5 or $30 \div 5$ or 6	M1	oe $2 \div 9$ or 0.22… $5 \div 30$ or 0.16… or 0.17
$5 \times$ their 4.5 or 22.5 or $7 \times$ their 4.5 or $2 \times$ their 6 or 12 or $7 \times$ their 6 or 42	M1dep	oe
31.5 or $31\dfrac{1}{2}$ or $\dfrac{63}{2}$	A1
Alternative method 3
$\dfrac{2}{7} \times$ purple = blue $\dfrac{5}{7} \times$ purple = red	M1	oe $\dfrac{2}{7} \times$ purple = 9 $\dfrac{5}{7} \times$ purple = 30
$9 \times \dfrac{7}{2}$ or $30 \times \dfrac{7}{5}$ or 42	M1dep	oe
31.5 or $31\dfrac{1}{2}$ or $\dfrac{63}{2}$	A1

Additional guidance

$28 + 3.5 = 31.5$	M1M1A1
$28 + 3.5$	M1M1A0
31.5, answer 31	M1M1A1
$31.5 + 42 = 73.5$	M1M1A0
10 4	M1M0A0
10, 4	M1M0A0
$10 + 4$	M1M0A0
‘He has 2.5 times more red than blue’	M1M0A0
2.5 : 1	M1M0A0
2.5	M0M0A0
28 on its own	M0M0A0

Foundation June 2017 Paper 2 Q19

AQACurrent spec2 marksRatio & Basic Proportion

19 Ben and Katy throw darts at a target.

Ben’s ratio of hits to misses is 5 : 1
Katy’s ratio of hits to misses is 3 : 1

Ben says,

“5 is bigger than 3, so I must have more hits than Katy.”

Give an example to show that this might not be true. [2 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 (hits and misses)
A counter example using both ratios or using numbers of hits and misses for both players	B2	eg Katy could be 6 : 2 and Ben hit 5 eg Ben 10 hits and 2 misses and Katy 12 hits and 4 misses B1 for a correct number of hits and misses (not 3 and 1) or a correct equivalent ratio for Katy
Alternative method 2 (hits and total throws or proportion of hits)
A counter example using total throws and number of hits for both players or using proportion of hits for both players	B2	eg Katy could have hit 6 out of 8, Ben hit 5 eg Katy could have $\dfrac{18}{24}$ and Ben $\dfrac{10}{12}$ B1 for a correct number of total throws and hits (not 3 out of 4) or a correct proportion of hits (not $\dfrac{3}{4}$) for Katy

Additional guidance

Must use the given ratios
(Ben) 5 : 1 (Katy) 6 : 2	B2
15 : 3 and 15 : 5 (so the same hits)	B2
(Katy) 6 : 2 or (Katy) 6 hits and 2 misses	B1
List of equivalent ratios for (Ben and) Katy with no counter example chosen	B1
15 : 3 and 9 : 3	B1
Fractions of hits out of total throws oe percentages or decimals or words eg $\dfrac{5}{6}$ and $\dfrac{3}{4}$ eg $\dfrac{20}{24}$ and $\dfrac{18}{24}$ eg $\dfrac{5}{6}$ and $\dfrac{6}{8}$	B0 B1 B2
Ben had (two) more throws – he had 6 and she had 4	B0

Higher June 2017 Paper 2 Q19

AQACurrent spec5 marksRatio & Basic Proportion

19 Rana sells 192 cakes in the ratio $\quad$ small : medium : large $\;=\; 7 : 6 : 11$

The profit for one medium cake is twice the profit for one small cake.
The profit for one large cake is three times the profit for one small cake.

Her total profit is £532.48

Work out the profit for one small cake. [5 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
$192 \div (7 + 6 + 11)$ or $192 \div 24$ or 8	M1	May be implied
$7 \times$ their 8 or 56 and $6 \times$ their 8 or 48 and $11 \times$ their 8 or 88	M1dep	56 : 48 : 88 is M2
their 56 $(\times 1)\ +$ their 48 $\times 2$ $+$ their 88 $\times 3$ or $56 + 96 + 264$ or 416	M1dep	May use $x$, $2x$ and $3x$ (any letter)
$532.48 \div$ their 416	M1dep	Allow working in pence
1.28	A1
Alternative method 2
$7\ (\times 1) + 6 \times 2 + 11 \times 3$ or $7 + 12 + 33$ or 52	M1	May use $x$, $2x$ and $3x$ (any letter)
$532.48 \div$ their 52 or 10.24	M1dep	Allow working in pence
$192 \div (7 + 6 + 11)$ or $192 \div 24$ or 8	M1	May be implied
their $10.24 \div$ their 8	M1dep	dep on M3 oe eg their $10.24 \times 7$ or 71.68 and their $71.68 \div (7 \times$ their 8)
1.28	A1

Additional guidance

(Alt 1) 56 : 96 : 264 with no subsequent addition	M1M1M0M0A0
$532.48 \div 24$ ($=$ 22.18… or 22.19) with no further valid working	Zero
$532.48 \div 192$ ($=$ 2.77…) with no further valid working	Zero
(Alt 1) 56 and 48 and 88 (or correct method leading to them) but not subsequently used	M2
(Alt 1) 8 (or correct method leading to it) but not subsequently used	M1
(Alt 2) 10.24 (or correct method leading to it)	M2
1.28 in working with Answer 71.68 (from $1.28 \times 56$)	M4A0

Foundation June 2017 Paper 3 Q18

AQACurrent spec5 marksRatio & Basic Proportion

18 A shop sells two brands of battery.

One brand A battery powers a toy for 5 hours.

One brand B battery powers the same toy for $5\dfrac{1}{2}$ hours.

Which brand is better value?

You must show your working. [5 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1 of 6 – cost per hour
$3.6(0) \div 8$ or (0).45 or $2.94 \div 6$ or (0).49	M1	$360 \div 8$ or 45 or $294 \div 6$ or 49
their (0).45 $\div$ 5 or (0).09 or their (0).49 $\div$ 5.5 or (0).08(9…)	M1dep	their 45 $\div$ 5 or 9 or their 49 $\div$ 5.5 or 8.(9…)
their (0).45 $\div$ 5 and their (0).49 $\div$ 5.5	M1dep	their 45 $\div$ 5 and their 49 $\div$ 5.5
(£)0.09 and (£)0.08(9…)	A1	9(p) and 8.(9…) (p)
brand B	A1ft	ft correct decision for their values with M3 scored
Alternative method 2 of 6 – cost per hour from price of pack
$8 \times 5$ or 40 or $6 \times 5.5$ or 33	M1
$3.6(0) \div$ their 40 or (0).09 or $2.94 \div$ their 33 or (0).08(9…)	M1dep	$360 \div$ their 40 or 9 or $294 \div$ their 33 or 8.(9…)
$3.6(0) \div$ their 40 and $2.94 \div$ their 33	M1dep	$360 \div$ their 40 and $294 \div$ their 33
(£)0.09 and (£)0.08(9…)	A1	9(p) and 8.(9…) (p)
brand B	A1ft	ft correct decision for their values with M3 scored
Alternative method 3 of 6 – number of hours per unit cost from number of batteries
$3.6(0) \div 8$ or (0).45 or $2.94 \div 6$ or (0).49	M1	$360 \div 8$ or 45 or $294 \div 6$ or 49
$5 \div$ their (0).45 or 11.1(…) or $5.5 \div$ their (0).49 or 11.2(…)	M1dep	$5 \div$ their 45 or (0).111(…) or $5.5 \div$ their 49 or (0).112(…)
$5 \div$ their (0).45 and $5.5 \div$ their (0).49	M1dep	$5 \div$ their 45 and $5.5 \div$ their 49
11.1(…) (hours) and 11.2(…) (hours)	A1	(0).111(…) (hours) and (0).112(…) (hours)
brand B	A1ft	ft correct decision for their values with M3 scored
Alternative method 4 of 6 – common number of batteries
Scaling towards a cost for a common number of batteries (eg 24 batteries) eg $8 \times 3 \times 5$ or 120 and $6 \times 4 \times 5.5$ or 132	M1
eg $3 \times 3.60$ or 10.8(0) and $4 \times 2.94$ or 11.76	M1	eg $3 \times 360$ or 1080 and $4 \times 294$ or 1176
eg their 10.8(0) $\div$ their 120 or (0).09 and their 11.76 $\div$ their 132 or (0).08(9…)	M1dep	eg their 1080 $\div$ their 120 or 9 and their 1176 $\div$ their 132 or 8.(9…) dependent on M1M1
(£)0.09 and (£)0.08(9…)	A1	9(p) and 8.(9…) (p)
brand B	A1ft	ft correct decision for their values with M3 scored
Alternative method 5 of 6 – number of hours per unit cost from batteries per unit cost
$8 \div 3.6(0)$ or 2.2(…) or $6 \div 2.94$ or 2.04(…)	M1	$8 \div 360$ or 0.022(…) or $6 \div 294$ or 0.0204(…)
their 2.2(…) $\times$ 5 or 11.1(…) or their 2.04(…) $\times$ 5.5 or 11.2(…)	M1dep	their 0.022(…) $\times$ 5 or 0.111(…) or their 0.0204(…) $\times$ 5.5 or 0.112(…)
their 2.2(…) $\times$ 5 and their 2.04(…) $\times$ 5.5	M1dep	their 0.022(…) $\times$ 5 and their 0.0204(…) $\times$ 5.5
11.1(…) (hours) and 11.2(…) (hours)	A1	(0).111(…) (hours) and (0).112(…) (hours)
brand B	A1ft	ft correct decision for their values with M3 scored
Alternative method 6 of 6 – cost for common number of battery hours
$3.6(0) \div 8$ or (0).45 or $2.94 \div 6$ or (0).49	M1	$360 \div 8$ or 45 or $294 \div 6$ or 49
Scaling towards a common number of battery hours (eg 55 hours) eg their (0).45 $\times$ 11 or their (0).49 $\times$ 10	M1dep	eg their 45 $\times$ 11 or their 49 $\times$ 10
eg their (0).45 $\times$ 11 and their (0).49 $\times$ 10	M1dep	eg their 45 $\times$ 11 and their 49 $\times$ 10
eg (£)4.95 and (£)4.9(0)	A1	eg 495(p) and 490(p)
brand B	A1ft	ft correct decision for their values with M3 scored

Additional guidance

For the first A mark the values must not be rounded to the same value
A1ft can be awarded after A0 for the same value for the correct decision eg 0.09 and 0.09 with decision ‘both the same’	M3A0A1ft
$8 \times 5 = 40$ and $40 \div 3.6(0)$ and $6 \times 5.5 = 33$ and $33 \div 2.94$ is equivalent to $8 \div 3.6(0) \times 5$ and $6 \div 2.94 \times 5.5$ on Alt 5	M3
$8 \times 5 = 40$ and $40 \div 3.6(0)$ is equivalent to $8 \div 3.6(0) \times 5$ on Alt method 5	M2
$6 \times 5.5 = 33$ and $33 \div 2.94$ is equivalent to $6 \div 2.94 \times 5.5$ on Alt method 5	M2
(0).45 $\div$ 5	M1M1
(0).45 $\div$ 5 and (0).49 $\div$ 5.5	M1M1M1
(0).45 $\div$ 5 and (0).415 $\div$ 5.5 $\qquad$ 0.415 is not from a correct method	M1M1M0
In Alt method 4 M1M1 can be awarded in either order
In Alt method 5 their 2.2(…) must be correct or from correct method their 2.04(…) must be correct or from correct method
Accept misread of 4 batteries (A) or 3 batteries (B) for up to M3A0A1ft
Accept working with minutes eg in Alt method 3 for 2^nd M1dep accept $\quad 300 \div 45 = 6.6(\ldots)$ or 6.7 or $\;330 \div 49 = 6.7(\ldots)$ for 3^rd M1dep accept $\quad 300 \div 45$ and $330 \div 49$ for first A mark must see 6.6(…) or 6.67 and 6.7(…) or 6.7 and 6.73(…)

Foundation June 2017 Paper 3 Q17

AQACurrent spec2 marksRatio & Basic Proportion

17 $CD$ and $PQ$ are lines of length 12 cm

(a) $CE : CD = 1 : 2$

Mark point $E$ on the line with a cross. [1 mark]

(b) $PR : RQ = 1 : 3$

Mark point $R$ on the line with a cross. [1 mark]

Foundation June 2017 Paper 2 Q16

AQACurrent spec1 markFractions Ratio & Basic Proportion

16 The counters in a bag are red or blue.

One fifth of the counters are red.

Work out the ratio $\qquad$ red counters : blue counters

Circle your answer. [1 mark]

1 : 4
1 : 5
4 : 5
1 : 6

Foundation June 2017 Paper 1 Q15

AQACurrent spec3 marksRatio & Basic Proportion

15 In a school show,

girls : boys = 1 : 1
girls who sing : girls who do not sing = 1 : 2

8 girls sing in the show.

How many students are in the show altogether? [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
$8 \times 2$ or 16	M1	implied by 8 : 16
their 16 $+$ 8 or 24	M1dep	$8 \times 3$
48	A1
Alternative method 2
$(1 + 2 = 3)$ $3 + 3$ or 6	M1
their $6 \times 8$	M1dep	their 6 must be from $\;3 + 3$
48	A1

Additional guidance

Beware 24 coming from incorrect working
eg Misread of 8 girls who do not sing in the show, leading to answer of 24

M1M1A0

Higher June 2017 Paper 3 Q11

AQACurrent spec3 marksRatio & Basic Proportion

11 Purple paint is made by mixing red paint and blue paint in the ratio 5 : 2

Yan has 30 litres of red paint and 9 litres of blue paint.

What is the maximum amount of purple paint he can make? [3 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
Any correct scaling of the ratio 5 : 2 eg 10 (:) 4 or 20 (:) 8 or 25 (:) 10	M1	oe
22.5 (:) 9 or 22.5 (red) or 30 (:) 12 or 12 (blue)	M1dep	oe
31.5 or $31\dfrac{1}{2}$ or $\dfrac{63}{2}$	A1
Alternative method 2
$9 \div 2$ or 4.5 or $30 \div 5$ or 6	M1	oe $2 \div 9$ or 0.22… $5 \div 30$ or 0.16… or 0.17
$5 \times$ their 4.5 or 22.5 or $7 \times$ their 4.5 or $2 \times$ their 6 or 12 or $7 \times$ their 6 or 42	M1dep	oe
31.5 or $31\dfrac{1}{2}$ or $\dfrac{63}{2}$	A1
Alternative method 3
$\dfrac{2}{7} \times$ purple = blue $\dfrac{5}{7} \times$ purple = red	M1	oe $\;\dfrac{2}{7} \times$ purple = 9 $\dfrac{5}{7} \times$ purple = 30
$9 \times \dfrac{7}{2}$ or $30 \times \dfrac{7}{5}$ or 42	M1dep	oe
31.5 or $31\dfrac{1}{2}$ or $\dfrac{63}{2}$	A1

Additional guidance

28 + 3.5 = 31.5	M1M1A1
28 + 3.5	M1M1A0
31.5, answer 31	M1M1A1
31.5 + 42 = 73.5	M1M1A0
10 4	M1M0A0
10, 4	M1M0A0
10 + 4	M1M0A0
‘He has 2.5 times more red than blue’	M1M0A0
2.5 : 1	M1M0A0
2.5	M0M0A0
28 on its own	M0M0A0

Higher June 2017 Paper 1 Q4

AQACurrent spec1 markRatio & Basic Proportion

4 $a : b = 4 : 3$

Circle the correct statement. [1 mark]

$b$ is $\dfrac{4}{7}$ of $a$
$b$ is $\dfrac{3}{7}$ of $a$
$b$ is $\dfrac{4}{3}$ of $a$
$b$ is $\dfrac{3}{4}$ of $a$

Cost per lb in \(\$\): \(2 \times 2.2\ (= 4.4)\) \(3.75 \times 1.2(0)\ (= 4.5)\) \(\text{``}4.5\text{''} \div \text{``}4.4\text{''}\ (= 1.02\ldots)\) and \(4.9(0) \div 5\ (= 0.98)\) New York and 1.02… and 0.98	Cost per lb in £: \(2 \times 2.2\ (= 4.4)\) \(4.9(0) \div 1.2(0)\ (= 4.08\ldots)\) \(3.75 \div \text{``}4.4\text{''}\ (= 0.85\ldots)\) and \(\text{``}4.08\ldots\text{''} \div 5\ (= 0.81(6\ldots))\) New York and 0.85… and 0.81(6…)
Cost per kg in \(\$\): \(5 \div 2.2\ (= 2.27\ldots)\) \(3.75 \times 1.2(0)\ (= 4.5)\) \(\text{``}4.5\text{''} \div 2\ (= 2.25)\) and \(4.9(0) \div \text{``}2.27\ldots\text{''}\ (= 2.15(6))\) New York and 2.25 and 2.15(6)	Cost per kg in £: \(5 \div 2.2\ (= 2.27\ldots)\) \(4.9(0) \div 1.2(0)\ (= 4.08\ldots)\) \(3.75 \div 2\ (= 1.87(5))\) and \(\text{``}4.08\ldots\text{''} \div \text{``}2.27\ldots\text{''}\ (= 1.79(6\ldots))\) New York and 1.87(5) and 1.79(6…)
Number of lbs per \(\$\): \(2 \times 2.2\ (= 4.4)\) \(3.75 \times 1.2(0)\ (= 4.5)\) \(\text{``}4.4\text{''} \div \text{``}4.5\text{''}\ (= 0.97(7\ldots))\) and \(5 \div 4.9(0)\ (= 1.02\ldots)\) New York and 0.97(7…) and 1.02…	Number of lbs per £: \(2 \times 2.2\ (= 4.4)\) \(4.9(0) \div 1.2(0)\ (= 4.08\ldots)\) \(\text{``}4.4\text{''} \div 3.75\ (= 1.17\ldots)\) and \(5 \div \text{``}4.08\ldots\text{''}\ (= 1.22\ldots)\) New York and 1.17… and 1.22…
Number of kg per \(\$\): \(5 \div 2.2\ (= 2.27\ldots)\) \(3.75 \times 1.2(0)\ (= 4.5)\) \(2 \div \text{``}4.5\text{''}\ (= 0.44\ldots)\) and \(\text{``}2.27\ldots\text{''} \div 4.9(0)\ (= 0.46\ldots)\) New York and 0.44… and 0.46…	Number of kg per £: \(5 \div 2.2\ (= 2.27\ldots)\) \(4.9(0) \div 1.2(0)\ (= 4.08\ldots)\) \(2 \div 3.75\ (= 0.53\ldots)\) and \(\text{``}2.27\ldots\text{''} \div \text{``}4.08\ldots\text{''}\ (= 0.55(6\ldots))\) New York and 0.53… and 0.55(6…)
Cost of 5 lb in \(\$\): \(2 \times 2.2\ (= 4.4)\) \(3.75 \times 1.2(0)\ (= 4.5)\) \(\text{``}4.5\text{''} \div \text{``}4.4\text{''} \times 5\ (= 5.11\ldots)\) New York and 5.11…	Cost of 5 lb in £: \(2 \times 2.2\ (= 4.4)\) \(4.9(0) \div 1.2(0)\ (= 4.08\ldots)\) \(3.75 \div \text{``}4.4\text{''} \times 5\ (= 4.26\ldots)\) New York and 4.08… and 4.26…
Cost of 2 kg in \(\$\): \(5 \div 2.2\ (= 2.27\ldots)\) \(3.75 \times 1.2(0)\ (= 4.5)\) \(4.9(0) \div \text{``}2.27\ldots\text{''} \times 2\ (= 4.31\ldots)\) New York and 4.5 and 4.31…	Cost of 2 kg in £: \(5 \div 2.2\ (= 2.27\ldots)\) \(4.9(0) \div 1.2(0)\ (= 4.08\ldots)\) \(\text{``}4.08\ldots\text{''} \div \text{``}2.27\ldots\text{''} \times 2\ (= 3.59\ldots)\) New York and 3.59…

Cost per lb in \(\$\): \(2 \times 2.2\ (= 4.4)\) \(3.75 \times 1.2(0)\ (= 4.5)\) \(\text{``}4.5\text{''} \div \text{``}4.4\text{''}\ (= 1.02\ldots)\) and \(4.9(0) \div 5\ (= 0.98)\) New York and 1.02… and 0.98	Cost per lb in £: \(2 \times 2.2\ (= 4.4)\) \(4.9(0) \div 1.2(0)\ (= 4.08\ldots)\) \(3.75 \div \text{``}4.4\text{''}\ (= 0.85\ldots)\) and \(\text{``}4.08\ldots\text{''} \div 5\ (= 0.81(6\ldots))\) New York and 0.85… and 0.81(6…)
Cost per kg in \(\$\): \(5 \div 2.2\ (= 2.27\ldots)\) \(3.75 \times 1.2(0)\ (= 4.5)\) \(\text{``}4.5\text{''} \div 2\ (= 2.25)\) and \(4.9(0) \div \text{``}2.27\ldots\text{''}\ (= 2.15(6))\) New York and 2.25 and 2.15(6)	Cost per kg in £: \(5 \div 2.2\ (= 2.27\ldots)\) \(4.9(0) \div 1.2(0)\ (= 4.08\ldots)\) \(3.75 \div 2\ (= 1.87(5))\) and \(\text{``}4.08\ldots\text{''} \div \text{``}2.27\ldots\text{''}\ (= 1.79(6\ldots))\) New York and 1.87(5) and 1.79(6…)
Number of lbs per \(\$\): \(2 \times 2.2\ (= 4.4)\) \(3.75 \times 1.2(0)\ (= 4.5)\) \(\text{``}4.4\text{''} \div \text{``}4.5\text{''}\ (= 0.97(7\ldots))\) and \(5 \div 4.9(0)\ (= 1.02\ldots)\) New York and 0.97(7…) and 1.02…	Number of lbs per £: \(2 \times 2.2\ (= 4.4)\) \(4.9(0) \div 1.2(0)\ (= 4.08\ldots)\) \(\text{``}4.4\text{''} \div 3.75\ (= 1.17\ldots)\) and \(5 \div \text{``}4.08\ldots\text{''}\ (= 1.22\ldots)\) New York and 1.17… and 1.22…
Number of kg per \(\$\): \(5 \div 2.2\ (= 2.27\ldots)\) \(3.75 \times 1.2(0)\ (= 4.5)\) \(2 \div \text{``}4.5\text{''}\ (= 0.44\ldots)\) and \(\text{``}2.27\ldots\text{''} \div 4.9(0)\ (= 0.46\ldots)\) New York and 0.44… and 0.46…	Number of kg per £: \(5 \div 2.2\ (= 2.27\ldots)\) \(4.9(0) \div 1.2(0)\ (= 4.08\ldots)\) \(2 \div 3.75\ (= 0.53\ldots)\) and \(\text{``}2.27\ldots\text{''} \div \text{``}4.08\ldots\text{''}\ (= 0.55(6\ldots))\) New York and 0.53… and 0.55(6…)
Cost of 5 lb in \(\$\): \(2 \times 2.2\ (= 4.4)\) \(3.75 \times 1.2(0)\ (= 4.5)\) \(\text{``}4.5\text{''} \div \text{``}4.4\text{''} \times 5\ (= 5.11\ldots)\) New York and 5.11…	Cost of 5 lb in £: \(2 \times 2.2\ (= 4.4)\) \(4.9(0) \div 1.2(0)\ (= 4.08\ldots)\) \(3.75 \div \text{``}4.4\text{''} \times 5\ (= 4.26\ldots)\) New York and 4.08… and 4.26…
Cost of 2 kg in \(\$\): \(5 \div 2.2\ (= 2.27\ldots)\) \(3.75 \times 1.2(0)\ (= 4.5)\) \(4.9(0) \div \text{``}2.27\ldots\text{''} \times 2\ (= 4.31\ldots)\) New York and 4.5 and 4.31…	Cost of 2 kg in £: \(5 \div 2.2\ (= 2.27\ldots)\) \(4.9(0) \div 1.2(0)\ (= 4.08\ldots)\) \(\text{``}4.08\ldots\text{''} \div \text{``}2.27\ldots\text{''} \times 2\ (= 3.59\ldots)\) New York and 3.59…

	London	Zurich
100g	\(3.5 \div 2 =\) £1.75 \(1.75 \times 1.25 =\) 2.1875 F	\(7.2 \div 360 =\) 2.00 F \(2.00 \div 1.25 =\) £1.60
200g	£3.50 \(3.5 \times 1.25 =\) 4.375 F	\(7.2 \div 360 \times 200 =\) 4.0 F \(4.0 \div 1.25 =\) £3.20
360g	\(3.5 \div 200 \times 360 =\) £6.30 \(6.30 \times 1.25 =\) 7.875 F	7.20 F \(7.20 \div 1.25 =\) £5.76
1g	£3.50 \(\div 200 =\) £0.0175 \(\times 1.25 =\) 0.021875 F	\(7.20 \div 360 =\) 0.02 F \(\div 1.25 =\) £0.016
40g	£3.50 \(\div 5 =\) £0.70 \(0.7 \times 1.25 =\) 0.875 F	\(7.20 \div 9 =\) 0.8 F \(0.8 \div 1.25 =\) £0.64
By weight	\(350 \div 200 =\) 1.75 p/g \(350 \times 1.25 = 4.375\) \(4.375 \div 200 =\) 0.021875 F/g	\(720 \div 360 =\) 0.02 F/g \(720 \div 1.25 = 576\) \(576 \div 360 =\) 1.6 p/g
By cost	\(200 \div 350 =\) 0.571 g/p \(350 \times 1.25 = 437.5\) \(200 \div 437.5 =\) 45.7 g/F	\(360 \div 720 =\) 50 g/F \(720 \div 1.25 = 576\) \(360 \div 576 =\) 0.625 g/P

	London	Zurich
100g	\(3.5 \div 2 =\) £1.75 \(1.75 \times 1.25 =\) 2.1875 F	\(7.2 \div 360 \times 100 =\) 2.00 F \(2.00 \div 1.25 =\) £1.60
200g	£3.50 \(3.5 \times 1.25 =\) 4.375 F	\(7.2 \div 360 \times 200 =\) 4.0 F \(4.0 \div 1.25 =\) £3.20
360g	\(3.5 \div 200 \times 360 =\) £6.30 \(6.30 \times 1.25 =\) 7.875 F	7.20 F \(7.20 \div 1.25 =\) £5.76
1g	£3.50 \(\div 200 =\) £0.0175 \(\times 1.25 =\) 0.021875 F	\(7.20 \div 360 =\) 0.02 F \(\div 1.25 =\) £0.016
40g	£3.50 \(\div 5 =\) £0.70 \(0.7 \times 1.25 =\) 0.875 F	\(7.20 \div 9 =\) 0.8 F \(0.8 \div 1.25 =\) £0.64
By weight	\(350 \div 200 =\) 1.75 p/g \(3.50 \times 1.25 = 4.375\) \(4.375 \div 200 =\) 0.021875 F/g	\(7.20 \div 360 =\) 0.02 F/g \(720 \div 1.25 = 576\) \(576 \div 360 =\) 1.6 p/g

Accept [0.66, 0.67] for \(\dfrac{2}{3}\)
\(2 : 3\)	B1B1
Answer \(1 : \dfrac{6}{4}\)	B1B0
\(12 : 30\) followed by \(2 : 5\)	B0B1ft