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Edexcel IGCSE 4MA1/2FR, Wednesday 4 June 2025: Worked Solutions and Mark Schemes

Sir Faraz Hassan

26 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)4MA1/2FR - Foundation Tier - Wednesday 4 June 2025100 marks · 2 hours · Calculator allowed

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/2FR (Foundation Tier), June 2025 series, sat Wednesday 4 June 2025 –100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

Both are PDF files hosted by Pearson: official question paper (PDF) and official mark scheme (PDF).

Try each question yourself first, then open the worked solution to check your method and see exactly where each method mark (M1) and accuracy mark (A1) is earned. The questions follow the same order as the original paper and carry the same marks.

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Every question with a full worked solution and mark scheme - free PDF

Worked solutions, questions 1 to 16 of 27

(a) Put these numbers in order of size, starting with the smallest.

$83 \qquad 148 \qquad 46 \qquad 229 \qquad 99$ [1 mark]

(b) Put these decimals in order of size, starting with the smallest.

$0.25 \qquad 0.5 \qquad 0.46 \qquad 0.08 \qquad 0.417$ [1 mark]

(d) Work out the number that is exactly halfway between $0.2$ and $0.3$ [1 mark]

(a)(b)(c)(d)

[Total 4 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

Five whole numbers:

83

148

46

229

99

Five decimals:

0.25

0.5

0.46

0.08

0.417

The decimal

0.81

The two decimals

0.2

and

0.3

None of the five whole numbers is repeated and no two of the five decimals are equal, so each list has one order and one order only.

Find

(a) the five whole numbers written out smallest first (b) the five decimals written out smallest first (c)

0.81

written as a fraction (d) the number exactly halfway between

0.2

and

0.3

Plan the Solution

Count the digits first. A three-digit number is bigger than any two-digit number, so that splits the list before a single digit is compared.
Fill the shorter decimals out with zeros until all five have three decimal places, the longest one there. Once they are all the same length they can be compared like whole numbers.
Count the decimal places in $0.81$ . Two places means hundredths, so the denominator is $100$ .
Halfway between two numbers is the midpoint: add the two numbers and halve the total.

Worked Solution [4 marks]

Rule - Place value: compare two numbers column by column from the left, and the first column where they differ decides which one is larger. Zeros written on the end of a decimal change nothing, and a decimal with two decimal places is a number of hundredths.

Step 1: (a) split the whole numbers by how many digits they have

46 < 83 < 99

148 < 229

46 < 83 < 99 < 148 < 229

(Reason: Every two-digit number is smaller than every three-digit number, so

46

83

and

99

all come before

148

and

229

. Inside each group the leading digit settles it:

4

, then

8

, then

9

in the tens column, and

1

before

2

in the hundreds column.)

Step 2: (b) give every decimal the same number of decimal places

0.25 = 0.250

0.5 = 0.500

0.46 = 0.460

0.08 = 0.080

0.417 = 0.417

(Reason:

0.417

has the most decimal places, three, so every decimal is filled out to three places. A zero written on the end of a decimal adds nothing to its value - it only records that there are no thousandths.)

Step 3: (b) compare the decimals as whole numbers of thousandths

80 < 250 < 417 < 460 < 500

0.08 < 0.25 < 0.417 < 0.46 < 0.5

(Reason: With three decimal places each decimal is a whole number of thousandths:

0.080

80

of them and

0.500

500

. Whole numbers are easy to put in order, and the order they come in is the order the decimals come in.)

Step 4: (c) read the decimal as a number of hundredths

0.81 = \dfrac{8}{10} + \dfrac{1}{100}

0.81 = \dfrac{80}{100} + \dfrac{1}{100} = \dfrac{81}{100}

(Reason: The

8

sits in the tenths column and the

1

in the hundredths column, so together they make

81

hundredths. Two decimal places always give a denominator of

100

, and since

81

and

100

share no factor, that fraction is already as simple as it goes.)

Step 5: (d) take the midpoint of the two decimals

\dfrac{0.2 + 0.3}{2} = \dfrac{0.5}{2} = 0.25

(Reason: Halfway between two numbers means the midpoint, and the midpoint is what you reach by adding the two numbers together and halving the total. That works whatever the two numbers are.)

(a)

46

83

99

148

229

(b)

0.08

0.25

0.417

0.46

0.5

(c)

\dfrac{81}{100}

(d)

0.25

Verification

Check 1: Read the answer to (a) backwards. If the list really runs smallest to largest, then read the other way it must run largest to smallest, and it must still hold the same five numbers the question printed. Backwards it reads

229

148

99

83

46

- largest first, the same five numbers, none repeated and none missing.

Check 2: Order the decimals a second way, without lining up any decimal places. Compare each one against a landmark instead:

0.08

is less than a tenth,

0.25

is a quarter,

0.417

is a little over four tenths,

0.46

is just under a half and

0.5

is exactly a half. Those landmarks fall in that same order, so the list matches the one the padding method gave.

Check 3: Turn the fraction back into a decimal. Dividing by

100

moves every digit two columns to the right.

\dfrac{81}{100} = 0.81

, which is the decimal the question printed.

Check 4: Measure the distance from the answer to each end of the pair.

0.25 - 0.2

and

0.3 - 0.25

should come out the same size as each other. Both gaps come to

0.05

, so

0.25

sits the same distance from

0.2

as it does from

0.3

, which is what halfway means.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $46$ , $83$ , $99$ , $148$ , $229$	B1	Correct answer only.	✓
(b) $0.08$ , $0.25$ , $0.417$ , $0.46$ , $0.5$	B1	Correct answer only.	✓
(c) $\dfrac{81}{100}$	B1	Any equivalent fraction is accepted.	✓
(d) $0.25$	B1	Allow $\dfrac{1}{4}$ or any equivalent.	✓

Full marks: 4/4

Marina recorded the ages, in years, of the seven cats at an animal rescue centre.
Here are her results.

$2 \qquad 9 \qquad 10 \qquad 6 \qquad 7 \qquad 9 \qquad 4$

(a) Write down the mode of the ages. [1 mark]

(b) Work out the median age. [2 marks]

(a)(b)(c)

[Total 4 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

The seven ages, in the order they were recorded:

2 \qquad 9 \qquad 10 \qquad 6 \qquad 7 \qquad 9 \qquad 4

Seven values, and

7

is odd, so once they are written in order of size exactly one of them sits in the middle.

Find

(a) the mode of the ages (b) the median age (c) the range of the ages

Plan the Solution

Write the seven ages in order of size first. The median cannot be read off until they are ordered, and ordering them also puts equal ages side by side for the mode and the smallest and largest ages at the two ends for the range.
The mode is the age that appears most often, so count how many times each age appears. The counts have to add back up to $7$ .
With $7$ ages in order, the middle one is the $4$ th, because $3$ ages sit below it and $3$ sit above it.
The range is one subtraction: the largest age take away the smallest age.

Worked Solution [4 marks]

Rule - Mode, median and range: the mode is the value that appears most often, the median is the middle value once the data is written in order of size, and the range is

\text{largest} - \text{smallest}

. Only the median needs the data ordered, but ordering it once serves all three.

Step 1: put the seven ages in order of size

2 \quad 4 \quad 6 \quad 7 \quad 9 \quad 9 \quad 10

(Reason: The same seven ages the question printed, smallest first, with nothing added, dropped or altered. Part (b) awards a method mark for this ordering on its own, and it sets the other two parts up as well: the two

9

s now sit side by side for part (a), and the smallest and largest ages sit at the two ends for part (c).)

Step 2: (a) count how many times each age appears

9 \text{ appears twice}

\text{mode} = 9

(Reason: Reading along the ordered list, the two

9

s are the only pair:

2

4

6

7

and

10

each appear once and

9

appears twice, which accounts for all

7

ages. The mode is the age that appears most often, so it is the age

9

itself, never the

2

that says how many times it was recorded.)

Step 3: (b) take the middle value of the ordered list

\text{middle position} = \dfrac{7 + 1}{2} = 4

2 \quad 4 \quad 6 \quad 7 \quad 9 \quad 9 \quad 10

\text{median} = 7

(Reason: There are

7

ages, an odd number, so one of them sits exactly in the middle. Counting in from both ends at once,

2

4

and

6

lie below it and

9

9

and

10

lie above it, which leaves the

4

th value of the ordered list on its own in the middle. The ordering has to come first: the middle of the list as it was recorded is not the median.)

Step 4: (c) take the smallest age from the largest

\text{range} = 10 - 2 = 8

(Reason: The ordered list has the largest age,

10

, at one end and the smallest,

2

, at the other, and the range is the gap between them. It measures how spread out the ages are, so it is a single number,

8

, and not the pair

2

10

(a)

9

(b)

7

(c)

8

Verification

Check 1: Go back to the list as the question printed it and tally the ages there, with nothing put in order. If the mode is right, one age appears more often than any other and the tallies add back up to seven.

9

appears twice and

2

4

6

7

and

10

appear once each, which is

2 + 5 = 7

ages altogether, so

9

is the only age recorded more than once.

Check 2: Write the ages in order the other way round, largest first, and count to the same position. The mark scheme accepts either ordering, so the middle value has to come out the same counting from either end. Largest first the list reads

10 \quad 9 \quad 9 \quad 7 \quad 6 \quad 4 \quad 2

, and the

4

th value from that end is

7

, with

3

ages on each side of it.

Check 3: Find the range without subtracting. Count up in ones from the smallest age to the largest, and separately add up the gaps between neighbouring ages in the ordered list, which must come to the same total. Counting

2

up to

10

takes

8

steps, and the six gaps

2 + 2 + 1 + 2 + 0 + 1

also come to

8

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $9$	B1	Correct answer only.	✓
(b) $2 \quad 4 \quad 6 \quad 7 \quad 9 \quad 9 \quad 10$ or $10 \quad 9 \quad 9 \quad 7 \quad 6 \quad 4 \quad 2$	M1	For ordering the numbers. Allow one error or one omission.	✓
(b) $7$	A1	A correct answer scores full marks, unless it comes from obviously incorrect working.	✓
(c) $8$	B1	Correct answer only.	✓

Full marks: 4/4

The diagram shows a 3-D shape.

(a) (i) Write down the mathematical name of this shape.
[1 mark]
(ii) Write down the number of edges of this shape. [1 mark]

The grid below shows six quadrilaterals.
Two of the quadrilaterals are congruent.

(b) Write down the letters of these two quadrilaterals.
[1 mark]
(c) Write down the letter of the quadrilateral that has no lines of symmetry. [1 mark]

(a) (i)(a) (ii)(b)(c)

[Total 4 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

A 3-D shape drawn in perspective, with the three edges at its hidden corner dashed.

Six quadrilaterals,

A

F

, drawn on a square grid.

Find

The mathematical name of the 3-D shape. The number of edges the 3-D shape has. The two quadrilaterals that are congruent. The quadrilateral that has no lines of symmetry.

Plan the Solution

Read the solid off the drawing: two equal square faces joined by four equal edges.
Count the edges in groups, so that the dashed ones are not left out.
Congruent means the same shape and the same size, so measure the sides of each quadrilateral and look for two that match.
A line of symmetry folds a shape exactly onto itself, so test each quadrilateral for a fold line.

Worked Solution [4 marks]

A cube has

6

square faces,

8

vertices and

12

edges. Congruent shapes have exactly the same size and shape, so one can be turned, or turned over, to fit exactly on the other. A line of symmetry folds a shape exactly onto itself.

Step 1: name the solid

(Reason: Every face of the drawing is a square and every edge is the same length, so the solid is a cube. A cube is the cuboid whose length, width and height are all equal, which is why a mark scheme will also accept cuboid or square prism here.)

Step 2: count the edges

4 + 4 = 8

8 + 4 = 12

(Reason: Count in groups so that none is missed. The square face at the front has

4

edges and the square face behind it has

4

more, which is

8

. Then

4

edges join the two faces, one at each corner. Only three of those four are drawn solid: the fourth runs back to the hidden corner and is dashed, so it is the one that gets forgotten.)

Step 3: find the congruent pair

\sqrt{1^{2} + 3^{2}} = \sqrt{10}

(Reason:

B

and

D

both have parallel sides of

3

squares and

1

square, a height of

3

squares, and two sloping sides that each run

1

square across and

3

squares up. Every side of one matches a side of the other, and

D

is simply

B

turned over, which changes nothing about its size.

F

has sloping sides of

\sqrt{10}

squares as well, but its parallel sides are

7

and

1

, so it is not congruent to either of them.)

Step 4: look for a line of symmetry

\sqrt{1^{2} + 2^{2}} = \sqrt{5}

\sqrt{2^{2} + 2^{2}} = \sqrt{8}

(Reason: Each of the six is a trapezium, and a trapezium folds onto itself only when its two sloping sides are equal. In

C

the left side rises

2

squares over

1

square, so it measures

\sqrt{5}

squares, while the right side falls

2

squares over

2

squares, so it measures

\sqrt{8}

squares. Those are not equal, so there is no fold line. The other five all have equal sloping sides, so each of them has one line of symmetry.)

(a) (i) cube(a) (ii)

12

(b)

B

and

D

(c)

C

Verification

Check 1: Count something else about the same solid. A cube has

8

corners and

6

faces, and for any solid of this kind

V - E + F = 2

8 - 12 + 6 = 2

, so

12

edges is right.

Check 2: Move

B

onto

D

instead of measuring. Turn

B

over in a horizontal line, then slide it

3

squares to the left, so that every corner

(x, y)

goes to

(x - 3, 11 - y)

(5, 9) \rightarrow (2, 2)

(6, 9) \rightarrow (3, 2)

(7, 6) \rightarrow (4, 5)

(4, 6) \rightarrow (1, 5)

and those four points are exactly the corners of

D

Check 3: Test

C

a second way. A fold line of a trapezium has to pass through the midpoint of both parallel sides, so compare the two midpoints. The long side runs from

9

13

, with midpoint

11

; the short side runs from

10

11

, with midpoint

10.5

. They are not above one another, so no fold line exists.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) (i) Cube	B1	Allow a misspelling. Allow cuboid. Allow prism or square prism or rectangular prism.	✓
(a) (ii) $12$	B1	cao	✓
(b) $B$ and $D$	B1	Allow $D$ and $B$ (or $b$ and $d$ etc).	✓
(c) $C$	B1	Allow $c$ .	✓

Full marks: 4/4

(a) Convert $5\,000$ millilitres into litres. [1 mark]

(b) Work out the difference between a length of $3$ metres and a length of $85$ centimetres.
Give your answer in centimetres. [2 marks]

(a) litres(b) centimetres

[Total 3 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

(a) A volume of

5\,000

millilitres.

(b) A length of

3

metres and a length of

85

centimetres.

Find

(a) That same volume written in litres. (b) The difference between the two lengths, given in centimetres.

Plan the Solution

One litre is $1\,000$ millilitres, so part (a) is a single division.
Part (b) gives the two lengths in different units, and lengths can only be subtracted once they are in the same unit.
The answer is wanted in centimetres, so change the metres into centimetres and do the subtraction there.

Worked Solution [3 marks]

Metric conversions:

1

litre is

1\,000

millilitres, and

1

metre is

100

centimetres. Multiply when moving to the smaller unit, divide when moving to the larger one.

Step 1: divide the millilitres by

1\,000

1 \text{ litre} = 1\,000 \text{ ml}

\dfrac{5\,000}{1\,000} = 5

(Reason: A litre is the larger unit, so the number must come out smaller.

5\,000

millilitres fills five one-litre bottles exactly.)

Step 2: write the

3

metres in centimetres

1 \text{ m} = 100 \text{ cm}

3 \times 100 = 300

(Reason: A centimetre is the smaller unit, so the number must come out bigger. Multiplying by

100

carries the metres into centimetres.)

Step 3: subtract the shorter length from the longer one

300 \text{ cm} - 85 \text{ cm} = 215 \text{ cm}

(Reason: Both lengths are now in centimetres, so they subtract directly and the difference is already in the unit the question asks for.)

(a)

5

litres(b)

215

centimetres

Verification

Check 1: Turn the answer to part (a) back into millilitres by multiplying by

1\,000

5 \times 1\,000 = 5\,000

millilitres, which is the volume the question gives.

Check 2: Do part (b) in metres instead.

85

centimetres is

0.85

metres, so subtract in metres and convert the difference afterwards.

3 - 0.85 = 2.15

metres, and

2.15 \times 100 = 215

centimetres, the same answer.

Check 3: Add the difference back on to the shorter length and see whether the longer length comes back.

85 + 215 = 300

centimetres, which is the

3

metres the question gives.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $5$	B1	For the correct answer.	✓
(b) $3 \times 100 \; (= 300)$ or $3 - 0.85 \; (= 2.15)$ or, with digits $300$ , either $300 - 85$ or $85 - 300$	M1	For conversion to centimetres, or a method to work out the difference in metres, or a method to find the difference in centimetres using their converted $300$ . For digits $300$ , allow $30$ , $300$ , $3\,000$ , $0.3$ , $0.03$ and so on, but not $3$ .	✓
$215$	A1	Allow $-215$ . A correct answer scores full marks unless it comes from obvious incorrect working.	✓

Full marks: 3/3

(a) Simplify the expression $a + a + a + a + a$ [1 mark]

(b) Simplify the expression $b \times c \times 7$ [1 mark]

(a)(b)(c) d =

[Total 3 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

(a) The sum

a + a + a + a + a

, which adds the same letter over and over.

(b) The product

b \times c \times 7

, which multiplies two letters and a number.

5d = 40

Find

(a) That sum written in its simplest form. (b) That product written in its simplest form. (c) The value of

d

that makes the equation true.

Plan the Solution

Part (a) adds one letter to itself several times, so count how many lots of $a$ there are and write that count in front of the letter.
Part (b) only multiplies, and multiplying can be done in any order, so move the $7$ to the front and write the letters straight after it.
Part (c) has $d$ multiplied by $5$ , so undo that multiplication by dividing both sides by $5$ .

Worked Solution [3 marks]

Algebra shorthand: repeated addition of one letter is written as a count in front of that letter, so

a + a = 2a

; a product is written with its number first and its letters after it, so

3 \times p \times q

is written

3pq

; and an equation stays balanced when the same thing is done to both sides.

Step 1: count the

a

terms in part (a)

a + a + a + a + a = 5a

(Reason: There are five separate lots of

a

being added. Adding a letter to itself is counted, not multiplied, so the count is written in front of the letter, where it is called the coefficient.)

Step 2: put the

7

in front in part (b)

b \times c \times 7 = 7 \times b \times c

7 \times b \times c = 7bc

(Reason: Multiplying can be done in any order, so the

7

moves to the front. A number multiplying letters is written next to them, so no multiplication signs are needed. Writing the letters the other way round, as

7cb

, is equally correct.)

Step 3: divide both sides by

5

in part (c)

5d = 40

\dfrac{5d}{5} = \dfrac{40}{5}

d = 8

(Reason:

5d

means five lots of

d

, so dividing both sides by

5

leaves

d

on its own and keeps the two sides equal.)

(a)

5a

(b)

7bc

(c)

d = 8

Verification

Check 1: Put

a = 3

into the sum the question gives, and into the answer, and see whether the two agree.

3 + 3 + 3 + 3 + 3 = 15

and

5 \times 3 = 15

, so the simplified form gives the same value as the sum it came from.

Check 2: Put

b = 2

and

c = 4

into the product the question gives, and into the answer.

2 \times 4 \times 7 = 56

and

7 \times 2 \times 4 = 56

, so reordering the factors has not changed the value.

Check 3: Put

d = 8

back into the equation the question gives.

5 \times 8 = 40

, which is the right-hand side, so

d = 8

is the solution.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $5a$	B1	For the correct answer.	✓
(b) $7bc$	B1	Or equivalent, for example $7cb$ .	✓
(c) $8$	B1	For the correct answer.	✓

Full marks: 3/3

The diagram shows a number machine.

(a) Work out the output when the input is $10$ [1 mark]

(b) Work out the output when the input is $-2$ [1 mark]

(a)(b)

[Total 2 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

A number machine with two boxes: the first multiplies by

4

, the second subtracts

15

The input enters the first box, and whatever leaves the second box is the output.

Find

(a) the output when the input is

10

(b) the output when the input is

-2

Plan the Solution

Follow the arrows from left to right. The number that leaves the first box is the number that enters the second.
Multiply by $4$ first, then subtract $15$ . The machine fixes that order, so the two boxes cannot be swapped.
Part (b) starts below zero, so carry the minus sign through the multiplication before subtracting.

Worked Solution [2 marks]

Rule - Read a number machine left to right: each box acts on the number that reaches it, so the output is

4 \times \text{input} - 15

Step 1: (a) The first box multiplies by

4

10 \times 4 = 40

(Reason: the input

10

reaches the first box, so that box acts on it before anything else happens)

Step 2: (a) The second box subtracts

15

40 - 15 = 25

(Reason: the number leaving the first box,

40

, is the number entering the second)

Step 3: (b) The first box multiplies by

4

again

-2 \times 4 = -8

(Reason: four lots of a negative number is negative, so the minus sign travels through the multiplication)

Step 4: (b) The second box subtracts

15

-8 - 15 = -23

(Reason: taking

15

away from a number that is already below zero moves it further below zero)

(a)

25

(b)

-23

Verification

Check 1: Run part (a) backwards through the machine: undo the subtraction by adding

15

, then undo the multiplication by dividing by

4

\dfrac{25 + 15}{4} = 10

, the input part (a) started from

Check 2: Do the same for part (b), starting from

-23

\dfrac{-23 + 15}{4} = -2

, the input part (b) started from

Check 3: Treat the whole machine as one calculation,

4n - 15

, and substitute each input into it.

4 \times 10 - 15 = 25

and

4 \times (-2) - 15 = -23

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$25$ cao	✓
(b)	B1	$-23$ cao	✓

Full marks: 2/2

Douglas sells $80$ pies for a total of £ $176$
$\dfrac{1}{5}$ of the pies are large.
Each large pie costs £ $3$
The remaining pies are small.

Work out the cost of one small pie. [4 marks]

[Total 4 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

Douglas sells

80

pies altogether and takes £

176

for them.

\dfrac{1}{5}

of the pies are large, and every large pie is priced at £

3

on its own.

The pies that are left over are the small ones, and they are all the same price as each other.

Find

the cost of one small pie, in pounds

Plan the Solution

Count the large pies first: $\dfrac{1}{5}$ of $80$ says how many there are, and every pie that is not large is small.
Work out what the large pies bring in on their own, then take that away from the total takings to leave what the small pies brought in.
Share the small pies' takings equally between the small pies. That is a division, and the divisor is the number of SMALL pies, not the number of pies on the stall.

Worked Solution [4 marks]

Rule - The takings split into two lots that make the whole:

\text{large takings} + \text{small takings} = 176

, so taking one lot away leaves the other. Sharing a lot equally is a division:

\text{price of one pie} = \dfrac{\text{takings}}{\text{number of those pies}}

Step 1: How many of the pies are large

\dfrac{1}{5} \times 80 = 16

(Reason: one fifth means splitting the

80

pies into five equal groups of

16

and taking one of those groups)

Step 2: How many of the pies are small

80 - 16 = 64

(Reason: every pie is either large or small, so the small ones are whatever is left of the

80

)

Step 3: What the large pies bring in

16 \times 3 = 48

(Reason:

16

large pies at £

3

each, so this part of the takings can be worked out without knowing anything about the small pies)

Step 4: What the small pies bring in

176 - 48 = 128

(Reason: the two lots together make the whole £

176

, so taking the large pies' £

48

away leaves the small pies' share of the takings)

Step 5: Share those takings between the small pies

\dfrac{128}{64} = 2

(Reason: the

64

small pies are all the same price, so the £

128

they brought in divides equally between them)

2

Verification

Check 1: Rebuild the takings from the two prices. If one small pie really costs £

2

, the

16

large pies and the

64

small pies must together bring in the total Douglas took.

16 \times 3 + 64 \times 2 = 48 + 128 = 176

, the total on the paper

Check 2: Use the average price instead of the counts. Douglas takes £

176

for

80

pies, so the average pie costs

\dfrac{176}{80} = 2.2

, and each large pie sits

0.8

above that average.

16 \times 0.8 = 12.8

above average altogether, spread over the

64

small pies gives

\dfrac{12.8}{64} = 0.2

below average each, so

2.2 - 0.2 = 2

Check 3: Work in batches of five pies, which is the smallest batch with the right mix: one large and four small. The

80

pies make

16

such batches.

\dfrac{176}{16} = 11

per batch, and

11 - 3 = 8

of that is the four small pies, so

\dfrac{8}{4} = 2

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{1}{5} \times 80 \; (= 16)$ or $\left(1 - \dfrac{1}{5}\right) \times 80 \; (= 64)$ oe	M1	for a method to find the number of large pies or the number of small pies	✓
“ $16$ ” $\times 3 \; (= 48)$	M1	for a method to find the total cost of the large pies. The quotation marks are the scheme's own: the candidate's number of large pies from the row above may be used here.	✓
$176$ − “ $48$ ” $(= 128)$	M1	for a method to find the total cost of the small pies. The quotation marks are the scheme's own: the candidate's total cost of the large pies from the row above may be used here.	✓
Correct answer scores full marks (unless from obvious incorrect working)	A1	$2$	✓

Full marks: 4/4

Gordon is going to book an apartment at a holiday resort for $7$ nights.

An apartment at the resort costs the same amount for each night.
Gordon knows that the apartment costs $150$ euros for $2$ nights.

The resort also charges an additional tourist tax of $5$ euros for each night.

Work out the total cost of the apartment and the tax for $7$ nights. [3 marks]

euros

[Total 3 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

Gordon is booking the apartment for

7

nights, and every night costs the same as every other night.

The apartment on its own costs

150

euros for

2

nights.

A tourist tax of

5

euros is charged for each night, on top of what the apartment costs.

Find

the total cost of the apartment and the tax for

7

nights, in euros

Plan the Solution

The $150$ euros pays for $2$ equal nights, so halving it gives what one night of the apartment costs. That is the only rate the question does not hand you.
Scale that nightly cost up to $7$ nights to get what the apartment costs on its own.
The tax is a separate charge made once for every night, so work it out over $7$ nights as well, then add the two amounts. Watch which count of nights each part uses: $2$ belongs to the price Gordon was quoted, and $7$ belongs to the stay.

Worked Solution [3 marks]

Rule - A cost that is the same every night is shared out equally:

\text{cost of one night} = \dfrac{\text{cost of several nights}}{\text{number of those nights}}

. A charge made once per night is multiplied by the number of nights, and a total made of two separate charges is their sum.

Step 1: What the apartment costs for one night

\dfrac{150}{2} = 75

(Reason: the

150

euros pays for

2

nights that cost the same as each other, so one night is half of it)

Step 2: What the apartment costs for the whole stay

75 \times 7 = 525

(Reason: every night of the stay costs the same

75

euros, so

7

nights cost seven times as much)

Step 3: The tax for the whole stay

5 \times 7 = 35

(Reason: the tax is charged once for every night that Gordon stays, and the stay is

7

nights long, not the

2

nights the quoted price happens to cover)

Step 4: Add the two charges together

525 + 35 = 560

(Reason: Gordon pays for the apartment and pays the tax, so the total is the

525

euros for the apartment plus the

35

euros of tax)

560

euros

Verification

Check 1: Price one whole night first, tax included, instead of keeping the two charges apart. One night costs

75 + 5 = 80

euros, and the stay is

7

of those nights.

80 \times 7 = 560

, the same total by a route that never forms

525

35

Check 2: Scale the quoted package instead of dividing it. Two nights cost

150 + 5 \times 2 = 160

euros once the tax on those two nights is added, and

7

nights is

3.5

such packages.

160 \times 3.5 = 560

, so no nightly rate is needed to reach the total

Check 3: Work backwards from the answer. Take the tax off the total, then share what is left between the nights and see whether it rebuilds the price the paper quotes.

560 - 35 = 525

, then

\dfrac{525}{7} = 75

a night, and

75 \times 2 = 150

euros for two nights, which is the figure Gordon was given

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{150}{2} \; (= 75)$ or $5 \times 7 \; (= 35)$ or $\dfrac{150 + 5 \times 2}{2} \; (= 80)$	M1	for a method to find the cost of the apartment without tax for $1$ night or for a method to find the total cost of the tax for $7$ nights or for a method to find the total cost of the apartment and the tax for $1$ night	✓
$7 \times$ “ $75$ ” $+$ “ $35$ ” $(= 525 + 35)$ or (“ $75$ ” $+ \, 5) \times 7$ or “ $80$ ” $\times 7$	M1	for a complete method to find the total cost of the apartment and the tax for $7$ nights. The quotation marks are the scheme's own: the candidate's own $75$ , $35$ or $80$ from the row above may be used here.	✓
Correct answer scores full marks (unless from obvious incorrect working)	A1	$560$	✓

Full marks: 3/3

Pierre chooses one food item and one drink at a hotel breakfast bar.

He can choose one food item from croissant ( $C$ ) or fruit ( $F$ ) or porridge ( $P$ )
He can choose one drink from milk ( $M$ ) or juice ( $J$ ) or tea ( $T$ )

(a) List all the possible combinations that Pierre can choose. [2 marks]

Yusuf and Rhys share some cherries in the ratio $3 : 4$

(b) What fraction of the cherries does Rhys get? [1 mark]

(a)(b)

[Total 3 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

(a) One food item from

C

F

P

, and one drink from

M

J

T

(b) Cherries shared between Yusuf and Rhys in the ratio

3 : 4

Find

(a) Every possible food-and-drink combination, with none missing and none repeated. (b) The fraction of the cherries that Rhys gets.

Plan the Solution

(a) Hold the food item fixed and run through all three drinks before moving on to the next food item. Working in that order is what stops a combination being missed.
(a) Then count what the finished list ought to hold: $3$ food items, each of them with $3$ drinks.
(b) Add the parts of the ratio to find how many equal shares the cherries are cut into, then put Rhys's parts over that total.

Worked Solution [3 marks]

Rule - Systematic listing: fix the first choice, run through every second choice, then move on to the next first choice. Rule - Ratio to fraction: in a share in the ratio

a : b

, the second person's fraction of the whole is

\dfrac{b}{a + b}

(a) Pair the croissant with each drink in turn

CM, \; CJ, \; CT

(Reason: The food item stays at

C

while only the drink changes, so all three croissant combinations come out together and none of them can be forgotten.)

(a) Repeat for the fruit, and then for the porridge

FM, \; FJ, \; FT

PM, \; PJ, \; PT

(Reason: The same three drinks are on offer whichever food item is picked, so every food item contributes

3

combinations of its own.)

(a) Check that nothing has been missed

3 \times 3 = 9

(Reason: There are

3

food items and, for each one of them,

3

drinks, so the finished list must hold that many combinations - with no repeats in it and nothing extra.)

(b) Count the equal parts in the ratio

3 + 4 = 7

(Reason: The ratio

3 : 4

cuts the cherries into

3

parts for Yusuf and

4

parts for Rhys, which is

7

equal parts altogether.)

(b) Put Rhys's parts over the total number of parts

\dfrac{4}{3 + 4} = \dfrac{4}{7}

(Reason: Rhys takes

4

of the

7

equal parts. A fraction of the whole always carries the number of equal parts underneath, not the other person's share.)

(a)

CM

CJ

CT

FM

FJ

FT

PM

PJ

PT

(b)

\dfrac{4}{7}

Verification

Check 1: Count the boxed list, and count how often each letter appears in it. The list holds

9

combinations, and each of

C

F

and

P

appears

3

times, as does each of

M

J

and

T

Check 2: Rebuild the list the other way round, drink first: all the

M

combinations, then all the

J

combinations, then all the

T

combinations.

CM

FM

PM

CJ

FJ

PJ

CT

FT

PT

- the same nine pairs in a different order, so none was missed and none was written twice.

Check 3: Test the fraction on a number of cherries that divides exactly: share

70

cherries in the ratio

3 : 4

. One part is

10

, so Yusuf gets

30

and Rhys gets

40

, and

\dfrac{40}{70} = \dfrac{4}{7}

Check 4: Add the two shares. Yusuf's fraction of the cherries and Rhys's fraction of the cherries must make one whole between them.

\dfrac{3}{7} + \dfrac{4}{7} = 1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $CM$ , $CJ$ , $CT$ , $FM$ , $FJ$ , $FT$ , $PM$ , $PJ$ , $PT$	B2	for correct outcomes listed with no repeats and no extras	✓
(a) partial credit	B1	for $5$ , $6$ , $7$ or $8$ correct outcomes listed, ignore repeats and ignore extras or for $9$ correct outcomes listed, ignore repeats and ignore extras	✓
(b) $\dfrac{4}{7}$	B1	oe	✓
(b) what oe accepts here	Note	oe is short for or equivalent: any fraction equal to $\dfrac{4}{7}$ scores the mark, for example $\dfrac{8}{14}$ or $\dfrac{40}{70}$ .	✓

Full marks: 3/3

The conversion graph below can be used to change between miles and kilometres.

(a) Use the graph to change $25$ miles into kilometres. [1 mark]

Claire delivers parcels for a courier firm.
She is paid $\$0.80$ for every kilometre she drives.

For one delivery round Claire was paid $\$19.20$ in total.

(b) Work out the number of miles she drove. [2 marks]

(a)(b)

[Total 3 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

(a) A conversion graph between miles and kilometres. The miles axis runs from

0

30

and the kilometres axis runs from

0

50

, and every small square is one unit.

(b) Claire is paid

\$0.80

for every kilometre she drives, and one delivery round paid her

\$19.20

altogether.

Find

(a)

25

miles, written in kilometres. (b) The number of miles Claire drove on that round.

Plan the Solution

(a) Go up from the miles axis to the line, then turn and go straight across to the kilometres axis, and read the value there.
(b) The money comes first. Every kilometre is worth the same $\$0.80$ , so the pay tells you how many kilometres she drove.
(b) Then use the same graph the other way round: across from the kilometres axis to the line, then down to the miles axis.

Worked Solution [3 marks]

Rule - Reading a conversion graph: start on the axis whose value you know, go to the line, then turn and go to the axis you want. Rule - A fixed rate: the number of kilometres is

\dfrac{\text{total pay}}{\text{pay for one kilometre}}

(a) Find 25 on the miles axis and go straight up to the line

25 \text{ miles}

(Reason: The miles axis is the horizontal one, so

25

miles is a point along the bottom. It is one of the numbered lines, so there are no squares to count before you start.)

(a) Turn and read across to the kilometres axis

(25, \; 40)

(Reason: Going straight across from the point on the line meets the kilometres axis at the numbered line

40

, so the point on the line is

(25, \; 40)

and

25

miles is

40

kilometres.)

(b) Turn the pay into kilometres

\dfrac{19.20}{0.80} = 24

(Reason: Every kilometre driven is worth

\$0.80

, so the number of kilometres is how many lots of

0.80

there are in

19.20

. That is a division, and it is the step the mark scheme gives the method mark for.)

(b) Go across from 24 on the kilometres axis to the line

24 \text{ km}

(Reason:

24

is not one of the numbered lines. Each small square is

1

kilometre, so

24

is the line one square below

25

. Go across from there until you meet the graph.)

(b) Read down to the miles axis

(15, \; 24)

(Reason: Dropping straight down from that point on the line lands on

15

on the miles axis, so

24

kilometres is

15

miles - the distance Claire drove.)

(a)

40

kilometres(b)

15

miles

Verification

Check 1: Build the pay back up from the kilometres:

24

kilometres at

\$0.80

for each one.

0.80 \times 24 = 19.20

, which is exactly what she was paid.

Check 2: Convert the answer forwards instead of backwards. Part (a) reads

25

miles as

40

kilometres, so one mile is

1.6

kilometres.

1.6 \times 15 = 24

kilometres, which is the number of kilometres the pay gave.

Check 3: Compare the two readings. Both points sit on the same straight line through the origin, so the miles and the kilometres must be in the same ratio at both of them.

\dfrac{15}{25} = \dfrac{24}{40}

, and both sides come to

0.6

Check 4: Check the size of the answer. A mile is longer than a kilometre, so the same journey must come to fewer miles than kilometres.

15

miles against

24

kilometres, which is the way round it should be.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $40$	B1	for $40$	✓
(b) $\dfrac{19.20}{0.80}$ $(= 24)$ oe	M1	for a method to find the number of kilometres she drove	✓
(b) $15$	A1	cao	✓
(b) what the printed scheme adds beside the answer	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓
(b) what oe and cao accept here	Note	oe is short for or equivalent, so any correct method for dividing the total pay by the pay for one kilometre earns the M1 - counting up in lots of $0.80$ as far as $19.20$ earns it just as the division does. cao is short for correct answer only, so the A1 needs $15$ itself.	✓

Full marks: 3/3

The bar chart gives information about the results of the matches played by a youth club's five-a-side football team.

The team gained
$3$ points for each match it won
$1$ point for each match it drew
$0$ points for each match it lost

The team drew $14$ matches.

Work out the total number of points the team gained. [3 marks]

[Total 3 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

The bar chart has three bars: won is

5

squares tall, drew is

7

squares tall and lost is

4

squares tall.

A win is worth

3

points, a draw is worth

1

point and a loss is worth

0

points.

The team drew

14

matches.

The vertical axis carries no numbers, so the scale is not given directly.

Find

The total number of points the team gained.

Plan the Solution

The vertical axis has no numbers on it, so the scale has to come from the one bar the question tells us about.
The drew bar is $7$ squares tall and stands for $14$ matches, which fixes what one square is worth.
Use that scale to read the won bar and the lost bar.
Multiply each number of matches by the points that result is worth, then add the three totals.

Worked Solution [3 marks]

Rule - Reading an unnumbered scale: divide the value you are given by the height of its own bar in squares to find what one square stands for, then read every other bar with that same scale.

Step 1: find what one square on the vertical axis stands for

\dfrac{14}{7} = 2

(Reason: The drew bar is

7

squares tall and the team drew

14

matches, so

7

squares stand for

14

matches and one square stands for

2

matches.)

Step 2: read the won bar and the lost bar

\text{won: } 5 \times 2 = 10

\text{lost: } 4 \times 2 = 8

(Reason: The won bar is

5

squares tall and the lost bar is

4

squares tall, and every square is worth

2

matches.)

Step 3: turn each result into points

\text{from wins: } 10 \times 3 = 30

\text{from draws: } 14 \times 1 = 14

\text{from losses: } 8 \times 0 = 0

(Reason: A win is worth

3

points, a draw is worth

1

point and a loss is worth

0

points, so each number of matches is multiplied by its own points value.)

Step 4: add the three totals

30 + 14 + 0 = 44

(Reason: Adding the points from the wins, the draws and the losses gives the total the team gained.)

44

points

Verification

Check 1: Count matches instead of points. The three bars are

5

7

and

4

squares tall, which is

16

squares in all at

2

matches a square.

16 \times 2 = 32

matches, and separately

10 + 14 + 8 = 32

Check 2: Price a whole square of each result instead. A square of wins is worth

3 \times 2 = 6

points, a square of draws is worth

1 \times 2 = 2

points, and a square of losses is worth nothing.

5 \times 6 + 7 \times 2 = 44

Check 3: Take the scale out of it, as the mark scheme allows. The won bar is

\dfrac{5}{7}

of the drew bar, so the wins come straight from the

14

matches.

\dfrac{5}{7} \times 14 = 10

wins, so

3 \times 10 + 14 = 44

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{14}{7} \, (= 2)$ or $\dfrac{5}{7} \times 14 \, (= 10)$ or $\dfrac{4}{7} \times 14 \, (= 8)$ or any correct value marked on the vertical axis	M1	for working with the scale	✓
$\text{their } 10 \times 3 \, (= 30)$ or $5 \times 3 + 7 \times 1 \, (= 15 + 7 = 22)$ or $5 \times 3 + 14 \, (= 15 + 14 = 29)$	M1	for a method to find the number of points won, or for a method to work out the total number of points using a vertical scale of $1$ match per cm. The candidate's own value from the first mark may be used here - that is what the printed scheme's quotation marks mean.	✓
M0M1 is possible	Note	The second method mark can be earned even when the first is not.	✓
$44$	A1	Correct answer scores full marks (unless from obvious incorrect working).	✓

Full marks: 3/3

The scale drawing shows the positions of two towns, Ashcombe and Redmere.

(a) Work out the real distance, in kilometres, between Ashcombe and Redmere. [3 marks]

On Sunday, Carla cycles directly from Ashcombe to Redmere.

Carla leaves Ashcombe at $12\ 50$
She arrives at Redmere at $17\ 15$

(b) Work out the time Carla takes to cycle directly from Ashcombe to Redmere. [2 marks]

(a) kilometres(b) hoursminutes

[Total 5 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

A scale drawing showing Ashcombe and Redmere, joined by a straight line.

Scale:

1

cm represents

5000

Carla leaves Ashcombe at

12\ 50

and arrives at Redmere at

17\ 15

Find

(a) The real distance between the two towns, in kilometres. (b) How long the journey takes, in hours and minutes.

Plan the Solution

Measure the line between the two crosses with a ruler. It is $9$ cm long.
Use the scale to turn that measurement into a real distance in metres.
Change the metres into kilometres, because part (a) asks for kilometres.
For part (b), count on from $12\ 50$ : first to the next whole hour, then in whole hours, then the last few minutes.

Worked Solution [5 marks]

Rule - Scale drawing: multiply the measured length by whatever

1

cm represents, and the answer is the real distance. Rule - Elapsed time: count on from the start time in stages, never subtracting clock times as if they were ordinary numbers.

Step 1: Measure the line on the scale drawing

9 \text{ cm}

(Reason: A ruler laid across the two crosses reads

9

cm. The mark scheme allows anything from

8.8

cm to

9.2

cm, so a reading a little either side of

9

still earns the mark.)

Step 2: Use the scale to get the real distance in metres

9 \times 5000 = 45\,000

(Reason: Every

1

cm on the drawing stands for

5000

m on the ground, so

9

cm stands for

9

lots of

5000

m.)

Step 3: Change the metres into kilometres

\dfrac{45\,000}{1000} = 45

(Reason: There are

1000

m in

1

km, so dividing the number of metres by

1000

turns the answer into kilometres, which is the unit part (a) asks for.)

Step 4: Count on from

12\ 50

to the next whole hour

60 - 50 = 10

(Reason: An hour is

60

minutes and the clock already reads

50

minutes past, so

10

more minutes reach

13\ 00

Step 5: Count on in whole hours to

17\ 00

17 - 13 = 4

(Reason: From

13\ 00

17\ 00

4

complete hours, with no odd minutes in them.)

Step 6: Add the minutes from each end

10 + 15 = 25

(Reason: The

10

minutes before

13\ 00

and the

15

minutes after

17\ 00

give

25

minutes, on top of the

4

whole hours.)

(a)

45

kilometres(b)

4

hours

25

minutes

Verification

Check 1: Take the other way through the scale.

5000

m is

5

km, so

1

cm on the drawing stands for

5

km, and the measured length is multiplied by that instead.

9 \times 5 = 45

kilometres, the same answer.

Check 2: Work backwards from the answer.

45

km is

45\,000

m, and each centimetre on the drawing stands for

5000

m, so dividing must give back the length that was measured.

\dfrac{45\,000}{5000} = 9

centimetres, which is the length on the drawing.

Check 3: Count part (b) a different way: in minutes after midnight.

12\ 50

770

minutes after midnight and

17\ 15

1035

minutes after midnight.

1035 - 770 = 265

minutes, and

4 \times 60 + 25 = 265

, so the two ways agree.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Measure the line and use the scale	M1	for measuring the line (a tolerance of $\pm 0.2$ cm, so $8.8$ cm to $9.2$ cm) and using the scale, that is a measurement from $8.8$ to $9.2$ times $5000$ , giving $44\,000$ to $46\,000$ , or $\dfrac{5000}{1000} = 5$ , or $\dfrac{\text{any seen value}}{1000}$ , or for an attempt to convert a value into kilometres. Do not allow $\dfrac{\text{any seen value}}{1000}$ if it is clearly linked to the wrong unit, eg mm.	✓
(a) Use of scale and conversion	M1	for use of scale and conversion: their $44\,000$ to $46\,000$ divided by $1000$ , or $8.8$ to $9.2$ times their $5$ . The printed scheme sets those values in quotation marks, which means the candidate's own earlier values may be used here.	✓
(a) The answer	A1	for $45$ . Allow $44$ to $46$ , and it must agree with their measurement. A correct answer scores full marks unless it comes from obviously incorrect working.	✓
(b) The time from $12\ 50$ to $17\ 15$	B2	for $4$ (hours) and $25$ (minutes).	✓
(b) Partial credit	(B1)	for $4$ (hours) or $25$ (minutes).	✓

Full marks: 5/5

(a) Factorise $15 - 5x$ [1 mark]

$d = 5p + 7r$

(b) Work out the value of $p$ when $d = 35$ and $r = 2$ [3 marks]

Beads are sold in small tubs and large tubs.

There are $12$ beads in a small tub.
There are $25$ beads in a large tub.

Oliver buys $m$ small tubs of beads and $n$ large tubs of beads.

(ii) Write an expression, in terms of $m$ and $n$ , for the total number of beads Oliver buys. [2 marks]

(a)(b) p =(c)(i)(c)(ii)

[Total 7 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

The expression

15 - 5x

The formula

d = 5p + 7r

, with

d = 35

and

r = 2

12

beads in a small tub, and

25

beads in a large tub.

Oliver buys

m

small tubs and

n

large tubs.

Find

(a)

15 - 5x

written as a factor outside a bracket. (b) The value of

p

. (c)(i) An expression for the total number of tubs. (c)(ii) An expression for the total number of beads.

Plan the Solution

For part (a), find the largest number that divides both $15$ and $5x$ , and take it outside a bracket.
For part (b), put $d = 35$ and $r = 2$ into the formula, then undo the operations around $p$ one at a time.
For part (c)(i), count the tubs themselves: a tub counts once whether it is small or large, so the sizes play no part.
For part (c)(ii), count the beads instead: each small tub carries $12$ and each large tub carries $25$ , so multiply first and add afterwards.
Both answers to part (c) stay as expressions, because $m$ and $n$ are never given a value.

Worked Solution [7 marks]

Rule - Factorising: take the highest common factor outside the bracket, so that

ab + ac = a(b + c)

. Rule - Substituting: replace every letter whose value is known, then rearrange until the letter you want stands alone. Rule - Forming an expression: an amount for each item is a multiplication, and separate groups are added.

Step 1: Find the highest common factor of

15

and

5x

15 = 5 \times 3

5x = 5 \times x

(Reason: Both terms contain a factor of

5

, and no larger number divides

15

, so

5

is the highest common factor of the two terms.)

Step 2: Write that common factor outside a bracket

15 - 5x = 5(3 - x)

(Reason: Dividing each term by

5

leaves

3

and

-x

, and those go inside the bracket in the same order and with the same signs. The subtraction sign stays where it was.)

Step 3: Substitute

d = 35

and

r = 2

into the formula

35 = 5p + 7 \times 2

35 = 5p + 14

(Reason: Putting the two known values into

d = 5p + 7r

turns the formula into an equation containing

p

alone, and

7 \times 2 = 14

is a number that can be worked out straight away.)

Step 4: Take the known number away from both sides

5p = 35 - 14 = 21

(Reason: The

14

is added to

5p

on the right-hand side, so subtracting it from both sides leaves

5p

on its own.)

Step 5: Divide both sides by

5

p = \dfrac{21}{5}

p = 4.2

(Reason:

5p

means

5

times

p

, so dividing both sides by

5

leaves

p

. The fraction

\dfrac{21}{5}

does not cancel, and as a decimal it is

4.2

. The mark scheme accepts either form.)

Step 6: Count the tubs for part (c)(i)

m + n

(Reason: Every tub counts as one tub whatever its size, so the

m

small tubs and the

n

large tubs give

m + n

tubs altogether. The

12

and the

25

are not used in this part.)

Step 7: Count the beads each size of tub contributes

12 \times m = 12m

25 \times n = 25n

(Reason: Each small tub holds

12

beads, so

m

of them hold

12m

beads; each large tub holds

25

beads, so

n

of them hold

25n

beads.)

Step 8: Add the two amounts together

12m + 25n

(Reason: The two sizes of tub are separate groups, so their bead counts add. Nothing further can be collected, because

12m

and

25n

are not like terms.)

(a)

5(3 - x)

(b)

p = \dfrac{21}{5}

, that is

4.2

(c)(i)

m + n

(c)(ii)

12m + 25n

Verification

Check 1: Multiply the bracket in part (a) back out. Multiplying the

5

by each term inside must rebuild the expression the question printed.

5(3 - x) = 15 - 5x

, which is the original expression.

Check 2: Put

p = 4.2

and

r = 2

back into

d = 5p + 7r

. If the value of

p

is right, the formula must return the

d = 35

the question gave.

5 \times 4.2 + 7 \times 2 = 21 + 14 = 35

, the value the question gave.

Check 3: Reach

p

the other way round. Rearrange the formula first, giving

p = \dfrac{d - 7r}{5}

, and only then put the numbers in, so the arithmetic happens in a different order.

p = \dfrac{35 - 14}{5} = \dfrac{21}{5} = 4.2

, the same value.

Check 4: Test part (c) on a case. If Oliver buys

4

small tubs and

3

large tubs, count directly: that is

4 + 3

tubs, holding

4 \times 12 = 48

beads and

3 \times 25 = 75

beads. The direct count gives

7

tubs and

48 + 75 = 123

beads, and the expressions give

m + n = 7

and

12 \times 4 + 25 \times 3 = 123

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The factorised expression	B1	for $5(3 - x)$ .	✓
(b) A correct substitution, or a correct first step in rearranging	M1	for $35 = 5 \times p + 7 \times 2$ , or for $5p = d - 7r$ , or for $\dfrac{d}{5} = p + \dfrac{7}{5}r$ , or equivalent. The mark is for a correct substitution in an equation, or for a correct first step in re-arranging.	✓
(b) Rearranging to reach $5p$ , $-5p$ , or a complete method for $p$	M1	for $5p = 35 - \text{"14"}$ or $5p = 21$ , or for $-5p = -35 + \text{"14"}$ or $-5p = -21$ , or for $(p =) \dfrac{35 - \text{"14"}}{5}$ or $(p =) \dfrac{-35 + \text{"14"}}{-5}$ , or equivalent. The mark is for rearranging to find the value of $5p$ , or the value of $-5p$ , or for a complete method to find the value of $p$ . The printed scheme sets $14$ in quotation marks, which means the candidate's own earlier value may be used here.	✓
(b) The value of $p$	A1	for $\dfrac{21}{5}$ , or equivalent, eg $4.2$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
(c)(i) The expression for the total number of tubs	B1	for $m + n$ , or equivalent. Condone $s + l$ .	✓
(c)(ii) The expression for the total number of beads	B2	for the final answer $12m + 25n$ or $25n + 12m$ .	✓
(c)(ii) Partial credit	(B1)	for $12m$ or $25n$ . For B2 or B1, condone use of $s$ and $l$ for $m$ and $n$ respectively.	✓

Full marks: 7/7

Here is a list of the ingredients needed to make $6$ muffins.

$\begin{array}{|c|}\hline \textbf{Chocolate muffins} \\ \text{Makes } 6 \text{ muffins} \\ 60 \text{ g sugar} \\ 75 \text{ g butter} \\ 1 \text{ egg} \\ 90 \text{ g flour} \\ 65 \text{ g chocolate} \\ \hline\end{array}$

Meera makes some of these muffins.
She uses $450$ g of flour.

(a) How many muffins does Meera make? [2 marks]

Claire makes $84$ of these muffins.

(b) What weight of butter does Claire use? [2 marks]

Meera and Claire pay a total of $624$ rupees for the ingredients.
The ratio of the amount Meera pays to the amount Claire pays is $3 : 5$

(a)(b) g(c) rupees

[Total 6 marks]

Show solution & mark schemeHide solution & mark scheme

Question 14 - Exam Solution

Understanding the Question

Given

One recipe makes

6

muffins from

60

g sugar,

75

g butter,

1

egg,

90

g flour and

65

g chocolate.

Meera uses

450

g of flour.

Claire makes

84

muffins.

Together they pay

624

rupees, shared in the ratio

3 : 5

, Meera to Claire.

Find

(a) The number of muffins Meera makes. (b) The weight of butter Claire uses. (c) The amount Claire pays.

Plan the Solution

Part (a) works from an ingredient, so count the recipes first: divide the flour used by the flour in one recipe, then multiply by the $6$ muffins each recipe makes.
Part (b) works the other way round, from the muffins. Divide by $6$ to count the recipes, then multiply by the butter in one recipe. Only the butter line of the card is needed.
Part (c) is a ratio share, not a recipe. Add the parts to see how many equal parts the money is cut into, find one part, then take Claire's parts.
In every part the same idea is doing the work: multiply or divide two matching quantities by the same number and they stay in proportion.

Worked Solution [6 marks]

Rule - Scaling a recipe: divide the amount used by the amount in one recipe to find the number of recipes, then multiply any other ingredient by that same number. Rule - Sharing in a ratio: add the parts to find how many equal parts there are, divide the total by that number to find one part, then multiply by the number of parts wanted, so that

a : b

shares a total

T

\dfrac{T}{a + b}

for each part.

Step 1: Work out how many recipes

450

g of flour makes

\dfrac{450}{90} = 5

(Reason: One recipe uses

90

g of flour, so dividing the flour used by the flour in one recipe counts the recipes. It divides exactly, giving

5

whole recipes, so nothing is left over and no rounding is needed.)

Step 2: Multiply by the

6

muffins one recipe makes

5 \times 6 = 30

(Reason: Every recipe makes the same

6

muffins, so

5

recipes make

5

lots of

6

. The other four ingredients play no part in this answer.)

Step 3: Work out how many recipes

84

muffins is

\dfrac{84}{6} = 14

(Reason: Part (b) starts from the muffins rather than from an ingredient, so this time divide by the

6

muffins in one recipe. Again it divides exactly, so Claire makes a whole number of recipes.)

Step 4: Multiply by the butter in one recipe

14 \times 75 = 1\,050

(Reason: Each recipe uses

75

g of butter, so

14

recipes use

14

lots of

75

g. The answer is a weight, so it is measured in grams.)

Step 5: Split the

624

rupees into equal parts

3 + 5 = 8

\dfrac{624}{8} = 78

(Reason: A ratio of

3 : 5

means the money is cut into

8

equal parts,

3

of them Meera's and

5

of them Claire's. Dividing the total by

8

gives the value of one part, in rupees.)

Step 6: Take Claire's share of those parts

5 \times 78 = 390

(Reason: Claire's share is

5

of the

8

equal parts, so her payment is that many lots of the value of one part. The question asks for Claire's amount, not Meera's.)

(a)

30

muffins(b)

1\,050

g(c)

390

rupees

Verification

Check 1: Scale part (a) back the other way. If

30

muffins is right, then it must be a whole number of recipes, and those recipes must use exactly the flour the question gave.

\dfrac{30}{6} = 5

recipes, and

5 \times 90 = 450

g of flour, the amount the question gives.

Check 2: Redo parts (a) and (b) per muffin instead of per recipe, so the arithmetic happens in a different order. One muffin takes

\dfrac{90}{6} = 15

g of flour and

\dfrac{75}{6} = 12.5

g of butter.

\dfrac{450}{15} = 30

muffins, and

84 \times 12.5 = 1\,050

g of butter, the same two answers.

Check 3: Work out Meera's payment instead and add the two together. Meera has

3

of the

8

parts, and the two payments must rebuild the total the question gave.

624 \times \dfrac{3}{8} = 234

rupees for Meera, and

234 + 390 = 624

, the total the question gives.

Check 4: Test that the two payments really are in the ratio the question states, by dividing each of them by the value of one part.

\dfrac{234}{78} = 3

and

\dfrac{390}{78} = 5

, so the payments are in the ratio

3 : 5

, Meera to Claire.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A method to find the number of recipes, or the flour needed for one muffin	M1	for $\dfrac{450}{90} \, (= 5)$ , or equivalent, or for $\dfrac{90}{6} \, (= 15)$ , or equivalent. The mark is for a method to find the number of batches of $6$ muffins made, or the amount of flour needed per muffin.	✓
(a) The number of muffins	A1	for $30$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
(b) A method to find the number of recipes, or the butter needed for one muffin	M1	for $\dfrac{84}{6} \, (= 14)$ , or equivalent, or for $\dfrac{75}{6} \, (= 12.5)$ , or equivalent. The mark is for a method to find the number of batches of $6$ muffins made, or the amount of butter needed per muffin.	✓
(b) The weight of butter	A1	for $1\,050$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
(c) A method to find one part of the ratio, three parts, or a complete method	M1	for $\dfrac{624}{3 + 5} \, (= 78)$ , or equivalent, or for $624 \times \dfrac{3}{3 + 5} \, (= 234)$ , or equivalent, or for $624 \times \dfrac{5}{3 + 5}$ , or equivalent. The mark is for a method to find the value of one part of the ratio, or for a method to find the value of three parts, or for a complete method.	✓
(c) The amount Claire pays	A1	for $390$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 6/6

Marcus has some wooden beads in a tin.

$7$ of the beads are yellow
$5$ of the beads are white
The rest of the beads are purple

Marcus is going to take at random a bead from the tin.

The probability that the bead is purple is $\dfrac{2}{5}$

Work out the number of purple beads in the tin. [3 marks]

[Total 3 marks]

Show solution & mark schemeHide solution & mark scheme

Question 15 - Exam Solution

Understanding the Question

Given

7

of the beads are yellow and

5

of them are white. Every other bead in the tin is purple.

One bead is taken at random, and the probability that it is purple is

\dfrac{2}{5}

The total number of beads in the tin is not given.

Find

The number of purple beads in the tin. It is a whole number, so the working should land on one - here it comes out at

8

Plan the Solution

Add the yellow and the white beads. That gives $12$ beads whose colour is already known, and they are exactly the beads that are not purple.
Turn the probability round. If $\dfrac{2}{5}$ of the tin is purple, work out what fraction of the tin is not purple.
Use that fraction to scale the known count up to the whole tin: find one fifth of the beads first, then all five fifths.
Subtract the beads that are not purple to leave the purple ones, then check that the probability comes back out.

Worked Solution [3 marks]

Rule - Probability of one random pick:

\dfrac{\text{number of beads of that colour}}{\text{total number of beads}}

. The probability that something happens and the probability that it does not happen always add to

1

Step 1: Count the beads that are not purple

7 + 5 = 12

(Reason: (Reason: yellow and white are the only other colours in the tin, so those

12

beads are precisely the ones that are not purple.))

Step 2: Work out the fraction that is not purple

1 - \dfrac{2}{5} = \dfrac{3}{5}

(Reason: (Reason: purple and not purple are the only two outcomes, so their probabilities add to

1

. Taking

\dfrac{2}{5}

away leaves the share of the tin that is not purple.))

Step 3: Scale up to the whole tin

\dfrac{12}{3} = 4

4 \times 5 = 20

(Reason: (Reason: think of the tin as

5

equal parts. The

12

beads that are not purple fill

3

of them, so dividing by

3

gives one part, and all

5

parts together make the whole tin.))

Step 4: Subtract to leave the purple beads

20 - 12 = 8

(Reason: (Reason: the tin holds

20

beads and

12

of them are yellow or white, so everything left over is purple.))

8

purple beads

Verification

Check 1: Put the answer back into the question. The tin would hold

8 + 12 = 20

beads altogether, of which

8

are purple, so the probability of taking a purple bead is

\dfrac{8}{20}

\dfrac{8}{20} = \dfrac{2}{5}

, which is the probability the question gives.

Check 2: Check the split a different way, as a ratio. The tin divides

8

purple to

12

not purple, and cancelling that ratio should give the

2 : 3

split that sits behind

\dfrac{2}{5}

\dfrac{8}{4} = 2

and

\dfrac{12}{4} = 3

, so the ratio is

2 : 3

as expected.

Check 3: Check the other colour instead. Out of

20

beads,

12

are not purple, and that share must be the fraction found in Step 2 and must complete the probability to

1

\dfrac{12}{20} = \dfrac{3}{5}

and

\dfrac{2}{5} + \dfrac{3}{5} = 1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Proportion, ratio or equation	M1	For finding the proportion of the beads that are not purple (this may be a percentage or a decimal), $1 - \dfrac{2}{5} \; (= \dfrac{3}{5})$ oe, or for a correct ratio (allow the ratio in any order), (purple : yellow + white =) $2 : 3$ Values in a ratio do not need labels, but if labels are used they must be correct. This is implied by, for example, $12$ beads being $3$ parts. OR for forming a correct equation in terms of the number of purple beads or the total number of beads (allow the use of any letter), eg $\dfrac{n}{n + 12} = \dfrac{2}{5}$ or $\dfrac{t - 12}{t} = \dfrac{2}{5}$	✓
A method to reach one fifth of the beads, or the total	M1	For a method to find the value of one fifth of the beads, $\dfrac{7 + 5}{3} \; (= 4)$ oe, or for a method to find the total number of beads in the tin, $\dfrac{7 + 5}{3} \times 5 \; (= 20)$ oe, or $\dfrac{7 + 5}{\dfrac{3}{5}} \; (= 20)$ oe. The printed scheme puts the $3$ and the $\dfrac{3}{5}$ in quotation marks, which means the candidate's own value from the first mark may be used here, so a follow-through on it is allowed. OR for a correct method to solve a correct equation, eg $n = \dfrac{12 \times 2}{5 - 2}$ or $t = \dfrac{5 \times 12}{5 - 2}$	✓
Answer $8$	A1	cao. A correct answer scores full marks unless it comes from obvious incorrect working. Note: an answer of $\dfrac{8}{20}$ scores M1M1A0 - the total has been found but the question asked for the number of purple beads, not the probability.	✓

Full marks: 3/3

The diagram shows a quadrilateral $ACDF$ .

In this quadrilateral, $ABEF$ is a trapezium and $BCDE$ is a parallelogram.

$FE = 26$ cm, $AB = 18$ cm, $BC = 12$ cm

Parallelogram $BCDE$ has an area of $96$ cm².

Find the area of trapezium $ABEF$ . [3 marks]

cm²

[Total 3 marks]

Show solution & mark schemeHide solution & mark scheme

Question 16 - Exam Solution

Understanding the Question

Given

A quadrilateral

ACDF

made of a trapezium

ABEF

and a parallelogram

BCDE

, joined along

BE

FE = 26

cm,

AB = 18

cm,

BC = 12

cm.

Area of parallelogram

BCDE

96

cm².

Find

The area of the trapezium

ABEF

, in cm².

Plan the Solution

$F$ , $E$ and $D$ sit on one straight line, and $A$ , $B$ and $C$ sit on another. The two lines are parallel, so the trapezium and the parallelogram have the same perpendicular height.
The parallelogram is the shape whose area is known, so use it to find that height first: its base is $BC = 12$ cm.
Then feed that height into the trapezium, whose parallel sides are $AB = 18$ cm and $FE = 26$ cm.
Watch which lengths belong to which shape: $BC$ is a side of the parallelogram only, and never goes inside the trapezium formula.

Worked Solution [3 marks]

Parallelogram:

\text{area} = \text{base} \times h

.
Trapezium:

\text{area} = \dfrac{1}{2}(a + b)h

, where

a

and

b

are the two parallel sides and

h

is the perpendicular distance between them.
Two shapes standing between the same pair of parallel lines share one value of

h

Step 1: Find the height from the parallelogram

12 \times h = 96

\dfrac{96}{12} = 8

(Reason: A parallelogram's area is base times perpendicular height, so

12

lots of

h

make

96

, giving

h = 8

cm. Because

FED

and

ABC

are parallel lines, this same

8

cm is the height of the trapezium.)

Step 2: Put that height into the trapezium formula

\dfrac{1}{2} \times (18 + 26) \times 8 = \dfrac{1}{2} \times 44 \times 8 = 176

(Reason: The trapezium's parallel sides are

AB

and

FE

, so those are the pair that goes inside the bracket. Half of

18 + 26

is the average width of the trapezium, and multiplying that average width by the height gives the area.)

176

cm²

Verification

Check 1: Compare the two shapes without finding the height at all. The trapezium's average width is

\dfrac{18 + 26}{2} = 22

cm against the parallelogram's

12

cm, and both stand on the same height, so the areas are in the ratio

22 : 12

96 \times \dfrac{22}{12} = 176

cm², which agrees.

Check 2: Cut the trapezium up instead of using its formula. Sliding

AB

up to the top line leaves a parallelogram

18

cm by

8

cm, plus a triangle whose base is

26 - 18 = 8

cm on the same height.

18 \times 8 + \dfrac{1}{2} \times 8 \times 8 = 144 + 32 = 176

cm², which agrees.

Check 3: Work out the whole quadrilateral a different way.

ED

equals

BC

, which is

12

cm, because

BCDE

is a parallelogram. So

FD

26 + 12 = 38

cm,

AC

18 + 12 = 30

cm, and

ACDF

is itself a trapezium on the same height.

\dfrac{1}{2} \times (38 + 30) \times 8 = 272

cm² for the whole quadrilateral, and taking the parallelogram off leaves

272 - 96 = 176

cm², which agrees.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{96}{12} \; (= 8)$	M1	for a complete method to find the height of the parallelogram	✓
$\dfrac{1}{2}(18 + 26) \times [8]$	M1	for a method to find the area of the trapezium (do not allow missing brackets unless recovered), where $[8]$ is what they believe to be the height of the trapezium and must be positive. Do not allow $[8]$ to be $12$ or $1$	✓
$176$	A1	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 3/3

Continue to questions 17 to 27

The remaining 11 questions, with the same full worked solutions and mark schemes

Frequently asked questions

There are 27 questions worth 100 marks in total, sat over 2 hours. It is Foundation tier and a calculator is allowed throughout, unlike UK GCSE Maths, where one paper is non-calculator.

Foundation tier targets grades 1 to 5, so grades 6 to 9 are only available on Higher tier. About 40 per cent of the questions are targeted at grades 4 and 5 and appear on both Paper 2FR and Paper 2HR, so the top of the Foundation paper overlaps with the bottom of the Higher paper.

Yes. The paper states in its own instructions that without sufficient working, correct answers may be awarded no marks. Several questions ask you to show your working clearly or to show clear algebraic working, and on those a bare answer scores nothing. That is why every solution here sets out the method mark by mark.

Yes, a Foundation tier formulae sheet is printed in the paper. It gives the area of a trapezium, the volume of a prism, the volume of a cylinder and the curved surface area of a cylinder. Everything else has to be recalled, so Pythagoras theorem, the angle facts and the percentage methods used on this paper are not provided. Nothing may be written on the formulae page.

Both are published by Pearson Edexcel and are linked directly from this page as PDF files. The solutions here are original: every question has been reworded, but all the numbers match the original paper, so the answers agree with the official mark scheme. This resource reproduces neither the exam paper nor the official mark scheme.

Keep revising

Once you have worked through this paper, read what the IGCSE is and how it is graded, or compare Edexcel 4MA1 with Cambridge 0580 if you are still choosing a board. Check the IGCSE grade boundaries to set your target, and if the exam is close, the four-week IGCSE Maths revision plan sets out what to do week by week.

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Edexcel IGCSE 4MA1/2FR, Wednesday 4 June 2025: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 16 of 27

Question 1, Calculator allowed

Question 1 - Exam Solution

Question 2, Calculator allowed

Question 2 - Exam Solution

Question 3, Calculator allowed

Question 3 - Exam Solution

Question 4, Calculator allowed

Question 4 - Exam Solution

Question 5, Calculator allowed

Question 5 - Exam Solution

Question 6, Calculator allowed

Question 6 - Exam Solution

Question 7, Calculator allowed

Question 7 - Exam Solution

Question 8, Calculator allowed

Question 8 - Exam Solution

Question 9, Calculator allowed

Question 9 - Exam Solution

Question 10, Calculator allowed

Question 10 - Exam Solution

Question 11, Calculator allowed

Question 11 - Exam Solution

Question 12, Calculator allowed

Question 12 - Exam Solution

Question 13, Calculator allowed

Question 13 - Exam Solution

Question 14, Calculator allowed

Question 14 - Exam Solution

Question 15, Calculator allowed

Question 15 - Exam Solution

Question 16, Calculator allowed

Question 16 - Exam Solution

Frequently asked questions

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