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Edexcel IGCSE 4MA1 Paper 2F, June 2024: Worked Solutions and Mark Schemes

Sir Faraz Hassan

4 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)Paper 2F - Foundation Tier - June 2024100 marks · 2 hours · Calculator allowed

Original worked solutions for Edexcel International GCSE Mathematics A (4MA1), Paper 2F (Foundation Tier), June 2024 – 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 14 of 28

The table shows the distance, in kilometres, each of $6$ volunteers travelled to get to a beach clean-up.

$\begin{array}{|l|c|c|c|c|c|c|} \hline \text{Name} & \text{Clive} & \text{Aoi} & \text{Rufus} & \text{Hinata} & \text{Cedric} & \text{Nora} \\ \hline \text{Distance (km)} & 7 & 12 & 10 & 5 & 4 & 9 \\ \hline \end{array}$

On the grid, draw a bar chart to show this information. [3 marks]

[Total 3 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

Six distances, in kilometres, one for each volunteer:

7, 12, 10, 5, 4, 9

A blank square grid to draw the chart on.

Find

A bar chart showing the six distances.

Plan the Solution

Read the six heights straight from the table - a bar chart of raw data shows the numbers that are already there, so nothing has to be worked out.
Choose a scale for the vertical axis that reaches the tallest distance, $12$ km, and keep it linear so that equal steps stand for equal distances.
Draw six bars of equal width, then label the scale and every bar.

Worked Solution [3 marks]

Rule - Bar chart: each bar's height is the value it stands for, the scale rises in equal steps, and every bar carries its own label.

Step 1: read the six distances from the table

7, 12, 10, 5, 4, 9

(Reason: these six numbers are the bar heights, so the tallest bar is Aoi's at

12

km and the shortest is Cedric's at

4

km)

Step 2: choose a scale for the vertical axis

1 \text{ square} = 1 \text{ km}

12 \times 1 = 12 \text{ squares}

(Reason: one square standing for

1

km keeps the scale linear and makes every bar as many squares tall as its distance, so the axis has to run at least as high as the tallest bar)

Step 3: draw the bars and label the chart

\text{Clive } 7 \quad \text{Aoi } 12 \quad \text{Rufus } 10

\text{Hinata } 5 \quad \text{Cedric } 4 \quad \text{Nora } 9

(Reason: equal widths and equal gaps keep the bars comparable, and the marks are for the heights and the labels, not for the width chosen)

A fully correct bar chart: bars of height

7, 12, 10, 5, 4, 9

kma linear scale on the distance axis, and every bar labelled with its name

Verification

Check 1: Read the finished chart back against its scale. One square is

1

km, so count the squares in each bar and compare with the table. Clive

7

, Aoi

12

, Rufus

10

, Hinata

5

, Cedric

4

, Nora

9

- all six agree with the table.

Check 2: Add the six bar heights and compare that total with the total of the six distances in the table. A bar read wrongly would change one total and not the other.

7 + 12 + 10 + 5 + 4 + 9 = 47

km both ways.

Check 3: The tallest bar should stand above the shortest by the range of the data.

12 - 4 = 8

squares between the top of Aoi's bar and the top of Cedric's, which is what the chart shows.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Bars drawn to the correct heights	B2	All six bars at the correct height for the scale used - $7, 12, 10, 5, 4, 9$ . Gaps or no gaps between the bars, and bars of different widths, are condoned.	✓
Partial credit for the bars	B1	Three, four or five bars at the correct height, or all six heights marked but no bars drawn.	✓
Labelling	B1	All labels correct: a linear scale on the distance axis, and an individual label for each of the six names.	✓

Full marks: 3/3

The diagram below shows shape $J$ on a square grid, then shape $K$ , then shape $L$ on a centimetre grid.

(a) Draw a shape on the grid that is congruent to shape $J$ [1 mark]

(b) Draw a shape on the grid that is an enlargement of shape $J$ with scale factor $2$ [2 marks]

Shape $K$ has exactly one line of symmetry.
(c) Draw that line of symmetry on shape $K$ [1 mark]

(d) Work out the perimeter of shape $L$ [1 mark]

(e) Work out the area of shape $L$ [1 mark]

(d) cm(e) cm²

[Total 6 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

Shape

J

on a square grid: an L-shape

3

squares wide and

2

squares tall, with a

2

1

rectangle missing from its top left corner.

Shape

K

: a six-sided shape that has exactly one line of symmetry.

Shape

L

on a centimetre grid: an L-shape

5

cm wide and

2

cm tall, with a

2

cm by

1

cm rectangle missing from its top right corner.

Find

(a) a shape congruent to shape

J

; (b) an enlargement of shape

J

with scale factor

2

; (c) the one line of symmetry of shape

K

; (d) the perimeter of shape

L

, in cm; (e) the area of shape

L

, in square centimetres.

Plan the Solution

Congruent means same size and same shape, so part (a) asks only for a copy: count the squares that make shape $J$ and lay them out again somewhere else on the grid.
Scale factor $2$ doubles every length, so for part (b) work round the outline of shape $J$ and draw each side twice as long.
A line of symmetry is a fold line. Shape $K$ has its two shortest sides meeting at one corner and its two medium sides meeting at the opposite corner, so for part (c) try the diagonal that joins those two corners.
For part (d), travel once round the outside of shape $L$ and add every side, the short ones at the step included.
For part (e), count the centimetre squares shape $L$ covers, or split it into two rectangles and add their areas.

Worked Solution [6 marks]

Rule - Congruent shapes have identical side lengths and identical angles, so only their position or their turn may differ. An enlargement of scale factor

2

multiplies every length by

2

and every area by

2^{2}

. A line of symmetry folds a shape exactly onto itself. Perimeter is the total distance round the outside; area is the number of unit squares covered.

Step 1 - Part (a): copy shape

J

without changing its size

3 \times 1 + 1 \times 1 = 4

(Reason: Shape

J

is a row of

3

squares with

1

more square sitting on the right-hand end, so it covers

4

squares altogether. A congruent shape has the same sides and the same angles, so drawing those

4

squares in the same arrangement anywhere on the grid earns the mark. Turning it or flipping it is allowed; making it bigger or smaller is not.)

Step 2 - Part (b): double every side of shape

J

3 \times 2 = 6

2 \times 2 = 4

1 \times 2 = 2

(Reason: Scale factor

2

multiplies every length by

2

, so the side of

3

squares becomes

6

, the side of

2

squares becomes

4

and each side of

1

square becomes

2

. The angles are unchanged, so the drawing is the same L-shape stretched equally in both directions.)

Step 3 - Part (c): find the fold line of shape

K

6 = 2 + 2 \times 2

(Reason: Shape

K

has

6

vertices. A fold line has to leave the shape looking the same, so it passes through

2

of them and swaps the other

4

2

matching pairs. The two shortest sides meet at the bottom left vertex and the two medium sides meet at the opposite vertex, so the fold line is the diagonal joining those two. Every other line leaves the shape looking different, so draw that one and no others.)

Step 4 - Part (d): add the sides of shape

L

5 + 1 + 2 + 1 + 3 + 2 = 14

(Reason: Start at the bottom left corner of shape

L

and travel once round the outside, writing down every side you walk along. Each square of the grid is

1

cm, and the two short sides at the step count in exactly the same way as the long ones.)

Step 5 - Part (e): count the centimetre squares of shape

L

5 \times 1 = 5

3 \times 1 = 3

5 + 3 = 8

(Reason: Split shape

L

into two rectangles: the bottom row is

5

cm by

1

cm and the row above it is

3

cm by

1

cm. Adding the two gives the number of centimetre squares the shape covers, which is its area.)

(a) a shape of the same size and shape as shape

J

(b) an L-shape with every side of shape

J

doubled(c) the diagonal joining the two opposite vertices where the equal sides meet(d)

14\text{ cm}

(e)

8\text{ cm}^{2}

Verification

Check 1: Shape

L

fits exactly inside a

5

cm by

2

cm rectangle, and the piece that is missing is cut from a corner, so the outline is exactly as long as the rectangle's.

2 \times (5 + 2) = 14

Check 2: Count the centimetre squares of shape

L

one row at a time: a row of

5

with a row of

3

above it.

5 + 3 = 8

Check 3: Enlarging by scale factor

2

multiplies area by

2^{2}

, so the shape drawn in part (b) should cover

4

times as many squares as shape

J

4 \times 2^{2} = 16

Check 4: Fold shape

K

along the drawn diagonal: the two shortest sides swap with each other, the two long slanting sides swap, and the two medium sides swap. every vertex lands on a vertex, so the fold works, and no other line does the same

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A shape drawn with the same side lengths and the same angles as shape $J$	B1	A shape of the same size and shape as the one given. It may be reflected or rotated.	✓
(b) An L-shape drawn with every side of shape $J$ doubled	B2	A shape that is an enlargement of the given shape. Any orientation is satisfactory. B1 for $2$ correctly enlarged sides.	✓
(c) The correct line of symmetry drawn on shape $K$	B1	The correct line, with no other lines drawn.	✓
(d) Perimeter of shape $L$	B1	$14$	✓
(e) Area of shape $L$	B1	$8$	✓

Full marks: 6/6

(a) Write these numbers in order of size.
Start with the smallest number.

$-77$ , $39$ , $-89$ , $43$ , $-6$ [1 mark]

(b) Write these decimals in order of size.
Start with the smallest decimal.

$0.134$ , $0.12$ , $0.145$ , $0.017$ , $0.3$ [1 mark]

(d) Change $\dfrac{27}{100}$ into a decimal. [1 mark]

There are $60$ cupcakes on a cake stall.
$\dfrac{7}{10}$ of the cupcakes are chocolate cupcakes.

(e) Work out how many of the cupcakes are not chocolate cupcakes. [2 marks]

(a)(b)(c) %(d)(e)

[Total 6 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

Five numbers:

-77

39

-89

43

-6

. Two of them are negative.

Five decimals:

0.134

0.12

0.145

0.017

0.3

. They do not all have the same number of decimal places.

The decimal

0.7

and the fraction

\dfrac{27}{100}

A cake stall holding

60

cupcakes, of which

\dfrac{7}{10}

are chocolate cupcakes.

Find

(a) the five numbers written in order, smallest first; (b) the five decimals written in order, smallest first; (c)

0.7

written as a percentage; (d)

\dfrac{27}{100}

written as a decimal; (e) how many of the cupcakes are not chocolate cupcakes.

Plan the Solution

For part (a), picture a number line. Everything negative sits to the left of $0$ , so both negative numbers come first, and among negatives the one furthest from zero is the smallest.
For part (b), give every decimal the same number of decimal places by filling the gaps with zeros. Once they all have three decimal places they can be compared as whole numbers of thousandths.
For part (c), a percentage counts hundredths, so multiply the decimal by $100$ .
For part (d), a fraction with denominator $100$ is already a number of hundredths, so the numerator gives the digits after the decimal point.
For part (e), the whole stall is $1$ whole. Subtract the fraction that are chocolate to get the fraction that are not, then take that fraction of $60$ .

Worked Solution [6 marks]

Rule - On a number line, further left means smaller, so a negative number with a bigger digit is worth less. Decimals are compared place by place from the left, and writing them all to the same number of decimal places makes that comparison safe. To turn a decimal into a percentage, multiply by

100

; a fraction over

100

is already a percentage, and its numerator gives the hundredths digits of the decimal. To find a fraction of an amount, divide by the denominator and multiply by the numerator.

Step 1 - Part (a): put the two negative numbers first

-89 < -77 < -6 < 0 < 39 < 43

(Reason: Every negative number is to the left of

0

, so

-89

-77

and

-6

all come before

39

and

43

. Among the negatives the order is the other way round from the digits:

89

is bigger than

77

, so

-89

is further left and therefore smaller. The two positive numbers keep their usual order.)

Step 2 - Part (b): give every decimal three decimal places, then compare

0.017,\ 0.120,\ 0.134,\ 0.145,\ 0.300

17 < 120 < 134 < 145 < 300

0.017 < 0.12 < 0.134 < 0.145 < 0.3

(Reason: Filling the short decimals with zeros does not change their value:

0.12

is the same as

0.120

and

0.3

is the same as

0.300

. Now every decimal is a number of thousandths, so they can be ordered as whole numbers. Without the zeros it is easy to be tricked by the length of a decimal and think

0.134

beats

0.3

Step 3 - Part (c): multiply the decimal by one hundred

0.7 \times 100 = 70

(Reason: Per cent means out of

100

, so a percentage counts hundredths.

0.7

7

tenths, and each tenth is

10

hundredths, so

0.7

70

hundredths.)

Step 4 - Part (d): read the hundredths straight off the fraction

\dfrac{27}{100} = 0.27

(Reason: The denominator

100

says the fraction counts hundredths, and the second place after the decimal point is the hundredths place.

27

hundredths is

2

tenths and

7

hundredths, which is written

0.27

Step 5 - Part (e): take the fraction that are not chocolate, then find it of sixty

1 - \dfrac{7}{10} = \dfrac{3}{10}

\dfrac{3}{10} \times 60 = 18

(Reason: The whole stall is

1

whole, so the cupcakes that are not chocolate make up the rest of it once the chocolate ones are taken away:

\dfrac{3}{10}

of the stall. To take tenths of

60

, split the

60

cupcakes into

10

equal groups of

6

and count three of those groups. The question asks for the cupcakes that are not chocolate, so it is the

\dfrac{3}{10}

that is wanted, never the

\dfrac{7}{10}

(a)

-89,\ -77,\ -6,\ 39,\ 43

(b)

0.017,\ 0.12,\ 0.134,\ 0.145,\ 0.3

(c)

70\%

(d)

0.27

(e)

18

cupcakes

Verification

Check 1: Walk along the answer to part (a) from left to right and subtract each number from the one after it. Every gap must come out positive, or two numbers are the wrong way round.

12,\ 71,\ 45,\ 4

Check 2: Multiply each decimal in the answer to part (b) by

1000

to clear the decimal point, and check the whole numbers climb.

17 < 120 < 134 < 145 < 300

Check 3: Work parts (c) and (d) backwards. A percentage is a count of hundredths, so put

70

over

100

; and read

0.27

as hundredths.

\dfrac{70}{100} = 0.7

and

0.27 = \dfrac{27}{100}

Check 4: The chocolate cupcakes and the ones that are not chocolate are the whole stall between them, so the two counts must add back to

60

. The chocolate ones number

\dfrac{7}{10} \times 60

42 + 18 = 60

Check 5: Do part (e) a different way, as a percentage:

\dfrac{3}{10}

30\%

, and

30\%

60

0.3 \times 60

0.3 \times 60 = 18

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The five numbers written in order, smallest first	B1	$-89,\ -77,\ -6,\ 39,\ 43$	✓
(b) The five decimals written in order, smallest first	B1	$0.017,\ 0.12,\ 0.134,\ 0.145,\ 0.3$ . Allow extra zeros, eg $0.017,\ 0.120,\ 0.134,\ 0.145,\ 0.300$ .	✓
(c) $0.7$ written as a percentage	B1	$70$	✓
(d) $\dfrac{27}{100}$ written as a decimal	B1	$0.27$	✓
(e) A complete method to reach the cupcakes that are not chocolate	M1	$1 - \dfrac{7}{10}$ ( $= \dfrac{3}{10} = 0.3$ , that is $30\%$ ) or $\dfrac{7}{10} \times 60$ ( $= 42$ ) oe.	✓
(e) The number of cupcakes that are not chocolate	A1	$18$ . A correct answer scores full marks, unless it comes from obvious incorrect working.	✓

Full marks: 6/6

The first five terms of a number sequence are shown below.

$106 \qquad 99 \qquad 92 \qquad 85 \qquad 78$

(a) (i) Write down the next term of the sequence.
[1 mark]
(ii) Explain how you found your answer to part (a)(i). [1 mark]

The $9\text{th}$ term of the sequence is $50$

(b) Work out the $12\text{th}$ term of the sequence. [1 mark]

Lorenzo says $7$ is a term in the sequence.
Lorenzo is wrong.

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

[Total 4 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

The first five terms of the sequence are

106, \ 99, \ 92, \ 85, \ 78

The

9\text{th}

term is

50

Lorenzo claims that

7

appears somewhere in the sequence.

Find

The next term, which is the

6\text{th}

term, and the rule used to get it. The

12\text{th}

term, starting from the

9\text{th}

term. A reason why

7

cannot be a term.

Plan the Solution

Subtract each term from the one after it. If every gap is the same, the sequence is linear and that gap is the common difference $d$ .
Add $d$ to the last printed term for part (a)(i). The value of $d$ is itself the explanation asked for in part (a)(ii).
For part (b), count the gaps between the $9\text{th}$ term and the $12\text{th}$ term, then apply $d$ that many times to $50$ .
For part (c), build the $n$ th term rule, then list the terms on either side of $7$ to show that the sequence steps straight over it.

Worked Solution [4 marks]

Rule, linear sequence: every term is the one before it plus a fixed common difference

d

, and the

n

th term is

dn + (a - d)

, where

a

is the first term.

Step 1: find the common difference

99 - 106 = -7

92 - 99 = -7

85 - 92 = -7

78 - 85 = -7

(Reason: (Reason: every gap comes to the same number, so the sequence is linear with

d = -7

. One gap on its own could be a coincidence, so all four are checked.))

Step 2: part (a)(i), take one more step

78 - 7 = 71

(Reason: (Reason:

78

is the

5\text{th}

term, so taking away

7

once gives the

6\text{th}

term.))

Step 3: part (a)(ii), say what the rule was

d = -7

(Reason: (Reason: the explanation the examiner wants is the common difference itself. Subtracting

7

, or adding

-7

, is what turns each term into the next one.))

Step 4: part (b), count the gaps, not the terms

12 - 9 = 3

(Reason: (Reason: the

9\text{th}

term is the starting point, so it is not a step. Only the moves to the

10\text{th}

11\text{th}

and

12\text{th}

terms count.))

Step 5: part (b), step on from the

9\text{th}

term to the

12\text{th}

50 - 3 \times 7 = 29

(Reason: (Reason: the

9\text{th}

term is

50

, and each further term takes away another

7

, so the number of gaps found in Step 4 is how many sevens come off.))

Step 6: part (c), build the

n

th term rule

a = 106

d = -7

dn + (a - d) = -7n + (106 + 7)

n\text{th term} = -7n + 113

(Reason: (Reason: the rule turns a question about a list into a question about one equation, which can then be tested for any value at all.))

Step 7: part (c), test whether

7

can be a term

-7n + 113 = 7

7n = 106

n = \dfrac{106}{7}

(Reason: (Reason:

7 \times 15 = 105

and

7 \times 16 = 112

, so

106

is not a multiple of

7

and

n

is not a whole number. A term position has to be a whole number.))

Step 8: part (c), show it in the list as well

29, \ 22, \ 15, \ 8, \ 1

(Reason: (Reason: after

15

the sequence goes to

8

and then to

1

, so it steps straight over

7

and never lands on it.))

(a)(i)

71

(a)(ii) subtract

7

from the term before(b)

29

15, \ 8, \ 1

, stepping over

7

Verification

Check 1: Put

n = 1

n = 5

and

n = 9

into

-7n + 113

106

78

and

50

, which are the first term, the last printed term and the

9\text{th}

term the question gives

Check 2: Run part (b) backwards: add

7

three times to the

12\text{th}

term and the

9\text{th}

term should come back.

29 + 3 \times 7 = 50

Check 3: Divide by

7

and compare remainders.

106 = 15 \times 7 + 1

, and taking

7

away never changes a remainder. every term leaves remainder

1

, while

7

leaves remainder

0

, so no term can equal

7

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) $71$	B1	cao	✓
(a)(ii) $-7$	B1	oe, for example 'take $7$ ', 'subtract $7$ ', 'each number is going down by $7$ ', $78 - 7$ , $-7n + 113$ , $-7 \times 6 + 113$ . If words are used, accept incorrect spelling where the meaning is clear.	✓
(b) $29$	B1	cao	✓
(c) correct reason	B1	oe, for example the sequence is $-7n + 113$ , or it goes $15, \ 8, \ 1$ , or the terms are not in the $7$ times table, or $8$ is in the sequence but $7$ is not. Not enough on its own: ' $7$ is not in the sequence', 'it will pass $7$ ', or 'Lorenzo is wrong'.	✓

Full marks: 4/4

Duncan has made two fair spinners for a game at his school fair.
Spinner $A$ has $4$ sides and can land on $1$ , $1$ , $2$ or $3$
Spinner $B$ has $5$ sides and can land on $5$ , $7$ , $8$ , $9$ or $11$

Duncan spins each spinner once.
He subtracts the number that spinner $A$ lands on from the number that spinner $B$ lands on to get his score.

(a) Complete the table to show all the possible scores.
The spinner $A$ numbers run across the top row and the spinner $B$ numbers run down the first column.
$\begin{array}{c|c|c|c|c} & 1 & 1 & 2 & 3 \\ \hline 5 & 4 & 4 & \phantom{00} & \phantom{00} \\ \hline 7 & \phantom{00} & \phantom{00} & 5 & 4 \\ \hline 8 & \phantom{00} & 7 & \phantom{00} & 5 \\ \hline 9 & \phantom{00} & 8 & \phantom{00} & \phantom{00} \\ \hline 11 & \phantom{00} & 10 & 9 & 8 \end{array}$ [2 marks]

(b) Find the probability that
(i) Duncan's score is an even number
[1 mark]
(ii) Duncan's score is greater than $7$ [1 mark]

(b)(i)(b)(ii)

[Total 4 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

Spinner

A

lands on

1

1

2

3

, each section equally likely

Spinner

B

lands on

5

7

8

9

11

, each section equally likely

Score

= B - A

, the spinner

B

number minus the spinner

A

number

Find

(a) the ten missing scores in the two-way table (b)(i)

P(\text{score is even})

(b)(ii)

P(\text{score} > 7)

Plan the Solution

Every cell of the table is one subtraction: the number at the start of the row minus the number at the top of the column, so fill each empty cell with $B - A$ .
The table has $4$ columns and $5$ rows, so there are $20$ equally likely outcomes. That $20$ is the denominator of both probabilities.
For each probability, count the cells that fit the description and put that count over the total. Count from the completed table, never from the spinners.
Read part (b)(ii) carefully: greater than $7$ does not include a score of $7$ itself.

Worked Solution [4 marks]

Rule - Equally likely outcomes:

P(\text{event}) = \dfrac{\text{number of successful outcomes}}{\text{total number of outcomes}}

Step 1: fill one cell, so the rule is clear

\text{score} = B - A

5 - 1 = 4

(Reason: The top left cell is spinner

B

5

with spinner

A

1

, and the table already prints

4

there. The same subtraction fills every other cell.)

Step 2: complete every cell of the table

\begin{array}{c|c|c|c|c} & 1 & 1 & 2 & 3 \\ \hline 5 & 4 & 4 & 3 & 2 \\ \hline 7 & 6 & 6 & 5 & 4 \\ \hline 8 & 7 & 7 & 6 & 5 \\ \hline 9 & 8 & 8 & 7 & 6 \\ \hline 11 & 10 & 10 & 9 & 8 \end{array}

(Reason: The two missing values on the top row are

5 - 2 = 3

and

5 - 3 = 2

. Across a row the score falls as the spinner

A

number rises; down a column it rises with the spinner

B

number. Both patterns are a quick way to spot a slip.)

Step 3: count the equally likely outcomes

4 \times 5 = 20

(Reason: Each of the

4

sections on spinner

A

can pair with each of the

5

sections on spinner

B

, so the completed table holds

20

outcomes and every one of them is equally likely.)

Step 4: part (b)(i), count the even scores

\text{Even scores}:\ 4,\,4,\,2,\,6,\,6,\,4,\,6,\,8,\,8,\,6,\,10,\,10,\,8 \implies 13\ \text{cells}

P(\text{even}) = \dfrac{13}{20}

(Reason: Work through the completed table one row at a time and count every even score.

13

of the

20

cells hold an even number, and

\dfrac{13}{20}

will not cancel.)

Step 5: part (b)(ii), count the scores greater than 7

\text{Scores} > 7:\ 8,\,8,\,10,\,10,\,9,\,8 \implies 6\ \text{cells}

P(\text{score} > 7) = \dfrac{6}{20} = \dfrac{3}{10}

(Reason: Greater than

7

means

8

and above, so a cell holding

7

does not count. Only the last two rows of the table reach that high.)

(a) the completed table above, with the ten missing scores in place(b)(i)

\dfrac{13}{20}

(b)(ii)

\dfrac{6}{20} = \dfrac{3}{10}

Verification

Check 1 - the scores the paper already printed: Test the printed cells against

B - A

. Three of them:

5 - 1 = 4

8 - 1 = 7

and

11 - 3 = 8

. All ten printed cells agree with the completed table, so the rule filling the empty cells is the right one.

Check 2 - the even count, without the table: A difference is even only when both numbers are odd or both are even. Spinner

B

shows

4

odd numbers and

1

even one; spinner

A

shows

3

odd and

1

even.

4 \times 3 + 1 \times 1 = 13

, the same count as reading the table.

Check 3 - the high scores, counted from the other end: Count the cells holding

7

or less instead. There are

14

of them, and the two counts must fill the table between them.

20 - 14 = 6

, which is the count part (b)(ii) needs.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) All ten missing scores correct in the table	B2	All correct. (B1 for $5$ , $6$ , $7$ , $8$ or $9$ values completed correctly.)	✓
(b)(i) $\dfrac{13}{20}$	B1ft	Or any equivalent: $0.65$ or $65\%$ . Follow through from a completed table.	✓
(b)(ii) $\dfrac{6}{20}$	B1ft	Or any equivalent: $\dfrac{3}{10}$ , $0.3$ or $30\%$ . Follow through from a completed table.	✓
Note on parts (b)(i) and (b)(ii)	note	Penalise incorrect notation only once. Penalise an incorrect denominator only once, as long as that denominator is greater than $13$ and is the same in both parts.	✓

Full marks: 4/4

There are $150$ vehicles in a car park.

Of these vehicles
$19$ are vans
$32$ are motorbikes
$3$ are lorries

The rest of the vehicles are cars.

Write the number of cars as a fraction of the total number of vehicles.
Give your fraction in its simplest form. [3 marks]

[Total 3 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

There are

150

vehicles in the car park altogether.

19

are vans,

32

are motorbikes and

3

are lorries.

Every vehicle that is left over is a car.

Find

The number of cars written as a fraction of

150

. The fraction must be given in its simplest form, so it has to be cancelled down.

Plan the Solution

Add the three groups that are not cars, so the vehicles already accounted for are known.
Subtract that total from $150$ to find how many cars there are.
Write the number of cars over the total number of vehicles.
Cancel the fraction by the highest common factor of its numerator and its denominator.

Worked Solution [3 marks]

Rule - a part as a fraction of a whole:

\dfrac{\text{part}}{\text{whole}}

, then divide the numerator and the denominator by their highest common factor.

Step 1: Count the vehicles that are not cars

19 + 32 + 3 = 54

(Reason: The vans, the motorbikes and the lorries are the three groups the question counts for you, so adding them gives the

54

vehicles that are already accounted for.)

Step 2: Subtract from the total to find the number of cars

150 - 54 = 96

(Reason: Every vehicle is a van, a motorbike, a lorry or a car, so whatever is left of the

150

after the other three groups are taken away must be the

96

cars.)

Step 3: Write the cars as a fraction of the total

\dfrac{96}{150}

(Reason: A fraction of a total puts the part on top and the whole underneath, so the number of cars goes over

150

and not over any smaller group.)

Step 4: Cancel down to the simplest form

\dfrac{96}{150} = \dfrac{6 \times 16}{6 \times 25} = \dfrac{16}{25}

(Reason: The highest common factor of

96

and

150

6

, so writing each of them as

6

times something lets that whole factor cancel in one go. Always test the result:

16

and

25

share no factor other than

1

, so nothing further will cancel and the fraction is genuinely in its simplest form.)

\dfrac{16}{25}

Verification

Check 1: Add all four groups back together. They must rebuild the

150

vehicles the car park started with.

19 + 32 + 3 + 96 = 150

Check 2: Turn the fraction before cancelling and the fraction after cancelling into decimals. Cancelling changes how a fraction is written, never how big it is, so the two decimals must be identical.

\dfrac{96}{150} = 0.64

and

\dfrac{16}{25} = 0.64

Check 3: Cross-multiply the two fractions. If

\dfrac{96}{150}

and

\dfrac{16}{25}

are equal then

96 \times 25

and

150 \times 16

must come to the same number.

96 \times 25 = 2400

and

150 \times 16 = 2400

Check 4: Break the final numerator and denominator into prime factors. If they have no prime factor in common, no further cancelling is possible and the fraction really is in its simplest form.

16 = 2^4

and

25 = 5^2

, with no prime factor in common

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$19 + 32 + 3 = 54$ or $150 - 19 - 32 - 3 = 96$	M1	For the number of vehicles that are not cars, or for the number of cars. Award also for the fraction $\dfrac{54}{150}$ .	✓
$\dfrac{150 - 54}{150} = \dfrac{96}{150}$	M1	This mark assumes the first one. Award also for any correct fraction that is not in its simplest form, or for $\dfrac{9}{25}$ .	✓
$\dfrac{16}{25}$	A1	A correct answer scores full marks, unless it comes from obviously incorrect working.	✓
Special case	SC	B2 for $0.64$ or $64\%$ if no other marks are scored - the right quantity, but given as a decimal or a percentage instead of as a fraction in its simplest form. B1 for $0.36$ or $36\%$ if no other marks are scored - the fraction of the vehicles that are not cars, so the wrong part of the total.	✓

Full marks: 3/3

$2$ flasks each contain $350$ millilitres of apple juice.
$5$ cartons each contain $y$ millilitres of apple juice.

Martin pours all the juice from the $2$ flasks and the $5$ cartons into a pan.
The total amount of juice that Martin pours into the pan is $2.8$ litres.

Work out the value of $y$ [4 marks]

y =

[Total 4 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

2

flasks, each holding

350

millilitres

5

cartons, each holding

y

millilitres

Poured together they make

2.8

litres

Find

The value of

y

, the number of millilitres in one carton

Plan the Solution

Change the total into millilitres, so that every amount is measured in the same unit.
Work out the juice that came from the $2$ flasks and take it off the total.
Share whatever is left equally between the $5$ cartons.

Worked Solution [4 marks]

Total juice

= 2 \times 350 + 5y

, with every volume measured in millilitres.

Step 1: Put both amounts into the same unit

2.8 \text{ litres} = 2.8 \times 1000 = 2800 \text{ ml}

(Reason: One litre is

1000

millilitres. Litres and millilitres cannot be mixed in one calculation, and this conversion is worth a mark on its own.)

Step 2: Work out the juice from the two flasks

2 \times 350 = 700 \text{ ml}

(Reason: Each flask holds

350

millilitres and there are

2

of them.)

Step 3: Take the flasks' juice off the total

2800 - 700 = 2100 \text{ ml}

(Reason: Everything in the pan came from either the flasks or the cartons, so whatever is left once the flasks are accounted for is the juice from the cartons.)

Step 4: Share what is left between the five cartons

5y = 2100

y = \dfrac{2100}{5} = 420

(Reason: The

5

cartons all hold the same amount, so divide the remaining juice by

5

y = 420

Verification

Check 1: Put

y = 420

back into the pan:

2 \times 350 + 5 \times 420 = 700 + 2100

2800 \text{ ml} = 2.8 \text{ litres}

, which is the total the question gives.

Check 2: Work the whole question in litres instead:

2 \times 0.35 = 0.7

, then

2.8 - 0.7 = 2.1

, then

\dfrac{2.1}{5} = 0.42

litres.

0.42

litres is

420

millilitres, so a different unit gives the same answer.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Convert so that both amounts are in one unit, e.g. $2.8 \text{ litres} = 2800 \text{ ml}$ or $350 \text{ ml} = 0.35 \text{ litres}$	B1	One correct conversion. It may be implied by later working that is consistently in one unit.	✓
$2800 - 2 \times 350 = 2100$ , or forming the equation $2 \times 350 + 5y = 2800$	M1	Method for the amount left for the five cartons, or an equivalent equation. Working in litres, $2.8 - 2 \times 0.35 = 2.1$ , earns the same mark.	✓
$\dfrac{2100}{5}$	M1	Divide their remaining amount by $5$ . Also earned by an answer of $0.42$ , which is the value in litres left unconverted.	✓
$y = 420$	A1	A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 4/4

Craft Corner and Palette House each have a special offer on pots of paint.

Ingrid buys $30$ pots of paint from Craft Corner using the special offer.
Bilal buys $30$ pots of paint from Palette House using the special offer.

Work out the difference between the amount that Ingrid pays and the amount that Bilal pays. [4 marks]

[Total 4 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

Craft Corner sells paint pots at

\$4.20

each, and its offer gives

1

free pot for every

2

pots paid for.

Palette House sells a pack of

5

paint pots for

\$18

, and its offer takes

25\%

off the price of each pack.

Ingrid and Bilal each buy

30

pots, each using their own shop's offer.

Find

The difference between the amount Ingrid pays at Craft Corner and the amount Bilal pays at Palette House.

Plan the Solution

Treat the two shops as two separate calculations, and only subtract at the very end.
Craft Corner first. The offer is really a $3$ for $2$ deal, so the useful question is how many of the $30$ pots are actually paid for.
Palette House next. Work out the full price of the packs, then reduce it by $25\%$ .
Finally subtract the smaller total from the larger one.

Worked Solution [4 marks]

Rule - a

3

for

2

offer: every group of

3

pots costs the price of

2

, so the number paid for is

\dfrac{2}{3}

of the number taken home. A reduction of

25\%

leaves

75\%

still to pay, so the amount paid is

0.75

of the full price.

Step 1: how many pots Ingrid actually pays for

\dfrac{30}{3} = 10

10 \times 2 = 20

(Reason: (Reason: the free pot comes with every

2

bought, so the pots come in groups of

3

. Thirty pots make

10

groups, and Ingrid pays for

2

pots in each group.))

Step 2: the amount Ingrid pays

20 \times 4.20 = 84

(Reason: (Reason: the

10

free pots cost nothing, so only the

20

paid-for pots are charged, at

\$4.20

each. Ingrid pays

\$84

.))

Step 3: the full price of Bilal's packs

\dfrac{30}{5} = 6

6 \times 18 = 108

(Reason: (Reason: Palette House only sells packs of

5

, and

30

pots is exactly

6

packs. Before the offer those packs would cost

\$108

.))

Step 4: the amount Bilal pays after the reduction

108 \times 0.75 = 81

(Reason: (Reason: taking

25\%

off leaves

75\%

of the price still to pay, and

75\%

written as a decimal is

0.75

. Multiplying by the decimal that is LEFT is quicker than working out the saving and subtracting it.))

Step 5: the difference between the two amounts

84 - 81 = 3

(Reason: (Reason: Ingrid pays the larger amount, so subtract Bilal's total from hers. The question asks for the difference, so the answer is the gap between them, not either total.))

\$3

Verification

Check 1: Cost Ingrid's pots a completely different way. Every

3

pots cost

2 \times 4.20

, so on average one pot costs

\dfrac{2 \times 4.20}{3}

, and all

30

pots cost thirty times that.

\dfrac{2 \times 4.20}{3} = 2.80

, and

30 \times 2.80 = 84

, which is Step 2 again.

Check 2: Cost Bilal's pots from the saving instead of the multiplier. Work out

25\%

of one pack, take it off the pack price, then multiply by the

6

packs.

0.25 \times 18 = 4.50

, so a pack costs

18 - 4.50 = 13.50

, and

6 \times 13.50 = 81

, which is Step 4 again.

Check 3: Is the size sensible? Both offers bring the price of one pot down to somewhere under

\$3

, so two totals in the eighties and a small gap between them is what we should expect. Ingrid pays

\$84

, Bilal pays

\$81

, and the gap is

\$3

- under

4\%

of either total, which is the sort of margin two rival offers should sit within.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A method that uses one of the two offers	M1	For working with the Craft Corner $3$ for $2$ offer, for example $\dfrac{2}{3} \times 30 \; (= 20)$ or $\dfrac{2}{3} \times 4.20 \; (= 2.80)$ , or for at least two multiples of $8.40$ listed against the number of pots bought. Or for working with the $25\%$ off offer at Palette House, for example $0.75 \times 18 \; (= 13.50)$ or $\dfrac{30}{5} \times 18 \times 0.75 \; (= 81)$ .	✓
One of the two amounts, correct	A1	For $84$ at Craft Corner, or $81$ at Palette House. Either one earns this mark on its own.	✓
A fully correct method for the difference	M1	For $84 - 81$ , where both amounts have come from correct working. Subtracting two values that were themselves wrong does not earn this.	✓
The difference	A1	For $3$ . The scheme also allows $-3$ , since a candidate who subtracts the other way round has still found the difference.	✓
A correct answer with no working shown	Note	A correct answer scores full marks, unless it has come from obviously incorrect working.	✓

Full marks: 4/4

A circle has a radius of $9$ cm.

Work out the area of the circle.
Give your answer correct to $3$ significant figures. [2 marks]

cm²

[Total 2 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

A circle whose radius is

9

cm.

Find

The area of the circle, correct to

3

significant figures.

Plan the Solution

The radius is known, so use the area formula $A = \pi r^{2}$ .
Square the radius first, then multiply by $\pi$ on the calculator.
Round only at the very end, so the third significant figure is safe.

Worked Solution [2 marks]

Area of a circle:

A = \pi r^{2}

, where

r

is the radius.

Step 1: Write down the rule

A = \pi r^{2}

(Reason: The question gives the radius, so the version of the formula written in terms of

r

is the one to reach for.)

Step 2: Square the radius

r = 9

r^{2} = 9 \times 9 = 81

(Reason: Squaring a length means multiplying it by itself, and it is the radius that is squared in the formula, never the whole of

\pi r

Step 3: Multiply by pi

A = \pi \times 81

A \approx 254.4690

(Reason: Use the calculator's own

\pi

key and keep every digit on the display. Rounding here instead of at the end can move the third significant figure.)

Step 4: Round to 3 significant figures

254.4690 \approx 254

(Reason: The first three significant figures are

2

5

and

4

. The digit after them is

4

, which is below

5

, so the last figure kept stays as it is.)

254 \text{ cm}^{2}

(to

3

significant figures)

Verification

Check 1: Run the calculation backwards: divide the unrounded area by

\pi

and square-root the result. It must give the radius back.

\sqrt{\dfrac{254.4690}{\pi}} \approx 9

, the radius the question gives.

Check 2: Trap the circle between two squares. The square drawn inside it has area

2r^{2}

and the square drawn round it has area

4r^{2}

, so the circle must sit between them.

2 \times 81 = 162

and

4 \times 81 = 324

, and

162 < 254 < 324

Check 3: Repeat the working with the approximations for

\pi

that the mark scheme allows,

3.14

and

\dfrac{22}{7}

3.14 \times 81 = 254.34

and

\dfrac{22}{7} \times 81 \approx 254.57

, which round to

254

and

255

, both inside the accepted band.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\pi \times 9^{2}$ or equivalent	M1	Any use of $\pi$ , $3.14$ , $3.142$ or $\dfrac{22}{7}$ multiplied by the radius squared. Writing $\pi \times 81$ earns it just as well.	✓
$254$	A1	Accept anything from $254$ to $255$ , which covers the approximations for $\pi$ allowed above. A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 2/2

Shape $A$ is drawn on the centimetre grid below.

On the second centimetre grid, draw a rectangle that has the same area as shape $A$ . [2 marks]

[Total 2 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

Shape

A

is drawn on a centimetre grid, so every small square measures

1

cm by

1

cm and covers an area of

1\text{ cm}^2

Shape

A

has eight sides, and every one of them runs along a grid line.

Find

A rectangle whose area is equal to the area of shape

A

Plan the Solution

Cut shape $A$ into rectangles whose sides lie along the grid lines.
Work out the area of each rectangle and add them, to get the area of shape $A$ .
Choose two side lengths whose product is that area, and draw that rectangle on the second grid.

Worked Solution [2 marks]

\text{area of a rectangle} = \text{length} \times \text{width}

Step 1: cut shape

A

into three rectangles

\text{left arm} = 3 \times 4 = 12\text{ cm}^2

\text{middle bar} = 6 \times 1 = 6\text{ cm}^2

\text{lower arm} = 2 \times 3 = 6\text{ cm}^2

(Reason: The tall arm on the left, the bar across the middle and the short arm at the lower right meet edge to edge and do not overlap. Each small square has area

1\text{ cm}^2

, so a piece's area is its width in squares multiplied by its height in squares.)

Step 2: add the three areas

12 + 6 + 6 = 24

(Reason: Together the three pieces make up the whole of shape

A

and nothing else, so shape

A

has area

24\text{ cm}^2

Step 3: choose a rectangle with the same area

6 \times 4 = 24

(Reason: Any two lengths whose product is

24

will do:

12

2

8

3

, or

6

4

. A rectangle

6

cm by

4

cm sits comfortably inside the second grid.)

Step 4: draw the rectangle on the grid

6\text{ squares across} \times 4\text{ squares down} = 24\text{ squares}

(Reason: Draw along the grid lines, counting

6

squares across and

4

squares down, so the finished rectangle really does cover

24

centimetre squares.)

A rectangle measuring

6

cm by

4

cm, drawn on the centimetre grid, area

24\text{ cm}^2

Verification

Check 1: Count the whole centimetre squares inside shape

A

one row at a time, working down from the top.

3 + 3 + 3 + 9 + 2 + 2 + 2 = 24

Check 2: Take the smallest rectangle that surrounds shape

A

, which is

9

cm by

7

cm, then subtract the two rectangular pieces missing from it:

6

3

at the top right and

7

3

at the bottom left.

63 - 18 - 21 = 24

Check 3: Multiply the drawn rectangle's own side lengths and compare the result with the area of shape

A

6 \times 4 = 24

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A rectangle of area $24\text{ cm}^2$ drawn on the centimetre grid	B2	For a correct rectangle, for example $2$ squares by $12$ squares, $3$ squares by $8$ squares, $4$ squares by $6$ squares, or $5$ squares by $4.8$ squares. The sides need not be whole numbers of squares.	✓
The area of shape $A$ stated as $24$ , or any rectangle drawn	B1	Partial credit, awarded independently: $1$ mark for stating $24$ , or for drawing any rectangle, where the two are not brought together into a correct rectangle of area $24\text{ cm}^2$ .	✓

Full marks: 2/2

(a) Simplify $e \times e \times e \times e \times e$
[1 mark]
(b) Simplify $m + m + m$
[1 mark]
(c) Simplify $3g^{2} + 7g^{2} - 4g^{2}$
[1 mark]
(d) Expand $a(a + 8)$
[1 mark]
(e) Factorise $15x + 20$ [1 mark]

Bridget sells $d$ packs of greetings cards and $h$ boxes of greetings cards.

Each pack contains $3$ greetings cards.
Each box contains $5$ greetings cards.

The total number of greetings cards that Bridget sells is $T$

(f) Write down a formula for $T$ in terms of $d$ and $h$ [3 marks]

(a)(b)(c)(d)(e)(f)

[Total 8 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

Five expressions to work on:

e \times e \times e \times e \times e

m + m + m

3g^{2} + 7g^{2} - 4g^{2}

a(a + 8)

and

15x + 20

For part (f): a pack holds

3

greetings cards and a box holds

5

greetings cards, and

d

packs and

h

boxes are sold.

Find

The simplest form of each expression in parts (a) to (e). A formula for the total number of cards,

T

, written in terms of

d

and

h

Plan the Solution

Part (a) is a product of powers of one letter, so add the indices. Parts (b) and (c) are sums of like terms, so combine the coefficients and leave the letter part alone.
Part (d) multiplies out: the term outside the bracket multiplies each term inside it.
Part (e) is the reverse of part (d): take out the highest common factor of $15$ and $20$ .
Part (f) counts the two kinds of container separately and adds the results, so the formula has one term for the packs and one for the boxes.

Worked Solution [8 marks]

Index law and the distributive law:

x^{m} \times x^{n} = x^{m+n}

and

p(q + r) = pq + pr

. Like terms are added by adding their coefficients only.

Step 1 (a): count how many times

e

is multiplied

e \times e \times e \times e \times e = e^{1+1+1+1+1} = e^{5}

(Reason: Every plain

e

means

e^{1}

, and multiplying powers of the same base adds the indices. There are

5

of them, so the index is

5

. Multiplying is not adding:

e \times e

e^{2}

, not

2e

Step 2 (b): add the like terms

m

m + m + m = (1 + 1 + 1)m = 3m

(Reason: There are three lots of the same term, so the coefficients

1

1

and

1

are added and the letter is unchanged. Adding is not multiplying:

m + m + m

3m

, while

m \times m \times m

would be

m^{3}

Step 3 (c): combine the coefficients of

g^{2}

3g^{2} + 7g^{2} - 4g^{2} = (3 + 7 - 4)g^{2} = 6g^{2}

(Reason: All three terms are lots of

g^{2}

, so they are like terms and only the numbers in front are combined. The letter part stays as

g^{2}

and is never squared again or doubled.)

Step 4 (d): multiply each term inside the bracket by

a

a(a + 8) = a \times a + a \times 8 = a^{2} + 8a

(Reason: The distributive law reaches both terms in the bracket, not just the first.

a \times a

a^{2}

and

a \times 8

8a

, so the

a

must still be there in the second term. The two terms are not like terms, so nothing further can be collected.)

Step 5 (e): take out the highest common factor of

15

and

20

15x + 20 = 5 \times 3x + 5 \times 4 = 5(3x + 4)

(Reason: The highest common factor of

15

and

20

5

, so

5

comes outside and what is left,

3x

and

4

, goes inside. The letter

x

appears in only one of the two terms, so it cannot be taken out as well.)

Step 6 (f): write one term for the packs and one for the boxes

\text{cards in the packs} = 3 \times d = 3d

\text{cards in the boxes} = 5 \times h = 5h

T = 3d + 5h

(Reason: Each of the

d

packs holds

3

cards, giving

3d

cards, and each of the

h

boxes holds

5

cards, giving

5h

cards. Every card sold is in a pack or in a box, so the two amounts are added. A formula is asked for, so the answer must start

T =

and not stop at the expression.)

(a)

e^{5}

(b)

3m

(c)

6g^{2}

(d)

a^{2} + 8a

(e)

5(3x + 4)

(f)

T = 3d + 5h

Verification

Check 1: Put numbers in. With

e = 2

m = 4

and

g = 2

, work out each original expression and each answer, and compare the two.

2 \times 2 \times 2 \times 2 \times 2 = 32 = 2^{5}

4 + 4 + 4 = 12 = 3 \times 4

12 + 28 - 16 = 24 = 6 \times 4

Check 2: Run parts (d) and (e) backwards. Factorise the answer to (d) and expand the answer to (e); each should give back the expression it came from.

a^{2} + 8a = a(a + 8)

and

5(3x + 4) = 15x + 20

Check 3: Test that

5

really is the highest common factor in part (e) by looking at what is left inside the bracket.

15 = 3 \times 5

and

20 = 4 \times 5

, and

3

and

4

have no common factor left

Check 4: Test the formula on a case that can be counted by hand:

4

packs and

2

boxes hold

4 \times 3 = 12

cards and

2 \times 5 = 10

cards.

T = 3 \times 4 + 5 \times 2 = 22

, and counting gives

12 + 10 = 22

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $e^{5}$	B1	For $e^{5}$ . The index must be written as a power of $e$ ; $5e$ earns nothing.	✓
(b) $3m$	B1	For $3m$ . $m^{3}$ earns nothing.	✓
(c) $6g^{2}$	B1	For $6g^{2}$ . The power must stay as $g^{2}$ .	✓
(d) $a^{2} + 8a$	B1	For $a^{2} + 8a$ . $a^{2} + 8$ earns nothing.	✓
(e) $5(3x + 4)$	B1	For $5(3x + 4)$ . A partial factorisation such as $5(3x + 20)$ or an unfactorised answer earns nothing.	✓
(f) $T = 3d + 5h$	B3	Allow $T = d3 + h5$ or $T = d \times 3 + h \times 5$ .	✓
(f) partial credit: the right expression without $T =$ , or one coefficient right	B2	For $3d + 5h$ with no $T =$ in front, or for $T = 3d + xh$ or $T = yd + 5h$ or $T = 5d + 3h$ , where $x \neq 0$ and $y \neq 0$ and either may be negative. The last of these is the swap: $5$ attached to the packs and $3$ to the boxes.	✓
(f) partial credit: one coefficient right with no $T =$ , or any expression in $d$ and $h$	B1	For $3d + xh$ or $yd + 5h$ where $x \neq 0$ and $y \neq 0$ and either may be negative, or for $5d + 3h$ , or for $T =$ an incorrect expression in $d$ and $h$ , for example $T = d + h$ or $T = d \times h$ , or for $T = 3d$ with or without a constant, or $T = 5h$ with or without a constant.	✓

Full marks: 8/8

Duncan buys a telescope in Australia.

The telescope costs $232$ Australian dollars.
In India, an identical telescope costs $12\,420$ Indian rupees.

The exchange rate is $1$ Australian dollar for $54$ Indian rupees.

The telescope costs more in Australia than in India.

Work out how much more.
You must give the units of your answer. [3 marks]

[Total 3 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

In Australia the telescope costs

232

Australian dollars.

In India the identical telescope costs

12\,420

Indian rupees.

The exchange rate is

1

Australian dollar for

54

Indian rupees.

Australia is the dearer of the two.

Find

How much more the telescope costs in Australia than in India, with the units written beside the number.

Plan the Solution

The two prices are in different currencies, so they cannot be compared yet. Turn the Indian price into Australian dollars by dividing by $54$ .
Subtract the smaller price from the larger one, now that both are in the same currency.
Write the units beside the number, because the question asks for them and the final mark is for them.

Worked Solution [3 marks]

Two prices can only be compared once they are in the same currency. Multiplying by the rate

54

turns Australian dollars into rupees, and dividing by

54

turns rupees back into Australian dollars.

Step 1: put the Indian price into Australian dollars

\dfrac{12\,420}{54} = 230

(Reason: Every Australian dollar is worth

54

rupees, so the number of dollars is found by dividing the rupee price by

54

. The telescope in India costs

230

Australian dollars.)

Step 2: subtract, now that both prices are in the same currency

232 - 230 = 2

(Reason: Both prices are Australian dollar prices, so subtracting the Indian one from the Australian one gives the extra amount Duncan pays by buying in Australia.)

Step 3: write the units beside the number

2 \text{ Australian dollars}

(Reason: A bare number would not say which currency the difference is in, and the same difference is

108

in rupees. The question asks for the units, so the currency must be written down.)

2

Australian dollars

Verification

Check 1: Work the whole question in rupees instead. The Australian price in rupees is

232 \times 54 = 12\,528

, and the gap between the two rupee prices is

12\,528 - 12\,420 = 108

rupees. Turning that gap back into Australian dollars gives

\dfrac{108}{54} = 2

, the same difference reached the other way round.

Check 2: Undo Step 1: multiply the Indian price in Australian dollars back by the rate, and add the difference back on to the Indian price.

230 \times 54 = 12\,420

, which is the rupee price the question gives, and

230 + 2 = 232

, which is the Australian price.

Check 3: Is the size sensible? Both telescopes cost about

230

Australian dollars, so the gap between them should be small. A gap of

2

Australian dollars on a price of

232

is under one per cent, which is what two almost equal prices should give.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Convert so that both prices are in one currency	M1	A complete method for one conversion, either way round: $\dfrac{12\,420}{54} \; (= 230)$ or $232 \times 54 \; (= 12\,528)$ .	✓
Subtract to find the difference	M1	Subtracting in whichever currency was chosen, using the candidate's own converted value: $232 - 230 = 2$ or $12\,528 - 12\,420 = 108$ . The quotation marks in the official scheme mark the converted value as the candidate's own.	✓
The answer, with its units	A1	cao $2$ Australian dollars, or $108$ rupees. The answer must carry the correct units, which may be shortened, for example $\$$ for dollars or r for rupees, and an incorrect spelling is allowed if the meaning is clear. A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

Owen wants to knit a gift for a newborn baby.

He can knit a jacket ( $J$ ) or a blanket ( $B$ ) or a hat ( $H$ ) or a scarf ( $S$ )
He can knit using white wool ( $W$ ) or using yellow wool ( $Y$ )

Owen chooses one item to knit and chooses one colour of wool.

Write down all the possible combinations for the item that Owen could knit. [2 marks]

[Total 2 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

Four items to choose from:

J

(jacket),

B

(blanket),

H

(hat),

S

(scarf)

Two wool colours to choose from:

W

(white) and

Y

(yellow)

Exactly one item is chosen, and exactly one colour is chosen.

Find

Every possible combination of one item with one colour, written as a list The list must be complete: no combination missing, none repeated, and nothing in it that is not a real pair

Plan the Solution

Work out how many combinations there should be before listing any, so the list can be checked against a target: $4 \times 2 = 8$ .
Be systematic. Take the items in the order they are given, and for each item write its white pair first and its yellow pair second.
Write the item letter first and the colour letter second every time, so no pair can accidentally be written twice in two different ways.

Worked Solution [2 marks]

Rule - Systematic listing: if there are

m

choices of one kind and

n

choices of another, then pairing every choice with every other choice gives

m \times n

combinations.

Step 1: work out how many combinations to expect

4 \times 2 = 8

(Reason: (Reason: each of the

4

items can be knitted in either of the

2

colours, so there are

8

pairs altogether. Knowing the target first is what stops the list being cut short.))

Step 2: pair the first item with each colour

J: JW, JY

(Reason: (Reason: the jacket

J

can be knitted in white

W

or in yellow

Y

, and there is no third choice, so the jacket contributes exactly

2

combinations.))

Step 3: repeat for the other three items

B: BW, BY

H: HW, HY

S: SW, SY

(Reason: (Reason: the same two colours are available whatever the item is, so each of the remaining

3

items contributes

2

more combinations, giving

6

here.))

Step 4: collect the four rows into one list

JW, JY, BW, BY, HW, HY, SW, SY

(Reason: (Reason: reading the rows in order gives

8

combinations, which matches Step 1. Because every row was built the same way, no pair is repeated and every pair is a real item-with-colour choice.))

JW, JY, BW, BY, HW, HY, SW, SY

Verification

Check 1: Count the entries in the finished list and compare with

4 \times 2 = 8

. The list holds

8

entries and all

8

are different, so nothing is missing and nothing is repeated.

Check 2: Count how many times each colour letter appears in the list.

W

appears

4

times and

Y

appears

4

times - once with each of the

4

items, as it must be.

Check 3: Read the list the other way round: count how many times each item letter appears.

J

B

H

and

S

each appear

2

times, once in white and once in yellow, so no item has been given a colour twice or left out.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
All $8$ combinations written down: $JW, JY, BW, BY, HW, HY, SW, SY$	B2	All correct combinations, with no repeats and no incorrect combinations. Any order is accepted.	✓
At least $4$ correct combinations written down	B1	Partial credit: at least $4$ correct combinations, ignoring any repeats and ignoring any incorrect combinations.	✓

Full marks: 2/2

Shape $A$ and shape $B$ are drawn on the grid below.

(a) Shape $A$ is mapped onto shape $B$ by a single transformation.
Describe this transformation fully. [2 marks]

(b) On the grid, draw the reflection of shape $A$ in the line with equation $y = -1$ [2 marks]

(a)

[Total 4 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

Shape

A

has vertices

(2,\,3)

(5,\,2)

(5,\,1)

and

(2,\,1)

Shape

B

has vertices

(-2,\,-3)

(-5,\,-2)

(-5,\,-1)

and

(-2,\,-1)

Both axes run from

-6

6

, so the whole grid is available for the image.

Find

(a) the single transformation that maps

A

onto

B

, described fully. (b) the reflection of shape

A

in the line

y = -1

, drawn on the grid.

Plan the Solution

Pair each vertex of $A$ with the vertex of $B$ it maps onto, and look for the pattern in the coordinates.
A full description needs THREE things for a rotation: the word rotation, the angle, and the centre. Two of the three earns only half the marks.
Find the centre from the data, not by eye: for a half turn it is the midpoint of every vertex and its image.
For (b), reflect one vertex at a time in the horizontal line $y = -1$ , then join the images in the same order round the shape.

Worked Solution [4 marks]

A rotation of

180^\circ

about the centre

(a,\,b)

maps

(x,\,y)

onto

(2a - x,\,2b - y)

, and the centre is the midpoint of every point and its image. A reflection in the horizontal line

y = c

maps

(x,\,y)

onto

(x,\,2c - y)

Step 1: pair up the vertices of

A

and

B

(2,\,3) \to (-2,\,-3)

(5,\,2) \to (-5,\,-2)

(5,\,1) \to (-5,\,-1)

(2,\,1) \to (-2,\,-1)

(Reason: Go round each shape in the same direction. The sloping side of

A

joins

(2,\,3)

(5,\,2)

and the sloping side of

B

joins

(-2,\,-3)

(-5,\,-2)

, so those vertices correspond.)

Step 2: read the rule that links every pair

(x,\,y) \to (-x,\,-y)

(Reason: Both coordinates change sign, and nothing else changes. A slide would add the same amount to every point, and a flip in one of the axes would change one coordinate only, so this is a turn.)

Step 3: find the centre of the turn

\dfrac{2 + (-2)}{2} = 0

\dfrac{3 + (-3)}{2} = 0

(Reason: The centre of a half turn is the midpoint of each point and its image. Working the midpoint of

(2,\,3)

and

(-2,\,-3)

coordinate by coordinate gives

(0,\,0)

, and the other three pairs give the same point, so the centre is the origin.)

Step 4: describe the transformation in full

180^\circ \text{ rotation about } (0,\,0)

(Reason: Three things are needed and all three are now known: the type (rotation), the angle

(180^\circ)

and the centre

((0,\,0))

. A half turn is the one angle where the direction does not matter, so clockwise or anticlockwise need not be stated.)

Step 5: reflect each vertex of

A

in the line

y = -1

(x,\,y) \to (x,\,-2 - y)

(2,\,3) \to (2,\,-5)

(5,\,2) \to (5,\,-4)

(5,\,1) \to (5,\,-3)

(2,\,1) \to (2,\,-3)

(Reason: A reflection in a horizontal line leaves every

x

where it is and sends

y

2c - y

. Here the mirror is

y = -1

, so

2c

-2

and each new

y

-2 - y

Step 6: check the distances, then join the four image points

3 - (-1) = 4

-1 - 4 = -5

1 - (-1) = 2

-1 - 2 = -3

(Reason: The top vertex is

4

above the mirror line, so its image is

4

below it, at

(2,\,-5)

. The bottom left vertex is

2

above, so its image is

2

below, at

(2,\,-3)

. Join the four images in the same order round the shape and the reflection is drawn.)

(a) Rotation of

180^\circ

about the origin

(0,\,0)

(b) Image drawn with vertices

(2,\,-3)

(2,\,-5)

(5,\,-3)

and

(5,\,-4)

Verification

Check 1: Turn shape

B

through

180^\circ

about the origin and see where it lands:

(-2,\,-3)

goes to

(2,\,3)

and

(-5,\,-2)

goes to

(5,\,2)

. Shape

B

maps back onto shape

A

, so the same rotation works in both directions.

Check 2: Compare the side lengths. Shape

A

has a horizontal side of

3

, a vertical side of

2

and a short vertical side of

1

, and shape

B

has exactly the same three. The two shapes are congruent, so the description is a rotation. An enlargement of scale factor

-1

about the origin sends the points to the same places, which is why the mark scheme allows that wording only as a special case.

Check 3: Test the reflection the other way round. Each image should sit as far below

y = -1

as its object sits above it, with the

x

-coordinate unmoved:

(5,\,2)

3

above the line and

(5,\,-4)

3

below it. All four

x

-coordinates are unchanged and all four distances match, so the drawn shape is a true reflection of

A

y = -1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Name the transformation	B1	rotation, or an equivalent word, with no mention of reflection, translation, enlargement, move, flip or any similar word	✓
(a) Give the angle and the centre	B1	$180^\circ$ , or a half turn, about $(0,\,0)$ , or $O$ , or 'the origin'. Ignore any reference to clockwise or anticlockwise.	✓
(a) Special case	SC B2	an answer of 'enlargement, centre the origin, scale factor $-1$ ' scores SC B2, because it moves every point to the right place but is not the description asked for	✓
(b) Draw the reflected shape	B2	a correct shape, with vertices $(2,\,-3)$ , $(2,\,-5)$ , $(5,\,-3)$ and $(5,\,-4)$	✓
(b) Partial credit	B1	a 'correct' shape reflected in any horizontal line, or a correct reflection in the line $x = -1$ , or shape $B$ reflected in $y = -1$	✓

Full marks: 4/4

Continue to questions 15 to 28

The remaining 14 questions, with the same full worked solutions and mark schemes

Frequently asked questions

There are 28 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 2F and Paper 2H, 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 Paper 2F, June 2024: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 14 of 28

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

Frequently asked questions

Keep revising

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