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Edexcel IGCSE 4MA1 Paper 2FR, November 2024: Worked Solutions and Mark Schemes

Sir Faraz Hassan

3 Aug 2026

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

Original worked solutions for Edexcel International GCSE Mathematics A (4MA1), Paper 2FR (Foundation Tier), November 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 15 of 26

(a) Arrange these five numbers in order of size, smallest first.
$122 \qquad 5 \qquad 150 \qquad 71 \qquad 39$ [1 mark]

(b) Arrange these five decimals in order of size, smallest first.
$0.7 \qquad 0.074 \qquad 3.77 \qquad 0.37 \qquad 0.13$ [1 mark]

(d) Write down what the digit $3$ is worth in the number $1324$ [1 mark]

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

[Total 4 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

Five whole numbers:

122, 5, 150, 71, 39

Five decimals:

0.7, 0.074, 3.77, 0.37, 0.13

The number five thousand and eighty four, written in words.

The number

1324

, with attention on its digit

3

Find

(a) the five numbers in order, smallest first (b) the five decimals in order, smallest first (c) five thousand and eighty four written in figures (d) what the

3

1324

is worth

Plan the Solution

All four parts are place value, so read every number column by column rather than at a glance.
(a) Count the digits first: a number with fewer digits is smaller. Only inside a group of the same length do you compare digits from the left.
(b) Give every decimal the same number of decimal places, then the comparison becomes a whole-number one.
(c) Write the thousands, hundreds, tens and units columns in turn, using $0$ for any column the words do not mention.
(d) Find which column the $3$ stands in, then multiply the digit by that column's place value.

Worked Solution [4 marks]

Rule - Place value: a digit is worth its face value multiplied by the column it stands in (units

1

, tens

10

, hundreds

100

, thousands

1000

), and numbers are compared by their highest column first.

Step 1: (a) sort the whole numbers by length, then by leading digit

5 < 39 < 71 < 122 < 150

(Reason:

5

has one digit,

39

and

71

have two, and

122

and

150

have three, so the one-digit number is smallest and the three-digit numbers are largest. Inside each group compare from the left:

3

tens is less than

7

tens, and

122

and

150

share a hundreds digit, so the tens decide it because

2

is less than

5

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

0.700, 0.074, 3.770, 0.370, 0.130

74 < 130 < 370 < 700 < 3770

0.074 < 0.13 < 0.37 < 0.7 < 3.77

(Reason: Filling the short ones with trailing zeros changes nothing, because a zero on the end of a decimal adds no value. With three decimal places each, the five decimals are

700

74

3770

370

and

130

thousandths, and ordering whole numbers of thousandths is easy. The commonest slip is reading

0.7

as smaller than

0.074

because it has fewer digits.)

Step 3: (c) build the number one column at a time

5000 + 84 = 5084

(Reason: Five thousand is a

5

in the thousands column, and eighty four is

8

tens and

4

units. The words never mention hundreds, so that column takes a

0

to hold the

5

in the thousands place. Leaving it out would give

584

, a different number.)

Step 4: (d) find the column the digit 3 stands in

1324 = 1000 + 300 + 20 + 4

3 \times 100 = 300

(Reason: Reading

1324

column by column gives

1

thousand,

3

hundreds,

2

tens and

4

units, so the

3

stands in the hundreds column. Its worth is the digit multiplied by that column's place value. The mark scheme accepts either the words or the number.)

(a)

5, 39, 71, 122, 150

(b)

0.074, 0.13, 0.37, 0.7, 3.77

(c)

5084

(d)

3

hundreds, that is

300

Verification

Check 1: Read both ordered lists backwards. They must run largest first and still hold all five of the given values, none lost and none invented:

150, 122, 71, 39, 5

and

3.77, 0.7, 0.37, 0.13, 0.074

. Five values in, five values out, in the opposite order

Check 2: Order the decimals a second way, without lining up any decimal points. Multiply every one by

1000

to get

700, 74, 3770, 370, 130

, order those whole numbers, then divide back by

1000

0.074, 0.13, 0.37, 0.7, 3.77

Check 3: Read the answer to (c) back out in words, column by column:

5

thousands,

0

hundreds,

8

tens,

4

units.

5084

reads as five thousand and eighty four

Check 4: Take the value of the

3

away from

1324

and look at what is left:

1324 - 300 = 1024

. The hundreds column of

1024

is now empty. Only the hundreds column emptied, so the

3

was worth

300

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $5, 39, 71, 122, 150$	B1	All five numbers in the correct order, smallest first.	✓
(b) $0.074, 0.13, 0.37, 0.7, 3.77$	B1	All five decimals in the correct order, smallest first.	✓
(c) $5084$	B1	The four-digit number, with the $0$ in the hundreds column.	✓
(d) $3$ hundreds	B1	Accept $300$ or the word hundreds on its own.	✓

Full marks: 4/4

An atlas records the total number of countries in each continent.
The bar chart shows this information for five of the continents.

Europe has a total of $44$ countries.

(a) Complete the bar chart by drawing the bar for Europe. [1 mark]

(b) Write down the name of the continent that has $23$ countries. [1 mark]

(d) Work out the total number of countries in Africa and Oceania altogether. [1 mark]

(b)(c)(d)

[Total 4 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

A bar chart of the number of countries in five continents: Asia

48

, North America

23

, Africa

54

, South America

12

and Oceania

14

Europe has

44

countries, and its bar has not been drawn.

The vertical scale runs from

0

56

and is labelled every

4

countries.

Find

(a) The Europe bar, drawn to the right height. (b) The continent whose bar reads

23

. (c) The continent whose total is

4

times the South America total. (d) The Africa total added to the Oceania total.

Plan the Solution

Read every bar against the vertical scale, remembering that one gridline to the next is $4$ countries, not $1$ .
For (a), find $44$ on the scale and draw a bar of the same width as the others up to that line.
For (b), look for the bar that stops just short of the line marked $24$ .
For (c), work out $4 \times 12$ first, then find the bar of that height.
For (d), add the two totals that have been read off.

Worked Solution [4 marks]

Rule - Reading a bar chart: the height of a bar against the vertical scale is the value it stands for, and a value between two gridlines is found by counting on from the gridline below it.

Step 1: draw the bar for Europe (part (a))

11 \times 4 = 44

(Reason: The scale rises in steps of

4

, so

44

is eleven whole steps above zero and is itself a labelled line. Draw a bar in the Europe position, the same width as the others, with its top level with that line.)

Step 2: find the bar that reads 23 (part (b))

23 = 20 + 3

(Reason: A bar at

23

sits three above the line marked

20

and just below the line marked

24

. Only the second bar does that, and it is North America.)

Step 3: find the continent with

4

times the South America total (part (c))

4 \times 12 = 48

(Reason: South America reads

12

, so the continent being asked for reads

48

. The only bar level with the line marked

48

is Asia.)

Step 4: add Africa and Oceania (part (d))

54 + 14 = 68

(Reason: Africa is the tallest bar and reads

54

. Oceania is the fifth bar and reads

14

. The word altogether means the two totals are added.)

(a) Europe bar drawn to a height of

44

(b) North America(c) Asia(d)

68

Verification

Check 1: Count the gridlines for part (a). The scale goes up in

4

s, and

11 \times 4 = 44

. The Europe bar is level with the eleventh gridline above zero, the one already labelled

44

Check 2: Do part (c) the other way round, by dividing instead of multiplying:

\dfrac{48}{12} = 4

. Asia has exactly

4

times as many countries as South America, which is what part (c) asked for.

Check 3: Do part (d) from the whole chart. All six continents come to

48 + 23 + 54 + 12 + 14 + 44 = 195

, and the four that part (d) does not use come to

48 + 23 + 12 + 44 = 127

, so

195 - 127 = 68

. Africa and Oceania together come to

68

, reached without adding those two bars directly.

Check 4: Make sure part (b) has only one answer by listing the other five totals:

48

54

12

14

and

44

. None of them is

23

, so North America is the only continent that fits.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A bar drawn in the Europe position of height $44$	B1	The bar must reach the line marked $44$ and be about the same width as the other bars.	✓
(b) North America	B1	The name of the continent, however it is spelled out.	✓
(c) Asia	B1	The name only. Working such as $4 \times 12 = 48$ earns nothing on its own.	✓
(d) $68$	B1	The answer $68$ . A misread of either bar loses the mark, as there is no method mark on this part.	✓

Full marks: 4/4

Part of a number line is shown below.

(a) Write down the number marked by the arrow. [1 mark]

(b) The diagram below shows the temperature dial on a pottery kiln.
On this dial, draw an arrow to show a temperature of $240^\circ \text{C}$ [1 mark]

(a)(c)

[Total 3 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

Part of a number line on which

57

58

and

59

are labelled, each labelled gap being cut into

5

equal parts by short marks.

A kiln dial labelled

50

100

150

200

250

300

, again with

5

equal parts in each labelled gap.

The number

0.786

Find

(a) the number the arrow on the number line points to (b) where on the dial an arrow showing

240^\circ \text{C}

must be drawn (c)

0.786

written correct to

2

decimal places

Plan the Solution

Both diagrams are scales, so treat them the same way: never read a scale until you know what one small division is worth.
Find the value of one small division by dividing a labelled gap by the number of equal parts it is cut into, then count divisions on from the nearest labelled mark.
For the rounding, look at the digit in the third decimal place of $0.786$ and decide which of the two neighbouring hundredths it sits nearer to.

Worked Solution [3 marks]

Rule - Reading a scale: one small division = the gap between two labelled marks divided by the number of equal parts between them. Rounding to

2

decimal places keeps

2

digits after the point, and the next digit decides whether the last one goes up.

Step 1: Work out one small division on the number line

\dfrac{58 - 57}{5} = \dfrac{1}{5} = 0.2

(Reason: (Reason: the gap from

57

58

is worth

1

and it is cut into

5

equal parts, so each short mark is worth

0.2

.))

Step 2: Count on from the

57

mark

57 + 3 \times 0.2 = 57.6

(Reason: (Reason: the arrow stands at the third short mark to the right of

57

, so add three small divisions to

57

.))

Step 3: Work out one small division on the kiln dial

\dfrac{250 - 200}{5} = \dfrac{50}{5} = 10

(Reason: (Reason: the dial is labelled every

50

degrees and each labelled gap carries

5

short marks, so one small division is

10

degrees.))

Step 4: Count on from the

200

mark to place the arrow

200 + 4 \times 10 = 240

(Reason: (Reason:

240

is four small divisions past

200

, so the arrow is drawn from the centre out to that mark.))

Step 5: Round

0.786

correct to

2

decimal places

0.786 \approx 0.79

(Reason: (Reason: keeping

2

decimal places means choosing between

0.78

and

0.79

. The next digit is

6

, which is

5

or more, so the hundredths digit goes up from

8

9

.))

(a)

57.6

(b) an arrow drawn on the dial pointing at

240

(c)

0.79

Verification

Check 1: Read the number line from the other end. The arrow is

2

small divisions to the left of

58

, and each is worth

0.2

58 - 2 \times 0.2 = 57.6

Check 2: Read the dial from the other end. Counting back from

250

, one small division is worth

10

degrees.

250 - 10 = 240

, so the arrow sits one short mark before the

250

label.

Check 3: Instead of a rounding rule, measure the distance from

0.786

to each of the two hundredths on either side of it.

0.786 - 0.78 = 0.006

but

0.79 - 0.786 = 0.004

, so

0.79

is the nearer of the two.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Reads the number line and gives $57.6$	B1	cao. $57.6$ only. $57.3$ scores nothing - it comes from treating a short mark as $0.1$ instead of $0.2$ .	✓
(b) Draws an arrow on the dial pointing at $240$	B1	Arrow pointing at $240$ , that is at the fourth short mark after $200$ . The mark is for the drawing, so there is no answer line for this part.	✓
(c) Gives $0.79$	B1	cao. $0.79$ only. $0.78$ scores nothing - it comes from cutting the number short rather than rounding it.	✓

Full marks: 3/3

The diagram below shows a quadrilateral.

(a) Write down the mathematical name of this type of quadrilateral. [1 mark]

The second diagram shows a solid 3-D shape.

(b) (i) Write down the mathematical name of this 3-D shape.
[1 mark]
(ii) Write down the number of edges this shape has. [1 mark]

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

[Total 3 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

A quadrilateral whose two shorter sides meet at one vertex and whose two longer sides meet at the opposite vertex

A solid with a triangular face at the front and an identical triangular face behind it

Find

the mathematical name of the quadrilateral the mathematical name of the 3-D shape the number of edges the 3-D shape has

Plan the Solution

Compare the four sides of the quadrilateral: are the equal sides next to each other, or opposite each other?
Count the lines of symmetry, because that is what tells a kite apart from a rhombus.
Look for the face that repeats all the way through the solid, then name the solid from the shape of that face.
Count the edges of the solid in three groups: the front face, the back face, and the edges that join the two faces.

Worked Solution [3 marks]

A kite has two pairs of equal adjacent sides and exactly one line of symmetry. A prism has the same cross-section from one end to the other, and a prism whose cross-section has

n

sides has

3n

edges.

Step 1: compare the sides and the symmetry of the quadrilateral

2 \text{ pairs of equal adjacent sides}

1 \text{ line of symmetry}

(Reason: The two shorter sides meet at the left-hand vertex and the two longer sides meet at the right-hand vertex, so the equal sides are next to each other and not opposite each other. Folding along the long diagonal maps each short side onto the other one, and that diagonal is the only fold that works. A rhombus would need all four sides equal and two lines of symmetry, and a parallelogram would need its equal sides parallel, so the quadrilateral is a kite.)

Step 2: look for the face that repeats through the solid

2 \text{ triangles} + 3 \text{ rectangles} = 5 \text{ faces}

(Reason: Cutting the solid anywhere between the two ends gives the same triangle every time, so the solid is a prism. The cross-section is a triangle, so it is a triangular prism.)

Step 3: count the edges in three groups

3 + 3 + 3 = 9

(Reason: There are three edges round the triangle at the front, three round the identical triangle at the back, and three more joining the two triangles, one at each corner. The dashed lines in the diagram are edges as well: they are the ones hidden behind the solid.)

(a) kite(b)(i) triangular prism(b)(ii)

9

edges

Verification

Check 1: A prism whose cross-section has

n

sides has

3n

edges. The cross-section here is a triangle, so

n = 3

3 \times 3 = 9

Check 2: Euler's formula for a solid says faces plus vertices equals edges plus

2

. This prism has

5

faces and

6

vertices.

5 + 6 = 9 + 2

Check 3: Test the other quadrilateral names against the drawing. A rhombus needs all four sides equal, a parallelogram needs both pairs of opposite sides parallel, and a trapezium needs one pair of parallel sides. This quadrilateral has

0

pairs of parallel sides and two different side lengths. No other name fits, so the quadrilateral is a kite.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Names the type of quadrilateral	B1	kite	✓
(b)(i) Names the 3-D shape	B1	prism, accept triangular prism	✓
(b)(ii) States how many edges the shape has	B1	$9$	✓

Full marks: 3/3

Amira has $5$ litres of orange squash in a large jug and a tray of empty glasses.
She fills as many glasses as possible with squash from the jug.
She puts $280$ millilitres of squash into each glass.

Work out how many glasses Amira completely fills. [3 marks]

[Total 3 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

The jug holds

5

litres of squash.

Every glass is filled with

280

millilitres of squash.

The glasses must be filled completely, so a part-filled glass does not count.

Find

The number of glasses that Amira can fill completely from the jug.

Plan the Solution

The two amounts are in different units, so change the litres into millilitres first.
Divide the total amount by the amount that goes into one glass.
That division does not come out exactly, so keep only the whole-number part: the leftover squash is not enough to fill another glass.

Worked Solution [3 marks]

Rule - Same units first:

1

litre =

1000

millilitres. Then the number of full glasses is

\dfrac{\text{total millilitres}}{\text{millilitres in one glass}}

, keeping only the whole-number part.

Step 1: Put both amounts into the same unit

5 \times 1000 = 5000 \text{ ml}

(Reason: The jug is measured in litres and each glass in millilitres, so one of them has to be changed before they can be compared. Millilitres is the easier choice here, because the glass is already measured in them.)

Step 2: Divide the total by the amount in one glass

\dfrac{5000}{280} = \dfrac{125}{7} = 17\dfrac{6}{7}

(Reason: Dividing counts how many lots of

280

millilitres fit inside

5000

millilitres. A calculator gives

17.857\ldots

, which is the same number written as a decimal.)

Step 3: Keep only the whole glasses

17 \times 280 = 4760

18 \times 280 = 5040

(Reason: Seventeen glasses take

4760

millilitres, which the jug can supply. Eighteen glasses would need

5040

millilitres and only

5000

millilitres are there, so the last

240

millilitres are left in the jug. The answer is rounded down, never up.)

17

glasses

Verification

Check 1: Multiply back:

17

glasses at

280

millilitres each.

17 \times 280 = 4760

millilitres, which is inside the

5000

millilitres available, with

240

millilitres to spare.

Check 2: Test one more glass, to be sure the answer is not

18

18 \times 280 = 5040

millilitres, which is

40

millilitres more than there is, so the eighteenth glass cannot be filled.

Check 3: Redo the whole question in litres instead, by converting the glass rather than the jug.

280

millilitres is

0.28

litres, and

\dfrac{5}{0.28} = 17\dfrac{6}{7}

as well, so the whole-number part is still

17

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$5 \times 1000 = 5000$ or $\dfrac{280}{1000} = 0.28$	M1	for a correct conversion between millilitres and litres	✓
$\dfrac{5000}{280}$ or $\dfrac{5}{0.28}$ or $\dfrac{125}{7}$ or $17\dfrac{6}{7}$	M1	for a complete method. Follow through an incorrect conversion, but an attempt to convert must have been made.	✓
$17$	A1	cao	✓
$17 \times 280 = 4760$ or $17 \times 0.28 = 4.76$	M2	Alternative complete method: building up to the largest multiple of $280$ that still fits inside $5000$ . This scores M2 in place of the two method marks above, and A1 still follows for $17$ .	✓
Working not required, so a correct answer scores full marks, unless it comes from obviously incorrect working.	Note	A guidance row. It carries no mark of its own and does not add to the total of three.	✓

Full marks: 3/3

(a) Work out the value of $3.4^2$
[1 mark]
(b) Work out the cube root of $373\,248$
[1 mark]
(c) Write $7 \times 7 \times 7 \times 7 \times 7 \div 7$ as a single power of $7$
[1 mark]
(d) Put one set of brackets into each calculation below to make the answer correct.
(i) $5 + 3 \times 2 = 16$
[1 mark]
(ii) $10 - 8 - 10 - 6 \div 2 = 0$ [1 mark]

(a)(b)(c)

[Total 5 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

(a) the square

3.4^2

(b) the cube root of

373\,248

7 \times 7 \times 7 \times 7 \times 7 \div 7

(d) two calculations, each of which needs one set of brackets:

5 + 3 \times 2

is to come to

16

, and

10 - 8 - 10 - 6 \div 2

is to come to

0

Find

The value of

3.4^2

, the cube root of

373\,248

, the single power of

7

, and where the one set of brackets goes in each calculation in part (d). Five separate answers, one mark each, and a calculator is allowed throughout.

Plan the Solution

(a) Squaring means multiplying the number by itself, so work out $3.4 \times 3.4$ .
(b) A cube root asks which number, multiplied by itself three times, gives $373\,248$ . Bracket it between $70$ and $80$ first, then test.
(c) Count the sevens. Five multiplied together make $7^5$ , and dividing by one more seven lowers the index by $1$ .
(d) Work each calculation out exactly as it is printed. Comparing that value with the target shows which operation has to be forced to happen first, and the brackets go round that operation.

Worked Solution [5 marks]

Brackets are worked out first, then powers and roots, then multiplication and division, and addition and subtraction last. Powers of the same base divide by subtracting the indices:

7^5 \div 7^1 = 7^{5-1}

Step 1 (part a): multiply

3.4

by itself

3.4^2 = 3.4 \times 3.4 = 11.56

(Reason: (Reason: the small

2

counts how many times the number appears in the multiplication, so

3.4^2

3.4

written down twice and multiplied.))

Step 2 (part b): find the number whose cube is

373\,248

70^3 = 343\,000

72^3 = 72 \times 72 \times 72

72 \times 72 = 5\,184

5\,184 \times 72 = 373\,248

(Reason: (Reason:

70^3

is a little under the target and

80^3

is far over it, so the root is in the low seventies. On a calculator the cube root key gives

72

straight away, but the cube is worth writing out as proof.))

Step 3 (part c): count the sevens

7 \times 7 \times 7 \times 7 \times 7 = 7^5

7^5 \div 7 = 7^{5-1}

7^{5-1} = 7^4

(Reason: (Reason: five sevens are multiplied together, and dividing cancels one of them, leaving four. In index form that is

5

take away

1

.))

Step 4 (part d(i)): force the addition to happen first

5 + 3 \times 2 = 11

(5 + 3) \times 2 = 8 \times 2 = 16

(Reason: (Reason: as printed, the multiplication is done before the addition and the calculation comes to

11

. The target

16

8

doubled, and

8

5 + 3

, so the brackets go round the addition.))

Step 5 (part d(ii)): halve the

10 - 6

instead

10 - 8 - 10 - 6 \div 2 = 10 - 8 - 10 - 3 = -11

10 - 8 - (10 - 6) \div 2 = 10 - 8 - 4 \div 2

10 - 8 - 4 \div 2 = 10 - 8 - 2 = 0

(Reason: (Reason: as printed, only the

6

is divided by

2

and the calculation comes to

-11

. Bracketing

10 - 6

makes

4

the thing that is halved, so

2

is taken from

10 - 8

and nothing is left.))

(a)

11.56

(b)

72

(c)

7^4

(d)(i)

(5 + 3) \times 2 = 16

(d)(ii)

10 - 8 - (10 - 6) \div 2 = 0

Verification

Check 1 (part a): Take the decimal point out and square

34

instead:

34 \times 34 = 1156

. Each factor carries one decimal place, so the answer carries two. Putting the point back two places gives

11.56

, which is the value found in step 1.

Check 2 (part b): Split the target into prime factors:

2^9 \times 3^6 = 512 \times 729 = 373\,248

. A cube root takes a third of each index.

2^3 \times 3^2 = 8 \times 9 = 72

, agreeing with step 2, and every index divides by

3

exactly, so the cube root is a whole number.

Check 3 (part c): Work the whole calculation out as ordinary numbers:

7 \times 7 \times 7 \times 7 \times 7 = 16\,807

, and then

16\,807 \div 7 = 2\,401

7^4 = 2\,401

, the same number, so the single power is right.

Check 4 (part d(i)): Try the only other place the brackets could go:

5 + (3 \times 2) = 11

, which is what the calculation already came to. Bracketing the multiplication changes nothing, so

(5 + 3) \times 2

is the only placement that reaches

16

Check 5 (part d(ii)): Try other placements:

(10 - 8) - 10 - 6 \div 2 = -11

10 - (8 - 10) - 6 \div 2 = 9

and

10 - 8 - (10 - 6 \div 2) = -5

. None of them is

0

, so the brackets must go round

10 - 6

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$11.56$ . Accept an exact equivalent, for example $\dfrac{289}{25}$ or $11\dfrac{14}{25}$ .	✓
(b)	B1	$72$ . Working is not required, so the correct answer alone scores the mark.	✓
(c)	B1	$7^4$ . The answer must be a single power of $7$ , so $2\,401$ on its own does not score.	✓
(d)(i)	B1	$(5 + 3) \times 2 = 16$	✓
(d)(ii)	B1	$10 - 8 - (10 - 6) \div 2 = 0$	✓

Full marks: 5/5

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

$-3 \quad 5 \quad 13 \quad 21 \quad 29$

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

(b) Explain why $326$ cannot be a term of this sequence. [1 mark]

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

[Total 3 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

A number sequence beginning

-3

5

13

21

29

The same amount is added to get from each term to the next, so this is an arithmetic sequence.

Find

(a)(i) the term that comes after

29

(a)(ii) an explanation of how that term was found (b) a reason why

326

can never appear in the sequence

Plan the Solution

Subtract each term from the one after it, to find the common difference.
Add that difference to the last term given, to get the next term.
Turn the pattern into an $n$ th term rule, so that part (b) can be settled by calculation rather than by guessing.
Test $326$ against the rule, then back the answer up with an odd and even argument.

Worked Solution [3 marks]

Rule - Arithmetic sequence: the terms go up by a fixed common difference

d

, so the

n

th term is

a + (n - 1)d

, where

a

is the first term.

Step 1: Find the common difference

5 - (-3) = 8

13 - 5 = 8

21 - 13 = 8

29 - 21 = 8

(Reason: All four gaps are the same, so the sequence is arithmetic with common difference

d = 8

. Subtracting a negative first term is what makes the first gap

8

and not

2

Step 2: Add the common difference to the last term given

29 + 8 = 37

(Reason: Part (a)(i) asks only for the next term, so one addition is enough: the sixth term is

37

Step 3: Say what the rule is, in words

\text{next term} = \text{previous term} + 8

(Reason: Part (a)(ii) asks how the answer was found. Every term is

8

more than the one before it, so the answer is simply that

8

was added.)

Step 4: Build the

n

th term rule

a = -3 \quad \text{and} \quad d = 8

a + (n - 1)d = -3 + (n - 1) \times 8 = 8n - 11

\text{check } n = 6: \quad 8 \times 6 - 11 = 37

(Reason: The rule is checked against a term that is already known before it is trusted. It gives

37

n = 6

, agreeing with Step 2, so it is safe to use in part (b).)

Step 5: Test whether

326

fits the rule

8n - 11 = 326

8n = 326 + 11 = 337

n = \dfrac{337}{8} = 42.125

(Reason: A term number counts positions, so

n

has to be a whole number. This value is not a whole number, so no position in the sequence produces

326

Step 6: Give the odd and even reason, and the two terms either side

8 \times 42 - 11 = 325

8 \times 43 - 11 = 333

(Reason:

8n

is always even, and taking away the odd number

11

leaves an odd number, so every term of the sequence is odd. The sequence steps from

325

straight to

333

, and the even number

326

is stepped over.)

(a)(i)

37

(a)(ii) Added

8

to the previous term(b) Every term is odd, but

326

is even, so it cannot be a term

Verification

Check 1: Write the sequence out one term further by adding

8

-3, \; 5, \; 13, \; 21, \; 29, \; 37

. The sixth term is

37

, which is what the rule gives at

n = 6

, so the two methods agree.

Check 2: Count in the other direction from the rule: an even answer would need

8n - 11

to be even, so

8n

would have to be odd.

8n

is a multiple of

8

and can never be odd, so no term of the sequence is even, and

326

is even.

Check 3: Work out the terms on either side of

326

: the

42

nd term is

325

and the

43

rd term is

333

326

lies between two consecutive terms without being either of them, so it is skipped.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) $37$	B1	$37$ written down as the next term. No working is needed for this mark.	✓
(a)(ii) Added $8$	B1	Any clear statement of the rule, e.g. 'added $8$ ', 'add $8$ ', ' $+8$ ', 'it goes up by $8$ ', 'the rule is $8n - 11$ ', or the calculation $8 \times 6 - 11 = 37$ .	✓
(b) Correct explanation	B1	Any correct explanation, e.g. the sequence is odd; $326$ is even; the $n$ th term $8n - 11$ is always odd; the sequence goes $325$ , $333$ ; the $42$ nd term is $325$ ; $\dfrac{337}{8} = 42.125$ is not a whole number; or $326$ is not $11$ less than a multiple of $8$ . Not accepted on its own: 'adding $8$ each time will not lead to $326$ ', 'it goes past $326$ ', or any argument that divides $326$ itself by $8$ instead of $337$ .	✓

Full marks: 3/3

Anish buys some screws and some washers.

Each box of screws costs £ $2.60$
Each pack of washers costs £ $3.94$

Anish buys $5$ boxes of screws and $4$ packs of washers.
Anish pays with a £ $50$ note.

Work out how much change Anish should get. [3 marks]

[Total 3 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

One box of screws costs £

2.60

, and one pack of washers costs £

3.94

5

boxes of screws and

4

packs of washers are bought

The whole purchase is paid for with a single £

50

note

Find

The change that should be given from the £

50

note

Plan the Solution

Cost the screws on their own: the price of one box multiplied by the number of boxes.
Cost the washers on their own, in the same way.
Add the two costs to get the total spent.
Take that total away from £ $50$ , because the change is whatever is left of the note.

Worked Solution [3 marks]

Rule - Change:

\text{change} = \text{amount paid} - \text{total cost}

, and each item's cost is its price multiplied by the number bought, so the two costs must both be found before anything is subtracted.

Step 1: Work out the cost of the screws

5 \times 2.60 = 13.00

(Reason:

5

boxes at £

2.60

each. Multiplying the price by the number of boxes is quicker and safer than adding £

2.60

five times over.)

Step 2: Work out the cost of the washers

4 \times 3.94 = 15.76

(Reason: The washers come in packs, so it is the number of packs that multiplies £

3.94

, and each item must be costed with its own quantity, never with the other item's.)

Step 3: Add the two costs to get the total spent

13.00 + 15.76 = 28.76

(Reason: The screws and the washers are paid for together, so the shop takes £

28.76

in one transaction.)

Step 4: Subtract the total from the amount handed over

50.00 - 28.76 = 21.24

(Reason: The £

50

note covers the whole cost, so what is left of it is the change. Writing £

50

50.00

lines the pence up under the pence before subtracting.)

21.24

Verification

Check 1: Redo the whole question in pence, where every value is a whole number and no decimal point can slip:

5 \times 260 = 1300

and

4 \times 394 = 1576

, giving

2876

pence spent.

5000 - 2876 = 2124

pence, which is £

21.24

, so the pounds working and the pence working agree exactly.

Check 2: Subtract one cost at a time instead of adding them first. Take the screws off the note:

50.00 - 13.00 = 37.00

, then take the washers off what is left.

37.00 - 15.76 = 21.24

, the same change, reached without the two costs ever being added together.

Check 3: Put the note back together: add the change to the total spent.

21.24 + 28.76 = 50.00

, exactly the amount handed over, so nothing has been counted twice and nothing has been dropped.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A method to find the cost of the screws, or of the washers, or of both	M1	Any one of: $5 \times 2.60$ or $13$ ; $50 - 5 \times 2.60$ or $37$ ; $4 \times 3.94$ or $15.76$ ; $50 - 4 \times 3.94$ or $34.24$ ; or $5 \times 2.60 + 4 \times 3.94$ ( $= 28.76$ ).	✓
A complete method for the change	M1	For example $50 - (5 \times 2.60) - (4 \times 3.94)$ , or $50 - 13 - 15.76$ , or $50 - 28.76$ , carried out on the candidate's own earlier values.	✓
The change, $21.24$	A1	£ $21.24$ on the answer line. Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

(a) Write $c \times c \times c$ in its simplest form. [1 mark]

(b) Simplify $12d \times 3e$ [1 mark]

(d) Solve the equation $2g - 3 = 6$ [2 marks]

(e) Expand the brackets $x(x - 4)$ [1 mark]

$P = 4y^2 + w$

(f) Work out the value of $P$ when $y = -3$ and $w = 2$ [2 marks]

(a)(b)(c) k =(d) g =(e)(f) P =

[Total 8 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

(a) the product

c \times c \times c

and (b) the product

12d \times 3e

\dfrac{k}{4} = 7

and (d) the equation

2g - 3 = 6

(e) the bracketed expression

x(x - 4)

(f) the formula

P = 4y^2 + w

, with

y = -3

and

w = 2

Find

(a) a single power of

c

, and (b) one term in

d

and

e

k

, and (d) the value of

g

(e)

x(x - 4)

written without brackets (f) the value of

P

(note that

y

is negative, so the square is positive)

Plan the Solution

(a) and (b) are simplifying: multiplying, so count how many of each letter there are and multiply the numbers separately.
(c) and (d) are solving: undo whatever has been done to the letter, doing the same to both sides.
(e) is expanding: multiply every term inside the bracket by the term outside.
(f) is substituting: replace $y$ and $w$ by their values, then work out the power before the multiplication.

Worked Solution [8 marks]

Rule - four separate skills, one habit each. Multiplying powers of the same letter adds the indices,

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

. Solving means applying the inverse operation to both sides. Expanding a bracket multiplies every term inside by the term outside. Substituting means replacing each letter by its value first, then following the order of operations - powers before multiplication.

Step 1 - part (a): add the indices

c \times c \times c = c^{1} \times c^{1} \times c^{1}

= c^{1+1+1} = c^3

(Reason: (Reason: a letter on its own is that letter to the power

1

, and multiplying powers of the same letter adds the indices, so three

c

s multiplied give

c^3

.))

Step 2 - part (b): numbers first, then letters

12d \times 3e = 12 \times 3 \times d \times e

= 36de

(Reason: (Reason: multiplication can be done in any order, so gather the numbers,

12 \times 3 = 36

, and the letters,

d \times e = de

. The letters are different, so nothing else combines.))

Step 3 - part (c): undo the division

\dfrac{k}{4} = 7

k = 7 \times 4 = 28

(Reason: (Reason:

k

has been divided by

4

, so the inverse is to multiply both sides by

4

.))

Step 4 - part (d): clear the number term, then divide

2g - 3 = 6

2g = 6 + 3 = 9

g = \dfrac{9}{2} = 4.5

(Reason: (Reason: add

3

to both sides to leave

2g

on its own, then divide both sides by

2

. An answer of

\dfrac{9}{2}

is equally acceptable.))

Step 5 - part (e): multiply both terms in the bracket

x(x - 4) = x \times x - x \times 4

= x^2 - 4x

(Reason: (Reason: the

x

outside multiplies the

x

and the

-4

inside. Missing the second term is the usual slip here.))

Step 6 - part (f): substitute, then square before multiplying

P = 4y^2 + w

P = 4 \times (-3)^2 + 2

P = 4 \times 9 + 2 = 38

(Reason: (Reason: replace

y

-3

and

w

2

, then follow the order of operations - the power is worked out before the multiplication by

4

, and only the

y

is squared, not the

4

.))

(a)

c^3

(b)

36de

(c)

k = 28

(d)

g = 4.5

(e)

x^2 - 4x

(f)

P = 38

Verification

Check 1 - put numbers in place of the letters: A simplified expression must give the same value as the original for every choice of letter. Try

c = 2

in (a),

d = 5

and

e = 2

in (b), and

x = 3

in (e).

2 \times 2 \times 2 = 8

and

2^3 = 8

;

60 \times 6 = 360

and

36 \times 5 \times 2 = 360

;

3(3 - 4) = -3

and

3^2 - 4 \times 3 = -3

Check 2 - put the solutions back into the equations: A solution is correct when it makes the original equation true, so substitute

k = 28

into

\dfrac{k}{4} = 7

and

g = 4.5

into

2g - 3 = 6

\dfrac{28}{4} = 7

and

2 \times 4.5 - 3 = 9 - 3 = 6

, so both equations balance.

Check 3 - part (f) a second way: Fold the

4

into the square, since

4y^2 = (2y)^2

. This avoids the

4 \times y^2

step altogether, and a negative multiplied by a negative is positive.

(2 \times -3)^2 + 2 = (-6)^2 + 2 = 36 + 2 = 38

, matching the value found by squaring first.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $c^3$	B1	Correct answer only.	✓
(b) $36de$	B1	Or equivalent, so $36ed$ is equally acceptable.	✓
(c) $k = 28$	B1	Correct answer only.	✓
(d) $2g = 6 + 3$ or $2g = 9$ or $g - \dfrac{3}{2} = \dfrac{6}{2}$ or $\dfrac{6 + 3}{2}$ or $\dfrac{9}{2}$	M1	For one correct first step - either clearing the $-3$ or dividing every term by $2$ . Or equivalent.	✓
(d) $4.5$	A1	Or equivalent, eg $\dfrac{9}{2}$ or $4\dfrac{1}{2}$ . Working is not required, so a correct answer scores full marks unless it follows obviously incorrect working.	✓
(e) $x^2 - 4x$	B1	Or $-4x + x^2$ .	✓
(f) $4 \times (-3)^2 + 2$ or $4(-3)^2 + 2$ or $4 \times 9 + 2$ or $4(9) + 2$ or $4 \times -3 \times -3 + 2$	M1	For substituting values for $y$ and $w$ .	✓
(f) $38$	A1	Working is not required, so a correct answer scores full marks unless it follows obviously incorrect working. A common wrong answer here is a negative one, from squaring the minus sign away.	✓

Full marks: 8/8

Using only a ruler and a pair of compasses, construct a square of side $7$ cm.
Two of the four sides, $PQ$ and $QR$ , have already been drawn for you.
You must leave in all your construction lines. [2 marks]

[Total 2 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

Two sides of the square,

PQ

and

QR

, already drawn and meeting at a right angle at

Q

PQ = QR = 7

cm, and every side of a square is the same length.

A ruler and a pair of compasses, and nothing else.

Find

The fourth vertex,

S

, and the two sides that complete the square. Every construction arc, left on the page - the arcs are what the marks are for.

Plan the Solution

The right angle at $Q$ is already drawn, so nothing has to be constructed for it.
The missing vertex is $7$ cm from $P$ and $7$ cm from $R$ , so it is where two arcs of radius $7$ cm cross.
Set the compasses once, across a side that is already drawn, and do not change them again.
Join the crossing point to the two free ends with a ruler.

Worked Solution [2 marks]

Rule - Equal arcs: the point that is

r

from

P

and

r

from

R

is where an arc of radius

r

about

P

crosses an arc of radius

r

about

R

Step 1: Open the compasses to one whole side

r = PQ = 7 \text{ cm}

(Reason: Every side of a square is the same length, so a side that is already drawn sets the radius exactly. Opening the compasses across

PQ

is more accurate than measuring with a ruler, and the compasses are not touched again.)

Step 2: Draw an arc from

P

PS = 7 \text{ cm}

(Reason: The missing vertex is

7

cm from

P

, so it lies somewhere on this arc. Sweep enough of the arc to be sure it will reach the second one.)

Step 3: Draw a second arc from

R

, with the compasses unchanged

RS = 7 \text{ cm}

(Reason: The missing vertex is also

7

cm from

R

. The two arcs cross at that vertex, and the crossing is the construction - the point is not measured off a ruler.)

Step 4: Rule

PS

and

SR

PS = SR = 7 \text{ cm}

\angle SPQ = \angle SRQ = 90^\circ

(Reason: Four sides of

7

cm, together with the right angle already given at

Q

, make the shape a square. Leave both arcs on the page: they are the construction lines the question asks for.)

Step 5: Why the two arcs are certain to cross

7^2 + 7^2 = 98

PR = \sqrt{98} \approx 9.9 \text{ cm}

7 + 7 = 14 \text{ cm}

(Reason: The two arc centres

P

and

R

are about

9.9

cm apart, and two radii of

7

cm reach

14

cm between them, which is further. So the arcs must overlap, and they cross at two points: the corner

Q

itself, and the vertex on the far side of

PR

. Compasses opened to only half a side would reach

7

cm in total and would never meet.)

The square

PQRS

, of side

7

cm, with both arcs of radius

7

cm left on the page

Verification

Check 1: Measure the two new sides.

PS

and

SR

should each come to

7

cm, the same as the two sides that were given.

PS = SR = 7

cm, so all four sides match

Check 2: A square has equal diagonals, each

\sqrt{7^2 + 7^2} = \sqrt{98}

cm. Measure

PR

and

QS

and compare them.

PR = QS \approx 9.9

cm, equal to one decimal place

Check 3: Check the reach of the compasses independently of the drawing. The centres are

9.9

cm apart and the two radii add to

7 + 7 = 14

cm.

14 > 9.9

, so a crossing point had to exist

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Arcs of radius $7$ cm drawn from $P$ and from $R$ , crossing at $S$ , with $PS$ and $SR$ ruled in	B2	For a fully correct square with arcs shown.	✓
Part of the construction correct, but not all of it	B1	For a correctly sized square with no arcs shown, or for an incorrect quadrilateral with arcs of equal radius shown, or for correct arcs not joined - all within or on the guidelines of the overlay.	✓

Full marks: 2/2

A tea room sells $4$ different types of tart.
cherry (C), fig (F), lemon (L), maple (M)
Amara is going to choose $2$ different types of tart.
(a) Write down all the possible combinations she can choose. [2 marks]

The two-way table gives some information about the flavours of iced tea sold by the tea room on Saturday and on Sunday.

(b) Complete the two-way table. [2 marks]

Amara asks $100$ visitors whether they prefer tarts or iced tea.
$33$ of the visitors prefer tarts.
One of the $100$ visitors is chosen at random.
(c) Find the probability that this visitor does not prefer tarts. [1 mark]

(a)(a)(a)(c)

[Total 5 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

A tea room sells

4

types of tart, coded cherry (C), fig (F), lemon (L) and maple (M).

Amara chooses

2

different types.

A two-way table of iced teas sold on Saturday and on Sunday, with four entries missing.

100

visitors are asked, and

33

of them prefer tarts.

Find

(a) every possible pair of tart types Amara can choose (b) the four missing entries of the two-way table (c) the probability that a visitor chosen at random does not prefer tarts

Plan the Solution

(a) List in a fixed order: take each type in turn and pair it only with the types that come after it. An ordered list is what makes a missed pair or a repeat impossible to hide.
(b) Every row adds to its row total and every column adds to its column total. Start at the line with only one gap in it, then use each new entry to open up the next line.
(c) Preferring tarts and not preferring tarts are complementary, so the two counts add to $100$ .

Worked Solution [5 marks]

Rule - Two-way table: every row adds to its row total, every column adds to its column total, and the grand total is the sum of the row totals and also the sum of the column totals. Complementary events: the probability that something does not happen is

1

minus the probability that it does.

Part (a): pair the first type with every later type

\text{CF, CL, CM}

(Reason: Fix cherry first and run through the types that come after it: fig, lemon, then maple. Working in a fixed order is what keeps the list complete.)

Part (a): the pairs that do not use cherry

\text{FL, FM}

\text{LM}

3 + 2 + 1 = 6

(Reason: Fig pairs with lemon and with maple, and lemon pairs with maple. That is

3 + 2 + 1 = 6

pairs in all, so nothing has been missed and nothing has been counted twice.)

Part (b): the Sunday row has only one gap

49 - 20 - 13 = 16

(Reason: The Sunday row adds to

49

, and two of its three flavours are already given, so the peach entry is whatever is left over.)

Part (b): the peach column now has only one gap

48 - 16 = 32

(Reason: The peach column adds to

48

, and the Sunday peach entry has just been found, so the Saturday peach entry is the difference.)

Part (b): the two totals that are left

32 + 11 + 28 = 71

71 + 49 = 120

(Reason: The Saturday row can now be added across, and the two row totals give the overall total. The column totals give the same figure,

48 + 31 + 41 = 120

Part (c): take the complement

100 - 33 = 67

\dfrac{100 - 33}{100} = \dfrac{67}{100}

(Reason: Every visitor either prefers tarts or does not, so the

67

who do not are out of the

100

who were asked.)

(a)

\text{CF, CL, CM, FL, FM, LM}

(b) Saturday peach

32

, Saturday total

71

, Sunday peach

16

, overall total

120

(c)

\dfrac{67}{100}

Verification

Check 1: Add each row across the completed table:

32 + 11 + 28 = 71

and

16 + 20 + 13 = 49

. Both row totals are the ones the table already carried.

Check 2: Add each column down:

32 + 16 = 48

11 + 20 = 31

and

28 + 13 = 41

. All three column totals agree with the printed ones.

Check 3: Reach the overall total the other way round:

48 + 31 + 41 = 120

, against

71 + 49 = 120

from the rows. The rows and the columns give the same overall total.

Check 4: Count the combinations with the pair count for

2

chosen from

4

\dfrac{4 \times 3}{2} = 6

. Six pairs were listed, and no pair appears twice.

Check 5: Add the two probabilities:

\dfrac{33}{100} + \dfrac{67}{100} = 1

. The two outcomes are complementary, as they must be.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) All six combinations listed: $\text{CF, CL, CM, FL, FM, LM}$	B2	for all $6$ combinations, with no extras and no repeats	✓
(a) A partial list	(B1)	for at least $3$ correct combinations, ignoring repeats	✓
(b) Correct two-way table: $32$ and $71$ on the Saturday row, $16$ on the Sunday row, $120$ as the overall total	B2	for all $4$ correct values	✓
(b) A partly completed table	(B1)	for $2$ or $3$ correct values	✓
(c) $\dfrac{67}{100}$	B1	oe, so $0.67$ and $67\%$ are equally acceptable	✓

Full marks: 5/5

A regular polygon has an exterior angle of $15^\circ$ at each of its vertices.
Work out how many sides this polygon has. [2 marks]

[Total 2 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

The polygon is regular, so all of its sides and all of its angles are equal.

Each exterior angle is

15^\circ

Find

The number of sides,

n

Plan the Solution

Use the fact that the exterior angles of any polygon add up to one full turn.
A regular polygon has $n$ equal exterior angles, so each one is $\dfrac{360}{n}$ degrees.
Set that equal to $15$ and solve for $n$ .

Worked Solution [2 marks]

Rule - Exterior angles: the exterior angles of any polygon add up to

360^\circ

, so for a regular polygon with

n

sides each exterior angle is

\dfrac{360^\circ}{n}

Step 1: Write down the exterior angle fact

\text{sum of the exterior angles} = 360^\circ

(Reason: (Reason: walking once around the outside of any polygon turns you through one complete turn, which is

360^\circ

. This is true whatever the number of sides.))

Step 2: Turn that into an equation

15n = 360

(Reason: (Reason: the polygon is regular, so all

n

of its exterior angles are the same size,

15^\circ

, and together they make

360^\circ

.))

Step 3: Solve for the number of sides

n = \dfrac{360}{15} = 24

(Reason: (Reason: divide both sides of

15n = 360

15

. A calculator is allowed here, but the division is worth doing in your head as a check.))

24

sides

Verification

Check 1: Multiply back. If there are

24

exterior angles of

15^\circ

, they must add to exactly one full turn.

24 \times 15 = 360

Check 2: Go the other way, through the interior angle. Each interior angle is

180 - 15 = 165

degrees, and the interior angles of an

n

-sided polygon add to

(n - 2) \times 180

degrees, which is

22 \times 180 = 3960

degrees here.

\dfrac{3960}{165} = 24

Check 3: Compare with a polygon whose sides you already know. A regular hexagon has

6

sides and an exterior angle of

60^\circ

. Our exterior angle is a quarter of that, and a smaller exterior angle means more sides, so expect four times as many.

4 \times 6 = 24

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Divide one full turn by the size of one exterior angle, or set the interior angle equal to $180 - 15$ .	M1	for $\dfrac{360}{15}$ oe, or for $\dfrac{(n - 2) \times 180}{n} = 180 - 15$	✓
State the number of sides.	A1	$24$	✓
Note	-	Working not required, so a correct answer scores full marks (unless it comes from obviously incorrect working).	✓

Full marks: 2/2

On the grid below, draw the graph of $y = 5 - 2x$ for all values of $x$ from $-2$ to $3$ . [3 marks]

[Total 3 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

The equation

y = 5 - 2x

The values of

x

to use: every whole number from

-2

3

A grid running from

-2

11

up the

y

axis

Find

The straight line

y = 5 - 2x

, drawn on the grid from

x = -2

x = 3

Plan the Solution

Work out $y$ for each whole number value of $x$ from $-2$ to $3$ .
Set the six pairs out as a table, so nothing is lost between the arithmetic and the plotting.
Plot the six points, check they lie in a straight line, then rule ONE line across the whole range.

Worked Solution [3 marks]

Rule - Table of values: substitute each value of

x

into

y = 5 - 2x

, plot each pair

(x, y)

, then join them with one ruled straight line.

Step 1: substitute each value of

x

5 - 2 \times (-2) = 5 + 4 = 9

5 - 2 \times (-1) = 5 + 2 = 7

5 - 2 \times 0 = 5 - 0 = 5

5 - 2 \times 1 = 5 - 2 = 3

5 - 2 \times 2 = 5 - 4 = 1

5 - 2 \times 3 = 5 - 6 = -1

(Reason: (Reason: multiply first, then subtract. Take care with a negative value of

x

: subtracting

2

lots of a negative number ADDS, so the

y

value goes up on the left of the axis.))

Step 2: set the results out as a table

\begin{array}{c|cccccc} x & -2 & -1 & 0 & 1 & 2 & 3 \\ y & 9 & 7 & 5 & 3 & 1 & -1 \end{array}

(Reason: (Reason: read the bottom row across - it falls

9, 7, 5, 3, 1, -1

, down by

2

every time. A constant step is the sign of a straight line, so a value that breaks the pattern is an arithmetic slip, caught before anything is plotted.))

Step 3: plot the points and rule the line

(-2, 9), (-1, 7), (0, 5), (1, 3), (2, 1), (3, -1)

-2 \leq x \leq 3

(Reason: (Reason: each column of the table is one point,

x

across then

y

up. Join them with a single ruled line that reaches both ends of the range - a line stopping short of

x = -2

x = 3

does not answer the question.))

The ruled straight line through

(-2, 9)

(0, 5)

and

(3, -1)

, drawn right across the grid from

x = -2

x = 3

Verification

Check 1 - the gradient: The equation is in the form

y = mx + c

, so the line must fall

2

for every

1

across. From

(-2, 9)

(3, -1)

the drawn line falls

10

over a run of

5

\dfrac{-10}{5} = -2

Check 2 - the intercept: Where the drawn line crosses the

y

axis,

x

0

, so the reading there should be the constant term of the equation.

5 - 2 \times 0 = 5

Check 3 - a point read back off the line: Pick a point in the middle of the range that was not used to set the ruler: at

x = 2

the drawn line should pass through

(2, 1)

5 - 2 \times 2 = 1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
State at least two correct points, or draw a line of negative gradient through $(0, 5)$ , or draw a line of gradient $-2$	B1	The first of the three marks. Any two correct pairs are enough, and they may be written in a table rather than plotted - for example $(0, 5)$ and $(1, 3)$ .	✓
Draw a correct straight line segment through at least three of the six points, or plot all six points without joining them	B2	The six points are $(-2, 9)$ , $(-1, 7)$ , $(0, 5)$ , $(1, 3)$ , $(2, 1)$ , $(3, -1)$ . A line of gradient $-2$ through $(0, 5)$ that does not run the full range also earns this.	✓
Draw the correct line right across, from $x = -2$ to $x = 3$	B3	Full marks. The line must be straight, ruled, and reach both ends of the stated range; a correct segment that stops short scores B2 only.	✓

Full marks: 3/3

A charity book sale raises $1200$ euros. Shalini, Marta and Lucas share the $1200$ euros.
Shalini gets $\dfrac{1}{5}$ of the $1200$ euros.
Marta gets $42\%$ of the $1200$ euros.
Lucas gets the rest of the $1200$ euros.

Given that
the amount of money Shalini gets : the amount of money Lucas gets $= 1 : n$
work out the value of $n$ [4 marks]

n =

[Total 4 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

A total of

1200

euros is shared between three people.

Shalini gets

\dfrac{1}{5}

of the

1200

euros.

Marta gets

42\%

of the

1200

euros.

Lucas gets whatever is left of the

1200

euros.

Shalini's money : Lucas's money

= 1 : n

, so the ratio has already been scaled to start at

1

Find

The value of

n

. Because the first number of the ratio is

1

n

is how many times Lucas's share is bigger than Shalini's, so expect a decimal rather than a whole number.

Plan the Solution

Work out Shalini's share, which is a fraction of the $1200$ euros.
Work out Marta's share, which is a percentage of the same $1200$ euros.
Take BOTH of those shares away from $1200$ to find Lucas's share. "The rest" means what is left after the other two have been paid, not after one of them.
Write Shalini's share : Lucas's share, then divide both parts by Shalini's share so the ratio starts at $1$ .

Worked Solution [4 marks]

Rule - to write a ratio

a : b

in the form

1 : n

, divide both parts by the first part, which gives

n = \dfrac{b}{a}

Step 1: Shalini's share

\dfrac{1}{5} \times 1200 = 240

(Reason: (Reason: a fifth of a quantity is that quantity split into

5

equal parts, so

1200

splits into parts of

240

euros.))

Step 2: Marta's share

\dfrac{42}{100} \times 1200 = 0.42 \times 1200

0.42 \times 1200 = 504

(Reason: (Reason:

42\%

means

42

hundredths, and

\dfrac{42}{100} = 0.42

, so the percentage becomes a multiplier. Both shares are taken from the SAME

1200

euros.))

Step 3: Lucas's share, which is the rest

240 + 504 = 744

1200 - 744 = 456

(Reason: (Reason: Shalini's and Marta's money has already gone out of the

1200

euros, so what is left over belongs to Lucas.))

Step 4: write the ratio in the form 1 : n

240 : 456

n = \dfrac{456}{240} = 1.9

(Reason: (Reason: dividing both parts of

240 : 456

240

leaves

1

on the left, which is the form the question asks for, so the right-hand part is

n

.))

n = 1.9

Verification

Check 1: Add the three shares back together. Nothing may be left over and nothing may be created, so they must come to exactly

1200

euros.

240 + 504 + 456 = 1200

Check 2: Redo it without any euros at all, in fractions of the whole. Shalini has

\dfrac{1}{5} = \dfrac{20}{100}

and Marta has

\dfrac{42}{100}

, so together they have

\dfrac{62}{100}

and Lucas is left with

\dfrac{38}{100}

. The ratio is then

38

hundredths against

20

hundredths.

n = \dfrac{38}{20} = 1.9

Check 3: Run the ratio forwards instead of backwards. If

1 : 1.9

is right, then multiplying Shalini's

240

euros by

1.9

must land on Lucas's share.

1.9 \times 240 = 456

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Find $\dfrac{1}{5}$ of $1200$	M1	For $\dfrac{1}{5} \times 1200 = 240$ or $\dfrac{1200}{5} = 240$ oe. Also awarded for $20\%$ or $0.2$ or $0.42$ or $\dfrac{42}{100}$ seen.	✓
Find $42\%$ of $1200$	M1	For $0.42 \times 1200 = 504$ or $\dfrac{42}{100} \times 1200 = 504$ oe, or for the combined share $\dfrac{1}{5} + \dfrac{42}{100} = \dfrac{62}{100}$ oe, or $0.2 + 0.42 = 0.62$ , or $20 + 42 = 62$ . Award both method marks (M2) for $744$ seen.	✓
Subtract both shares from $1200$	M1	For $1200 - (240 + 504) = 456$ oe, or $1200 - 744 = 456$ . Follow through on the candidate's own two shares. Also awarded in fraction form for $1 - \dfrac{62}{100} = \dfrac{38}{100} = \dfrac{19}{50}$ oe, or $1 - 0.62 = 0.38$ , or $100 - 62 = 38$ .	✓
State the value of $n$	A1	$1.9$ oe, so $\dfrac{19}{10}$ and $1\dfrac{9}{10}$ are equally acceptable. Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 4/4

The table gives information about the number of plums in each of $30$ punnets.

(a) Write down the mode of the number of plums in a punnet. [1 mark]

(b) Work out the mean number of plums in a punnet. [3 marks]

(a)(b)

[Total 4 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

A frequency table of the number of plums in a punnet, taking the values

11

12

13

14

15

and

16

The matching frequencies are

7

8

7

5

1

and

2

There are

30

punnets altogether, so the frequencies must add to

30

Find

(a) The mode of the number of plums in a punnet. (b) The mean number of plums in a punnet.

Plan the Solution

For the mode, read down the frequency column and find the largest frequency. The mode is the value beside it, not the frequency itself.
For the mean, multiply each number of plums by its frequency, add those products to get the total number of plums, then divide by the total frequency.
Check the frequencies add to $30$ before dividing, because that total is the divisor.

Worked Solution [4 marks]

Rule - Averages from a frequency table: the mode is the value with the largest frequency, and

\text{mean} = \dfrac{\Sigma fx}{\Sigma f}

, where

x

is a value and

f

is its frequency.

Part (a): find the largest frequency

\text{largest frequency} = 8

\text{mode} = 12

(Reason: The frequency column counts how often each value occurs, so the value that occurs most often sits beside the largest frequency. That frequency is

8

, on the

12

row, so the mode is

12

plums. The mode is the value

12

, never the count

8

Part (b): multiply each number of plums by its frequency

11 \times 7 = 77

12 \times 8 = 96

13 \times 7 = 91

14 \times 5 = 70

15 \times 1 = 15

16 \times 2 = 32

(Reason: Each product is the total number of plums held by the punnets on that row:

7

punnets with

11

plums in each hold

77

plums between them. Take the frequency from the same row as the value.)

Add the products to find the total number of plums

77 + 96 + 91 + 70 + 15 + 32 = 381

(Reason: Adding the six products counts every plum in all

30

punnets, giving

381

plums altogether.)

Divide the total by the number of punnets

\text{mean} = \dfrac{381}{30} = 12.7

(Reason: The

381

plums are shared between

30

punnets. A mean does not have to be a whole number, so

12.7

is a proper final answer and should not be rounded unless the question asks for it.)

(a)

12

(b)

12.7

Verification

Check 1: Add the frequency column on its own:

7 + 8 + 7 + 5 + 1 + 2 = 30

. The frequencies total

30

, the number of punnets the question gives, so

30

is the right divisor.

Check 2: Ignore the table and think of the

30

punnets listed one by one. The reading

12

appears

8

times, while

11

and

13

each appear only

7

times. The most common reading is

12

, which agrees with part (a).

Check 3: Work the mean out a different way. Take

11

off every value and average what is left:

0 \times 7 + 1 \times 8 + 2 \times 7 + 3 \times 5 + 4 \times 1 + 5 \times 2 = 51

, then

\dfrac{51}{30} = 1.7

11 + 1.7 = 12.7

, which agrees with part (b). The mean also lies between

11

and

16

, as any mean of these values must.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Write down the value with the largest frequency	B1	$12$	✓
(b) Multiply each number of plums by its frequency	M1	For at least four correct products, e.g. $11 \times 7 + 12 \times 8 + 13 \times 7 + 14 \times 5 + 15 \times 1 + 16 \times 2$ or $77 + 96 + 91 + 70 + 15 + 32$ , leading to $381$ .	✓
Divide the total by the total frequency	M1	Their total divided by $30$ , e.g. $\dfrac{381}{30}$ .	✓
Give the mean	A1	$12.7$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working. Allow $13$ if both method marks are scored, since that is $12.7$ rounded to the nearest whole plum.	✓

Full marks: 4/4

Continue to questions 16 to 26

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

Frequently asked questions

There are 26 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 1H, 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 2FR, November 2024: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 15 of 26

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

Frequently asked questions

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