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Edexcel IGCSE 4MA1 Paper 1FR, November 2024: Worked Solutions and Mark Schemes

Sir Faraz Hassan

1 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)Paper 1FR - Foundation Tier - November 2024100 marks · 2 hours · Calculator allowed

Original worked solutions for Edexcel International GCSE Mathematics A (4MA1), Paper 1FR (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 13 of 24

Seven numbers are printed in the box below.

Each of your answers must be one of these seven numbers.

Write down

(a) a number that is odd [1 mark]

(b) a number that is a multiple of $12$ [1 mark]

(d) a number that is a factor of $24$ [1 mark]

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

[Total 4 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

The seven numbers in the box:

6

15

19

28

38

44

48

Every answer has to be taken from that list, so nothing is worked out from scratch.

Find

(a) one of them that is odd (b) one of them that is a multiple of

12

24

Plan the Solution

Take one part at a time and read the whole list against that one property.
Odd or even is settled by the last digit: a number ending in $0$ , $2$ , $4$ , $6$ or $8$ is even.
A multiple of $12$ sits in the $12$ times table, so look for the number that $12$ divides exactly.
A prime has exactly two factors, itself and $1$ , so any number that splits into a product of two smaller numbers is out.
A factor of $24$ divides into $24$ with nothing left over. Watch the direction here: $48$ is a multiple of $24$ , not a factor of it.

Worked Solution [4 marks]

Rule: a multiple of

12

12

times a whole number, and a factor of

24

is a whole number that divides

24

exactly. A prime has exactly two factors, and an odd number is one that

2

does not divide.

Step 1: pick out an odd number

15 = 2 \times 7 + 1

19 = 2 \times 9 + 1

(Reason: The other five numbers end in

6

8

8

4

and

8

, so halving each one leaves nothing over and they are all even. Only

15

and

19

leave

1

over, and either of them earns the mark.)

Step 2: pick out a multiple of

12

48 = 12 \times 4

44 = 12 \times 3 + 8

(Reason: The

12

times table runs

12

24

36

48

and only the last of those is in the box. The second line shows the closest miss:

44

8

past a multiple of

12

Step 3: pick out a prime number

6 = 2 \times 3

15 = 3 \times 5

28 = 2 \times 14

38 = 2 \times 19

44 = 2 \times 22

48 = 2 \times 24

(Reason: Every number in the box except

19

splits into two smaller whole numbers, so each has more than two factors. Nothing from

2

4

divides

19

, and

5 \times 5 = 25

is already past it, so the testing stops there and

19

is prime.)

Step 4: pick out a factor of

24

24 = 6 \times 4

48 = 24 \times 2

(Reason: The factors of

24

are

1

2

3

4

6

8

12

and

24

, and the only one of those in the box is

6

. The second line is the trap:

48

divides BY

24

, which makes it a multiple, not a factor.)

(a)

15

(or

19

)(b)

48

(c)

19

(d)

6

Verification

Check 1: Read the four answers back against the box:

15

48

19

and

6

. Each one is a number that is actually printed in the box, which is what the question demands.

Check 2: Test parts (b) and (d) the other way round, by dividing instead of multiplying.

\dfrac{48}{12} = 4

and

\dfrac{24}{6} = 4

, so both divisions come out as whole numbers.

Check 3: Count the factors of

19

directly, testing

2

3

and

4

, and stopping there because

5 \times 5 = 25

is bigger than

19

. None of them divides it, so

19

has only the factors

1

and

19

, which is exactly two.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) an odd number written down	B1	Accept $15$ or $19$ . Both values given together, and no other value, also scores.	✓
(b) a multiple of $12$ written down	B1	$48$ , and no other number from the box. Any extra value written alongside it loses the mark.	✓
(c) a prime number written down	B1	$19$ , and no other number from the box.	✓
(d) a factor of $24$ written down	B1	$6$ , and no other number from the box. $48$ is a common wrong answer here because it is a multiple of $24$ rather than a factor of it.	✓

Full marks: 4/4

(a) Express the decimal $0.13$ as a fraction. [1 mark]

(b) Fill in the empty box so that the statement below is correct.
$\dfrac{16}{20} = \dfrac{\boxed{\phantom{4}}}{5}$ [1 mark]

(a)

[Total 2 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

The decimal

0.13

The statement

\dfrac{16}{20} = \dfrac{\boxed{\phantom{4}}}{5}

, with the numerator on the right missing.

Find

(a)

0.13

written as a fraction. (b) The number that belongs in the empty box.

Plan the Solution

(a) Count the digits after the decimal point. Two decimal places means the number is a count of hundredths, so the denominator is $100$ .
(b) Look at what has happened to the denominator: $20$ has become $5$ . Do exactly the same thing to the numerator.

Worked Solution [2 marks]

Place value: a decimal with two decimal places is a number of hundredths, so it is written over

100

. Equivalent fractions: whatever you divide the denominator by, you must divide the numerator by as well, or the fraction changes value.

Step 1: part (a), count the decimal places

0.13 = \dfrac{13}{100}

(Reason: There are two digits after the decimal point, so the last digit sits in the hundredths column. Thirteen hundredths is written as

\dfrac{13}{100}

Step 2: part (a), check whether the fraction cancels

\dfrac{13}{100} = \dfrac{130}{1000}

(Reason:

13

is a prime number and

100

is not a multiple of

13

, so nothing cancels and

\dfrac{13}{100}

is already in its simplest form. An equivalent fraction such as

\dfrac{130}{1000}

is accepted as well.)

Step 3: part (b), see what has happened to the denominator

\dfrac{20}{5} = 4

(Reason: The denominator has gone from

20

down to

5

, and

20

shared into groups of

5

gives

4

, so the denominator has been divided by

4

Step 4: part (b), do the same to the numerator

\dfrac{16}{4} = 4

(Reason: Whatever the denominator is divided by, the numerator must be divided by as well, otherwise the two fractions are not equal.)

Step 5: part (b), write out the completed statement

\dfrac{16}{20} = \dfrac{4}{5}

(Reason: The numerator on the right is

4

, so

4

is the number that goes in the box.)

(a)

\dfrac{13}{100}

(b)

4

Verification

Check 1: Turn the part (a) answer back into a decimal. Dividing

13

100

moves every digit two places to the right.

\dfrac{13}{100} = 0.13

Check 2: Cross-multiply the two fractions in part (b). They are equivalent only if the two products are equal.

16 \times 5 = 80

and

20 \times 4 = 80

Check 3: Write both fractions in part (b) as decimals on the calculator and compare them.

\dfrac{16}{20} = 0.8

and

\dfrac{4}{5} = 0.8

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $\dfrac{13}{100}$	B1	Or any equivalent fraction, for example $\dfrac{130}{1000}$ . The fraction does not have to be given in its simplest form.	✓
(b) $4$	B1	For $4$ written in the box.	✓

Full marks: 2/2

(a) Complete this statement by writing a number in the box.
$\dfrac{5073}{\boxed{\phantom{00}}} = 19$ [1 mark]

(b) Complete this statement by writing a number in the box.
The cube root of $\boxed{\phantom{00}}$ is $14$ [1 mark]

Here is a list of five numbers.
$973 \quad 987 \quad 393 \quad 151 \quad 139$

(c)

[Total 4 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

(a) The statement

\dfrac{5073}{\boxed{\phantom{00}}} = 19

, with one number missing.

(b) A number whose cube root is

14

973 \quad 987 \quad 393 \quad 151 \quad 139

Find

(a) The number that belongs in the box. (b) The number that belongs in the box. (c) The difference between the largest and the smallest number in the list.

Plan the Solution

(a) Read the statement backwards. If $5073$ divided by the box gives $19$ , then $19$ lots of the box make $5073$ , so the box is $\dfrac{5073}{19}$ .
(b) Cubing undoes a cube root, so cube the $14$ .
(c) Write the five numbers in order of size, read off the two ends, then subtract the smaller from the larger.

Worked Solution [4 marks]

Inverse operations - if

\dfrac{a}{b} = c

then

b = \dfrac{a}{c}

, and if

\sqrt[3]{n} = k

then

n = k^{3}

. A difference is

\text{largest} - \text{smallest}

(a) Turn the division round

\dfrac{5073}{\boxed{\phantom{00}}} = 19 \implies \boxed{\phantom{00}} = \dfrac{5073}{19}

\dfrac{5073}{19} = 267

(Reason: The box is what

5073

is divided by, so the box is

5073

shared into

19

equal parts. A calculator is allowed on this paper, and the division comes out exactly, with nothing left over.)

(b) Undo the cube root

\sqrt[3]{\boxed{\phantom{00}}} = 14 \implies \boxed{\phantom{00}} = 14^{3}

14^{3} = 14 \times 14 \times 14 = 2744

(Reason: Cubing is the inverse of taking a cube root, so cube the

14

. Working in two stages,

14 \times 14 = 196

and then

196 \times 14 = 2744

139 \quad 151 \quad 393 \quad 973 \quad 987

\text{largest} = 987, \quad \text{smallest} = 139

(Reason: Ordering the numbers makes the two ends of the list obvious, and this is where the method mark is earned - the mark scheme awards it for identifying

987

and

139

, or simply for writing the numbers in order of size. Take care here:

973

and

987

differ only in the middle digit.)

987 - 139 = 848

(Reason: The difference between two numbers is what is left when the smaller is taken away from the larger.)

(a)

267

(b)

2744

(c)

848

Verification

Check 1: Multiply the answer to (a) back into the statement:

19

lots of

267

19 \times 267 = 5073

, the number the statement starts with.

Check 2: Test the answer to (b) by prime factors instead of by cubing:

2744 = 8 \times 343

, where

8 = 2^{3}

and

343 = 7^{3}

\sqrt[3]{2744} = 2 \times 7 = 14

, so

2744

really is the number whose cube root is

14

Check 3: Add the difference found in (c) back on to the smallest number in the list.

139 + 848 = 987

, the largest number in the list.

Check 4: Do the subtraction in (c) a second way, with no borrowing: take off

140

and then put the extra

1

back.

987 - 140 = 847

and

847 + 1 = 848

, which agrees.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The number in the box	B1	For $267$ written in the box.	✓
(b) The number in the box	B1	For $2744$ written in the box.	✓
(c) Identify the largest and the smallest number	M1	For identifying $987$ and $139$ - they may be shown or circled - or for $987 - n$ where $n$ is a number from the list, or for $m - 139$ where $m$ is a number from the list, or for writing the numbers in order of size.	✓
(c) The difference	A1	For $848$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 4/4

The pictogram shows information about the numbers of different types of vehicle that Emily counted passing the school gate.

(a) How many Vans did Emily count? [1 mark]

Emily counted $10$ Motorbikes.
(b) Show this information on the pictogram. [1 mark]

Emily counted more Lorries than Buses.
(c) How many more? [1 mark]

(a)(c)

[Total 3 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

The key: one whole symbol represents

8

vehicles.

Each whole symbol is drawn as

4

small squares of equal size.

Car:

3

whole symbols. Van:

3

whole symbols and a half symbol.

Lorry:

4

whole symbols and one small square. Bus: one whole symbol and

3

small squares.

Taxi:

4

whole symbols. The Motorbike row is empty.

Emily counted

10

Motorbikes.

Find

(a) The number of Vans shown in the pictogram. (b) What must be drawn in the Motorbike row to show

10

Motorbikes. (c) How many more Lorries than Buses Emily counted.

Plan the Solution

Start with the key. It fixes the value of a whole symbol, and the way the symbol is divided fixes the value of one small square.
For (a), count the whole symbols in the Van row, work out the value of the part symbol, and add.
For (b), change $10$ vehicles into small squares first, then group the small squares back into a symbol.
For (c), total the Lorry row and the Bus row separately, then subtract.

Worked Solution [3 marks]

Rule - Pictogram: a part of a symbol is worth the same part of the key. The symbol here is split into

4

equal small squares and the whole symbol is

8

vehicles, so one small square is

\dfrac{8}{4} = 2

vehicles.

Step 1: work out what one small square is worth

\dfrac{8}{4} = 2

(Reason: The key gives the value of a whole symbol only. Every reading in this question uses part symbols as well, so the value of one small square is the number to have ready first.)

Step 2: read the Van row for part (a)

3 \times 8 = 24

\dfrac{1}{2} \times 8 = 4

24 + 4 = 28

(Reason: The Van row shows

3

whole symbols and a half symbol. Half a symbol is half of

8

, which is the same as the two small squares that are drawn.)

Step 3: turn

10

into small squares for part (b)

\dfrac{10}{2} = 5

5 = 4 + 1

(Reason: Working in small squares avoids guessing how much of a symbol to draw.

5

small squares are needed, and

4

of them make one whole symbol, so one small square is left over.)

Step 4: total the Lorry row and the Bus row

4 \times 8 + 2 = 34

8 + 3 \times 2 = 14

(Reason: The Lorry row is

4

whole symbols and one small square. The Bus row is one whole symbol and

3

small squares, so the part symbol there is worth three small squares, not half a symbol.)

Step 5: subtract for part (c)

34 - 14 = 20

(Reason: The question asks how many more, so take the smaller row total away from the larger one.)

(a)

28

Vans(b) One whole symbol and one small square drawn in the Motorbike row(c)

20

more Lorries

Verification

Check 1: Count the whole pictogram two different ways once the Motorbike row is complete. There are

71

small squares altogether, each worth

2

vehicles.

71 \times 2 = 142

, and adding the six row totals gives

24 + 28 + 34 + 14 + 10 + 32 = 142

Check 2: Do part (c) without totalling either row. The Lorry row has

17

small squares and the Bus row has

7

, so subtract the small squares first and convert once.

17 - 7 = 10

small squares, and

10 \times 2 = 20

Check 3: Work part (a) backwards. Turn

28

back into symbols and see whether that is what the Van row actually shows.

\dfrac{28}{8} = 3.5

symbols, which is the three whole symbols and the half symbol that are drawn

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $28$	B1	cao. From $3.5 \times 8 = 28$ , or from $14$ small squares worth $2$ each.	✓
(b) One whole symbol and one small square drawn in the Motorbike row	B1	oe, eg $5$ small squares from the symbol shown. Whatever is drawn must come to $10$ .	✓
(c) $20$	B1	cao. From $34 - 14 = 20$ .	✓

Full marks: 3/3

A travel guide lists the average January temperature in five cities.

Here are the same five temperatures, in °C
$-8 \quad 24 \quad -3 \quad 12 \quad -14$

(a) Write these numbers in order of size.
Start with the smallest number. [1 mark]

(b) Work out the difference between the January temperature in Moscow and the January temperature in Lisbon. [1 mark]

The January temperature in Novosibirsk is $13$ °C lower than the January temperature in Riga.

The January temperature in Marrakesh is $25$ °C higher than the January temperature in Regina.

(d) Work out the January temperature in Marrakesh. [1 mark]

(a)(b) °C(c) °C(d) °C

[Total 4 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

Five January temperatures, in °C:

-8

(Moscow),

24

(Chennai),

-3

(Riga),

12

(Lisbon),

-14

(Regina)

Novosibirsk is

13

°C lower than Riga

Marrakesh is

25

°C higher than Regina

Find

(a) the five temperatures written out in order, smallest first (b) the difference between

-8

°C and

12

°C (c) the January temperature in Novosibirsk (d) the January temperature in Marrakesh

Plan the Solution

Picture a number line. Temperatures below zero sit to the left of zero, and the further left a number is, the smaller it is.
For (a), the below-zero temperatures come first, coldest first, then the above-zero ones.
For (b), the difference is the size of the gap between the two temperatures on that line.
For (c), lower means move to the left, so subtract. For (d), higher means move to the right, so add.

Worked Solution [4 marks]

Rule - Numbers increase from left to right on a number line, so a temperature further below zero is the smaller one. A difference is the size of the gap between two temperatures, and subtracting a negative is the same as adding.

Step 1: put the five temperatures in order

-14 < -8 < -3 < 12 < 24

(Reason: Three of the temperatures are below zero, so those three come first. Of those,

-14

is furthest below zero, so it is the smallest, and

-3

is the closest to zero, so it is the largest of the three.)

Step 2: work out the difference for part (b)

12 - (-8) = 12 + 8 = 20

(Reason: Moscow is

8

degrees below zero and Lisbon is

12

degrees above zero, so the gap is the climb up to zero followed by the climb above it.)

Step 3: work out the temperature in Novosibirsk for part (c)

-3 - 13 = -16

(Reason: Lower means subtract. Riga is already

3

degrees below zero, and dropping a further

13

degrees puts Novosibirsk

16

degrees below zero.)

Step 4: work out the temperature in Marrakesh for part (d)

-14 + 25 = 11

(Reason: Higher means add. The first

14

degrees of the rise only bring Regina up to zero, so the remaining

11

degrees are above zero.)

(a)

-14, -8, -3, 12, 24

(b)

20

°C(c)

-16

°C(d)

11

°C

Verification

Check 1: Read the ordered list back: every temperature must be warmer than the one before it, and all five must appear once. The three below-zero temperatures come first, coldest first, then the two above zero, and no temperature is repeated or missing.

Check 2: Count part (b) along the number line instead:

8

degrees from Moscow up to zero, then

12

degrees from zero up to Lisbon.

8 + 12 = 20

, which is the difference found in part (b).

Check 3: Reverse part (c): put the

13

degrees back on the Novosibirsk temperature and Riga should come back.

-16 + 13 = -3

, the January temperature in Riga.

Check 4: Reverse part (d): take the

25

degrees back off the Marrakesh temperature and Regina should come back.

11 - 25 = -14

, the January temperature in Regina.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $-14, -8, -3, 12, 24$	B1	All five temperatures written out in order, smallest first. No working is needed - the list alone earns the mark.	✓
(b) $20$	B1	The answer alone earns the mark. A difference is a size, so it is not negative.	✓
(c) $-16$	B1	The answer alone earns the mark. The minus sign is part of the answer.	✓
(d) $11$	B1	The answer alone earns the mark.	✓

Full marks: 4/4

Here is a regular octagon.

(a) Write down how many lines of symmetry a regular octagon has. [1 mark]

The diagram below shows a different shape.

(b) On the shape, mark a right angle with the letter $R$ .
[1 mark]
(c) On the shape, mark an obtuse angle with the letter $O$ .
[1 mark]
(d) By measuring, work out the size of the angle marked $x$ [1 mark]

(a)(d) °

[Total 4 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

A regular octagon.

A four-sided shape whose left-hand edge is vertical and whose bottom edge is horizontal, with one angle marked

x

Find

(a) how many lines of symmetry a regular octagon has (b) a right angle on the shape, marked with the letter

R

O

(d) the size of

x

, found by measuring

Plan the Solution

Part (a) is a rule, not a drawing job: a regular polygon with $n$ sides has $n$ lines of symmetry.
Parts (b) and (c) are the two definitions. A right angle is exactly $90^\circ$ ; an obtuse angle is bigger than $90^\circ$ but smaller than a straight angle.
Part (d) says by measuring, so the method is a protractor: centre on the point, base line along one arm, then read the scale that starts at zero on that arm.
Finish by checking the reading against the angle sum of a quadrilateral, which must come to $360^\circ$ .

Worked Solution [4 marks]

Regular polygons and quadrilaterals: a regular

n

-gon has

n

lines of symmetry and rotational symmetry of order

n

, and the four angles of any quadrilateral add up to

360^\circ

Step 1: count the octagon's lines of symmetry

\text{regular polygon with } n \text{ sides} \rightarrow n \text{ lines of symmetry}

n = 8

(Reason: An octagon has

8

sides, so it has

8

mirror lines:

4

joining opposite vertices, and

4

joining the midpoints of opposite sides. Nothing needs to be drawn or measured.)

Step 2: mark the right angle with

R

\text{vertical edge} \perp \text{horizontal edge}

\text{angle at the bottom left vertex} = 90^\circ

(Reason: The left-hand edge runs straight up and the bottom edge runs straight across, so the corner where they meet is square. It is the only square corner on the shape, so the letter

R

goes at the bottom left vertex. A clear right-angle symbol drawn in that corner earns the mark just as well.)

Step 3: mark an obtuse angle with

O

\text{obtuse: bigger than } 90^\circ \text{ and smaller than a straight angle}

\text{angle at the bottom right vertex} \approx 148^\circ

(Reason: At the bottom right vertex the base stops and the edge turns back upwards, opening that corner out well past square but not as far as a straight line. That is the obtuse angle, so the letter

O

goes there.)

Step 4: measure the angle marked

x

x = 42^\circ

\text{check: } 360^\circ - 90^\circ - 148^\circ - 80^\circ = 42^\circ

(Reason: Put the centre of the protractor on the point and lay its base line along one of the two arms, then read up from zero on that arm. The reading is

42^\circ

. Taking the other three angles from

360^\circ

gives the same value, which is why a hand measurement anywhere from

40^\circ

44^\circ

is accepted.)

(a)

8

lines of symmetry(b)

R

at the bottom left vertex(c)

O

at the bottom right vertex(d)

x = 42^\circ

Verification

Check 1: Add all four angles of the shape, in degrees. Any quadrilateral must total

360^\circ

, so a measurement that breaks this is a measurement to take again.

90 + 80 + 148 + 42 = 360

Check 2: Count the octagon's mirror lines in two families instead of using the rule: those through pairs of opposite vertices, and those through the midpoints of opposite sides.

4 + 4 = 8

, which also matches its rotational symmetry of order

8

Check 3: Sanity-check the size before writing it down. The marked corner is the sharpest on the shape, so

x

has to come out acute.

42^\circ

is less than

90^\circ

, and it sits inside the accepted

40^\circ

44^\circ

band.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $8$	B1	Award for $8$ only. No working is required and none is expected.	✓
(b) $R$ marked at the bottom left vertex	B1	Accept a clear, unambiguous right-angle symbol drawn in that corner instead of the letter $R$ .	✓
(c) $O$ marked at the bottom right vertex	B1	Award for a mark that clearly identifies that vertex and no other.	✓
(d) $42$	B1	Allow $40$ to $44$ , which is the tolerance for reading a protractor by hand.	✓

Full marks: 4/4

At a community centre games evening, everyone played draughts or dominoes or backgammon or cribbage.
Each person played only one of these games.

Marina starts to draw a pie chart to show information about the games played.

$24$ people played dominoes.

(a) Work out the number of people who played draughts. [2 marks]

$40$ people played cribbage.

(b) Work out the size of the angle on the pie chart for the sector representing cribbage.
You do not need to complete the pie chart. [2 marks]

(a)(b)

[Total 4 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

The sector for dominoes is

72^\circ

, and

24

people played dominoes.

The sector for draughts is

96^\circ

40

people played cribbage.

Each person played one game only, so the whole circle stands for everybody at the games evening.

Find

(a) the number of people who played draughts (b) the angle of the sector for cribbage

Plan the Solution

A pie chart shares one full turn between the groups, so the same number of degrees stands for one person in every sector.
The dominoes sector is the one we know completely: $72^\circ$ for $24$ people. Use it to find the angle for one person.
Then (a) turns an angle into people, and (b) turns people into an angle. Both use that one figure.

Worked Solution [4 marks]

Rule - Pie chart:

\text{angle for one person} = \dfrac{\text{angle of a known sector}}{\text{number of people in that sector}}

, and every sector on the chart uses that same figure.

Step 1: find the angle that stands for one person

\dfrac{72}{24} = 3

(Reason: the dominoes sector is

72^\circ

and it stands for

24

people, so one person is worth

3^\circ

of the pie chart)

Step 2: turn the draughts angle into people

\dfrac{96}{3} = 32

(Reason: the draughts sector is

96^\circ

, and every

3^\circ

of it is one more person)

Step 3: turn the cribbage people into an angle

40 \times 3 = 120

(Reason: there are

40

people, and each one takes

3^\circ

of the pie chart)

(a)

32

people(b)

120^\circ

Verification

Check 1: Turn the answer to (a) back into an angle. If

32

people played draughts, their sector must be

32

lots of

3^\circ

32 \times 3 = 96

, which is the draughts angle printed on the pie chart

Check 2: Count the whole games evening two ways. One person is

3^\circ

, so the full turn holds

\dfrac{360}{3} = 120

people. The three known sectors use

72 + 96 + 120 = 288

degrees, leaving

360 - 288 = 72

degrees for backgammon, which is

24

people.

24 + 32 + 40 + 24 = 120

, the same total, so the four sectors close the circle

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A method that fixes the size of one person or one degree, for example $\dfrac{72}{24} = 3$ , $\dfrac{24}{72} = \dfrac{1}{3}$ , $\dfrac{96}{72} = \dfrac{4}{3}$ or $24 \times \dfrac{360}{72} = 120$	M1	Any method that finds the degrees per person, the people per degree, the total number of people represented by the pie chart, or the scale factor between the two sectors.	✓
(a) $32$	A1	Correct answer only. A correct answer scores both marks unless it comes from obviously incorrect working.	✓
(b) $3 \times 40$ , or an equivalent such as $\dfrac{40}{24} \times 72$	M1ft	Follow through their own degrees-per-person figure. The working may already have been seen in part (a), for example $\dfrac{72}{24} = 3$ .	✓
(b) $120$	A1	Accept anything from $119.5$ to $120.3$ .	✓

Full marks: 4/4

Elise buys
$3$ scented candles at $\$1.80$ each
$4$ packets of flower seeds at $\$1.20$ a packet
$2$ identical jars of honey

The total cost is $\$17.10$

Work out the cost of one jar of honey. [4 marks]

[Total 4 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

3

scented candles at

\$1.80

each, and

4

packets of flower seeds at

\$1.20

a packet

2

jars of honey, and the word identical tells us the two jars cost the same as each other

The whole order comes to

\$17.10

Three different items, and only one of the three has an unknown price

Find

The cost of ONE jar of honey The order pays for

2

jars, so the last move will be to split what the jars cost between them

Plan the Solution

Price the two groups we can price: $3$ candles at a known price each, and $4$ packets at a known price each
Add those two amounts to find how much of the $\$17.10$ is already accounted for
Take that away from the total, which leaves the cost of the two jars together
Halve what is left, because the two jars are identical

Worked Solution [4 marks]

Rule - when a total is made up of several items and only one item's price is unknown, take off everything you CAN price and whatever is left belongs to the rest:

\text{total} - \text{priced items} = \text{unpriced items}

Step 1: price the two groups whose prices are given

1.80 \times 3 = 5.40

1.20 \times 4 = 4.80

5.40 + 4.80 = 10.20

(Reason: every candle costs the same and every packet costs the same, so each group is a multiplication rather than an addition, and the two group costs added together show that

\$10.20

of the order is already accounted for)

Step 2: take the priced items off the total

17.10 - 10.20 = 6.90

(Reason: the candles and the seed packets are paid for out of the same

\$17.10

, so whatever is left of the total after they are removed has to be the cost of the two jars of honey together)

Step 3: split what is left between the two jars

\dfrac{6.90}{2} = 3.45

(Reason: the jars are identical, so the

\$6.90

they cost between them divides equally in two)

\$3.45

Verification

Check 1: Put the answer back into the shopping list and rebuild the whole order:

3 \times 1.80 = 5.40

for the candles,

4 \times 1.20 = 4.80

for the seeds and

2 \times 3.45 = 6.90

for the honey.

5.40 + 4.80 + 6.90 = 17.10

, which is exactly the total the question states

Check 2: Run the question backwards. If one jar really costs

\$3.45

, then two jars cost

2 \times 3.45 = 6.90

, so taking that off the total should leave precisely the candles and the seeds.

17.10 - 6.90 = 10.20

, and the candles and seeds do come to

5.40 + 4.80 = 10.20

Check 3: Do the whole question again in cents, where every value is a whole number and no decimal point can slip: the candles cost

3 \times 180 = 540

cents and the seeds cost

4 \times 120 = 480

cents.

1710 - 540 - 480 = 690

cents for two jars, so one jar is

\dfrac{690}{2} = 345

cents, which is

\$3.45

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $3 \times 1.80 = 5.40$ or $4 \times 1.20 = 4.80$ or $3 \times 1.80 + 4 \times 1.20 = 10.20$	M1	for a method to find the cost of the candles, or the cost of the seed packets, or the total cost of the candles and the seed packets together	✓
eg $17.10 - (5.40 + 4.80) = 6.90$ or $17.10 - 10.20 = 6.90$	M1	for a method to find the cost of the $2$ jars of honey. Follow through on the candidate's own group costs from the first mark	✓
eg $\dfrac{6.90}{2}$	M1	for a complete method: the cost of the two jars, halved. Follow through on the candidate's own value for the two jars	✓
$3.45$	A1	cao. A correct answer scores full marks on its own, unless it comes from obviously incorrect working	✓
If no other marks are awarded, an answer of $7.05$ scores SCB2, and an answer of $14.10$ scores SCB1	SC	both come from one nameable slip: taking off the price of a single candle and a single packet instead of all $3$ candles and all $4$ packets. That leaves $14.10$ for the SCB1, and halving it gives the $7.05$ for the SCB2	✓

Full marks: 4/4

The grid shows three points, $A$ , $B$ and $C$ .

(a) Write down the coordinates of $A$ . [1 mark]

(b) Work out the coordinates of the midpoint of $AB$ . [2 marks]

The point $D$ is marked on the grid so that $ABCD$ is a rectangle.

(a)(b)(c)

[Total 5 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

Three points marked on a numbered grid:

A

B

and

C

ABCD

is a rectangle, so

D

is its fourth vertex - the corner opposite

B

Find

(a) The coordinates of

A

. (b) The coordinates of the midpoint of

AB

. (c) The coordinates of

D

Plan the Solution

Read $A$ straight off the grid: count across from the $y$ -axis first, then up from the $x$ -axis, and write the across value first.
For the midpoint, average the two $x$ -coordinates and average the two $y$ -coordinates. Averaging keeps the sign, so a negative coordinate is added, not ignored.
For $D$ , use the shape rather than a formula: in a rectangle $AD$ and $BC$ are opposite sides, so the move from $A$ to $D$ is the same as the move from $B$ to $C$ .
The letters $ABCD$ go round the rectangle in order, so $D$ is next to $A$ and $C$ , never next to $B$ .

Worked Solution [5 marks]

Rule - Midpoint:

x_M = \dfrac{x_1 + x_2}{2}

and

y_M = \dfrac{y_1 + y_2}{2}

. Rule - Rectangle: opposite sides are equal and parallel, so the step from

B

C

is also the step from

A

D

Step 1: Read the coordinates of

A

off the grid

A = (-4,\ 2)

(Reason: From

O

, the cross at

A

4

squares to the left and

2

squares up. Left of the

y

-axis makes the

x

-coordinate negative, and the across value is always written first.)

Step 2: Average the coordinates of

A

and

B

x_M = \dfrac{-4 + 2}{2} = -1

y_M = \dfrac{2 + 4}{2} = 3

M = (-1,\ 3)

(Reason: The midpoint sits halfway, so each of its coordinates is the mean of the two end values: add the pair and halve it. The

x

-coordinates are

-4

and

2

, and the

y

-coordinates are

2

and

4

Step 3: Step from

A

by the same move that takes

B

C

(3 - 2,\ 1 - 4) = (1,\ -3)

D = (-4 + 1,\ 2 - 3) = (-3,\ -1)

(Reason:

BC

and

AD

are opposite sides of the rectangle, so they are the same length and point the same way. The move from

B

C

1

right and

3

down, so the same move from

A

lands on

D

(a)

(-4,\ 2)

(b)

(-1,\ 3)

(c)

(-3,\ -1)

Verification

Check 1: Step from

A

M

, then from

M

B

. If

M

really is the midpoint, the two steps are identical.

(3,\ 1)

both times, so

M

lies halfway along

AB

Check 2: A rectangle needs a right angle at

B

. Multiply the two side steps

AB = (6,\ 2)

and

BC = (1,\ -3)

across and add:

6 \times 1 + 2 \times (-3)

6 - 6 = 0

, so

AB

and

BC

are perpendicular.

Check 3: The diagonals of a rectangle bisect each other, so the midpoint of

AC

must come out the same as the midpoint of

BD

. Both give

(-0.5,\ 1.5)

, which also confirms

D = (-3,\ -1)

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The coordinates of $A$ written down	B1	$(-4,\ 2)$ , correct answer only. The $x$ -coordinate must come first.	✓
(b) The coordinates of the midpoint of $AB$	B2	$(-1,\ 3)$ . B1 for $(-1,\ y)$ or $(x,\ 3)$ or $(3,\ -1)$ , or for the midpoint of $AB$ shown unambiguously on the diagram.	✓
(c) The coordinates of $D$	B2	$(-3,\ -1)$ . B1 for $(-3,\ y)$ or $(x,\ -1)$ or $(-1,\ -3)$ , or for $D$ marked correctly on the diagram.	✓

Full marks: 5/5

Ivy is baking from an American cookery book, so its oven temperatures are given in degrees Fahrenheit (°F) rather than in degrees Celsius (°C)
She uses this rule to change a temperature from °C to °F

(a) Change an oven temperature of $175^\circ\text{C}$ to a temperature in °F [2 marks]

(b) Change an oven temperature of $482^\circ\text{F}$ to a temperature in °C [2 marks]

(a) °F(b) °C

[Total 4 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

A function machine that changes a temperature in degrees Celsius into a temperature in degrees Fahrenheit: multiply by

1.8

, then add

32

Part (a) starts with

175^\circ\text{C}

, and part (b) starts with

482^\circ\text{F}

Find

(a) the same temperature written in degrees Fahrenheit. (b) the same temperature written in degrees Celsius. One machine, used forwards in (a) and backwards in (b).

Plan the Solution

(a) Feed the temperature in at the left and work through the two boxes in the order they are drawn.
(b) The temperature is coming out of the right-hand end, so travel the other way and undo each box: subtract $32$ first, then divide by $1.8$ .
Undoing a machine reverses the ORDER of the steps as well as the operations themselves, so the subtraction comes before the division.
A calculator is allowed, so type each stage separately rather than one long expression - that is where marks are lost here.

Worked Solution [4 marks]

Rule - forwards, follow the boxes as drawn: multiply by

1.8

, then add

32

.
Backwards, use the inverse operations in the opposite order: subtract

32

, then divide by

1.8

Step 1 (a) - multiply by

1.8

175 \times 1.8 = 315

(Reason: The temperature going in is

175

, and the first box it meets multiplies by

1.8

. This product on its own earns the method mark.)

Step 2 (a) - then add

32

315 + 32 = 347

(Reason: What comes out of the first box goes straight into the second, so the

32

is added to

315

and never to

175

Step 3 (b) - undo the

+ 32

first

482 - 32 = 450

(Reason: Part (b) walks back through the machine, and the last thing the machine did was add

32

, so the first thing to undo is that addition.)

Step 4 (b) - then undo the

\times 1.8

\dfrac{450}{1.8} = 250

(Reason: The opposite of multiplying by

1.8

is dividing by

1.8

, and it is the

450

from step 3 that is divided, not the

482

the part started with.)

(a)

347^\circ\text{F}

(b)

250^\circ\text{C}

Verification

Check 1: Put part (a)'s answer back through the machine in reverse:

347 - 32 = 315

, then

\dfrac{315}{1.8} = 175

. It lands on

175^\circ\text{C}

, the temperature part (a) began with.

Check 2: Send part (b)'s answer forwards through the machine as drawn:

250 \times 1.8 = 450

, then

450 + 32 = 482

. It lands on

482^\circ\text{F}

, the temperature part (b) began with.

Check 3: Is the machine itself being used properly? Water boils at

100^\circ\text{C}

, and the machine gives

100 \times 1.8 + 32 = 212

. The familiar boiling point

212^\circ\text{F}

comes out, so the rule is being applied the right way round.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $175 \times 1.8 = 315$	M1	for a correct first step	✓
(a) $315 + 32 = 347$	A1	cao. A correct answer scores full marks unless it comes from obviously incorrect working.	✓
(b) $482 - 32 = 450$	M1	for $-32$ , or for dividing by $1.8$	✓
(b) $\dfrac{450}{1.8} = 250$	A1	cao. A correct answer scores full marks unless it comes from obviously incorrect working.	✓
(b) special case - an answer of $464.2$ and so on	SC B1	This row carries no mark of its own; it records what one particular wrong answer is worth. It comes from typing the reverse machine into a calculator without brackets: the calculator works out $\dfrac{32}{1.8}$ first and subtracts only that from $482$ , instead of subtracting $32$ and then dividing.	✓

Full marks: 4/4

(a) Simplify $c + c + c + c + c$ [1 mark]

(b) Simplify $7w + 10y - 9w + 2y$ [2 marks]

The $n\text{th}$ term of a sequence is given by $7n - 4$

(d) Solve $7(x + 5) = 8 - 3x$
You must show clear algebraic working. [3 marks]

(a)(b)(c) 1st term(c) 5th term(d) x =

[Total 8 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

(a) the sum

c + c + c + c + c

(b) the expression

7w + 10y - 9w + 2y

n\text{th}

term is

7n - 4

(d) the equation

7(x + 5) = 8 - 3x

Find

(a) the sum written as a single term in

c

(b) the expression with its like terms collected (c) the

1\text{st}

term and the

5\text{th}

term of the sequence (d) the value of

x

, with algebraic working shown

Plan the Solution

(a) count how many $c$ terms are being added.
(b) handle the $w$ terms and the $y$ terms separately, because they are not like terms.
(c) substitute the position into the rule: $n = 1$ for the first term, then $n = 5$ .
(d) expand the bracket, gather the $x$ terms on one side and the numbers on the other, then divide.

Worked Solution [8 marks]

Like terms collect by adding their number parts only, and the letter is unchanged; a term of a sequence comes from substituting its position

n

into the

n\text{th}

term rule; and a linear equation is solved by expanding, collecting like terms, then dividing by the number in front of

x

Step 1: part (a), add the five

c

terms

c + c + c + c + c = 5c

(Reason: Adding

c

five times is the same as

5

lots of

c

. Only how many there are changes; the letter itself is unchanged.)

Step 2: part (b), collect the terms in

w

7w - 9w = -2w

(Reason: The terms in

w

are like terms, so their number parts combine into one number part and the letter

w

is carried through unchanged.)

Step 3: part (b), collect the terms in

y

and write the answer

10y + 2y = 12y

7w + 10y - 9w + 2y = -2w + 12y

(Reason: A term in

w

and a term in

y

are not like terms, so they cannot be combined with each other and both are kept in the answer.)

Step 4: part (c), the

1\text{st}

term, where

n = 1

7 \times 1 - 4 = 7 - 4 = 3

(Reason: The

1\text{st}

term sits in position

n = 1

, so

1

is substituted for

n

7n - 4

Step 5: part (c), the

5\text{th}

term, where

n = 5

7 \times 5 - 4 = 35 - 4 = 31

(Reason: The same rule is used with

n = 5

. Multiplying comes before subtracting, so the

35

is found first and the

4

is taken off afterwards.)

Step 6: part (d), expand the bracket

7(x + 5) = 7x + 35

7x + 35 = 8 - 3x

(Reason: Everything inside the bracket is multiplied by

7

7 \times x = 7x

and

7 \times 5 = 35

. The right-hand side is untouched.)

Step 7: part (d), gather the

x

terms and the numbers

7x + 3x = 8 - 35

10x = -27

(Reason: Adding

3x

to both sides removes the

-3x

from the right, and subtracting

35

from both sides removes it from the left, leaving the letters on one side and the numbers on the other.)

Step 8: part (d), divide to find

x

x = \dfrac{-27}{10}

x = -2.7

(Reason: Both sides are divided by

10

, the number multiplying

x

. The answer is negative because

-27

is negative, and a decimal answer is perfectly acceptable here.)

(a)

5c

(b)

-2w + 12y

(c)

1\text{st}

term

3

5\text{th}

term

31

(d)

x = -2.7

Verification

Check 1: Part (a): put

c = 2

into both forms. The sum gives

2 + 2 + 2 + 2 + 2

and the single term gives

5 \times 2

10 = 10

, so the two forms agree

Check 2: Part (b): put

w = 1

and

y = 1

into both forms. The original gives

7 + 10 - 9 + 2

and the collected form gives

-2 + 12

10 = 10

, so nothing was lost or gained in collecting

Check 3: Part (c): list the sequence from

7n - 4

. The terms should rise by

7

each time, so the

5\text{th}

term should be the

1\text{st}

term plus four steps of

7

3, 10, 17, 24, 31

and

3 + 4 \times 7 = 31

Check 4: Part (d): substitute

x = -2.7

into each side of the original equation. The left gives

7(-2.7 + 5) = 7 \times 2.3

and the right gives

8 - 3 \times (-2.7) = 8 + 8.1

16.1 = 16.1

, so both sides balance

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $5c$	B1	cao	✓
(b) $-2w + 12y$	B2	oe (B1 for $-2w$ or $12y$ )	✓
(c) $3$ for the $1\text{st}$ term	B1	cao	✓
(c) $31$ for the $5\text{th}$ term	B1	cao	✓
(d) $7x + 35$	M1	a correct expansion of the bracket (need not be in an equation)	✓
(d) $7x + 3x = 8 - 35$ oe, eg $10x = 8 - 35$ or $10x = -27$	M1	for isolating terms in $x$ and number terms; ft from an incorrect expansion in the form $7x + c$ or $x + 35$	✓
(d) $x = -2.7$	A1	dep on M1, oe. Working required.	✓

Full marks: 8/8

The diagram shows a quadrilateral $ABCD$ joined along $CD$ to a triangle $CDE$ .
In triangle $CDE$ , $CD = CE$ .
$D$ , $E$ and $F$ lie on one straight line.

Work out the size of angle $BCE$ .
Give a reason for each stage of your working. [5 marks]

angle BCE =

[Total 5 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

ABCD

is a quadrilateral, and triangle

CDE

is joined to it along

CD

CD = CE

, so triangle

CDE

is isosceles.

D

E

and

F

lie on one straight line.

Marked on the figure:

\angle DAB = 78^\circ

\angle ABC = 128^\circ

\angle ADC = 86^\circ

and

\angle CEF = 125^\circ

Find

The size of

\angle BCE

, the whole angle at

C

between

CB

and

CE

. A reason is wanted at every stage, so each line of working names the angle fact it uses.

Plan the Solution

Start at $E$ . The straight line $DEF$ turns the $125^\circ$ into $\angle CED$ .
$CD = CE$ copies that angle across to $\angle CDE$ , and the triangle then gives $\angle DCE$ .
The four angles of $ABCD$ give the other piece, $\angle BCD$ .
$CD$ lies inside the angle being asked for, so the two pieces at $C$ add.

Worked Solution [5 marks]

Angles on a straight line add to

180^\circ

, angles in a triangle add to

180^\circ

, and angles in a quadrilateral add to

360^\circ

Step 1: the angle of the triangle at

E

\angle CED = 180^\circ - 125^\circ = 55^\circ

(Reason: Angles on a straight line add to

180^\circ

, and

DEF

is a straight line, so

\angle CED

and the marked

125^\circ

together make

180^\circ

Step 2: the other base angle of the triangle

\angle CDE = \angle CED = 55^\circ

(Reason:

CD = CE

, so triangle

CDE

is isosceles and the two base angles, at

D

and at

E

, are equal.)

Step 3: the angle of the triangle at

C

\angle DCE = 180^\circ - 55^\circ - 55^\circ = 70^\circ

(Reason: Angles in a triangle add to

180^\circ

, so take both base angles of triangle

CDE

away from

180^\circ

Step 4: the fourth angle of the quadrilateral

\angle BCD = 360^\circ - (78^\circ + 128^\circ + 86^\circ)

\angle BCD = 360^\circ - 292^\circ = 68^\circ

(Reason: Angles in a quadrilateral add to

360^\circ

, and the other three angles of

ABCD

are

78^\circ

128^\circ

and

86^\circ

Step 5: put the two angles at

C

together

\angle BCE = \angle BCD + \angle DCE

\angle BCE = 68^\circ + 70^\circ = 138^\circ

(Reason:

CD

lies inside

\angle BCE

, so

\angle BCD

and

\angle DCE

together make the whole angle at

C

\angle BCE = 138^\circ

Verification

Check 1: Put the

68^\circ

back into the quadrilateral and add all four of its angles.

78^\circ + 128^\circ + 68^\circ + 86^\circ = 360^\circ

Check 2: Add the three angles of triangle

CDE

, then check the straight line at

E

as well.

55^\circ + 55^\circ + 70^\circ = 180^\circ

and

55^\circ + 125^\circ = 180^\circ

Check 3: A different route to the same angle.

ABCED

is a pentagon, so its five angles add to

540^\circ

. Its angle at

D

is the whole of

\angle ADE = 86^\circ + 55^\circ = 141^\circ

, and its angle at

C

is the angle being found.

540^\circ - 78^\circ - 128^\circ - 55^\circ - 141^\circ = 138^\circ

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\angle CED = 180^\circ - 125^\circ = 55^\circ$ or $\angle BCD = 360^\circ - (86^\circ + 78^\circ + 128^\circ) = 68^\circ$	M1	Either one of these two starting angles earns the mark. Angles may be marked on the diagram or labelled unambiguously in the working.	✓
$\angle DCE = 180^\circ - (2 \times 55^\circ) = 70^\circ$	M1	The apex angle of the isosceles triangle, from the candidate value for $\angle CED$ . Angles may be marked on the diagram or labelled unambiguously in the working.	✓
$\angle BCE = 68^\circ + 70^\circ = 138^\circ$	A1	For $138$ . A correct answer of $138$ scores these 3 marks unless it comes from obvious incorrect working.	✓
Reasons: angles on a straight line add to $180^\circ$ ; isosceles triangle; angles in a triangle add to $180^\circ$ ; angles in a quadrilateral add to $360^\circ$ .	B2	Dependent on both method marks, for all correct reasons for the method used. Award B1 instead, dependent on one method mark, for one correct reason for the method used.	✓

Full marks: 5/5

Rory has some marbles in a box.
He has $n$ red marbles.
He has twice as many blue marbles as red marbles.
He has $7$ more green marbles than red marbles.

(a) Write an expression, in terms of $n$ , for the total number of red marbles, blue marbles and green marbles that Rory has.
Write your answer in its simplest form. [2 marks]

The total number of marbles that Rory has is $T$

(b) Write an expression, in terms of $T$ and $n$ , for the number of marbles that Rory has that are not red marbles, blue marbles or green marbles. [1 mark]

(a)(b)

[Total 3 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

Red marbles:

n

Blue marbles: twice as many as red, so

2n

Green marbles:

7

more than red, so

n + 7

The total number of marbles in the box is

T

Find

(a) An expression in terms of

n

for the number of red, blue and green marbles altogether, in its simplest form (b) An expression in terms of

T

and

n

for how many of the marbles are not red, blue or green

Plan the Solution

Write each colour as an expression in $n$ before adding anything.
Read "twice as many" as $2 \times n$ and " $7$ more than" as $n + 7$ , so the green marbles carry an $n$ as well as the $7$ .
Add the three expressions, then collect the $n$ terms together and the number terms together.
For part (b), the marbles of any other colour are what is left when the three colours are taken away from $T$ , so subtract the whole of the part (a) expression.

Worked Solution [3 marks]

Rule - Collecting like terms: write every quantity in terms of the same letter, add them, then add the

n

terms together and the number terms together. Only like terms may be combined.

Step 1: Write each colour in terms of

n

\text{red} = n

\text{blue} = 2 \times n = 2n

\text{green} = n + 7

(Reason: (Reason: "twice as many blue as red" doubles the red amount, and "

7

more green than red" adds

7

to it. Notice that green is

n + 7

, not just

7

- the

n

is still there.))

Step 2: Add the three amounts

\text{total} = n + 2n + (n + 7)

(Reason: (Reason: the total of the three colours is red plus blue plus green. The brackets keep the green expression together while it is written down.))

Step 3: Collect the like terms

n + 2n + n = 4n

\text{total} = 4n + 7

(Reason: (Reason: three lots of

n

come from red, blue and green, so their coefficients

1

2

and

1

add up. The

7

has no

n

with it, so it cannot be combined with them and stays as it is.))

Step 4: (b) Take the three colours away from

T

\text{other colours} = T - (4n + 7)

(Reason: (Reason: every marble is either one of the three colours or it is not, so the rest are

T

minus the number that are red, blue or green. The brackets matter: the whole of

4n + 7

is subtracted, which is the same as

T - 4n - 7

.))

(a)

4n + 7

(b)

T - (4n + 7)

Verification

Check 1: Put

n = 5

into the words of the question:

5

red, twice as many blue so

10

, and

7

more green than red so

12

. Add them, then compare with the expression.

5 + 10 + 12 = 27

and

4 \times 5 + 7 = 27

Check 2: Repeat with a different value,

n = 10

, so that a rule which only happens to work once cannot pass:

10

red,

20

blue,

17

green.

10 + 20 + 17 = 47

and

4 \times 10 + 7 = 47

Check 3: Test part (b) on real numbers. Take

n = 5

and a box holding

T = 60

marbles. Count the three colours, then count what is left.

60 - 27 = 33

marbles of other colours, and

T - (4n + 7) = 60 - (20 + 7) = 33

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Write the blue marbles as $2n$ , or the green marbles as $n + 7$	M1	for $2n$ oe or $n + 7$ oe	✓
(a) Add and collect the like terms	A1	for $4n + 7$ oe, eg $7 + 4n$ . A correct answer scores full marks unless it comes from obviously incorrect working	✓
(b) Subtract the total of the three colours from $T$	B1ft	for $T - (4n + 7)$ oe, eg $T - 4n - 7$ . Follow through $T$ minus the candidate's own answer to part (a)	✓

Full marks: 3/3

Continue to questions 14 to 24

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

Frequently asked questions

There are 24 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 1FR 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 1FR, November 2024: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 13 of 24

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

Frequently asked questions

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