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

Sir Faraz Hassan

31 Jul 2026

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

Original worked solutions for Edexcel International GCSE Mathematics A (4MA1), Paper 1F (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 14 of 26

The table gives information about the total length, in kilometres, of the road network on each of seven islands.

(a) Which of these seven islands has the longest road network? [1 mark]

(b) Write the number $6124$ in words. [1 mark]

The road network on Corvel is longer than the road network on Northwick.

(d) Write down the value of the $6$ in the number $4763$ [1 mark]

Two numbers in the table round to $4000$ when written correct to the nearest thousand.

(e) Write down these two numbers. [1 mark]

(a)(b)(c) km(d)(e)and

[Total 5 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

Seven islands, and the length in kilometres of the road network on each:

8947

3808

13\,600

4763

6124

2758

and

4361

Corvel holds the

8947

km network and Northwick the

2758

km one.

Find

(a) The island with the longest road network. (b) The number

6124

written out in words. (c) How much longer Corvel's network is than Northwick's. (d) What the

6

4763

is worth. (e) The two lengths that round to

4000

to the nearest thousand.

Plan the Solution

(a) Compare by place value. A five-digit number beats every four-digit one, so count digits first.
(b) Split $6124$ into thousands, hundreds, tens and ones, then read the columns from the left.
(c) How much longer is a difference, so subtract the smaller length from the larger.
(d) Find which column the $6$ stands in and give it that column's value.
(e) Round each length to the nearest thousand and keep the ones that land on four thousand.

Worked Solution [5 marks]

Every digit is worth its face value times the column it stands in, so

4763 = 4000 + 700 + 60 + 3

. Comparing, subtracting and rounding whole numbers all work column by column, starting from the largest column.

(a) Pick out the longest network

2758, \; 3808, \; 4361, \; 4763, \; 6124, \; 8947, \; 13\,600

(Reason: Put the seven lengths in order. Six of them have four digits and only

13\,600

has five, so it is the largest. That row is Eskvale.)

(b) Read 6124 out of its columns

6124 = 6000 + 100 + 20 + 4

(Reason: Six thousands, one hundred, two tens and four ones. Reading the columns from the left gives six thousand one hundred and twenty-four.)

8947 - 2758 = 6189

(Reason: Corvel's network is

8947

km and Northwick's is

2758

km, so take the smaller from the larger. Both the units column and the tens column need a borrow.)

(d) Give the 6 the value of its column

4763 = 4000 + 700 + 60 + 3

(Reason: The

6

stands in the tens column, so it is worth six tens. That is

60

, not

6

(e) Round every length to the nearest thousand

3808 \to 4000

4361 \to 4000

4763 \to 5000

2758 \to 3000

(Reason: The hundreds digit decides the rounding, so a length rounds to

4000

exactly when it lies from

3500

up to

4500

. Only

3808

and

4361

do;

4763

is past

4500

, so it rounds to

5000

(a) Eskvale(b) six thousand one hundred and twenty-four(c)

6189

km(d)

60

(e)

3808

and

4361

Verification

Check 1: Add the difference back on to Northwick's length. It should come back to Corvel's.

2758 + 6189 = 8947

Check 2: Rebuild

4763

from its columns with the

6

put back in as

60

4000 + 700 + 60 + 3 = 4763

Check 3: Read the words of part (b) back as a number: six thousand, one hundred, twenty-four.

6000 + 100 + 24 = 6124

Check 4: Count how many of the seven lengths lie from

3500

up to

4500

. Exactly two do,

3808

and

4361

, which is the number of answers part (e) asks for.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	Eskvale. Accept $13\,600$ .	✓
(b)	B1	Six thousand one hundred and twenty four. A hyphen in twenty-four is accepted.	✓
(c)	B1	$6189$ and nothing else. Units are not required.	✓
(d)	B1	$60$ . Accept $6$ tens, or sixty.	✓
(e)	B1	$3808$ and $4361$ , in either order. Accept Tarnholm and Greymoor.	✓

Full marks: 5/5

(a) Write $10p - 4p + 9p$ in its simplest form. [1 mark]

(b) Simplify fully $9 \times 4q$ . [1 mark]

(a)(b)(c) r =

[Total 3 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

10p - 4p + 9p

- three terms, every one of them a multiple of

p

9 \times 4q

- a number multiplied by a term in

q

4r = 15

- a one-step equation, with

r

multiplied by

4

Find

The simplest form of

10p - 4p + 9p

. The simplest form of

9 \times 4q

. The value of

r

Plan the Solution

Part (a) is a collecting exercise: the terms are like terms, so combine the coefficients and keep the $p$ .
Part (b) is a multiplication: multiply the two numbers and keep the $q$ . There is no adding to do here.
Part (c) is a one-step equation: apply the inverse of multiplying by $4$ to both sides.
Each part is worth one mark, so each answer is the whole of the work - no method marks are on offer.

Worked Solution [3 marks]

Rule - Like terms are collected by combining their coefficients and keeping the letter; a term is multiplied by a number by multiplying its coefficient; and an equation is undone by the inverse operation, applied to both sides.

Step 1: (a) Combine the coefficients

10p - 4p + 9p = (10 - 4 + 9)p

(10 - 4 + 9)p = 15p

(Reason: Every term is a multiple of

p

, so they are like terms. Only the coefficients

10

-4

and

9

are combined; the letter is carried through unchanged, so the answer is a term in

p

and not a number.)

Step 2: (b) Multiply the numbers, keep the letter

9 \times 4q = 9 \times 4 \times q

9 \times 4 \times q = 36q

(Reason: Multiplication can be done in any order, so the two numbers

9

and

4

are multiplied first. Nothing is added here:

9 \times 4q

is nine lots of

4q

, which is

36q

Step 3: (c) Undo the multiplication

4r = 15

r = \dfrac{15}{4} = 3.75

(Reason: The

4

and the

r

are multiplied together, so the inverse of that multiplication is applied to both sides to leave

r

on its own. The answer does not have to be a whole number.)

(a)

15p

(b)

36q

(c)

r = 3.75

Verification

Check 1: Put

p = 2

into the original expression and into the answer. The original gives

10(2) - 4(2) + 9(2) = 20 - 8 + 18

, and the answer gives

15(2)

30 = 30

, so the two expressions agree.

Check 2: Put

q = 5

into the original expression and into the answer. The original gives

9 \times 4(5) = 9 \times 20

, and the answer gives

36(5)

180 = 180

, so the two expressions agree.

Check 3: Substitute the solution back into the equation: work out

4 \times 3.75

and compare it with

15

4 \times 3.75 = 15

, which is the right-hand side.

Check 4: Read the answer as a division with a remainder instead:

15

divided by

4

3

remainder

3

, so

r = 3\dfrac{3}{4}

3\dfrac{3}{4} = \dfrac{15}{4} = 3.75

, the same value in all three forms the mark scheme accepts.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Collect the like terms in $p$	B1	$15p$ with no further working needed. The letter must be present: $15$ on its own is not the simplified expression.	✓
(b) Multiply the coefficient by $9$	B1	$36q$ . Adding instead of multiplying gives $13q$ , which earns nothing.	✓
(c) Divide both sides by $4$	B1	$3.75$ , or any equivalent form: $\dfrac{15}{4}$ and $3\dfrac{3}{4}$ are both accepted.	✓

Full marks: 3/3

A tin contains $10$ marbles.
$5$ marbles are black
$3$ marbles are green
$2$ marbles are red

Priya is going to take at random a marble from the tin.

(a) On the probability scale, mark with a cross the probability that the marble is black. [1 mark]

(b) On the probability scale, mark with a cross the probability that the marble is orange. [1 mark]

Lukas has three tubs of beads, $A$ , $B$ and $C$
He tries to find the probability of taking at random a white bead from each tub.
He writes his probabilities in a table.

The probability that Lukas writes for tub $C$ is incorrect.

(c)

[Total 3 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

A tin of

10

marbles:

5

black,

3

green and

2

red. One marble is taken at random, so every marble is equally likely.

Two probability scales, each running from

0

1

in ten equal steps, to be marked with a cross.

Three probabilities written for three different tubs of beads:

0.7

0.45

and

1.2

Find

The probability that the marble is black, shown as a cross on the first scale. The probability that the marble is orange, shown as a cross on the second scale. A reason why

1.2

cannot be a probability at all.

Plan the Solution

Parts (a) and (b) are answered ON the scale, so each answer is a cross rather than a written number. Work the probability out first, then find where it sits between $0$ and $1$ .
For (a), count the black marbles out of the total and write the fraction in its simplest form.
For (b), check whether the tin holds any orange marbles at all. The three colours already account for all $10$ marbles, so the event is impossible.
For (c), compare $1.2$ with the two ends of the probability scale. One sentence earns the mark, and it must say what is wrong with the value itself.

Worked Solution [3 marks]

Rule - When the outcomes are equally likely, the probability of an event is the number of ways it can happen divided by the total number of outcomes, and the answer is always a number from

0

, which means impossible, to

1

, which means certain.

Step 1: (a) Write the probability of black as a fraction

P(\text{black}) = \dfrac{5}{10}

\dfrac{5}{10} = \dfrac{1}{2} = 0.5

(Reason: The tin holds

10

marbles and one is taken at random, so each marble is equally likely. The probability of black is therefore the number of black marbles over the total number of marbles, and that fraction is then written in its simplest form.)

Step 2: (a) Mark the cross halfway along the scale

\dfrac{1}{2} = 0.5

(Reason: A probability of

\dfrac{1}{2}

sits exactly halfway between

0

and

1

, which is the tick already labelled

\dfrac{1}{2}

in the middle of the scale. The cross goes on the scale line itself, not above it or below it.)

Step 3: (b) Count the orange marbles

5 + 3 + 2 = 10

P(\text{orange}) = \dfrac{0}{10}

(Reason: The black, green and red marbles already account for every one of the

10

marbles in the tin, so there is no orange marble to take. The number of ways the event can happen is

0

Step 4: (b) Mark the cross at the left-hand end

\dfrac{0}{10} = 0

(Reason: An event that cannot happen has probability

0

, and

0

is the left-hand end of the scale. An impossible event is still marked on the scale, not off it.)

Step 5: (c) Test the value written for tub C against the scale

1.2 - 1 = 0.2

1.2 \times 100 = 120

(Reason: The largest a probability can be is

1

, which means certain, so

1.2

lies

0.2

beyond the right-hand end of the scale and no event can have it. Read as a percentage it is

120

per cent, and nothing is more than

100

per cent certain. The values

0.7

and

0.45

both sit on the scale, so those two are not the problem.)

(a) a cross on the scale at

\dfrac{1}{2}

(b) a cross on the scale at

0

(c)

1.2

is greater than

1

, and a probability can never be greater than

1

Verification

Check 1: Add the probabilities of the three colours that are in the tin. If every marble has been counted once, they must total

1

\dfrac{5}{10} + \dfrac{3}{10} + \dfrac{2}{10} = 1

, so nothing has been missed and there is no room left for an orange marble.

Check 2: Work part (a) the other way round. The marbles that are not black are the

3

green and the

2

red, so subtract that probability from

1

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

, the same probability, so black and not-black are equally likely.

Check 3: Change part (a) into a decimal and a percentage, then read the position off the scale: halfway along ought to be

50

per cent.

\dfrac{5}{10} = 0.5

, which is

50

per cent, and the middle tick of a scale from

0

1

is exactly where the cross was put.

Check 4: Try to mark all three of Lukas's values on a probability scale and see which one has nowhere to go.

0.7

and

0.45

both land on the scale, while

1.2 - 1 = 0.2

shows that the third lands past the end of it.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A cross on the probability scale at $\dfrac{1}{2}$	B1	The cross must be at $\dfrac{1}{2}$ , the middle tick, and on the scale line. Correct answer only, so a cross anywhere else scores nothing and no working can earn the mark.	✓
(b) A cross on the probability scale at $0$	B1	The cross must be at $0$ , the left-hand end. Correct answer only. Leaving the scale blank because the event cannot happen scores nothing: $0$ is itself a probability.	✓
(c) A correct reason	B1	Any statement that a probability cannot be more than $1$ or equivalent: it is over $1$ ; it is more than $100$ per cent; probability runs from $0$ to $1$ ; $1.2$ is impossible; it has to be $1$ or less. A contradictory answer scores nothing, and so does saying only that the value is too high. Arguing that the probabilities should add up to $1$ also scores nothing - these three are for three different tubs, so their total means nothing at all.	✓

Full marks: 3/3

The diagram shows a polygon.

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

(b) Write down the number that the arrow points to on the number line below. [1 mark]

Here are four clock faces, labelled $A$ , $B$ , $C$ and $D$ .

(d) Complete this sentence by writing a suitable metric unit on the answer line.
The length of a large cruise ship is $300$ [1 mark]

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

[Total 4 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

A polygon drawn from eight straight sides of equal length.

A number line whose labelled marks are

100

200

300

and

400

, with five equal small divisions between one labelled mark and the next, and an arrow pointing at a mark between

200

and

300

Four clock faces,

A

B

C

and

D

, each showing a different time.

A sentence giving the length of a large cruise ship as

300

of a metric unit that has been left out.

Find

The mathematical name of the polygon. The number the arrow points to. The letter of the clock face that shows quarter to five. The metric unit that completes the sentence.

Plan the Solution

Name the polygon by counting its sides. The name comes from the number of sides, so count them one at a time round the outline rather than judging by the look of the shape.
On a number line, work out what ONE small division is worth before reading anything off it: divide the gap between two labelled marks by the number of divisions inside that gap, then count on from the nearer label.
Read a clock in two halves. The long hand gives the minutes and the short hand gives the hour. Quarter to means the long hand is at $9$ , and the hour is the one the short hand is heading towards, not the one it has just left.
For the unit, test each candidate in turn: turn $300$ of it into metres and ask whether a ship could really be that long.

Worked Solution [4 marks]

Rule - On a number line one small division is worth the gap between two labelled marks divided by the number of divisions in it; on a clock face the long hand counts

5

minutes for every number it passes; and a polygon is named by how many sides it has.

Step 1: (a) Count the sides of the polygon

4 + 4 = 8

(Reason: Four of the edges run straight across or straight up and down - the top, the bottom and the two ends - and four more cut across the corners, so the outline is made of

8

straight sides. A polygon with

8

sides is an octagon.)

Step 2: (b) Work out what one small division is worth

\dfrac{100}{5} = 20

(Reason: From

100

200

the line climbs by

100

, and that gap is split into

5

equal divisions. Count the GAPS and not the ticks: between the two labels there are

5

gaps but only

4

ticks, so each division is worth

20

Step 3: (b) Count on from 200 to the arrow

200 + 2 \times 20 = 240

(Reason: The arrow stands

2

small divisions to the right of

200

, and Step 2 found each division to be worth

20

, so add

2

lots of

20

on to

200

Step 4: (c) Read the four clock faces

9 \times 5 = 45

60 - 45 = 15

(Reason: Every number on the dial stands for

5

minutes, so a long hand at

9

means

45

minutes past the hour, which leaves

15

minutes - a quarter of an hour - still to run. Quarter to five therefore needs the long hand at

9

AND the short hand between

4

and

5

, on its way to

5

. Only clock

C

has both. Clock

A

has its long hand at

9

but its short hand between

5

and

6

, so that face is quarter to six, and clocks

B

and

D

have the long hand at

3

, which is quarter PAST.)

Step 5: (d) Test each metric unit for length

\dfrac{300}{1000} = 0.3

\dfrac{300}{100} = 3

300 \times 1000 = 300\,000

(Reason: Turn

300

of each unit into metres.

300

millimetres is only

0.3

of a metre and

300

centimetres is only

3

metres, both far too short for a ship, while

300

kilometres is

300\,000

metres, which is longer than many countries are wide. Metres is the one unit left, and

300

metres is about right for a large cruise ship.)

(a) octagon(b)

240

(c)

C

(d) metres

Verification

Check 1: Test the count of sides a different way. The angles inside a polygon with

n

sides add up to

(n - 2) \times 180

degrees, so work that out for

8

sides and share it equally between the corners.

(8 - 2) \times 180 = 1080

and

\dfrac{1080}{8} = 135

, so each corner of the figure is a

135

degree angle, which is exactly what the corners of a regular octagon measure.

Check 2: Read the number line from the other end. Counting back, the arrow stands

3

small divisions short of

300

300 - 3 \times 20 = 240

, the same number, so the reading does not depend on which label it was counted from.

Check 3: Come at the time from the words instead of from the dial. A quarter of an hour is

60

minutes shared into

4

, and quarter TO means that many minutes before the next hour, so work out which number on the dial the long hand must reach.

\dfrac{60}{4} = 15

minutes before the hour, which is

60 - 15 = 45

minutes past it, and

\dfrac{45}{5} = 9

puts the long hand on the

9

- which is where clock

C

has it, with its short hand not yet at

5

Check 4: Measure the ship in people rather than in units. A grown adult is about

2

metres tall, so divide the length by

2

for each candidate unit and see which count is believable.

\dfrac{300}{2} = 150

adults lying end to end if the unit is metres, which is a believable ship, against

\dfrac{300\,000}{2} = 150\,000

adults if the unit is kilometres, which is not.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) octagon	B1	Correct answer only. The word octagon on its own earns the mark, and regular octagon earns it too. Naming any other polygon scores nothing, and so does describing the shape without naming it.	✓
(b) $240$	B1	Correct answer only. $240$ with no working still earns the mark. The two wrong readings this question is built to catch are $220$ , from counting one division instead of two, and $250$ , from sharing the $100$ between $4$ divisions instead of $5$ . Neither earns anything.	✓
(c) $C$	B1	A lower case c is accepted as well as a capital $C$ . The question asks for a letter, so it is the letter that must be written down. Clock $A$ is the near miss, since it also has its long hand at $9$ .	✓
(d) metres	B1	The abbreviation m is accepted as well as the word metres. Any other unit scores nothing: the sentence measures a length, so a unit of mass or of capacity is wrong outright, and centimetres or kilometres would make the ship $3$ metres or $300\,000$ metres long.	✓

Full marks: 4/4

Here is a list of six numbers.

$2 \qquad 8 \qquad 9 \qquad 18 \qquad 24 \qquad 28$

(a) Using only the numbers in this list, write down
(i) an odd number
[1 mark]
(ii) a number that is a multiple of $4$ and also a multiple of $6$
[1 mark]
(iii) a cube number
[1 mark]
(iv) a prime number [1 mark]

(b) Work out the value of
$6^2 + 2^3 \times 5$ [1 mark]

(a)(i)(a)(ii)(a)(iii)(a)(iv)(b)

[Total 5 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

A list of six numbers:

2, \; 8, \; 9, \; 18, \; 24, \; 28

For part (b), the calculation

6^2 + 2^3 \times 5

, which mixes powers, a multiplication and an addition.

Find

(a) One number from the list for each of four properties: odd, a multiple of both numbers, a cube, and prime. Each answer must be taken from the list, not invented. (b) The value of

6^2 + 2^3 \times 5

Plan the Solution

Part (a) is answered by testing the list, not by recalling a fact. Take one property at a time and run every number in the list past it.
Odd means it will not divide by $2$ . A multiple of both $4$ and $6$ is a multiple of $12$ , their lowest common multiple. A cube is a whole number times itself times itself. A prime has exactly two factors.
Each property is matched by exactly one number in this list, so each part of (a) has a single right answer.
Part (b) is order of operations: powers first, then the multiplication, then the addition last.

Worked Solution [5 marks]

Rule - Number properties, then order of operations: a number is odd if it leaves a remainder of

1

on division by

2

; it is a cube if it equals

k \times k \times k

for some whole number

k

; it is prime if it has exactly two factors. In a calculation, work out powers first, then multiplication, and addition last.

Step 1: (a)(i) test the list for an odd number

2, 8, 18, 24, 28 \text{ all divide by } 2

9 = 2 \times 4 + 1

(Reason: An odd number leaves a remainder of

1

when it is divided by

2

. Five of the six numbers are even, so

9

is the only odd number available.)

Step 2: (a)(ii) test the list for a multiple of both

4

and

6

\text{LCM of } 4 \text{ and } 6 = 12

24 = 12 \times 2

24 = 4 \times 6 = 6 \times 4

(Reason: A number in both times tables is a multiple of the lowest common multiple,

12

. Only

24

is a multiple of

12

here. The near misses are worth naming:

8

and

28

are multiples of

4

but not of

6

, and

18

is a multiple of

6

but not of

4

Step 3: (a)(iii) test the list for a cube number

8 = 2 \times 2 \times 2 = 2^3

1^3 = 1

3^3 = 27

(Reason: A cube number is a whole number multiplied by itself and then by itself again. The cubes around this list are

1

8

and

27

, and only

8

appears in the list.)

Step 4: (a)(iv) test the list for a prime number

2 \text{ has factors } 1 \text{ and } 2 \text{ only}

9 = 3 \times 3

8, 18, 24, 28 \text{ are even and larger than } 2

(Reason: A prime number has exactly two factors, itself and

1

. Any even number larger than

2

has

2

as a third factor, and

9

has the extra factor

3

, so

2

is the only prime in the list. It is the one even prime, which is why it is easy to miss.)

Step 5: (b) work out

6^2 + 2^3 \times 5

6^2 = 36

2^3 \times 5 = 8 \times 5 = 40

6^2 + 2^3 \times 5 = 36 + 40 = 76

(Reason: Powers are worked out first, then the multiplication, and the addition comes last. Adding first would give

44 \times 5 = 220

, a completely different value, which is why the order matters.)

(a)(i)

9

(a)(ii)

24

(a)(iii)

8

(a)(iv)

2

(b)

76

Verification

Check 1: Go back to the list and test all six numbers against all four properties, rather than only checking the four answers chosen. Odd: only

9

. Multiple of

12

: only

24

. Cube: only

8

. Prime: only

2

. Each property is matched once, so every answer in part (a) is forced.

Check 2: Redo part (b) a different way. Both terms are multiples of

4

, since

36 = 4 \times 9

and

40 = 4 \times 10

, so the

4

can be taken out.

4 \times (9 + 10) = 4 \times 19 = 76

, the same value as before.

Check 3: Replace the multiplication with repeated addition:

2^3 \times 5

means five lots of

8

8 + 8 + 8 + 8 + 8 = 40

and

36 + 40 = 76

, so the answer to (b) is confirmed without multiplying.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) writes down $9$	B1	cao. $9$ is the only odd number in the list. Working is not required, so a correct answer alone scores the mark.	✓
(a)(ii) writes down $24$	B1	cao. Accept $24$ only. $8$ or $28$ (a multiple of $4$ alone) and $18$ (a multiple of $6$ alone) score no mark.	✓
(a)(iii) writes down $8$	B1	cao. $8 = 2^3$ . A square number such as $9$ scores no mark.	✓
(a)(iv) writes down $2$	B1	cao. $2$ is the only prime in the list; $9$ has the factor $3$ and the rest are even and larger than $2$ .	✓
(b) writes down $76$	B1	cao. Working is not required, so a correct answer scores full marks. $220$ (adding before multiplying) and $66$ (reading $2^3$ as $2 \times 3$ ) score no mark.	✓

Full marks: 5/5

The first diagram shows two straight lines that meet at a point.
The second diagram shows the quadrilateral $ABDE$ together with the isosceles triangle $BCD$ , in which $BD = DC$ .
$ABC$ is a straight line.

(a) (i) Find the value of $x$ .
[1 mark]
(ii) Give the reason for your answer. [1 mark]

(b) Find the value of $y$ . [3 marks]

(a)(i) x =(a)(ii)(b) y =

[Total 5 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

Two straight lines meet at a point. One angle at that point is

235^{\circ}

and the other is

x^{\circ}

ABDE

is a quadrilateral,

BCD

is an isosceles triangle with

BD = DC

, and

ABC

is a straight line.

In the second figure, angle

EAB = 90^{\circ}

, angle

AED = 98^{\circ}

and angle

DCB = 54^{\circ}

Find

The value of

x

, the angle fact that gives it, and the value of

y

Plan the Solution

Part (a): the two marked angles are the only angles at that point, so together they must make one complete turn of $360^{\circ}$ .
Part (b): the isosceles triangle gives the angle at $B$ inside the triangle; the straight line $ABC$ turns that into $\angle ABD$ , which is the quadrilateral's angle at $B$ .
Then use the four angles of $ABDE$ . Keep to that order: $y$ cannot be found until $\angle ABD$ is known.

Worked Solution [5 marks]

Angles around a point add up to

360^{\circ}

. Angles on a straight line add up to

180^{\circ}

. The base angles of an isosceles triangle are equal, and the four angles of a quadrilateral add up to

360^{\circ}

Step 1: one complete turn at the point

x + 235 = 360

360 - 235 = 125

(Reason: the two marked angles are the only angles at that point, and a complete turn is

360^{\circ}

)

Step 2: say which angle fact was used

235 + 125 = 360

(Reason: part (a)(ii) wants the fact, not the arithmetic: angles around a point add up to

360^{\circ}

)

Step 3: the angles of isosceles triangle BCD

\angle DBC = \angle DCB = 54^{\circ}

180 - 54 - 54 = 72

(Reason:

BD = DC

, so the angles opposite those two equal sides are equal, and the apex angle

\angle BDC

is whatever is left of

180^{\circ}

)

Step 4: the quadrilateral's angle at B

\angle ABD = 180 - \angle DBC

180 - 54 = 126

(Reason:

ABC

is a straight line, so

\angle ABD

and

\angle DBC

together make

180^{\circ}

. This obtuse angle is the one inside

ABDE

, not the

54^{\circ}

inside the triangle)

Step 5: the four angles of quadrilateral ABDE

\angle EAB + \angle ABD + \angle BDE + \angle DEA = 360

90 + 126 + 98 = 314

360 - 314 = 46

(Reason: the four angles of any quadrilateral add up to

360^{\circ}

, so

y

is what the other three leave)

(a)(i)

x = 125

(a)(ii) Angles around a point add up to

360^{\circ}

(b)

y = 46

Verification

Check 1: Put the answer to part (a) back beside the given angle and add them

235 + 125 = 360

, exactly one complete turn

Check 2: Add all four angles of

ABDE

with

y = 46

in place

90 + 126 + 46 + 98 = 360

Check 3: Take a different route entirely.

AEDC

is also a quadrilateral, since its fourth side lies along the straight line

ABC

. Its angle at

D

y

plus the apex angle

72^{\circ}

, and

46 + 72 = 118

, so its four angles are

90^{\circ}

98^{\circ}

118^{\circ}

and

54^{\circ}

90 + 98 + 118 + 54 = 360

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) $x = 125$	B1	cao. No working is required, so $125$ on its own scores the mark.	✓
(a)(ii) correct reason	B1	Angles around a point add up to $360^{\circ}$ . The words 'around a point' (or 'at a point') and the value $360$ must both appear.	✓
(b) $\angle ABD = 180 - 54 = 126$ or $\angle BDC = 180 - 2 \times 54 = 72$ or $\angle BDC = 180 - 108 = 72$	M1	Any one of these three. If angles are written on the diagram they must be correctly assigned, and if angle notation is used it must be correctly assigned.	✓
(b) $360 - (98 + 90 + 126)$ or $360 - (98 + 90 + 54 + 72)$ or $360 - 314$	M1	For a complete method that reaches $y$ . The quotation marks the mark scheme puts round $126$ and $72$ mean the candidate's own value from the previous mark may be used here.	✓
(b) $y = 46$	A1	cao	✓
Note	(no mark)	Working is not required in part (b), so a correct answer scores full marks unless it comes from obviously incorrect working. A common wrong answer is $100$ , which is what putting the apex angle $72^{\circ}$ into the quadrilateral in place of $\angle ABD = 126^{\circ}$ gives.	✓

Full marks: 5/5

Douglas works at a bottling plant.

His normal hourly rate of pay is £ $14$
His overtime hourly rate of pay is £ $21$

Douglas is paid the normal hourly rate of pay for $35$ hours in one week.
His total pay for this week is £ $679$

Work out the number of hours of overtime he works in this week. [4 marks]

[Total 4 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

Normal rate of pay: £

14

for each hour

Overtime rate of pay: £

21

for each hour

35

hours of the week are paid at the normal rate

Total pay for the week: £

679

Find

The number of hours of overtime worked in this week

Plan the Solution

Work out the pay earned at the normal rate for the $35$ hours.
Take that away from the total pay, so that what is left is the pay earned at the overtime rate.
Divide the overtime pay by the overtime rate to turn it back into a number of hours.

Worked Solution [4 marks]

Rule - Pay in two parts:

\text{total pay} = \text{normal hours} \times \text{normal rate} + \text{overtime hours} \times \text{overtime rate}

Step 1: The pay earned at the normal rate

35 \times 14 = 490

(Reason: (Reason: Douglas is paid £

14

for each of the

35

hours at the normal rate, so £

490

of his pay has nothing to do with overtime.))

Step 2: The pay that is left for overtime

679 - 490 = 189

(Reason: (Reason: the week's pay is only these two parts, so everything above the £

490

must have been earned at the overtime rate. That leaves £

189

of overtime pay.))

Step 3: Turn the overtime pay back into hours

\dfrac{189}{21} = 9

(Reason: (Reason: each hour of overtime is worth £

21

, so dividing the overtime pay by

21

counts how many overtime hours it took to earn it.))

9

hours of overtime

Verification

Check 1: Build the week's pay forwards from the answer:

35

hours at £

14

gives

490

, and

9

hours at £

21

gives

189

490 + 189 = 679

, the total pay the question gives

Check 2: Count the week a different way. Douglas works

35 + 9 = 44

hours altogether, and every hour earns at least £

14

, while each overtime hour earns an extra

21 - 14 = 7

pounds on top:

44 \times 14 = 616

and

9 \times 7 = 63

616 + 63 = 679

, the same total by a route that never uses the £189

Check 3: Check the answer is sensible. Overtime is paid in whole hours here, and

189

is a multiple of

21

, so the division has to come out exactly.

9

is a whole number of hours, and

9 \times 21 = 189

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$35 \times 14 = 490$	M1	Method for the pay earned at the normal hourly rate.	✓
$679 - 490 = 189$	M1	Method for the part of the total pay earned at the overtime rate, allowing their own value in place of $490$ . The official scheme also awards this mark and the next one together, as M2, for $679 - 9 \times 21$ or $490 + 9 \times 21$ or equivalent.	✓
$\dfrac{189}{21}$	M1	Method for turning the overtime pay into hours, allowing their own value in place of $189$ , divided by the overtime rate.	✓
$9$	A1	Working not required, so a correct answer scores full marks, unless it comes from obvious incorrect working.	✓

Full marks: 4/4

(a) Simplify the expression $10x - 7y - 6x + 4y$ [2 marks]

You are given the formula $T = 4d - 6e$

(b) Work out the value of $T$ when $d = 13$ and $e = 7$ [2 marks]

(a)(b) T =(c) p =

[Total 6 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

(a) The expression

10x - 7y - 6x + 4y

, which has two

x

terms and two

y

terms.

(b) The formula

T = 4d - 6e

, together with

d = 13

and

e = 7

5p + 11 = 28

, which has one unknown letter.

Find

(a) The expression written with as few terms as possible. (b) The value of

T

. (c) The value of

p

Plan the Solution

(a) Group the terms that carry the same letter, then combine their number parts, keeping the sign that sits in front of each term.
(b) Replace $d$ and $e$ by their values, work out the two products, then subtract.
(c) Undo the $+11$ first, then undo the multiplication by $5$ , doing the same to both sides each time.

Worked Solution [6 marks]

Rule - Collect, substitute, then undo: like terms are terms carrying exactly the same letter, and only those may be combined; a letter is replaced by its value before any arithmetic is done; and an equation is solved by undoing its operations in reverse order, doing the same to both sides.

Step 1: Sort part (a) into like terms

10x - 7y - 6x + 4y

(10x - 6x) + (-7y + 4y)

(Reason: Like terms carry exactly the same letter, so the two

x

terms go together and the two

y

terms go together. The sign written in front of a term belongs to that term, so

-7y

keeps its minus sign when it is moved.)

Step 2: Combine each pair

10x - 6x = 4x

-7y + 4y = -3y

10x - 7y - 6x + 4y = 4x - 3y

(Reason:

10

of something with

6

of it taken away leaves

4

of it. For the

y

terms, start at

-7

and add

4

, which moves up the number line to

-3

, so that term is still negative.)

Step 3: Substitute the given values in part (b)

T = 4d - 6e

T = 4 \times 13 - 6 \times 7

(Reason: Each letter is replaced by its own value, so

d

becomes

13

and

e

becomes

7

. Writing

4d

means

4

multiplied by

d

, so the multiplication signs are put back in.)

Step 4: Work out the two products, then subtract

4 \times 13 = 52

6 \times 7 = 42

T = 52 - 42 = 10

(Reason: Multiplication is done before subtraction, so both products are worked out first. The formula then takes

42

away from

52

Step 5: Undo the addition in part (c)

5p + 11 = 28

5p + 11 - 11 = 28 - 11

5p = 17

(Reason: On the left,

11

has been added, so

11

is taken off both sides. Doing the same to each side keeps the equation balanced and leaves

5p

on its own.)

Step 6: Undo the multiplication

\dfrac{5p}{5} = \dfrac{17}{5}

p = \dfrac{17}{5} = 3.4

(Reason:

5p

means

5

multiplied by

p

, so both sides are divided by

5

. Since

17

is not a multiple of

5

, the answer is a fraction, and

\dfrac{17}{5}

and

3.4

are the same number.)

(a)

4x - 3y

(b)

T = 10

(c)

p = \dfrac{17}{5} = 3.4

Verification

Check 1: Part (a): put

x = 2

and

y = 1

into both forms. The original gives

20 - 7 - 12 + 4

and the simplified form gives

8 - 3

, so the two expressions must agree.

5 = 5

Check 2: Part (b): work backwards. Adding back the

42

that was subtracted must return the first product,

4 \times 13

10 + 42 = 52 = 4 \times 13

Check 3: Part (c): put

p = 3.4

back into the left-hand side of the original equation and compare it with the right-hand side.

5 \times 3.4 + 11 = 17 + 11 = 28

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $4x - 3y$	B2	Both terms correct, in either order, so $-3y + 4x$ also scores B2. If not B2, award B1 for $4x$ alone or for $-3y$ alone.	✓
(b) $4 \times 13$ and $6 \times 7$ , or $52$ and $42$	M1	Method mark for substituting both values and forming the two products. The sign attached to $6 \times 7$ may be either way round.	✓
(b) $10$	A1	Working not required, so a correct answer scores full marks (unless it comes from obviously incorrect working).	✓
(b) $-50$	SC B1	Special case, awarded only when no other marks are earned. It names one specific error: substituting the two values the wrong way round gives $4e - 6d = 28 - 78 = -50$ .	✓
(c) $5p = 28 - 11$ or $5p = 17$ or $p + \dfrac{11}{5} = \dfrac{28}{5}$ oe	M1	Method mark for a correct first step towards getting $p$ on its own. Also allow $11 - 28 = -5p$ or $\dfrac{28 - 11}{5}$ .	✓
(c) $\dfrac{17}{5}$	A1	Or equivalent, for example $3.4$ or the mixed number $3\dfrac{2}{5}$ . Working not required, so a correct answer scores full marks.	✓

Full marks: 6/6

Triangle $ABC$ is equilateral. Each of its three sides is $9$ cm long.

Using only a ruler and a pair of compasses, construct triangle $ABC$
The side $AB$ has already been drawn for you.
Every construction line you draw must be left showing. [2 marks]

[Total 2 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

Triangle

ABC

is equilateral, so

AB = BC = CA = 9

The side

AB

is already drawn, and it is exactly

9

cm long

A ruler and a pair of compasses only, with every construction line left showing

Find

The position of the third vertex

C

, and then the completed triangle

Plan the Solution

The third vertex has to be $9$ cm from $A$ and $9$ cm from $B$ at the same time.
Every point $9$ cm from $A$ lies on one arc, and every point $9$ cm from $B$ lies on another.
The two arcs cross at only one point above $AB$ , so that crossing point must be $C$ .
Draw both arcs first, then join the crossing point to each end of the given side. The arcs stay on the page: they are the working, and the marks are for them.

Worked Solution [2 marks]

Rule - Constructing an equilateral triangle: the third vertex is the point that is the same distance from both ends of the given side. An arc of radius

9

cm centred on

A

and an arc of radius

9

cm centred on

B

cross at exactly that point.

Step 1: Open the compasses to the length of a side

\text{compass radius} = AB = 9\text{ cm}

(Reason: (Reason: every side of the triangle is

9

cm, and

AB

is already drawn

9

cm long. Resting the compass point on

A

and the pencil on

B

therefore sets the radius without measuring anything. Do not change that setting again.))

Step 2: Draw an arc centred on A

AC = 9\text{ cm}

(Reason: (Reason: with the point on

A

, every place the pencil can reach is

9

cm from

A

. Sweep a generous arc right across the space above

AB

, so there is plenty of it for the second arc to cross.))

Step 3: Draw an arc centred on B

BC = 9\text{ cm}

(Reason: (Reason: keeping the same setting, lift the compasses, put the point on

B

and sweep a second arc. Where the two arcs cross, the point is

9

cm from

A

and

9

cm from

B

at once, so that crossing is

C

.))

Step 4: Join the triangle and leave the arcs showing

AB = BC = CA = 9\text{ cm}

\angle A = \angle B = \angle C = 60^\circ

(Reason: (Reason: use the ruler to join

C

A

and

C

B

. All three sides now measure

9

cm, so the triangle is equilateral and every angle is

60^\circ

. Do not rub the arcs out: without them the answer scores only one of the two marks.))

Triangle

ABC

constructed with two arcs of radius

9

cm, one centred on

A

and one centred on

B

, crossing at

C

, with all construction arcs left showing

Verification

Check 1: Measure

AC

and

BC

with the ruler. Both were drawn with the compasses fixed at one setting, so both must come out the same as

AB

AC = BC = AB = 9\text{ cm}

Check 2: Measure straight up from

AB

C

. An equilateral triangle of side

9

cm stands

9\sin 60^\circ

high, so a badly set pair of compasses shows up at once.

9\sin 60^\circ = 7.79\text{ cm (2 d.p.)}

Check 3: Put a protractor on each corner in turn. All three sides are equal, so all three angles are equal, and the three of them add up to

180^\circ

\dfrac{180^\circ}{3} = 60^\circ

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Two arcs of radius $9$ cm drawn, one centred on $A$ and one centred on $B$ , crossing above $AB$	B1	Awarded for two intersecting arcs within or on the guidelines of the overlay. Also awarded for an accurate triangle drawn with no arcs.	✓
Triangle $ABC$ completed from the crossing point, with the correct intersecting arcs $9$ cm from $A$ and $9$ cm from $B$ still showing	B2	Full marks. The arcs must lie within or on the guidelines of the overlay.	✓
Note	note	Working required. Here the arcs are the working: a triangle measured out with a ruler and a protractor, with no arcs left on the page, is worth B1 only.	✓

Full marks: 2/2

There are $29$ kayaks on the rack at a lake hire centre.

$10$ of the kayaks are yellow.
The rest of the kayaks are green or blue.

Erin takes one of these kayaks at random.

(a) Write down the probability that she takes a yellow kayak. [1 mark]

The probability that Erin takes a green kayak is $\dfrac{7}{29}$

(b) Work out the probability that she takes a blue kayak. [2 marks]

(a)(b)

[Total 3 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

There are

29

kayaks on the rack, and one is taken at random, so every kayak is equally likely

10

of the kayaks are yellow

The rest are green or blue, and

P(\text{green}) = \dfrac{7}{29}

Find

(a)

P(\text{yellow})

(b)

P(\text{blue})

Plan the Solution

Part (a) is a straight count: the number of yellow kayaks over the $29$ kayaks on the rack.
The green probability is already written over $29$ , so its numerator, $7$ , is the number of green kayaks.
Every kayak is yellow, green or blue, so taking the yellow ones and the green ones away from $29$ leaves the blue ones.
Put that count over the same total. There is a second route for part (b) - take both known probabilities away from 1 - and it is used as a check.

Worked Solution [3 marks]

Rule - For equally likely outcomes,

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

, and the probabilities of all the separate colours add up to

1

Step 1: (a) Write the yellow probability straight from the counts

P(\text{yellow}) = \dfrac{10}{29}

(Reason: (Reason:

10

of the

29

kayaks are yellow, and each one is equally likely to be taken. There is nothing to cancel here:

29

is a prime number, so

\dfrac{10}{29}

is already in its simplest form.))

Step 2: Turn the green probability back into a number of kayaks

\dfrac{7}{29} \times 29 = 7

(Reason: (Reason: the denominator of

\dfrac{7}{29}

is the total number of kayaks, so the numerator is already a count.

7

of the kayaks on the rack are green.))

Step 3: (b) Subtract to find how many kayaks are blue

29 - 10 - 7 = 12

(Reason: (Reason: every kayak on the rack is yellow, green or blue. Take away the

10

yellow ones and the

7

green ones, and only the blue ones are left.))

Step 4: Write the blue probability over the same total

P(\text{blue}) = \dfrac{12}{29}

(Reason: (Reason: the rack still holds

29

kayaks, so the total underneath does not change - only the count on top does. As a decimal this is

0.4137\ldots

, which the mark scheme accepts anywhere from

0.41

0.42

.))

(a)

\dfrac{10}{29}

(b)

\dfrac{12}{29}

Verification

Check 1: The kayak taken is certainly yellow, green or blue, so the three probabilities must add to

1

\dfrac{10}{29} + \dfrac{7}{29} + \dfrac{12}{29} = \dfrac{29}{29} = 1

Check 2: Work part (b) the other way round: instead of counting kayaks, take both known probabilities away from

1

1 - \dfrac{10}{29} - \dfrac{7}{29} = \dfrac{12}{29}

Check 3: Count the kayaks instead of the probabilities: the three colours must account for the whole rack.

10 + 7 + 12 = 29

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $P(\text{yellow}) = \dfrac{10}{29}$	B1	Or any equivalent. Accept the decimal 0.34(48275...) or the percentage 34.(48275...)%, truncated or rounded.	✓
(b) A complete method for the blue kayaks: $29 - 10 - 7$ , or $\dfrac{29 - 10 - 7}{29}$ , or $1 - \dfrac{10 + 7}{29}$	M1	Also awarded for the blue count 12 on its own, or for the decimal method 1 - 0.34(482...) - 0.24(137...).	✓
(b) $P(\text{blue}) = \dfrac{12}{29}$	A1	Or any equivalent. Accept 0.41(37931...) to 0.42, or 41.(37931...)% to 42%.	✓
Note	note	Working is not required in part (b), so a correct answer scores full marks on its own, unless it comes from obvious incorrect working. Incorrect probability notation is penalised only once across the question.	✓

Full marks: 3/3

Oliver is going to bake some scones.

Here is a list of ingredients for making $24$ scones.

Ingredients for $24$ scones
$120$ g butter
$60$ g sugar
$200$ g flour

Oliver has
five $250$ g packs of butter
$750$ g of sugar
$1.4$ kg of flour

Work out the maximum number of scones that Oliver can make.
Show your working clearly. [4 marks]

[Total 4 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

A recipe that makes

24

scones uses

120

g of butter,

60

g of sugar and

200

g of flour.

Oliver has five

250

g packs of butter,

750

g of sugar and

1.4

kg of flour.

One of the three amounts is written in kilograms while the recipe is written in grams.

Find

The largest number of scones Oliver can bake before one of the ingredients runs out.

Plan the Solution

Put every amount into grams first, so the store cupboard and the recipe can be compared.
Take each ingredient in turn and work out how many batches of $24$ that ingredient alone would allow.
The smallest of the three answers is the limit, because that ingredient runs out first. The other two are not the answer, however large they are.
Turn batches back into scones by multiplying the limiting number of batches by $24$ .

Worked Solution [4 marks]

Rule - The ingredient that runs out first sets the limit: divide the amount held by the amount one batch needs, do this for every ingredient, take the smallest result, then multiply by

24

Step 1: Write every amount in grams

1.4 \times 1000 = 1400 \text{ g of flour}

5 \times 250 = 1250 \text{ g of butter}

(Reason: (Reason: the recipe is written in grams, so the

1.4

kg of flour has to become

1400

g and the five packs have to be added together before anything can be compared. The sugar is already in grams at

750

g.))

Step 2: How many batches does the butter allow?

\dfrac{1250}{120} = 10.4\ldots \text{ batches}

(Reason: (Reason: one batch takes

120

g of butter, so dividing the butter held by

120

counts how many whole batches the butter could supply.))

Step 3: How many batches does the sugar allow?

\dfrac{750}{60} = 12.5 \text{ batches}

(Reason: (Reason: one batch takes

60

g of sugar, so the sugar he has is divided by

60

in exactly the same way as the butter.))

Step 4: How many batches does the flour allow?

\dfrac{1400}{200} = 7 \text{ batches}

(Reason: (Reason: one batch takes

200

g of flour, and

1400

divides by

200

exactly, so the flour allows precisely

7

batches with nothing left over.))

Step 5: Take the smallest, then count the scones

\text{smallest} = 7 \text{ batches, from the flour}

7 \times 24 = 168

(Reason: (Reason: the flour is the first ingredient to run out, so it decides the total. Only

7

batches can be baked, and each batch makes

24

scones.))

168

scones

Verification

Check 1: Rebuild the shopping list.

168

scones is

7

full batches, so multiply each recipe amount by

7

and compare with what Oliver actually has.

7 \times 200 = 1400

g of flour, which is all of it;

7 \times 120 = 840

g of butter out of

1250

7 \times 60 = 420

g of sugar out of

750

g - every amount is available

Check 2: Test that

168

really is the maximum by trying one more batch. An eighth batch would need

8 \times 200

g of flour.

1600 > 1400

, so an eighth batch is impossible and

192

scones cannot be made

Check 3: Reach the answer a different way, working per single scone instead of per batch. One scone takes

\dfrac{200}{24}

g of flour, so the flour allows

1400 \times \dfrac{24}{200}

scones.

1400 \times \dfrac{24}{200} = 168

, the same limit found by a completely different route

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Convert the flour into grams	M1	for $1.4 \times 1000 = 1400$ oe	✓
Work out how many batches ONE ingredient allows	M1	for $\dfrac{5 \times 250}{120} = 10.4\ldots$ or $\dfrac{750}{60} = 12.5$ or $\dfrac{1400}{200} = 7$ oe; or for the grams needed by one scone, $\dfrac{120}{24} = 5$ or $\dfrac{60}{24} = 2.5$ or $\dfrac{200}{24} = 8.3\ldots$ oe. A correct list of multiples of $120$ , $60$ or $200$ scores this mark	✓
Work out how many scones ALL THREE ingredients allow	M1	for $250$ and $300$ and $168$ oe; or for the three batch counts $10.4\ldots$ and $12.5$ and $7$ each multiplied by $24$	✓
Choose the smallest and state the number of scones	A1	for $168$ . Working is required, and this mark is not awarded unless all three method marks are earned	✓

Full marks: 4/4

The table gives information about the number of training sessions each of $30$ members of a swimming club completed in one week.

$\begin{array}{|c|c|}\hline \text{Number of training sessions} & \text{Frequency} \\ \hline 0 & 2 \\ 1 & 5 \\ 2 & 11 \\ 3 & 7 \\ 4 & 4 \\ 5 & 1 \\ \hline \end{array}$

(a) Work out the mean number of training sessions. [3 marks]

Erin is a member of the swimming club.
The probability that Erin travels to the pool by bus is $0.79$

(b) Work out the probability that Erin does not travel to the pool by bus. [1 mark]

(a)(b)

[Total 4 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

A frequency table for the

30

members:

0

1

2

3

4

5

training sessions, with frequencies

2

5

11

7

4

1

The frequencies total

30

, which is the number of members the question states

The probability that Erin travels to the pool by bus is

0.79

Find

(a) The mean number of training sessions for the members of the club (b) The probability that Erin does not travel to the pool by bus

Plan the Solution

Part (a): a frequency table is a shorthand for a long list. Turn it back into a total by multiplying each number of sessions by how many members did that number.
Add those products to get the total number of sessions swum by the whole club, then divide by how many members there are.
Part (b): travelling by bus and not travelling by bus are the only two outcomes, so their probabilities add to $1$ . Subtract to get the one that is missing.

Worked Solution [4 marks]

Rule - Mean from a frequency table:

\text{mean} = \dfrac{\text{total of all the values}}{\text{total frequency}}

. Rule - Complementary events: an event and its opposite have probabilities that add to

1

Step 1: Multiply each number of sessions by its frequency

0 \times 2 = 0

1 \times 5 = 5

2 \times 11 = 22

3 \times 7 = 21

4 \times 4 = 16

5 \times 1 = 5

(Reason: each row stands for a whole group of members, so the

7

members who trained

3

times contribute

21

sessions between them, not

3

)

Step 2: Add the products to get the total number of sessions

0 + 5 + 22 + 21 + 16 + 5 = 69

(Reason: adding the products puts the whole club back together as one running total, which is the top of the mean)

Step 3: Add the frequencies to get the total number of members

2 + 5 + 11 + 7 + 4 + 1 = 30

(Reason: this is the bottom of the mean, and it agrees with the

30

members the question states, so no row has been missed)

Step 4: Divide the total number of sessions by the total number of members

\dfrac{69}{30} = 2.3

(Reason: the mean shares the

69

sessions equally between the

30

members)

Step 5: Part (b) - take the probability of travelling by bus away from

1

1 - 0.79 = 0.21

(Reason: Erin either travels by bus or does not, and nothing else is possible, so the two probabilities must add to

1

)

(a)

2.3

training sessions(b)

0.21

Verification

Check 1: Multiply the mean back by the number of members. If

2.3

really is the fair share,

30

shares of it must rebuild the total.

2.3 \times 30 = 69

, the total found in step 2

Check 2: A mean must lie between the smallest and the largest value in the table, and it should sit near the tallest bar. Most members,

11

of them, trained

2

times.

0

is the least and

5

is the most, and

2.3

lies between them, just above

2

Check 3: Add the two probabilities in part (b). Two outcomes that cover every possibility must account for the whole of the probability between them.

0.79 + 0.21 = 1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Multiply each value by its frequency and add the products	M1	for at least $4$ products added, which need not be evaluated, or for $71$ (the total reached when the top row $0 \times 2$ is copied down as $2$ )	✓
(a) Divide the total by the total frequency	M1	for their $69$ divided by $30$ , that is $\dfrac{69}{30}$	✓
(a) Answer	A1	cao $2.3$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working	✓
(b) Answer	B1	$0.21$ or an equivalent, for example $\dfrac{21}{100}$ or $21$ per cent	✓

Full marks: 4/4

Draw the graph of $y = 2x - 3$ on the grid below.
Use values of $x$ from $-2$ to $4$ [3 marks]

[Total 3 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

The equation

y = 2x - 3

Values of

x

from

-2

4

A printed grid to draw the graph on

Find

The graph of

y = 2x - 3

, drawn across the whole of that interval

Plan the Solution

$y = 2x - 3$ is of the form $y = mx + c$ , so its graph is a straight line and a short table of values is enough.
Put each whole value of $x$ from $-2$ to $4$ into the equation and work out $y$ .
Plot each pair on the grid and rule one line through them.
Take the line all the way from $x = -2$ to $x = 4$ : the marks are for a line across the whole interval, not for the points on their own.

Worked Solution [3 marks]

Rule - A straight line

y = mx + c

m

is the gradient, the steps up for each step across, and

c

is where the line crosses the

y

-axis. Two points fix the line and a third is a check, so a table of values is the safest way to draw one.

Step 1: Work out

y

at the two ends,

x = -2

and

x = 4

2 \times (-2) - 3 = -4 - 3 = -7

2 \times 4 - 3 = 8 - 3 = 5

(Reason: (Reason: the ends of the interval are the two points the question names, and between them they fix the whole line. Multiply by

2

first and take the

3

off afterwards; a negative value of

x

still has to go through that multiplication before the subtraction.))

Step 2: Fill in the values in between

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

2 \times 0 - 3 = -3

2 \times 1 - 3 = -1

2 \times 2 - 3 = 1

2 \times 3 - 3 = 3

(Reason: (Reason: only two points are strictly needed for a straight line, but a full table catches a slip. Each

y

should be

2

more than the one before it, and here they run

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

- a constant step, which is exactly what a straight line means.))

Step 3: Plot the seven points and rule one line through them

(-2, -7),\ (-1, -5),\ (0, -3),\ (1, -1),\ (2, 1),\ (3, 3),\ (4, 5)

(Reason: (Reason: read each pair as across first, then up or down. Every point should land on one straight line; if one is off, that value of

y

is worth checking again before the line is ruled. Use a ruler and take the line right from

x = -2

x = 4

.))

Step 4: Read the gradient back off the drawing as a check

\dfrac{5 - (-7)}{4 - (-2)} = \dfrac{12}{6} = 2

(Reason: (Reason: from one end of the drawn line to the other is

12

squares up for

6

squares across, so the drawing has gradient

2

- the number in front of the

x

. A line drawn with the right gradient through the right point on the

y

-axis cannot be anything else.))

The straight line through

(-2, -7)

and

(4, 5)

, ruled right across the grid from

x = -2

x = 4

Verification

Check 1: Take a point off the middle of the drawn line,

(3, 3)

, and put

x = 3

back into the equation. It has to give the same

y

2 \times 3 - 3 = 3

Check 2: Count the squares between the two ends of the drawn line:

12

up for

6

across. The gradient of the drawing must equal the

2

in the equation.

\dfrac{12}{6} = 2

Check 3: The line should cut the

y

-axis at

-3

, the constant term, and the middle point of the seven should sit halfway between the two ends.

\dfrac{-7 + 5}{2} = -1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
At least $2$ correct points stated (a table counts) or plotted, or a line with a positive gradient through $(0, -3)$ , or a line with a gradient of $2$	B1	The first band. Enough correct work to show the equation has been used, but no correct line yet.	✓
A correct straight line segment through at least $3$ of $(-2, -7), (-1, -5), (0, -3), (1, -1), (2, 1), (3, 3), (4, 5)$ , or all seven of them plotted but not joined	B2	The second band. A correct short line, or seven correct points with no line ruled through them.	✓
A correct line drawn between $x = -2$ and $x = 4$	B3	Full marks. The line is correct and it covers the whole interval the question asks for.	✓
Note	note	The three marks are one B3 award, not three marks collected in turn: B1 and B2 say what a partly correct answer is worth. A correct line that stops short of either end drops to the B2 band, so the last thing to check is that the line reaches both ends.	✓

Full marks: 3/3

Yaniv makes $490$ clay pots each week for a pottery studio.
$86$ of the pots are glazed.

(a) Write $86$ as a percentage of $490$
Give your answer correct to one decimal place. [2 marks]

A glazed pot weighs $375$ grams before it is fired.
The pot loses $12\%$ of its weight when it is fired.

(b) Work out the weight of the pot after it is fired. [3 marks]

(a) %(b) grams

[Total 5 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

490

pots are made each week.

86

of those pots are glazed.

A glazed pot weighs

375

grams before it is fired.

Firing takes away

12\%

of that weight.

Find

(a)

86

written as a percentage of

490

, correct to one decimal place. (b) The weight of one pot after firing, in grams.

Plan the Solution

Part (a): put the glazed pots over the total made, $\dfrac{86}{490}$ , then multiply by $100$ .
Keep the full decimal on the calculator and round only at the very end, or the last digit can come out wrong.
Part (b): find $12\%$ of $375$ grams, then take that loss away from $375$ grams.
A one-step alternative for part (b): losing $12\%$ leaves $88\%$ , so multiply by $0.88$ .

Worked Solution [5 marks]

Rule - Percentage of a total:

\dfrac{\text{part}}{\text{whole}} \times 100

. Rule - Percentage decrease: take the loss off the original, or multiply the original by

\dfrac{100 - 12}{100}

Step 1: Write the glazed pots as a fraction of all the pots

\dfrac{86}{490} = 0.175510\ldots

(Reason: A percentage compares a part with a whole, so the

86

glazed pots go on top and the

490

pots made altogether go underneath.)

Step 2: Turn the decimal into a percentage

0.175510\ldots \times 100 = 17.5510\ldots\%

(Reason: Multiplying by

100

converts a decimal to a percentage, in the same way that

0.5

becomes

50\%

Step 3: Round to one decimal place

17.5510\ldots \approx 17.6

(Reason: The digit in the second decimal place is

5

, so the first decimal place rounds up from

5

6

. Rounding the answer, not the working, is what keeps the last digit safe.)

Step 4: Work out the weight the pot loses in the kiln

\dfrac{12}{100} = 0.12

0.12 \times 375 = 45

(Reason: The pot loses

12\%

of the weight it had before firing, so the multiplier goes on the weight of one pot,

375

grams.)

Step 5: Take the loss away from the starting weight

375 - 45 = 330

(Reason: What comes out of the kiln is what went in minus what was lost, so the baked pot weighs

330

grams.)

(a)

17.6\%

(b)

330

grams

Verification

Check 1: Put the rounded percentage back.

17.6\%

0.176

, so take that fraction of the

490

pots.

0.176 \times 490 = 86.24

, within a quarter of a pot of

86

, which is all a rounded percentage can give.

Check 2: Test the rounding boundary. Half way between

17.5

and

17.6

17.55

17.5510\ldots > 17.55

, so one decimal place gives

17.6

and not

17.5

Check 3: Do part (b) by the other method. Losing

12\%

means keeping

88\%

, so multiply by

0.88

0.88 \times 375 = 330

grams, the same weight reached a different way.

Check 4: Compare the fired weight with the raw weight as a fraction.

\dfrac{330}{375} = 0.88 = 88\%

, so exactly

12\%

of the weight has gone.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $\dfrac{86}{490} \times 100$ or $0.175510\ldots \times 100$	M1	A correct method for writing one number as a percentage of another. The multiplication by $100$ may be implied.	✓
(a) $17.6$	A1	awrt $17.6$ . Working is not required, so a correct answer scores both marks unless it comes from obviously incorrect working.	✓
(b) $\dfrac{12}{100} \times 375$ or $0.12 \times 375$ $(= 45)$	M1	A calculation must be seen. Writing $12\%$ of $375$ in words is not enough unless $45$ is seen. The split $37.5 + 3.75 + 3.75$ earns the same mark.	✓
(b) $375 - 45$	M1	Awarded on the candidate's own loss, so $375 - 37.5 - 3.75 - 3.75$ and $375 - 37.5 - 7.5$ both score it.	✓
(b) $330$	A1	Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓
Alternative for (b): $\dfrac{88}{100} \times 375$	M2	The one-step multiplier method earns both method marks at once, because $100\% - 12\% = 88\%$ does the subtraction inside the multiplier.	✓

Full marks: 5/5

Continue to questions 15 to 26

The remaining 12 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 1F 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 1F, November 2024: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 14 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

Frequently asked questions

Keep revising

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