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Cambridge IGCSE 0580/21, May/June 2024: Worked Solutions and Mark Schemes

Q: How many questions are on Cambridge 0580/21, May/June 2024, and how long is the exam?

There are 22 questions worth 70 marks in total, sat over 1 hour 30 minutes. It is Extended tier and a calculator is allowed throughout - the paper's own instructions say you should use a calculator where appropriate.

Q: What grades can you get on the Extended route?

Extended is graded A* to E. An Extended candidate sits two papers - Paper 2 and Paper 4 - marked out of 70 and 130, so 200 in total, and the grade comes from the combined mark rather than from either paper alone. In the June 2024 series the Extended thresholds were A* 175, A 150, B 117, C 85, D 66 and E 48 out of 200.

Q: Do you have to show your working?

Yes. The paper states in its own instructions that you must show all necessary working clearly, and that answers should be given to three significant figures, or one decimal place for angles in degrees, unless the question specifies otherwise. The mark scheme awards method marks for working that is shown, so a bare answer can score less than the question is worth. That is why every solution here sets out the method mark by mark.

Q: Is a formulae sheet given in the exam?

No. There is no formulae sheet in this paper. The June 2024 Extended papers print no formula list of any kind, so every formula has to be recalled - including the ones a formulae sheet would normally give, such as the area of a trapezium, the volume of a prism and the sine and cosine rules. A printed list of formulas arrived with the 2025 syllabus and is not part of this series.

Q: Where can I get the official question paper and mark scheme?

Both are published by Cambridge Assessment International Education 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.

Sir Faraz Hassan

14 Sept 2026

Table of Contents▾

Cambridge IGCSE Mathematics (0580)0580/21 - Extended - May/June 202470 marks · 1 hour 30 minutes · Calculator allowed

Original worked solutions for Cambridge IGCSE Mathematics, Paper 0580/21 (Extended), May/June 2024 series, sat Friday 3 May 2024 in Cambridge administrative zone 2 –70 marks, 1 hour 30 minutes, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Cambridge Assessment International Education. This resource reproduces neither the exam paper nor the official mark scheme.

Cambridge sits this paper on different days in different administrative zones. The date above is the one published for Zone 2; timetables for the other zones of this series are no longer available from Cambridge.

Both are PDF files hosted by Cambridge: 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 independent mark (B1) is earned. The questions follow the same order as the original paper and carry the same marks.

Download printable PDF

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

Worked solutions, questions 1 to 11 of 22

Two sides of parallelogram $ABCD$ are drawn on the grid.

Work out the coordinates of the point $D$ . [2 marks]

[Total 2 marks]

Show solution & mark schemeHide solution & mark scheme

Question 1 - Exam Solution

Understanding the Question

Given

Two sides of a parallelogram

ABCD

are drawn on the grid.

Reading the three plotted vertices off the grid:

A = (-7, 5)

B = (-1, 1)

and

C = (3, 3)

Find

The coordinates of the fourth vertex

D

Plan the Solution

The letters run round the shape in order, so the four sides are $AB$ , $BC$ , $CD$ and $DA$ . That makes $AD$ the side opposite $BC$ .
Opposite sides of a parallelogram are parallel and the same length, so the move from $A$ to $D$ is the very same move as from $B$ to $C$ .
So work that move out from the two points that are given, then start at $A$ and make the same move.
Confirm the answer a different way with the diagonals, which cut each other in half in any parallelogram.

Worked Solution [2 marks]

Rule - Opposite sides of a parallelogram: in

ABCD

the move from

A

D

equals the move from

B

C

, so

D = A + (C - B)

Step 1: read the three given vertices off the grid

A = (-7, 5)

B = (-1, 1)

C = (3, 3)

(Reason: Every one of the three points sits on a crossing of the grid lines, so each coordinate is a whole number.

A

is seven squares left of the

y

-axis and five up,

B

is one left and one up, and

C

is three right and three up.)

Step 2: work out the move from B to C

3 - (-1) = 4

3 - 1 = 2

(Reason: Subtract

B

from

C

, the across coordinates first and then the up ones. Take care with the first line:

B

has a negative

x

-coordinate, and subtracting

-1

adds one.)

Step 3: make that same move, starting from A

-7 + 4 = -3

5 + 2 = 7

(Reason: The move runs from

A

D

in the same direction it runs from

B

C

, which is four squares to the right and two squares up. Apply it to each of

A

's coordinates in turn.)

Step 4: write down the coordinates of D

D = (-3, 7)

(Reason: Plotting the point three squares left of the

y

-axis and seven up closes the shape, and

AD

comes out parallel to

BC

with

CD

parallel to

AB

D = (-3, 7)

Verification

Check 1: Test the OTHER pair of opposite sides. The move from

A

B

must equal the move from

D

C

B - A = (6, -4)

and

C - D = (6, -4)

, the same move

Check 2: The diagonals of a parallelogram cut each other in half, so

AC

and

BD

must have the same midpoint. This uses

D

in a completely different way from the working. the midpoint of

AC

(-2, 4)

and the midpoint of

BD

(-2, 4)

Check 3: Opposite sides must also come out the same LENGTH. Use Pythagoras on the across and up steps of each side.

AD = BC = \sqrt{20}

and

AB = DC = \sqrt{52}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The coordinates of D	Note	The Answer column gives $(-3, 7)$ , and the Marks column gives $2$ for it.	✓
Partial Marks	B1	For correct diagram, or correct coordinates for their point $D$ , or for $(-3, k)$ or $(k, 7)$ .	✓

Full marks: 2/2

A jar of beads belongs to Meera.
From the jar she takes one bead at random.
There is a probability of $0.6$ that the bead she takes is glass.

(a) Find the probability that the bead she takes is not glass. [1 mark]

(b) Three types of bead are in the jar: glass, clay and metal.

$\begin{array}{|c|c|c|c|}\hline \textbf{Type of bead} & \text{Glass} & \text{Clay} & \text{Metal} \\ \hline \textbf{Number of beads} & & 14 & 14 \\ \hline \textbf{Probability} & 0.6 & & \\ \hline \end{array}$

Fill in the missing values in the table. [2 marks]

[Total 3 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

One bead is taken at random from Meera's jar.

The probability that the bead is glass is

0.6

Of the three types, the jar holds

14

clay beads and

14

metal beads.

Find

(a) The probability that the bead is not glass. (b) The number of glass beads, and the probability for clay and the probability for metal.

Plan the Solution

Part (a) is a complement. Glass and not glass are the only two things that can happen, so the two probabilities add up to $1$ .
In part (b), clay and metal are the only types that are not glass, so the $14$ and the $14$ together are the beads that are not glass. Part (a) has already given the probability of that outcome.
Knowing how many beads carry a probability of $0.4$ is enough to scale back up to the whole jar, and the glass beads are then whatever is left.
Each missing probability is that type's count out of the total number of beads.

Worked Solution [3 marks]

Rule - Probability from a count: the probabilities of all the outcomes add up to

1

, and the probability of one type is

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

Step 1: part (a), take the given probability away from 1

1 - 0.6 = 0.4

(Reason: Glass and not glass are the only two things that can happen, so their probabilities add up to

1

. Taking

0.6

away from

1

leaves

0.4

Step 2: part (b), count the beads that are not glass

14 + 14 = 28

(Reason: Clay and metal are the only other types, so every bead that is not glass is one of these

28

. Part (a) has already priced that outcome at

0.4

Step 3: scale up to the whole jar

\dfrac{28}{0.4} = 70

(Reason: Those

28

beads make up

0.4

of the jar. Dividing by

0.4

undoes that scaling and gives the whole jar, so there are

70

beads in it altogether.)

Step 4: find the number of glass beads

70 - 28 = 42

(Reason: Every bead in the jar is glass, clay or metal, so the glass beads are whatever is left once the other two types are taken away from the total.)

Step 5: find the two missing probabilities

\dfrac{14}{70} = 0.2

(Reason: Clay and metal each have

14

beads out of the

70

in the jar, so each probability is

14

out of

70

. The two come out the same because the two counts are the same.)

(a)

0.4

(b) number of glass beads

42

(b) probability of clay

0.2

, probability of metal

0.2

Verification

Check 1: Add the three counts together. They must come to the total worked out in step 3.

42 + 14 + 14 = 70

Check 2: Add the three probabilities. Every bead is one of the three types, so they must come to

1

0.6 + 0.2 + 0.2 = 1

Check 3: Read the given probability back off the completed table. The glass beads out of the whole jar must return the

0.6

the question states, which uses the finished table in the opposite direction from the working.

\dfrac{42}{70} = 0.6

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The probability of not glass	Note	The Answer column gives $0.4$ oe, and the Marks column gives $1$ for it.	✓
(b) The completed table	Note	The Answer column gives $42$ , then $0.2$ and $0.2$ , and the Marks column gives $2$ for it.	✓
Partial Marks	B1	For $42$ .	✓
Partial Marks	B1	For $0.2$ and $0.2$ .	✓
Partial Marks	SC1	If B0 scored, for their two probabilities being half their (a).	✓
Where the special case comes from	Note	Both of the other two types have the same count, so each of their probabilities comes out as half of the part (a) answer. A candidate who halves their own part (a) has used that method, whatever their part (a) was.	✓

Full marks: 3/3

Some information about two sequences is given in the table.

$\begin{array}{|c|c|c|}\hline & n\text{th term} & 5\text{th term} \\ \hline \text{Sequence } A & 60 - 4n & \\ \hline \text{Sequence } B & n^2 - 300 & \\ \hline \end{array}$

(a) Fill in the missing values in the table. [2 marks]

(b) Work out the smallest positive number in sequence $B$ . [2 marks]

[Total 4 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

Sequence

A

has

n

th term

60 - 4n

Sequence

B

has

n

th term

n^2 - 300

The table leaves the

5

th term of each sequence blank.

Find

(a) The

5

th term of sequence

A

and the

5

th term of sequence

B

. (b) The smallest positive number in sequence

B

Plan the Solution

Part (a) is substitution. The $5$ th term is the term at $n = 5$ , so $5$ goes in place of $n$ in each rule.
Sequence $B$ starts far below zero and climbs, because $n^2$ grows while the $300$ stays put. Its terms stay negative until $n^2$ passes $300$ .
So part (b) is asking where that crossing happens. Find the smallest whole number $n$ whose square is greater than $300$ , then work out that term.
The terms only get bigger after the crossing, so the first positive term is also the smallest positive one.

Worked Solution [4 marks]

Rule - The

n

th term rule gives any term you ask it for: put that term's position in for

n

. A number is positive when it is greater than zero, and a position is always one of

1, 2, 3, \ldots

Step 1: the

5

th term of sequence

A

60 - 4 \times 5 = 60 - 20 = 40

(Reason: The

5

th term is the one at

n = 5

, so

5

replaces

n

. Four lots of

5

20

, and

20

taken from

60

leaves

40

Step 2: the

5

th term of sequence

B

5^2 - 300 = 25 - 300 = -275

(Reason: The same position goes into the second rule, but here it is squared first.

25

is far smaller than

300

, so the term lands well below zero and the answer is negative.)

Step 3: part (b), say what positive means here

n^2 - 300 > 0

n^2 > 300

(Reason: A positive term is one greater than zero. Adding

300

to both sides leaves

n^2

on its own, so the question has turned into: which squares beat

300

Step 4: find the first whole number

n

that works

17^2 = 289

18^2 = 324

(Reason: The square root marks the crossing point:

\sqrt{300} = 17.32\ldots

, so the first whole number past it is

18

. The squares either side settle it -

289

is still below

300

and

324

is above it.)

Step 5: work out that term of sequence B

18^2 - 300 = 324 - 300 = 24

(Reason: Putting that value of

n

into the rule for sequence

B

gives the first term above zero. Every term after it is bigger, because a bigger

n

has a bigger square, so this one is the smallest positive number in the sequence.)

(a)

5

th term of sequence

A

40

(a)

5

th term of sequence

B

-275

(b)

24

Verification

Check 1: Count sequence

A

out term by term instead of using the rule. It starts at

56

and drops by

4

every time, so the fifth number in the list is part (a)'s first answer.

56, 52, 48, 44, 40

Check 2: Reach the

5

th term of sequence

B

from the

4

th rather than from the rule. Squaring

5

instead of

4

adds

9

, so the

5

th term must sit

9

above the

4

th.

4^2 - 300 = -284

, then

-284 + 9 = -275

Check 3: List the terms of sequence

B

on either side of the answer. If the term before it were positive too,

24

could not be the smallest positive one.

16^2 - 300 = -44

17^2 - 300 = -11

18^2 - 300 = 24

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The two $5$ th terms	Note	The Answer column gives $40$ and $-275$ , and the Marks column gives $2$ for it.	✓
Partial Marks	B1	For each of the two correct terms.	✓
(b) The smallest positive number	Note	The Answer column gives $24$ , and the Marks column gives $2$ for it.	✓
Partial Marks	B1	For $324$ or $289$ or $\sqrt{300}$ or $17.3\ldots$ .	✓
Where those four values come from	Note	The four values the Partial Marks cell names are the landmarks on the way to the answer: the square just below $300$ , the square just above it, the square root of $300$ itself, and that root written as a decimal. Any one of them shows the candidate has located where the sequence turns positive.	✓

Full marks: 4/4

An odd number is a factor of $140$ and a factor of $210$ .
What is the greatest value this number can have? [2 marks]

[Total 2 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

The number is a factor of

140

and a factor of

210

The number is odd.

Find

The greatest number that meets both of those conditions.

Plan the Solution

Write $140$ and $210$ as products of prime factors.
A number that divides both can be built only from primes that appear in both lists.
Odd means no factor of $2$ , so multiply the shared primes with every $2$ left out.

Worked Solution [2 marks]

Rule - Common factors from primes: a number divides both only if every prime inside it appears in both factorisations, and an odd number contains no factor of

2

Step 1: Write

140

as a product of primes

140 = 2 \times 2 \times 5 \times 7

(Reason: Halving twice gives

140 = 2 \times 70

then

70 = 2 \times 35

, and

35 = 5 \times 7

Step 2: Write

210

as a product of primes

210 = 2 \times 3 \times 5 \times 7

(Reason: Halving once gives

210 = 2 \times 105

, then

105 = 3 \times 35

and

35 = 5 \times 7

Step 3: Multiply the primes that appear in both lists

2 \times 5 \times 7 = 70

(Reason: A factor of both numbers can use only the primes the two lists share, so that product is the highest common factor.)

Step 4: Leave the

2

out to make it odd

70 = 2 \times 5 \times 7

5 \times 7 = 35

(Reason: Any factor that still contains a

2

is even, so the

2

is dropped. What is left is the largest odd number that divides both.)

35

Verification

Check 1: Divide each of the two numbers by

35

\dfrac{140}{35} = 4

and

\dfrac{210}{35} = 6

, both whole numbers, and

35

is odd.

Check 2: List every odd factor of

140

and test the largest of them.

1

5

7

and

35

. The largest is

35

, and it divides

210

as well, so nothing bigger can work.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The greatest odd number	Note	The Answer column gives $35$ , and the Marks column gives $2$ for it.	✓
An answer of $5$ , $7$ or $70$	B1	For answer $5$ , $7$ or $70$ .	✓
Both prime factorisations, or factor trees or tables, or the shared odd primes	M1	Or for $2 \times 2 \times 5 \times 7$ and $2 \times 3 \times 5 \times 7$ , or two correct factor trees or tables, or $5 \times 7 \times k$ seen.	✓
Where those values come from	Note	$70$ is the highest common factor of the two numbers, and it is even. $5$ and $7$ are the odd primes shared by the two factorisations, and $5 \times 7 = 35$ .	✓

Full marks: 2/2

Work out the value of each of the following.

(a) $\sqrt[3]{343} - \sqrt{40.96}$ [1 mark]

(b) $(192 + 4 \times 16)^{1.25}$ [1 mark]

(a)(b)

[Total 2 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

Part (a) is

\sqrt[3]{343} - \sqrt{40.96}

Part (b) is

(192 + 4 \times 16)^{1.25}

A calculator is allowed on this paper. Every number here is exact, so each part also comes out by hand.

Find

(a) The value of

\sqrt[3]{343} - \sqrt{40.96}

. (b) The value of

(192 + 4 \times 16)^{1.25}

Plan the Solution

Take part (a) one root at a time. $343$ is a cube number and $40.96$ is a square, so both roots are whole or exact and the subtraction is exact too.
In part (b) the bracket is worked out first, and inside the bracket the multiplication comes before the addition.
Then turn the index into a fraction: $1.25 = \dfrac{5}{4}$ . A denominator of $4$ is a fourth root and a numerator of $5$ is a fifth power.

Worked Solution [2 marks]

Rule - A fractional index is a root and a power together:

a^{\dfrac{m}{n}} = \left(\sqrt[n]{a}\right)^{m}

, where the denominator gives the root and the numerator gives the power.

Step 1: part (a), the cube root

7 \times 7 \times 7 = 343

\sqrt[3]{343} = 7

(Reason: A cube root asks which number multiplied by itself three times gives

343

. Trying

7

gives

343

exactly, so this root is a whole number.)

Step 2: part (a), the square root

6.4 \times 6.4 = 40.96

\sqrt{40.96} = 6.4

(Reason: Because

40.96

has two decimal places, its square root has one.

64 \times 64 = 4096

fixes the digits, and placing the decimal point gives

6.4

Step 3: part (a), subtract the square root from the cube root

7 - 6.4 = 0.6

(Reason: Both roots are exact, so no rounding enters the subtraction and the answer is exact as well.)

Step 4: part (b), work out the bracket

4 \times 16 = 64

192 + 64 = 256

(Reason: Inside the bracket the multiplication is done before the addition, so

4 \times 16

becomes

64

first. Adding that to

192

gives

256

. Adding first would give

3136

instead.)

Step 5: part (b), write the index as a fraction

1.25 = \dfrac{5}{4}

256^{1.25} = \left(\sqrt[4]{256}\right)^{5}

(Reason: A fractional index splits into a root and a power. The denominator

4

names the fourth root and the numerator

5

names the fifth power.)

Step 6: part (b), take the root, then the power

\sqrt[4]{256} = 4

4^{5} = 1024

(Reason: Rooting first keeps the numbers small:

4 \times 4 \times 4 \times 4 = 256

, and then

4^{5}

1024

(a)

0.6

(b)

1024

Verification

Check 1 - part (a), cube and square back: Reverse both roots.

7^{3}

should be the number under the cube root, and

6.4^{2}

the number under the square root.

7^{3} = 343

and

6.4^{2} = 40.96

, so both roots are right and

7 - 6.4 = 0.6

Check 2 - part (a), the same subtraction in tenths: Write both roots over

10

, so nothing is a decimal:

7 = \dfrac{70}{10}

and

6.4 = \dfrac{64}{10}

\dfrac{70}{10} - \dfrac{64}{10} = \dfrac{6}{10} = \dfrac{3}{5}

, which is the second spelling of this value in the Answer column.

Check 3 - part (b), in powers of two: The bracket is a power of two,

256 = 2^{8}

, and raising a power to a power multiplies the indices.

(2^{8})^{1.25} = 2^{10} = 1024

, reached without taking a root at all.

Check 4 - part (b), work backwards: Start from the answer. If

1024 = 4^{5}

, then the fourth root used in step 6 must have come from

4^{4}

4^{4} = 256

, which is the value of the bracket, so the chain closes.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The two roots, subtracted	Note	The Answer column gives $0.6$ or $\dfrac{3}{5}$ , and the Marks column gives $1$ for it.	✓
(b) The bracket, raised to the power	Note	The Answer column gives $1024$ , and the Marks column gives $1$ for it.	✓
Where the marks sit	Note	Neither part has an entry in the Partial Marks column, so the whole of each mark is for the answer itself.	✓

Full marks: 2/2

In the diagram, $5$ kites are congruent to kite $ABCD$ .
Each pair of neighbouring kites shares one edge.
$DCE$ is a straight line, and angle $DAB = 40^{\circ}$ .

What is the value of $x$ ? [3 marks]

x =

[Total 3 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

In the diagram,

5

kites are congruent to kite

ABCD

, and each pair of neighbouring kites shares one edge.

Angle

DAB = 40^{\circ}

, and

DCE

is a straight line.

The angle marked

x^{\circ}

is inside one of the kites, at a corner where two of them meet.

Find

The value of

x

Plan the Solution

Count the kites with a corner at $C$ : the kites congruent to $ABCD$ , and $ABCD$ itself.
Those corners lie along the straight line $DCE$ , and congruent kites have equal angles, so divide $180^{\circ}$ by that count to get the angle one kite has at $C$ .
Kite $ABCD$ then has two known angles, one at $A$ and one at $C$ . Its other two are equal to each other, because a kite is symmetrical about the diagonal joining those two.
Use the angle sum of a quadrilateral to find that equal pair, then match the marked angle to it.

Worked Solution [3 marks]

Rule - Angles on a straight line add to

180^{\circ}

, and the four angles of a quadrilateral add to

360^{\circ}

. Congruent shapes have equal angles, and a kite is symmetrical about one diagonal, so the angles at the two vertices that diagonal does not join are equal.

Step 1: count the kites that meet at

C

5 + 1 = 6

(Reason: Every kite has one corner at

C

. The

5

congruent kites supply

5

of those corners, and kite

ABCD

supplies one more.)

Step 2: share the straight angle

DCE

between them

\dfrac{180}{6} = 30

(Reason: Angles on a straight line add to

180^{\circ}

, so the corners at

C

fill

180^{\circ}

between them. The kites are congruent, so their angles at

C

are all equal and each one takes the same share.)

Step 3: use the angle sum of kite

ABCD

360 - 40 - 30 = 290

\dfrac{290}{2} = 145

(Reason: The four angles of a quadrilateral add to

360^{\circ}

. Taking off the

40^{\circ}

A

and the

30^{\circ}

C

leaves the angles at

B

and

D

. The diagonal

AC

is the kite's line of symmetry, so those two are equal and each takes half of what is left.)

Step 4: read off

x

x = 145

(Reason: The angle marked

x^{\circ}

sits at a corner of one of the congruent kites, matching

B

and

D

in kite

ABCD

. Congruent kites have equal angles, so it takes the same value.)

x = 145

Verification

Check 1: Add the four angles of kite

ABCD

and see whether they come to

360^{\circ}

40 + 30 + 145 + 145 = 360

Check 2: Put the angles at

C

back on the straight line:

6

kites, each giving the same angle.

6 \times 30 = 180

Check 3: Reach the answer a second way. The diagonal

AC

halves both the angle at

A

and the angle at

C

, so work inside triangle

ACD

instead of inside the whole kite.

180 - 20 - 15 = 145

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The answer	Note	The Answer column gives $145$ , and the Marks column gives $3$ for it.	✓
Partial Marks	M1	For $\dfrac{180}{6}$ , or any angle congruent to $BCD = 30$ .	✓
Partial Marks	M1	For $\dfrac{360 - 40 - \text{their } 30}{2}$ oe.	✓
Where the first Partial Marks row's two values come from	Note	Kite $ABCD$ has a corner at $C$ as well as the kites congruent to it, so $6$ equal angles fill the straight line $DCE$ and each of them is $30$ . An angle congruent to $BCD$ is that same angle in one of the other kites.	✓
What the second Partial Marks row allows	Note	The second method mark is for the working rather than for the $30$ . A candidate who found a different angle at $C$ still earns it by using their own value in the same way.	✓

Full marks: 3/3

Triangle $JKL$ and a semicircle with diameter $JL$ are joined to make the shape in the diagram.
$JK = JL = 12.8$ cm, and triangle $JKL$ is both isosceles and right-angled.

(a) What is the area of this shape? [3 marks]

(b) What is the perimeter of this shape? [4 marks]

(a) cm²(b) cm

[Total 7 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

Triangle

JKL

is right-angled and isosceles, with

JK = JL = 12.8

cm.

A semicircle stands on

JL

as its diameter.

Find

(a) the area of the shape (b) the perimeter of the shape

Plan the Solution

(a) Add the area of triangle $JKL$ to the area of the semicircle.
(b) Walk round the outside: $JK$ , then $KL$ , then the curved edge. $JL$ is inside the shape, so it is left out.

Worked Solution [7 marks]

For a triangle, area

\dfrac{1}{2} \times \text{base} \times \text{height}

. For a semicircle of radius

r

on a diameter

d

, area

\dfrac{1}{2} \pi r^{2}

and curved edge

\dfrac{1}{2} \pi d

Part (a), step 1: the area of the triangle

\dfrac{1}{2} \times 12.8 \times 12.8 = 81.92

(Reason: The two equal sides of a right-angled isosceles triangle are the ones holding the right angle, so the right angle is at

J

and

JK

and

JL

are the base and the height.)

Part (a), step 2: the radius of the semicircle

\dfrac{12.8}{2} = 6.4

(Reason:

JL

is the diameter, so the radius is half of it.)

Part (a), step 3: the area of the semicircle

\dfrac{1}{2} \times \pi \times 6.4^{2} = 64.3398\ldots

(Reason: A whole circle of radius

6.4

has area

\pi \times 6.4^{2}

, and a semicircle is half of that. Keep the unrounded value for the next step.)

Part (a), step 4: add the two areas

81.92 + 64.3398\ldots = 146.2598\ldots

(Reason: The triangle and the semicircle meet along

JL

and do not overlap, so the two areas simply add. To 3 significant figures this is

146

Part (b), step 1: the length of KL

12.8^{2} + 12.8^{2} = 327.68

\sqrt{327.68} = 18.1019\ldots

(Reason:

KL

is the hypotenuse of the right-angled triangle, so Pythagoras gives it from the two sides of

12.8

cm.)

Part (b), step 2: the length of the curved edge

\dfrac{1}{2} \times \pi \times 12.8 = 20.1061\ldots

(Reason: The whole circle on

JL

has circumference

\pi \times 12.8

, and the curved edge of this shape is half of it.)

Part (b), step 3: add the three edges of the boundary

12.8 + 18.1019\ldots + 20.1061\ldots = 51.0081\ldots

(Reason: The boundary is

JK

, then

KL

, then the arc back to

J

JL

is a line inside the shape, so it is not walked along. To 3 significant figures this is

51.0

(a)

146 \text{ cm}^{2}

(b)

51.0 \text{ cm}

Verification

Check 1: A semicircle's area is also

\dfrac{\pi d^{2}}{8}

, which works from the diameter and never touches the radius. Here

12.8^{2} = 163.84

and

\dfrac{163.84}{8} = 20.48

. The semicircle comes to

20.48 \times \pi = 64.3398\ldots

, so the shape is

81.92 + 64.3398\ldots = 146.2598\ldots

again.

Check 2: The right angle is at

J

, so the other two angles are

45^\circ

each and trigonometry gives

KL = \dfrac{12.8}{\sin 45^\circ}

without using Pythagoras at all.

KL = 18.1019\ldots

once more, and with the arc at

20.1061\ldots

the boundary is

12.8 + 18.1019\ldots + 20.1061\ldots = 51.0081\ldots

as before.

Check 3: A size check on the curved edge. An arc joining the two ends of a diameter has to be longer than the diameter itself, and half of

\pi \times 12.8

is roughly

1.57 \times 12.8

. The arc,

20.1061\ldots

, is comfortably longer than

12.8

, which is what an arc on a diameter must be.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The area of the shape	Note	The Answer column gives $146$ or $146.2$ to $146.3$ , and the Marks column gives $3$ for it.	✓
(a) Area of the triangle	M1	For $\dfrac{1}{2} \times 12.8 \times 12.8$ .	✓
(a) Area of the semicircle	M1	For $\left[\dfrac{1}{2} \times\right] \pi \times \left(\dfrac{12.8}{2}\right)^{2}$ .	✓
(b) The perimeter of the shape	Note	The Answer column gives $51[.0]$ or $51.00$ to $51.01\ldots$ , and the Marks column gives $4$ for it.	✓
(b) The semicircular arc	M1	For $\dfrac{1}{2} \times \pi \times 12.8$ .	✓
(b) The length of KL	M2	For $\sqrt{12.8^{2} + 12.8^{2}}$ or $\dfrac{12.8}{\sin 45}$ oe.	✓
(b) Part-way towards KL	M1	Or for $12.8^{2} + 12.8^{2}$ oe, or $\sin 45 = \dfrac{12.8}{KL}$ oe.	✓

Full marks: 7/7

A sequence starts with the five terms below.

$11 \qquad 18 \qquad 25 \qquad 32 \qquad 39$

What is an expression for the $n$ th term of this sequence? [2 marks]

[Total 2 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

The first five terms of a sequence are

11

18

25

32

and

39

Find

An expression for the

n

th term.

Plan the Solution

Check that the step from one term to the next is the same all the way along the list.
Use that step as the number in front of $n$ , then work out what has to be added to it.

Worked Solution [2 marks]

A sequence that goes up by the same amount every time has

n

th term

kn + j

, where

k

is that common difference and

j

is what is left when the matching multiple of

k

is taken away from a term.

Step 1: is the step between terms always the same?

18 - 11 = 7

25 - 18 = 7

32 - 25 = 7

39 - 32 = 7

(Reason: The same

7

every time, so the terms climb at a steady rate and the expression is a multiple of

n

with a number added on. The multiplier is that common difference, so the expression begins

7n

Step 2: write the 7 times table under the sequence

7 \times 1 = 7

7 \times 2 = 14

7 \times 3 = 21

(Reason:

7n

runs

7

14

21

28

35

n

runs from

1

5

. These are not the sequence itself, but each one sits directly under a term of it.)

Step 3: how far above the times table each term sits

11 - 7 = 4

(Reason: Take the multiple of

7

away from the term standing over it. The same amount is left at every one of the five terms, so one pair settles it, and that amount is the number added to

7n

Step 4: the nth term

7n + 4

(Reason: The two pieces go together:

7

for every step along the list, and

4

added on. At

n = 1

that gives

11

, which is where the sequence starts.)

7n + 4

Verification

Check 1: Put

n = 1

and

n = 5

into the expression. They must give the first and the fifth numbers on the paper,

11

and

39

7 \times 1 + 4 = 11

and

7 \times 5 + 4 = 39

, the two ends of the printed list.

Check 2: Take one term of the expression away from the next one. Whatever

n

is, the answer has to be the sequence's own step of

7

7(n + 1) + 4 - (7n + 4) = 7

, so the expression climbs by

7

a term, exactly as the list does.

Check 3: Build the number to add from the other end of the list instead. The fifth term is

39

and the fifth multiple of

7

35

39 - 35 = 4

, the same number to add as the first term gave.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
An expression for the $n$ th term	Note	The Answer column gives $7n + 4$ oe final answer, and the Marks column gives $2$ for it.	✓
A partly correct expression	B1	For $7n + j$ or $kn + 4$ , $k \neq 0$ , or $7n + 4$ seen then spoilt.	✓

Full marks: 2/2

A delivery van is worth $\$8000$ .
Its value falls exponentially by $25\%$ each year.

What is the value of the van after $3$ years? [2 marks]

[Total 2 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

A delivery van is worth

\$8000

Its value falls exponentially by

25\%

each year.

Find

The value of the van after

3

years.

Plan the Solution

Turn the $25\%$ fall into the fraction of the value that is still there after one year.
Use that multiplier once for each of the $3$ years, then apply it to the starting value.

Worked Solution [2 marks]

Exponential decay: a starting value

V

that falls by

r\%

of itself every year is worth

V \times \left(1 - \dfrac{r}{100}\right)^{n}

after

n

years. Each year's fall is taken from the value at the start of that year, not from the original value, which is why the multiplier is used again and again rather than added up.

Step 1: what is left after one year

1 - \dfrac{25}{100} = 0.75

(Reason: Losing

25\%

of the value leaves

75\%

of it, and

75\%

as a decimal is

0.75

. So after any one year the van is worth

0.75

of what it was worth a year earlier.)

Step 2: what is left after three years

0.75 \times 0.75 \times 0.75 = 0.421875

(Reason: The fall happens once a year for

3

years, and each one acts on the value left by the year before, so the multiplier is used

3

times over. That is

0.75^{3}

, the three-year multiplier.)

Step 3: apply the multiplier to the starting value

8000 \times 0.421875 = 3375

(Reason: The van started at

\$8000

, so multiplying by the three-year multiplier gives what it is worth once the

3

years have passed.)

\$3375

Verification

Check 1: Do it in three separate steps, with no power anywhere. A quarter of

8000

2000

, and each later year loses a quarter of whatever is left at the time.

8000 - 2000 = 6000

, then

6000 - 1500 = 4500

, then

4500 - 1125 = 3375

, the same value the multiplier gave.

Check 2: Work in fractions instead of decimals. Taking a quarter off leaves

\dfrac{3}{4}

, so three years leave

\left(\dfrac{3}{4}\right)^{3} = \dfrac{27}{64}

of the value.

\dfrac{8000}{64} = 125

, and

125 \times 27 = 3375

, reached without a decimal at any point.

Check 3: Run the years backwards. Dividing by

0.75

undoes one year's fall, so doing it three times has to bring the original value back.

\dfrac{3375}{0.75} = 4500

\dfrac{4500}{0.75} = 6000

\dfrac{6000}{0.75} = 8000

, the van's value at the start.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The value after three years	Note	The Answer column gives $3375$ , and the Marks column gives $2$ for it.	✓
The method	M1	For $8000 \times \left(1 - \dfrac{25}{100}\right)^{3}$ oe.	✓
Where the bracket in the Partial Marks row comes from	Note	A fall of $25$ per cent leaves $1 - \dfrac{25}{100}$ of the value, so the bracket is one year's multiplier, and the index $3$ is the number of years it is applied over.	✓

Full marks: 2/2

Bilal's $\$1500$ is paid into a savings account.
Compound interest is added each year at a rate of $r$ %.
His investment is worth $\$1656.73$ when $8$ years have passed.

What is the value of $r$ ? [3 marks]

r =

[Total 3 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

Paid into the account:

\$1500

Compound interest each year:

r

Value after

8

years:

\$1656.73

Find

The value of

r

Plan the Solution

Set the $8$ -year compound interest expression equal to $\$1656.73$ .
Divide by $1500$ so that the growth factor stands on its own.
Take the eighth root for one year's multiplier, then read $r$ off it.

Worked Solution [3 marks]

Compound interest multiplies the amount by the same factor every year, so after

n

years an investment of

P

is worth

P \times \left(1 + \dfrac{r}{100}\right)^{n}

Step 1: write the

8

-year equation

1500 \times \left(1 + \dfrac{r}{100}\right)^{8} = 1656.73

(Reason: Each year the balance is multiplied by the same factor, so

8

years multiply by it

8

times over.)

Step 2: divide by

1500

\left(1 + \dfrac{r}{100}\right)^{8} = \dfrac{1656.73}{1500} = 1.1044866\ldots

(Reason: Dividing the final value by the starting value leaves the whole

8

-year growth factor and nothing else.)

Step 3: take the eighth root

1 + \dfrac{r}{100} = \sqrt[8]{\dfrac{1656.73}{1500}} = 1.0125000\ldots

(Reason: The eighth root undoes the eighth power, so what is left is one year's multiplier.)

Step 4: turn the multiplier into a percentage

1.0125 - 1 = 0.0125

0.0125 \times 100 = 1.25

(Reason: Taking

1

away leaves the interest part of the multiplier, and multiplying by

100

writes that decimal as a percentage, which is

r

r = 1.25

Verification

Check 1: Put the answer back into the question and grow the money with the power: multiply

\$1500

1.0125

eight times over.

1500 \times 1.0125^{8} = 1656.73

to the nearest cent, which is the value the question gives.

Check 2: Drop the power and do it the long way, one year at a time:

1500 \to 1518.75 \to 1537.73 \to 1556.96 \to 1576.42 \to 1596.12 \to 1616.07 \to 1636.28 \to 1656.73

. The eighth multiplication lands on

\$1656.73

, so eight years at

1.25

% really does turn

\$1500

into that amount.

Check 3: Weigh the answer against simple interest at the same rate, which would give

1500 \times 0.0125 \times 8 = 150

. The account actually gained

1656.73 - 1500 = 156.73

, a little more than the

150

simple interest would give, which is what compounding should do. A much larger rate would overshoot badly.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The value of $r$	Note	The Answer column gives $1.25$ or $1.250\ldots$ , and the Marks column gives $3$ for it.	✓
Taking the eighth root	M2	For $\sqrt[8]{\dfrac{1656.73}{1500}}$ oe.	✓
Setting up the equation	M1	Or for $1656.73 = 1500(k)^{8}$ oe for any $k$ .	✓
What $k$ stands for	Note	Here $k$ is the yearly multiplier, the number the amount is multiplied by each year, and the $8$ is the number of years.	✓

Full marks: 3/3

Which inequalities define $R$ , the region that is not shaded? [4 marks]

[Total 4 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

The grid is shaded everywhere except inside

R

The region has four straight edges. Two of them are drawn solid and two are drawn dashed.

Find

Every inequality a point has to satisfy in order to lie inside

R

Plan the Solution

Read the four corners of the region off the grid.
Turn each edge into the equation of the line it lies along.
Fix the direction of each inequality by testing one point taken from well inside the region.
Fix the strictness from the drawing, then gather the four statements together.

Worked Solution [4 marks]

Rule - A boundary drawn solid belongs to its region, so it gives

\leqslant

\geqslant

. A boundary drawn dashed does not belong to it, so it gives

<

>

. Which way an inequality points is settled by one test point taken from inside the region.

Step 1: read the corners of the region

(1, 1), \quad (5, 5), \quad (6, 5), \quad (6, 1)

(Reason: Every corner sits on a crossing of the grid lines, so all four can be read straight off the diagram. One edge slopes, one is upright, and the other two are level.)

Step 2: the equation of the sloping edge

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

y = x

(Reason: The sloping edge runs from

(1, 1)

(5, 5)

. Its gradient is the rise divided by the run, and it passes through

(1, 1)

, so every point on it has its two coordinates equal.)

Step 3: which side of the sloping edge the region is on

(4, 2)

2 < 4

y < x

(Reason: The point

(4, 2)

lies well inside

R

, and there the

y

coordinate is the smaller of the two. So the region is the side on which

y

is below

x

. This edge is dashed, so the line itself is left out and the inequality is strict.)

Step 4: the upright edge

x = 6

4 < 6

x < 6

(Reason: The upright edge joins

(6, 1)

(6, 5)

, so every point on it has

x = 6

. The test point has

x = 4

, which puts the region on the left of that line, and this edge is dashed as well.)

Step 5: the two level edges

y = 1

y = 5

1 \leqslant y \leqslant 5

(Reason: The lower edge is

y = 1

and the upper edge is

y = 5

, and the region lies between them, since the test point has

y = 2

. Both are drawn solid, so points on either line do belong to

R

, giving

y \geqslant 1

and

y \leqslant 5

, which are written together as one double inequality.)

y < x

x < 6

1 \leqslant y \leqslant 5

Verification

Check 1: Put a point that is plainly inside the region,

(5, 2)

, into all four statements, then do the same with a point that is plainly outside,

(2, 4)

(5, 2)

satisfies every one of them, while

(2, 4)

already fails

y < x

Check 2: Test the middle of each edge in turn:

(3, 3)

on the sloping edge,

(6, 3)

on the upright edge,

(3, 1)

on the lower edge and

(5.5, 5)

on the upper edge. Each one should pass exactly when the paper draws that edge solid. The two dashed midpoints fail, on

y < x

and on

x < 6

, and the two solid midpoints pass

Check 3: Sweep the whole grid at quarter-unit steps. For each point decide from the picture alone whether it is in the unshaded region, then decide again from the four inequalities, and compare the two answers. The picture and the inequalities agree at all

1089

points tested

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The answer	Note	The Answer column gives $y < x$ , $x < 6$ and $1 \leqslant y \leqslant 5$ oe, and the Marks column gives $4$ for it.	✓
Partial Marks	B1	For $y < x$ .	✓
Partial Marks	B1	For $x < 6$ .	✓
Partial Marks	B2	For $1 \leqslant y \leqslant 5$ .	✓
Partial Marks	B1	Or for $y \geqslant 1$ or $y \leqslant 5$ .	✓
Partial Marks	SC2	If B0 scored, for $y \leqslant x$ , $x \leqslant 6$ and $1 < y < 5$ oe.	✓
Partial Marks	SC1	If B0 scored, or for three correct from $y = x$ , $x = 6$ , $y = 1$ and $y = 5$ .	✓
Where the special case comes from	Note	The special case is the two kinds of boundary read the wrong way round: each dashed edge treated as part of the region and each solid edge treated as outside it. All four lines are then right and only the inequality signs are wrong.	✓

Full marks: 4/4

Continue to questions 12 to 22

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

Frequently asked questions

There are 22 questions worth 70 marks in total, sat over 1 hour 30 minutes. It is Extended tier and a calculator is allowed throughout - the paper's own instructions say you should use a calculator where appropriate.

Extended is graded A* to E. An Extended candidate sits two papers - Paper 2 and Paper 4 - marked out of 70 and 130, so 200 in total, and the grade comes from the combined mark rather than from either paper alone. In the June 2024 series the Extended thresholds were A* 175, A 150, B 117, C 85, D 66 and E 48 out of 200.

Yes. The paper states in its own instructions that you must show all necessary working clearly, and that answers should be given to three significant figures, or one decimal place for angles in degrees, unless the question specifies otherwise. The mark scheme awards method marks for working that is shown, so a bare answer can score less than the question is worth. That is why every solution here sets out the method mark by mark.

No. There is no formulae sheet in this paper. The June 2024 Extended papers print no formula list of any kind, so every formula has to be recalled - including the ones a formulae sheet would normally give, such as the area of a trapezium, the volume of a prism and the sine and cosine rules. A printed list of formulas arrived with the 2025 syllabus and is not part of this series.

Both are published by Cambridge Assessment International Education 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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Cambridge IGCSE 0580/21, May/June 2024: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 11 of 22

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

Frequently asked questions

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