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Cambridge IGCSE 0580/41, May/June 2024: Worked Solutions and Mark Schemes

Q: How many questions are on Cambridge 0580/41, May/June 2024, and how long is the exam?

There are 9 questions worth 130 marks in total, sat over 2 hours 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

16 Sept 2026

Table of Contents▾

Cambridge IGCSE Mathematics (0580)0580/41 - Extended - May/June 2024130 marks · 2 hours 30 minutes · Calculator allowed

Original worked solutions for Cambridge IGCSE Mathematics, Paper 0580/41 (Extended), May/June 2024 series, sat Tuesday 7 May 2024 in Cambridge administrative zone 2 –130 marks, 2 hours 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 4 of 9

(a) The areas, in km², of the world's four largest rainforests are given in this table.

$\begin{array}{|l|c|}\hline \textbf{Rainforest} & \textbf{Area}\ (\text{km}^2) \\ \hline \text{Amazon} & 5\,500\,000 \\ \hline \text{Congo} & 2\,000\,000 \\ \hline \text{Atlantic} & 1\,315\,000 \\ \hline \text{Valdivian} & 250\,000 \\ \hline \end{array}$

(i) What is the area of the Valdivian rainforest, as a percentage of the area of the Amazon rainforest? [1 mark]

(ii) Give the ratio of the rainforest areas Valdivian : Atlantic : Congo in its simplest form. [2 marks]

(iii) Of the area of the Amazon rainforest, $60\%$ lies in Brazil and $10\%$ lies in Colombia.
Peru holds $43\dfrac{1}{3}\%$ of the remaining area of the rainforest.
What percentage of the Amazon rainforest is in Brazil, Colombia and Peru? [3 marks]

(iv) $\dfrac{27}{50}$ of the total area of rainforest in the world is the area of the Amazon rainforest.
What is the total area of rainforest in the world?
Give your answer correct to the nearest $100\,000$ km². [3 marks]

(v) Every minute, the world loses $60.7$ hectares of rainforest.
What is the total area, in hectares, of rainforest lost in $365$ days?
Give your answer in standard form. [3 marks]

(b) Correct to the nearest $10$ km, the Amazon river is $6440$ km long.
Correct to the nearest $100$ km, the Congo river is $4400$ km long.
What is the upper bound of the difference between the lengths of the Amazon river and the Congo river? [3 marks]

(a)(i) %(a)(ii)(a)(iii) %(a)(iv) km²(a)(v) hectares(b) km

[Total 15 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

Areas in km²: Amazon

5\,500\,000

, Congo

2\,000\,000

, Atlantic

1\,315\,000

, Valdivian

250\,000

Of the Amazon:

60\%

in Brazil,

10\%

in Colombia, and

43\dfrac{1}{3}\%

of what is left in Peru.

\dfrac{27}{50}

of the world's rainforest area is the Amazon.

60.7

hectares are lost every minute.

Amazon river

6440

km to the nearest

10

km, Congo river

4400

km to the nearest

100

km.

Find

(a)(i) The Valdivian area as a percentage of the Amazon area. (a)(ii) Valdivian : Atlantic : Congo in its simplest form. (a)(iii) The percentage of the Amazon in Brazil, Colombia and Peru. (a)(iv) The world total, to the nearest

100\,000

km². (a)(v) The hectares lost in

365

days, in standard form. (b) The upper bound of the difference between the two river lengths.

Plan the Solution

A part as a percentage of a whole is $\dfrac{\text{part}}{\text{whole}} \times 100$ .
A ratio is simplest once every term has been divided by the highest common factor of all three.
For (iii), find what is left after Brazil and Colombia, take Peru's share of that, then add the three shares.
For (iv), dividing by $\dfrac{27}{50}$ is the same as multiplying by $\dfrac{50}{27}$ .
For (v), turn $365$ days into minutes, multiply by the rate, then write the result in standard form.
For (b), a difference is largest when the first length is as big as it can be and the second as small as it can be.

Worked Solution [15 marks]

Rule - Part of a whole, and bounds: a part as a percentage of a whole is

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

; a ratio is in its simplest form once every term has been divided by their highest common factor; and the largest value of

A - C

is the upper bound of

A

minus the lower bound of

C

Step 1: Write the Valdivian area over the Amazon area, then multiply by

100

\dfrac{250\,000}{5\,500\,000} \times 100 = \dfrac{100}{22}

\dfrac{100}{22} = 4.545

(Reason: Both areas divide by

250\,000

, so the fraction cancels to

\dfrac{1}{22}

. The division gives the recurring decimal

4.5454

…, which is

4.55

3

significant figures.)

Step 2: Divide all three areas by their highest common factor

\dfrac{250\,000}{5000} = 50

\dfrac{1\,315\,000}{5000} = 263

\dfrac{2\,000\,000}{5000} = 400

(Reason: The largest number that divides all three areas is

5000

. The three answers cannot cancel again, because

263

is prime.)

Step 3: Find the percentage of the Amazon left after Brazil and Colombia

100 - 60 - 10 = 30

(Reason: Brazil and Colombia take

70\%

between them, so the remaining area Peru's share is measured against is

30\%

of the rainforest.)

Step 4: Take Peru's share of that

30\%

\dfrac{130}{3} \times \dfrac{30}{100} = 13

(Reason:

43\dfrac{1}{3}

\dfrac{130}{3}

, and a percentage of a percentage is found by multiplying. So Peru holds

13\%

of the whole rainforest.)

Step 5: Add the three shares

60 + 10 + 13 = 83

(Reason: Brazil, Colombia and Peru together, as a percentage of the whole Amazon rainforest.)

Step 6: Divide the Amazon area by

\dfrac{27}{50}

\dfrac{5\,500\,000 \times 50}{27} = 10\,185\,185.19

(Reason: The Amazon area is

\dfrac{27}{50}

of the world total, so the world total is the Amazon area divided by

\dfrac{27}{50}

, which is the same as multiplying by

\dfrac{50}{27}

Step 7: Round to the nearest

100\,000

10\,200\,000

(Reason:

10\,185\,185.19

lies between

10\,100\,000

and

10\,200\,000

, and it is much nearer the second of those, so that is the answer in km².)

Step 8: Count the minutes in

365

days

365 \times 24 \times 60 = 525\,600

(Reason: Each day holds

24

hours and each hour holds

60

minutes.)

Step 9: Multiply the minutes by the rate

60.7 \times 525\,600 = 31\,903\,920

(Reason:

60.7

hectares go every minute, so this is the whole year's loss in hectares.)

Step 10: Write the total in standard form

31\,903\,920 = 3.190392 \times 10^{7}

(Reason: Standard form needs one non-zero digit in front of the point, and the point moves

7

places. To

3

significant figures the number in front of the point is

3.19

Step 11: Write the bounds of the Amazon river length

6440 - 5 = 6435

6440 + 5 = 6445

(Reason: A length given to the nearest

10

km is within

5

km of the value stated, so the largest the Amazon river can be is

6445

km.)

Step 12: Write the bounds of the Congo river length

4400 - 50 = 4350

4400 + 50 = 4450

(Reason: A length given to the nearest

100

km is within

50

km of the value stated, so the smallest the Congo river can be is

4350

km.)

Step 13: Subtract the smallest Congo length from the largest Amazon length

6445 - 4350 = 2095

(Reason: The difference is greatest when the first length is as large as it can be and the second as small as it can be, so this takes the upper bound of the Amazon length with the lower bound of the Congo length.)

(a)(i)

4.55\%

(a)(ii)

50 : 263 : 400

(a)(iii)

83\%

(a)(iv)

10\,200\,000

km²(a)(v)

3.19 \times 10^{7}

hectares(b)

2095

Verification

Check 1: Divide the Amazon area by

22

and see whether the Valdivian area comes back.

\dfrac{5\,500\,000}{22} = 250\,000

, the Valdivian area exactly, so the percentage really is one part in

22

Check 2: Multiply every term of the simplified ratio back up by

5000

50 \times 5000 = 250\,000

263 \times 5000 = 1\,315\,000

and

400 \times 5000 = 2\,000\,000

, the three areas in the table.

Check 3: Work out the percentage of the Amazon that is in none of the three countries, two ways. From the answer,

100 - 83 = 17

; from the remaining area,

30 - 13 = 17

. The two agree.

Check 4: Take

\dfrac{27}{50}

of the rounded world total and compare it with the Amazon area.

10\,200\,000 \times \dfrac{27}{50} = 5\,508\,000

, which is

5\,500\,000

to the nearest

100\,000

, as it should be after rounding.

Check 5: Divide the year's loss by the number of minutes in the year, and compare the standard form with the total.

\dfrac{31\,903\,920}{525\,600} = 60.7

, the rate given, and

3.19 \times 10^{7} = 31\,900\,000

, which matches

31\,903\,920

3

significant figures.

Check 6: Reach the upper bound a different way: take the difference of the two stated lengths and add both rounding slacks.

6440 - 4400 = 2040

and

2040 + 5 + 50 = 2095

, the same upper bound.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) The percentage	Note	The Answer column gives $4.55$ or $4.545$ …, and the Marks column gives $1$ for it.	✓
(a)(ii) The simplest ratio	Note	The Answer column gives $50 : 263 : 400$ cao, and the Marks column gives $2$ for it.	✓
A correct simplification of the three areas	M1	For a correct simplification from $250\,000 : 1\,315\,000 : 2\,000\,000$ .	✓
(a)(iii) The percentage in the three countries	Note	The Answer column gives $83$ cao, and the Marks column gives $3$ for it.	✓
Peru's share of the area left, in one expression	M2	For $\dfrac{43\dfrac{1}{3}}{100} \times (100 - 60 - 10)$ oe.	✓
The area left after Brazil and Colombia, seen	M1	Or for $100 - 60 - 10$ seen.	✓
(a)(iv) The world total	Note	The Answer column gives $10\,200\,000$ cao, and the Marks column gives $3$ for it.	✓
The total before rounding	B2	For $10\,185\,185$ to $10\,185\,200$ .	✓
The division that gives the total	M1	Or for $\dfrac{5\,500\,000}{27}\ [\times 50]$ .	✓
Where the partial range comes from	Note	The range $10\,185\,185$ to $10\,185\,200$ is the total before it is rounded, and it rounds to the answer $10\,200\,000$ .	✓
(a)(v) The area lost, in standard form	Note	The Answer column gives $3.19 \times 10^{7}$ or $3.190\ldots \times 10^{7}$ , and the Marks column gives $3$ for it.	✓
The total before it is written in standard form	B2	For $31\,903\,920$ .	✓
The product that gives the total	M1	Or for $60.7 \times 60 \times 24 \times 365$ .	✓
A candidate's own number converted to standard form	SC1	If B0 scored, for correctly converting their number seen to standard form to $3\text{sf}$ or better.	✓
(b) The upper bound of the difference	Note	The Answer column gives $2095$ nfww, and the Marks column gives $3$ for it.	✓
The subtraction, with the other length inside its stated range	M2	For $6445 - C$ where $4300 \leqslant C < 4400$ oe, or $A - 4350$ where $6440 < A \leqslant 6450$ oe.	✓
One of the four bounds seen	M1	Or for $6440 + 5$ or $6440 - 5$ or $4400 + 50$ or $4400 - 50$ seen oe.	✓
Where the two bounds in the M2 row come from	Note	The Amazon length is at most $6445$ and the Congo length is at least $4350$ , and $6445 - 4350 = 2095$ .	✓

Full marks: 15/15

(a) Draw each of these images on the grid.

(i) The image of triangle $T$ under a reflection in the $x$ -axis
[1 mark]
(ii) The image of triangle $T$ under a translation by the vector $\binom{-5}{-2}$
[2 marks]
(iii) The image of triangle $T$ under an enlargement with scale factor $-\dfrac{1}{2}$ and centre $(-1,\ 1)$ . [2 marks]

(b) An enlargement with scale factor $3$ maps shape $P$ onto shape $Q$ .
Shape $Q$ is then mapped onto shape $R$ by an enlargement with scale factor $\dfrac{2}{5}$ .
Shape $P$ has an area of $10$ cm².
What is the area of shape $R$ ? [3 marks]

(b) cm²

[Total 8 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

Triangle

T

is drawn on the grid, with vertices

(2,\ -1)

(1,\ -3)

and

(5,\ -3)

An enlargement with scale factor

3

maps

P

onto

Q

, and one with scale factor

\dfrac{2}{5}

maps

Q

onto

R

Shape

P

has an area of

10

cm².

Find

(a)(i) The image of

T

after a reflection in the

x

-axis. (a)(ii) The image of

T

after a translation by

\binom{-5}{-2}

. (a)(iii) The image of

T

after an enlargement with scale factor

-\dfrac{1}{2}

and centre

(-1,\ 1)

. (b) The area of shape

R

Plan the Solution

Transform one vertex at a time, then join the three images up in the same order.
The $x$ -axis is the line $y = 0$ , so a reflection in it leaves $x$ alone and turns $y$ into $-y$ .
A translation adds the top number of the vector to every $x$ and the bottom number to every $y$ .
For an enlargement, measure the step from the centre to the vertex, multiply that step by the scale factor, and set off from the centre again.
For (b), an enlargement with scale factor $k$ multiplies every area by $k^2$ , so do it once for each enlargement.

Worked Solution [8 marks]

Rule - Transformations of points, and area: a reflection in the

x

-axis sends

(x,\ y)

(x,\ -y)

; a translation by

\binom{a}{b}

sends it to

(x + a,\ y + b)

; an enlargement with scale factor

k

and centre

C

sends a point

X

C + k(X - C)

; and that enlargement multiplies every area by

k^2

Step 1: Read the three vertices of triangle

T

off the grid

(2,\ -1) \qquad (1,\ -3) \qquad (5,\ -3)

(Reason: The apex sits one square below the

x

-axis and two to the right of the

y

-axis; the flat edge runs along

y = -3

from

x = 1

x = 5

Step 2: Reflect each vertex in the

x

-axis

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

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

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

(Reason: A point

3

below the mirror goes

3

above it, so each

x

-coordinate stays where it is and each

y

-coordinate changes sign.)

Step 3: Add the translation vector to each vertex

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

(1,\ -3) \to (-4,\ -5)

(5,\ -3) \to (0,\ -5)

(Reason: The vector

\binom{-5}{-2}

moves every point

5

squares left and

2

squares down, so

5

comes off each

x

-coordinate and

2

off each

y

-coordinate.)

Step 4: Measure the step from the centre

(-1,\ 1)

to each vertex

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

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

(5,\ -3) \to (6,\ -4)

(Reason: Each step is the vertex take away the centre. For the first one that is

2 - (-1) = 3

across and

-1 - 1 = -2

up.)

Step 5: Multiply each step by

-\dfrac{1}{2}

and set off from the centre again

(3,\ -2) \to (-2.5,\ 2)

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

(6,\ -4) \to (-4,\ 3)

(Reason: The minus sign turns the step round, so the image lands on the far side of the centre, and the half makes it half as long. For the first vertex that gives

-1 - \dfrac{1}{2} \times 3 = -2.5

and

1 - \dfrac{1}{2} \times (-2) = 2

Step 6: Turn the first scale factor into an area scale factor

10 \times 3^2 = 90

(Reason: An enlargement with scale factor

3

makes every length

3

times as long, so every area is multiplied by

3^2 = 9

Step 7: Do the same for the second enlargement

90 \times \left(\dfrac{2}{5}\right)^2 = 90 \times \dfrac{4}{25} = 14.4

(Reason: Squaring

\dfrac{2}{5}

gives the area scale factor

\dfrac{4}{25}

, and that is applied to the area of

Q

to reach the area of

R

in cm².)

(a)(i)

(2,\ 1)

(1,\ 3)

(5,\ 3)

(a)(ii)

(-3,\ -3)

(-4,\ -5)

(0,\ -5)

(a)(iii)

(-2.5,\ 2)

(-2,\ 3)

(-4,\ 3)

(b)

14.4

cm²

Verification

Check 1: A reflection and a translation both move a shape without changing it, so compare the squared side lengths of

T

with those of the two images. Triangle

T

gives

5

13

and

16

, and so do both images, so neither one has been stretched or turned into a different shape.

Check 2: Take the far vertex

(5,\ -3)

and check that its image really is on the opposite side of the centre and half as far away. The step from the centre is

(6,\ -4)

and the step to the image is

(-3,\ 2)

, because

-\dfrac{1}{2} \times 6 = -3

and

-\dfrac{1}{2} \times (-4) = 2

. The two steps point opposite ways, and the second is half the length of the first.

Check 3: Count the area of

T

and of its enlarged image by the base-and-height rule, and compare them with the area scale factor.

T

has base

4

and height

2

, so its area is

4

; the image has base

2

and height

1

, so its area is

1

. That is a quarter, which is what

\left(-\dfrac{1}{2}\right)^2 = \dfrac{1}{4}

demands.

Check 4: Reach (b) the other way round: combine the two enlargements into one, then square only once. The two together multiply every length by

3 \times \dfrac{2}{5} = 1.2

, so the area is multiplied by

1.2^2 = 1.44

and

10 \times 1.44 = 14.4

, the same answer.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) The image after the reflection	Note	The Answer column gives Triangle at $(2,\ 1)$ $(1,\ 3)$ $(5,\ 3)$ , and the Marks column gives $1$ for it.	✓
(a)(ii) The image after the translation	Note	The Answer column gives Triangle at $(-4,\ -5)$ $(-3,\ -3)$ $(0,\ -5)$ , and the Marks column gives $2$ for it.	✓
A translation by a vector with one component correct	B1	For translation by $\binom{-5}{k}$ or $\binom{k}{-2}$ .	✓
(a)(iii) The image after the enlargement	Note	The Answer column gives Triangle at $(-2.5,\ 2)$ $(-4,\ 3)$ $(-2,\ 3)$ , and the Marks column gives $2$ for it.	✓
An enlargement with the correct scale factor from any centre	B1	For enlargement by sf $-\dfrac{1}{2}$ with any centre.	✓
(b) The area of shape R	Note	The Answer column gives $14.4$ , and the Marks column gives $3$ for it.	✓
Both area scale factors applied to the area of P, in one expression	M2	For $[10 \times]\ 3^2 \times \left(\dfrac{2}{5}\right)^2$ oe.	✓
One of the two area scale factors seen or implied	M1	Or for $3^2$ or $\left(\dfrac{2}{5}\right)^2$ soi.	✓

Full marks: 8/8

(a) $C = \dfrac{1}{4}xy^{2}$
(i) When $x = 5$ and $y = 8$ , what is the value of $C$ ?
[2 marks]
(ii) Given that $C = 15$ and $x = 2.4$ , what positive value does $y$ take? [2 marks]

(b) Write the expression below as a single fraction, giving it in its simplest form.
$\dfrac{4}{x - 1} - \dfrac{3}{2x + 5}$ [3 marks]

(d) Give the expression below in its simplest form.
$\left(\dfrac{y^{8}}{16x^{16}}\right)^{-\dfrac{3}{4}}$ [3 marks]

(a)(i) C =(a)(ii) y =(b)(c)(d)

[Total 13 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

C = \dfrac{1}{4}xy^{2}

Part (b):

\dfrac{4}{x - 1} - \dfrac{3}{2x + 5}

Part (c):

(2x + 3)(4 - x)^{2}

Part (d):

\left(\dfrac{y^{8}}{16x^{16}}\right)^{-\dfrac{3}{4}}

Find

(a)(i) the value of

C

when

x = 5

and

y = 8

(a)(ii) the positive value of

y

when

C = 15

and

x = 2.4

(b) one fraction, in its simplest form (c) the expansion, simplified (d) the expression in its simplest form

Plan the Solution

(a) is substitution both ways round: put the numbers straight in for (i), and undo the formula one operation at a time for (ii).
(b) needs one common denominator. The two denominators share no factor, so their product $(x - 1)(2x + 5)$ is the smallest one available.
(c) squares the bracket first, then multiplies the two brackets together and collects like terms.
(d) is index laws: a negative index turns the fraction over, and a power of a power multiplies the indices.

Worked Solution [13 marks]

Index laws and a common denominator:

\left(\dfrac{a}{b}\right)^{-n} = \left(\dfrac{b}{a}\right)^{n}

(a^{m})^{n} = a^{mn}

, and two fractions are subtracted over the product of their denominators when those denominators share no factor.

(a)(i) Put the numbers into the formula

C = \dfrac{1}{4} \times 5 \times 8^{2}

8^{2} = 64

C = \dfrac{1}{4} \times 5 \times 64 = 80

(Reason: The index is worked out before the multiplication, so

8^{2}

becomes

64

first, and a quarter of

320

80

(a)(ii) Substitute, then clear the quarter

15 = \dfrac{1}{4} \times 2.4 \times y^{2}

y^{2} = \dfrac{15 \times 4}{2.4} = 25

(Reason: Multiplying both sides by

4

and then dividing by

2.4

leaves

y^{2}

by itself.)

(a)(ii) Take the positive square root

y = \sqrt{25} = 5

(Reason:

25

has two square roots,

5

and

-5

, and only the positive one is asked for.)

(b) Write both fractions over one denominator

\dfrac{4}{x - 1} - \dfrac{3}{2x + 5} = \dfrac{4(2x + 5) - 3(x - 1)}{(x - 1)(2x + 5)}

(Reason: The denominators share no factor, so the common denominator is their product. Each numerator is multiplied by the denominator that the other fraction brought.)

(b) Expand the numerator and collect

4(2x + 5) - 3(x - 1) = 8x + 20 - 3x + 3 = 5x + 23

(Reason: Each bracket is multiplied out, and then the

x

terms and the number terms are collected separately.)

(b) Write the single fraction

\dfrac{4}{x - 1} - \dfrac{3}{2x + 5} = \dfrac{5x + 23}{(x - 1)(2x + 5)}

(Reason: Nothing cancels:

5x + 23

shares no factor with

x - 1

or with

2x + 5

, so this is already in its simplest form. The denominator may be left expanded as

2x^{2} + 3x - 5

instead.)

(4 - x)^{2} = (4 - x)(4 - x) = 16 - 8x + x^{2}

(Reason: A squared bracket is the bracket multiplied by itself, never each term squared on its own.)

(2x + 3)(16 - 8x + x^{2}) = 32x - 16x^{2} + 2x^{3} + 48 - 24x + 3x^{2}

(Reason: Every term in the first bracket multiplies every term in the second, which gives six terms.)

2x^{3} - 16x^{2} + 3x^{2} + 32x - 24x + 48 = 2x^{3} - 13x^{2} + 8x + 48

(Reason:

-16x^{2} + 3x^{2} = -13x^{2}

and

32x - 24x = 8x

, and the

48

has nothing to pair with.)

(d) Use the negative index to turn the fraction over

\left(\dfrac{y^{8}}{16x^{16}}\right)^{-\dfrac{3}{4}} = \left(\dfrac{16x^{16}}{y^{8}}\right)^{\dfrac{3}{4}}

(Reason: A negative index means the reciprocal, so turning the fraction upside down makes the index positive.)

(d) Apply the three-quarter power to each part

16^{\dfrac{3}{4}} = \left(\sqrt[4]{16}\right)^{3} = 2^{3} = 8

\left(x^{16}\right)^{\dfrac{3}{4}} = x^{12}

\left(y^{8}\right)^{\dfrac{3}{4}} = y^{6}

(Reason: A power of

\dfrac{3}{4}

is the fourth root followed by the cube. For the letters the indices multiply:

16 \times \dfrac{3}{4} = 12

and

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

(d) Put the three results back together

\left(\dfrac{16x^{16}}{y^{8}}\right)^{\dfrac{3}{4}} = \dfrac{8x^{12}}{y^{6}}

(Reason: Each part stays where it started, so the answer is

\dfrac{8x^{12}}{y^{6}}

, which may also be written

8x^{12}y^{-6}

(a)(i)

C = 80

(a)(ii)

y = 5

(b)

\dfrac{5x + 23}{(x - 1)(2x + 5)}

(c)

2x^{3} - 13x^{2} + 8x + 48

(d)

\dfrac{8x^{12}}{y^{6}}

Verification

Check 1: Part (a) run backwards. Put

x = 2.4

and

y = 5

into the formula and see whether

C

comes back out.

\dfrac{1}{4} \times 2.4 \times 25 = 15

, the value the question gives.

Check 2: Part (b) at a number. Put

x = 3

into the printed expression and into the single fraction.

\dfrac{4}{2} - \dfrac{3}{11} = \dfrac{19}{11}

and

\dfrac{38}{2 \times 11} = \dfrac{19}{11}

Check 3: Part (c) at a number. Put

x = 1

into

(2x + 3)(4 - x)^{2}

and into the expansion.

5 \times 9 = 45

and

2 - 13 + 8 + 48 = 45

Check 4: Part (d) at numbers. Put

x = 1

and

y = 2

into the printed expression and into the simplified one.

\left(\dfrac{256}{16}\right)^{-\dfrac{3}{4}} = 16^{-\dfrac{3}{4}} = \dfrac{1}{8}

and

\dfrac{8 \times 1}{64} = \dfrac{1}{8}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) The value of C	Note	The Answer column gives 80, and the Marks column gives 2 for it.	✓
(a)(i) Substituting into the formula	M1	For $\dfrac{1}{4} \times 5 \times 8^{2}$ .	✓
(a)(ii) The positive value of y	Note	The Answer column gives 5, and the Marks column gives 2 for it.	✓
(a)(ii) Rearranging to reach y squared	M1	For $[y^{2} =] \dfrac{15 \times 4}{2.4}$ oe.	✓
(b) The single fraction	Note	The Answer column gives $\dfrac{5x + 23}{(x - 1)(2x + 5)}$ or $\dfrac{5x + 23}{2x^{2} + 3x - 5}$ as the final answer, and the Marks column gives 3 for it.	✓
(b) The numerator over the common denominator	B1	For $4(2x + 5) - 3(x - 1)$ oe isw.	✓
(b) The common denominator	B1	For common denominator $= (x - 1)(2x + 5)$ oe isw.	✓
(c) The expansion, simplified	Note	The Answer column gives $2x^{3} - 13x^{2} + 8x + 48$ as the final answer, and the Marks column gives 3 for it.	✓
(c) Expanding all three brackets	B2	For correct expansion of 3 brackets but unsimplified, or for simplified four-term expression of correct form with 3 terms correct.	✓
(c) Expanding two brackets only	B1	Or for correct expansion of two brackets with at least 3 terms out of 4 correct.	✓
(d) The expression in its simplest form	Note	The Answer column gives $\dfrac{8x^{12}}{y^{6}}$ or $8x^{12}y^{-6}$ as the final answer, and the Marks column gives 3 for it.	✓
(d) Two elements of the final answer correct	B2	For two elements correct in final answer, or for correct answer seen then spoiled, or for correct expression where all parts of the power have been dealt with, or for $(\;)^{-1}$ or $\left(\dfrac{2x^{4}}{y^{2}}\right)^{3}$ .	✓
(d) One element of the final answer correct	B1	Or for $8$ or $y^{6}$ or $y^{-6}$ or $x^{12}$ correct in final answer, or for $\left(\dfrac{16x^{16}}{y^{8}}\right)^{\dfrac{3}{4}}$ or $\left(\dfrac{y^{2}}{2x^{4}}\right)^{-3}$ .	✓

Full marks: 13/13

(a) Some motorbikes each travel $195$ m, and Haoran records the time, in seconds, that each one takes.
This information is shown in the box and whisker plot.

(i) What is the median time? [1 mark]

(ii) What is the interquartile range? [1 mark]

(iii) How much greater is the average speed of the fastest motorbike than the average speed of the slowest motorbike?
Give your answer in kilometres per hour. [5 marks]

(b) For each of $80$ different vans, Rosalind records the distance it can travel on a full tank of fuel.
These distances are shown in the table.

$\begin{array}{|l|c|c|c|c|c|}\hline \text{Distance } (d\ \text{km}) & 250 < d \leqslant 300 & 300 < d \leqslant 400 & 400 < d \leqslant 420 & 420 < d \leqslant 450 & 450 < d \leqslant 500 \\ \hline \text{Frequency} & 7 & 13 & 19 & 21 & 20 \\ \hline \end{array}$

(i) Which class interval contains the median? [1 mark]

(ii) What is an estimate of the mean? [4 marks]

(iii) This information is drawn as a histogram.
On that histogram the bar for $250 < d \leqslant 300$ has a height of $2.8$ cm.
What is the height of the bar for each of the intervals below? [3 marks]

(iv) From the $80$ vans, two are chosen at random.
Work out the probability that, on a full tank of fuel, one of these vans can travel more than $450$ km while the other can travel not more than $300$ km. [3 marks]

(a)(i) s(a)(ii) s(a)(iii) km/h(b)(i)(b)(ii) km(b)(iii) 300 < d ⩽ 400 cm(b)(iii) 400 < d ⩽ 420 cm(b)(iii) 420 < d ⩽ 450 cm(b)(iv)

[Total 18 marks]

Show solution & mark schemeHide solution & mark scheme

Question 4 - Exam Solution

Understanding the Question

Given

(a) Each motorbike travels

195

m, and the box and whisker plot shows the times in seconds.

(b)

80

vans, with the distances grouped in the table, and a histogram whose bar for

250 < d \leqslant 300

2.8

cm tall.

Find

(a)(i) The median time. (a)(ii) The interquartile range. (a)(iii) The difference between the two average speeds, in km/h. (b)(i) The class interval that the median falls in. (b)(ii) An estimate of the mean. (b)(iii) The heights of three more bars. (b)(iv) The probability for the two vans chosen.

Plan the Solution

Read the five values off the box plot: the two whisker ends, the two box edges and the line inside the box.
Average speed is distance divided by time, so the shortest time gives the fastest motorbike.
To turn m/s into km/h, multiply by $\dfrac{3600}{1000}$ .
For grouped data, treat every value in an interval as the midpoint of that interval.
On a histogram the height of a bar is the frequency density, $\dfrac{\text{frequency}}{\text{class width}}$ , drawn to one fixed scale.
For two vans chosen without replacement, the second is chosen from $79$ , and the two orders both count.

Worked Solution [18 marks]

Rule - Reading a box plot and summarising grouped data: the interquartile range is the upper quartile minus the lower quartile; average speed is

\dfrac{\text{distance}}{\text{time}}

, and

1

m/s is

\dfrac{3600}{1000}

km/h; an estimate of the mean of grouped data is

\dfrac{\Sigma fx}{\Sigma f}

with

x

the midpoint of each class; and the height of a histogram bar is its frequency density,

\dfrac{\text{frequency}}{\text{class width}}

Step 1: Read the five values off the box and whisker plot

\text{minimum} = 6

\text{lower quartile} = 8.2

\text{median} = 9.3

\text{upper quartile} = 11.6

\text{maximum} = 13

(Reason: The two end bars give the smallest and largest times, the two edges of the box give the quartiles, and the line inside the box gives the median. The scale runs from

5

14

with five small squares to the second, so each square is

0.2

s.)

Step 2: Write down the median

\text{median} = 9.3

(Reason: The line drawn inside the box marks the median, and it stands

4.3

s to the right of the

5

on the scale.)

Step 3: Take the lower quartile from the upper quartile

11.6 - 8.2 = 3.4

(Reason: The interquartile range is the width of the box: the upper quartile minus the lower quartile.)

Step 4: Turn the shortest and the longest times into speeds in metres per second

\dfrac{195}{6} = 32.5

\dfrac{195}{13} = 15

(Reason: The fastest motorbike is the one that takes the least time,

6

s, and the slowest takes

13

s. Average speed is distance divided by time, and both cover

195

m.)

Step 5: Change each speed to kilometres per hour

32.5 \times \dfrac{3600}{1000} = 117

15 \times \dfrac{3600}{1000} = 54

(Reason: There are

3600

seconds in an hour and

1000

metres in a kilometre, so multiplying by

\dfrac{3600}{1000}

turns metres per second into kilometres per hour.)

Step 6: Subtract the two speeds

117 - 54 = 63

(Reason: The question asks how much greater the faster average speed is, so it is the larger speed minus the smaller.)

Step 7: Build the frequencies up until the middle of the data is reached

7,\ 20,\ 39,\ 60,\ 80

\dfrac{80}{2} = 40

(Reason: With

80

distances the median lies between the

40

th and the

41

st. Only

39

of them are at most

420

km, while

60

are at most

450

km, so both of those two lie in

420 < d \leqslant 450

Step 8: Take the midpoint of each interval

\dfrac{250 + 300}{2} = 275

275,\ 350,\ 410,\ 435,\ 475

(Reason: The individual distances are not known, so every van in an interval is treated as travelling the midpoint of that interval. The other four midpoints are found the same way.)

Step 9: Multiply each midpoint by its frequency and add

275 \times 7 + 350 \times 13 + 410 \times 19 + 435 \times 21 + 475 \times 20 = 32\,900

(Reason: This is

\Sigma fx

: the total distance the

80

vans would travel if every one of them managed its own interval's midpoint.)

Step 10: Divide by the number of vans

\dfrac{32\,900}{80} = 411.25

(Reason: It is only an estimate because each distance has been replaced by the midpoint of the interval it falls in.)

Step 11: Use the bar that is given to find the scale of the histogram

\dfrac{7}{50} = 0.14

\dfrac{2.8}{0.14} = 20

(Reason: The interval

250 < d \leqslant 300

has width

50

and frequency

7

, so its frequency density is

0.14

. That bar is drawn

2.8

cm tall, so one unit of frequency density is drawn as

20

cm.)

Step 12: Work out the other three frequency densities and scale each one

\dfrac{13}{100} = 0.13

0.13 \times 20 = 2.6

\dfrac{19}{20} = 0.95

0.95 \times 20 = 19

\dfrac{21}{30} = 0.7

0.7 \times 20 = 14

(Reason: The three class widths are

100

20

and

30

. The narrow interval

400 < d \leqslant 420

holds almost as many vans as the wide one, which is why its bar is so much taller.)

Step 13: Count the vans in the two intervals the last part asks about

450 < d \leqslant 500 \rightarrow 20

250 < d \leqslant 300 \rightarrow 7

(Reason: More than

450

km is the top interval, which holds

20

vans. Not more than

300

km is the bottom interval, which holds

7

Step 14: Multiply the two probabilities, then double for the two orders

2 \times \dfrac{20}{80} \times \dfrac{7}{79} = \dfrac{7}{158}

(Reason: The first van is not put back, so the second is chosen from

79

. The van that travels furthest could be picked first or second, so there are two ways round and the product is doubled.)

(a)(i)

9.3

s(a)(ii)

3.4

s(a)(iii)

63

km/h(b)(i)

420 < d \leqslant 450

(b)(ii)

411.25

km(b)(iii)

2.6

cm,

19

cm,

14

cm(b)(iv)

\dfrac{7}{158}

Verification

Check 1: Subtract the two speeds while they are still in metres per second, and convert only once at the end.

\dfrac{195}{6} - \dfrac{195}{13} = 17.5

m/s, and

17.5 \times 3.6 = 63

km/h, the same answer by a different order of operations.

Check 2: Estimate the mean from an assumed mean of

400

instead. The weighted total of the deviations is

7 \times (-125) + 13 \times (-50) + 19 \times 10 + 21 \times 35 + 20 \times 75 = 900

400 + \dfrac{900}{80} = 411.25

, the same estimate, and it sits inside the table's range of

250

500

km as it must.

Check 3: On a histogram the AREAS of the bars are in the same ratio as the frequencies, so compare the given bar with the tall one.

2.8 \times 50 = 140

and

19 \times 20 = 380

, and

\dfrac{380}{140} = \dfrac{19}{7}

- exactly the ratio of the frequencies

19

and

7

Check 4: Count ordered pairs of vans instead of using probabilities:

20 \times 7 + 7 \times 20 = 280

favourable ordered pairs out of

80 \times 79 = 6320

\dfrac{280}{6320} = \dfrac{7}{158}

, which is the same probability.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) The median time	Note	The Answer column gives 9.3, and the Marks column gives 1 for it.	✓
(a)(ii) The interquartile range	Note	The Answer column gives 3.4, and the Marks column gives 1 for it.	✓
(a)(iii) The difference between the two average speeds	Note	The Answer column gives 63, and the Marks column gives 5 for it.	✓
(a)(iii) Both speeds converted, then subtracted	M4	For $\dfrac{195}{6} \times \dfrac{3600}{1000} - \dfrac{195}{13} \times \dfrac{3600}{1000}$ oe.	✓
(a)(iii) One speed converted, or the difference taken before converting	M3	Or for $\dfrac{195}{6} \times \dfrac{3600}{1000}$ oe or $\dfrac{195}{13} \times \dfrac{3600}{1000}$ oe, or for $\left( \dfrac{195}{6} - \dfrac{195}{13} \right)[\times k]$ oe.	✓
(a)(iii) Second method: a speed with the conversion seen	M1	Or for $\dfrac{195}{6}$ or $\dfrac{195}{13}$ or their speed $\times \dfrac{3600}{1000}$ seen.	✓
(a)(iii) Second method: selecting the two times	M1	For selecting 6 and 13.	✓
(b)(i) The class interval containing the median	Note	The Answer column gives $420 < d \leqslant 450$ , and the Marks column gives 1 for it.	✓
(b)(ii) The estimate of the mean	Note	The Answer column gives 411.25, and the Marks column gives 4 for it.	✓
(b)(ii) The midpoints of the five intervals	M1	For 275, 350, 410, 435, 475 soi.	✓
(b)(ii) The total of frequency times midpoint	M1	For $\Sigma fx$ .	✓
(b)(ii) Dividing that total by the number of vans	M1 dep	For their $\dfrac{\Sigma fx}{80}$ .	✓
(b)(iii) The three bar heights	Note	The Answer column gives 2.6, 19 and 14, and the Marks column gives 3 for it.	✓
(b)(iii) Each correct height	B1	For each.	✓
(b)(iii) Three correct frequency densities	SC1	If 0 scored, for 3 of 0.14, 0.13, 0.95 or 0.7 oe.	✓
(b)(iv) The probability	Note	The Answer column gives $\dfrac{7}{158}$ oe, and the Marks column gives 3 for it.	✓
(b)(iv) Both fractions multiplied, doubled for the two orders	M2	For $[2\times] \dfrac{20}{80} \times \dfrac{7}{79}$ oe.	✓
(b)(iv) One of the four fractions seen	M1	Or for $\dfrac{20}{80}$ or $\dfrac{7}{79}$ or $\dfrac{7}{80}$ or $\dfrac{20}{79}$ oe seen.	✓
(b)(iv) The answer worked out with replacement	SC1	After 0 scored, for $\dfrac{7}{160}$ oe.	✓

Full marks: 18/18

Continue to questions 5 to 9

The remaining 5 questions, with the same full worked solutions and mark schemes

Frequently asked questions

There are 9 questions worth 130 marks in total, sat over 2 hours 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 0580/41, May/June 2024: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 4 of 9

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

Frequently asked questions

Keep revising

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