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Cambridge IGCSE 0580/21, May/June 2025: Worked Solutions and Mark Schemes

Q: How many questions are on Cambridge 0580/21, May/June 2025, and how long is the exam?

There are 23 questions worth 100 marks in total, sat over 2 hours. It is Extended tier and a calculator is not allowed: the paper's own instructions say that calculators must not be used in this paper, so every answer is worked by hand.

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 - each marked out of 100, so 200 in total, and the grade comes from the combined mark rather than from either paper alone. In the June 2025 series the thresholds for candidates who sat Paper 21 with Paper 41 were A* 156, A 131, B 106, C 81, D 68 and E 55 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. 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?

Yes. The question paper prints a List of formulas. It gives the area of a triangle and of a circle, the circumference of a circle, the curved surface areas of a cylinder and a cone, the surface area of a sphere, and the volumes of a prism, a pyramid, a cylinder, a cone and a sphere; the quadratic formula; and, for a triangle with sides a, b, c opposite angles A, B, C, the sine rule, the cosine rule and the area as one half of ab sin C. Anything not on the list, such as the area of a trapezium, still has to be recalled.

Sir Faraz Hassan

22 Sept 2026

Table of Contents▾

Cambridge IGCSE Mathematics (0580)0580/21 - Extended - May/June 2025100 marks · 2 hours · Non-calculator

Original worked solutions for Cambridge IGCSE Mathematics, Paper 0580/21 (Extended), May/June 2025 series, sat Friday 2 May 2025 in Cambridge administrative zone 2 – 100 marks, 2 hours, calculator not allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are Cambridge Assessment International Education's. This resource reproduces neither the exam paper nor the official mark scheme.

Cambridge publishes a timetable for each administrative zone, and the date above is the one published for Zone 2.

Cambridge has not yet published this paper on its public website. We will link the official question paper and mark scheme here as soon as they are released.

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 14 of 23

What is the simplest form of the expression below?

$7c - 5d + c + 3d$ [2 marks]

[Total 2 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

The expression

7c - 5d + c + 3d

It has four terms: two in

c

and two in

d

Find

The expression in its simplest form.

Plan the Solution

Put the $c$ terms together and the $d$ terms together, each term keeping the sign in front of it.
Add the coefficients within each group, remembering that $c$ on its own means $1c$ .
Stop when one $c$ term and one $d$ term are left, because unlike terms do not combine.

Worked Solution [2 marks]

Rule - Collecting like terms: terms with the same letter combine by adding their coefficients, and the sign in front of a term travels with it.

Step 1: Group the like terms

7c - 5d + c + 3d = 7c + c - 5d + 3d

(Reason: Each term moves together with the sign in front of it, so the

5d

keeps its minus sign while

c

and

3d

stay positive.)

Step 2: Combine the

c

terms

7c + c = 8c

(Reason: A

c

on its own means

1c

, so the coefficients add to

7 + 1 = 8

Step 3: Combine the

d

terms

-5d + 3d = -2d

(Reason: Start at

-5

and add

3

-5 + 3 = -2

. The minus sign belongs to the

5d

alone, so the

3d

is added, not taken away.)

Step 4: Write the simplified expression

7c - 5d + c + 3d = 8c - 2d

(Reason:

8c

and

2d

have different letters, so they are unlike terms and cannot be combined. This is the final answer: going on to merge them into one term would spoil it.)

8c - 2d

Verification

Check 1: Put

c = 2

and

d = 3

into the original expression and into the answer.

7 \times 2 - 5 \times 3 + 2 + 3 \times 3 = 10

and

8 \times 2 - 2 \times 3 = 10

, so the two forms give the same value.

Check 2: Put

c = 1

and

d = 0

into the original, then

c = 0

and

d = 1

. Each choice leaves only one letter's terms.

7 + 1 = 8

is the coefficient of

c

and

-5 + 3 = -2

is the coefficient of

d

, matching

8c - 2d

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The simplified expression	Note	The Answer column gives $8c - 2d$ final answer, and the Marks column gives $2$ for it.	✓
An answer of $8c - kd$ or $kc - 2d$ , or the correct answer spoilt	B1	For answer $8c - kd$ or $kc - 2d$ or for correct answer seen and spoilt.	✓

Full marks: 2/2

In the diagram, a pair of parallel lines is crossed by two straight lines.

What are the values of $w$ , $x$ and $y$ ? [4 marks]

w =x =y =

[Total 4 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

Two parallel lines, each marked with an arrow, are crossed by a long straight line and a steep straight line.

The long line makes an angle of

158^\circ

where it crosses the right-hand parallel line, and the steep line makes an angle of

76^\circ

where it crosses the left-hand parallel line.

The diagram is not to scale, so no angle can be measured from it.

Find

The values of

w

x

and

y

Plan the Solution

$w$ sits in the same position as the $158^\circ$ angle, but at the other parallel line: corresponding angles.
$x$ and the $76^\circ$ angle make a Z shape with the steep line: alternate angles.
For $y$ , use the triangle made by the long line, the steep line and the right-hand parallel line. Its angles are $x$ , $y$ and the angle beside the $158^\circ$ on the straight parallel line.

Worked Solution [4 marks]

Rule - Angles and parallel lines: corresponding angles are equal and alternate angles are equal; angles on a straight line add up to

180^\circ

, and so do the angles in a triangle.

Step 1: Find

w

with corresponding angles

w = 158

(Reason: The long line crosses both parallel lines. At each crossing the marked angle is in the same position: below the crossing, between the part of the long line running to the left and the part of the parallel line running down to the right. So

w

and the

158^\circ

angle are corresponding angles, and corresponding angles are equal.)

Step 2: Find

x

with alternate angles

x = 76

(Reason: The steep line crosses both parallel lines.

x

is at the upper crossing, to the right of the steep line; the

76^\circ

angle is at the lower crossing, to the left of it. Both lie between the parallel lines, on opposite sides of the steep line, so they form a Z shape: they are alternate angles, and alternate angles are equal.)

Step 3: Use the straight line at the right-hand crossing

180 - 158 = 22

(Reason: The

158^\circ

angle and the angle beside it, above the long line, together make the straight right-hand parallel line, so they add up to

180^\circ

. This

22^\circ

angle is one corner of the triangle made by the long line, the steep line and the right-hand parallel line.)

Step 4: Find

y

from the angles in the triangle

y = 180 - 76 - 22

y = 82

(Reason: The triangle's corners are the three points where its lines cross. Its angles are

x = 76^\circ

at the top,

22^\circ

on the right, and

y

where the steep line meets the long line. Angles in a triangle add up to

180^\circ

w = 158

x = 76

y = 82

Verification

Check 1: The

158^\circ

angle lies outside the triangle, beside its

22^\circ

corner, so it is an exterior angle of the triangle. An exterior angle equals the sum of the two interior angles opposite it, which are

x

and

y

76 + 82 = 158

, which is the marked angle.

Check 2: Use the other triangle, made by the long line, the steep line and the left-hand parallel line. Its corners are the given

76^\circ

, the angle beside

w

on the straight long line, which is

180 - 158 = 22

degrees, and the angle vertically opposite

y

180 - 76 - 22 = 82

, and vertically opposite angles are equal, so

y = 82

again.

Check 3: Co-interior angles between parallel lines add up to

180^\circ

. At the upper crossing of the steep line, the angle beside

x

on the right-hand parallel line is co-interior with the

76^\circ

angle: both are to the left of the steep line, between the parallels. The angle beside

x

180 - 76 = 104

, and

104 + 76 = 180

, as co-interior angles must.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The values of $w$ , $x$ and $y$	Note	The Answer column gives $[w =]\ 158$ , $[x =]\ 76$ and $[y =]\ 82$ , and the Marks column gives $4$ for it.	✓
The value of $w$	B1	For $w$ correct.	✓
The value of $x$	B1	For $x$ correct.	✓
The value of $y$	B2FT	For $y = 158 - \textit{their}\ x$ correctly evaluated.	✓
The value of $y$ : the alternative to the B2FT	B1	Or for $22$ (identified), or $82$ in position vertically opposite to $y$ , or for $y = 158 - \textit{their}\ x$ .	✓

Full marks: 4/4

Laura puts $\$1500$ into a savings account that pays simple interest at $3\%$ per year.

How much are her savings worth altogether at the end of $6$ years? [3 marks]

[Total 3 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

Laura puts

\$1500

into a savings account.

The account pays simple interest at

3\%

per year.

The value is wanted at the end of

6

years. No calculator: every step is done by hand.

Find

The total value of her savings, in dollars, at the end of

6

years: the

\$1500

she put in together with all the interest.

Plan the Solution

Work out $3\%$ of $\$1500$ by finding $1\%$ first and then multiplying by $3$ . That is the interest for one year.
Simple interest is the same amount every year, so multiply one year's interest by $6$ .
Add the interest to the $\$1500$ : the question asks for the total value, not just the interest.

Worked Solution [3 marks]

Rule - Simple interest:

I = \dfrac{P \times R \times T}{100}

, where

P

is the amount invested,

R

is the rate per cent per year and

T

is the number of years. The total value is

P + I

Step 1: Find

3\%

\$1500

\dfrac{1500}{100} = 15

3 \times 15 = 45

(Reason: Per cent means out of

100

, so

1\%

1500

1500

divided by

100

, which is

15

. Three lots of that make

3\%

, so one year's interest is

\$45

Step 2: Find the interest for all

6

years

45 \times 6 = 270

(Reason: Simple interest is always worked out on the original

\$1500

, so every year earns the same

\$45

. Interest already paid into the account earns nothing more; that would be compound interest.)

Step 3: Add the interest to the amount she put in

1500 + 270 = 1770

(Reason: The total value at the end of

6

years is the

\$1500

she put in plus the interest from every one of the

6

years.)

\$1770

Verification

Check 1: Use the formula in one go, with

P = 1500

R = 3

and

T = 6

. Cancel the

100

into the

1500

first, which leaves

15

, and

3 \times 6 = 18

\dfrac{1500 \times 3 \times 6}{100} = 15 \times 18 = 270

, and

1500 + 270 = 1770

Check 2: Over

6

years the interest adds up to

6 \times 3 = 18

per cent of the

\$1500

, so the total value is

118\%

of it. Since

1\%

1500

15

118 \times 15 = 1770

Check 3: Work backwards from the answer: take away the

\$1500

to leave the interest, share it over the

6

years, and compare one year's interest with the

\$1500

1770 - 1500 = 270

\dfrac{270}{6} = 45

and

\dfrac{45}{1500} = \dfrac{3}{100}

, which is

3\%

per year.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The total value at the end of $6$ years	Note	The Answer column gives $1770$ , and the Marks column gives $3$ for it.	✓
The interest for the $6$ years	B2	For $270$ .	✓
The total value: the alternative to the B2	M2	Or for $1500 + \dfrac{1500 \times 3 \times 6}{100}$ .	✓
The interest calculation: the alternative to the B2 and the M2	M1	Or for $\dfrac{1500 \times 3[\times 6]}{100}$ .	✓

Full marks: 3/3

What is the value of this calculation?

$\dfrac{5}{6} - \dfrac{2}{3} \times \dfrac{3}{8}$ [3 marks]

[Total 3 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

The calculation

\dfrac{5}{6} - \dfrac{2}{3} \times \dfrac{3}{8}

, which has no brackets.

No calculator: every step is done by hand.

Find

The value of

\dfrac{5}{6} - \dfrac{2}{3} \times \dfrac{3}{8}

Plan the Solution

There are no brackets, so the order of operations (BIDMAS) says the multiplication $\dfrac{2}{3} \times \dfrac{3}{8}$ is done before the subtraction.
Multiply the two fractions: numerator times numerator, and denominator times denominator.
Write $\dfrac{5}{6}$ over the same denominator as the product, subtract the numerators, then simplify.

Worked Solution [3 marks]

Rule - Order of operations with fractions: multiply before subtracting. To multiply,

\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}

. To subtract, write both fractions over a common denominator first, then subtract the numerators.

Step 1: Do the multiplication first

\dfrac{2}{3} \times \dfrac{3}{8} = \dfrac{2 \times 3}{3 \times 8} = \dfrac{6}{24}

(Reason: Multiplication comes before subtraction, so

\dfrac{2}{3} \times \dfrac{3}{8}

is worked out first: multiply the numerators together and the denominators together. Leave

\dfrac{6}{24}

as it is for now, because

24

is a multiple of

6

, which makes the subtraction easier. Working from left to right instead gives

\dfrac{5}{6} - \dfrac{2}{3} = \dfrac{1}{6}

and then

\dfrac{1}{6} \times \dfrac{3}{8} = \dfrac{1}{16}

, which is the wrong answer the mark scheme's SC1 names.)

Step 2: Write

\dfrac{5}{6}

over the denominator

24

\dfrac{5}{6} = \dfrac{5 \times 4}{6 \times 4} = \dfrac{20}{24}

(Reason: The smallest number that both

6

and

24

divide into is

24

, because

6 \times 4 = 24

. Multiplying the top and the bottom of

\dfrac{5}{6}

by the same number,

4

, changes how the fraction looks but not its value.)

Step 3: Subtract the numerators

\dfrac{20}{24} - \dfrac{6}{24} = \dfrac{14}{24}

(Reason: Both fractions are now in twenty-fourths, so take the numerators away,

20 - 6 = 14

, and keep the denominator

24

Step 4: Simplify

\dfrac{14}{24} = \dfrac{7 \times 2}{12 \times 2} = \dfrac{7}{12}

(Reason: Both

14

and

24

are even, so the common factor

2

cancels from the top and the bottom.

7

and

12

have no common factor other than

1

, so

\dfrac{7}{12}

is in its simplest form.)

\dfrac{7}{12}

Verification

Check 1: Cancel before multiplying: the

3

on top of

\dfrac{3}{8}

cancels with the

3

underneath

\dfrac{2}{3}

. Then subtract over the common denominator

12

instead of

24

\dfrac{2}{3} \times \dfrac{3}{8} = \dfrac{2}{8} = \dfrac{1}{4}

, then

\dfrac{5}{6} - \dfrac{1}{4} = \dfrac{10}{12} - \dfrac{3}{12} = \dfrac{7}{12}

Check 2: Adding undoes subtracting, so adding the product back on to the answer must give

\dfrac{5}{6}

. The product

\dfrac{6}{24}

\dfrac{1}{4}

, which is

\dfrac{3}{12}

\dfrac{7}{12} + \dfrac{3}{12} = \dfrac{10}{12} = \dfrac{5}{6}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The value of the calculation	Note	The Answer column gives $\dfrac{7}{12}$ oe, and the Marks column gives $3$ for it.	✓
The product of the two fractions	M1	For $\dfrac{6}{24}$ oe.	✓
The subtraction over a common denominator	M1	For correct use of common denominator in subtraction $\dfrac{5}{6} - \textit{their}\ \dfrac{6}{24}$ , e.g. $\dfrac{20}{24}$ and $\dfrac{6}{24}$ oe.	✓
Special case: answer $\dfrac{1}{16}$	SC1	If $0$ scored, for answer $\dfrac{1}{16}$ oe.	✓

Full marks: 3/3

A regular polygon has interior angles of $150^\circ$ .

How many sides does this polygon have? [2 marks]

[Total 2 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

A regular polygon, so all of its sides are equal and all of its interior angles are equal.

Each interior angle is

150^\circ

Find

The number of sides of the polygon,

n

Plan the Solution

At each corner the interior angle and the exterior angle sit on a straight line, so one exterior angle is $180^\circ$ take away the interior angle.
The exterior angles of any polygon add up to $360^\circ$ . In a regular polygon they are all equal, so $360^\circ$ divided by one exterior angle gives the number of sides.

Worked Solution [2 marks]

Rule - Exterior angles of a regular polygon:

\text{number of sides} = \dfrac{360^\circ}{\text{one exterior angle}}

Find one exterior angle

180 - 150 = 30

(Reason: The interior and exterior angles at a corner make a straight line, and angles on a straight line add up to

180^\circ

, so each exterior angle is

30^\circ

Divide

360^\circ

by one exterior angle

\dfrac{360}{180 - 150} = \dfrac{360}{30} = 12

(Reason: Going once round the polygon turns through

360^\circ

in total, a turn of

30^\circ

at every corner. That takes

12

corners, and a polygon has as many sides as corners.)

12

sides

Verification

Check 1 - the interior angle sum: The interior angles of a

12

-sided polygon add up to

(12 - 2) \times 180 = 1800

degrees. Shared equally between its

12

angles, each one is

\dfrac{1800}{12} = 150

degrees, the interior angle given.

Check 2 - the interior angle equation: Set the interior angle formula equal to

150

and solve:

\dfrac{180(n - 2)}{n} = 150

gives

180n - 360 = 150n

, so

30n = 360

n = 12

, the same number of sides.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The number of sides	Note	The Answer column gives $12$ , and the Marks column gives $2$ for it.	✓
Method for the number of sides	M1	For $\dfrac{360}{180 - 150}$ or for $\dfrac{180(n - 2)}{n} = 150$ .	✓

Full marks: 2/2

The grid shows the line $x + y = 7$ .

(a) Draw the line $y = 2x + 1$ on the grid. [2 marks]

(b) Solve this pair of simultaneous equations using your graph.
$x + y = 7$
$y = 2x + 1$ [1 mark]

(b) x =(b) y =

[Total 3 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

The line

x + y = 7

is already drawn on a grid numbered from

0

8

across and from

0

9

up.

Part (a) adds a second line,

y = 2x + 1

, to the same grid.

Find

(a) The line

y = 2x + 1

, drawn on the grid. (b) The values of

x

and

y

that make both equations true at once, read from the graph.

Plan the Solution

(a) Choose some values of $x$ , work out $y = 2x + 1$ for each one, then plot the points and rule one straight line through them.
The grid stops at $y = 9$ , so use $x$ from $0$ to $4$ : over that range $y$ runs from $1$ up to $9$ .
(b) Every point on a line fits that line's equation, so the one point that is on both lines fits both equations. Read off the coordinates of the point where the lines cross.

Worked Solution [3 marks]

Rule - Solving simultaneous equations with a graph: draw both lines on the same axes. The point where they cross is the solution, its

x

-coordinate giving

x

and its

y

-coordinate giving

y

Step 1: work out five points on

y = 2x + 1

2 \times 0 + 1 = 1

2 \times 1 + 1 = 3

2 \times 2 + 1 = 5

2 \times 3 + 1 = 7

2 \times 4 + 1 = 9

(Reason: Put each chosen value of

x

into

2x + 1

: multiply it by

2

, then add

1

. The lines take

x = 0

1

2

3

and

4

in turn, and each answer is the

y

-value that goes with that

x

Step 2: plot the points and rule the line

\begin{array}{|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 \\ \hline y & 1 & 3 & 5 & 7 & 9 \\ \hline \end{array}

(Reason: Plot

(0, 1)

(1, 3)

(2, 5)

(3, 7)

and

(4, 9)

, then join them with a ruler right across the grid, from the

y

-axis to the top edge. The points must lie in one straight line, which is the check that none of them is wrong: each step of

1

square to the right goes

2

squares up, so the gradient is

2

, and the line meets the

y

-axis at

1

. It is the gold line in the figure at Step 3.)

Step 3: read off where the two lines cross

x = 2

y = 5

(Reason: The gold line crosses the navy line

x + y = 7

at a grid point two squares to the right of the

y

-axis and five squares up. The dashed lines from that point, down to the

x

-axis and across to the

y

-axis, give the two readings.)

(a) the line

y = 2x + 1

, ruled from

(0, 1)

(4, 9)

(b)

x = 2

y = 5

Verification

Check 1: Test the drawn line against the mark scheme's description. From

(0, 1)

(4, 9)

it rises

8

squares while it runs

4

across.

\dfrac{8}{4} = 2

, so the gradient is

2

, and the line meets the

y

-axis at

1

Check 2: Put

x = 2

and

y = 5

back into the first equation,

x + y = 7

2 + 5 = 7

Check 3: Put the same values into the second equation,

y = 2x + 1

2 \times 2 + 1 = 5

, which is the value of

y

Check 4: Solve the pair without the graph. Replace

y

in the first equation with

2x + 1

x + 2x + 1 = 7

, so

3x = 6

, giving

x = 2

and then

y = 2 \times 2 + 1 = 5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The line drawn	Note	The Answer column gives “Correct line drawn”, and the Marks column gives $2$ for it.	✓
(a) Partial Marks	M1	For a line with gradient $2$ or for a line with positive gradient and intercept at $y = 1$ .	✓
(b) The solution	Note	The Answer column gives $[x =]\ 2$ , $[y =]\ 5$ , and the Marks column gives $1$ for it.	✓
(b) Follow through	Note	FT the intersection of their (a) with the given line.	✓

Full marks: 3/3

What fraction is equal to the recurring decimal $0.2\dot{6}$ ?
Give your answer in its simplest form. [3 marks]

[Total 3 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

The recurring decimal

0.2\dot{6}

, whose dot sits over the

6

alone, so only the

6

repeats

Find

The decimal written as a fraction in its simplest form

Plan the Solution

Call the decimal $x$ .
Multiply it by $10$ and by $100$ , which gives two numbers with exactly the same recurring part after the point.
Take the smaller from the larger. The recurring parts cancel and a whole number is left.
Divide to get a fraction, then cancel it until nothing more will cancel.

Worked Solution [3 marks]

Rule - shift, then subtract: multiply a recurring decimal by two powers of

10

that leave the same recurring tail after the point. Subtracting one line from the other removes the tail, and dividing turns what is left into a fraction.

Step 1: Give the decimal a name

x = 0.2\dot{6} = 0.2666\ldots

(Reason: The dot is over the

6

only, so the

2

appears once and every digit after it is a

6

. Calling the decimal

x

lets it be handled with algebra.)

Step 2: Multiply by

10

and by

100

10x = 2.\dot{6}

100x = 26.\dot{6}

(Reason: One digit comes before the repeating part, so multiplying by

10

moves the point just past it. The repeating part is one digit long, so multiplying by

100

moves the point one place further. Both lines now end in the same tail,

0.666\ldots

, after the point.)

Step 3: Subtract to clear the recurring tail

100x - 10x = 26.\dot{6} - 2.\dot{6}

90x = 24

(Reason: The two tails match digit for digit, so they cancel exactly and only the whole numbers are left on the right,

26

take away

2

. On the left,

100x - 10x

collects to

90x

Step 4: Divide, then cancel to the simplest form

x = \dfrac{24}{90}

\dfrac{24}{90} = \dfrac{4}{15}

(Reason: Dividing both sides by

90

gives the fraction. The highest common factor of

24

and

90

6

, because

24 = 6 \times 4

and

90 = 6 \times 15

. After dividing it out,

4 = 2 \times 2

and

15 = 3 \times 5

have no factor in common, so the fraction is in its simplest form.)

\dfrac{4}{15}

Verification

Check 1 - put the fraction back into the subtraction: The subtraction ends at

90x = 24

, so multiplying the answer by

90

has to give

24

again.

\dfrac{4}{15} \times 90 = 24

Check 2 - divide it out by hand: Short division of

4

15

15

goes into

40

twice with

10

left over, then into

100

six times with

10

left over again. The same remainder keeps coming back, so every digit after that is a

6

. The division gives

0.2666\ldots

, which is

0.2\dot{6}

Check 3 - split the decimal into two parts: Write the decimal as

0.2

plus

0.0\dot{6}

. The first part is

\dfrac{2}{10}

. The second is one tenth of

0.\dot{6}

, and a single repeating digit is that digit over

9

, so it is one tenth of

\dfrac{6}{9}

, which is

\dfrac{6}{90}

\dfrac{2}{10} + \dfrac{6}{90} = \dfrac{18}{90} + \dfrac{6}{90} = \dfrac{24}{90} = \dfrac{4}{15}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The fraction in its simplest form	Note	The Answer column gives $\dfrac{4}{15}$ , and the Marks column gives $3$ for it.	✓
The right fraction, not yet cancelled	B2	For $\dfrac{24}{90}$ oe.	✓
A subtraction, an equation or a sum that leads to that fraction	M1	Or for $26.66\ldots - 2.66\ldots$ oe, or for $90x = 24$ oe, or for $\dfrac{2}{10} + \dfrac{6}{90}$ oe.	✓
Where the B2 value comes from	Note	The $\dfrac{24}{90}$ in the B2 row is $\dfrac{4}{15}$ before it is cancelled.	✓

Full marks: 3/3

$\mathbf{m} = \begin{pmatrix} 11 \\ 5 \end{pmatrix}$ and $\mathbf{n} = \begin{pmatrix} 8 \\ -3 \end{pmatrix}$

(a) What is the column vector $2\mathbf{m} - \mathbf{n}$ ? [2 marks]

(b) The magnitude of the vector $\begin{pmatrix} 5 \\ \sqrt{y} \end{pmatrix}$ is $7$ .
What is the value of $y$ ? [2 marks]

(a)(b) y =

[Total 4 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

The column vectors

\mathbf{m} = \begin{pmatrix} 11 \\ 5 \end{pmatrix}

and

\mathbf{n} = \begin{pmatrix} 8 \\ -3 \end{pmatrix}

(b) The vector

\begin{pmatrix} 5 \\ \sqrt{y} \end{pmatrix}

has a magnitude of

7

Find

(a) The column vector

2\mathbf{m} - \mathbf{n}

(b) The value of

y

Plan the Solution

(a) Double each component of $\mathbf{m}$ , then take away each component of $\mathbf{n}$ , top from top and bottom from bottom.
(b) The magnitude is the length of the vector. Its two components are the two shorter sides of a right-angled triangle, so Pythagoras' theorem links them to the length $7$ .
Square both sides so that no square root is left, then solve for $y$ .

Worked Solution [4 marks]

Rule - Vectors work component by component: a number multiplies every component, and subtracting pairs the top entries together and the bottom entries together. The magnitude of

\begin{pmatrix} a \\ b \end{pmatrix}

\sqrt{a^2 + b^2}

Step 1: Double

\mathbf{m}

2\mathbf{m} = 2\begin{pmatrix} 11 \\ 5 \end{pmatrix} = \begin{pmatrix} 22 \\ 10 \end{pmatrix}

(Reason: Multiplying a vector by

2

multiplies each of its components by

2

2 \times 11 = 22

on top and

2 \times 5 = 10

underneath.)

Step 2: Take

\mathbf{n}

away, one component at a time

2\mathbf{m} - \mathbf{n} = \begin{pmatrix} 22 \\ 10 \end{pmatrix} - \begin{pmatrix} 8 \\ -3 \end{pmatrix}

\begin{pmatrix} 22 - 8 \\ 10 - (-3) \end{pmatrix} = \begin{pmatrix} 14 \\ 13 \end{pmatrix}

(Reason: The top entry of

\mathbf{n}

comes off the top entry of

2\mathbf{m}

, and the bottom entry off the bottom entry. The bottom entry of

\mathbf{n}

is negative, and taking away a negative number is the same as adding the positive one, so

10 - (-3)

10 + 3

Step 3: Write the magnitude with Pythagoras' theorem

\sqrt{5^2 + (\sqrt{y})^2} = 7

(Reason: Drawn as an arrow, the vector goes

5

across and

\sqrt{y}

up. Those two moves are the shorter sides of a right-angled triangle, and the arrow itself is the hypotenuse, whose length is the magnitude

7

Step 4: Square both sides and solve for

y

5^2 + (\sqrt{y})^2 = 7^2

25 + y = 49

y = 49 - 25 = 24

(Reason: Squaring both sides removes the outer square root. Squaring

\sqrt{y}

gives back

y

, and

5^2 = 25

and

7^2 = 49

, so taking

25

from both sides leaves

y

. The square root of

y

only exists when

y

is at least

0

, and

24

is.)

(a)

\begin{pmatrix} 14 \\ 13 \end{pmatrix}

(b)

y = 24

Verification

Check 1 - (a) add n back on: If the answer is

2\mathbf{m} - \mathbf{n}

, then adding

\mathbf{n}

to it has to give

2\mathbf{m}

again.

\begin{pmatrix} 14 \\ 13 \end{pmatrix} + \begin{pmatrix} 8 \\ -3 \end{pmatrix} = \begin{pmatrix} 22 \\ 10 \end{pmatrix}

, which is

2\mathbf{m}

Check 2 - (a) group it a different way:

2\mathbf{m} - \mathbf{n}

is also

\mathbf{m} + (\mathbf{m} - \mathbf{n})

. First

\mathbf{m} - \mathbf{n} = \begin{pmatrix} 11 - 8 \\ 5 + 3 \end{pmatrix} = \begin{pmatrix} 3 \\ 8 \end{pmatrix}

, then add

\mathbf{m}

to it.

\begin{pmatrix} 11 \\ 5 \end{pmatrix} + \begin{pmatrix} 3 \\ 8 \end{pmatrix} = \begin{pmatrix} 14 \\ 13 \end{pmatrix}

Check 3 - (b) put y = 24 back in: With

y = 24

the vector is

\begin{pmatrix} 5 \\ \sqrt{24} \end{pmatrix}

. Work out its magnitude from the start.

\sqrt{5^2 + (\sqrt{24})^2} = \sqrt{25 + 24} = \sqrt{49} = 7

Check 4 - (b) difference of two squares: From

y + 5^2 = 7^2

, the value of

y

7^2 - 5^2

, and a difference of two squares factorises, so neither square has to be worked out.

7^2 - 5^2 = (7 - 5)(7 + 5) = 2 \times 12 = 24

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The column vector $2\mathbf{m} - \mathbf{n}$	Note	The Answer column gives $\begin{pmatrix} 14 \\ 13 \end{pmatrix}$ , and the Marks column gives $2$ for it.	✓
(a) One entry right, or $2\mathbf{m}$	B1	For $\begin{pmatrix} 14 \\ k \end{pmatrix}$ , or for $\begin{pmatrix} k \\ 13 \end{pmatrix}$ , or for $\begin{pmatrix} 22 \\ 10 \end{pmatrix}$ .	✓
(a) Where the B1 forms come from	Note	In the first two forms one entry matches the answer, $14$ on top or $13$ underneath, with $k$ in the other place. The third form, $\begin{pmatrix} 22 \\ 10 \end{pmatrix}$ , is $2\mathbf{m}$ before $\mathbf{n}$ is taken away.	✓
(b) The value of $y$	Note	The Answer column gives $24$ , and the Marks column gives $2$ for it.	✓
(b) Pythagoras' theorem with the magnitude	M1	For $y + 5^2 = 7^2$ or better.	✓
(b) Why the M1 has $y$ and no square root	Note	The $y$ in the M1 form is $(\sqrt{y})^2$ , the square of the vector's lower entry, so the square root has already gone.	✓

Full marks: 4/4

Some information about the scores a group of friends got in a quiz is given in the table.

$\begin{array}{|l||c|c|c|}\hline \text{Score} & 4 & 5 & 8 \\ \hline \text{Frequency} & 2 & 4 & n \\ \hline \end{array}$

The mean of their scores is $6$ .

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

n =

[Total 3 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

A frequency table of quiz scores: the scores

4

5

and

8

have frequencies

2

4

and

n

The mean of the scores is

6

. No calculator: every step is done by hand.

Find

The value of

n

, the number of friends who scored

8

Plan the Solution

The total of all the scores is each score times its frequency, added up: $2 \times 4 + 4 \times 5 + n \times 8$ .
The number of friends is the total of the frequencies, $2 + 4 + n$ .
The total divided by the number of friends is the mean, so set that fraction equal to $6$ .
Multiply both sides by the number of friends to clear the fraction, then solve the linear equation that is left.

Worked Solution [3 marks]

Rule - mean from a frequency table:

\text{mean} = \dfrac{\text{total of all the scores}}{\text{number of people}}

, where the total is each score times its frequency, added up, and the number of people is the total of the frequencies.

Step 1: Write the mean as a fraction

\text{mean} = \dfrac{2 \times 4 + 4 \times 5 + n \times 8}{2 + 4 + n}

\dfrac{2 \times 4 + 4 \times 5 + n \times 8}{2 + 4 + n} = 6

(Reason: Each score counts as many times as its frequency:

2

friends scored

4

4

scored

5

and

n

scored

8

, so the top is the total of all the scores. The bottom adds up the frequencies, which counts the friends. The mean is given as

6

Step 2: Simplify the top and the bottom

2 \times 4 + 4 \times 5 + n \times 8 = 8 + 20 + 8n = 28 + 8n

2 + 4 + n = 6 + n

(Reason:

2 \times 4 = 8

and

4 \times 5 = 20

, while

n \times 8

stays as

8n

because

n

is not known yet. Collecting the numbers gives

28

on the top and

6

on the bottom.)

Step 3: Multiply both sides by the number of friends

8 + 20 + 8n = 6(2 + 4 + n)

28 + 8n = 36 + 6n

(Reason: The total divided by the number of friends gives the mean, so the total equals the mean times the number of friends. Multiplying both sides by

2 + 4 + n

clears the fraction. On the right the

6

multiplies every term in the bracket:

6 \times 2 + 6 \times 4 + 6 \times n = 12 + 24 + 6n

Step 4: Solve for

n

8n - 6n = 36 - 28

2n = 8

n = 4

(Reason: Taking

6n

and

28

from both sides puts the

n

terms on the left and the numbers on the right. Dividing both sides by

2

then leaves

n

on its own.)

n = 4

Verification

Check 1 - put $n = 4$ back into the table: With

4

friends scoring

8

, the total of the scores is

8 + 20 + 4 \times 8

and there are

2 + 4 + 4

friends. Dividing the total by the number of friends must give the mean.

\dfrac{8 + 20 + 32}{2 + 4 + 4} = \dfrac{60}{10} = 6

Check 2 - balance the scores around the mean: At the mean, the amounts by which scores fall below it and rise above it are equal. Each score of

4

2

below

6

and each score of

5

1

below; each score of

8

2

above, and there are

4

of them. Below:

2 \times 2 + 4 \times 1 = 8

. Above:

4 \times 2 = 8

. The two balance.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The value of $n$	Note	The Answer column gives $4$ nfww, and the Marks column gives $3$ for it.	✓
The equation for the total of the scores	M2	For $8 + 20 + 8n = 6(2 + 4 + n)$ or better.	✓
The mean as a fraction, or one side of the equation, instead of the M2	M1	Or for $\dfrac{2 \times 4 + 4 \times 5 + n \times 8}{2 + 4 + n} [= 6]$ oe, or $8 + 20 + 8n$ or $6(2 + 4 + n)$ .	✓
Where the M2 equation comes from	Note	$8 + 20 + 8n$ is the total of all the scores, $2 \times 4 + 4 \times 5 + n \times 8$ , and $6(2 + 4 + n)$ is the mean times the number of people, so the equation says that the two ways of finding the total agree.	✓
What nfww means	Note	nfww is the scheme's abbreviation for not from wrong working: the answer $4$ must not come from wrong working.	✓

Full marks: 3/3

The three points $A$ , $B$ and $C$ lie on a circle with centre $O$ .
The tangent to the circle at $A$ is the line $DE$ .
The diagram marks two angles at $C$ : angle $ACO = 35^{\circ}$ and angle $BCO = 40^{\circ}$ .

What is the size of each of these angles?

(a) Angle $AOC$ [1 mark]

(b) Angle $ABC$ [1 mark]

(d) Angle $OAB$ [1 mark]

(a) Angle AOC =(b) Angle ABC =(c) Angle DAC =(d) Angle OAB =

[Total 4 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

A

B

and

C

lie on a circle with centre

O

, so

OA = OB = OC

, because all three are radii

The line

DE

is the tangent to the circle at

A

Angle

ACO = 35^{\circ}

and angle

BCO = 40^{\circ}

The diagram is not to scale, so no angle can be measured off it, and there is no calculator: every step is done by hand.

Find

(a) Angle

AOC

(b) Angle

ABC

DAC

(d) Angle

OAB

Plan the Solution

(a) $OA$ and $OC$ are radii, so triangle $AOC$ is isosceles. Its base angles are equal, and the angle sum of the triangle gives the angle at $O$ .
(b) Angle $AOC$ at the centre and angle $ABC$ at the circumference stand on the same arc $AC$ , so halve the answer to (a).
(c) Angle $DAC$ lies between the tangent and the chord $AC$ , so the alternate segment theorem makes it equal to angle $ABC$ .
(d) Add the two marked angles to get angle $ACB$ , use the angle sum of triangle $ABC$ to find angle $BAC$ , then take away angle $OAC$ from (a).

Worked Solution [4 marks]

Rule - Circle theorems: two radii make an isosceles triangle; the angle at the centre is twice the angle at the circumference standing on the same arc; the angle between a tangent and a chord equals the angle in the alternate segment.

Step 1: (a) Angle

AOC

from the isosceles triangle

AOC

OA = OC

OAC = OCA = 35^{\circ}

AOC = 180 - 2 \times 35

180 - 70 = 110

(Reason: Both

OA

and

OC

are radii, so they are the same length and triangle

AOC

is isosceles. The angles opposite the equal sides are equal, so angle

OAC

is also

35^{\circ}

. The three angles of a triangle add to

180^{\circ}

, and the two base angles take

70^{\circ}

of that, which leaves

110^{\circ}

O

Step 2: (b) Angle

ABC

, half the angle at the centre

ABC = \dfrac{1}{2} \times AOC

\dfrac{110}{2} = 55

(Reason: Angle

AOC

is at the centre and angle

ABC

is at the circumference, and both stand on the same arc

AC

, the arc that does not contain

B

. The angle at the centre is twice the angle at the circumference, so angle

ABC

is half of the

110^{\circ}

found in (a).)

Step 3: (c) Angle

DAC

, by the alternate segment theorem

DAC = ABC

DAC = 55^{\circ}

(Reason: The line

DA

is part of the tangent at

A

, and

AC

is a chord. The angle between a tangent and a chord equals the angle in the alternate segment: the angle the chord makes at any point of the circle on the other side of

AC

from

D

B

is such a point, so angle

DAC

equals angle

ABC

, which is

55^{\circ}

from (b).)

Step 4: (d) Angle

OAB

, from the angles of triangle

ABC

ACB = ACO + BCO

35 + 40 = 75

BAC = 180 - ACB - ABC

180 - 75 - 55 = 50

OAB = BAC - OAC

50 - 35 = 15

(Reason: The radius

OC

splits angle

ACB

into the two marked angles, so angle

ACB

75^{\circ}

. The angles of triangle

ABC

add to

180^{\circ}

, which leaves

50^{\circ}

for angle

BAC

. The radius

OA

splits angle

BAC

into angle

OAC

, which is

35^{\circ}

from step 1, and angle

OAB

, the angle wanted. The figure at this step marks every angle the four parts found.)

(a)

AOC = 110^{\circ}

(b)

ABC = 55^{\circ}

(c)

DAC = 55^{\circ}

(d)

OAB = 15^{\circ}

Verification

Check 1 - (c) by the tangent and the radius: The radius

OA

meets the tangent at

A

at a right angle, so angle

DAO

90^{\circ}

C

is on the same side of

OA

D

, so angle

DAC

is what is left of that right angle after angle

OAC

90 - 35 = 55

, the same as (c)

Check 2 - (b) and (d) from the three triangles at $O$ : Join

OB

. Triangle

OBC

is isosceles too, so angle

OBC

40^{\circ}

and angle

BOC

180 - 80 = 100

. The three angles at

O

make a full turn, so angle

AOB

360 - 110 - 100 = 150

, and the isosceles triangle

OAB

shares what is left of its angle sum equally between its two base angles.

\dfrac{180 - 150}{2} = 15

for angle

OAB

and for angle

OBA

, so angle

ABC

15 + 40 = 55

Check 3 - (d) by the tangent at $A$ : The angle between the tangent

AE

and the chord

AB

equals the angle in the alternate segment, angle

ACB

, which is

75^{\circ}

. Angle

OAE

is a right angle, and angle

OAB

is what is left of it.

90 - 75 = 15

, the same as (d)

Check 4 - the angle sum of triangle $ABC$ : Angle

BAC

35 + 15

, angle

ABC

55

and angle

ACB

35 + 40

. The three must add to

180

50 + 55 + 75 = 180

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Angle $AOC$	Note	The Answer column gives $110$ , and the Marks column gives $1$ for it.	✓
(b) Angle $ABC$	Note	The Answer column gives $55$ , and the Marks column gives $1$ for it.	✓
(b) Follow-through from (a)	Note	FT $\dfrac{\text{their } 110}{2}$ .	✓
(c) Angle $DAC$	Note	The Answer column gives $55$ , and the Marks column gives $1$ for it.	✓
(c) Follow-through from (b)	Note	FT their (b).	✓
(c) The condition on that follow-through	Note	Provided $\text{their (b)} < 90$ .	✓
(d) Angle $OAB$	Note	The Answer column gives $15$ , and the Marks column gives $1$ for it.	✓
(d) Follow-through from (b)	Note	FT $70 - \text{their (b)}$ .	✓
(d) The condition on that follow-through	Note	Provided $\text{their (b)} < 70$ .	✓
What FT and their mean	Note	FT is the scheme's abbreviation for follow through after error. Their (b) means the candidate's own answer to (b), and their $110$ means the candidate's own answer to (a), used where the $110$ would be.	✓
Where the 70 comes from	Note	The angle sum of triangle $ABC$ , less angle $ACB$ (the two marked angles at $C$ ) and less angle $OAC$ (equal to angle $OCA$ ), leaves $70$ for angle $OAB$ and angle $ABC$ together. So angle $OAB$ is $70$ minus angle $ABC$ , which is why (d) follows through from (b).	✓

Full marks: 4/4

The graph of $y = \mathrm{f}(x)$ and the point $P(-2, 11)$ are drawn on the grid below.

The graph of $y = \mathrm{f}(x)$ is touched at the point $(a, b)$ by the tangent from $P$ .
Both $a$ and $b$ are integers.

(a) Draw this tangent, and use it to find the value of $a$ and the value of $b$ . [2 marks]

(b) What is the equation of the tangent?
Give your answer in the form $y = mx + c$ . [3 marks]

(a) a =(a) b =(b) y =

[Total 5 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

The graph of

y = \mathrm{f}(x)

, drawn on a grid with

x

from

-2

3

and

y

from

-10

15

The point

P(-2, 11)

, marked on the left edge of the grid, above the curve.

The tangent from

P

touches the curve at

(a, b)

, where

a

and

b

are integers.

Find

(a) The values of

a

and

b

, by drawing the tangent. (b) The equation of the tangent, in the form

y = mx + c

Plan the Solution

(a) Lay a ruler through $P$ and turn it until it just touches the curve: it meets the curve at one point, with the whole curve on one side of it. Rule the line and read off the point where it touches.
(b) The gradient $m$ is rise over run between two points on the ruled line, $P$ and the touching point. Then put the touching point into $y = mx + c$ to find $c$ , and check it against where the line crosses the $y$ -axis.

Worked Solution [5 marks]

Rule - Straight line through two points:

y = mx + c

, where the gradient

m = \dfrac{\text{rise}}{\text{run}}

and

c

is where the line crosses the

y

-axis. A tangent touches a curve at one point without crossing it there.

Step 1: rule the tangent from

P

and find where it touches

(a, b) = (1, 2)

(Reason: Put the ruler's edge on

P

and turn it about

P

until it rests against the curve, touching it at one point with the whole curve below the ruler. Rule that line. It touches the curve at the grid point where

x = 1

and

y = 2

, which fits the question's clue that

a

and

b

are integers. The solution figure shows the ruled line in gold.)

Step 2: write down

a

and

b

a = 1

b = 2

(Reason: The touching point is

(a, b)

, so

a

is its

x

-coordinate and

b

is its

y

-coordinate. A correct ruled tangent together with both values earns both marks for (a).)

Step 3: find the gradient

m

\dfrac{2 - 11}{1 - (-2)} = \dfrac{-9}{3} = -3

m = -3

(Reason: Use rise over run from

P(-2, 11)

to the touching point

(1, 2)

. The rise is the change in

y

and the run is the change in

x

, both taken in the same order. The line falls from left to right, so its gradient is negative.)

Step 4: find the

y

-intercept

c

2 = -3 \times 1 + c

c = 2 + 3 = 5

(Reason: Put the touching point

(1, 2)

into

y = -3x + c

and solve for

c

. On the graph the ruled line crosses the

y

-axis at

5

, which agrees.)

Step 5: write the equation

y = -3x + 5

y = 5 - 3x

(Reason: Put

m = -3

and

c = 5

into

y = mx + c

. The two lines are the same equation, and the mark scheme writes it as

5 - 3x

(a)

a = 1

b = 2

(b)

y = 5 - 3x

Verification

Check 1: Put

x = -2

into

y = 5 - 3x

. The tangent must pass through

P

5 - 3 \times (-2) = 11

, the

y

-coordinate of

P

Check 2: Put

x = 1

into

y = 5 - 3x

. The tangent must pass through the touching point.

5 - 3 \times 1 = 2

, which is

b

Check 3: Compare the line with the curve one unit either side of the touching point. The curve passes through the grid points

(0, 3)

and

(2, -3)

5 - 3 \times 0 = 5

and

5 - 3 \times 2 = -1

, so the line is above the curve on both sides: it touches at

(1, 2)

without crossing

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The tangent and the values	Note	The Answer column gives “For correct ruled tangent and $[a =]\ 1$ , $[b =]\ 2$ ”, and the Marks column gives $2$ for it.	✓
(a) Partial Marks	B1	For correct ruled tangent or both values correct without a correct tangent.	✓
(b) The equation	Note	The Answer column gives $[y =]\ 5 - 3x$ , and the Marks column gives $3$ for it.	✓
(b) Full marks FT their (a)	3FT	Full marks FT their (a) provided $m < 0$ , $c \neq 0$ .	✓
(b) Without the 3FT: the gradient term	B1	Otherwise, for $(\text{their } {-3})x + c$ .	✓
(b) Without the 3FT: the gradient term, by rise over run	M1	Or for correct $\dfrac{\text{rise}}{\text{run}}$ for their line.	✓
(b) Without the 3FT: the intercept	B1	For $mx + c$ where $c$ is the correct intercept for their graph, $m \neq 0$ .	✓

Full marks: 5/5

For one day, a record was kept of the time each of $120$ office workers spent working at a computer.
The results are displayed in the cumulative frequency diagram.

(a) What estimate of the interquartile range does the cumulative frequency diagram give? [2 marks]

(b) For $70\%$ of the office workers, the time spent at a computer was less than $k$ hours.
What estimate of the value of $k$ does the cumulative frequency diagram give? [2 marks]

(a) h(b) k =

[Total 4 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

A cumulative frequency curve for

120

office workers, showing the time each of them spent working at a computer on one day.

Time runs from

0

10

hours along the bottom, and cumulative frequency runs from

0

120

up the side.

One small square of the grid is

0.2

hours across and

2

workers up.

(b)

70\%

of the workers spent less than

k

hours at a computer.

Find

(a) An estimate of the interquartile range, in hours. (b) An estimate of

k

Plan the Solution

(a) The lower quartile is a quarter of the way through the $120$ workers, and the upper quartile three quarters of the way. Work out those two positions, read each one across to the curve and down to the time axis, then subtract.
(b) Turn $70\%$ into a number of workers, then read across from that cumulative frequency to the curve and down in the same way.

Worked Solution [4 marks]

Rule - Quartiles from a cumulative frequency curve: for

n

values, read the lower quartile at

\dfrac{1}{4}n

and the upper quartile at

\dfrac{3}{4}n

on the cumulative frequency axis. Then

\text{IQR} = \text{UQ} - \text{LQ}

Step 1: find where the quartiles are on the cumulative frequency axis

\dfrac{1}{4} \times 120 = 30

\dfrac{3}{4} \times 120 = 90

(Reason: The lower quartile belongs to the worker a quarter of the way up the ordered list, and the upper quartile to the worker three quarters of the way up. With

120

workers, these are at cumulative frequencies

30

and

90

Step 2: read both quartiles off the curve

\text{LQ} = 4.2

\text{UQ} = 5.6

(Reason: From

30

on the cumulative frequency axis, go across to the curve and straight down to the time axis: the line lands one small square past

4

, at

4.2

hours. From

90

the line lands two small squares short of

6

, at

5.6

hours. The solution figure shows both reading lines. Either quartile read correctly is worth a mark on its own.)

Step 3: subtract the lower quartile from the upper quartile

\text{IQR} = 5.6 - 4.2 = 1.4

(Reason: The interquartile range is the whole gap between the two quartiles, in hours. Halving it gives

0.7

, which the mark scheme credits only with a special-case mark, SC1.)

Step 4: find

70\%

of the

120

workers

\dfrac{70}{100} \times 120 = 84

(Reason: The cumulative frequency axis counts workers, not percentages, and it runs up to

120

, so

70\%

of the workers means

70\%

120

Step 5: read across from

84

k = 5.4

(Reason: From

84

on the cumulative frequency axis, go across to the curve and straight down: the line lands two small squares past

5

, at

5.4

hours. So

84

workers, which is

70\%

of them, spent less than

5.4

hours at a computer. The second solution figure shows this reading.)

(a)

1.4

hours(b)

k = 5.4

Verification

Check 1: (a) Count the small squares along the time axis between the two reading lines, from

4.2

5.6

hours. Each small square is

0.2

hours wide.

7 \times 0.2 = 1.4

hours, the same interquartile range

Check 2: (b) Work backwards from

5.4

hours: go up to the curve and across, and the line meets the cumulative frequency axis at

84

. Then write

84

as a fraction of the

120

workers.

\dfrac{84}{120} = \dfrac{7}{10}

, which is

70\%

Check 3: (b)

70\%

is more than

25\%

and less than

75\%

, so

k

must lie between the lower and upper quartiles found in (a).

4.2 < 5.4 < 5.6

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The interquartile range	Note	The Answer column gives $1.4$ , and the Marks column gives $2$ for it.	✓
(a) One quartile correct	B1	For $[\text{UQ} =]\ 5.6$ or for $[\text{LQ} =]\ 4.2$	✓
(a) Special case, instead of the B1	SC1	Or for $0.7$	✓
(b) The value of $k$	Note	The Answer column gives $5.4$ , and the Marks column gives $2$ for it.	✓
(b) $70\%$ of the workers	B1	For $84$ seen.	✓

Full marks: 4/4

Solid $A$ and solid $B$ are drawn above.
A cone and a hemisphere, each of radius $6$ cm, together make up solid $A$ .
The sloping edge of the cone is $10$ cm long.
Solid $B$ is a cylinder. Its radius is $4$ cm and its height is $h$ cm.

Solid $A$ and solid $B$ have the same $\textbf{total}$ surface area.

(a) What is the value of $h$ ? [5 marks]

(b) What is the height of solid $A$ ? [3 marks]

(a) h =(b) cm

[Total 8 marks]

Show solution & mark schemeHide solution & mark scheme

Question 13 - Exam Solution

Understanding the Question

Given

Solid

A

: a cone and a hemisphere, both of radius

6

cm, with the cone's sloping edge

10

cm long

Solid

B

: a cylinder of radius

4

cm and height

h

The total surface areas of

A

and

B

are equal

No calculator is allowed, so every area stays a multiple of

\pi

, and the

\pi

cancels by hand

Find

(a) The value of

h

(b) The height of solid

A

, from the tip of the cone to the bottom of the hemisphere

Plan the Solution

(a) List the surfaces that make up each total. For $A$ : the curved surface of the cone and the curved surface of the hemisphere, because the flat circle where they are joined is inside the solid. For $B$ : two circular ends and the curved surface. Set the two totals equal and solve for $h$ .
(b) The height of $A$ is the height of the cone plus the radius of the hemisphere. The cone's height, its radius and its sloping edge form a right-angled triangle, so Pythagoras gives the height of the cone.

Worked Solution [8 marks]

Rule - Surface areas: curved surface of a cone

\pi r l

; surface of a sphere

4\pi r^2

, so the curved surface of a hemisphere is

2\pi r^2

; curved surface of a cylinder

2\pi r h

; circle

\pi r^2

. Pythagoras:

a^2 + b^2 = c^2

Step 1: (a) the curved surface of the cone

\pi \times 6 \times 10 = 60\pi

(Reason: The sloping edge is the cone's slant height

l

, so

r = 6

and

l = 10

go into

\pi r l

. Only the curved surface counts: the cone's flat base is the circle where it is joined to the hemisphere.)

Step 2: (a) the curved surface of the hemisphere

\dfrac{1}{2} \times 4\pi r^2 = 2\pi r^2

2 \times \pi \times 6^2 = 72\pi

(Reason: A hemisphere is half of a sphere, and a sphere's surface is

4\pi r^2

, so the curved surface of the hemisphere is

2\pi r^2

. Its radius is the same

6

cm as the cone's. Its flat face is the same joined circle, inside the solid, so that does not count either.)

Step 3: (a) the total surface area of solid

A

60\pi + 72\pi = 132\pi

(Reason: The two curved surfaces are the whole outside of solid

A

. The circle where the cone meets the hemisphere is covered on both sides, so it is not part of the surface.)

Step 4: (a) the total surface area of solid

B

2 \times \pi \times 4^2 = 32\pi

2 \times \pi \times 4 \times h = 8\pi h

32\pi + 8\pi h

(Reason: A cylinder has two circular ends, each

\pi r^2

, and a curved surface

2\pi r h

. With

r = 4

the two ends make

32\pi

and the curved surface is

8\pi h

. A total surface area includes both ends.)

Step 5: (a) set the two totals equal and solve

32\pi + 8\pi h = 132\pi

8\pi h = 100\pi

8h = 100

h = \dfrac{100}{8} = 12.5

(Reason: Take

32\pi

from both sides, then divide both sides by

\pi

. Every term carries

\pi

, so its value is never needed. Leaving out the cylinder's two ends would give

8\pi h = 132\pi

and the answer

16.5

instead, which is the special case in the mark scheme.)

Step 6: (b) the height of the cone, by Pythagoras

x^2 + 6^2 = 10^2

x = \sqrt{10^2 - 6^2} = \sqrt{64} = 8

(Reason: Call the height of the cone

x

cm. It runs straight down the middle of the cone, at right angles to the radius, so the height, the radius

6

and the sloping edge

10

make a right-angled triangle with the sloping edge as its hypotenuse. By hand,

100 - 36 = 64

, and

64 = 8^2

Step 7: (b) the height of solid

A

8 + 6 = 14

(Reason: Solid

A

stands from the tip of the cone to the lowest point of the hemisphere. Below the joined circle the hemisphere goes down by its radius,

6

cm, so the height is the cone's

8

cm plus

6

cm. The figure here is a cross-section through the middle of solid

A

, drawn to scale.)

(a)

h = 12.5

(b)

14

Verification

Check 1 - (a) back into solid $B$ : Put

h = 12.5

into the total surface area of solid

B

32\pi + 8\pi \times 12.5 = 132\pi

, the total surface area of solid

A

Check 2 - (a) dividing by $2\pi$ first: Every term of

60\pi + 72\pi = 32\pi + 8\pi h

is a multiple of

2\pi

. Divided through by

2\pi

, the equation becomes

30 + 36 = 16 + 4h

4h = 66 - 16 = 50

, so

h = 12.5

again

Check 3 - (b) Pythagoras the other way round: Square the height of the cone and its radius and add them. The sum must be the square of the sloping edge.

8^2 + 6^2 = 64 + 36 = 100 = 10^2

Check 4 - (b) the $3$ , $4$ , $5$ triangle: The radius

6

and the sloping edge

10

are

2 \times 3

and

2 \times 5

, so the triangle is the

3

4

5

right-angled triangle doubled, and its third side is

2 \times 4

2 \times 4 = 8

for the cone, and

8 + 6 = 14

for solid

A

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The value of $h$	Note	The Answer column gives $12.5$ oe, and the Marks column gives $5$ for it.	✓
(a) First method: the equation simplified	M4	For $8[\pi]h = 100[\pi]$ .	✓
(a) Second method: the two totals set equal	M3	Or for $\pi \times 6 \times 10 + 2 \times \pi \times 6^2 = 2 \times \pi \times 4^2 + 2 \times \pi \times 4 \times h$ oe.	✓
(a) Third method: the surface area of solid A	M1	Or for $\pi \times 6 \times 10 + 2 \times \pi \times 6^2$ oe.	✓
(a) Third method: the surface area of solid B	M1	For $2 \times \pi \times 4^2 + 2 \times \pi \times 4 \times h$ oe.	✓
(a) In place of the three methods: special case	SC2	Or for answer $16.5$ .	✓
(a) The square brackets in the first method	Note	The square brackets in $8[\pi]h = 100[\pi]$ mark the $\pi$ as optional: $8\pi h = 100\pi$ and $8h = 100$ both earn that mark.	✓
(a) Where the special case comes from	Note	The special case is the answer $16.5$ . It comes from leaving out the two circular ends of the cylinder, so that only its curved surface is set equal to the surface area of solid $A$ .	✓
(b) The height of solid $A$	Note	The Answer column gives $14$ , and the Marks column gives $3$ for it.	✓
(b) The height of the cone	M2	For $\sqrt{10^2 - 6^2}$ .	✓
(b) The height of the cone, by Pythagoras	M1	Or for $6^2 + x^2 = 10^2$ .	✓
(b) What the letter stands for	Note	In the Pythagoras line, $x$ is the height of the cone.	✓

Full marks: 8/8

The functions $\mathrm{f}$ and $\mathrm{g}$ are given by $\mathrm{f}(x) = 3x - 4$ and $\mathrm{g}(x) = 4x + 1$ .

(a) What is the value of $\mathrm{f}(-2)$ ? [1 mark]

(b) What is $\mathrm{f}^{-1}(x)$ ? [2 marks]

(d) The expression below is to be simplified.
$\dfrac{2}{\mathrm{f}(x)} - \dfrac{5}{\mathrm{g}(x)}$
Give your answer as a single fraction in terms of $x$ . [3 marks]

(a)(b) f⁻¹(x) =(c) a =(c) b =(d)

[Total 8 marks]

Show solution & mark schemeHide solution & mark scheme

Question 14 - Exam Solution

Understanding the Question

Given

\mathrm{f}(x) = 3x - 4

and

\mathrm{g}(x) = 4x + 1

, two linear functions

There is no calculator, so every step is done by hand, and every value stays exact.

Find

(a)

\mathrm{f}(-2)

(b)

\mathrm{f}^{-1}(x)

, the inverse of

\mathrm{f}

(c)

a

and

b

, where

\mathrm{fg}(x) = ax + b

(d)

\dfrac{2}{\mathrm{f}(x)} - \dfrac{5}{\mathrm{g}(x)}

as a single fraction in terms of

x

Plan the Solution

(a) Put $-2$ in place of every $x$ in $3x - 4$ .
(b) Write $y = 3x - 4$ , make $x$ the subject, then write the result in terms of $x$ .
(c) $\mathrm{fg}(x)$ means $\mathrm{g}$ first and then $\mathrm{f}$ , so put $4x + 1$ into $\mathrm{f}$ , expand, and read off the number of $x$ and the constant.
(d) Replace $\mathrm{f}(x)$ and $\mathrm{g}(x)$ by their expressions, put both fractions over the common denominator $(3x - 4)(4x + 1)$ , then expand and simplify the numerator.

Worked Solution [8 marks]

Rule - Functions:

\mathrm{f}(-2)

means put

-2

in place of every

x

; the inverse

\mathrm{f}^{-1}

comes from making

x

the subject of

y = \mathrm{f}(x)

;

\mathrm{fg}(x) = \mathrm{f}(\mathrm{g}(x))

, with

\mathrm{g}

applied first; and two fractions are subtracted over a common denominator.

Step 1: (a) Substitute

x = -2

into

\mathrm{f}(x)

\mathrm{f}(-2) = 3 \times (-2) - 4

-6 - 4 = -10

(Reason: Every

x

3x - 4

is replaced by

-2

, in a bracket so that the sign stays with it. Three lots of

-2

-6

, and taking away a further

4

goes further below zero, to

-10

Step 2: (b) Make

x

the subject of

y = 3x - 4

y = 3x - 4

y + 4 = 3x

x = \dfrac{y + 4}{3}

\mathrm{f}^{-1}(x) = \dfrac{x + 4}{3}

(Reason: The inverse undoes

\mathrm{f}

. The function

\mathrm{f}

multiplies by

3

and then subtracts

4

, so its inverse adds

4

first and then divides by

3

, the operations in reverse order. Adding

4

to both sides is the first step. Once

x

is the subject, the letter

y

is replaced by

x

so that

\mathrm{f}^{-1}

is written in terms of

x

, as the question asks.)

Step 3: (c) Put

\mathrm{g}(x)

into

\mathrm{f}

\mathrm{fg}(x) = \mathrm{f}(4x + 1)

\mathrm{f}(4x + 1) = 3(4x + 1) - 4

3(4x + 1) - 4 = 12x + 3 - 4

12x + 3 - 4 = 12x - 1

(Reason: In

\mathrm{fg}(x)

the function written next to the

x

acts first, so

\mathrm{g}

turns

x

into

4x + 1

, and then

\mathrm{f}

is applied to that whole bracket: three times it, then subtract

4

. Matching

12x - 1

with

ax + b

term by term gives

a = 12

and

b = -1

. The other order is a different function:

\mathrm{gf}(x) = 4(3x - 4) + 1 = 12x - 15

, which gets

a

right and

b

wrong.)

Step 4: (d) Write both fractions in terms of

x

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

(Reason: The expression is in terms of

\mathrm{f}(x)

and

\mathrm{g}(x)

, and the answer has to be in terms of

x

, so each is replaced by the expression it stands for.)

Step 5: (d) Put both fractions over

(3x - 4)(4x + 1)

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

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

(Reason: The two denominators have no factor in common, so the simplest common denominator is their product. The first fraction is multiplied top and bottom by

4x + 1

and the second by

3x - 4

, which changes neither fraction's value. With one denominator the numerators can be subtracted, and the minus sign applies to the whole of

5(3x - 4)

, which is why it stays in its bracket.)

Step 6: (d) Expand and simplify the numerator

2(4x + 1) = 8x + 2

-5(3x - 4) = -15x + 20

8x + 2 - 15x + 20 = 22 - 7x

(Reason: Each bracket is multiplied out with the sign in front of it. In the second,

-5

times

3x

-15x

, and

-5

times

-4

is positive

20

, because two negatives multiply to give a positive. Then the

x

terms are collected,

8x - 15x = -7x

, and the numbers,

2 + 20 = 22

Step 7: (d) Write the answer as a single fraction

\dfrac{2}{3x - 4} - \dfrac{5}{4x + 1} = \dfrac{22 - 7x}{(3x - 4)(4x + 1)}

(3x - 4)(4x + 1) = 12x^2 + 3x - 16x - 4

12x^2 + 3x - 16x - 4 = 12x^2 - 13x - 4

(Reason: The numerator

22 - 7x

is zero only at

x = \dfrac{22}{7}

, where neither bracket of the denominator is zero, so it shares no factor with either bracket and nothing cancels. The denominator may be left as two brackets or multiplied out to

12x^2 - 13x - 4

; both give the same single fraction.)

(a)

\mathrm{f}(-2) = -10

(b)

\mathrm{f}^{-1}(x) = \dfrac{x + 4}{3}

(c)

a = 12

b = -1

(d)

\dfrac{22 - 7x}{(3x - 4)(4x + 1)}

Verification

Check 1 - (b) undoes (a): The inverse must send the answer to (a) back to where it started, so

\mathrm{f}^{-1}(-10)

should be

-2

\dfrac{-10 + 4}{3} = \dfrac{-6}{3} = -2

Check 2 - (b) applying $\mathrm{f}$ to the inverse: Put

\dfrac{x + 4}{3}

into

\mathrm{f}

: multiply by

3

and subtract

4

. The result should be

x

itself.

3 \times \dfrac{x + 4}{3} - 4 = x + 4 - 4 = x

Check 3 - (c) with numbers: Take

x = 1

\mathrm{g}(1) = 5

and

\mathrm{f}(5) = 11

. Take

x = 0

\mathrm{g}(0) = 1

and

\mathrm{f}(1) = -1

. The expression

12x - 1

should give the same two values.

12 \times 1 - 1 = 11

and

12 \times 0 - 1 = -1

Check 4 - (d) at $x = 1$ : At

x = 1

\mathrm{f}(1) = -1

and

\mathrm{g}(1) = 5

, so the original expression is

\dfrac{2}{-1} - \dfrac{5}{5} = -3

. The single fraction should give the same value.

\dfrac{22 - 7}{(-1)(5)} = \dfrac{15}{-5} = -3

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The value of $\mathrm{f}(-2)$	Note	The Answer column gives $-10$ , and the Marks column gives $1$ for it.	✓
(b) The inverse function $\mathrm{f}^{-1}(x)$	Note	The Answer column gives $\dfrac{x + 4}{3}$ oe, and the Marks column gives $2$ for it.	✓
(b) The first step of the rearrangement	M1	For correct first step, $y + 4 = 3x$ or $x = 3y - 4$ or $\dfrac{y}{3} = x - \dfrac{4}{3}$ .	✓
(c) The values of $a$ and $b$	Note	The Answer column gives $a = 12$ , $b = -1$ , and the Marks column gives $2$ for it.	✓
(c) One of the two values	B1	For either $a$ or $b$ correct.	✓
(c) In place of the B1: $\mathrm{fg}(x)$ written out	M1	Or for $3(4x + 1) - 4$ .	✓
(d) The single fraction	Note	The Answer column gives $\dfrac{22 - 7x}{(3x - 4)(4x + 1)}$ or $\dfrac{22 - 7x}{12x^2 - 13x - 4}$ final answer, and the Marks column gives $3$ for it.	✓
(d) The numerator over the common denominator	B1	For $2(4x + 1) - 5(3x - 4)$ oe or better isw.	✓
(d) The common denominator	B1	For common denominator $(3x - 4)(4x + 1)$ oe isw.	✓
(d) What oe, or better and isw mean	Note	oe is the scheme's abbreviation for or equivalent, and isw for ignore subsequent working: once the expression has been written, a slip in the working after it does not take that mark away. Or better means that a simplified form of the same numerator, such as $22 - 7x$ , is credited as well.	✓

Full marks: 8/8

Continue to questions 15 to 23

The remaining 9 questions, with the same full worked solutions and mark schemes

Frequently asked questions

There are 23 questions worth 100 marks in total, sat over 2 hours. It is Extended tier and a calculator is not allowed: the paper's own instructions say that calculators must not be used in this paper, so every answer is worked by hand.

Extended is graded A* to E. An Extended candidate sits two papers - Paper 2 and Paper 4 - each marked out of 100, so 200 in total, and the grade comes from the combined mark rather than from either paper alone. In the June 2025 series the thresholds for candidates who sat Paper 21 with Paper 41 were A* 156, A 131, B 106, C 81, D 68 and E 55 out of 200.

Yes. The paper states in its own instructions that you must show all necessary working clearly. 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.

Yes. The question paper prints a List of formulas. It gives the area of a triangle and of a circle, the circumference of a circle, the curved surface areas of a cylinder and a cone, the surface area of a sphere, and the volumes of a prism, a pyramid, a cylinder, a cone and a sphere; the quadratic formula; and, for a triangle with sides a, b, c opposite angles A, B, C, the sine rule, the cosine rule and the area as one half of ab sin C. Anything not on the list, such as the area of a trapezium, still has to be recalled.

Cambridge has not yet published this paper on its public website. We will link the official question paper and mark scheme here as soon as they are released. 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 2025: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 14 of 23

Question 1, Non-calculator

Question 1 - Exam Solution

Question 2, Non-calculator

Question 2 - Exam Solution

Question 3, Non-calculator

Question 3 - Exam Solution

Question 4, Non-calculator

Question 4 - Exam Solution

Question 5, Non-calculator

Question 5 - Exam Solution

Question 6, Non-calculator

Question 6 - Exam Solution

Question 7, Non-calculator

Question 7 - Exam Solution

Question 8, Non-calculator

Question 8 - Exam Solution

Question 9, Non-calculator

Question 9 - Exam Solution

Question 10, Non-calculator

Question 10 - Exam Solution

Question 11, Non-calculator

Question 11 - Exam Solution

Question 12, Non-calculator

Question 12 - Exam Solution

Question 13, Non-calculator

Question 13 - Exam Solution

Question 14, Non-calculator

Question 14 - Exam Solution

Frequently asked questions

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