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Cambridge IGCSE 0580/21, May/June 2024: Worked Solutions, Questions 12 to 22

Sir Faraz Hassan

14 Sept 2026

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

This is the rest of the paper. Questions 1 to 11, the paper's overview and the frequently asked questions are on the first page.

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

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

Both are PDF files hosted by Cambridge: official question paper (PDF) and official mark scheme (PDF).

Try each question yourself first, then open the worked solution to check your method and see exactly where each method mark (M1) and independent mark (B1) is earned. The questions follow the same order as the original paper and carry the same marks.

Download printable PDF

All 22 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 12 to 22 of 22

What values of $x$ and $y$ satisfy both of these equations?
You must show all your working.

$\dfrac{3x}{2} + 5y = 5$
$4x - 3y = 46$ [4 marks]

x =y =

[Total 4 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

First equation:

\dfrac{3x}{2} + 5y = 5

Second equation:

4x - 3y = 46

Two linear equations in the same two unknowns,

x

and

y

, one of them carrying a fraction.

Find

The value of

x

and the value of

y

. One pair of values has to fit both equations at once, so expect a single solution pair.

Plan the Solution

Multiply the first equation by $2$ so that no fraction is left in it.
Scale both equations until one letter carries the same coefficient in each.
Subtract to remove that letter, then solve the single equation that is left.
Substitute the value found back into one of the equations to get the other letter.

Worked Solution [4 marks]

Rule - Elimination: multiply one or both equations so that one letter has the same size of coefficient in both, then add or subtract the equations to remove that letter.

Step 1: Clear the fraction

\dfrac{3x}{2} + 5y = 5

2 \times \dfrac{3x}{2} + 2 \times 5y = 2 \times 5

3x + 10y = 10

(Reason: Multiplying every term by

2

clears the denominator, so both equations now have whole-number coefficients and are easier to scale.)

Step 2: Match the coefficients of

x

\text{Cleared equation} \times 4: \quad 12x + 40y = 40

\text{Second equation} \times 3: \quad 12x - 9y = 138

(Reason: The lowest common multiple of

3

and

4

12

, so both equations can be written starting with

12x

. Every term on both sides is multiplied, the constant included.)

Step 3: Subtract to eliminate

x

(12x + 40y) - (12x - 9y) = 40 - 138

49y = -98

y = \dfrac{-98}{49} = -2

(Reason: Both equations now start with

12x

, so subtracting one from the other removes

x

completely. Subtracting

-9y

adds

9y

, which is where the

49y

comes from.)

Step 4: Substitute back to find

x

3x + 10 \times (-2) = 10

3x - 20 = 10

3x = 30

x = \dfrac{30}{3} = 10

(Reason: Putting

y = -2

into the cleared equation leaves one unknown, which solves in one line.)

x = 10

y = -2

Verification

Check 1: Put

x = 10

and

y = -2

into the first equation in the form the question prints it, fraction and all.

\dfrac{3 \times 10}{2} + 5 \times (-2) = 15 - 10 = 5

Check 2: Put the same pair into the second equation.

4 \times 10 - 3 \times (-2) = 40 + 6 = 46

Check 3: Reach the pair a different way: rearrange the second equation for

y

and put

x = 10

into it.

\dfrac{4 \times 10 - 46}{3} = \dfrac{-6}{3} = -2

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Equating a set of coefficients	M1	Correctly equating one set of coefficients.	✓
Eliminating a variable	M1	Correct method to eliminate one variable.	✓
Both values	A2	For $x = 10$ and $y = -2$ .	✓
The value of $x$	A1	For $x = 10$ .	✓
The value of $y$	A1	For $y = -2$ .	✓
Special case	SC1	If M0 scored, for $2$ values satisfying one of the original equations.	✓
About the special case	Note	M0 records that neither method mark was earned. The special case then covers a pair of values that satisfies one of the two printed equations.	✓

Full marks: 4/4

A cyclic quadrilateral is drawn in the diagram.

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

p =

[Total 3 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

A quadrilateral drawn inside a circle with all four of its vertices on the circle, so it is a cyclic quadrilateral.

Going round the shape the marked angles are

p^\circ

5m^\circ

(4m + 38)^\circ

and

4m^\circ

Find

The value of

p

Plan the Solution

Opposite angles of a cyclic quadrilateral add to $180^\circ$ . Read off which vertex faces which before writing anything down.
Start with the pair that carries only $m$ , the $5m^\circ$ and $4m^\circ$ angles, because that equation has one unknown in it.
Then bring $m$ to the other pair, $p^\circ$ and $(4m + 38)^\circ$ , which is where $p$ lives.

Worked Solution [3 marks]

Rule - Cyclic quadrilateral: opposite angles add to

180^\circ

Step 1: use the pair of opposite angles that carries only

m

5m + 4m = 180

9m = 180

m = 20

(Reason: The

5m^\circ

and

4m^\circ

angles are at opposite vertices, so they add to

180

. No other letter appears, so this equation gives

m

on its own.)

Step 2: work out the angle that faces

p^\circ

4 \times 20 + 38 = 118

(Reason: The bottom vertex is the one opposite

p^\circ

, and its angle is

(4m + 38)^\circ

. Putting

m = 20

into that expression turns it into a number.)

Step 3: use the second pair of opposite angles

p + 118 = 180

p = 180 - 118 = 62

(Reason: Opposite angles add to

180

, so taking the facing angle away from

180

is what is left for

p

p = 62

Verification

Check 1: Test BOTH pairs of opposite angles, not just the one that was solved. With

m = 20

the four angles are

62

100

118

and

80

62 + 118 = 180

and

100 + 80 = 180

Check 2: The angles of any quadrilateral add to

360

, cyclic or not, so adding all four is a test that does not use the circle at all.

62 + 100 + 118 + 80 = 360

Check 3: Work backwards. If

p = 62

then the angle facing it has to be

118

, which means

4m + 38 = 118

118 - 38 = 80

, so

4m = 80

and

m = 20

, the value Step 1 found.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The value of $p$	Note	The Answer column gives $62$ , and the Marks column gives $3$ for it.	✓
Finding $m$	B2	For $m = 20$ .	✓
Method for $m$ , or method for $p$	M1	Or for $5m + 4m = 180$ soi, or $p + 4m + 38 = 180$ soi.	✓
Where the scheme's value comes from	Note	The scheme's $m = 20$ is the value that makes the $5m$ and $4m$ angles total $180$ .	✓

Full marks: 3/3

A circle of radius $9$ cm is shown in the diagram.

What is the area of the shaded major sector? [3 marks]

cm²

[Total 3 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

A circle of radius

9

An angle of

48^{\circ}

at the centre, marking off the unshaded minor sector

The shaded region is the major sector, the larger of the two

Find

The area of the shaded major sector

Plan the Solution

Take the $48^{\circ}$ minor sector away from a full turn, to get the angle of the major sector.
Write that angle over $360$ : that is the fraction of the circle the sector covers.
Multiply the fraction by the area of the whole circle, $\pi r^{2}$ .
Round to $3$ significant figures.

Worked Solution [3 marks]

Rule - Area of a sector:

\dfrac{\theta}{360} \times \pi r^{2}

, where

\theta

is the angle at the centre in degrees.

Step 1: Find the angle of the major sector

360^{\circ} - 48^{\circ} = 312^{\circ}

(Reason: The major sector and the

48^{\circ}

minor sector together make a full turn, so the major sector takes whatever the minor one leaves.)

Step 2: Write the sector as a fraction of the whole circle

\dfrac{312}{360}

(Reason: An angle of

312^{\circ}

out of a full turn of

360^{\circ}

is that fraction of the circle, so the sector's area is the same fraction of the circle's area.)

Step 3: Multiply by the area of the whole circle

9^{2} = 81

\dfrac{312}{360} \times \pi \times 81 = 220.539\ldots

(Reason: The whole circle has area

\pi r^{2}

, and squaring the radius gives

81

, so the circle is

\pi \times 81

square centimetres.)

Step 4: Round the answer

220.539\ldots = 221 \text{ (3 s.f.)}

(Reason: An answer is given to

3

significant figures unless the question asks for something else, and the units are square centimetres because this is an area.)

221 \text{ cm}^{2}

Verification

Check 1 - take the minor sector away instead: Work out the whole circle and the

48^{\circ}

minor sector separately. The circle is

\pi \times 81 = 254.469\ldots

and the minor sector is

\dfrac{48}{360} \times 254.469 = 33.929\ldots

254.469 - 33.929 = 220.54

Check 2 - the radian form of the sector formula: The same sector by

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

, with the angle in radians.

312^{\circ}

5.4454

radians, and this formula never forms the fraction used above.

0.5 \times 81 \times 5.4454 = 220.54

Check 3 - a size check: The shaded sector is more than half the circle, because

312^{\circ}

is more than

180^{\circ}

, and it must be less than the whole circle. Half the circle is about

127 \text{ cm}^{2}

and the whole circle about

254 \text{ cm}^{2}

, and

221

sits between them.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The area of the shaded major sector	Note	The Answer column gives $221$ or $220.5$ to $220.6$ , and the Marks column gives 3 for it.	✓
The full method: the major sector's fraction of the circle, times the circle's area	M2	For $\dfrac{360 - 48}{360} \times \pi \times 9^{2}$	✓
Partial method: the sector formula with an angle that is not a full turn	M1	Or for $\dfrac{k}{360} \times \pi \times 9^{2}$ where $k < 360$	✓
The angle of the major sector on its own	B1	Or for $312$	✓
Where the partial award's value comes from	Note	The $312$ in the B1 row is $360 - 48$ , the angle of the shaded major sector.	✓

Full marks: 3/3

What fraction, in its simplest form, is equal to $0.1\dot{4}\dot{6}$ ?
You must show all your working. [3 marks]

[Total 3 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

The recurring decimal

0.1\dot{4}\dot{6}

, in which the block

46

repeats without end

All the working must be shown, so the fraction has to be reached by algebra

Find

The decimal written as a fraction in its simplest form

Plan the Solution

Give the decimal a name, $x$ .
Multiply by $1000$ and by $10$ , so that both lines end in the same recurring tail.
Subtract the smaller line from the larger, which cancels the tail and leaves a whole number.
Divide to reach a fraction, then cancel that fraction to its simplest form.

Worked Solution [3 marks]

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

10

chosen so that both results carry the same recurring tail. Subtracting one from the other cancels that tail and leaves a whole number, and dividing then turns the decimal into a fraction.

Step 1: Give the decimal a name

x = 0.1\dot{4}\dot{6} = 0.1464646\ldots

(Reason: Naming the decimal turns the question into algebra, which is what the working has to be. Written out, the

1

happens once and then the block

46

repeats without end.)

Step 2: Multiply by

1000

and by

10

1000x = 146.\dot{4}\dot{6}

10x = 1.\dot{4}\dot{6}

(Reason: One digit stands before the repeating block, so multiplying by

10

moves the point just past it. The block is two digits long, so multiplying by

1000

moves the point two places further on. Both lines then break off at the same place in the pattern, leaving the same tail

0.4646\ldots

on each.)

Step 3: Subtract to clear the recurring tail

1000x - 10x = 146.\dot{4}\dot{6} - 1.\dot{4}\dot{6}

990x = 145

(Reason: The two tails are identical, so subtracting cancels them exactly and what is left is the whole number

146 - 1

. On the left,

1000x - 10x

collects to

990x

Step 4: Divide, then cancel to the simplest form

x = \dfrac{145}{990}

\dfrac{145}{990} = \dfrac{29}{198}

(Reason: Dividing both sides by

990

gives the fraction. Both parts are multiples of

5

, since

145 = 5 \times 29

and

990 = 5 \times 198

. After that cancelling,

29

is prime and

198

is not a multiple of it, so nothing is left to cancel.)

\dfrac{29}{198}

Verification

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

990x = 145

, so multiplying the answer by

990

must give that same whole number back.

\dfrac{29}{198} \times 990 = 145

Check 2 - divide it out again: A calculator divides

29

198

. The digits it shows must be the decimal the question prints, not merely something close to it. The display reads

0.1464646\ldots

, which is

0.1\dot{4}\dot{6}

Check 3 - the fraction really is in its simplest form: Break the denominator into prime factors and look for a factor it shares with

29

198 = 2 \times 3^{2} \times 11

, and

29

is prime, so the only factor they share is

1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Two multiples of the decimal, subtracted so the recurring tails cancel	M1	For $146.4̇6̇ - 1.4̇6̇$ oe	✓
The fraction in its simplest form	A2	For $\dfrac{29}{198}$ cao	✓
The right fraction, not yet cancelled	A1	For $\dfrac{145}{990}$ oe	✓
A fraction over 990 reached without enough working	SC1	If M0 scored, for $\dfrac{k}{990}$ with insufficient working	✓
Where the partial award's value comes from	Note	The $\dfrac{145}{990}$ in the A1 row is $\dfrac{29}{198}$ before it is cancelled.	✓

Full marks: 3/3

(a) Shade the region $M' \cap N'$ on the Venn diagram. [1 mark]

(b) What is $\mathrm{n}(B \cap (A' \cup C))$ ? [1 mark]

(b)

[Total 2 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

Part (a): a rectangle for the universal set holding two overlapping circles,

M

and

N

Part (b): a rectangle holding three overlapping circles,

A

B

and

C

, with a number written in every one of its eight regions.

Those numbers are

33

A

alone,

16

A

and

B

only,

3

B

alone,

10

in all three,

18

A

and

C

only,

4

B

and

C

only,

9

C

alone, and

20

outside all three.

Find

(a) The region

M' \cap N'

, shaded on the first diagram. (b) The value of

\mathrm{n}(B \cap (A' \cup C))

Plan the Solution

A dash is the complement sign, so $M'$ means everything outside $M$ . Two complements joined by $\cap$ means outside both at once.
In part (b) work from inside the brackets outwards. Build $A' \cup C$ first, then throw away anything that is not also in $B$ .
Do it one region at a time. Each of the eight regions is wholly inside a circle or wholly outside it, so every region gets a plain yes or no.

Worked Solution [2 marks]

Rule - a dash is the complement, everything outside the set;

\cap

means in both;

\cup

means in one or the other or both; and

\mathrm{n}(X)

is how many members

X

has.

Step 1: read the two dashes in

M' \cap N'

M' \cap N' = (M \cup N)'

(Reason: A point in

M'

is outside

M

, and a point in

N'

is outside

N

. To be in both at once a point has to miss both circles, which is the same as sitting outside the whole of

M \cup N

. So the shading is every part of the rectangle that no circle covers, and the overlap stays clear along with the rest.)

Step 2: build

A' \cup C

A' \cup C \text{ holds } 3, 10, 18, 4, 9, 20

(Reason:

A'

is the four regions with no

A

over them, holding

3

4

9

and

20

C

is the four regions inside the bottom circle, holding

18

10

4

and

9

. A union keeps anything on either list, and the only two regions on neither are the ones holding

33

and

16

Step 3: keep only the regions that are also in

B

B \text{ holds } 16, 3, 10, 4

B \cap (A' \cup C) \text{ holds } 3, 10, 4

(Reason: There are four regions inside

B

. Of those, the one holding

16

is the only one Step 2 left out, because it lies inside

A

and outside

C

and so fails both halves of the union. The other three survive.)

Step 4: count the members in those three regions

3 + 10 + 4 = 17

(Reason:

\mathrm{n}

counts members, so add together the numbers written in the three regions Step 3 kept.)

(a) Every part of the rectangle outside both circles is shaded(b)

17

Verification

Check 1: Take the unwanted slice off the whole of

B

instead of building the answer up. All four regions of

B

together give

16 + 3 + 10 + 4 = 33

, and the only one of them that fails the condition is the region inside

A

and outside

C

33 - 16 = 17

Check 2: Split the union instead, so that

B \cap (A' \cup C)

becomes

B \cap A'

together with

B \cap C

. The first holds

7

members and the second holds

14

, and the region holding

4

sits in both of them, so it must not be counted twice.

7 + 14 - 4 = 17

Check 3: For part (a), test one point from each of the four areas of the first diagram. A point in

M

alone fails

M'

; a point in the overlap fails both dashes; a point in

N

alone fails

N'

. Only a point that misses both circles passes, so the shading is the outside area and nothing else.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The shaded region	Note	The Answer column gives a diagram with the whole rectangle shaded except inside the two circles, and the Marks column gives $1$ for it.	✓
(b) The number of members	Note	The Answer column gives $17$ , and the Marks column gives $1$ for it.	✓
Neither part has a Partial Marks entry	Note	Both parts leave the Partial Marks column empty, so on each part the whole mark is for the answer itself and nothing is printed for a method.	✓

Full marks: 2/2

What is the area of triangle $ABC$ ? [2 marks]

cm²

[Total 2 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

In triangle

ABC

, side

AB

6.7

cm and side

BC

5.9

cm.

The angle where those two sides meet, at vertex

B

, is marked

81

degrees.

Find

The area of triangle

ABC

, in

\text{cm}^2

Plan the Solution

Two sides and the angle between them are given, so the area comes straight from $\dfrac{1}{2}ab\sin C$ and no perpendicular height has to be found first.
Confirm that the marked angle is the one where the two given sides actually meet. That is the only angle this rule accepts.
Multiply everything in one go, keep the unrounded value on the calculator, and round once at the very end.

Worked Solution [2 marks]

Rule - Area from two sides and the angle between them:

\text{Area} = \dfrac{1}{2}ab\sin C

, where

a

and

b

are the two sides and

C

is the angle between them.

Choose the rule that fits what is given

\text{Area} = \dfrac{1}{2} \times AB \times BC \times \sin B

(Reason: Two sides and the angle between them are known, and that is exactly what this rule takes. A half-base-times-height calculation would need a perpendicular length the question never gives.)

Put in the two sides and the angle between them

\text{Area} = \dfrac{1}{2} \times 6.7 \times 5.9 \times \sin 81^\circ

(Reason: The sides

6.7

cm and

5.9

cm both meet at

B

, and the angle marked there is

81^\circ

Work the product out

\dfrac{1}{2} \times 6.7 \times 5.9 = 19.765

19.765 \times \sin 81^\circ = 19.5216\ldots

(Reason: Halve the product of the two sides first, then multiply by the sine of the angle. Leave the value on the calculator display rather than rounding partway through.)

Round the answer

\text{Area} = 19.5 \text{ cm}^2

(Reason:

19.5216\ldots

is not exact, so it is given to three significant figures. That is the

19.5

the mark scheme gives, and the scheme also accepts the longer

19.52\ldots

19.5 \text{ cm}^2

Verification

Check 1 - undo the multiplication: Divide the area back by

\dfrac{1}{2} \times 6.7 \times 5.9

. If the working is right, what is left has to be the sine of the marked angle.

\dfrac{19.5216\ldots}{19.765} = 0.98768\ldots = \sin 81^\circ

Check 2 - build it from a perpendicular height: Drop a perpendicular from

A

on to

BC

. Its length is

6.7 \times \sin 81^\circ = 6.6175\ldots

cm, so half the base times that height gives the area a second way.

\dfrac{1}{2} \times 5.9 \times 6.6175\ldots = 19.52\ldots

Check 3 - is the size sensible: With these two sides the area is at its largest when the angle between them is

90^\circ

, which would give

\dfrac{1}{2} \times 6.7 \times 5.9 = 19.765

. An angle of

81^\circ

is close to a right angle, so the area should land just below that ceiling. The answer

19.5

sits a little under

19.765

, as expected.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The area of triangle $ABC$	Note	The Answer column gives $19.5$ or $19.52\ldots$ , and the Marks column gives $2$ for it.	✓
Method: the two sides and the angle between them	M1	For $\dfrac{1}{2} \times 6.7 \times 5.9 \times \sin 81$ oe.	✓
Note: which angle the rule needs	Note	The rule needs the angle between the two sides being used. The sides $6.7$ and $5.9$ meet at $B$ , and the angle marked there is $81$ , so both sides and the angle drop straight into the formula with nothing to work out first.	✓

Full marks: 2/2

The graph of $y = x^3 + 4x^2 - 2$ is drawn in the diagram for $-3 \leqslant x \leqslant 1.5$ .

Draw a suitable straight line to solve the equation $x^3 + 4x^2 - 2 = 2x$ for $-3 \leqslant x \leqslant 1.5$ . [3 marks]

x =x =

[Total 3 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

The curve

y = x^3 + 4x^2 - 2

is already drawn for

-3 \leqslant x \leqslant 1.5

Find

Every value of

x

-3 \leqslant x \leqslant 1.5

that satisfies

x^3 + 4x^2 - 2 = 2x

, found by ruling one straight line.

Plan the Solution

Read the equation as two graphs: the curve that is already there, and one straight line.
Rule that line right across the grid.
Take the $x$ -coordinate of each point where the line meets the curve.

Worked Solution [3 marks]

Two graphs meet where their

y

-values are equal, so the

x

-coordinate of every crossing point solves the equation made by putting the two expressions equal to each other.

Step 1: Decide which straight line to draw

x^3 + 4x^2 - 2 = 2x

(Reason: The left-hand side is exactly the expression that has already been plotted, so the equation holds wherever the curve's

y

-value is equal to

2x

. The line to add is therefore

y = 2x

Step 2: Work out two points on that line

2 \times (-1) = -2

2 \times 1.5 = 3

(Reason: Two points fix a straight line. Take

x = -1

and

x = 1.5

, which are both inside the range shown. The line also passes through the origin, so rule it right across the grid.)

Step 3: Read the x-coordinate of each crossing

x = -0.52

x = 0.88

(Reason: The ruled line cuts the curve at two points inside

-3 \leqslant x \leqslant 1.5

. Dropping from each crossing to the

x

-axis gives a reading a small part of a square to the left of

-0.5

, and another the same distance to the left of

0.9

x = -0.52

x = 0.88

Verification

Check 1: Rearranged, the equation is

x^3 + 4x^2 - 2x - 2 = 0

. Substitute the two ends of the square that the first crossing sits in.

(-0.6)^3 + 4 \times (-0.6)^2 - 2 \times (-0.6) - 2 = 0.424

and

(-0.5)^3 + 4 \times (-0.5)^2 - 2 \times (-0.5) - 2 = -0.125

. The sign changes, so a solution lies between

-0.6

and

-0.5

Check 2: The same test on the square that the second crossing sits in.

0.85^3 + 4 \times 0.85^2 - 2 \times 0.85 - 2 = -0.195875

and

0.9^3 + 4 \times 0.9^2 - 2 \times 0.9 - 2 = 0.169

. The sign changes again, so a solution lies between

0.85

and

0.9

Check 3: This cubic has three solutions, and they add to

-4

. Test whether the one that was not read off really is out of range. The two crossings add to about

0.35

, so the third solution is about

-4.35

, well outside

-3 \leqslant x \leqslant 1.5

. The two read off the graph are all there are in the range.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The straight line	B1	For the line $y = 2x$ ruled.	✓
Both solutions	B2	For $x = -0.5$ to $-0.55$ and $x = 0.85$ to $0.9$ .	✓
The negative solution only	B1	For $-0.5$ to $-0.55$ .	✓
The positive solution only	B1	For $0.85$ to $0.9$ .	✓

Full marks: 3/3

Each expression below is to be factorised completely.

(a) $12m^2 - 75t^2$ [3 marks]

(b) $xy + 15 + 3y + 5x$ [2 marks]

(a)(b)

[Total 5 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

(a)

12m^2 - 75t^2

(b)

xy + 15 + 3y + 5x

Two expressions, neither of them already written as a product.

Find

(a)

12m^2 - 75t^2

written as a product, factorised completely. (b)

xy + 15 + 3y + 5x

written as a product, factorised completely.

Plan the Solution

(a) Two terms. Take out the number that divides both of them, and look at what is left inside the bracket.
(a) $4m^2 - 25t^2$ is one square subtract another, so it splits into two brackets.
(b) Four terms, with nothing dividing all four. Put them in pairs and take a factor out of each pair.
(b) The same bracket should come out of both pairs; that bracket is the last factor to take out.
Completely means keep going until no bracket has a factor left inside it.

Worked Solution [5 marks]

Rule - Factorise completely: take out every common factor first, then factorise what is left. A difference of two squares splits as

a^2 - b^2 = (a + b)(a - b)

(a) Take out the factor common to both terms

12m^2 - 75t^2 = 3(4m^2 - 25t^2)

(Reason: The largest number dividing both

12

and

75

3

, and no letter appears in both terms, so only the number comes out.)

(a) Write the bracket as a difference of two squares

4m^2 - 25t^2 = (2m)^2 - (5t)^2

(Reason: Both terms inside the bracket are perfect squares, so the bracket is one square subtract another.)

(a) Split the difference of two squares

3(4m^2 - 25t^2) = 3(2m + 5t)(2m - 5t)

(Reason: Using

a^2 - b^2 = (a + b)(a - b)

with

a = 2m

and

b = 5t

. Neither bracket has a factor left inside it, so the factorising stops here.)

(b) Put the four terms into pairs

xy + 15 + 3y + 5x = xy + 5x + 3y + 15

(Reason: Nothing divides all four terms, so they are paired off instead:

xy

with

5x

, and

3y

with

15

(b) Take a factor out of each pair

xy + 5x + 3y + 15 = x(y + 5) + 3(y + 5)

(Reason:

x

comes out of the first pair and

3

out of the second, and each pair leaves

y + 5

behind.)

(b) Take out the bracket both parts share

x(y + 5) + 3(y + 5) = (x + 3)(y + 5)

(Reason:

y + 5

is a factor of both parts, so it comes out and

x + 3

is what is left.)

(a)

3(2m + 5t)(2m - 5t)

(b)

(x + 3)(y + 5)

Verification

Check 1: Multiply part (a)'s brackets out, then put the

3

back.

(2m + 5t)(2m - 5t) = 4m^2 - 25t^2

, so

3(2m + 5t)(2m - 5t) = 12m^2 - 75t^2

, which is the expression in part (a).

Check 2: Put

m = 2

and

t = 1

into part (a)'s expression and into its answer.

12 \times 4 - 75 = -27

and

3 \times 9 \times (-1) = -27

Check 3: Multiply

(x + 3)(y + 5)

out, term by term.

xy + 5x + 3y + 15

, the same four terms as in part (b).

Check 4: Put

x = 2

and

y = 4

into part (b)'s expression and into its answer.

8 + 15 + 12 + 10 = 45

and

5 \times 9 = 45

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The complete factorisation	Note	The Answer column gives $3(2m + 5t)(2m - 5t)$ final answer, and the Marks column gives $3$ for it.	✓
(a) Factorised into two brackets, one of them still holding a common factor	B2	For $(6m + 15t)(2m - 5t)$ or $(2m + 5t)(6m - 15t)$ .	✓
(a) The common factor taken out, or the two brackets without it	B1	Or for $3(4m^2 - 25t^2)$ or $(2m + 5t)(2m - 5t)$ .	✓
(b) The complete factorisation	Note	The Answer column gives $(x + 3)(y + 5)$ final answer, and the Marks column gives $2$ for it.	✓
(b) Either grouping of the four terms into pairs	B1	For $x(y + 5) + 3(y + 5)$ or $y(x + 3) + 5(x + 3)$ .	✓

Full marks: 5/5

Find every value of $x$ in the interval $0^\circ \leqslant x \leqslant 360^\circ$ that satisfies $8 \sin x + 6 = 1$ . [3 marks]

x =or x =

[Total 3 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

The equation

8 \sin x + 6 = 1

The interval

0^\circ \leqslant x \leqslant 360^\circ

, so the calculator is set to degrees

Find

Every value of

x

in that interval that satisfies the equation Expect two of them: one full turn takes every sine value strictly between

-1

and

1

twice

Plan the Solution

Rearrange the equation until $\sin x$ stands alone.
Read the acute angle $a$ from $\sin^{-1}\left(\dfrac{5}{8}\right)$ , leaving the minus sign aside for the moment.
The sine is negative, so take the two angles of the turn whose sine is negative: $180^\circ + a$ and $360^\circ - a$ .
Write each angle to one decimal place.

Worked Solution [3 marks]

Rule - Negative sine over one turn: if

\sin x = -k

with

0 < k < 1

and

a = \sin^{-1} k

, then the solutions in

0^\circ \leqslant x \leqslant 360^\circ

are

x = 180^\circ + a

and

x = 360^\circ - a

Step 1: get

\sin x

on its own

8 \sin x + 6 = 1

8 \sin x = 1 - 6 = -5

\sin x = -\dfrac{5}{8}

(Reason: Take

6

from both sides, then divide both sides by

8

. The result sits between

-1

and

1

, so the equation does have solutions.)

Step 2: find the acute angle behind the answer

\sin^{-1}\left(\dfrac{5}{8}\right) = 38.682\ldots^\circ

(Reason: The inverse sine of a positive value gives the acute angle

a

. The minus sign is not needed here - it settles which two angles are wanted, not how big

a

is.)

Step 3: the solution between

180^\circ

and

270^\circ

x = 180^\circ + 38.682^\circ = 218.682^\circ

(Reason: Sine is negative for every angle between

180^\circ

and

270^\circ

, and the one wanted sits

a

past

180^\circ

Step 4: the solution between

270^\circ

and

360^\circ

x = 360^\circ - 38.682^\circ = 321.318^\circ

(Reason: Sine is negative across that quarter of the turn as well, and the second solution sits

a

short of

360^\circ

Step 5: write each angle to one decimal place

x = 218.7^\circ

x = 321.3^\circ

(Reason: Neither decimal terminates, so each angle is given to one decimal place.)

x = 218.7^\circ

x = 321.3^\circ

Verification

Check 1: Both angles have the same sine,

-0.625

. Put that back into

8 \sin x + 6

8 \times (-0.625) + 6 = 1

, which is the right-hand side

Check 2: One solution sits

a

past

180^\circ

and the other

a

short of

360^\circ

, so the pair must add to

540^\circ

whatever

a

turns out to be.

218.682 + 321.318 = 540

Check 3: Count the crossings over the whole turn:

8 \sin x + 6

runs from

6

up to

14

, down to

-2

270^\circ

and back to

6

. It passes

1

once on the way down and once on the way up, so there are exactly two solutions - the two found above.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The two values of $x$	Note	The Answer column gives $218.7$ , $321.3$ , and the Marks column gives 3 for it.	✓
One value correct	B2	For one correct.	✓
Method: $\sin x$ on its own	M1	Or for $\sin x = -\dfrac{5}{8}$ oe.	✓
Special case	SC1	If M1 or 0 scored, for two reflex angles with a sum of $540$ or two non-reflex angles with a sum of $180$ .	✓

Full marks: 3/3

The solid in the diagram is a cuboid.
In it, $EF = 9.1$ cm, $EH = 6.5$ cm and $HD = 4$ cm.

What is the angle between $CE$ and the base $CDHG$ ? [4 marks]

[Total 4 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

A cuboid with

EF = 9.1

cm,

EH = 6.5

cm and

HD = 4

EH

is a vertical edge, so

E

is directly above

H

The base

CDHG

is a rectangle, with

HG = EF = 9.1

cm and

GC = HD = 4

Find

The angle between the line

CE

and the plane of the base

CDHG

Plan the Solution

The line $CE$ meets the base at $C$ . Drop $E$ straight down onto the base and it lands on $H$ , so $CH$ is the shadow of $CE$ on the base.
The angle wanted is therefore the angle at $C$ in triangle $ECH$ , and that triangle has its right angle at $H$ because $EH$ is perpendicular to the base.
So find $CH$ with Pythagoras in the base rectangle first, then use the tangent ratio.

Worked Solution [4 marks]

Rule - the angle between a line and a plane: project the line onto the plane, then work in the right-angled triangle made by the line, its projection and the perpendicular between them.

Step 1: find the base diagonal

CH

HG = EF = 9.1 \text{ cm}

GC = HD = 4 \text{ cm}

CH^2 = HG^2 + GC^2

9.1^2 + 4^2 = 82.81 + 16 = 98.81

CH = \sqrt{98.81} = 9.9403\ldots \text{ cm}

(Reason: (Reason:

CDHG

is a rectangle measuring

9.1

cm by

4

cm, so its diagonal comes straight from Pythagoras. Keep the surd rather than a rounded value, because it is used again in the next step.))

Step 2: name the angle

\angle ECH

(Reason:

E

sits directly above

H

, so

H

is the point where

E

lands on the base. That makes

CH

the projection of

CE

, and the angle between

CE

and the base is the angle between

CE

and

CH

. (Reason: picking the right triangle is the whole difficulty here, and the mark scheme has a mark for this identification on its own.))

Step 3: use the tangent ratio in triangle

ECH

\tan(\angle ECH) = \dfrac{EH}{CH} = \dfrac{6.5}{\sqrt{98.81}}

(Reason: (Reason: triangle

ECH

has its right angle at

H

. Seen from

C

, the vertical

EH

is the opposite side and the projection

CH

is the adjacent side, and opposite over adjacent is the tangent.))

Step 4: work out the angle

\angle ECH = \tan^{-1}\!\left(\dfrac{6.5}{\sqrt{98.81}}\right) = 33.18\ldots^{\circ}

\angle ECH = 33.2^{\circ} \text{ (3 s.f.)}

(Reason: (Reason: the calculator must be in degree mode. The unrounded value is

33.18\ldots

, which rounds to

33.2

to three significant figures.))

33.2^{\circ}

Verification

Check 1: Use the space diagonal instead of the base diagonal.

CE = \sqrt{4^2 + 9.1^2 + 6.5^2} = \sqrt{141.06}

, and from

C

the side

EH

is opposite while

CE

is the hypotenuse, so use the sine.

\sin^{-1}\!\left(\dfrac{6.5}{\sqrt{141.06}}\right) = 33.2^{\circ}

to three significant figures, the same angle

Check 2: Use the cosine on the same triangle: adjacent

CH = \sqrt{98.81}

over hypotenuse

CE = \sqrt{141.06}

. This uses neither of the other two ratios.

\cos^{-1}\!\left(\dfrac{\sqrt{98.81}}{\sqrt{141.06}}\right) = 33.2^{\circ}

to three significant figures, the same angle again

Check 3: A size check. The height

EH = 6.5

cm is smaller than the base diagonal

CH = 9.94\ldots

cm, so the opposite side is shorter than the adjacent side and the angle has to be under

45^{\circ}

33.2^{\circ}

is under

45^{\circ}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The angle between $CE$ and the base $CDHG$	Note	The Answer column gives $33.2$ or $33.18\ldots$ , and the Marks column gives $4$ for it.	✓
Method - the tangent of angle $ECH$	M3	For $\tan = \dfrac{6.5}{\sqrt{4^2 + 9.1^2}}$ oe.	✓
Method - the squaring that gives a diagonal	M2	Or for $4^2 + 9.1^2$ oe, or $4^2 + 9.1^2 + 6.5^2$ oe.	✓
Method - identifying the angle	M1	Or for recognising the angle $ECH$ .	✓

Full marks: 4/4

Only red beads and blue beads are kept in tin $A$ and in tin $B$ .
At random, Curtis takes one bead from tin $A$ and Heather takes one bead from tin $B$ .

For Curtis, the probability of taking a red bead is $0.4$ .
There is a probability of $0.25$ that both beads taken are red.

What is the probability that both beads taken are blue? [4 marks]

[Total 4 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

Tin

A

and tin

B

hold red beads and blue beads only, and one bead is taken at random from each

\mathrm{P}(\text{Curtis red}) = 0.4

\mathrm{P}(\text{both red}) = 0.25

Find

\mathrm{P}(\text{both blue})

, the probability that Curtis's bead and Heather's bead are both blue

Plan the Solution

Each tin holds two colours only, so a blue probability is $1$ minus the matching red probability.
The two beads come from different tins and are taken at random, so the picks are independent and their probabilities multiply.
Heather's red probability is not given, so take it out of the $0.25$ first. Then change both reds into blues and multiply.

Worked Solution [4 marks]

Rule - Independent events:

\mathrm{P}(X \text{ and } Y) = \mathrm{P}(X) \times \mathrm{P}(Y)

. Complement:

\mathrm{P}(\text{not } X) = 1 - \mathrm{P}(X)

Step 1: Heather's probability of a red bead

\mathrm{P}(\text{Heather red}) = \dfrac{0.25}{0.4} = 0.625

(Reason: The picks are independent, so the two red probabilities multiply to give

0.25

. Dividing by Curtis's

0.4

undoes that multiplication.)

Step 2: Curtis's probability of a blue bead

\mathrm{P}(\text{Curtis blue}) = 1 - 0.4 = 0.6

(Reason: Tin

A

holds red beads and blue beads only, so its two probabilities add to

1

Step 3: Heather's probability of a blue bead

\mathrm{P}(\text{Heather blue}) = 1 - 0.625 = 0.375

(Reason: Tin

B

holds two colours only as well, so the same subtraction from

1

applies to it.)

Step 4: Multiply the two blue probabilities

\mathrm{P}(\text{both blue}) = 0.6 \times 0.375 = 0.225

(Reason: Independent again: the probability that both blue picks happen is the product of the two separate probabilities.)

0.225

Verification

Check 1: Rebuild the figure the question gives. Multiply Curtis's red probability

0.4

by Heather's red probability

0.625

0.4 \times 0.625 = 0.25

, which is the both-red probability the question states.

Check 2: There are four colour pairs and they must total

1

. Red then blue is

0.4 \times 0.375 = 0.15

, and blue then red is

0.6 \times 0.625 = 0.375

0.25 + 0.15 + 0.375 + 0.225 = 1

Check 3: Count whole beads instead of using probabilities. Tin

A

with

2

red and

3

blue gives

0.4

, and tin

B

with

5

red and

3

blue gives both-red

0.25

. That makes

5 \times 8 = 40

equally likely pairs, of which

3 \times 3

are blue with blue.

\dfrac{9}{40} = 0.225

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The probability that both beads taken are blue	Note	The Answer column gives $0.225$ oe, and the Marks column gives $4$ for it.	✓
Complete method in one expression	M3	For $\left(1 - \dfrac{0.25}{0.4}\right) \times (1 - 0.4)$ oe.	✓
Second method: Heather's probability of a red bead	M2	Or for $\dfrac{0.25}{0.4}$ .	✓
Second method: an equation for Heather's probability of a red bead	M1	Or for $0.4 \times p = 0.25$ oe.	✓
Second method: both blue from their probability of a red bead	M1	For $(1 - \text{their P(Heather red)}) \times (1 - 0.4)$ oe.	✓
What the letter p stands for	Note	The letter $p$ in the second method stands for the probability that Heather takes a red bead.	✓

Full marks: 4/4

Keep revising

That is the whole paper. Read what the IGCSE is and how it is graded, or compare Edexcel 4MA1 with Cambridge 0580 if you are still choosing a board. Check the IGCSE grade boundaries to set your target, and if the exam is close, the four-week IGCSE Maths revision plan sets out what to do week by week.

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Cambridge IGCSE 0580/21, May/June 2024: Worked Solutions, Questions 12 to 22

Worked solutions, questions 12 to 22 of 22

Question 12, Calculator allowed

Question 12 - Exam Solution

Question 13, Calculator allowed

Question 13 - Exam Solution

Question 14, Calculator allowed

Question 14 - Exam Solution

Question 15, Calculator allowed

Question 15 - Exam Solution

Question 16, Calculator allowed

Question 16 - Exam Solution

Question 17, Calculator allowed

Question 17 - Exam Solution

Question 18, Calculator allowed

Question 18 - Exam Solution

Question 19, Calculator allowed

Question 19 - Exam Solution

Question 20, Calculator allowed

Question 20 - Exam Solution

Question 21, Calculator allowed

Question 21 - Exam Solution

Question 22, Calculator allowed

Question 22 - Exam Solution

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