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Cambridge IGCSE 0580/41, May/June 2024: Worked Solutions, Questions 5 to 9

Sir Faraz Hassan

16 Sept 2026

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

This is the rest of the paper. Questions 1 to 4, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Cambridge IGCSE Mathematics, Paper 0580/41 (Extended), May/June 2024 series, sat Tuesday 7 May 2024 in Cambridge administrative zone 2 –130 marks, 2 hours 30 minutes, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Cambridge Assessment International Education. This resource reproduces neither the exam paper nor the official mark scheme.

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

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

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

Download printable PDF

All 9 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 5 to 9 of 9

(a) The point $P$ has coordinates $(1, 7)$ .
The point $Q$ has coordinates $(5, -5)$ .

(i) What is $\overrightarrow{PQ}$ ? [2 marks]

(ii) Show that $|\overrightarrow{OP}|$ and $|\overrightarrow{OQ}|$ are equal. [3 marks]

(iii) A circle with centre $O$ has $PQ$ as a chord.
Work out this circle's circumference. [2 marks]

(iv) A different circle, with centre $R$ , has $PQ$ as its diameter.
What are the coordinates of $R$ ? [2 marks]

(v) What is the equation of the perpendicular bisector of $PQ$ ?
Give your answer in the form $y = mx + c$ . [4 marks]

(b) Point $A$ has position vector $\mathbf{a}$ .
Point $B$ has position vector $\mathbf{b}$ .

$M$ lies on $AB$ , and $AM : MB = 2 : 3$ .

In terms of $\mathbf{a}$ and $\mathbf{b}$ , what is the position vector of $M$ ?
Give your answer in its simplest form. [4 marks]

(a)(i)(a)(iii)(a)(iv)(a)(v) y =(b)

[Total 17 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

P(1, 7)

and

Q(5, -5)

, with

O

the origin.

A circle centred on

O

with

PQ

as a chord, and a second circle with

PQ

as its diameter and

R

as its centre.

Position vectors

\mathbf{a}

and

\mathbf{b}

, with

M

AB

and

AM : MB = 2 : 3

Find

\overrightarrow{PQ}

, and a demonstration that

|\overrightarrow{OP}|

and

|\overrightarrow{OQ}|

are the same length. The circumference of the circle centred on

O

, the coordinates of

R

, and the perpendicular bisector of

PQ

in the form

y = mx + c

\overrightarrow{OM}

in terms of

\mathbf{a}

and

\mathbf{b}

Plan the Solution

Subtract position vectors for $\overrightarrow{PQ}$ , then use $\sqrt{x^2 + y^2}$ for each distance from the origin.
$P$ and $Q$ both lie on the circle centred $O$ , so that common distance is its radius and the circumference follows.
The centre of a circle is halfway along any diameter, so $R$ is the midpoint of $PQ$ .
The perpendicular bisector goes through that midpoint with the negative reciprocal of the gradient of $PQ$ .
For part (b), go from $O$ to $A$ and then part of the way along $AB$ ; a ratio of $2 : 3$ makes that part $\dfrac{2}{5}$ .

Worked Solution [17 marks]

Rule - Vectors and coordinate geometry:

\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}

, a point's distance from the origin is

\sqrt{x^2 + y^2}

, the midpoint of a segment is the average of its ends, and perpendicular gradients multiply to

-1

(a)(i) Subtract the position vectors

\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}

\overrightarrow{PQ} = \begin{pmatrix} 5 \\ -5 \end{pmatrix} - \begin{pmatrix} 1 \\ 7 \end{pmatrix} = \begin{pmatrix} 4 \\ -12 \end{pmatrix}

(Reason: The vector from

P

Q

is where you finish minus where you start, taken one component at a time.)

(a)(ii) The distance of

P

from the origin

|\overrightarrow{OP}| = \sqrt{1^2 + 7^2} = \sqrt{1 + 49} = \sqrt{50}

(Reason: Pythagoras on the right-angled triangle whose two shorter sides are the coordinates of

P

(a)(ii) The distance of

Q

from the origin

|\overrightarrow{OQ}| = \sqrt{5^2 + (-5)^2} = \sqrt{25 + 25} = \sqrt{50}

(Reason: Squaring removes the sign, so a

y

-coordinate of

-5

contributes

25

just as

5

would. Both magnitudes come out as

\sqrt{50}

, which is what this part asks you to show.)

(a)(iii) The radius of the circle centred

O

r = |\overrightarrow{OP}| = |\overrightarrow{OQ}| = \sqrt{50}

(Reason:

PQ

is a chord, so

P

and

Q

sit on the circle and

OP

and

OQ

are both radii. Part (ii) has already shown they are equal.)

(a)(iii) Apply the circumference formula

C = 2 \times \pi \times r

C = 2 \times \pi \times \sqrt{50} = 44.4288\ldots

(Reason: Keep the surd in the calculator instead of rounding the radius first, or the last figure drifts.)

(a)(iii) Round the circumference

C = 44.4

(Reason: A calculated length is given to three significant figures unless the question says otherwise, and

44.4288\ldots

rounds to

44.4

(a)(iv)

R

is the midpoint of

PQ

R = \left( \dfrac{1 + 5}{2}, \dfrac{7 + (-5)}{2} \right) = (3, 1)

(Reason: The centre of a circle sits halfway along any diameter, so average the

x

-coordinates and average the

y

-coordinates.)

(a)(v) The gradient of

PQ

\dfrac{7 - (-5)}{1 - 5} = \dfrac{12}{-4} = -3

(Reason: Rise over run, taking

P

first on the top and on the bottom so the two subtractions match.)

(a)(v) The perpendicular gradient

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

(Reason: Perpendicular gradients multiply to

-1

, so turn the gradient upside down and change its sign.)

(a)(v) Fit the line through

R

1 = \dfrac{1}{3} \times 3 + c

c = 0

y = \dfrac{1}{3}x

(Reason: A bisector cuts

PQ

at its midpoint, so

R(3, 1)

lies on it. Here the constant works out as zero, so the line passes through the origin.)

(b) The step from

A

M

\overrightarrow{AB} = \mathbf{b} - \mathbf{a}

\overrightarrow{AM} = \dfrac{2}{2 + 3}\overrightarrow{AB} = \dfrac{2}{5}(\mathbf{b} - \mathbf{a})

(Reason: A ratio of

2 : 3

splits

AB

into

5

equal parts, and

M

stands after

2

of them.)

(b) Travel from

O

M

\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM}

\overrightarrow{OM} = \mathbf{a} + \dfrac{2}{5}(\mathbf{b} - \mathbf{a}) = \dfrac{3}{5}\mathbf{a} + \dfrac{2}{5}\mathbf{b}

(Reason: Collecting the

\mathbf{a}

terms gives

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

, and that collecting is what makes it the simplest form.)

(a)(i)

\overrightarrow{PQ} = \begin{pmatrix} 4 \\ -12 \end{pmatrix}

(a)(ii)

|\overrightarrow{OP}| = |\overrightarrow{OQ}| = \sqrt{50}

(a)(iii)

44.4

(a)(iv)

R = (3, 1)

(a)(v)

y = \dfrac{1}{3}x

(b)

\overrightarrow{OM} = \dfrac{3}{5}\mathbf{a} + \dfrac{2}{5}\mathbf{b}

Verification

Check 1: Put

x = 3

into the bisector and see whether it reaches

R

\dfrac{3}{3} = 1

, so

(3, 1)

is on the line.

Check 2: Measure

R

against both ends of the diameter.

RP = \sqrt{2^2 + 6^2} = \sqrt{40}

and

RQ = \sqrt{2^2 + 6^2} = \sqrt{40}

, equal as a centre requires.

Check 3: Work the circumference the other way round, from the diameter.

2 \times \sqrt{50} = 14.1421\ldots

and

\pi \times 14.1421\ldots = 44.4288\ldots

, which matches.

Check 4: A point ON the line

AB

has coefficients adding to

1

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

, so

M

is on

AB

and not off it.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) $\overrightarrow{PQ}$ as a column vector	Note	The Answer column gives the column vector with $4$ above $-12$ , and the Marks column gives $2$ for it.	✓
(a)(i) Each component of $\overrightarrow{PQ}$	B1	For each.	✓
(a)(ii) The sum of squares for $\overrightarrow{OP}$	M1	For $1^2 + 7^2$ .	✓
(a)(ii) The sum of squares for $\overrightarrow{OQ}$	M1	For $5^2 + ([-]5)^2$ .	✓
(a)(ii) Both magnitudes equal $\sqrt{50}$	A1	Both $\sqrt{50}$ oe.	✓
(a)(ii) The condition on the accuracy mark	Note	With no errors seen.	✓
(a)(ii) The special case when nothing else is scored	SC1	If M0M0A0 scored, for $\sqrt{50}$ oe for each.	✓
(a)(iii) The circumference	Note	The Answer column gives $44.4$ or $44.42[8\ldots]$ to $44.435$ , and the Marks column gives $2$ for it.	✓
(a)(iii) The follow-through allowed	Note	FT their (a)(ii) correct to 3sf or better.	✓
(a)(iii) The method for the circumference	M1	For $2 \times \pi \times \text{their } \sqrt{50}$ oe.	✓
(a)(iv) The centre $R$	Note	The Answer column gives $(3, 1)$ , and the Marks column gives $2$ for it.	✓
(a)(iv) Each coordinate of $R$	B1	For each.	✓
(a)(v) The equation of the perpendicular bisector	Note	The Answer column gives $[y =] \dfrac{1}{3}x$ , and the Marks column gives $4$ for it.	✓
(a)(v) A correct equation in the wrong form	B3	For a correct equation in the wrong form as final answer.	✓
(a)(v) The perpendicular gradient on its own	B2	Or for $\dfrac{1}{3}$ stated or used as perpendicular gradient.	✓
(a)(v) Second method: the gradient of $PQ$	M1	Or for $[\text{grad } PQ] = \dfrac{7 - -5}{1 - 5}$ oe.	✓
(a)(v) Second method: the negative reciprocal	M1	For $\dfrac{-1}{\text{their grad } PQ}$ .	✓
(a)(v) Second method: substituting into $y = mx + c$	M1dep	For substituting their (a)(iv) or $(0, 0)$ into $y = \text{their } mx + c$ oe, dep on the 2nd M1 or B2.	✓
(b) The position vector of $M$	Note	The Answer column gives $\dfrac{3}{5}\mathbf{a} + \dfrac{2}{5}\mathbf{b}$ final answer, and the Marks column gives $4$ for it.	✓
(b) An unsimplified correct answer	B3	For an unsimplified correct answer.	✓
(b) $\overrightarrow{AM}$ or $\overrightarrow{BM}$ in terms of $\mathbf{a}$ and $\mathbf{b}$	B2	Or for $AM = \dfrac{2}{5}(\mathbf{b} - \mathbf{a})$ soi, or $BM = \dfrac{3}{5}(\mathbf{a} - \mathbf{b})$ soi.	✓
(b) $\overrightarrow{AB}$ , a correct route, or a correct diagram	B1	Or for $AB = \mathbf{b} - \mathbf{a}$ or $BA = \mathbf{a} - \mathbf{b}$ , or for a correct route for $OM$ , or for correct diagram.	✓

Full marks: 17/17

The positions of two beacons $A$ and $B$ , a yacht $C$ and a marina $H$ are marked on the diagram.
$B$ lies due west of $C$ .

(a) From yacht $C$ , what is the bearing of the marina? [1 mark]

(b) (i) Show that the size of angle $CBH$ is $100^\circ$ . [1 mark]

(ii) Show that the length of $BH$ is $13.7$ km, correct to $1$ decimal place. [3 marks]

(d) Yacht $C$ leaves at $1$ pm and sails the $32$ km straight to the marina, holding a speed of $10$ knots.

(i) At what time does yacht $C$ reach the marina?
Give this time correct to the nearest minute.
[ $1 \text{ knot} = 1.852 \text{ km/h}$ ] [4 marks]

(ii) How far is yacht $C$ from the marina at the moment when it is closest to beacon $B$ ? [3 marks]

(a)(c)(d)(i)(d)(ii) km

[Total 17 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

Two beacons

A

and

B

, a yacht

C

and a marina

H

, with

B

due west of

C

AB = 11

km,

AH = 14

km and

HC = 32

km.

Angle

BCH = 25^\circ

and angle

BHC = 55^\circ

The yacht leaves at

1

pm at

10

knots, and

1

knot is

1.852

km/h.

Find

The bearing of the marina from the yacht, and a demonstration that angle

CBH

100^\circ

and that

BH

13.7

km. The bearing of

A

from

B

. The arrival time, and the distance still to run when the yacht is nearest to

B

Plan the Solution

A bearing is measured clockwise from north, so start (a) from the $270^\circ$ line that points from $C$ back to $B$ .
The angles of triangle $BCH$ add to $180^\circ$ , which leaves the angle at $B$ .
Two angles and a side opposite one of them make triangle $BCH$ a sine-rule triangle.
Triangle $ABH$ has all three sides known, so the cosine rule gives the angle at $B$ . Adding it to the bearing of $H$ from $B$ turns it into a bearing.
A knot is a nautical mile an hour, so convert the speed once and then divide distance by speed.
The shortest distance from $B$ to the yacht's track is the perpendicular, and that makes a right-angled triangle with $BH$ as its hypotenuse.

Worked Solution [17 marks]

Rule - Bearings, the sine rule and the cosine rule: a bearing is measured clockwise from north and written with three figures,

\dfrac{a}{\sin A} = \dfrac{b}{\sin B}

, and

a^2 = b^2 + c^2 - 2bc\cos A

(a) Turn round the line from

C

B

270 - 25 = 245

(Reason:

B

is due west of

C

, so from

C

the beacon lies on a bearing of

270^\circ

. The marina sits

25^\circ

on the anticlockwise side of that line, so its bearing is

25

less.)

(b)(i) The angles of triangle

BCH

180 - (55 + 25) = 100

(Reason: The angles of any triangle add to

180^\circ

. The angles at

H

and

C

are given, so the one at

B

is whatever is left.)

(b)(ii) Set up the sine rule in triangle

BCH

\dfrac{BH}{\sin 25} = \dfrac{32}{\sin 100}

(Reason:

BH

is opposite the

25^\circ

angle and

HC

is opposite the

100^\circ

angle, so those are the two pairs that go together.)

(b)(ii) Rearrange and work it out

BH = \dfrac{32 \times \sin 25}{\sin 100}

BH = 13.7324\ldots

(Reason: Multiply both sides by

\sin 25

. Keep every figure in the calculator instead of rounding here, or the first decimal place is not safe.)

(b)(ii) Round to

1

decimal place

BH = 13.7

(Reason: The digit after the first decimal place is

3

, so the

7

stays as it is, and that is the

13.7

km the question asks to be shown.)

H

from

B

90 + 100 = 190

(Reason:

C

is due east of

B

, so

BC

points along

090^\circ

. Turning clockwise through the

100^\circ

of angle

CBH

lands on

BH

ABH

\cos B = \dfrac{11^2 + 13.7^2 - 14^2}{2 \times 11 \times 13.7}

(Reason: All three sides are known, so the rearranged cosine rule gives the cosine of the angle at

B

. The side that is subtracted is the one opposite that angle.)

11^2 + 13.7^2 - 14^2 = 112.69

2 \times 11 \times 13.7 = 301.4

(Reason: Square each side first, then combine the squares the way the rearranged rule sets them out.)

\dfrac{112.69}{301.4} = 0.3739

\cos^{-1}(0.3739) = 68.0\ldots

(Reason: The calculator must be in degree mode. Angle

ABH

comes out as

68.0^\circ

1

decimal place.)

H

190 + 68.0 = 258.0

(Reason:

A

lies on the far side of

BH

from

C

, so the turn carries on clockwise past

190^\circ

. A bearing is written with three figures, so the answer is

258^\circ

(d)(i) Change the speed into km/h

10 \times 1.852 = 18.52

(Reason: A knot is one nautical mile an hour, and the question gives

1

knot as

1.852

km/h.)

(d)(i) Divide the distance by the speed

\dfrac{32}{18.52} \times 60 = 103.67

(Reason: Distance over speed gives the time in hours, and the

60

turns those hours straight into minutes.)

(d)(i) Change the minutes into hours and minutes

60 + 44 = 104

(Reason:

103.67

minutes is

104

minutes to the nearest minute, which is

1

hour and

44

minutes. Starting at

1

pm, the yacht ties up at

2\ 44

pm.)

(d)(ii) Find where the yacht passes closest to

B

\dfrac{x}{13.7} = \cos 55

(Reason: The shortest distance from a point to a line is the perpendicular, so drop one from

B

onto

CH

and call the foot

X

. In right-angled triangle

BXH

the side

HX

is adjacent to the

55^\circ

angle and

BH

is the hypotenuse.)

(d)(ii) Work out that distance

x = 13.7 \times \cos 55 = 7.858\ldots

x = 7.86

(Reason: That is the stretch of

CH

still lying between the yacht and the marina, given to

3

significant figures.)

(a)

245^\circ

(b)(i) angle

CBH

100^\circ

(b)(ii)

BH

13.7

km(c)

258^\circ

(d)(i)

2\ 44

pm(d)(ii)

7.86

Verification

Check 1: Walk the bearings round the figure. The reverse of the bearing of

H

from

C

should lead back through the

55^\circ

angle to the reverse of

190^\circ

245 - 180 = 65

, then

65 - 55 = 10

, and the reverse of

190^\circ

190 - 180 = 10

, which agrees.

Check 2: Add all three angles of triangle

BCH

25 + 55 + 100 = 180

, as a triangle requires.

Check 3: Put the unrounded

BH

back into the sine rule and compare the two ratios.

\dfrac{13.7324}{\sin 25} = 32.49\ldots

and

\dfrac{32}{\sin 100} = 32.49\ldots

, the same ratio.

Check 4: Run the cosine rule the other way round, using the angle just found to rebuild

AH

11^2 + 13.7^2 - 2 \times 11 \times 13.7 \times \cos 68.04^\circ = 196.0\ldots

, and

\sqrt{196} = 14

, which is

AH

Check 5: Multiply the sailing time back by the speed.

18.52 \times 1.7279 = 32.0

km, which is

HC

Check 6: Test right-angled triangle

BXH

with Pythagoras, where

BX = 13.7 \times \sin 55

11.2224^2 + 7.858^2 = 187.69

, and

13.7^2 = 187.69

, so the two sides do make up the hypotenuse.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The bearing of the marina from the yacht	Note	The Answer column gives $245$ , and the Marks column gives $1$ for it.	✓
(b)(i) The angle at $B$ from the angle sum	M1	For $180 - (55 + 25)\ [=100]$ .	✓
(b)(ii) The sine rule rearranged for $BH$	M2	For $\dfrac{32 \times \sin 25}{\sin 100}$ oe.	✓
(b)(ii) The sine rule before it is rearranged	M1	For $\dfrac{\sin 25}{BH} = \dfrac{\sin 100}{32}$ oe.	✓
(b)(ii) The unrounded length of $BH$	A1	For $13.73\ldots$ .	✓
(c) The bearing of $A$ from $B$	Note	The Answer column gives $258$ or $257.9$ to $258.0\ldots$ , and the Marks column gives $5$ for it.	✓
(c) The angle at $B$ on its own	B4	For $67.9$ to $68.0\ldots$ .	✓
(c) Second method: the cosine rule rearranged for $\cos B$	M2	Or for $[\cos =] \dfrac{11^2 + 13.7^2 - 14^2}{2 \times 13.7 \times 11}$ .	✓
(c) Second method: the value of the cosine	A1	For $0.3738$ to $0.376$ .	✓
(c) Second method: the cosine rule before it is rearranged	M1	Or for $14^2 = 11^2 + 13.7^2 - 2 \times 11 \times 13.7 \times \cos B$ .	✓
(c) Turning the angle at $B$ into a bearing	M1dep	For $190 + \text{their angle } B$ , dep on at least M1.	✓
(d)(i) The arrival time	Note	The Answer column gives $2\ 44$ pm or $14\ 44$ cao, and the Marks column gives $4$ for it.	✓
(d)(i) The journey time, as hours and minutes or as minutes	B3	For $1 \text{ hour } 44$ or $1 \text{ hour } 43.6$ to $1 \text{ hour } 43.8$ or $104$ or $103.6$ to $103.8$ .	✓
(d)(i) The journey time in hours	B2	Or for $1.727$ to $1.73$ .	✓
(d)(i) The method, taken as far as minutes	M2	Or for $\dfrac{32}{10 \times 1.852} \times 60$ .	✓
(d)(i) The method, taken only as far as hours	M1	Or for $\dfrac{32}{10 \times 1.852}$ .	✓
(d)(ii) The distance still to run	Note	The Answer column gives $7.857$ to $7.88$ , and the Marks column gives $3$ for it.	✓
(d)(ii) The trigonometry in the right-angled triangle	M2	For $\dfrac{x}{13.7} = \cos 55$ oe.	✓
(d)(ii) Recognising where the shortest distance falls	M1	Or for dist to $H$ occurs when perpendicular from $B$ meets $CH$ soi.	✓

Full marks: 17/17

A crate in the shape of a cuboid is drawn above.
The crate has no top.

(a) (i) What is the surface area of the inside of the open crate? [3 marks]

(ii) Cylindrical tins, each of height $20$ cm and diameter $15$ cm, are put into the crate.
What is the greatest number of these tins that fit completely inside the crate? [3 marks]

(b) The mass of a solid bronze cone is $750$ g.
The bronze has a density of $8.9$ g/cm³.
The radius and the height of the cone are in the ratio $1 : 3$ .

(i) Show that the cone has a radius of $2.99$ cm, correct to $3$ significant figures.
[ $\text{Density} = \dfrac{\text{mass}}{\text{volume}}$ ]
[The volume, $V$ , of a cone with radius $r$ and height $h$ is $V = \dfrac{1}{3}\pi r^2 h$ .] [4 marks]

(ii) What is the total surface area of the cone?
[The curved surface area, $A$ , of a cone with radius $r$ and slant height $l$ is $A = \pi r l$ .] [5 marks]

(a)(i) cm²(a)(ii)(b)(ii) cm²

[Total 15 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

An open crate in the shape of a cuboid,

70

cm long,

30

cm deep and

40

cm high, with no top.

Tins that are cylinders of height

20

cm and diameter

15

cm.

A solid bronze cone of mass

750

g, made of bronze of density

8.9

g/cm³, whose radius and height are in the ratio

1 : 3

Find

The surface area of the inside of the open crate, and the greatest number of tins that fit inside it. A demonstration that the cone has radius

2.99

cm, and then the total surface area of that cone.

Plan the Solution

An open box has five faces, not six: a base and two matching pairs of walls.
A tin cannot be cut, so divide each edge of the crate by the measurement of the tin that lies along it and round every count down before multiplying.
Density ties mass to volume, so turn the $750$ g into a volume first.
The ratio makes the height $3r$ , and $\dfrac{1}{3}\pi r^2 \times 3r$ collapses to $\pi r^3$ , which leaves one unknown in one equation.
The radius, the height and the slant height make a right-angled triangle, so Pythagoras gives $l$ before $A = \pi r l$ can be used.
A cone standing on its base has a flat circle as well as a curved surface, so the total is $\pi r l + \pi r^2$ .

Worked Solution [15 marks]

Rule - Surface area, density and the cone: an open box has five faces,

\text{density} = \dfrac{\text{mass}}{\text{volume}}

V = \dfrac{1}{3}\pi r^2 h

and

A = \pi r l

(a)(i) The base of the crate

70 \times 30 = 2100

(Reason: The crate stands on a rectangle

70

cm by

30

cm, and that face is inside the crate, so it is one of the faces counted.)

(a)(i) The two end walls

2 \times 30 \times 40 = 2400

(Reason: The two ends stand on the

30

cm edges and rise the full

40

cm. They are a matching pair, so one area is doubled.)

(a)(i) The two long walls

2 \times 40 \times 70 = 5600

(Reason: The other two walls stand on the

70

cm edges and rise to the same height as the ends. They are a matching pair as well.)

(a)(i) Add the five faces

2100 + 2400 + 5600 = 10\,100

(Reason: A base and four walls, with no lid to add. Every term is a length times a length, so the answer is in square centimetres.)

(a)(ii) How many tins lie along the

70

cm edge

\dfrac{70}{15} = 4.66\ldots

(Reason: Each tin needs

15

cm of that edge. Four tins use

60

cm and a fifth would need

75

cm, so the count is

4

- part of a tin is no use.)

(a)(ii) Across the crate, and up it

\dfrac{30}{15} = 2

\dfrac{40}{20} = 2

(Reason: The tins stand upright, so it is the diameter that has to fit across the

30

cm and the height of a tin that has to fit up the

40

cm. Both of these divide exactly.)

(a)(ii) Multiply the three counts

4 \times 2 \times 2 = 16

(Reason: Eight tins cover the base in a

4

2

block, and a second layer of eight stands on top of them.)

(b)(i) Turn the mass into a volume

V = \dfrac{750}{8.9} = 84.2696\ldots

(Reason: Density is mass divided by volume, so volume is mass divided by density. Keep every figure in the calculator here - rounding now would move the third significant figure of the radius.)

(b)(i) Put the ratio into the cone formula

h = 3r

V = \dfrac{1}{3}\pi r^2 \times 3r = \pi r^3

(Reason: The ratio

1 : 3

makes the height three times the radius. The

3

and the

\dfrac{1}{3}

cancel, which leaves one unknown.)

(b)(i) Solve for

r

\pi r^3 = 84.2696\ldots

r^3 = \dfrac{84.2696\ldots}{\pi} = 26.8238\ldots

r = \sqrt[3]{26.8238\ldots} = 2.9934\ldots

(Reason: Divide by

\pi

, then take the cube root. Every figure is carried until the last line.)

(b)(i) Round to

3

significant figures

r = 2.99

(Reason: The fourth significant figure is

4

, so the third stays as it is. That is the

2.99

cm the question asks to be shown.)

(b)(ii) The height and the slant height

h = 3 \times 2.99 = 8.97

2.99^2 + 8.97^2 = 89.401

l = \sqrt{89.401} = 9.4552\ldots

(Reason: The radius, the height and the slant height make a right-angled triangle, with the slant height as its hypotenuse.)

(b)(ii) The curved surface

\pi \times 2.99 \times 9.4552 = 88.816

(Reason: This is the formula the question gives,

A = \pi r l

, with the radius and the slant height just found.)

(b)(ii) The base circle

\pi \times 2.99^2 = 28.086

(Reason: The cone is solid and stands on its base, so there is a flat circle of radius

2.99

cm to add as well.)

(b)(ii) Add the two areas

88.816 + 28.086 = 116.902

A = 117

(Reason: Rounded to

3

significant figures, the total surface area is

117

cm².)

(a)(i)

10\,100

cm²(a)(ii)

16

tins(b)(i) the radius is

2.99

cm(b)(ii)

117

cm²

Verification

Check 1: Build the four walls as one band round the base instead of adding them face by face.

2 \times (70 + 30) = 200

cm round the base,

200 \times 40 = 8000

cm² of wall, and

8000 + 2100 = 10\,100

cm² once the base is added.

Check 2: Measure the space the tins take up inside the crate.

4 \times 15 = 60

cm of the

70

cm,

2 \times 15 = 30

cm of the

30

cm and

2 \times 20 = 40

cm of the

40

cm. A fifth tin along the longest edge would need

75

cm.

Check 3: Put the unrounded radius back and rebuild the mass it came from. With

r = 2.9934\ldots

the volume

\pi r^3

84.2696\ldots

cm³, and

8.9 \times 84.2696\ldots = 750

Check 4: Rebuild the surface area a different way. For this cone

l = r\sqrt{10}

, so the total is

\pi r^2 (1 + \sqrt{10})

1 + \sqrt{10} = 4.1623

and

28.086 \times 4.1623 = 116.902

Check 5: Is the size sensible? This cone is three times as tall as its radius, so the curved surface should be a few times the base circle.

\dfrac{88.816}{28.086} = 3.162

, which is

\sqrt{10}

4

significant figures - the ratio

\dfrac{l}{r}

for this cone.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) The surface area of the inside of the crate	Note	The Answer column gives $10\,100$ , and the Marks column gives $3$ for it.	✓
(a)(i) The five faces, all of them	M2	For $30 \times 70 + 2 \times 30 \times 40 + 2 \times 40 \times 70$ .	✓
(a)(i) One face of the crate worked out	M1	Or for $30 \times 40$ or $30 \times 70$ or $40 \times 70$ .	✓
(a)(ii) The greatest number of tins	Note	The Answer column gives $16$ , and the Marks column gives $3$ for it.	✓
(a)(ii) How many tins fit along each edge	M2	For $2$ fit width, $2$ fit height and $4$ fit length soi.	✓
(a)(ii) One edge divided by one tin	M1	Or for $70$ , $30$ or $40$ divided by $15$ or $20$ .	✓
(b)(i) The volume equation, with the ratio already used	M2	For $\dfrac{1}{3}\pi r^2 \times 3r = \text{their } \dfrac{750}{8.9}$ oe.	✓
(b)(i) The density formula used on the two given values	M1	For using $750$ and $8.9$ correctly in $v = \dfrac{m}{d}$ oe or $\dfrac{750}{8.9}$ .	✓
(b)(i) The cube of the radius made the subject	M1dep	For $r^3 = \dfrac{\text{their }\left(\dfrac{750}{8.9}\right)}{\pi}$ oe.	✓
(b)(i) The unrounded radius	A1	For $r = 2.993\ldots$ .	✓
(b)(ii) The total surface area of the cone	Note	The Answer column gives $117$ or $116.9$ to $117.2$ , and the Marks column gives $5$ for it.	✓
(b)(ii) Both surfaces in one expression	M4	For $\pi \times 2.99^2 + \pi \times 2.99 \times \sqrt{2.99^2 + (3 \times 2.99)^2}$ oe.	✓
(b)(ii) The curved surface area on its own	M3	Or for $\pi \times 2.99 \times \sqrt{2.99^2 + (3 \times 2.99)^2}$ .	✓
(b)(ii) The slant height	M2	Or for $\sqrt{2.99^2 + (3 \times 2.99)^2}$ .	✓
(b)(ii) Pythagoras started, or the base circle alone	M1	Or for $2.99^2 + (3 \times 2.99)^2$ , or for $\pi \times 2.99^2$ .	✓

Full marks: 15/15

(a) Sketch the graph of $y = x^2 + 7x - 18$ on the axes below.
The values where the graph meets the $x$ -axis and the $y$ -axis must be written on your sketch. [4 marks]

(b) (i) What is the derivative of $y = x^2 - 3x - 28$ ? [2 marks]

(ii) For $y = x^2 - 3x - 28$ , give the coordinates of the turning point. [3 marks]

What are the coordinates of $P$ and $Q$ ?
You must show all your working and give your answers correct to $2$ decimal places. [6 marks]

(b)(i)(b)(ii)(c)(c)

[Total 15 marks]

Show solution & mark schemeHide solution & mark scheme

Question 8 - Exam Solution

Understanding the Question

Given

The curve

y = x^2 + 7x - 18

, and a blank pair of axes to sketch it on.

The curve

y = x^2 - 3x - 28

, used in parts (b) and (c).

The line

y = 5 - 2x

, which cuts that curve at

P

and

Q

Find

A sketch of

y = x^2 + 7x - 18

carrying the values where it meets each axis. The derivative of

y = x^2 - 3x - 28

, and the coordinates of its turning point. The coordinates of

P

and

Q

, each correct to

2

decimal places.

Plan the Solution

A sketch needs three things: which way up the curve is, where it cuts each axis, and roughly where its lowest point sits.
The $x^2$ term is positive, so the curve is U shaped, and it meets the $x$ -axis where the right-hand side factorises to zero.
Putting $x = 0$ in gives the $y$ -axis value in one line, because it leaves only the constant term.
The lowest point of a U shaped quadratic sits halfway between its two $x$ -axis values.
A turning point is where the gradient is zero, so differentiate first and then solve.
Two graphs meet where their $y$ -values agree, so set the two right-hand sides equal and collect every term on one side.
The quadratic that comes out has no factors, so the formula is the way through, and the y-values then come from the simpler of the two equations.

Worked Solution [15 marks]

Rule - Sketching a quadratic, and where two graphs meet: factorise for the

x

-axis values, put

x = 0

for the

y

-axis value, solve

\dfrac{dy}{dx} = 0

for a turning point, and set the two expressions for

y

equal where two graphs cross.

(a) Factorise the right-hand side

x^2 + 7x - 18 = (x + 9)(x - 2)

(Reason: Two numbers that multiply to

-18

and add to

7

are

9

and

-2

(a) The values where the graph meets the

x

-axis

(x + 9)(x - 2) = 0

x = -9 \quad \text{or} \quad x = 2

(Reason: The graph meets the

x

-axis where

y = 0

, and a product is zero only when one of its brackets is zero.)

(a) The value where the graph meets the

y

-axis

y = 0^2 + 7(0) - 18 = -18

(Reason: Every point on the

y

-axis has

x = 0

, so only the constant term survives.)

(a) The lowest point, so the curve can be placed

x = \dfrac{-9 + 2}{2} = -3.5

y = (-3.5)^2 + 7(-3.5) - 18 = -30.25

(Reason: A U shaped quadratic is symmetrical, so its lowest point is halfway between the two

x

-axis values. Both coordinates are negative, which puts the lowest point in the third quadrant.)

(a) Draw the curve

y = x^2 + 7x - 18

(Reason: The coefficient of

x^2

is positive, so the curve is U shaped. Mark

-9

and

2

on the

x

-axis and

-18

on the

y

-axis, write those three values beside the marks, and draw one smooth curve through them.)

(b)(i) Differentiate term by term

y = x^2 - 3x - 28

\dfrac{dy}{dx} = 2x - 3

(Reason: Bring each power down in front and drop it by one:

x^2

gives

2x

-3x

gives

-3

, and a constant has no gradient of its own.)

(b)(ii) The gradient is zero at a turning point

2x - 3 = 0

x = 1.5

(Reason: The curve is level for an instant at its turning point, so set the derivative from part (b)(i) equal to zero and solve.)

(b)(ii) The

y

-coordinate of the turning point

y = 1.5^2 - 3(1.5) - 28

y = 2.25 - 4.5 - 28 = -30.25

(Reason: The turning point lies on the curve, so put the

x

-value back into the equation of the curve rather than into the derivative.)

y

-value where they cross

x^2 - 3x - 28 = 5 - 2x

(Reason: At

P

and at

Q

the point lies on both graphs, so the two expressions for

y

must be equal there.)

x^2 - 3x - 28 - 5 + 2x = 0

x^2 - x - 33 = 0

(Reason: Take

5

from both sides and add

2x

to both sides. The

-3x

and the

2x

leave

-x

, and

-28 - 5

leaves

-33

x = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-33)}}{2(1)}

x = \dfrac{1 \pm \sqrt{133}}{2}

(Reason: Here

a = 1

b = -1

and

c = -33

. Under the root,

1 + 132 = 133

, which is not a square number, so the answers will not be exact.)

x

-values

x = \dfrac{1 + 11.5325\ldots}{2} = 6.2662\ldots

x = \dfrac{1 - 11.5325\ldots}{2} = -5.2662\ldots

(Reason: Keep the whole calculator display at this stage. Rounding here would move the second decimal place of the

y

-values.)

y

-values

y = 5 - 2(6.2662\ldots) = -7.5325\ldots

y = 5 - 2(-5.2662\ldots) = 15.5325\ldots

(Reason: Use the line, not the curve:

y = 5 - 2x

is the shorter calculation and gives the same point. Watch the second one, where subtracting a negative adds.)

2

decimal places

(-5.27,\ 15.53)

(6.27,\ -7.53)

(Reason: The third decimal place is

6

in both

x

-values, so each of those rounds up, and

2

in both

y

-values, so each of those stays as it is.)

(a) a U shaped curve through

(-9,\ 0)

(2,\ 0)

and

(0,\ -18)

(b)(i)

\dfrac{dy}{dx} = 2x - 3

(b)(ii)

(1.5,\ -30.25)

(c)

(-5.27,\ 15.53)

and

(6.27,\ -7.53)

Verification

Check 1: Expand the brackets again and see the original expression come back.

(x + 9)(x - 2) = x^2 - 2x + 9x - 18 = x^2 + 7x - 18

Check 2: Put both

x

-axis values back into

y = x^2 + 7x - 18

; each must give zero.

81 - 63 - 18 = 0

and

4 + 14 - 18 = 0

Check 3: Find the turning point of

y = x^2 - 3x - 28

without differentiating, by completing the square. Completing the square gives

(x - 1.5)^2 - 30.25

, which is smallest when the bracket is zero, so the turning point is

(1.5,\ -30.25)

. Part (a)'s curve reaches the same depth, since

-18 - 12.25 = -30.25

Check 4: Both points must sit on the line, so put the rounded

x

-values into

y = 5 - 2x

5 - 2(-5.27) = 15.54

and

5 - 2(6.27) = -7.54

, each within

0.01

of the rounded

y

-values, which is why the mark scheme accepts either figure.

Check 5: The two solutions of

x^2 - x - 33 = 0

must add to

1

, because the coefficient of

x

-1

-5.2662\ldots + 6.2662\ldots = 1

, and the two values sit either side of

0.5

, the line of symmetry of that quadratic.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The sketch	Note	The Answer column gives correct sketch with roots indicated at $x = -9$ and $x = 2$ and $y$ intercept = $-18$ , minimum should be in 3rd quadrant, and the Marks column gives $4$ for it.	✓
(a) The shape of the curve	B1	For a U shaped parabola.	✓
(a) The two values on the $x$ -axis	B2	For roots at $-9$ and $2$ on diagram.	✓
(a) The factorisation instead	M1	Or for $(x + 9)(x - 2) \; [= 0]$ .	✓
(a) The value on the $y$ -axis	B1	For $y$ – intercept at $-18$ on diagram.	✓
(a) The cap on the sketch	Note	Maximum 3 marks if sketch not fully correct.	✓
(b)(i) The derivative	Note	The Answer column gives $2x - 3$ , and the Marks column gives $2$ for it.	✓
(b)(i) One term differentiated correctly	B1	For $2x + k$ or $kx^{[p]} - 3$ .	✓
(b)(ii) The turning point	Note	The Answer column gives $(1.5,\ -30.25)$ oe, and the Marks column gives $3$ for it.	✓
(b)(ii) The $x$ -coordinate alone	B2	For $x = 1.5$ .	✓
(b)(ii) The derivative set to zero, or the completed square	M1	Or for $\text{their (b)(i)} = 0$ , or for $(x - 1.5)^2$ .	✓
(c) The rearranged quadratic	B1	For $x^2 - x - 33 \; [= 0]$ seen.	✓
(c) How the formula row follows through	Note	FT their quadratic dep on no factors.	✓
(c) The quadratic formula with the right values in it	B2FT	For $\dfrac{[--]1 \pm \sqrt{([-]1)^2 - 4(1)(-33)}}{2 \times 1}$ oe.	✓
(c) The discriminant under the root	B1	For $\sqrt{([-]1)^2 - 4(1)(-33)}$ or better.	✓
(c) One branch of the formula	B1	Or for $\dfrac{[--]1 + \sqrt{q}}{2(1)}$ oe or $\dfrac{[--]1 - \sqrt{q}}{2(1)}$ oe.	✓
(c) The two $x$ -values	B2	For $-5.27$ or $-5.267$ to $-5.266$ and $6.27$ or $6.266$ to $6.267$ .	✓
(c) One of the two $x$ -values	B1	For each.	✓
(c) The special case for two swapped signs	SC1	If 0 scored, for $-6.27$ and $5.27$ .	✓
(c) The two points	B1	For $(-5.27,\ 15.53 \text{ or } 15.54)$ and $(6.27,\ -7.53 \text{ or } -7.54)$ .	✓

Full marks: 15/15

$\mathrm{f}(x) = 4x + 1 \qquad \mathrm{g}(x) = 6 - 2x \qquad \mathrm{h}(x) = 3^{x-2}$

(a) Work out the value of each of the following.
(i) $\mathrm{f}(3)$ [1 mark]

(ii) $\mathrm{gf}(3)$ [1 mark]

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

(d) What is the value of $\mathrm{hh}(2)$ ? [2 marks]

(e) For which value of $x$ is $\mathrm{h}^{-1}(x) = 10$ ? [2 marks]

(a)(i)(a)(ii)(b) g⁻¹(x) =(c) x =(d)(e) x =

[Total 12 marks]

Show solution & mark schemeHide solution & mark scheme

Question 9 - Exam Solution

Understanding the Question

Given

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

\mathrm{g}(x) = 6 - 2x

\mathrm{h}(x) = 3^{x-2}

Two linear functions and one function with the variable in the index.

Find

(a)(i)

\mathrm{f}(3)

and (ii)

\mathrm{gf}(3)

(b) the inverse

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

(c)

x

when

\mathrm{f}(x) = \mathrm{g}(2x - 7)

(d)

\mathrm{hh}(2)

(e)

x

when

\mathrm{h}^{-1}(x) = 10

Plan the Solution

To evaluate a function at a number, replace every $x$ in its formula by that number.
In $\mathrm{gf}(3)$ the function written nearest the number acts first, so work out $\mathrm{f}(3)$ and feed the result into $\mathrm{g}$ .
For the inverse, write $y = \mathrm{g}(x)$ , make $x$ the subject, then rename the letters.
For (c), put $2x - 7$ into $\mathrm{g}$ first, then solve the linear equation that is left.
For (d), work outwards: $\mathrm{h}(2)$ first, then $\mathrm{h}$ of that answer.
For (e), an inverse undoes its function, so $\mathrm{h}^{-1}(x) = 10$ is the same statement as $x = \mathrm{h}(10)$ .

Worked Solution [12 marks]

Rule - Composite and inverse functions:

\mathrm{gf}(x)

means apply

\mathrm{f}

first and then

\mathrm{g}

; and an inverse reverses a function, so

\mathrm{g}^{-1}(y) = x

exactly when

\mathrm{g}(x) = y

Step 1: (a)(i) Put

3

in place of

x

\mathrm{f}

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

4 \times 3 + 1 = 12 + 1 = 13

(Reason: the formula

4x + 1

is a rule for what to do to its input, so the

3

is multiplied by

4

before the

1

is added)

Step 2: (a)(ii) Apply

\mathrm{f}

first, then

\mathrm{g}

\mathrm{gf}(3) = \mathrm{g}(13)

\mathrm{g}(13) = 6 - 2 \times 13

6 - 2 \times 13 = 6 - 26 = -20

(Reason: the function written nearest the number acts first, so the

13

from part (i) becomes the input of

\mathrm{g}

)

Step 3: (b) Make

x

the subject of

y = 6 - 2x

y = 6 - 2x

2x = 6 - y

x = \dfrac{6 - y}{2}

\mathrm{g}^{-1}(x) = \dfrac{6 - x}{2}

(Reason: the inverse sends an output of

\mathrm{g}

back to the input it came from, so the working is that rearrangement; the letters are renamed at the end because an inverse is written as a function of

x

)

Step 4: (c) Feed

2x - 7

into

\mathrm{g}

\mathrm{g}(2x - 7) = 6 - 2(2x - 7)

6 - 2(2x - 7) = 6 - 4x + 14

6 - 4x + 14 = 20 - 4x

(Reason: both terms inside the bracket are multiplied by

-2

, and a negative multiplied by a negative is positive, so the constant term grows rather than shrinks)

Step 5: (c) Collect the terms and solve

4x + 1 = 20 - 4x

4x + 4x = 20 - 1

8x = 19

x = \dfrac{19}{8} = 2.375

(Reason: the

x

terms go on one side and the numbers on the other, which leaves a single division by

8

)

Step 6: (d) Work outwards for

\mathrm{hh}(2)

\mathrm{h}(2) = 3^{2-2} = 3^{0} = 1

\mathrm{hh}(2) = \mathrm{h}(1)

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

(Reason: the inner

\mathrm{h}

acts first and gives

1

, and that

1

is then the input of the outer

\mathrm{h}

)

Step 7: (e) Turn the inverse statement round

\mathrm{h}^{-1}(x) = 10

x = \mathrm{h}(10) = 3^{10-2}

3^{10-2} = 3^{8} = 6561

(Reason: an inverse undoes its function, so saying that

\mathrm{h}^{-1}

sends

x

10

is the same as saying that

\mathrm{h}

sends

10

x

)

(a)(i)

13

(a)(ii)

-20

(b)

\mathrm{g}^{-1}(x) = \dfrac{6 - x}{2}

(c)

x = 2.375

(d)

\dfrac{1}{3}

(e)

x = 6561

Verification

Check 1 - part (a)(ii): Compose the two functions into one formula first,

\mathrm{gf}(x) = 6 - 2(4x + 1) = 4 - 8x

, then put

x = 3

into that instead.

4 - 8 \times 3 = 4 - 24 = -20

Check 2 - part (b):

\mathrm{g}(5) = 6 - 10 = -4

, so the inverse must send

-4

back to

5

\dfrac{6 - (-4)}{2} = \dfrac{10}{2} = 5

Check 3 - part (c): Put

x = 2.375

into each side of

\mathrm{f}(x) = \mathrm{g}(2x - 7)

separately. The bracket gives

2 \times 2.375 - 7 = -2.25

4 \times 2.375 + 1 = 10.5

and

6 - 2 \times (-2.25) = 10.5

Check 4 - part (d): A negative index is a reciprocal, not a negative number, so multiplying the answer by the base must give

1

\dfrac{1}{3} \times 3 = 1

Check 5 - part (e): Divide

6561

3

over and over: it takes

8

divisions to reach

1

6561 = 3^{8}

, so the index

x - 2

8

and

x = 10

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) The value of $\mathrm{f}(3)$	Note	The Answer column gives $13$ , and the Marks column gives $1$ for it.	✓
(a)(ii) The value of $\mathrm{gf}(3)$	Note	The Answer column gives $-20$ , and the Marks column gives $1$ for it.	✓
(a)(ii) The instruction printed in the Partial Marks column	Note	FT $6 - 2(\textit{their} \text{ (a)(i)})$ .	✓
(b) The expression for $\mathrm{g}^{-1}(x)$	Note	The Answer column gives $\dfrac{6 - x}{2}$ oe final answer, and the Marks column gives $2$ for it.	✓
(b) The first step of the rearrangement	M1	For the correct first step: $x = 6 - 2y$ , $y - 6 = -2x$ , $\dfrac{y}{2} = 3 - x$ .	✓
(c) The value of $x$	Note	The Answer column gives $2.375$ oe, and the Marks column gives $4$ for it.	✓
(c) Substituting $2x - 7$ into $\mathrm{g}$	B1	For $6 - 2(2x - 7)$ oe.	✓
(c) Expanding the bracket	B1	For $4x + 1 = 6 - 4x + 14$ .	✓
(c) Rearranging to a single term in x	M1	For $8x = 19$ FT their linear equation rearranged correctly from $ax + b = cx + d$ to form $ex = f$ .	✓
(d) The value of $\mathrm{hh}(2)$	Note	The Answer column gives $\dfrac{1}{3}$ or $0.333\ldots$ , and the Marks column gives $2$ for it.	✓
(d) Reaching the inner value, or writing the whole index expression	M1	For $\mathrm{h}(1)$ or $3^{(3^{x-2} - 2)}$ or $3^{(3^{2-2} - 2)}$ , or better.	✓
(e) The value of $x$	Note	The Answer column gives $6561$ , and the Marks column gives $2$ for it.	✓
(e) Turning the inverse statement round	M1	For $3^{10-2}$ or $x = \mathrm{h}(10)$ .	✓

Full marks: 12/12

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 0580/41, May/June 2024: Worked Solutions, Questions 5 to 9

Worked solutions, questions 5 to 9 of 9

Question 5, Calculator allowed

Question 5 - Exam Solution

Question 6, Calculator allowed

Question 6 - Exam Solution

Question 7, Calculator allowed

Question 7 - Exam Solution

Question 8, Calculator allowed

Question 8 - Exam Solution

Question 9, Calculator allowed

Question 9 - Exam Solution

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