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Edexcel IGCSE 4MA1 Paper 1H, November 2024: Worked Solutions and Mark Schemes

Sir Faraz Hassan

29 Jul 2026

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Edexcel International GCSE Mathematics A (4MA1)Paper 1H - Higher Tier - November 2024100 marks · 2 hours · Calculator allowed

Original worked solutions for Edexcel International GCSE Mathematics A (4MA1), Paper 1H (Higher Tier), November 2024 – 100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

Both are PDF files hosted by Pearson: 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 accuracy mark (A1) is earned. The questions follow the same order as the original paper and carry the same marks.

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

Worked solutions, questions 1 to 15 of 25

The table gives information about the hourly rates of pay of the $60$ workers at a fruit-packing plant.

(a) Write down the modal class. [1 mark]

(b) Work out an estimate for the mean hourly rate of pay, in dollars, of the $60$ workers. [4 marks]

(a)(b) dollars

[Total 5 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

The hourly rates of pay, in dollars, of

60

workers, grouped into five classes of width

5

18 + 16 + 14 + 8 + 4 = 60

Find

(a) the modal class (b) an estimate for the mean hourly rate of pay of the

60

workers

Plan the Solution

Part (a) needs no arithmetic. The modal class is the class with the greatest frequency, so it is read straight from the table.
Part (b) can only be an estimate, because a grouped table does not record what any one worker earns. Assume every worker in a class is paid the midpoint of that class.
Multiply each midpoint by its frequency, add the five products, then divide by the total frequency of $60$ .

Worked Solution [5 marks]

Rule - Estimated mean of grouped data: the estimate is

\dfrac{\sum fx}{\sum f}

, taking

x

as the midpoint of each class.

Part (a): compare the five frequencies

18, \; 16, \; 14, \; 8, \; 4

10 < p \le 15

(Reason: The modal class is the class that occurs most often, so it is the class with the greatest frequency. That frequency is

18

, in the first class, and the answer is written as the class itself rather than as the frequency.)

Part (b): take the midpoint of each class

\dfrac{10 + 15}{2} = 12.5

\dfrac{15 + 20}{2} = 17.5

\dfrac{20 + 25}{2} = 22.5

\dfrac{25 + 30}{2} = 27.5

\dfrac{30 + 35}{2} = 32.5

(Reason: The table says how many workers fall in each class, not what any one of them is paid, so the midpoint stands in for every rate in that class. Using the midpoint is what makes the estimate a fair one, because rates below it and above it balance out.)

Multiply each midpoint by its frequency, then add

12.5 \times 18 + 17.5 \times 16 + 22.5 \times 14 + 27.5 \times 8 + 32.5 \times 4 = 225 + 280 + 315 + 220 + 130

225 + 280 + 315 + 220 + 130 = 1170

(Reason: Each product is an estimate of the total hourly pay of one class, so adding the five products estimates the total hourly pay of all

60

workers together.)

Divide by the total frequency

18 + 16 + 14 + 8 + 4 = 60

\dfrac{1170}{60} = 19.5

(Reason: A mean is a total shared between everybody it covers. Adding the frequency column first confirms the

60

workers the question states, which is the number the total must be divided by.)

(a)

10 < p \le 15

; (b)

19.5

dollars

Verification

Check 1 - the frequency column: Add the frequencies:

18 + 16 + 14 + 8 + 4

60

, which is the number of workers the question states, so nothing has been missed from the table

Check 2 - the assumed-mean method, counting each midpoint in steps of $5$ from $22.5$ :

18 \times (-2) + 16 \times (-1) + 14 \times 0 + 8 \times 1 + 4 \times 2 = -36

22.5 + \dfrac{5 \times (-36)}{60} = 19.5

Check 3 - the estimate must sit between the two extreme estimates: Paying every worker the lower bound of their class gives

\dfrac{1020}{60}

, and paying every worker the upper bound gives

\dfrac{1320}{60}

17

dollars and

22

dollars, and the estimate

19.5

dollars lies between them

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Write down the class with the greatest frequency	B1	$10 < p \le 15$ . Accept $10 \le p \le 15$ , $10 < p < 15$ , $10 \le p < 15$ or $10 - 15$	✓
(b) At least four correct products, added. They need not be evaluated	M2	$12.5 \times 18 + 17.5 \times 16 + 22.5 \times 14 + 27.5 \times 8 + 32.5 \times 4 \; (= 1170)$	✓
(b) If M2 is not earned: at least four products, each using a consistent value from within its class (end points allowed), added - for example the lower bound of every class. Correct midpoints for at least four products, not added, also earns this mark	M1	$10 \times 18 + 15 \times 16 + 20 \times 14 + 25 \times 8 + 30 \times 4 \; (= 1020)$	✓
(b) Divide the total by the total frequency	dM1	$\dfrac{1170}{60} \; (= 19.5)$	✓
(b) The estimated mean hourly rate of pay	A1	$19.5$ , or any equivalent form such as $19.50$	✓
(b) Note on the two most common wrong methods	note	Paying every worker the lower bound of their class totals $1020$ , and the upper bound totals $1320$ . Neither can reach the correct estimate, and each scores M1 at most. A fully correct answer scores all four marks unless it clearly follows from incorrect working.	✓

Full marks: 5/5

Use a ruler and a pair of compasses only to construct the bisector of angle $ABC$ .
You must leave in all your construction arcs. [2 marks]

[Total 2 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

Angle

ABC

, drawn as the two arms

BA

and

BC

meeting at the point

B

A ruler and a pair of compasses, and nothing else - no protractor, and no measuring of the angle

Find

The bisector of angle

ABC

, drawn accurately, with every construction arc left on the page

Plan the Solution

The bisector is the line through $B$ that splits the angle into two equal halves, so the way to find it without measuring is to build a shape that is symmetrical about it.
Mark one point on each arm, both the same distance from $B$ . One compass setting does this in a single sweep.
Find a point that is the same distance from each of those two points. A second compass setting, used twice, does this.
Join $B$ to that point. The two halves of the figure are then mirror images, so the angle has been bisected.
Leave every arc on the page. Half the marks here are for the arcs, not for the line.

Worked Solution [2 marks]

Rule - Bisecting an angle: with one compass setting, cut both arms from

B

P

and at

Q

; with a second setting, draw an arc from

P

and an arc from

Q

so that they cross at

R

; the bisector is the straight line through

B

and

R

Cut both arms with one compass setting

BP = BQ

(Reason: Put the compass point on

B

, open the compasses to any convenient width, and sweep an arc that crosses both arms - at

P

BA

and at

Q

BC

. Because the setting is not touched between the two crossings,

BP

and

BQ

are the same length, and that single fact is what makes everything drawn after it symmetrical about the line being looked for.)

Cross a pair of arcs from

P

and from

Q

PR = QR

(Reason: Open the compasses to a second setting, comfortably more than half of

PQ

, and draw a short arc from

P

into the angle. Without changing the setting, draw a second arc from

Q

. They meet at

R

. A setting smaller than half of

PQ

leaves the two arcs short of each other and there is no

R

to find, which is the one thing that can go wrong at this stage.)

Join

B

R

and extend the line

\angle ABR = \angle RBC

(Reason: Line the ruler up on

B

and on

R

, draw the line, and carry it past

R

so that it clearly cuts across the angle. Leave the arcs exactly where they are - rubbing them out throws away the mark they earn.)

Why the two halves have to be equal

BP = BQ, \; PR = QR, \; BR = BR

\dfrac{46}{2} = 23

(Reason: Triangles

BPR

and

BQR

have three pairs of equal sides, so they are congruent, and congruent triangles have equal angles at

B

. That argument uses neither compass setting, which is why the construction works on any angle at all. Halving the angle drawn at

B

on this page gives the size each half must come out at, and a protractor laid on the finished drawing agrees with it.)

The straight line through

B

and

R

, with the arc from

B

and the two arcs from

P

and

Q

all left on the diagram

Verification

Check 1 - measure the two halves: The angle

ABC

printed on this page measures

46

degrees, so measure angle

ABR

and angle

RBC

with a protractor each half measures

23

degrees, and

23 + 23 = 46

, so the whole angle has been accounted for

Check 2 - the shape $BPRQ$ is a kite:

BP = BQ

came from the first compass setting and

PR = QR

came from the second, so

BPRQ

is a kite whose long diagonal is

BR

a kite is symmetrical about that diagonal, so

BR

is a mirror line of the figure and the two parts of the angle at

B

have to match

Check 3 - the distance from $R$ to each arm: Measure the perpendicular distance from

R

to the arm

BA

, then from

R

to the arm

BC

the two distances are the same, which is the property that defines every point lying on the bisector

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A fully correct bisector of angle $ABC$ , with all the relevant construction arcs left on the diagram	B2	The bisector and the arcs may cross anywhere on or inside the overlay guidelines	✓
All the construction arcs drawn, but no bisector drawn	B1	The arcs on their own show the method, even though the line that answers the question is missing	✓
A correct bisector, on or within the guidelines, but with no arcs, or with too few arcs	B1	A line in the right place with no evidence of how it was found. Measuring the angle with a protractor and halving it lands here at best	✓
Note on the overlay	note	An overlay is supplied with the official mark scheme. It carries the guidelines that both the B2 row and the B1 rows refer to, and it is what decides whether a drawn bisector is close enough to be correct	✓

Full marks: 2/2

(a) Write $(p^3)^5$ as a single power of $p$ . [1 mark]

(b) Multiply out the brackets and simplify $2n(4n + 3) + n(n - 4)$ [2 marks]

You must show clear algebraic working. [3 marks]

(a)(b)(c) x =

[Total 6 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

(p^3)^5

- a power raised to another power

2n(4n + 3) + n(n - 4)

- two brackets, each with a term outside it

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

- a linear equation with the fraction on the left

Find

(a)

(p^3)^5

written as a single power of

p

. (b) the expansion in its simplest form. (c) the value of

x

, with the algebra shown.

Plan the Solution

(a) A power raised to another power keeps its base, and the two indices multiply.
(b) Multiply every term inside each bracket by the term outside it, then collect the squared terms and the terms in $n$ separately.
(c) Clear the fraction first by multiplying both sides by $3$ , then gather the terms in $x$ on one side and the numbers on the other.
In (c) every line keeps the two sides balanced, which is what the instruction to show clear algebraic working is asking for.

Worked Solution [6 marks]

Rule - index law, expanding a bracket, and balance:

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

a(b + c) = ab + ac

, and an equation stays true when both sides are treated in exactly the same way.

Part (a) - multiply the indices

(p^3)^5 = p^{3 \times 5} = p^{15}

(Reason: A power raised to another power is that power used over and over, so

p

cubed used five times uses the index

3

five times. The base stays as

p

and the two indices multiply.)

Part (b) - multiply out each bracket

2n(4n + 3) = 8n^2 + 6n

n(n - 4) = n^2 - 4n

(Reason: Each term inside a bracket is multiplied by the term outside it, so

2n

multiplies the

4n

and the

3

, and

n

multiplies the

n

and the minus

4

. Four terms come out of the two brackets.)

Part (b) - collect the like terms

8n^2 + 6n + n^2 - 4n = 9n^2 + 2n

(Reason: The two squared terms add to give

9

lots of

n

squared, and the two terms in

n

add to give

2

lots of

n

. A term in

n

squared and a term in

n

are unlike, so nothing further combines.)

Part (c) - clear the fraction

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

2x + 5 = 3(4 - x)

(Reason: Multiplying both sides by

3

undoes the division on the left. The whole of the right hand side is multiplied by

3

, so it is written inside a bracket first.)

Part (c) - multiply out the right hand side

2x + 5 = 12 - 3x

(Reason: The

3

outside the bracket multiplies both terms inside it, not only the first one. This is the step the mark scheme singles out, and it is where marks are most often lost.)

Part (c) - gather the terms in

x

2x + 3x = 12 - 5

5x = 7

(Reason: Adding

3x

to both sides and taking

5

from both sides puts every term in

x

on the left and every number on the right. That rearrangement is what the second method mark is for.)

Part (c) - divide to finish

x = \dfrac{7}{5} = 1.4

(Reason: Dividing both sides by

5

leaves

x

on its own. The fraction is exact and

1.4

is the same value written as a decimal, and the mark scheme accepts either.)

(a)

p^{15}

(b)

9n^2 + 2n

(c)

x = 1.4

Verification

Check 1 - part (a): Put

p = 2

into each form and work the two out as plain numbers:

(2^3)^5 = 8^5 = 32\,768

and

2^{15} = 32\,768

, so the two forms are the same expression.

Check 2 - part (b): Put

n = 3

into the expression given and into the answer:

2 \times 3 \times 15 + 3 \times (-1) = 87

and

9 \times 9 + 6 = 87

, so the two agree.

Check 3 - part (c): Put

x = 1.4

back into the equation given and work out each side separately: the left side is

\dfrac{2 \times 1.4 + 5}{3} = \dfrac{7.8}{3} = 2.6

and the right side is

4 - 1.4 = 2.6

, so the two sides are equal.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Multiply the indices	B1	cao for $p^{15}$ . No other form earns this mark.	✓
(b) Expand both brackets	M1	for expanding with at least $3$ correct terms, for example $8n^2 + 6n + n^2 - 4n$ . Seeing $8n^2$ is what counts, not $2n \times 4n$ left unworked. Where the terms are written in a list or a table, a term with no sign in front of it may be taken as a plus.	✓
(b) Collect the like terms	A1	oe for $9n^2 + 2n$ . The forms $2n + 9n^2$ , $n(9n + 2)$ and $n(2 + 9n)$ are all accepted.	✓
(c) Clear the fraction	M1	for removal of the fraction and multiplying out the right hand side correctly by $3$ , giving $2x + 5 = 12 - 3x$ , or for separating the fraction on the left in an equation, giving $\dfrac{2}{3}x + \dfrac{5}{3} = 4 - x$ oe.	✓
(c) Rearrange for the terms in $x$	M1ft	dep on $4$ terms, for correctly rearranging their four term equation so that the terms in $x$ are on one side and the number terms on the other, for example $2x + 3x = 12 - 5$ or $5x = 7$ .	✓
(c) Divide to finish	A1	dep on M2 for $\dfrac{7}{5}$ oe, for example $1.4$ or $1\dfrac{2}{5}$ . Working is required, so an answer standing on its own does not score.	✓

Full marks: 6/6

The Venn diagram shows the universal set ℰ and the two sets $\mathit{A}$ and $\mathit{B}$ .

(a) Write down the members of the set $\mathit{B}$ [1 mark]

(b) Write down the members of the set $\mathit{A} \cap \mathit{B}$ [1 mark]

(a)(b)(c)

[Total 3 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

The region inside

\mathit{A}

only holds

9

The overlap of

\mathit{A}

and

\mathit{B}

holds

3

and

7

The region inside

\mathit{B}

only holds

5

and

2

The rectangle holds

4

6

and

8

outside both circles

So the universal set holds the eight numbers

\{2, 3, 4, 5, 6, 7, 8, 9\}

Find

The members of

\mathit{B}

, of

\mathit{A} \cap \mathit{B}

and of

\mathit{A}^\prime

, read straight off the diagram

Plan the Solution

Read the diagram one region at a time: $\mathit{A}$ only, the overlap, $\mathit{B}$ only, and the numbers outside both circles.
A set is everything drawn inside its own circle, so $\mathit{B}$ is the overlap together with the $\mathit{B}$ -only region.
An intersection is the overlap region on its own, where the two circles cross.
The complement $\mathit{A}^\prime$ is everything in the universal set that is not inside $\mathit{A}$ , so it is the $\mathit{B}$ -only region together with the numbers outside both circles.
Each part is worth one mark for the complete listing, so every member must appear, with no repeats and nothing extra.

Worked Solution [3 marks]

Rule - reading a Venn diagram: a set is the union of every region inside its own circle, an intersection is the overlapping region, and

\mathit{A}^\prime

is everything in the universal set that lies outside

\mathit{A}

Part (a) - take both of the regions inside circle

\mathit{B}

\mathit{B} = \{3, 7\} \cup \{5, 2\}

\mathit{B} = \{2, 3, 5, 7\}

(Reason: Circle

\mathit{B}

covers two regions - its overlap with

\mathit{A}

, and the part of

\mathit{B}

on its own - so a number in either region is a member of

\mathit{B}

. The

3

and the

7

are in

\mathit{B}

even though they are in

\mathit{A}

as well.)

Part (b) - take the overlap only

\mathit{A} \cap \mathit{B} = \{3, 7\}

(Reason:

\mathit{A} \cap \mathit{B}

is the region where the two circles cross, so only a number inside both circles qualifies. The

9

is inside

\mathit{A}

alone and the

5

and the

2

are inside

\mathit{B}

alone, so none of them belongs to the intersection.)

Part (c) - take everything outside circle

\mathit{A}

\mathit{A} = \{3, 7, 9\}

\mathit{A}^\prime = \{2, 4, 5, 6, 8\}

(Reason:

\mathit{A}^\prime

is everything in the universal set that is not inside circle

\mathit{A}

. The

5

and the

2

sit inside

\mathit{B}

, but they are still outside

\mathit{A}

, so they join the

4

6

and

8

from outside both circles.)

(a)

2, 3, 5, 7

; (b)

3, 7

; (c)

2, 4, 5, 6, 8

Verification

Check 1 - the four regions account for every number: Count the numbers region by region - one in

\mathit{A}

only, two in the overlap, two in

\mathit{B}

only and three outside both circles - and compare the total with the eight numbers printed inside the rectangle

1 + 2 + 2 + 3 = 8

Check 2 - a set and its complement: Every number in the universal set is either inside

\mathit{A}

or in

\mathit{A}^\prime

and never in both, so the three members of

\mathit{A}

and the five members of the answer to part (c) must account for the whole universal set

3 + 5 = 8

Check 3 - the overlap is counted once in each set: Adding the size of

\mathit{A}

to the size of

\mathit{B}

counts the two members of the overlap twice, so taking the overlap off again must leave the five numbers that are drawn inside the two circles

3 + 4 - 2 = 5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The members of the set $\mathit{B}$	B1	cao $2$ $3$ $5$ $7$ . All four numbers must be present, with no repeats and no other numbers. Any order is accepted, and so is any separator between them (commas, colons and so on)	✓
(b) The members of the set $\mathit{A} \cap \mathit{B}$	B1	cao $3$ $7$ . Both numbers must be present, with no repeats and no other numbers. Any order, any separator	✓
(c) The members of the set $\mathit{A}^\prime$	B1	cao $2$ $4$ $5$ $6$ $8$ . All five numbers must be present, with no repeats and no other numbers. Any order, any separator. Because nothing may be missing, a listing of $4$ $6$ $8$ alone - the numbers outside both circles, with the members of $\mathit{B}$ that lie outside $\mathit{A}$ forgotten - scores zero	✓

Full marks: 3/3

A cylinder is shown in the diagram below.

The cylinder has radius $70$ cm and height $18$ cm.

Find the volume of the cylinder.

Give your answer in litres, correct to the nearest litre. [4 marks]

litres

[Total 4 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

A cylinder with radius

r = 70 \text{ cm}

and height

h = 18 \text{ cm}

Both lengths are in centimetres, so the volume comes out in cubic centimetres

The conversion the answer needs at the end:

1 \text{ litre} = 1000 \text{ cm}^3

Find

The volume of the cylinder, in litres, correct to the nearest litre

Plan the Solution

Work out the volume in cubic centimetres first, from the radius and the height.
Leave the volume as a multiple of $\pi$ while the working goes on, so that no rounding creeps in early.
There are $1000$ cubic centimetres in one litre, so divide the volume in cubic centimetres by $1000$ to turn it into litres.
Round to the nearest whole litre only at the very end, after the calculator has done the last multiplication.
The four marks split two and two: two for the volume in cubic centimetres, and two for the conversion and the rounded answer.

Worked Solution [4 marks]

Rule - Volume of a cylinder:

V = \pi r^2 h

, where

r

is the radius of the circular face and

h

is the height. A litre is

1000 \text{ cm}^3

, so

\dfrac{\text{volume in cm}^3}{1000}

is the volume in litres

Write down the rule, then put in the radius and the height

V = \pi r^2 h

V = \pi \times 70^2 \times 18

(Reason: The volume of a cylinder is the area of its circular face multiplied by its height. Only the radius is squared, never the height, so the

70

is squared and the

18

is not.)

Work out the volume in cubic centimetres

70^2 = 4900

4900 \times 18 = 88\,200

V = 88\,200\pi \text{ cm}^3

(Reason: Squaring the radius and then multiplying by the height leaves the volume as a multiple of

\pi

. Keeping it as

88\,200\pi

means nothing has been rounded yet, and the mark scheme awards this second mark for

88\,200\pi

or for the decimal

277\,088.472

that it comes to.)

Change the cubic centimetres into litres

1 \text{ litre} = 1000 \text{ cm}^3

V = \dfrac{88\,200\pi}{1000} = 88.2\pi \text{ litres}

(Reason: A litre is

1000

cubic centimetres, so the number of litres is the volume in cubic centimetres divided by

1000

. Dividing the exact form

88\,200\pi

keeps the working exact, and the mark scheme gives this third mark on the volume the candidate actually found, so an earlier slip does not stop it.)

Work out the decimal, then round to the nearest litre

V = 88.2\pi = 277.088\ldots \text{ litres}

V \approx 277 \text{ litres}

(Reason: The calculator turns

88.2\pi

into

277.088

and so on. The first decimal place is a

0

, which is below

5

, so the value rounds down to

277

whole litres.)

277

litres

Verification

Check 1 - work backwards from the answer: Turn the answer in litres back into cubic centimetres, then divide by the height and by

\pi

, and see whether the square of the radius comes back:

277.088\ldots \times 1000 = 277\,088.472\ldots \text{ cm}^3

, and

\dfrac{277\,088.472\ldots}{18\pi} = 4900 = 70^2

, so the radius is

70

cm again

Check 2 - units that need no conversion at all: Work in decimetres instead. One cubic decimetre is exactly one litre, so this route never divides by

1000

anywhere. The radius is

7

dm and the height is

1.8

dm:

\pi \times 7^2 \times 1.8 = 88.2\pi = 277.088\ldots

, which is

277

litres to the nearest litre

Check 3 - the two usual approximations for $\pi$ : Replace

\pi

3.14

, then by

\dfrac{22}{7}

, which is where the band in the mark scheme comes from:

3.14 \times 88\,200 = 276\,948 \text{ cm}^3

and

\dfrac{22}{7} \times 88\,200 = 277\,200 \text{ cm}^3

, that is

276.948

litres and

277.2

litres, and both round to

277

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Use of the volume of a cylinder	M1	for use of $\pi r^2 h$ , for example $\pi \times 70^2 \times 18$ oe, or $\pi \times 0.7^2 \times 0.18$ oe where the lengths are changed into metres first	✓
The volume, before any conversion	A1	for $88\,200\pi$ or $277\,088.472$ in cubic centimetres, or for $0.0882\pi$ or $\dfrac{441}{5000}\pi$ or $0.277088472$ in cubic metres. Allow $276\,948$ to $277\,200$ in cubic centimetres, or $0.276948$ to $0.277200$ in cubic metres	✓
Change the volume into litres	M1	for dividing their volume in cubic centimetres by $1000$ , for example $\dfrac{277\,088.472}{1000}$ , or $\dfrac{88\,200\pi}{1000}$ giving $88.2\pi$ or $\dfrac{441}{5}\pi$ , or for multiplying their volume in cubic metres by $1000$ . Allow any value for their volume that contains $\pi$ , $70$ and $18$ to be divided by $1000$ , and any value that contains $\pi$ , $0.7$ and $0.18$ to be multiplied by $1000$	✓
The volume in litres, to the nearest litre	A1	awrt $277$ . Any value that rounds to $277$ scores, so $277.088\ldots$ left on the answer line is not penalised here, although the question asks for the nearest litre and $277$ is the answer wanted	✓
Note - a correct answer standing on its own	note	A correct answer scores full marks, unless it comes from obviously incorrect working	✓

Full marks: 4/4

Three numbers are written here as products of their prime factors.

$A = 2^3 \times 5^4 \times 7 \times 11$

$B = 2^2 \times 5^2 \times 7^2$

$C = 2^2 \times 5^3 \times 7^4$

Work out the highest common factor (HCF) of $A$ , $B$ and $C$

Give your answer as a product of prime factors. [2 marks]

[Total 2 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

A = 2^3 \times 5^4 \times 7 \times 11

, which is built from the primes

2

5

7

and

11

B = 2^2 \times 5^2 \times 7^2

, which has no

11

in it

C = 2^2 \times 5^3 \times 7^4

, which has no

11

in it either

All three numbers are already written as products of prime factors, so there is no factorising to do first

Find

The highest common factor of

A

B

and

C

The answer left as a product of prime factors, not turned into a single number

Plan the Solution

A factor shared by all three numbers can only be built from primes that appear in all three, so the first job is to see which primes those are.
The prime $11$ appears in $A$ alone, so it cannot be part of any factor of $B$ or of $C$ .
For each prime that does appear in all three, take the lowest power of it that appears - going any higher would stop it dividing the number that has the fewest of them.
Multiply those lowest powers together and leave the answer in that form, because a product of prime factors is what the question asks for.
Both marks are for the answer itself. There is no method mark in this question, so the product has to be right.

Worked Solution [2 marks]

Rule - HCF from prime factors: with each number written as a product of prime powers, the HCF is the product of the LOWEST power of every prime that appears in all of them. A prime missing from one of the numbers counts there as that prime to the power

0

, and since

p^0 = 1

it multiplies nothing into the answer

Line the three numbers up prime by prime

A = 2^3 \times 5^4 \times 7^1 \times 11^1

B = 2^2 \times 5^2 \times 7^2 \times 11^0

C = 2^2 \times 5^3 \times 7^4 \times 11^0

(Reason: Writing all three numbers with the same four primes lets the powers be read straight down a column. The single

7

A

is written as

7

to the power

1

, and the

11

that is missing from

B

and from

C

is written as

11

to the power

0

, so that no prime is left out of the comparison by accident.)

Take the lowest power of each prime

2^3, 2^2, 2^2 \implies 2^2

5^4, 5^2, 5^3 \implies 5^2

7^1, 7^2, 7^4 \implies 7^1

11^1, 11^0, 11^0 \implies 11^0

11^0 = 1

(Reason: A common factor has to divide all three numbers, so it can use each prime only as many times as the number with the fewest of them has it.

B

has just two

2

s and just two

5

s, and

A

has just one

7

. The lowest power of

11

is the power

0

, and

11

to the power

0

1

, so

11

contributes nothing at all.)

Multiply the lowest powers together

\text{HCF} = 2^2 \times 5^2 \times 7

(Reason: Multiplying the lowest powers gives the largest number that divides all three. This product is the answer the question asks for, so it is left exactly as it stands, and no

1

is written into it.)

See what that product comes to as an ordinary number

2^2 \times 5^2 \times 7 = 4 \times 25 \times 7 = 700

(Reason: Multiplying the product out is a useful size check, and the mark scheme is content to see the number in the working space beside the product. It is not the form the answer line wants, though. The prime

11

belongs to

A

alone, so it takes no part in a factor shared by all three numbers, and a bare number left on its own with no product anywhere beside it scores one mark instead of two.)

2^2 \times 5^2 \times 7

Verification

Check 1 - divide each number by the answer: The product comes to

700

. Divide each of the three numbers by

700

and see whether every answer is a whole number:

A = 385\,000 = 700 \times 550

B = 4900 = 700 \times 7

and

C = 1\,200\,500 = 700 \times 1715

, so

700

divides all three exactly

Check 2 - nothing larger can work: Any common factor must be a factor of the smallest of the three numbers,

B = 4900

. Its factors above

700

are

980

1225

2450

and

4900

, so test each of those against

A = 385\,000

: the four divisions leave remainders of

840

350

350

and

2800

, so not one of them divides

A

, and no common factor above

700

exists

Check 3 - what is left after dividing by $700$ : Divide all three numbers by

700

and factorise what is left:

550 = 2 \times 5^2 \times 11

, then

7

, then

1715 = 5 \times 7^3

: those three share no prime at all, so their own highest common factor is

1

, which means

700

cannot be scaled up any further and is the highest common factor of

A

B

and

C

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The HCF as a product of prime factors	B2	for $2^2 \times 5^2 \times 7$ . Allow $2 \times 2 \times 5 \times 5 \times 7$ , and the factors may be written in any order, so $2^2 . 5^2 . 7$ is accepted. The answer must be a product of prime factors, and $1$ is not allowed in the final answer	✓
A product that is nearly right, or the value $700$	B1	for $2^p \times 5^q \times 7^r$ where two of $p$ , $q$ , $r$ are correct, or for one mistake in their product, for example $2^2 \times 5^2$ oe, $2^2 \times 7$ oe, $5^2 \times 7$ oe, $2^2 \times 5 \times 7$ oe, $2 \times 5^2 \times 7$ oe or $2^2 \times 5^2 \times 7 \times 11$ oe, or for $700$	✓
Note - $700$ on the answer line	note	$2 \times 2 \times 5 \times 5 \times 7$ in the working space with $700$ on the answer line is awarded B2, and $2^2 \times 5^2 \times 7$ in the working space with $700$ on the answer line is awarded B2. It is $700$ with no product anywhere that drops to B1	✓

Full marks: 2/2

Shop A and Shop B are both selling the same model of bicycle, and each shop has an offer on it.

The normal price of the bicycle in Shop A is not the same as the normal price of the bicycle in Shop B.

Which shop takes more money off its normal price?
Show your working clearly. [4 marks]

[Total 4 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

Shop A's notice prints its normal price, £

475

, and takes

16\%

off that price

Shop B's notice prints no normal price at all - only

15\%

off, and the £

408

you pay

The normal price in Shop A is not the same as the normal price in Shop B, so the two percentages are percentages of different amounts

Find

Which shop takes more money off its own normal price The money off in each shop, in pounds, because two percentages of two different prices cannot be compared as percentages Both amounts written down: the mark scheme awards the final mark only when

72

and

76

are both seen, so the letter on its own scores nothing

Plan the Solution

Shop A is one step. Its normal price is printed on the notice, so the money off is $16\%$ of £ $475$ .
Shop B is a reverse percentage. The £ $408$ is not its normal price - it is what is left after $15\%$ has gone, so £ $408$ is $85\%$ of the normal price.
Divide the £ $408$ by $0.85$ to get back to the full $100\%$ , then take $15\%$ of that normal price to get the money off. Subtracting £ $408$ from it does the same job.
Compare the two amounts of money, not the two percentages. A bigger percentage taken off a smaller price is not always more money.
Say which shop, and leave both amounts in the working where the examiner can see them.

Worked Solution [4 marks]

Rule - Reverse percentage: taking

p\%

off a price multiplies it by

\dfrac{100 - p}{100}

, so dividing the price paid by that multiplier gives the normal price back. The money off is always that percentage of the NORMAL price, never of the price paid.

Shop A: take

16\%

of the price on its notice

0.16 \times 475 = 76

(Reason: Shop A is the easy one, because it prints its normal price. Taking

16\%

off means multiplying that normal price by

0.16

to find the money off, and

0.16

475

76

. So Shop A takes £

76

off, and a customer there pays £

399

Shop B: work out what the £

408

is a percentage of

x - 0.15x = 408

0.85x = 408

(Reason: Shop B never prints its normal price, so call it

x

. Taking

15\%

off leaves

85\%

behind, which means the £

408

on the notice is

0.85

of the normal price rather than the normal price itself. Everything else in this question depends on reading the £

408

that way.)

Divide by the multiplier to get Shop B's normal price

\dfrac{408}{0.85} = 480

(Reason: Multiplying by

0.85

took the price from

100\%

down to

85\%

, so dividing by

0.85

undoes it and brings the price back up to the full

100\%

. Shop B's normal price is £

480

, which is £

5

more than Shop A's - exactly the sort of difference the question warns about in its second line.)

Shop B: take

15\%

of that normal price

0.15 \times 480 = 72

(Reason: The

15\%

comes off Shop B's own normal price of £

480

, not off the £

408

a customer actually hands over, so the money off is

15\%

480

. Subtracting gives the same amount: £

480

less the £

408

paid leaves £

72

Compare the two amounts of money

76 > 72

76 - 72 = 4

(Reason: The question asks which shop takes more MONEY off, so the comparison is between £

76

and £

72

and not between

16\%

and

15\%

. Shop A takes £

4

more off. The percentages alone could not have settled it:

16\%

is the bigger percentage but it is taken off the smaller normal price, so the two effects pull against each other.)

Shop A - it takes £

76

off, while Shop B takes £

72

off

Verification

Check 1 - put Shop B's normal price back through its own offer: Take

15\%

off £

480

and see whether the £

408

printed on the notice comes back:

0.85 \times 480 = 408

the notice's £

408

is reproduced exactly, so £

480

is the right normal price, and

480 - 408 = 72

gives the money off a second way, by subtraction rather than by taking a percentage

Check 2 - Shop B by the $1\%$ method: The £

408

85\%

of the normal price, so

1\%

of that price is

\dfrac{408}{85} = 4.8

, which makes

15\%

equal to

4.8 \times 15 = 72

72

again, this time without dividing by

0.85

anywhere, and

100\%

comes to

4.8 \times 100 = 480

, which agrees with Shop B's normal price

Check 3 - Shop A from the other end: Taking

16\%

off leaves

84\%

, so the price paid in Shop A is

0.84 \times 475 = 399

, and the money off is

475 - 399 = 76

76

again, reached without multiplying by

0.16

, so £

76

against £

72

holds and Shop A is £

4

the better offer

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Working for Shop A	M1	for $475 \times 0.16$ $(= 76)$ oe, or $475 \times (1 - 0.16)$ $(= 399)$ oe	✓
Working for Shop B - reading what the £ $408$ is a percentage of	M1	for $1 - 0.15$ $(= 0.85)$ , or $x - 0.15x = 408$ , or $100(\%) - 15(\%)$ $(= 85(\%))$ , or $\dfrac{408}{85}$ $(= 4.8)$ oe	✓
Working for Shop B - the normal price, or the money off	M1	for $\dfrac{408}{0.85}$ $(= 480)$ , or $\dfrac{408}{85} \times 100$ $(= 480)$ , or $408 \times \dfrac{100}{85}$ $(= 480)$ oe, or their own $1\%$ value $4.8 \times 100$ $(= 480)$ , or $\dfrac{408}{85} \times 15$ $(= 72)$	✓
The decision, with both amounts shown	A1	dep on M2, for A, with correct working and with $72$ and $76$ both seen	✓
Note - working required	note	The answer column reads Working required, so A written on its own, with no working, earns nothing at all. Both amounts must appear: $76$ for Shop A and $72$ for Shop B	✓

Full marks: 4/4

(a) (i) Write $x^2 + 5x - 24$ as a product of two brackets.
[2 marks]
(ii) Hence solve the equation $x^2 + 5x - 24 = 0$ .
[1 mark]
(b) Find the range of values of $y$ for which $3y + 5 > 7y - 10$ .
You must show clear algebraic working. [3 marks]

(a)(i)(a)(ii)(b)

[Total 6 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

The quadratic expression

x^2 + 5x - 24

, with

1

as the coefficient of

x^2

The equation

x^2 + 5x - 24 = 0

, the same expression set equal to zero

The inequality

3y + 5 > 7y - 10

, which is linear in

y

Find

Part (a)(i): the expression written as a product of two brackets, each of the form

(x + a)

with

a

an integer Part (a)(ii): the values of

x

that satisfy the equation - a quadratic, so expect two solutions Part (b): every value of

y

that makes the inequality true, given as a single inequality in

y

Plan the Solution

Factorise by hunting for two integers whose product is the constant term $-24$ and whose sum is $5$ , the coefficient of $x$
The word hence means part (a)(ii) is free once part (a)(i) is done: set each bracket equal to zero and read off the two solutions
Treat the inequality exactly like an equation, but reverse the inequality sign at the moment both sides are divided by a negative number
As a safeguard, collect the $y$ terms on the side that keeps their coefficient positive, where no reversal is needed, and confirm the same answer appears

Worked Solution [6 marks]

Factorising

x^2 + bx + c

: find integers

p

and

q

with

pq = c

and

p + q = b

, so that

x^2 + bx + c = (x + p)(x + q)

. A product of two brackets equals zero only when one of the brackets equals zero. For an inequality, dividing or multiplying both sides by a negative number reverses the direction of the inequality sign.

Set up the pair of numbers the brackets need

x^2 + 5x - 24 = (x + p)(x + q)

p \times q = -24

p + q = 5

(Reason: Expanding

(x + p)(x + q)

gives

x^2 + (p + q)x + pq

, so the constant term fixes the product of

p

and

q

and the coefficient of

x

fixes their sum.)

Test the integer pairs whose product is

-24

12 \times (-2) = -24 \text{ but } 12 + (-2) = 10

6 \times (-4) = -24 \text{ but } 6 + (-4) = 2

8 \times (-3) = -24 \text{ and } 8 + (-3) = 5

(Reason: Every pair here has the right product, so the sum is what decides. Only

8

and

-3

give the

5

that the coefficient of

x

demands.)

Write down the factorised form

x^2 + 5x - 24 = (x + 8)(x - 3)

(Reason: With

p = 8

and

q = -3

, the bracket

(x + (-3))

is written as

(x - 3)

. Both numbers inside the brackets are integers, which is the form the mark scheme requires.)

Use the factors to solve the equation

(x + 8)(x - 3) = 0

x + 8 = 0 \text{ or } x - 3 = 0

x = -8 \text{ or } x = 3

(Reason: A product is zero only when at least one factor is zero, so each bracket is set to zero in turn and the resulting one-step equation is solved.)

Collect the

y

terms on one side of the inequality

3y + 5 > 7y - 10

3y - 7y > -10 - 5

-4y > -15

(Reason: Subtracting

7y

from both sides and subtracting

5

from both sides keeps the inequality balanced, exactly as it would keep an equation balanced.)

Divide both sides by

-4

-4y > -15

y < \dfrac{15}{4}

(Reason: Dividing by a negative number reverses the direction of the inequality, so the greater-than sign becomes a less-than sign. The two negatives in the division also cancel, leaving a positive

15

over

4

Check it another way, keeping the coefficient of

y

positive

3y + 5 > 7y - 10

5 + 10 > 7y - 3y

15 > 4y

\dfrac{15}{4} > y

(Reason: Moving the

y

terms to the right instead leaves

4y

, whose coefficient is positive, so dividing by

4

needs no reversal. Reading

15

over

4

is greater than

y

from the other end gives the same solution as before.)

(a)(i)

(x + 8)(x - 3)

(ii)

x = -8

x = 3

(b)

y < 3.75

Verification

Check 1: Expand the two brackets and compare with the expression in the question

(x + 8)(x - 3) = x^2 - 3x + 8x - 24 = x^2 + 5x - 24

, the original expression

Check 2: Substitute each solution into

x^2 + 5x - 24

(-8)^2 + 5(-8) - 24 = 64 - 40 - 24 = 0

and

3^2 + 5(3) - 24 = 9 + 15 - 24 = 0

Check 3: Compare the sum and the product of the two solutions with the coefficients of the quadratic

-8 + 3 = -5

, the negative of the coefficient of

x

, and

(-8) \times 3 = -24

, the constant term

Check 4: Test one value inside the solution set and one outside it in

3y + 5 > 7y - 10

y = 0

gives

5 > -10

, which is true, and

y = 4

gives

17 > 18

, which is false

Check 5: Test the boundary value

y = 3.75

in both sides of the inequality

3(3.75) + 5 = 16.25

and

7(3.75) - 10 = 16.25

, so the two sides are equal rather than one being greater, and

3.75

itself is not part of the solution

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Start to factorise the quadratic	M1	$(x \pm 8)(x \pm 3)$ , or a partial factorisation $x(x - 3) + 8(x - 3)$ or $x(x + 8) - 3(x + 8)$ , or any $(x + a)(x + b)$ with $ab = -24$ or $a + b = 5$ , where $a$ and $b$ are integers	✓
Give the fully correct factorisation	A1	$(x + 8)(x - 3)$ . Any letter may be used in place of $x$ , and the answer must be in the form $(x + a)(x + b)$ with $a$ and $b$ integers	✓
Guidance on a bare correct answer in part (a)(i)	note	A correct answer scores full marks unless it follows obvious incorrect working	✓
Solve the equation from the factorised form	B1ft	$-8$ and $3$ , followed through from the candidate's own answer to part (a)(i) provided their factors are of the form $(x + a)(x + b)$ . Award B0 for $-8$ and $3$ if no marks were scored in part (a)(i)	✓
Collect the $y$ terms and the number terms	M1	$3y - 7y > -10 - 5$ , or $5 + 10 > 7y - 3y$ . The use of $=$ is allowed and an incorrect inequality sign is condoned at this stage	✓
Simplify to a single inequality in $y$	M1	$-4y > -15$ , or $15 > 4y$ , or $y = \dfrac{15}{4}$ or equivalent. Again $=$ is allowed and an incorrect inequality sign is condoned	✓
State the solution with the correct inequality sign	A1	$y < \dfrac{15}{4}$ or equivalent, for example $y < 3.75$ or $\dfrac{15}{4} > y$ or $3.75 > y$ . Algebraic working is required, and the answer line must carry the correct inequality sign	✓
Guidance on an answer line that drops the inequality	note	Sight of the correct answer in the working space, with just $(y =) \dfrac{15}{4}$ or equivalent on the answer line, gains M2 only	✓

Full marks: 6/6

(a) The number $\mathrm{8.4 \times 10^{-5}}$ is written in standard form. Write it as an ordinary number. [1 mark]

(b) Work out the product $\mathrm{(6.5 \times 10^{-40}) \times (8 \times 10^{185})}$ . Give your answer in standard form. [2 marks]

(a)(b)

[Total 3 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

Part (a) gives the number

\mathrm{8.4 \times 10^{-5}}

, already written in standard form

Part (b) gives the product

\mathrm{(6.5 \times 10^{-40}) \times (8 \times 10^{185})}

, with both numbers written in standard form

A number is in standard form when its number part is at least

1

and less than

10

, multiplied by a whole-number power of

10

Find

Part (a): the same value written out in full as an ordinary number, with no power of

10

left in it Part (b): the value of the product, written in standard form, so its number part must again be at least

1

and less than

10

Plan the Solution

Part (a): a negative index means a division, so read the power of $10$ as dividing by $100\,000$ , then move the decimal point five places to the left.
Part (b): multiply the two number parts, and separately add the two indices, because powers of the same base add when they are multiplied.
Then test the result against the definition of standard form. The number part comes out as $52$ , which is too big, so it has to be adjusted.
Adjust it by making the number part ten times smaller and the power of $10$ one higher, which leaves the value unchanged.
Check part (b) by dividing the final answer by one of the two numbers in the question and confirming the other one comes back.

Worked Solution [3 marks]

Multiplying in standard form: multiply the two number parts together, and add the two indices, because powers of the same base add when they are multiplied. Then look at the number part. If it is

10

or more the answer is not yet in standard form, so move its decimal point one place to the left and add

1

to the index. That makes the number part ten times smaller and the power of

10

ten times bigger, so the value itself does not change.

Read what the negative index in part (a) means

8.4 \times 10^{-5} = \dfrac{8.4}{10^{5}}

10^{5} = 100\,000

(Reason: A negative index means the reciprocal of the matching positive power, so multiplying by

10

to the power negative

5

is the same as dividing by

10

to the power

5

. That power of

10

100\,000

Divide

8.4

100\,000

\dfrac{8.4}{100\,000} = 0.000084

(Reason: Dividing by

100\,000

moves the decimal point five places to the left:

8.4

becomes

0.84

, then

0.084

, then

0.0084

, then

0.00084

, and finally

0.000084

. Counting the five places one at a time is what stops a zero being lost, and the

8

finishes in the fifth decimal place.)

Split part (b) into number parts and powers of

10

(6.5 \times 10^{-40}) \times (8 \times 10^{185}) = (6.5 \times 8) \times (10^{-40} \times 10^{185})

6.5 \times 8 = 52

(Reason: Multiplication can be carried out in any order, so the two number parts are collected together and the two powers of

10

are collected together. The number parts give

52

Add the two indices

10^{-40} \times 10^{185} = 10^{-40 + 185}

-40 + 185 = 145

(Reason: Powers of the same base are multiplied by adding their indices, not by multiplying them. Adding a negative index is a subtraction, so this is

185

take away

40

, which is

145

Put the two halves back together

(6.5 \times 10^{-40}) \times (8 \times 10^{185}) = 52 \times 10^{145}

(Reason: This is the line the mark scheme awards the method mark for. The value is already right, but it is not yet an answer, because standard form needs a number part between

1

and

10

and

52

is not one.)

Rewrite

52

in standard form

52 = 5.2 \times 10

52 \times 10^{145} = 5.2 \times 10^{146}

(Reason: The number part has to be at least

1

and less than

10

, so the decimal point in

52

moves one place to the left to give

5.2

. That makes the number part ten times smaller, so the power of

10

must be ten times bigger for the value to stay the same: the index goes up by one, from

145

146

(a)

0.000084

(b)

5.2 \times 10^{146}

Verification

Check 1: Multiply the ordinary number from part (a) back by

100\,000

and see whether

8.4

returns

0.000084 \times 100\,000 = 8.4

Check 2: Reach the same ordinary number a second way, as a fraction over a power of ten, with no decimal point moved at all

\dfrac{84}{1\,000\,000} = 0.000084

Check 3: Undo part (b) by dividing the answer by the second number in the question, number parts first

\dfrac{5.2}{8} = 0.65

Check 4: Now the indices of that same division. With the

0.65

above it, this rebuilds the first number of the question, since

0.65

times ten to the power negative

39

6.5

times ten to the power negative

40

146 - 185 = -39

Check 5: Add the two indices the other way round, reading the negative one as a subtraction, and confirm the same total appears

185 - 40 = 145

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Write the standard-form number in part (a) as an ordinary number	B1	$0.000084$ , which the paper prints with a digit space as $0.000\,084$	✓
Reach a correct product in part (b), in any form	M1	Sight of $\mathrm{52 \times 10^{145}}$ , or of $5.2$ multiplied by any power of $10$ , or of any number from $1$ up to but not including $10$ multiplied by $\mathrm{10^{146}}$	✓
Give the answer in standard form	A1	$\mathrm{5.2 \times 10^{146}}$ , with the number part between $1$ and $10$ and the index correct	✓
Guidance on a bare correct answer in part (b)	note	A correct answer scores full marks unless it comes from obviously incorrect working	✓

Full marks: 3/3

The two triangles below are similar.

Work out the value of $\mathit{x}$

Show your working clearly. [5 marks]

x =

[Total 5 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

Triangle

\mathit{ABC}

and triangle

\mathit{DEF}

are similar, and each one has a right angle: at

\mathit{A}

in the small triangle and at

\mathit{D}

in the large one

The small triangle has

\mathit{AB} = 7.5

cm and

\mathit{AC} = \mathit{x}

The large triangle has

\mathit{DF} = 24

cm and hypotenuse

\mathit{EF} = 51

The marked angles pair the vertices up: the single arc puts

\mathit{B}

with

\mathit{E}

, and the double arc puts

\mathit{C}

with

\mathit{F}

Find

The value of

\mathit{x}

, which is the length of

\mathit{AC}

in centimetres. Only one length of the large triangle is given alongside its hypotenuse, so the side that matches

\mathit{AB}

has to be worked out before the two triangles can be compared.

Plan the Solution

Pair the vertices up first, using the right angles and the arcs. That is what decides which side matches which, and every later step depends on it.
The side matching $AB$ is $DE$ , and $DE$ is not given, so find it from Pythagoras theorem in triangle $DEF$ , where $EF$ is the hypotenuse and $DF$ is the other short side.
Divide $DE$ by $AB$ to get the scale factor from the small triangle to the large one.
The side $x$ matches $DF$ , and $x$ is on the small triangle, so divide $DF$ by that scale factor to come back down.
Check the answer on the pair of sides that has not been used, and again on the angles, which must match if the triangles really are similar.

Worked Solution [5 marks]

Similar triangles: matching sides are all in the same ratio, so the scale factor from the small triangle to the large one is

\mathit{k} = \dfrac{DE}{AB}

, and dividing a length of the large triangle by

\mathit{k}

gives the length that matches it on the small one.

Match the two triangles up, vertex by vertex

AB \text{ matches } DE

AC \text{ matches } DF

BC \text{ matches } EF

(Reason: The right angles are at

A

and at

D

, so those two vertices go together. The single arc pairs

B

with

E

, and the double arc pairs

C

with

F

. Reading the letters in that order pairs the sides as well. The known side of the small triangle is

AB

, so the length it must be compared with is

DE

, and

DE

is not given.)

Find

DE

using Pythagoras theorem

DE^2 = 51^2 - 24^2

51^2 - 24^2 = 2601 - 576 = 2025

DE = \sqrt{2025} = 45

(Reason: In triangle

DEF

the right angle is at

D

, so

EF

is the hypotenuse and

DE

is one of the two shorter sides. Squaring the hypotenuse and taking away the square of the other short side leaves the square of

DE

2025

is a perfect square, so

DE

is exactly

45

cm and nothing has to be rounded.)

Work out the scale factor from the small triangle to the large one

k = \dfrac{DE}{AB} = \dfrac{45}{7.5}

\dfrac{45}{7.5} = 6

(Reason: A scale factor is a length on the large triangle divided by the length that matches it on the small triangle.

DE

and

AB

are such a pair, so

45

divided by

7.5

gives it. Every side of the large triangle is

6

times the matching side of the small one.)

Divide

DF

by the scale factor to reach

x

x = \dfrac{DF}{k}

\dfrac{24}{6} = 4

(Reason:

AC

matches

DF

, and

AC

belongs to the small triangle, so the large triangle length is divided by the scale factor rather than multiplied by it. Multiplying instead is the commonest slip on this question: it would make the small triangle bigger than the large one, which the figure rules out at a glance.)

State the value of

x

x = 4

(Reason: The question asks for the value of

x

, so the answer is the number

4

, and the length

AC

4

cm. It is sensible:

x

is the shorter of the two sides meeting at the right angle in triangle

ABC

, exactly as

24

is shorter than

45

in triangle

DEF

x = 4

Verification

Check 1: Scale the answer back up. A length on the small triangle multiplied by

6

must give the length that matches it on the large one, so

4

must return the

24

the question gives

4 \times 6 = 24

Check 2: Test the pair of sides that has not been used. The hypotenuse of the small triangle is

8.5

cm, because

8.5

squared is

72.25

and that is

7.5

squared added to

4

squared. Scaled up by

6

it must give the hypotenuse of the large triangle

8.5 \times 6 = 51

Check 3: Compare the two triangles as a ratio instead of a scale factor, which never forms

6

at all. Each small side divided by the large side matching it must give the same fraction

\dfrac{7.5}{45} = \dfrac{4}{24}

Check 4: Check the angles rather than the sides. The tangent of the angle at

F

DE

over

DF

, and the tangent of the angle at

C

, which is marked as equal to it, is

AB

over

AC

, so the two must come to the same value

\dfrac{45}{24} = \dfrac{7.5}{4}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Apply Pythagoras theorem to triangle $DEF$	M1	For example $51^2 = \mathit{DE}^2 + 24^2$ , or $\mathit{DE}^2 = 51^2 - 24^2$ , or $2601 = \mathit{DE}^2 + 576$ . A trigonometric start earns the same mark: $\cos \mathit{DFE} = \dfrac{24}{51}$ or $\sin \mathit{DEF} = \dfrac{24}{51}$	✓
Square root to reach the missing side of the large triangle	M1	$\mathit{DE} = \sqrt{51^2 - 24^2} = \sqrt{2025} = 45$ , or, from the trigonometric start, angle $\mathit{DFE} = 61.9$ or angle $\mathit{DEF} = 28.0$ , both as the mark scheme prints them, cut off after the first decimal place	✓
A correct method for the scale factor	M1	For example $\dfrac{45}{7.5} = 6$ from their own $\mathit{DE}$ , or $\dfrac{7.5}{45} = \dfrac{1}{6}$ , for which a decimal of $0.17$ or better is accepted, or the ratio written as $\dfrac{x}{24} = \dfrac{7.5}{45}$ . Correct use of the sine rule, the cosine rule or Pythagoras theorem is allowed here. Special case: a candidate who adds the squares instead of subtracting them reaches $\sqrt{3177} = 3\sqrt{353}$ , about $56.36$ , and may still earn this mark by using that value in place of $45$	✓
Use the scale factor to find $x$	dM1	Dependent on the previous method mark. For example $\dfrac{24}{6}$ with their own scale factor, or $24 \times \dfrac{1}{6}$ , or $\dfrac{7.5}{45} \times 24$ . Finding the small hypotenuse first earns it too: $\dfrac{51}{6} = 8.5$ followed by $\sqrt{8.5^2 - 7.5^2}$	✓
The value of $x$	A1	$4$ , and it must come from correct figures	✓
Working must be shown	note	The mark scheme prints Working required beside the answer, and the question itself asks for the working to be shown clearly, so an answer of $4$ standing on its own scores nothing	✓

Full marks: 5/5

(a) The table below is for the function $y = 2(x + \dfrac{1}{x})$ . Work out the four missing values of $y$ . [2 marks]

$\displaystyle{\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & 0.5 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline y & \boxed{\phantom{00.0}} & \boxed{\phantom{00.0}} & 5 & 6.7 & \boxed{\phantom{00.0}} & \boxed{\phantom{00.0}} & 12.3 \\ \hline \end{array}}$

(b) Using your completed table, draw the graph of $y = 2(x + \dfrac{1}{x})$ on the grid for $0.5 \leqslant x \leqslant 6$ [2 marks]

[Total 4 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

y = 2(x + \dfrac{1}{x})

A table of values with

x = 0.5

1

2

3

4

5

and

6

Three values of

y

are already printed:

5

x = 2

6.7

x = 3

and

12.3

x = 6

A grid running from

0

6

across and

0

13

Find

the four missing values of

y

in the table, then the graph of

y = 2(x + \dfrac{1}{x})

for

0.5 \leqslant x \leqslant 6

Plan the Solution

Read the function as an instruction: take the reciprocal of $x$ , add $x$ to it, then double the total.
Work out the four missing values one at a time. All four come out exactly, so nothing has to be rounded.
Use the three values the table already prints as a running check on the method.
Plot all seven points, then join them with one smooth curve that starts at $x = 0.5$ and stops at $x = 6$ .

Worked Solution [4 marks]

Rule - Substitution into

y = 2(x + \dfrac{1}{x})

: work out

\dfrac{1}{x}

first, add

x

to it, then multiply the total by

2

Substitute

x = 0.5

2 \times (0.5 + 2) = 2 \times 2.5 = 5

(Reason: One divided by

0.5

2

, so the bracket comes to

2.5

. Doubling that gives

y = 5

. Notice how large

y

is at this end: the reciprocal term is what pushes the curve back up for small

x

Substitute

x = 1

2 \times (1 + 1) = 2 \times 2 = 4

(Reason: At

x = 1

the reciprocal is

1

as well, so the bracket is just

2

. This is the smallest value anywhere in the table, and it is where the curve turns.)

Substitute

x = 4

2 \times (4 + 0.25) = 2 \times 4.25 = 8.5

(Reason: A quarter is

0.25

, so the bracket is

4.25

and doubling gives

8.5

exactly. From here on the reciprocal is small and

y

is only a little more than

2x

Substitute

x = 5

2 \times (5 + 0.2) = 2 \times 5.2 = 10.4

(Reason: Work out the reciprocal of

5

first, add it on, then double the bracket. This value is exact too, so no rounding is needed.)

The completed table

\displaystyle{\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & 0.5 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline y & 5 & 4 & 5 & 6.7 & 8.5 & 10.4 & 12.3 \\ \hline \end{array}}

(Reason: The four values worked out above are

5

4

8.5

and

10.4

. The other three were already printed, and

6.7

and

12.3

are the only rounded entries in the row.)

Plot the seven points and join them

(0.5, 5), \; (1, 4), \; (2, 5), \; (3, 6.7)

(4, 8.5), \; (5, 10.4), \; (6, 12.3)

(Reason: Plot every point from the completed row, then join them with a single smooth curve. The curve dips to its lowest point at

(1, 4)

, turns, and climbs steadily to

(6, 12.3)

. Two things lose the accuracy mark: joining the points with straight line segments, and letting the curve run on past

x = 0.5

x = 6

Missing values of

y

5

4

8.5

and

10.4

- then a smooth curve through all seven points

Verification

Check 1: Rewrite the function as the single fraction

\dfrac{2(x^2 + 1)}{x}

, which needs no bracket at all, and test it on

x = 5

\dfrac{2 \times (5^2 + 1)}{5} = \dfrac{52}{5} = 10.4

Check 2: The values the table already prints have to come out of the same rule, correct to

1

decimal place

2 \times (3 + \dfrac{1}{3}) = 6.7

Check 3: The smallest number in the completed row should sit exactly where the curve turns, at

x = 1

, and every other value in the row is larger than it

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

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The four missing values of $y$	B2	B2 oe for all four values: $5$ , $4$ , $8.5$ and $10.4$ . B1 for $2$ or $3$ correct values of $y$ . The marks may also be awarded if the values are plotted correctly on the graph.	✓
(b) The seven points plotted	M1ft	M1 ft their table, dep on B1, for at least $6$ points plotted correctly, within or on the circles on the overlay.	✓
(b) The curve itself	A1	A1 for a correct smooth curve between $x = 0.5$ and $x = 6$ .	✓
Notes	Note	A correct answer scores full marks unless it comes from obviously incorrect working. If a fully correct graph is drawn but the table in part (a) is left incomplete, the marks for part (a) are still awarded. Any curve drawn for $x$ below $0.5$ or above $6$ is ignored.	✓

Full marks: 4/4

$ABC$ is a triangle with a right angle at $B$ , and $D$ is a point on the side $BC$ .

$AB = 4\,250$ m, angle $BAD = 47^\circ$ and angle $BCA = 24^\circ$ .

Work out the length of $DC$ . Give your answer correct to the nearest integer. [4 marks]

[Total 4 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

ABC

is a triangle with a right angle at

B

D

is a point on

BC

, so

BD

and

DC

together make up

BC

AB = 4\,250 \text{ m}

angle

BAD = 47^\circ

angle

BCA = 24^\circ

Find

The length of

DC

, correct to the nearest integer

Plan the Solution

The right angle at $B$ belongs to two triangles at once: $ABD$ and the whole triangle $ABC$ . $AB$ is the side next to the known angle in each of them, so the tangent ratio reaches everything.
Use the $47^\circ$ angle at $A$ to find $BD$ , the near part of the base.
Use the $24^\circ$ angle at $C$ to find $BC$ , the whole base.
$D$ lies between $B$ and $C$ , so $DC$ is what is left when $BD$ is taken away from $BC$ .
Keep the unrounded values on the calculator and round only at the very end.

Worked Solution [4 marks]

Tangent ratio in a right-angled triangle:

\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}

Find

BD

in triangle

ABD

\tan 47^\circ = \dfrac{BD}{4\,250}

BD = 4\,250 \times \tan 47^\circ = 4\,557.567\ldots

(Reason: Angle

ABD

is a right angle,

BD

is opposite the

47^\circ

angle at

A

, and

AB

is adjacent to it. So

BD

and

AB

are linked by the tangent of

47^\circ

Find the whole of

BC

in triangle

ABC

\tan 24^\circ = \dfrac{4\,250}{BC}

BC = \dfrac{4\,250}{\tan 24^\circ} = 9\,545.656\ldots

(Reason: Angle

ABC

is the same right angle. This time the known angle is the

24^\circ

C

, and

AB

is the side opposite it while

BC

is the side adjacent to it, so

BC

is reached by dividing by the tangent instead of multiplying by it.)

Subtract to leave

DC

DC = BC - BD

9\,545.656 - 4\,557.567 = 4\,988.089

(Reason:

D

is between

B

and

C

, so

BD

and

DC

make up the whole of

BC

. Taking

BD

away from

BC

leaves the part that is asked for.)

Round to the nearest integer

DC = 4\,988 \text{ m}

(Reason:

4988.089...

lies between

4988

and

4989

, and the first digit after the decimal point is

0

, so it rounds down to

4988

DC = 4988

Verification

Check 1: Put the two parts of the base back together and compare the total with

BC

4\,557.567 + 4\,988.089 = 9\,545.656 \text{ m}

Check 2: Work

DC

out a second way, from triangle

ADC

alone. Its angles are

19^\circ

A

, with

137^\circ

D

and

24^\circ

C

, and

AD = \dfrac{4\,250}{\cos 47^\circ} = 6\,231.686\ldots

The sine rule gives

DC = AD \times \dfrac{\sin 19^\circ}{\sin 24^\circ} = 4\,988.089\ldots

, which is the same length.

Check 3: The two angles at

A

must fill angle

BAC

, and angle

BAC

is the complement of the

24^\circ

angle at

C

47 + 19 = 66 = 90 - 24

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct trigonometric statement in one of the right-angled triangles	M1	eg $\tan 47^\circ = \dfrac{BD}{4\,250}$ or $\tan 24^\circ = \dfrac{4\,250}{BC}$ or $\tan 66^\circ = \dfrac{BC}{4\,250}$ or $AD = \dfrac{4\,250}{\cos 47^\circ}$ or $AC = \dfrac{4\,250}{\sin 24^\circ}$	✓
A correct value for one of the lengths the method needs	M1	eg $BD = 4\,250 \times \tan 47^\circ = 4\,557.567\ldots$ or $BC = \dfrac{4\,250}{\tan 24^\circ} = 9\,545.656\ldots$ or $AD = 6\,231.686\ldots$ or $AC = 10\,449.021\ldots$	✓
A complete method for $DC$	M1	eg the subtraction $9\,545.656 - 4\,557.567$ , or $DC = \dfrac{6\,231.686 \times \sin 19^\circ}{\sin 24^\circ}$ , or $DC = \dfrac{10\,449.021 \times \sin 19^\circ}{\sin 137^\circ}$ , or the cosine rule $6\,231.686^2 + 10\,449.021^2 - 2 \times 6\,231.686 \times 10\,449.021 \times \cos 19^\circ$ followed by a square root	✓
The answer	A1	The answer $4\,988 \text{ m}$ , and anything in the range $4\,932$ to $4\,990$ is allowed, which absorbs answers built from figures rounded at an earlier stage. A correct answer scores full marks unless it comes from obvious incorrect working.	✓

Full marks: 4/4

Martin drives a delivery van to a bakery. The probability that Martin arrives on time on Friday is $0.7$ , the probability that he arrives on time on Saturday after arriving on time on Friday is $0.9$ , and the probability that he arrives on time on Saturday after arriving late on Friday is $0.6$

(a) Use this information to complete the probability tree diagram. [2 marks]

(b) Work out the probability that Martin arrives on time on both Friday and Saturday. [2 marks]

(b)

[Total 4 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

The probability that Martin arrives on time on Friday is

0.7

If he arrives on time on Friday, the probability that he arrives on time on Saturday is

0.9

If he arrives late on Friday, the probability that he arrives on time on Saturday is

0.6

A two stage probability tree: Friday first, then Saturday

Find

Part (a): the five probabilities missing from the tree. Part (b): the probability that Martin arrives on time on both days.

Plan the Solution

Write the given Saturday probabilities on the pairs they belong to: $0.9$ follows an on time Friday, $0.6$ follows a late Friday.
Every pair of branches leaving the same point covers all the outcomes, so subtract each given probability from $1$ to get its partner.
For part (b), pick out the single path that is on time on Friday and then on time on Saturday.
Multiply along that path, then check the four complete paths add up to $1$ .

Worked Solution [4 marks]

Rule - Tree diagrams: the branches leaving any one point add up to

1

, and the probability of a whole path is the product of the probabilities along it.

Complete the tree

1 - 0.7 = 0.3

1 - 0.9 = 0.1

1 - 0.6 = 0.4

(Reason: On time and late are the only two ways each day can go, so each pair of branches adds up to

1

. The

0.9

belongs to the upper Saturday pair because that pair follows an on time Friday, and the

0.6

belongs to the lower pair, which follows a late Friday.)

Read the completed tree back

0.7 + 0.3 = 1

0.9 + 0.1 = 1

0.6 + 0.4 = 1

(Reason: A pair that does not add up to

1

is the quickest sign that a probability has been written on the wrong branch, so this is worth doing before any multiplying starts.)

Multiply along the on time, on time path

0.7 \times 0.9 = 0.63

(Reason: Following a single path through the tree means both things happen, so the probabilities along that path are multiplied together.)

Give the answer in the forms the mark scheme accepts

0.63 = \dfrac{63}{100}

(Reason: The accuracy mark is awarded for any equivalent form, so the fraction and

63\%

earn it too.)

0.63

Verification

Check 1: Work out the other three paths and add all four path probabilities:

0.63 + 0.07 + 0.18 + 0.12 = 1

Check 2: Count

1000

weekends instead of using decimals:

700

arrive on time on Friday, and

9

out of every

10

of those arrive on time on Saturday as well, which is

630

weekends out of

1000

\dfrac{630}{1000} = 0.63

Check 3: Sanity check the size of the answer: arriving on time on both days must be harder than arriving on time on Friday alone.

0.63

is smaller than

0.7

, as it has to be

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) All three pairs of probabilities correct	B2	All $3$ correct pairs of probabilities on the correct branches: $0.7$ and $0.3$ on Friday, $0.9$ and $0.1$ on the upper Saturday pair, $0.6$ and $0.4$ on the lower Saturday pair. Equivalent fractions are allowed.	✓
(a) One correct pair of probabilities	B1	If B2 is not earned, one correct pair of probabilities on a correct branch. The given $0.7$ together with $0.3$ counts as a pair.	✓
(b) A complete method involving one product	M1ft	A complete method involving one product, $0.7 \times 0.9$ or equivalent.	✓
(b) The probability	A1ft	$0.63$ , or any equivalent form, for example $\dfrac{63}{100}$ or $63\%$ .	✓
(b) Note	note	A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 4/4

$B$ is inversely proportional to the square of $d$

$B$ takes the value $0.25$ when $d = 12$

Find a formula for $B$ in terms of $d$ [3 marks]

[Total 3 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

B

is inversely proportional to the square of

d

B

takes the value

0.25

when

d = 12

, and that is the only pair of values the question supplies

Find

a formula for

B

in terms of

d

- an equation in

d

with no unknown constant left in it

Plan the Solution

Write inverse square proportion as an equation, using a letter for the constant of proportionality. Without that letter there is nothing for the given values to fix.
Substitute the one pair of values the question supplies, squaring $d$ first so the equation carries a single number rather than a power.
Solve for the constant, then put its value back into the formula and leave $d$ as a letter.

Worked Solution [3 marks]

Rule - Inverse proportion to a square: when one quantity is inversely proportional to the square of another, it equals a constant divided by that square, so the formula has the shape

\dfrac{k}{d^{2}}

for some constant. Substitute the pair of values given to find that constant, then write the formula out with the constant in place of the letter.

Write the proportion as an equation with a constant

B = \dfrac{k}{d^{2}}

(Reason: Inversely proportional to the square of

d

means

B

is a constant divided by

d

squared. The constant has to appear as a letter at this stage; a version with no constant in it cannot be made to fit the pair of values, and the mark scheme awards nothing for it.)

Substitute the pair of values the question gives

0.25 = \dfrac{k}{12^{2}}

0.25 = \dfrac{k}{144}

(Reason: The value of

B

goes on the left and the value of

d

goes into the denominator. Squaring the

12

before going any further leaves a single number under the constant, so what is left to solve is a one-step equation.)

Clear the fraction and read off the constant

k = 0.25 \times 144

k = 36

(Reason: Multiplying both sides by

144

clears the fraction and leaves the constant on its own. A quarter of

144

36

Put the constant back into the formula

B = \dfrac{36}{d^{2}}

(Reason: The question asks for a formula for

B

in terms of

d

, so

d

stays a letter and only the constant is replaced by its value. Stopping at the value of the constant answers a different question from the one asked.)

B = \dfrac{36}{d^{2}}

Verification

Check 1: Put

d = 12

back into the finished formula. Squaring the

12

makes the denominator

144

, and the fraction has to return the value of

B

the question gives.

\dfrac{36}{144} = 0.25

Check 2: Halve

d

from

12

6

. Inverse square proportion must then multiply

B

4

, so the formula has to return

4

lots of

0.25

, which is

1

\dfrac{36}{6^{2}} = \dfrac{36}{36} = 1

Check 3: Work out

B

multiplied by the square of

d

d = 12

, at

d = 6

and at

d = 3

. For an inverse square relationship that product has to come to the same number every time, and that number is the constant in the formula.

0.25 \times 144 = 1 \times 36 = 4 \times 9 = 36

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Write the relationship as a formula carrying a constant	M1	for $B = \dfrac{k}{d^{2}}$ oe, or the alternative form $Bk = \dfrac{1}{d^{2}}$ oe. The constant of proportionality must be written as a symbol - $k$ is the usual choice, and $D$ or $b$ are allowed instead - so a version with no constant in it, $B = \dfrac{1}{d^{2}}$ , earns nothing here.	✓
Substitute the given pair into a correct formula	M1	for substituting $B$ and $d$ into a correct formula: $0.25 = \dfrac{k}{12^{2}}$ oe, or $0.25 = \dfrac{k}{144}$ oe, or $k = 36$ . Both method marks are available together for $0.25 = \dfrac{k}{12^{2}}$ oe. In the alternative form the right-hand side is $\dfrac{1}{12^{2}}$ oe or $\dfrac{1}{144}$ oe, and the constant comes to $\dfrac{1}{36}$ .	✓
The finished formula	A1	for the finished formula $B = \dfrac{36}{d^{2}}$ , or any equivalent - $B = 36 \times \dfrac{1}{d^{2}}$ and the negative-index form both count. Full marks are also awarded when $B = \dfrac{k}{d^{2}}$ is written on the answer line and $k = 36$ is clearly given in the body of the working.	✓
General guidance	note	A correct answer scores full marks unless it comes from obviously incorrect working.	✓
Rearranged, but not a formula for $B$	SC	M2A0 where the constant has been found but the answer is left as a rearrangement rather than as a formula for $B$ - for example an answer that gives the square of $d$ as $\dfrac{36}{B}$ , or $d$ as $\dfrac{6}{\sqrt{B}}$ . Both method marks are earned and the accuracy mark is lost.	✓

Full marks: 3/3

Expand and simplify $3x(2x - 1)(5x + 4)$ , giving your answer in the form $ax^3 + bx^2 + cx$ where $a$ , $b$ and $c$ are integers. [3 marks]

[Total 3 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

The expression

3x(2x - 1)(5x + 4)

, a single term multiplied by two linear brackets

a

b

and

c

are integers

Find

The expansion, simplified, in the form

ax^3 + bx^2 + cx

the values of

a

b

and

c

Plan the Solution

Multiply the two brackets together first, so that only one bracket is left to deal with.
Collect the two $x$ terms straight away, so three terms are carried forward instead of four.
Multiply each of those three terms by $3x$ , adding the indices.
Never share the $3x$ out between the two brackets - that answers a different question and the mark scheme refuses the first method mark for it.
Check by substituting a value of $x$ into the original product and into the expansion.

Worked Solution [3 marks]

Expanding a triple product: expand one pair of brackets first,

(px + q)(rx + s) = prx^2 + (ps + qr)x + qs

, then multiply every term of that quadratic by the single term outside, using

x^m \times x^n = x^{m+n}

Expand the two brackets

(2x - 1)(5x + 4) = 10x^2 + 8x - 5x - 4

10x^2 + 8x - 5x - 4 = 10x^2 + 3x - 4

(Reason: Each term in the first bracket multiplies each term in the second, so there are four products; the two

x

terms are then collected into a single term. Taking this pair of brackets first is the shorter road, because it leaves only one bracket for the

3x

to multiply.)

Multiply every term by

3x

3x \times 10x^2 = 30x^3

3x \times 3x = 9x^2

3x \times (-4) = -12x

3x(10x^2 + 3x - 4) = 30x^3 + 9x^2 - 12x

(Reason: The

3x

multiplies all three terms of the quadratic, and the indices add, so

3x

times

10x^2

gives

30x^3

. The

3x

is not shared out between the two original brackets: doing that gives

6x^2 - 3x + 15x^2 + 12x = 21x^2 + 9x

, which is not even a cubic, and the mark scheme names it as an expansion that earns nothing.)

Read off

a

b

and

c

30x^3 + 9x^2 - 12x

a = 30, b = 9, c = -12

(Reason: The three terms are already in descending powers of

x

and there is no constant term, which is exactly the form the question asks for, so the three integers can be read straight off.)

30x^3 + 9x^2 - 12x

Verification

Check 1: Substitute

x = 1

into the original product,

3 \times 1 \times (2 - 1) \times (5 + 4)

, and into the expansion the product gives

27

, and

30 + 9 - 12 = 27

Check 2: Substitute

x = 2

, so the product is

6 \times 3 \times 14

and the expansion is

240 + 36 - 24

both give

252

Check 3: Expand in the other order:

3x(2x - 1) = 6x^2 - 3x

first, then multiply that by

(5x + 4)

30x^3 + 24x^2 - 15x^2 - 12x = 30x^3 + 9x^2 - 12x

, the same expansion

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Expand one pair of brackets	M1	For an expansion with only one error, eg $(2x - 1)(5x + 4) = 10x^2 + 8x - 5x - 4$ , or $3x(2x - 1) = 6x^2 - 3x$ , or $3x(5x + 4) = 15x^2 + 12x$ . This mark is not awarded for $6x^2 - 3x + 15x^2 + 12x$ , which shares the $3x$ out between the two brackets instead of multiplying all three factors together.	✓
Complete the expansion	M1ft	Follow through on the candidate's own quadratic, dependent on the first M1, allowing one further error, eg $3x(10x^2 + 3x - 4) = 30x^3 + 9x^2 - 12x$ , or $(6x^2 - 3x)(5x + 4) = 30x^3 + 24x^2 - 15x^2 - 12x$ , or $(15x^2 + 12x)(2x - 1) = 30x^3 - 15x^2 + 24x^2 - 12x$ .	✓
The simplified cubic	A1	cao $30x^3 + 9x^2 - 12x$ . Terms may be in any order but must be simplified, dependent on the first M1. Accept $a = 30$ , $b = 9$ , $c = -12$ . Ignore subsequent working for a correct factorisation, eg $3(10x^3 + 3x^2 - 4x)$ ; do not ignore an incorrect simplification.	✓
Partial credit when the two brackets containing the $3x$ are multiplied out	note	Expanding $(6x^2 - 3x)(5x + 4)$ has four products to get right: M2 for three of the four correct, giving $30x^3 + 24x^2 - 15x^2 - 12x$ , and M1 for two of the four correct.	✓

Full marks: 3/3

Continue to questions 16 to 25

The remaining 10 questions, with the same full worked solutions and mark schemes

Frequently asked questions

There are 25 questions worth 100 marks in total, sat over 2 hours. It is Higher tier and a calculator is allowed throughout, unlike UK GCSE Maths, where one paper is non-calculator.

Higher tier targets grades 4 to 9, so the lower grades 1 to 3 are only reachable on the tier below. About 40 per cent of the questions are targeted at grades 4 and 5 and appear on both Paper 1F and Paper 1H, so the lowest grades on this Higher paper are the ones the two tiers share.

Yes. The paper states in its own instructions that without sufficient working, correct answers may be awarded no marks. Several questions ask you to show your working clearly or to show clear algebraic working, and on those a bare answer scores nothing. That is why every solution here sets out the method mark by mark.

Yes, a Higher tier formulae sheet is printed in the paper. It gives the area of a trapezium, the volume of a prism, the volume and curved surface area of a cylinder, the volume and curved surface area of a cone, the volume and surface area of a sphere, the area of a triangle from two sides and the included angle, the sine rule, the cosine rule, the sum of an arithmetic series and the quadratic formula. Other results, such as Pythagoras theorem and the trigonometric ratios for right-angled triangles, still have to be recalled. Nothing may be written on the formulae page.

Both are published by Pearson Edexcel and are linked directly from this page as PDF files. The solutions here are original: every question has been reworded, but all the numbers match the original paper, so the answers agree with the official mark scheme. This resource reproduces neither the exam paper nor the official mark scheme.

Keep revising

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

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Edexcel IGCSE 4MA1 Paper 1H, November 2024: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 15 of 25

Question 1, Calculator allowed

Question 1 - Exam Solution

Question 2, Calculator allowed

Question 2 - Exam Solution

Question 3, Calculator allowed

Question 3 - Exam Solution

Question 4, Calculator allowed

Question 4 - Exam Solution

Question 5, Calculator allowed

Question 5 - Exam Solution

Question 6, Calculator allowed

Question 6 - Exam Solution

Question 7, Calculator allowed

Question 7 - Exam Solution

Question 8, Calculator allowed

Question 8 - Exam Solution

Question 9, Calculator allowed

Question 9 - Exam Solution

Question 10, Calculator allowed

Question 10 - Exam Solution

Question 11, Calculator allowed

Question 11 - Exam Solution

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

Frequently asked questions

Keep revising

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