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Edexcel IGCSE 4MA1 Paper 2H, November 2024: Worked Solutions and Mark Schemes

Q: How many questions are on Edexcel 4MA1 Paper 2H, November 2024, and how long is the exam?

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.

Q: What grades can you get from the Higher paper?

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 2F and Paper 2H, so the lowest grades on this Higher paper are the ones the two tiers share.

Q: Do you have to show your working?

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.

Q: Is a formulae sheet given in the exam?

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.

Q: Where can I get the official question paper and mark scheme?

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.

Sir Faraz Hassan

30 Jul 2026

Table of Contents▾

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

Original worked solutions for Edexcel International GCSE Mathematics A (4MA1), Paper 2H (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.

Download printable PDF

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

Worked solutions, questions 1 to 15 of 25

Show that $1\dfrac{5}{7} \times 2\dfrac{3}{16} = 3\dfrac{3}{4}$
Show every stage of your working. [3 marks]

[Total 3 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

The product of two mixed numbers,

1\dfrac{5}{7}

and

2\dfrac{3}{16}

The result that has to be reached:

3\dfrac{3}{4}

Find

Show that the left-hand side works out to exactly

3\dfrac{3}{4}

. The working is the answer, so every stage has to be written down.

Plan the Solution

Write each mixed number as an improper fraction, because a mixed number cannot be multiplied a piece at a time.
Multiply the numerators together and the denominators together to get one fraction, $\dfrac{420}{112}$ .
Cancel that fraction to lowest terms, then turn it back into a mixed number and compare it with $3\dfrac{3}{4}$ .

Worked Solution [3 marks]

Rule - Multiplying mixed numbers: write every mixed number as an improper fraction first, then

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

. Common factors may be cancelled before multiplying or after.

Step 1: Write both mixed numbers as improper fractions

1\dfrac{5}{7} = \dfrac{1 \times 7 + 5}{7} = \dfrac{12}{7}

2\dfrac{3}{16} = \dfrac{2 \times 16 + 3}{16} = \dfrac{35}{16}

(Reason: Multiply the whole number by the denominator and add the numerator; the denominator itself never changes. So

1\dfrac{5}{7}

has numerator

1 \times 7 + 5 = 12

and

2\dfrac{3}{16}

has numerator

2 \times 16 + 3 = 35

Step 2: Multiply the numerators, and multiply the denominators

\dfrac{12}{7} \times \dfrac{35}{16} = \dfrac{12 \times 35}{7 \times 16} = \dfrac{420}{112}

(Reason: The product of two fractions is the product of the numerators over the product of the denominators:

12 \times 35 = 420

on top and

7 \times 16 = 112

underneath.)

Step 3: Cancel the fraction to its lowest terms

\dfrac{420}{112} = \dfrac{15}{4}

(Reason: Both parts divide by

28

, since

\dfrac{420}{28} = 15

and

\dfrac{112}{28} = 4

. Cancelling before multiplying gives the same fraction:

\dfrac{12}{7} \times \dfrac{35}{16} = \dfrac{3}{4} \times \dfrac{5}{1} = \dfrac{15}{4}

Step 4: Write the improper fraction as a mixed number

\dfrac{15}{4} = \dfrac{3 \times 4 + 3}{4} = 3\dfrac{3}{4}

(Reason:

4

goes into

15

three times with

3

left over, so

\dfrac{15}{4} = 3\dfrac{3}{4}

, which is exactly the result the question asked for.)

1\dfrac{5}{7} \times 2\dfrac{3}{16} = \dfrac{420}{112} = \dfrac{15}{4} = 3\dfrac{3}{4}

as required

Verification

Check 1: Compare the two sides over the same denominator. The target is

3\dfrac{3}{4} = \dfrac{15}{4}

, and scaling top and bottom by

28

gives

\dfrac{15 \times 28}{4 \times 28}

\dfrac{15 \times 28}{4 \times 28} = \dfrac{420}{112}

, the unsimplified product from Step 2, so both sides are the same number

Check 2: Work backwards. Dividing the result by one factor must return the other, so

\dfrac{15}{4}

divided by

\dfrac{35}{16}

should give

1\dfrac{5}{7}

\dfrac{15}{4} \times \dfrac{16}{35} = \dfrac{240}{140} = \dfrac{12}{7} = 1\dfrac{5}{7}

, the first factor recovered

Check 3: Use decimals, which avoids fractions altogether:

2\dfrac{3}{16} = 2.1875

and

3\dfrac{3}{4} = 3.75

, so

\dfrac{3.75}{2.1875}

must come to

1\dfrac{5}{7}

\dfrac{3.75}{2.1875} = 1.714285\ldots = \dfrac{12}{7} = 1\dfrac{5}{7}

, agreeing with the fraction working

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Both mixed numbers written as improper fractions: $\dfrac{12}{7} \times \dfrac{35}{16}$	M1	For both fractions written as improper fractions. The multiplication sign may be implied, and any equivalent pair of improper fractions scores this mark.	✓
Multiplying across, or cancelling: $\dfrac{12 \times 35}{7 \times 16} = \dfrac{420}{112}$	M1	For multiplying the numerators and multiplying the denominators, or for cancelling the fractions fully, or for cancelling partially and then multiplying across. Writing both fractions over a common denominator also scores it, for example $\dfrac{192}{112} \times \dfrac{245}{112} = \dfrac{47\,040}{12\,544}$ .	✓
Completion: $\dfrac{420}{112} = \dfrac{15}{4} = 3\dfrac{3}{4}$	A1	Completion to the given result. Working is required, so an unsupported statement of the answer scores nothing here.	✓
Note	note	If the working shows clearly that $3\dfrac{3}{4} = \dfrac{15}{4}$ , it is enough to show that the left-hand side comes to $\dfrac{15}{4}$ ; the mixed number need not be written again.	✓

Full marks: 3/3

A carpenter measures the width of a bookshelf as $1.4$ metres, correct to one decimal place.

(a) Write down the upper bound of the width of the bookshelf. [1 mark]

(b) Write down the lower bound of the width of the bookshelf. [1 mark]

(a) metres(b) metres

[Total 2 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

A measured width of

1.4

metres, correct to one decimal place.

So the true width is any value that rounds to

1.4

at one decimal place.

Find

(a) The upper bound of the width, in metres. (b) The lower bound of the width, in metres.

Plan the Solution

Find the rounding unit. One decimal place means the width was rounded to the nearest $0.1$ of a metre.
Halve that unit. Half of $0.1$ is $0.05$ , the greatest amount the true width can differ from the measurement.
Add the half-unit to $1.4$ for the upper bound, and subtract it for the lower bound.
Keep the two the right way round: the upper bound is the larger of the pair.

Worked Solution [2 marks]

Rule - Bounds of a rounded measurement: if

x

is given correct to the nearest

u

, the lower bound is

x - \dfrac{u}{2}

and the upper bound is

x + \dfrac{u}{2}

Step 1: find the rounding unit, and half of it

\text{one decimal place} \implies u = 0.1

\dfrac{u}{2} = \dfrac{0.1}{2} = 0.05

(Reason: Rounding to one decimal place means the width was rounded to the nearest

0.1

m, so the true width lies within half of

0.1

m of the

1.4

m that was written down.)

Step 2: (a) add the half-unit to get the upper bound

1.4 + 0.05 = 1.45

(Reason: Every width below

1.45

m rounds down to

1.4

m, while

1.45

m itself rounds up to

1.5

m, so this is where the interval stops.)

Step 3: (b) subtract the half-unit to get the lower bound

1.4 - 0.05 = 1.35

(Reason: Going the same distance the other way from the measurement gives the smallest width that still rounds to

1.4

m. Anything smaller would be measured as

1.3

m instead.)

(a)

1.45

metres(b)

1.35

metres

Verification

Check 1: Round each bound back to one decimal place.

1.35

rounds up to

1.4

, and any width just below

1.45

rounds down to

1.4

1.35 \leq \text{width} < 1.45

, and every width in that interval is measured as

1.4

Check 2: The gap between the two bounds should be exactly one rounding unit, that is one whole step at this degree of accuracy.

1.45 - 1.35 = 0.1

, which is one step of one decimal place

Check 3: The measurement should sit exactly halfway between the two bounds.

\dfrac{1.35 + 1.45}{2} = 1.4

, the width the carpenter wrote down

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The upper bound of the width	B1	$1.45$ . Allow $1.44\dot{9}$ or $1.44999(9\ldots)$ , which are the same number as $1.45$ .	✓
(b) The lower bound of the width	B1	$1.35$ , cao. No working is needed for either part, so nothing else earns this mark.	✓
Special case - the two bounds interchanged	SCB1	If (a) is given as $1.35$ and (b) as $1.45$ , the pair of bounds is right but the labels are swapped. Score B0 then B1, so one mark in total.	✓

Full marks: 2/2

Triangle $PQR$ is shown in the diagram below.

Calculate the value of $x$ .
Give your answer correct to one decimal place. [3 marks]

x =

[Total 3 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

Triangle

PQR

, with the right angle at

Q

PR = 8.6

cm - the side facing the right angle, so it is the hypotenuse.

The angle at

R

43^\circ

QR = x

cm - the side running along one arm of that angle.

Find

The value of

x

, correct to one decimal place.

Plan the Solution

Stand at the $43^\circ$ angle and name the two sides the question involves.
$QR$ lies along an arm of that angle, so it is the adjacent side, and $PR$ faces the right angle, so it is the hypotenuse.
Adjacent with hypotenuse is the cosine ratio, so no third side has to be found first.
Rearrange for $x$ , evaluate with the calculator in degree mode, and round only at the very end.

Worked Solution [3 marks]

Rule - Cosine ratio in a right-angled triangle:

\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}

Step 1: Name the two sides at the

43^\circ

angle

\text{hypotenuse} = PR = 8.6

\text{adjacent} = QR = x

(Reason: The hypotenuse is always the side opposite the right angle, and the right angle is at

Q

, so it is

PR

. The side

QR

runs from

R

along the angle rather than across from it, so it is the adjacent side.)

Step 2: Write down the cosine ratio

\cos 43^\circ = \dfrac{x}{8.6}

(Reason: Adjacent over hypotenuse is cosine. The side

PQ

never appears, which is exactly why cosine is the efficient choice here.)

Step 3: Rearrange to make

x

the subject

x = 8.6 \times \cos 43^\circ

(Reason: Multiplying both sides by

8.6

clears the fraction. This single line is worth both method marks on its own.)

Step 4: Evaluate, with the calculator in degree mode

\cos 43^\circ = 0.7313537\ldots

x = 8.6 \times 0.7313537\ldots = 6.2896418\ldots

(Reason: A calculator left in radian mode gives a completely different value, so check the DEG indicator before pressing cosine. Keep the full display value and do not round yet.)

Step 5: Round to one decimal place

x = 6.2896418\ldots

x = 6.3 \text{ (to 1 d.p.)}

(Reason: The second decimal digit is

8

, so the first decimal digit rounds up from

2

3

x = 6.3

Verification

Check 1 - Pythagoras closes the triangle: Work out the third side a different way:

PQ = 8.6\sin 43^\circ = 5.8651858\ldots

cm. Squaring both legs and adding them must give the square of the hypotenuse.

6.2896418^2 + 5.8651858^2 \approx 73.96

, and

8.6^2 = 73.96

Check 2 - run the trigonometry backwards: Put the answer back into the ratio. If

x

is right, the angle at

R

must come back out as

43^\circ

\cos^{-1}\!\left(\dfrac{6.2896418}{8.6}\right) = 43.0^\circ

Check 3 - is the size sensible: No trigonometry needed. A leg is shorter than the hypotenuse, and because

43^\circ

is less than

45^\circ

the side along the angle is the longer of the two legs, so

x

must sit between

\dfrac{8.6}{\sqrt{2}}

and

8.6

6.0811\ldots < 6.3 < 8.6

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\cos 43^\circ = \dfrac{x}{8.6}$	M1	A correct trig statement for $x$ or for $QR$ , or a correct Pythagoras statement for $x^2$ . Also accept $\tan 43^\circ = \dfrac{8.6\sin 43^\circ}{x}$ , $\sin(90 - 43) = \dfrac{x}{8.6}$ , $\dfrac{x}{\sin(90 - 43)} = \dfrac{8.6}{\sin 90}$ or $x^2 = 8.6^2 - (8.6\sin 43^\circ)^2$ .	✓
$x = 8.6\cos 43^\circ$	M1	A fully correct calculation to find $x$ . Also accept $x = \dfrac{8.6\sin 43^\circ}{\tan 43^\circ}$ , $x = 8.6\sin(90 - 43)$ , $x = \dfrac{8.6\sin 47^\circ}{\sin 90^\circ}$ or $x = \sqrt{8.6^2 - (8.6\sin 43^\circ)^2}$ . A student who goes straight to this line scores both method marks.	✓
$x = 6.3$	A1	Award for anything which rounds to $6.3$ , seen even if it is then rounded incorrectly afterwards. A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

$N$ is a number.
$17\%$ of $N$ is $357$

(a) Work out the value of $N$ [2 marks]

In $2019$ , the number of members of a cycling club was $650$
In $2020$ , the number of members of the club was $806$

(b) Work out the percentage increase in the number of members. [3 marks]

(a) N =(b) %

[Total 5 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

17\%

N

357

The club had

650

members in

2019

and

806

members in

2020

Two separate percentage skills: one works backwards from a percentage, the other measures a change

Find

(a) The value of

N

(b) The percentage increase from

650

members to

806

members

Plan the Solution

Part (a) is a reverse percentage. $17\%$ of $N$ means $0.17 \times N$ , so the equation is $0.17N = 357$ .
Undo the multiplication by dividing $357$ by $0.17$ . Do not take $17\%$ of $357$ - that would make the answer smaller, when it must be far larger.
Part (b) is a percentage change. Find the increase first, then write it as a fraction of the original amount.
The original amount is the $2019$ figure, so $650$ goes on the bottom of that fraction. Multiply by $100$ to turn it into a percentage.

Worked Solution [5 marks]

Rule - Reverse percentage: if

p\%

N

A

, then

N = \dfrac{100A}{p}

.
Rule - Percentage change:

\text{percentage change} = \dfrac{\text{change}}{\text{original amount}} \times 100

Step 1 - Part (a): write

17\%

N

as an equation

17\% = \dfrac{17}{100} = 0.17

0.17 \times N = 357

(Reason: Per cent means out of

100

, so

17\%

N

0.17 \times N

. Writing the equation first makes it clear that

N

has been multiplied, so a division will undo it.)

Step 2 - Part (a): divide to undo the multiplication

N = \dfrac{357}{0.17}

N = 2100

(Reason: Dividing both sides by

0.17

leaves

N

on its own. Without a calculator the same value comes from

\dfrac{357 \times 100}{17}

, or from

1\%

N

being

\dfrac{357}{17} = 21

Step 3 - Part (b): find the increase

806 - 650 = 156

(Reason: The increase is what has been added on: the

2020

figure minus the

2019

figure.

156

extra members is the change the percentage has to describe.)

Step 4 - Part (b): write the increase as a percentage of the original

\dfrac{156}{650} \times 100 = 24

(Reason: A percentage increase is always measured against the original amount, so the

2019

figure goes on the bottom of the fraction. Multiplying by

100

turns the decimal into a percentage.)

(a)

N = 2100

(b)

24\%

Verification

Check 1: Put

N = 2100

back into part (a) and take

17\%

of it.

0.17 \times 2100 = 357

, the amount the question gives.

Check 2: Reach part (a) a second way, with no decimal at all.

1\%

N

\dfrac{357}{17} = 21

, so

N

100

times that.

21 \times 100 = 2100

, the same value as the division gave.

Check 3: Test part (b) forwards. An increase of

24\%

multiplies the original by

1.24

1.24 \times 650 = 806

, the

2020

membership.

Check 4: Compare the two figures directly instead of using the increase:

\dfrac{806}{650} = 1.24

, so

2020

124\%

2019

124\% - 100\% = 24\%

, so the increase is

24\%

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Part (a): $\dfrac{357}{0.17}$ or $0.17N = 357$ or $\dfrac{17}{100} \times N = 357$ or $\dfrac{357 \times 100}{17}$	M1	A correct calculation for $N$ , or a correct equation in $N$ , or equivalent. Not $17\% \times N = 357$ .	✓
$N = 2100$	A1	cao. A correct answer scores both marks, unless it comes from obviously incorrect working.	✓
Part (b): $806 - 650 = 156$ or $\dfrac{806}{650} = 1.24$	M1	Either the increase, or the multiplier from $2019$ to $2020$ , or equivalent.	✓
$\dfrac{806 - 650}{650} \times 100$ or $1.24 \times 100 = 124$ or $1.24 - 1 = 0.24$	M1	A correct calculation for the percentage increase, or $124$ or $0.24$ seen as the answer or in part of the working.	✓
$24$ per cent	A1	cao. A correct answer scores all three marks, unless it comes from obviously incorrect working.	✓
An answer between $19.3$ and $19.4$ , when no other mark in part (b) has been earned	SCB1	This special case is for one specific error: dividing the increase by the $2020$ figure, $806$ , instead of by the original $650$ . The method is right and only the base is wrong, so it is worth one mark.	✓

Full marks: 5/5

Dylan has made a biased five-sided spinner to use in a board game.
The five sections of the spinner are numbered $1$ , $2$ , $3$ , $4$ , $5$

The table gives the probabilities that, when the spinner is spun, it will land on $2$ or on $3$ or on $5$

The probability that the spinner will land on $1$ is the same as the probability that the spinner will land on $4$

Dylan is going to spin the spinner $400$ times.

Work out an estimate for the number of times the spinner will land on $4$ [4 marks]

[Total 4 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

A biased five-sided spinner, its sections numbered

1

2

3

4

5

Probability of landing on

2

0.14

Probability of landing on

3

0.17

Probability of landing on

5

0.21

Landing on

1

and landing on

4

are equally likely, so those two probabilities are equal

The spinner is going to be spun

400

times

Find

An estimate for the number of times the spinner will land on

4

Plan the Solution

The five sections are the only outcomes there are, so the five probabilities must total $1$ .
Add the three probabilities the table gives, then subtract from $1$ to see how much is left for $1$ and $4$ together.
Halve that leftover, because landing on $1$ and landing on $4$ are equally likely.
Multiply the probability by $400$ to turn it into an expected number of spins.

Worked Solution [4 marks]

Rule - Total probability and expected frequency: the probabilities of all the possible outcomes add up to

1

, and an estimate for the number of times one outcome happens in

n

trials is

p \times n

Step 1: add the three probabilities the table gives

0.14 + 0.17 + 0.21 = 0.52

(Reason: landing on

2

, on

3

and on

5

cannot happen together, so their probabilities simply add)

Step 2: find the probability left over for

1

and

4

1 - 0.52 = 0.48

(Reason: the five sections are the only places the spinner can stop, so all five probabilities must total

1

, and whatever is not already used belongs to

1

and

4

)

Step 3: split that leftover into two equal parts

\dfrac{0.48}{2} = 0.24

(Reason: the spinner is just as likely to land on

1

as on

4

, so the probability left over is shared equally between the two of them)

Step 4: turn that probability into a number of spins

0.24 \times 400 = 96

(Reason: an estimate for how often an outcome happens is its probability multiplied by the number of trials, and here there are

400

trials)

96

times

Verification

Check 1: Put the answer back into the table and add all five probabilities. A complete set of outcomes must total

1

0.14 + 0.17 + 0.24 + 0.21 + 0.24 = 1

Check 2: Work in spins rather than probabilities. Multiply each given probability by

400

, then share the spins that are left between

1

and

4

56 + 68 + 84 + 96 + 96 = 400

Check 3: Run the question backwards: turn the answer back into a probability by dividing by the number of spins, and see whether it matches step 3.

\dfrac{96}{400} = 0.24

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$1 - (0.14 + 0.17 + 0.21) \; (= 0.48)$	M1	Correct use of the fact that the probabilities total $1$ . Accept instead $0.14 + 0.17 + 0.21 + x + x = 1$ oe, or a correct calculation for an estimate for the number of times the spinner lands on $2$ or on $3$ or on $5$ , eg $0.14 \times 400 \; (= 56)$ or $0.52 \times 400 \; (= 208)$ .	✓
$\dfrac{0.48}{2} \; (= 0.24)$	M1	A completely correct method to find the probability that the spinner lands on $4$ - the value may be written straight into the table. Accept instead a completely correct method for the number of times it lands on $1$ or on $4$ , eg $400 - 56 - 68 - 84 \; (= 192)$ oe or $0.48 \times 400 \; (= 192)$ .	✓
$0.24 \times 400$	M1	A correct calculation to find the estimate required, or an answer leading from $96$ seen, eg $\dfrac{96}{400}$ . Accept instead $\dfrac{192}{2}$ .	✓
Correct answer scores full marks, unless it comes from obviously incorrect working	A1	$96$ cao	✓
Special case - halving the total that was just added instead of the total that is left	SC	$B1$ for an answer of $104$ if no other marks have been awarded. It comes from $\dfrac{0.52}{2} = 0.26$ and then $0.26 \times 400 = 104$ , which halves the probability already accounted for rather than the probability still to be shared.	✓

Full marks: 4/4

The diagram shows a solid prism.
The cross-section of the prism is a right-angled triangle.

Work out the total surface area of the prism. [3 marks]

cm²

[Total 3 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

A solid prism whose cross-section is a right-angled triangle with shorter sides

6

cm and

8

cm, and hypotenuse

10

cm.

The prism is

15

cm long.

The figure is not accurately drawn, so every length must come from the labels and not from measuring.

Find

The total surface area of the prism, in square centimetres. Total means every face: the prism is solid, so there are

5

faces on the outside - two triangular ends and three rectangles.

Plan the Solution

Confirm the cross-section really is right-angled, because half base times height only applies then: test whether $6^2 + 8^2 = 10^2$ .
Work out the area of one triangular end, then double it for the two ends.
Sweep each side of the triangle along the prism. Each gives a rectangle $15$ cm long and as wide as that side, so the three widths are $8$ cm, $6$ cm and $10$ cm.
Add the five face areas.
Finish in square centimetres, because an area is being asked for.

Worked Solution [3 marks]

Rule - Surface area of a prism: add the areas of every face. A triangular prism has five of them - two identical triangular ends, and one rectangle for each side of the triangle, each rectangle as long as the prism.

Step 1: check that the cross-section is right-angled

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

10^2 = 100

(Reason: The three sides satisfy Pythagoras, so the

6

cm and

8

cm sides meet at the right angle. That is what makes half base times height the correct area rule here.)

Step 2: the two triangular ends

\dfrac{1}{2} \times 8 \times 6 = 24

2 \times 24 = 48

(Reason: The base and the height are the two sides that meet at the right angle,

8

cm and

6

cm, never the

10

cm hypotenuse. The two ends are identical, so one area is worked out and then doubled.)

Step 3: the three rectangular faces

15 \times 8 = 120

15 \times 6 = 90

15 \times 10 = 150

(Reason: Every rectangle runs the whole length of the prism, so each is

15

cm by the width of the side it came from. The

10

cm side gives the sloping face, which is easy to forget because it faces away in the drawing.)

Step 4: add the areas of all five faces

48 + 120 + 90 + 150 = 408

(Reason: Two triangular ends and three rectangles, each counted exactly once. Adding gives the total surface area in

\text{cm}^2

, because every term is an area.)

408 \text{ cm}^2

Verification

Check 1 - factor the length out: All three rectangles are

15

cm long, so together they cover the perimeter of the triangle multiplied by the length:

(6 + 8 + 10) \times 15

24 \times 15 = 360

for the rectangles, then

360 + 48 = 408

with the two ends - the same total, reached by grouping the faces a different way.

Check 2 - half of a cuboid: Two of these prisms fit together along their sloping faces to make a cuboid

6

cm by

8

cm by

15

cm, whose surface area is

2 \times (6 \times 8) + 2 \times (6 \times 15) + 2 \times (8 \times 15) = 516

. Cutting it in half gives each prism half of that surface, plus the new sloping face the cut creates,

10 \times 15

\dfrac{516}{2} = 258

, then

258 + 150 = 408

- the same total, with no face area added one at a time.

Check 3 - is the size sensible? The whole cuboid that contains the prism has surface area

516

\text{cm}^2

, and the prism is that cuboid with a corner sliced off, so its surface must be smaller than

516

but comfortably larger than the three rectangles alone,

360

360 < 408 < 516

, so the answer sits exactly where a surface area for this solid should.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct method for the areas of two different faces, for example $15 \times 8$ together with $\dfrac{1}{2} \times 8 \times 6$	M1	For a correct method to find the areas of $2$ different faces (that is, not $2$ triangles). Allow $8 \times 6$ as one area, and allow this mark even when the areas themselves are incorrect.	✓
$0.5 \times 8 \times 6 = 24$ (and $\times 2 = 48$ ) oe, with $15 \times 8 = 120$ , $15 \times 6 = 90$ , $15 \times 10 = 150$ , written with the intention of adding them	M1	For adding together $4$ or $5$ values for area (condone $48$ as $1$ or $2$ areas), at least $3$ of which come from a correct method.	✓
$408$	A1	cao. A correct answer scores full marks unless it comes from obviously incorrect working.	✓
An answer of $456$	SC	Special case: B2 for an answer of $456$ if no other marks are awarded. This is the total with the triangular ends left unhalved - each end counted as the whole rectangle, $2 \times 8 \times 6 = 96$ in place of $48$ .	✓
Note - three faces only	Note	$(6 + 8 + 10) \times 15$ (the sum must be seen) accounts for $3$ faces. Award the second method mark for it only when it is clearly not intended as the volume - for example when the area of a triangular end is then added.	✓

Full marks: 3/3

(a) On the grid below, draw each of these three straight lines.
(i) $x = 3$
(ii) $y = 1$
(iii) $x + y = 7$
Write the equation of each line beside the line you have drawn. [3 marks]

(b) On the same grid, shade the one region where all three of these inequalities hold.
$x \geq 3$ , $y \geq 1$ and $x + y \leq 7$
Mark this region with the letter $R$ . [1 mark]

[Total 4 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

A grid on which

x

and

y

both run from

0

8

Three equations to draw:

x = 3

y = 1

and

x + y = 7

Three inequalities to satisfy at the same time:

x \geq 3

y \geq 1

and

x + y \leq 7

Find

(a) The three straight lines, drawn on the grid, each labelled with its equation. (b) The one region where all three inequalities hold, shaded and marked

R

Plan the Solution

An equation with only $x$ in it gives a vertical line and an equation with only $y$ in it gives a horizontal line, so draw those two straight away.
For $x + y = 7$ , work out where it crosses each axis and join those two points with a ruler.
Decide which side of each line to keep by testing one point that is clearly on one side of it.
The region wanted is where the three kept sides overlap. Find the corners where the boundaries cross, shade the overlap, and write $R$ inside it.

Worked Solution [4 marks]

Rule - draw the boundary from the equation, then choose the side with a test point. An inequality written with

\geq

\leq

includes its own boundary, so the lines themselves belong to the region.

Step 1: Draw

x = 3

x = 3 \text{ passes through } (3, 0) \text{ and } (3, 8)

(Reason: Every point whose

x

-coordinate is

3

lies on this line, whatever

y

is, so the line is vertical and runs the full height of the grid.)

Step 2: Draw

y = 1

y = 1 \text{ passes through } (0, 1) \text{ and } (8, 1)

(Reason: This time the

y

-coordinate is fixed and

x

is free, so the line is horizontal, one square above the

x

-axis.)

Step 3: Draw

x + y = 7

x = 0 \text{ gives } y = 7

y = 0 \text{ gives } x = 7

\text{join } (0, 7) \text{ to } (7, 0)

(Reason: Two points fix a straight line, and the two axis crossings are the easiest pair to read off. The gradient is

-1

, so the line drops one square for every square it moves across, which is a quick way to check it.)

Step 4: Choose the side of each line to keep

x \geq 3 \text{: keep the side to the right of } x = 3

y \geq 1 \text{: keep the side above } y = 1

x + y \leq 7 \text{: keep the side below } x + y = 7

(Reason: Test the point

(4, 2)

4 \geq 3

is true,

2 \geq 1

is true, and

4 + 2 = 6

, which is not more than

7

. One point on the correct side of all three lines is enough to fix which side each inequality keeps.)

Step 5: Find the corners of the region

x = 3 \text{ and } y = 1 \text{ meet at } (3, 1)

y = 1 \text{ and } x + y = 7 \text{ meet where } x = 6, \text{ so } (6, 1)

x = 3 \text{ and } x + y = 7 \text{ meet where } y = 4, \text{ so } (3, 4)

(Reason: A corner sits where two of the boundaries cross, so take the three lines in pairs. All three inequalities allow equality, so the edges and the corners are part of the region rather than excluded from it.)

Step 6: Shade the region and label it

R

\text{the triangle } (3, 1), (6, 1), (3, 4)

(Reason: The three kept sides overlap in a single triangle. Shade it and write

R

inside, so there is no doubt which region is meant.)

(a)

x = 3

vertical,

y = 1

horizontal,

x + y = 7

from

(0, 7)

(7, 0)

(b)

R

is the triangle with corners

(3, 1)

(6, 1)

and

(3, 4)

Verification

Check 1: Take a point inside the shading,

(4, 2)

, and test all three inequalities:

4 \geq 3

2 \geq 1

and

4 + 2 = 6

, which is at most

7

. All three hold, so

(4, 2)

does belong to

R

Check 2: Take a point just outside the sloping edge,

(5, 3)

. Now

5 \geq 3

and

3 \geq 1

still hold, but

5 + 3 = 8

8

is more than

7

, so

(5, 3)

is correctly left outside the shading.

Check 3: Count the shape instead. The base runs from

(3, 1)

(6, 1)

, so it is

3

squares long, and the height from

(3, 1)

up to

(3, 4)

3

squares. Area

\dfrac{1}{2} \times 3 \times 3 = 4.5

squares, which is what counting the squares inside the shading gives.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) The line $x = 3$ drawn	B1	A vertical line through $3$ on the $x$ -axis. Dashed or solid is accepted, of minimum length $2$ squares.	✓
(a)(ii) The line $y = 1$ drawn	B1	A horizontal line through $1$ on the $y$ -axis. Dashed or solid is accepted, of minimum length $2$ squares.	✓
(a)(iii) The line $x + y = 7$ drawn	B1	The line joining $(0, 7)$ to $(7, 0)$ . Dashed or solid is accepted, of minimum length $2$ squares.	✓
Note on labelling in part (a)	-	A missing label is condoned when the line is unambiguous. Where the lines are unlabelled the scheme awards on what has been drawn: $x = 3$ and $y = 3$ score B1 B0; $y = 1$ and $x = 1$ score B0 B1; $x = 3$ , $x = 1$ and $y = 1$ score B0 B1; $x = 3$ , $y = 1$ and $y = 3$ score B1 B0; and $x = 3$ , $x = 1$ , $y = 1$ and $y = 3$ score B0 B0.	✓
(b) The correct region shaded and labelled $R$	B1	Shaded in or shaded out, labelled $R$ or with a clear intention that it is the region wanted. Follow through is allowed only for one vertical line (not $x = 0$ ), one horizontal line (not $y = 0$ ) and one line of negative gradient, for example $x = 1$ , $y = 3$ and $x + y = 7$ .	✓
Note on the shading in part (b)	-	Lines must be at least $2$ cm long, and the shaded area must be fully enclosed before the mark in part (b) can be given.	✓

Full marks: 4/4

Rosie puts $4$ apples in a basket.
The mean weight of the $4$ apples in the basket is $145$ grams.

Martin puts one more apple into the basket.
The mean weight of the $5$ apples in the basket is $142$ grams.

Work out the weight of the apple that Martin puts into the basket. [3 marks]

grams

[Total 3 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

A basket holds

4

apples whose mean weight is

145

grams.

One more apple is put in, and the mean weight of the

5

apples is then

142

grams.

Nothing is said about any individual apple. The two means are the only facts given.

Find

The weight, in grams, of the apple that Martin adds. A mean on its own says nothing about one single item, so each mean has to be turned back into a TOTAL weight before the extra apple can be picked out.

Plan the Solution

Rearrange the mean formula into $\text{total} = \text{mean} \times \text{number of apples}$ .
Use it once with $4$ apples and a mean of $145$ grams, then again with $5$ apples and a mean of $142$ grams. The number of apples changes as well as the mean.
Subtract the first total from the second. The only difference between the two baskets is the apple Martin added, so that difference is its weight.
Sanity-check the size at the end: the mean fell when the apple went in, so the new apple must be lighter than $142$ grams.

Worked Solution [3 marks]

Rule - Mean and total:

\text{mean} = \dfrac{\text{total weight}}{\text{number of apples}}

, so

\text{total weight} = \text{mean} \times \text{number of apples}

. A mean is a total in disguise, and going back to totals is what makes two groups of different sizes comparable.

Step 1: the total weight of the first 4 apples

4 \times 145 = 580

(Reason: Four apples averaging

145

grams weigh the same altogether as four apples of exactly

145

grams each, so the total is

580

grams however the four are shared out.)

Step 2: the total weight of all 5 apples

5 \times 142 = 710

(Reason: The second mean is a mean of

5

apples, not

4

, so the count in the multiplication changes as well as the mean itself.)

Step 3: subtract the two totals

710 - 580 = 130

(Reason: The two baskets hold the same first four apples, so everything they share cancels in the subtraction. What is left over is exactly the apple Martin added, in grams.)

130 \text{ grams}

Verification

Check 1 - put the answer back in: Add the answer to the first total and take the mean of the

5

apples:

580 + 130 = 710

, then divide by

5

\dfrac{710}{5} = 142

, which is the mean the question states for the

5

apples.

Check 2 - balance the weights about the new mean: Distances from the new mean must cancel out. Each of the first four apples averages

145 - 142 = 3

grams ABOVE the new mean, which is

4 \times 3 = 12

grams of surplus between them, so the new apple must sit

12

grams BELOW it.

142 - 12 = 130

grams - the same answer, with no total worked out anywhere.

Check 3 - is the size sensible? Adding the apple pulled the mean DOWN, from

145

grams to

142

grams, so the new apple has to weigh less than the new mean - and it must of course be a positive weight.

0 < 130 < 142

, so the answer sits exactly where an apple that drags the mean down should sit. An answer above

142

would have been impossible.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$4 \times 145$ $(= 580)$ or $5 \times 142$ $(= 710)$ , or $\dfrac{145 + 145 + 145 + 145 + x}{5} = 142$ oe	M1	For one correct product, or for a correct equation for the weight of the last apple.	✓
$5 \times 142 - 4 \times 145$ or $710 - 580$ using their own two totals, or $145 + 145 + 145 + 145 + x = 5 \times 142$	M1	For a fully correct method to find the weight of the fifth apple, or a fully correct equation to find the missing weight with no denominator.	✓
$130$	A1	cao	✓
Note - a correct answer on its own	Note	A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

Meera puts $20\,000$ euros into a savings bond for $3$ years.
The bond pays $3.5\%$ per year compound interest.

Work out how much money Meera will have in the bond at the end of the $3$ years.
Give your answer correct to the nearest euro. [3 marks]

euros

[Total 3 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

Amount put into the bond:

20\,000

euros

Compound interest:

3.5\%

per year

Time in the bond:

3

years

Find

The value of the bond after

3

years, correct to the nearest euro

Plan the Solution

Compound interest adds the same percentage each year to whatever is in the bond at the time, so turn $3.5\%$ growth into a single multiplier for one year.
One year is one multiplication, so $3$ years is that multiplier to the power $3$ .
Multiply the $20\,000$ euros by the three-year multiplier, then round the result to the nearest whole euro.

Worked Solution [3 marks]

Compound interest:

\text{final amount} = \text{starting amount} \times \text{multiplier}^{n}

, where

n

is the number of years and the multiplier is

1

plus the yearly rate as a decimal.

Step 1: turn the percentage into a one-year multiplier

100\% + 3.5\% = 103.5\%

\text{multiplier} = \dfrac{103.5}{100} = 1.035

(Reason: (Reason: the bond keeps what it already had and adds

3.5\%

on top, so each year's value is

103.5\%

of the year before. Writing that as

1.035

replaces a two-stage percentage calculation with one multiplication.))

Step 2: raise the multiplier to the power 3

1.035^{3} = 1.108717875

(Reason: (Reason: the money stays in the bond for

3

years, so it is multiplied by

1.035

three times. Keep every decimal place here; rounding this multiplier is what produces the wrong answers

22\,160

and

22\,180

.))

Step 3: multiply the amount invested by the three-year multiplier

20\,000 \times 1.035^{3} = 20\,000 \times 1.108717875 = 22\,174.3575

(Reason: (Reason: the multiplier scales the whole balance, so the

20\,000

euros is multiplied by it once for the whole three-year period.))

Step 4: round to the nearest euro

22\,174.3575 \approx 22\,174

(Reason: (Reason: the question asks for the nearest euro, and the first decimal place is

3

, which is less than

5

, so the euros digit stays as it is.))

22\,174

euros

Verification

Check 1: Grow the bond one year at a time instead of using a power:

20\,000 \times 1.035 = 20\,700

, then

20\,700 \times 1.035 = 21\,424.5

, then

21\,424.5 \times 1.035 = 22\,174.3575

. the same total,

22\,174.3575

, so the power of

3

was applied the right number of times

Check 2: Look at the interest on its own. The bond has earned

22\,174.3575 - 20\,000 = 2\,174.3575

euros. Simple interest at the same rate would earn only

3 \times 700 = 2\,100

euros, so compound interest must come out a little higher.

2\,174.3575 > 2\,100

, and the extra

74.3575

euros is the interest earned on earlier interest

Check 3: Work backwards. Dividing the final amount by the three-year multiplier must return the amount that was put in:

\dfrac{22\,174.3575}{1.108717875} = 20\,000

. the original

20\,000

euros comes back exactly, so nothing was multiplied one time too many or too few

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$20\,000 \times 1.035 = 20\,700$ oe, or $20\,000 \times 0.035 = 700$ oe	M1	For finding $103.5\%$ of $20\,000$ , or $3.5\%$ of $20\,000$ . Accept $(1 + \dfrac{3.5}{100})$ for $1.035$ , but not $(1 + 3.5\%)$ .	✓
$20\,700 \times 1.035 = 21\,424.5$ , then $21\,424.5 \times 1.035$ oe	M1	Dependent on the first method mark, for a complete method that carries the multiplier through all three years.	✓
The value of the bond after $3$ years, to the nearest euro	A1	$22\,174$ . Allow $22\,174$ to $22\,175$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
Both method marks earned in one line	Note	M2 for $20\,000 \times 1.035^{3}$ , where $1.035^{3} = 1.108717875$ . M2 is also earned by $20\,000 \times 1.035^{4} = 22\,950.4600125$ , which is a complete method run for four years instead of three, so the A mark is lost.	✓
The interest given instead of the total	Note	If the correct answer is seen and $20\,000$ is then subtracted to leave $2\,174$ to $2\,175$ , award full marks. The same range with no working gains 2 marks.	✓
Misread: $2000$ invested in place of $20\,000$	SC	B2 for $2000 \times 1.035^{3} = 2217.43575$ . The compound-interest method is completely right and only the amount invested was misread.	✓
The three-year multiplier cut short to $3$ decimal places	SC	B2 for $22\,160$ , from $20\,000 \times 1.108$ , or B2 for $22\,180$ , from $20\,000 \times 1.109$ . Both come from rounding $1.108717875$ before multiplying.	✓
One recognisable piece of the method only	SC	B1 if no other marks are awarded and any of these is seen, whether or not it is given as the answer: $20\,000 \times 1.035^{n}$ , $20\,000 \times 0.965^{3} = 17\,972.6425$ (a decrease instead of an increase), $20\,000 \times 0.105 = 2\,100$ , $20\,000 \times 1.105 = 22\,100$ , or $20\,000 \times 1.105^{2}$ .	✓

Full marks: 3/3

A company has two offices, Riverside and Eastgate.
Everyone who works at the Riverside office and everyone who works at the Eastgate office named their preferred way of travelling to work from bus, bicycle and walking.

The pie chart shows information about the results for the Riverside office.
The table shows information about the results for the Eastgate office.

There are $300$ people who work at the Riverside office.
There are $320$ people who work at the Eastgate office.

More people at the Riverside office than at the Eastgate office said bus was their preferred way of travelling to work.

How many more? [5 marks]

[Total 5 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

Riverside office:

300

people, shown on a pie chart with sector angles bus

126

, bicycle

102

and walking

132

degrees.

Eastgate office:

320

people, given in a table as bus

3x + 6

, bicycle

5x + 8

and walking

7x - 9

The three ways of travelling are the only choices, so in each office they account for everybody.

Find

How many more people chose the bus at the Riverside office than at the Eastgate office. Two figures are needed before that subtraction can happen: one read from the table, one read from the pie chart.

Plan the Solution

The three Eastgate expressions must add up to $320$ , so form an equation and solve it for $x$ .
Substitute that value back into the bus row, $3x + 6$ , to get the Eastgate bus figure.
For Riverside, the bus sector is $126$ degrees out of $360$ , so take that fraction of $300$ people.
Subtract the smaller figure from the larger one.

Worked Solution [5 marks]

Rule - Pie chart: a category's frequency is its angle out of

360

degrees, multiplied by the total. And when the parts of a total are given as expressions in

x

, add the parts and set the sum equal to that total.

Step 1: add the three Eastgate expressions and set the total to

320

(3x + 6) + (5x + 8) + (7x - 9) = 320

15x + 5 = 320

(Reason: Everyone at the Eastgate office chose exactly one way of travelling, so the three expressions must account for all

320

people. Collecting like terms gives

3x + 5x + 7x = 15x

and

6 + 8 - 9 = 5

Step 2: solve the equation for

x

15x = 320 - 5 = 315

x = \dfrac{315}{15} = 21

(Reason: Take

5

from both sides, then divide both sides by

15

. A whole number is expected here, because it counts people.)

Step 3: work out how many at the Eastgate office chose the bus

3 \times 21 + 6 = 69

(Reason: The bus row of the table reads

3x + 6

, so it is that expression, and no other row, that

x = 21

goes into.)

Step 4: work out how many at the Riverside office chose the bus

\dfrac{126}{360} \times 300 = 105

(Reason: The whole pie chart is

360

degrees and stands for all

300

people, so the bus sector of

126

degrees stands for the same fraction of those

300

people.)

Step 5: subtract to find how many more

105 - 69 = 36

(Reason: The question asks how many more chose the bus at Riverside, so the Eastgate figure comes off the Riverside figure.)

36

more people

Verification

Check 1: With

x = 21

the three Eastgate rows come to

69

113

and

138

people.

69 + 113 + 138 = 320

, the number of people at the Eastgate office.

Check 2: Turn the other two Riverside sectors into people the same way:

\dfrac{102}{360} \times 300 = 85

and

\dfrac{132}{360} \times 300 = 110

105 + 85 + 110 = 300

, the number of people at the Riverside office.

Check 3: Do the Riverside step a completely different way. The whole chart is

360

degrees for

300

people, which is

1.2

degrees per person, so divide the bus angle by that.

\dfrac{126}{1.2} = 105

, the same Riverside bus figure as before.

Check 4: Put the difference back on to the smaller of the two groups.

69 + 36 = 105

, which is the Riverside bus group, so the subtraction was the right way round.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$3x + 6 + 5x + 8 + 7x - 9 = 320$ oe, for example $15x + 5 = 320$	M1	A correct method to find the correct value of $x$ for the Eastgate office, for example an equation. It can be implied by $\dfrac{320 - 5}{15}$ oe.	✓
$(x =) 21$ or $(3x =) 63$	A1	For the correct value of $x$ or of $3x$ . A correct answer of $21$ or $63$ on its own scores these first two marks, unless it comes from obviously incorrect working.	✓
$3 \times 21 + 6 = 69$ or $63 + 6 = 69$	M1ft	Dependent on the first M1. A correct method for the number at the Eastgate office who chose the bus, following through their value of $x$ as long as only one value is offered and it is clearly intended as $x$ . Look for $69$ written beside the table.	✓
$\dfrac{126}{360} \times 300 = 105$ oe, for example $\dfrac{300}{360} = \dfrac{5}{6}$ then $\dfrac{5}{6} \times 126 = 105$ , or $\dfrac{360}{300} = 1.2$ then $\dfrac{126}{1.2} = 105$	M1	Independent of the marks above. A correct method for the number at the Riverside office whose preferred way of travelling is the bus. Because $\dfrac{300}{360}$ is $0.83$ recurring, a rounded $0.83$ is allowed here.	✓
$105 - 69 = 36$	A1	cao. Dependent on the A1 already scored, so the $36$ must come from a correct $x$ .	✓

Full marks: 5/5

The diagram shows a regular pentagon $ABCDE$ joined to a regular hexagon $DEFGHI$ along the edge $DE$ that the two shapes share.

$AF$ is a straight line.

Work out the size of angle $EAF$ . [5 marks]

[Total 5 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

ABCDE

is a regular pentagon

DEFGHI

is a regular hexagon

The two shapes are joined along the edge

DE

, so every edge in the figure is the same length

AF

is a straight line

Find

The size of angle

EAF

, in degrees

Plan the Solution

Work out the interior angle of a regular pentagon, then the interior angle of a regular hexagon.
Three angles meet at $E$ : the pentagon's angle $AED$ , the hexagon's angle $DEF$ , and angle $AEF$ underneath them. They fill one full turn, so subtracting gives angle $AEF$ .
Angle $EAF$ is a base angle of triangle $AEF$ , which is isosceles. Take angle $AEF$ out of the triangle's angle sum, then halve what is left.

Worked Solution [5 marks]

Rule - interior angle of a regular polygon with

n

sides:

\dfrac{(n - 2) \times 180}{n}

degrees. Angles at a point add to

360

degrees, the angles of a triangle add to

180

degrees, and the two base angles of an isosceles triangle are equal.

Step 1: the interior angle of the regular pentagon

ABCDE

\dfrac{(5 - 2) \times 180}{5} = \dfrac{540}{5} = 108

(Reason: A pentagon can be cut into three triangles from one vertex, so its five interior angles add to

(5 - 2) \times 180 = 540

degrees. The pentagon is regular, so those five angles are equal, and angle

AED

is one of them.)

Step 2: the interior angle of the regular hexagon

DEFGHI

\dfrac{(6 - 2) \times 180}{6} = \dfrac{720}{6} = 120

(Reason: A hexagon can be cut into four triangles, so its interior angles add to

(6 - 2) \times 180 = 720

degrees. The hexagon is regular, so divide by the six equal angles it has to get the size of angle

DEF

Step 3: the angle

AEF

that is left at

E

360 - 108 - 120 = 132

(Reason: Angle

AED

in the pentagon, angle

DEF

in the hexagon and angle

AEF

below them fit round the point

E

with no gap and no overlap, so all three together make one full turn of

360

degrees. What is left is obtuse, which is what the figure suggests.)

Step 4: what is left for the other two angles of triangle

AEF

180 - 132 = 48

(Reason: The three angles of triangle

AEF

add to

180

degrees, so angle

EAF

and angle

EFA

share the

48

degrees that are left over.)

Step 5: halve it, because the triangle is isosceles

\dfrac{48}{2} = 24

(Reason:

AE

is an edge of the pentagon and

EF

is an edge of the hexagon, and the two shapes are joined along the edge

DE

, so every edge in the figure is the same length. That makes

AE

and

EF

equal, so triangle

AEF

is isosceles and its base angles

EAF

and

EFA

are equal. Each one is half of what was left.)

Angle

EAF

24

Verification

Check 1: Reach angle

AEF

a second way, from exterior angles. A regular pentagon has exterior angle

\dfrac{360}{5} = 72

degrees and a regular hexagon has exterior angle

\dfrac{360}{6} = 60

degrees. Taking each interior angle off the full turn leaves exactly those two exterior angles.

72 + 60 = 132

Check 2: Put the three angles at

E

back together. If they do not close the full turn, one of them is wrong.

108 + 120 + 132 = 360

Check 3: Put the answer back into triangle

AEF

. Both base angles are

24

degrees, so with the obtuse angle they must come to the angle sum of a triangle.

24 + 24 + 132 = 180

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Interior angle of the pentagon $\dfrac{3 \times 180}{5} = 108$ , or exterior angle of the pentagon $\dfrac{360}{5} = 72$	M1	Allow this in the working, but not if it is labelled in the wrong place on the diagram, unless the candidate clearly starts again. Any equivalent method scores.	✓
Interior angle of the hexagon $\dfrac{4 \times 180}{6} = 120$ , or exterior angle of the hexagon $\dfrac{360}{6} = 60$	M1	Allow this in the working, but not if it is labelled in the wrong place on the diagram, unless the candidate clearly starts again. Any equivalent method scores.	✓
A complete method for the obtuse angle $AEF$ : $360 - (108 + 120) = 132$ , or $72 + 60 = 132$ , or $(180 - 108) + (180 - 120) = 132$	M1	A fully correct method for the size of the obtuse angle $AEF$ , but not if it is labelled in the wrong place on the diagram. A value carried forward must come from correct working.	✓
A complete method for angle $EAF$ : $\dfrac{180 - 132}{2} = 24$ , or $\dfrac{180 - (72 + 60)}{2} = 24$	M1	A fully correct method for the size of angle $EAF$ . A value carried forward must come from correct working.	✓
Answer $24$	A1	Correct answer only. A correct answer scores full marks unless it clearly comes from incorrect working.	✓

Full marks: 5/5

(a) Solve $\dfrac{3x + 2}{5} - \dfrac{2x + 1}{3} = x$
You must show clear algebraic working. [3 marks]

(b) Rearrange the formula $f = \sqrt{\dfrac{a + bc}{c - d}}$ to make $c$ the subject. [4 marks]

(a) x =(b)

[Total 7 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

(a) The equation

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

, whose denominators are

5

and

3

(b) The formula

f = \sqrt{\dfrac{a + bc}{c - d}}

, in which

c

appears twice, once on the top of the fraction and once on the bottom.

Find

(a) The value of

x

, with the algebra shown. (b) A formula for

c

in terms of

a

b

d

and

f

Plan the Solution

(a) Multiply every term by $15$ , the lowest common multiple of $5$ and $3$ . The $x$ on the right must be multiplied too.
(a) Expand the brackets, collect the $x$ terms on one side and the numbers on the other, then divide.
(b) Square both sides first. The square root is the outermost operation, so undoing it is the opening move.
(b) Multiply by $c - d$ to clear the fraction, expand, then gather every term containing $c$ on one side and everything else on the other.
(b) Factorise the $c$ out of that side and divide by the bracket. Because $c$ appears twice, it can only be made the subject after factorising.

Worked Solution [7 marks]

Rule - Clearing denominators: multiply every term by the lowest common denominator (or, for a square root, square first and then multiply by the denominator); then collect every term in the unknown on one side and factorise it out.

Step 1 (a): multiply every term by

15

15 \times \dfrac{3x + 2}{5} - 15 \times \dfrac{2x + 1}{3} = 15 \times x

3(3x + 2) - 5(2x + 1) = 15x

(Reason: (Reason:

15

is the lowest common multiple of

5

and

3

, so each denominator cancels.

\dfrac{15}{5} = 3

is what the first bracket is multiplied by, and

\dfrac{15}{3} = 5

is what the second bracket is multiplied by.))

Step 2 (a): expand both brackets

9x + 6 - 10x - 5 = 15x

(Reason: (Reason: the second bracket is being subtracted, so

-5

multiplies both terms inside it:

-5 \times 2x = -10x

and

-5 \times 1 = -5

. Losing that second sign is the commonest slip here.))

Step 3 (a): collect like terms on the left

-x + 1 = 15x

(Reason: (Reason:

9x - 10x = -x

and

6 - 5 = 1

. The equation now has no brackets and no fractions, which is what the second method mark is for.))

Step 4 (a): gather the

x

terms

1 = 15x + x

1 = 16x

(Reason: (Reason: adding

x

to both sides keeps every

x

term positive, so

15x + x = 16x

.))

Step 5 (a): divide by

16

x = \dfrac{1}{16}

(Reason: (Reason: dividing both sides by

16

leaves

x

on its own. As a decimal this is

0.0625

, which the mark scheme also accepts.))

Step 6 (b): square both sides

f^{2} = \dfrac{a + bc}{c - d}

(Reason: (Reason: squaring is the inverse of taking a square root, so the root disappears and the fraction underneath is left untouched.))

Step 7 (b): multiply by

c - d

and expand

f^{2}(c - d) = a + bc

cf^{2} - df^{2} = a + bc

(Reason: (Reason: multiplying both sides by the denominator clears the fraction. Expanding puts the

c

term into the open, which is what the next step needs.))

Step 8 (b): put the

c

terms on one side

cf^{2} - bc = a + df^{2}

(Reason: (Reason: subtract

bc

from both sides and add

df^{2}

to both sides. Everything containing

c

is now on the left and everything else is on the right.))

Step 9 (b): factorise and divide

c(f^{2} - b) = a + df^{2}

c = \dfrac{a + df^{2}}{f^{2} - b}

(Reason: (Reason:

c

is a common factor of both terms on the left, so it can be taken outside a bracket. Dividing by that bracket makes

c

the subject.))

(a)

x = \dfrac{1}{16}

(b)

c = \dfrac{a + df^{2}}{f^{2} - b}

Verification

Check 1: (a) Put

x = \dfrac{1}{16}

back into the left-hand side. The numerators become

3x + 2 = \dfrac{35}{16}

and

2x + 1 = \dfrac{9}{8}

, so the two fractions are

\dfrac{7}{16}

and

\dfrac{6}{16}

\dfrac{7}{16} - \dfrac{6}{16} = \dfrac{1}{16}

, which is exactly the value of

x

Check 2: (a) Work the same equation in decimals with

x = 0.0625

: the numerators are

2.1875

and

1.125

, giving

\dfrac{2.1875}{5} = 0.4375

and

\dfrac{1.125}{3} = 0.375

0.4375 - 0.375 = 0.0625

, the same answer reached without fractions

Check 3: (b) Test the rearrangement on numbers. Take

a = 2

b = 3

d = 1

and

f = 4

. The new formula gives

c = \dfrac{2 + 1 \times 16}{16 - 3} = \dfrac{18}{13}

. Feeding that back into the original gives

a + bc = \dfrac{80}{13}

and

c - d = \dfrac{5}{13}

\sqrt{\dfrac{80}{5}} = \sqrt{16} = 4

, the value of

f

that was chosen at the start

Check 4: (b) Multiply the numerator and the denominator of the answer by

-1

. A fraction is unchanged when both parts change sign, so the two forms must be the same formula.

c = \dfrac{-a - df^{2}}{b - f^{2}}

, which is the alternative form the mark scheme allows

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Writing both fractions over the common denominator $15$ , or multiplying every term by $15$ to remove the denominators, or splitting the left-hand side into $\dfrac{3}{5}x + \dfrac{2}{5} - \dfrac{2}{3}x - \dfrac{1}{3}$	M1	A correct method for dealing with the two denominators. If the student expands at this stage, allow one of the four terms on the left to be wrong. If the denominators are removed here, a correct method must be shown or implied.	✓
(a) An equation with no brackets and no fractions, e.g. $9x + 6 - 10x - 5 = 15x$ or $1 - x = 15x$	M1	Or an equation with a common denominator for all terms and the numerators simplified, e.g. $\dfrac{-x + 1}{15} = \dfrac{15x}{15}$ . Allow one error in the four terms on the left across the two method marks, and follow through for the simplifying.	✓
(a) $x = \dfrac{1}{16}$	A1	Or equivalent, e.g. $0.0625$ (allow $0.062$ or $0.063$ ). Working is required.	✓
(b) Squaring both sides: $f^{2} = \dfrac{a + bc}{c - d}$	M1	For squaring both sides in a correct equation.	✓
(b) Multiplying by the denominator and expanding: $cf^{2} - df^{2} = a + bc$	M1	For multiplying by the denominator and expanding in a correct equation.	✓
(b) Isolating the terms in $c$ : $cf^{2} - bc = a + df^{2}$	M1	For isolating the terms in $c$ on one side and the other terms on the other side, in a correct equation.	✓
(b) $c = \dfrac{a + df^{2}}{f^{2} - b}$	A1	Or equivalent, e.g. $c = \dfrac{-a - df^{2}}{b - f^{2}}$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 7/7

The frequency table gives information about the weights, in kilograms, of $60$ bags of flour on a market stall.

$\begin{array}{|c|c|}\hline \textbf{Weight } (w \textbf{ kilograms}) & \textbf{Frequency} \\ \hline 0 < w \leq 1 & 4 \\ \hline 1 < w \leq 2 & 15 \\ \hline 2 < w \leq 3 & 20 \\ \hline 3 < w \leq 4 & 11 \\ \hline 4 < w \leq 5 & 6 \\ \hline 5 < w \leq 6 & 4 \\ \hline\end{array}$

(a) Complete the cumulative frequency table.

$\begin{array}{|c|c|}\hline \textbf{Weight } (w \textbf{ kilograms}) & \textbf{Cumulative frequency} \\ \hline 0 < w \leq 1 & \phantom{000} \\ \hline 0 < w \leq 2 & \phantom{000} \\ \hline 0 < w \leq 3 & \phantom{000} \\ \hline 0 < w \leq 4 & \phantom{000} \\ \hline 0 < w \leq 5 & \phantom{000} \\ \hline 0 < w \leq 6 & \phantom{000} \\ \hline\end{array}$ [1 mark]

(b) On the grid below, draw a cumulative frequency graph for your table. [2 marks]

(d) Use your graph to find an estimate for the number of these bags that weigh more than $3.7$ kilograms. [2 marks]

[Total 6 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

The weights, in kilograms, of

60

bags of flour, grouped into six classes each

1

kilogram wide.

The frequencies

4

15

20

11

6

4

, which total

60

Find

(a) The cumulative frequency for each class. (b) The cumulative frequency graph. (c) An estimate of the median weight, read from the graph. (d) An estimate of how many bags weigh more than

3.7

kilograms.

Plan the Solution

Add the frequencies down the column. Each running total counts every bag at or below that weight, so the last total must be the whole sample.
Plot each total at the upper end of its class, then join the points. A class is only complete at its top boundary, so that is where its total belongs.
The median is the $30$ th value, because $\dfrac{60}{2} = 30$ . Read across from $30$ and down to the weight axis.
For part (d), read up from $3.7$ to the graph and across. That gives how many bags are $3.7$ kilograms or less, so subtract it from $60$ .

Worked Solution [6 marks]

Rule - Cumulative frequency: each entry is a running total of the frequencies, plotted at the UPPER end of its class. For

n

values the median is read at

\dfrac{n}{2}

on the cumulative frequency axis.

Step 1: add the frequencies down the column

0 + 4 = 4

4 + 15 = 19

19 + 20 = 39

39 + 11 = 50

50 + 6 = 56

56 + 4 = 60

(Reason: Each row of the cumulative table asks how many bags weigh at most that much, so every frequency so far is included. The column reads

4

19

39

50

56

60

Step 2: plot each total at the upper end of its class

(1, 4) \quad (2, 19) \quad (3, 39)

(4, 50) \quad (5, 56) \quad (6, 60)

(Reason: The class

0 < w \leq 1

is only complete at

w = 1

, so its total of

4

belongs at

1

kilogram and not in the middle of the class. The graph is taken back to

(0, 0)

because no bag weighs

0

kilograms or less, then the points are joined with a smooth curve.)

Step 3: read the median at a cumulative frequency of

30

\dfrac{60}{2} = 30

2 + \dfrac{11}{20} = 2.55

(Reason:

19

bags weigh

2

kilograms or less and

39

weigh

3

kilograms or less, so the

30

th value lies in the class

2 < w \leq 3

. Reading across at

30

and down gives about

2.55

kilograms; the mark scheme accepts anything from

2.3

2.7

Step 4: read the graph at

3.7

kilograms

39 + 0.7 \times 11 = 46.7

(Reason: Reading up from

3.7

to the graph and across to the cumulative frequency axis gives about

47

bags weighing

3.7

kilograms or less. The mark scheme allows any reading from

45

48

, which is one square either way.)

Step 5: subtract from the total

60 - 47 = 13

(Reason: The graph gives the number of bags at or below

3.7

kilograms, so take it from

60

to get the number above. The answer counts bags, so it must be a whole number - the mark scheme accepts

12

13

14

15

(a)

4, 19, 39, 50, 56, 60

(b) the six points plotted at the upper end of each class, joined with a curve(c)

2.55

kilograms(d)

13

bags

Verification

Check 1: Add the whole frequency column. The last cumulative total has to be the entire sample, and nothing else can be.

4 + 15 + 20 + 11 + 6 + 4 = 60

, which is the last entry in the cumulative column

Check 2: Test the median against the class it should sit in.

19

bags weigh

2

kilograms or less and

39

weigh

3

kilograms or less, so the

30

th value must lie between

2

and

3

2.55

lies between

2

and

3

, and it is just over halfway, as

30

is just over halfway from

19

39

Check 3: Work part (d) from the frequency column instead of the graph:

0.3

of the

11

bags in

3 < w \leq 4

, plus all

6

and all

4

in the two classes above it.

3.3 + 6 + 4 = 13.3

, which is

13

bags to the nearest whole bag

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The cumulative frequencies $4, 19, 39, 50, 56, 60$	B1	All six entries correct.	✓
(b) The cumulative frequency graph	B2	Fully correct graph: the points plotted at the ends of the intervals and joined with a curve or line segments. If not B2 then B1 for 5 or 6 of their points at the ends of the intervals and joined, or for 5 or 6 points plotted correctly at the ends of the intervals but not joined, or for 5 or 6 points from a table with only one arithmetic error plotted consistently within each interval at their correct heights and joined. Ignore the part from 0 to the first plotted point.	✓
(c) Read across at $30$ and down to the weight axis: $2.55$	B1ft	Any value from $2.3$ to $2.7$ . Follow through an increasing graph, reading across at $30$ or $30.5$ .	✓
(d) A reading taken at $3.7$ kilograms, about $47$	M1	A correct method to take a reading at $3.7$ kg, for example $45$ to $48$ , or a correct value for their own graph, within one square.	✓
$60 - 47 = 13$	A1ft	$12$ , $13$ , $14$ or $15$ . Follow through their increasing graph if the value falls outside that range. It must be a whole number: a non-integer answer earns the M1 only.	✓

Full marks: 6/6

Given that

$\dfrac{3^{2n+3}}{3^{4}} = 3^{3} \times 3^{1-2n}$

work out the value of $n$ .
You must show clear working. [3 marks]

n =

[Total 3 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

\dfrac{3^{2n+3}}{3^{4}} = 3^{3} \times 3^{1-2n}

Every power on the page has the same base,

3

, and every index is linear in

n

Find

The value of

n

Plan the Solution

Use $\dfrac{a^{p}}{a^{q}} = a^{p-q}$ to turn the left-hand side into one power of $3$ .
Use $a^{p} \times a^{q} = a^{p+q}$ to turn the right-hand side into one power of $3$ .
With a single power of $3$ on each side, the two indices must be equal. That gives a linear equation in $n$ to solve.
The answer is not a whole number here, so leave it as an exact fraction rather than rounding.

Worked Solution [3 marks]

Rule - Index laws:

\dfrac{a^{p}}{a^{q}} = a^{p-q}

and

a^{p} \times a^{q} = a^{p+q}

. If

a^{p} = a^{q}

and

a

is a positive number other than

1

, then

p = q

Step 1: make the left-hand side a single power of

3

\dfrac{3^{2n+3}}{3^{4}} = 3^{2n+3-4} = 3^{2n-1}

(Reason: dividing two powers of the same base subtracts the indices, so the

4

comes off the index

2n+3

)

Step 2: make the right-hand side a single power of

3

3^{3} \times 3^{1-2n} = 3^{3+1-2n} = 3^{4-2n}

(Reason: multiplying two powers of the same base adds the indices, and

3+1 = 4

)

Step 3: equate the two indices

3^{2n-1} = 3^{4-2n}

2n - 1 = 4 - 2n

(Reason: the base

3

is now the same on both sides, so the equation can only hold when the indices match)

Step 4: solve the equation for

n

2n + 2n = 4 + 1

4n = 5

n = \dfrac{5}{4}

(Reason: collect the

n

terms on one side and the numbers on the other, then divide by

4

)

n = \dfrac{5}{4} = 1.25

Verification

Check 1: Put

n = \dfrac{5}{4}

into each index. The left-hand index is

2n-1

and the right-hand index is

4-2n

2 \times \dfrac{5}{4} - 1 = \dfrac{3}{2}

and

4 - 2 \times \dfrac{5}{4} = \dfrac{3}{2}

, so the two indices agree.

Check 2: Work each side of the original equation out on a calculator with

n = 1.25

, so the numerator index is

2 \times 1.25 + 3 = 5.5

and the second factor index is

1 - 2 \times 1.25 = -1.5

\dfrac{3^{5.5}}{3^{4}} = 3^{1.5}

and

3^{3} \times 3^{-1.5} = 3^{1.5}

, which is

5.196

3

decimal places on both sides.

Check 3: Reach the answer a second way. Multiply both sides by

3^{4}

first, which clears the fraction and gives one power of

3

on each side straight away.

3^{2n+3} = 3^{8-2n}

, so

2n+3 = 8-2n

, giving

4n = 5

and the same

n = \dfrac{5}{4}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $3^{2n+3-4}\left(= 3^{3} \times 3^{1-2n}\right)$ or $3^{2n-1}\left(= 3^{3} \times 3^{1-2n}\right)$ or $\left(3^{2n+3} =\right)3^{7} \times 3^{1-2n}$ or $\left(\dfrac{3^{2n+3}}{3^{4}} =\right)3^{4-2n}$ or $\left(3^{2n+3} =\right)3^{3} \times 3^{5-2n}$ or $\dfrac{3^{2n}}{3^{4}} = 3^{1-2n}$ (division by $3^{3}$ ). This is not an exhaustive list. Or one of the indices $2n+3-4$ , $3+1-2n$ , $2n-1$ , $4-2n$ .	M1	For one rule of indices used to correctly combine two or more of the given expressions. The part in brackets is not needed, and the working must include algebra. Or for one of the four indices listed, provided it is clear that it applies to the left-hand side or to the right-hand side.	✓
eg $3^{2n+3-4} = 3^{3+1-2n}$ or $3^{2n+3} = 3^{8-2n}$ or $3^{2n-1} = 3^{4-2n}$ or $2n+3-4 = 3+1-2n$ oe eg $2n-1 = 4-2n$	M1	A correct single power of $3$ on both sides, or a correct equation in $n$ without indices. Some students go straight to this line and gain M2.	✓
$n = \dfrac{5}{4}$ , with working shown	A1	oe, dep on M1. Working is required, so an answer of $\dfrac{5}{4}$ on its own does not earn this mark.	✓

Full marks: 3/3

Using algebra, show that the recurring decimal $0.7\dot{6}\dot{3}$ is equal to the fraction $\dfrac{42}{55}$
You must show your working clearly. [2 marks]

[Total 2 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

The recurring decimal

0.7\dot{6}\dot{3}

, which is written out in full as

0.7636363\ldots

The dots sit on the

6

and the

3

, so the block

63

repeats for ever, while the

7

does not repeat.

Find

An algebraic proof that this decimal is exactly

\dfrac{42}{55}

This is a show that question, so the working is the answer: every line has to be written down, and a calculator display on its own earns nothing.

Plan the Solution

Give the decimal a letter, $x$ , so that it can be worked with algebraically.
Multiply by $10$ so that the repeating block starts immediately after the decimal point.
Multiply by $1000$ as well, so a second line has its repeating block in exactly the same position.
Subtract one line from the other. The two recurring tails are identical, so they cancel and leave a whole number.
Solve the equation that is left, then cancel the fraction to its lowest terms.

Worked Solution [2 marks]

Rule - Recurring decimals: let

x

be the decimal, then multiply it by two powers of ten that place the repeating block in the same position and subtract. Multipliers

10^{n}

and

10^{m}

line the tails up when

n - m

is a whole number of blocks and

10^{m}x

has already passed the digits that do not repeat.

Step 1: give the decimal a letter

x = 0.7\dot{6}\dot{3}

x = 0.7636363\ldots

(Reason: Naming the decimal

x

is what turns this into algebra, and the mark scheme insists on algebra for the final mark.)

Step 2: move the repeating block up to the decimal point

10x = 7.636363\ldots

(Reason: Multiplying by

10

shifts every digit one place left, so the

7

that does not repeat is now in front of the point and the tail begins with the block.)

Step 3: make a second line with the block in the same position

1000x = 763.636363\ldots

(Reason: The block

63

is two digits long, so multiplying by a further

100

shifts it by exactly one whole block and the two tails now match digit for digit.)

Step 4: subtract, so the recurring tails cancel

1000x - 10x = 763.636363\ldots - 7.636363\ldots

990x = 756

(Reason: Both decimals end in the same never-ending tail, so the tails subtract away and a whole number is left. This is the whole point of choosing those two multipliers.)

Step 5: solve for x

x = \dfrac{756}{990}

(Reason: Dividing both sides by

990

leaves

x

on its own as a single fraction. The decimal has already become a fraction here, but it is not yet in its lowest terms.)

Step 6: cancel to lowest terms

756 = 18 \times 42

990 = 18 \times 55

\dfrac{756}{990} = \dfrac{18 \times 42}{18 \times 55} = \dfrac{42}{55}

(Reason: The highest common factor of

756

and

990

18

, so cancelling that factor from top and bottom gives the fraction the question asked for.)

0.7\dot{6}\dot{3} = \dfrac{756}{990} = \dfrac{42}{55}

as required

Verification

Check 1: Go the other way. Turn

\dfrac{42}{55}

back into a decimal by long division and read the digits.

\dfrac{42}{55} = 0.7\dot{6}\dot{3}

, the decimal the question started from

Check 2: Use a different pair of multipliers,

100x

and

x

, which the mark scheme also allows. The subtraction now leaves a terminating decimal instead of a whole number.

99x = 75.6

, so

x = \dfrac{75.6}{99} = \dfrac{756}{990} = \dfrac{42}{55}

Check 3: Build the decimal from place value instead of by subtracting:

0.7

plus the recurring part

0.0\dot{6}\dot{3}

, which is

\dfrac{63}{990}

because a two digit block over

99

is shifted one place further along.

\dfrac{7}{10} + \dfrac{63}{990} = \dfrac{693 + 63}{990} = \dfrac{756}{990} = \dfrac{42}{55}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Two multiples of $x$ whose recurring tails match, written down with the intention of subtracting, eg $1000x = 763.63\ldots$ and $10x = 7.63\ldots$	M1	Any pair that subtracts to a whole number or a terminating decimal, eg $10\,000x = 7636.36\ldots$ with $100x = 76.36\ldots$ , or $100x = 76.363\ldots$ with $x = 0.763\ldots$ . If the recurring is not shown, at least one of the numbers must be given to at least $5$ significant figures. The difference itself may be $75.6$ , $756$ , $7560$ and so on.	✓
Complete the working to $\dfrac{42}{55}$ , eg $990x = 756$ leading to $x = \dfrac{756}{990} = \dfrac{42}{55}$	A1	Dependent on the M1, and algebra must be used for this final mark to be awarded. Allow, for instance, $99x = 75.6$ and then $\dfrac{756}{990} = \dfrac{42}{55}$ .	✓
Note - no algebra used	-	A response that reaches $\dfrac{42}{55}$ without using algebra scores a maximum of $1$ mark. Working is required throughout.	✓

Full marks: 2/2

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 2F and Paper 2H, 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 2H, 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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