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Edexcel IGCSE 4MA1/1H, Thursday 15 May 2025: Worked Solutions and Mark Schemes

Q: How many questions are on Edexcel 4MA1/1H, sat Thursday 15 May 2025, 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 1F and Paper 1H, 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

27 Aug 2026

Table of Contents▾

Edexcel International GCSE Mathematics A (4MA1)4MA1/1H - Higher Tier - Thursday 15 May 2025100 marks · 2 hours · Calculator allowed

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/1H (Higher Tier), June 2025 series, sat Thursday 15 May 2025 –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 11 of 25

The table gives information about the distances $100$ members of a swimming club travel to reach the pool.

$\begin{array}{|c|c|}\hline \textbf{Distance } (d \textbf{ km}) & \textbf{Frequency} \\ \hline 0 < d \leq 5 & 26 \\ \hline 5 < d \leq 10 & 40 \\ \hline 10 < d \leq 15 & 16 \\ \hline 15 < d \leq 20 & 10 \\ \hline 20 < d \leq 25 & 8 \\ \hline \end{array}$

(a) Write down the modal class.
[1 mark]
(b) Work out an estimate for the mean distance. [4 marks]

(a)(b) km

[Total 5 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

The distances

d

km travelled by

100

members, grouped into five classes that are each

5

km wide.

The frequencies are

26

40

16

10

and

8

, and they add to

100

Only the class each distance falls in is known, never the distance itself.

Find

(a) The modal class. (b) An estimate for the mean distance, in km.

Plan the Solution

(a) Pick out the class with the highest frequency, and give the class interval as the answer, not the frequency.
(b) Let the midpoint of each class stand for every distance in that class. Multiply each midpoint by its frequency, add the five products, then divide by the total frequency $100$ .
The word estimate is the warning: midpoints have replaced the real distances, so the answer cannot be exact.

Worked Solution [5 marks]

Estimated mean of grouped data: the midpoint

x

of a class stands for every value in it, so the estimate is the total of the

fx

products divided by the total of the frequencies

f

Step 1: Compare the five frequencies

26 \quad 40 \quad 16 \quad 10 \quad 8

(Reason: The mode is what occurs most often, so with grouped data the modal class is the class holding the highest frequency. Here

40

is the largest of the five.)

Step 2: Name the class that frequency belongs to

5 < d \leq 10

(Reason: The answer is the class interval itself, not the frequency

40

and not a single distance. Every class here is

5

km wide, so no class can be top merely by being wider than the rest.)

Step 3: Find the midpoint of each class

\dfrac{0 + 5}{2} = 2.5

\dfrac{5 + 10}{2} = 7.5

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

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

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

(Reason: The table never says where inside a class a distance falls, so the midpoint is used as the best single stand-in for all of them. This is exactly what makes the mean an estimate.)

Step 4: Multiply each midpoint by its frequency

2.5 \times 26 = 65

7.5 \times 40 = 300

12.5 \times 16 = 200

17.5 \times 10 = 175

22.5 \times 8 = 180

(Reason: Each product is the distance one whole class contributes:

26

members at roughly

2.5

km each contribute roughly

65

km between them.)

Step 5: Add the five products

65 + 300 + 200 + 175 + 180 = 920

(Reason: This total is an estimate of the distance travelled by all

100

members together, in km.)

Step 6: Divide by the total frequency

\dfrac{920}{100} = 9.2

(Reason: A mean is a total shared out equally, so the estimated total distance is divided by the number of members,

100

. Divide by the total frequency, never by the

5

classes.)

(a)

5 < d \leq 10

(b)

9.2

Verification

Check 1: Trap the answer between two totals that need no midpoints. Put every distance at the bottom of its class and the estimate is

\dfrac{670}{100} = 6.7

; put every distance at the top of its class and it is

\dfrac{1170}{100} = 11.7

. A midpoint estimate has to land between the two.

6.7 < 9.2 < 11.7

Check 2: Rebuild the mean without ever forming the total

920

. Take

12.5

off every midpoint, leaving

-10

-5

0

5

and

10

. Weighted by the frequencies these total

-330

, and the

12.5

is then added back.

12.5 + \dfrac{-330}{100} = 12.5 - 3.3 = 9.2

Check 3: Test part (a) by the definition that survives unequal classes. The frequency density of

5 < d \leq 10

\dfrac{40}{5} = 8

members per km, and the nearest rival,

0 < d \leq 5

, has

\dfrac{26}{5} = 5.2

members per km. The same class,

5 < d \leq 10

, comes top whether frequencies or densities are compared.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $5 < d \leq 10$	B1	Allow $5$ to $10$ , or $5 < d < 10$ , or $5 \leq d \leq 10$ , or $5 \leq d < 10$ .	✓
(b) $2.5 \times 26 + 7.5 \times 40 + 12.5 \times 16 + 17.5 \times 10 + 22.5 \times 8 \; (= 920)$ or $65 + 300 + 200 + 175 + 180 \; (= 920)$	M2	For at least $4$ correct products added. They need not be evaluated, ie the work may be left in the form $2.5 \times 26 + 7.5 \times 40 + \ldots$ If not M2 then award M1 for consistent use of values within the interval, including end points, for at least $4$ products added. Again they need not be evaluated, ie the work may be left in the form $5 \times 26 + 10 \times 40 + \ldots$ Or award M1 for correct midpoints used for at least $4$ products and not added.	✓
$\dfrac{920}{100}$ , where either value may be the candidate's own	M1	Dependent on at least M1. Allow division by their $\Sigma f$ provided the addition, or a total under the frequency column, is seen.	✓
$9.2$	A1	Or equivalent, eg $9\dfrac{1}{5}$ or $\dfrac{46}{5}$ . A correct answer scores full marks, unless it comes from obvious incorrect working.	✓
Answer of $6.7$ or $9.7$ or $11.7$	SC B2	For an answer of 6.7 or 9.7 or 11.7	✓
Where those three special-case answers come from	Note	Each is one of the totals in the working column divided by $100$ . Lower class bounds give products $0$ , $200$ , $160$ , $150$ , $160$ and a total of $670$ , so $6.7$ . The values $3$ , $8$ , $13$ , $18$ , $23$ give a total of $970$ , so $9.7$ . Upper class bounds give a total of $1170$ , so $11.7$ .	✓

Full marks: 5/5

Elise makes candles.
Each candle costs $6$ euros to make.

Elise packs the candles into boxes to sell.
Each box contains $4$ candles.

Elise sells $80$ boxes of candles for a total of $2160$ euros.

(a) Work out the percentage profit Elise makes.
Show your working clearly. [4 marks]

The height of each candle is $9$ cm, correct to the nearest cm

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

The weight of each candle is $120$ g, correct to the nearest $10$ g

(a) %(b) cm(c) g

[Total 6 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

Each candle costs

6

euros to make.

Each box holds

4

candles.

80

boxes are sold for

2160

euros in total.

The height of a candle is

9

cm, correct to the nearest cm.

The weight of a candle is

120

g, correct to the nearest

10

Find

(a) the percentage profit (b) the lower bound of the height (c) the upper bound of the weight

Plan the Solution

Work out what the whole order cost to make, then compare it with the money taken in.
Percentage profit is measured against the cost, never against the money taken in - so the profit is divided by the cost.
A measurement given to the nearest $1$ cm or the nearest $10$ g can be up to half of that step away from the value written down. Halve the step, then move that far from the stated value.

Worked Solution [6 marks]

Rule - Percentage profit:

\text{profit} = \text{income} - \text{cost}

, then

\text{percentage profit} = \dfrac{\text{profit}}{\text{cost}} \times 100

.
Rule - Bounds: a measurement given to the nearest

u

lies within

\dfrac{u}{2}

of the value written down.

Step 1: what the whole order cost to make

4 \times 6 = 24

24 \times 80 = 1920

(Reason: Each box holds

4

candles at

6

euros each, so filling one box costs

24

euros, and there are

80

boxes.)

Step 2: the profit

2160 - 1920 = 240

(Reason: Profit is what is left of the

2160

euros taken in once the

1920

euros the candles cost to make has been taken off.)

Step 3: the profit as a fraction of the cost

\dfrac{240}{1920} = 0.125

(Reason: Percentage profit compares the profit with what was spent, so the

240

euros of profit is divided by the

1920

euros of cost - not by the money taken in.)

Step 4: write that decimal as a percentage

0.125 \times 100 = 12.5

(Reason: A decimal is turned into a percentage by multiplying it by

100

Step 5: the lower bound of the height

9 - 0.5 = 8.5

(Reason: The height is given to the nearest cm, and half of

1

cm is

0.5

cm. Every height from

8.5

cm up to

9.5

cm is recorded as

9

cm, so

8.5

cm is the smallest it can be.)

Step 6: the upper bound of the weight

120 + 5 = 125

(Reason: The weight is given to the nearest

10

g, and half of

10

g is

5

g. Every weight from

115

g up to

125

g is recorded as

120

g, so

125

g is the upper bound.)

(a)

12.5

%(b)

8.5

cm(c)

125

Verification

Check 1: Do part (a) one box at a time instead of all

80

at once. One box brings in

\dfrac{2160}{80} = 27

euros and costs

24

euros to fill, so it makes

3

euros of profit.

\dfrac{3}{24} \times 100 = 12.5

Check 2: Compare the money taken in with the cost as one multiplier:

\dfrac{2160}{1920} = 1.125

. That says the income is

112.5

per cent OF the cost, and a percentage profit measures how far ABOVE the cost that is.

112.5 - 100 = 12.5

Check 3: Round both bounds back. To the nearest cm,

8.5

cm goes up to

9

cm while

8.4

cm goes down to

8

cm. To the nearest

10

124.9

g goes to

120

g while

125.1

g goes to

130

g. The height cannot be below

8.5

cm, and

125

g is where the weight stops being recorded as

120

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$4 \times 6 \times 80 \ (= 1920)$ or $4 \times 6 \ (= 24)$ or $\dfrac{2160}{80} \ (= 27)$ or $\dfrac{2160}{80 \times 4} \ (= 6.75)$	M1	(a) for method to work out total income or income from one box or expenditure for one box or income per candle	✓
$2160 - 1920 \ (= 240)$ or $\dfrac{2160}{1920} \ (= 1.125)$ or $27 - 24 \ (= 3)$ or $\dfrac{27}{24} \ (= 1.125)$ or $6.75 - 6 \ (= 0.75)$ or $\dfrac{6.75}{6} \ (= 1.125)$	M1	for working out the profit or income divided by expenditure. The candidate's own value from the previous mark may be used in place of $1920$ , $27$ , $24$ or $6.75$	✓
$\dfrac{240}{1920} \ (\times 100)$ or $0.125 \ (\times 100)$ or $\left(\dfrac{2160}{1920} - 1\right)(\times 100)$ or $(1.125 - 1) \ (\times 100)$ or $1.125 \times 100 \ (= 112.5)$	M1	for a method to reach one step from the answer ie getting to $\dfrac{1}{8}$ oe or $0.125$ or $112.5$ . The candidate's own earlier values may be used in place of $240$ , $1920$ , $3$ , $24$ , $0.75$ or $1.125$	✓
$12.5$	A1	Working required	✓
$8.5$	B1	(b) cao	✓
$125$	B1	(c) allow $124.9$ or $124.99\ldots$	✓

Full marks: 6/6

$AB$ and $BC$ are two sides of a regular polygon that has $n$ sides.

Find the value of $n$ .
You must show your working clearly. [4 marks]

n =

[Total 4 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

Three straight lines meet at the point

B

The angle between

BA

and the third line is

148^\circ

The angle between that third line and

BC

50^\circ

AB

and

BC

are two sides of a regular polygon with

n

sides.

Find

The value of

n

, the number of sides of the polygon.

Plan the Solution

The three lines leaving $B$ fill a whole turn, so taking the two marked angles away from $360^\circ$ leaves the polygon's own angle at $B$ .
That angle is an interior angle of the regular polygon. Its exterior angle is whatever is left over to $180^\circ$ .
Every polygon's exterior angles add to $360^\circ$ , and in a regular polygon they are all the same size, so dividing counts how many there are.
The figure is not accurately drawn, so nothing may be measured off it. Only the two marked values are usable.

Worked Solution [4 marks]

Rule - Regular polygon:

\text{interior angle} + \text{exterior angle} = 180^\circ

, and

n = \dfrac{360^\circ}{\text{exterior angle}}

Step 1: Find the polygon's angle at

B

\angle ABC = 360^\circ - (148^\circ + 50^\circ)

\angle ABC = 360^\circ - 198^\circ = 162^\circ

(Reason: The three lines drawn from

B

use up a full turn, and angles at a point add to

360^\circ

. What is left after the two marked angles is the angle inside the polygon.)

Step 2: Turn that interior angle into the exterior angle

180^\circ - 162^\circ = 18^\circ

(Reason: An interior angle and the exterior angle beside it lie along a straight line, so the two of them together make

180^\circ

Step 3: Divide a whole turn by one exterior angle

n = \dfrac{360^\circ}{18^\circ} = 20

(Reason: The exterior angles of any polygon add to

360^\circ

. The polygon is regular, so all

n

of them are equal and the division counts them.)

n = 20

Verification

Check 1: Work back from the answer. A regular polygon with

20

sides has exterior angle

\dfrac{360^\circ}{20} = 18^\circ

, so its interior angle is what is left to

180^\circ

180^\circ - 18^\circ = 162^\circ

Check 2: Use the interior-angle sum instead, which never mentions exterior angles. The angles of a

20

-sided polygon add to

180^\circ \times (20 - 2) = 3240^\circ

, shared equally between

20

corners.

\dfrac{3240^\circ}{20} = 162^\circ

Check 3: Put the three angles at

B

back together and see whether they close the turn.

148^\circ + 50^\circ + 162^\circ = 360^\circ

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $360 - (148 + 50) \; (= 162)$ or $180 - 50 \; (= 130)$ or $180 - 148 \; (= 32)$	M1	for method to interior angle of the polygon or start to the method of finding the exterior angle of the polygon	✓
eg $180 - 162 \; (= 18)$ or $148 - 130 \; (= 18)$ or $50 - 32 \; (= 18)$ or $180(n - 2) = 162n$ or $\dfrac{180(n - 2)}{n} = 162$ The printed scheme puts the first row's value in quotation marks in each of these, so the candidate's own $162$ , $130$ or $32$ is accepted in its place.	M1	for method to find the exterior angle or for setting up an equation using sum of interior angles formula	✓
eg $\dfrac{360}{18}$ or $(n =) \; \dfrac{360}{180 - 162}$ The printed scheme quotes the $18$ and the $162$ here as well, so the candidate's own values are accepted.	M1	for a complete method	✓
$20$ Working required	A1	dep on M1	✓

Full marks: 4/4

$x^{5} \times x^{7} = x^{m}$

(a) Work out the value of $m$ [1 mark]

$\dfrac{y^{8}}{y^{3}} = y^{n}$

(b) Work out the value of $n$ [1 mark]

(a) m =(b) n =(c)

[Total 4 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

x^{5} \times x^{7} = x^{m}

\dfrac{y^{8}}{y^{3}} = y^{n}

The expression

(5a^{4}r^{2})^{3}

: a number and two powers inside one bracket, all raised to the power

3

Find

The value of

m

. The value of

n

(5a^{4}r^{2})^{3}

written as a single simplified term.

Plan the Solution

Parts (a) and (b) use the two index laws for powers of the same letter: a product adds the indices, a quotient subtracts them.
In part (c) three things sit inside the bracket, so cube each of them in turn: the number $5$ , the power $a^{4}$ and the power $r^{2}$ .
Raising a power to a power multiplies the two indices, so the $3$ outside the bracket multiplies the $4$ and the $2$ . It is not added to them.

Worked Solution [4 marks]

Rule - Index laws:

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

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

, and

(kx^{p})^{q} = k^{q}x^{pq}

Step 1: part (a), a product of two powers of

x

x^{5} \times x^{7} = x^{5 + 7} = x^{12}

m = 12

(Reason: Multiplying powers of the same letter adds the indices:

5

factors of

x

followed by

7

more gives

12

factors of

x

altogether.)

Step 2: part (b), a quotient of two powers of

y

\dfrac{y^{8}}{y^{3}} = y^{8 - 3} = y^{5}

n = 5

(Reason: Dividing powers of the same letter subtracts the indices:

3

of the

8

factors of

y

on top cancel with the

3

underneath, leaving

5

Step 3: part (c), cube every factor inside the bracket

(5a^{4}r^{2})^{3} = 5^{3} \times (a^{4})^{3} \times (r^{2})^{3}

(Reason: The bracket holds three factors multiplied together, and the power

3

applies to every one of them, the number

5

included.)

Step 4: part (c), work out each of the three factors

5^{3} = 125

(a^{4})^{3} = a^{4 \times 3} = a^{12}

(r^{2})^{3} = r^{2 \times 3} = r^{6}

(Reason: Cubing the number means

5 \times 5 \times 5

, and raising a power to a power multiplies the two indices.)

Step 5: part (c), put the three factors back together

(5a^{4}r^{2})^{3} = 125a^{12}r^{6}

(Reason: The number and the two powers are multiplied, so they are written side by side as one term with the number in front.)

(a)

m = 12

(b)

n = 5

(c)

125a^{12}r^{6}

Verification

Check 1: Part (a) with a number instead of a letter. Put

x = 2

: the left-hand side is

2^{5} \times 2^{7} = 32 \times 128 = 4096

, and

2^{12} = 4096

m = 12

Check 2: Part (b) the same way. Put

y = 2

: the left-hand side is

\dfrac{256}{8} = 32

, and

2^{5} = 32

n = 5

Check 3: Part (c) with numbers. Put

a = 2

and

r = 3

. Inside the bracket,

5 \times 16 \times 9 = 720

, so the question gives

720^{3} = 373\,248\,000

, while the answer gives

125 \times 4096 \times 729 = 373\,248\,000

. Both sides give

373\,248\,000

Check 4: Part (c) counted rather than calculated. Writing the bracket out three times gives

5 \times 5 \times 5

for the number,

4 + 4 + 4

factors of

a

and

2 + 2 + 2

factors of

r

125a^{12}r^{6}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$12$ . Accept $x^{12}$ .	✓
(b)	B1	$5$ . Accept $y^{5}$ .	✓
(c)	B2	For $125a^{12}r^{6}$ .	✓
(c)	B1	For a product in the form $ka^{p}r^{q}$ where $2$ from $k$ , $p$ or $q$ are correct, eg $5a^{12}r^{6}$ . Allow $125a^{12}$ or $125r^{6}$ or $a^{12}r^{6}$ so as long as not added to any other terms.	✓

Full marks: 4/4

A music shop is having a sale.
In the sale, normal prices are reduced by $28\%$
The sale price of a guitar is $198$ euros.

Work out the normal price of the guitar. [3 marks]

euros

[Total 3 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

In the sale, normal prices are reduced by

28\%

The sale price of the guitar is

198

euros.

Find

The normal price of the guitar - the price before the reduction.

Plan the Solution

The reduction is measured against the normal price, not against the sale price, so start by asking what percentage of the normal price a shopper actually pays: $100\% - 28\%$ .
Turn that percentage into a multiplier, so the sale price is the normal price multiplied by it.
Undo that multiplication by dividing $198$ by the multiplier. This is a reverse percentage, so never take $28\%$ off $198$ and never add it back on.

Worked Solution [3 marks]

Rule - Reverse percentage: after a reduction, sale price = normal price

\times

multiplier, so normal price =

\dfrac{\text{sale price}}{\text{multiplier}}

Step 1: Find what percentage of the normal price is actually paid

100\% - 28\% = 72\%

1 - 0.28 = 0.72

(Reason: The

28\%

comes off the normal price, so the sale price is the

72\%

that is left, and

0.72

is that percentage written as a multiplier.)

Step 2: Write the sale price as a multiplication

0.72n = 198

(Reason: Let

n

be the normal price in euros. Taking

72\%

of it means multiplying it by

0.72

, and the result is the sale price of

198

euros.)

Step 3: Divide by the multiplier to get the normal price

n = \dfrac{198}{0.72} = 275

(Reason: Dividing by the multiplier undoes the multiplication in Step 2, taking the sale price back up to the price the shop charged before the reduction.)

275

euros

Verification

Check 1: Put the answer back into the question and take the reduction off it:

0.28 \times 275 = 77

euros comes off.

275 - 77 = 198

, which is the sale price the question gives.

Check 2: Reach it a different way, one per cent at a time. If

198

euros is

72\%

, then one per cent is

\dfrac{198}{72} = 2.75

euros.

2.75 \times 100 = 275

euros, the same normal price by a method that never uses the multiplier.

Check 3: Is the size sensible? A reduction makes a price smaller, so the normal price must be the larger of the two, and the amount taken off should be a little over a quarter of it.

275

is larger than

198

, and

\dfrac{77}{275} = 0.28

, which is the reduction the question states.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $1 - 0.28 \ (= 0.72)$ oe or $0.72x = 198$ or $100(\%) - 28(\%) \ (= 72(\%))$ or $\dfrac{198}{72} \ (= 2.75)$ oe	M1	for a correct first step	✓
eg $(x =) \ \dfrac{198}{0.72}$ oe or $\dfrac{198}{72} \times 100$ oe or $2.75 \times 100$ . The printed scheme writes the first step's value in quotation marks, so the candidate's own value from the first step may be used here.	M1	for a complete method	✓
$275$ - correct answer scores full marks (unless from obvious incorrect working)	A1	cao	✓

Full marks: 3/3

(a) Solve the equation $x - 4 = \dfrac{3 + 2x}{6}$
You must show clear algebraic working. [3 marks]

(b) (i) Factorise $y^2 - 11y + 30$
[2 marks]
(ii) Hence solve $y^2 - 11y + 30 = 0$ [1 mark]

(a) x =(b)(i)(b)(ii)

[Total 6 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

(a)

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

, a linear equation with a fraction on the right-hand side

(b)

y^2 - 11y + 30

, a quadratic expression whose

y^2

coefficient is

1

Find

(a) the value of

x

(b)(i)

y^2 - 11y + 30

written as a product of two brackets (b)(ii) the values of

y

that make that product zero. A quadratic factorising into two different brackets has two of them.

Plan the Solution

(a) Multiply every term on both sides by $6$ so the fraction disappears, then collect the $x$ terms on one side and the numbers on the other.
(b)(i) Because the $y^2$ coefficient is $1$ , look for two numbers with product $30$ and sum $-11$ . Each becomes the constant in its own bracket.
(b)(ii) The word "Hence" means reuse part (i). A product is zero only when one of the factors is zero, so set each bracket to zero in turn.

Worked Solution [6 marks]

Rule - Clear a fraction by multiplying EVERY term by its denominator. Factorise

y^2 + by + c

by finding two numbers with product

c

and sum

b

Step 1 - Multiply every term by

6

6(x - 4) = 3 + 2x

6x - 24 = 3 + 2x

(Reason: Multiplying both sides by

6

cancels the denominator on the right. The whole of the left-hand side is multiplied, so it is bracketed first and then expanded; the

4

is inside that bracket, so it is multiplied too.)

Step 2 - Collect the

x

terms on one side

6x - 2x = 3 + 24

4x = 27

(Reason: Subtracting

2x

from both sides and adding

24

to both sides leaves one

x

term on the left and one number on the right. This is the rearrangement the second method mark is for.)

Step 3 - Divide by

4

x = \dfrac{27}{4} = 6.75

(Reason:

27

does not divide exactly by

4

, so the answer is left as the fraction

\dfrac{27}{4}

. The decimal

6.75

is the same value and is equally acceptable.)

Step 4 - Find two numbers with product

30

and sum

-11

(-6) \times (-5) = 30

(-6) + (-5) = -11

(Reason: The product is positive and the sum is negative, so both numbers are negative. The negative pairs multiplying to

30

are

-1

and

-30

-2

and

-15

-3

and

-10

-5

and

-6

. Only the last pair adds to

-11

Step 5 - Write the two brackets

y^2 - 11y + 30 = (y - 6)(y - 5)

(Reason: Each number found becomes the constant in its own bracket, and each bracket starts with

y

because the

y^2

coefficient is

1

Step 6 - Set each bracket to zero

(y - 6)(y - 5) = 0

y - 6 = 0 \quad \text{or} \quad y - 5 = 0

y = 6 \quad \text{or} \quad y = 5

(Reason: Two numbers multiply to zero only when at least one of them is zero, so each bracket is taken to be zero in turn. Solving the two one-step equations gives the two solutions.)

(a)

x = \dfrac{27}{4} = 6.75

(b)(i)

(y - 6)(y - 5)

(b)(ii)

y = 6

y = 5

Verification

Check 1: Put

x = 6.75

back into the original equation and work out each side on its own. They must come out the same.
On the right,

2 \times 6.75 = 13.5

6.75 - 4 = 2.75

and

\dfrac{3 + 13.5}{6} = 2.75

, so both sides agree.

Check 2: Expand the brackets from (b)(i) and compare the result with the expression printed in the question.

y^2 - 5y - 6y + 30 = y^2 - 11y + 30

, which is the expression given.

Check 3: Substitute each solution from (b)(ii) into

y^2 - 11y + 30

. A genuine solution makes it zero.

36 - 66 + 30 = 0

and

25 - 55 + 30 = 0

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $6x - 24 = 3 + 2x$ or $x - 4 = \dfrac{3}{6} + \dfrac{2}{6}x$ oe	M1	for correct removal of fraction and expansion of bracket in a correct equation or separating fraction (RHS) in an equation	✓
(a) $6x - 2x = 3 + 24$ or $4x = 27$ or $-24 - 3 = 2x - 6x$ or $-27 = -4x$ oe or $x - \dfrac{2}{6}x = \dfrac{3}{6} + 4$ oe or $-4 - \dfrac{3}{6} = \dfrac{2}{6}x - x$ oe	M1ft	(dep on 4 terms) correctly rearranging their 4 term equation for terms in $x$ on one side of equation and number terms on the other	✓
(a) $\dfrac{27}{4}$ Working required	A1	oe eg $6.75$ or $6\dfrac{3}{4}$ , dep on M1	✓
(b)(i) $(y \pm 6)(y \pm 5)$ or $(6 \pm y)(5 \pm y)$ or $y(y - 6) - 5(y - 6)$ or $y(y - 5) - 6(y - 5)$	M1	for $(y \pm 6)(y \pm 5)$ or $(6 \pm y)(5 \pm y)$ or for $(y + a)(y + b)$ where $ab = 30$ or $a + b = -11$ or $y(y + a) + b(y + a)$ or $y(y + b) + a(y + b)$ where $ab = 30$ or $a + b = -11$	✓
(b)(i) $(y - 6)(y - 5)$ Correct answer scores full marks (unless from obvious incorrect working)	A1	oe, allow any letter for $y$	✓
(b)(ii) $(y =) 6$ , $(y =) 5$	B1	must ft from their answer in (b)(i) ft from their factors in the form $(y + a)(y + b)$	✓

Full marks: 6/6

The diagram shows a solid cylinder.

The radius of the cylinder is $8$ cm.
The height of the cylinder is $h$ cm.
The volume of the cylinder is $3892$ cm³.

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

h =

[Total 3 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

A solid cylinder with radius

8

cm and height

h

Volume of the cylinder:

3892

cm³

Find

The value of

h

, correct to one decimal place

Plan the Solution

Start from the volume formula for a cylinder, $V = \pi r^{2} h$ .
Substitute $r = 8$ and $V = 3892$ , which leaves $h$ as the only unknown.
Work out the circular cross-section area $\pi \times 8^{2}$ first, then divide the volume by it.
Keep the full calculator value all the way through, and round only at the end.

Worked Solution [3 marks]

Rule - Volume of a cylinder:

V = \pi r^{2} h

- the area of the circular cross-section multiplied by the height.

Step 1: Put the given values into the volume formula

V = \pi r^{2} h

3892 = \pi \times 8^{2} \times h

(Reason: The radius is

8

cm and the volume is

3892

cm³, so

h

is the only letter left in the equation.)

Step 2: Work out the area of the circular cross-section

8^{2} = 64

\pi \times 64 = 201.0619\ldots

(Reason: Square the radius first, then multiply by

\pi

. Keep the full calculator value - rounding at this point would move the final answer.)

Step 3: Rearrange to make h the subject

h = \dfrac{3892}{\pi \times 8^{2}}

h = \dfrac{3892}{201.0619\ldots}

(Reason: Dividing both sides by the cross-section area leaves

h

on its own.)

Step 4: Divide, then round to one decimal place

h = 19.3572\ldots

h = 19.4 \text{ (to 1 d.p.)}

(Reason: The digit after the first decimal place is a

5

, so

19.35\ldots

rounds up to

19.4

h = 19.4

(to one decimal place)

Verification

Check 1: Put the unrounded

h

back into

\pi r^{2} h

and rebuild the volume.

\pi \times 8^{2} \times 19.3572\ldots = 3892

, the volume the question gives.

Check 2: Divide by the

8^{2}

and by the

\pi

as two separate steps, in the other order - the alternative the mark scheme prints.

\dfrac{3892}{64} = 60.8125

and

\dfrac{60.8125}{\pi} = 19.3572\ldots

, the same value as before.

Check 3: Repeat the division with the two approximations for

\pi

that the mark scheme allows.

3.14

gives

19.36\ldots

and

\dfrac{22}{7}

gives

19.34\ldots

, so both land inside the

19.3

19.4

the scheme accepts.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$3892 = \pi \times 8^{2} \times h$ or $\pi \times 8^{2} \; (= 64\pi = 201.0\ldots)$	M1	Allow the use of $3.14\ldots$ or $\dfrac{22}{7}$ for $\pi$ .	✓
$(h =) \dfrac{3892}{\pi \times 8^{2}}$ oe, for example $\dfrac{3892}{64} = 60.8\ldots$ and $\dfrac{60.8\ldots}{\pi}$	M1	Allow the use of $3.14\ldots$ or $\dfrac{22}{7}$ for $\pi$ .	✓
$19.4$	A1	Allow $19.3$ to $19.4$ .	✓
A correct answer scores full marks.	Note	Unless it comes from obviously incorrect working.	✓

Full marks: 3/3

(a) Express $520$ million in standard form. [1 mark]

(b) Convert $8.79 \times 10^{-5}$ to an ordinary number. [1 mark]

(a)(b)(c)

[Total 4 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

(a) the number

520

million

(b)

8.79 \times 10^{-5}

(c)

(5 \times 10^{42}) \times (7 \times 10^{-180})

Three separate pieces of standard-form work: going into it, coming back out of it, and multiplying inside it.

Find

(a) the same number written as

a \times 10^{n}

(b) the same number written out in full, with no power of ten (c) the product, given in standard form

Plan the Solution

Standard form is one front number times a power of ten, and the front number must be at least $1$ and less than $10$ . Every part below comes down to that one condition.
For (a), write the number out in full first. Then the index is a matter of counting how far the decimal point travels, not of guessing.
For (b), a negative index sends the point the other way, so expect an answer smaller than $1$ .
For (c), multiply the front numbers, add the indices, and only then ask whether the front number still obeys the condition. It will not, and fixing it changes the index.

Worked Solution [4 marks]

Rule - Standard form: a number is written as

a \times 10^{n}

, where

a

is at least

1

and less than

10

, and

n

is a whole number. To multiply two such numbers, multiply the front numbers and add the indices:

10^{m} \times 10^{n} = 10^{m+n}

Part (a): write

520

million out in full

520 \times 1\,000\,000 = 520\,000\,000

(Reason: One million is

1\,000\,000

, so

520

million is

520

lots of it. Writing the number out in full turns the next step into counting rather than guessing.)

Part (a): move the decimal point until one digit stands in front of it

520\,000\,000 = 5.2 \times 10^{8}

(Reason: The point starts after the last zero and moves

8

places to the left, to sit between the

5

and the

2

. Each place is one power of ten, so the index is

8

. The front number

5.2

is at least

1

and less than

10

, which is what standard form asks for.)

Part (b): read what the negative index is telling you

8.79 \times 10^{-5} = \dfrac{8.79}{10^{5}}

(Reason: A negative index means divide, not subtract - the answer is not a negative number. Dividing by

10^{5}

sends every digit

5

places to the right, so the result must be smaller than

1

Part (b): move the point

5

places and fill the gaps with zeros

\dfrac{8.79}{100\,000} = 0.0000879

(Reason: The

8

starts in the units column and finishes five columns to its right, so it lands in the fifth decimal place and four zeros sit between the point and it. Count them in the answer:

0.0000879

Part (c): multiply the front numbers

5 \times 7 = 35

(Reason: Multiplication can be done in any order, so gather the ordinary numbers together and the powers of ten together, and deal with each pair once.)

Part (c): add the indices

10^{42} \times 10^{-180} = 10^{42 + (-180)} = 10^{-138}

(Reason: Multiplying powers of the same base adds the indices. Adding a negative index is a subtraction, so

42 - 180 = -138

, and the answer so far is

35 \times 10^{-138}

Part (c): put the answer back into standard form

35 \times 10^{-138} = 3.5 \times 10 \times 10^{-138}

3.5 \times 10 \times 10^{-138} = 3.5 \times 10^{-137}

(Reason: The front number has to be less than

10

, and

35

is not, so the work is not finished at the previous line. Moving the point of

35

one place to the left divides the front number by

10

, so the power of ten has to go up by one to leave the value unchanged.)

(a)

5.2 \times 10^{8}

(b)

0.0000879

(c)

3.5 \times 10^{-137}

Verification

Check 1: Work part (a) backwards. Move the point of

5.2

eight places to the right and count the digits that come out.

5.2 \times 10^{8} = 520\,000\,000

, which reads as

520

million.

Check 2: Work part (b) backwards. Moving the point five places the other way must give the front number the question started from.

0.0000879 \times 10^{5} = 8.79

Check 3: Compare the two ways of writing part (c). The unfinished form and the standard form must be the same number, and the index must differ by exactly the one place the front number moved.

35 \times 10^{-138} = 3.5 \times 10^{-137}

, and

-138 + 1 = -137

Check 4: Is the size sensible? The

10^{-180}

is far stronger than the

10^{42}

, so the product has to be a tiny positive number rather than a large one, and the indices on their own say roughly where it lands.

42 - 180 = -138

, and two front numbers below

10

can only shift that by one place, so an index of

-137

is the size to expect.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$5.2 \times 10^{8}$	✓
(b)	B1	$0.0000879$	✓
(c)	M1	$35 \times 10^{-138}$ or $3.5 \times 10 \times 10^{-138}$ or $3.5 \times 10^{n}$ where $n \neq -137$	✓
(c)	A1	$3.5 \times 10^{-137}$ . A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 4/4

The diagram shows a trapezium $ABCD$ that has one line of symmetry.

angle $ADC = 60^{\circ} \qquad AD = 12 \text{ cm} \qquad DC = 47 \text{ cm}$

Find the area of the trapezium.
Give your answer correct to $3$ significant figures.
You must show your working clearly. [5 marks]

cm²

[Total 5 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

A trapezium

ABCD

with one line of symmetry, so

AB

is parallel to

DC

and

AD = BC

angle

ADC = 60^{\circ}

AD = 12

cm and

DC = 47

Find

The area of the trapezium, correct to

3

significant figures

Plan the Solution

Drop a perpendicular from $A$ down to $DC$ . It cuts off a right-angled triangle whose hypotenuse is $AD = 12$ cm, with the $60^{\circ}$ angle at $D$ .
Use $\sin 60^{\circ}$ on that triangle for the height of the trapezium, and $\cos 60^{\circ}$ for the piece of $DC$ it stands on.
The one line of symmetry makes the triangle at the $C$ end congruent to it, so the same piece is used up at each end of $DC$ .
Take both pieces off $47$ to get $AB$ , then put the two parallel sides and the height into the trapezium formula.
Keep the full calculator value of the height all the way through, and round only at the very end.

Worked Solution [5 marks]

Rule - Area of a trapezium:

\text{Area} = \dfrac{1}{2}(a + b)h

- half the sum of the two parallel sides, multiplied by the perpendicular height between them.

Step 1: Work out the perpendicular height of the trapezium

h = 12 \sin 60^{\circ}

h = 6\sqrt{3} = 10.3923\ldots

(Reason: In the right-angled triangle the height is the side opposite the

60^{\circ}

angle and

12

cm is the hypotenuse, so the height is

12 \sin 60^{\circ}

. This value is not exact as a decimal, so carry it in the calculator rather than rounding it now.)

Step 2: Work out how much of DC that triangle stands on

12 \cos 60^{\circ} = 6

(Reason: The base of the same triangle lies along

DC

and is the side next to the

60^{\circ}

angle, so it is

12 \cos 60^{\circ}

. Because

\cos 60^{\circ} = \dfrac{1}{2}

, this length is exactly

6

cm.)

Step 3: Work out the length of AB

AB = 47 - 6 - 6 = 35

(Reason: The line of symmetry makes the triangle at the

C

end congruent to the one at the

D

end, so

6

cm of

DC

is used up at each end. What is left between the two feet is

AB

Step 4: Put both parallel sides and the height into the formula

\text{Area} = \dfrac{1}{2} \times (47 + 35) \times 10.3923\ldots

\dfrac{1}{2} \times (47 + 35) = 41

\text{Area} = 41 \times 10.3923\ldots

(Reason: The parallel sides are

DC = 47

cm and

AB = 35

cm. Half of their sum is

41

, so the area is

41

lots of the height.)

Step 5: Multiply out, then round to three significant figures

\text{Area} = 426.0844\ldots

\text{Area} = 426 \text{ cm}^{2} \text{ (to 3 s.f.)}

(Reason: The first three significant figures are

4

2

and

6

, and the digit after them is a

0

, so the

426

is left as it stands.)

426 \text{ cm}^{2}

(to

3

significant figures)

Verification

Check 1: Rebuild the shape as a rectangle with one triangle at each end - the second complete method the mark scheme prints - and add the three areas.

35 \times 10.3923\ldots = 363.73\ldots

and each end triangle is

\dfrac{1}{2} \times 6 \times 10.3923\ldots = 31.17\ldots

, so the total is

363.73\ldots + 2 \times 31.17\ldots = 426.08\ldots

Check 2: Cut the trapezium along the diagonal

AC

instead and use

\dfrac{1}{2}ab \sin C

on each triangle. This route never works out a height at all. Triangle

ADC

\dfrac{1}{2} \times 12 \times 47 \times \sin 60^{\circ} = 244.21\ldots

and triangle

ABC

\dfrac{1}{2} \times 12 \times 35 \times \sin 120^{\circ} = 181.86\ldots

, giving

426.08\ldots

again.

Check 3: Find the height a second way, by Pythagoras on the same right-angled triangle rather than by the sine.

\sqrt{12^{2} - 6^{2}} = \sqrt{108} = 10.3923\ldots

, the same height as in Step 1.

Check 4: Check the size is sensible. The trapezium must be larger than a rectangle on the short parallel side and smaller than one on the long parallel side, both at the same height.

35 \times 10.3923\ldots = 363.7\ldots

and

47 \times 10.3923\ldots = 488.4\ldots

, and

426

lies between them.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$12 \sin 60 \; (= 6\sqrt{3} = 10.3(9\ldots))$ or $\sqrt{12^{2} - 6^{2}} \; (= 6\sqrt{3} = 10.3(9\ldots))$ or (Area $ADC$ =) $\dfrac{1}{2} \times 12 \times 47 \times \sin 60 \; (= 244.2\ldots)$	M1	for a method to find the height of the trapezium or the area of triangle $ADC$ . The printed scheme puts the $6$ in quotation marks, so a candidate's own value from the base row may be used. The first two M1 marks can be awarded in either order.	✓
$12 \cos 60 \; (= 6)$ or $\sqrt{12^{2} - (6\sqrt{3})^{2}} \; (= 6)$	M1	(indep) for a method to find the base of the triangle, condoning missing brackets around the $6\sqrt{3}$ , which the printed scheme puts in quotation marks so a candidate's own height may be used. The first two M1 marks can be awarded in either order.	✓
( $AB$ =) $47 - 6 - 6 \; (= 35)$	M1	(dep on previous M1) for a method to find the length of $AB$ . The printed scheme puts each $6$ in quotation marks, so a candidate's own value may be used.	✓
(Trapezium =) $\dfrac{1}{2} \times (47 + 35) \times 10.3(9\ldots)$ or (Rectangle + two triangles =) $35 \times 10.3(9\ldots) + 2 \times \dfrac{1}{2} \times 6 \times 10.3(9\ldots)$ or $35 \times 10.3(9\ldots) + 2 \times \dfrac{1}{2} \times 6 \times 12 \times \sin 60$ or (Triangle $ADC$ + Triangle $ABC$ =) $244.2\ldots + \dfrac{1}{2} \times 12 \times 35 \times \sin 120$ oe, for example $(47 - 6) \times 10.3(9\ldots)$	M1	for a complete method. Each value the printed scheme puts in quotation marks may be a candidate's own. There are other methods, and marks should be awarded for a complete method that should give the correct area.	✓
$426$	A1	(dep on M1) allow $420$ to $427$ from correct working. The printed scheme also states that working is required.	✓

Full marks: 5/5

(a) The table below gives values of $y = x^3 + 2x + 3$ for values of $x$ from $-3$ to $3$
$\begin{array}{|c|c|c|c|c|c|c|c|}\hline x & -3 & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline y & \phantom{-30} & -9 & 0 & 3 & 6 & \phantom{-30} & 36 \\ \hline \end{array}$
Work out the two missing values and complete the table. [1 mark]

(b) Draw the graph of $y = x^3 + 2x + 3$ on the grid below, for $-3 \leq x \leq 3$ [2 marks]

[Total 3 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

The curve

y = x^3 + 2x + 3

A table of values running from

x = -3

x = 3

, with the cells for

x = -3

and

x = 2

left blank

A grid running from

-3

3

across and from

-30

40

Find

The two missing

y

values The curve, drawn on the grid for

-3 \leq x \leq 3

Plan the Solution

Put each missing $x$ value into $y = x^3 + 2x + 3$ and work the answer out.
Cube first, then double, then add $3$ . An odd power of a negative number stays negative, so the sign matters at $x = -3$ .
Plot all seven points from the finished table.
Join them with one smooth curve - a cubic has no corners and no straight stretches.

Worked Solution [3 marks]

Rule - Table of values: work out

y

for every

x

the table asks for, plot each pair

(x, y)

, then join the points with a single smooth curve.

Step 1: substitute

x = -3

(-3)^3 + 2 \times (-3) + 3 = -27 - 6 + 3 = -30

(Reason: Reason: take the cube first, then the

2x

term, then the

3

. Cubing is an odd power, so a negative number stays negative when it is cubed.)

Step 2: substitute

x = 2

2^3 + 2 \times 2 + 3 = 8 + 4 + 3 = 15

(Reason: Reason:

2^3

means

2 \times 2 \times 2

, which is

8

- it is not

2 \times 3

Step 3: write the completed table out as seven points

(-3, -30) \quad (-2, -9) \quad (-1, 0) \quad (0, 3)

(1, 6) \quad (2, 15) \quad (3, 36)

(Reason: Reason: the five values the table already carries are used exactly as printed. Only the two blanks were ours to fill.)

Step 4: plot the points and join them

-3 \leq x \leq 3

(Reason: Reason: one mark is for the plotting and one is for the curve. Draw it only between

x = -3

and

x = 3

, and keep it curved throughout - joining the points with straight lines loses the accuracy mark.)

(a)

-30

and

15

(b) a smooth curve through all seven points, from

(-3, -30)

(3, 36)

Verification

Check 1: Test the reading of the formula on a value the table already gives. Put

x = -1

in and see whether it returns the printed value.

(-1)^3 + 2 \times (-1) + 3 = -1 - 2 + 3 = 0

, and the table prints

0

there

Check 2: Take the

3

off every

y

. What is left is

x^3 + 2x

, which is an odd expression, so the values at

x

and

-x

must be equal and opposite.

-30 - 3 = -33

against

36 - 3 = 33

, and

15 - 3 = 12

against

-9 - 3 = -12

- both pairs match

Check 3: The gradient is

3x^2 + 2

, which is never smaller than

2

, so

y

has to rise all the way across the table and the curve can have no turning point. the

y

row reads

-30

-9

0

3

6

15

36

- rising throughout, and every value sits between

-30

and

40

, so the whole curve fits on the grid

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $-30$ and $15$ written in the two empty cells	B1	for $-30$ and $15$ in the correct place. This may be awarded if they are plotted correctly on the graph.	✓
(b) at least $6$ points plotted correctly	M1ft	for at least $6$ points plotted correctly (within the circles on the overlay). Follow through their incorrect table.	✓
(b) correct curve between $x = -3$ and $x = 3$	A1	for a correct curve between $x = -3$ and $x = 3$ (clear intention to go through all the points, and it must be curved). Ignore to the left of $x = -3$ and to the right of $x = 3$ .	✓
Guidance on a blank table	Note	If a fully correct graph is shown but a blank table is shown in (a), then award the mark for (a).	✓
Guidance on the answer alone	Note	Correct answer scores full marks (unless from obvious incorrect working).	✓

Full marks: 3/3

$OABC$ is a sector of a circle with centre $O$ .

In the sector, angle $AOC = 140^\circ$ and $OA = OC = 16$ cm.

Work out the area of the sector.
Give your answer correct to $3$ significant figures. [2 marks]

cm²

[Total 2 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

A sector

OABC

of a circle with centre

O

angle

AOC = 140^\circ

OA = OC = 16

cm, so the radius is

r = 16

Find

The area of the sector, correct to

3

significant figures.

Plan the Solution

Write the angle at the centre as a fraction of a full turn of $360^\circ$ .
Work out the area of the whole circle from the radius.
Multiply the two, and round only on the last line.

Worked Solution [2 marks]

Rule - Area of a sector:

\text{Area} = \dfrac{\theta}{360} \times \pi r^2

, where

\theta

is the angle at the centre, in degrees.

Step 1: Write the sector as a fraction of the whole circle

\dfrac{140}{360} = \dfrac{7}{18}

(Reason: The angle at the centre is

140^\circ

out of a full turn of

360^\circ

, and

140

and

360

both divide by

20

Step 2: Work out the area of the whole circle

\pi \times 16^2 = 256\pi

256\pi \approx 804.2477

(Reason:

OA

and

OC

are radii, so

r = 16

and

r^2 = 256

. Keeping the

\pi

here means nothing is rounded early.)

Step 3: Take that fraction of the whole circle

\dfrac{7}{18} \times 256\pi \approx 312.763

(Reason: The sector is that share of the whole circle, so the two multiply. On a calculator the whole line goes in at once.)

Step 4: Round to 3 significant figures

312.763\ldots \approx 313

\text{Area} \approx 313 \text{ cm}^2

(Reason: The first three significant figures are

3

1

and

2

. The next digit is

7

, so the

2

rounds up to

3

. Area is measured in

\text{cm}^2

313 \text{ cm}^2

Verification

Check 1: Turn

140^\circ

into radians,

\dfrac{7\pi}{9}

, and use the other sector formula,

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

. A different formula must give the same area.

\dfrac{1}{2} \times 16^2 \times \dfrac{7\pi}{9} \approx 312.763

Check 2: Work backwards. Divide the sector area by the area of the whole circle and the fraction of the turn should come back.

\dfrac{312.763}{804.2477} \approx 0.3889

, and

\dfrac{140}{360} \approx 0.3889

Check 3: The mark scheme allows

3.14

\dfrac{22}{7}

in place of

\pi

, so neither may change the answer.

3.14

gives

312.60\ldots

and

\dfrac{22}{7}

gives

312.88\ldots

, both

313

3

significant figures.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $\pi \times 16^2 \times \dfrac{140}{360}$ oe eg $256\pi \times \dfrac{7}{18}$	M1	allow use of $3.14\ldots$ or $\dfrac{22}{7}$ for $\pi$ . $\dfrac{140}{360}$ may be seen as an equivalent fraction or decimal eg $\dfrac{7}{18}$ or $0.3\dot{8}$ , $0.388(8\ldots)$ or $0.389$	✓
$313$	A1	accept $311.8$ to $313$	✓
Correct answer scores full marks (unless from obvious incorrect working)	Note	Printed on the official scheme for this question, and it applies to the whole question.	✓

Full marks: 2/2

Continue to questions 12 to 20

The remaining 9 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 4MA1/1H, Thursday 15 May 2025: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 11 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

Frequently asked questions

Keep revising

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