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Edexcel IGCSE 4MA1 Paper 2FR, November 2024: Worked Solutions, Questions 16 to 26

Sir Faraz Hassan

3 Aug 2026

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

Back to questions 1 to 15

This is the rest of the paper. Questions 1 to 15, the paper's overview and the frequently asked questions are on the first page.

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

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

Try each question yourself first, then open the worked solution to check your method and see exactly where each method mark (M1) and accuracy mark (A1) is earned. The questions follow the same order as the original paper and carry the same marks.

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All 26 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 16 to 26 of 26

Glass jars are made at a bottling plant.

On Wednesday, $375$ jars per hour were made at the plant.
On Wednesday, jars were made at the plant for $8$ hours.

On Thursday, $20\%$ more jars were made than on Wednesday.
On Thursday, $300$ jars per hour were made at the plant.

Work out for how many hours the plant made jars on Thursday. [4 marks]

hours

[Total 4 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

Wednesday:

375

jars per hour, for

8

hours

Thursday:

20\%

more jars in total than on Wednesday

Thursday:

300

jars per hour

Find

The number of hours the plant made jars on Thursday

Plan the Solution

Multiply Wednesday's hourly rate by Wednesday's hours to get Wednesday's total.
Increase that total by $20\%$ to get Thursday's total. Notice that the two hourly rates are different, so Thursday's total cannot be found from Wednesday's rate.
Divide Thursday's total by $300$ jars per hour to get the number of hours.

Worked Solution [4 marks]

Rule - Rate: number made = hourly rate multiplied by number of hours, so number of hours =

\dfrac{\text{number made}}{\text{hourly rate}}

Step 1: Work out how many jars were made on Wednesday

375 \times 8 = 3000

(Reason: The plant made

375

jars in each of

8

hours, so the total is the hourly rate multiplied by the number of hours.)

Step 2: Increase Wednesday's total by 20%

100\% + 20\% = 120\% = 1.2

3000 \times 1.2 = 3600

(Reason:

20\%

more means

120\%

of Wednesday's total, and

120\%

as a decimal multiplier is

1.2

. Adding

20\%

3000

, which is

600

, to

3000

gives the same total.)

Step 3: Divide Thursday's total by Thursday's hourly rate

\dfrac{3600}{300} = 12

(Reason: Every hour on Thursday produced

300

jars, so the number of hours is Thursday's total shared into groups of

300

12

hours

Verification

Check 1: Work forwards from the answer. Make

300

jars an hour for

12

hours, then compare that total with Wednesday's

3000

300 \times 12 = 3600

, and

\dfrac{3600 - 3000}{3000} = 0.2

, which is exactly

20\%

Check 2: Reach the answer a different way, by proportion, without finding either total. Thursday's rate is

\dfrac{300}{375} = \dfrac{4}{5}

of Wednesday's rate, while Thursday's output is

\dfrac{6}{5}

of Wednesday's output, so the time must scale by

\dfrac{6}{5} \times \dfrac{5}{4}

\dfrac{6}{5} \times \dfrac{5}{4} = \dfrac{3}{2}

, and

\dfrac{3}{2} \times 8 = 12

hours, which agrees

Check 3: Sense check. Thursday's rate of

300

an hour is slower than Wednesday's

375

an hour, yet more jars were made, so Thursday must have taken longer than Wednesday's

8

hours.

12

hours is longer than

8

hours, as expected

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$375 \times 8$ (= $3000$ )	M1	For a complete method to find the number of jars made on Wednesday. Award for $3000$ seen anywhere.	✓
$375 \times 8 \times 1.2$ (= $3600$ )	M1	For a complete method to find the number of jars made on Thursday, using their Wednesday total. Also award for $3000 \times \left(1 + \dfrac{20}{100}\right)$ or $1.2 \times 3000$ or $(0.2 \times 3000) + 3000$ or $600 + 3000$ .	✓
$\dfrac{3600}{300}$	M1	For dividing their Thursday total by $300$ jars per hour.	✓
$12$	A1	Working not required, so a correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 4/4

Express $1400$ as a product of powers of its prime factors.
Show your working clearly. [3 marks]

[Total 3 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

The number

1400

, to be broken down into primes.

Find

The product of powers of primes that makes

1400

, written in index form.

Plan the Solution

Divide by the smallest prime that goes in, and keep dividing by that same prime for as long as it goes in.
Move up to the next prime and repeat, stopping when the number left is itself prime.
Count how many times each prime was used and write that count as a power, which is what index form means.

Worked Solution [3 marks]

Rule - Prime factorisation: keep dividing by the smallest prime that goes in until

1

is reached. The divisors used, collected as powers, are the answer.

Step 1: divide by the smallest prime,

2

, for as long as it goes in

\dfrac{1400}{2} = 700

\dfrac{700}{2} = 350

\dfrac{350}{2} = 175

(Reason:

1400

is even, so

2

divides it. Carry on until the number left is odd:

175

is odd, so

2

will not go again.)

Step 2: move up to the next prime that goes in,

5

\dfrac{175}{5} = 35

\dfrac{35}{5} = 7

(Reason:

3

is skipped because

1 + 7 + 5 = 13

is not a multiple of

3

, and

175

ends in a

5

, so

5

is the next prime that divides.)

Step 3: stop when the number left is itself prime

\dfrac{7}{7} = 1

1400 = 2 \times 2 \times 2 \times 5 \times 5 \times 7

(Reason:

7

is prime, so the ladder ends there. The divisors used down the ladder are the prime factors of

1400

Step 4: collect the repeated primes as powers

1400 = 2^{3} \times 5^{2} \times 7

(Reason: There are three

2

s, two

5

s and one

7

, and writing those counts as powers is what puts the answer in index form.)

2^{3} \times 5^{2} \times 7

Verification

Check 1: Multiply the powers back together.

2^{3} = 8

and

5^{2} = 25

8 \times 25 \times 7 = 1400

Check 2: Split

1400

a completely different way, as

14 \times 100

, and factorise each piece on its own.

(2 \times 7) \times (2^{2} \times 5^{2}) = 2^{3} \times 5^{2} \times 7

Check 3: Divide

1400

by the powers one at a time. A correct factorisation lands exactly on

1

\dfrac{1400}{8} = 175 \text{, then } \dfrac{175}{25} = 7 \text{, then } \dfrac{7}{7} = 1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Start the prime factorisation, e.g. $1400 = 2 \times 2 \times 350$	M1	For two correct stages of prime factorisation with no incorrect stage, or for at least three stages with no more than one incorrect stage. Each stage gives two factors, and it may be shown in a factor tree, a ladder or a list, e.g. $2, 2, 350$ . Other openings such as $2 \times 7 \times 100$ or $5 \times 7 \times 40$ score the same.	✓
Reach the complete list of primes, $2 \times 2 \times 2 \times 5 \times 5 \times 7$	M1	Dependent on the first M1. Awarded for $2 \times 2 \times 2 \times 5 \times 5 \times 7$ , or for the primes listed as $2^{3}, 5^{2}, 7$ . Ignore any $1$ s. It may be seen in a fully correct factor tree or ladder rather than written out.	✓
Write the answer in index form, $2^{3} \times 5^{2} \times 7$	A1	Dependent on both method marks. The factors may be given in any order, but the answer must be in index form as the question asks, and a $1$ in the final answer is not allowed.	✓
Note - working required	-	The question asks for the working to be shown, so the answer on its own earns nothing. This row carries no mark of its own.	✓

Full marks: 3/3

The grid shows two shapes, $A$ and $B$ .

(a) Describe fully the single transformation that maps shape $A$ onto shape $B$ . [2 marks]

(b) On the same grid, rotate shape $A$ through $180^\circ$ about the point $(-1, 0)$ .
Label your new shape $C$ . [2 marks]

(a)

[Total 4 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

Shape

A

has vertices

(2, 1)

(2, 2)

(3, 2)

and

(4, 1)

Shape

B

has vertices

(5, -4)

(5, -3)

(6, -3)

and

(7, -4)

Both are trapeziums of the same size, drawn on a grid running from

-8

8

on each axis.

Find

(a) The single transformation that maps

A

onto

B

, named in full and with its vector. (b) The image of

A

after a rotation of

180^\circ

about

(-1, 0)

, drawn on the grid and labelled

C

Plan the Solution

Compare the two shapes first. $B$ is the same size as $A$ and the same way up, so nothing has been turned, flipped or resized. That leaves a translation.
Pick one vertex, and the vertex in the matching position on the other shape, then subtract to get the vector. Repeating that for all four vertices is what proves it really is a translation.
For part (b), a rotation of $180^\circ$ is a half turn: every point finishes on the opposite side of the centre, the same distance away.
Rotate the four vertices one at a time, join them in the same order, and write $C$ inside the new shape.

Worked Solution [4 marks]

Rule - Translation: the vector from

(x_1, y_1)

(x_2, y_2)

\binom{x_2 - x_1}{y_2 - y_1}

. Rotation of

180^\circ

about

(a, b)

(x, y) \to (2a - x, 2b - y)

Step 1: Decide what kind of transformation part (a) is

\text{same size, same way up} \implies \text{a translation}

(Reason: Shape

B

is not turned, not flipped and not a different size, so it cannot be a rotation, a reflection or an enlargement. A translation is the only single transformation left, and naming any of the others alongside it loses the mark.)

Step 2: Match a pair of corresponding vertices

(2, 1) \to (5, -4)

(Reason: Take the bottom left vertex of

A

and the bottom left vertex of

B

. Matching by position on the shape is what matters; pairing the wrong two vertices gives the wrong vector.)

Step 3: Subtract to find the translation vector

5 - 2 = 3

-4 - 1 = -5

\binom{3}{-5}

(Reason: Image coordinate minus object coordinate, the

x

values first and then the

y

values. A positive top number means right, and a negative bottom number means down, so this is

3

right and

5

down.)

Step 4: Write down the half turn rule for part (b)

(x, y) \to (2a - x, 2b - y)

(x, y) \to (-2 - x, -y)

(Reason: The centre is

(a, b) = (-1, 0)

, so

2a = -2

and

2b = 0

. A half turn needs no protractor: each image point sits the same distance from the centre, on the opposite side of it.)

Step 5: Rotate each vertex of shape A

(2, 1) \to (-4, -1)

(2, 2) \to (-4, -2)

(3, 2) \to (-5, -2)

(4, 1) \to (-6, -1)

(Reason: Substitute the vertices into the rule one at a time. Working down the list in order keeps the image vertices in the same order as the original ones, which is what makes joining them up straightforward.)

Step 6: Join the image points in order and label the shape

(-6, -1) \quad (-4, -1) \quad (-4, -2) \quad (-5, -2)

(Reason: Join the four image points in the same order as the vertices of

A

, so the sloping edge still sits opposite the vertical edge. Then write

C

inside the shape, because the question asks for it to be labelled.)

(a) Translation by the vector

\binom{3}{-5}

(b)

C

drawn at

(-6, -1)

(-4, -1)

(-4, -2)

and

(-5, -2)

Verification

Check 1: Undo the translation. Take

3

off the

x

coordinate of a vertex of

B

, and add

5

to its

y

coordinate.

5 - 3 = 2

and

-4 + 5 = 1

, so

(5, -4)

goes back to

(2, 1)

, which is a vertex of

A

Check 2: Subtract all four corresponding pairs, not just the one used in the working. A translation moves every point by the same vector. All four pairs give

3

across and

5

down, so one single vector carries the whole shape and the transformation really is a translation.

Check 3: The centre of a half turn is the midpoint of every point and its image, so test the midpoints instead of repeating the rule. The midpoint of

(2, 1)

and

(-4, -1)

(-1, 0)

, and the other three pairs give

(-1, 0)

as well.

Check 4: A rotation cannot change the size or shape of anything, so compare the edges of

C

with the edges of

A

. Each shape has a vertical edge of

1

square, a short horizontal edge of

1

square, a sloping edge across one square, and a long horizontal edge of

2

squares, so

C

is congruent to

A

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Name the transformation	B1	for translation. No mark if reflection, rotation, enlargement, mirrored, turned, move or flipped is stated as well. Move together with translation is acceptable, and a misspelling such as translat is allowed.	✓
(a) Give the vector	B1	for (vector =) $\binom{3}{-5}$	✓
(b) Draw the image and label it	B2	for the shape drawn at $(-6, -1)$ $(-4, -1)$ $(-4, -2)$ $(-5, -2)$ . Condone a missing label.	✓
(b) Partial credit	B1	if not B2, for a correct trapezium drawn with the correct orientation in the wrong position, or for $3$ points plotted correctly. The commonest cause of both is rotating about the origin instead of about $(-1, 0)$ .	✓

Full marks: 4/4

Four numbers are written below in ascending order of size.

$x \qquad x \qquad y \qquad 15$

In this list $x$ and $y$ are integers.

The four numbers have
a median of $12.5$
a range of $4$

Work out the value of $x$ and the value of $y$ . [2 marks]

x =y =

[Total 2 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

The list

x, x, y, 15

, written in ascending order of size, with

x

and

y

both integers.

The four numbers have a median of

12.5

and a range of

4

Find

The value of

x

and the value of

y

Plan the Solution

The range uses only the smallest and the largest number. Ascending order fixes which is which, so the range gives an equation containing $x$ on its own - start there.
With four numbers there is no single middle one, so the median is the mean of the middle two, $x$ and $y$ . That gives a second equation linking the two letters.
Put the value of $x$ into that second equation to get $y$ , then read the finished list back to confirm it is still ascending and both numbers are integers.

Worked Solution [2 marks]

Rule - Range and median:

\text{range} = \text{largest} - \text{smallest}

, and for an even amount of numbers in order the median is the mean of the middle two.

Step 1: use the range, which involves only the two ends of the list

15 - x = 4

x = 15 - 4 = 11

(Reason: The list is in ascending order, so the smallest number is

x

and the largest is

15

. The range is the largest take away the smallest, and it is given as

4

, so

x

is the only unknown left in that equation.)

Step 2: use the median, which for four numbers is the mean of the middle two

\dfrac{x + y}{2} = 12.5

x + y = 25

(Reason: Four is an even amount, so no single number sits in the middle. The middle two are

x

and

y

, and their mean is

12.5

. Multiplying both sides by

2

clears the fraction and leaves the sum of the pair.)

Step 3: substitute the value of

x

to find

y

11 + y = 25

y = 25 - 11 = 14

(Reason: Step 1 gives

x = 11

, and putting that into

x + y = 25

leaves a one-step equation in

y

Step 4: write the finished list and confirm it obeys everything the question said

11, 11, 14, 15

11 \leq 11 \leq 14 \leq 15

(Reason: The list is still in ascending order, and

11

and

14

are both integers, so nothing the question stated has been broken. This matters here: the median and the range on their own do not pin the letters down, and the order is what decides which value belongs to which letter.)

x = 11

y = 14

Verification

Check 1: Work the median back out of the finished list

11, 11, 14, 15

. The middle two numbers are

11

and

14

\dfrac{11 + 14}{2} = \dfrac{25}{2} = 12.5

Check 2: Work the range back out of the same list, largest take away smallest. This uses the two numbers Check 1 did not.

15 - 11 = 4

Check 3: Test the swap,

x = 14

and

y = 11

. It gives

14, 14, 11, 15

, whose median and range both still come out as

12.5

and

4

, so the only thing that rules it out is the ascending order.

11 < 14

, so

14, 14, 11, 15

is not in ascending order

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Both values correct: $x = 11$ and $y = 14$	B2	Full marks for both values. These are B marks, so they are for the values themselves and no working is required; a correct pair with no method shown still scores B2.	✓
Exactly one value correct: $x = 11$ or $y = 14$	B1	Awarded when one of the two values is right and the other is not - for example the range is used correctly to reach $x = 11$ but the median is then treated as the mean of all four numbers, or $y = 14$ is right while $x$ is not. The mark scheme prints this row in brackets beneath the B2, as the partial award.	✓
Special case - the letters the wrong way round: $x = 14$ and $y = 11$	-	SC B1. This row carries no mark of its own; it records what that one specific wrong answer is worth. The error is naming the letters the wrong way round, and it is worth a mark because the arithmetic behind it is sound: $14, 14, 11, 15$ has a median of $12.5$ and a range of $4$ , so both stated conditions are met and only the ascending order is broken.	✓

Full marks: 2/2

$\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$
$A = \{\text{factors of }6\}$
$B = \{\text{prime numbers}\}$

(a) List the members of the set
(i) $A \cup B$
[1 mark]
(ii) $A'$ [1 mark]

Anjali claims that $A \cap B = \emptyset$
Anjali is wrong.
(b) Explain why. [1 mark]

$C$ is a set with $4$ members such that
the set $A \cap C$ has $2$ members
the set $B \cap C$ has $2$ members
Set $A \cap C$ and set $B \cap C$ have no members in common.
(c) List the $4$ members of set $C$ [2 marks]

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

[Total 5 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

The universal set

\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

, so every set here is taken from the numbers

1

10

A = \{\text{factors of }6\}

and

B = \{\text{prime numbers}\}

C

has

4

members, with

A \cap C

and

B \cap C

each having

2

members and sharing no member

Find

(a)(i) the members of

A \cup B

(a)(ii) the members of

A'

(b) a reason why

A \cap B = \emptyset

is wrong (c) the

4

members of

C

Plan the Solution

Write out $A$ and $B$ as lists first. Every part of the question is read off those two lists.
For (a)(i) collect everything that is in either list, writing a repeated member once only.
For (a)(ii) go through the universal set and keep whatever is missing from $A$ .
For (b) find the members the two lists share. A set that has a member in it cannot be the empty set.
For (c) work out where each pair of members of $C$ is allowed to come from, then check the answer against all three conditions.

Worked Solution [5 marks]

Rule - Set notation:

A \cup B

(union) is everything in

A

or in

B

or in both;

A \cap B

(intersection) is everything in both;

A'

(complement) is everything in

\mathcal{E}

that is not in

A

; and

\emptyset

is the set with no members at all.

Step 1: List the members of

A

6 = 1 \times 6 = 2 \times 3

A = \{1, 2, 3, 6\}

(Reason: (Reason: the factors of

6

are the whole numbers that divide into

6

exactly. The two factor pairs

1 \times 6

and

2 \times 3

give all four of them, and all four are inside the universal set.))

Step 2: List the members of

B

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

(Reason: (Reason: a prime number has exactly two factors, itself and

1

. So

1

is not prime, and

4

6

8

9

and

10

all have more than two factors.))

Step 3: (a)(i) Take the union of the two lists

A \cup B = \{1, 2, 3, 6\} \cup \{2, 3, 5, 7\}

A \cup B = \{1, 2, 3, 5, 6, 7\}

(Reason: (Reason: the union collects everything that appears in either list.

2

and

3

appear in both lists, but a set never repeats a member, so each is written once.))

Step 4: (a)(ii) Take the complement of

A

\mathcal{E} = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}

A' = \{4, 5, 7, 8, 9, 10\}

(Reason: (Reason:

A'

is everything in the universal set that is not a factor of

6

. Every number in

\mathcal{E}

is in exactly one of

A

and

A'

, so the two lists together must account for all

10

of them.))

Step 5: (b) Work out

A \cap B

A \cap B = \{1, 2, 3, 6\} \cap \{2, 3, 5, 7\}

A \cap B = \{2, 3\}

(Reason: (Reason:

2

and

3

are factors of

6

and they are also prime, so they sit in both sets. The intersection therefore has members in it, and a set with members in it is not the empty set.))

Step 6: (c) Decide where the members of

C

may come from

A \cap B = \{2, 3\}

A \setminus B = \{1, 6\}

B \setminus A = \{5, 7\}

(Reason: (Reason: if a member of

A \cap C

were also prime, it would sit in

B \cap C

as well, and those two sets are told to share no member. So

A \cap C

cannot use

2

3

, and by the same argument neither can

B \cap C

.))

Step 7: (c) List the

4

members of

C

A \cap C = \{1, 6\}

B \cap C = \{5, 7\}

C = \{1, 5, 6, 7\}

(Reason: (Reason: only

1

and

6

are left for

A \cap C

and only

5

and

7

are left for

B \cap C

, and each of those sets needs exactly

2

members, so both are forced. That is already

4

members, which is all the room

C

has.))

(a)(i)

A \cup B = \{1, 2, 3, 5, 6, 7\}

(a)(ii)

A' = \{4, 5, 7, 8, 9, 10\}

(b)

2

and

3

are members of both

A

and

B

, so

A \cap B = \{2, 3\}

, which is not empty(c)

C = \{1, 5, 6, 7\}

Verification

Check 1: Count the complement.

A

has

4

members and the universal set has

10

, so

A'

must have

10 - 4 = 6

members.

4, 5, 7, 8, 9, 10

6

members, and putting

A

and

A'

together gives every number from

1

10

exactly once

Check 2: Count the union a different way, using

|A \cup B| = |A| + |B| - |A \cap B|

, which subtracts the shared members so they are not counted twice.

4 + 4 - 2 = 6

, and the list

1, 2, 3, 5, 6, 7

has

6

members

Check 3: Test

C = \{1, 5, 6, 7\}

against all three conditions in the question.

A \cap C = \{1, 6\}

has

2

members,

B \cap C = \{5, 7\}

has

2

members, those two sets share nothing, and

C

has

4

members

Check 4: Check that no other answer works, by testing every set of

4

numbers chosen from

1

12

against the same three conditions. Going past

10

also tests whether a number from outside the universal set could be used. Exactly one of the

495

sets passes, and it is

\{1, 5, 6, 7\}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) List the members of $A \cup B$	B1	$1, 2, 3, 5, 6, 7$ - in any order, with no repeats	✓
(a)(ii) List the members of $A'$	B1	$4, 5, 7, 8, 9, 10$ - in any order, with no repeats	✓
(b) Explain why the intersection is not empty	B1	For identifying $2$ , or $3$ , or $2$ and $3$ , as a member of both $A$ and $B$ , together with an explanation showing the meaning of intersection and of the empty set. Accept equivalent wording, for example that the two sets have a common member or that $A \cap B = \{2, 3\}$ . If a common number is quoted, it must be correct.	✓
(c) List the $4$ members of set $C$	B2	$1, 5, 6, 7$ - in any order. Partial credit: B1 for three correct values with no more than one incorrect, or for four correct values with no more than one incorrect.	✓

Full marks: 5/5

The diagram shows the design for a pendant, which will be made using wire.
The design is a circle inside a square $ABCD$

The circle touches the square at the points $E$ , $F$ , $G$ and $H$

The area of the square is $81$ cm²

Calculate the total length of wire that will be needed to make the square and the circle.
Give your answer correct to $3$ significant figures. [4 marks]

[Total 4 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

The design is a circle inside a square

ABCD

, and the circle touches the square at

E

F

G

and

H

The area of the square is

81

cm², which is a square number.

The wire has to go all the way round the square and all the way round the circle.

Find

The total length of wire: the perimeter of the square added to the circumference of the circle. The answer is wanted correct to

3

significant figures, so the full decimal is kept until the very last line.

Plan the Solution

The area of a square is the side multiplied by itself, so square-rooting $81$ gives the side.
Four equal sides then give the perimeter of the square.
A circle that touches all four sides fits exactly across the square, so its diameter is the same length as a side. That is the link the whole question turns on.
Use $C = \pi d$ for the circumference, add the two lengths, and round only at the end.

Worked Solution [4 marks]

Rule - Square:

A = s^2

and

P = 4s

. Circle:

C = \pi d

, and

d = 2r

Step 1: Find the length of one side of the square

s^2 = 81

s = \sqrt{81} = 9

(Reason: The area of a square is one side multiplied by itself, so the side is the square root of the area.

81

is a square number and

9 \times 9 = 81

, so the side is exactly

9

cm with nothing to round.)

Step 2: Work out the perimeter of the square

P = 4 \times 9 = 36

(Reason: The wire round the square runs along all four sides, and every side of a square is the same length, so multiply by

4

. That is

36

cm of wire before the circle is even considered.)

Step 3: Work out the circumference of the circle

d = 9

C = \pi d

C = \pi \times 9 = 28.274\ldots

(Reason: The circle touches all four sides, so it stretches right across the square and its diameter is the same as the side:

9

cm. Keep the decimal running rather than writing

28.3

here, because rounding twice drags the final answer off.)

Step 4: Add the two lengths and round

36 + 9\pi = 64.274\ldots

64.274\ldots \approx 64.3

(Reason: One length of wire makes both outlines, so the perimeter and the circumference are added. The first

3

significant figures are

6

4

and

2

; the next digit is

7

, so the

2

rounds up to

3

64.3

Verification

Check 1: Work backwards: take the square's perimeter off the total, then divide what is left by

\pi

. That should hand back the diameter.

\dfrac{64.274\ldots - 36}{\pi} = 9

, which is the side of the square, exactly as the inscribed circle requires.

Check 2: Keep the answer exact rather than decimal, then let the calculator do the whole sum in one go:

36 + 9\pi

36 + 9\pi = 64.2743\ldots

, which rounds to

64.3

and sits inside the mark scheme's band of

64.26

64.3

Check 3: A rough size check, done without a calculator.

\pi

is a little over

3

, so the circumference is a little over

27

cm. That gives roughly

36 + 27 = 63

cm, so an answer near

64

cm is the right size. An answer near

50

cm would mean the radius had been used where the diameter belongs, and an answer near

93

cm would mean the side had been used as the radius.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A method for the length of a side of the square	M1	for $\sqrt{81}$ $(= 9)$ or $9$ or $9 \times 9$ $(= 81)$ . This may be seen on the diagram rather than in the working.	✓
The perimeter of the square	M1	for $4 \times 9$ $(= 36)$ oe. The first M mark can be implied by $36$ .	✓
A correct expression for the circumference of the circle	M1	for using $2\pi r$ or $\pi D$ , eg $\pi \times 9$ $(= 28.2(743\ldots))$ or $9\pi$ . The first M mark can be implied by $28.2(743\ldots)$ rounded or truncated to $1$ dp, or by $9\pi$ .	✓
The total length of wire	A1	for $64.3$ . Accept anything from $64.26$ to $64.3$ , which is the band a candidate lands in when the circumference is rounded or truncated part way through.	✓
Note	-	Working not required, so a correct answer scores full marks (unless it comes from obviously incorrect working). This row carries no mark of its own and does not add to the total of four.	✓

Full marks: 4/4

(a) Solve $\dfrac{2f}{3} = 4f - 17$
You must show clear algebraic working. [3 marks]

(b) Simplify $(e + 12)^{0}$ where $e > 0$ [1 mark]

(d) Factorise fully $20x^{5}y + 12x^{3}y^{4}$ [2 marks]

(a) f =(b)(c)(d)

[Total 8 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

(a) An equation with a fraction on one side:

\dfrac{2f}{3} = 4f - 17

(b) A bracket raised to the power

0

(e + 12)^{0}

, and we are told

e > 0

\dfrac{12a^{4}h^{6}}{4ah^{2}}

(d) A two term expression to factorise:

20x^{5}y + 12x^{3}y^{4}

Find

(a) The value of

f

, with the algebra written out - the mark scheme says working is required. (b) The simplified value of the power. (c) The quotient written as a single term, simplified fully. (d) The expression written as a product, factorised fully.

Plan the Solution

(a) The fraction is the awkward part, so remove it first: multiply every term on both sides by $3$ . Then gather the $f$ terms on one side to reach the form $af = b$ , and divide.
(b) Check the bracket cannot be zero, then apply the zero index law.
(c) Deal with the numbers and each letter separately: divide the numbers, and subtract the indices of $a$ and of $h$ .
(d) Build the highest common factor from the numbers and the LOWEST power of each letter, take it outside a bracket, then divide each term by it.

Worked Solution [8 marks]

Rule - Linear equations and index laws: multiply out to clear a fraction, then collect like terms;

\dfrac{x^{m}}{x^{n}} = x^{m-n}

x^{0} = 1

for any

x

that is not zero, and to factorise you take out the highest common factor of every term.

Step 1: (a) Multiply both sides by

3

to clear the fraction

\dfrac{2f}{3} \times 3 = (4f - 17) \times 3

2f = 12f - 51

(Reason: Multiplying by

3

cancels the denominator on the left. On the right BOTH terms inside the bracket are multiplied by

3

, so

4f

becomes

12f

and

17

becomes

51

Step 2: (a) Collect the

f

terms on one side

2f - 12f = -51

-10f = -51

(Reason: Subtracting

12f

from both sides leaves a two term equation in the form

af = b

, which is what the second method mark is for.)

Step 3: (a) Divide both sides by

-10

f = \dfrac{-51}{-10}

f = \dfrac{51}{10} = 5.1

(Reason: A negative divided by a negative is positive, so

f = \dfrac{51}{10}

. The mark scheme accepts the equivalent decimal

5.1

Step 4: (b) Use the zero index law

e > 0 \implies e + 12 > 12

(e + 12)^{0} = 1

(Reason: Anything except zero raised to the power

0

equals

1

. Because

e > 0

, the bracket is at least

12

and so can never be zero - the answer is

1

whatever

e

is.)

Step 5: (c) Divide the numbers, and subtract the indices letter by letter

\dfrac{12}{4} = 3

\dfrac{a^{4}}{a} = a^{4-1} = a^{3}

\dfrac{h^{6}}{h^{2}} = h^{6-2} = h^{4}

\dfrac{12a^{4}h^{6}}{4ah^{2}} = 3a^{3}h^{4}

(Reason: A quotient of terms splits into three separate divisions: the numbers, the

a

powers and the

h

powers. Dividing powers of the same letter means SUBTRACTING the indices, so

a

has index

4 - 1

and

h

has index

6 - 2

Step 6: (d) Build the highest common factor of the two terms

20x^{5}y = 4 \times 5 \times x^{3} \times x^{2} \times y

12x^{3}y^{4} = 4 \times 3 \times x^{3} \times y \times y^{3}

\text{HCF} = 4x^{3}y

(Reason: The largest number that divides both

20

and

12

4

. For each letter take the LOWEST power that appears in both terms:

x^{3}

and

y

Step 7: (d) Divide each term by the highest common factor and write the bracket

\dfrac{20x^{5}y}{4x^{3}y} = 5x^{2}

\dfrac{12x^{3}y^{4}}{4x^{3}y} = 3y^{3}

20x^{5}y + 12x^{3}y^{4} = 4x^{3}y(5x^{2} + 3y^{3})

(Reason: What is left inside the bracket is each original term divided by the common factor. The bracket

5x^{2} + 3y^{3}

has no common factor left -

5

and

3

share no factor and the two terms share no letter - so the factorisation is full.)

(a)

f = \dfrac{51}{10} = 5.1

(b)

1

(c)

3a^{3}h^{4}

(d)

4x^{3}y(5x^{2} + 3y^{3})

Verification

Check 1: (a) Put

f = 5.1

back into the original equation and work out each side on its own.

\dfrac{2 \times 5.1}{3} = \dfrac{10.2}{3} = 3.4

and

4 \times 5.1 - 17 = 20.4 - 17 = 3.4

, so the two sides agree.

Check 2: (b) Test the rule with a number: taking

e = 1

makes the bracket

13

13^{0} = 1

, and the same happens for

e = 0.5

e = 100

- the bracket is never zero, so the value is always

1

Check 3: (c) Multiply the answer back by the denominator; it must rebuild the numerator.

4ah^{2} \times 3a^{3}h^{4} = 12a^{4}h^{6}

, which is the term we started with, so the indices are

3

and

4

Check 4: (d) Expand the factorised form and compare it with the original expression.

4x^{3}y \times 5x^{2} = 20x^{5}y

and

4x^{3}y \times 3y^{3} = 12x^{3}y^{4}

, which add to give the expression in the question.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A correct first step: both sides multiplied by $3$ and expanded, e.g. $2f = 12f - 51$ or $2f = -51 + 12f$	M1	Also allowed: the right hand side written as two terms each over $2$ , e.g. $\dfrac{f}{3} = 2f - \dfrac{17}{2}$ , or $0.3f = 2f - 8.5$ , or $f = 6f - 25.5$ . Decimals to $1$ decimal place or better are accepted, rounded or truncated.	✓
(a) A correct two term equation in the form $af = b$ , e.g. $-10f = -51$ or $10f = 51$ or $5f = \dfrac{51}{2}$	M1	Follow through is allowed from these equations only: $2f = 12f - 17$ , $2f = 4f - 51$ and $6f = 12f - 51$ or equivalent. Decimals to $1$ decimal place or better, rounded or truncated.	✓
(a) $f = \dfrac{51}{10}$ or equivalent, with the algebra shown	A1	Dependent on at least the first method mark. Working is required, so an answer with no algebra scores no accuracy mark. Accept $5.1$ as an equivalent.	✓
(b) $1$	B1	Independent of the rest of the question. Any bracket that is not zero raised to the power $0$ gives $1$ , and $e > 0$ guarantees the bracket is not zero.	✓
(c) $3a^{3}h^{4}$ or equivalent	B2	B1 for a product in the form $ka^{p}h^{q}$ where $2$ of $k$ , $p$ , $q$ are correct, allowing multiplication signs, e.g. $5a^{3}h^{4}$ or $\dfrac{12a^{3}h^{4}}{4}$ . Allow $3a^{3}$ , $a^{3}h^{4}$ or $3h^{4}$ as long as it is not added to any other term.	✓
(d) $4x^{3}y(5x^{2} + 3y^{3})$	B2	B1 for any correct factorisation with at least a two term factor outside the bracket, e.g. $2x^{3}y(10x^{2} + 6y^{3})$ , $x^{3}y(20x^{2} + 12y^{3})$ , $2x(10x^{4}y + 6x^{2}y^{4})$ , $4y(5x^{5} + 3x^{3}y^{3})$ or $4x^{3}(5x^{2}y + 3y^{4})$ . B1 also for the correct highest common factor together with a two term expression containing at most one incorrect term, e.g. $4x^{3}y(5x^{2} + \text{...})$ .	✓

Full marks: 8/8

$\dfrac{3^{-2} \times 3^{5}}{3^{10}} = 3^{n}$

Work out the value of $n$ . [2 marks]

n =

[Total 2 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

One statement about powers,

\dfrac{3^{-2} \times 3^{5}}{3^{10}} = 3^{n}

Every power in it has the same base,

3

, so the index laws apply straight away.

Find

The value of the index

n

, which is a whole number here.

Plan the Solution

Tidy the numerator first: $3^{-2} \times 3^{5}$ is a multiplication of powers of the same base, so the indices add.
Then handle the fraction: dividing powers of the same base subtracts the indices.
That leaves a single power of $3$ . Compare it with $3^{n}$ and read off $n$ .
A calculator is allowed, so the whole thing can be checked as an ordinary fraction at the end.

Worked Solution [2 marks]

Rule - Index laws for the same base:

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

and

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

Step 1: Combine the two powers in the numerator

3^{-2} \times 3^{5} = 3^{-2 + 5} = 3^{3}

(Reason: The base is

3

on both, so the two powers combine into a single power of

3

and only the indices are worked with. A negative index is used exactly like any other index here.)

Step 2: Divide by the denominator

\dfrac{3^{3}}{3^{10}} = 3^{3 - 10} = 3^{-7}

(Reason: Dividing powers of the same base subtracts the index of the denominator from the index of the numerator. The denominator index

10

is the larger, so the result is a negative index.)

Step 3: Compare the single power with

3^{n}

3^{-7} = 3^{n}

n = -7

(Reason: Two powers of the same base are equal only when their indices are equal, so the index on the left is the value of

n

n = -7

Verification

Check 1: Work the left-hand side out as an ordinary number instead of using any index law. The numerator is

3^{-2} \times 3^{5} = 27

and the denominator is

3^{10} = 59049

, so the fraction is

\dfrac{27}{59049} = \dfrac{1}{2187}

3^{-7} = \dfrac{1}{3^{7}} = \dfrac{1}{2187}

, the same number, so

n = -7

is right.

Check 2: Take the two steps in the other order. Divide first:

\dfrac{3^{-2}}{3^{10}} = 3^{-12}

, then multiply by

3^{5}

3^{-12} \times 3^{5} = 3^{-12 + 5} = 3^{-7}

, so

n = -7

again, from a different order of working.

Check 3: A sign sanity check. Adding the indices above the line gives

-2 + 5 = 3

, and the index below the line is

10

, which is bigger. A smaller index on top than underneath must leave a negative index, and

-7

is negative. A positive answer such as

7

could be rejected on sight.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct application of an index rule as a first step, for example $3^{3}$ or $(3^{-2}) \times 3^{-5}$ or $\dfrac{3^{3}}{3^{10}}$ or $\dfrac{(3^{5})}{3^{12}}$ or $\dfrac{(3^{-2})}{3^{5}}$ or $3^{-12} (\times 3^{5})$ or equivalent. A correct calculation for $n$ also earns it: $-2 + 5 - 10$ or $-12 + 5$ or $3 - 10$ or equivalent.	M1	Method: one index law applied correctly to the same base, or the index arithmetic written out in full. Either route into the question scores this mark.	✓
$n = -7$	A1	Accuracy: the value $-7$ . Allow $3^{-7}$ as the answer.	✓
Working is not required on this question, so a correct answer alone scores full marks, unless it follows obviously incorrect working.	Note	Guidance row, carrying no mark of its own. It tells the marker that an unsupported $-7$ is worth the full $2$ marks.	✓

Full marks: 2/2

In a clearance sale at an electrical store, all normal prices are reduced by $17\%$
The sale price of a washing machine is $6225$ rupees.
Work out the normal price of the washing machine. [3 marks]

rupees

[Total 3 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

Every normal price in the sale is reduced by

17\%

The sale price of the washing machine is

6225

rupees

Find

The normal price of the washing machine, in rupees

Plan the Solution

The normal price is the whole amount, so it counts as $100\%$ . The reduction is taken off the NORMAL price, not off the sale price - that is the whole difficulty of the question.
Work out what percentage of the normal price the shopper actually pays, and write that percentage as a decimal multiplier.
The sale price is the normal price multiplied by that decimal, so divide the sale price by it to get back to the normal price.
Check by taking $17\%$ off the answer and confirming that $6225$ comes back.

Worked Solution [3 marks]

Rule - Reverse percentage: a reduction of

r\%

leaves

(100 - r)\%

of the original price, so

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

, where the multiplier is that remaining percentage written as a decimal. You divide by the multiplier - you never add the percentage back on.

Step 1: Work out the percentage that is left

100\% - 17\% = 83\%

83\% = \dfrac{83}{100} = 0.83

(Reason: (Reason: the normal price is the whole amount,

100\%

. Taking

17\%

off it leaves

83\%

of the normal price, and

83\%

written as a decimal is

0.83

.))

Step 2: Write the sale price in terms of the normal price

\text{normal price} \times 0.83 = 6225

(Reason: (Reason:

83\%

of the normal price is what the shopper pays, and what the shopper pays is

6225

rupees, so those two amounts are equal.))

Step 3: Divide to undo the multiplication

\text{normal price} = \dfrac{6225}{0.83} = 7500

(Reason: (Reason: the normal price was multiplied by

0.83

to give the sale price, so dividing the sale price by

0.83

takes it back to the normal price. Dividing by a number smaller than

1

makes the amount bigger, which is what a reversed reduction should do.))

Step 4: The same answer from

1\%

of the normal price

1\% \text{ of the normal price} = \dfrac{6225}{83} = 75

100\% \text{ of the normal price} = 75 \times 100 = 7500

(Reason: (Reason:

6225

rupees is

83\%

of the normal price, so dividing it by

83

gives

1\%

of the normal price. Multiplying that by

100

rebuilds the full

100\%

. The mark scheme allows this route as well, and it gives the same answer.))

7500

rupees

Verification

Check 1: Put the answer through the sale. Multiply it by

0.83

and see whether the sale price comes back.

7500 \times 0.83 = 6225

, which is exactly the sale price the question gives

Check 2: Work the reduction out as a percentage, which is a different calculation from Check 1: subtract to find how much came off, then compare that with the normal price.

7500 - 6225 = 1275

and

\dfrac{1275}{7500} = 0.17

, so the reduction really is

17\%

of the normal price

Check 3: Test the tempting wrong method, which is to add

17\%

back onto the sale price. If that worked,

17\%

of the sale price would equal the amount that came off.

6225 \times 0.17 = 1058.25

, but the amount that came off is

1275

, so

6225 + 1058.25 = 7283.25

is not the normal price

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Find the percentage, or the multiplier, that is left after the $17\%$ reduction	M1	For $1 - 0.17$ or $0.83$ or $\dfrac{83}{100}$ , or for $100(\%) - 17(\%)$ or $83(\%)$ , or for $\dfrac{6225}{83}\ (= 75)$ oe	✓
Divide the sale price by that multiplier	M1	For $\dfrac{6225}{0.83}$ or $\dfrac{6225}{83} \times 100$ or $\dfrac{6225 \times 100}{83}$ oe, or for $75 \times 100$ . The official scheme writes the multiplier in quotation marks, which means the candidate's own value from the first mark is followed through.	✓
State the normal price	A1	$7500$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

(a) Change $6.04 \times 10^{5}$ into an ordinary number. [1 mark]

(b) Express $0.000\,07$ in standard form. [1 mark]

(c) Work out $\dfrac{7.6 \times 10^{10}}{4 \times 10^{5} - 2 \times 10^{4}}$
Give your answer in standard form. [2 marks]

(a)(b)(c)

[Total 4 marks]

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Question 25 - Exam Solution

Understanding the Question

Given

A number written in standard form,

6.04 \times 10^{5}

An ordinary number,

0.000\,07

The fraction

\dfrac{7.6 \times 10^{10}}{4 \times 10^{5} - 2 \times 10^{4}}

, whose denominator is a subtraction.

Find

(a) The same value written as an ordinary number. (b) The same value written in standard form. (c) The value of the fraction, in standard form.

Plan the Solution

Parts (a) and (b) are one move each: multiplying by $10^{5}$ shifts the digits left, dividing by $10^{5}$ shifts them right.
Part (c) has an order to it. The denominator $4 \times 10^{5} - 2 \times 10^{4}$ is a subtraction, so it must be worked out as one number before any dividing happens.
Write both terms as ordinary numbers first, because $10^{5}$ and $10^{4}$ are different powers and cannot be subtracted directly.
Put the denominator back into standard form, then divide the front numbers and subtract the indices.
Finish by checking the front number of the answer is at least $1$ and less than $10$ .

Worked Solution [4 marks]

Rule - Standard form: every number is written

a \times 10^{n}

where

1 \leq a < 10

and

n

is a whole number. To divide two numbers in standard form, divide the values of

a

and subtract the indices:

\dfrac{a \times 10^{m}}{b \times 10^{n}} = \dfrac{a}{b} \times 10^{m-n}

Step 1: Part (a) - multiply

6.04

10^{5}

10^{5} = 100\,000

6.04 \times 100\,000 = 604\,000

(Reason: Multiplying by

10^{5}

moves every digit

5

places to the left, which is the same as moving the decimal point in

6.04

five places to the right. Zeros fill the empty places.)

Step 2: Part (b) - count the places in

0.000\,07

0.000\,07 = \dfrac{7}{100\,000}

\dfrac{7}{100\,000} = \dfrac{7}{10^{5}} = 7 \times 10^{-5}

(Reason: Standard form needs a front number with

1 \leq a < 10

, so the

7

must sit in the units place. Getting it there means moving the point

5

places, and because the number is smaller than

1

the index is negative.)

Step 3: Part (c) - work out the denominator first

4 \times 10^{5} = 400\,000

2 \times 10^{4} = 20\,000

400\,000 - 20\,000 = 380\,000

(Reason: The bottom of the fraction is a subtraction, and a fraction bar means the whole of the bottom is divided into the whole of the top.

10^{5}

and

10^{4}

are not the same power, so the two terms are turned into ordinary numbers before subtracting.)

Step 4: Part (c) - put the denominator in standard form and divide

380\,000 = 3.8 \times 10^{5}

\dfrac{7.6 \times 10^{10}}{3.8 \times 10^{5}} = \dfrac{7.6}{3.8} \times 10^{10 - 5}

= 2 \times 10^{5}

(Reason: Divide the front numbers and subtract the indices:

\dfrac{7.6}{3.8} = 2

and

10 - 5 = 5

. Since

1 \leq 2 < 10

, the answer is already in standard form and needs no adjusting.)

(a)

604\,000

(b)

7 \times 10^{-5}

(c)

2 \times 10^{5}

Verification

Check 1: Part (a) backwards: count how many places the point in

604\,000

has to move to leave a front number between

1

and

10

604\,000 = 6.04 \times 10^{5}

Check 2: Part (b) backwards:

10^{-5}

means one part in

100\,000

, so the answer should turn back into the number given.

\dfrac{7}{100\,000} = 0.000\,07

Check 3: Part (c) backwards: multiplying the answer by the denominator must rebuild the numerator.

2 \times 10^{5} \times 3.8 \times 10^{5} = 7.6 \times 10^{10}

Check 4: Part (c) a second way, with ordinary numbers only and no index laws at all.

\dfrac{76\,000\,000\,000}{380\,000} = 200\,000 = 2 \times 10^{5}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Write $6.04 \times 10^{5}$ as an ordinary number	B1	Answer of $604\,000$ . No working is needed for this mark.	✓
(b) Write $0.000\,07$ in standard form	B1	Answer of $7 \times 10^{-5}$ . The negative index is essential; $7 \times 10^{5}$ earns nothing.	✓
(c) Simplify the denominator	M1	For $380\,000$ or $3.8 \times 10^{5}$ or $38 \times 10^{4}$ or equivalent seen anywhere.	✓
(c) Divide and give the answer in standard form	A1	Answer of $2 \times 10^{5}$ . Accept $2.0 \times 10^{5}$ or $2.00 \times 10^{5}$ and so on. Accept a dot or a comma in place of the multiplication sign, for example $2 \, . \, 10^{5}$ .	✓
(c) Special case - the division is right but the form is not	SC B1	Awarded only when no other mark is earned in part (c), for a final answer of $200\,000$ or $20 \times 10^{4}$ or $0.2 \times 10^{6}$ or equivalent, or $2 \times 10^{n}$ with $n \neq 5$ . The student has divided correctly and then written the result in a form that is not standard form. It is not awarded when the denominator was simplified incorrectly.	✓
(c) Note on working	(no mark)	Working is not required in part (c), so a correct answer on its own scores both marks, unless it clearly follows from incorrect working.	✓

Full marks: 4/4

The diagram shows the hexagon $ABCDEF$ .

Angle $BCF = 30^\circ$
$AB$ , $FC$ and $ED$ are parallel.

Work out the total area of $ABCDEF$ .
You must show all of your working. [5 marks]

cm²

[Total 5 marks]

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Question 26 - Exam Solution

Understanding the Question

Given

ED = 23

cm,

FE = 4.7

cm, and the angles at

E

and

D

are right angles

AB = 11

cm and

BC = 5

Angle

BCF = 30^\circ

AB

FC

and

ED

are parallel, so the hexagon is a rectangle with a trapezium on top

Find

The area of hexagon

ABCDEF

, in

\text{cm}^2

Plan the Solution

Cut the hexagon along the dotted line $FC$ into rectangle $FEDC$ and trapezium $ABCF$ .
Use the right angles at $E$ and $D$ to say that $FEDC$ is a rectangle, so $FC = ED = 23$ cm.
The trapezium's own height is not given, so get it from the $5$ cm side and the $30^\circ$ angle at $C$ .
Work out the two areas separately, then add them.

Worked Solution [5 marks]

Rectangle:

\text{area} = \text{length} \times \text{width}

. Trapezium:

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

, where

a

and

b

are the parallel sides and

h

is the perpendicular distance between them. In a right-angled triangle,

\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}

Step 1: cut the hexagon along

FC

ABCDEF = FEDC + ABCF

(Reason: The hexagon has no area formula of its own, but the two pieces do.

FC

is parallel to

ED

and the angles at

E

and

D

are right angles, so

FEDC

is a rectangle;

AB

is parallel to

FC

, so

ABCF

is a trapezium.)

Step 2: the area of rectangle

FEDC

23 \times 4.7 = 108.1

(Reason: Opposite sides of a rectangle are equal, so

FC = ED = 23

cm and

DC = FE = 4.7

cm. The rectangle is therefore

23

cm by

4.7

cm.)

Step 3: the perpendicular height of trapezium

ABCF

\sin 30^\circ = \dfrac{h}{5}

h = 5 \sin 30^\circ

5 \times 0.5 = 2.5

(Reason: Drop a perpendicular from

B

onto

FC

. That makes a right-angled triangle whose hypotenuse is

BC = 5

cm, with the

30^\circ

angle at

C

. The height is the side opposite that angle, so it is the sine that is needed, and the height is

2.5

cm. The

5

cm is the slope, not the height.)

Step 4: the area of trapezium

ABCF

\text{area} = \dfrac{1}{2}(AB + FC) \times h

\dfrac{1}{2} \times (11 + 23) \times 2.5 = 42.5

(Reason: The two parallel sides are

AB = 11

cm and

FC = 23

cm, and

h

is the perpendicular distance between them found in Step 3.)

Step 5: add the two areas

108.1 + 42.5 = 150.6

(Reason: The rectangle and the trapezium meet along

FC

and do not overlap, so their areas simply add to give the area of the hexagon. The units are

\text{cm}^2

because an area is being measured.)

150.6\text{ cm}^2

Verification

Check 1: Build the shape the opposite way round. The hexagon sits inside a rectangle

23

cm by

4.7 + 2.5 = 7.2

cm, and what has to come off are the two corner triangles above

FC

. Their bases add to

23 - 11 = 12

cm and both have height

2.5

cm.

23 \times 7.2 = 165.6

and

\dfrac{1}{2} \times 12 \times 2.5 = 15

, so the hexagon is

165.6 - 15 = 150.6

Check 2: Do the trapezium without its formula. Split

ABCF

into a parallelogram

11

cm wide and a triangle of base

12

cm, both of height

2.5

cm.

11 \times 2.5 = 27.5

and

\dfrac{1}{2} \times 12 \times 2.5 = 15

, so the trapezium is

27.5 + 15 = 42.5

and the hexagon is

108.1 + 42.5 = 150.6

Check 3: Get the height without using sine at all. The horizontal step from

B

across to

C

5 \cos 30^\circ \approx 4.33

, and squaring it gives exactly

18.75

. Pythagoras in the same right-angled triangle then gives the height.

5^2 - 18.75 = 6.25

and

2.5^2 = 6.25

, so the height really is

2.5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The area of the rectangle: $23 \times 4.7 = 108.1$ oe	B1	Awarded whatever else happens in the question. It may be embedded in a single calculation for the whole shape, such as $23 \times (4.7 + 2.5) = 165.6$ .	✓
A correct method for the height of the trapezium: $\sin 30^\circ = \dfrac{x}{5}$ oe, where $x$ is that height	M1	The sine rule form $\dfrac{x}{\sin 30^\circ} = \dfrac{5}{\sin 90^\circ}$ scores this too, as does the Pythagoras route $5 \cos 30^\circ \approx 4.33$ together with $x^2 = 5^2 - (5 \cos 30^\circ)^2$ .	✓
The height itself: $x = 5 \sin 30^\circ = 2.5$ oe	M1	Or the same value from the other route, $x = \sqrt{5^2 - (5 \cos 30^\circ)^2} = 2.5$ .	✓
A correct method for the area of the trapezium, or for the whole shape: $\dfrac{1}{2} \times (11 + 23) \times 2.5 = 42.5$ oe	M1	Any correct complete method earns it, for example $(11 \times 2.5) + (\dfrac{1}{2} \times 12 \times 2.5) = 42.5$ , or the whole shape at once as $165.6 - 15 = 150.6$ . The candidate's own height carries through.	✓
Working required. $150.6$	A1	Not awarded without at least one method mark. Accept anything which rounds to $150.6$ ; allow $151$ ; accept $\dfrac{753}{5}$ .	✓

Full marks: 5/5

Keep revising

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

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