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Edexcel IGCSE 4MA1/2FR, Wednesday 4 June 2025: Worked Solutions, Questions 17 to 27

Sir Faraz Hassan

26 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)4MA1/2FR - Foundation Tier - Wednesday 4 June 2025100 marks · 2 hours · Calculator allowed

Back to questions 1 to 16

This is the rest of the paper. Questions 1 to 16, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/2FR (Foundation Tier), June 2025 series, sat Wednesday 4 June 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

All 27 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 17 to 27 of 27

(a) Factorise fully $18c - 45cd$ [2 marks]

(b) Solve the equation $\dfrac{5 - 2x}{6} = 3x - 4$
You must show clear algebraic working. [3 marks]

(a)(b) x =

[Total 5 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

Part (a): the expression

18c - 45cd

, which has two terms.

Part (b): the equation

\dfrac{5 - 2x}{6} = 3x - 4

, with a fraction on the left-hand side only.

Find

(a)

18c - 45cd

written as a product, with the highest common factor outside the brackets. (b) The value of

x

, with the algebraic working shown.

Plan the Solution

(a) Factorising fully means taking out the highest common factor, so deal with the numbers and the letters separately: first the HCF of $18$ and $45$ , then any letter that appears in both terms.
(a) Finish by checking that nothing inside the brackets could still come out. That is what the word fully is asking for.
(b) The fraction is the awkward part, so clear it first: multiply both sides by $6$ . The whole of the right-hand side is multiplied, not just its first term.
(b) Then gather the $x$ terms on one side and the numbers on the other, and divide.

Worked Solution [5 marks]

Factorising fully: take out the highest common factor of the numbers, together with every letter that appears in every term, then write what is left inside the brackets.
Solving an equation that carries a fraction: multiply both sides by the denominator, so

\dfrac{a}{6} = b

becomes

a = 6b

.
Whatever is done to one side of an equation is done to the whole of the other side.

Step 1 (a): Find the highest common factor of the numbers

18 = 2 \times 3 \times 3

45 = 3 \times 3 \times 5

3 \times 3 = 9

(Reason: Writing each number as a product of primes shows what the two share: two threes, and nothing else. So the largest number that divides both

18

and

45

9

Step 2 (a): Take out the common letter, then divide each term

\dfrac{18c}{9c} = 2

\dfrac{45cd}{9c} = 5d

18c - 45cd = 9c(2 - 5d)

(Reason:

c

is in both terms, so it comes out alongside the

9

, while

d

is in the second term only and has to stay inside. Dividing each term by

9c

gives what goes in the brackets, and the minus sign between the two terms is kept. Nothing more can come out:

2

and

5

share no factor, and no letter is in both parts of the bracket.)

Step 3 (b): Multiply both sides by 6 to clear the fraction

6 \times \dfrac{5 - 2x}{6} = 6(3x - 4)

5 - 2x = 18x - 24

(Reason: Multiplying by

6

cancels the denominator on the left. On the right the bracket is multiplied by

6

as a whole, so every term inside it is multiplied, not just the first one.)

Step 4 (b): Collect the x terms on one side and the numbers on the other

5 + 24 = 18x + 2x

29 = 20x

(Reason: Adding

2x

to both sides clears the

-2x

from the left, and adding

24

to both sides clears the

-24

from the right. The four terms are then sorted: numbers on one side,

x

terms on the other.)

Step 5 (b): Divide both sides by 20

x = \dfrac{29}{20}

x = 1.45

(Reason:

20

lots of

x

make

29

, so one

x

\dfrac{29}{20}

. The mark scheme accepts that fraction, the mixed number

1\dfrac{9}{20}

or the decimal

1.45

(a)

9c(2 - 5d)

(b)

x = 1.45

Verification

Check 1: Part (a): multiply the brackets out again. Every term inside is multiplied by

9c

9c \times 2 = 18c

and

9c \times (-5d) = -45cd

, giving

18c - 45cd

, the expression the question gives.

Check 2: Part (a) again, with numbers instead of letters, which tests the factorising without repeating the algebra. Put

c = 2

and

d = 3

into each form.

18 \times 2 - 45 \times 2 \times 3 = 36 - 270 = -234

, and

9 \times 2 \times (2 - 15) = 18 \times (-13) = -234

. The two forms agree.

Check 3: Part (b): put

x = 1.45

back into each side of the original equation separately, and see whether they meet. Left-hand side

\dfrac{5 - 2.9}{6} = \dfrac{2.1}{6} = 0.35

, right-hand side

3 \times 1.45 - 4 = 4.35 - 4 = 0.35

. The two sides are equal.

Check 4: Part (b) by a second method, never clearing the fraction at all. Split the left-hand side into

\dfrac{5}{6} - \dfrac{2}{6}x

, then collect:

\dfrac{5}{6} + 4 = 3x + \dfrac{1}{3}x

, so

\dfrac{29}{6} = \dfrac{10}{3}x

x = \dfrac{29}{6} \times \dfrac{3}{10} = \dfrac{29}{20} = 1.45

, the same value, reached without multiplying through.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$9c(2 - 5d)$	B2	for $9c(2 - 5d)$ or $-9c(5d - 2)$	✓
$9(2c - 5cd)$ or $c(18 - 45d)$ or $3c(6 - 15d)$ or $3(6c - 15cd)$	B1	for $9(2c - 5cd)$ or $c(18 - 45d)$ or $3c(6 - 15d)$ or $3(6c - 15cd)$ or $9c(p + qd)$ where $p$ and $q$ are non-zero integers or $(2 - 5d)$ as a factor	✓
eg $5 - 2x = 18x - 24$ or $\dfrac{5}{6} - \dfrac{2}{6}x = 3x - 4$	M1	for removal of the fraction and correctly multiplying out RHS by $6$ in an equation or separating fractions on the LHS in an equation	✓
$5 + 24 = 18x + 2x$ oe or $29 = 20x$ oe or $\dfrac{5}{6} + 4 = \dfrac{2}{6}x + 3x$ oe	M1ft	dep on $4$ terms, for correctly rearranging their $4$ term equation for terms in $x$ on one side of the equation and number terms on the other	✓
$1.45$ , working required	A1	dep on M1, oe eg $\dfrac{29}{20}$ or $1\dfrac{9}{20}$	✓

Full marks: 5/5

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 18 - Exam Solution

Understanding the Question

Given

The number

1400

Working is required, so an answer written down on its own earns nothing.

Find

1400

written as a product of powers of its prime factors.

Plan the Solution

Divide by the smallest prime, $2$ , for as long as the result is a whole number.
Move up through the primes - $3$ , then $5$ , then $7$ - dividing whenever one goes exactly.
Stop once what is left is prime, then write each repeated prime as a power.

Worked Solution [3 marks]

Rule - Prime factorisation: keep dividing by the smallest prime that goes exactly until the quotient is

1

. Every whole number greater than

1

is a product of primes in exactly one way, so the order the factors are found in cannot change the answer.

Step 1: Take out every factor of

2

\dfrac{1400}{2} = 700

\dfrac{700}{2} = 350

\dfrac{350}{2} = 175

(Reason:

1400

700

and

350

are all even, so

2

divides three times.

175

is odd, so

2

will not go again.)

Step 2: Move up to the next prime that divides,

5

\dfrac{175}{5} = 35

\dfrac{35}{5} = 7

(Reason:

175

is not a multiple of

3

- its digits add to

13

- but it ends in

5

, so

5

goes twice. What is left is

7

, which is prime, so the dividing stops there.)

Step 3: Collect the repeated factors into powers

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

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

(Reason: The three

2

s and the two

5

s are written as powers; the single

7

needs no index.)

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: Start somewhere else entirely. Split

1400

14 \times 100

and break each part down separately.

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

Check 3: Every base in the answer must itself be prime:

2

5

and

7

divide by nothing except

1

and themselves. All three bases are prime, so the answer really is a product of primes.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Prime factorisation started	M1	for finding $2$ prime factors after at least $2$ stages of prime factorisation with $0$ incorrect stages, or for finding $2$ prime factors after at least $3$ stages of prime factorisation with no more than $1$ incorrect stage. Examples of the amount of work needed: $2 \times 2 \times 350$ , $2 \times 7 \times 100$ , $2 \times 5 \times 140$ , $5 \times 5 \times 56$ , $7 \times 5 \times 40$ , or $2 \times 7 \times 25 \times 4$ reached from $14 \times 100 = 14 \times 25 \times 4$ .	✓
Note	Note	Each stage gives $2$ factors and may be shown in a factor tree, a table or a list, but $2$ prime factors must be seen. Example of finding $2$ prime factors after at least $3$ stages with $1$ incorrect stage: $1400 = 10 \times 14 = 2 \times 5 \times 2 \times 7$ .	✓
All six prime factors identified	M1 dep	dep on M1 - for the factors $2, 2, 2, 5, 5, 7$ identified with no others, in any form: listed, multiplied or added, for example $2 \times 2 \times 2 \times 5 \times 5 \times 7$ . Ignore $1$ s. May be seen in a fully correct factor tree or ladder.	✓
The factorisation written in index form	A1 dep on M2	dep on M2 - for $2^{3} \times 5^{2} \times 7$ . Working is required. May be in any order, and the multiplication signs may be written as dots.	✓

Full marks: 3/3

Work out the values of $x$ and $y$ that satisfy both of the simultaneous equations below.

$3x + 2y = 10$
$3x - 4y = 16$

You must show clear algebraic working. [3 marks]

x =y =

[Total 3 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

3x + 2y = 10

3x - 4y = 16

Two linear equations in the same two letters, and the

x

-terms are already identical.

Find

The one pair of values,

x

and

y

, that makes both equations true at once.

Plan the Solution

Both equations begin with $3x$ , so nothing has to be multiplied first: subtracting one equation from the other removes $x$ straight away.
Solve the single-letter equation that is left, to get $y$ .
Put that value back into one of the original equations, to get $x$ .
Test the pair in both original equations, not only the one used for the substitution.

Worked Solution [3 marks]

Rule - Elimination: when one letter has the same coefficient in both equations, subtracting one equation from the other removes that letter and leaves a single-letter equation to solve.

Step 1: Label the equations and compare the

x

-terms

3x + 2y = 10 \qquad (1)

3x - 4y = 16 \qquad (2)

(Reason: Both equations contain

3x

, so the coefficients of

x

already match and one subtraction will remove it.)

Step 2: Subtract equation (2) from equation (1)

(3x + 2y) - (3x - 4y) = 10 - 16

6y = -6

(Reason: The

x

-terms cancel because

3x - 3x = 0

, and subtracting

-4y

adds

4y

, so the left side is

6y

. On the right,

10 - 16 = -6

Step 3: Divide by

6

to find

y

y = \dfrac{-6}{6} = -1

(Reason: Dividing both sides of

6y = -6

6

leaves

y

on its own.)

Step 4: Substitute

y = -1

into equation (1) to find

x

3x + 2 \times (-1) = 10

3x - 2 = 10

3x = 12

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

(Reason: With

y

known, equation (1) has only one letter left in it. Adding

2

to both sides gives

3x = 12

, and dividing by

3

gives

x

x = 4

y = -1

Verification

Check 1: Put

x = 4

and

y = -1

into equation (1).

3 \times 4 + 2 \times (-1) = 12 - 2 = 10

Check 2: The same pair must also fit equation (2), which was not the equation used for the substitution.

3 \times 4 - 4 \times (-1) = 12 + 4 = 16

Check 3: Solve a second way: double equation (1) to give

6x + 4y = 20

, then add equation (2) so the

y

-terms cancel instead.

9x = 36

, so

x = 4

again, and equation (2) then gives

y = -1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $3x + 2y = 10$ subtract $3x - 4y = 16$ $(6y = -6)$ or eg $6x + 4y = 20$ add $3x - 4y = 16$ $(9x = 36)$ or eg $3\left(\dfrac{10 - 2y}{3}\right) - 4y = 16$ or $3\left(\dfrac{16 + 4y}{3}\right) + 2y = 10$ or eg $3x - 4\left(\dfrac{10 - 3x}{2}\right) = 16$ or $3x + 2\left(\dfrac{3x - 16}{4}\right) = 10$ or $6y = -6$ oe or $9x = 36$ oe	M1	a correct method to eliminate $x$ or $y$ : coefficients of $x$ or $y$ are the same and the correct operation to eliminate is selected; if operator not written, the correct operation can be implied by $2$ out of $3$ terms correct. Allow one arithmetic error if multiplying to equate coefficients, or for a correct substitution of one variable into the other equation. NB: the mark is for the method and not for the result of the method. However, if the correct result of this method is seen, the mark can be awarded.	✓
$3x + 2 \times (-1) = 10$ or $3x - 4 \times (-1) = 16$ or $x = \dfrac{10 - 2 \times (-1)}{3}$ or $x = \dfrac{16 + 4 \times (-1)}{3}$ or $3 \times 4 + 2y = 10$ or $3 \times 4 - 4y = 16$ or $y = \dfrac{10 - 3 \times 4}{2}$ or $y = \dfrac{3 \times 4 - 16}{4}$	M1	dep on M1 a correct substitution to find the value of the second variable using their value, or for starting again with elimination or substitution (as above). The printed scheme puts the substituted value in quotation marks, so the candidate's own value from the first mark is accepted here, not only $-1$ and $4$ .	✓
$x = 4$ and $y = -1$ , with working required	A1	dep on M1	✓

Full marks: 3/3

(a) On the grid above, draw the reflection of shape $A$ in the line $y = x$ [2 marks]

(b) On the grid above, draw the enlargement of shape $B$ with scale factor $2$ and centre $(1, 1)$ [2 marks]

[Total 4 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

Shape

A

is the trapezium with vertices

(5, 2)

(8, 2)

(7, 3)

and

(6, 3)

Shape

B

is the quadrilateral with vertices

(3, 2)

(5, 2)

(4, 4)

and

(3, 4)

The mirror line for part (a) is

y = x

. The centre for part (b) is

(1, 1)

and the scale factor is

2

Find

(a) The image of shape

A

after a reflection in

y = x

, drawn on the grid. (b) The image of shape

B

after an enlargement of scale factor

2

with centre

(1, 1)

, drawn on the grid.

Plan the Solution

(a) Work vertex by vertex. Reflecting in $y = x$ swaps the two coordinates of a point, so map all four vertices and then join them up in the same order.
(b) Work out from the centre. Count the steps across and up from $(1, 1)$ to a vertex, double both steps, and count that far again from the centre.
Each answer is the drawing itself, so there is no answer line to fill in - the marks are for the shape on the grid.

Worked Solution [4 marks]

Reflection in

y = x

(a, b) \rightarrow (b, a)

. Enlargement of scale factor

k

with centre

(p, q)

(a, b) \rightarrow (p + k(a - p), q + k(b - q))

Step 1: Read shape A off the grid

(5, 2), (8, 2), (7, 3), (6, 3)

(Reason: Every vertex sits on a grid intersection, so each coordinate can be counted straight off the axes. The trapezium has a base

3

squares long and a top edge

1

square long.)

Step 2: Swap the coordinates of every vertex

(5, 2) \rightarrow (2, 5)

(8, 2) \rightarrow (2, 8)

(7, 3) \rightarrow (3, 7)

(6, 3) \rightarrow (3, 6)

(Reason: The mirror line

y = x

makes the two axes change places, so every point

(a, b)

lands on

(b, a)

. Nothing is stretched, so the image is exactly the same size and shape as shape

A

Step 3: Join the four image points in the same order

(2, 5) \rightarrow (2, 8) \rightarrow (3, 7) \rightarrow (3, 6) \rightarrow (2, 5)

(Reason: Going round the object and round the image in the same order keeps the trapezium the right way round. Drawing the mirror line first is worth a mark on its own, even if the image then goes wrong.)

Step 4: Read shape B off the grid

(3, 2), (5, 2), (4, 4), (3, 4)

(Reason: The centre of enlargement

(1, 1)

lies outside shape

B

, so the image will sit further away from that point than the object does, on the same side of it.)

Step 5: Work out the image of the vertex (3, 4)

3 - 1 = 2

4 - 1 = 3

1 + 2 \times 2 = 5

1 + 2 \times 3 = 7

(3, 4) \rightarrow (5, 7)

(Reason: From the centre

(1, 1)

this vertex is

2

squares across and

3

squares up. Scale factor

2

doubles both steps, to

4

across and

6

up, and those doubled steps are then counted from the centre itself.)

Step 6: Repeat for the other three vertices

(3, 2) \rightarrow (5, 3)

(5, 2) \rightarrow (9, 3)

(4, 4) \rightarrow (7, 7)

(Reason: The same counting works every time. The vertex

(5, 2)

4

across and

1

up from the centre, so its image is

8

across and

2

up from the centre, at

(9, 3)

Step 7: Join the image points and check along the ray lines

(5, 3) \rightarrow (9, 3) \rightarrow (7, 7) \rightarrow (5, 7) \rightarrow (5, 3)

(Reason: A straight line from the centre through any vertex of shape

B

carries on to that vertex's image, and the image is twice as far out. Every edge of the image is twice the length of the edge it came from.)

(a) Image of shape

A

drawn with vertices

(2, 5)

(2, 8)

(3, 6)

and

(3, 7)

(b) Image of shape

B

drawn with vertices

(5, 3)

(5, 7)

(7, 7)

and

(9, 3)

Verification

Check 1: (a) A mirror line is the perpendicular bisector of the join, so the midpoint of

(5, 2)

and its image

(2, 5)

has to lie on

y = x

\dfrac{5 + 2}{2} = 3.5

across and

\dfrac{2 + 5}{2} = 3.5

up, so the midpoint

(3.5, 3.5)

sits on the line

Check 2: (a) A reflection cannot change the area. Both the trapezium and its image have parallel sides of

3

and

1

, with those two sides

1

square apart.

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

square units for each of them

Check 3: (b) The centre, a vertex and that vertex's image must lie on one straight line, with the image twice as far from the centre. Test it on

(5, 2)

. The vertex is

4

across and

1

up from

(1, 1)

, and the image

(9, 3)

8

across and

2

up, so

\dfrac{8}{4} = 2

agrees with

\dfrac{2}{1} = 2

Check 4: (b) An enlargement of scale factor

k

multiplies an area by

k^2

. Shape

B

covers

3

square units and its image covers

12

square units.

\dfrac{12}{3} = 4

and

2^2 = 4

, so the two areas agree with a scale factor of

2

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Vertices at $(2, 5)$ , $(2, 8)$ , $(3, 6)$ , $(3, 7)$	B2	for correct shape in correct position	✓
(a) Partial credit	B1	for correct orientation of shape but wrong position, or for $3$ out of $4$ vertices correct, or for $y = x$ drawn	✓
(b) Vertices at $(5, 3)$ , $(5, 7)$ , $(7, 7)$ , $(9, 3)$	B2	for correct shape in correct position	✓
(b) Partial credit	B1	for correct size and orientation of shape but wrong position, or for $3$ out of $4$ vertices correct	✓

Full marks: 4/4

Nathan is going to lay paving slabs on a patio for a customer.
The area of the patio is $45$ m²

Nathan buys one pack of paving slabs for each $1.5$ m² of patio area.
Each pack of paving slabs costs £ $64$

Nathan also buys $5$ bags of jointing sand.
Each bag of jointing sand costs £ $12$

Nathan charges the customer £ $3\,000$

Work out his percentage profit.
Give your answer correct to one decimal place. [5 marks]

[Total 5 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

The patio has an area of

45

m².

One pack of paving slabs covers

1.5

m², and one pack costs £

64

5

bags of jointing sand are bought, and each bag costs £

12

The customer is charged £

3\,000

Find

Nathan's percentage profit, correct to one decimal place.

Plan the Solution

Count the packs first: divide the area of the patio by the area one pack covers.
Turn everything bought into money and add it up, so the total cost is a single figure.
Profit is the money charged less that total cost.
Percentage profit measures the profit against what was spent, so divide the profit by the total cost - not by the amount charged - and multiply by $100$ .

Worked Solution [5 marks]

Rule - Percentage profit: take the total cost away from the money charged to find the profit, then divide the profit by the total cost and multiply by

100

. The cost is what the profit is being compared with, so the cost goes underneath in the fraction.

Step 1: Work out how many packs of slabs are needed

\dfrac{45}{1.5} = 30

(Reason: One pack covers

1.5

m², so the number of packs is the area of the patio divided by the area one pack covers. It divides exactly, giving

30

whole packs, so there is no part pack to round up.)

Step 2: Work out what the packs of slabs cost

30 \times 64 = 1\,920

(Reason: Every pack costs the same £

64

, so

30

packs cost

30

lots of £

64

Step 3: Work out what the jointing sand costs

5 \times 12 = 60

(Reason: The sand is priced by the bag, so

5

bags at £

12

each is

5

lots of £

12

. This cost does not depend on the area of the patio.)

Step 4: Add the two costs to find the total cost

1\,920 + 60 = 1\,980

(Reason: The slabs and the sand are the only things Nathan buys, so the two costs added together are everything he spends. This total is the figure the percentage will be measured against.)

Step 5: Take the total cost away from the money charged

3\,000 - 1\,980 = 1\,020

(Reason: Profit is what is left of the customer's money once everything Nathan has paid out has been taken off, so the whole of the total cost comes off the £

3\,000

Step 6: Write the profit as a percentage of the total cost

\dfrac{1\,020}{1\,980} \times 100 = 51.5151\ldots

(Reason: Percentage profit compares the profit with the money spent, so the profit goes on top and the total cost £

1\,980

goes underneath. Multiplying by

100

turns that fraction into a percentage. The division never stops: the digits

51

repeat for ever.)

Step 7: Round to one decimal place

51.5

(Reason: The question asks for one decimal place. The digit after the first decimal place is

1

, which is below

5

, so the

5

in the first decimal place stays as it is. The answer is a percentage, so it needs a per cent sign.)

51.5\%

Verification

Check 1: Put the percentage back. The profit is

\dfrac{17}{33}

of the cost, so adding that profit on to the total cost must rebuild the money the customer is charged.

1\,980 \times \dfrac{17}{33} = 1\,020

of profit, and

1\,980 + 1\,020 = 3\,000

, the amount charged.

Check 2: Cost the job a different way, by the square metre instead of by the pack. One pack covers

1.5

m² for £

64

, so one square metre of slabs costs

\dfrac{64}{1.5}

pounds.

\dfrac{64}{1.5} \times 45 = 1\,920

for the slabs, and

1\,920 + 60 = 1\,980

altogether, the same total cost as before.

Check 3: Compare the money charged with the money spent in one go. The charge is some percentage of the cost, and the first

100

per cent of that only replaces what was spent, so the rest is the profit.

\dfrac{3\,000}{1\,980} \times 100 - 100 = 51.5

to one decimal place, the same answer by a different road.

Check 4: Test the size of the answer. A profit of exactly

50

per cent would mean charging half as much again as the job cost.

1\,980 \times 1.5 = 2\,970

, and the customer is charged £

3\,000

, a little more than that, so a percentage profit just above

50

is the right size.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A method to find the number of packs needed, or the cost of the sand, or the cost of the slabs per square metre	M1	for $\dfrac{45}{1.5} \, (= 30)$ , or for $5 \times 12 \, (= 60)$ , or for $\dfrac{64}{1.5} \, \left( = \dfrac{128}{3} = 42.6(6\ldots) \right)$ . The mark is for a method to find the number of packs needed, or the cost of the jointing sand, or the cost of the slabs per m².	✓
A method to find the cost of the packs of slabs	M1	for $\text{their } 30 \times 64 \, (= 1\,920)$ , or for $\text{their } 42.6(6\ldots) \times 45 \, (= 1\,920)$ . The printed scheme puts quotation marks round the $30$ and the $42.6(6\ldots)$ , so a candidate's own earlier value may be used here.	✓
A method to find the total cost, or the profit	M1	for $\text{their } 1\,920 + \text{their } 60 \, (= 1\,980)$ , or for $3\,000 - 1\,920 - 60 \, (= 1\,020)$ . The first alternative is printed in quotation marks, so a candidate's own two costs may be used; the second is printed without them.	✓
A method to find the percentage profit, or to be one step away from it	M1	for example $\dfrac{3\,000 - \text{their } 1\,980}{\text{their } 1\,980} \, (= 0.515\ldots)$ , or $\dfrac{3\,000 - \text{their } 1\,980}{\text{their } 1\,980} \times 100$ , or $\dfrac{3\,000}{\text{their } 1\,980} \, (= 1.515\ldots)$ , or $\dfrac{3\,000}{\text{their } 1\,980} \times 100 \, (= 151.5\ldots)$ , or $\dfrac{3\,000}{\text{their } 1\,980} \times 100 - 100$ . Every printed alternative puts the total cost in quotation marks, so the candidate's own total cost may be used throughout.	✓
The percentage profit	A1	for $51.5$ , awrt $51.5$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
Special case - the jointing sand left out of the total cost: an answer of $56.3$ or $56.25$	SC	Special case $\text{B3}$ for an answer of $56.3$ or $56.25$ , which comes from using $1\,920$ instead of $1\,980$ as the total cost.	✓

Full marks: 5/5

(a) Write down the value of $5^{0}$ [1 mark]

$\dfrac{5^{9} \times 5^{-3}}{5^{-2}} = 5^{k}$

(b) Work out the value of $k$ [2 marks]

(a)(b) k =(c)

[Total 5 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

Part (a) gives the power

5^{0}

, a base with index

0

Part (b) gives

\dfrac{5^{9} \times 5^{-3}}{5^{-2}} = 5^{k}

, so both sides are powers of the same base.

Part (c) gives

\left(2d^{4}e^{5}\right)^{3}

, a number and two letters multiplied together inside one bracket, and the whole bracket is cubed.

Find

The value of

5^{0}

. The value of

k

\left(2d^{4}e^{5}\right)^{3}

written in its simplest form.

Plan the Solution

Part (a) is the zero-index rule on its own. Any base except $0$ raised to the index $0$ is $1$ .
Part (b): tidy the top of the fraction first with the multiplication rule, then divide by the bottom with the division rule, so the left-hand side becomes one single power of $5$ .
Once each side is a single power of the same base, the two indices must be equal, and that gives $k$ without ever working out a large number.
Part (c): a power outside a bracket reaches every factor inside it, so cube the $2$ and multiply each letter's index by $3$ .

Worked Solution [5 marks]

Rule - The index laws:

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

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

\left(a^{m}\right)^{n} = a^{mn}

and

a^{0} = 1

for every

a

that is not

0

. A power outside a bracket applies to every factor inside it, so

\left(ab\right)^{n} = a^{n}b^{n}

Step 1: Part (a) - use the zero-index rule

\dfrac{5^{3}}{5^{3}} = 5^{3-3} = 5^{0}

\dfrac{125}{125} = 1

5^{0} = 1

(Reason: A power divided by itself is

1

, and the division rule turns that same division into an index of

3-3

. So an index of

0

has to mean

1

. This is the value of the power, not the value of the base, so the

5

does not appear in the answer.)

Step 2: Part (b) - multiply the two powers on the top

5^{9} \times 5^{-3} = 5^{9 + (-3)} = 5^{6}

(Reason: Multiplying powers of the same base adds the indices. Adding

-3

is the same as taking

3

away, so the index on top becomes

6

Step 3: Part (b) - divide by the power underneath

\dfrac{5^{6}}{5^{-2}} = 5^{6 - (-2)} = 5^{8}

(Reason: Dividing powers of the same base takes the bottom index away from the top index. The index underneath is negative here, and taking a negative number away is the same as adding it on, so watch the two minus signs meeting.)

Step 4: Part (b) - match the indices on the two sides

5^{8} = 5^{k}

k = 8

(Reason: Both sides are now a single power of the same base

5

, and one power of

5

can only equal another when their indices are equal. The question asks for

k

itself, so the answer is the index, not the power.)

Step 5: Part (c) - give every factor inside the bracket the power 3

\left(2d^{4}e^{5}\right)^{3} = 2^{3} \times \left(d^{4}\right)^{3} \times \left(e^{5}\right)^{3}

(Reason: The bracket holds three factors multiplied together, and cubing the bracket cubes each one of them. Leaving the

2

untouched is the usual slip here, because it carries no visible index.)

Step 6: Part (c) - work out each of the three factors

2^{3} = 8

\left(d^{4}\right)^{3} = d^{4 \times 3} = d^{12}

\left(e^{5}\right)^{3} = e^{5 \times 3} = e^{15}

(Reason: The number is cubed, not multiplied by

3

2 \times 2 \times 2 = 8

. A power raised to a power multiplies the indices, so

4

becomes

12

and

5

becomes

15

Step 7: Part (c) - write the three factors as one expression

\left(2d^{4}e^{5}\right)^{3} = 8d^{12}e^{15}

(Reason: The number and the two letters are all different, so nothing here will combine or cancel any further. That is what fully simplified means for an expression like this one.)

(a)

1

(b)

k = 8

(c)

8d^{12}e^{15}

Verification

Check 1: Test part (a) without using the zero-index rule at all. Step down the powers of

5

, where each step divides by

5

125

25

5

. One more step reaches

5^{0}

\dfrac{5}{5} = 1

, so

5^{0}

1

, the same as part (a).

Check 2: Work the left-hand side of part (b) out as an ordinary number instead of as indices.

5^{-3}

means

\dfrac{1}{125}

, and dividing by

5^{-2}

is the same as multiplying by

25

5^{9} \times 5^{-3} = 15\,625

, then

15\,625 \times 25 = 390\,625

, and

5^{8} = 390\,625

, so the index really is

8

Check 3: Reach part (b) a third way, by writing an equation in the indices alone and never touching a power. The top index is

9 - 3

and the bottom index is

-2

9 - 3 = k - 2

, and

6 + 2 = 8

, so

k = 8

again.

Check 4: Test part (c) with numbers. Put

d = 1

and

e = 2

into the question and into the answer. Two different expressions cannot agree for every substitution, so agreement is real evidence, and this pair tests the index on

e

\left(2 \times 1 \times 32\right)^{3} = 64^{3} = 262\,144

, and

8 \times 1 \times 32\,768 = 262\,144

Check 5: Now swap the two numbers over,

d = 2

and

e = 1

, which tests the index on

d

and leaves

e

out of the way.

\left(2 \times 16 \times 1\right)^{3} = 32^{3} = 32\,768

, and

8 \times 4\,096 \times 1 = 32\,768

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The value of $5^{0}$	B1	for $1$ , cao.	✓
(b) One correct application of an index rule, or a correct equation in the indices, or a complete method for $k$	M1	for one correct application of an index rule (must be seen in powers of $5$ ), eg $(5^{9} \times 5^{-3} =) \, 5^{6}$ or $\left( \dfrac{5^{9}}{5^{-2}} = \right) \, 5^{11}$ or $\left( \dfrac{5^{-3}}{5^{-2}} = \right) \, 5^{-1}$ or $(5^{k} \times 5^{-2} =) \, 5^{k-2}$ ; this could be after an initial mistake - working will need to be clearly seen. Or for forming a correct equation in the indices alone, $9 - 3 = k - 2$ oe. Or for a complete method for the value of $k$ , $9 - 3 - (-2)$ or $9 - 3 + 2$ .	✓
(b) The value of $k$	A1	for $8$ , condone $5^{8}$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
(c) The expression simplified fully	B2	for a correct answer, $8d^{12}e^{15}$ .	✓
(c) Partial credit inside that B2	(B1)	(B1 for answer of the form $kd^{m}e^{n}$ where at least two of $k = 8$ , $m = 12$ and $n = 15$ are correct)	✓

Full marks: 5/5

A silver pendant has a mass of $48.3$ g
Silver has a density of $10.5$ g/cm³

Calculate the volume of the pendant. [2 marks]

cm³

[Total 2 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

The pendant has a mass of

48.3

Silver has a density of

10.5

g/cm³

Density is the mass of one cubic centimetre of the material.

Find

The volume of the pendant, in cubic centimetres.

Plan the Solution

Start from the density formula, $\text{density} = \dfrac{\text{mass}}{\text{volume}}$ .
Substitute the two values the question gives, then rearrange so the volume is on its own.
Divide the mass by the density, then multiply back to check the mass returns to $48.3$ g.

Worked Solution [2 marks]

Rule - Density:

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

, so

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

Step 1: Write down the density formula

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

(Reason: Density compares mass with volume, so this formula is the link between the two values the question gives.)

Step 2: Put in the values given

10.5 = \dfrac{48.3}{v}

(Reason: The letter

v

stands for the volume in cm³. Substituting

10.5

and

48.3

correctly into a correct density formula is what earns the method mark.)

Step 3: Rearrange to make the volume the subject

10.5 \times v = 48.3

v = \dfrac{48.3}{10.5}

(Reason: Multiply both sides by

v

, then divide both sides by

10.5

. The mark scheme accepts any one of these three forms for the method mark.)

Step 4: Work out the division

\dfrac{48.3}{10.5} = 4.6

(Reason: A calculator is allowed on this paper, so the volume is one division. The same value as a fraction is

\dfrac{23}{5}

, or

4\dfrac{3}{5}

written as a mixed number.)

4.6

cm³

Verification

Check 1: Multiply the volume back by the density:

10.5 \times 4.6

. This must return the mass the question states.

10.5 \times 4.6 = 48.3

, the mass in the question.

Check 2: Estimate the size of the answer.

48.3

is close to

50

and

10.5

is close to

10

\dfrac{50}{10} = 5

, so a volume a little under

5

cm³ is the right size.

Check 3: Clear the decimals instead of using a calculator: multiply the top and the bottom of the fraction by

10

\dfrac{483}{105} = \dfrac{23}{5} = 4.6

, one of the equivalent forms the mark scheme allows.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$10.5 = \dfrac{48.3}{v}$ or $10.5v = 48.3$ or $(v =) \dfrac{48.3}{10.5}$	M1	oe for substituting $10.5$ and $48.3$ correctly into a correct formula for density; may use any letter for the volume	✓
$4.6$ (a correct answer scores full marks, unless it comes from obvious incorrect working)	A1	allow $\dfrac{23}{5}$ or $4\dfrac{3}{5}$ oe	✓

Full marks: 2/2

A list of $7$ numbers has a mean of $60$

A group of $3$ of these numbers has a mean of $46$

Work out the mean of the remaining $4$ numbers. [3 marks]

[Total 3 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

A list of

7

numbers has a mean of

60

A group of

3

of those numbers has a mean of

46

The two groups together make up the whole list, because

3 + 4 = 7

Find

The mean of the remaining

4

numbers.

Plan the Solution

A mean is a total shared out equally, so a mean can be turned back into a total: $\text{total} = \text{mean} \times \text{how many}$ .
Turn both means into totals: the total of all $7$ numbers, and the total of the $3$ numbers.
Subtract to leave the total of the other $4$ numbers, then share that total between $4$ .

Worked Solution [3 marks]

Rule - Mean:

\text{mean} = \dfrac{\text{total}}{\text{how many}}

, so

\text{total} = \text{mean} \times \text{how many}

Step 1: Turn the overall mean back into a total

60 \times 7 = 420

(Reason: A mean of

60

across

7

numbers means the

7

numbers share out to

60

each, so together they come to

60 \times 7

. Either this total or the one in Step 2 earns the first method mark.)

Step 2: Turn the second mean back into a total the same way

46 \times 3 = 138

(Reason: The same rule applied to the smaller group:

3

numbers with a mean of

46

add up to

46 \times 3

Step 3: Subtract to leave the total of the other 4 numbers

420 - 138 = 282

(Reason: The

3

numbers and the other

4

numbers together are the whole list, so taking the smaller total away from the whole total leaves exactly the total of the numbers that remain. This is the second method mark.)

Step 4: Share that total between the numbers that are left

\dfrac{282}{4} = 70.5

(Reason: There are

4

numbers left, so their total is divided by

4

to give their mean. A calculator is allowed on this paper, so this is one division; the same value written as a fraction is

\dfrac{141}{2}

, or

70\dfrac{1}{2}

as a mixed number.)

70.5

Verification

Check 1: Put the list back together. Add the total of the

3

numbers to the total of the other

4

, then divide by

7

. That must return the mean the question starts from.

138 + 282 = 420

and

\dfrac{420}{7} = 60

, the mean in the question.

Check 2: Balance the list about its overall mean. The

3

numbers average

14

below

60

, a shortfall of

3 \times 14 = 42

, and the other

4

numbers have to carry that whole shortfall between them.

\dfrac{42}{4} = 10.5

above the overall mean, and

60 + 10.5 = 70.5

Check 3: Write down a list that fits both statements. Take

44

46

48

, whose mean is

46

, and take

68

70

71

73

as the other four. All seven add to

420

, so their mean is

60

, and the four that are left add to

282

, giving a mean of

70.5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$60 \times 7 \, (= 420)$ or $46 \times 3 \, (= 138)$	M1	may be embedded within an equation	✓
$\text{"}420\text{"} - \text{"}138\text{"} \, (= 282)$	M1	for a method to find the sum of the $4$ numbers. Allow this mark if they do further incorrect work using $282$ . The quotation marks mean the candidate's own earlier totals may be used.	✓
$70.5$ (a correct answer scores full marks, unless it comes from obvious incorrect working)	A1	allow $\dfrac{141}{2}$ oe eg $\dfrac{282}{4}$ or $70\dfrac{1}{2}$	✓

Full marks: 3/3

In its winter sale, a kitchen shop in Lucerne reduces the normal price of every appliance by $15\%$

The sale price of an espresso machine is $612$ Swiss francs.

Work out the normal price of the espresso machine. [3 marks]

Swiss francs

[Total 3 marks]

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Question 25 - Exam Solution

Understanding the Question

Given

In the sale, the normal price of every appliance is reduced by

15\%

The sale price of the espresso machine is

612

Swiss francs.

The reduction is a percentage of the normal price, not of the sale price, and the normal price is the thing that has not been given.

Find

The normal price of the espresso machine, in Swiss francs.

Plan the Solution

The normal price is the whole amount, so it counts as $100\%$ . Taking $15\%$ off leaves $85\%$ of it, so the sale price is the normal price multiplied by $0.85$ .
That multiplication has already happened, so it has to be undone. Dividing the sale price by $0.85$ gets back to the price the shop started from.
Finish by taking $15\%$ off the answer and looking for $612$ again.

Worked Solution [3 marks]

Rule - Reverse percentage:

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

, so

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

, where the multiplier for a reduction of

15\%

0.85

Step 1: Work out what percentage of the normal price is left

100\% - 15\% = 85\%

1 - 0.15 = 0.85

(Reason: The normal price is the whole amount, so it is

100\%

. A reduction of

15\%

leaves

85\%

of it, and

85\%

written as a decimal is

0.85

. Either form earns the first method mark, and so does going straight to

\dfrac{612}{85}

Step 2: Say what the shop did to the normal price

\text{normal price} \times 0.85 = 612

(Reason: The shop started at the normal price and multiplied it by

0.85

to get the ticket in the window. This is the sentence the whole question turns on: the

15\%

is a share of the normal price, which is bigger than

612

, so it is worth more than

15\%

612

Step 3: Undo the multiplication

\text{normal price} = \dfrac{612}{0.85}

(Reason: Dividing by

0.85

reverses multiplying by

0.85

, so it takes the sale price back to the price it came from. Writing this down is the complete method, and it earns the second method mark.)

Step 4: Work out the division

\dfrac{612}{0.85} = 720

(Reason: A calculator is allowed on this paper, so this is one division. Dividing by

0.85

is the same as multiplying by

\dfrac{100}{85}

, which cancels to

\dfrac{20}{17}

, and

612 \times \dfrac{20}{17} = 720

the same way.)

720

Swiss francs

Verification

Check 1: Run the sale forwards. Take

15\%

off the answer; the sale price in the question must come back.

0.15 \times 720 = 108

comes off, and

720 - 108 = 612

, the sale price the question gives.

Check 2: Work the answer out a different way. The sale price is

85\%

of the normal price, so divide it by

85

to find

1\%

, then take

100

of those.

\dfrac{612}{85} = 7.2

for

1\%

, and

7.2 \times 100 = 720

for the whole price.

Check 3: Is the size sensible? The money taken off is

15\%

of the larger normal price, so it has to be more than

15\%

of the smaller sale price.

0.15 \times 612 = 91.8

, while the money actually taken off is

720 - 612 = 108

, which is the larger of the two, as it has to be.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$1 - 0.15 \, (= 0.85)$ or $100(\%) - 15(\%) \, (= 85(\%))$ or $\dfrac{612}{85} \, (= 7.2)$ oe	M1	may be seen embedded. Do not allow $(1 - 15\%)$ unless processed correctly.	✓
$\dfrac{612}{\text{"}0.85\text{"}}$ oe or $\dfrac{612}{\text{"}85\text{"}} \times 100$ oe or $\text{"}7.2\text{"} \times 100$	M1	for a complete method. The quotation marks mean the candidate's own earlier value may be used here, so a wrong multiplier carried forward correctly still earns this mark.	✓
$720$	A1	A correct answer scores full marks, unless it comes from obvious incorrect working.	✓

Full marks: 3/3

The straight line $L$ is parallel to the line with equation $y = 2 - 5x$
$L$ passes through the point $(0, 6)$

Work out an equation of $L$ [2 marks]

[Total 2 marks]

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Question 26 - Exam Solution

Understanding the Question

Given

The line with equation

y = 2 - 5x

L

is parallel to that line.

L

passes through

(0, 6)

Find

An equation of

L

Plan the Solution

Rewrite the given line with the $x$ term first, so its gradient can be read straight off.
Parallel lines have equal gradients, so give $L$ that same gradient.
Substitute $(0, 6)$ into $y = mx + c$ to find the constant term.

Worked Solution [2 marks]

Rule - Parallel lines: two lines are parallel exactly when their gradients are equal. In

y = mx + c

m

is the gradient and

c

is the

y

-intercept.

Step 1: Write the given line as

y = mx + c

y = 2 - 5x

y = -5x + 2

(Reason: Swapping the two terms round does not change the line, and it puts the gradient in front of the

x

. The gradient of the given line is

-5

, because the minus sign belongs to the

5

, and its

y

-intercept is

2

Step 2: Give

L

the same gradient

m = -5

(Reason: Parallel lines rise at exactly the same rate, so being parallel is the same statement as having equal gradients.

L

therefore takes the gradient of the line it is parallel to, and its equation so far is

y = mx + c

with that gradient in place of

m

Step 3: Substitute

(0, 6)

to find

c

6 = -5 \times 0 + c

c = 6

(Reason: The point lies on

L

, so

x = 0

and

y = 6

must fit the equation.

-5 \times 0 = 0

, which leaves

c = 6

. An

x

-coordinate of

0

puts the point on the

y

-axis, so here the constant is the

y

-intercept itself.)

Step 4: Write the equation of the line

y = -5x + 6

(Reason: Putting the gradient and the intercept into

y = mx + c

gives the equation of

L

. Any equivalent rearrangement earns the mark, for example

y = 6 - 5x

y + 5x = 6

y = -5x + 6

Verification

Check 1: Put

x = 0

into the answer.

y = -5 \times 0 + 6 = 6

, so

(0, 6)

does lie on

L

Check 2: Compare the two lines written the same way round:

y = -5x + 2

and

y = -5x + 6

. The gradients are both

-5

, so the lines are parallel, and the intercepts

2

and

6

differ, so

L

is not the line it is parallel to.

Check 3: Measure the vertical gap between the two lines at

x = 1

and again at

x = 4

. At

x = 1

the gap is

1 - (-3) = 4

and at

x = 4

it is

-14 - (-18) = 4

. A gap that never changes is what parallel means.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $y = -5x$ $(+ k)$ or $y - a = -5(x - b)$ or eg $y = mx + 6$ or $y - 6 = m(x - 0)$ or eg $-5x + 6$ or $L = -5x + 6$	M1	for the equation of any line with gradient $-5$ other than $y = 2 - 5x$ or for the equation of any line passing through the point $(0, 6)$ or the correct line missing $y =$ or with the wrong subject	✓
$y = -5x + 6$	A1	oe equation eg $y = 6 - 5x$ or $y - 6 = -5(x - 0)$ or $y + 5x = 6$ Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 2/2

The diagram shows triangle $ADE$ and triangle $CDB$ .

$ABD$ is a straight line.
$AE = 28 \text{ cm}$ , $ED = 45 \text{ cm}$ , $AB = 21 \text{ cm}$ , $CD = 35 \text{ cm}$
$\text{angle } AED = \text{angle } CBD = 90^\circ$

Calculate the area of triangle $CDB$
Give your answer correct to $3$ significant figures. [5 marks]

cm²

[Total 5 marks]

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Question 27 - Exam Solution

Understanding the Question

Given

ABD

is a straight line, and

B

is the point on it with

AB = 21 \text{ cm}

Triangle

ADE

has a right angle at

E

, with

AE = 28 \text{ cm}

and

ED = 45 \text{ cm}

Triangle

CDB

has a right angle at

B

, with

CD = 35 \text{ cm}

Find

The area of triangle

CDB

, correct to

3

significant figures. Neither side of that triangle is given, so both have to be built out of the numbers that are.

Plan the Solution

Triangle $ADE$ is right-angled at $E$ , so Pythagoras turns $AE$ and $ED$ into the hypotenuse $AD$ .
Because $ABD$ is a straight line, $BD$ is simply $AD$ with $AB$ taken off it.
Triangle $CDB$ is right-angled at $B$ , so $CD$ is its hypotenuse and Pythagoras gives the shorter side $BC$ - a subtraction this time, not an addition.
$BD$ and $BC$ meet at a right angle, so they are the base and the height of triangle $CDB$ and the area is half their product.

Worked Solution [5 marks]

Rule - Pythagoras: in a right-angled triangle

a^2 + b^2 = c^2

, where

c

is the hypotenuse, so a shorter side comes from

a^2 = c^2 - b^2

. Area of a right-angled triangle

= \dfrac{1}{2} \times \text{base} \times \text{height}

Step 1: Find

AD

with Pythagoras in triangle

ADE

AD^2 = 28^2 + 45^2 = 784 + 2\,025 = 2\,809

AD = \sqrt{2\,809} = 53 \text{ cm}

(Reason: Angle

AED

90^\circ

, so

AD

is the hypotenuse of triangle

ADE

and the two given sides are squared and added.

2\,809

is a perfect square, so

AD

is exactly

53

Step 2: Take

AB

off

AD

to reach

BD

BD = AD - AB = 53 - 21 = 32 \text{ cm}

(Reason:

ABD

is a straight line, so

B

lies on

AD

and

AB + BD = AD

. This is the step that joins the two triangles:

BD

belongs to both of them.)

Step 3: Find

BC

with Pythagoras in triangle

CDB

BC^2 = 35^2 - 32^2 = 1\,225 - 1\,024 = 201

BC = \sqrt{201} = 14.177 \ldots \text{ cm}

(Reason: Angle

CBD

90^\circ

, so

CD = 35 \text{ cm}

is the hypotenuse and

BC

is one of the shorter sides. A shorter side is found by subtracting the squares, not by adding them. Keep

\sqrt{201}

exact so nothing is lost to rounding before the last line.)

Step 4: Work out the area of triangle

CDB

\text{Area of } CDB = \dfrac{1}{2} \times BD \times BC = \dfrac{1}{2} \times 32 \times \sqrt{201} = 16\sqrt{201}

16 \times 14.177446 = 226.8391

\text{Area} = 227 \text{ cm}^2 \text{ to 3 significant figures}

(Reason: The right angle at

B

makes

BD

and

BC

the base and the height, so the area is half their product.

\dfrac{1}{2} \times 32 = 16

, which is why the exact answer is

16\sqrt{201}

; rounding

226.8391 \ldots

gives

227

227 \text{ cm}^2

Verification

Check 1: Reach the area by trigonometry instead. In triangle

CDB

\cos BDC = \dfrac{32}{35}

, so angle

BDC \approx 23.8955^\circ

, and the area is

\dfrac{1}{2} \times CD \times BD \times \sin BDC

\dfrac{1}{2} \times 35 \times 32 \times \sin 23.8955^\circ = 226.839 \ldots

, which is

227

3

significant figures.

Check 2: Put

BC

back into Pythagoras. The squares of the two shorter sides of triangle

CDB

must add up to the square of the hypotenuse

CD

32^2 + 201 = 1\,024 + 201 = 1\,225 = 35^2

Check 3: Test

AD

the same way:

28

45

and

53

must be a Pythagorean triple.

28^2 + 45^2 = 784 + 2\,025 = 2\,809 = 53^2

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct method to find $AD^2$ , or angle $EDA$ , or angle $EAD$	M1	For example $28^2 + 45^2 = 784 + 2\,025 = 2\,809$ . Or, for angle $EDA$ , $\tan^{-1} \dfrac{28}{45} = 31.8 \ldots$ . Or, for angle $EAD$ , $\tan^{-1} \dfrac{45}{28} = 58.1 \ldots$ .	✓
A correct method to find $AD$	M1	For example $AD = \sqrt{28^2 + 45^2} = \sqrt{784 + 2\,025} = \sqrt{2\,809} = 53$ , or equivalent. Or, from the angle above, $\dfrac{28}{\sin 31.8 \ldots} = 53 \ldots$ or $\dfrac{45}{\cos 31.8 \ldots} = 53 \ldots$ or $\dfrac{45}{\sin 58.1 \ldots} = 53 \ldots$ or $\dfrac{28}{\cos 58.1 \ldots} = 53 \ldots$ .	✓
A correct method to find $BC$ , or angle $BDC$ , or angle $BCD$	M1	For example $BC = \sqrt{35^2 - (53 - 21)^2} = \sqrt{1\,225 - 1\,024} = \sqrt{201} = 14.1 \ldots$ . Or, for angle $BDC$ , $\cos^{-1} \dfrac{53 - 21}{35} = 23.8 \ldots$ . Or, for angle $BCD$ , $\sin^{-1} \dfrac{53 - 21}{35} = 66.1 \ldots$ . The printed scheme puts the $53$ in quotation marks, so the candidate's own value for $AD$ may be used in its place.	✓
A correct method to find the area of triangle $CDB$	M1	For example $\dfrac{1}{2} \times 14.1 \ldots \times (53 - 21)$ , giving $16\sqrt{201}$ . Or $\dfrac{1}{2} \times 35 \times (53 - 21) \times \sin 23.8 \ldots$ . Or $\dfrac{1}{2} \times 35 \times 14.1 \ldots \times \sin 66.1 \ldots$ . The printed scheme puts each of these values in quotation marks, so the candidate's own $AD$ , $BC$ and angle may be used.	✓
The answer	A1	Anything which rounds to $227$ . Accept $16\sqrt{201}$ .	✓
Guidance for the whole question	Note	A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

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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Edexcel IGCSE 4MA1/2FR, Wednesday 4 June 2025: Worked Solutions, Questions 17 to 27

Worked solutions, questions 17 to 27 of 27

Question 17, Calculator allowed

Question 17 - Exam Solution

Question 18, Calculator allowed

Question 18 - Exam Solution

Question 19, Calculator allowed

Question 19 - Exam Solution

Question 20, Calculator allowed

Question 20 - Exam Solution

Question 21, Calculator allowed

Question 21 - Exam Solution

Question 22, Calculator allowed

Question 22 - Exam Solution

Question 23, Calculator allowed

Question 23 - Exam Solution

Question 24, Calculator allowed

Question 24 - Exam Solution

Question 25, Calculator allowed

Question 25 - Exam Solution

Question 26, Calculator allowed

Question 26 - Exam Solution

Question 27, Calculator allowed

Question 27 - Exam Solution

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