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Edexcel IGCSE 4MA1/2HR, Wednesday 4 June 2025: Worked Solutions and Mark Schemes

Sir Faraz Hassan

2 Sept 2026

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

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/2HR (Higher 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.

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

Worked solutions, questions 1 to 12 of 26

(a) Fully factorise $18c - 45cd$ [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 1 - Exam Solution

Understanding the Question

Given

(a) The expression

18c - 45cd

(b) The equation

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

Find

(a)

18c - 45cd

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

x

, with clear algebraic working shown.

Plan the Solution

(a) Split each term into its number and its letters. The highest common factor is the largest number that divides both numbers, together with every letter that appears in both terms.
(a) Divide each term by that factor, put what is left inside one bracket, then expand to check that nothing is still waiting to come out.
(b) Multiply both sides by $6$ to clear the fraction. Every term on the right is multiplied, not only the first one.
(b) Gather the $x$ terms on one side and the numbers on the other, then divide by the coefficient of $x$ .

Worked Solution [5 marks]

Rule - Factorise fully:

ab - ac = a(b - c)

, where

a

is the highest common factor, so the bracket has nothing left to take out. Rule - Solve: an equation stays balanced only when whatever is done to one side is done to the whole of the other side.

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

18 = 2 \times 3^{2}

45 = 3^{2} \times 5

(Reason: The largest number dividing both

18

and

45

9

. The letter

c

is in both terms and

d

is in the second term only, so the highest common factor is

9c

Step 2: (a) Divide each term by 9c and write the bracket

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

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

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

(Reason: Taking

9c

out of each term leaves

2

and

5d

, and the minus sign between the terms is kept. Inside the bracket

2

and

5

share no factor and

d

is not in both terms, so the factorisation is full.)

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

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

5 - 2x = 18x - 24

(Reason: The fraction has denominator

6

, so multiplying both sides by

6

clears it. On the right, every term must be multiplied, not only the

3x

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 moves the

x

terms to the right, and adding

24

to both sides moves the numbers to the left. That leaves four terms rearranged into two.)

Step 5: (b) Divide by the coefficient of x

x = \dfrac{29}{20}

\dfrac{29}{20} = 1.45

(Reason: Dividing both sides by

20

leaves

x

on its own. The scheme accepts the fraction

\dfrac{29}{20}

and the mixed number

1\dfrac{9}{20}

as well as the decimal

1.45

(a)

9c(2 - 5d)

(b)

x = 1.45

Verification

Check 1: Multiply

9c

by each term inside the bracket and compare with the expression the question prints.

9c \times 2 = 18c

and

9c \times 5d = 45cd

, so the bracket expands back to

18c - 45cd

Check 2: Look inside the bracket for anything else that could come out: a number dividing both terms, or a letter appearing in both terms.

2

and

5

share no factor above

1

, and

d

is missing from the first term, so nothing is left to take out.

Check 3: Put

x = 1.45

into the left-hand side and the right-hand side of the original equation separately.

\dfrac{5 - 2 \times 1.45}{6} = 0.35

and

3 \times 1.45 - 4 = 0.35

, so the two sides agree.

Check 4: Turn the fraction answer and the mixed number into decimals and compare all three forms.

\dfrac{29}{20} = 1.45

and

1 + \dfrac{9}{20} = 1.45

, so

\dfrac{29}{20}

1\dfrac{9}{20}

and

1.45

are the same number.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B2	for $9c(2 - 5d)$ or $-9c(5d - 2)$	✓
(a)	(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	✓
(b)	M1	eg $5 - 2x = 18x - 24$ or $\dfrac{5}{6} - \dfrac{2}{6}x = 3x - 4$ . For removal of the fraction and correctly multiplying out RHS by $6$ in an equation, or separating fractions on the LHS in an equation.	✓
(b)	M1ft	$5 + 24 = 18x + 2x$ oe or $29 = 20x$ oe or $\dfrac{5}{6} + 4 = \dfrac{2}{6}x + 3x$ oe. 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.	✓
(b)	A1	Working required. $1.45$ , 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.
You must show your working clearly. [3 marks]

[Total 3 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

The number

1400

Nothing else is given, so every prime factor has to be found by dividing.

Find

1400

written as a product of powers of its prime factors, with the working shown.

Plan the Solution

Divide by the smallest prime that goes into the number, then keep dividing each quotient the same way.
Stop as soon as the quotient left is itself a prime number.
Count how many times each prime was used, and write that count as an index.

Worked Solution [3 marks]

Rule - Prime factorisation: divide by the smallest prime that goes in, repeat on each quotient until what is left is prime, then write a prime used

n

times as that prime to the power

n

Step 1: Divide by

2

while the number stays even

\dfrac{1400}{2} = 700

\dfrac{700}{2} = 350

\dfrac{350}{2} = 175

(Reason:

1400

is even, and so are

700

and

350

, so

2

divides three times.

175

is odd, so

2

will not go again.)

Step 2: Move on to the next prime that divides, which is

5

\dfrac{175}{5} = 35

\dfrac{35}{5} = 7

(Reason: The digits of

175

add to

13

, so

3

does not divide it, but it ends in

5

5

does.

35

ends in

5

as well, so

5

divides a second time.)

Step 3: Stop at a prime, and list every factor used

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

(Reason:

7

is prime, so the dividing stops there. The factors used are the three

2

s, the two

5

s and the

7

that is left, which is the list of six prime factors the question is asking to be written more neatly.)

Step 4: Collect equal primes into powers

2 \times 2 \times 2 = 2^{3}

5 \times 5 = 5^{2}

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

(Reason: A repeated prime is written as a power, so the number of times a prime was used becomes its index. A prime used once keeps no index, which is why the

7

is written on its own.)

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

Verification

Check 1: Multiply the powers back together:

2^{3} = 8

and

5^{2} = 25

, so

8 \times 25 = 200

and

200 \times 7 = 1400

1400

, the number the question started with.

Check 2: Start the factorising somewhere completely different:

1400 = 14 \times 100

, with

14 = 2 \times 7

and

100 = 2^{2} \times 5^{2}

. The same three primes and the same three indices,

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

, so the starting split does not change the answer.

Check 3: Confirm every base really is prime:

2

5

and

7

each have no factors apart from

1

and themselves. All three bases are prime, so this is a product of powers of prime factors and not just any factorisation.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Two prime factors reached from the start of the factorisation	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. Each stage gives $2$ factors and may be in a factor tree, in a table or listed; examples of the amount of work needed for the award of this mark are $2 \times 2 \times 350$ , $2 \times 7 \times 100$ , $2 \times 5 \times 140$ , $5 \times 5 \times 56$ , $7 \times 5 \times 40$ and $2 \times 7 \times 25 \times 4$ , but we want to see $2$ prime factors. Example of finding $2$ prime factors after at least $3$ stages with $1$ incorrect stage: $1400 = 10 \times 14 = 2 \times 5 \times 2 \times 7$ .	✓
Every prime factor identified, and nothing else	M1	Dependent on the first M1. For factors $2,\ 2,\ 2,\ 5,\ 5,\ 7$ identified with no others, in any form, for example listed, multiplied or added, such as $2 \times 2 \times 2 \times 5 \times 5 \times 7$ . Ignore $1$ s. It may be seen in a fully correct factor tree or ladder.	✓
The product written in index form	A1	Dependent on both method marks. $2^{3} \times 5^{2} \times 7$ . It may be in any order, and the multiplication signs may be written as dots. Working is required.	✓

Full marks: 3/3

Solve the simultaneous equations

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

You must show clear algebraic working. [3 marks]

x =y =

[Total 3 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

Two equations that must both be true at the same time:

3x + 2y = 10

3x - 4y = 16

The coefficient of

x

3

in each of them.

Find

The value of

x

and the value of

y

that satisfy both equations at once. Both equations are linear, so expect exactly one solution pair - not two.

Plan the Solution

Write the two equations one under the other and compare the coefficients.
The $x$ terms are already identical, so subtracting one equation from the other removes $x$ with no multiplying needed first.
Solve the single equation in $y$ that is left.
Substitute that value back into one of the original equations to find $x$ .
Test the pair in both original equations, because a pair that fits only one of them is not a solution.

Worked Solution [3 marks]

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

Step 1: line the equations up and compare the

x

terms

3x + 2y = 10 \quad \text{(1)}

3x - 4y = 16 \quad \text{(2)}

(Reason: The coefficient of

x

3

in both equations, so neither equation has to be multiplied to make the coefficients match.)

Step 2: subtract equation (2) from equation (1) to eliminate

x

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

6y = -6

y = -1

(Reason: Subtracting reverses the sign of every term of equation (2), so the

3x

terms cancel and an equation in

y

alone is left.)

Step 3: 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) holds one unknown only. Equation (2) would give the same value, and it is used as the second check below.)

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: Put the same pair into equation (2), which was not the equation used to find

x

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

Check 3: Solve again by the other elimination: double equation (1) to get

6x + 4y = 20

, then add equation (2) so that the

y

terms cancel.

9x = 36

, so

x = 4

, and equation (1) then gives

y = -1

again.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $3x + 2y = 10$ and $3x - 4y = 16$ subtracted, giving $(6y = -6)$ or eg $6x + 4y = 20$ and $3x - 4y = 16$ added, giving $(9x = 36)$ 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.	✓
eg $3x + 2 \times (-1) = 10$ or $3x - 4 \times (-1) = 16$ or $x = \dfrac{10 - 2 \times (-1)}{3}$ or $x = \dfrac{16 + 4 \times (-1)}{3}$ or eg $3 \times 4 + 2y = 10$ or $3 \times 4 - 4y = 16$ or $y = \dfrac{10 - 3 \times 4}{2}$ or $y = \dfrac{3 \times 4 - 16}{4}$ The printed scheme sets each of these values in quotation marks, so the candidate's own earlier value may be used in place of $-1$ or $4$ .	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)	✓
$x = 4$ $y = -1$ Working required	A1	dep on M1. The working column of the printed scheme states that working is required.	✓

Full marks: 3/3

Nathan is going to cover a studio wall with acoustic panels for a customer.
The area of the wall is $45$ m²

Nathan buys one pack of panels for each $1.5$ m² of wall area.
Each pack of panels costs £ $64$

Nathan also buys $5$ tubes of panel adhesive.
Each tube of panel adhesive costs £ $12$

Nathan charges the customer £ $3000$

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

[Total 5 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

The wall has area

45

m²

One pack of acoustic panels covers

1.5

m² and costs £

64

5

tubes of panel adhesive are bought, at £

12

each

The customer is charged £

3000

Find

The percentage profit, correct to one decimal place.

Plan the Solution

Divide the wall area by the area one pack covers, to get the number of packs.
Cost the packs, cost the adhesive, and add the two together to get what the job cost Nathan.
Take the total cost away from the amount charged, to get the profit.
Write the profit as a percentage of the COST. Percentage profit is always measured against what was spent, never against what was charged.

Worked Solution [5 marks]

Percentage profit:

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

, where

\text{profit} = \text{amount charged} - \text{total cost}

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

\dfrac{45}{1.5} = 30

(Reason: One pack covers

1.5

m², so the number of packs is the wall area divided by

1.5

. It comes out whole, and it fits the wall exactly:

30 \times 1.5 = 45

Step 2: Work out the cost of the panels

30 \times 64 = 1920

(Reason: Every pack costs the same £

64

, so the cost of the panels is the number of packs multiplied by the price of one pack.)

Step 3: Work out the total cost of the job

5 \times 12 = 60

1920 + 60 = 1980

(Reason: The adhesive is bought as well as the panels, so its cost is added on. The job costs Nathan £

1980

altogether.)

Step 4: Work out the profit

3000 - 1980 = 1020

(Reason: Profit is what the customer is charged minus what the job cost.)

Step 5: Write the profit as a percentage of the cost

\dfrac{1020}{1980} \times 100 = \dfrac{1700}{33}

\dfrac{1700}{33} \approx 51.5

(Reason: A calculator gives

51.5151\ldots

, and one decimal place is asked for, so it rounds to

51.5

. Notice what the profit is divided by: the cost

1980

, not the

3000

charged.)

51.5\%

Verification

Check 1: Put the profit back on to the cost. The total cost and the profit must rebuild the amount charged.

1980 + 1020 = 3000

, which is exactly what the customer pays.

Check 2: Reach the total cost by the other route the mark scheme allows: the cost per square metre first, then the whole wall.

\dfrac{64}{1.5} \times 45 = 1920

, and

1920 + 60 = 1980

, the same total cost by a different road.

Check 3: A size check. Half of the cost is

990

, so a fifty per cent profit would be a profit of £

990

. The actual profit of £

1020

is a little above £

990

, so the percentage must be a little above fifty per cent, and

51.5\%

is.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{45}{1.5}$ (= $30$ ) or $5 \times 12$ (= $60$ ) or $\dfrac{64}{1.5}$ (= $\dfrac{128}{3}$ , that is $42.6(6\ldots)$ )	M1	for a method to find the number of packs needed or the cost of adhesive or cost of panels per m²	✓
$30 \times 64$ (= $1920$ ) or $42.6(6\ldots) \times 45$ (= $1920$ )	M1	for a method to find the cost of the packs of panels. The printed scheme puts $30$ and $42.6(6\ldots)$ in quotation marks, so the candidate's own value from the row above may be used.	✓
$1920 + 60$ (= $1980$ ) or $3000 - 1920 - 60$ (= $1020$ )	M1	for a method to find the total cost or the profit. The printed scheme puts $1920$ and $60$ in quotation marks on the addition route, so the candidate's own earlier values may be used there; the subtraction alternative is printed unquoted.	✓
eg $\dfrac{3000 - 1980}{1980}$ (= $0.515\ldots$ ) or $\dfrac{3000 - 1980}{1980} \times 100$ or $\dfrac{3000}{1980}$ (= $1.515\ldots$ ) or $\dfrac{3000}{1980} \times 100$ (= $151.5\ldots$ ) or $\dfrac{3000}{1980} \times 100 - 100$	M1	for a method to find the percentage profit or be one step away. The printed scheme puts $1980$ in quotation marks, so the candidate's own total cost may be used.	✓
Correct answer scores full marks (unless from obvious incorrect working). Answer $51.5$	A1	awrt $51.5$	✓
answer $56.3$ or $56.25$ (from use of $1920$ instead of $1980$ as total cost)	SC B3	a special case in the printed scheme: this answer scores $3$ marks	✓

Full marks: 5/5

(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 5 - Exam Solution

Understanding the Question

Given

Shape

A

is a trapezium with vertices

(5, 2)

(8, 2)

(7, 3)

and

(6, 3)

Shape

B

is a trapezium with vertices

(3, 2)

(3, 4)

(4, 4)

and

(5, 2)

Part (a) reflects shape

A

in the line

y = x

Part (b) enlarges shape

B

by scale factor

2

with centre

(1, 1)

Find

The image of shape

A

after the reflection, drawn on the first grid. The image of shape

B

after the enlargement, drawn on the second grid.

Plan the Solution

Read the four vertices of each shape off the grid before doing anything else. Both parts are then done one vertex at a time.
For part (a), use the rule for a reflection in $y = x$ : the two coordinates of a point change places. Drawing the mirror line first is worth doing, and the mark scheme gives a mark for it on its own.
For part (b), measure each vertex FROM the centre $(1, 1)$ , double that step, then step out again from the centre. Doubling the coordinates themselves would enlarge from the origin, which gives a shape of the right size in the wrong place.
Plot the four image points, join them in the same order as the original, and label the image.

Worked Solution [4 marks]

Reflection in the line

y = x

sends

(x, y) \to (y, x)

. Enlargement with centre

(a, b)

and scale factor

k

sends

(x, y) \to (a + k(x - a), \; b + k(y - b))

Step 1: Read the vertices of shape A off the grid

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

(Reason: Shape

A

is a trapezium: its base is

3

squares long, its top is

1

square long, and both sloping sides rise one square as they move one square across. Everything in part (a) is done to these four points.)

Step 2: Reflect each vertex by swapping its coordinates

(5, 2) \to (2, 5)

(8, 2) \to (2, 8)

(7, 3) \to (3, 7)

(6, 3) \to (3, 6)

(Reason: The mirror line

y = x

is the line through

(0, 0)

and

(10, 10)

where the two coordinates are equal, and reflecting in it exchanges the roles of

x

and

y

. So

(5, 2)

goes to

(2, 5)

: the point that was

5

across and

2

up becomes

2

across and

5

up.)

Step 3: Plot the four image points and join them up

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

(Reason: Join them in the same order as the original, so the

3

-square side still faces the

1

-square side. The image is congruent to shape

A

: a reflection moves a shape and turns it over, and never changes its size.)

Step 4: Read the vertices of shape B off the grid

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

(Reason: Shape

B

is a trapezium with a vertical left-hand side

2

squares tall, a horizontal top

1

square long and a horizontal base

2

squares long.)

Step 5: Work out the image of the first vertex

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

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

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

(Reason: The vertex

(3, 2)

2

squares right of the centre and

1

square above it. Scale factor

2

doubles those steps to

4

and

2

, and they are stepped out from the centre

(1, 1)

, which is why the centre's own coordinate opens each line. Measuring from the origin instead is the classic slip: it gives a shape of exactly the right size and the right way round, sitting in the wrong place.)

Step 6: Do the same for the other three vertices

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

(4, 4) \to (7, 7)

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

(Reason: Each one is treated the same way: double the step from the centre, then step that doubled amount out from

(1, 1)

. For

(4, 4)

the step is

3

right and

3

up, which doubles to

6

and

6

, giving

(7, 7)

Step 7: Plot the four image points and join them up

(5, 3), (5, 7), (7, 7), (9, 3)

(Reason: Join them in the original order. Every side has doubled in length and stays parallel to the side it came from, which is what an enlargement of scale factor

2

looks like. The centre

(1, 1)

itself does not move, and a ray drawn from it through any vertex passes straight through that vertex's image.)

(a)

(2, 5)

(2, 8)

(3, 6)

(3, 7)

(b)

(5, 3)

(5, 7)

(7, 7)

(9, 3)

Verification

Check 1: Measure the sides of the image in part (a). A reflection is a congruence, so every side must keep the length it had. Both trapeziums have sides of

3

\sqrt{2}

1

and

\sqrt{2}

squares, so the image really is congruent to shape

A

Check 2: Take a vertex and its image and find the midpoint of the line joining them. If the reflection is right, that midpoint sits on the mirror line. For

(5, 2)

and

(2, 5)

the midpoint is

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

across and

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

up, and

(3.5, 3.5)

has equal coordinates, so it lies on

y = x

Check 3: Step from the centre

(1, 1)

to a vertex, and from the centre to that vertex's image. The second step must be exactly

2

times the first, or the image is not on the ray. To

(3, 2)

the step is

2

right and

1

up; to

(5, 3)

it is

2 \times 2 = 4

right and

2 \times 1 = 2

up, so the image vertex sits on the same ray, twice as far out.

Check 4: Count the area of shape

B

and the area of its image. An enlargement of scale factor

2

multiplies area by

2^2

, not by

2

. Shape

B

covers

3

squares and its image covers

12

squares, and

12 = 4 \times 3

, which is the area scale factor it should be.

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 (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 (B1 for correct size and orientation of shape but wrong position or for $3$ out of $4$ vertices correct)	✓

Full marks: 4/4

(a) State 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 6 - Exam Solution

Understanding the Question

Given

(a)

5^{0}

(b)

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

(c)

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

Three separate demands on the index laws. They share no working, so each is finished on its own.

Find

The value of

5^{0}

. The value of

k

, which is an index and not a power of

5

. The simplified form of

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

, with no bracket left in it.

Plan the Solution

Part (a) is the zero index read straight off. Any non-zero number raised to the power $0$ is $1$ , so there is nothing to calculate.
Part (b): every term is a power of $5$ , so leave the base alone and work in the indices only. Multiplying adds them, dividing subtracts them.
The index on the bottom is $-2$ , and subtracting a negative adds, so that division makes the index bigger rather than smaller. That is the whole difficulty of the part.
Part (c): the bracket holds a product, so the outside power $3$ lands on all three factors, the $2$ included. Then a power of a power multiplies the indices.

Worked Solution [5 marks]

Rule - 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}

(ab)^{n} = a^{n}b^{n}

, and

a^{0} = 1

for every non-zero

a

Step 1: part (a), read off the zero index

5^{0} = 1

(Reason: Any non-zero number raised to the power

0

1

. The division law forces it:

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

, and that fraction is

\dfrac{125}{125} = 1

Step 2: part (b), collapse the multiplication on the top

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

(Reason: The base is the same in both, so multiplying adds the indices.

9 + (-3) = 6

, so the whole top is a single power,

5^{6}

Step 3: part (b), divide by the power on the bottom

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

(Reason: The base is the same again, so dividing subtracts the indices. The index on the bottom is

-2

, so it is that whole negative number that gets subtracted, and subtracting a negative adds.)

Step 4: part (b), match the indices on the two sides

5^{8} = 5^{k}

k = 8

(Reason: Two equal powers of the same base must have equal indices. The question asks for

k

, so the answer is

8

and not

5^{8}

Step 5: part (c), put the outside power on to every factor

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

= 8 \times d^{4 \times 3} \times e^{5 \times 3} = 8d^{12}e^{15}

(Reason: A power outside a bracket applies to each factor inside it, so the

2

is cubed as well and becomes

8

, not

6

. A power of a power then multiplies the indices:

4 \times 3 = 12

and

5 \times 3 = 15

(a)

5^{0} = 1

(b)

k = 8

(c)

8d^{12}e^{15}

Verification

Check 1 - part (a) from the pattern of powers: Go down the powers of

5

, dividing by

5

each time:

5^{3} = 125

5^{2} = 25

5^{1} = 5

. One more division continues the pattern to

5^{0}

\dfrac{5}{5} = 1

, so the pattern lands on

5^{0} = 1

Check 2 - part (b) by working both sides out as numbers: Evaluate the left-hand side without any index law.

5^{9} \times 5^{-3} = \dfrac{1953125}{125} = 15625

, and dividing by

5^{-2}

multiplies by

25

15625 \times 25 = 390625

and

5^{8} = 390625

, so the index is

k = 8

Check 3 - part (c) by substituting numbers: Put

d = 2

and

e = 1

into the original bracket and into the simplified answer, and compare the two numbers.

\left(2 \times 2^{4} \times 1^{5}\right)^{3} = 32^{3} = 32768

and

8 \times 2^{12} \times 1^{15} = 8 \times 4096 = 32768

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $1$	B1	cao	✓
(b) eg $\left(5^{9} \times 5^{-3} =\right) 5^{6}$ or $\left(\dfrac{5^{9}}{5^{-2}} =\right) 5^{11}$ or $\left(\dfrac{5^{-3}}{5^{-2}} =\right) 5^{-1}$ or $\left(5^{k} \times 5^{-2} =\right) 5^{k-2}$ or $9 - 3 = k - 2$ oe or $9 - 3 - -2$ or $9 - 3 + 2$	M1	for one correct application of an index rule (must be seen in powers of $5$ ) this could be after an initial mistake - working will need to be clearly seen or for forming a correct equation in the indices alone or for a complete method for the value of $k$	✓
$8$ Correct answer scores full marks (unless from obvious incorrect working)	A1	condone $5^{8}$	✓
(c) $8d^{12}e^{15}$	B2	for a correct answer (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

The mass of a solid silver pendant is $48.3$ g
The density of silver is $10.5$ g/cm³

Work out the volume of the silver pendant. [2 marks]

cm³

[Total 2 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

Mass of the pendant:

48.3

Density of silver:

10.5

g/cm³

The pendant is solid silver right through, so that one density applies to the whole of it.

Find

The volume of the pendant, in cm³. Density ties mass and volume together, so this is the third quantity of a set of three with the other two given.

Plan the Solution

Density is a compound measure: it says how much mass sits in each cm³ of the material. One formula ties the three quantities together, so any two of them give the third.
Write that formula down first and put the two given values into it. The volume is the unknown, and in the formula it starts underneath, so it cannot simply be read off.
Rearrange to make the volume the subject, then divide. Grams divided by grams per cm³ leaves cm³, so the unit looks after itself and nothing needs converting.
A calculator is allowed, so the division is done directly. The numbers are chosen so that it terminates, and an answer that runs on and on is the sign that the mass and the density went into the fraction the wrong way up.

Worked Solution [2 marks]

Rule - Density:

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

, which rearranges to

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

Step 1: write the density formula and substitute the two given values

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

10.5 = \dfrac{48.3}{V}

(Reason: Density is mass per unit volume, so the mass goes on top and the volume underneath. Writing

V

for the volume in cm³ puts the unknown exactly where the formula puts it, and the letter itself does not matter.)

Step 2: rearrange to make the volume the subject

10.5 \times V = 48.3

V = \dfrac{48.3}{10.5}

(Reason: Multiplying both sides by

V

lifts it out of the denominator, and dividing both sides by

10.5

then leaves it alone. The mass finishes on top, which is the sensible way round: a heavier pendant of the same metal takes up more room.)

Step 3: work out the division

V = \dfrac{48.3}{10.5} = 4.6

(Reason: Grams divided by grams per cm³ leaves cm³, so the answer is already a volume with no conversion needed. Silver is dense, so a mass of nearly fifty grams is packed into only a few cm³.)

4.6 \text{ cm}^{3}

Verification

Check 1 - put the answer back into the density formula: Density multiplied by volume must return the mass the question gives. Multiply

10.5

by the answer and compare.

10.5 \times 4.6 = 48.3

, which is the mass in the question

Check 2 - do the same division as a fraction instead: Clear both decimals by multiplying top and bottom by

10

, then cancel by

21

. This is the form the mark scheme also accepts.

\dfrac{48.3}{10.5} = \dfrac{483}{105} = \dfrac{23}{5}

, and

\dfrac{23}{5} = 4.6

Check 3 - trap the answer between two whole numbers: One cm³ of silver has a mass of

10.5

g, so work out the mass of

4

cm³ and of

5

cm³ and see where the pendant falls between them.

10.5 \times 4 = 42

and

10.5 \times 5 = 52.5

, and

42 < 48.3 < 52.5

, so the volume lies between

4

and

5

cm³

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	✓
$4.6$ Correct answer scores full marks (unless 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 $60$

The mean of $3$ of the numbers in the list is $46$

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

[Total 3 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

A list of

7

numbers whose mean is

60

A group of

3

of those numbers whose mean is

46

The other

4

numbers are whatever is left of the list once that group of

3

is taken out, so the two groups do not overlap and between them they are the whole list.

Not one individual number is given, and none is needed.

Find

The mean of the remaining

4

numbers. A mean is asked for, not a total, so whatever else happens the last move is a division by

4

Plan the Solution

A mean on its own says nothing about how many numbers went into it, so the two means given here cannot be combined directly. Totals can be combined, and a mean turns into a total in one multiplication.
Multiply each mean by how many numbers it covers. The whole list gives one total; the group of $3$ gives another.
The group and the rest make up the list between them and share no numbers, so subtracting the smaller total from the larger leaves exactly the total of the remaining $4$ numbers.
Divide that total by $4$ , because that is how many numbers it covers. Nothing here forces a whole-number answer, and this one is not a whole number.

Worked Solution [3 marks]

Rule - Mean:

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

, which rearranges to

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

. Totals may be added and subtracted; means may not.

Step 1: turn the overall mean into the total of all 7 numbers

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

60 \times 7 = 420

(Reason: A mean of

60

spread over

7

numbers carries exactly the same information as a total of

420

shared equally between them, so nothing is lost by multiplying. The numbers themselves stay unknown and the question never needs them.)

Step 2: turn the mean of the smaller group into that group's total

46 \times 3 = 138

(Reason: The same rule applied to the group on its own: its mean multiplied by how many numbers are in that group gives the total those numbers contribute to the list. Which count belongs to which mean is the thing to be careful about here, because the question hands over two of them.)

Step 3: subtract to leave the total of the remaining 4 numbers

420 - 138 = 282

(Reason: The group of

3

and the remaining

4

together account for all

7

numbers and share none of them, so their two totals must add up to

420

. Taking one total away therefore leaves the other.)

Step 4: divide that total by how many numbers it covers

\dfrac{282}{4} = 70.5

(Reason: The total

282

belongs to

4

numbers, so dividing by

4

turns it back into a mean. It lands above

60

, which is what has to happen: the group taken out averaged below

60

, so what is left has to sit above

60

to pull the whole list back up to it.)

70.5

Verification

Check 1 - rebuild a list that fits both facts and take its mean directly: Nothing says the numbers are all different, so take the

3

46

each and the other

4

70.5

each. Both groups then have the right mean, so the mean of all

7

must come back to

60

3 \times 46 + 4 \times 70.5 = 138 + 282 = 420

, and

\dfrac{420}{7} = 60

, which is the mean the question gives

Check 2 - balance the distances from the overall mean: Measure every number against the overall mean of

60

instead of totalling anything. The

3

numbers average

14

below it, so they are short by

42

between them, and the other

4

have to be

42

above it between them to balance that out.

(60 - 46) \times 3 = 42

, then

\dfrac{42}{4} = 10.5

above the mean each, so

60 + 10.5 = 70.5

Check 3 - read the last division as a fraction: Leave the final division uncalculated and cancel it instead, which also gives the answer in the other forms the mark scheme accepts.

\dfrac{282}{4} = \dfrac{141}{2}

, which is

70.5

, also written

70\dfrac{1}{2}

, and

46 < 60 < 70.5

as it must be

Mark Scheme Breakdown

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

Full marks: 3/3

An electrical store reduces all of its normal prices by $15\%$ in a clearance sale.
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 9 - Exam Solution

Understanding the Question

Given

Every normal price in the store is reduced by

15\%

in the sale.

The sale price of the espresso machine is

612

Swiss francs.

Find

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

Plan the Solution

The $15\%$ comes off the NORMAL price, so the ticket in the sale shows $85\%$ of it.
That makes the sale price the normal price multiplied by $0.85$ , and the sale price is the value we already know.
So reverse the multiplication: divide $612$ by $0.85$ .
Finish by taking $15\%$ off the answer, which must bring it back to $612$ .

Worked Solution [3 marks]

Reverse percentage:

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

, so

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

. Dividing by the multiplier is what undoes the reduction.

Step 1: find the multiplier for a sale price

100\% - 15\% = 85\%

\dfrac{85}{100} = 0.85

(Reason: Taking

15\%

off leaves

85\%

of the normal price, and a percentage becomes a decimal multiplier when it is divided by

100

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

0.85 \times n = 612

(Reason: Let

n

be the normal price. The store multiplies every normal price by

0.85

, and for this espresso machine that gives the

612

Swiss francs on the sale ticket.)

Step 3: reverse the multiplication

n = \dfrac{612}{0.85} = 720

(Reason: Multiplying by

0.85

made the price smaller, so dividing by

0.85

undoes the reduction and returns the price before the sale.)

720

Swiss francs

Verification

Check 1: Take the reduction off the answer.

15\%

720

0.15 \times 720 = 108

, and that must leave the price on the sale ticket.

720 - 108 = 612

, the sale price the question gives

Check 2: Work it out a second way, by the unitary method. If

85\%

612

Swiss francs, then

1\%

\dfrac{612}{85} = 7.2

Swiss francs.

7.2 \times 100 = 720

, the same normal price by a different method

Check 3: Sanity check the size and the share. A sale price must be smaller than the normal price, and the amount taken off must be

15\%

of the normal price, not of the sale price.

720 > 612

and

\dfrac{108}{720} = 0.15

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 it is processed correctly.	✓
$\dfrac{612}{0.85}$ oe, where the $0.85$ may be the candidate's own value from the row above or $\dfrac{612}{85} \times 100$ oe, again with their own $85$ or $7.2 \times 100$ , with their own $7.2$	M1	For a complete method.	✓
Correct answer scores full marks (unless from obvious incorrect working)	A1	The answer is $720$ .	✓

Full marks: 3/3

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

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

[Total 2 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

A line with equation

y = 2 - 5x

The line

L

is parallel to that line

The line

L

passes through

(0, 6)

Find

An equation of the line

L

Plan the Solution

Write $y = 2 - 5x$ in the form $y = mx + c$ so its gradient can be read straight off.
Parallel means equal gradients, so give $L$ the same $m$ and leave $c$ unknown.
Substitute the coordinates of $(0, 6)$ to pin down $c$ , then write the equation out.

Worked Solution [2 marks]

Rule - Parallel lines: two straight lines are parallel exactly when their gradients are equal. So a line parallel to

y = mx + c

keeps the same

m

and only its

c

changes.

Step 1: Read the gradient of the given line

y = 2 - 5x

y = -5x + 2

(Reason: In the form

y = mx + c

the number multiplying

x

is the gradient, so putting the

x

-term first shows the given line has gradient

-5

. The

2

is that line's intercept and plays no part in the gradient.)

Step 2: Give

L

the same gradient

y = -5x + c

(Reason: Parallel lines have equal gradients, so

L

also has gradient

-5

. Only the intercept

c

is still unknown, and that is what the point is for.)

Step 3: Use the point

(0, 6)

to find

c

6 = -5 \times 0 + c

c = 6

(Reason: The point lies on

L

, so its coordinates satisfy

y = -5x + c

: substitute

x = 0

and

y = 6

. The

x

-term vanishes, which is why a point on the

y

-axis hands you the intercept directly.)

Step 4: Write the equation of

L

y = -5x + 6

(Reason: Putting the gradient

-5

and the intercept

6

back into

y = mx + c

gives the equation of

L

y = -5x + 6

Verification

Check 1 - the line really passes through the point: Put

x = 0

into

y = -5x + 6

-5 \times 0 + 6 = 6

, which is the

y

-coordinate of

(0, 6)

Check 2 - the gradient read back from two points: Take

x = 1

and

x = 3

L

-5 \times 1 + 6 = 1

and

-5 \times 3 + 6 = -9

. The gradient between those two points is

\dfrac{-9 - 1}{3 - 1} = -5

, the same gradient as

y = 2 - 5x

Check 3 - the gap between the two lines: The given line at those same two values gives

2 - 5 \times 1 = -3

and

2 - 5 \times 3 = -13

. The vertical gap is

1 - (-3) = 4

and

-9 - (-13) = 4

- the same at both values, so the two lines stay a constant distance apart and never meet.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $y = -5x \; (+ k)$ or $y - a = -5(x - b)$ or $y = mx + 6$ or $y - 6 = m(x - 0)$ or $-5x + 6$ 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 with $y =$ missing or with the wrong subject	✓
$y = -5x + 6$	A1	oe equation eg $y = 6 - 5x$ or $y - 6 = -5(x - 0)$ or $y + 5x = 6$	✓
Guidance printed beside the answer row	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 2/2

The diagram shows two triangles, $ADE$ and $CDB$ .

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

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

cm²

[Total 5 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

Two triangles,

ADE

and

CDB

, with

ABD

a straight line

AE = 28

cm,

ED = 45

cm,

AB = 21

cm,

CD = 35

angle

AED

= angle

CBD = 90^{\circ}

, so both triangles are right-angled

Find

The area of triangle

CDB

, correct to

3

significant figures

Plan the Solution

Triangle $AED$ is right-angled at $E$ and $AD$ is its hypotenuse, so Pythagoras gives $AD$ .
$B$ lies on $AD$ , so $BD$ is what is left of $AD$ once $AB$ is taken off.
In triangle $CBD$ the right angle is at $B$ , so this time $CD$ is the hypotenuse and Pythagoras gives $BC$ .
$BC$ meets $BD$ at a right angle, so $BD$ is a base of triangle $CDB$ and $BC$ is the height that belongs to it.

Worked Solution [5 marks]

Rule - Pythagoras: in a right-angled triangle

h^2 = a^2 + b^2

, where

h

is the hypotenuse, so a shorter side is found by subtracting:

a^2 = h^2 - b^2

. Area of a triangle

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

, with the height measured perpendicular to that base.

Step 1: find

AD

from triangle

AED

AD^2 = AE^2 + ED^2

28^2 + 45^2 = 784 + 2025 = 2809

AD = \sqrt{2809} = 53

(Reason: Angle

AED = 90^{\circ}

, so

AD

is the hypotenuse of triangle

AED

and the two given sides are the shorter pair.

2809

is a perfect square, so

AD

comes out whole.)

Step 2: find

BD

BD = AD - AB

53 - 21 = 32

(Reason:

ABD

is a straight line, so

B

sits on

AD

and the two pieces

AB

and

BD

add to

AD

. That one length is what carries the first triangle into the second.)

Step 3: find

BC

from triangle

CBD

BC^2 = CD^2 - BD^2

35^2 - 32^2 = 1225 - 1024 = 201

BC = \sqrt{201} \approx 14.1774

(Reason: Angle

CBD = 90^{\circ}

, so here it is

CD

that is the hypotenuse and

BC

is one of the shorter sides. Its square is therefore the DIFFERENCE of the other two squares, not the sum. Leaving it as

\sqrt{201}

keeps the working exact.)

Step 4: find the area of triangle

CDB

\text{Area} = \dfrac{1}{2} \times BD \times BC

\dfrac{1}{2} \times 32 \times \sqrt{201} = 16\sqrt{201} = 226.8391...

(Reason: The base

BD

and the height

BC

are perpendicular, which is exactly what the area formula asks for. Halving the

32

first turns

\dfrac{1}{2} \times 32

into

16

and leaves the exact answer

16\sqrt{201}

Step 5: round to

3

significant figures

226.8391... = 227 \text{ (to 3 significant figures)}

(Reason: The fourth significant figure of

226.8391...

8

, which is

5

or more, so the third figure rounds up and

226

becomes

227

. The unit is

\text{cm}^2

because an area is being measured.)

227

cm² (exact value

16\sqrt{201}

cm²)

Verification

Check 1: Put both results back through Pythagoras. In triangle

AED

28^2 + 45^2 = 2809

and

53^2 = 2809

. In triangle

CBD

32^2 + 201 = 1225

and

35^2 = 1225

2809 = 2809

and

1225 = 1225

, so both right angles are consistent with the lengths used

Check 2: Work the area out from the angle at

D

instead, so no height is used at all.

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

gives angle

BDC = 23.8955...^{\circ}

, and the area is

\dfrac{1}{2} \times 35 \times 32 \times \sin BDC

. This gives

226.8391...

as well, from a formula that never mentions

BC

Check 3: Is the size sensible?

\sqrt{201}

lies between

\sqrt{196} = 14

and

\sqrt{225} = 15

, so the area

16\sqrt{201}

lies between

16 \times 14 = 224

and

16 \times 15 = 240

227

sits inside that range, near the lower end because

201

is much closer to

196

than to

225

Mark Scheme Breakdown

Step	Mark	Description	Got it?
( $AD^2 =$ ) $28^2 + 45^2$ ( $784 + 2025 = 2809$ ) or ( $EDA =$ ) $\tan^{-1}\left(\dfrac{28}{45}\right)$ ( $31.8(9...)$ ) or ( $EAD =$ ) $\tan^{-1}\left(\dfrac{45}{28}\right)$ ( $58.1(09...)$ )	M1	for a correct method to find $AD^2$ or angle $EDA$ or angle $EAD$	✓
eg ( $AD =$ ) $\sqrt{28^2 + 45^2}$ ( $\sqrt{784 + 2025} = \sqrt{2809} = 53$ ) oe or ( $AD =$ ) $\dfrac{28}{\sin 31.8(9...)}$ ( $53...$ ) or ( $AD =$ ) $\dfrac{45}{\cos 31.8(9...)}$ ( $53...$ ) oe or ( $AD =$ ) $\dfrac{45}{\sin 58.1...}$ ( $53...$ ) or ( $AD =$ ) $\dfrac{28}{\cos 58.1...}$ ( $53...$ ) oe	M1	for a correct method to find $AD$	✓
eg ( $BC =$ ) $\sqrt{35^2 - (53 - 21)^2}$ ( $\sqrt{1225 - 1024} = \sqrt{201} = 14.1(7...)$ ) or ( $BDC =$ ) $\cos^{-1}\left(\dfrac{53 - 21}{35}\right)$ ( $23.8(9...)$ ) or ( $BCD =$ ) $\sin^{-1}\left(\dfrac{53 - 21}{35}\right)$ ( $66.1...$ ) The printed scheme puts the $53$ in quotation marks, so the candidate's own value for $AD$ may be used here.	M1	for correct method to find $BC$ or angle $BDC$ or angle $BCD$	✓
eg $\dfrac{1}{2} \times 14.1(7...) \times (53 - 21)$ ( $16\sqrt{201}$ ) or $\dfrac{1}{2} \times 35 \times (53 - 21) \times \sin 23.8(9...)$ or $\dfrac{1}{2} \times 35 \times 14.1(7...) \times \sin 66.1...$ The $53$ , the $14.1(7...)$ , the $23.8(9...)$ and the $66.1...$ are all in quotation marks in the printed scheme, so the candidate's own earlier values may be used.	M1	for a correct method to find the area of triangle $CDB$	✓
$227$	A1	awrt $227$ ; accept $16\sqrt{201}$	✓
Correct answer scores full marks (unless from obvious incorrect working)	Note	Guidance printed alongside the scheme. It awards nothing of its own and does not change the total of $5$ marks.	✓

Full marks: 5/5

Zubair buys a motorboat for $\$16\,000$

The motorboat depreciates at a rate of $12\%$ each year for the first $2$ years.
In the third year, the motorboat depreciates at a rate of $x\%$

At the end of $3$ years, the value of the motorboat is $\$11\,461.12$

Work out the value of $x$ [3 marks]

[Total 3 marks]

Show solution & mark schemeHide solution & mark scheme

Question 12 - Exam Solution

Understanding the Question

Given

Value when new:

\$16\,000

12\%

depreciation in each of the first

2

years

x\%

depreciation in the third year

Value after

3

years:

\$11\,461.12

Find

The value of

x

Plan the Solution

A loss of $12\%$ leaves $88\%$ , so multiply by $0.88$ once for each of the first two years.
Divide the value after three years by the value after two years. What is left is the multiplier for the third year on its own.
Turn that multiplier into a percentage loss to read off $x$ .

Worked Solution [3 marks]

Rule - a decrease of

r\%

multiplies a value by

1 - \dfrac{r}{100}

, and a run of yearly changes multiplies those multipliers together.

Step 1: write the

12\%

loss as a multiplier

1 - \dfrac{12}{100} = 0.88

(Reason: Losing

12\%

leaves

88\%

of the value standing, and

88\%

as a decimal is

0.88

Step 2: apply that multiplier for the first

2

years

0.88^{2} = 0.7744

16\,000 \times 0.7744 = 12\,390.4

(Reason: Two years of the same loss multiply together, so the value after two years is

16\,000 \times 0.88^{2}

. Squaring the multiplier is not the same as doubling the loss.)

Step 3: find the multiplier for the third year

\dfrac{11\,461.12}{12\,390.4} = 0.925

(Reason: The third year acts on the value at the start of that year, not on the price when new. Dividing the final value by

12\,390.4

leaves the third year's multiplier on its own.)

Step 4: turn that multiplier into a percentage loss

1 - 0.925 = 0.075

0.075 \times 100 = 7.5

(Reason: A multiplier of

0.925

is a decrease of

0.075

, which is

7.5\%

. Subtracting the other way round,

0.925 - 1

, gives

-0.075

and the answer

-7.5

- the size is right but the sign says the boat gained value.)

x = 7.5

Verification

Check 1: Work all three years forward, using

0.925

for the third year, and compare with the value the question gives.

16\,000 \times 0.88 \times 0.88 \times 0.925 = 11\,461.12

Check 2: Take the multiplier for all three years together and divide out the two years that are known. This is the mark scheme's own alternative and it uses the price when new, so it does not reuse Step 2's value.

\dfrac{11\,461.12}{16\,000} = 0.71632

and

\dfrac{0.71632}{0.7744} = 0.925

Check 3: Work out the money lost in the third year and write it as a percentage of the value at the start of that year.

12\,390.4 - 11\,461.12 = 929.28

and

\dfrac{929.28}{12\,390.4} \times 100 = 7.5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$16\,000 \times \left(1 - \dfrac{12}{100}\right)^{2} \; (= 12\,390.4)$ or $16\,000 \times 0.7744 \; (= 12\,390.4)$ oe or $16\,000 \times \left(1 - \dfrac{12}{100}\right) \; (= 14\,080)$ and $14\,080 \times \left(1 - \dfrac{12}{100}\right) \; (= 12\,390.4)$ oe or $\left(1 - \dfrac{12}{100}\right)^{2} \; (= 0.7744)$ and $\dfrac{11\,461.12}{16\,000} \; \left(= \dfrac{4477}{6250} = 0.71632\right)$	M1	for a method to find the value of the motorboat after two years or a method to find the overall percentage multiplier after two years and the overall percentage multiplier for the three years May be seen embedded, eg in an equation Do not allow $\left(1 - 12\%\right)$ unless processed correctly The printed scheme quotes $14\,080$ in the second alternative, so the candidate's own value may be used there	✓
eg $\dfrac{12\,390.4 - 11\,461.12}{12\,390.4} \; (\times 100) \; (= 0.075)$ or $1 - \dfrac{11\,461.12}{12\,390.4} \; (\times 100)$ or $1 - 0.925 \; (= 0.075)$ or $\dfrac{11\,461.12}{12\,390.4} \; (\times 100) \; (= 0.925)$ or $\dfrac{0.71632}{0.7744} \; (\times 100) \; (= 0.925)$	M1	for a complete method to find the value of the decimal equivalent of $x\%$ or for a complete method to find the percentage multiplier for the third year May be seen embedded, eg within a correct equation rearranged to one of these equivalent forms The printed scheme quotes $12\,390.4$ , $0.71632$ and $0.7744$ here, so the candidate's own values from the first mark may be used	✓
$7.5$	A1	oe	✓
Correct answer only	Note	Correct answer only scores full marks (unless from obviously incorrect working)	✓
for an answer of $-7.5$	SC B2	SCB2 for answer $-7.5$	✓

Full marks: 3/3

Continue to questions 13 to 18

The remaining 6 questions, with the same full worked solutions and mark schemes

Frequently asked questions

There are 26 questions worth 100 marks in total, sat over 2 hours. It is Higher tier and a calculator is allowed throughout, unlike UK GCSE Maths, where one paper is non-calculator.

Higher tier targets grades 4 to 9, so the lower grades 1 to 3 are only reachable on the tier below. About 40 per cent of the questions are targeted at grades 4 and 5 and appear on both Paper 2FR and Paper 2HR, so the lowest grades on this Higher paper are the ones the two tiers share.

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

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

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

Keep revising

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

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Edexcel IGCSE 4MA1/2HR, Wednesday 4 June 2025: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 12 of 26

Question 1, Calculator allowed

Question 1 - Exam Solution

Question 2, Calculator allowed

Question 2 - Exam Solution

Question 3, Calculator allowed

Question 3 - Exam Solution

Question 4, Calculator allowed

Question 4 - Exam Solution

Question 5, Calculator allowed

Question 5 - Exam Solution

Question 6, Calculator allowed

Question 6 - Exam Solution

Question 7, Calculator allowed

Question 7 - Exam Solution

Question 8, Calculator allowed

Question 8 - Exam Solution

Question 9, Calculator allowed

Question 9 - Exam Solution

Question 10, Calculator allowed

Question 10 - Exam Solution

Question 11, Calculator allowed

Question 11 - Exam Solution

Question 12, Calculator allowed

Question 12 - Exam Solution

Frequently asked questions

Keep revising

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