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Edexcel IGCSE 4MA1/2HR, Wednesday 4 June 2025: Worked Solutions, Questions 13 to 18

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

Back to questions 1 to 12

This is part two of three. Questions 1 to 12, the paper's overview and the frequently asked questions are on the first page.

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

Download printable PDF

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

Worked solutions, questions 13 to 18 of 26

(a) Factorise the expression $4x^2 - 25y^2$ [2 marks]

(b) Show that $4x(x + 3)(2x - 5)$ can be written in the form $ax^3 + bx^2 + cx$ ,
where $a$ , $b$ and $c$ are integers whose values you must find. [3 marks]

[Total 5 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

(a) The expression

4x^2 - 25y^2

- two terms with a minus sign between them.

(b) The product

4x(x + 3)(2x - 5)

- a single term multiplied by two brackets.

Find

(a)

4x^2 - 25y^2

written as a product. (b) The expanded, simplified form, and with it the integers

a

b

and

c

Plan the Solution

(a) Test each term for being a perfect square: $4x^2 = (2x)^2$ and $25y^2 = (5y)^2$ , with a subtraction between them.
(a) Apply the difference of two squares with $A = 2x$ and $B = 5y$ .
(b) Expand two of the three factors, then multiply that result by the factor left over. Any pair may go first.
(b) Collect like terms, then compare with $ax^3 + bx^2 + cx$ to read the three integers off.

Worked Solution [5 marks]

Rule - Difference of two squares:

A^2 - B^2 = (A + B)(A - B)

. Rule - Expanding a triple product: expand any two of the factors first, then multiply that answer by the third, and collect like terms at the end.

Step 1: write each term of part (a) as a square

4x^2 = (2x)^2

25y^2 = (5y)^2

(Reason: Both

4

and

25

are square numbers, and

x^2

and

y^2

are already squares, so each term is a perfect square and the subtraction between them makes it a difference of two squares.)

Step 2: use the difference of two squares

4x^2 - 25y^2 = (2x)^2 - (5y)^2

(2x)^2 - (5y)^2 = (2x + 5y)(2x - 5y)

(Reason: With

A = 2x

and

B = 5y

, the rule

A^2 - B^2 = (A + B)(A - B)

gives the two brackets. One sign is a plus and the other a minus: two plus signs would give

4x^2 + 20xy + 25y^2

instead.)

Step 3: part (b) - expand the first two factors

4x(x + 3) = 4x^2 + 12x

(Reason: Multiply

4x

by each term inside the bracket:

4x \times x = 4x^2

and

4x \times 3 = 12x

Step 4: multiply that by the remaining bracket

(4x^2 + 12x)(2x - 5) = 8x^3 - 20x^2 + 24x^2 - 60x

(Reason: Every term in the first bracket multiplies every term in the second:

4x^2 \times 2x = 8x^3

4x^2 \times (-5) = -20x^2

12x \times 2x = 24x^2

and

12x \times (-5) = -60x

Step 5: collect the two

x^2

terms

8x^3 - 20x^2 + 24x^2 - 60x = 8x^3 + 4x^2 - 60x

(Reason: Only the two

x^2

terms are like terms. The

x^3

term and the

x

term each stand alone, so they are carried down unchanged.)

Step 6: read off

a

b

and

c

a = 8

b = 4

c = -60

(Reason: Comparing

8x^3 + 4x^2 - 60x

with

ax^3 + bx^2 + cx

matches the coefficients term by term. All three are integers, which is what the question asks for. There is no constant term in either, so nothing is left over.)

(a)

(2x + 5y)(2x - 5y)

(b)

8x^3 + 4x^2 - 60x

(b)

a = 8

b = 4

c = -60

Verification

Check 1: Expand

(2x + 5y)(2x - 5y)

term by term and watch what happens to the two middle terms.

4x^2 - 10xy + 10xy - 25y^2 = 4x^2 - 25y^2

, which is the expression the question printed.

Check 2: Put

x = 3

and

y = 1

into the printed expression, and into the factorised form.

4 \times 3^2 - 25 \times 1^2 = 36 - 25 = 11

and

(2 \times 3 + 5) \times (2 \times 3 - 5) = 11 \times 1 = 11

Check 3: Put

x = 4

into the original product

4x(x + 3)(2x - 5)

, and into the cubic.

4 \times 4 \times 7 \times 3 = 336

and

8 \times 64 + 4 \times 16 - 60 \times 4 = 336

Check 4: Run part (b) backwards: take out the common factor

4x

, then factorise the quadratic that is left inside.

8x^3 + 4x^2 - 60x = 4x(2x^2 + x - 15) = 4x(2x - 5)(x + 3)

, the product the question started from.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Write it as a difference of two squares	M1	$(2x \pm 5y)(2x \pm 5y)$ or $(2x)^2 - (5y)^2$	✓
(a) The factorised answer	A1	$(2x + 5y)(2x - 5y)$	✓
(a)	Note	Correct answer only scores full marks, unless it comes from obviously incorrect working.	✓
(b) Expand one pair of factors	M1	$4x(x + 3) = 4x^2 + 12x$ or $4x(2x - 5) = 8x^2 - 20x$ or $(x + 3)(2x - 5) = 2x^2 - 5x + 6x - 15$ $(= 2x^2 + x - 15)$ . An expansion with only one error. Do not award this mark for $4x^2 + 12x + 8x^2 - 20x$ .	✓
(b) Multiply by the factor that is left	M1	$(4x^2 + 12x)(2x - 5) = 8x^3 - 20x^2 + 24x^2 - 60x$ or $(8x^2 - 20x)(x + 3) = 8x^3 + 24x^2 - 20x^2 - 60x$ or $4x(2x^2 - 5x + 6x - 15) = 8x^3 - 20x^2 + 24x^2 - 60x$ or $4x(2x^2 + x - 15) = 8x^3 + 4x^2 - 60x$ . Follow through, dependent on the first M1, allowing one further error.	✓
(b) The simplified cubic	A1	$8x^3 + 4x^2 - 60x$ . Working required. Correct answer only, dependent on the first M1. Terms may be in any order but must be simplified.	✓
(b) Alternative to the two method marks above	M2	For $3$ terms, out of a maximum of $4$ terms, from $8x^3 - 20x^2 + 24x^2 - 60x$ . If not M2, then M1 for $2$ correct out of a maximum of $4$ .	✓
(b)	Note	Ignore subsequent working after a correct factorisation: $8x^3 + 4x^2 - 60x$ must be seen previously to award 3 marks, for example $4(2x^3 + x^2 - 15x)$ or $x(8x^2 + 4x - 60)$ . Do not ignore subsequent working after an incorrect simplification, or after further incorrect work following $8x^3 + 4x^2 - 60x$ : for example $8x^3 + 4x^2 - 60x = 2x^3 + x^2 - 15x$ gets M2A0.	✓

Full marks: 5/5

Hana has two tins of counters, tin $A$ and tin $B$ .
In tin $A$ there are only $5$ red counters and $4$ green counters.
In tin $B$ there are only $7$ red counters and $3$ green counters.

Hana takes at random a counter from tin $A$
She then takes at random a counter from tin $B$

(a) Use this information to complete the probability tree diagram. [2 marks]

(b) Work out the probability that Hana takes two red counters. [2 marks]

Hana puts the counters back into the tins they came from.

Hana also has a jar of counters.
In the jar, there are only red counters and green counters.
When a counter is taken at random from the jar, the probability that it is a green counter is $\dfrac{2}{11}$

Hana takes at random a counter from tin $A$
She then takes at random a counter from tin $B$
She then takes at random a counter from the jar.

(b)(c)

[Total 7 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

Tin

A

5

red counters and

4

green counters

Tin

B

7

red counters and

3

green counters

The jar: the probability of taking a green counter is

\dfrac{2}{11}

One counter is taken at random from each container, so the three choices are independent

Find

(a) the six missing probabilities on the tree diagram (b) the probability of two red counters (c) the probability of more red counters than green counters when one counter is taken from tin

A

, one from tin

B

and one from the jar

Plan the Solution

Write each colour as a fraction of the counters in that tin, then fill the six branches in.
For two red counters, multiply along the pair of red branches.
For the jar, take the green probability away from $1$ to get the red probability.
More red than green out of three counters means two red or three red, so list those four outcomes, work each one out, and add them.

Worked Solution [7 marks]

Rule - Tree diagrams: multiply the probabilities along a path to find the probability of that path, then add the probabilities of the separate paths that make up the event. The two branches leaving any one point always add to

1

Step 1: The two branches for tin

A

5 + 4 = 9

\dfrac{5}{9} \text{ red} \qquad \dfrac{4}{9} \text{ green}

(Reason: Tin

A

holds

9

counters altogether, so

5

of the

9

are red and

4

of the

9

are green. These two go on the first pair of branches.)

Step 2: The four branches for tin

B

7 + 3 = 10

\dfrac{7}{10} \text{ red} \qquad \dfrac{3}{10} \text{ green}

(Reason: Tin

B

is untouched by what came out of tin

A

, so the same pair of probabilities goes on both of the second-stage pairs. That completes part (a).)

Step 3: Part (b), two red counters

\dfrac{5}{9} \times \dfrac{7}{10} = \dfrac{35}{90} = \dfrac{7}{18}

(Reason: Two red counters is the single path red then red, so multiply along it. Cancelling

\dfrac{35}{90}

5

gives

\dfrac{7}{18}

Step 4: The jar's red probability

1 - \dfrac{2}{11} = \dfrac{9}{11}

(Reason: The jar holds red counters and green counters only, so the two probabilities add to

1

. The question gives the green one, so the red one is what is left.)

Step 5: Which outcomes have more red than green

\text{RRR} \qquad \text{RRG} \qquad \text{RGR} \qquad \text{GRR}

(Reason: Three counters are taken, so more red than green means either three red or exactly two red. Writing the colours in the order tin

A

, tin

B

, jar, that is one all-red outcome and three outcomes with a single green, which is

4

paths in all.)

Step 6: Work out each of the four paths

\dfrac{5}{9} \times \dfrac{7}{10} \times \dfrac{9}{11} = \dfrac{315}{990}

\dfrac{5}{9} \times \dfrac{7}{10} \times \dfrac{2}{11} = \dfrac{70}{990}

\dfrac{5}{9} \times \dfrac{3}{10} \times \dfrac{9}{11} = \dfrac{135}{990}

\dfrac{4}{9} \times \dfrac{7}{10} \times \dfrac{9}{11} = \dfrac{252}{990}

(Reason: Each path is multiplied straight through, in the order RRR, RRG, RGR, GRR. The denominators are always

9 \times 10 \times 11

, so writing every path over

990

makes the next step a single addition.)

Step 7: Add the four paths

\dfrac{315}{990} + \dfrac{70}{990} + \dfrac{135}{990} + \dfrac{252}{990} = \dfrac{772}{990}

\dfrac{772}{990} = \dfrac{386}{495}

(Reason: The four paths cannot happen together, so their probabilities add. Both

772

and

990

are even, so the fraction cancels by

2

(a)

\dfrac{5}{9}

and

\dfrac{4}{9}

for tin

A

, then

\dfrac{7}{10}

and

\dfrac{3}{10}

on each tin

B

pair(b)

\dfrac{7}{18}

(c)

\dfrac{386}{495}

Verification

Check 1: The two branches leaving any one point must add to

1

, so test each pair on the completed diagram.

\dfrac{5}{9} + \dfrac{4}{9} = 1

and

\dfrac{7}{10} + \dfrac{3}{10} = 1

Check 2: Do part (b) the other way round. The three paths that are not two reds come to

\dfrac{15}{90}

\dfrac{28}{90}

and

\dfrac{12}{90}

, which total

\dfrac{55}{90}

1 - \dfrac{55}{90} = \dfrac{35}{90} = \dfrac{7}{18}

Check 3: Do part (c) the other way round. The four outcomes with more green than red are RGG, GRG, GGR and GGG, and over

990

they come to

30 + 56 + 108 + 24 = 218

1 - \dfrac{218}{990} = \dfrac{772}{990} = \dfrac{386}{495}

Check 4: All eight outcomes of the three-stage tree must total

1

. Over

990

their numerators are

315 + 70 + 135 + 30 + 252 + 56 + 108 + 24 = 990

\dfrac{990}{990} = 1

, and as a decimal the answer to part (c) is

\dfrac{386}{495} = 0.7798

4

decimal places, which is a sensible size for an event that is more likely than not.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $\dfrac{5}{9}$ and $\dfrac{4}{9}$ on the tin $A$ pair; $\dfrac{7}{10}$ and $\dfrac{3}{10}$ on each tin $B$ pair	B2	for all 3 correct pairs of probabilities on the correct branches	✓
(a) 1 or 2 of the three pairs completed correctly	(B1)	for 1 or 2 correct pairs of probabilities on the correct branches	✓
(a) the forms accepted on a branch	Note	Accept decimals or percentages rounded or truncated to at least 2sf. NB $\dfrac{5}{9} = 0.55(55\ldots)$ and $\dfrac{4}{9} = 0.44(44\ldots)$	✓
(b) $\dfrac{5}{9} \times \dfrac{7}{10}$ or $1 - \left(\dfrac{5}{9} \times \dfrac{3}{10} + \dfrac{4}{9} \times \dfrac{7}{10} + \dfrac{4}{9} \times \dfrac{3}{10}\right)$	M1ft	ft diagram, oe. Allow ft their tree diagram provided the relevant probabilities are less than 1 in each case	✓
(b) $\dfrac{7}{18}$	A1ft	ft diagram, oe fraction, decimal or percentage. NB $\dfrac{7}{18} = \dfrac{35}{90} = 0.38(88\ldots)$ . For A1, allow decimals or percentages that round or truncate correctly to at least 2sf. ISW any attempt to convert to other form once correct probability seen	✓
(b) an answer written down with no working	Note	Correct answer only scores full marks (unless from obviously incorrect working)	✓
(c) (RRR =) $\dfrac{5}{9} \times \dfrac{7}{10} \times \left(1 - \dfrac{2}{11}\right)$ $\left(= \dfrac{315}{990} = \dfrac{7}{22}\right)$ or (RRG =) $\dfrac{5}{9} \times \dfrac{7}{10} \times \dfrac{2}{11}$ $\left(= \dfrac{70}{990} = \dfrac{7}{99}\right)$ or (RGR =) $\dfrac{5}{9} \times \dfrac{3}{10} \times \left(1 - \dfrac{2}{11}\right)$ $\left(= \dfrac{135}{990} = \dfrac{3}{22}\right)$ or (GRR =) $\dfrac{4}{9} \times \dfrac{7}{10} \times \left(1 - \dfrac{2}{11}\right)$ $\left(= \dfrac{252}{990} = \dfrac{14}{55}\right)$ OR (GGG =) $\dfrac{4}{9} \times \dfrac{3}{10} \times \dfrac{2}{11}$ $\left(= \dfrac{24}{990} = \dfrac{4}{165}\right)$ or (GGR =) $\dfrac{4}{9} \times \dfrac{3}{10} \times \left(1 - \dfrac{2}{11}\right)$ $\left(= \dfrac{108}{990} = \dfrac{6}{55}\right)$ or (GRG =) $\dfrac{4}{9} \times \dfrac{7}{10} \times \dfrac{2}{11}$ $\left(= \dfrac{56}{990} = \dfrac{28}{495}\right)$ or (RGG =) $\dfrac{5}{9} \times \dfrac{3}{10} \times \dfrac{2}{11}$ $\left(= \dfrac{30}{990} = \dfrac{1}{33}\right)$	M1ft	ft diagram, for a correct calculation to find the probability of one relevant outcome, eg RRR or RRG or RGR or GRR, OR eg GGG or GGR or GRG or RGG. Allow ft their tree diagram provided the relevant probabilities are less than 1 in each case	✓
(c) (RRR =) $\left[\dfrac{7}{18}\right] \times \left(1 - \dfrac{2}{11}\right)$ and (RRG =) $\left[\dfrac{7}{18}\right] \times \dfrac{2}{11}$	Note	May see RRR or RRG found using their answer to part (b), where $\left[\dfrac{7}{18}\right]$ stands for their answer to part (b) and must be less than 1	✓
(c) " $\dfrac{7}{22}$ " + " $\dfrac{7}{99}$ " + " $\dfrac{3}{22}$ " + " $\dfrac{14}{55}$ " oe OR $1$ − (" $\dfrac{4}{165}$ " + " $\dfrac{6}{55}$ " + " $\dfrac{28}{495}$ " + " $\dfrac{1}{33}$ ")	M1ft	ft diagram, for a method to find the probability required. Condone one error in one of the four relevant outcomes or omission of one outcome. The quotation marks mean the candidate's own values, taken from their own tree diagram, may be used	✓
(c) $\dfrac{5}{9} \times \dfrac{7}{10}$ $\left(= \dfrac{7}{18}\right)$	Note	Note that P(RR) is P(RRR) added to P(RRG), so $\dfrac{5}{9} \times \dfrac{7}{10}$ may be seen in place of " $\dfrac{7}{22}$ " + " $\dfrac{7}{99}$ " for this mark, and similarly for P(GG)	✓
(c) $\dfrac{386}{495}$	A1ft	ft diagram, correct probability, oe fraction, decimal or percentage. NB $\dfrac{386}{495} = \dfrac{772}{990} = 0.77(979\ldots)$ . For A1, allow decimals or percentages that round or truncate correctly to at least 2sf. ISW any attempt to convert to other form once correct probability seen	✓
(c) an answer written down with no working	Note	Correct answer only scores full marks (unless from obviously incorrect working)	✓

Full marks: 7/7

$A$ , $B$ and $C$ are points on a circle with centre $O$ .

Angle $BAO = 36^{\circ}$
Angle $BCO = 21^{\circ}$

Find the size of angle $ACO$ . [3 marks]

[Total 3 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

A

B

and

C

lie on a circle with centre

O

Angle

BAO = 36^{\circ}

Angle

BCO = 21^{\circ}

OA = OB = OC

, because all three are radii of the same circle

Find

The size of angle

ACO

Plan the Solution

$OB$ splits the figure into three triangles that meet at $O$ . Every one of them has two radii for sides, so every one of them is isosceles.
Use the equal base angles to write angle $ABO$ and angle $CBO$ , then add them to get angle $ABC$ .
Double angle $ABC$ to get angle $AOC$ , the angle at the centre standing on the same arc.
Finish inside triangle $OAC$ , where the two base angles share whatever is left of $180^{\circ}$ .

Worked Solution [3 marks]

Rule - Circle angles: two radii are equal, so the triangle they make is isosceles and its base angles are equal; and the angle at the centre is twice the angle at the circumference standing on the same arc,

AOC = 2 \times ABC

Step 1: Name the equal sides

OA = OB = OC

(Reason: Every radius of a circle is the same length, so triangle

OAB

, triangle

OBC

and triangle

OAC

are all isosceles. That single fact is what the two marked angles are there to be used with, and nothing else in the question is needed.)

Step 2: The base angles of the two triangles the question marks

ABO = BAO = 36^{\circ}

CBO = BCO = 21^{\circ}

(Reason: In triangle

OAB

the equal sides are

OA

and

OB

, so the angles opposite them, at

A

and at

B

, are equal. Triangle

OBC

works the same way, which turns the

21^{\circ}

C

into an angle at

B

Step 3: Add them to get angle

ABC

ABC = ABO + CBO

36 + 21 = 57

(Reason:

O

lies inside triangle

ABC

, so

OB

runs across the inside of angle

ABC

and cuts it into exactly the two pieces found in step 2.)

Step 4: Turn it into the angle at the centre on arc

AC

AOC = 2 \times ABC

2 \times 57 = 114

(Reason: Angle

ABC

stands at the circumference and angle

AOC

stands at the centre, and both stand on the same arc

AC

, so the one at the centre is twice the one at the circumference.)

Step 5: Finish inside the isosceles triangle

OAC

ACO = \dfrac{180 - AOC}{2}

\dfrac{180 - 114}{2} = 33

(Reason: The three angles of triangle

OAC

add to

180^{\circ}

. Taking the angle at the centre away leaves the two base angles together, and

OA = OC

makes those two equal, so one of them is half of what is left.)

ACO = 33^{\circ}

Verification

Check 1: Put the answer back into triangle

ABC

and test its angle sum. Angle

BAC

36 + 33

, angle

BCA

21 + 33

, and angle

ABC

57

from step 3.

69 + 54 + 57 = 180

Check 2: The three angles at

O

must fill a full turn. Each one is the apex of its own isosceles triangle, so they are

180 - 2 \times 36

180 - 2 \times 21

and

180 - 2 \times 33

108 + 138 + 114 = 360

Check 3: Reach the answer a completely different way, with the tangent at

A

. The radius meets a tangent at a right angle, so the angle between that tangent and

AB

90 - 36 = 54

, and the alternate segment theorem makes that equal to angle

ACB

. Angle

ACO

is then what is left after angle

BCO

is taken off.

54 - 21 = 33

, the same answer, from a route that never uses the angle at the centre

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$ABC$ = $21 + 36 \; (= 57)$ or angle $ABO = 36$ and angle $CBO = 21$ and $21 + 36 \; (= 57)$ or (reflex) $AOC$ = $360 - (21 + 21 + 36 + 36) \; (= 246)$ or (obtuse) $AOC$ = $2 \times (21 + 36) \; (= 114)$ or $BAX$ = $90 - 36 \; (= 54)$ with the tangent at $A$ drawn oe or $BCY$ = $90 - 21 \; (= 69)$ with the tangent at $C$ drawn oe or angle $ADB$ = $180 - 90 - 36 \; (= 54)$ OR eg $x$ = angle $ACO$ and $y$ = angle $ABC$ and $x + x + 2y = 180$ oe and $y + (x + 36) + (x + 21) = 180$ oe or eg $180 - 2x = 2(180 - 2x - 21 - 36)$	M1	for a method to find one of the angles on the scheme; for this mark, values and calculations must be linked to the correct angle by notation or by being marked on the diagram. May also be awarded for a correct method to set up and solve an equation to find one of these angles; the variable must be clearly defined, where $X$ is a point on the tangent at $A$ , where $Y$ is a point on the tangent at $C$ , where $AD$ is a diameter. OR for any correct pair of simultaneous equations with clearly defined variables one of which must be angle $ACO$ , or any correct equation in terms of $ACO$ only	✓
( $ACO$ =) $\dfrac{180 - 2 \times 57}{2}$ or ( $ACO$ =) $\dfrac{180 - (360 - 246)}{2}$ or ( $ACO$ =) $\dfrac{180 - 57 - 21 - 36}{2}$ or ( $ACO$ =) $54 - 21$ or ( $ACO$ =) $69 - 36$ or ( $ACO$ =) $90 - 57$ OR eg $180 - 2x = 2(180 - 2x - 21 - 36) \implies x = \ldots$	M1	for a complete method to find angle $ACO$ OR forms a correct equation in terms of $ACO$ only and solves to get a value (condone arithmetic errors). Implies the 1st M mark (provided no incorrect working seen). The printed scheme puts $57$ , $246$ , $54$ and $69$ in quotation marks in this row, which means the candidate's own earlier value may be used in place of each of them	✓
$33$	A1	Correct answer only scores full marks (unless from obviously incorrect working)	✓

Full marks: 3/3

Show that $\dfrac{4}{3\sqrt{5} + 7}$ can be written in the form $a - \sqrt{b}$ , where $a$ and $b$ are integers.
Show every stage of your working. [3 marks]

[Total 3 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

The fraction

\dfrac{4}{3\sqrt{5} + 7}

, which has a surd in its denominator.

The form the answer must take:

a - \sqrt{b}

, with

a

and

b

both integers.

Find

The values of

a

and

b

, with every stage of the working shown.

Plan the Solution

Multiply the numerator and the denominator by the conjugate $3\sqrt{5} - 7$ . That fraction is worth $1$ , so it changes how the expression looks and not what it is worth.
Expand the numerator, and expand the denominator as a difference of two squares so that the surd terms cancel and a whole number is left below the line.
Divide each term of the numerator by that whole number.
Turn the multiple of a surd into a single surd, so the answer reads $a - \sqrt{b}$ .

Worked Solution [3 marks]

Rationalising a denominator: a denominator of the form

p\sqrt{q} + r

is cleared by multiplying the numerator and the denominator by its conjugate

p\sqrt{q} - r

. The denominator becomes

(p\sqrt{q})^2 - r^2

, and both of those are whole numbers, so no surd is left below the line. A multiple of a surd is then written as one surd using

k\sqrt{m} = \sqrt{k^2 m}

Step 1: Multiply the numerator and the denominator by the conjugate

\dfrac{4}{3\sqrt{5} + 7} \times \dfrac{3\sqrt{5} - 7}{3\sqrt{5} - 7}

(Reason: The conjugate of

3\sqrt{5} + 7

3\sqrt{5} - 7

- the same two terms with the sign between them reversed. The fraction

\dfrac{3\sqrt{5} - 7}{3\sqrt{5} - 7}

is equal to

1

, so multiplying by it leaves the value of the expression unchanged.)

Step 2: Expand the numerator

4(3\sqrt{5} - 7) = 12\sqrt{5} - 28

(Reason: Multiply each term inside the bracket by

4

4 \times 3\sqrt{5} = 12\sqrt{5}

and

4 \times 7 = 28

Step 3: Expand the denominator

(3\sqrt{5} + 7)(3\sqrt{5} - 7) = (3\sqrt{5})^2 - 7^2

(3\sqrt{5})^2 = 3^2 \times 5 = 45

45 - 49 = -4

(Reason: The two brackets are a difference of two squares, so the two

\sqrt{5}

terms cancel and no surd survives. Squaring

3\sqrt{5}

squares the

3

as well as the

\sqrt{5}

, and

7^2 = 49

, so the denominator is left as a whole number. Expanding all four terms instead is equally valid, as long as every one of them is correct.)

Step 4: Divide the numerator by the denominator

\dfrac{12\sqrt{5} - 28}{-4} = -3\sqrt{5} + 7

-3\sqrt{5} + 7 = 7 - 3\sqrt{5}

(Reason: Divide both terms of the numerator by

-4

12\sqrt{5}

gives

-3\sqrt{5}

and

-28

gives

7

. Writing the whole number first puts the expression the right way round for the form the question asks for.)

Step 5: Write the multiple of a surd as a single surd

3\sqrt{5} = \sqrt{3^2 \times 5} = \sqrt{45}

7 - 3\sqrt{5} = 7 - \sqrt{45}

(Reason: Take the

3

inside the square root by squaring it. The expression now reads

a - \sqrt{b}

with

a = 7

and

b = 45

, and both of those are integers, which is what the question asked to be shown.)

\dfrac{4}{3\sqrt{5} + 7} = 7 - \sqrt{45}

a = 7

and

b = 45

, both integers

Verification

Check 1: Work both forms out as decimals on the calculator and compare them. The original fraction gives

0.2917960675

and

7 - \sqrt{45}

gives

0.2917960675

, agreeing to ten decimal places, so the two forms are the same number.

Check 2: Work backwards. If the answer is right, multiplying it by the denominator we started with must give the numerator we started with:

(7 - 3\sqrt{5})(7 + 3\sqrt{5}) = 49 - 45

. That comes to

4

, which is the numerator of the original fraction, so the rationalising was done correctly.

Check 3: Check the form itself. The question demands

a - \sqrt{b}

with

a

and

b

integers, and a subtraction sign between them.

7 - \sqrt{45}

has

a = 7

and

b = 45

, both whole numbers, and a minus sign between the two terms, so it is in the required form.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{4}{3\sqrt{5} + 7} \times \dfrac{3\sqrt{5} - 7}{3\sqrt{5} - 7}$ or $\dfrac{4}{3\sqrt{5} + 7} \times \dfrac{-3\sqrt{5} + 7}{-3\sqrt{5} + 7}$ oe	M1	for multiplying the numerator and denominator by $3\sqrt{5} - 7$ or $-3\sqrt{5} + 7$ (may be implied)	✓
eg $\dfrac{4(3\sqrt{5} - 7)}{45 - 21\sqrt{5} + 21\sqrt{5} - 49}$ or $\dfrac{4(3\sqrt{5} - 7)}{45 - 7^2}$ or $\dfrac{12\sqrt{5} - 28}{45 - 21\sqrt{5} + 21\sqrt{5} - 49}$ or $\dfrac{12\sqrt{5} - 28}{45 - 7^2}$ or $\dfrac{4(3\sqrt{5} - 7)}{45 - 49}$ or $\dfrac{4(3\sqrt{5} - 7)}{-4}$ or $\dfrac{12\sqrt{5} - 28}{45 - 49}$ or $\dfrac{12\sqrt{5} - 28}{-4}$	M1	for expanding the denominator in a correct fraction the denominator may be $4$ terms which all need to be correct $\dfrac{4}{3\sqrt{5} + 7} \times \dfrac{3\sqrt{5} - 7}{3\sqrt{5} - 7} = 7 - 3\sqrt{5}$ scores M1M0 Implies the 1st mark	✓
Working required $7 - \sqrt{45}$	A1 dep on M2	dep on M2	✓
Working required	Note	The answer column of the scheme is marked $7 - \sqrt{45}$ with working required, so the rationalising must be seen for the accuracy mark. The two special cases below say what an unsupported answer is worth instead.	✓
$7 - \sqrt{45}$ written down with no method mark earned	SC B1	SC B1 for answer $7 - \sqrt{45}$ with no method marks awarded	✓
$7 - \sqrt{45}$ with the 1st M1 earned but not the 2nd	SC B2	SC B2 for $7 - \sqrt{45}$ if you would award the 1st M1 but not the 2nd M1 (total 2 marks)	✓

Full marks: 3/3

Rearrange the formula
$p = \dfrac{8k^{2} + 5}{7 - 3k^{2}}$
to make $k$ the subject. [4 marks]

[Total 4 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

The formula

p = \dfrac{8k^{2} + 5}{7 - 3k^{2}}

, which gives

p

in terms of

k

k

appears twice, once above the line and once below it, and every time it appears it is squared.

Find

The same formula rearranged so that

k

stands alone on one side, written in terms of

p

. Only

k^{2}

can be reached directly, so expect a

\pm

in the answer.

Plan the Solution

Multiply both sides by $7 - 3k^{2}$ to clear the fraction, then expand the bracket.
Collect every term containing $k^{2}$ on one side and every term without it on the other.
Factorise, so that $k^{2}$ appears once only, then divide by the bracket.
Square-root both sides, keeping both the positive and the negative root.

Worked Solution [4 marks]

Changing the subject when the new subject appears more than once: clear the fraction first, then gather every term containing the new subject on one side and everything else on the other, factorise so that the new subject appears once only, divide by the bracket, and finally undo the power. Undoing a square gives two roots, so a

\pm

is kept.

Step 1: Multiply both sides by the denominator

p(7 - 3k^{2}) = 8k^{2} + 5

7p - 3k^{2}p = 8k^{2} + 5

(Reason: Multiplying both sides by

7 - 3k^{2}

clears the fraction. Expand the left-hand side by multiplying each term inside the bracket by

p

p \times 7 = 7p

and

p \times (-3k^{2}) = -3k^{2}p

. Multiplying the

7

but leaving the

3k^{2}

untouched is the commonest slip here, and expanding both terms correctly is exactly what the first method mark is for.)

Step 2: Collect the

k^{2}

terms on one side

7p - 5 = 8k^{2} + 3k^{2}p

(Reason: Add

3k^{2}p

to both sides and subtract

5

from both sides. Every term containing

k^{2}

is now on the right, and every term without it is on the left. A term changes sign as it crosses the equals sign, so the

+5

on the right becomes

-5

on the left.)

Step 3: Factorise, so that

k^{2}

appears once only

7p - 5 = k^{2}(8 + 3p)

(Reason:

k^{2}

is a factor of both

8k^{2}

and

3k^{2}p

, so it comes outside a bracket and leaves

8 + 3p

inside. This is the step the question turns on: while

k^{2}

is written in two separate terms it cannot be made the subject, and once it is written once it can.)

Step 4: Divide, then take the square root

k^{2} = \dfrac{7p - 5}{8 + 3p}

k = \pm\sqrt{\dfrac{7p - 5}{8 + 3p}}

(Reason: Divide both sides by

8 + 3p

to leave

k^{2}

on its own, then square-root both sides. Squaring hides the sign, so undoing it gives two roots, one positive and one negative, and both of them are genuine values of

k

k = \pm\sqrt{\dfrac{7p - 5}{8 + 3p}}

Verification

Check 1: Put one value through both formulas. Take

k = 1

. The formula the question gives makes

p = \dfrac{8 \times 1 + 5}{7 - 3 \times 1} = \dfrac{13}{4} = 3.25

. Now feed that value of

p

into the rearrangement.

\dfrac{7 \times 3.25 - 5}{8 + 3 \times 3.25} = \dfrac{17.75}{17.75} = 1

, and

\sqrt{1} = 1

, which is the value of

k

it started from.

Check 2: Repeat with a value that makes

p

negative, so the rearrangement is tested below zero as well as above it. Take

k = 2

, which gives

\dfrac{8 \times 4 + 5}{7 - 3 \times 4} = \dfrac{37}{-5} = -7.4

\dfrac{7 \times (-7.4) - 5}{8 + 3 \times (-7.4)} = \dfrac{-56.8}{-14.2} = 4

, and

\sqrt{4} = 2

. Putting

k = -2

into the original formula gives the same

p = -7.4

, which is why the answer keeps both signs.

Check 3: Check it with no numbers at all. Put

k^{2} = \dfrac{7p - 5}{8 + 3p}

back into the top and the bottom of the original formula separately, over the common denominator

8 + 3p

. The top becomes

\dfrac{8(7p - 5) + 5(8 + 3p)}{8 + 3p} = \dfrac{71p}{8 + 3p}

and the bottom becomes

\dfrac{7(8 + 3p) - 3(7p - 5)}{8 + 3p} = \dfrac{71}{8 + 3p}

. Dividing the top by the bottom leaves

\dfrac{71p}{71} = p

, which is the formula the question started from, so the rearrangement holds for every value of

p

and not just for the two tested above.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$7p - 3k^{2}p = 8k^{2} + 5$	M1	for correctly multiplying both sides by the denominator and expanding the brackets	✓
$7p - 5 = 8k^{2} + 3k^{2}p$ or $-3k^{2}p - 8k^{2} = 5 - 7p$	M1ft	dep on 2 terms in $k^{2}$ and 2 other terms for correctly collecting their $k^{2}$ terms on one side and their other terms on the other side Note: eg $8k^{2} + 3k^{2}$ does not count as 2 terms in $k^{2}$	✓
eg $7p - 5 = k^{2}(8 + 3p)$ or $k^{2}(-3p - 8) = 5 - 7p$	M1ft	dep on previous M1 for correctly factorising for $k^{2}$ or $-k^{2}$ in their equation	✓
$k = (\pm)\sqrt{\dfrac{7p - 5}{8 + 3p}}$	A1	oe eg $k = (\pm)\sqrt{\dfrac{5 - 7p}{-3p - 8}}$ or $k = (\pm)\left(\dfrac{7p - 5}{8 + 3p}\right)^{\dfrac{1}{2}}$ or $k = (\pm)\left(\dfrac{7p - 5}{8 + 3p}\right)^{0.5}$ (condone omission of $\pm$ ) NB: to award A1 we must see $k = (\pm)\sqrt{\dfrac{7p - 5}{8 + 3p}}$ in working if $(\pm)\sqrt{\dfrac{7p - 5}{8 + 3p}}$ alone is given as an answer	✓
Working not required	Note	The working column of the printed scheme reads: Working not required, so correct answer scores full marks (unless from obvious incorrect working).	✓

Full marks: 4/4

The diagram shows triangle $OAB$ .

$\overrightarrow{OA} = 4\mathbf{a}$
$\overrightarrow{OB} = 4\mathbf{b}$

The point $P$ lies on $AB$ so that $AP : PB = 1 : 3$

(a) Write down an expression for $\overrightarrow{AB}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ [1 mark]

(b) Find $\overrightarrow{OP}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .
Give your answer in its simplest form. [2 marks]

(a)(b)

[Total 3 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

A triangle

OAB

with

\overrightarrow{OA} = 4\mathbf{a}

and

\overrightarrow{OB} = 4\mathbf{b}

P

AB

, dividing it so that

AP : PB = 1 : 3

No lengths and no angles are given, so every step is vector algebra.

Find

(a)

\overrightarrow{AB}

in terms of

\mathbf{a}

and

\mathbf{b}

(b)

\overrightarrow{OP}

in terms of

\mathbf{a}

and

\mathbf{b}

, simplified

Plan the Solution

Only $\overrightarrow{OA}$ and $\overrightarrow{OB}$ are known, so travel from $A$ to $B$ the long way, back through $O$ .
Turn the ratio into a fraction: $1 : 3$ cuts $AB$ into equal parts, and $AP$ is one of them.
Reach $P$ as $\overrightarrow{OA} + \overrightarrow{AP}$ , then collect the $\mathbf{a}$ terms and the $\mathbf{b}$ terms.

Worked Solution [3 marks]

Rule - Triangle law:

\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = -\overrightarrow{OA} + \overrightarrow{OB}

. And a point dividing

AB

in the ratio

m : n

sits

\dfrac{m}{m + n}

of the way from

A

B

Step 1: Go from

A

B

through

O

\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB}

\overrightarrow{AB} = -4\mathbf{a} + 4\mathbf{b}

(Reason: Reversing a vector changes its sign, so

\overrightarrow{AO} = -\overrightarrow{OA} = -4\mathbf{a}

. That is part (a). Factorised as

4(\mathbf{b} - \mathbf{a})

it says the same thing.)

Step 2: Turn

AP : PB = 1 : 3

into a fraction of

\overrightarrow{AB}

1 + 3 = 4

\overrightarrow{AP} = \dfrac{1}{4}\overrightarrow{AB}

(Reason: The two numbers in the ratio count parts, so

AB

is cut into

4

equal parts and

AP

1

of them. The

3

is the other piece, not the number of parts.)

Step 3: Build

\overrightarrow{OP}

starting from

O

\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}

\overrightarrow{OP} = 4\mathbf{a} + \dfrac{1}{4}(-4\mathbf{a} + 4\mathbf{b})

(Reason: Travel out along

\overrightarrow{OA}

, then a quarter of the way along

\overrightarrow{AB}

. This substitution alone earns the method mark.)

Step 4: Collect the terms

\overrightarrow{OP} = 4\mathbf{a} - \mathbf{a} + \mathbf{b}

\overrightarrow{OP} = 3\mathbf{a} + \mathbf{b}

(Reason: A quarter of

-4\mathbf{a}

-\mathbf{a}

, and a quarter of

4\mathbf{b}

\mathbf{b}

. Then

4\mathbf{a} - \mathbf{a} = 3\mathbf{a}

, and nothing else will combine.)

(a)

\overrightarrow{AB} = -4\mathbf{a} + 4\mathbf{b}

(b)

\overrightarrow{OP} = 3\mathbf{a} + \mathbf{b}

Verification

Check 1: Follow

\overrightarrow{OA}

and then

\overrightarrow{AB}

. Together they have to land on

B

4\mathbf{a} + (-4\mathbf{a} + 4\mathbf{b}) = 4\mathbf{b} = \overrightarrow{OB}

Check 2: Reach

P

from the far end instead, using

\overrightarrow{PB} = \dfrac{3}{4}\overrightarrow{AB}

\overrightarrow{OB} - \overrightarrow{PB} = 4\mathbf{b} - \dfrac{3}{4}(-4\mathbf{a} + 4\mathbf{b}) = 3\mathbf{a} + \mathbf{b}

Check 3: Put numbers on the vectors. With

\mathbf{a}

one unit right and

\mathbf{b}

one unit up,

A

is at

(4, 0)

and

B

is at

(0, 4)

. A quarter of the way from

A

B

is the point

(3, 1)

3\mathbf{a} + \mathbf{b}

points to

(3, 1)

, the same point

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $-4\mathbf{a} + 4\mathbf{b}$	B1	oe eg $4(\mathbf{b} - \mathbf{a})$	✓
(b) $4\mathbf{a} + \dfrac{1}{4}(-4\mathbf{a} + 4\mathbf{b})$ oe or $4\mathbf{a} + \dfrac{1}{4}[\overrightarrow{AB}]$ or $4\mathbf{b} - \dfrac{3}{4}(-4\mathbf{a} + 4\mathbf{b})$ oe or $4\mathbf{b} - \dfrac{3}{4}[\overrightarrow{AB}]$	M1ft	for a correct expression ft their (a), where $[\overrightarrow{AB}]$ is their answer to (a) of the form $m\mathbf{a} + n\mathbf{b}$ , $m, n \neq 0$	✓
$3\mathbf{a} + \mathbf{b}$	A1	allow $\mathbf{b} + 3\mathbf{a}$	✓
Part (b), from the printed scheme	Note	Correct answer only scores full marks (unless from obviously incorrect working).	✓

Full marks: 3/3

Continue to questions 19 to 26

The remaining 8 questions, with the same full worked solutions and mark schemes

Keep revising

That is part two of three. 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, Questions 13 to 18

Worked solutions, questions 13 to 18 of 26

Question 13, Calculator allowed

Question 13 - Exam Solution

Question 14, Calculator allowed

Question 14 - Exam Solution

Question 15, Calculator allowed

Question 15 - Exam Solution

Question 16, Calculator allowed

Question 16 - Exam Solution

Question 17, Calculator allowed

Question 17 - Exam Solution

Question 18, Calculator allowed

Question 18 - Exam Solution

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