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Edexcel IGCSE 4MA1/1H, Thursday 15 May 2025: Worked Solutions, Questions 21 to 25

Sir Faraz Hassan

27 Aug 2026

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

Back to questions 12 to 20

This is the rest of the paper. Questions 1 to 20, the paper's overview and the frequently asked questions are on the first two pages.

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/1H (Higher Tier), June 2025 series, sat Thursday 15 May 2025 –100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

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

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

Download printable PDF

All 25 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 21 to 25 of 25

A jar contains $20$ beads.

$9$ of the beads are red
$7$ of the beads are yellow
$4$ of the beads are green

Callum takes at random three beads from the jar.

Work out the probability that exactly two of the three beads are the same colour. [3 marks]

[Total 3 marks]

Show solution & mark schemeHide solution & mark scheme

Question 21 - Exam Solution

Understanding the Question

Given

A jar of

20

beads:

9

red,

7

yellow and

4

green

Three beads are taken at random, and none is put back, so each draw changes what is left in the jar

Find

The probability that exactly two of the three beads are the same colour

Plan the Solution

Exactly two the same means one matching pair and one bead of a different colour. All three alike does not count, and neither does all three different.
There are only three colours, so there are only three cases: two reds, two yellows, two greens. Work each one out and add.
Nothing is put back, so the denominators count down $20$ , $19$ , $18$ . Keeping every fraction over $6840$ makes the adding easy.
In each case the odd bead can be drawn first, second or third. Those $3$ orders have the same probability, so work out one and multiply by $3$ .

Worked Solution [3 marks]

Rule - Taking without replacement: each bead taken is gone, so the totals count down

20

19

18

. A matching pair and one odd bead can come out in

3

orders of equal probability, so one product multiplied by

3

covers the whole case.

Step 1: check the colours, and pin down what the question allows

9 + 7 + 4 = 20

(Reason: The three colours account for every bead, so no fourth colour is hiding. Exactly two the same means one matching pair plus one bead of a different colour, which leaves three cases to work out: two reds, two yellows, two greens.)

Step 2: two reds and one bead that is not red

\dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{11}{18} = \dfrac{792}{6840}

3 \times \dfrac{792}{6840} = \dfrac{2376}{6840}

(Reason: There are

9

reds out of

20

. The first red is gone, so the second is

8

out of

19

, and the third bead must be one of the

20 - 9 = 11

that are not red, out of the

18

still in the jar. That product is only the order red, red, other. The odd bead could just as easily come first or second, and each of those orders has the same probability, so multiply by

3

Step 3: two yellows and one bead that is not yellow

\dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{13}{18} = \dfrac{546}{6840}

3 \times \dfrac{546}{6840} = \dfrac{1638}{6840}

(Reason: The same shape with new numbers:

7

yellows, then

6

out of

19

, then one of the

20 - 7 = 13

beads that are not yellow out of the

18

left. Leaving it over

6840

rather than cancelling now means the three cases can simply be added at the end.)

Step 4: two greens and one bead that is not green

\dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{16}{18} = \dfrac{192}{6840}

3 \times \dfrac{192}{6840} = \dfrac{576}{6840}

(Reason: Green is the scarce colour: only

4

to begin with, so the second green is

3

out of

19

. The third bead is one of the

20 - 4 = 16

beads that are not green. Take the complement from the colour you have just paired, never from one of the other colours - that swap is the commonest way this question is lost.)

Step 5: add the three cases, then cancel

\dfrac{2376}{6840} + \dfrac{1638}{6840} + \dfrac{576}{6840} = \dfrac{4590}{6840}

\dfrac{4590}{6840} = \dfrac{51}{76}

(Reason: Two reds, two yellows and two greens cannot happen at the same time, so the three probabilities add. The highest common factor of

4590

and

6840

90

, and dividing both by it leaves

\dfrac{51}{76}

, which is

0.671

3

significant figures.)

\dfrac{51}{76}

(

0.671

3

significant figures)

Verification

Check 1: Count the other way round. Every one of the

20 \times 19 \times 18 = 6840

ordered picks falls into exactly one of three camps: all three the same colour, all three different, or exactly two the same. Work out the first two camps and take them away.

9 \times 8 \times 7 + 7 \times 6 \times 5 + 4 \times 3 \times 2 = 738

6 \times 9 \times 7 \times 4 = 1512

, and

6840 - 738 - 1512 = 4590

Check 2: Throw order away completely and count selections instead. There are

1140

ways of choosing three beads from

20

. A pair of reds with one other bead can be chosen in

36 \times 11

ways, a pair of yellows in

21 \times 13

ways and a pair of greens in

6 \times 16

ways.

396 + 273 + 96 = 765

and

\dfrac{765}{1140} = \dfrac{51}{76}

Check 3: The three camps have to account for everything, so their probabilities must add to

1

. Put the answer back beside the other two and see whether they close.

\dfrac{4590}{6840} + \dfrac{738}{6840} + \dfrac{1512}{6840} = 1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $(\text{P}(RRY) =) \dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{7}{18} \left( = \dfrac{504}{6840} = \dfrac{7}{95} \right)$ oe or $(\text{P}(RRG) =) \dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{4}{18} \left( = \dfrac{288}{6840} = \dfrac{4}{95} \right)$ oe or $(\text{P}(RRR) =) \dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{7}{18} \left( = \dfrac{504}{6840} = \dfrac{7}{95} \right)$ oe or $(\text{P}(RRR') =) \dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{11}{18} \left( = \dfrac{792}{6840} = \dfrac{11}{95} \right)$ oe or $(\text{P}(YYR) =) \dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{9}{18} \left( = \dfrac{378}{6840} = \dfrac{21}{380} \right)$ oe or $(\text{P}(YYG) =) \dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{4}{18} \left( = \dfrac{168}{6840} = \dfrac{7}{285} \right)$ oe or $(\text{P}(YYY) =) \dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{5}{18} \left( = \dfrac{210}{6840} = \dfrac{7}{228} \right)$ oe or $(\text{P}(YYY') =) \dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{13}{18} \left( = \dfrac{546}{6840} = \dfrac{91}{1140} \right)$ oe or $(\text{P}(GGR) =) \dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{9}{18} \left( = \dfrac{108}{6840} = \dfrac{3}{190} \right)$ oe or $(\text{P}(GGY) =) \dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{7}{18} \left( = \dfrac{84}{6840} = \dfrac{7}{570} \right)$ oe or $(\text{P}(GGG) =) \dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{2}{18} \left( = \dfrac{24}{6840} = \dfrac{1}{285} \right)$ oe or $(\text{P}(GGG') =) \dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{16}{18} \left( = \dfrac{192}{6840} = \dfrac{8}{285} \right)$ oe or $(\text{P}(RGY) =) \dfrac{9}{20} \times \dfrac{7}{19} \times \dfrac{4}{18} \left( = \dfrac{252}{6840} = \dfrac{7}{190} \right)$ oe	M1	for finding one correct product, does not need to be labelled, or for an answer of $\dfrac{17}{76}$ oe eg $0.22(3\ldots)$ or $22(.3\ldots)\%$ or $\dfrac{65}{76}$ oe eg $0.85(5\ldots)$ or $85(.5\ldots)\%$	✓
$(\text{P}(RRR' \text{ or } YYY' \text{ or } GGG') =)$ $\left( 3 \times \dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{11}{18} \right)$ + $\left( 3 \times \dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{13}{18} \right)$ + $\left( 3 \times \dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{16}{18} \right)$ oe or $(\text{P}(RRY \text{ or } RRG \text{ or } YYR \text{ or } YYG \text{ or } GGR \text{ or } GGY) =)$ $\left( 3 \times \dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{7}{18} \right)$ + $\left( 3 \times \dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{4}{18} \right)$ + $\left( 3 \times \dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{9}{18} \right)$ + $\left( 3 \times \dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{4}{18} \right)$ + $\left( 3 \times \dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{9}{18} \right)$ + $\left( 3 \times \dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{7}{18} \right)$ oe or $(1 - \text{P}(RRR \text{ or } YYY \text{ or } GGG \text{ or } RGY) =)$ $1 - \left( \left( \dfrac{9}{20} \times \dfrac{8}{19} \times \dfrac{7}{18} \right) + \left( \dfrac{7}{20} \times \dfrac{6}{19} \times \dfrac{5}{18} \right) + \left( \dfrac{4}{20} \times \dfrac{3}{19} \times \dfrac{2}{18} \right) + \left( 6 \times \dfrac{9}{20} \times \dfrac{7}{19} \times \dfrac{4}{18} \right) \right)$ oe	M1	for a complete calculation	✓
$\dfrac{51}{76}$	A1	oe eg $0.67(1\ldots)$ or $67(.1\ldots)\%$	✓
Correct answer scores full marks (unless from obvious incorrect working).	Note	The printed scheme carries this in italics in the row with the final A1, and it awards nothing of its own.	✓
for an answer of $\dfrac{669}{1000}$ oe eg $0.66(9)$ or $66(.9)\%$	SCB1	a special case, awarded only when no other marks are earned in this question	✓
Where the three named wrong answers come from.	Note	$\dfrac{17}{76}$ is what is left when the three orders are never counted - it is $\dfrac{792 + 546 + 192}{6840}$ , one order per case. $\dfrac{65}{76}$ comes from the subtraction route with the six orders of three different colours counted once instead of six times - it is $\dfrac{6840 - 738 - 252}{6840}$ . $\dfrac{669}{1000}$ is the answer with the beads put back each time, $3 \times \left( \left( \dfrac{9}{20} \right)^{2} \times \dfrac{11}{20} + \left( \dfrac{7}{20} \right)^{2} \times \dfrac{13}{20} + \left( \dfrac{4}{20} \right)^{2} \times \dfrac{16}{20} \right)$ , where every denominator stays at $20$ . This row awards nothing; it is here so that each special case has a nameable error behind it.	✓

Full marks: 3/3

Solve the simultaneous equations

$x^2 + 3y + y^2 = 7$
$y = x + 2$

You must show clear algebraic working. [5 marks]

[Total 5 marks]

Show solution & mark schemeHide solution & mark scheme

Question 22 - Exam Solution

Understanding the Question

Given

x^2 + 3y + y^2 = 7

, an equation with a squared term in each letter, so it is a curve

y = x + 2

, a straight line, with

y

already on its own

Algebraic working is asked for, so the values on their own would earn nothing.

Find

Every pair of values of

x

and

y

that fits both equations at the same time. A quadratic appears on the way, so expect two solution pairs.

Plan the Solution

The line gives $y$ in terms of $x$ , so replace every $y$ in the curve equation by $x + 2$ . That leaves one letter instead of two.
Expand, then collect everything on one side, to reach a three term quadratic in $x$ alone.
Factorise that quadratic and read off the two values of $x$ .
Put each value of $x$ back into $y = x + 2$ , the simpler of the two equations, and keep each $y$ beside the $x$ it came from.

Worked Solution [5 marks]

Rule - Substitution (line into curve): where a linear equation gives one letter on its own, write that letter's expression everywhere the letter appears in the other equation. One letter is eliminated, and the quadratic left behind has two roots - each root carrying its own partner.

Step 1: substitute the line into the curve

x^2 + 3(x + 2) + (x + 2)^2 = 7

(Reason: The second equation says

y

and

x + 2

are the same thing, so every

y

in the first equation may be written as

x + 2

. Both of them have to go - the

y

inside

3y

as well as the

y^2

. Replacing only one of them leaves both letters in the equation, and nothing can be solved from there.)

Step 2: expand, then collect into a three term quadratic

x^2 + 3x + 6 + x^2 + 4x + 4 = 7

2x^2 + 7x + 10 = 7

2x^2 + 7x + 3 = 0

(Reason:

(x + 2)^2

means

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

, not

x^2 + 4

. Collecting then gives

x^2 + x^2 = 2x^2

3x + 4x = 7x

and

6 + 4 = 10

. Taking

7

from both sides leaves zero on the right, which is the form a quadratic must be in before it can be factorised.)

Step 3: factorise, and read off the two values of

x

2x^2 + 7x + 3 = (2x + 1)(x + 3)

2x + 1 = 0 \quad \text{or} \quad x + 3 = 0

x = -\dfrac{1}{2} \quad \text{or} \quad x = -3

(Reason: The

2x^2

can only come from

2x \times x

, and the

+3

only from

1 \times 3

. What decides which bracket takes the

1

and which takes the

3

is the middle term: multiply the brackets out again and the two

x

terms must add to

7x

. A product is zero only when one of its brackets is zero, and that is what turns one factorised line into two small equations.)

Step 4: put each value of

x

back into the line

x = -\dfrac{1}{2} \implies y = -\dfrac{1}{2} + 2 = \dfrac{3}{2}

x = -3 \implies y = -3 + 2 = -1

(Reason: Use

y = x + 2

rather than the curve: it is one addition each, with no squaring to go wrong, and the mark scheme allows either equation. Keep each

y

with the

x

that produced it - the marks are for pairs, not for four loose numbers.)

Step 5: write the two solution pairs

x = -\dfrac{1}{2}, \; y = \dfrac{3}{2}

x = -3, \; y = -1

(Reason: Each root of the quadratic carries one value of

y

, so the answer is two pairs and not four separate values. Written as coordinates,

\left( -\dfrac{1}{2}, \dfrac{3}{2} \right)

and

(-3, -1)

are the two points where the line crosses the curve, and the mark scheme allows that form as well.)

x = -\dfrac{1}{2}, \; y = \dfrac{3}{2}

x = -3, \; y = -1

Verification

Check 1, the pair $\left( -\dfrac{1}{2}, \dfrac{3}{2} \right)$ : Put both values into the curve equation

x^2 + 3y + y^2

, working the three terms out one at a time, then test the line as well. A pair has to fit both equations, not just the one it was found from.

\left( -\dfrac{1}{2} \right)^2 + 3 \times \dfrac{3}{2} + \left( \dfrac{3}{2} \right)^2 = \dfrac{1}{4} + \dfrac{9}{2} + \dfrac{9}{4} = 7

and

-\dfrac{1}{2} + 2 = \dfrac{3}{2}

Check 2, the pair $(-3, -1)$ : The same test on the second pair. Squaring a negative gives a positive, so

(-3)^2

9

and

(-1)^2

1

, while the middle term

3y

stays negative.

(-3)^2 + 3 \times (-1) + (-1)^2 = 9 - 3 + 1 = 7

and

-3 + 2 = -1

Check 3: Start from the other end. Rearranging the line gives

x = y - 2

, and substituting that into the curve leaves a quadratic in

y

instead. This is the mark scheme's own second method, and it must arrive at the same two pairs.

(y - 2)^2 + 3y + y^2 = 7

gives

2y^2 - y - 3 = 0

, so

(y + 1)(2y - 3) = 0

and

y = -1

y = \dfrac{3}{2}

, and

x = y - 2

turns those into

-3

and

-\dfrac{1}{2}

Check 4: Test the two roots against the quadratic without factorising anything. Whatever method solves

ax^2 + bx + c = 0

, its roots must add to

-\dfrac{b}{a}

and multiply to

\dfrac{c}{a}

. Here

a = 2

b = 7

and

c = 3

, so the sum should be

-\dfrac{7}{2}

and the product

\dfrac{3}{2}

-\dfrac{1}{2} + (-3) = -\dfrac{7}{2}

and

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

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $x^2 + 3(x + 2) + (x + 2)^2 = 7$ or eg $(y - 2)^2 + 3y + y^2 = 7$	M1	for substitution of $y = x + 2$ (or $x = \pm y \pm 2$ ) into $x^2 + 3y + y^2 = 7$ to obtain an equation in $x$ only (or $y$ only)	✓
eg $2x^2 + 7x + 3 (= 0)$ or $2x^2 + 7x = -3$ or eg $2y^2 - y - 3 (= 0)$ or $2y^2 - y = 3$	M1ft	dep on previous M1 for multiplying out and collecting terms, forming a three term quadratic in any form of $ax^2 + bx + c \, (= 0)$ where at least $2$ coefficients ( $a$ or $b$ or $c$ ) are correct	✓
eg $(2x + 1)(x + 3) (= 0)$ or $x = \dfrac{-7 \pm \sqrt{(7)^2 - 4 \times 2 \times 3}}{2 \times 2}$ or $\left( x + \dfrac{7}{4} \right)^2 - \left( \dfrac{7}{4} \right)^2 = -\dfrac{3}{2}$ $\left( x = -\dfrac{1}{2} \text{ and } x = -3 \right)$ or eg $(y + 1)(2y - 3) (= 0)$ or $y = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 2 \times -3}}{2 \times 2}$ or $\left( y - \dfrac{1}{4} \right)^2 - \left( \dfrac{1}{4} \right)^2 = \dfrac{3}{2}$ $\left( y = -1 \text{ and } y = \dfrac{3}{2} \right)$	M1ft	dep on first M1 method to solve their $3$ term quadratic using any correct method (allow one sign error and some simplification - allow as far as eg $\dfrac{-7 \pm \sqrt{49 - 24}}{4}$ or $\dfrac{1 \pm \sqrt{1 + 24}}{4}$ or if factorising allow brackets which expanded give $2$ out of $3$ terms correct) or correct values for $x$ or correct values for $y$	✓
$y = \text{"}-\dfrac{1}{2}\text{"} + 2 \left( = \dfrac{3}{2} \right)$ and $y = \text{"}-3\text{"} + 2 \left( = -1 \right)$ or $x = \text{"}-1\text{"} - 2 \left( = -3 \right)$ and $x = \text{"}\dfrac{3}{2}\text{"} - 2 \left( = -\dfrac{1}{2} \right)$	M1ft	dep on previous M1 for substituting their $2$ found values of $x$ or $y$ into one of the two given equations or fully correct values for the other variable (correct labels for $x$ / $y$ ) or for one correct pair of values. The printed row puts the substituted values in quotation marks, so a candidate's own earlier values may be used here.	✓
Working required $x = -\dfrac{1}{2}$ and $y = \dfrac{3}{2}$ $x = -3$ and $y = -1$	A1	oe dep on M2 (allow coordinates)	✓
If they find the values of $y$ but think they are the values of $x$ then the maximum mark is $3$	Note	The printed scheme carries this line beneath the table, and it awards nothing of its own.	✓

Full marks: 5/5

In the diagram, $OAB$ is a triangle.
$P$ is the midpoint of $OA$
$Q$ is a point on $OB$

$ABR$ and $PQR$ are straight lines.

$\overrightarrow{OA} = 12\mathbf{a} \qquad \overrightarrow{OB} = 8\mathbf{b}$

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

$AB : BR = 1 : 2 \qquad \overrightarrow{OQ} = n\mathbf{b}$

(b) Using a vector method, work out the value of $n$ [4 marks]

(a)(b) n =

[Total 5 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

OAB

is a triangle with

\overrightarrow{OA} = 12\mathbf{a}

and

\overrightarrow{OB} = 8\mathbf{b}

P

is the midpoint of

OA

, and

Q

lies on

OB

ABR

and

PQR

are straight lines, with

AB : BR = 1 : 2

and

\overrightarrow{OQ} = n\mathbf{b}

Find

(a)

\overrightarrow{AB}

in terms of

\mathbf{a}

and

\mathbf{b}

(b) the value of

n

Plan the Solution

For (a), travel from $A$ to $B$ the long way round, through $O$ .
For (b), build $\overrightarrow{PR}$ out of $\overrightarrow{PA}$ and $\overrightarrow{AR}$ . Read the ratio carefully: $AB : BR = 1 : 2$ makes $AR$ three parts long, not two.
Write $\overrightarrow{PQ}$ using $n$ , then use the fact that $P$ , $Q$ and $R$ lie on one straight line.
Compare the $\mathbf{a}$ parts to find the multiplier, then the $\mathbf{b}$ parts give $n$ .

Worked Solution [5 marks]

Rule - Vectors on a straight line: if

P

Q

and

R

lie on one line, then

\overrightarrow{PQ} = \lambda \overrightarrow{PR}

for some number

\lambda

. Because

\mathbf{a}

and

\mathbf{b}

are not parallel, the

\mathbf{a}

parts and the

\mathbf{b}

parts must then match separately.

Step 1: (a) Travel from

A

B

through

O

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

\overrightarrow{AB} = -12\mathbf{a} + 8\mathbf{b}

(Reason: Going from

A

O

\overrightarrow{OA} = 12\mathbf{a}

travelled backwards, which is

-12\mathbf{a}

. Adding

\overrightarrow{OB} = 8\mathbf{b}

finishes the journey at

B

Step 2: (b) Find

\overrightarrow{PA}

, and count how many parts long

AR

\overrightarrow{PA} = \dfrac{1}{2} \times 12\mathbf{a} = 6\mathbf{a}

AB : BR = 1 : 2 \implies \overrightarrow{AR} = 3\overrightarrow{AB}

(Reason:

P

is the midpoint of

OA

, so

\overrightarrow{PA}

is half of

12\mathbf{a}

. Along

ABR

AB

1

part and

BR

2

more parts, so the whole of

AR

3

parts.)

Step 3: Build

\overrightarrow{PR}

from those two pieces

\overrightarrow{PR} = \overrightarrow{PA} + \overrightarrow{AR}

\overrightarrow{PR} = 6\mathbf{a} + 3(-12\mathbf{a} + 8\mathbf{b}) = -30\mathbf{a} + 24\mathbf{b}

(Reason: Walk from

P

back to

A

, then straight along

ABR

all the way to

R

Step 4: Write

\overrightarrow{PQ}

using

n

\overrightarrow{PQ} = \overrightarrow{PO} + \overrightarrow{OQ}

\overrightarrow{PQ} = -6\mathbf{a} + n\mathbf{b}

(Reason:

\overrightarrow{PO}

reverses

\overrightarrow{OP} = 6\mathbf{a}

, and

\overrightarrow{OQ} = n\mathbf{b}

is given.)

Step 5:

PQR

is a straight line, so

\overrightarrow{PQ} = \lambda \overrightarrow{PR}

-6\mathbf{a} + n\mathbf{b} = \lambda(-30\mathbf{a} + 24\mathbf{b})

-6 = -30\lambda \implies \lambda = \dfrac{1}{5}

(Reason:

\mathbf{a}

and

\mathbf{b}

are not parallel, so the two sides can only be equal if the

\mathbf{a}

parts match and the

\mathbf{b}

parts match. The

\mathbf{a}

parts pin down

\lambda

Step 6: Compare the

\mathbf{b}

parts to get

n

n = 24\lambda = 24 \times \dfrac{1}{5}

n = \dfrac{24}{5} = 4.8

(Reason: The

\mathbf{b}

part of

\overrightarrow{PR}

24

, and

\lambda = \dfrac{1}{5}

scales it down to

\dfrac{24}{5}

(a)

\overrightarrow{AB} = -12\mathbf{a} + 8\mathbf{b}

(b)

n = 4.8

Verification

Check 1: Put real vectors in:

\mathbf{a} = (3, 1)

and

\mathbf{b} = (-1, 4)

. Then

A

(36, 12)

B

(-8, 32)

P

(18, 6)

and

R

(-96, 72)

. The line

PR

crosses

OB

(-4.8, 19.2)

, which is

4.8\mathbf{b}

n = 4.8

Check 2: Take a different journey to

Q

, namely

O

P

Q

n\mathbf{b} = 6\mathbf{a} + k(-30\mathbf{a} + 24\mathbf{b})

. There is no

\mathbf{a}

on the left, so

0 = 6 - 30k

and

k = \dfrac{1}{5}

n = 24 \times \dfrac{1}{5} = 4.8

Check 3: Ask where this puts

Q

. Since

\overrightarrow{OQ} = 4.8\mathbf{b}

and

\overrightarrow{OB} = 8\mathbf{b}

, the rest of the side is

\overrightarrow{QB} = 3.2\mathbf{b}

, giving

OQ : QB = 3 : 2

8 - 4.8 = 3.2

, so

Q

sits on

OB

nearer

B

, which is where the diagram puts it

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\overrightarrow{AB} = -12\mathbf{a} + 8\mathbf{b}$	B1	oe	✓
$(\overrightarrow{PR} =) \; 6\mathbf{a} + 3(-12\mathbf{a} + 8\mathbf{b}) \; (= -30\mathbf{a} + 24\mathbf{b})$ oe or $(\overrightarrow{RP} =) \; -3(-12\mathbf{a} + 8\mathbf{b}) - 6\mathbf{a} \; (= 30\mathbf{a} - 24\mathbf{b})$	M1	for method to find $\overrightarrow{PR}$ or $\overrightarrow{RP}$ , ft their $\overrightarrow{AB}$ . The printed scheme puts $-12\mathbf{a} + 8\mathbf{b}$ in quotation marks, so a candidate's own answer to part (a) may be used here.	✓
$(\overrightarrow{PQ} =) \; -6\mathbf{a} + n\mathbf{b}$ oe eg $(\overrightarrow{PQ} =) \; -6\mathbf{a} + m \times 8\mathbf{b}$ or $(\overrightarrow{OQ} =) \; 6\mathbf{a} + k(-30\mathbf{a} + 24\mathbf{b})$ or $(\overrightarrow{AQ} =) \; -12\mathbf{a} + n\mathbf{b}$	M1	for method to find $\overrightarrow{PQ}$ or $\overrightarrow{OQ}$ or $\overrightarrow{AQ}$ . The printed scheme puts $-30\mathbf{a} + 24\mathbf{b}$ in quotation marks in the $\overrightarrow{OQ}$ form, so a candidate's own earlier value may be used.	✓
$\overrightarrow{PQ} = \lambda \overrightarrow{PR}$ eg $-6\mathbf{a} + n\mathbf{b} = \lambda(-30\mathbf{a} + 24\mathbf{b})$ or $-6 = -30\lambda$ oe or $\lambda = \dfrac{1}{5}$ oe OR $\mu \overrightarrow{PQ} = \overrightarrow{PR}$ eg $\mu(-6\mathbf{a} + n\mathbf{b}) = -30\mathbf{a} + 24\mathbf{b}$ or $-6\mu = -30$ oe or $\mu = 5$ OR $\overrightarrow{OQ} = \overrightarrow{OP} + k\overrightarrow{PR}$ eg $n\mathbf{b} = 6\mathbf{a} + k(-30\mathbf{a} + 24\mathbf{b})$ or $0 = 6 - 30k$ or $k = \dfrac{1}{5}$ oe OR $\overrightarrow{AQ} = \overrightarrow{AR} + x\overrightarrow{RP}$ eg $-12\mathbf{a} + n\mathbf{b} = 3(-12\mathbf{a} + 8\mathbf{b}) + x(30\mathbf{a} - 24\mathbf{b})$ or $-36 + 30x = -12$ or $x = \dfrac{4}{5}$ oe	M1	for setting up an equation to find the value of the unknown coefficient(s)	✓
Working required. $n = 4.8$	A1	oe eg $\dfrac{24}{5}$ , dep on M1	✓

Full marks: 5/5

An arithmetic series has $30$ terms.

The first term of the series is $a$
The common difference between consecutive terms is $d$

The $20$ th term of the series is $123$
The sum of all $30$ terms of the series is $2880$

Find the value of $a$ and the value of $d$
You must show clear algebraic working. [5 marks]

a =d =

[Total 5 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

An arithmetic series with

30

terms, first term

a

and common difference

d

The

20

th term:

U_{20} = 123

The sum of all

30

terms:

S_{30} = 2880

Find

The value of the first term

a

The value of the common difference

d

Two unknowns, so two equations are needed and they are solved together

Plan the Solution

Turn the $20$ th term into an equation using $U_n = a + (n - 1)d$
Turn the sum of the $30$ terms into a second equation using $S_n = \dfrac{n}{2}\left(2a + (n - 1)d\right)$
Make the coefficients of $a$ match, then subtract to leave an equation in $d$ only
Substitute that value of $d$ back into the simpler equation to find $a$

Worked Solution [5 marks]

Rule - Arithmetic series: the

n

th term is

U_n = a + (n - 1)d

, and the sum of the first

n

terms is

S_n = \dfrac{n}{2}\left(2a + (n - 1)d\right)

Step 1: turn the

20

th term into an equation

U_{20} = a + (20 - 1)d

a + 19d = 123

(Reason: the

n

th term of an arithmetic series is

U_n = a + (n - 1)d

, so the

20

th term sits

19

common differences after the first term, not

20

)

Step 2: turn the sum of the

30

terms into a second equation

S_{30} = \dfrac{30}{2}\left(2a + (30 - 1)d\right)

2880 = 15(2a + 29d)

2a + 29d = 192

(Reason: the sum of the first

n

terms is

S_n = \dfrac{n}{2}\left(2a + (n - 1)d\right)

, and here

\dfrac{30}{2} = 15

, so dividing both sides by

15

clears the bracket and leaves a second equation in

a

and

d

)

Step 3: match the coefficients of

a

and subtract

2 \times (a + 19d) = 2 \times 123

2a + 38d = 246

(2a + 38d) - (2a + 29d) = 246 - 192

9d = 54

d = \dfrac{54}{9} = 6

(Reason: both equations now begin with

2a

, so subtracting one from the other removes

a

completely and leaves an equation in

d

alone)

Step 4: substitute

d = 6

back to find

a

a + 19 \times 6 = 123

a + 114 = 123

a = 123 - 114 = 9

(Reason: either equation will do, and

a + 19d = 123

is the simpler one;

19 \times 6 = 114

, which leaves

a

on its own)

Step 5: state both values clearly

a = 9

d = 6

(Reason: the final mark is awarded only when

a

and

d

are clearly identified, so both are written out and labelled rather than left inside the working)

a = 9

d = 6

Verification

Check 1: Put

a = 9

and

d = 6

into the

20

th term,

a + 19d

9 + 19 \times 6 = 9 + 114 = 123

Check 2: Put both values into

S_{30} = \dfrac{30}{2}\left(2a + 29d\right)

, where

2 \times 9 = 18

and

29 \times 6 = 174

15 \times (18 + 174) = 15 \times 192 = 2880

Check 3: Total the series a different way. The last term is

9 + 29 \times 6 = 183

, and an arithmetic series also sums as

\dfrac{n}{2}(U_1 + U_n)

15 \times (9 + 183) = 15 \times 192 = 2880

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$123 = a + (20 - 1)d$ or $123 = a + 19d$	M1	for using $U_n = a + (n - 1)d$	✓
$2880 = \dfrac{30}{2}\left(2a + (30 - 1)d\right)$ or $2880 = \dfrac{30}{2}\left(2a + 29d\right)$ or $192 = 2a + 29d$	M1	for using $S_n = \dfrac{n}{2}\left(2a + (n - 1)d\right)$	✓
eg $192 = 2a + 29d$ , $123 = a + 19d$ $(\times 2)$ , $246 = 2a + 38d$ , subtracting $54 = 9d$ or $192 = 2(123 - 19d) + 29d$ oe or $d = 6$ or $192 = 2a + 29d$ $(\times 19)$ , $123 = a + 19d$ $(\times 29)$ , $3648 = 38a + 551d$ , $3567 = 29a + 551d$ , subtracting $81 = 9a$ or $192 = 2a + 29\left(\dfrac{123 - a}{19}\right)$ oe or $a = 9$	M1	(dep on M2) for a correct method to find $a$ or $d$ : coefficients of $a$ or $d$ the same in correct equations and correct operator to eliminate the selected variable, resulting in an equation in $a$ only or in $d$ only, or writing $a$ or $d$ in terms of the other variable and correctly substituting (condone missing brackets)	✓
eg $192 = 2a + 29(\text{"6"})$ oe or $123 = a + 19(\text{"6"})$ oe or $192 = 2(\text{"9"}) + 29d$ oe or $123 = \text{"9"} + 19d$ oe	M1	(dep on M3) for substituting their found value of $a$ or $d$ into a correct equation	✓
Working required $a = 9$ $d = 6$	A1	dep on M2, and $a$ and $d$ must be clearly identified	✓
The quotation marks in the substitution row	Note	a value in quotation marks is the candidate's own earlier value, so the substitution mark is available on a follow-through even when that value is wrong	✓

Full marks: 5/5

$PQRS$ is a square.
$PR$ is a diagonal of the square.

The vertex $P$ has coordinates $(4, 7)$
The vertex $R$ has coordinates $(8, -5)$

Work out an equation of the straight line that passes through $Q$ and $S$
Give your answer in the form $ay = bx + c$ where $a$ , $b$ and $c$ are integers. [5 marks]

[Total 5 marks]

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Question 25 - Exam Solution

Understanding the Question

Given

PQRS

is a square, and

PR

is a diagonal of it

The vertex

P

is at

(4,\ 7)

The vertex

R

is at

(8,\ -5)

Q

and

S

are the ends of the other diagonal, and their coordinates are not given

Find

An equation of the straight line through

Q

and

S

It must be written as

ay = bx + c

with

a

b

and

c

integers, so any fraction has to be cleared before the answer is stated

Plan the Solution

The diagonals of a square bisect each other at right angles, so $QS$ is the perpendicular bisector of $PR$
Find the midpoint of $PR$ : that centre point lies on $QS$ as well
Find the gradient of $PR$ , then take its negative reciprocal for the gradient of $QS$
Put the gradient and the centre into $y - y_1 = m(x - x_1)$ , then multiply through to clear the fraction

Worked Solution [5 marks]

Rule - Diagonals of a square: they bisect each other at right angles, so

QS

passes through the midpoint of

PR

and

m_{PR} \times m_{QS} = -1

Step 1: find the midpoint of

PR

\left(\dfrac{4 + 8}{2},\ \dfrac{7 + (-5)}{2}\right) = (6,\ 1)

(Reason: the diagonals of a square bisect each other, so the midpoint of

PR

is the centre of the square and therefore lies on

QS

too; it is the one point of

QS

the question hands you)

Step 2: find the gradient of

PR

m_{PR} = \dfrac{7 - (-5)}{4 - 8} = \dfrac{12}{-4} = -3

(Reason: gradient is the change in

y

divided by the change in

x

; taking the two points in the other order gives

\dfrac{-5 - 7}{8 - 4} = \dfrac{-12}{4}

, the same value)

Step 3: turn that into the gradient of

QS

m_{PR} \times m_{QS} = -1

-3 \times m_{QS} = -1

m_{QS} = \dfrac{-1}{-3} = \dfrac{1}{3}

(Reason: the diagonals of a square meet at right angles, and perpendicular gradients multiply to

-1

, so the gradient of

QS

is the negative reciprocal of

-3

; dividing

-1

by a negative number gives a positive gradient)

Step 4: build the equation through the centre

y - 1 = \dfrac{1}{3}(x - 6)

y - 1 = \dfrac{1}{3}x - 2

y = \dfrac{1}{3}x - 1

(Reason: a line through

(x_1,\ y_1)

with gradient

m

y - y_1 = m(x - x_1)

, and here the point is the centre

(6,\ 1)

with gradient

\dfrac{1}{3}

)

Step 5: clear the fraction to get integer coefficients

3 \times y = 3 \times \left(\dfrac{1}{3}x - 1\right)

3y = x - 3

(Reason: the question asks for

ay = bx + c

with

a

b

and

c

integers, and multiplying every term by

3

removes the fraction, leaving

a = 3

b = 1

and

c = -3

)

3y = x - 3

Verification

Check 1: Find the other two vertices by turning the half-diagonal a quarter turn each way about the centre

(6,\ 1)

. They are

(12,\ 3)

and

(0,\ -1)

. Put each one into

3y = x - 3

3 \times 3 = 9

and

12 - 3 = 9

; then

3 \times (-1) = -3

and

0 - 3 = -3

Check 2: Multiply the two diagonal gradients together. Perpendicular lines must give

-1

-3 \times \dfrac{1}{3} = -1

Check 3: Every point of

QS

must be the same distance from

P

as from

R

. Test the centre

(6,\ 1)

: squared distance to

P

(6 - 4)^2 + (1 - 7)^2

, and to

R

it is

(6 - 8)^2 + (1 + 5)^2

4 + 36 = 40

and

4 + 36 = 40

Check 4: The two diagonals of a square are equal in length. Compare their squared lengths,

PR

first and then

QS

(8 - 4)^2 + (-5 - 7)^2 = 16 + 144 = 160

and

(12 - 0)^2 + (3 - (-1))^2 = 144 + 16 = 160

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\left(\dfrac{4 + 8}{2},\ \dfrac{7 - 5}{2}\right)$ oe or $(6,\ 1)$	M1	for finding the midpoint of $PR$	✓
$\dfrac{7 - -5}{4 - 8}\left(= -\dfrac{12}{4} = -3\right)$ oe	M1	for method to find the gradient of $PR$	✓
$\text{"}-3\text{"} \times m = -1$ oe or $(m =) \dfrac{-1}{\text{"}-3\text{"}}$ or $(m =) \dfrac{1}{3}$	M1ft	for finding the gradient of $QS$ , may be seen embedded in an equation, ft their gradient of $PR$	✓
$\text{"}1\text{"} = \text{"}\dfrac{1}{3}\text{"}(\text{"}6\text{"}) + c$ oe or $c = -1$ or $y - 1 = \dfrac{1}{3}(x - 6)$ or $y = \dfrac{1}{3}x - 1$	M1ft	(dep on previous M1) for finding the equation through $QS$ , ft their gradient of $PR$ and their midpoint of $PR$ , do not allow $(4,\ 7)$ or $(8,\ -5)$ as midpoint $PR$	✓
$3y = x - 3$	A1	oe eg $6y = 2x - 6$ or $3y - x + 3 = 0$ etc but must be integer coefficients, accept $a = 3$ , $b = 1$ , $c = -3$	✓
Correct answer scores full marks (unless from obvious incorrect working)	Note	the printed scheme carries this line beside the method rows, so a correct equation earns all five marks even when the midpoint and gradient working is not shown	✓
The quotation marks in the two follow-through rows	Note	a value in quotation marks is the candidate's own earlier value, so the gradient row and the equation row are both available on a follow-through even when the gradient of $PR$ was found wrongly	✓

Full marks: 5/5

Keep revising

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

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Edexcel IGCSE 4MA1/1H, Thursday 15 May 2025: Worked Solutions, Questions 21 to 25

Worked solutions, questions 21 to 25 of 25

Question 21, Calculator allowed

Question 21 - Exam Solution

Question 22, Calculator allowed

Question 22 - Exam Solution

Question 23, Calculator allowed

Question 23 - Exam Solution

Question 24, Calculator allowed

Question 24 - Exam Solution

Question 25, Calculator allowed

Question 25 - Exam Solution

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