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Edexcel IGCSE 4MA1/2HR, Wednesday 4 June 2025: Worked Solutions, Questions 19 to 26

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 13 to 18

This is the rest of the paper. Questions 1 to 18, 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 4MA1/2HR (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 19 to 26 of 26

Two functions $\text{f}$ and $\text{g}$ are defined as follows.

$\text{f}(x) = 3x - 4$ where $x > 2$
$\text{g}(x) = \dfrac{x}{2x + 1}$

(a) Write down the value of $x$ that must be left out of every domain of $\text{g}$
[1 mark]
(b) Work out $\text{gf}(x)$
Give your answer in its simplest form. [2 marks]

(a)(b) gf(x) =

[Total 3 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

\text{f}(x) = 3x - 4

where

x > 2

\text{g}(x) = \dfrac{x}{2x + 1}

One linear function and one algebraic fraction.

Find

(a) the value of

x

that no domain of

\text{g}

may contain (b)

\text{gf}(x)

, in its simplest form

Plan the Solution

For (a), a fraction has no value when its denominator is zero, so solve $2x + 1 = 0$ .
For (b), $\text{gf}(x)$ means do $\text{f}$ first and then $\text{g}$ , so write $3x - 4$ wherever an $x$ appears in the formula for $\text{g}$ .
Expand the bracket in the denominator, collect the constants, then check whether anything cancels.

Worked Solution [3 marks]

Rule - Composite functions:

\text{gf}(x) = \text{g}(\text{f}(x))

, so the inner function

\text{f}

is applied first. And a fraction has no value wherever its denominator is

0

Step 1: find where the denominator of

\text{g}

is zero

2x + 1 = 0

2x = -1

x = -\dfrac{1}{2}

(Reason: (Reason: dividing by

0

has no value, so this one value of

x

has to be left out of every domain of

\text{g}

. Every other value of

x

is allowed, which is why the answer is a value and not an inequality.))

Step 2: put the whole of

\text{f}(x)

into

\text{g}

\text{gf}(x) = \text{g}(3x - 4)

\text{gf}(x) = \dfrac{3x - 4}{2(3x - 4) + 1}

(Reason: (Reason:

\text{gf}

means do

\text{f}

first, so every

x

in the formula for

\text{g}

is replaced by the whole of

3x - 4

, brackets and all. This unsimplified fraction already earns the method mark.))

Step 3: expand the bracket and collect the denominator

2(3x - 4) + 1 = 6x - 8 + 1 = 6x - 7

(Reason: (Reason: multiply both terms inside the bracket by

2

, then add the

1

. Only the denominator changes here; the numerator is left exactly as it is.))

Step 4: check the fraction is already in its simplest form

\text{gf}(x) = \dfrac{3x - 4}{6x - 7}

(Reason: (Reason:

3x - 4

and

6x - 7

share no common factor, and one is not a multiple of the other, so nothing cancels and the fraction is as simple as it goes. Multiplying top and bottom by

-1

gives

\dfrac{4 - 3x}{7 - 6x}

, which is the same answer written another way.))

(a)

-\dfrac{1}{2}

(b)

\text{gf}(x) = \dfrac{3x - 4}{6x - 7}

Verification

Check 1: Put

x = -\dfrac{1}{2}

into the denominator of

\text{g}

2 \times \left(-\dfrac{1}{2}\right) + 1 = -1 + 1 = 0

, so

\text{g}

has no value there, while every other

x

gives a denominator that is not zero.

Check 2: Take

x = 3

, which is inside the domain of

\text{f}

. Then

3 \times 3 - 4 = 5

, so the composite is

\text{g}(5)

. Working forwards,

\dfrac{5}{2 \times 5 + 1} = \dfrac{5}{11}

. The answer gives

\dfrac{3 \times 3 - 4}{6 \times 3 - 7} = \dfrac{5}{11}

, the same value.

Check 3: The new denominator is zero when

6x - 7 = 0

, that is at

x = \dfrac{7}{6}

\dfrac{7}{6}

is smaller than

2

, so no value in the given domain

x > 2

is lost. This agrees with part (a):

\text{f}(x) > 2

for every

x > 2

, so

\text{f}(x)

never lands on

-\dfrac{1}{2}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $-\dfrac{1}{2}$	B1	oe. Accept $x = -\dfrac{1}{2}$ , accept $x \neq -\dfrac{1}{2}$ . Do not allow inequalities, eg $x > -\dfrac{1}{2}$ or $x \leq -\dfrac{1}{2}$ .	✓
(b) $\dfrac{3x - 4}{2(3x - 4) + 1}$	M1	for a correct unsimplified expression for $\text{gf}(x)$	✓
(b) $\dfrac{3x - 4}{6x - 7}$	A1	oe eg $\dfrac{4 - 3x}{7 - 6x}$ . Correct answer seen followed by incorrect subsequent working scores M1A0. Correct answer only scores full marks (unless from obviously incorrect working).	✓

Full marks: 3/3

The histogram gives some information about the weights, in grams, of some pebbles.

$75$ pebbles weigh less than $100$ grams.

A pebble is chosen at random.

Find an estimate for the probability that this pebble weighs between $400$ grams and $600$ grams. [4 marks]

[Total 4 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

A histogram of the weights, in grams, of some pebbles. The frequency density axis carries no numbers, so its scale has to be worked out.

75

pebbles weigh less than

100

grams.

The bars meet at

0

100

200

450

500

and

700

grams.

Find

An estimate for the probability that a pebble chosen at random weighs between

400

grams and

600

grams.

Plan the Solution

The first bar is the only one whose frequency is given, so use it to fix the scale: its class width is $100$ and its frequency is $75$ .
Read the height of every bar off that scale and turn each one into a frequency, to get the total number of pebbles.
Do the same for the strip between $400$ grams and $600$ grams, which cuts two of the bars part way.
The probability is that strip's total over the whole total.

Worked Solution [4 marks]

On a histogram the frequency of a class is the AREA of its bar, not its height:

\text{frequency} = \text{frequency density} \times \text{class width}

Step 1: Use the first bar to fix the frequency density scale

\dfrac{75}{100} = 0.75

\dfrac{0.75}{3} = 0.25

(Reason: The first bar covers

0

100

grams, so its class width is

100

, and the question gives its frequency as

75

. Frequency density is frequency divided by class width, so that bar has a frequency density of

0.75

. It stands

3

large squares high, so one large square up the axis is

0.25

Step 2: Work out how many pebbles there are altogether

0.75 \times 100 = 75

2.25 \times 100 = 225

1 \times 250 = 250

2.5 \times 50 = 125

0.25 \times 200 = 50

75 + 225 + 250 + 125 + 50 = 725

(Reason: The five bars are

3

9

4

10

and

1

large squares high, so at

0.25

a square their frequency densities are

0.75

2.25

1

2.5

and

0.25

. Multiply each one by its own class width to get its frequency, then add. The first bar gives back the

75

the question states, which is the scale confirming itself.)

Step 3: Work out how many pebbles weigh between

400

grams and

600

grams

1 \times 50 = 50

2.5 \times 50 = 125

0.25 \times 100 = 25

50 + 125 + 25 = 200

(Reason: The strip runs across three bars. It takes the last

50

grams of the third bar (

400

450

), the whole of the fourth bar (

450

500

, also

50

grams wide) and the first

100

grams of the fifth bar (

500

600

). Only the part of a bar inside the strip counts, so use each piece's own width, not the whole class width.)

Step 4: Write the estimate as a probability

\dfrac{200}{725} = \dfrac{8}{29}

(Reason: The probability is the number of pebbles in the strip out of the number of pebbles altogether. Both

200

and

725

divide by

25

, which cancels the fraction to

\dfrac{8}{29}

, or

0.28

2

significant figures.)

\dfrac{200}{725} = \dfrac{8}{29}

Verification

Check 1: Count large squares instead. One large square is

50

grams wide and

0.25

high, so it stands for

50 \times 0.25 = 12.5

pebbles. The bars cover

6 + 18 + 20 + 10 + 4 = 58

large squares, of which

4 + 10 + 2 = 16

lie between

400

and

600

grams.

58 \times 12.5 = 725

and

16 \times 12.5 = 200

, so the probability is

\dfrac{16}{58} = \dfrac{8}{29}

Check 2: Count small squares, which is a finer measure of the same areas. One small square is

10

grams wide and

0.05

high, so it stands for

10 \times 0.05 = 0.5

pebbles. The bars cover

150 + 450 + 500 + 250 + 100 = 1450

small squares, and

100 + 250 + 50 = 400

of them lie in the strip.

1450 \times 0.5 = 725

and

400 \times 0.5 = 200

, giving

\dfrac{400}{1450} = \dfrac{8}{29}

Check 3: Is the size sensible? A quarter of the

58

large squares would be

\dfrac{58}{4} = 14.5

, and the strip covers

16

of them, so the answer must be a little over a quarter.

\dfrac{8}{29} = 0.28

2

significant figures, which is a little over

0.25

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(frequency density $=$ ) $\dfrac{75}{100} \; (= 0.75)$ or any one correct value marked on the frequency density axis or uses their own linear frequency density scale to find the area of at least one of the third, fourth or fifth bars OR eg $1$ (large) box / $1$ (large) square $= 12.5$ pebbles oe or $5$ (sml) squares $= 2.5$ pebbles or $1$ (sml) square / $1$ box $= 0.5$ pebbles	M1	for a correct calculation of frequency density or a correct value on the frequency density axis or use of their own linear frequency density scale to find the area of at least one of the third, fourth or fifth bars OR a correct measure of scale Implied by a correct frequency for the third ( $250$ ), fourth ( $125$ ) or fifth ( $50$ ) bar	✓
eg using frequency densities $75 + (2.25 \times 100) + (1 \times 250) + (2.5 \times 50) + (0.25 \times 200)$ $(= 75 + 225 + 250 + 125 + 50 = 725)$ OR eg using a measure of scale (eg large squares) $(6 + 18 + 20 + 10 + 4) \times 12.5 \; (= 725)$ oe OR eg total no. of large squares $= 6 + 18 + 20 + 10 + 4 \; (= 58)$ or total no. of sml squares $= 150 + 450 + 500 + 250 + 100 \; (= 1450)$	M1	for a method to work out the total number of pebbles OR the total number of squares. Allow one error in a frequency density value, class width value or number of squares but not an omission. Allow ft their own linear frequency density scale. The frequency density values in the row beside this one are the candidate's own values read off their scale, which is what the printed scheme marks by putting them in quotation marks.	✓
eg using frequency densities $(1 \times 50) + (2.5 \times 50) + (0.25 \times 100)$ $(= 50 + 125 + 25 = 200)$ OR eg using a measure of scale (eg large squares) $(4 + 10 + 2) \times 12.5 \; (= 200)$ oe OR eg no. of large squares $= 4 + 10 + 2 \; (= 16)$ or no. of sml squares $= 100 + 250 + 50 \; (= 400)$	M1	for a method to estimate the number of pebbles between $400$ g and $600$ g OR the total number of squares between $400$ g and $600$ g Allow one error in a frequency density value, class width value or number of squares but not an omission Allow ft their own linear frequency density scale	✓
$\dfrac{200}{725}$	A1	oe eg $\dfrac{8}{29}$ or any correct decimal or correct percentage rounded or truncated to $2$ sf. NB $\dfrac{200}{725} = 0.2758\ldots$	✓
Correct answer only scores full marks (unless from obviously incorrect working)	Note	A correct answer on its own earns all four marks, unless the working shown with it is obviously incorrect.	✓
Counting squares alone, eg counting large squares $\dfrac{4 + 10 + 2}{6 + 18 + 20 + 10 + 4}$	Note	M3 for any complete method that relies only on counting squares. Allow one error in the number of squares in each of the numerator and the denominator, but do not allow any omissions. A consistent method must be used for the award of more than one mark.	✓

Full marks: 4/4

Solve the inequality $2x^2 - 7x - 15 > 0$
You must show clear algebraic working. [3 marks]

[Total 3 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

The inequality

2x^2 - 7x - 15 > 0

The working has to be algebraic, so the critical values must come from factorising, from the formula or from completing the square.

Find

Every value of

x

for which

2x^2 - 7x - 15

is greater than

0

Plan the Solution

Factorise $2x^2 - 7x - 15$ into two brackets.
Set each bracket equal to $0$ to get the two critical values.
The $x^2$ coefficient is positive, so the curve is a U-shape. Test one value from each of the three regions to see which ones sit above the axis.
Write the answer as two separate inequalities.

Worked Solution [3 marks]

When

a > 0

, the graph of

y = ax^2 + bx + c

is a U-shape, so

ax^2 + bx + c

is negative between the two roots and positive outside them.

Step 1: Factorise the quadratic

2 \times (-15) = -30

3 + (-10) = -7

2x^2 + 3x - 10x - 15 = x(2x + 3) - 5(2x + 3)

2x^2 - 7x - 15 = (2x + 3)(x - 5)

(Reason: Multiply the

x^2

coefficient by the constant term:

2 \times (-15) = -30

. Two numbers with product

-30

and sum

-7

are

3

and

-10

, so the middle term splits into

3x - 10x

and each pair of terms leaves the same bracket.)

Step 2: Find the critical values

(2x + 3)(x - 5) = 0

2x + 3 = 0 \Rightarrow x = -\dfrac{3}{2}

x - 5 = 0 \Rightarrow x = 5

(Reason: A product is

0

only when one of its factors is

0

, so each bracket is solved on its own. These two values are the critical values: they are where the curve meets the

x

-axis, and they are the only places the sign of the expression can change.)

Step 3: Test one value in each region

2(-2)^2 - 7(-2) - 15 = 8 + 14 - 15 = 7

2(0)^2 - 7(0) - 15 = 0 - 0 - 15 = -15

2(6)^2 - 7(6) - 15 = 72 - 42 - 15 = 15

(Reason: The critical values

-\dfrac{3}{2}

and

5

cut the number line into three regions. Take

x = -2

from the left region,

x = 0

from the middle one and

x = 6

from the right region. Only the two outer regions give a positive value, which is exactly what a U-shaped curve does.)

Step 4: Write the solution as two inequalities

x < -\dfrac{3}{2}

x > 5

(Reason: The solution is two separate stretches of the number line, so it is written as two separate inequalities. A single chain such as

-\dfrac{3}{2} > x > 5

earns nothing, because no number is both smaller than

-\dfrac{3}{2}

and larger than

5

at once. A comma, the word or, or

\cup

may be used to link them.)

x < -\dfrac{3}{2}

x > 5

Verification

Check 1: Expand the brackets again:

(2x + 3)(x - 5) = 2x^2 - 10x + 3x - 15

2x^2 - 7x - 15

, which is the expression the question gives, so the factorisation is right

Check 2: Find the critical values a second way, with the quadratic formula:

x = \dfrac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 2 \times (-15)}}{2 \times 2} = \dfrac{7 \pm \sqrt{169}}{4}

\dfrac{7 + 13}{4} = 5

and

\dfrac{7 - 13}{4} = -\dfrac{3}{2}

, the same two critical values

Check 3: Substitute one value from each region into

2x^2 - 7x - 15

and read off the signs.

x = -2

gives

7

x = 0

gives

-15

and

x = 6

gives

15

, so only the two outer regions are positive

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $(2x + 3)(x - 5)$ oe or $\dfrac{-(-7) \pm \sqrt{(-7)^2 - 4 \times 2 \times (-15)}}{2 \times 2}$ oe or $2\left[\left(x - \dfrac{7}{4}\right)^2 - \left(\dfrac{7}{4}\right)^2\right] - 15$ oe	M1	for a correct method to find the critical values. Minimum evidence for the quadratic formula is a two-term discriminant, eg $\dfrac{7 \pm \sqrt{49 + 120}}{4}$ (must have the $\pm$ ). Allow $(x + 1.5)(2x - 10)$ as a correct factorisation, but do not allow $\left(x + \dfrac{3}{2}\right)(x - 5)$ unless preceded by division of the quadratic by $2$ . Allow M1A1 for the correct critical values and evidence of another algebraic method that has led to these, eg $x(2x + 3) - 5(2x + 3)$ and the correct critical values, or $2x(x - 5) + 3(x - 5)$ and the correct critical values, or $(2x + 3)(2x - 10)$ and the correct critical values.	✓
$(x =) -\dfrac{3}{2}$ and $(x =) 5$	A1	dep on M1 for correct critical values oe	✓
$x < -\dfrac{3}{2}$ , $x > 5$ Working required	A1	dep on M1 for correct inequalities (must be separate inequalities). Allow interval notation, eg $\left(-\infty, -\dfrac{3}{2}\right) \cup (5, \infty)$ or $\left(-\infty, -\dfrac{3}{2}\right), (5, \infty)$ or $]-\infty, -\dfrac{3}{2}[ \cup ]5, \infty[$ or $]-\infty, -\dfrac{3}{2}[ , ]5, \infty[$ . Acceptable notation: allow a comma, a space, the word or, the word and, or $\cup$ to link the two regions. Do not allow as a single inequality $-\dfrac{3}{2} > x > 5$ .	✓

Full marks: 3/3

The diagram shows a solid wooden block in the shape of a cuboid.

The volume of the block is $V$ cm³
Work out the maximum value of $V$ [5 marks]

[Total 5 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

A solid cuboid measuring

x

cm by

x

cm by

(15 - 4x)

Its volume is

V

cm³, so

V

depends only on

x

Find

The maximum value of

V

Plan the Solution

Multiply the three edge lengths together to write $V$ in terms of $x$ .
Multiply out, so that every term is a power of $x$ and can be differentiated term by term.
A maximum is a turning point, so differentiate and solve $\dfrac{dV}{dx} = 0$ .
Two roots come out. Keep the one that gives a real block, confirm it is a maximum, then substitute it back into $V$ .

Worked Solution [5 marks]

Rule - Turning points:

V

has a turning point where

\dfrac{dV}{dx} = 0

, and that turning point is a maximum when

\dfrac{d^2V}{dx^2}

is negative there. Differentiate term by term with

\dfrac{d}{dx}(ax^n) = nax^{n-1}

Step 1: Write the volume in terms of

x

V = x \times x \times (15 - 4x)

V = x^2(15 - 4x)

(Reason: The cross-section is a square of side

x

cm and the block is

(15 - 4x)

cm long, so the volume is the three edge lengths multiplied together.)

Step 2: Multiply out to get a cubic in

x

V = 15x^2 - 4x^3

(Reason:

x^2 \times 15 = 15x^2

and

x^2 \times (-4x) = -4x^3

. Differentiating is only tidy once every term is a single power of

x

Step 3: Differentiate

\dfrac{dV}{dx} = 30x - 12x^2

(Reason: Bring each power down and reduce it by one:

15x^2

gives

2 \times 15x = 30x

, and

-4x^3

gives

3 \times (-4)x^2 = -12x^2

Step 4: Set the gradient to zero and solve

30x - 12x^2 = 0

6x(5 - 2x) = 0

x = 0 \quad \text{or} \quad x = 2.5

(Reason: At a turning point the graph is momentarily flat, so the gradient is

0

. Taking out the common factor

6x

gives the two roots.

x = 0

would leave a block with no width at all, so the root that matters here is

x = 2.5

Step 5: Confirm that this root gives the maximum

\dfrac{d^2V}{dx^2} = 30 - 24x

30 - 24 \times 2.5 = -30

(Reason: Differentiating a second time gives

30 - 24x

. At

x = 2.5

this is negative, so the gradient is falling as it passes through zero and the turning point is a maximum. At the rejected root

x = 0

the same expression is

+30

, which is a minimum.)

Step 6: Work out

V

when

x = 2.5

V = 15 \times (2.5)^2 - 4 \times (2.5)^3

15 \times 6.25 - 4 \times 15.625 = 93.75 - 62.5 = 31.25

(Reason: Squaring gives

2.5^2 = 6.25

and cubing gives

2.5^3 = 15.625

. Substituting both into

15x^2 - 4x^3

gives the largest volume the block can have.)

V = 31.25

Verification

Check 1: Work out

V

x = 2.4

and at

x = 2.6

, one on each side of

2.5

. Both should come out smaller.

2.4^2 \times (15 - 4 \times 2.4) = 31.104

and

2.6^2 \times (15 - 4 \times 2.6) = 31.096

, both below

31.25

Check 2: At

x = 2.5

the block measures

2.5

cm by

2.5

cm by

5

cm. Multiply those three edges directly, without using the expanded cubic.

2.5 \times 2.5 \times 5 = 31.25

Check 3: Multiply

V

4

to get

4V = (2x)(2x)(15 - 4x)

. Those three brackets add up to

15

whatever

x

is, and a product of positive numbers with a fixed sum is largest when they are all equal, which needs

2x = 15 - 4x

, so

x = 2.5

again. Each bracket is then

5

, and

5 \times 5 \times 5 = 125

, so

4V

can never exceed

125

and

V

can never exceed

\dfrac{125}{4}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$(V =) \; x(x)(15 - 4x) \quad (= 15x^2 - 4x^3)$	M1	for a correct expression for the volume (condone missing $V = \ldots$ )	✓
$\left(\dfrac{dV}{dx} = \right) \; 30x - 12x^2$	M1ft	ft dep on a $V$ of the form $ax^3 + bx^2$ where $a, b \neq 0$ , for a two-term derivative with at least one term correct for their $V$ , eg $15(2)x$ or $30x$ or $-3(4)x^2$ or $-12x^2$	✓
$30x - 12x^2 = 0 \implies x = \ldots$ or $(x =) \; 2.5$ oe	M1ft	dep on previous M mark, for equating their two-term first derivative to $0$ and solving to get a value of $x$ (the value of $x$ obtained must be greater than $0$ and must not be where their second derivative is equal to $0$ ) or the correct value of $x$	✓
$(V =) \; 2.5 \times 2.5 \times (15 - 4 \times 2.5)$ or $(V =) \; 15(2.5)^2 - 4(2.5)^3$	M1	dep on M3, for a full and correct substitution of their value of $x$ . The printed scheme writes that value in quotation marks, so a candidate's own earlier value may be used throughout this row.	✓
$31.25$	A1	oe eg $\dfrac{125}{4}$ . Ignore any units on the answer line.	✓
Correct answer only scores full marks (unless from obviously incorrect working)	Note	Guidance printed beside the final row of the scheme. It awards no mark of its own.	✓

Full marks: 5/5

The diagram shows a triangular prism $ABCDEF$ .
The base $ABCD$ is a horizontal rectangle.

$M$ is the midpoint of the line $AC$

$AC = 40$ cm
angle $CAE = 35^\circ$
angle $ABF = 90^\circ$

Find the size of angle $CME$
Give your answer correct to $3$ significant figures. [3 marks]

[Total 3 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

A triangular prism

ABCDEF

whose base

ABCD

is a horizontal rectangle

AC = 40

cm, angle

CAE = 35^\circ

, angle

ABF = 90^\circ

M

is the midpoint of

AC

, so it sits half way along the base's diagonal

Find

The size of angle

CME

, correct to

3

significant figures

Plan the Solution

Angle $ABF = 90^\circ$ is the fact that makes $BF$ stand vertically on the base, and $CE$ is the matching edge of the other triangular face, so $CE$ is vertical too.
A vertical edge is perpendicular to every line of the horizontal base drawn through its foot, so triangle $ACE$ and triangle $MCE$ are both right-angled at $C$ .
Use the $35^\circ$ angle with $AC$ to get the height $CE$ , halve $AC$ to get $CM$ , then take the inverse tangent of $CE$ over $CM$ .

Worked Solution [3 marks]

Rule - Right-angled trigonometry:

\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}

, so a missing opposite side is

\text{adjacent} \times \tan \theta

and a missing angle is

\tan^{-1}\left(\dfrac{\text{opposite}}{\text{adjacent}}\right)

Step 1: Show that

CE

is perpendicular to the base

\angle ABF = 90^\circ

BF \perp BA \text{ and } BF \perp BC

CE \parallel BF \text{, so } CE \perp AC \text{ and } CE \perp CM

(Reason: The base is a rectangle, so

BA

and

BC

already meet at right angles at

B

. The given angle

ABF = 90^\circ

makes

BF

perpendicular to

BA

as well, and

BF

is a side of the rectangular face

BCEF

, so it is perpendicular to

BC

too. Perpendicular to two different directions of the base means perpendicular to the base itself. The two triangular ends are congruent, so

CE

is parallel and equal to

BF

and stands vertically as well.)

Step 2: Work out the height

CE

\tan 35^\circ = \dfrac{CE}{40}

CE = 40 \tan 35^\circ = 28.0083\ldots \text{ cm}

(Reason: In triangle

ACE

the right angle is at

C

, so

AC

is the side adjacent to the

35^\circ

angle at

A

and

CE

is the side opposite it. Any correct route to

CE

earns the first method mark:

\dfrac{40}{\tan 55^\circ}

and

\dfrac{40 \sin 35^\circ}{\sin 55^\circ}

both give the same length.)

Step 3: Work out

CM

CM = \dfrac{40}{2} = 20 \text{ cm}

(Reason:

M

is the midpoint of

AC

, so

CM

is exactly half of

AC

. Halving is the whole of what the word midpoint is doing here, and it is the step that is easiest to skip.)

Step 4: Put the two lengths into the tangent ratio

\tan(\angle CME) = \dfrac{CE}{CM}

\dfrac{28.0083}{20} = 1.400415

(Reason: Triangle

MCE

has its right angle at

C

as well, so

CE

is opposite the angle at

M

and

CM

is adjacent to it. Reaching this ratio is the second method mark.)

Step 5: Take the inverse tangent, then round

\angle CME = \tan^{-1}(1.400415)

\angle CME = 54.4703\ldots^\circ

\angle CME = 54.5^\circ \text{ (3 s.f.)}

(Reason: The calculator must be in degree mode. The third significant figure of

54.4703\ldots

is the

4

in the first decimal place and the digit after it is

7

, so it rounds up to

54.5

\angle CME = 54.5^\circ

Verification

Check 1: Reach the angle with sine instead of tangent. The hypotenuse of triangle

MCE

ME = \sqrt{20^{2} + 28.0083^{2}} = 34.4161\ldots

, and

CE

is opposite the angle at

M

\sin^{-1}\left(\dfrac{28.0083}{34.4161}\right) = 54.47^\circ

Check 2: Do the whole thing in letters first.

CE = AC \tan 35^\circ

and

CM = \dfrac{AC}{2}

, so

\tan(\angle CME) = \dfrac{AC \tan 35^\circ}{\dfrac{AC}{2}} = 2 \tan 35^\circ

and the

40

cancels out completely.

\tan^{-1}(2 \tan 35^\circ) = 54.47^\circ

, the same angle, so the answer never depended on

AC

Check 3: Is the size sensible?

M

is only half as far from

C

A

is, and both look up at the same height

CE

, so the angle at

M

has to be the larger of the two.

54.5^\circ > 35^\circ

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $(CE =) \; 40 \times \tan 35 \; (= 28(.0\ldots))$ or $(CE =) \; \dfrac{40}{\tan(90 - 35)} \; (= 28(.0\ldots))$ or $(CE =) \; \dfrac{40 \sin 35}{\sin(90 - 35)} \; (= 28(.0\ldots))$	M1	for a correct method to find $CE$	✓
eg $(\angle CME =) \; \tan^{-1}\left(\dfrac{[CE]}{\dfrac{40}{2}}\right)$ or $(\angle CME =) \; \sin^{-1}\left(\dfrac{[CE]}{\sqrt{\left(\dfrac{40}{2}\right)^{2} + [CE]^{2}}}\right)$ or $(\angle CME =) \; \cos^{-1}\left(\dfrac{\dfrac{40}{2}}{\sqrt{\left(\dfrac{40}{2}\right)^{2} + [CE]^{2}}}\right)$	M1	for a complete method to find angle $CME$ . The printed scheme puts the height in quotation marks, which means the candidate's own value from the first mark may be used; it is written $[CE]$ here.	✓
$54.5$	A1	awrt $54.5$	✓
From the printed scheme, beside the answer	Note	Correct answer only scores full marks (unless from obviously incorrect working).	✓

Full marks: 3/3

$PQ$ is a straight line drawn on a square grid, where $1$ unit on each axis represents $1$ cm.

The point $P$ has coordinates $(-5,\ a)$ and the point $Q$ has coordinates $(7,\ 3a)$ , where $a > 0$

The length of $PQ$ is $4\sqrt{10}$ cm.

Work out an equation of the perpendicular bisector of $PQ$ .
Give your answer in the form $y = mx + c$ where $m$ and $c$ are integers. [6 marks]

[Total 6 marks]

Show solution & mark schemeHide solution & mark scheme

Question 24 - Exam Solution

Understanding the Question

Given

P(-5,\ a)

and

Q(7,\ 3a)

, where

a > 0

The length

PQ = 4\sqrt{10}

The grid scale is

1

unit to

1

cm, so a distance in grid units is that same number of centimetres

Find

An equation of the perpendicular bisector of

PQ

It must come out in the form

y = mx + c

with

m

and

c

integers, so expect whole numbers at the end

Plan the Solution

The length is given, so square it and use the distance formula to build an equation in $a$ alone.
Solve that equation, take the positive root, and write down $P$ and $Q$ as ordinary number pairs.
A perpendicular bisector needs two things: a gradient and a point. The gradient is the negative reciprocal of $m_{PQ}$ .
The point is the midpoint of $PQ$ , because the bisector cuts $PQ$ in half.
Finish with the point-gradient form and rearrange into the form asked for.

Worked Solution [6 marks]

Rule - Perpendicular bisector: it passes through the midpoint

\left(\dfrac{x_1 + x_2}{2},\ \dfrac{y_1 + y_2}{2}\right)

, and its gradient

m_{\perp}

satisfies

m \times m_{\perp} = -1

, where

m

is the gradient of the segment.

Step 1: Turn the given length into an equation in

a

PQ^2 = (7 - (-5))^2 + (3a - a)^2 = 12^2 + (2a)^2

(4\sqrt{10})^2 = 4^2 \times 10 = 160

144 + 4a^2 = 160

(Reason: Squaring both sides clears the square root, so a length with a surd in it becomes an ordinary equation. The vertical gap is

3a - a = 2a

, and squaring that gives

4a^2

, not

2a^2

Step 2: Solve for

a

and write down the two points

4a^2 = 160 - 144 = 16

a^2 = 4

a = 2

P(-5,\ 2) \qquad Q(7,\ 6)

(Reason: Both

2

and

-2

square to

4

, but the question states

a > 0

, so only

a = 2

is allowed. Then

3a = 6

Step 3: Work out the gradient of

PQ

m_{PQ} = \dfrac{6 - 2}{7 - (-5)} = \dfrac{4}{12} = \dfrac{1}{3}

(Reason: The gradient is the change in

y

divided by the change in

x

. Subtracting

-5

adds

5

, so the horizontal change is

12

and the vertical change is

4

Step 4: Turn that into the gradient of the perpendicular bisector

m_{PQ} \times m_{\perp} = -1

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

m_{\perp} = -3

(Reason: Perpendicular gradients multiply to

-1

, so

m_{\perp}

is the negative reciprocal of

\dfrac{1}{3}

: turn the fraction upside down and change the sign.)

Step 5: Find the midpoint of

PQ

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

(Reason: The bisector cuts

PQ

in half, so it goes through the midpoint. Average the

x

coordinates, then average the

y

coordinates.)

Step 6: Put the gradient and the midpoint into a line

y - 4 = -3(x - 1)

y - 4 = -3x + 3

y = -3x + 7

(Reason: Point-gradient form through

(1,\ 4)

with gradient

-3

. This gives

m = -3

and

c = 7

, both integers, which is the form the question asks for.)

y = -3x + 7

Verification

Check 1 - the midpoint lies on the line: Put

x = 1

into

y = -3x + 7

and compare with the midpoint's

y

coordinate.

-3(1) + 7 = 4

, which is the

y

coordinate of

M(1,\ 4)

Check 2 - the midpoint is the same distance from each end: Work out

MP

and

MQ

with the distance formula, then double one of them and compare with the length the question gives.

MP = \sqrt{6^2 + 2^2} = \sqrt{40}

and

MQ = \sqrt{6^2 + 2^2} = \sqrt{40}

, and

2\sqrt{40} = 4\sqrt{10}

, the given length

Check 3 - the two lines really are perpendicular: Multiply the gradient of

PQ

by the gradient of the bisector.

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

, so they meet at right angles

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\sqrt{(3a - a)^2 + (7 - (-5))^2} = 4\sqrt{10}$ oe or $(2a)^2 + 12^2 = (4\sqrt{10})^2$ oe or $4a^2 + 144 = 160$ oe	M1	for forming a correct equation in terms of $a$ ; brackets must be used correctly, but allow recovery from missing or incorrect brackets to be recovered; condone $4\sqrt{10}^{\,2}$ in place of $(4\sqrt{10})^2$	✓
$a = \sqrt{\dfrac{160 - 144}{4}}\ \left(= \sqrt{4} = 2\right)$	M1	dep on M1; for a complete method to solve a correct equation for $a$ ; condone inclusion of $\pm$ . The printed row puts $160$ and $144$ in quotation marks, so the candidate's own earlier values may be used here	✓
$m_{PQ} = \dfrac{3a - a}{7 - (-5)}\ \left(= \dfrac{2a}{12} = \dfrac{a}{6}\right)$ oe or $m_{PQ} = \dfrac{3 \times [2] - [2]}{7 - (-5)}\ \left(= \dfrac{4}{12} = \dfrac{1}{3}\right)$ oe	M1ft	for a method to find the gradient of $PQ$ , where $[2]$ is what they believe the value of $a$ to be; must be positive and clearly identified	✓
$\left[\dfrac{a}{6}\right] \times m_{\perp} = -1$ or $m_{\perp} = \dfrac{-1}{\left[\dfrac{a}{6}\right]}\ \left(= -\dfrac{6}{a}\right)$ oe or $\left[\dfrac{1}{3}\right] \times m_{\perp} = -1$ or $m_{\perp} = \dfrac{-1}{\left[\dfrac{1}{3}\right]}\ (= -3)$ oe	M1ft	for a method to find the gradient of the perpendicular bisector, where $\left[\dfrac{a}{6}\right]$ or $\left[\dfrac{1}{3}\right]$ is what they believe to be the gradient of $PQ$ ; must be clearly identified	✓
$\dfrac{-5 + 7}{2}\ (= 1)$ and $\dfrac{3a + a}{2}\ (= 2a)$ or $\dfrac{-5 + 7}{2}\ (= 1)$ and $\dfrac{3 \times [2] + [2]}{2}\ (= 4)$	M1ft	for a method to find the $x$ coordinate and $y$ coordinate of the midpoint of $PQ$ ; condone if the coordinates are the wrong way around, where $[2]$ is what they believe the value of $a$ to be; must be positive and clearly identified	✓
$y = -3x + 7$ . Correct answer only scores full marks (unless from obviously incorrect working)	A1	oe correct equation in required form eg $y = 7 - 3x$	✓

Full marks: 6/6

The grid below shows the graph of $y = a\sin(x + b)^{\circ} + c$ for $0 \leq x \leq 360$

Work out a suitable value for $a$ , a suitable value for $b$ and a suitable value for $c$ [3 marks]

a =b =c =

[Total 3 marks]

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Question 25 - Exam Solution

Understanding the Question

Given

The graph of

y = a\sin(x + b)^{\circ} + c

drawn on a grid for

0 \leq x \leq 360

A highest point at

(45,\ 4)

and a lowest point at

(225,\ -2)

The curve passes through

y = 1

on the way down at

x = 135

and on the way back up at

x = 315

Find

A suitable value for each of

a

b

and

c

Plan the Solution

Read the highest and the lowest value of $y$ off the grid. Every letter comes out of those two numbers.
The curve waves about the line halfway between them, and the height of that line is $c$ .
Half the drop from the highest point to the lowest point is $a$ .
Get $b$ from where the peak sits, because a sine is at its highest when the angle inside it is $90^{\circ}$ .

Worked Solution [3 marks]

Sine curves:

y = a\sin(x + b)^{\circ} + c

waves about the line

y = c

, reaching

c + a

at its highest and

c - a

at its lowest, and it peaks where

x + b = 90

Step 1: Read the highest and the lowest point off the grid

y_{\text{max}} = 4

y_{\text{min}} = -2

(Reason: The peak sits on the gridline

y = 4

and the trough on

y = -2

. These are the only two numbers the graph really has to give, and the rest of the question is what you can build from them.)

Step 2: The value halfway between them is

c

c = \dfrac{4 + (-2)}{2} = \dfrac{2}{2} = 1

(Reason: A sine curve is symmetrical about its middle line, so that line sits exactly halfway between the highest and the lowest value. Its height is the number added on at the end, which is

c

Step 3: Half the drop between them is

a

a = \dfrac{4 - (-2)}{2} = \dfrac{6}{2} = 3

(Reason: The peak is

a

above the middle line and the trough is

a

below it, so the whole drop from peak to trough is

2a

. Halving it leaves

a

Step 4: Where the peak sits gives

b

45 + b = 90

b = 90 - 45 = 45

(Reason: A sine reaches its highest value when the angle inside it is

90^{\circ}

. This peak is at

x = 45

, so the angle inside there is

45 + b

, and that has to be

90

Step 5: Write the equation the three values give

y = 3\sin(x + 45)^{\circ} + 1

(Reason: Reading the graph has fixed all three letters. Turning

a

negative works just as well:

y = -3\sin(x + 225)^{\circ} + 1

draws exactly the same curve, which is why the mark scheme accepts that pair too.)

a = 3

b = 45

c = 1

Verification

Check 1: Put

x = 225

into

y = 3\sin(x + 45)^{\circ} + 1

. The angle inside becomes

270^{\circ}

, and

\sin 270^{\circ} = -1

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

, which is the lowest point the grid shows.

Check 2: Test the start of the curve instead, which is a reading the working never used. At

x = 0

the angle inside is

45^{\circ}

, and

\sin 45^{\circ} = 0.7071

to four decimal places.

3 \times 0.7071 + 1 = 3.1213

, which is why the curve leaves the

y

-axis a little above

3

rather than on it.

Check 3: Rebuild both turning points from the answer: the highest value should come to

c + a

and the lowest to

c - a

1 + 3 = 4

and

1 - 3 = -2

, matching the peak and the trough on the grid.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The value of $a$	B1	for $a = 3$ or $a = -3$	✓
The value of $b$	B1	for $a > 0$ and $b = 45$ or for $a < 0$ and $b = 225$ if no answer for $a$ is seen allow this mark for $b = 45$	✓
The value of $c$	B1	for $c = 1$	✓
Alternative angles	Note	Allow correct alternative angles for the value of $b$ , eg $45$ or $-315$ , $225$ or $-135$	✓

Full marks: 3/3

The diagram shows a solid hemisphere, $H$

The radius of $H$ is $x$ cm
The volume of $H$ is $6174\pi \text{ cm}^{3}$

A dish is made by removing a solid hemisphere from $H$ , so that the dish has a uniform thickness of $2$ cm

Work out the total surface area of the dish.
Give your answer in terms of $\pi$ [5 marks]

cm²

[Total 5 marks]

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Question 26 - Exam Solution

Understanding the Question

Given

A solid hemisphere

H

whose radius is

x

The volume of

H

6174\pi \text{ cm}^{3}

A dish made by removing a solid hemisphere from

H

, leaving a uniform thickness of

2

Find

The total surface area of the dish, in terms of

\pi

Plan the Solution

A hemisphere is half a sphere, so its volume is half of $\dfrac{4}{3}\pi x^{3}$ . Put that equal to $6174\pi$ and the radius comes out of it.
The $\pi$ sits on both sides of that equation, so it cancels and every number in the working stays whole.
Taking a hemisphere out of the middle leaves a wall $2$ cm thick all the way round, so the hollow has a radius $2$ cm smaller than $H$ .
The dish has three surfaces and no others: the curved outside, the curved inside, and the flat ring round the top. Work out all three, then add them.

Worked Solution [5 marks]

Hemisphere: volume

= \dfrac{1}{2} \times \dfrac{4}{3}\pi r^{3} = \dfrac{2}{3}\pi r^{3}

and curved surface area

= \dfrac{1}{2} \times 4\pi r^{2} = 2\pi r^{2}

Step 1: Turn the volume into an equation

\dfrac{1}{2} \times \dfrac{4}{3}\pi x^{3} = 6174\pi

\dfrac{2}{3}\pi x^{3} = 6174\pi

(Reason: A hemisphere is half a sphere, so its volume is half of

\dfrac{4}{3}\pi x^{3}

. Half of

\dfrac{4}{3}

\dfrac{2}{3}

, which is where the

\dfrac{2}{3}\pi x^{3}

comes from.)

Step 2: Solve for the radius

x

x^{3} = \dfrac{6174 \times 3}{2} = 9261

x = \sqrt[3]{9261} = 21

(Reason: Dividing both sides by

\pi

clears it completely, so no decimal ever enters the working. Multiplying by

\dfrac{3}{2}

undoes the

\dfrac{2}{3}

, and

9261

is a perfect cube:

21 \times 21 \times 21 = 9261

Step 3: Find the radius of the hollow

21 - 2 = 19

(Reason: The wall is

2

cm thick all the way round, so the hemisphere that was taken out has the same centre and a radius

2

cm smaller. The dish therefore runs between radius

19

cm on the inside and radius

21

cm on the outside.)

Step 4: The two curved surfaces

2\pi \times 21^{2} = 2\pi \times 441 = 882\pi

2\pi \times 19^{2} = 2\pi \times 361 = 722\pi

(Reason: The curved surface of a hemisphere is half a sphere's

4\pi r^{2}

, which is

2\pi r^{2}

. The dish has two of them, because it is curved on the outside and curved on the inside: one at radius

21

and one at radius

19

Step 5: The flat ring round the top

\pi(21)^{2} - \pi(19)^{2} = 441\pi - 361\pi = 80\pi

(Reason: The top of the dish is a flat ring. Its area is the whole circle of radius

21

with the circle of the opening, radius

19

, taken out of it.)

Step 6: Add the three surfaces

882\pi + 722\pi + 80\pi = 1684\pi

(Reason: The outside, the inside and the ring make up the whole of the dish's surface, with nothing left out and nothing counted twice, so adding them gives the total. The question asks for the answer in terms of

\pi

, so it is left exactly as it stands.)

1684\pi \text{ cm}^{2}

Verification

Check 1: Put

x = 21

back into the volume of a hemisphere and see whether the figure the paper gives comes back.

\dfrac{2}{3}\pi \times 21^{3} = \dfrac{2}{3}\pi \times 9261 = 6174\pi

, which is the volume of

H

Check 2: Work the ring out a second way.

441 - 361

is a difference of two squares, so it factorises as

(21 - 19)(21 + 19)

and neither radius has to be squared at all.

\pi \times 2 \times 40 = 80\pi

, the ring's area again, from arithmetic that shares no step with the first way.

Check 3: Regroup the same three pieces. The curved outside and the whole top circle together make

3\pi r^{2}

at radius

21

; the curved inside with the opening taken off it leaves

\pi r^{2}

at radius

19

3\pi \times 441 + \pi \times 361 = 1323\pi + 361\pi = 1684\pi

, the same total by a different grouping, and it is the form the mark scheme's alternative method uses.

Check 4: Check which hemisphere is which, because this is the one place the question can be misread. If the

2

cm were added on rather than taken off, the dish would run from radius

21

on the inside out to radius

23

2\pi \times 529 + 2\pi \times 441 + \pi \times 88 = 2028\pi

, which is the value the mark scheme names as a special case. The question says the hemisphere of volume

6174\pi \text{ cm}^{3}

H

itself and the dish is cut out of it, so

21

is the outside radius and

1684\pi

is the answer.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Form a correct equation from the volume of $H$	M1	oe for forming a correct equation; allow use of any letter eg $\dfrac{1}{2} \times \dfrac{4}{3}\pi x^{3} = 6174\pi$	✓
Find the radius of the hemisphere	M1	for a correct method to find the radius of the hemisphere $(x =) \sqrt[3]{\dfrac{6174\pi \times 3}{2\pi}}$ $(= \sqrt[3]{9261} = 21)$	✓
The area of the top of the dish, or the two curved surfaces	M1ft	oe for a method to find the area of the top of the dish eg $\pi(21)^{2} - \pi(21 - 2)^{2}$ $(= 441\pi - 361\pi = 80\pi)$ or the total area of the two curved surfaces of the dish eg $2\pi(21)^{2} + 2\pi(21 - 2)^{2}$ $(= 882\pi + 722\pi = 1604\pi)$ where $21$ is what they believe to be the radius of the hemisphere	✓
A complete method for the total	M1ft	ft their $21$ for a complete method: their $1604\pi$ added to their $80\pi$	✓
The total surface area of the dish	A1	cao $1684\pi$	✓
Alternative method: the formula for a hemispherical shell	M2	oe for use of the formula for the total surface area of a hemispherical shell in a complete method eg $3\pi(21)^{2} + \pi(21 - 2)^{2}$ This M2 stands in place of the two M1ft marks above. If not M2, allow M1 for this formula used with omission of $\pi$ eg $3(21)^{2} + (21 - 2)^{2}$	✓
Special case	SC	SC B4 for $2028\pi$ (use of $23$ as the outer radius)	✓
Correct answer only	Note	A correct answer only scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 5/5

Keep revising

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

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Edexcel IGCSE 4MA1/2HR, Wednesday 4 June 2025: Worked Solutions, Questions 19 to 26

Worked solutions, questions 19 to 26 of 26

Question 19, Calculator allowed

Question 19 - Exam Solution

Question 20, Calculator allowed

Question 20 - Exam Solution

Question 21, Calculator allowed

Question 21 - Exam Solution

Question 22, Calculator allowed

Question 22 - Exam Solution

Question 23, Calculator allowed

Question 23 - Exam Solution

Question 24, Calculator allowed

Question 24 - Exam Solution

Question 25, Calculator allowed

Question 25 - Exam Solution

Question 26, Calculator allowed

Question 26 - Exam Solution

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