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Edexcel IGCSE 4MA1/1H, Thursday 15 May 2025: Worked Solutions, Questions 12 to 20

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 1 to 11

This is part two of three. Questions 1 to 11, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/1H (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 12 to 20 of 25

Write $\dfrac{5}{4} + \dfrac{x - 3}{6x}$ as a single fraction in its simplest form. [3 marks]

[Total 3 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

The sum of two fractions,

\dfrac{5}{4}

and

\dfrac{x - 3}{6x}

The denominators

4

and

6x

are different, and one of them contains

x

Find

One single fraction, written in its simplest form. So the final numerator and denominator must share no factor, not even

x

Plan the Solution

Find the lowest common denominator of $4$ and $6x$ .
Rewrite each fraction over that denominator, multiplying the top by whatever the bottom was multiplied by.
Add the two numerators, expanding the bracket in full.
Check whether anything cancels before writing the answer down.

Worked Solution [3 marks]

Rule - Adding algebraic fractions: rewrite both fractions over the lowest common denominator, add the numerators, then cancel any factor common to the top and the bottom.

Step 1: Find the lowest common denominator

4 = 2 \times 2

6x = 2 \times 3 \times x

2 \times 2 \times 3 \times x = 12x

(Reason:

12x

is the smallest expression that both

4

and

6x

divide into exactly, so it is the lowest common denominator)

Step 2: Rewrite each fraction with denominator

12x

\dfrac{5}{4} = \dfrac{5 \times 3x}{4 \times 3x} = \dfrac{15x}{12x}

\dfrac{x - 3}{6x} = \dfrac{2(x - 3)}{12x} = \dfrac{2x - 6}{12x}

(Reason: multiplying the top and the bottom by the same amount leaves a fraction's value unchanged, and the bracket must be multiplied out in full)

Step 3: Add the numerators

\dfrac{15x}{12x} + \dfrac{2x - 6}{12x} = \dfrac{15x + 2x - 6}{12x}

\dfrac{15x + 2x - 6}{12x} = \dfrac{17x - 6}{12x}

(Reason: the denominators now match, so only the numerators are added, and

15x + 2x = 17x

)

Step 4: Check that nothing cancels

\dfrac{17x - 6}{12x}

(Reason: no whole number divides both

17

and

6

except

1

, so nothing can be taken out of the numerator, and

x

is not a factor of

17x - 6

because of the

-6

)

\dfrac{17x - 6}{12x}

Verification

Check 1 - substitute $x = 3$ : At

x = 3

the second fraction is zero, so the original sum is just

\dfrac{5}{4}

. The answer must give the same value.

\dfrac{5}{4} + \dfrac{3 - 3}{6 \times 3} = \dfrac{5}{4}

and

\dfrac{17 \times 3 - 6}{12 \times 3} = \dfrac{45}{36} = \dfrac{5}{4}

Check 2 - substitute $x = 2$ : A second value, chosen so the second fraction is negative and the two sides cannot agree by accident.

\dfrac{5}{4} + \dfrac{2 - 3}{6 \times 2} = \dfrac{5}{4} - \dfrac{1}{12} = \dfrac{7}{6}

and

\dfrac{17 \times 2 - 6}{12 \times 2} = \dfrac{28}{24} = \dfrac{7}{6}

Check 3 - split the answer back apart: Divide each term of the numerator by

12x

and rebuild the original sum the other way round.

\dfrac{17x - 6}{12x} = \dfrac{17}{12} - \dfrac{1}{2x}

and

\dfrac{5}{4} + \dfrac{x - 3}{6x} = \dfrac{5}{4} + \dfrac{1}{6} - \dfrac{1}{2x} = \dfrac{17}{12} - \dfrac{1}{2x}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $\dfrac{5(6x)}{24x} + \dfrac{4(x - 3)}{24x}$ oe, or $\dfrac{5(6x)}{4(6x)} + \dfrac{4(x - 3)}{4(6x)}$ oe, or $\dfrac{30x}{24x} + \dfrac{4(x - 3)}{24x}$ oe, or $\dfrac{30x + 4(x - 3)}{24x}$ oe, or $\dfrac{15x}{12x} + \dfrac{2(x - 3)}{12x}$ oe, or $\dfrac{15x + 2(x - 3)}{12x}$ oe	M1	for two correct fractions with common denominator or a single correct fraction	✓
eg $\dfrac{30x + 4x - 12}{24x}$ oe, or $\dfrac{30x}{24x} + \dfrac{4x - 12}{24x}$ oe, or $\dfrac{30x}{24x} + \dfrac{4x}{24x} - \dfrac{12}{24x}$ oe, or $\dfrac{34x}{24x} - \dfrac{12}{24x}$ oe, or $\dfrac{34x - 12}{24x}$ oe, or $\dfrac{15x + 2x - 6}{12x}$ oe	M1	for correct fraction(s) with bracket(s) expanded correctly	✓
$\dfrac{17x - 6}{12x}$	A1	oe but must be simplified, eg $\dfrac{-6 + 17x}{12x}$ . Do not ISW incorrect simplification, eg $\dfrac{17x - 6}{12x} = \dfrac{11}{12}$ is M2A0	✓
Correct answer scores full marks (unless from obvious incorrect working)	Note	The printed scheme awards all three marks for a correct simplified single fraction on the answer line even where no working is shown, unless the working that is shown is obviously incorrect. This row carries no mark of its own.	✓

Full marks: 3/3

Beatrice bought a lakeside cabin for $\$750\,000$

In the first year, the value of the cabin fell by $4\%$
In the second year, the value of the cabin fell by $6.5\%$

In the third year, the value of the cabin rose by $x\%$

At the end of the third year, the value of Beatrice's cabin was $\$698\,445$

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

x =

[Total 3 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

The cabin was bought for

\$750\,000

First year: a fall of

4\%

Second year: a fall of

6.5\%

Third year: a rise of

x\%

After the three years the cabin was worth

\$698\,445

Find

The value of

x

, the percentage rise in the third year. The three changes act one after another, so expect a chain of multipliers rather than one percentage added on at the end.

Plan the Solution

Write each fall as a multiplier: a fall of $4\%$ leaves $96\%$ and a fall of $6.5\%$ leaves $93.5\%$ .
Apply the two falls in order to get the value the third year starts from.
Divide the final value by that value to get the third year's multiplier.
Turn that multiplier back into a percentage to get $x$ .

Worked Solution [3 marks]

Rule - percentage change as a multiplier: a fall of

p\%

multiplies by

1 - \dfrac{p}{100}

and a rise of

p\%

multiplies by

1 + \dfrac{p}{100}

. Changes in successive years multiply together, so the whole chain is

750\,000 \times 0.96 \times 0.935 \times \left(1 + \dfrac{x}{100}\right) = 698\,445

Step 1: Write each fall as a multiplier

1 - \dfrac{4}{100} = 0.96

1 - \dfrac{6.5}{100} = 0.935

(Reason: (Reason: a fall of

4\%

leaves

96\%

of the value and a fall of

6.5\%

leaves

93.5\%

of it, so each year becomes a single multiplication instead of a subtraction.))

Step 2: Work down to the value at the end of the second year

750\,000 \times 0.96 = 720\,000

720\,000 \times 0.935 = 673\,200

(Reason: (Reason: the second fall is taken off the

\$720\,000

that is left after the first year, never off the original

\$750\,000

. Adding the two percentages to get

10.5\%

is the standard slip here.))

Step 3: Find the multiplier for the third year

\dfrac{698\,445}{673\,200} = 1.0375

(Reason: (Reason: dividing the value at the END of a year by the value at the START of that same year leaves exactly the multiplier for that year, because the start value cancels.))

Step 4: Turn the multiplier back into a percentage

1.0375 - 1 = 0.0375

0.0375 \times 100 = 3.75

(Reason: (Reason: the rise multiplier is

1 + \dfrac{x}{100}

, so taking away

1

leaves

\dfrac{x}{100}

and multiplying by

100

gives

x

itself.))

x = 3.75

Verification

Check 1: Put the answer back on to the value the third year started from: raise

673\,200

3.75\%

673\,200 \times 1.0375 = 698\,445

, which is the value the question gives.

Check 2: Do all three years in one go, as a single multiplier:

0.96 \times 0.935 \times 1.0375 = 0.93126

750\,000 \times 0.93126 = 698\,445

, so the chain agrees with the year-by-year working.

Check 3: Check the cash rise instead of the multiplier: work out

3.75\%

673\,200

and add it on.

0.0375 \times 673\,200 = 25\,245

and

673\,200 + 25\,245 = 698\,445

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$750\,000 \times (1 - 0.04) \ (= 720\,000)$ oe and $720\,000 \times (1 - 0.065) \ (= 673\,200)$ or $750\,000 \times (1 - 0.04) \times (1 - 0.065) \ (= 673\,200)$ or $750\,000 \times 0.96 \times 0.935 \ (= 673\,200)$	M1	NB: accept $\left(1 - \dfrac{4}{100}\right)$ for $0.96$ but not $(1 - 4\%)$ , and accept $\left(1 - \dfrac{6.5}{100}\right)$ for $0.935$ but not $(1 - 6.5\%)$ . Calculations may be seen as part of an equation, eg $750\,000 \times 0.96 \times 0.935 \times \left(1 + \dfrac{x}{100}\right) = 698\,445$ $1.0375$ or $25\,245$ imply M1.	✓
eg $\dfrac{698\,445 - 673\,200}{673\,200} \ (\times 100) \ (= 0.0375)$ or $\dfrac{698\,445}{673\,200} - 1 \ (= 0.0375)$ or $1.0375 - 1 \ (= 0.0375)$ oe or $\dfrac{698\,445}{673\,200} \times 100 \ (= 103.75)$	M1	For a method to reach a value one step away from $x$ , ie a method leading to $0.0375$ or $103.75$ . The printed scheme writes $673\,200$ in quotation marks in this row, so the candidate's own value from the first mark may be used in its place.	✓
$3.75$	A1	oe eg $3\dfrac{3}{4}$ or $\dfrac{15}{4}$ . Correct answer scores full marks (unless from obvious incorrect working).	✓

Full marks: 3/3

Lorenzo is going to play one game of squash and one game of table tennis.

He can either win or lose each game.

The probability that he will win the game of squash is $0.7$
The probability that he will win the game of table tennis is $0.4$

(a) Complete the probability tree diagram. [2 marks]

(b) Work out the probability that Lorenzo loses both games. [2 marks]

(b)

[Total 4 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

Lorenzo plays one game of squash and then one game of table tennis.

He either wins or loses each game - neither game can be drawn.

The probability that he wins the squash is

0.7

The probability that he wins the table tennis is

0.4

Find

(a) The three pairs of probabilities that belong on the branches of the tree. (b) The probability that he loses both games. Losing both games is one single path through the tree, so expect one multiplication rather than a sum.

Plan the Solution

The two branches leaving any one point cover everything that can happen there, so their probabilities add to $1$ . Subtract to fill in the missing one of each pair.
The squash result does not change how likely he is to win the table tennis, so the same pair of probabilities goes on both table tennis pairs.
For (b), find the one path that is lose and then lose, and multiply the probabilities along it.

Worked Solution [4 marks]

Rule - a probability tree: the branches leaving any one point cover everything that can happen there, so their probabilities add to

1

; and the probability of a whole path is the probabilities along that path multiplied together.

Step 1: Complete the squash pair

1 - 0.7 = 0.3

(Reason: (Reason: he either wins or loses, so those two branches take up the whole of the probability at that point. Taking

0.7

away from

1

leaves the probability that he loses the squash.))

Step 2: Complete both table tennis pairs

1 - 0.4 = 0.6

(Reason: (Reason: the same subtraction fills a table tennis pair - and because the squash result does not change how likely he is to win the table tennis,

0.4

and

0.6

go on the lower pair as well as the upper one. That is the third pair the mark scheme is looking for.))

Step 3: Multiply along the lose, lose path

0.3 \times 0.6 = 0.18

(Reason: (Reason: losing both games is one single route through the tree - down at the first pair of branches, then down again at the second - and the probability of a whole route is the probabilities along it multiplied together.))

(a)

0.3

on the lose squash branch, and

0.4

and

0.6

on each table tennis pair(b)

0.18

Verification

Check 1: Multiply along all four paths of the completed tree and add the four results. Every way the two games can turn out is one of those four paths, so they must total

1

0.28 + 0.42 + 0.12 + 0.18 = 1

, and the last of those four,

0.18

, is the lose then lose path.

Check 2: Do the same multiplication in fractions instead of decimals:

0.3

\dfrac{3}{10}

and

0.6

\dfrac{6}{10}

\dfrac{3}{10} \times \dfrac{6}{10} = \dfrac{18}{100} = \dfrac{9}{50} = 0.18

, which is the same answer in two of the forms the mark scheme accepts.

Check 3: Count instead of multiplying. Imagine

1\,000

repeats of this pair of games: he loses the squash in about

300

of them, and he goes on to lose the table tennis in

60\%

of those.

0.6 \times 300 = 180

repeats out of

1\,000

, and

\dfrac{180}{1\,000} = 0.18

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) All three pairs of probabilities on the correct branches: $0.7$ and $0.3$ on the squash pair, $0.4$ and $0.6$ on each table tennis pair	B2	For all $3$ correct pairs of probabilities on the correct branches.	✓
For $1$ or $2$ correct pairs of probabilities on the correct branches	(B1)	The printed scheme gives this row in brackets: it is the partial credit inside part (a)'s two marks, not a mark on top of them.	✓
Allow equivalent fractions or percentages	Note	Printed on the official scheme under part (a), so it covers both of the rows above it: a branch labelled with a fraction or a percentage is marked exactly as a decimal one is.	✓
$0.3 \times 0.6$	M1ft	(b) The printed scheme writes both values in quotation marks, so a candidate's own two probabilities, read off their own tree, may be used in place of $0.3$ and $0.6$ . Both probabilities must be less than $1$ .	✓
$0.18$	A1ft	oe eg $\dfrac{18}{100}$ or $\dfrac{9}{50}$ or $\dfrac{0.18}{1}$ or $18\%$ .	✓
Correct answer scores full marks (unless from obvious incorrect working)	Note	Printed in italics on the official scheme beside part (b)'s answer.	✓

Full marks: 4/4

$\left(\sqrt{3}\right)^{5} = k\sqrt{3}$ where $k$ is an integer.

(a) Work out the value of $k$ [1 mark]

(b) Show that $\dfrac{21}{3 - \sqrt{2}}$ can be written in the form $c + \sqrt{d}$
where $c$ and $d$ are integers.
Set out each stage of your working clearly. [3 marks]

k =

[Total 4 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

\left(\sqrt{3}\right)^{5} = k\sqrt{3}

, with

k

an integer

Part (b) starts from

\dfrac{21}{3 - \sqrt{2}}

, a fraction carrying a surd in its denominator

Find

(a) the integer

k

(b) integers

c

and

d

for which

\dfrac{21}{3 - \sqrt{2}} = c + \sqrt{d}

Plan the Solution

(a) Split the power. $\left(\sqrt{3}\right)^{5}$ is $\left(\sqrt{3}\right)^{4}$ multiplied by one spare $\sqrt{3}$ , and an even power of a square root is a whole number.
(b) Clear the surd out of the denominator by multiplying the top and the bottom by $3 + \sqrt{2}$ . That pairing is a difference of two squares, so the roots cancel and the denominator becomes a whole number.
(b) Cancel the fraction, expand the bracket, then write the surd term as one square root, because the form $c + \sqrt{d}$ carries no number in front of the root.

Worked Solution [4 marks]

Rule - Rationalising a denominator: multiply by

\dfrac{a + \sqrt{b}}{a + \sqrt{b}}

, because

(a - \sqrt{b})(a + \sqrt{b}) = a^{2} - b

has no surd left in it. A coefficient enters a root squared:

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

Step 1: split the fifth power of the root

\left(\sqrt{3}\right)^{5} = \left(\sqrt{3}\right)^{4} \times \sqrt{3}

\left(\sqrt{3}\right)^{4} = \left(\left(\sqrt{3}\right)^{2}\right)^{2} = 3^{2} = 9

(Reason: Squaring a square root undoes it, so

\left(\sqrt{3}\right)^{2} = 3

. Taking the even part of the power first leaves a whole number multiplied by one single

\sqrt{3}

Step 2: read off k

\left(\sqrt{3}\right)^{5} = 9\sqrt{3}

9\sqrt{3} = k\sqrt{3} \implies k = 9

(Reason: Both sides are now a whole number multiplied by

\sqrt{3}

, so those whole numbers must be the same.)

Step 3: multiply the top and the bottom by the conjugate

\dfrac{21}{3 - \sqrt{2}} = \dfrac{21}{3 - \sqrt{2}} \times \dfrac{3 + \sqrt{2}}{3 + \sqrt{2}}

(Reason:

\dfrac{3 + \sqrt{2}}{3 + \sqrt{2}}

is equal to

1

, so the value of the fraction is untouched. The conjugate is the multiplier to reach for because

3 + \sqrt{2}

and

3 - \sqrt{2}

multiply to give a whole number.)

Step 4: expand the denominator

(3 - \sqrt{2})(3 + \sqrt{2}) = 9 + 3\sqrt{2} - 3\sqrt{2} - 2 = 7

\dfrac{21}{3 - \sqrt{2}} = \dfrac{21(3 + \sqrt{2})}{7}

(Reason: The two middle terms are opposites and cancel, which is the difference of two squares:

3^{2} - \left(\sqrt{2}\right)^{2} = 9 - 2 = 7

. The denominator now holds no surd.)

Step 5: cancel, then expand the numerator

\dfrac{21(3 + \sqrt{2})}{7} = 3(3 + \sqrt{2})

3(3 + \sqrt{2}) = 9 + 3\sqrt{2}

(Reason:

\dfrac{21}{7} = 3

, so cancelling before the bracket is expanded keeps the numbers small. Expanding first would give

\dfrac{63 + 21\sqrt{2}}{7}

, which reaches the same place.)

Step 6: write the surd term as a single square root

3\sqrt{2} = \sqrt{3^{2} \times 2} = \sqrt{18}

(Reason: The form

c + \sqrt{d}

carries no number in front of the root, so the

3

has to be taken inside. A coefficient enters a square root squared.)

Step 7: state the result in the form the question asks for

\dfrac{21}{3 - \sqrt{2}} = 9 + \sqrt{18}

(Reason: Comparing with

c + \sqrt{d}

gives

c = 9

and

d = 18

, and both of those are integers, which is what the question demands.)

(a)

k = 9

(b)

\dfrac{21}{3 - \sqrt{2}} = 9 + \sqrt{18}

c = 9

and

d = 18

Verification

Check 1: Square both sides of part (a). Squaring clears the root, so the check becomes whole-number arithmetic.

\left(\sqrt{3}\right)^{10} = 3^{5} = 243

and

\left(9\sqrt{3}\right)^{2} = 81 \times 3 = 243

Check 2: Work both sides of part (a) out as decimals on the calculator.

\left(\sqrt{3}\right)^{5} \approx 15.5885

and

9\sqrt{3} \approx 15.5885

Check 3: Multiply the answer to part (b) back by the original denominator. If the rationalising is sound, the product is the original numerator.

\left(9 + \sqrt{18}\right)\left(3 - \sqrt{2}\right) = 27 - 9\sqrt{2} + 9\sqrt{2} - 6 = 21

Check 4: Work the original fraction and the final answer out as decimals.

\dfrac{21}{3 - \sqrt{2}} \approx 13.2426

and

9 + \sqrt{18} \approx 13.2426

Check 5: Read the answer back against the form the question demands,

c + \sqrt{d}

with both letters integers.

c = 9

and

d = 18

, and both are integers

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $9$	B1	Allow $9\sqrt{3}$ .	✓
(b) $\dfrac{21}{3 - \sqrt{2}} \times \dfrac{3 + \sqrt{2}}{3 + \sqrt{2}}$ or $\dfrac{21}{3 - \sqrt{2}} \times \dfrac{-3 - \sqrt{2}}{-3 - \sqrt{2}}$	M1	For explicitly multiplying the numerator and the denominator by $3 + \sqrt{2}$ or $-3 - \sqrt{2}$ .	✓
eg $\dfrac{21\left(3 + \sqrt{2}\right)}{9 - 3\sqrt{2} + 3\sqrt{2} - 2}$ or $\dfrac{21\left(3 + \sqrt{2}\right)}{3^{2} - 2}$ or $\dfrac{21\left(3 + \sqrt{2}\right)}{9 - 2}$ or $\dfrac{21\left(3 + \sqrt{2}\right)}{7}$ or $\dfrac{63 + 21\sqrt{2}}{9 - 2}$ or $\dfrac{63 + 21\sqrt{2}}{7}$	M1	Dependent on the first M1. The denominator may be $4$ terms, which all need to be correct. $\dfrac{21}{3 - \sqrt{2}} \times \dfrac{3 + \sqrt{2}}{3 + \sqrt{2}} = 9 + 3\sqrt{2}$ scores M1M0.	✓
Working required $9 + \sqrt{18}$	A1	Dependent on M2.	✓
$9 + \sqrt{18}$ gained with no method marks awarded	SC B1	Special case.	✓
$9 + \sqrt{18}$ gained if you would award the first M1 but not the second M1	SC B2	Special case, total $2$ marks.	✓

Full marks: 4/4

$\text{f}(x) = 9 - \sqrt{x}$ where $x \geqslant 0$
$\text{g}(x) = 4x^2$

(a) Work out $\text{f}(9)$ [1 mark]

(b) Solve $\text{fg}(x) < 0$
You must show clear algebraic working. [3 marks]

(a)(b)

[Total 4 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

\text{f}(x) = 9 - \sqrt{x}

for

x \geqslant 0

\text{g}(x) = 4x^2

Two functions: one built on a square root, one a quadratic.

Find

(a) The value of

\text{f}(9)

(b) Every value of

x

for which

\text{fg}(x) < 0

. The answer is a range, not a single number.

Plan the Solution

(a) Replace $x$ with $9$ in $\text{f}(x)$ and work out the square root before subtracting.
(b) $\text{fg}(x)$ means do $\text{g}$ first, so put $4x^2$ in place of $x$ in $\text{f}(x)$ .
Simplify $\sqrt{4x^2}$ , which turns the expression into a linear one, then solve the inequality.
Watch the inequality sign: adding to both sides never turns it, and dividing by a positive number never turns it either.

Worked Solution [4 marks]

Rule - Composite functions:

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

. Work

\text{g}

out first, then feed that result into

\text{f}

. The letter standing nearest the

x

acts first.

Step 1: put

9

into

\text{f}(x)

\text{f}(9) = 9 - \sqrt{9}

\text{f}(9) = 9 - 3 = 6

(Reason: The square root sign means the positive square root, so

\sqrt{9} = 3

and not

-3

. The root is worked out first, then subtracted from

9

Step 2: build

\text{fg}(x)

by putting

\text{g}(x)

inside

\text{f}

\text{fg}(x) = \text{f}(4x^2)

\text{fg}(x) = 9 - \sqrt{4x^2}

(Reason:

\text{fg}(x)

says do

\text{g}

first, then

\text{f}

, so every

x

9 - \sqrt{x}

is replaced by

4x^2

. Nothing is squared or rooted yet - this is only the substitution.)

Step 3: take the square root out

\sqrt{4x^2} = 2x

\text{fg}(x) = 9 - 2x

(Reason: A root over a product splits into the root of each factor, so

\sqrt{4x^2} = \sqrt{4} \times \sqrt{x^2}

. Here

\sqrt{4} = 2

, and the root sign means the positive square root, so

\sqrt{x^2}

x

for

x \geqslant 0

. The square root has gone, and what is left is linear.)

Step 4: solve the inequality

9 - 2x < 0

9 < 2x

\dfrac{9}{2} < x

x > \dfrac{9}{2}

(Reason: Add

2x

to both sides so the

x

term is positive, then halve both sides. Neither move turns the sign round. Reading

\dfrac{9}{2} < x

from right to left gives

x > \dfrac{9}{2}

, the same statement written the usual way round, and

\dfrac{9}{2} = 4.5

(a)

\text{f}(9) = 6

(b)

x > \dfrac{9}{2}

Verification

Check 1: Run part (a) backwards. If

9 - \sqrt{x} = 6

then

\sqrt{x} = 3

, so

x = 3^2 = 9

. The input comes back as

9

, which is the number the question put in.

Check 2: Test a value inside the answer. Take

x = 5

, which is bigger than

4.5

. Then

4 \times 5^2 = 100

and

9 - \sqrt{100} = 9 - 10 = -1

\text{fg}(5) = -1

, which is less than

0

, so

x = 5

does belong to the solution set.

Check 3: Test a value outside it and the boundary itself. For

x = 4

4 \times 4^2 = 64

and

9 - \sqrt{64} = 9 - 8 = 1

. For

x = 4.5

4 \times 4.5^2 = 81

and

9 - \sqrt{81} = 9 - 9 = 0

\text{fg}(4) = 1

and

\text{fg}(4.5) = 0

, and neither is less than

0

, so the boundary is excluded and the inequality is strict.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $6$	B1	cao	✓
(b) $9 - \sqrt{4x^2}\ (<0)$	M1	for substituting $\text{g}(x)$ in $\text{f}(x)$ , allow incorrect inequality sign or $=$ sign	✓
(b) eg $9 - 2x\ (<0)$ or $9 < 2x$ or $81 < 4x^2$ or $9^2 < 4x^2$	M1	for removing the square root, allow incorrect inequality sign or $=$ sign	✓
(b) Working required. $x > \dfrac{9}{2}$	A1	(dep on M1) oe eg $x > 4.5$ , $\dfrac{9}{2} < x$ , $4.5 < x$	✓

Full marks: 4/4

The histogram gives information about the times, in minutes, that some visitors spent in an art gallery.

Work out an estimate for the proportion of these visitors who spent more than $40$ minutes in the gallery. [3 marks]

[Total 3 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

A histogram of frequency density against time

t

, in minutes.

Five bars:

0 \leq t < 10

at frequency density

2.5

;

10 \leq t < 15

4

;

15 \leq t < 30

3.6

;

30 \leq t < 50

0.5

;

50 \leq t < 60

1.7

Find

An estimate for the proportion of these visitors with

t > 40

Plan the Solution

On a histogram the frequency is the area of the bar, so multiply each frequency density by its class width.
Add the five frequencies to get the total number of visitors.
The bar from $30$ to $50$ has to be split at $40$ , because only the part above $40$ counts.
Write the number above $40$ over the total, then give the fraction in its simplest form.

Worked Solution [3 marks]

Rule - Histogram:

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

. The area of a bar is the frequency for that class.

Step 1: Turn each bar into a frequency

\text{0 to 10 minutes: } 10 \times 2.5 = 25

\text{10 to 15 minutes: } 5 \times 4 = 20

\text{15 to 30 minutes: } 15 \times 3.6 = 54

\text{30 to 50 minutes: } 20 \times 0.5 = 10

\text{50 to 60 minutes: } 10 \times 1.7 = 17

(Reason: The width of a bar is its class width in minutes and its height is the frequency density, so

\text{width} \times \text{height}

is the number of visitors in that class.)

Step 2: Add the frequencies to find the total

25 + 20 + 54 + 10 + 17 = 126

(Reason: There were

126

visitors altogether, and that total is the denominator of the proportion.)

Step 3: Split the

30

50

bar at

40

\text{40 to 50 minutes: } 10 \times 0.5 = 5

\text{50 to 60 minutes: } 10 \times 1.7 = 17

5 + 17 = 22

(Reason: The bar from

30

50

has a constant height, so the part from

40

50

10

minutes wide; the last bar lies entirely above

40

Step 4: Write the proportion and simplify

\dfrac{22}{126} = \dfrac{11}{63}

(Reason: Both

22

and

126

divide by

2

\dfrac{11}{63}

Verification

Check 1: Measure in centimetre squares on the printed grid instead. Across,

1

cm is

5

minutes; up,

1

cm is

0.5

of a unit of frequency density. The five bars are

10

8

21.6

4

and

6.8

square centimetres, and above

40

minutes there are

2

and

6.8

\dfrac{8.8}{50.4} = \dfrac{11}{63}

Check 2: Work out the other part instead. The visitors with

t \leq 40

number

25 + 20 + 54 + 5 = 104

, and the two parts must make the whole.

104 + 22 = 126

and

\dfrac{52}{63} + \dfrac{11}{63} = 1

Check 3: Count in small squares instead. One small square is

1

minute across and

0.1

of a unit of frequency density up, so the bars hold

250

200

540

100

and

170

small squares, and above

40

minutes there are

50 + 170 = 220

\dfrac{220}{1260} = \dfrac{11}{63}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$10 \times 2.5 = 25$ , $5 \times 4 = 20$ , $15 \times 3.6 = 54$ , $20 \times 0.5 = 10$ , $10 \times 1.7 = 17$ , $10 \times 0.5 = 5$ (area of the $40$ to $50$ bar)	M1	for finding the area of at least $2$ bars, either using $\text{freq density} \times \text{mins}$ or use of counting small squares or $\text{cm}^2$ . Values may be seen on the diagram. $22$ or $220$ or $8.8$ implies M1	✓
$(10 \times 2.5) + (5 \times 4) + (15 \times 3.6) + (20 \times 0.5) + (10 \times 1.7) = 126$ or $25 + 20 + 54 + 10 + 17 = 126$ or $250 + 200 + 540 + 100 + 170 = 1260$ or $10 + 8 + 21.6 + 4 + 6.8 = 50.4$	M1	for method to find total number of visitors (allow one error or omission) or total number of small squares or $\text{cm}^2$ for the method used (allow one error or omission). The printed scheme puts those five values in quotation marks, so a candidate's own earlier frequencies may be used.	✓
$\dfrac{11}{63}$	A1	oe eg $\dfrac{22}{126}$ or $\dfrac{220}{1260}$ or $0.174(60\ldots)$ or $0.175$ or $17.4(60\ldots)\%$ or $17.5\%$ or $22$ out of $126$ . If $\dfrac{22}{126}$ is seen in the workings and $22$ is on the answer line, award M2A0	✓
On the answer line	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 3/3

$y$ is inversely proportional to $x^3$

When $x = 3$ , $y = 4$

Find the value of $x$ when $y = 864$ [4 marks]

x =

[Total 4 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

y

is inversely proportional to

x^3

One pair of values:

y = 4

when

x = 3

Find

The value of

x

when

y = 864

Plan the Solution

Turn the words into an equation. Inversely proportional to $x^3$ means $y = \dfrac{k}{x^3}$ for some constant $k$ .
Use the pair of values the question gives to work out $k$ .
Put $y = 864$ into the finished formula and rearrange to leave $x^3$ on its own.
Take the cube root to get $x$ .

Worked Solution [4 marks]

Rule - Inverse proportion:

y \propto \dfrac{1}{x^3}

means

y = \dfrac{k}{x^3}

, where

k

is a constant. Find

k

from one pair of values, and the same formula then holds for every other pair.

Step 1: Write the statement as an equation

y \propto \dfrac{1}{x^3}

y = \dfrac{k}{x^3}

(Reason: Inversely proportional means

y

is a constant divided by

x^3

, not multiplied by it. The constant has to be written as a symbol such as

k

until its value is known.)

Step 2: Substitute

x = 3

and

y = 4

to find

k

4 = \dfrac{k}{3^3}

k = 4 \times 3^3 = 108

(Reason: Multiplying both sides by

3^3

leaves

k

on its own, so the constant comes from the one pair of values the question gives.)

Step 3: Put

y = 864

into the formula

864 = \dfrac{108}{x^3}

x^3 = \dfrac{108}{864}

\dfrac{108}{864} = \dfrac{1}{8}

(Reason: Multiply both sides by

x^3

and then divide by

864

. Both

108

and

864

divide by

108

, which is what cancels the fraction down to

\dfrac{1}{8}

Step 4: Take the cube root

x = \sqrt[3]{\dfrac{1}{8}}

x = \dfrac{1}{2} = 0.5

(Reason: Cube rooting undoes the cube. Since

2^3 = 8

, the cube root of

\dfrac{1}{8}

\dfrac{1}{2}

, and a bigger

y

gives a smaller

x

, which is what inverse proportion should do.)

x = 0.5

Verification

Check 1: Put

x = 0.5

back into

y = \dfrac{108}{x^3}

. Cubing

0.5

gives

0.125

, and the

y

that comes out has to be

864

\dfrac{108}{0.5^3} = \dfrac{108}{0.125} = 864

Check 2: Work in multipliers and never find the constant at all. Here

y

is multiplied by

\dfrac{864}{4} = 216

, so

x^3

is divided by

216

and

x

is divided by the cube root of

216

6^3 = 216

and

\dfrac{3}{6} = 0.5

Check 3: For inverse proportion to

x^3

the product

y \times x^3

is the same for every pair, so the given pair and the answer pair must produce the same number.

4 \times 27 = 108

and

864 \times 0.5^3 = 108

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$y = \dfrac{k}{x^3}$ or $hy = \dfrac{1}{x^3}$	M1	do not award for $y = \dfrac{1}{x^3}$ . Constant of proportionality must be a symbol such as $k$ . Condone use of $\propto$ for method marks	✓
$4 = \dfrac{k}{3^3}$ or $k = 4 \times 3^3 \; (= 108)$ or $h \times 4 = \dfrac{1}{3^3}$ or $h = \dfrac{1}{4 \times 3^3} \; \left(= \dfrac{1}{108}\right)$	M1	for substitution of $x$ and $y$ into a correct formula. Condone use of $\propto$ for method marks	✓
eg $(x^3 =) \dfrac{4 \times 3^3}{864} \; \left(= \dfrac{1}{8}\right)$ or $(x^3 =) \dfrac{108}{864} \; \left(= \dfrac{1}{8}\right)$ oe	M1	for method to find $x^3$ . Condone use of $\propto$ for method marks. The printed scheme writes the second form with $108$ in quotation marks, so a candidate's own earlier constant may be used	✓
$0.5$	A1	oe	✓
On the answer line	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 4/4

A curve has equation $y = \text{f}(x)$
The curve has exactly one minimum point.
The coordinates of that minimum point are $(8, -12)$

Write down the coordinates of the minimum point on the curve with equation

(i) $y = \text{f}(x) + 3$ [1 mark]

(ii) $y = \text{f}(2x)$ [1 mark]

(i)(ii)

[Total 2 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

A curve

y = \text{f}(x)

with exactly one minimum point.

That minimum point is

(8, -12)

No formula for

\text{f}(x)

is given, so the answer has to come from the transformation itself.

Find

(i) The minimum point of

y = \text{f}(x) + 3

. (ii) The minimum point of

y = \text{f}(2x)

Plan the Solution

For each new equation, ask whether the change happens after $\text{f}$ has acted, or to the input before it acts.
A change after $\text{f}$ can only move the output, so only the $y$ -coordinate shifts.
A change to the input can only move the $x$ -coordinate, so the height stays the same.
Then apply that movement to the one point the question gives, $(8, -12)$ , because the lowest point of the new curve is the image of the lowest point of the old one.

Worked Solution [2 marks]

Rule - transforming

y = \text{f}(x)

: the curve

y = \text{f}(x) + a

is the curve translated

a

units up, and the curve

y = \text{f}(kx)

is the curve stretched parallel to the

x

-axis with scale factor

\dfrac{1}{k}

Step 1: See where the

+3

acts

y = \text{f}(x) + 3

(Reason: The

3

is added after

\text{f}

has produced its output, so every height on the curve goes up by

3

and no

x

-coordinate changes.)

Step 2: Move the minimum point up

3

y_{\text{new}} = -12 + 3 = -9 \implies (8, -9)

(Reason: The whole curve slides up, so the lowest point of the new curve is the image of the lowest point of the old one. Its

x

-coordinate stays at

8

and only its height moves.)

Step 3: See where the

2

\text{f}(2x)

acts

y = \text{f}(2x)

(Reason: Here the input is doubled before

\text{f}

acts, so the new curve reaches at

x

whatever the old curve reached at

2x

. Nothing is done to the output, so no height changes.)

Step 4: Halve the

x

-coordinate of the minimum

2x = 8 \implies x = \dfrac{8}{2} = 4 \implies (4, -12)

(Reason: The lowest value of

\text{f}

is still

-12

, and it happens when the input is

8

. The input is now

2x

, so that is

x = 4

, with the height unchanged.)

(i)

(8, -9)

(ii)

(4, -12)

Verification

Check 1: Test it on a curve that fits the question.

\text{f}(x) = (x - 8)^2 - 12

has one minimum, at

(8, -12)

, and

\text{f}(x) + 3 = (x - 8)^2 - 9

, which is smallest when

(x - 8)^2 = 0

. Lowest at

x = 8

, height

-9

, giving

(8, -9)

Check 2: The same curve for the second part:

\text{f}(2x) = (2x - 8)^2 - 12

, which is smallest when

2x - 8 = 0

. Lowest at

x = 4

, height

-12

, giving

(4, -12)

Check 3: Ask which coordinate each change is even able to move. Adding

3

happens to the output of

\text{f}

, while doubling happens to its input, so one answer must keep its

x

-coordinate and the other must keep its height. (i) keeps

x = 8

; (ii) keeps

y = -12

. Both answers do.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(i) Write down the coordinates	B1	for $(8, -9)$	✓
(ii) Write down the coordinates	B1	for $(4, -12)$	✓

Full marks: 2/2

The diagram shows a quadrilateral $ABCD$ .
The diagonal $BD$ is drawn.

Find the length of $BC$ .
Give your answer correct to $3$ significant figures. [5 marks]

[Total 5 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

In triangle

ABD

AB = 9.4

cm,

AD = 12.8

cm, and the angle

BAD

between them is

72^\circ

In triangle

BCD

: the angle

BDC

39^\circ

and the angle

BCD

54^\circ

The diagonal

BD

is a side of both triangles, so it is the only link between them

Find

The length of

BC

, correct to

3

significant figures

Plan the Solution

Triangle $ABD$ hands you two sides and the angle BETWEEN them, which is the cosine rule's shape. It gives $BD$ .
$BD$ is the bridge. Once it is known, triangle $BCD$ has two angles and one side, which is the sine rule's shape.
Pair each side with the angle FACING it: $BC$ faces $39^\circ$ and $BD$ faces $54^\circ$ .
Keep the unrounded $BD$ in the calculator and round once, right at the end.

Worked Solution [5 marks]

Rule - Cosine rule:

a^2 = b^2 + c^2 - 2bc\cos A

, where

A

is the angle between the sides

b

and

c

. Sine rule:

\dfrac{a}{\sin A} = \dfrac{b}{\sin B}

, where every side sits over the sine of the angle facing it.

Step 1: put triangle

ABD

into the cosine rule

BD^2 = 9.4^2 + 12.8^2 - 2 \times 9.4 \times 12.8 \times \cos 72^\circ

(Reason: Two sides and the angle between them is exactly what the cosine rule takes. The

72^\circ

sits between

AB

and

AD

, so the side it finds is the third one,

BD

Step 2: work out

BD

9.4^2 + 12.8^2 = 88.36 + 163.84 = 252.2

2 \times 9.4 \times 12.8 \times \cos 72^\circ = 240.64 \times \cos 72^\circ = 74.3618

252.2 - 74.3618 = 177.8382

BD = \sqrt{177.8382} = 13.3356

(Reason: The cosine rule gives

BD^2

, never

BD

, so the square root is a step in its own right and is easy to forget. Writing

BD

down as

13.3

here would drag the final answer with it, so carry the full display value forward.)

Step 3: match each side of triangle

BCD

to the angle facing it

\dfrac{BC}{\sin 39^\circ} = \dfrac{BD}{\sin 54^\circ}

(Reason: The angle at

D

39^\circ

and it looks straight across the triangle at

BC

. The angle at

C

54^\circ

and it looks across at

BD

. Pairing a side with the angle it touches, instead of the one facing it, is the commonest way to lose this question.)

Step 4: rearrange and work out

BC

BC = \dfrac{BD}{\sin 54^\circ} \times \sin 39^\circ

BC = \dfrac{13.3356}{\sin 54^\circ} \times \sin 39^\circ = 10.3735

(Reason: Multiplying both sides of the sine rule by

\sin 39^\circ

leaves

BC

on its own. Type it as one calculation so that nothing in the middle gets rounded.)

Step 5: round to

3

significant figures

BC = 10.3735\ldots

BC = 10.4 \text{ cm}

(Reason: The first three significant figures are

1

0

and

3

. The next digit is

7

, so the

3

rounds up to

4

BC = 10.4

Verification

Check 1: The sine rule says every side divided by the sine of the angle facing it gives the SAME number. Work the two out separately and compare them.

\dfrac{\sin 39^\circ}{10.3735} = 0.06067

and

\dfrac{\sin 54^\circ}{13.3356} = 0.06067

Check 2: Work backwards through a different route. The third angle of triangle

BCD

180^\circ - 39^\circ - 54^\circ = 87^\circ

, and the sine rule then gives

DC = 16.4611

. Putting

BC

and

DC

into the cosine rule must hand

BD^2

straight back.

10.3735^2 + 16.4611^2 - 2 \times 10.3735 \times 16.4611 \times \cos 54^\circ = 177.84

and

13.3356^2 = 177.84

Check 3: A sanity check with no calculator. In any triangle the larger angle faces the longer side, and in triangle

BCD

the angle facing

BC

39^\circ

while the angle facing

BD

54^\circ

. So

BC

has to come out SHORTER than

BD

, and

10.4

is shorter than

13.3

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$(BD^2 =) 9.4^2 + 12.8^2 - 2 \times 9.4 \times 12.8 \times \cos 72 \quad (= 177.8\ldots)$	M1	for applying the cosine rule. Also allow $(BD^2 =) 88.36 + 163.84 - 2 \times 9.4 \times 12.8 \times \cos 72 \quad (= 177.8\ldots)$ oe, or $(BD =) \sqrt{9.4^2 + 12.8^2 - 2 \times 9.4 \times 12.8 \times \cos 72}$ .	✓
$(BD =) 13.3$	A1	allow $13.3$ to $13.342$ , or $\sqrt{177.8\ldots}$ , or $\sqrt{178}$ .	✓
eg $\dfrac{BC}{\sin 39} = \dfrac{BD}{\sin 54}$ or $\dfrac{\sin 39}{BC} = \dfrac{\sin 54}{BD}$ , where $BD$ stands for the candidate's own earlier value	M1ft	for applying the sine rule. Allow use of their $BD$ .	✓
$(BC =) \dfrac{BD}{\sin 54} \times \sin 39$ , where $BD$ stands for the candidate's own earlier value	M1ft	for a method to find $BC$ using the sine rule. Allow use of their $BD$ .	✓
$10.4$	A1	allow $10.3$ to $10.4$ .	✓
A correct answer scores full marks, unless it comes from obviously incorrect working.	Note	The printed scheme carries this in italics beside the final A1, and it awards nothing of its own.	✓

Full marks: 5/5

Continue to questions 21 to 25

The remaining 5 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 4MA1/1H, Thursday 15 May 2025: Worked Solutions, Questions 12 to 20

Worked solutions, questions 12 to 20 of 25

Question 12, Calculator allowed

Question 12 - Exam Solution

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

Question 19, Calculator allowed

Question 19 - Exam Solution

Question 20, Calculator allowed

Question 20 - Exam Solution

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