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Edexcel IGCSE 4MA1/2H, Wednesday 4 June 2025: Worked Solutions, Questions 11 to 19

Sir Faraz Hassan

31 Aug 2026

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

Back to questions 1 to 10

This is part two of three. Questions 1 to 10, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/2H (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 11 to 19 of 26

Yusuf has a jar of glass beads.

$17$ of the beads are green
$28$ of the beads are yellow
the rest of the beads are purple

Yusuf is going to take at random a bead from the jar.

The probability that Yusuf will take a purple bead is $\dfrac{4}{9}$

Work out the number of purple beads that are in the jar. [3 marks]

[Total 3 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

The jar holds

17

green beads,

28

yellow beads, and some purple beads.

One bead is taken at random, and the probability that it is purple is

\dfrac{4}{9}

Find

The number of purple beads in the jar. The total is unknown as well, so expect to find the whole jar on the way to the

4

ninths that are purple.

Plan the Solution

The green and the yellow beads are exactly the beads that are not purple, so add them: that is the only part of the jar the question actually counts.
Purple and not purple are the only two outcomes, so their probabilities add to $1$ . That turns the counted beads into a known fraction of the jar.
Split that fraction into its $5$ equal ninths to find one ninth, then build the whole jar back up and take the purple share.

Worked Solution [3 marks]

Rule - the probability of taking a purple bead is the number of purple beads over the total number of beads, and the probabilities of all the outcomes add to

1

Step 1: count the beads that are not purple

17 + 28 = 45

(Reason: Every bead is green, yellow or purple, so the

17

green and the

28

yellow beads together are all the beads that are not purple.)

Step 2: find the probability of not purple

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

(Reason: Purple and not purple are the only two outcomes, so their probabilities add to

1

. Those

45

beads therefore make up

\dfrac{5}{9}

of the jar.)

Step 3: work out one ninth of the jar

\dfrac{45}{5} = 9

(Reason: Those

45

beads are

5

of the

9

equal ninths, so one ninth is

45

shared into

5

equal groups.)

Step 4: build the whole jar, then take the purple share

9 \times 9 = 81

4 \times 9 = 36

(Reason: The whole jar is

9

ninths, so it holds

81

beads, and the purple beads are the

4

ninths that are left once the counted beads are taken out.)

36

purple beads

Verification

Check 1: Put the answer back in. The jar would then hold

17

green,

28

yellow and

36

purple beads, so work out the probability of purple from those counts.

\dfrac{36}{17 + 28 + 36} = \dfrac{36}{81} = \dfrac{4}{9}

Check 2: Cross-multiply instead of dividing: nine lots of the purple beads should equal four lots of all the beads.

9 \times 36 = 324

and

4 \times 81 = 324

Check 3: A different quantity altogether. The beads that are not purple should outnumber the purple ones in the ratio five to four, because their probabilities are five ninths and four ninths.

\dfrac{45}{36} = \dfrac{5}{4}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{17 + 28}{9 - 4}$ or $\dfrac{45}{5}$ or $1 - \dfrac{4}{9} \left( = \dfrac{5}{9} \right)$ or $(17 + 28 =) \dfrac{5}{9}$ oe or $\dfrac{p}{p + 28 + 17} = \dfrac{4}{9}$ oe or $\dfrac{p}{p + 45} = \dfrac{4}{9}$ or $\dfrac{m - 45}{m} = \dfrac{4}{9}$	M1	Allow $0.55(555\ldots)$ or $55(.555\ldots)\%$ truncated or rounded	✓
$\dfrac{17 + 28}{5} \times 4$ or $\dfrac{45}{5} \times 4$ or $\dfrac{17 + 28}{\text{their } \dfrac{5}{9}} (= 81)$ or $\dfrac{45}{\text{their } \dfrac{5}{9}} (= 81)$ or $(17 + 28) \times \text{their } \dfrac{9}{5} (= 81)$ or $45 \times \text{their } \dfrac{9}{5} (= 81)$ or $\text{their } \dfrac{5}{9} = \dfrac{17 + 28}{n}$ oe or $n = 81$ or $9p = 4(p + 28 + 17)$ or $9p - 4p = 180$ oe or $5p = 180$ oe or $9(m - 45) = 4m$ or $9m - 4m = 405$ oe or $5m = 405$ oe or $m = 81$	M1	for the correct calculation for the total number of beads or for the correct calculation for the number of purple beads or for the correct equation for the total number of beads (removing the denominators) or for the correct equation for the number of purple beads (removing the denominators). A value in quotation marks in the printed scheme is written here as their value, meaning the candidate's own earlier value may be used.	✓
$36$	A1	cao. Working not required, so correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 3/3

Multiply out and simplify $3x(2x + 5)(7x - 4)$ [3 marks]

[Total 3 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

The expression

3x(2x + 5)(7x - 4)

Three factors multiplied together: the single term

3x

, the bracket

(2x + 5)

and the bracket

(7x - 4)

Find

The same expression written without brackets, with every pair of like terms collected.

Plan the Solution

Multiplication can be done in any order, so take the factors two at a time.
Expand $(2x + 5)(7x - 4)$ first: every term in one bracket multiplies every term in the other, which gives four products.
Collect the two $x$ terms, so the bracket becomes a quadratic with three terms.
Multiply that quadratic by $3x$ , one term at a time.
Check the result by expanding in a different order, and by substituting a number.

Worked Solution [3 marks]

Rule - Expanding brackets: multiply every term in one bracket by every term in the other, then collect like terms. When two single terms are multiplied, multiply the numbers and add the indices:

x^a \times x^b = x^{a+b}

Step 1: expand

(2x + 5)(7x - 4)

(2x + 5)(7x - 4) = 2x \times 7x + 2x \times (-4) + 5 \times 7x + 5 \times (-4)

= 14x^2 - 8x + 35x - 20

(Reason: Each term in the first bracket multiplies each term in the second, so there are four products. The signs travel with the terms:

2x \times (-4) = -8x

and

5 \times (-4) = -20

Step 2: collect the like terms inside the bracket

-8x + 35x = 27x

(2x + 5)(7x - 4) = 14x^2 + 27x - 20

(Reason: Only the two

x

terms are alike.

14x^2

is the only squared term and

-20

is the only number, so neither of them changes.)

Step 3: multiply the quadratic by

3x

3x(14x^2 + 27x - 20) = 3x \times 14x^2 + 3x \times 27x - 3x \times 20

= 42x^3 + 81x^2 - 60x

(Reason: Multiply the numbers and add the indices each time:

3 \times 14 = 42

and

x \times x^2 = x^3

, so

3x \times 14x^2 = 42x^3

. Nothing can be collected now, because the three terms carry different powers of

x

42x^3 + 81x^2 - 60x

Verification

Check 1 - expand in a different order: Take the single term into the first bracket instead:

3x(2x + 5) = 6x^2 + 15x

, then multiply that by

(7x - 4)

(6x^2 + 15x)(7x - 4) = 42x^3 - 24x^2 + 105x^2 - 60x

, which collects to

42x^3 + 81x^2 - 60x

Check 2 - substitute a value for $x$ : Put

x = 2

into the original expression. Then

3x = 6

2x + 5 = 9

and

7x - 4 = 10

. The original gives

6 \times 9 \times 10 = 540

, and the answer gives

42 \times 8 + 81 \times 4 - 60 \times 2 = 540

Check 3 - is the answer the right shape? Each factor has degree

1

, so the product has degree

3

. Every term carries a factor of

x

, so there is no number on its own, and the highest coefficient is

3 \times 2 \times 7 = 42

. The answer is a cubic, its highest term is

42x^3

, and it has no constant term.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
One pair of factors expanded: $3x(2x + 5) = 6x^2 + 15x$ or $3x(7x - 4) = 21x^2 - 12x$ or $(2x + 5)(7x - 4) = 14x^2 - 8x + 35x - 20$ $(14x^2 + 27x - 20)$	M1	An expansion with only one error. Do not award this mark for $6x^2 + 15x + 21x^2 - 12x$ or $(6x^2 + 15x)(21x^2 - 12x)$ .	✓
$(6x^2 + 15x)(7x - 4) = 42x^3 - 24x^2 + 105x^2 - 60x$ or $(21x^2 - 12x)(2x + 5) = 42x^3 + 105x^2 - 24x^2 - 60x$ or $3x(14x^2 - 8x + 35x - 20) = 42x^3 - 24x^2 + 105x^2 - 60x$ or $3x(14x^2 + 27x - 20) = 42x^3 + 81x^2 - 60x$	M1	ft dep on M1. Allow one further error.	✓
$42x^3 + 105x^2 - 24x^2 - 60x$ with $3$ of the terms correct	M2	An alternative to the two method marks above, not an addition to them. M2 for $3$ (out of a maximum of $4$ ) of $42x^3 + 105x^2 - 24x^2 - 60x$ . M1 for $2$ correct out of a maximum of $4$ .	✓
$42x^3 + 81x^2 - 60x$	A1	cao (terms may be in any order but must be simplified) dep on M1. ISW correct factorisation, eg $3(14x^3 + 27x^2 - 20x)$ . Do not ISW incorrect simplification, eg $14x^3 + 27x^2 - 20x$ .	✓
Working not required, so a correct answer scores full marks (unless from obvious incorrect working).	Note	Guidance for the whole question. This row awards nothing of its own.	✓

Full marks: 3/3

The frequency table gives information about the times, in minutes, that $60$ visitors took to find their way through a hedge maze.

$\begin{array}{|c|c|}\hline \textbf{Time}\ (t\ \textbf{minutes}) & \textbf{Frequency} \\ \hline 0 < t \leq 10 & 8 \\ \hline 10 < t \leq 20 & 13 \\ \hline 20 < t \leq 30 & 12 \\ \hline 30 < t \leq 40 & 17 \\ \hline 40 < t \leq 50 & 7 \\ \hline 50 < t \leq 60 & 3 \\ \hline \end{array}$

(a) Complete the cumulative frequency table.

$\begin{array}{|c|c|}\hline \textbf{Time}\ (t\ \textbf{minutes}) & \textbf{Cumulative frequency} \\ \hline 0 < t \leq 10 & \phantom{000} \\ \hline 0 < t \leq 20 & \phantom{000} \\ \hline 0 < t \leq 30 & \phantom{000} \\ \hline 0 < t \leq 40 & \phantom{000} \\ \hline 0 < t \leq 50 & \phantom{000} \\ \hline 0 < t \leq 60 & \phantom{000} \\ \hline \end{array}$ [1 mark]

(b) On the grid below, draw a cumulative frequency graph for your table. [2 marks]

(d) Use your graph to find an estimate for the number of these visitors who took more than $45$ minutes to find their way through the maze. [2 marks]

[Total 6 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

The times, in minutes, that

60

visitors took, grouped into six classes each of width

10

minutes.

The frequencies, in order, are

8

13

12

17

7

and

3

Find

(a) the cumulative frequency for each class (b) the cumulative frequency graph (c) an estimate of the median time (d) an estimate of how many visitors took more than

45

minutes

Plan the Solution

Add the frequencies down the table. Each running total counts every visitor who had finished by the END of that class.
Plot each running total against the UPPER end of its class, so the first point sits at $10$ minutes and not in the middle of the class, then join the points.
The median is read at half of $60$ , so go across from $30$ on the cumulative frequency axis and down to the time axis.
For part (d), read UP from $45$ minutes to find how many had already finished, then take that away from $60$ .

Worked Solution [6 marks]

Rule - Cumulative frequency: plot each running total against the UPPER end of its class, and estimate the median by reading across from

\dfrac{n}{2}

on the cumulative frequency axis.

Step 1: add the frequencies down the table

8

8 + 13 = 21

21 + 12 = 33

33 + 17 = 50

50 + 7 = 57

57 + 3 = 60

(Reason: Each cumulative frequency counts every visitor who had finished by the END of that class, so it is the running total of the frequency column. The last entry has to come out as

60

, the number of visitors, and that is the check that no frequency has been dropped.)

Step 2: plot each total at the end of its class

(10,\ 8) \qquad (20,\ 21) \qquad (30,\ 33)

(40,\ 50) \qquad (50,\ 57) \qquad (60,\ 60)

(Reason: Every visitor in the first class had finished by

10

minutes, so that point goes at

10

and not at the middle of the class. Joining the points gives the graph, and it may be started at

(0,\ 0)

because nobody had finished after

0

minutes.)

Step 3: read the median off the graph

\dfrac{60}{2} = 30

20 + \dfrac{30 - 21}{33 - 21} \times 10 = 27.5

(Reason: Half of

60

30

, so go across from

30

on the cumulative frequency axis. The graph passes that height on the segment from

(20,\ 21)

(30,\ 33)

, and the working finds how far along that segment it happens.)

Step 4: read up from 45 minutes, then subtract

50 + \dfrac{45 - 40}{50 - 40} \times (57 - 50) = 53.5

60 - 54 = 6

(Reason: Reading up from

45

minutes to the graph and across to the cumulative frequency axis gives

53.5

. A cumulative frequency counts whole visitors, so about

54

had finished within

45

minutes, and everyone else out of the

60

took longer than that.)

(a)

8

21

33

50

57

60

(b) the six totals plotted at the ends of the classes and joined(c)

27.5

minutes(d)

6

visitors

Verification

Check 1: Add the six frequencies straight up. The last cumulative frequency has to be the number of visitors, because by

60

minutes every visitor has finished.

8 + 13 + 12 + 17 + 7 + 3 = 60

Check 2: Find the median from the other end of its class. The graph reaches

33

30

minutes, which overshoots by

3

visitors out of the

12

in that class, so come back that fraction of the

10

minutes.

30 - \dfrac{33 - 30}{33 - 21} \times 10 = 27.5

Check 3: Count the visitors still going instead of subtracting from

60

. Of the

7

in the class

40 < t \leq 50

the graph gives

54 - 50 = 4

finished by

45

minutes, and all

3

in the last class were still going.

(57 - 54) + 3 = 6

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	for $8$ , $21$ , $33$ , $50$ , $57$ , $60$	✓
(b)	M1ft	for at least $5$ points plotted correctly at the end of the interval, or follow through from an ascending table (follow through from a table with only one arithmetic error that may be continued through the table) for all $6$ points plotted consistently within each interval in the frequency table at the correct height	✓
(b)	A1	for fully correct plotting with points joined; accept a curve or line segments; accept a curve that is not joined at $(0, 0)$	✓
(b)	Note	A histogram or bar chart type graph scores zero marks unless a cumulative frequency diagram is drawn over the histogram or bar chart. Ignore any part of the graph before $(10, 8)$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓
(c)	B1	accept an answer in the range $27$ to $28$ , or follow through an ascending graph	✓
(d)	M1ft	follow through for a line going up from the $x$ -axis at $45$ to the line and across to the $y$ -axis, or for a mark on the line at the correct point, or for a correct reading from the vertical scale, for example $54$ or $53.5$	✓
(d)	A1	accept integer value $5$ or $6$ or $7$ , or follow through from their ascending graph for an integer value	✓
(d)	Note	Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 6/6

The five graphs below are labelled A to E.

(a) State the letter of the graph that could have the equation $y = 5 - x^2$ [1 mark]

(b) State the letter of the graph that could have the equation $y = 2x^3$ [1 mark]

(a)(b)

[Total 2 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

Five sketch graphs on separate axes, labelled A to E. No scales are marked, so only the shape of each curve can be read.

Two equations to match to a sketch:

y = 5 - x^2

and

y = 2x^3

Find

(a) The letter of the graph that could have the equation

y = 5 - x^2

. (b) The letter of the graph that could have the equation

y = 2x^3

Plan the Solution

Read the highest power of $x$ in each equation. That alone decides which family of curve it belongs to.
Read the sign in front of that highest power. It decides which way up a parabola sits, and which way a cubic runs.
Sort the five sketches into families first. Each family appears exactly once, so once the family is named the letter is forced.

Worked Solution [2 marks]

The highest power of

x

fixes the shape. A term in

x^2

gives a parabola, opening upwards when its coefficient is positive and downwards when it is negative. A term in

x^3

gives a cubic, running from bottom left to top right when its coefficient is positive. A term in

x

alone gives a straight line, and

\dfrac{1}{x}

gives a curve in two branches with a break at

x = 0

Step 1: read the shape of

y = 5 - x^2

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

(Reason: The highest power is

2

, so the curve is a parabola. The number in front of

x^2

-1

, which is negative, so the parabola opens downwards: it turns at a maximum, not a minimum.)

Step 2: test three values, then match

5 - (-3)^2 = -4

5 - 0^2 = 5

5 - 3^2 = -4

(Reason: The curve is below the

x

-axis at

x = -3

, above it at

x = 0

and below it again at

x = 3

- up in the middle and down at both ends. Graph B is the only sketch that does that. Graph D is a parabola as well, but it opens upwards, which needs a positive

x^2

term.)

Step 3: read the shape of

y = 2x^3

2 \times (-2)^3 = -16

2 \times 2^3 = 16

(Reason: The highest power is

3

, so the curve is a cubic. The two values either side of the

y

-axis are the same size but opposite in sign, so the curve sits below the

x

-axis on the left and above it on the right. The coefficient

2

is positive, so it runs upwards from left to right.)

Step 4: match the cubic to a sketch

2 \times 0^3 = 0

(Reason: The curve must pass through the origin, which rules out graph A - it has a break at

x = 0

and never meets either axis - and rules out graph E, which is straight and crosses the

y

-axis above the origin. Both parabolas are out too, because a parabola gives the same value at

x

and at

-x

while a cubic changes sign. Graph C is the one left: it climbs from the bottom left, flattens at the origin and rises steeply to the top right.)

(a) Graph B(b) Graph C

Verification

Check 1: Test the symmetry for part (a). Put

x = 1

and

x = -1

into

y = 5 - x^2

5 - 1^2 = 4

and

5 - (-1)^2 = 4

- the same height either side of the

y

-axis. Only B and D have that symmetry, and of those two only B turns at a maximum.

Check 2: Test part (b) the same way. Put

x = 1

and

x = -1

into

y = 2x^3

2 \times 1^3 = 2

and

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

- equal in size but opposite in sign, so the curve crosses from below the

x

-axis to above it as it passes through the origin. Graph C is the only sketch shaped that way.

Check 3: Count the families. The five sketches are five different shapes: a two-branch curve, a downward parabola, a cubic, an upward parabola and a straight line. A downward parabola and a cubic each appear exactly once, so the two answers are forced and no letter is used twice.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	B - cao	✓
(b)	B1	C - cao	✓

Full marks: 2/2

Simplify fully
$\left( \dfrac{2a^{3}}{10a^{7}c^{2}} \right)^{-3}$ [3 marks]

[Total 3 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

The expression

\left( \dfrac{2a^{3}}{10a^{7}c^{2}} \right)^{-3}

One fraction in

a

and

c

, raised to the power

-3

Find

The same expression written as simply as possible. Expect one number multiplied by one power of

a

and one power of

c

Plan the Solution

Tidy the inside of the bracket first: cancel the numbers, then subtract the indices of $a$ with $\dfrac{x^{m}}{x^{n}} = x^{m-n}$ .
Leave the index outside the bracket until last. It is negative, so it turns the fraction over: raising to $-3$ is the same as flipping the fraction and raising to $3$ .
Cube each factor separately. The number in front is a factor too, so it is cubed as well as the letters.

Worked Solution [3 marks]

Rule - Index laws:

\dfrac{x^{m}}{x^{n}} = x^{m-n}

(x^{m})^{n} = x^{mn}

and

x^{-n} = \dfrac{1}{x^{n}}

Step 1: Simplify inside the bracket

\dfrac{2a^{3}}{10a^{7}c^{2}} = \dfrac{2}{10} \times \dfrac{a^{3}}{a^{7}} \times \dfrac{1}{c^{2}}

\dfrac{2}{10} = \dfrac{1}{5}

\dfrac{a^{3}}{a^{7}} = a^{3-7} = a^{-4} = \dfrac{1}{a^{4}}

\dfrac{2a^{3}}{10a^{7}c^{2}} = \dfrac{1}{5a^{4}c^{2}}

(Reason: Dividing powers of the same letter subtracts the indices, and

a^{-4}

is another way of writing

\dfrac{1}{a^{4}}

, so both letters end up in the denominator.)

Step 2: Turn the fraction over to deal with the negative index

\left( \dfrac{1}{5a^{4}c^{2}} \right)^{-3} = \left( 5a^{4}c^{2} \right)^{3}

(Reason: A negative index means the reciprocal:

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

, so a fraction raised to

-3

is that fraction turned upside down and raised to

3

. Flipping

\dfrac{1}{5a^{4}c^{2}}

leaves

5a^{4}c^{2}

Step 3: Cube every factor inside the bracket

\left( 5a^{4}c^{2} \right)^{3} = 5^{3} \times a^{4 \times 3} \times c^{2 \times 3}

5^{3} \times a^{4 \times 3} \times c^{2 \times 3} = 125a^{12}c^{6}

(Reason: A power outside a bracket multiplies every index inside it, so

a^{4}

becomes

a^{12}

and

c^{2}

becomes

c^{6}

. The number in front is a factor of the bracket like the letters are, so it is raised to the power too.)

125a^{12}c^{6}

Verification

Check 1: Put

a = 2

and

c = 3

into the original expression, and into the answer, and compare the two values. Inside the bracket

\dfrac{2 \times 8}{10 \times 128 \times 9} = \dfrac{16}{11\,520} = \dfrac{1}{720}

, so the original is

720^{3} = 373\,248\,000

, and the answer gives

125 \times 4096 \times 729 = 373\,248\,000

Check 2: Work backwards. Take the cube root of the answer and check it is the reciprocal of the bracket that Step 1 produced.

\sqrt[3]{125a^{12}c^{6}} = 5a^{4}c^{2}

, and

\dfrac{1}{5a^{4}c^{2}}

is exactly the simplified bracket, so the negative index has been undone correctly.

Check 3: Take a different order of working: turn the fraction over first, cube the top and the bottom without simplifying anything, and only then cancel.

\dfrac{(10a^{7}c^{2})^{3}}{(2a^{3})^{3}} = \dfrac{1000a^{21}c^{6}}{8a^{9}} = 125a^{12}c^{6}

, since

\dfrac{1000}{8} = 125

and

a^{21-9} = a^{12}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Fully simplified: $125a^{12}c^{6}$	B3	for $125a^{12}c^{6}$ oe, eg $\dfrac{125a^{12}}{c^{-6}}$ or $\dfrac{125c^{6}}{a^{-12}}$ or $\dfrac{125}{a^{-12}c^{-6}}$ or $\dfrac{a^{12}}{125^{-1}c^{-6}}$ or $\dfrac{1}{125^{-1}a^{-12}c^{-6}}$ or $\dfrac{a^{12}c^{6}}{125^{-1}}$	✓
Two correct terms	B2	for $2$ correct terms (Allow eg $125a^{12}$ or $125c^{6}$ or $ka^{12}c^{6}$ as long as not added to any other terms)	✓
One correct term	B1	for one correct term (allow eg $a^{12}$ or $a^{-12}c^{6}$ or $\dfrac{1}{125^{-1}}$ or $\dfrac{1000a^{21}c^{6}}{8a^{9}}$ as long as not added to any other terms)	✓

Full marks: 3/3

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

[Total 4 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

The formula

c = \dfrac{t^2 + 3}{7 - 8t^2}

The letter

t

appears in the numerator and in the denominator, and only ever as

t^2

Find

A formula for

t

in terms of

c

Plan the Solution

Multiply both sides by $7 - 8t^2$ so that no $t$ is left underneath a fraction bar.
Gather every term containing $t^2$ on one side and everything else on the other.
Take $t^2$ out as a common factor, then divide by the bracket that is left.
Square root last, and keep both signs.

Worked Solution [4 marks]

Rule - Changing the subject: clear the fraction, collect every term containing the new subject on one side, factorise it out, then undo what is left around it. Here the new subject appears as

t^2

, so the last step is a square root and both signs are kept.

Step 1: multiply both sides by the denominator

c(7 - 8t^2) = t^2 + 3

7c - 8ct^2 = t^2 + 3

(Reason: (Reason: while

t

sits underneath a fraction bar nothing can be collected, so the fraction is cleared first. Multiplying out the bracket turns the left-hand side into the two terms

7c

and

-8ct^2

.))

Step 2: collect the

t^2

terms on one side

7c - 3 = t^2 + 8ct^2

(Reason: (Reason: add

8ct^2

to both sides and take the

3

across, so that the two

t^2

terms end up together on the right and the terms with no

t

in them end up on the left.))

Step 3: factorise

t^2

out

7c - 3 = t^2(1 + 8c)

(Reason: (Reason:

t^2

is a common factor of

t^2

and

8ct^2

, and taking it out is what turns two

t^2

terms into one.))

Step 4: divide, then square root

t^2 = \dfrac{7c - 3}{1 + 8c}

t = \pm\sqrt{\dfrac{7c - 3}{1 + 8c}}

(Reason: (Reason: dividing by the bracket leaves

t^2

on its own. Squaring hides a sign, so undoing it gives two values: both

t

and

-t

lead back to the same

c

.))

t = \pm\sqrt{\dfrac{7c - 3}{1 + 8c}}

Verification

Check 1 - put $t = 1$ through both formulas: The formula on the paper gives

\dfrac{1 + 3}{7 - 8} = \dfrac{4}{-1} = -4

, so

c = -4

. Feeding

c = -4

into the rearranged formula gives

\dfrac{7 \times (-4) - 3}{1 + 8 \times (-4)} = \dfrac{-31}{-31} = 1

, and the square root of that is

1

again.

Check 2 - work the other way, from $c$ back to $t$ : Put

c = 1

into the rearranged formula:

\dfrac{7 - 3}{1 + 8} = \dfrac{4}{9}

, so

t = \pm\dfrac{2}{3}

. Putting

t^2 = \dfrac{4}{9}

back into the formula on the paper, and multiplying the top and the bottom by

9

, gives

\dfrac{4 + 27}{63 - 32} = \dfrac{31}{31} = 1

, the value of

c

we started with.

Check 3 - why both signs are kept: The formula uses

t

only as

t^2

, so

t

and

-t

have to give the same

c

t = -1

gives

\dfrac{1 + 3}{7 - 8} = -4

, the same value as

t = 1

, so an answer carrying one sign only would lose half of it.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$7c - 8ct^2 = t^2 + 3$ oe	M1	For multiplying both sides by the denominator and expanding the brackets.	✓
$7c - 3 = t^2 + 8ct^2$ oe or $-8ct^2 - t^2 = 3 - 7c$ oe	M1	Follow through, dependent on 2 terms in $t^2$ and 2 other terms. For collecting the $t^2$ terms on one side and the other terms on the other side.	✓
$7c - 3 = t^2(1 + 8c)$ oe or $t^2(-8c - 1) = 3 - 7c$ oe	M1	Follow through, dependent on the previous M1. For factorising for $t^2$ .	✓
$t = (\pm)\sqrt{\dfrac{7c - 3}{1 + 8c}}$	A1	Or equivalent, for example $t = (\pm)\sqrt{\dfrac{3 - 7c}{-8c - 1}}$ or $t = (\pm)\left(\dfrac{7c - 3}{1 + 8c}\right)^{\dfrac{1}{2}}$ or $t = (\pm)\left(\dfrac{7c - 3}{1 + 8c}\right)^{0.5}$ .	✓
NB	Note	To award the A1 we must see $t = (\pm)\sqrt{\dfrac{7c - 3}{1 + 8c}}$ in the working if $(\pm)\sqrt{\dfrac{7c - 3}{1 + 8c}}$ alone is given as the answer.	✓
Working	Note	Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 4/4

$y = 4x^3 + 5x^2 + 2x$

(a) Find $\dfrac{dy}{dx}$ [2 marks]

(b) Work out the coordinates of the turning points on the curve with equation $y = 4x^3 + 5x^2 + 2x$
You must show clear algebraic working. [4 marks]

(a)(b)

[Total 6 marks]

Show solution & mark schemeHide solution & mark scheme

Question 17 - Exam Solution

Understanding the Question

Given

y = 4x^3 + 5x^2 + 2x

A cubic, so the derivative is a quadratic and the curve can have up to two turning points.

Find

(a)

\dfrac{dy}{dx}

(b) the coordinates of the turning points, so an

x

and a

y

for each one

Plan the Solution

Differentiate term by term: multiply each term by its power, then take one off the power.
A turning point is where the curve is momentarily flat, so set $\dfrac{dy}{dx} = 0$ and solve.
Every coefficient of the derivative is even, so take the factor $2$ out before factorising the quadratic.
Put each root back into the original cubic to get its $y$ value. The derivative gives gradients, not heights.

Worked Solution [6 marks]

Rule - Turning points: differentiate, set

\dfrac{dy}{dx} = 0

, solve for

x

, then substitute each

x

back into the original equation to get

y

Step 1: Differentiate term by term

y = 4x^3 + 5x^2 + 2x

\dfrac{dy}{dx} = 3 \times 4x^{2} + 2 \times 5x + 2

\dfrac{dy}{dx} = 12x^2 + 10x + 2

(Reason: The power rule: multiply by the power, then take one off it. So

4x^3

gives

12x^2

5x^2

gives

10x

, and

2x

gives

2

. That is part (a) answered.)

Step 2: Set the gradient equal to zero

12x^2 + 10x + 2 = 0

2(6x^2 + 5x + 1) = 0

6x^2 + 5x + 1 = 0

(Reason: At a turning point the tangent is horizontal, so the gradient is

0

. Every coefficient is even, so dividing through by

2

leaves smaller numbers to factorise.)

Step 3: Factorise and solve for

x

6x^2 + 5x + 1 = (3x + 1)(2x + 1)

(3x + 1)(2x + 1) = 0

3x + 1 = 0 \implies x = -\dfrac{1}{3}

2x + 1 = 0 \implies x = -\dfrac{1}{2}

(Reason: Split the

6

3 \times 2

and the

1

1 \times 1

; the cross terms

3x + 2x

rebuild the middle term

5x

. A product is zero only when one of its brackets is zero.)

Step 4: Substitute

x = -\dfrac{1}{2}

into the original equation

y = 4\left(-\dfrac{1}{2}\right)^{3} + 5\left(-\dfrac{1}{2}\right)^{2} + 2\left(-\dfrac{1}{2}\right)

y = -\dfrac{1}{2} + \dfrac{5}{4} - 1 = -\dfrac{1}{4}

(Reason: The height comes from the original cubic, never from the derivative. Cubing a negative keeps it negative, while squaring it makes it positive, so the first term is

-\dfrac{1}{2}

and the second is

+\dfrac{5}{4}

Step 5: Substitute

x = -\dfrac{1}{3}

into the original equation

y = 4\left(-\dfrac{1}{3}\right)^{3} + 5\left(-\dfrac{1}{3}\right)^{2} + 2\left(-\dfrac{1}{3}\right)

y = -\dfrac{4}{27} + \dfrac{5}{9} - \dfrac{2}{3} = -\dfrac{7}{27}

(Reason: The same substitution again. Writing all three terms over a denominator of

27

keeps the fraction arithmetic exact.)

(a)

\dfrac{dy}{dx} = 12x^2 + 10x + 2

(b)

\left(-\dfrac{1}{2},\ -\dfrac{1}{4}\right)

and

\left(-\dfrac{1}{3},\ -\dfrac{7}{27}\right)

Verification

Check 1: Expand the factorisation back out.

(3x + 1)(2x + 1)

should rebuild

6x^2 + 5x + 1

, and doubling every term should rebuild part (a).

2(6x^2 + 5x + 1) = 12x^2 + 10x + 2

Check 2: Put each root back into the derivative. The gradient must come out as exactly

0

at a turning point.

12\left(-\dfrac{1}{2}\right)^{2} + 10\left(-\dfrac{1}{2}\right) + 2 = 0

and

12\left(-\dfrac{1}{3}\right)^{2} + 10\left(-\dfrac{1}{3}\right) + 2 = 0

Check 3: The two roots of

6x^2 + 5x + 1 = 0

must add to

-\dfrac{5}{6}

and multiply to

\dfrac{1}{6}

-\dfrac{1}{2} - \dfrac{1}{3} = -\dfrac{5}{6}

and

\left(-\dfrac{1}{2}\right)\left(-\dfrac{1}{3}\right) = \dfrac{1}{6}

Check 4: Redo the second turning point in decimals on the calculator.

x \approx -0.3333

gives a height just below the first turning point, which is what a local minimum sitting to the right of a local maximum should do.

-\dfrac{7}{27} \approx -0.2593

, lower than

-\dfrac{1}{4} = -0.25

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Two of $12x^2 + 10x + 2$	M1	for differentiating 2 or 3 terms correctly	✓
(a) $12x^2 + 10x + 2$	A1	for all 3 terms correct	✓
(b) $(3x + 1)(4x + 2)\ (= 0)$ or $(6x + 2)(2x + 1)(= 0)$ or $2(3x + 1)(2x + 1)\ (= 0)$ or $(3x + 1)(2x + 1)(= 0)$ or $\dfrac{-10 \pm \sqrt{10^2 - 4 \times 12 \times 2}}{2 \times 12}$ or $\dfrac{-5 \pm \sqrt{5^2 - 4 \times 6 \times 1}}{2 \times 6}$ or $12\left[\left(x + \dfrac{10}{24}\right)^{2} - \left(\dfrac{10}{24}\right)^{2}\right] + 2\ (= 0)$ oe or $6\left[\left(x + \dfrac{5}{12}\right)^{2} - \left(\dfrac{5}{12}\right)^{2}\right] + 1\ (= 0)$ oe	M1	ft dep on M1 for a correct method to solve their 3 term quadratic equation (with at least 2 correct coefficients) using any correct method (if factorising, allow brackets which expanded give 2 out of 3 terms correct) (if using formula allow one sign error and some simplification - allow as far as $\dfrac{-10 \pm \sqrt{100 - 96}}{24}$ ) Derivative must be a 3 term quadratic for this M mark NB Can be implied by answers of $(x =) -\dfrac{1}{2}$ and $(x =) -\dfrac{1}{3}$	✓
(b) $-\dfrac{1}{2}$ , $-\dfrac{1}{3}$	A1	oe dep on previous M1. Allow $-0.33(333)$ or $-0.\dot{3}$ for correct $x$ values	✓
(b) $(y =) 4 \times \left(-\dfrac{1}{2}\right)^{3} + 5 \times \left(-\dfrac{1}{2}\right)^{2} + 2\left(-\dfrac{1}{2}\right)\ \left(= -\dfrac{1}{4}\right)$ or $(y =) 4 \times \left(-\dfrac{1}{3}\right)^{3} + 5 \times \left(-\dfrac{1}{3}\right)^{2} + 2\left(-\dfrac{1}{3}\right)\ \left(= -\dfrac{7}{27}\right)$	M1	ft dep on previous M1 for substituting at least one $x$ value into $y$ NB Can be implied by one correct value of $y$	✓
(b) Working required. $\left(-\dfrac{1}{2},\ -\dfrac{1}{4}\right)$ , $\left(-\dfrac{1}{3},\ -\dfrac{7}{27}\right)$	A1	oe dep on M1 for correct coordinates $(-0.5,\ -0.25)$ , $(-0.33,\ -0.25(9 \ldots))$	✓

Full marks: 6/6

Show, using algebra, that $0.\dot{7}0\dot{2} = \dfrac{26}{37}$ [2 marks]

[Total 2 marks]

Show solution & mark schemeHide solution & mark scheme

Question 18 - Exam Solution

Understanding the Question

Given

The recurring decimal

0.\dot{7}0\dot{2}

. The dots sit on the

7

and on the

2

, so the block

702

repeats for ever:

0.702702702\ldots

The fraction it is claimed to equal:

\dfrac{26}{37}

Find

Show that

0.\dot{7}0\dot{2}

is exactly

\dfrac{26}{37}

. The working is the answer here, so there is nothing to write on an answer line, and the working must be algebra - a solution with no algebra is capped at 1 mark.

Plan the Solution

Give the recurring decimal a letter, so that there is something to do algebra with.
Count the digits in the repeating block: $702$ is $3$ digits long, so the multiplier is $10^{3} = 1000$ .
Write the two lines under each other and subtract. The endless tails are identical, so they cancel and leave a whole number.
Divide to reach a fraction, then cancel that fraction to its lowest terms.

Worked Solution [2 marks]

Rule - Recurring decimal to fraction: call the decimal

x

. If the repeating block is

n

digits long, multiply by

10^{n}

and subtract

x

. The identical tails cancel, and

(10^{n} - 1)x

is left equal to a whole number.

Step 1: give the decimal a letter

x = 0.\dot{7}0\dot{2} = 0.702702702\ldots

(Reason: Algebra needs something to hold on to, so the decimal is called

x

. The dots say the block

702

repeats without end, so the digits after the point never stop and never change pattern.)

Step 2: multiply by

1000

1000x = 702.\dot{7}0\dot{2} = 702.702702\ldots

(Reason: The repeating block is

3

digits long, so multiplying by

10^{3} = 1000

slides the point past exactly one whole block. The tail left after the point is then the same tail as before, digit for digit.)

Step 3: subtract the two lines

1000x - x = 702.702702\ldots - 0.702702\ldots

999x = 702

(Reason: Both lines carry exactly the same endless tail, so subtracting wipes it out and leaves a whole number. Choosing the multiplier to match the block length is what makes that happen, and it is what earns the first mark.)

Step 4: divide, then cancel

x = \dfrac{702}{999}

702 = 27 \times 26

999 = 27 \times 37

\dfrac{702}{999} = \dfrac{26}{37}

(Reason: Divide both sides by

999

. The highest common factor of

702

and

999

27

, so dividing the top and the bottom by

27

puts the fraction in its lowest terms. Cancelling by anything smaller leaves the fraction unfinished, and the second mark is for the completion.)

0.\dot{7}0\dot{2} = \dfrac{26}{37}

as required

Verification

Check 1: Two fractions are equal exactly when their cross products match. Multiply

26

999

, then multiply

37

702

26 \times 999 = 25\,974

and

37 \times 702 = 25\,974

Check 2: Start again with a different pair of lines,

10000x

and

10x

. Their tails match too, so subtracting gives

9990x = 7020

\dfrac{7020}{9990} = \dfrac{26}{37}

- the same fraction, from different multipliers

Check 3: Go the other way and divide

26

37

by hand. The remainders come out

1

10

26

- and

26

is where the division started. Once a remainder repeats the digits must repeat too, so the quotient is

0.702702702\ldots

, which is

0.\dot{7}0\dot{2}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $(1000x =) 702.702\ldots$ and $(x =) 0.702\ldots$	M1	for $2$ recurring decimals that when subtracted give a whole number or terminating decimal with intention to subtract (ie give $702$ or $7020$ or $70200$ etc) eg $(1000x =) 702.702\ldots$ and $(x =) 0.702\ldots$ or $(10000x =) 7027.02\ldots$ and $(10x =) 7.02\ldots$ or $(100000x =) 70270.20\ldots$ and $(100x =) 70.20\ldots$ with intention to subtract $x$ is not required to award this mark (if recurring dots not shown in both numbers then showing at least one of the numbers to at least $6$ sf) NB Accept bar notation for dot notation to indicate recurring decimals	✓
eg $1000x - x = 702.702\ldots - 0.702\ldots = 702$ $(999x = 702)$ and $\dfrac{702}{999} = \dfrac{26}{37}$ or $10000x - 10x = 7027.02\ldots - 7.02\ldots = 7020$ $(9990x = 7020)$ and $\dfrac{7020}{9990} = \dfrac{26}{37}$ or $100000x - 100x = 70270.20\ldots - 70.20\ldots = 70200$ $(99900x = 70200)$ and $\dfrac{70200}{99900} = \dfrac{26}{37}$ Answer column: shown	A1	for completion to $\dfrac{26}{37}$ dep on M1 and must use algebra for this final mark to be awarded No algebra used gets a maximum of 1 mark	✓
Working required	Note	The fraction is printed in the question, so a solution that only restates it earns nothing: the working is what is being marked.	✓

Full marks: 2/2

Solve the inequality $4x^2 + 4x - 15 < 0$
You must show clear algebraic working. [3 marks]

[Total 3 marks]

Show solution & mark schemeHide solution & mark scheme

Question 19 - Exam Solution

Understanding the Question

Given

The quadratic inequality

4x^2 + 4x - 15 < 0

The coefficient of

x^2

4

, which is positive, so

y = 4x^2 + 4x - 15

is a parabola opening upwards: it dips below the axis once and comes back up.

The sign is strict.

< 0

means negative, so a value of

x

that makes the expression exactly

0

is not part of the solution.

Find

Every value of

x

for which

4x^2 + 4x - 15

is negative, written as one inequality. Clear algebraic working is asked for, so the two critical values must be reached by factorising, by the formula or by completing the square. Both answer marks depend on that method mark, so critical values written down with nothing behind them score nothing at all.

Plan the Solution

Solve the equation $4x^2 + 4x - 15 = 0$ first. Its two solutions are the critical values, where the curve crosses the $x$ -axis and where the expression can change sign.
Factorise by splitting the middle term: look for two numbers with product $4 \times (-15) = -60$ and sum $4$ .
Decide which side of the critical values the curve sits below the axis. With a positive $x^2$ coefficient that is the region between them, and one test value confirms it.
Write the answer as a single double inequality, with strict signs at both ends.

Worked Solution [3 marks]

Rule - Quadratic inequality: solve the equation to find the critical values, then choose the region. For

ax^2 + bx + c < 0

with

a > 0

the expression is negative between the two roots and positive outside them.

Step 1: solve the equation

4x^2 + 4x - 15 = 0

4x^2 + 4x - 15 = 0

4 \times (-15) = -60

10 \times (-6) = -60

10 + (-6) = 4

4x^2 + 10x - 6x - 15 = 0

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

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

(Reason: The expression can only change sign where it is zero, so the equation comes first. To factorise it, look for two numbers whose product is

4 \times (-15) = -60

and whose sum is

4

: they are

10

and

-6

. Splitting the middle term with them leaves the common bracket

(2x + 5)

to take out. This is the method mark, and the quadratic formula or completing the square would earn it just as well.)

Step 2: read the critical values off the brackets

2x + 5 = 0 \implies x = -\dfrac{5}{2} = -2.5

2x - 3 = 0 \implies x = \dfrac{3}{2} = 1.5

(Reason: A product is zero only when one of its factors is zero, so each bracket gives one critical value. These are the two places where the curve

y = 4x^2 + 4x - 15

meets the

x

-axis, and they are what the first answer mark is for.)

Step 3: decide which side of the critical values is negative

4 \times 0^2 + 4 \times 0 - 15 = -15

4 \times 2^2 + 4 \times 2 - 15 = 9

4 \times (-3)^2 + 4 \times (-3) - 15 = 9

(Reason: The critical values

-2.5

and

1.5

cut the number line into three pieces, and the expression keeps one sign right across each piece. Testing

x = 0

from between them gives

-15

, which is negative. Testing

x = 2

and

x = -3

from outside gives

9

both times, which is positive. That is exactly the shape a positive

x^2

coefficient forces: below the axis between the roots, above it either side.)

Step 4: write the solution as one inequality

-2.5 < x < 1.5

(Reason: The negative piece is the one between the critical values, so

x

has to be greater than

-2.5

and less than

1.5

at the same time, and those two conditions are written as one double inequality. Both signs stay strict, because at

-2.5

and at

1.5

the expression is

0

, and the question asks for less than

0

. Turning the answer inside out, as

x < -2.5

x > 1.5

, would describe where the expression is positive instead.)

-2.5 < x < 1.5

(equivalently

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

)

Verification

Check 1: Take two values from inside the interval,

x = 1

and

x = -2

, and work the expression out for each.

4 + 4 - 15 = -7

and

16 - 8 - 15 = -7

, both negative, so both values belong

Check 2: Take one value from beyond each end,

x = 2

and

x = -3

, so that both sides outside the interval are tested.

16 + 8 - 15 = 9

and

36 - 12 - 15 = 9

, both positive, so neither value belongs

Check 3: Multiply the brackets out again and compare the result with the expression the question prints.

(2x + 5)(2x - 3) = 4x^2 - 6x + 10x - 15 = 4x^2 + 4x - 15

Check 4: Reach the two critical values a completely different way, with the quadratic formula, and see whether they come back the same.

\dfrac{-4 \pm \sqrt{16 + 240}}{8} = \dfrac{-4 \pm 16}{8}

, giving

1.5

and

-2.5

again

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$(2x - 3)(2x + 5)$ or $\dfrac{-4 \pm \sqrt{4^2 - 4 \times 4 \times -15}}{2 \times 4}$ or $4\left[\left(x + \dfrac{1}{2}\right)^2 - \left(\dfrac{1}{2}\right)^2\right] - 15 \, (= 0)$ or $4\left(x + \dfrac{4}{2 \times 4}\right)^2 - \dfrac{4^2}{4 \times 4} + -15 \, (= 0)$	M1	for a correct method to solve the quadratic equation $4x^2 + 4x - 15 = 0$ Allow $(4x - 6)(x + 2.5)$ or $(4x + 10)(x - 1.5)$ or $(4x - 6)(4x + 10)$ leading to $(x - 1.5)(x + 2.5)$ or $(4x - 6)(4x + 10)$ leading to correct values of $x$ Do not allow $(x - 1.5)(x + 2.5)$ without previous working (If using formula allow some simplification - allow as far as $\dfrac{-4 \pm \sqrt{16 + 240}}{8}$ )	✓
$1.5, -2.5$ oe	A1	oe dep on M1	✓
Answer column: $-2.5 < x < 1.5$	A1	oe dep on M1 Allow $x > -2.5$ (and) $x < 1.5$ oe Allow any variable as long as used all the way through	✓
Where the three marks fall	Note	The two answer marks sit in separate rows and both depend on the method mark, so a solution that states the critical values with no method behind them - the $(x - 1.5)(x + 2.5)$ case the first row rules out - scores nothing rather than one. The second answer mark is for the region and not for the numbers, so a candidate who reaches $1.5$ and $-2.5$ and then writes $x < -2.5$ or $x > 1.5$ keeps the first answer mark and loses the second: that is the region where the expression is positive.	✓

Full marks: 3/3

Continue to questions 20 to 26

The remaining 7 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 4MA1/2H, Wednesday 4 June 2025: Worked Solutions, Questions 11 to 19

Worked solutions, questions 11 to 19 of 26

Question 11, Calculator allowed

Question 11 - Exam Solution

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

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