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Edexcel IGCSE 4MA1/1HR, Thursday 15 May 2025: Worked Solutions, Questions 12 to 19

Sir Faraz Hassan

31 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)4MA1/1HR - 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 4MA1/1HR (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 26 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 12 to 19 of 26

The table gives information about the ages of the $80$ people who visited an art gallery one Saturday.
$\begin{array}{|c|c|}\hline \textbf{Age } (n \textbf{ years}) & \textbf{Frequency} \\ \hline 10 < n \leq 20 & 12 \\ \hline 20 < n \leq 30 & 15 \\ \hline 30 < n \leq 40 & 20 \\ \hline 40 < n \leq 50 & 18 \\ \hline 50 < n \leq 60 & 9 \\ \hline 60 < n \leq 70 & 6 \\ \hline \end{array}$
(a) Complete the cumulative frequency table.
$\begin{array}{|c|c|}\hline \textbf{Age } (n \textbf{ years}) & \textbf{Cumulative frequency} \\ \hline 10 < n \leq 20 & \phantom{000} \\ \hline 10 < n \leq 30 & \phantom{000} \\ \hline 10 < n \leq 40 & \phantom{000} \\ \hline 10 < n \leq 50 & \phantom{000} \\ \hline 10 < n \leq 60 & \phantom{000} \\ \hline 10 < n \leq 70 & \phantom{000} \\ \hline \end{array}$
[1 mark]
(b) On the grid below, draw a cumulative frequency graph for your completed table. [2 marks]

(c) Use your graph to work out an estimate for the percentage of these $80$ people who are more than $46$ years of age.
Give your answer correct to the nearest whole number. [3 marks]

[Total 6 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

A grouped frequency table for the ages of the

80

visitors, in six classes of width

10

years running from

10

70

The frequencies are

12

15

20

18

9

and

6

, which add to

80

Find

(a) the six cumulative frequencies. (b) the cumulative frequency graph, drawn on the grid. (c) the percentage of the

80

visitors who are more than

46

, to the nearest whole number.

Plan the Solution

Add the frequencies downwards. Each cumulative frequency is a running total, so it counts everyone up to the TOP of that class.
Plot every cumulative frequency against the UPPER end of its class, and start at $(10,\ 0)$ because nobody is aged $10$ or under.
Draw a line up from $46$ to the graph and across to the cumulative frequency axis. That reading estimates how many people are $46$ or under.
Take the reading away from $80$ to count the people OVER $46$ , then write that count as a percentage of $80$ .

Worked Solution [6 marks]

Rule - Cumulative frequency: each entry is the running total of the frequencies so far, and every point is plotted at the UPPER end of its class. Reading up from a value on the horizontal axis estimates how many are at or below it.

Step 1: add the frequencies downwards

12

12 + 15 = 27

27 + 20 = 47

47 + 18 = 65

65 + 9 = 74

74 + 6 = 80

(Reason: Each row of the cumulative frequency table counts everyone up to the top of that class, so every row is the row above it plus the next frequency. The last row must come out as

80

, because by the top class every visitor has been counted.)

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

(10,\ 0)

(20,\ 12)

(30,\ 27)

(40,\ 47)

(50,\ 65)

(60,\ 74)

(70,\ 80)

(Reason: A cumulative frequency is only complete at the TOP of its class, so the

12

is plotted at

20

and not at the midpoint

15

. The graph starts at

(10,\ 0)

because nobody is aged

10

or under. Join the points with a smooth curve or with straight line segments.)

Step 3: read the graph at

46

47 + \dfrac{6}{10} \times (65 - 47) = 57.8

57.8 \approx 58

(Reason: The value

46

lies between

40

and

50

, where the graph climbs from

47

65

. Six tenths of the way across gives

57.8

, so the graph is read as about

58

people aged

46

or under. The mark scheme accepts any reading from

57

59

Step 4: count the people over

46

80 - 58 = 22

(Reason: Every visitor is either

46

or under, or over

46

, so taking the reading away from the total

80

leaves the number who are over

46

Step 5: write that count as a percentage of

80

\dfrac{22}{80} \times 100 = 27.5

27.5 \approx 28

(Reason: A percentage is the part over the whole, multiplied by

100

. Here the part is the

22

people over

46

and the whole is

80

. The question asks for the nearest whole number, so

27.5

is written as

28

(a)

12

27

47

65

74

80

(b) the seven points joined in order, rising from

(10,\ 0)

(70,\ 80)

(c)

28\%

Verification

Check 1: Add the six frequencies straight down the table. The last cumulative frequency has to equal that total, because by the top class everybody has been counted.

12 + 15 + 20 + 18 + 9 + 6 = 80

, and the last entry of the completed table is

80

Check 2: Work out the percentage aged

46

or under instead. Everybody sits on one side or the other, so the two percentages must add to

100

\dfrac{58}{80} \times 100 = 72.5

and

27.5 + 72.5 = 100

Check 3: Sense-check the reading against the table. Because

46

sits inside the class

40 < n \leq 50

, the reading must lie between the cumulative frequencies at the two ends of that class.

47 < 57.8 < 65

. Also

22

out of

80

is a little over a quarter, and

28\%

is a little over

25\%

Check 4: Test whether rounding the reading up to

58

changed the answer, by redoing the percentage from the unrounded

57.8

\dfrac{80 - 57.8}{80} \times 100 = 27.75

, which is still

28

to the nearest whole number.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$12$ , $27$ , $47$ , $65$ , $74$ , $80$	✓
(b)	M1	ft from table for at least $5$ points plotted correctly at end of interval or ft from sensible table (ft from a table with only one arithmetic error that may be continued through table) for all $6$ points plotted consistently within each interval in the freq table at the correct height	✓
(b)	A1	correct cf graph. Accept curve or line segments. Accept curve that is not joined at $(10,\ 0)$	✓
(b)	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓
(c)	M1ft	a line up from $46$ to their graph and a line across to the vertical axis or a mark on the curve at the correct point and a mark on the vertical axis at the correct point or a reading of $57$ to $59$ from their cf graph or a value of $21$ to $23$ or a correct value for their graph. Must be ascending (could be a line of best fit).	✓
(c)	M1ft	method to find the fraction or percentage of people aged over or under $46$ , ft from their graph, or a value in the range $0.26$ to $0.29$ or $0.71$ to $0.74$ or $71\%$ to $74\%$ . For example, over $46$ : $\dfrac{80 - 58}{80} = 0.275$ , or under $46$ : $\dfrac{58}{80} = 0.725$ . The printed scheme sets the $58$ in quotation marks, so the candidate's own reading may be used in place of it.	✓
(c)	A1ft	$28$ . Accept $26$ to $29$ , ft their cf graph.	✓
(c)	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 6/6

Here are the numbers of bicycles hired from a seaside kiosk on each of $11$ days in July.

$9 \qquad 10 \qquad 12 \qquad 15 \qquad 17 \qquad 18 \qquad 19 \qquad 19 \qquad 20 \qquad 25 \qquad 26$

Work out the interquartile range of the numbers of bicycles hired. [2 marks]

[Total 2 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

The numbers of bicycles hired on each of

11

days, listed as the kiosk recorded them.

The eleven values are

9

10

12

15

17

18

19

19

20

25

and

26

, already written from smallest to largest.

Find

The interquartile range of the

11

values.

Plan the Solution

Check the list is in order and count how many values there are. A quartile is a position in an ordered list, so the order and the count decide everything that follows.
Work out the position of the lower quartile and the position of the upper quartile. With $11$ values both positions land on a whole number, so no in-between value is needed.
Count along the ordered list to read the value sitting at each of those two positions.
Take the lower quartile away from the upper quartile.

Worked Solution [2 marks]

Rule - Interquartile range: for

n

values written in order, the lower quartile is the value in position

\dfrac{n + 1}{4}

, the upper quartile is the value in position

\dfrac{3(n + 1)}{4}

, and the interquartile range is the upper quartile minus the lower quartile.

Step 1: check the order, and count the values

9 \qquad 10 \qquad 12 \qquad 15 \qquad 17 \qquad 18 \qquad 19 \qquad 19 \qquad 20 \qquad 25 \qquad 26

(Reason: Quartiles are read from a list that runs from smallest to largest, so the first job is always to check the order. Here the values are already in order, so nothing moves. Counting along gives

11

values, so

n = 11

Step 2: find the lower quartile

\dfrac{11 + 1}{4} = 3

3\text{rd value} = 12

(Reason: The lower quartile sits at position

\dfrac{n + 1}{4}

, which here is position

3

. Counting three along the ordered list gives

9

, then

10

, then

12

, so the third value is

12

. Count the first value as position one, not as position zero.)

Step 3: find the upper quartile

\dfrac{3 \times (11 + 1)}{4} = 9

9\text{th value} = 20

(Reason: The upper quartile sits at position

\dfrac{3(n + 1)}{4}

, which here is position

9

. Counting nine along the ordered list lands on

20

. A quick way to check the position is that it is three times the lower quartile's position,

3 \times 3 = 9

Step 4: subtract

20 - 12 = 8

(Reason: The interquartile range is the upper quartile minus the lower quartile. It measures how spread out the middle half of the data is, so it is a subtraction and not an average. This is the line the mark scheme wants to see: the two quartiles

20

and

12

identified without any doubt about which is which.)

8

Verification

Check 1: Find the quartiles a different way, without any position formula. Split the ordered list at the median and take the middle value of each half. With

11

values the median is the sixth one, and it belongs to neither half. The median is

18

. The lower half is

9

10

12

15

17

, whose middle value is

12

. The upper half is

19

19

20

25

26

, whose middle value is

20

. Again

20 - 12 = 8

Check 2: Count the values outside the two quartiles. A quarter of the data should sit below the lower quartile and a quarter above the upper quartile, so the two tails should hold about the same number of values. Below

12

there are two values,

9

and

10

. Above

20

there are two values,

25

and

26

. The tails match, so the quartiles are sitting where a quarter of the way in and a quarter of the way back should put them.

Check 3: Compare the answer with the full range. The interquartile range covers only the middle half of the values, so it has to come out smaller than the range of all

11

. The range is

26 - 9 = 17

, and

8 < 17

. The interquartile range is a little under half the range, which is what a list with one low value and two high ones should give.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Working	M1	$20 \; (-) \; 12$ for both values unambiguously identified. The printed scheme brackets the minus sign, so the subtraction itself need not be written down.	✓
Answer	A1	$8$	✓
Note	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 2/2

Find the value of $6.7 \times 10^{135} + 3 \times 10^{134}$
Give your answer in standard form. [2 marks]

[Total 2 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

Two numbers written in standard form,

6.7 \times 10^{135}

and

3 \times 10^{134}

, to be added.

The two powers of ten are not the same: one term carries

10^{135}

and the other carries

10^{134}

, so they are one power apart.

Find

The total, written in standard form.

Plan the Solution

Compare the two powers of ten. They are different, so the front numbers cannot be added straight away.
Rewrite one term so that both terms carry the same power of ten. Moving up one power divides the front number by $10$ ; moving down one power multiplies it by $10$ .
Add the two front numbers, keeping the shared power of ten as a common factor.
Check the total is in standard form: the front number must be at least $1$ and less than $10$ .

Worked Solution [2 marks]

Rule - Adding in standard form: two terms can be added only when they carry the same power of ten. Match the powers first, then add the front numbers and keep the shared power. Moving up one power divides the front number by

10

; moving down one power multiplies it by

10

. The answer is written as

a \times 10^{n}

with

1 \leq a < 10

and

n

a whole number.

Step 1: make the two powers of ten match

3 \times 10^{134} = 0.3 \times 10^{135}

(Reason: The two terms carry different powers of ten, so nothing can be added yet. Writing the smaller term at the larger power raises the power by one, from

10^{134}

10^{135}

, and the front number must then be divided by

10

so that the value itself does not change. Dividing

3

10

gives

0.3

, so the second term becomes

0.3 \times 10^{135}

and is ready to be added to the first.)

Step 2: add the front numbers

6.7 \times 10^{135} + 0.3 \times 10^{135} = (6.7 + 0.3) \times 10^{135}

(6.7 + 0.3) \times 10^{135} = 7 \times 10^{135}

(Reason: Both terms now carry

10^{135}

, so it is a common factor and only the front numbers are added. This is the working the method mark is for: the mark scheme awards it for any line that gets the two terms on to one power of ten, and

(6.7 + 0.3) \times 10^{135}

written on its own is enough.)

Step 3: check the answer is in standard form

7 \times 10^{135}

(Reason: Standard form needs a front number of at least

1

and less than

10

, multiplied by a whole-number power of ten. The front number here is

7

, which is inside that range, so the answer is already in standard form and nothing has to be adjusted. Had the powers been matched at

10^{134}

instead, the total would have come out as

70 \times 10^{134}

- the same value, but with a front number too big for standard form, so one more step would still be needed.)

7 \times 10^{135}

Verification

Check 1: Match the powers the other way round. Carry both terms at

10^{134}

instead of

10^{135}

. A correct total cannot depend on which power was chosen. Moving down one power multiplies the front number by

10

, so

6.7 \times 10^{135} = 67 \times 10^{134}

. Adding gives

67 \times 10^{134} + 3 \times 10^{134} = 70 \times 10^{134}

, and converting that back to standard form gives

70 \times 10^{134} = 7 \times 10^{135}

. The same total, reached without ever writing

0.3

Check 2: Undo the addition. Take the second term back off the answer; the first term should reappear exactly, because addition and subtraction are inverses.

7 \times 10^{135} - 3 \times 10^{134} = 6.7 \times 10^{135}

, which is the term the question started with, so nothing has been gained or lost in the rewriting.

Check 3: Sanity-check the size. The second term is far smaller than the first, so the total must be a little more than the first term and nowhere near double it. Doubling the first term gives

2 \times 6.7 \times 10^{135} = 1.34 \times 10^{136}

, and the answer

7 \times 10^{135}

sits just above the first term and well below that. The first term is a little over

22

times the second, so a small rise is exactly what adding it should produce.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Working	M1	eg $0.3 \times 10^{135}$ or $67 \times 10^{134}$ or $(6.7 + 0.3) \times 10^{135}$ or $70 \times 10^{134}$ or $0.7 \times 10^{136}$ or $7 \times 10^{n}$ with $n \neq 135$	✓
Answer	A1	$7 \times 10^{135}$	✓
Note	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 2/2

(a) Solve the equation
$\dfrac{5a + 8}{3} - \dfrac{2a + 5}{4} = 23$
Show clear algebraic working. [4 marks]

(b) Write $\left( \dfrac{\sqrt{y}}{3} \right)^{-1}$ in the form $cy^{n}$ , where $c$ and $n$ are numbers to be found. [2 marks]

(a) a =(b)

[Total 6 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

An equation made of two algebraic fractions:

\dfrac{5a + 8}{3} - \dfrac{2a + 5}{4} = 23

An expression raised to the power

-1

\left( \dfrac{\sqrt{y}}{3} \right)^{-1}

Find

(a) The value of

a

, with clear algebraic working shown. (b) The numbers

c

and

n

that turn the expression into

cy^{n}

Plan the Solution

(a) Multiply every term by $12$ , the lowest common multiple of $3$ and $4$ , so both denominators disappear.
(a) Expand the two brackets, collect the $a$ terms on one side and the numbers on the other, then divide.
(b) A power of $-1$ means the reciprocal, so start by turning the fraction upside down.
(b) Replace the square root by a half power, move it to the numerator, then read off $c$ and $n$ .

Worked Solution [6 marks]

Rule - Clear fractions first, then use the index laws: multiply every term by the lowest common multiple of the denominators, and use

\left( \dfrac{p}{q} \right)^{-1} = \dfrac{q}{p}

with

\sqrt{y} = y^{\dfrac{1}{2}}

and

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

Step 1: Multiply every term by

12

12 \times \dfrac{5a + 8}{3} - 12 \times \dfrac{2a + 5}{4} = 12 \times 23

4(5a + 8) - 3(2a + 5) = 276

(Reason: The lowest common multiple of

3

and

4

12

. Multiplying every term by it cancels both denominators, leaving

12

divided by

3

, which is

4

, in front of the first bracket and

3

in front of the second. On the right,

12 \times 23 = 276

Step 2: Expand the brackets

20a + 32 - 6a - 15 = 276

(Reason: Multiply out each bracket:

4(5a + 8)

gives

20a + 32

, and

3(2a + 5)

gives

6a + 15

. The minus sign in front of the second bracket applies to both of its terms, so both are taken away.)

Step 3: Collect like terms

14a + 17 = 276

(Reason: The two terms in

a

give

20a - 6a = 14a

, and the two numbers on the left give

32 - 15 = 17

Step 4: Solve for

a

14a = 276 - 17 = 259

a = \dfrac{259}{14} = \dfrac{37}{2} = 18.5

(Reason: Subtract

17

from both sides, then divide both sides by

14

. Both

259

and

14

have a factor of

7

, so the fraction cancels to

\dfrac{37}{2}

, which is

18.5

. The mark scheme accepts any of these three forms.)

Step 5: Deal with the power of

-1

\left( \dfrac{\sqrt{y}}{3} \right)^{-1} = \dfrac{3}{\sqrt{y}}

(Reason: A power of

-1

means the reciprocal, and the reciprocal of a fraction is that fraction turned upside down:

\left( \dfrac{p}{q} \right)^{-1} = \dfrac{q}{p}

Step 6: Write the square root as an index

\dfrac{3}{\sqrt{y}} = \dfrac{3}{y^{\dfrac{1}{2}}} = 3y^{-\dfrac{1}{2}}

(Reason: A square root is a half power, so

\sqrt{y} = y^{\dfrac{1}{2}}

. Moving a power up from the denominator changes the sign of the index, which gives

c = 3

and

n = -\dfrac{1}{2}

, that is

n = -0.5

(a)

a = 18.5

(b)

3y^{-0.5}

, so

c = 3

and

n = -0.5

Verification

Check 1: Put

a = 18.5

back into the left-hand side of the original equation. The numerators become

100.5

and

42

\dfrac{100.5}{3} - \dfrac{42}{4} = 33.5 - 10.5 = 23

Check 2: Check the fraction forms the mark scheme also accepts really are the same number: cancel

\dfrac{259}{14}

7

, since

259 = 7 \times 37

and

14 = 7 \times 2

\dfrac{259}{14} = \dfrac{37}{2} = 18.5

Check 3: Test part (b) with

y = 9

. Then

\sqrt{9} = 3

, so the original expression is the reciprocal of

1

, which is

1

\dfrac{3}{\sqrt{9}} = \dfrac{3}{3} = 1

Check 4: Test part (b) with

y = 36

. Then

\sqrt{36} = 6

, so the original expression is the reciprocal of

2

, which is

0.5

\dfrac{3}{\sqrt{36}} = \dfrac{3}{6} = 0.5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) For example $12 \times \dfrac{5a + 8}{3} - 12 \times \dfrac{2a + 5}{4} = 12 \times 23$ or $4(5a + 8) - 3(2a + 5) = 12 \times 23 \,(= 276)$ or $\dfrac{4(5a + 8)}{12} - \dfrac{3(2a + 5)}{12} \,(= 23)$ or $\dfrac{4(5a + 8) - 3(2a + 5)}{12} \,(= 23)$	M1	for clear intention to multiply all terms by $12$ or a multiple of $12$ , or to express the left-hand side as two fractions over $12$ or a multiple of $12$ , or as a single fraction with a denominator of $12$ or a multiple of $12$ . If the numerator is expanded, allow one sign error or one numerical error but not both. Accept $\dfrac{20a + 32}{12} - \dfrac{6a + 15}{12} \,(= 23)$ or $\dfrac{20a + 32 - 6a + 15}{12} \,(= 23)$ .	✓
(a) For example $20a + 32 - 6a - 15 = 12 \times 23 \,(= 276)$ or $14a + 17 = 276$	M1	follow through, for expanding the brackets and multiplying both sides by the denominator with no more than one error in total, leading to a linear equation. Accept a linear equation leading to $14a - 17 = 276$ oe or $14a + 47 = 276$ oe or $26a + 17 = 276$ . This mark implies the previous M mark if that has not already been awarded.	✓
(a) For example $20a - 6a = 276 - 32 + 15$ oe or $14a = 259$	M1	follow through, dependent on the previous M1, for correctly rearranging so that the terms in $a$ are on one side and the number terms are on the other side.	✓
(a) $a = 18.5$ - working required	A1	or equivalent, dependent on both method marks, for example $\dfrac{259}{14}$ or $\dfrac{37}{2}$ .	✓
(b) For example $\dfrac{3}{\sqrt{y}} \left( = \dfrac{3\sqrt{y}}{y} \right)$ or $\dfrac{3}{y^{0.5}}$ or $\dfrac{3}{y^{\dfrac{1}{2}}}$ or $\left( \dfrac{y^{0.5}}{3} \right)^{-1}$ or $\left( \dfrac{y^{\dfrac{1}{2}}}{3} \right)^{-1}$ oe	M1	for a correct first step by applying one of the following index rules: $\sqrt{x} = x^{\dfrac{1}{2}} = x^{0.5}$ or $\left( \dfrac{a}{b} \right)^{-1} = \dfrac{b}{a}$ .	✓
(b) $3y^{-0.5}$	A1	or equivalent, for example $3y^{-\dfrac{1}{2}}$ ; accept $c = 3$ and $n = -0.5$ oe. A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 6/6

Show, using algebra, that the recurring decimal $0.6\dot{1}\dot{2}$ is equal to $\dfrac{101}{165}$ . [2 marks]

[Total 2 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

The recurring decimal

0.6\dot{1}\dot{2}

, with a dot over the

1

and a dot over the

2

Written out in full that is

0.6121212\ldots

, so the

6

does not recur and the block

12

repeats forever.

Find

Show that this decimal is exactly

\dfrac{101}{165}

. The working is the answer here, and the second mark is only given if algebra is used, so a calculator display proves nothing on its own.

Plan the Solution

Give the decimal a letter, $x$ , so that what follows is algebra and not arithmetic.
Multiply by $10$ to move the non-recurring $6$ in front of the decimal point.
The block is $2$ digits long, so multiply by a further $100$ to get a second number carrying exactly the same tail.
Subtract the two. The endless tails are identical, so they cancel and leave a whole number.
Divide by the coefficient, then cancel the fraction down to $\dfrac{101}{165}$ .

Worked Solution [2 marks]

Recurring decimals: multiply

x

by two powers of ten that leave the same recurring tail immediately after the decimal point, then subtract to remove that tail.

Step 1: Give the decimal a letter

x = 0.6\dot{1}\dot{2} = 0.6121212\ldots

(Reason: Naming the decimal is what turns this into algebra, and the final mark depends on it. Everything after this line is ordinary equation solving.)

Step 2: Multiply by two powers of ten

10x = 6.121212\ldots

1\,000x = 612.121212\ldots

(Reason: One digit sits between the point and the block, so

10x

starts the block immediately after the point. The block is

2

digits long, so a further

100

gives

1\,000x

with exactly the same tail.)

Step 3: Subtract to remove the recurring tail

1\,000x - 10x = 612.121212\ldots - 6.121212\ldots

990x = 606

(Reason: Both numbers end in the same endless

0.121212\ldots

, so subtracting takes it away exactly and leaves the whole number

606

Step 4: Divide, then cancel

x = \dfrac{606}{990}

\dfrac{606}{990} = \dfrac{101 \times 6}{165 \times 6} = \dfrac{101}{165}

(Reason: Both

606

and

990

are multiples of

6

. Nothing is left to cancel afterwards, because

101

is prime and is not a factor of

165

0.6\dot{1}\dot{2} = \dfrac{101}{165}

as required

Verification

Check 1: Go the other way. Divide

101

165

and read the digits off the display.

\dfrac{101}{165} = 0.6121212\ldots

, which is the decimal the question started from, dots and all.

Check 2: Use a different pair of multipliers. Subtract

x

from

100x

instead:

61.212121\ldots - 0.612121\ldots

99x = 60.6

, and

\dfrac{60.6}{99} = \dfrac{606}{990} = \dfrac{101}{165}

again. This pair leaves a terminating decimal rather than a whole number, which the mark scheme allows just as readily.

Check 3: Split the decimal instead of shifting it:

0.6\dot{1}\dot{2} = 0.6 + 0.0\dot{1}\dot{2}

, and the recurring part on its own is

\dfrac{12}{990}

\dfrac{6}{10} + \dfrac{12}{990} = \dfrac{594 + 12}{990} = \dfrac{606}{990} = \dfrac{101}{165}

, reached without any subtraction at all.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Two multiples of $x$ carrying the same recurring tail, written down so that one can be subtracted from the other	M1	For $2$ recurring decimals that when subtracted give a whole number or terminating decimal, with intention to subtract (ie give $60.6$ or $606$ or $6\,060$ etc), eg $(1\,000x =) \; 612.12\ldots$ and $(10x =) \; 6.12\ldots$ , or $(100\,000x =) \; 61\,212.12\ldots$ and $(1\,000x =) \; 612\ldots$ , or $(100x =) \; 61.212\ldots$ and $(x =) \; 0.612\ldots$ , with intention to subtract. $x$ is not required to award this mark. If recurring dots are not shown in both numbers, then at least one of the numbers must be shown to at least $5$ significant figures. Or $\dfrac{6}{10} + 1\,000x \, (12.12\ldots) - 10x \, (0.12\ldots)$ .	✓
Complete the algebra to $\dfrac{101}{165}$	A1	For completion to $\dfrac{101}{165}$ , dep on M1, and algebra must be used for this final mark to be awarded. Eg $10\,000x - 100x = 6\,121.21\ldots - 61.21\ldots = 6\,060$ , $(9\,900x = 6\,060)$ and $\dfrac{6\,060}{9\,900} = \dfrac{101}{165}$ , or $1\,000x - 10x = 612.12\ldots - 6.12\ldots = 606$ , $(990x = 606)$ and $\dfrac{606}{990} = \dfrac{101}{165}$ , or $100x - x = 61.212\ldots - 0.612\ldots = 60.6$ , $(99x = 60.6)$ and $\dfrac{60.6}{99} = \dfrac{101}{165}$ oe, or $0.6 + \ldots$ and $(1\,000x - 10x = 990x = 12)$ and $0.6 + \dfrac{12}{990} = \dfrac{0.6 \times 990 + 12}{990} = \dfrac{101}{165}$ oe. Allow for instance $99x = 60.6$ and then $\dfrac{606}{990} = \dfrac{101}{165}$ .	✓
Working required	Note	No algebra used gets a maximum of 1 mark.	✓

Full marks: 2/2

Three numbers are consecutive multiples of $4$ .
Prove that the difference between the square of the largest number and the square of the smallest number is always a multiple of $64$ .

You must show clear algebraic working. [3 marks]

[Total 3 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

Three numbers that are consecutive multiples of

4

Consecutive multiples of

4

rise in steps of

4

, so once the smallest is named the other two follow from it

Find

A proof that the square of the largest number minus the square of the smallest number is always a multiple of

64

The proof has to be algebraic: it must cover every such set of three numbers at once, not a few chosen examples.

Plan the Solution

Name the three numbers with one letter: $4n$ , $4n + 4$ and $4n + 8$ , where $n$ is any integer.
Square the largest and square the smallest, then subtract one from the other.
Factorise what is left and show that $64$ comes out as a factor, with an integer beside it.

Worked Solution [3 marks]

Rule - Consecutive multiples of

4

: every multiple of

4

4n

for some integer

n

, and the next two are

4n + 4

and

4n + 8

. A number is a multiple of

64

exactly when it can be written as

64 \times (\text{an integer})

Step 1: name the three numbers algebraically

4n, \; 4n + 4, \; 4n + 8

(Reason:

n

stands for any integer, so

4n

stands for any multiple of

4

. Adding

4

each time gives the next two consecutive multiples, so the smallest of the three is

4n

and the largest is

4n + 8

Step 2: square the largest and square the smallest

(4n + 8)^2 = 16n^2 + 64n + 64

(4n)^2 = 16n^2

(Reason: Expand the bracket in full - square the first term, double the product of the two terms, then square the second term. Only the largest and the smallest are squared, so the middle number

4n + 4

plays no part in the proof.)

Step 3: subtract, then take out the common factor

(4n + 8)^2 - (4n)^2 = 16n^2 + 64n + 64 - 16n^2

= 64n + 64

= 64(n + 1)

(Reason: The

16n^2

terms cancel, which is what makes this difference so tidy. Both terms left over have

64

as a factor, since

64n = 64 \times n

and

64 = 64 \times 1

, so

64

can be taken outside a bracket.)

Step 4: read the conclusion off the factorised form

64(n + 1)

(Reason:

n

is an integer, so

n + 1

is an integer as well. The difference is therefore

64

multiplied by an integer, which is precisely what it means to be a multiple of

64

. Nothing in the working depended on which integer

n

is, so this holds for any three consecutive multiples of

4

(4n + 8)^2 - (4n)^2 = 64n + 64 = 64(n + 1)

n + 1

is an integer, so the difference is always a multiple of

64

Verification

Check 1: Test the result on real numbers. Take

n = 3

, which gives the three consecutive multiples

12

16

and

20

, then square the largest and the smallest and subtract.

20^2 - 12^2 = 400 - 144 = 256

, and

256 = 64 \times 4

, which is the

64(n + 1)

above with

n = 3

Check 2: Test a second, larger set so the check is not a coincidence. Take

n = 7

, giving

28

32

and

36

36^2 - 28^2 = 1296 - 784 = 512

, and

512 = 64 \times 8

, again matching

64(n + 1)

with

n = 7

Check 3: Redo the algebra a different way, so no expansion is repeated. Use the difference of two squares,

a^2 - b^2 = (a + b)(a - b)

, with

a = 4n + 8

and

b = 4n

(4n + 8 + 4n)(4n + 8 - 4n) = (8n + 8) \times 8 = 64n + 64

, the same expression the expansion gave

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $4n, \; 4n + 4, \; 4n + 8$ or $4n, \; 4(n + 1), \; 4(n + 2)$ or $4n - 4, \; 4n, \; 4n + 4$ or $4(n - 1), \; 4n, \; 4(n + 1)$	M1	for correct expressions for $3$ consecutive multiples of $4$ (any letter can be used) may just see the first and third multiple for this mark	✓
eg $(4n + 8)^2 - (4n)^2 \; (= 16n^2 + 64n + 64 - 16n^2)$ or $(4(n + 2))^2 - (4n)^2 \; (= 16n^2 + 64n + 64 - 16n^2)$ or $(4n + 4)^2 - (4n - 4)^2 \; (= 16n^2 + 32n + 16 - 16n^2 + 32n - 16)$ or $(4(n + 1))^2 - (4(n - 1))^2 \; (= 16n^2 + 32n + 16 - 16n^2 + 32n - 16)$	M1	for squaring the largest and smallest multiple of $4$ and subtracting (no need to expand or simplify for this mark)	✓
eg $(4n + 8)^2 - (4n)^2 = 16n^2 + 64n + 64 - 16n^2 = 64n + 64$ or $(4n + 8)^2 - (4n)^2 = (4n + 8 + 4n)(4n + 8 - 4n) = 8(8n + 8) = 64n + 64$ or $(4n + 4)^2 - (4n - 4)^2 = 16n^2 + 32n + 16 - 16n^2 + 32n - 16 = 64n$ or $(4(n + 1))^2 - (4(n - 1))^2 = (4(n + 1) + 4(n - 1))(4(n + 1) - 4(n - 1)) = 8n \times 8 = 64n$ Answer given as correctly shown	A1	dep on M2, for use of algebra to show correct conclusion	✓
Working required	Note	algebraic working must be seen. Trying particular multiples of $4$ , however many of them, shows the statement in those cases and does not prove it in every case	✓

Full marks: 3/3

$F$ varies inversely with the cube of $r$
When $r = 2$ , $F = 6$

(a) Work out a formula for $F$ in terms of $r$ [3 marks]

(b) Work out the value of $r$ when $F = 3072$ [2 marks]

(a)(b) r =

[Total 5 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

F

varies inversely with the cube of

r

One pair of values:

F = 6

when

r = 2

Find

(a) A formula for

F

in terms of

r

(b) The value of

r

when

F = 3072

Plan the Solution

Turn the words into an equation. Inverse proportion to the cube of $r$ means $F = \dfrac{k}{r^{3}}$ for some constant $k$ .
Put the one pair of values the question gives into that equation to pin $k$ down, then write the formula out in full. That is part (a).
For part (b), put $F = 3072$ into the finished formula, rearrange to leave $r^{3}$ on its own, and take the cube root.

Worked Solution [5 marks]

Rule - Inverse proportion:

F \propto \dfrac{1}{r^{3}}

means

F = \dfrac{k}{r^{3}}

, where

k

is a constant. One pair of values fixes

k

, and that same

k

then holds for every other pair.

Step 1: Write the statement as an equation

F \propto \dfrac{1}{r^{3}}

F = \dfrac{k}{r^{3}}

(Reason: Inversely proportional to the cube means

F

is a constant divided by

r^{3}

, never multiplied by it, and the cube is on the

r

alone. The constant has to be carried as a letter such as

k

until a pair of values fixes it; the mark scheme allows any letter here but not

1

Step 2: Substitute

F = 6

and

r = 2

6 = \dfrac{k}{2^{3}}

6 = \dfrac{k}{8}

(Reason: The pair of values is the only thing that can pin

k

down, so it goes into the equation just written. Cubing the

2

gives

8

, so the equation now has one unknown in it instead of three.)

Step 3: Work out

k

, then write the formula

k = 6 \times 8 = 48

F = \dfrac{48}{r^{3}}

(Reason: Multiplying both sides by

8

leaves

k

on its own. Putting

k = 48

back into

F = \dfrac{k}{r^{3}}

gives the formula part (a) asks for, with

F

as the subject.)

Step 4: Put

F = 3072

into the formula

3072 = \dfrac{48}{r^{3}}

r^{3} = \dfrac{48}{3072} = \dfrac{1}{64}

(Reason: Multiply both sides by

r^{3}

and then divide by

3072

, so that

r^{3}

is left on its own. Watch which number ends up on top:

F

has gone up a long way from

6

, so

r

must come out smaller than the

2

that produced it.)

Step 5: Take the cube root

r = \sqrt[3]{\dfrac{1}{64}}

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

(Reason: Cube rooting undoes the cube. Since

4^{3} = 64

, the cube root of

\dfrac{1}{64}

\dfrac{1}{4}

. The mark scheme prints

\dfrac{1}{4}

and accepts any equivalent form, so

0.25

scores the mark just as well.)

(a)

F = \dfrac{48}{r^{3}}

(b)

r = \dfrac{1}{4}

, or

0.25

Verification

Check 1: Put

r = 2

back into the formula from part (a). Cubing

2

gives

8

, and the

F

that comes out has to be the

6

the question states.

\dfrac{48}{8} = 6

Check 2: Put

r = \dfrac{1}{4}

into the same formula. Cubing

\dfrac{1}{4}

gives

\dfrac{1}{64}

, and dividing by

\dfrac{1}{64}

is the same as multiplying by

64

48 \times 64 = 3072

Check 3: Work in multipliers and never use the constant at all.

F

is multiplied by

\dfrac{3072}{6} = 512

, so

r^{3}

is divided by

512

and

r

is divided by the cube root of

512

8^{3} = 512

and

\dfrac{2}{8} = \dfrac{1}{4}

Check 4: For inverse proportion to the cube, the product

F \times r^{3}

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

6 \times 2^{3} = 48

and

3072 \times \left( \dfrac{1}{4} \right)^{3} = 48

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$F = \dfrac{k}{r^{3}}$ or $Fr^{3} = k$ or $kF = \dfrac{1}{r^{3}}$	M1	oe. $k$ can be any letter (must be a letter and not $1$ )	✓
$6 = \dfrac{k}{2^{3}}$ oe or $k = 48$ or $6k = \dfrac{1}{2^{3}}$ oe or $k = \dfrac{1}{48}$	M1	for substitution of $F$ and $r$ into a correct formula, implies the first M1 if you see this stage. Condone use of $\propto$ for method marks	✓
$F = \dfrac{48}{r^{3}}$	A1	oe with $F$ the subject eg $F = 48 \times \dfrac{1}{r^{3}}$ or $F = 48 \times r^{-3}$ . Award $3$ marks if answer is $F = \dfrac{k}{r^{3}}$ and $k = 48$ clearly given in the body of the script. M2A0 for $Fr^{3} = 48$ or $r = \sqrt[3]{\dfrac{48}{F}}$ or $r^{3} = \dfrac{48}{F}$	✓
$(r^{3} =) \dfrac{48}{3072}$ oe eg $\dfrac{1}{64}$ or $0.01(5625)$ rounded or truncated	M1ft	allow use of their $48$ as long as M2 was gained in part (a). The printed scheme puts that $48$ in quotation marks, which is what grants the follow-through from a candidate's own constant	✓
$\dfrac{1}{4}$	A1	oe	✓
On the answer row of part (a) and of part (b)	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 5/5

Malee has $9$ tiles.
There is a number on each tile.

Malee puts the $9$ tiles in a box.
She takes a tile from the box at random and does not put the tile back.
She then takes a second tile from the box at random.

Work out the probability that the sum of the numbers on the two tiles is less than $5$ [3 marks]

[Total 3 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

Nine tiles, carrying the numbers

1

1

2

3

3

3

5

6

and

6

Two tiles are taken at random, and the first one is NOT put back before the second is taken.

Find

The probability that the two numbers taken add up to less than

5

Plan the Solution

Work out which pairs of numbers actually add to less than $5$ . Only the small numbers can manage it, so that list is short.
Count the outcomes. There are $9$ tiles for the first draw but only $8$ for the second, because the first tile is not put back.
Take each possible FIRST tile in turn, write down the probability of that branch, and add the branches up. Two different first tiles cannot both happen, so adding is allowed.

Worked Solution [3 marks]

Rule - Without replacement: the second fraction has a denominator one smaller than the first, because one tile has gone. For two draws in a row,

P(A \text{ then } B) = P(A) \times P(B \text{ after } A)

, and outcomes that cannot happen together are added.

Step 1: which pairs add to less than

5

1 + 1 = 2

1 + 2 = 3

1 + 3 = 4

2 + 3 = 5

1 + 5 = 6

(Reason: Start with the smallest numbers on the tiles and work upwards. The first three sums are under

5

, the fourth is exactly

5

and so does not count, and every remaining pair is larger still. So one of the two tiles has to be a

1

, and the other has to be a

1

, a

2

or a

3

Step 2: count the outcomes

9 \times 8 = 72

(Reason: There are

9

tiles to choose from first and only

8

left for the second draw, so there are

72

outcomes in order. If the tile were put back there would be

9 \times 9 = 81

outcomes instead and the answer would come out as

\dfrac{21}{81}

- that is exactly the wrong answer the mark scheme carries a special case for, so read the words does not put the tile back carefully.)

Step 3: take each possible first tile in turn

\dfrac{2}{9} \times \dfrac{5}{8} = \dfrac{10}{72}

\dfrac{1}{9} \times \dfrac{2}{8} = \dfrac{2}{72}

\dfrac{3}{9} \times \dfrac{2}{8} = \dfrac{6}{72}

(Reason: The first tile is a

1

with probability

\dfrac{2}{9}

, and then the other

1

, the

2

and all three

3

s still finish the job, which is

5

of the

8

tiles left. The first tile is the

2

with probability

\dfrac{1}{9}

, and then only a

1

works, which is

2

of the

8

left. The first tile is a

3

with probability

\dfrac{3}{9}

, and again only a

1

works. A first tile of

5

6

can never work, so there is no fourth line.)

Step 4: add the branches

\dfrac{10}{72} + \dfrac{2}{72} + \dfrac{6}{72} = \dfrac{18}{72}

\dfrac{18}{72} = \dfrac{1}{4}

(Reason: The three branches start with different tiles, so they cannot happen together and their probabilities simply add. Dividing top and bottom by

18

turns

\dfrac{18}{72}

into

\dfrac{1}{4}

. The mark scheme prints

\dfrac{18}{72}

and accepts any equivalent form, so

\dfrac{9}{36}

\dfrac{1}{4}

0.25

and

25\%

all score the accuracy mark.)

\dfrac{18}{72} = \dfrac{1}{4}

, or

0.25

Verification

Check 1: Count pairs instead of ordered draws. Order does not change a sum, so there are

{}^{9}C_{2} = 36

pairs altogether. The pairs that work are the two

1

s together, which is

1

pair, a

1

with the

2

, which is

2

pairs, and a

1

with a

3

, which is

6

pairs.

1 + 2 + 6 = 9

and

\dfrac{9}{36} = \dfrac{1}{4}

Check 2: Count the draws that FAIL instead. Going through the value pairs the other way - a

1

with a

5

or a

6

, the

2

with a

3

, a

5

or a

6

, two

3

s together, and so on - gives

54

ordered draws whose sum is

5

or more. The two counts have to fill the

72

between them.

\dfrac{54}{72} = \dfrac{3}{4}

and

\dfrac{18}{72} + \dfrac{54}{72} = 1

Check 3: Is the size sensible? Divide

18

72

. One draw in four working is a believable size for an event that needs one of only two

1

s to turn up among the two tiles taken.

\dfrac{18}{72} = 0.25

, which is

25\%

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{2}{9} \times \dfrac{1}{8} \left( = \dfrac{2}{72} \right)$ oe or $\dfrac{1}{9} \times \dfrac{2}{8} \left( = \dfrac{2}{72} \right)$ oe or $\dfrac{2}{9} \times \dfrac{3}{8} \left( = \dfrac{6}{72} \right)$ oe or $\dfrac{3}{9} \times \dfrac{2}{8} \left( = \dfrac{6}{72} \right)$ oe or $\dfrac{2}{9} \times \dfrac{5}{8} \left( = \dfrac{10}{72} \right)$ oe or $\dfrac{5}{9} \times \dfrac{2}{8} \left( = \dfrac{10}{72} \right)$ oe, or ${}^{9}C_{2}$ or $\dfrac{9!}{2!7!}$ or $\dfrac{9 \times 8}{2}$ or $36$ or $1 + 2 + 6 \left( = 9 \right)$	M1	for finding one correct product, or for the correct number of total outcomes, or for the correct number of outcomes when the sum $< 5$ . NB if using decimals allow $2$ decimal places truncated or rounded	✓
$3 \times \dfrac{2}{72} + 2 \times \dfrac{6}{72}$ oe or $\dfrac{2}{72} + \dfrac{6}{72} + \dfrac{10}{72}$ oe, or ${}^{9}C_{2}$ or $\dfrac{9!}{2!7!}$ or $\dfrac{9 \times 8}{2}$ or $36$ and $1 + 2 + 6 \left( = 9 \right)$	M1	for a complete correct method, or for the correct number of total outcomes AND for the correct number of outcomes when the sum $< 5$ . The printed scheme puts each of those fractions in quotation marks, which is what allows a candidate's own products from the first M1 to be used here	✓
$\dfrac{18}{72}$	A1	oe eg $\dfrac{9}{36}$ or $0.25$ or $25\%$	✓
$\dfrac{21}{81}$	SC B1	oe eg $\dfrac{7}{27}$ or $0.259(25\ldots)$ or $25.9(25\ldots)\%$ truncated or rounded	✓
ALT method - $\dfrac{2}{9} \times \dfrac{1}{8} \left( = \dfrac{2}{72} \right)$ oe or $\dfrac{1}{9} \times \dfrac{2}{8} \left( = \dfrac{2}{72} \right)$ or $\dfrac{1}{9} \times \dfrac{1}{8} \left( = \dfrac{1}{72} \right)$ oe or $\dfrac{2}{9} \times \dfrac{2}{8} \left( = \dfrac{4}{72} \right)$ or $\dfrac{3}{9} \times \dfrac{1}{8} \left( = \dfrac{3}{72} \right)$ oe or $\dfrac{1}{9} \times \dfrac{3}{8} \left( = \dfrac{3}{72} \right)$ oe or $\dfrac{2}{9} \times \dfrac{3}{8} \left( = \dfrac{6}{72} \right)$ oe or $\dfrac{3}{9} \times \dfrac{2}{8} \left( = \dfrac{6}{72} \right)$ oe or $\dfrac{1}{9} \times \dfrac{6}{8} \left( = \dfrac{6}{72} \right)$ oe or $\dfrac{6}{9} \times \dfrac{1}{8} \left( = \dfrac{6}{72} \right)$ oe or $\dfrac{3}{9} \times \dfrac{6}{8} \left( = \dfrac{18}{72} \right)$ oe or $\dfrac{6}{9} \times \dfrac{3}{8} \left( = \dfrac{18}{72} \right)$ oe or $\dfrac{1}{9} \times \dfrac{8}{8} \left( = \dfrac{8}{72} \right)$ oe or $\dfrac{8}{9} \times \dfrac{1}{8} \left( = \dfrac{8}{72} \right)$ or $\dfrac{2}{9} \times \dfrac{8}{8} \left( = \dfrac{16}{72} \right)$ oe or $\dfrac{8}{9} \times \dfrac{2}{8} \left( = \dfrac{16}{72} \right)$ oe	Note	The mark scheme prints a full ALT method on its next page, carrying the same three marks and working from the draws that fail instead. This is that method's first method mark: for finding one correct product. NB if using decimals allow $2$ decimal places truncated or rounded	✓
ALT method - $1 - \left( 2 \times \dfrac{1}{72} + 7 \times \dfrac{2}{72} + 4 \times \dfrac{3}{72} + 2 \times \dfrac{4}{72} + 3 \times \dfrac{6}{72} \right)$ or $1 - \left( \dfrac{6}{72} + \dfrac{6}{72} + \dfrac{18}{72} + \dfrac{8}{72} + \dfrac{16}{72} \right)$ or $1 - \dfrac{54}{72}$ oe	Note	The ALT method's second method mark: for a complete correct method. The printed scheme puts each of those fractions in quotation marks, which is what allows a candidate's own products from the ALT first mark to be used here	✓
On the answer row	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓
On the answer row	Note	Do not allow $\dfrac{6}{9} \times \dfrac{3}{8} = \dfrac{18}{72}$ or $\dfrac{3}{9} \times \dfrac{6}{8} = \dfrac{18}{72}$ as this an incorrect method (M1M0A0)	✓

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 4MA1/1HR, Thursday 15 May 2025: Worked Solutions, Questions 12 to 19

Worked solutions, questions 12 to 19 of 26

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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