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Edexcel IGCSE 4MA1 Paper 2F, June 2024: Worked Solutions, Questions 15 to 28

Sir Faraz Hassan

4 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)Paper 2F - Foundation Tier - June 2024100 marks · 2 hours · Calculator allowed

Back to questions 1 to 14

This is the rest of the paper. Questions 1 to 14, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Edexcel International GCSE Mathematics A (4MA1), Paper 2F (Foundation Tier), June 2024 – 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 28 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 15 to 28 of 28

$\mathcal{E} = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}$

$A = \{11, 14, 16, 19\}$
$B = \{12, 14, 15, 16, 18, 20\}$

Show this information on the Venn diagram below. [3 marks]

[Total 3 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

The universal set

\mathcal{E} = \{11, 12, 13, 14, 15, 16, 17, 18, 19, 20\}

- the ten whole numbers from

11

20

inclusive

A = \{11, 14, 16, 19\}

and

B = \{12, 14, 15, 16, 18, 20\}

A Venn diagram with two overlapping circles inside a rectangle, so there are four regions to fill: circle

A

alone, the overlap, circle

B

alone, and the space inside the rectangle but outside both circles.

Find

Which of the four regions each of the ten numbers belongs to Every member of

\mathcal{E}

is written on the diagram exactly once, so the region outside both circles has to be filled in as well - it is one of the four parts being marked.

Plan the Solution

Start with the overlap. A number is written once only, so the numbers that are in $A$ and in $B$ have to be settled before anything else is placed.
Then take the overlap away from each circle in turn: what is left of $A$ goes in the left crescent, and what is left of $B$ goes in the right crescent.
Anything in $\mathcal{E}$ that has not been used yet goes in the rectangle, outside both circles.
Finish by counting. The four regions must hold $10$ numbers between them, because that is how many members $\mathcal{E}$ has.

Worked Solution [3 marks]

Rule - Every member of

\mathcal{E}

goes in exactly one of the four regions: in both sets, in

A

only, in

B

only, or in neither set.

Step 1: find the numbers that are in both sets

A = \{11, 14, 16, 19\}

B = \{12, 14, 15, 16, 18, 20\}

A \cap B = \{14, 16\}

(Reason: (Reason:

14

and

16

are the only numbers that appear in both lists, so they belong in the overlap where the two circles cross. Settling the overlap first is what stops a number being written twice.))

Step 2: take the overlap out of circle A

A \text{ only} = \{11, 19\}

(Reason: (Reason:

A

has

4

members and

2

of them are already in the overlap, so

2

are left for the crescent of

A

that lies outside

B

.))

Step 3: take the overlap out of circle B

B = \{12, 14, 15, 16, 18, 20\}

B \text{ only} = \{12, 15, 18, 20\}

(Reason: (Reason:

14

and

16

have already been placed in the overlap, so they are not written a second time in the crescent that belongs to

B

alone.))

Step 4: place what is left outside both circles

4 + 6 - 2 = 8

10 - 8 = 2

\text{Outside both circles} = \{13, 17\}

(Reason: (Reason: the two circles hold

8

different numbers between them, because the

2

shared numbers would otherwise be counted twice. That leaves

2

members of

\mathcal{E}

in neither set -

13

and

17

- and they go inside the rectangle but outside both circles.))

A

only:

11, 19

In both

A

and

B

14, 16

B

only:

12, 15, 18, 20

Outside both circles:

13, 17

Verification

Check 1: Count the numbers written in the four regions and compare the total with the size of

\mathcal{E}

. The regions hold

2

2

4

and

2

numbers, and

2 + 2 + 4 + 2 = 10

, which is exactly how many members

\mathcal{E}

has - nothing missing and nothing written twice.

Check 2: Read circle

A

back off the finished diagram - the crescent and the overlap together - and compare it with the given set.

11, 19

from the crescent and

14, 16

from the overlap give

\{11, 14, 16, 19\}

, which is

A

exactly.

Check 3: Do the same for circle

B

: its crescent and the overlap together must rebuild

B

12, 15, 18, 20

from the crescent and

14, 16

from the overlap give

\{12, 14, 15, 16, 18, 20\}

, which is

B

exactly.

Check 4: Scan the ten numbers on the diagram for a repeat, and check that every member of the universal set appears somewhere. Each of

11

20

appears once and once only, so no number has been left out and none has been placed in two regions.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
All four parts of the Venn diagram correct: $11, 19$ in $A$ only, $14, 16$ in the overlap, $12, 15, 18, 20$ in $B$ only, and $13, 17$ outside both circles	B3	Full marks for a completely correct Venn diagram - all $4$ parts right, including the numbers outside both circles.	✓
Two or three of the four parts of the Venn diagram correct	B2	Partial credit. A candidate who fills the two circles correctly but leaves $13$ and $17$ off the diagram has three parts right, so this is where that answer lands.	✓
Exactly one of the four parts of the Venn diagram correct	B1	Partial credit. Writing the whole of $B$ into the $B$ crescent and the whole of $A$ into the $A$ crescent - that is, never removing the shared numbers - typically leaves only one region right.	✓

Full marks: 3/3

Use a calculator to find the value of

$\dfrac{17.9}{8.61 + 2.36} - 1.2^{2}$

Write your answer as a decimal.
Copy down every figure shown on your calculator display. [2 marks]

[Total 2 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

The calculation

\dfrac{17.9}{8.61 + 2.36} - 1.2^{2}

, to be worked out on a calculator.

A fraction bar, so the whole of

8.61 + 2.36

is the divisor.

Find

The value as a decimal, written with every figure the calculator display shows.

Plan the Solution

Total the denominator first: $8.61 + 2.36$ .
Divide $17.9$ by that total, keeping the full display.
Square $1.2$ , then subtract it.
Copy the display figure for figure. The question asks for all of them, so nothing is rounded.

Worked Solution [2 marks]

Rule - order of operations: a fraction bar groups like a bracket, so the whole of

8.61 + 2.36

is worked out before dividing, and the power

1.2^{2}

is worked out before the subtraction.

Step 1: total the denominator

8.61 + 2.36 = 10.97

(Reason: The fraction bar acts as a bracket, so everything under it is added before any dividing is done.)

Step 2: divide

\dfrac{17.9}{10.97} = 1.631722880583\ldots

(Reason: The full display is kept. Rounding to

1.63

here would change the figures the final answer is judged on.)

Step 3: work out the power

1.2^{2}

1.2^{2} = 1.44

(Reason: A power is evaluated before the subtraction, so this value is ready to take away in the next step.)

Step 4: subtract

1.631722880583\ldots - 1.44 = 0.191722880583\ldots

(Reason: Taking the square away from the quotient completes the calculation.)

Step 5: read the display

0.1917228806

(Reason: A ten-figure display rounds the last figure it can show, so it reads

0.1917228806

. Every figure is written down, exactly as it appears.)

0.1917228806

Verification

Check 1: Do it as one fraction instead. Take

1.44 \times 10.97 = 15.7968

from

17.9

, then divide by

10.97

\dfrac{2.1032}{10.97} = 0.191722880583\ldots

, the same figures as Step 4.

Check 2: Work it exactly, as a fraction with no decimals:

\dfrac{1790}{1097} - \dfrac{36}{25}

\dfrac{5258}{27425} = 0.191722880583\ldots

, which agrees to every figure shown.

Check 3: Reverse the calculation: add

1.44

back on, then multiply by

10.97

(0.191722880583\ldots + 1.44) \times 10.97 = 17.9

, the numerator we started from.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A start to the calculation, for example $8.61 + 2.36 = 10.97$ or $1.63(17\ldots)$ or $1.43(17\ldots)$ , or an answer given as the fraction $\dfrac{5258}{27425}$ , or an answer rounded too soon, such as $0.19$ , $0.191$ , $0.192$ or $0.1917$ .	M1	Method: the denominator totalled, or the division carried out, or the value left as a fraction or rounded before the answer line.	✓
$0.19172(28806)$	A1	The value written to at least $5$ decimal places. A correct answer scores both marks unless it follows obviously incorrect working.	✓

Full marks: 2/2

Eight numbers are listed in order of size, from smallest to largest.

$h \quad 6 \quad 7 \quad 8 \quad j \quad 16 \quad k \quad k$

Each of $h$ , $j$ and $k$ is an integer.

The median of the eight numbers is $10$
The mode of the eight numbers is $18$
The range of the eight numbers is $13$

Work out the value of $h$ , the value of $j$ and the value of $k$ . [3 marks]

h =j =k =

[Total 3 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

Eight numbers, already in order of size:

h \quad 6 \quad 7 \quad 8 \quad j \quad 16 \quad k \quad k

h

j

and

k

are integers, so no fractions or decimals are involved.

Median

= 10

, mode

= 18

, range

= 13

Find

The value of

h

, the value of

j

and the value of

k

Plan the Solution

Take the three clues one at a time, and start with the clue that pins down a letter on its own.
The mode is the value written most often. Every number in the list appears once except $k$ , which appears twice, so the mode has to be $k$ . That gives $k$ straight away.
With $8$ numbers there is no single middle number, so the median is the mean of the $4$ th and $5$ th numbers, which are $8$ and $j$ . That gives $j$ .
The list is in order of size, so the largest number is $k$ and the smallest is $h$ . The range clue then gives $h$ .

Worked Solution [3 marks]

Rule - For a list already in order of size: the mode is the value written most often; the median of

8

values is

\dfrac{4\text{th value} + 5\text{th value}}{2}

; and the range is

\text{largest} - \text{smallest}

Step 1: Use the mode to find

k

\text{mode} = k

k = 18

(Reason: The mode is the value that occurs most often. The numbers

6

7

8

and

16

are each written once, so the only value that can occur twice is

k

. The mode is

18

, so

k

must be

18

Step 2: Use the median to find

j

\dfrac{8 + j}{2} = 10

8 + j = 2 \times 10 = 20

j = 20 - 8 = 12

(Reason: There are

8

numbers, an even amount, so the median is the mean of the middle two. Counting along the list, the

4

th number is

8

and the

5

th number is

j

. Their mean is given as

10

, so their total must be

20

Step 3: Use the range to find

h

k - h = 13

18 - h = 13

h = 18 - 13 = 5

(Reason: The eight numbers are already written in order of size, so the largest is

k

and the smallest is

h

. The range is largest minus smallest, and

k

was found to be

18

in Step 1.)

h = 5

j = 12

k = 18

Verification

Check 1: Put the three values back into the list and work out all three statistics from scratch. The eight numbers become

5, \; 6, \; 7, \; 8, \; 12, \; 16, \; 18, \; 18

, whose middle two are

8

and

12

. Median

\dfrac{8 + 12}{2} = 10

, mode

18

(the only repeat), range

18 - 5 = 13

- all three given facts come back out.

Check 2: Check the answers keep the list in order of size, which the question states it is in. Reading along:

5 \leq 6

at the start,

8 \leq 12 \leq 16

in the middle, and

16 \leq 18

at the end. Every number sits where the question places it, so

h = 5

j = 12

and

k = 18

are allowed values and not just arithmetic that happens to work.

Check 3: Check no other set of integers fits. A search over every integer triple that keeps the list in order gives exactly one that has median

10

, a single mode of

18

and range

13

. Only

(h, j, k) = (5, 12, 18)

survives, so the answer is unique - each clue pins down exactly one letter.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Finds one of the three letters, or writes a correct statement for one of them, e.g. $k = 18$ or $\dfrac{8 + j}{2} = 10$ or $k - h = 13$ .	M1	For a correct value for $h$ , $j$ or $k$ , or for a correct statement for one of these.	✓
Finds a second letter, e.g. both $k = 18$ and $j = 12$ , or both correct statements for them.	M1	For $2$ correct values from $h$ , $j$ or $k$ , or for $2$ correct statements for them.	✓
All three values correct: $h = 5$ , $j = 12$ , $k = 18$ .	A1	All correct. A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

(a) On the grid, draw the straight line with equation
(i) $y = 2$
(ii) $x = 6$
(iii) $y = x + 1$
Write the equation beside each line you draw. [3 marks]

(b) Show, by shading on the grid, the region that satisfies all three of these inequalities
$y \geq 2 \qquad x \leq 6 \qquad y \leq x + 1$
Label this region $R$ [1 mark]

[Total 4 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

A square grid running from

0

8

on the

x

-axis and from

0

8

on the

y

-axis.

Three lines to draw:

y = 2

x = 6

and

y = x + 1

Three inequalities to satisfy at the same time:

y \geq 2

x \leq 6

and

y \leq x + 1

Find

(a) the three straight lines, drawn on the grid, each with its equation written beside it. (b) the one region that satisfies all three inequalities at once, shaded and labelled

R

Plan the Solution

Draw the two easy lines first. $y = 2$ is horizontal and $x = 6$ is vertical, so neither needs a table.
For $y = x + 1$ , work out two points that are far apart, plot them, join them with a ruler, and keep a third point as a check that the line is straight.
Turn each inequality into a side of its own line, and test a single point to settle which side that is.
The region wanted is the part of the grid on the correct side of all three lines at once. Its corners are where the boundary lines cross, so work those out before shading, and label the region $R$ .

Worked Solution [4 marks]

A line

y = a

is horizontal and a line

x = b

is vertical. A line

y = mx + c

has gradient

m

and passes through

(0,\,c)

. A set of inequalities is satisfied by the OVERLAP of the regions the separate inequalities allow, and the corners of that overlap are the points where the boundary lines cross.

Step 1: draw

y = 2

y = 2

(0,\,2) \text{ and } (8,\,2)

(Reason: Every point on this line has

y

-coordinate

2

, whatever

x

is, so it is a horizontal line

2

squares above the

x

-axis. Rule it right across the grid, from

(0,\,2)

(8,\,2)

Step 2: draw

x = 6

x = 6

(6,\,0) \text{ and } (6,\,8)

(Reason: Here it is the

x

-coordinate that is fixed and

y

that is free, so the line is vertical,

6

squares to the right of the

y

-axis. It runs from

(6,\,0)

up to

(6,\,8)

Step 3: draw

y = x + 1

x = 0 : \ y = 0 + 1 = 1

x = 7 : \ y = 7 + 1 = 8

x = 3 : \ y = 3 + 1 = 4

(Reason: Two points far apart give the most accurate line, so use the ends of the grid: at

x = 0

the line is at

(0,\,1)

, and it reaches the top of the grid at

(7,\,8)

. The third point

(3,\,4)

should land exactly on the ruler - if it does not, one of the three is plotted wrongly.)

Step 4: decide which side of each line to keep

y \geq 2 : \text{ on or above } y = 2

x \leq 6 : \text{ on or left of } x = 6

y \leq x + 1 : \text{ on or below } y = x + 1

(Reason: Test one point rather than guessing. Take

(6,\,2)

: it gives

2 \geq 2

(true),

6 \leq 6

(true) and

2 \leq 6 + 1

(true), so

(6,\,2)

is on the correct side of all three lines and the region is the one containing it.)

Step 5: find the corners of the region

y = 2 \text{ and } y = x + 1 : \ x + 1 = 2 \Rightarrow x = 1 \Rightarrow (1,\,2)

y = 2 \text{ and } x = 6 : \ (6,\,2)

x = 6 \text{ and } y = x + 1 : \ y = 6 + 1 = 7 \Rightarrow (6,\,7)

(Reason: A corner is a point where two of the boundaries cross, so take the lines two at a time and solve them as a pair. The three corners are what tell you the shape to shade is a triangle rather than a strip that runs off the grid.)

Step 6: shade the triangle and label it

R

\text{corners } (1,\,2) , \ (6,\,2) , \ (6,\,7)

(Reason: The three lines cut exactly one region out of the grid that is on the right side of all of them at once. Shade that triangle and write

R

inside it, so it is clear which region has been chosen.)

(a)

y = 2

horizontal through

(0,\,2)

;

x = 6

vertical through

(6,\,0)

;

y = x + 1

through

(0,\,1)

and

(7,\,8)

(b)

R

is the shaded triangle with corners

(1,\,2)

(6,\,2)

and

(6,\,7)

Verification

Check 1: Take a point well inside the shaded triangle,

(5,\,4)

, and put it into all three inequalities:

4 \geq 2

5 \leq 6

and

4 \leq 5 + 1

. All three are true, so

(5,\,4)

really does lie in the region.

Check 2: Step just across each boundary in turn.

(5,\,1)

is below

y = 2

;

(7,\,4)

is to the right of

x = 6

;

(2,\,5)

is above

y = x + 1

. Each of the three points breaks exactly one inequality and keeps the other two, so every boundary is being used, and on the correct side.

Check 3: Each corner must satisfy BOTH equations that meet there.

(1,\,2)

y = 2

and

1 + 1 = 2

(6,\,2)

y = 2

and

x = 6

(6,\,7)

x = 6

and

6 + 1 = 7

. Every corner lies on both of the lines that cross there, so the triangle drawn is the overlap of the three regions.

Check 4: Measure the triangle two ways. Its horizontal side runs from

(1,\,2)

(6,\,2)

and its vertical side from

(6,\,2)

(6,\,7)

, so

\dfrac{1}{2} \times 5 \times 5 = 12.5

square units; counting whole and part squares on the grid gives the same. Both give

12.5

square units, which is what a triangle with two sides of

5

squares at right angles must have.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) The line y = 2 drawn	B1	a horizontal line through $(0,\,2)$ and $(8,\,2)$ . It may be solid, dotted or dashed, must be at least $2$ cm long, and need not be labelled	✓
(a)(ii) The line x = 6 drawn	B1	a vertical line through $(6,\,0)$ and $(6,\,8)$ . It may be solid, dotted or dashed, must be at least $2$ cm long, and need not be labelled	✓
(a)(iii) The line y = x + 1 drawn	B1	a straight line of gradient $1$ through $(0,\,1)$ and $(7,\,8)$ . It may be solid, dotted or dashed, must be at least $2$ cm long, and need not be labelled	✓
(b) Correct region indicated	B1ft	the triangle with corners $(1,\,2)$ , $(6,\,2)$ and $(6,\,7)$ , shaded and labelled $R$ . Follow through on the candidate's own lines, dependent on at least B2 already scored in part (a) and on there being a vertical line, a horizontal line and a diagonal line with a positive gradient	✓
(b) Special case	SC B1	the lines $y = x + 1$ , $y = 6$ and $x = 2$ with the matching area shaded scores SC B1. That is the candidate who has swapped the letters over on the two simple lines, drawing $y = 6$ for $x = 6$ and $x = 2$ for $y = 2$ , and then shaded correctly for the lines drawn	✓
Note on part (a)	note	the mark scheme awards the three marks in (a) for the lines alone - they need not be labelled - although the question does ask for a label on each line	✓

Full marks: 4/4

An aircraft takes $9$ hours $36$ minutes to fly from Doha to Cape Town.
The aircraft flies at an average speed of $820$ km/h.
Work out the total distance the aircraft flies. [3 marks]

[Total 3 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

Time for the flight:

9

hours

36

minutes

Average speed:

820

km/h

Find

The total distance the aircraft flies, in kilometres.

Plan the Solution

The speed is given in kilometres per HOUR, so the time has to be in hours before anything is multiplied. $9$ hours $36$ minutes is not $9.36$ hours.
Turn the $36$ minutes into a part of an hour by dividing by $60$ , then add it to the $9$ whole hours.
Then use $\text{distance} = \text{speed} \times \text{time}$ and write the answer in kilometres.

Worked Solution [3 marks]

Speed, distance and time:

\text{distance} = \text{speed} \times \text{time}

, where the time is measured in hours because the speed is in kilometres per hour.

Step 1: Change the minutes into hours

\dfrac{36}{60} = 0.6

9 \text{ hours } 36 \text{ minutes} = 9.6 \text{ hours}

(Reason: There are

60

minutes in an hour, so

36

minutes is

\dfrac{36}{60}

of an hour, which is

0.6

of an hour. Adding that to the

9

whole hours gives

9.6

hours.)

Step 2: Multiply the speed by the time in hours

\text{distance} = 820 \times 9.6 = 7872

(Reason: Distance is speed multiplied by time. The speed counts kilometres for every hour, so multiplying by the number of hours gives the number of kilometres.)

Step 3: Write the distance in kilometres

\text{distance} = 7872 \text{ km}

(Reason: The speed was in kilometres per hour and the time was in hours, so the distance comes out in kilometres and the unit on the answer line is already km.)

7872

Verification

Check 1: Split the journey. Work out the distance for the

9

whole hours and the distance for the

36

minutes separately, then add them.

820 \times 9 = 7380

and

\dfrac{820}{60} \times 36 = 492

, and

7380 + 492 = 7872

Check 2: Work backwards. Dividing the distance by the time must give the speed back.

\dfrac{7872}{9.6} = 820

, which is the speed the question gives.

Check 3: Work in minutes instead of hours. The flight is

9 \times 60 + 36 = 576

minutes, and the aircraft flies

\dfrac{820}{60}

km each minute.

576 \times \dfrac{820}{60} = 7872

, the same distance from a method that never forms

9.6

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Convert the time into hours (or into minutes)	M1	For a correct conversion of the time into hours or into minutes, eg $9.6$ hours, $9\dfrac{36}{60}$ hours, $9\dfrac{3}{5}$ hours or $576$ minutes.	✓
Use $\text{distance} = \text{speed} \times \text{time}$ with the time in hours	M1	eg $820 \times 9.6$ , $820 \times \dfrac{576}{60}$ , $576 \times \dfrac{820}{60}$ or $576 \times \dfrac{41}{3}$ (allow $13.7$ for $\dfrac{41}{3}$ ) or equivalent. Use of $9.36$ is allowed for this mark only. Award M2 for $820 \times 9 + \dfrac{820}{60} \times 36$ $(= 7380 + 492)$ , or for $\dfrac{34\,560}{60 \times 60} \times 820$ or equivalent.	✓
Correct answer	A1	$7872$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
Special case: the time read as a decimal	SCB1	An answer of $7675.2$ earns SCB1 if no other marks are awarded. It comes from treating $9$ hours $36$ minutes as $9.36$ hours instead of $9.6$ hours, so the $36$ minutes is never divided by $60$ .	✓

Full marks: 3/3

Show that $2\dfrac{4}{7} \times 3\dfrac{1}{9} = 8$
You must show all your working. [3 marks]

[Total 3 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

The two mixed numbers

2\dfrac{4}{7}

and

3\dfrac{1}{9}

The result that has to be proved:

2\dfrac{4}{7} \times 3\dfrac{1}{9} = 8

Find

A complete piece of working that ends at

8

. The answer is printed in the question, so the marks are for the working, not for the number.

Plan the Solution

Write each mixed number as an improper fraction: multiply the whole number by the denominator, then add the numerator.
Look for common factors across the multiplication sign. The $9$ underneath cancels with the $18$ on top, and the $7$ underneath cancels with the $28$ on top.
Multiply the numerators together and the denominators together.
Simplify and state that the result is $8$ , as required.

Worked Solution [3 marks]

Rule - Mixed numbers and fractions:

n\dfrac{p}{q} = \dfrac{n \times q + p}{q}

, and

\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}

Step 1: Write each mixed number as an improper fraction

2\dfrac{4}{7} = \dfrac{2 \times 7 + 4}{7} = \dfrac{18}{7}

3\dfrac{1}{9} = \dfrac{3 \times 9 + 1}{9} = \dfrac{28}{9}

(Reason: Multiply the whole number by the denominator and then add the numerator on top. The denominator itself never changes, so the

7

and the

9

stay where they are.)

Step 2: Cancel the common factors before multiplying

\dfrac{18}{7} \times \dfrac{28}{9} = \dfrac{18 \times 28}{7 \times 9}

\dfrac{18 \times 28}{7 \times 9} = \dfrac{2 \times 4}{1 \times 1}

(Reason: A factor on top may be cancelled with a factor underneath, whichever fraction it came from. Here

9

goes into

18

exactly

2

times and

7

goes into

28

exactly

4

times, so both denominators cancel away completely.)

Step 3: Multiply the cancelled fractions

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

(Reason: The numerators multiply to

8

and the denominators multiply to

1

, and a fraction whose denominator is

1

is just its numerator. This is the required result.)

Step 4: The same answer without cancelling first

\dfrac{18}{7} \times \dfrac{28}{9} = \dfrac{504}{63} = 8

(Reason: Multiplying straight across gives

\dfrac{504}{63}

, and since

63 \times 8 = 504

that fraction is exactly

8

. Either route earns all three marks, so cancelling is a shortcut rather than a requirement.)

2\dfrac{4}{7} \times 3\dfrac{1}{9} = \dfrac{18}{7} \times \dfrac{28}{9} = 8

as required

Verification

Check 1: Work backwards. If the product really is

8

, then dividing

8

3\dfrac{1}{9}

must give back

2\dfrac{4}{7}

. Dividing by a fraction means multiplying by its reciprocal.

8 \times \dfrac{9}{28} = \dfrac{72}{28} = \dfrac{18}{7} = 2\dfrac{4}{7}

Check 2: Do it without improper fractions at all. Treat it as two brackets,

(2 + \dfrac{4}{7})(3 + \dfrac{1}{9})

, and expand, putting every term over

63

6 + \dfrac{2}{9} + \dfrac{12}{7} + \dfrac{4}{63} = \dfrac{378 + 14 + 108 + 4}{63} = \dfrac{504}{63} = 8

Check 3: A decimal check on the calculator, which is allowed on this paper:

\dfrac{18}{7} \approx 2.571428571

and

\dfrac{28}{9} \approx 3.111111111

2.571428571 \times 3.111111111 \approx 8

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Write both mixed numbers as improper fractions: $\dfrac{18}{7}$ and $\dfrac{28}{9}$ .	M1	for correct improper fractions	✓
Cancel fully, eg $\dfrac{18}{7} \times \dfrac{28}{9} = \dfrac{2 \times 4}{1 \times 1}$ , or multiply without cancelling, eg $\dfrac{18 \times 28}{7 \times 9} = \dfrac{504}{63}$ .	M1dep	for cancelling fractions fully, or for cancelling partially with a clear intention to multiply, or for not cancelling but with a clear intention to multiply. An arithmetic error in the multiplication is allowed here.	✓
Reach the printed result from fully correct working, eg $\dfrac{504}{63} = 8$ or $2 \times 4 = 8$ .	A1	Shown. Dependent on both method marks, for a correct answer from fully correct working.	✓
Alternative presentation the scheme also accepts	note	A candidate may write $8 = \dfrac{8}{1}$ underneath the printed $8$ , and then need only show that the given product comes to $\dfrac{8}{1}$ .	✓

Full marks: 3/3

Triangle $ABC$ is shown in the diagram below.
The angle at $B$ is a right angle.

Calculate the value of $x$ .
Give your answer correct to one decimal place. [3 marks]

x =

[Total 3 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

Triangle

ABC

with a right angle at

B

The angle at

A

34^{\circ}

The side

AC = 6.5

cm, which is the hypotenuse

The side

BC = x

Find

The value of

x

, correct to one decimal place

Plan the Solution

Stand at the $34^{\circ}$ angle and name the two sides that matter: $BC$ faces that angle, so it is the opposite; $AC$ faces the right angle, so it is the hypotenuse.
Opposite and hypotenuse together means the sine ratio, not the cosine and not the tangent.
Write the ratio, rearrange it to make $x$ the subject, then evaluate with the calculator in degree mode.
Round only at the very end, to one decimal place.

Worked Solution [3 marks]

Rule - Sine ratio in a right-angled triangle:

\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}

Step 1: Name the sides from the marked angle

\text{opposite} = BC = x

\text{hypotenuse} = AC = 6.5

(Reason: (Reason: the hypotenuse is always the side facing the right angle, and the opposite is the side facing the angle you are using, here the

34^{\circ}

angle at

A

.))

Step 2: Choose the ratio and write it down

\sin 34^{\circ} = \dfrac{x}{6.5}

(Reason: (Reason: the two sides in play are the opposite and the hypotenuse, and that pair belongs to sine. This single line is worth the first method mark.))

Step 3: Rearrange to make x the subject

x = 6.5 \times \sin 34^{\circ}

(Reason: (Reason: multiplying both sides by

6.5

clears the fraction and leaves

x

alone. This is the fully correct method the second mark is for.))

Step 4: Evaluate on the calculator

x = 6.5 \times \sin 34^{\circ} = 3.63475\ldots

(Reason: (Reason: the calculator must be in degree mode, or the answer will be wildly out. Keep the full display and do not round yet.))

Step 5: Round to one decimal place

x \approx 3.6

(Reason: (Reason: the digit after the first decimal place is

3

, which is below

5

, so the

6

stays as it is. The mark scheme accepts anything that rounds to

3.6

.))

x = 3.6

cm (to 1 decimal place)

Verification

Check 1: Find the third side and test Pythagoras.

AB = 6.5 \cos 34^{\circ} = 5.38874\ldots

, so

AB^{2} + BC^{2} = 29.0385\ldots + 13.2114\ldots

AB^{2} + BC^{2} = 42.25 = 6.5^{2}

, exactly the square of the hypotenuse

Check 2: Come at it from the other acute angle instead. The angles give

180 - 90 - 34 = 56

, and

BC

is adjacent to the

56^{\circ}

angle, so

x = 6.5 \times \cos 56^{\circ}

6.5 \times \cos 56^{\circ} = 3.63475\ldots

, the same value, reached without using sine at all

Check 3: A size check. The hypotenuse is the longest side of any right-angled triangle, so

x

must be smaller than

6.5

. Also

34^{\circ}

is less than

45^{\circ}

, so the side facing it must be the shorter of the two legs.

3.6 < 5.4 < 6.5

, so the answer sits where it should

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct trig statement for $x$	M1	$\sin 34 = \dfrac{x}{6.5}$ or $\dfrac{x}{\sin 34} = \dfrac{6.5}{\sin 90}$ or $6.5^{2} - (6.5 \times \cos 34)^{2}$ or $\cos 56 = \dfrac{x}{6.5}$ oe	✓
A fully correct method to find $x$	M1	$(x =) \; 6.5 \times \sin 34$ or $x = \dfrac{6.5 \times \sin 34}{\sin 90}$ or $(x =) \sqrt{6.5^{2} - (6.5 \times \cos 34)^{2}}$ or $(x =) \; 6.5 \times \cos 56$ oe	✓
The value of x	A1	awrt $3.6$	✓
Note	note	A correct answer scores full marks unless it comes from obvious incorrect working.	✓

Full marks: 3/3

A moving walkway at an airport carries passengers at a steady speed of $w$ metres per second.
Change this speed to a speed in kilometres per hour.
Give your answer in terms of $w$ in its simplest form. [3 marks]

kilometres per hour

[Total 3 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

A speed of

w

metres per second.

There are

60

seconds in a minute and

60

minutes in an hour.

There are

1000

metres in a kilometre.

Find

The same speed written in kilometres per hour, in terms of

w

, as simply as possible.

Plan the Solution

A speed in kilometres per hour is a distance in kilometres travelled in one hour, so start by finding how far the walkway carries you in $1$ hour.
Turn that distance from metres into kilometres by dividing by $1000$ .
Tidy the number that is multiplying $w$ , because the question asks for the simplest form.

Worked Solution [3 marks]

Rule - to change metres per second into kilometres per hour, multiply by

3600

to turn seconds into hours and divide by

1000

to turn metres into kilometres.

Step 1: how far does it travel in one hour?

60 \times 60 = 3600

3600 \times w = 3600w

(Reason: (Reason: one hour is

60

minutes and each minute is

60

seconds, so one hour is

3600

seconds. The walkway covers

w

metres in every one of them, so in an hour it covers

3600w

metres.))

Step 2: change those metres into kilometres

3600w \text{ metres}

\dfrac{3600w}{1000} \text{ kilometres}

(Reason: (Reason: a kilometre is

1000

metres, so a distance in metres is divided by

1000

to give the same distance in kilometres. The speed is now a number of kilometres in one hour, which is exactly what kilometres per hour means.))

Step 3: simplify the number in front of w

\dfrac{3600}{1000} = 3.6

\dfrac{3600w}{1000} = 3.6w

(Reason: (Reason: the

3600

and the

1000

are just numbers, so they can be worked out on their own. Dividing

3600

1000

gives the multiplier that turns any speed in metres per second into kilometres per hour. In fraction form it is

\dfrac{18}{5}

, which is the same number.))

3.6w

kilometres per hour

Verification

Check 1 - put a number in: Take

w = 5

, a speed of

5

metres per second. In one hour that is

5 \times 3600 = 18000

metres, and

\dfrac{18000}{1000} = 18

kilometres. The formula gives

3.6 \times 5 = 18

kilometres per hour, the same speed

Check 2 - convert back the other way: A speed of

3.6

kilometres per hour is

3600

metres travelled in

3600

seconds, so it should come back to

1

metre per second.

\dfrac{3600}{3600} = 1

, so the conversion undoes itself

Check 3 - is the size sensible? Kilometres are long and hours are long, but an hour is

3600

times a second while a kilometre is only

1000

times a metre, so the number counting kilometres per hour must come out bigger than the number counting metres per second. The multiplier

3.6

is greater than

1

, so the speed in kilometres per hour is the larger number, as expected

Mark Scheme Breakdown

Step	Mark	Description	Got it?
One correct conversion step	M1	For any one of $\dfrac{w}{1000}$ , $w \times 10^{-3}$ , $0.001w$ , $w \times 60 \times 60$ or $w \times 3600$ oe. Also award this mark for $\dfrac{3600}{1000}$ , $\dfrac{18}{5}$ or $3.6$ oe written on its own, without a link to $w$ .	✓
A fully correct method, including w	M1	For $\dfrac{w \times 60 \times 60}{1000}$ oe, for example $w \times \dfrac{3600}{1000}$ . Both conversions must be present and applied to $w$ .	✓
The simplified answer	A1	For $3.6w$ . Accept $\dfrac{18}{5}w$ or $3\dfrac{3}{5}w$ ; allow $3.6 \times w$ .	✓
Note	note	A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

The diagram shows a hexagon $ABCDEF$

$AF = 21$ cm, $CD = 15$ cm, $AB = FE = 13$ cm

$CD$ is parallel to $AF$
The perpendicular height of the hexagon is $h$ cm

The area of the hexagon is $390$ cm²

Work out the value of $h$ [4 marks]

h =

[Total 4 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

A hexagon

ABCDEF

with

AF = 21

cm,

CD = 15

cm and

AB = FE = 13

CD

is parallel to

AF

, and the angles at

A

and at

F

are right angles

The area of the whole hexagon is

390

cm², and its perpendicular height is

h

Find

The value of

h

, the perpendicular height of the hexagon in cm

Plan the Solution

Cut the hexagon along $BE$ into two familiar shapes: a rectangle underneath and a trapezium on top.
Find the rectangle's area from the numbers given, then subtract it from $390$ to leave the trapezium's area.
Write the trapezium's area with the trapezium formula. Its parallel sides are known, so the only unknown left is its height, $h - 13$ .
Solve that equation, then add the $13$ cm of the rectangle back on to reach the full height $h$ .

Worked Solution [4 marks]

Rule - Composite shape: the total area is the sum of its parts, and a trapezium with parallel sides

a

and

b

a distance

d

apart has area

\dfrac{1}{2}(a + b)d

Step 1: Split the hexagon and find the rectangle underneath

BE = AF = 21 \text{ cm}

\text{Area of } ABEF = 21 \times 13 = 273 \text{ cm}^2

(Reason: The angles at

A

and

F

are right angles and

AB = FE

, so

ABEF

is a rectangle. That makes

BE

the same length as

AF

, and a rectangle's area is its length times its width.)

Step 2: Find the area left over for the trapezium

\text{Area of } BCDE = 390 - 273 = 117 \text{ cm}^2

(Reason: The rectangle and the trapezium together make the whole hexagon, so whatever is left of the

390

cm² after the rectangle belongs to the trapezium

BCDE

Step 3: Write the trapezium's area in terms of h

\dfrac{1}{2} \times (21 + 15) \times (h - 13) = 117

18(h - 13) = 117

(Reason: The trapezium's parallel sides are

BE = 21

cm and

CD = 15

cm. It sits on top of the

13

cm rectangle, so its own height is

h - 13

cm, and

\dfrac{1}{2} \times (21 + 15) = 18

Step 4: Solve for h

h - 13 = \dfrac{117}{18} = 6.5

h = 6.5 + 13 = 19.5

(Reason: Dividing both sides by

18

gives the trapezium's height,

6.5

cm. That is only the part above

BE

, so the

13

cm of the rectangle is added back to give the height of the whole hexagon.)

h = 19.5

Verification

Check 1: Put

h = 19.5

back into the original split. The trapezium's height is

19.5 - 13 = 6.5

cm, so its area is

18 \times 6.5 = 117

cm², and the rectangle is

273

cm².

273 + 117 = 390

cm², which is the area the question gives

Check 2: Cut the shape a different way. Start from the whole

21

cm by

19.5

cm rectangle that encloses it and take off the two corner triangles, each with base

\dfrac{21 - 15}{2} = 3

cm and height

6.5

cm, so

2 \times \dfrac{1}{2} \times 3 \times 6.5 = 19.5

cm².

409.5 - 19.5 = 390

cm², the same area from a completely different split

Check 3: Expand instead of dividing. From

273 + 18(h - 13) = 390

the left side becomes

18h - 234 + 273 = 18h + 39

, which is one of the equations the mark scheme lists.

18h + 39 = 390

, so

18h = 351

and

h = 19.5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct area for one part of the shape, eg the rectangle $ABEF$ : $13 \times 21 = 273$	M1	A correct calculation for an area linked to the shape. $h - 13$ may be written as $x$ or $y$ , and even as $h$ ; brackets may be missing for this mark only.	✓
$390 - 273 = 117$ , or $13 \times 21$ together with $\dfrac{1}{2}(15 + 21)(h - 13)$	M1	For considering the area of all parts of the shape. The parts need not be added or subtracted, but the brackets must be used correctly.	✓
$\dfrac{117}{0.5 \times (15 + 21)} = 6.5$ , or the equation $273 + 18(h - 13) = 390$	M1	A correct calculation for the height of the trapezium or for the height of the shape, or a correct equation involving either. Typical equations simplify to $18y = 117$ , $18h - 234 = 117$ or $18h + 39 = 390$ .	✓
$h = 19.5$	A1	oe, eg $\dfrac{39}{2}$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 4/4

Rohan buys $600$ balloons for a school fair.
He buys round balloons, heart balloons and star balloons so that

$\text{number of round balloons} : \text{number of heart balloons} : \text{number of star balloons} = 9 : 4 : 2$

$45\%$ of the round balloons are red.

$\dfrac{5}{8}$ of the heart balloons are red.

All of the star balloons are red.

Work out the number of balloons that are red. [5 marks]

[Total 5 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

A total of

600

balloons.

Round : heart : star

= 9 : 4 : 2

45\%

of the round balloons are red.

\dfrac{5}{8}

of the heart balloons are red.

Every one of the star balloons is red.

Find

The total number of balloons that are red.

Plan the Solution

Add the ratio parts to see how many equal parts the $600$ balloons are split into.
Divide to find the size of one part, then multiply to find how many balloons there are of each shape.
Take the red fraction of each shape separately - a percentage for the round, a fraction for the heart, all of the star.
Add the three red amounts to get the total.

Worked Solution [5 marks]

Rule - Sharing in a ratio: one part is

\dfrac{\text{total amount}}{\text{sum of the ratio parts}}

, and each share is that one part multiplied by its own number of parts.

Step 1: Add the ratio parts

9 + 4 + 2 = 15

(Reason: (Reason: the ratio

9 : 4 : 2

splits the whole order into

15

equal parts, so the parts must be added before anything can be shared.))

Step 2: Work out one part

\dfrac{600}{15} = 40

(Reason: (Reason: the

600

balloons are shared equally between the

15

parts, so one part is

40

balloons.))

Step 3: Work out how many of each shape

\text{round} = 9 \times 40 = 360

\text{heart} = 4 \times 40 = 160

\text{star} = 2 \times 40 = 80

(Reason: (Reason: each shape gets its own number of parts. As a check,

360 + 160 + 80 = 600

, which is the whole order.))

Step 4: Work out how many of each shape are red

0.45 \times 360 = 162

\dfrac{5}{8} \times 160 = 100

\text{star: all } 80

(Reason: (Reason:

45\%

is the same as

0.45

, so multiply the round balloons by

0.45

; multiply the heart balloons by their red fraction; and every one of the

80

star balloons is red, so all of them count.))

Step 5: Add the three red amounts

162 + 100 + 80 = 342

(Reason: (Reason: the red balloons are the red round ones, the red heart ones and the star ones together - no balloon is counted twice, because each balloon has only one shape.))

342

balloons are red

Verification

Check 1: Work out what fraction of the whole order is red in one go:

0.45 \times \dfrac{9}{15} = \dfrac{27}{100}

for the round,

\dfrac{5}{8} \times \dfrac{4}{15} = \dfrac{1}{6}

for the heart, and

\dfrac{2}{15}

for the star.

\left(\dfrac{27}{100} + \dfrac{1}{6} + \dfrac{2}{15}\right) \times 600 = \dfrac{57}{100} \times 600 = 342

Check 2: Count the balloons that are NOT red instead:

0.55 \times 360 = 198

round and

\dfrac{3}{8} \times 160 = 60

heart, with none of the star balloons left out.

198 + 60 + 0 = 258

are not red, and

600 - 258 = 342

are red.

Check 3: Work in ratio parts rather than balloons:

0.45 \times 9 = 4.05

parts of round,

\dfrac{5}{8} \times 4 = 2.5

parts of heart and

2

parts of star are red, out of

15

parts.

4.05 + 2.5 + 2 = 8.55

, and

\dfrac{8.55}{15} \times 600 = 342

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct method to find one share	M1	e.g. $\dfrac{600}{9 + 4 + 2} = 40$ , or any one of $360$ , $160$ or $80$ seen. Also allow $45\%$ of $600$ , or $\dfrac{5}{8}$ of $600$ , or the fraction of the share that is round balloons.	✓
A correct method to find the number of star balloons	M1	e.g. $2 \times 40 = 80$ or $\dfrac{2}{15} \times 600 = 80$ , or the fraction of the share that is heart balloons. This implies the first M1.	✓
A correct method to find the number of red round balloons	M1	e.g. $0.45 \times (9 \times 40) = 162$ or $0.45 \times 600 \times \dfrac{9}{15} = 162$ , or the total of the parts that are red, $4.05 + 2.5 + 2 = 8.55$ . This implies the first M1.	✓
A correct method to find the number of red heart balloons	M1	e.g. $\dfrac{5}{8} \times (4 \times 40) = 100$ or $\dfrac{5}{8} \times 600 \times \dfrac{4}{15} = 100$ , or multiplying the total of the correct red parts by $600$ , e.g. $\dfrac{8.55}{9 + 4 + 2} \times 600$ , which implies all previous M marks.	✓
Correct answer	A1	$342$ cao. A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 5/5

Lorenzo invests $4500$ koruna in a savings bond for $4$ years.
The bond pays $2.4\%$ per year compound interest.

Work out how much money Lorenzo will have in the savings bond at the end of $4$ years.
Give your answer correct to the nearest koruna. [3 marks]

koruna

[Total 3 marks]

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Question 25 - Exam Solution

Understanding the Question

Given

Amount invested:

4500

koruna

Time in the bond:

4

years

Compound interest at

2.4\%

per year - the interest is added to the bond, so the next year's interest is worked out on a bigger amount

Find

The value of the bond at the end of

4

years, to the nearest koruna

Plan the Solution

Turn the yearly increase into a single multiplier: $100\% + 2.4\% = 102.4\%$ , which is $1.024$ .
Compound interest applies that multiplier once for every year, so $4$ years means the multiplier is used $4$ times - that is $1.024^{4}$ .
Multiply the $4500$ koruna by that power, then round the result to the nearest whole koruna.

Worked Solution [3 marks]

Rule - Compound growth:

A = P \times \left(1 + \dfrac{r}{100}\right)^{n}

, where

P

is the starting amount,

r

is the percentage rate per year and

n

is the number of years.

Step 1: Write the yearly increase as one multiplier

100\% + 2.4\% = 102.4\%

102.4\% = \dfrac{102.4}{100} = 1.024

(Reason: Adding

2.4\%

to an amount leaves you with

102.4\%

of it, so one multiplication by

1.024

does a whole year in a single step.)

Step 2: Raise the multiplier to the power of the number of years

1.024^{4} = 1.099511627776

(Reason: Interest is added at the end of each of the

4

years, and every time the running total is multiplied by

1.024

, so the four multiplications collapse into one power.)

Step 3: Multiply the amount invested by that power

4500 \times 1.099511627776 = 4947.802324992

(Reason: The starting amount is scaled by the same factor every year, so multiplying the

4500

koruna by the power gives the value of the bond at the end.)

Step 4: Round to the nearest koruna

4947.802324992 \approx 4948

(Reason: The first digit after the decimal point is

8

, which is

5

or more, so the whole-number part rounds up from

4947

4948

4948

koruna

Verification

Check 1: Build the total up one year at a time instead of using a power: multiply by

1.024

four separate times, starting from

4500

. The year-end totals are

4608

4718.592

4831.838208

and

4947.802324992

- the same final amount, so the power method agrees with the year-by-year method.

Check 2: Work backwards. Dividing the final amount by the multiplier four times must return the amount that was invested.

\dfrac{4947.802324992}{1.024^{4}} = 4500

, which is exactly what Lorenzo put in.

Check 3: Compare with simple interest, which would pay

4500 \times 0.024 \times 4 = 432

koruna. Compound interest must pay a little more, because later years earn interest on the earlier interest. The interest here is

4947.802324992 - 4500 = 447.802324992

koruna, which is more than

432

but not by much over

4

years - exactly the size of gap to expect.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$4500 \times 1.024$ $(= 4608)$ or equivalent, or $4500 \times 0.024$ $(= 108)$	M1	One year of compound growth started correctly - either the multiplier method, or working out one year's interest of $108$ koruna.	✓
$4608 \times 1.024$ $(= 4718.592)$ and $4718.592 \times 1.024$ $(= 4831.838208)$ and $4831.838208 \times 1.024$ $(= 4947.802324992)$	M1	The growth carried through all four years, each year multiplying the previous total. The mark scheme prints these figures in quotation marks, which means the candidate's own running totals are followed through.	✓
$4948$	A1	Accept anything from $4947$ to $4948$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
Single-step alternative: $4500 \times 1.024^{4}$	M2	Doing all four years in one power earns both method marks together, with no year-by-year working needed.	✓
If no other mark has been awarded	SCB1	Award $1$ mark for any of $4500 \times 0.024 \times 4 = 432$ or $0.096 \times 4500 = 432$ (simple interest instead of compound), $4500 + 4500 \times 0.024 \times 4 = 4932$ or $4500 \times 1.096 = 4932$ (the simple-interest total), $0.976 \times 4500 = 4392$ (a decrease of $2.4\%$ for one year), $0.904 \times 4500 = 4068$ (a simple decrease over $4$ years), $0.976^{4} \times 4500 = 4083.304660992$ (a compound decrease), or $4500 \times 1.024^{3} = 4831.838208$ (stopping a year early).	✓

Full marks: 3/3

Solve the simultaneous equations

$6x + 4y = 1$
$3x + 5y = 8$

You must show clear algebraic working. [3 marks]

x =y =

[Total 3 marks]

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Question 26 - Exam Solution

Understanding the Question

Given

Equation 1:

6x + 4y = 1

Equation 2:

3x + 5y = 8

Two linear equations in the same two unknowns, so exactly one pair of values fits both.

Find

The value of

x

and the value of

y

that satisfy both equations at the same time. Algebraic working is demanded, so a pair found by trial and improvement earns nothing.

Plan the Solution

The coefficient of $x$ is $6$ in equation 1 and $3$ in equation 2, so doubling equation 2 makes both of them $6x$ .
Subtract the two equations to remove $x$ and leave one equation in $y$ alone.
Put that value of $y$ back into one of the original equations to get $x$ .
Test the pair in both original equations, and repeat the solve by eliminating $y$ instead, as an independent check.

Worked Solution [3 marks]

Rule - Elimination: multiply one or both equations until one letter has the same coefficient in both, then subtract when those coefficients have the same sign, or add when they have opposite signs, to remove that letter.

Step 1: Match the

x

terms

3x + 5y = 8

2 \times (3x + 5y) = 2 \times 8

6x + 10y = 16

(Reason: Multiplying every term of equation 2 by

2

turns its

3x

into

6x

, matching equation 1. The right-hand side must be doubled as well, so

8

becomes

16

Step 2: Subtract to eliminate

x

(6x + 10y) - (6x + 4y) = 16 - 1

6y = 15

y = \dfrac{15}{6} = \dfrac{5}{2} = 2.5

(Reason: Both equations now begin with

6x

, so subtracting removes

x

completely. That leaves

10y - 4y = 6y

on the left and

16 - 1 = 15

on the right, and dividing by

6

gives

y

Step 3: Substitute back to find

x

6x + 4 \times 2.5 = 1

6x + 10 = 1

6x = 1 - 10 = -9

x = \dfrac{-9}{6} = -\dfrac{3}{2} = -1.5

(Reason: Putting

y = 2.5

into the first original equation leaves

x

as the only unknown. Take the

10

across to the right, then divide by

6

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

y = \dfrac{5}{2} = 2.5

Verification

Check 1: Put

x = -1.5

and

y = 2.5

into the left-hand side of equation 1,

6x + 4y

6 \times (-1.5) + 4 \times 2.5 = -9 + 10 = 1

, which is the right-hand side of equation 1.

Check 2: Put the same pair into the left-hand side of equation 2,

3x + 5y

. A pair that only fits one equation is not a solution.

3 \times (-1.5) + 5 \times 2.5 = -4.5 + 12.5 = 8

, which is the right-hand side of equation 2.

Check 3: Solve it again the other way round, eliminating

y

first:

5

times equation 1 gives

30x + 20y = 5

, and

4

times equation 2 gives

12x + 20y = 32

. Subtract.

18x = 5 - 32 = -27

, so

x = \dfrac{-27}{18} = -1.5

, the same value reached by a different elimination.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Multiply equation 2 by $2$ to give $6x + 10y = 16$ , then subtract equation 1 from it.	M1	A correct method to eliminate $x$ or $y$ : multiplying one or both equations so that one value can be eliminated, and the correct operation to eliminate it, which can be shown by $2$ out of $3$ terms correct for subtraction or addition (allow one arithmetic error in multiplying). Or a correct substitution of one variable into the other equation. Note: the mark is for the method and not for the result, although a correct result also earns it.	✓
Substitute the letter already found back into either original equation to reach the other letter.	M1	A correct method to calculate the value of the other letter: substitution of the found variable into an equation (the equation does not need to be solved), or starting again with elimination or substitution.	✓
Both values stated, with algebraic working shown.	A1	Working required. $x = -1.5$ and $y = 2.5$ , or equivalent: the answer must be a vulgar fraction, a mixed number or a decimal, so an unsimplified $y = \dfrac{12.5}{5}$ is not accepted.	✓

Full marks: 3/3

(i) Write $x^2 + 9x - 22$ as a product of two brackets. [2 marks]

(ii) Hence solve the equation $x^2 + 9x - 22 = 0$ [1 mark]

(i)(ii)

[Total 3 marks]

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Question 27 - Exam Solution

Understanding the Question

Given

The expression

x^2 + 9x - 22

The equation

x^2 + 9x - 22 = 0

, which is the same expression set equal to zero

A quadratic with no coefficient in front of

x^2

, so it factorises into two simple brackets if it factorises at all

Find

(i) The two brackets whose product is the given expression (ii) The values of

x

that make the expression zero. A quadratic can have two, so expect two.

Plan the Solution

Compare with $x^2 + bx + c$ : here $b = 9$ and $c = -22$ .
Hunt for two numbers whose product is $-22$ and whose sum is $9$ . Because the product is negative, one number is positive and the other is negative.
List the factor pairs of $-22$ and test each one on the sum. Every pair passes the product test, so the sum is what decides.
Drop the winning pair straight into $(x + p)(x + q)$ .
For part (ii), the word "hence" means use those brackets: a product is zero only when one of its factors is zero, so set each bracket to $0$ in turn.

Worked Solution [3 marks]

Rule - Factorising

x^2 + bx + c

: find two numbers

p

and

q

with

pq = c

and

p + q = b

. Then

x^2 + bx + c = (x + p)(x + q)

Step 1: Read off the product and the sum

x^2 + 9x - 22 \text{ gives } b = 9 \text{ and } c = -22

(Reason: The two numbers in the brackets must multiply to give

c = -22

and add to give

b = 9

Step 2: Test the factor pairs of

-22

1 \times (-22) = -22 \text{ and } 1 + (-22) = -21

2 \times (-11) = -22 \text{ and } 2 + (-11) = -9

(-2) \times 11 = -22 \text{ and } (-2) + 11 = 9

(Reason: Every pair here multiplies to

-22

, so the product test cannot separate them. Only

-2

and

11

also add to

9

, so that is the pair to use.)

Step 3: Write the two brackets

x^2 + 9x - 22 = (x + (-2))(x + 11)

x^2 + 9x - 22 = (x - 2)(x + 11)

(Reason: The winning pair goes straight into

(x + p)(x + q)

, and

x + (-2)

is tidied to

x - 2

Step 4: Use the factorisation on the equation

(x - 2)(x + 11) = 0

x - 2 = 0 \text{ or } x + 11 = 0

(Reason: Two numbers multiply to zero only when at least one of them is zero, so each bracket is taken in turn.)

Step 5: Solve each bracket

x = 2 \text{ or } x = -11

(Reason: Add

2

to both sides of the first bracket, and subtract

11

from both sides of the second.)

(i)

(x - 2)(x + 11)

(ii)

x = 2

x = -11

Verification

Check 1: Expand the brackets again and collect the two

x

terms.

(x - 2)(x + 11) = x^2 + 11x - 2x - 22 = x^2 + 9x - 22

Check 2: Put

x = 2

back into

x^2 + 9x - 22

. Squaring the

2

gives

4

4 + 18 - 22 = 0

Check 3: Put

x = -11

into the same expression. Squaring the

-11

gives

121

, and

9 \times (-11)

-99

121 - 99 - 22 = 0

Check 4: The two solutions of

x^2 + bx + c = 0

must add to

-b

and multiply to

c

, which tests both answers at once without expanding anything.

2 - 11 = -9 \text{ and } 2 \times (-11) = -22

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(i) Brackets of the form $(x \pm 2)(x \pm 11)$	M1	Or $(x + a)(x + b)$ where $ab = -22$ or $a + b = 9$ . The method mark is for finding the right pair of numbers, even if the signs are not yet settled.	✓
(i) $(x - 2)(x + 11)$	A1	Correct answer scores full marks, unless it comes from obvious incorrect working.	✓
(ii) $2, -11$	B1ft	Must follow through from the candidate's own factors in part (i): a candidate who wrote $(x + 2)(x - 11)$ in part (i) earns this mark for $-2, 11$ .	✓

Full marks: 3/3

Bilal uses a smartwatch to count the number of steps he walks each day for $7$ days.

For the first $4$ days, his mean number of steps is $11\,800$
For the next $3$ days, his mean number of steps is $13\,207$

Work out his mean number of steps for the $7$ days. [3 marks]

[Total 3 marks]

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Question 28 - Exam Solution

Understanding the Question

Given

A step count for each of

7

days, split into a block of

4

days and a block of

3

days

Mean for the first

4

days

= 11\,800

Mean for the next

3

days

= 13\,207

Find

The mean number of steps for all

7

days Note: this is a weighted mean, so it is not the halfway value

\dfrac{11\,800 + 13\,207}{2}

Plan the Solution

A mean on its own cannot be added to another mean. Turn each mean back into a total first.
Multiply each block's mean by the number of days in that block to get the steps in that block.
Add the two block totals to get the steps for the whole $7$ days.
Divide that grand total by $7$ to get the mean for the $7$ days.

Worked Solution [3 marks]

Mean:

\text{mean} = \dfrac{\text{total}}{\text{number of values}}

, which rearranges to

\text{total} = \text{mean} \times \text{number of values}

. Totals can be added; means cannot.

Step 1: turn the first mean back into a total

4 \times 11\,800 = 47\,200

(Reason: A mean of

11\,800

over

4

days means those four days add up to

4 \times 11\,800

steps.)

Step 2: turn the second mean back into a total

3 \times 13\,207 = 39\,621

(Reason: The same rearrangement for the shorter block:

3

days at a mean of

13\,207

steps come to

3 \times 13\,207

steps altogether.)

Step 3: add the two block totals

47\,200 + 39\,621 = 86\,821

(Reason: Totals may be added because every one of the

7

days is counted exactly once:

4

days from the first block and

3

from the second.)

Step 4: divide the grand total by the number of days

\dfrac{86\,821}{7} = 12\,403

(Reason: Back to the mean itself:

86\,821

steps shared over

7

days. The division is exact, so the mean is a whole number of steps.)

12\,403

steps

Verification

Check 1: Work backwards. If the mean for the

7

days is

12\,403

, the seven days must add up to

7 \times 12\,403

, and that must be the grand total from Step 3.

7 \times 12\,403 = 86\,821

, which is

47\,200 + 39\,621

Check 2: Reach the answer a different way. Start at the lower mean and add the share of the gap that belongs to the

3

busier days: the gap between the two means is

13\,207 - 11\,800 = 1407

11\,800 + \dfrac{3}{7} \times 1407 = 11\,800 + 603 = 12\,403

Check 3: Sense check the position of the answer. It must lie between

11\,800

and

13\,207

, and nearer the first because more days are in that block. The two gaps should be in the reverse ratio of the day counts,

3 : 4

12\,403 - 11\,800 = 603

and

13\,207 - 12\,403 = 804

, and

603 : 804 = 3 : 4

Mark Scheme Breakdown

Step	Mark	Description	Got it?
One correct block total, or the total for all $7$ days	M1	$4 \times 11\,800 \ (= 47\,200)$ or $3 \times 13\,207 \ (= 39\,621)$ or $86\,821$	✓
A fully correct method for the mean of the $7$ days	M1	$\dfrac{47\,200 + 39\,621}{7}$ , that is $\dfrac{86\,821}{7}$ . Their two block totals may be followed through here.	✓
The mean for the $7$ days	A1	$12\,403$ . A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 3/3

Keep revising

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

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Edexcel IGCSE 4MA1 Paper 2F, June 2024: Worked Solutions, Questions 15 to 28

Worked solutions, questions 15 to 28 of 28

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