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Edexcel IGCSE 4MA1/2F, Wednesday 4 June 2025: Worked Solutions and Mark Schemes

Sir Faraz Hassan

25 Aug 2026

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

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/2F (Foundation Tier), June 2025 series, sat Wednesday 4 June 2025 –100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

Both are PDF files hosted by Pearson: official question paper (PDF) and official mark scheme (PDF).

Try each question yourself first, then open the worked solution to check your method and see exactly where each method mark (M1) and accuracy mark (A1) is earned. The questions follow the same order as the original paper and carry the same marks.

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Every question with a full worked solution and mark scheme - free PDF

Worked solutions, questions 1 to 17 of 28

Mei is finding out the areas, in square kilometres, of five countries in South America.
The table gives this information.

$\begin{array}{|l|c|}\hline \textbf{Country} & \textbf{Area (square kilometres)} \\ \hline \text{Paraguay} & 406\,752 \\ \hline \text{Suriname} & 163\,800 \\ \hline \text{Uruguay} & 173\,626 \\ \hline \text{Venezuela} & 926\,690 \\ \hline \text{Chile} & 756\,100 \\ \hline \end{array}$

(a) Which of these countries has the largest area? [1 mark]

(b) Write the number $163\,800$ in words. [1 mark]

(c) Circle the word that describes the value of the $7$ in $406\,752$
$7 \text{ tenths} \qquad 7 \text{ tens} \qquad 7 \text{ hundreds} \qquad 7 \text{ hundredths} \qquad 7 \text{ thousandths}$ [1 mark]

(d) Write the number $173\,626$ correct to the nearest thousand. [1 mark]

(e) Work out the difference between the area of Chile and the area of Uruguay. [1 mark]

(a)(b)(d)(e) square kilometres

[Total 5 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

Paraguay,

406\,752

square kilometres

Suriname,

163\,800

square kilometres

Uruguay,

173\,626

square kilometres

Venezuela,

926\,690

square kilometres

Chile,

756\,100

square kilometres

Every area is a whole number with six digits, so the same place-value columns line up across all five.

Find

(a) the country with the largest area (b)

163\,800

written out in words (c) what the

7

406\,752

is worth (d)

173\,626

to the nearest thousand (e) the difference between the areas of Chile and Uruguay

Plan the Solution

Compare the areas from the left. The first column where they differ settles which is largest.
Split $163\,800$ at the thousands, say each piece, then join them.
Count the columns from the right in $406\,752$ until you reach the $7$ .
For the nearest thousand, find the two multiples of $1\,000$ the number sits between and see which one it is closer to.
Difference means subtract, so take the smaller area away from the larger one.

Worked Solution [5 marks]

Rule - Place value: reading a whole number from the right, the columns are units, tens, hundreds, thousands, ten thousands and hundred thousands. Line two numbers up by those columns and you can compare them, round them and subtract them.

Step 1: (a) compare the five areas

926\,690 > 756\,100 > 406\,752 > 173\,626 > 163\,800

(Reason: All five areas have six digits, so the hundred thousands column decides it straight away:

9

beats

7

4

1

and

1

. Nothing below that column can change the order.)

Step 2: (b) split the number at the thousands

163\,800 = 163\,000 + 800

(Reason: The digits above the thousands column read

163

, so that part is one hundred and sixty three thousand. What is left is

800

, which is eight hundred.)

Step 3: (c) write the number in expanded form

406\,752 = 400\,000 + 6\,000 + 700 + 50 + 2

7 \times 100 = 700

(Reason: Counting from the right, the

2

is units, the

5

is tens and the

7

is the third column, the hundreds. So the

7

stands for seven hundreds.)

Step 4: (d) find the nearest multiple of one thousand

173\,000 < 173\,626 < 174\,000

173\,626 - 173\,000 = 626

174\,000 - 173\,626 = 374

(Reason: The number sits between those two multiples of

1\,000

. It is

374

from the one above and

626

from the one below, so the one above is nearer.)

Step 5: (e) subtract the smaller area from the larger

756\,100 - 173\,626 = 582\,474

(Reason: Chile has the larger area, so Chile goes on top. Take the area of Uruguay away from it, lining the two numbers up by their place-value columns.)

(a) Venezuela(b) one hundred and sixty three thousand and eight hundred(c)

7

hundreds(d)

174\,000

(e)

582\,474

square kilometres

Verification

Check 1: Read only the hundred thousands column of the five areas:

4

1

1

9

7

. The

9

belongs to Venezuela, and no other area even reaches nine hundred thousand square kilometres.

Check 2: Count the columns of

406\,752

from the right instead of from the left:

2

units,

5

tens, then the

7

. The

7

lands in the third column, which is the hundreds column, worth

700

Check 3: Test it against the halfway point instead of the two gaps. Halfway between

173\,000

and

174\,000

173\,500

173\,626

is above the halfway point, so it rounds up to

174\,000

Check 4: Reverse the subtraction. Add the difference back on to the area of Uruguay as the table gives it, and the area of Chile should come back.

173\,626 + 582\,474 = 756\,100

, which is exactly Chile's area.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Venezuela	B1	Allow incorrect spelling if the meaning is clear. Allow V. Accept the numerical answer $926\,690$ , and accept it written with a comma separating the thousands from the hundreds.	✓
(b) one hundred (and) sixty three thousand (and) eight hundred	B1	Allow incorrect spelling if the meaning is clear, eg hunder or hudred for hundred, sixety or sisty for sixty etc.	✓
(c) (7) hundreds circled	B1	(7) hundreds clearly indicated with no other words indicated.	✓
(d) $174\,000$	B1	Allow 'one hundred and seventy-four thousand' or $174$ thousand.	✓
(e) $582\,474$	B1	The printed scheme carries no notes beside this row.	✓

Full marks: 5/5

Nadia asks $20$ adults which garden bird they most like to see.
Here are her results.
$\begin{array}{lllll} \text{robin} & \text{blackbird} & \text{sparrow} & \text{wren} & \text{robin} \\ \text{blackbird} & \text{magpie} & \text{robin} & \text{sparrow} & \text{blackbird} \\ \text{wren} & \text{robin} & \text{blackbird} & \text{robin} & \text{sparrow} \\ \text{magpie} & \text{blackbird} & \text{wren} & \text{robin} & \text{sparrow} \end{array}$
(a) Complete the frequency table to show this information.
$\begin{array}{|l|c|c|}\hline \textbf{Bird} & \textbf{Tally} & \textbf{Frequency} \\ \hline \text{robin} & \phantom{000000} & \phantom{000} \\ \hline \text{sparrow} & \phantom{000000} & \phantom{000} \\ \hline \text{wren} & \phantom{000000} & \phantom{000} \\ \hline \text{blackbird} & \phantom{000000} & \phantom{000} \\ \hline \text{magpie} & \phantom{000000} & \phantom{000} \\ \hline \end{array}$
[2 marks]
(b) Complete the bar chart for the information in your table. [3 marks]

[Total 5 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

A list of

20

results, one for each adult Nadia asked.

Five birds appear in the list: robin, sparrow, wren, blackbird and magpie.

An empty frequency table, and a grid whose vertical axis is numbered from

0

8

in ones.

Find

(a) The frequency of each bird, written into the table. (b) A bar chart of those frequencies, with every bar labelled.

Plan the Solution

Go along the list in the order it is written and put one tally mark against a bird each time it appears.
Total each row of tally marks. That total is the frequency.
Add the five frequencies. They must come to $20$ , because every adult is counted exactly once.
Draw a bar for each bird up to its frequency, keeping the widths equal and the gaps equal, and write the bird's name under its own bar.

Worked Solution [5 marks]

Rule - Frequency: a frequency is a count of how many times something appears, so the frequencies must add up to the number of results. On a bar chart the height of a bar is that frequency.

Step 1: Tally the results, one mark at a time

\begin{array}{|l|c|}\hline \textbf{Bird} & \textbf{Tally} \\ \hline \text{robin} & \vert\,\vert\,\vert\,\vert\,\vert\quad \vert \\ \hline \text{sparrow} & \vert\,\vert\,\vert\,\vert \\ \hline \text{wren} & \vert\,\vert\,\vert \\ \hline \text{blackbird} & \vert\,\vert\,\vert\,\vert\,\vert \\ \hline \text{magpie} & \vert\,\vert \\ \hline \end{array}

(Reason: Working through the list in the order it is written means no result is missed and none is counted twice. Grouping the marks in fives makes them quick to total.)

Step 2: Total each row of tally marks

\begin{array}{|l|c|c|}\hline \textbf{Bird} & \textbf{Tally} & \textbf{Frequency} \\ \hline \text{robin} & \vert\,\vert\,\vert\,\vert\,\vert\quad \vert & 6 \\ \hline \text{sparrow} & \vert\,\vert\,\vert\,\vert & 4 \\ \hline \text{wren} & \vert\,\vert\,\vert & 3 \\ \hline \text{blackbird} & \vert\,\vert\,\vert\,\vert\,\vert & 5 \\ \hline \text{magpie} & \vert\,\vert & 2 \\ \hline \end{array}

(Reason: The frequency is simply how many tally marks that bird has, so each row is counted and the total written in the last column.)

Step 3: Check the frequencies against the number of adults asked

6 + 4 + 3 + 5 + 2 = 20

(Reason: Every one of the

20

answers is counted exactly once, so the five frequencies have to add up to

20

. They do, so nothing has been missed and nothing counted twice.)

Step 4: Read the bar heights off the table

6 \quad 4 \quad 3 \quad 5 \quad 2

(Reason: The height of each bar is that bird's frequency, taken in the same order as the table. The scale is numbered in ones, so every bar top sits on a grid line.)

Step 5: Draw the bars and label them

(Reason: Bars of equal width with equal gaps, each drawn up to its frequency, and the bird's name written under its own bar. The labels earn a mark of their own, so a chart with five correct bars and nothing written underneath does not score full marks.)

(a) robin

6

, sparrow

4

, wren

3

, blackbird

5

, magpie

2

(b) bars of height

6

4

3

5

and

2

, each labelled with its bird

Verification

Check 1 - count down the columns instead of along the rows: Counting the

20

results column by column uses a completely different order, so a result missed the first time is not missed in the same place again. robin

6

, sparrow

4

, wren

3

, blackbird

5

, magpie

2

- the same five frequencies

Check 2 - the frequencies add to the number of adults asked: Add the five frequencies and compare with the number of adults.

6 + 4 + 3 + 5 + 2 = 20

, and Nadia asked

20

adults

Check 3 - read the finished bar chart back: Read each bar height off the grid and compare it with the table it came from. The bars measure

6

4

3

5

and

2

, and the tallest bar is the robin, which is the bird that appears most often in the list

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Frequencies $6$ , $4$ , $3$ , $5$ , $2$	B2	for all correct frequencies (ignore tally column)	✓
(a) Part marks	B1	for $3$ or $4$ correct frequencies, or for $4$ or $5$ tallies correct but not totalled, or for frequencies written as probabilities with $3$ or $4$ or $5$ correct numerators	✓
(b) A correctly labelled bar chart	B2	ft for all $5$ correct bars, follow through their figures (do not follow through $0$ )	✓
(b) Part marks	B1ft	for $3$ or $4$ correct bars, follow through their figures (do not follow through $0$ ), or for $5$ correct indications of heights, eg cross, dot, etc (do not follow through $0$ )	✓
(b) What is condoned	Note	Condone gaps of different widths or no gaps between bars, and also bars of different widths	✓
(b) Labels	B1	for all $5$ bars labelled - use of initials or a key is acceptable	✓

Full marks: 5/5

(a) Write $d + d + d + d$ in a simpler form. [1 mark]

(b) Simplify $w \times w \times w \times w \times w$ [1 mark]

(d) Simplify $7r + 9y + 3r - 4y$ [2 marks]

(a)(b)(c) x =(d)

[Total 5 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

(a)

d + d + d + d

, the same letter added four times

(b)

w \times w \times w \times w \times w

, the same letter multiplied five times

(c)

6x = 42

, an equation with one operation in it

(d)

7r + 9y + 3r - 4y

, four terms sharing two different letters

Find

(a) the sum written as one term (b) the product written as one power of

w

x

(d) the expression with its like terms collected

Plan the Solution

Parts (a) and (b) look alike and are not. One is repeated addition, the other is repeated multiplication, so settle which before writing anything down.
Repeated addition of a letter gives a coefficient, so count the terms in (a): four lots of $d$ .
Repeated multiplication of a letter gives an index, so count the factors in (b): five copies of $w$ .
Part (c) has one operation on $x$ , so undo it with the opposite operation.
Part (d) mixes two letters. Collect the $r$ terms and the $y$ terms separately, and carry each sign with the term in front of which it sits.

Worked Solution [5 marks]

Like terms only: repeated addition gives a coefficient,

d + d + d + d = 4d

; repeated multiplication gives an index,

w \times w \times w \times w \times w = w^{5}

; and terms in different letters, such as

10r

and

5y

, can never be added into one term.

(a) Count the terms

d + d + d + d = 4 \times d = 4d

(Reason: Adding the same letter four times is the same as multiplying it by

4

. The number in front records how many there are, so the answer is one term, not four. It is not

d^{4}

: nothing here is being multiplied.)

(b) Count the factors

w \times w \times w \times w \times w = w^{5}

(Reason: There are

5

copies of

w

multiplied together, and the index is exactly that count. It is not

5w

, which would mean five copies added.)

6x = 42

x = \dfrac{42}{6}

x = 7

(Reason:

x

has been multiplied by

6

, so dividing both sides by

6

leaves

x

on its own. Whatever is done to one side is done to the other.)

(d) Collect the

r

terms

7r + 3r = 10r

(Reason:

7r

and

3r

are like terms, because they carry the same letter, so their coefficients add:

7 + 3 = 10

(d) Collect the

y

terms

9y - 4y = 5y

(Reason: The same letter again, so the coefficients combine. The sign in front of the

4y

belongs to that term and travels with it, so read the operation before deciding what to do with the

9

and the

4

(d) Write the two collected terms together

7r + 9y + 3r - 4y = 10r + 5y

(Reason:

10r

and

5y

are unlike terms, so there is nothing further to collect. Two terms is the finished answer, and writing

15ry

would be inventing a multiplication the expression never had.)

(a)

4d

(b)

w^{5}

(c)

x = 7

(d)

10r + 5y

Verification

Check 1: Put

d = 3

into both forms in (a), then

w = 2

into both forms in (b). A correct simplification agrees with the original at every value.

3 + 3 + 3 + 3 = 12

and

4 \times 3 = 12

;

2 \times 2 \times 2 \times 2 \times 2 = 32

and

2^{5} = 32

Check 2: Put

x = 7

back into the equation given in (c) and read the left-hand side.

6 \times 7 = 42

, which is the right-hand side

Check 3: Put

r = 2

and

y = 5

into the expression in (d) and into the collected answer.

14 + 45 + 6 - 20 = 45

and

10 \times 2 + 5 \times 5 = 45

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Simplify $d + d + d + d$	B1	$4d$ . Allow $d4$ .	✓
(b) Simplify $w \times w \times w \times w \times w$	B1	$w^{5}$	✓
(c) Solve $6x = 42$	B1	$7$	✓
(d) Simplify $7r + 9y + 3r - 4y$	B2	For $10r + 5y$ .	✓
(d) Partial credit	B1	For $10r$ or for $(+)5y$ .	✓

Full marks: 5/5

The diagram below shows a number line.

(a) Write down the number that the arrow is pointing to. [1 mark]

(b) On the number line below, mark the number $265$ with an arrow ( $\uparrow$ ). [1 mark]

(a)

[Total 2 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

A number line with

74

75

and

76

labelled, and

5

equal small divisions between one labelled number and the next. An arrow points at the first mark to the right of

75

A second number line with

260

280

and

300

labelled, and

4

equal small divisions between one labelled number and the next. Nothing is marked on it yet.

Find

(a) The number the arrow is pointing to. (b) Which mark to draw the arrow on, so that it shows

265

Plan the Solution

Work out what one small division is worth on each line. Both parts turn on that single value, and the two lines do not have the same one.
Share the gap between two labelled numbers equally between the SPACES on the line. The marks and the spaces are different things, and there is always one fewer mark than there are spaces.
For part (a), start at the labelled number just below the arrow and count on. For part (b), work out how many small divisions the target is above the labelled number just below it, then draw the arrow on that mark.

Worked Solution [2 marks]

Rule - one small division: share the gap between two labelled numbers equally between the spaces on the line. Count the spaces between the marks, never the marks themselves.

Part (a): find what one small division is worth

75 - 74 = 1

\dfrac{1}{5} = 0.2

(Reason: The gap from

74

75

1

, and it is split into

5

equal spaces, so one small division is one fifth of

1

. There are only

4

marks between the two labels, which is exactly why counting marks instead of spaces goes wrong.)

Part (a): count on from

75

75 + 0.2 = 75.2

(Reason: The arrow stands on the first mark to the right of

75

, which is one small division past it, so add one small division to

75

Part (b): find what one small division is worth on the second line

280 - 260 = 20

\dfrac{20}{4} = 5

(Reason: This line is scaled differently. The gap from

260

280

20

, split into

4

equal spaces, so here each small division is worth

5

Part (b): count how many divisions past

260

the arrow goes

265 - 260 = 5

\dfrac{5}{5} = 1

(Reason:

265

5

more than

260

, and one small division is worth

5

, so the arrow goes on the first mark to the right of

260

. Draw it pointing up at that mark.)

(a)

75.2

(b) The arrow on the first mark to the right of

260

, the mark that stands for

265

Verification

Check 1 - part (a), read the line from the other end: From

74

to the arrow is

6

small divisions, and from

74

76

10

small divisions covering

2

. This never uses the value of one division.

74 + \dfrac{6}{10} \times 2 = 75.2

Check 2 - part (a), as a mixed number: One small division is a fifth, so the first mark after

75

75

and a fifth. The mark scheme accepts that form as well as the decimal, so the two must agree.

75 + \dfrac{1}{5} = 75.2

Check 3 - part (b), count back from the far end:

300

8

small divisions above

260

. Counting back

7

of them must land on the same mark the arrow was drawn on.

300 - 7 \times 5 = 265

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$75.2$ . Allow $75\dfrac{1}{5}$ or equivalent.	✓
(b)	B1	A clear indication of marking the first notch after $260$ . Correct answer only.	✓

Full marks: 2/2

The grid below shows three points, $P$ , $Q$ and $R$ .

(a) Write down the coordinates of the point $P$ [1 mark]

(b) On the grid, mark with a cross (×) the point with coordinates $(3, -1)$
Label this point $T$ [1 mark]

(a)(c)

[Total 4 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

Three points marked with crosses on a square grid:

P

Q

and

R

The grid is numbered from

-4

4

across and from

-3

5

up.

Find

(a) The coordinates of

P

. (b) Where a cross for

(3, -1)

goes, labelled

T

. (c) The coordinates of the midpoint of

QR

Plan the Solution

Read every coordinate as a pair, across first and then up. A point left of the vertical axis has a negative first number, and a point below the horizontal axis has a negative second number.
For (b), start at the origin, count $3$ squares right and $1$ square down, and put the cross where those two gridlines meet.
For (c), a midpoint sits halfway between the two ends, so average the two across-numbers and then average the two up-numbers.

Worked Solution [4 marks]

Rule - Midpoint: the midpoint of the segment joining

(x_1,\ y_1)

and

(x_2,\ y_2)

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

Step 1: read

P

off the grid

P = (-2,\ 3)

(Reason: From the origin,

P

2

squares to the left, so the across-number is

-2

, and

3

squares up, so the up-number is

3

. A coordinate pair is always written across first, then up.)

Step 2: put the cross for

T

(3, -1)

T = (3,\ -1)

(Reason: Start at

O

. Count

3

squares right, then

1

square down, and mark that grid point with a cross. Write

T

beside it so the examiner can see which point is meant.)

Step 3: average the across-numbers of

Q

and

R

Q = (3,\ 4) \qquad R = (1,\ -2)

\dfrac{3 + 1}{2} = \dfrac{4}{2} = 2

(Reason: The midpoint is halfway along

QR

, so its across-number is halfway between

3

and

1

Step 4: average the up-numbers of

Q

and

R

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

(Reason: Adding

-2

is the same as taking away

2

, so work out the top of the fraction first and then halve it.)

Step 5: write the midpoint as a coordinate pair

\text{midpoint of } QR = (2,\ 1)

(Reason: Put the two averages together, across first. On the grid,

(2, 1)

sits exactly halfway along the line from

Q

R

(a)

(-2,\ 3)

(b)

T

marked at

(3,\ -1)

on the grid(c)

(2,\ 1)

Verification

Check 1: From

Q(3, 4)

(2, 1)

1

square left and

3

squares down. Take that same step once more from

(2, 1)

. It lands on

(1, -2)

, which is

R

, so

(2, 1)

is exactly halfway.

Check 2: A midpoint is the same distance from both ends. From

(2, 1)

Q

1

across and

3

up; from

(2, 1)

R

1

across and

3

down.

1^2 + 3^2 = 10

both times, so the two distances are equal.

Check 3:

P

is left of the vertical axis and above the horizontal axis, so its pair must read negative then positive.

T

is right of the vertical axis and below the horizontal axis, so its pair must read positive then negative.

(-2, 3)

and

(3, -1)

carry exactly those signs.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$(-2, 3)$	✓
(b)	B1	$T$ clearly marked. A point clearly marked at $(3, -1)$ need not be labelled if meaning is clear.	✓
(c)	B2	$(2, 1)$	✓
(c)	B1	for $(2, X)$ or $(Y, 1)$ or the midpoint unambiguously marked in the correct place on the diagram	✓

Full marks: 4/4

(a) Change $\dfrac{19}{5}$ into a mixed number. [1 mark]

(b) Circle the two fractions in the list below that are equivalent.

$\dfrac{19}{20} \qquad \dfrac{4}{5} \qquad \dfrac{20}{24} \qquad \dfrac{28}{35} \qquad \dfrac{5}{9}$ [1 mark]

(a)(c)

[Total 3 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

The improper fraction

\dfrac{19}{5}

A list of five fractions:

\dfrac{19}{20}

\dfrac{4}{5}

\dfrac{20}{24}

\dfrac{28}{35}

and

\dfrac{5}{9}

The decimal

0.3

Find

(a)

\dfrac{19}{5}

written as a mixed number. (b) The two fractions in the list that are equivalent. (c)

0.3

written as a fraction.

Plan the Solution

For (a), find how many whole ones are hidden inside $\dfrac{19}{5}$ . Every $5$ fifths make one whole, so divide the top by the bottom and keep the remainder in fifths.
For (b), put every fraction in the list into its simplest form. Two fractions are equivalent exactly when they simplify to the same thing, so once they are all cancelled down the pair stands out.
For (c), read the decimal off its place value. The first place after the point is tenths, so the denominator is $10$ .

Worked Solution [3 marks]

Rule - Equivalent fractions: multiplying or dividing the top and the bottom by the same number leaves the value unchanged, so

\dfrac{a}{b} = \dfrac{a \times k}{b \times k}

. A mixed number splits an improper fraction into the whole ones it contains plus the part left over.

Step 1: count the whole ones inside

\dfrac{19}{5}

19 = 3 \times 5 + 4

(Reason: Every

5

fifths make one whole one.

3 \times 5 = 15

, so three whole ones use up

15

of the fifths and

4

fifths are left over.)

Step 2: keep the leftover over the same denominator

\dfrac{19}{5} = \dfrac{15}{5} + \dfrac{4}{5}

\dfrac{19}{5} = 3\dfrac{4}{5}

(Reason: The

4

that is left over is still being counted in fifths, so it keeps the denominator

5

. A mixed number is written with the whole number and the fraction side by side, with no plus sign between them.)

Step 3: put each fraction in the list into its simplest form

\dfrac{20}{24} = \dfrac{5}{6}

\dfrac{28}{35} = \dfrac{4}{5}

(Reason: Cancel wherever the top and the bottom share a factor:

20

and

24

share a factor of

4

, and

28

and

35

share a factor of

7

. The other three will not cancel at all -

19

is prime, and

\dfrac{4}{5}

and

\dfrac{5}{9}

are already as simple as they go.)

Step 4: pick out the two that match

\dfrac{4}{5} = \dfrac{4 \times 7}{5 \times 7} = \dfrac{28}{35}

(Reason: Two fractions in the list simplify to the same thing, and those are the two to circle:

\dfrac{4}{5}

and

\dfrac{28}{35}

. Multiplying the top and the bottom of

\dfrac{4}{5}

7

is the same statement read the other way round.)

Step 5: read

0.3

off its place value

0.3 = \dfrac{3}{10}

(Reason: The first place after the decimal point is the tenths place, so

0.3

3

tenths, which is

3

over

10

. Nothing divides into both

3

and

10

, so it will not cancel any further.)

(a)

3\dfrac{4}{5}

(b)

\dfrac{4}{5}

and

\dfrac{28}{35}

(c)

\dfrac{3}{10}

Verification

Check 1: Turn the mixed number in (a) back into an improper fraction. Three whole ones are

15

fifths, and the

4

fifths are added on.

3 \times 5 + 4 = 19

, which is the numerator the question started with.

Check 2: Test the pair in (b) without cancelling anything. Two fractions are equal exactly when the cross products match, so multiply the top of each by the bottom of the other.

4 \times 35 = 140

and

5 \times 28 = 140

, so the two are equal. No other pair in the list gives a match.

Check 3: Turn the answer to (c) back into a decimal by writing it in hundredths.

\dfrac{3}{10} = \dfrac{30}{100} = 0.3

, which is the decimal the question gave.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$3\dfrac{4}{5}$	✓
(b)	B1	$\dfrac{4}{5}$ , $\dfrac{28}{35}$ . Both fractions and no other fraction circled or clearly indicated in the list	✓
(c)	B1	$\dfrac{3}{10}$ oe fraction	✓

Full marks: 3/3

$2$ muffins and $5$ biscuits cost $\$4.50$
$3$ muffins cost $\$2.25$

Work out the cost of $1$ biscuit. [4 marks]

[Total 4 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

One order of

2

muffins and

5

biscuits costs

\$4.50

altogether.

A second order of

3

muffins costs

\$2.25

with no biscuits in it at all.

Every muffin costs the same as every other muffin, and every biscuit costs the same as every other biscuit.

Find

The cost of one biscuit.

Plan the Solution

The second order is the way in, because it holds only one kind of item. An order of muffins and nothing else prices a muffin.
Once one muffin has a price, the $2$ muffins hiding inside the first order can be costed and taken off the $\$4.50$ , and what is left is the price of the $5$ biscuits between them.
Share that remainder between the $5$ biscuits to reach one biscuit.
Work in dollars all the way through, and keep the two orders apart until the subtraction.

Worked Solution [4 marks]

Rule - one kind of item at a time: an order holding a single kind of item gives the price of that item by division, that price then lifts the same item out of a mixed order by subtraction, and the remainder is shared out by division again.

Step 1: price one muffin from the muffins-only order

\dfrac{2.25}{3} = 0.75

(Reason: (Reason: the second order is

3

muffins and nothing else, so its

\$2.25

is shared equally between

3

identical muffins. Dividing by

3

gives

\$0.75

for one of them.))

Step 2: cost the muffins that sit inside the first order

2 \times 0.75 = 1.50

(Reason: (Reason: the first order contains

2

of those same muffins, and a muffin costs the same wherever it is bought, so the muffin part of that order comes to

\$1.50

in all.))

Step 3: take the muffins off the first order

4.50 - 1.50 = 3.00

(Reason: (Reason: the first order is the

2

muffins and the

5

biscuits together. Remove everything the muffins account for and what remains is exactly what the

5

biscuits cost between them.))

Step 4: share that remainder between the five biscuits

\dfrac{3.00}{5} = 0.60

(Reason: (Reason: the

5

biscuits are identical, so the money left over splits equally between them. One biscuit costs

\$0.60

, which is

60

cents.))

\$0.60

Verification

Check 1: Rebuild both orders from the two prices. A muffin at

\$0.75

and a biscuit at

\$0.60

have to reproduce the two totals the question gives, not just one of them.

3 \times 0.75 = 2.25

for the muffins-only order, and

2 \times 0.75 + 5 \times 0.60 = 4.50

for the mixed one. Both are the printed totals.

Check 2: Reach the biscuit price without ever pricing a muffin. Take

3

copies of the first order, which is

6

muffins and

15

biscuits, and

2

copies of the second, which is

6

muffins on their own. Subtracting removes the muffins outright.

3 \times 4.50 - 2 \times 2.25 = 9

for

15

biscuits, so one biscuit is

\dfrac{9}{15} = 0.60

, and the muffin price was never used at any point.

Check 3: Is the size sensible? There are more biscuits than muffins in the first order, so the biscuits should take the larger share of the

\$4.50

, and a biscuit should come out cheaper than a muffin rather than dearer. The biscuits account for

3.00

of the

4.50

and the muffins

1.50

, and a biscuit at

\$0.60

sits just under a muffin at

\$0.75

- both of which are what a bakery counter should look like.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct method to find the cost of 1 muffin	M1	For $\dfrac{2.25}{3} \; (= 0.75)$ or $\dfrac{225}{3} \; (= 75)$ or $\dfrac{2}{3} \times 2.25 \; (= 1.50)$ or $\dfrac{2}{3} \times 225 \; (= 150)$ . Allow $\dfrac{1}{3}$ as $0.33\ldots$ or $\dfrac{2}{3}$ as $0.66\ldots$ , truncated or rounded. Or for setting up two equations in which the coefficients of $m$ are equal: $6m + 15b = 13.50$ and $6m = 4.50$ oe, or $6m + 15b = 1350$ and $6m = 450$ oe, with $m$ for muffins and $b$ for biscuits, or the use of 2 different letters (which may not be defined).	✓
A correct method to find the cost of the 5 biscuits	M1	For $4.50 - 2 \times 0.75 \; (= 3)$ or $450 - 2 \times 75 \; (= 300)$ , where the printed scheme puts the $0.75$ and the $75$ in quotation marks, so the candidate's own earlier value for one muffin may be used in their place. Or for finding an equation for $b$ : $15b = 9$ oe or $15b = 900$ oe.	✓
A correct method to find the cost of 1 biscuit	M1	For $\dfrac{3}{5}$ or $\dfrac{300}{5} \; (= 60)$ , where the printed scheme again puts the $3$ and the $300$ in quotation marks, so the candidate's own value for the $5$ biscuits may be used. Or for $\dfrac{9}{15}$ or $\dfrac{900}{15} \; (= 60)$ .	✓
The cost of 1 biscuit	A1	For $0.6$ or $0.60$ , or for the dollar sign crossed out and $60$ cents given instead.	✓
An answer given with no working	Note	Working is not required here, so a correct answer scores full marks unless it has come from obviously incorrect working.	✓
Special case - the muffin division turned upside down, carried through both steps	SC B2	Awarded only when no other marks are earned, for $4.50 - 2 \times \dfrac{3}{2.25}$ , which comes to $1.83\ldots$ , and then that value divided by $5$ , which comes to $0.36\ldots$ dollars. Or, in cents, for $450 - 2 \times \dfrac{3}{225}$ , which comes to $449.9\ldots$ , and then that value divided by $5$ , which comes to $89.9\ldots$ cents. The error behind it is the first division taken the wrong way round, $\dfrac{3}{2.25}$ in place of $\dfrac{2.25}{3}$ , after which the method is followed correctly.	✓
Special case - the same upside-down division, first step only	SC B1	Awarded only when no other marks are earned, for $4.50 - 2 \times \dfrac{3}{2.25}$ on its own, which comes to $1.83\ldots$ , or for $450 - 2 \times \dfrac{3}{225}$ on its own, which comes to $449.9\ldots$ cents.	✓

Full marks: 4/4

In the diagram, $ABC$ and $EBD$ are straight lines.
Triangle $BCD$ is isosceles, with $BD = BC$

Angle $ABE = 104^\circ$

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

x =

[Total 3 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

ABC

and

EBD

are straight lines, crossing at

B

Triangle

BCD

is isosceles, with

BD = BC

Angle

ABE = 104^\circ

The angle marked

x^\circ

is angle

BDC

, the angle of the triangle at

D

Find

The value of

x

, which is the size of angle

BDC

in degrees

Plan the Solution

The two straight lines cross at $B$ , so the angle of the triangle at $B$ is vertically opposite the $104^\circ$ that is marked above it. That carries the given angle down into the triangle.
The equal sides $BD$ and $BC$ make the angles at $D$ and at $C$ equal, so those two share whatever the angle at $B$ leaves.
Take the angle at $B$ away from $180^\circ$ , then halve what is left.

Worked Solution [3 marks]

Rule - Vertically opposite angles are equal; the three angles of a triangle add up to

180^\circ

; and in an isosceles triangle the two angles opposite the equal sides are equal.

Step 1: Carry the

104^\circ

down into the triangle

\text{angle } DBC = \text{angle } ABE = 104^\circ

(Reason:

ABC

and

EBD

are straight lines that cross at

B

, so angle

DBC

and angle

ABE

are vertically opposite, and vertically opposite angles are equal.)

Step 2: Find what the angles at

D

and

C

\text{angle } BDC + \text{angle } BCD = 180^\circ - 104^\circ = 76^\circ

(Reason: The three angles of triangle

BCD

add up to

180^\circ

, so taking the

104^\circ

B

away leaves

76^\circ

for the other two angles together.)

Step 3: Halve it, because the two angles are equal

x = \dfrac{76}{2} = 38

(Reason:

BD = BC

, so the angles opposite those two sides, at

D

and at

C

, are equal. Each one is therefore half of the

76^\circ

they share.)

x = 38

Verification

Check 1: Put the three angles of triangle

BCD

back together:

104

B

, and

38

at each of

D

and

C

104 + 38 + 38 = 180

Check 2: Work along the straight line

ABC

instead. Angle

ABD

180 - 104 = 76

, and it is the exterior angle of the triangle at

B

, so it must equal the two interior angles opposite it added together.

38 + 38 = 76

Check 3: Split the isosceles triangle down the middle. The line from

B

to the midpoint of

DC

is perpendicular to

DC

and halves the angle at

B

, giving a right-angled triangle with

52^\circ

B

90 - 52 = 38

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\text{angle } DBC = 104$ or $180 - 104 \, (= 76)$	M1	for correctly finding $DBC$ or $EBC$ or $ABD$ . May be seen on the diagram. This is not awarded if the angles are incorrectly assigned. (Ignore incorrect angles on the diagram if a student shows $38$ on the answer line)	✓
$\dfrac{180 - 104}{2}$ oe or $\dfrac{76}{2}$ oe	M1	The printed scheme gives no further notes against this mark. oe means any equivalent working is accepted.	✓
Working not required, so correct answer scores full marks (unless from obvious incorrect working). Answer $38$	A1	cao - correct answer only.	✓

Full marks: 3/3

Neil has $2$ litres of lemonade in a large bottle and some empty glasses.
He pours lemonade from the bottle to completely fill as many glasses as possible.
He pours $180$ millilitres of lemonade into each glass.

Work out how much lemonade Neil has left in the bottle after he has completely filled as many glasses as possible.
Give your answer in millilitres. [4 marks]

millilitres

[Total 4 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

2

litres of lemonade in the bottle at the start

Every completely filled glass takes

180

millilitres

A conversion is needed:

1

litre is

1000

millilitres

Find

The amount of lemonade left in the bottle, in millilitres, once no further glass can be completely filled.

Plan the Solution

Put both amounts into the same unit by changing the $2$ litres into millilitres.
Divide to see how many $180$ millilitre glasses fit into that amount.
The division does not come out as a whole number, so keep the whole-number part only - a part-filled glass does not count.
Multiply that number of glasses by $180$ to find how much lemonade leaves the bottle, then subtract from the starting amount.

Worked Solution [4 marks]

Rule - Convert first, then count whole glasses:

1 \text{ litre} = 1000 \text{ millilitres}

, and lemonade left

= \text{start} - (\text{number of full glasses} \times 180)

Step 1: Put both amounts into the same unit

2 \times 1000 = 2000

(Reason: The glasses are measured in millilitres, so the bottle must be measured in millilitres too. There are

1000

millilitres in

1

litre, so the bottle holds

2000

millilitres.)

Step 2: Divide to see how many glasses can be completely filled

\dfrac{2000}{180} = \dfrac{100}{9}

(Reason: Each full glass takes

180

millilitres, so dividing counts the glasses. Now

\dfrac{100}{9}

is a little more than

11

but less than

12

, so eleven glasses can be filled completely and a twelfth cannot.)

Step 3: Work out how much lemonade is poured out

11 \times 180 = 1980

(Reason: Only completely filled glasses are poured, and there are

11

of them, each taking

180

millilitres.)

Step 4: Subtract to find what is left in the bottle

2000 - 1980 = 20

(Reason: The bottle started with

2000

millilitres and

1980

millilitres went into the glasses, so what stays behind is the difference. The question asks for millilitres, so no converting back is needed.)

20

millilitres

Verification

Check 1: Add the glasses up instead of multiplying:

180, 360, 540, 720

and so on reaches

1800

after ten glasses, and one more glass makes

1980

1980 + 20 = 2000

, so every millilitre of lemonade is accounted for

Check 2: Test that a twelfth glass really is impossible. Filling one needs another

180

millilitres, and only

20

millilitres are left in the bottle.

20 < 180

, so no further glass can be completely filled and the pouring stops here

Check 3: Work in litres all the way through instead. Each glass is

0.18

litres, so eleven glasses take

1.98

litres, leaving

0.02

litres in the bottle. Multiplying by

1000

turns that into

20

millilitres, the same answer by a different unit

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Writes $2$ litres as $2000$ (millilitres), or $180$ millilitres as $0.18$ (litres)	B1	Can be implied from correct working	✓
eg $\dfrac{2000}{180}$ oe or $\dfrac{2}{0.18}$ oe or $11(.111....)$ or $\dfrac{100}{9}$ or $11\dfrac{1}{9}$ or $180, 360, 540, 720, ...., 1800, 1980$ or $2000, 1820, 1640, 1460, ...., 200, 20$	M1	ft their millilitres over $180$ , or $2$ over their litres, for this mark. NB Repeated subtraction must continue to a number less than $180$ . eg $\dfrac{200}{180}$ or $200 - 180$ gains this mark only. eg $\dfrac{500}{180}$ or $500 - 180 - 180$ gains this mark only. eg $\dfrac{2}{1.8}$ or $2 - 1.8$ gains this mark only. Allow one arithmetic error for repeated addition or repeated subtraction for this mark	✓
$11 \times 180$ ( $= 1980$ ) or $11 \times 0.18$ ( $= 1.98$ ) or $180, 360, 540, 720, ...., 1800, 1980$ or $2000, 1820, 1640, 1460, ...., 200, 20$	M1	dep on B1, or for an answer of $0.02$ . No errors allowed for repeated addition or repeated subtraction for this mark	✓
Answer: $20$	A1	Working not required, so correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 4/4

Wei is asked to work out the value of $30 - 4x$ when $x = -5$
Here is his working and his answer.

$30 - 4x = 30 - 4 \times -5$
$= 30 - 20$
$= 10$

Wei's answer is wrong.

(a) Explain the mistake Wei has made in his working. [1 mark]

Amara is thinking of a number.
She calls her number $Y$
$Y$ is a positive, odd number and $Y \leqslant 9$

(b) Write down all the possible values of $Y$ [2 marks]

(a)(b)

[Total 3 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

Wei must work out

30 - 4x

when

x = -5

, and his working ends at

10

Y

is positive,

Y

is odd, and

Y \leqslant 9

Find

(a) what Wei has done wrong, said in words (b) every value

Y

can take

Plan the Solution

(a) Substitute $x = -5$ , then follow the order of operations: the multiplication is done before the subtraction.
(a) Work the correct third line out, compare it with Wei's third line, and name the difference.
(b) Take the three conditions one at a time. Positive removes $0$ and the negative numbers, odd removes the even numbers, and $Y \leqslant 9$ sets the ceiling.

Worked Solution [3 marks]

Rule - a negative multiplied by a negative gives a positive, so

-4 \times -5 = +20

, and a term worth

-20

that is being subtracted adds to the total.

Part (a), Step 1: substitute

x = -5

30 - 4x = 30 - 4 \times -5

(Reason:

4x

means

4 \times x

, so

x

is replaced by

-5

and nothing else on the line changes. Wei's first line is right.)

Part (a), Step 2: multiply before subtracting

-4 \times -5 = +20

(Reason: The term being taken away is

4 \times -5

, which is

-20

. Reading the expression as

30 + (-4) \times (-5)

makes the sign rule visible: two negatives multiply to a positive, so this is

+20

Part (a), Step 3: finish the line

30 - 4 \times -5 = 30 + 20

30 + 20 = 50

(Reason: Taking away a negative amount adds to the total, so the third line should read

30 + 20

and the value of the expression is

50

Part (a), Step 4: say what went wrong

30 - 20 = 10

(Reason: Wei's own third line is correct arithmetic, but it starts from the wrong sign: he treated

-4 \times -5

-20

instead of

+20

, so he subtracted where he should have added. Naming that one sign is what the mark is for.)

Part (b), Step 5: take the conditions one at a time

Y > 0, \quad Y \text{ odd}, \quad Y \leqslant 9

(Reason: Positive rules out

0

and every negative number. Odd rules out

2

4

6

and

8

. The third condition sets the ceiling, and

Y \leqslant 9

lets

9

itself in.)

Part (b), Step 6: write the list

Y = 1, 3, 5, 7, 9

(Reason: Start at

1

and count on in twos. The next odd number after

9

11

, which breaks

Y \leqslant 9

, so the list stops at

9

(a)

-4 \times -5 = +20

, not

-20

, so the third line should be

30 + 20

and the answer should be

50

(b)

Y = 1

3

5

7

9

Verification

Check 1: Work the value out the other way round: build the term

4x

first, then subtract it from

30

4 \times -5 = -20

and

30 - (-20) = 50

, which is the value Step 3 reached

Check 2: Find where Wei's

10

really comes from by substituting the positive value

x = 5

instead.

30 - 4 \times 5 = 10

, so

10

is the value at

x = 5

, not at

x = -5

Check 3: Count what the conditions allow: from

1

9

every other whole number is odd. Five values are allowed, and the list holds five

Check 4: Test the three numbers just outside the list, one for each condition.

0

is not positive,

2

is even, and

11

is greater than

9

, so nothing is missing and nothing extra belongs

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A correct reason	B1	He should have got $30 + 20$ oe, or the answer should be $50$ oe, or $-4 \times -5 = (+)20$ oe, or minus and minus gives a positive oe, etc	✓
(b) $1, 3, 5, 7, 9$	B2	For $1, 3, 5, 7, 9$ with no extras (in any order)	✓
(b) part marks	B1	For four correct values with no more than one incorrect, or for five correct values with no more than one incorrect	✓

Full marks: 3/3

The members of a summer activity camp take part in one session, either on Thursday or on Friday.
They each pick one activity from badminton or swimming or archery.
The two-way table gives some information about their choices.

$\begin{array}{|l|c|c|c|c|}\hline & \textbf{badminton} & \textbf{swimming} & \textbf{archery} & \textbf{Total} \\ \hline \textbf{Thursday} & 29 & & 58 & 110 \\ \hline \textbf{Friday} & & & & \\ \hline \textbf{Total} & 62 & & 105 & 240 \\ \hline \end{array}$

(a) Complete the two-way table. [3 marks]

One of the members who picks archery is chosen at random.

(b) Write down the probability that this member takes part in archery on Thursday. [1 mark]

(b)

[Total 4 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

Every member takes part on exactly one of the two days, and picks exactly one of the three activities.

The Thursday row reads

29

for badminton and

58

for archery, and adds to

110

The Total row reads

62

for badminton and

105

for archery, and the grand total is

240

Six cells of the table are empty: the whole Friday row, one cell in the Thursday row and one cell in the Total row.

Find

(a) The six missing entries in the two-way table. (b) The probability that a member picked at random from those who chose archery is one of the Thursday members.

Plan the Solution

Start with a row or a column that has only ONE empty cell, because that cell is forced: it is whatever is left when the entries already there are taken from the total.
Each entry filled in opens up another row or another column with a single gap, so keep working outwards until every cell is full.
Part (b) is a conditional probability. Only the $105$ members who picked archery are in the running, so the archery total is the denominator, not the grand total of $240$ .

Worked Solution [4 marks]

Rule - two-way table: every row adds to its own row total, every column adds to its own column total, and the row totals and the column totals both add to the same grand total of

240

Step 1: close the Thursday row

110 - 29 - 58 = 23

(Reason: The Thursday row has only one empty cell, so that cell is forced. Take the two Thursday entries already printed from the row total of

110

, which leaves

23

for swimming on Thursday.)

Step 2: the Friday total

240 - 110 = 130

(Reason: The Total column holds only the two day totals, and they add to the grand total, so the Friday total is

240

take away the Thursday total of

110

Step 3: the Friday badminton and archery entries

62 - 29 = 33

105 - 58 = 47

(Reason: Each of those two columns has a single gap, so take the Thursday entry from the column total. Badminton leaves

33

and archery leaves

47

Step 4: close the Friday row

130 - 33 - 47 = 50

(Reason: The Friday row now has one empty cell left, so take the two Friday entries just worked out in step 3 from the Friday total of

130

Step 5: the swimming column total

240 - 62 - 105 = 73

(Reason: The three column totals add to the grand total, so the swimming total is what is left when the badminton total of

62

and the archery total of

105

are taken from

240

. Reading it down the swimming column instead gives the same thing:

23 + 50 = 73

Step 6: the completed table

\begin{array}{|l|c|c|c|c|}\hline & \textbf{badminton} & \textbf{swimming} & \textbf{archery} & \textbf{Total} \\ \hline \textbf{Thursday} & 29 & 23 & 58 & 110 \\ \hline \textbf{Friday} & 33 & 50 & 47 & 130 \\ \hline \textbf{Total} & 62 & 73 & 105 & 240 \\ \hline \end{array}

(Reason: Every row now adds to the total at the end of it and every column adds to the total at the foot of it, which is what part (a) is marked on. The marks are for the entries, so a table that is right in only some cells still scores.)

Step 7: the probability for part (b)

\text{P(Thursday)} = \dfrac{58}{105}

(Reason: The member is picked from those who chose archery, so the archery total of

105

is the denominator. Of those

105

, the

58

sitting in the Thursday row are the ones that count. The fraction will not cancel, because

58 = 2 \times 29

and

105 = 3 \times 5 \times 7

share no factor.)

(a) Thursday swimming

23

; Friday badminton

33

, swimming

50

, archery

47

, total

130

; swimming total

73

(b)

\dfrac{58}{105}

Verification

Check 1: Add each row across and compare it with the total printed at the end of that row:

29 + 23 + 58

33 + 50 + 47

and

62 + 73 + 105

110

130

and

240

, which are the three row totals the table already carries.

Check 2: Now add down the columns instead, which uses each entry a second time in a different pairing:

29 + 33

23 + 50

58 + 47

and

110 + 130

62

73

105

and

240

, which are the four column totals the table already carries.

Check 3: The archery column has only two entries, so the Thursday share and the Friday share of it must add to

1

. Work out the Friday share on its own, from the cell part (b) never touches.

\dfrac{58}{105} + \dfrac{47}{105} = \dfrac{105}{105} = 1

, so nobody who picked archery has been counted twice or left out.

Check 4: Test the denominator by picking the wrong one on purpose. Using the grand total would give

\dfrac{58}{240}

, which is the chance that a member picked from the WHOLE camp does archery on Thursday. The two fractions are nowhere near each other:

\dfrac{58}{240}

is under a quarter, while

\dfrac{58}{105}

is over a half. The two questions have different answers, and the archery total is the denominator this part needs.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A correctly completed table	B3	B3 for a correctly completed table. (B2 for $4$ or $5$ correct entries. B1 for $2$ or $3$ correct entries.)	✓
(b) $\dfrac{58}{105}$	B1	B1 for $\dfrac{58}{105}$ oe, or $0.55(238\ldots)$ or $55(.238\ldots)\%$ , truncated or rounded.	✓

Full marks: 4/4

(a) Rearrange $w = 3y - d$ to make $y$ the subject. [2 marks]

Marco buys $b$ bags of marbles and $p$ pouches of marbles.

There are $12$ marbles in each bag.
There are $3$ marbles in each pouch.

In total, Marco buys $T$ marbles.

(b) Write down a formula for $T$ in terms of $b$ and $p$ [3 marks]

(a)(b)

[Total 5 marks]

Show solution & mark schemeHide solution & mark scheme

Question 12 - Exam Solution

Understanding the Question

Given

(a) The formula

w = 3y - d

(b)

b

bags holding

12

marbles each, and

p

pouches holding

3

marbles each

(b)

T

is the total number of marbles bought

Find

(a)

y

on its own, written in terms of

w

and

d

(b) A formula for

T

in terms of

b

and

p

Plan the Solution

(a) $y$ is buried inside $3y - d$ . Undo the operations around it in reverse order: the subtraction first, then the multiplication.
(a) Whatever is done to one side is done to the other, so the formula stays true on every line.
(b) Work out the marbles that come from the bags and the marbles that come from the pouches separately, then add the two amounts.
(b) There is no number to find here. The answer is a formula, so the letters stay in it.

Worked Solution [5 marks]

Rule - Changing the subject: undo each operation around the wanted letter, doing the same thing to both sides, until that letter stands alone. Rule - Building a formula: multiply the number in one item by how many of those items there are, then add the separate amounts together.

Part (a), Step 1 - undo the subtraction

w = 3y - d

w + d = 3y

(Reason: Reason: on the right-hand side

d

is being taken away, so adding

d

to both sides cancels it and leaves

3y

standing alone.)

Part (a), Step 2 - undo the multiplication

3y = w + d

y = \dfrac{w + d}{3}

(Reason: Reason:

y

is multiplied by

3

, so dividing both sides by

3

leaves

y

by itself. The formula is then written with

y

on the left, which is what making

y

the subject means.)

Part (b), Step 1 - the marbles in the bags

12 \times b = 12b

(Reason: Reason: one bag holds

12

marbles and there are

b

bags, so the bags account for

12b

marbles.)

Part (b), Step 2 - the marbles in the pouches

3 \times p = 3p

(Reason: Reason: one pouch holds

3

marbles and there are

p

pouches, so the pouches account for

3p

marbles.)

Part (b), Step 3 - add the two amounts

T = 12b + 3p

(Reason: Reason: every marble comes either from a bag or from a pouch, so the total

T

is the two amounts added together. This cannot be tidied any further, because

12b

and

3p

are not like terms.)

(a)

y = \dfrac{w + d}{3}

(b)

T = 12b + 3p

Verification

Check 1 - put numbers through part (a): Choose

y = 4

and

d = 5

. The formula in the question gives

3 \times 4 - 5 = 7

, so

w = 7

. Now feed

w = 7

and

d = 5

into the rearranged formula.

\dfrac{7 + 5}{3} = \dfrac{12}{3} = 4

, which is the value of

y

that was started with.

Check 2 - rearrange the answer back again: Multiply both sides of the answer by

3

to get

3y = w + d

, then take

d

away from both sides.

3y - d = w

, which is the formula the question started from, so nothing has been lost.

Check 3 - count part (b) a different way: Take

b = 5

bags and

p = 4

pouches. Counting them separately gives

5 \times 12 = 60

marbles from the bags and

4 \times 3 = 12

marbles from the pouches, so

72

marbles altogether. The formula gives

12 \times 5 + 3 \times 4 = 60 + 12 = 72

, the same total.

Check 4 - test the two numbers in front of the letters: One bag and no pouches must come to

12

marbles, and no bags and one pouch must come to

3

marbles. Put those two cases into the formula.

12 \times 1 + 3 \times 0 = 12

and

12 \times 0 + 3 \times 1 = 3

, so neither number in front of a letter has been swapped.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A correct first step	M1	For a correct first step: $w + d = 3y$ oe or $-3y = -w - d$ oe or $\dfrac{w}{3} = y - \dfrac{d}{3}$ oe or $\dfrac{w + d}{3}$ oe or $y = w + d$ divided by $3$ .	✓
(a) The rearranged formula	A1	$y = \dfrac{w + d}{3}$ oe, eg $y = \dfrac{-d - w}{-3}$ or $y = \dfrac{w}{3} + \dfrac{d}{3}$ oe, or $y = (w + d)$ divided by $3$ . Must see $y = \ldots$ on the answer line or in the working.	✓
(a) Working	Note	Working is not required, so a correct answer scores full marks, unless it comes from obviously incorrect working.	✓
(b) The formula	B3	For $T = 12b + 3p$ oe. Accept $T = 12 \times b + 3 \times p$ .	✓
(b) Almost the full formula	B2	For $12b + 3p$ or $T = 12b + xp$ or $T = yb + 3p$ or a correct equation with other letters, eg $T = 12m + 3n$ .	✓
(b) One correct part of the formula	B1	For $12b + xp$ or $12b$ or $yb + 3p$ or $3p$ or $12p + 3b$ or $T = kb + cp$ where $k \neq 0$ or $k \neq 12$ and $c \neq 0$ or $c \neq 3$ .	✓
(b) Letters	Note	Accept upper or lower case for $T$ , $b$ and $p$ , including a mixture of these, for B3, B2 and B1.	✓

Full marks: 5/5

Find the value of $\dfrac{9.75 - 1.4^{2}}{9.2 + \sqrt{6.3}}$
Write down all the figures on your calculator display. [2 marks]

[Total 2 marks]

Show solution & mark schemeHide solution & mark scheme

Question 13 - Exam Solution

Understanding the Question

Given

One fraction, with

9.75 - 1.4^{2}

on the top line and

9.2 + \sqrt{6.3}

on the bottom line.

A calculator is allowed, and the answer is to be copied from its display.

Find

The value of the fraction, written with every figure the display shows.

Plan the Solution

Work out the top line on its own, squaring before subtracting.
Work out the bottom line on its own, adding $\sqrt{6.3}$ to $9.2$ without rounding the root.
Divide the top line by the bottom line, then copy the display exactly as it stands.

Worked Solution [2 marks]

Rule - the fraction bar groups: work out the whole of the top line and the whole of the bottom line first, and only then divide the one by the other.

Step 1: work out the top line

9.75 - 1.4^{2} = 9.75 - 1.96 = 7.79

(Reason: The power is worked out before the subtraction, so square the

1.4

first and only then take it away from

9.75

Step 2: work out the bottom line

\sqrt{6.3} = 2.50998\ldots

9.2 + 2.50998\ldots = 11.70998\ldots

(Reason: The root does not stop, so keep it as the calculator holds it. Every figure of the bottom line is needed, because the question asks for every figure of the answer.)

Step 3: divide, and copy the whole display

\dfrac{7.79}{11.70998\ldots} = 0.6652445134

(Reason: Do the division in one calculation. Retyping a rounded

11.70998

agrees to seven figures and then drifts, so the end of the display would be wrong; and nothing is rounded at the finish, because all the figures are what is asked for.)

0.6652445134

Verification

Check 1: Estimate with easier numbers. The top line is about

7.8

and the bottom line about

11.7

, and that ratio is exactly two thirds.

\dfrac{7.8}{11.7} = \dfrac{2}{3}

, so the answer must sit just below

0.6667

, and

0.6652445134

does.

Check 2: Work backwards. Multiplying the displayed value by the bottom line must return the top line,

7.79

0.6652445134 \times 11.70998008 = 7.79

, the top line back again.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$7.79$ or $11.709\ldots$ or $0.67$ or $0.665$ or $0.6652$	M1	Any one of these values seen scores the method mark.	✓
Working not required, so correct answer scores full marks (unless from obvious incorrect working)	A1	Answer $0.66524(4\ldots)$ , to at least $5$ significant figures. The calculator value is $0.6652445134$ .	✓

Full marks: 2/2

A factory makes bicycles.

The factory makes $125$ bicycles each hour for $9$ hours a day.
The factory makes bicycles for $5$ days each week.

$24\%$ of the bicycles the factory makes are blue.

Work out the number of blue bicycles that the factory makes each week. [3 marks]

[Total 3 marks]

Show solution & mark schemeHide solution & mark scheme

Question 14 - Exam Solution

Understanding the Question

Given

The factory makes

125

bicycles each hour.

It makes bicycles for

9

hours a day, on

5

days each week.

24\%

of the bicycles it makes are blue.

Find

The number of blue bicycles the factory makes in one week.

Plan the Solution

Build the weekly total up in stages: one hour, then one day, then one week.
Write $24\%$ as a decimal, then multiply it by that weekly total.
A calculator is allowed, but write the multiplications down: the method marks are for the chain, not for the final number.

Worked Solution [3 marks]

Percentage of an amount: write the percentage as a decimal, then multiply by the amount.

24\%

of an amount is

0.24

times that amount.

Step 1: Find the bicycles made in one day

125 \times 9 = 1125

(Reason: Each hour produces

125

bicycles and the factory runs for

9

hours, so a day is nine lots of

125

Step 2: Find the bicycles made in one week

1125 \times 5 = 5625

(Reason: A week is

5

of those days, so multiply the daily total by

5

Step 3: Write the percentage as a decimal

\dfrac{24}{100} = 0.24

(Reason: Per cent means out of

100

, so

24\%

24

parts out of

100

Step 4: Take that share of the weekly total

0.24 \times 5625 = 1350

(Reason: The blue bicycles are

0.24

of every bicycle made that week, and the answer comes out as a whole number, which is what a count of bicycles must be.)

1350

blue bicycles

Verification

Check 1: Take the percentage first instead of last.

24\%

125

30

blue bicycles an hour, and the factory runs

9 \times 5 = 45

hours a week.

30 \times 45 = 1350

Check 2: Count the bicycles that are not blue instead. They are the other

76\%

of the week, and the two groups must add back to the weekly total.

0.76 \times 5625 = 4275

and

4275 + 1350 = 5625

Check 3: Avoid decimals altogether.

\dfrac{24}{100}

cancels to

\dfrac{6}{25}

, and

5625

splits into

25

equal parts of

225

\dfrac{6}{25} \times 5625 = 6 \times 225 = 1350

Mark Scheme Breakdown

Step	Mark	Description	Got it?
One correct product	M1	$125 \times 9\ (= 1125)$ or $125 \times 5\ (= 625)$ or $9 \times 5\ (= 45)$ or $125 \times 9 \times 5\ (= 5625)$ or $0.24 \times 125\ (= 30)$ or $0.24 \times 9\ (= 2.16)$ or $0.24 \times 5\ (= 1.2)$	✓
A complete method	M1	For a complete method: $0.24 \times (125 \times 9 \times 5)$ oe or $0.24 \times \text{"5625"}$ oe or $\text{"30"} \times 9 \times 5$ oe or $125 \times \text{"2.16"} \times 5$ oe or $125 \times 9 \times \text{"1.2"}$ oe. A value in quotation marks is the candidate's own earlier value, so this mark follows through from it.	✓
The answer	A1	$1350$	✓
Unsupported answers	Note	Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓
Special case	SC B2	SCB2 for $4275$	✓
Where that special case comes from	Note	$4275$ is $76\%$ of $5625$ , so it is the count of the bicycles that are not blue. The whole weekly chain has been built correctly and the wrong share taken at the last step, which is why the special case is worth two marks and not none.	✓

Full marks: 3/3

A circle has a radius of $16$ cm.

Work out the circumference of the circle.
Give your answer correct to $3$ significant figures.
[2 marks]

[Total 2 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

A circle whose radius is

r = 16

cm.

The length given is the radius, so the distance right across the circle is twice it.

Find

The circumference of the circle - the distance all the way round the edge. The answer is wanted correct to

3

significant figures.

Plan the Solution

Double the radius to get the diameter, because $C = \pi d$ is built on the distance right across the circle.
Multiply that diameter by $\pi$ , using the calculator's own value of $\pi$ and not a rounded one.
Round at the very end, and only then, to $3$ significant figures.

Worked Solution [2 marks]

Rule - Circumference of a circle:

C = \pi d

, where

d

is the diameter. The diameter is twice the radius, so this is the same rule as

C = 2 \pi r

Step 1: Double the radius to get the diameter

d = 2 \times 16 = 32

(Reason: The circumference formula multiplies

\pi

by the distance right across the circle, and that distance is two radii, so

32

cm.)

Step 2: Multiply the diameter by

\pi

C = \pi \times 32 = 100.5309649\ldots

(Reason: The calculator's own value for

\pi

is used, so every figure is still there when the rounding comes. Nothing is written down to

2

3

figures at this stage.)

Step 3: Round to

3

significant figures

C = 101 \text{ cm}

(Reason: The first three significant figures of

100.5309649\ldots

are

1

0

and

0

, and the digit after them is

5

, so the last of them rounds up and

100

becomes

101

101 \text{ cm}

Verification

Check 1: Work the calculation backwards. Dividing the circumference by

\pi

must give the diameter back, and halving that must return the radius the question started with.

\dfrac{100.5309649\ldots}{\pi} = 32

, and half of

32

16

cm, which is the radius given.

Check 2: Squeeze it.

\pi

lies between

3

and

3.2

, so the circumference has to lie between those two multiples of the diameter.

3 \times 32 = 96

and

3.2 \times 32 = 102.4

, so the circumference is somewhere between

96

cm and

102.4

cm, and

101

cm sits inside that.

Check 3: Redo it with

\dfrac{22}{7}

in place of

\pi

, one of the approximations the mark scheme allows.

\dfrac{22}{7} \times 32 = 100.5714\ldots

, which still rounds to

101

3

significant figures.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $2 \times \pi \times 16$ or $\pi \times 32$	M1	allow $3.14\ldots$ or $\dfrac{22}{7}$ for $\pi$ . The printed row writes the $32$ in quotation marks, so the candidate's own value for $2 \times 16$ may be used here.	✓
$101$ Working not required, so correct answer scores full marks (unless from obvious incorrect working)	A1	accept $100$ to $101$	✓
NB, printed in the A1 row's notes	Note	An answer of $101$ reached from working written as $\pi \times 16^{2}$ scores M0A0. The printed row glosses that working as $32\pi$ , the value such a candidate has evidently keyed in. This is the 'unless from obvious incorrect working' exception in the row above doing its job: the answer on the line matches, but the method written beside it is the area formula, so neither mark is earned. Separately, the range that row accepts is what the approximations allowed by the M1 need - $3.14 \times 32 = 100.48$ gives $100$ , and $\dfrac{22}{7}$ gives $101$ .	✓

Full marks: 2/2

The diagram shows the positions of two lighthouses, $A$ and $B$

The bearing of $B$ from $A$ is $073^{\circ}$

Work out the bearing of $A$ from $B$ [2 marks]

[Total 2 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

Two lighthouses,

A

and

B

, with a north line drawn at each

The bearing of

B

from

A

073^{\circ}

The two north lines point the same way, so they are parallel

Find

The bearing of

A

from

B

, measured clockwise from the north line at

B

Plan the Solution

A bearing is the clockwise turn from north, and it is always written with three figures.
Use the parallel north lines to find the angle at $B$ between north and the line $BA$ .
That angle opens the wrong way, so take it off a full turn of $360^{\circ}$ to get the clockwise bearing.

Worked Solution [2 marks]

Rule - Back bearing: the north lines at

A

and

B

are parallel, so the bearing of

A

from

B

and the bearing of

B

from

A

always differ by

180^{\circ}

Step 1: Mark the angle the diagram gives you at

A

\angle NAB = 73^{\circ}

(Reason: The bearing

073^{\circ}

is the clockwise turn from the north line at

A

round to the line

AB

, which is exactly the angle printed on the diagram.)

Step 2: Use the parallel north lines to get the angle at

B

180^{\circ} - 73^{\circ} = 107^{\circ}

(Reason: The north lines are parallel and

AB

cuts across both of them, so the angle at

A

and the angle at

B

between north and

BA

are allied (co-interior) angles and add to

180^{\circ}

Step 3: Turn clockwise from north at

B

360^{\circ} - 107^{\circ} = 253^{\circ}

(Reason: The

107^{\circ}

opens anticlockwise from the north line at

B

, and a bearing must be measured clockwise, so take that angle off the full turn of

360^{\circ}

253^{\circ}

Verification

Check 1: Reverse it. Taking

180^{\circ}

off a back bearing must return the bearing you started with.

253^{\circ} - 180^{\circ} = 073^{\circ}

Check 2: Test the parallel-line step on its own: allied angles across a transversal must add to

180^{\circ}

73^{\circ} + 107^{\circ} = 180^{\circ}

Check 3: Sense check the direction.

B

is north-east of

A

, so

A

must be south-west of

B

, and every south-west bearing lies between

180^{\circ}

and

270^{\circ}

180^{\circ} < 253^{\circ} < 270^{\circ}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$180 + 73$ or $360 - (180 - 73)$ or $360 - 107$ or $270 - (90 - 73)$ or $270 - 17$ or $270 - (180 - 90 - 73)$	M1	or for $73$ or for $107$ or for $253$ seen in the correct place on the diagram by point $B$ or correctly identified by labelling	✓
$253$	A1	Working not required, so correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 2/2

Here is a list of ingredients to make a mushroom pasta bake for $4$ people.

$\begin{array}{|c|}\hline \textbf{Mushroom pasta bake} \\ \textbf{Ingredients for 4 people} \\ 100 \text{ g mushrooms} \\ 50 \text{ g grated Cheddar} \\ 300 \text{ g penne} \\ 40 \text{ g butter} \\ 2 \text{ onions} \\ \hline \end{array}$

Owen is going to make the mushroom pasta bake for $12$ people.

(a) Work out how much grated Cheddar he needs. [2 marks]

Nadia makes the mushroom pasta bake.
She uses $180$ g of butter.

(b) Work out how many people Nadia makes the mushroom pasta bake for. [2 marks]

(a)(b)

[Total 4 marks]

Show solution & mark schemeHide solution & mark scheme

Question 17 - Exam Solution

Understanding the Question

Given

The recipe makes enough for

4

people.

It uses

50

g of grated Cheddar and

40

g of butter.

Owen is cooking for

12

people.

Nadia uses

180

g of butter.

Find

(a) The mass of grated Cheddar needed for

12

people. (b) The number of people

180

g of butter is enough for.

Plan the Solution

Both parts are the same idea. A recipe scales up and down in one ratio, so every ingredient is multiplied by the same number.
(a) Compare the two numbers of people to get that multiplier, then apply it to the $50$ g of Cheddar.
(b) Compare the two masses of butter to get the multiplier, then apply it to the $4$ people the recipe serves.
The unitary method - working out one person's share first - reaches the same answers and is used in the checks.

Worked Solution [4 marks]

Rule - Scaling a recipe: every ingredient, and the number of people, is multiplied by the same scale factor, and

\text{scale factor} = \dfrac{\text{new amount}}{\text{recipe amount}}

Step 1: part (a) - find the scale factor

\dfrac{12}{4} = 3

(Reason: the recipe serves

4

people and Owen is cooking for

12

, so every ingredient is multiplied by

3

Step 2: part (a) - scale the grated Cheddar

50 \times 3 = 150

(Reason: the recipe uses

50

g of Cheddar for

4

people, so three times the recipe needs three times the Cheddar, in grams.)

Step 3: part (b) - find the scale factor from the butter

\dfrac{180}{40} = 4.5

(Reason: here the multiplier has to come from the butter, because that is the ingredient Nadia measured out.)

Step 4: part (b) - scale the number of people

4 \times 4.5 = 18

(Reason: the recipe serves

4

people, and the number of people is multiplied by the same scale factor as every ingredient.)

(a)

150

g(b)

18

people

Verification

Check 1: Part (a) by the unitary method. One person needs

\dfrac{50}{4} = 12.5

g of grated Cheddar, so

12

people need

12.5 \times 12

150

g, which is the answer to part (a).

Check 2: Part (b) by the unitary method. One person needs

\dfrac{40}{4} = 10

g of butter, so

180

g of butter is enough for

\dfrac{180}{10}

people.

18

people, which is the answer to part (b).

Check 3: Work backwards from part (b). Serving

18

people is

\dfrac{18}{4} = 4.5

recipes, and that much butter is

4.5 \times 40 = 180

g. The butter comes back to the

180

g Nadia used, and

18

is larger than

12

, as it must be, because

12

people would use only

120

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $\dfrac{12}{4}\ (= 3)$ or $\dfrac{50}{4}\ (= 12.5)$ oe	M1	For a correct method to find the scale factor, or a correct calculation to find the amount of grated Cheddar for one person.	✓
(a) $150$	A1	Working not required, so a correct answer scores full marks (unless from obvious incorrect working).	✓
(b) $\dfrac{180}{40}\ (= 4.5)$ oe or $\dfrac{40}{4}\ (= 10)$	M1	For a correct method to find the scale factor, or a correct calculation to find the amount of butter for one person.	✓
(b) $18$	A1	Working not required, so a correct answer scores full marks (unless from obvious incorrect working).	✓

Full marks: 4/4

Continue to questions 18 to 28

The remaining 11 questions, with the same full worked solutions and mark schemes

Frequently asked questions

There are 28 questions worth 100 marks in total, sat over 2 hours. It is Foundation tier and a calculator is allowed throughout, unlike UK GCSE Maths, where one paper is non-calculator.

Foundation tier targets grades 1 to 5, so grades 6 to 9 are only available on Higher tier. About 40 per cent of the questions are targeted at grades 4 and 5 and appear on both Paper 2F and Paper 2H, so the top of the Foundation paper overlaps with the bottom of the Higher paper.

Yes. The paper states in its own instructions that without sufficient working, correct answers may be awarded no marks. Several questions ask you to show your working clearly or to show clear algebraic working, and on those a bare answer scores nothing. That is why every solution here sets out the method mark by mark.

Yes, a Foundation tier formulae sheet is printed in the paper. It gives the area of a trapezium, the volume of a prism, the volume of a cylinder and the curved surface area of a cylinder. Everything else has to be recalled, so Pythagoras theorem, the angle facts and the percentage methods used on this paper are not provided. Nothing may be written on the formulae page.

Both are published by Pearson Edexcel and are linked directly from this page as PDF files. The solutions here are original: every question has been reworded, but all the numbers match the original paper, so the answers agree with the official mark scheme. This resource reproduces neither the exam paper nor the official mark scheme.

Keep revising

Once you have worked through this 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 4MA1/2F, Wednesday 4 June 2025: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 17 of 28

Question 1, Calculator allowed

Question 1 - Exam Solution

Question 2, Calculator allowed

Question 2 - Exam Solution

Question 3, Calculator allowed

Question 3 - Exam Solution

Question 4, Calculator allowed

Question 4 - Exam Solution

Question 5, Calculator allowed

Question 5 - Exam Solution

Question 6, Calculator allowed

Question 6 - Exam Solution

Question 7, Calculator allowed

Question 7 - Exam Solution

Question 8, Calculator allowed

Question 8 - Exam Solution

Question 9, Calculator allowed

Question 9 - Exam Solution

Question 10, Calculator allowed

Question 10 - Exam Solution

Question 11, Calculator allowed

Question 11 - Exam Solution

Question 12, Calculator allowed

Question 12 - Exam Solution

Question 13, Calculator allowed

Question 13 - Exam Solution

Question 14, Calculator allowed

Question 14 - Exam Solution

Question 15, Calculator allowed

Question 15 - Exam Solution

Question 16, Calculator allowed

Question 16 - Exam Solution

Question 17, Calculator allowed

Question 17 - Exam Solution

Frequently asked questions

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