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Edexcel IGCSE 4MA1/2F, Wednesday 4 June 2025: Worked Solutions, Questions 18 to 28

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

Back to questions 1 to 17

This is the rest of the paper. Questions 1 to 17, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/2F (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.

Download printable PDF

All 28 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 18 to 28 of 28

Callum has six tiles.
He writes a number on each tile so that

the range of the numbers is $10$
the median of the numbers is $7.5$
the mode of the numbers is $6$

Callum arranges the tiles so that the numbers are in order of size.

Three of the numbers are hidden.

Complete the tiles above to show the three numbers that are hidden. [3 marks]

[Total 3 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

Six numbers, one on each tile, laid out in order of size.

The tiles that can be read are the third,

6

, the fifth,

10

, and the sixth,

14

The range is

10

, the median is

7.5

and the mode is

6

Find

The number on the first tile, the number on the second tile and the number on the fourth tile.

Plan the Solution

The tiles are in order of size, so the smallest number is on the first tile and the largest is on the sixth. That turns the range into one subtraction.
Six numbers have no single middle value, so the median is the mean of the third and the fourth. The third is already on the page, so the fourth follows.
The mode is the only condition left, and the second tile is the only tile left, so the mode fixes it.
Finish by testing all three conditions on the completed row, and by checking the row is still in order of size.

Worked Solution [3 marks]

Rule - Range:

\text{largest} - \text{smallest}

. Median of six values in order: the mean of the third and the fourth,

\dfrac{\text{3rd} + \text{4th}}{2}

. Mode: the value that occurs most often.

Step 1: use the range to find the first tile

14 - 10 = 4

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

14

on the last tile and the smallest is on the first. The range is the largest take away the smallest, so the smallest sits

10

below

14

Step 2: use the median to find the fourth tile

2 \times 7.5 = 15

15 - 6 = 9

(Reason: Six numbers have no single middle value, so the median is the mean of the third and the fourth. Those two must therefore total twice

7.5

, and the third tile already reads

6

, so take that away from the total.)

Step 3: use the mode to find the second tile

4, 6, 6, 9, 10, 14

(Reason: The second tile lies between

4

and

6

in the order. A

5

would leave every value appearing once, so there would be no mode at all, and a

4

would make

4

the most common value instead. Only a second

6

makes

6

the mode.)

First tile:

4

Second tile:

6

Fourth tile:

9

The completed row:

4, 6, 6, 9, 10, 14

Verification

Check 1 - the range: Take the smallest number on the completed row away from the largest.

14 - 4 = 10

Check 2 - the median: Six numbers, so average the third and the fourth.

\dfrac{6 + 9}{2} = 7.5

Check 3 - the mode: Count how many times each value appears on the completed row.

6

appears twice and every other value appears once, so the mode is

6

Check 4 - the order: Read the completed row from left to right. Every number is at least as large as the one before it, so the tiles are still in order of size.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
First tile	B1	First tile $4$	✓
Second tile	B1	Second tile $6$ , or a list of $6$ numbers with a mode of $6$	✓
Fourth tile	B1	Fourth tile $9$	✓
Special case	SC B2	for $4$ , $6$ and $9$ in the incorrect order	✓

Full marks: 3/3

(a) On the grid, draw the enlargement of shape $A$ with scale factor $3$ and centre $(0, 1)$ . [2 marks]

(b) Give a full description of the single transformation that maps triangle $P$ onto triangle $Q$ . [3 marks]

(b)

[Total 5 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

Shape

A

on the grid has vertices

(1, 1)

(3, 1)

(3, 4)

(2, 4)

(2, 2)

and

(1, 2)

The enlargement has scale factor

3

and centre

(0, 1)

Triangle

P

has vertices

(2, 2)

(4, 2)

and

(2, 6)

Triangle

Q

has vertices

(6, 8)

(8, 8)

and

(8, 4)

Find

(a) the six vertices of the enlarged shape, so that it can be drawn on the grid (b) the single transformation that maps

P

onto

Q

, described fully

Plan the Solution

(a) Work from the centre $(0, 1)$ . For each vertex of shape $A$ , count how far across and how far up it is from that centre.
(a) Multiply both counts by $3$ , then count out again from the centre to reach the image vertex.
(a) Join the six image points in the same order as shape $A$ .
(b) The two triangles are the same size, so look for a rotation, a reflection or a translation rather than an enlargement that changes the size.
(b) Match each vertex of $P$ with its image on $Q$ , join the pairs, and use the joins to find the centre.

Worked Solution [5 marks]

Rule - Enlargement: if a point is

a

across and

b

up from the centre, its image is

3a

across and

3b

up from that same centre. Rule - Describing a rotation: give the word rotation, the angle, the direction (not needed for

180^\circ

) and the centre.

Step 1: Count from the centre

(0, 1)

to the first vertex

1 - 0 = 1

1 - 1 = 0

(Reason: The vertex

(1, 1)

1

across and

0

up from the centre. In an enlargement every distance is measured from the centre, never from the origin, and here the centre is not the origin.)

Step 2: Multiply both counts by the scale factor

3

3 \times 1 = 3

3 \times 0 = 0

(1, 1) \rightarrow (3, 1)

(Reason: The image is

3

across and

0

up from

(0, 1)

, which lands on

(3, 1)

Step 3: Do the same for the other five vertices

(3, 1) \rightarrow (9, 1)

(3, 4) \rightarrow (9, 10)

(2, 4) \rightarrow (6, 10)

(2, 2) \rightarrow (6, 4)

(1, 2) \rightarrow (3, 4)

(Reason: Take

(3, 4)

as the pattern: it is

3

across and

3

up from the centre, so its image is

9

across and

9

up, and

9

up from

y = 1

y = 10

Step 4: Plot the six image points and join them in order

3 \times 2 = 6

3 \times 3 = 9

(Reason: Joining the images in the same order as shape

A

keeps the shape itself and changes only its size. Shape

A

2

squares wide and

3

squares tall, so the image is

6

wide and

9

tall.)

Step 5: Compare triangle

P

with triangle

Q

6 - 2 = 4

8 - 4 = 4

4 - 2 = 2

8 - 6 = 2

(Reason: Triangle

P

has a vertical side of

4

and a horizontal side of

2

, and so does triangle

Q

. The triangles are congruent, so the transformation does not change the size.)

Step 6: Match each vertex of

P

with its image on

Q

(2, 2) \rightarrow (8, 8)

(4, 2) \rightarrow (6, 8)

(2, 6) \rightarrow (8, 4)

(Reason: The right angle of

P

is at

(2, 2)

and the right angle of

Q

is at

(8, 8)

, so those two corners match. The short side and the long side then fix the other two.)

Step 7: Join each vertex to its image and find the midpoint

\dfrac{2 + 8}{2} = 5

\dfrac{4 + 6}{2} = 5

\dfrac{6 + 4}{2} = 5

(Reason: All three joins have the same midpoint

(5, 5)

: halfway between

2

and

8

5

, and halfway between

4

and

6

5

. A transformation that sends every point to the opposite side of one fixed point, the same distance away, is a half turn about that point.)

Step 8: Write the description in full

180^\circ

(Reason: A half turn is a rotation of

180^\circ

. A full description of a rotation needs the word rotation, the angle and the centre; at

180^\circ

no direction is needed, because clockwise and anticlockwise give the same image.)

(a) Enlargement drawn with vertices

(3, 1)

(9, 1)

(9, 10)

(6, 10)

(6, 4)

and

(3, 4)

(b) Rotation of

180^\circ

about

(5, 5)

Verification

Check 1: (a) Every image vertex must be

3

times as far from the centre as its object vertex. The vertex

(3, 4)

3

across and

3

up from

(0, 1)

, and its image

(9, 10)

9

across and

9

up.

9 = 3 \times 3

, so the image vertex is on the ray and three times as far out

Check 2: (a) An enlargement of scale factor

3

makes every length three times as long. Shape

A

2

squares wide and

3

squares tall; the shape drawn is

6

wide and

9

tall.

\dfrac{6}{2} = 3

and

\dfrac{9}{3} = 3

Check 3: (b) A half turn about

(5, 5)

sends

(x, y)

(10 - x, 10 - y)

. Test it on the vertex

(2, 6)

P

10 - 2 = 8

and

10 - 6 = 4

, which is the vertex

(8, 4)

Q

Check 4: (b) A translation would leave every side pointing the same way. In

P

the short side points right from

(2, 2)

; in

Q

it points left from

(8, 8)

. The sides point opposite ways, so the transformation is a turn and not a slide

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Correct shape drawn in correct position	B2	Shape drawn with coordinates $(3, 1)$ , $(9, 1)$ , $(9, 10)$ , $(6, 10)$ , $(6, 4)$ , $(3, 4)$	✓
(a) Shape of the correct size, wrong position	B1	for a shape of the correct size but in the wrong position	✓
(a) Marking note	Note	NB Overlay is available	✓
(b) The transformation is named	B1	Rotation, with no mention of any other transformation words or move, flip, transform, up, right etc	✓
(b) The angle	B1	$180^\circ$ , allow half turn	✓
(b) The centre	B1	(centre) $(5, 5)$ , must be a coordinate and not a vector	✓
(b) Alternative full description	B2	for enlargement scale factor $-1$ (Ignore any reference to clockwise or anticlockwise)	✓
(b) Marking note	Note	Turn is not sufficient	✓

Full marks: 5/5

Show that $7\dfrac{1}{3} - 3\dfrac{4}{7} = 3\dfrac{16}{21}$
You must show all your working. [3 marks]

[Total 3 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

The subtraction of two mixed numbers,

7\dfrac{1}{3} - 3\dfrac{4}{7}

The result that has to be proved:

7\dfrac{1}{3} - 3\dfrac{4}{7} = 3\dfrac{16}{21}

Find

A complete piece of working that ends at

3\dfrac{16}{21}

. 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.
The denominators $3$ and $7$ share no factor, so the lowest common denominator is $3 \times 7 = 21$ .
Rewrite both fractions over that denominator, then subtract the numerators.
Turn the improper fraction back into a mixed number and compare it with the printed result.
A second route avoids improper fractions altogether: keep the whole numbers, borrow one whole from the $7$ , and subtract in two parts. The mark scheme awards full marks for either route.

Worked Solution [3 marks]

Rule - Mixed numbers, then a common denominator:

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

, and

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

Step 1: Write each mixed number as an improper fraction

7\dfrac{1}{3} = \dfrac{7 \times 3 + 1}{3} = \dfrac{22}{3}

3\dfrac{4}{7} = \dfrac{3 \times 7 + 4}{7} = \dfrac{25}{7}

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

3

and the

7

stay where they are. This alone earns the first method mark.)

Step 2: Write both fractions over the common denominator

21

\dfrac{22}{3} = \dfrac{22 \times 7}{21} = \dfrac{154}{21}

\dfrac{25}{7} = \dfrac{25 \times 3}{21} = \dfrac{75}{21}

(Reason: The denominators

3

and

7

have no common factor, so the lowest common denominator is

3 \times 7 = 21

. Multiply the top and the bottom of the first fraction by

7

, and the top and the bottom of the second fraction by

3

. Whatever is done underneath must be done on top, or the value of the fraction changes.)

Step 3: Subtract the numerators

\dfrac{154}{21} - \dfrac{75}{21} = \dfrac{154 - 75}{21} = \dfrac{79}{21}

(Reason: Once both fractions are over the same denominator, only the numerators are subtracted. The denominator

21

is carried straight through, and

154 - 75 = 79

Step 4: Write the improper fraction back as a mixed number

79 = 3 \times 21 + 16

\dfrac{79}{21} = 3\dfrac{16}{21}

(Reason:

21

goes into

79

three times with

16

left over, so the whole-number part is

3

and the remainder

16

sits over

21

. That is the printed result, as required.)

Step 5: The same result without improper fractions

7\dfrac{1}{3} = 7\dfrac{7}{21}

3\dfrac{4}{7} = 3\dfrac{12}{21}

7\dfrac{7}{21} - 3\dfrac{12}{21} = 6\dfrac{28}{21} - 3\dfrac{12}{21} = 3\dfrac{16}{21}

(Reason: This time only the fraction parts are rewritten. Since

\dfrac{7}{21}

is smaller than

\dfrac{12}{21}

, take one whole from the

7

and add it to the fraction as

\dfrac{21}{21}

, which turns

7\dfrac{7}{21}

into

6\dfrac{28}{21}

. The whole numbers and the fractions then subtract separately, and this route is worth full marks too.)

7\dfrac{1}{3} - 3\dfrac{4}{7} = \dfrac{22}{3} - \dfrac{25}{7} = \dfrac{79}{21} = 3\dfrac{16}{21}

as required

Verification

Check 1: Add the answer back on. If the difference really is

3\dfrac{16}{21}

, then adding

3\dfrac{4}{7}

to it must return the number the question started from,

7\dfrac{1}{3}

3\dfrac{16}{21} + 3\dfrac{12}{21} = 6\dfrac{28}{21} = 7\dfrac{7}{21} = 7\dfrac{1}{3}

Check 2: Take the whole numbers apart first, which is the mark scheme's other route.

\dfrac{7}{21}

\dfrac{5}{21}

short of

\dfrac{12}{21}

, so the answer sits

\dfrac{5}{21}

below

4

4 - \dfrac{5}{21} = \dfrac{84 - 5}{21} = \dfrac{79}{21} = 3\dfrac{16}{21}

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

7\dfrac{1}{3} \approx 7.333333333

and

3\dfrac{4}{7} \approx 3.571428571

7.333333333 - 3.571428571 \approx 3.761904762

and

3\dfrac{16}{21} \approx 3.761904762

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Correct improper fractions, $\dfrac{22}{3}$ and $\dfrac{25}{7}$ , or the fractional parts written over a common denominator, $(7)\dfrac{7}{21}$ and $(3)\dfrac{12}{21}$ , or $(7)\dfrac{7a}{21a}$ and $(3)\dfrac{12a}{21a}$ .	M1	for correct improper fractions or fractional part of numbers written correctly over a common denominator	✓
$\dfrac{154}{21} - \dfrac{75}{21}$ or $\dfrac{22 \times 7}{21} - \dfrac{25 \times 3}{21}$ or $\dfrac{22 \times 7 - 25 \times 3}{21}$ or $\dfrac{154a}{21a} - \dfrac{75a}{21a}$ or $7\dfrac{7}{21} - 3\dfrac{12}{21} = 4 - \dfrac{5}{21}$ oe or $7\dfrac{7}{21} - 3\dfrac{12}{21} = 6\dfrac{28}{21} - 3\dfrac{12}{21}$ .	M1	for correct fractions with a common denominator with minus sign or mixed numbers to the stage shown. $\dfrac{154}{21} - \dfrac{75}{21}$ or $\dfrac{22 \times 7}{21} - \dfrac{25 \times 3}{21}$ implies the first M1	✓
$\dfrac{154}{21} - \dfrac{75}{21} = \dfrac{79}{21} = 3\dfrac{16}{21}$ or $4 - \dfrac{5}{21} = 3\dfrac{16}{21}$ or $7\dfrac{7}{21} - 3\dfrac{12}{21} = 6\dfrac{28}{21} - 3\dfrac{12}{21} = 3\dfrac{16}{21}$ . Working required.	A1	Dep on M2 for a correct answer from fully correct working. If a student shows that $3\dfrac{16}{21} = \dfrac{79}{21}$ then they must show correct working to $\dfrac{79}{21}$ and can gain full marks for this	✓
What the answer column asks for	Note	The answer column reads: a fully correct solution shown. The result $3\dfrac{16}{21}$ is printed in the question, so there is no separate answer to write on an answer line and every mark here is for the working.	✓

Full marks: 3/3

Using a ruler and a pair of compasses only, construct the perpendicular bisector of the line $AB$
Show all your construction lines. [2 marks]

[Total 2 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

A line segment

AB

, drawn on the page.

A ruler and a pair of compasses, and nothing else.

Find

The perpendicular bisector of

AB

: the straight line that cuts

AB

in half and crosses it at

90^{\circ}

. Every construction arc left showing on the diagram.

Plan the Solution

Open the compasses to a radius bigger than half of $AB$ , and do not change that setting again.
Stand the point on $A$ and draw an arc on each side of the line, then move the point to $B$ and draw two more.
The two pairs of arcs cross at two points. Rule the straight line through them.
Leave every arc on the page: the arcs are half of what is being marked.

Worked Solution [2 marks]

Rule - Perpendicular bisector: with the same radius from

A

and from

B

, the arcs meet at two points

P

and

Q

with

AP = BP = AQ = BQ

. A point that is the same distance from

A

as from

B

lies on the perpendicular bisector, so the line

PQ

is that bisector.

Step 1: Open the compasses to more than half of

AB

, then draw an arc from

A

on each side of the line

r > \dfrac{1}{2} AB

(Reason: If the radius

r

is smaller than half of

AB

, an arc drawn from

A

and an arc drawn from

B

never reach each other and nothing crosses. Any setting past halfway works, and it must not be touched again.)

Step 2: Keep the same radius and draw two arcs from

B

AP = BP = AQ = BQ = r

(Reason: The radius has not changed, so all four of those lengths are the same

r

. The arcs cross at two points:

P

on one side of

AB

and

Q

on the other.)

Step 3: Rule the straight line through

P

and

Q

PQ \text{ is the perpendicular bisector of } AB

(Reason:

P

and

Q

are each the same distance from

A

as from

B

, and every point with that property lies on the perpendicular bisector. Two points fix a line, so ruling

PQ

draws it.)

Step 4: Read off what the ruled line has done to

AB

APBQ \text{ is a rhombus}

AM = MB \quad \text{and} \quad \angle AMQ = 90^{\circ}

(Reason: All four sides of

APBQ

are the compass radius

r

, so it is a rhombus, and the diagonals of a rhombus cut each other in half at right angles. Those diagonals are

AB

and

PQ

, so

M

is the midpoint of

AB

and the angle there is a right angle.)

The perpendicular bisector of

AB

: the straight line ruled through the two points where the pairs of arcs cross, with all four arcs left on the diagram.

Verification

Check 1: Measure

AM

and

MB

along

AB

with the ruler. They come out equal, so the ruled line has cut

AB

in half.

Check 2: Fold the paper along the ruled line.

A

lands exactly on

B

, which happens only along the perpendicular bisector.

Check 3: Look at the quadrilateral

APBQ

: every side is one compass radius, because the setting never changed. A rhombus, and the diagonals of a rhombus bisect each other at right angles, which is exactly what this construction claims.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A fully correct perpendicular bisector of $AB$ , with 2 pairs of intersecting arcs shown	B2	B2 for a fully correct perpendicular bisector with 2 pairs of intersecting arcs shown (the line and the arcs can intersect on or within the overlay guidelines)	✓
Partial credit	(B1)	B1 for 2 pairs of intersecting arcs and no perpendicular bisector drawn, or for a correct perpendicular bisector drawn within or on guidelines but no arcs or insufficient arcs, or one pair of intersecting arcs and perpendicular bisector drawn on just one side of $AB$	✓
Marking note	Note	NB Overlay is available	✓

Full marks: 2/2

$ABCD$ and $EFGH$ are similar quadrilaterals.
The diagram shows both of them.

$AB = 5$ cm, $BC = 4$ cm, $CD = y$ cm
$EF = x$ cm, $FG = 10$ cm, $GH = 24$ cm

(a) Find the value of $x$ . [2 marks]

(b) Find the value of $y$ . [2 marks]

x =y =

[Total 4 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

ABCD

and

EFGH

are similar, so one is an enlargement of the other.

AB = 5

cm,

BC = 4

cm,

CD = y

EF = x

cm,

FG = 10

cm,

GH = 24

The two names are written in matching order, so the sides pair off in that order.

Find

(a) the value of

x

, which is the length of

EF

(b) the value of

y

, which is the length of

CD

Plan the Solution

Pair the sides off by their letters: $AB$ with $EF$ , $BC$ with $FG$ , and $CD$ with $GH$ .
Only $BC$ and $FG$ have a number at both ends, so that is the pair the scale factor has to come from.
Part (a) asks for a side of the larger quadrilateral, so multiply by the scale factor.
Part (b) asks for a side of the smaller one, so divide by the scale factor instead.

Worked Solution [4 marks]

Rule - Similar shapes: matching sides are in the same ratio, so

\dfrac{EF}{AB} = \dfrac{FG}{BC} = \dfrac{GH}{CD}

Step 1: pair off the sides that match

AB \to EF, \quad BC \to FG, \quad CD \to GH

(Reason: The names are written in matching order, so the side named by the first two letters of one is the side named by the first two letters of the other.)

Step 2: work out the scale factor

\text{scale factor} = \dfrac{FG}{BC} = \dfrac{10}{4} = 2.5

(Reason:

BC

and

FG

are the only matching pair with a number on both quadrilaterals, so they are the only pair that can give the scale factor.)

Step 3: (a) multiply, to cross to the larger quadrilateral

x = 5 \times 2.5 = 12.5

(Reason:

EF

is on the larger quadrilateral and the side that matches it is

AB = 5

cm, so

5

is multiplied by the scale factor.)

Step 4: (b) divide, to cross back to the smaller quadrilateral

y = \dfrac{24}{2.5} = 9.6

(Reason:

CD

is on the smaller quadrilateral and the side that matches it is

GH = 24

cm, so

24

is divided by the scale factor.)

(a)

x = 12.5

(b)

y = 9.6

Verification

Check 1: Divide each side of

EFGH

by the side of

ABCD

that matches it. Similar shapes have to give the same ratio every time.

\dfrac{12.5}{5} = 2.5

\dfrac{10}{4} = 2.5

\dfrac{24}{9.6} = 2.5

Check 2: Work the other way round. Shrinking the larger quadrilateral by the scale factor has to give the smaller one's own numbers back.

\dfrac{12.5}{2.5} = 5

and

9.6 \times 2.5 = 24

Check 3: Size sense.

EFGH

is the larger quadrilateral, so every side of it must be longer than the side of

ABCD

it matches.

12.5 > 5

and

24 > 9.6

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A correct scale factor, or a correct equation in $x$	M1	For a correct scale factor, which may be expressed as a fraction, a decimal or a ratio, and may or may not be used: $\dfrac{10}{4} = \dfrac{5}{2} = 2.5$ or $\dfrac{4}{10} = \dfrac{2}{5} = 0.4$ , or for a correct equation in $x$ : $\dfrac{x}{5} = \dfrac{10}{4}$ oe or $\dfrac{x}{10} = \dfrac{5}{4}$ oe. Allow any letter for $x$ .	✓
(a) The value of $x$	A1	$12.5$ oe, eg $\dfrac{50}{4}$ or $\dfrac{25}{2}$ or $12\dfrac{1}{2}$ or $12\dfrac{2}{4}$ . Working is not required, so a correct answer scores full marks unless it comes from obvious incorrect working.	✓
(b) A correct method, or a correct equation in $y$	M1	$\dfrac{24}{[2.5]}$ oe, or a correct equation in $y$ : $\dfrac{y}{24} = \dfrac{4}{10}$ oe or $\dfrac{y}{24} = \dfrac{5}{[12.5]}$ oe or $\dfrac{y}{4} = \dfrac{24}{10}$ oe or $\dfrac{y}{5} = \dfrac{24}{[12.5]}$ oe. Follow through: $[2.5]$ is their scale factor from (a), and $[12.5]$ is their answer to (a). Allow any letter for $y$ .	✓
(b) The value of $y$	A1	$9.6$ oe, eg $\dfrac{48}{5}$ or $9\dfrac{3}{5}$ . Working is not required, so a correct answer scores full marks unless it comes from obvious incorrect working. If (a) gives $x = 9.6$ and (b) gives $y = 12.5$ , the marks are M1 A0 M1 A0.	✓

Full marks: 4/4

Nico, Ella and Rosa run a craft stall together.
One Saturday the stall takes £ $240$ , which they share in the ratios $3 : 4 : 5$

Ella and Rosa each give £ $10$ of their share to Nico.

Work out the ratio of the amounts of money that Nico, Ella and Rosa now have.
Give your answer in its simplest form. [4 marks]

[Total 4 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

Nico, Ella and Rosa share £

240

in the ratios

3 : 4 : 5

, so Nico takes

3

parts, Ella takes

4

parts and Rosa takes

5

parts.

Ella and Rosa each give £

10

of their share to Nico, so Nico is given £

10

twice.

Find

The ratio

\text{Nico} : \text{Ella} : \text{Rosa}

of the amounts of money they hold after the two gifts, in its simplest form.

Plan the Solution

Add the three parts of the ratio to find how many equal shares the £ $240$ is split into.
Divide by that number to find one share, then multiply to find what each of them starts with.
Move the money: Nico gains £ $10$ from Ella and £ $10$ from Rosa, and each of those two is £ $10$ worse off.
Write the three new amounts as a ratio, then divide every part by their highest common factor.

Worked Solution [4 marks]

Rule - Sharing in a given ratio:

\text{one share} = \dfrac{\text{total}}{\text{sum of the parts}}

, and each person receives their own number of parts lots of one share.

Step 1: count the shares the money is split into

3 + 4 + 5 = 12

(Reason: The ratio

3 : 4 : 5

splits the money into

12

equal shares. It is the parts added together that the total is divided by, never one part on its own.)

Step 2: work out one share

\dfrac{240}{12} = 20

(Reason: The £

240

is split into

12

equal shares, so one share is worth £

20

Step 3: work out what each of them starts with

\text{Nico: } 3 \times 20 = 60

\text{Ella: } 4 \times 20 = 80

\text{Rosa: } 5 \times 20 = 100

(Reason: Each of them takes their own number of parts lots of one share, and the three amounts add back to £

240

Step 4: move the two £

10

gifts

\text{Nico: } 60 + 10 + 10 = 80

\text{Ella: } 80 - 10 = 70

\text{Rosa: } 100 - 10 = 90

(Reason: Nico is given £

10

by Ella and another £

10

by Rosa, so he gains £

20

altogether, while Ella and Rosa are each £

10

worse off.)

Step 5: write the new amounts as a ratio and simplify

80 : 70 : 90

\dfrac{80}{10} : \dfrac{70}{10} : \dfrac{90}{10} = 8 : 7 : 9

(Reason: The highest common factor of

80

70

and

90

10

, so dividing every part by

10

leaves the ratio in its simplest form.)

8 : 7 : 9

Verification

Check 1: Add the three new amounts. The £

20

that leaves Ella and Rosa arrives with Nico, so the total has to still be £

240

80 + 70 + 90 = 240

Check 2: Work backwards. Share £

240

in the ratio

8 : 7 : 9

, whose parts add to

24

, and the three amounts must come back.

\dfrac{240}{24} = 10

, then

8 \times 10 = 80

7 \times 10 = 70

9 \times 10 = 90

Check 3: Test that nothing is left to cancel, because the answer has to be in its simplest form.

7

is prime and divides neither

8

nor

9

. The only whole number that divides all three parts is

1

, so

8 : 7 : 9

cannot be cancelled any further.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct method to find the value of one share	M1	For a correct method to find the value of one share: $\dfrac{240}{3 + 4 + 5} = 20$ or $240 \times \dfrac{3}{3 + 4 + 5} = 60$ oe or $240 \times \dfrac{4}{3 + 4 + 5} = 80$ oe or $240 \times \dfrac{5}{3 + 4 + 5} = 100$ oe. NB $\dfrac{240}{3} = 80$ , $\dfrac{240}{4} = 60$ and $\dfrac{240}{5} = 48$ scores M0.	✓
The correct values for two of the three people after the gifts	M1	For the correct values for $2$ of the people after Ella and Rosa give Nico £ $10$ . For two of: (Nico) $3 \times [20] + 10 + 10 = 80$ or $[60] + 10 + 10 = 80$ ; (Ella) $4 \times [20] - 10 = 70$ or $[80] - 10 = 70$ ; (Rosa) $5 \times [20] - 10 = 90$ or $[100] - 10 = 90$ . A value in square brackets is the candidate's own value from the first method mark, so this mark follows through on it.	✓
All three new amounts, or the final ratio in the wrong order, or the final ratio unsimplified	M1	For all $3$ of $80$ , $70$ and $90$ correct (ignore units), or for the correct values for the final ratio in the wrong order (ignore units), eg $9 : 7 : 8$ oe, or for the correct values for the final ratio unsimplified (ignore units), eg $4 : 3.5 : 4.5$ oe.	✓
The final ratio in its simplest form	A1	$8 : 7 : 9$ . Working is not required, so a correct answer scores full marks unless it comes from obvious incorrect working. Other orders are acceptable if they are labelled correctly on the answer line or in the working.	✓

Full marks: 4/4

Lukas buys a vintage wristwatch for $4000$ Swiss francs.
The value of the wristwatch increases by $7\%$ each year.

Work out the value of the wristwatch at the end of $3$ years.
Give your answer correct to the nearest Swiss franc. [3 marks]

Swiss francs

[Total 3 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

Lukas pays

4000

Swiss francs for the wristwatch.

The value of the wristwatch increases by

7\%

each year, and each year's increase is worked out on the value at the start of that year, not on the price he paid.

Find

The value of the wristwatch at the end of

3

years, correct to the nearest Swiss franc.

Plan the Solution

Turn the $7\%$ rise into a single decimal multiplier, so that one multiplication does a whole year.
Use that multiplier once for each of the $3$ years, which means raising it to the power $3$ .
Multiply the $4000$ Swiss francs by it to get the value at the end of the third year.
Round only at the very end, so that no accuracy is lost part-way through.

Worked Solution [3 marks]

Rule - Repeated percentage change:

\text{final value} = \text{starting value} \times (\text{multiplier})^{n}

, where the multiplier for a rise of

p\%

1 + \dfrac{p}{100}

and

n

is the number of years.

Step 1: write the increase as one multiplier

100\% + 7\% = 107\%

\dfrac{107}{100} = 1.07

(Reason: Adding

7\%

to a value leaves

107\%

of it, and

107\%

written as a decimal is

1.07

. Multiplying by

1.07

does the whole year in one move, so there is no separate addition to make.)

Step 2: use the multiplier once for each year

4000 \times 1.07 \times 1.07 \times 1.07 = 4000 \times 1.07^{3}

(Reason: The value is multiplied by

1.07

at the end of every year, so over

3

years it is multiplied by

1.07

three times. Three of the same factor is written as a power, and the power is the number of years.)

Step 3: work out the value at the end of the third year

4000 \times 1.07^{3} = 4900.172

(Reason: A calculator gives

1.07^{3} = 1.225043

, and the

4000

Swiss francs multiplied by that is the value at the end of the third year, before any rounding.)

Step 4: round to the nearest Swiss franc

4900.172 \approx 4900

(Reason: The first digit after the decimal point is

1

, which is less than

5

, so the whole-number part does not change. The rounding is done once, at the end, so the

4579.6

and the other yearly values are never rounded on the way through.)

4900

Swiss francs

Verification

Check 1: Do the three rises one at a time instead, working each year's

7\%

out on the value at the start of that year.

4000 + 280 = 4280

, then

4280 + 299.6 = 4579.6

, then

4579.6 + 320.572 = 4900.172

Check 2: Undo the three rises. Dividing by

1.07

three times has to give back the price Lukas paid.

\dfrac{4900.172}{1.07} = 4579.6

\dfrac{4579.6}{1.07} = 4280

\dfrac{4280}{1.07} = 4000

Check 3: Test the size against simple interest. Taking

7\%

of the original

4000

three times gives

3 \times 280 = 840

, and compound growth must beat that, because the second and third rises are worked out on larger amounts. The rise here is

4900.172 - 4000 = 900.172

, which is

900.172 - 840 = 60.172

more than the flat

840

. Bigger, as compounding requires, and only a little bigger over

3

years.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct method to find $7\%$ of $4000$ or $107\%$ of $4000$	M1	For finding $7\%$ of $4000$ or $107\%$ of $4000$ : $0.07 \times 4000\,(= 280)$ or $1.07 \times 4000\,(= 4280)$ .	✓
The three yearly rises taken in turn, or the three rises added together	M1	For $4000 + [280]\,(= 4280)$ oe and $0.07 \times [4280]\,(= 299.6)$ and $[4280] + [299.6]\,(= 4579.6)$ and $0.07 \times [4579.6]\,(= 320.572)$ , or for $[280] + [299.6] + [320.572]\,(= 900.172)$ . A value in square brackets is the candidate's own value from earlier in the question, so this mark follows through on it. The printed row gives that last total as $900$ with its decimal part in brackets, so either form is accepted.	✓
The power method, worth both method marks on its own	M2	For $1.07^{3} \times 4000$ or $1.07^{4} \times 4000\,(= 5243.18404)$ .	✓
The value at the end of $3$ years, to the nearest Swiss franc	A1	$4900$ . Allow answers in the range $4900$ to $4901$ . Working is not required, so a correct answer scores full marks unless it comes from obvious incorrect working.	✓
Special case, if no other mark is awarded	SCB1	If no other mark is awarded, SCB1 for $4000 \times 0.07 \times 3\,(= 840)$ or $4000 \times 0.21\,(= 840)$ or $4000 + 4000 \times 0.07 \times 3\,(= 4840)$ or $4000 \times 1.21\,(= 4840)$ or $0.93 \times 4000\,(= 3720)$ or $0.79 \times 4000\,(= 3160)$ or $0.93^{3} \times 4000\,(= 3217.428)$ or $4000 \times 1.07^{2}\,(= 4579.6)$ .	✓
Where the special cases come from, and how the M2 row is used	Note	This row awards nothing. Each special case above is one nameable slip. $4000 \times 0.07 \times 3$ , $4000 \times 0.21$ and $4000 \times 1.21$ all take $7\%$ of the original $4000$ three times, which is simple interest rather than a value that grows each year. $0.93 \times 4000$ , $0.79 \times 4000$ and $0.93^{3} \times 4000$ treat the change as a fall instead of a rise. $4000 \times 1.07^{2}$ uses the multiplier one year too few. On the printed scheme the M2 row sits beside both M1 rows and replaces them, so a candidate who writes it has the two method marks outright.	✓

Full marks: 3/3

Here is a pair of simultaneous equations.

$3x + 5y = 8$
$4x + y = -3.5$

Work out the value of $x$ and the value of $y$ .
You must show clear algebraic working. [3 marks]

x =y =

[Total 3 marks]

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Question 25 - Exam Solution

Understanding the Question

Given

The first equation,

3x + 5y = 8

The second equation,

4x + y = -3.5

. Both are linear, so exactly one pair of values fits them both.

Find

The value of

x

and the value of

y

that satisfy both equations at once. Algebraic working is required, so the pair has to be reached by elimination or substitution, never by trial.

Plan the Solution

The $y$ term in the second equation is just $y$ , so multiplying that equation by $5$ is enough to make both equations carry $5y$ .
Subtract one equation from the other to remove $y$ and leave a single equation in $x$ .
Solve that for $x$ , then substitute the value back into the simpler equation to find $y$ .
Test the pair in both original equations. A sign slip often satisfies one equation and fails the other, so one substitution is not a check.

Worked Solution [3 marks]

Rule - Elimination: multiply one or both equations so that the coefficients of the same letter match, then add or subtract the equations to remove that letter.

Step 1: label the two equations

3x + 5y = 8 \qquad \text{(1)}

4x + y = -3.5 \qquad \text{(2)}

(Reason: Numbering them keeps the working clear, so every line below can say which equation it came from. That is part of what showing clear algebraic working means.)

Step 2: match the

y

terms

5 \times (4x + y) = 5 \times (-3.5)

20x + 5y = -17.5 \qquad \text{(3)}

(Reason: Equation (2) carries just

y

, so multiplying every term of it by

5

makes both equations carry

5y

. The right-hand side is multiplied as well:

5 \times (-3.5) = -17.5

Step 3: subtract to eliminate

y

, then solve for

x

(3x + 5y) - (20x + 5y) = 8 - (-17.5)

-17x = 25.5

x = \dfrac{25.5}{-17} = -1.5

(Reason: Both equations now contain

5y

, so subtracting removes

y

completely and leaves

3x - 20x = -17x

. On the right, subtracting a negative adds:

8 - (-17.5) = 8 + 17.5 = 25.5

. Dividing a positive by a negative then gives a negative value.)

Step 4: substitute back to find

y

4 \times (-1.5) + y = -3.5

-6 + y = -3.5

y = -3.5 + 6 = 2.5

(Reason: Equation (2) is the simpler one to put

x

back into, and

4 \times (-1.5) = -6

. Moving that

-6

across the equals sign changes its sign, so

6

is added to

-3.5

x = -1.5

y = 2.5

Verification

Check 1: Put both values into equation (1), which was not the equation used to find

y

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

Check 2: Put both values into equation (2) as well. A sign slip usually satisfies one equation and fails the other, so both have to be tested.

4 \times (-1.5) + 2.5 = -6 + 2.5 = -3.5

Check 3: Solve the pair a second way, eliminating

x

instead of

y

4

times equation (1) and

3

times equation (2) both give

12x

12x + 20y = 32

minus

12x + 3y = -10.5

gives

17y = 42.5

, so

y = \dfrac{42.5}{17} = 2.5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct method to eliminate $x$ or $y$	M1	For a correct method to eliminate $x$ or $y$ : coefficients of $x$ or $y$ the same and correct operator to eliminate the selected variable (condone any one arithmetic error in multiplication), or writing $x$ or $y$ in terms of the other variable and correctly substituting (condone missing brackets). For example $3x + 5y = 8$ with $20x + 5y = -17.5$ and subtracting ( $3x - 20x = 8 - (-17.5)$ or $-17x = 25.5$ ), or $12x + 20y = 32$ with $12x + 3y = -10.5$ and subtracting ( $20y - 3y = 32 - (-10.5)$ or $17y = 42.5$ ), or $3x + 5(-3.5 - 4x) = 8$ , or $4x + \dfrac{8 - 3x}{5} = -3.5$ , or $3\left(\dfrac{-3.5 - y}{4}\right) + 5y = 8$ , or $4\left(\dfrac{8 - 5y}{3}\right) + y = -3.5$ . NB The mark is for the method and not for the result of the method. However, if the correct result of the method is seen, the mark can be awarded.	✓
A correct method to find the other variable	M1	Dependent on the first M1. For a correct method to find the other variable by substitution of the found variable into one equation, or for repeating the above method to find the second variable. For example $3 \times [-1.5] + 5y = 8$ , or $4 \times [-1.5] + y = -3.5$ , or $y = -3.5 - 4 \times [-1.5]$ , or $y = \dfrac{8 - 3 \times [-1.5]}{5}$ , or $3x + 5 \times [2.5] = 8$ , or $4x + [2.5] = -3.5$ , or $x = \dfrac{-3.5 - [2.5]}{4}$ , or $x = \dfrac{8 - 5 \times [2.5]}{3}$ . A value in square brackets is the candidate's own value from the first method mark, so this mark follows through on it.	✓
Both values, with working shown	A1	$x = -1.5$ and $y = 2.5$ , oe, dependent on the first M1. The printed working column says that working is required, so an answer given with no algebraic working scores nothing here.	✓
Marking note	Note	oe means or equivalent: the same two values written another way, such as $x = -\dfrac{3}{2}$ and $y = \dfrac{5}{2}$ , are the same answer. The second method mark depends on the first, so a candidate who never reaches a correct elimination or substitution cannot pick it up by finding a second value from a wrong first one.	✓

Full marks: 3/3

(a) Solve the inequality $7 - 3t < 2t + 15$ [2 marks]

The region $R$ , shaded on the grid below, is bounded by three straight lines.

(b) Write down three inequalities that together describe the region $R$ . [3 marks]

(a)(b)(b)(b)

[Total 5 marks]

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Question 26 - Exam Solution

Understanding the Question

Given

Part (a): the inequality

7 - 3t < 2t + 15

, with

t

on both sides.

Part (b): a shaded region

R

on a grid, bounded by three straight lines.

Read off the grid: one line is upright through

2

on the

x

axis, one is flat through

3

on the

y

axis, and the sloping one joins

(0, 9)

(9, 0)

Find

Part (a): every value of

t

that makes the inequality true. Part (b): three inequalities that between them describe

R

and nothing outside it.

Plan the Solution

Part (a): rearrange it exactly as you would an equation. Move the $t$ terms to one side and the numbers to the other, then divide.
Keep an eye on the symbol. It turns round only when both sides are multiplied or divided by a negative number, or when the two sides are swapped over.
Part (b): write each boundary line as an equation first. Two of them are read straight off the axes; the sloping one comes from the two points where it meets them.
Then take one point well inside $R$ and see which way each symbol has to point.
Last, look at whether the edges belong to $R$ . The lines are drawn solid and the corners are part of the region, so each symbol carries an equals sign as well.

Worked Solution [5 marks]

Rule - Inequalities and regions: rearrange an inequality exactly as you would an equation, remembering that multiplying or dividing both sides by a negative number turns the symbol round; and describe a region by writing each boundary line as an equation first, then fixing the direction of every symbol by testing one point inside the region.

(a) Get the

t

terms on one side

7 - 3t < 2t + 15

7 - 3t + 3t < 2t + 15 + 3t

7 < 5t + 15

(Reason: Adding

3t

to both sides clears the

-3t

on the left and leaves

5t

on the right. Adding the same amount to both sides never changes which side is the bigger, so the symbol is untouched.)

(a) Get the number terms on the other side

7 - 15 < 5t + 15 - 15

-8 < 5t

(Reason: Taking

15

from both sides leaves the

5t

on its own, and on the left

7 - 15 = -8

. Subtracting the same amount from both sides leaves the symbol alone too.)

(a) Divide both sides by

5

\dfrac{-8}{5} < \dfrac{5t}{5}

-1.6 < t

(Reason: The divisor is

5

, which is positive, so the symbol stays exactly as it is. It is only dividing (or multiplying) by a negative number that turns it round. And

-\dfrac{8}{5} = -1.6

(a) Write the answer with

t

first

t > -1.6

(Reason: Reading

-1.6 < t

from the other end says that

t

is the greater of the two. Swapping the two sides over is the other move that turns the symbol round, and the two statements say the same thing.)

(b) Write down the equation of each boundary line

x = 2

y = 3

x + y = 9

(Reason: The upright line crosses the

x

axis at

2

and never moves left or right, so every point on it has

x = 2

. The flat line has

y = 3

all the way along. The sloping line joins

(0, 9)

(9, 0)

, and the two coordinates of every point on it add up to

9

(b) Test one point inside

R

to fix each direction

3 \geq 2

4 \geq 3

3 + 4 \leq 9

(Reason: The point

(3, 4)

sits well inside the shaded triangle. Its

x

is to the right of

x = 2

, its

y

is above

y = 3

, and

3 + 4 = 7

, which is below

9

. One point inside settles which way all three symbols point.)

(b) Decide whether the edges belong to the region

2 + 7 = 9

6 + 3 = 9

(Reason: The corners

(2, 7)

and

(6, 3)

lie on the sloping line, and the corner

(2, 3)

lies on both of the other two. The lines are drawn solid, so the edges count as part of

R

and each symbol carries an equals sign as well.)

(b) Write the three inequalities

x \geq 2

y \geq 3

x + y \leq 9

(Reason: Each boundary line has become an inequality pointing towards

R

. Together the three of them describe the shaded triangle and nothing outside it.)

(a)

t > -1.6

(b)

x \geq 2

y \geq 3

x + y \leq 9

Verification

Check 1: Part (a). Try a value on each side of the boundary. Put

t = 0

into both sides, then put

t = -2

into both sides.

7 - 0 = 7

and

2 \times 0 + 15 = 15

, and

7 < 15

is true. But

7 - 3 \times (-2) = 13

and

2 \times (-2) + 15 = 11

, and

13 < 11

is false. So the values above

-1.6

work and the values below it do not.

Check 2: Part (a). Put the boundary value

t = -1.6

into both sides. If it really is the boundary, the two sides must come out equal there.

7 - 3 \times (-1.6) = 11.8

and

2 \times (-1.6) + 15 = 11.8

, so the two sides meet exactly at that value and nowhere else.

Check 3: Part (b). Test the three corners of

R

against all three inequalities. At

(2, 3)

the two coordinates add to

5

, at

(2, 7)

they add to

9

, and at

(6, 3)

they add to

9

. Every corner has

x \geq 2

and

y \geq 3

and a total of at most

9

, so all three corners belong to

R

Check 4: Part (b). Take one point just outside each edge and confirm that exactly one inequality fails each time. That is what shows no inequality is missing and none of them is cutting into the region.

(1, 5)

fails

x \geq 2

and nothing else,

(4, 1)

fails

y \geq 3

and nothing else, and

(7, 5)

fails

x + y \leq 9

and nothing else.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$-3t - 2t < 15 - 7$ or $-5t < 8$ oe, or $7 - 15 < 2t + 3t$ or $-8 < 5t$ oe, or $t = -1.6$ or $t < -1.6$	M1	for correctly isolating terms in $t$ on one side and number terms on the other side (use of $=$ or any inequality symbol or variable is permitted)	✓
(a) $t > -1.6$	A1	oe eg $-1.6 < t$ or $t > -\dfrac{8}{5}$ or $-\dfrac{8}{5} < t$ oe. Must have correct inequality symbol on answer line. Working not required, so a correct answer scores full marks (unless it comes from obvious incorrect working). NB sight of the correct answer in the working space and just $(t =) -1.6$ oe on the answer line gains M1 only.	✓
(b) $x \geq 2$	B1	oe, allow $x > 2$ or $2 < x$	✓
(b) $y \geq 3$	B1	oe, allow $y > 3$ or $3 < y$	✓
(b) $x + y \leq 9$	B1	oe, allow $x + y < 9$ or $y < 9 - x$ or $9 > x + y$	✓
(b) special case	SC B2	for all of $x \leq 2$ , $y \leq 3$ , $x + y \geq 9$ oe, or $x < 2$ , $y < 3$ , $x + y > 9$	✓
(b) special case	SC B1	for all of $x = 2$ , $y = 3$ , $x + y = 9$ oe	✓
(b) what the two special cases are	Note	The first is the script of a student who found the right three lines but pointed every symbol the wrong way, so the answer describes the region outside the triangle. The second is the script of a student who wrote the three boundary lines as equations and never turned them into inequalities at all.	✓

Full marks: 5/5

In the diagram, $ABD$ and $ABC$ are right-angled triangles.

$AB = 16$ cm, $BC = 12$ cm, $AD = 1.5 \times AC$

Work out the length of $CD$ .
Give your answer correct to $3$ significant figures. [5 marks]

[Total 5 marks]

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Question 27 - Exam Solution

Understanding the Question

Given

Triangles

ABD

and

ABC

are both right-angled at

B

AB = 16

cm and

BC = 12

AD = 1.5 \times AC

B

C

and

D

lie on one straight line, with

C

between

B

and

D

Find

The length of

CD

The answer correct to

3

significant figures

Plan the Solution

Triangle $ABC$ has both short sides given, so Pythagoras gives its hypotenuse $AC$ .
Scale that by $1.5$ to get $AD$ , which is the hypotenuse of triangle $ABD$ .
In triangle $ABD$ the hypotenuse and one short side are then known, so Pythagoras gives $BD$ .
$C$ sits on $BD$ , so $CD$ is what is left when $BC$ is taken off $BD$ .
Keep the full decimal for $BD$ and round only at the very end.

Worked Solution [5 marks]

Pythagoras' theorem. In a right-angled triangle the square on the hypotenuse equals the sum of the squares on the other two sides, so

h^2 = a^2 + b^2

when the hypotenuse is the unknown, and

a^2 = h^2 - b^2

when the hypotenuse is the side already known.

Step 1: Pythagoras in triangle

ABC

AC^2 = BC^2 + AB^2

12^2 + 16^2 = 144 + 256 = 400

AC = \sqrt{400} = 20 \text{ cm}

(Reason: The right angle is at

B

, so

AC

is the hypotenuse of this triangle and the two squares are added.)

Step 2: Scale

AC

up to get

AD

AD = 1.5 \times AC

1.5 \times 20 = 30

(Reason: The question states that

AD

1.5

times

AC

, and

AC

has just been found to be

20

cm.)

Step 3: Pythagoras in triangle

ABD

BD^2 = AD^2 - AB^2

30^2 - 16^2 = 900 - 256 = 644

BD = \sqrt{644} = 25.3771\ldots

(Reason: This time the hypotenuse

AD

is the side that is known, so

AB^2

is taken away rather than added. Keep the full decimal for the next step.)

Step 4: Subtract to reach

CD

CD = BD - BC

25.3771\ldots - 12 = 13.3771\ldots

(Reason:

B

C

and

D

lie on one straight line with

C

between the other two, so the two lengths simply subtract.)

Step 5: Round to

3

significant figures

CD = 13.3771\ldots \approx 13.4 \text{ cm}

(Reason: The first three significant figures of

13.3771\ldots

are

1

3

and

3

. The next digit is

7

, so the third one rounds up to

4

CD = 13.4

Verification

Check 1: Work backwards. If

BD = 25.3771\ldots

and

AB = 16

, then Pythagoras must rebuild the hypotenuse

AD

644 + 256 = 900

and

\sqrt{900} = 30

, which is exactly

1.5 \times 20

Check 2: Reach

BD

by trigonometry instead. In triangle

ABD

\cos BAD = \dfrac{16}{30}

, and then

BD = 30 \sin BAD

BAD = 57.769\ldots^{\circ}

and

30 \sin 57.769\ldots^{\circ} = 25.377\ldots

, giving

CD = 13.4

cm again

Check 3: Do it in surd form, with no decimal until the last line.

BD^2 = 2.25 \times 400 - 256

, so

BD = 2\sqrt{161}

2.25 \times 400 - 256 = 900 - 256 = 644

and

644 = 4 \times 161

, so

CD = 2\sqrt{161} - 12 = 13.377\ldots

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$(AC^2 =)\; 12^2 + 16^2 \; (= 144 + 256 = 400)$ or $(BAC =) \tan^{-1}\left(\dfrac{12}{16}\right) \; (= 36.8(698\ldots))$ or $36.9$ or $(BCA =) \tan^{-1}\left(\dfrac{16}{12}\right) \; (= 53.1(301\ldots))$	M1	for a correct method using triangle $ABC$	✓
$(AC =) \sqrt{12^2 + 16^2} \; (= \sqrt{144 + 256} = \sqrt{400} = 20)$ or $(AC =) \dfrac{16}{\cos 36.8^{\circ}} \; (= 20)$ or $(AC =) \dfrac{12}{\sin 36.8^{\circ}} \; (= 20)$ or $(AC =) \dfrac{16}{\sin 53.1^{\circ}} \; (= 20)$ or $(AC =) \dfrac{12}{\cos 53.1^{\circ}} \; (= 20)$	M1	for a correct method to find $AC$ . The printed scheme puts $36.8$ and $53.1$ in quotation marks, so a candidate's own angle from the first row may be used in their place.	✓
$(BD^2 =) (1.5 \times 20)^2 - 16^2 \; (= 644)$ or $(BD^2 =) 30^2 - 16^2 \; (= 900 - 256 = 644)$ or $(BAD =) \cos^{-1}\left(\dfrac{16}{30}\right) \; (= 57.7(690\ldots))$ or $57.8$ or $(BDA =) \sin^{-1}\left(\dfrac{16}{30}\right) \; (= 32.2(309\ldots))$ or $(BCA =) \sin^{-1}\left(\dfrac{16}{20}\right) \; (= 53.1(301\ldots))$ and $\dfrac{CD}{\sin(180 - 126.9 - 32.2)} = \dfrac{30}{\sin(180 - 53.1)}$ or equivalent	M1	for a correct method using a triangle to find $BD^2$ or angle $BAD$ or angle $BDA$ , or for a correct equation for side $CD$ . The printed scheme puts $20$ and $30$ in quotation marks, so the candidate's own earlier values may be used.	✓
$(BD =) \sqrt{(1.5 \times 20)^2 - 16^2} \; (= 25.3(771\ldots))$ or $(BD =) \sqrt{30^2 - 16^2} \; (= \sqrt{900 - 256} = \sqrt{644} = 2\sqrt{161} = 25.3(771\ldots))$ or $(BD =) 16 \tan 57.7^{\circ} \; (= 25.3(771\ldots))$ or $(BD =) 30 \sin 57.7^{\circ} \; (= 25.3(771\ldots))$ or $(BD =) \sqrt{16^2 + 30^2 - 2 \times 16 \times 30 \cos 57.7^{\circ}} \; (= 25.3(771\ldots))$ or $(BD =) 30 \cos 32.2^{\circ} \; (= 25.3(771\ldots))$ or $(BD =) \dfrac{16}{\tan 32.2^{\circ}} \; (= 25.3(771\ldots))$ or $(BD =) \dfrac{16}{\sin 32.2^{\circ}} \times \sin 57.7^{\circ} \; (= 25.3(771\ldots))$ or $(CD =) \dfrac{30}{\sin 126.9^{\circ}} \times \sin 20.9^{\circ}$ or equivalent	M1	for a correct method to find $BD$ or $CD$ . Here too the printed scheme quotes $20$ , $30$ and the angles, so the candidate's own earlier values are allowed in their place.	✓
the final answer	A1	awrt $13.4$	✓
Working not required, so a correct answer scores full marks	Note	unless it comes from obviously incorrect working	✓

Full marks: 5/5

Bilal has a jar of glass beads.

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

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

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

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

[Total 3 marks]

Show solution & mark schemeHide solution & mark scheme

Question 28 - Exam Solution

Understanding the Question

Given

A jar of glass beads:

17

green,

28

yellow, and the rest purple.

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

\dfrac{4}{9}

Find

The number of purple beads in the jar. The jar's total is not given, so it has to come out of the

\dfrac{4}{9}

as well.

Plan the Solution

Green and yellow are the only other colours, so adding them gives the number of beads that are not purple.
Take $\dfrac{4}{9}$ away from $1$ to get the probability that a bead is not purple.
Those beads make up $\dfrac{5}{9}$ of the jar, so dividing by $5$ gives one ninth of it.
Purple takes four of the nine ninths, so multiply one ninth by $4$ .

Worked Solution [3 marks]

Rule - The probabilities of all the outcomes add to

1

, so the beads that are not purple make up

1 - \dfrac{4}{9}

of the jar. Ninths are equal shares, so every ninth holds the same number of beads.

Step 1: Count the beads that are not purple

17 + 28 = 45

(Reason: Green and yellow are the only other colours in the jar, so these

45

beads are exactly the ones that are not purple.)

Step 2: Work out the probability that a bead is not purple

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

(Reason: A bead taken at random is either purple or it is not, so the two probabilities add to

1

Step 3: Find how many beads make one ninth of the jar

\dfrac{45}{5} = 9

(Reason: The

45

beads that are not purple are

\dfrac{5}{9}

of the jar, which is five equal ninths of it.)

Step 4: Scale one ninth up to the purple share

4 \times 9 = 36

(Reason: Purple takes four of the nine equal ninths, and each ninth holds

9

beads.)

36

purple beads

Verification

Check 1: Put the purple beads back with the others:

45 + 36 = 81

beads in the jar altogether. Now read the probability straight off that jar.

\dfrac{36}{81} = \dfrac{4}{9}

Check 2: Test the other colours the same way. Out of

81

beads,

45

are green or yellow, and that share should be the

\dfrac{5}{9}

from Step 2.

\dfrac{45}{81} = \dfrac{5}{9}

, and

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

Check 3: Reach it by algebra instead of by ninths. With

p

purple beads the jar holds

p + 45

beads, so

\dfrac{p}{p + 45} = \dfrac{4}{9}

. Multiplying out the denominators gives

9p = 4p + 180

5p = 180

, so

p = 36

Mark Scheme Breakdown

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

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 4MA1/2F, Wednesday 4 June 2025: Worked Solutions, Questions 18 to 28

Worked solutions, questions 18 to 28 of 28

Question 18, Calculator allowed

Question 18 - Exam Solution

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Question 21 - Exam Solution

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Question 24, Calculator allowed

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Question 28, Calculator allowed

Question 28 - Exam Solution

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