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

Sir Faraz Hassan

31 Aug 2026

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

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

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

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

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

Worked solutions, questions 1 to 10 of 26

Theo has six tiles.
He writes one 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 $6$

Theo lays the tiles in a row so that the numbers are in order of size.

The numbers on three of the tiles are hidden.

Write the three hidden numbers on the tiles above. [3 marks]

[Total 3 marks]

Show solution & mark schemeHide solution & mark scheme

Question 1 - Exam Solution

Understanding the Question

Given

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

a, b, 6, c, 10, 14

The range of the numbers is

10

The median of the numbers is

7.5

The mode of the numbers is

6

Find

The three hidden numbers

a

b

and

c

- the first, second and fourth tiles.

Plan the Solution

The tiles are in order of size, so $14$ is the largest number and $a$ is the smallest. The range gives $a$ straight away.
Six numbers have two middle ones, the third and the fourth. The median gives $c$ .
Once $a$ and $c$ are known, every settled number appears just once, so the mode is what forces a second $6$ . That gives $b$ .
Finish by testing all three statistics on the completed row.

Worked Solution [3 marks]

Rule - Range

= \text{largest} - \text{smallest}

. For six numbers in order of size the median is

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

. The mode is the value that appears most often.

Step 1: Name the three hidden numbers

a, b, 6, c, 10, 14

(Reason: The tiles are already in order of size, so

a

is the smallest number and

14

is the largest.)

Step 2: Use the range to find the first tile

14 - a = 10

a = 14 - 10 = 4

(Reason: The range is the largest number take away the smallest, and the row is in order, so those two are

14

and

a

Step 3: Use the median to find the fourth tile

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

c = 2 \times 7.5 - 6 = 9

(Reason: Six numbers have two middle ones. Doubling the median and taking away the other middle number leaves

c

Step 4: Use the mode to find the second tile

4, b, 6, 9, 10, 14

b = 6

(Reason:

4

9

10

and

14

each appear once, so the mode can only be

6

if a second

6

is written. The row is in order of size, so

b

lies between

4

and

6

, and only

6

itself makes

6

the most common number.)

Step 5: Write the three numbers on the tiles

4, 6, 6, 9, 10, 14

(Reason: The hidden numbers are

4

6

and

9

, on the first, second and fourth tiles.)

First tile:

4

Second tile:

6

Fourth tile:

9

Completed row:

4

6

6

9

10

14

Verification

Check 1: Take the smallest number of the completed row away from the largest.

14 - 4 = 10

, which is the range given.

Check 2: The two middle numbers of the completed row are

6

and

9

. Work out their mean.

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

, which is the median given.

Check 3: Count how many times each number appears in the completed row, and read the row from left to right.

6

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

6

, and the row is still in order of size.

Mark Scheme Breakdown

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

Full marks: 3/3

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

(b) Describe fully the single transformation that takes triangle $P$ to triangle $Q$ . [3 marks]

(b)

[Total 5 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

(a) Shape

A

has vertices

(1, 1)

(3, 1)

(3, 4)

(2, 4)

(2, 2)

and

(1, 2)

(a) Scale factor

3

, centre

(0, 1)

(b) Triangle

P

has vertices

(2, 2)

(2, 6)

and

(4, 2)

(b) Triangle

Q

has vertices

(6, 8)

(8, 8)

and

(8, 4)

Find

(a) The image of shape

A

, drawn on the grid. (b) The single transformation that takes triangle

P

to triangle

Q

, described fully.

Plan the Solution

(a) Take one vertex at a time. Count the step from the centre to the vertex, multiply that step by $3$ , then count that far out from the centre again.
(a) Join the six images in the same order as the original, so the image comes out the same shape the right way round.
(b) Check the size first. $P$ and $Q$ are congruent, so nothing that changes the size can do it.
(b) Join each vertex of $P$ to the matching vertex of $Q$ . A half turn puts the centre at the midpoint of every join, so all three midpoints must agree.
(b) A full description of a rotation needs three things: the word rotation, the angle, and the centre.

Worked Solution [5 marks]

Rule - Enlargement with scale factor

k

about a centre

C

: count the step from

C

to the point, multiply that step by

k

, and count out from

C

again. Rule - A half turn about

(a, b)

sends

(x, y)

(2a - x, 2b - y)

, so the centre is the midpoint of every point and its image.

Step 1: Read the vertices of shape

A

off the grid

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

(Reason: (Reason: the enlargement is worked out corner by corner, so the six vertices have to be read off before anything is multiplied. Going round the outline in order is what keeps the L the right way round later.))

Step 2: Enlarge each vertex from the centre

(0, 1)

(1, 1) \rightarrow (0, 1) + 3(1, 0) = (3, 1)

(3, 1) \rightarrow (0, 1) + 3(3, 0) = (9, 1)

(3, 4) \rightarrow (0, 1) + 3(3, 3) = (9, 10)

(2, 4) \rightarrow (0, 1) + 3(2, 3) = (6, 10)

(2, 2) \rightarrow (0, 1) + 3(2, 1) = (6, 4)

(1, 2) \rightarrow (0, 1) + 3(1, 1) = (3, 4)

(Reason: (Reason: the bracket is the step from the centre to the vertex, not the vertex itself. Multiplying that step by

3

and adding it back on to the centre lands on the image.))

Step 3: Plot the six images and join them in the same order

(3, 1),\; (9, 1),\; (9, 10),\; (6, 10),\; (6, 4),\; (3, 4)

(Reason: (Reason: joining them in the original order keeps the L the right way round. The image lands directly to the right of

A

and shares the whole edge from

(3, 1)

(3, 4)

with it, which is a quick way to see the drawing is in the right place.))

Step 4: Compare triangle

P

with triangle

Q

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

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

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

(Reason: (Reason: both triangles have a short side of

2

and a long side of

4

, so they are congruent and nothing that changes the size can be involved. The two legs of

P

point up and right; the matching legs of

Q

point down and left, which is a turn of

180^{\circ}

.))

Step 5: Join matching vertices and take the midpoint of each join

\left( \dfrac{2 + 8}{2}, \dfrac{2 + 8}{2} \right) = (5, 5)

\left( \dfrac{2 + 8}{2}, \dfrac{6 + 4}{2} \right) = (5, 5)

\left( \dfrac{4 + 6}{2}, \dfrac{2 + 8}{2} \right) = (5, 5)

(Reason: (Reason: a half turn carries every point to the far side of the centre and the same distance away, so the centre is the midpoint of every join. All three midpoints agree, and that agreement is the check that the centre is right.))

Step 6: Test the rule about

(5, 5)

(x, y) \rightarrow (2 \times 5 - x, 2 \times 5 - y) = (10 - x, 10 - y)

(2, 2) \rightarrow (10 - 2, 10 - 2) = (8, 8)

(2, 6) \rightarrow (10 - 2, 10 - 6) = (8, 4)

(4, 2) \rightarrow (10 - 4, 10 - 2) = (6, 8)

(Reason: (Reason: the rule sends every vertex of

P

to a vertex of

Q

, so the description is complete - a rotation,

180^{\circ}

, about

(5, 5)

. No direction is needed: a half turn clockwise and a half turn anticlockwise land in the same place.))

(a) Image 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: Every length in the image should be

3

times the matching length in shape

A

. The bottom edge of

A

runs from

(1, 1)

(3, 1)

, so it is

2

squares long.

3 \times 2 = 6

, and the image edge from

(3, 1)

(9, 1)

6

squares long.

Check 2: Area multiplies by the square of the scale factor, which is a test the side lengths cannot give on their own. Shape

A

covers

4

squares.

3^{2} \times 4 = 36

, and the image covers

36

squares.

Check 3: Apply the half-turn rule

(x, y) \rightarrow (10 - x, 10 - y)

to all three vertices of

P

and see whether triangle

Q

comes out. It gives

(8, 8)

(8, 4)

and

(6, 8)

, which is triangle

Q

exactly.

Check 4: A rotation has only one centre, so every join must have the same midpoint. Working the three out separately tests

(5, 5)

a second way. All three joins have midpoint

(5, 5)

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) A shape of the correct size but in the wrong position	B1	B1 for a shape of the correct size but in the wrong position.	✓
(a) Marking aid	Note	NB Overlay is available.	✓
(a) The error behind the B1	Note	A shape of the correct size in the wrong position is what enlarging from the origin gives instead of from $(0, 1)$ : every image lands $2$ squares too high.	✓
(b) Rotation	B1	With no mention of any other transformation words or move, flip, transform, up, right etc. Turn on its own is not sufficient.	✓
(b) $180^{\circ}$	B1	Allow half turn.	✓
(b) (centre) $(5, 5)$	B1	Must be a coordinate and not a vector.	✓
(b) Alternative: enlargement, scale factor $-1$	B2	B2 for enlargement SF $-1$ in place of the first two marks. Ignore any reference to clockwise or anticlockwise.	✓

Full marks: 5/5

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

[Total 3 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

A subtraction of two mixed numbers:

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

The denominators

3

and

7

are different.

The fraction being taken away,

\dfrac{4}{7}

, is the bigger of the two fractional parts.

Find

Show that the difference comes to exactly

3\dfrac{16}{21}

Plan the Solution

Turn each mixed number into an improper fraction, so the whole numbers and the fractions travel together instead of being handled separately.
Rewrite both fractions over the common denominator $21$ , the lowest common multiple of $3$ and $7$ .
Subtract the numerators, then turn the improper fraction back into a mixed number and compare it with the result the question states.

Worked Solution [3 marks]

Rule - Subtracting mixed numbers: write each mixed number as an improper fraction, put both fractions over a common denominator, subtract the numerators, then write the result back as a mixed number.

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 add the numerator:

7 \times 3 + 1 = 22

and

3 \times 7 + 4 = 25

. The whole numbers are now inside the fractions, so nothing has to be borrowed later.)

Step 2: Put both fractions over the denominator

21

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

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

(Reason: The lowest common multiple of

3

and

7

21

, so the first fraction is multiplied top and bottom by

7

and the second top and bottom by

3

. Multiplying top and bottom by the same number does not change the value.)

Step 3: Subtract the numerators

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

(Reason: Once both fractions share a denominator the denominator is left alone and only the numerators are subtracted, because the two fractions are now counted in the same sized pieces.)

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

79 = 3 \times 21 + 16

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

(Reason: There are

3

whole lots of

21

inside

79

, with

16

twenty-firsts left over. That is the result the question gives, so the statement is shown.)

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

as required

Verification

Check 1: Add

3\dfrac{4}{7}

back on to the result. If the subtraction is right the total must return to

7\dfrac{1}{3}

\dfrac{79}{21} + \dfrac{75}{21} = \dfrac{154}{21} = \dfrac{22}{3}

Check 2: Work both sides out as decimals on the calculator and compare them.

\dfrac{22}{3} - \dfrac{25}{7} = 3.7619

and

3 + \dfrac{16}{21} = 3.7619

Check 3: Do the subtraction the other way the mark scheme allows, keeping the whole numbers apart:

7\dfrac{7}{21} - 3\dfrac{12}{21}

leaves

4 - \dfrac{5}{21}

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

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Improper fractions, or the fractional parts over a common denominator	M1	for correct improper fractions or fractional part of numbers written correctly over a common denominator: $\dfrac{22}{3} \; (-) \; \dfrac{25}{7}$ or $(7)\dfrac{7}{21} \; (-) \; (3)\dfrac{12}{21}$ or $(7)\dfrac{7a}{21a} \; (-) \; (3)\dfrac{12a}{21a}$	✓
Correct fractions over a common denominator, with the subtraction shown	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 $\dfrac{22 \times 7}{21} - \dfrac{25 \times 3}{21}$ 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 $7\dfrac{7}{21} - 3\dfrac{12}{21} = 6\dfrac{28}{21} - 3\dfrac{12}{21}$ . $\dfrac{154}{21} - \dfrac{75}{21}$ or $\dfrac{22 \times 7}{21} - \dfrac{25 \times 3}{21}$ implies the first M1	✓
A fully correct solution reaching the given result	A1	Dep on M2 for a correct answer from fully correct working: $\dfrac{154}{21} - \dfrac{75}{21} = \dfrac{79}{21} = 3\dfrac{16}{21}$ or $4 - \dfrac{5}{21} = 3\dfrac{16}{21}$ or $7\dfrac{7}{21} - 3\dfrac{12}{21} = 6\dfrac{28}{21} - 3\dfrac{12}{21} = 3\dfrac{16}{21}$ . 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	✓
Working required	Note	The result is printed in the question, so it is the working that is being marked. A solution that writes the given result down again, with nothing between the two mixed numbers and it, earns none of the three marks.	✓

Full marks: 3/3

Use a ruler and a pair of compasses only to construct the perpendicular bisector of the line $AB$
You must leave all of your construction lines on the diagram. [2 marks]

[Total 2 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

A straight line segment

AB

, already drawn on the page.

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

Find

The perpendicular bisector of

AB

: the line that cuts

AB

into two equal halves and meets it at a right angle. Every construction line must still be on the page at the end. Nothing is rubbed out.

Plan the Solution

Open the compasses to a radius bigger than half of $AB$ , so that arcs swung from the two ends can reach each other.
Swing a pair of arcs from $A$ , one either side of the line, then a matching pair from $B$ without touching the compass setting in between.
Rule the straight line through the two points where the arcs cross, and take it a little way past the line on both sides.

Worked Solution [2 marks]

Rule - Perpendicular bisector: a point is the same distance from

A

as it is from

B

exactly when it lies on the perpendicular bisector of

AB

, so any two such points fix that line.

Step 1: Set the compasses and swing two arcs from

A

r > \dfrac{1}{2}AB

(Reason: The radius

r

is chosen once and then left alone for the whole construction. If

r

were smaller than half of

AB

, arcs swung from the two ends could never reach each other and there would be nothing to cross. Swing one arc into the space above the line and one into the space below it.)

Step 2: Swing two matching arcs from

B

AP = BP = AQ = BQ = r

(Reason: The compasses are not adjusted between the two pairs, so all four of these lengths are the one radius. That is what makes

P

and

Q

each exactly as far from

A

as they are from

B

Step 3: Rule the line through

P

and

Q

PQ \perp AB

(Reason: Two points fix a straight line, and both of these are equidistant from

A

and

B

, so the line through them is the perpendicular bisector. Take it past

AB

on both sides so the crossing is clear, and leave all four arcs on the page, because the mark scheme asks to see them.)

Step 4: Why the ruled line is the perpendicular bisector

AP = PB = BQ = QA

PQ \perp AB \quad \text{and} \quad AM = MB

(Reason: Four equal sides make

APBQ

a rhombus, and the diagonals of a rhombus cross at right angles and cut each other in half. The diagonals here are

PQ

and

AB

, so

PQ

meets

AB

90^\circ

and at its midpoint

M

. That is both halves of what perpendicular bisector means.)

The perpendicular bisector of

AB

is the line ruled through the two arc crossings

P

and

Q

, with both pairs of arcs left showing on the diagram.

Verification

Check 1: Look at the four radii. The compasses were opened once and never touched again, so

AP

BP

AQ

and

BQ

are all the same length. Both arc crossings are the same distance from

A

as from

B

, so both of them lie on the bisector.

Check 2: Measure the finished figure where the ruled line cuts

AB

. A protractor should read

90^\circ

, and a ruler should give the same length either side of the crossing. The line meets

AB

at a right angle and

AM = MB

, which is exactly what the question asked for.

Check 3: Fold the paper along the line that has just been ruled.

A

lands on top of

B

, which only happens when the fold is the perpendicular bisector of

AB

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Construct the perpendicular bisector of $AB$	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).	✓
A partly correct construction	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 for one pair of intersecting arcs and the perpendicular bisector drawn on just one side of $AB$ .	✓
Marking note	Note	An overlay is available.	✓

Full marks: 2/2

The diagram shows quadrilateral $ABCD$ and quadrilateral $EFGH$ .
The two quadrilaterals are similar.

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

(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 5 - Exam Solution

Understanding the Question

Given

Two similar quadrilaterals,

ABCD

and

EFGH

, with the letters matching in order.

AB = 5\text{ cm}

BC = 4\text{ cm}

CD = y\text{ cm}

EF = x\text{ cm}

FG = 10\text{ cm}

GH = 24\text{ cm}

Find

(a) the value of

x

, which is the length

EF

(b) the value of

y

, which is the length

CD

Plan the Solution

Read the correspondence off the letters: $A$ goes with $E$ , $B$ with $F$ , $C$ with $G$ and $D$ with $H$ . So $AB$ matches $EF$ , $BC$ matches $FG$ , and $CD$ matches $GH$ .
Only one matching pair has both of its lengths given, $BC$ and $FG$ , so that pair is what fixes the scale factor.
Multiply by the scale factor to go from the smaller quadrilateral to the larger one, and divide by it to come back the other way.
Finish by checking the size of each answer: $x$ is a side of the larger quadrilateral, so it must be bigger than $5$ , and $y$ is a side of the smaller one, so it must be smaller than $24$ .

Worked Solution [4 marks]

Rule - Similar shapes: every pair of matching sides is in the same ratio. Writing

k

for the scale factor from the smaller shape to the larger one,

\text{large side} = k \times \text{matching small side}

and

\text{small side} = \dfrac{\text{matching large side}}{k}

Step 1 (a): pick the matching pair whose lengths are both known

BC = 4\text{ cm}

FG = 10\text{ cm}

(Reason: The letters match in order, so

BC

in the smaller quadrilateral corresponds to

FG

in the larger one. Every other pair has an unknown in it, so this is the only pair that can give the scale factor.)

Step 2 (a): work out the scale factor

k = \dfrac{FG}{BC} = \dfrac{10}{4} = 2.5

(Reason: Dividing a large side by its matching small side gives the scale factor from

ABCD

EFGH

. A value bigger than

1

is the sign that

EFGH

is the enlargement.)

Step 3 (a): scale

AB

up to

EF

x = EF = k \times AB = 2.5 \times 5 = 12.5

(Reason: Going from the smaller quadrilateral to the larger one multiplies every length by

k

, and

EF

is the side that matches

AB

Step 4 (b): scale

GH

back down to

CD

y = CD = \dfrac{GH}{k} = \dfrac{24}{2.5} = 9.6

(Reason: Coming back the other way undoes the enlargement, so divide by the same scale factor, and

CD

is the side that matches

GH

(a)

x = 12.5

(b)

y = 9.6

Verification

Check 1: Every matching pair must give the same scale factor. Test the other two pairs,

AB

with

EF

and

CD

with

GH

\dfrac{12.5}{5} = 2.5

and

\dfrac{24}{9.6} = 2.5

Check 2: Work inside each quadrilateral instead of across the pair. The ratio

AB : BC

ABCD

must equal the ratio

EF : FG

EFGH

\dfrac{5}{4} = 1.25

and

\dfrac{12.5}{10} = 1.25

Check 3: Check that each answer has landed on the right shape.

x

is a side of the larger quadrilateral and

y

is a side of the smaller one, so swapping the two answers over is the slip to rule out.

12.5 > 5

and

9.6 < 24

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $\dfrac{10}{4}$ ( $= \dfrac{5}{2} = 2.5$ ) or $\dfrac{4}{10}$ ( $= \dfrac{2}{5} = 0.4$ ) or $\dfrac{x}{5} = \dfrac{10}{4}$ oe or $\dfrac{x}{10} = \dfrac{5}{4}$ oe	M1	For a correct scale factor, which can be expressed as a fraction, a decimal or a ratio (it may or may not be used), or for a correct equation in $x$ . Allow any letter for $x$ .	✓
(a) Working not required, so a correct answer scores full marks (unless it comes from obviously incorrect working)	A1	$12.5$ oe, eg $\dfrac{50}{4}$ or $\dfrac{25}{2}$ or $12\dfrac{1}{2}$ or $12\dfrac{2}{4}$	✓
(b) $\dfrac{24}{[2.5]}$ oe or $\dfrac{y}{24} = \dfrac{4}{10}$ 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	M1ft	Follow through, ie $[2.5]$ is their scale factor from (a); or for a correct equation in $y$ . Allow any letter for $y$ . Follow through their answer to (a), ie $[12.5]$ is their answer to (a).	✓
(b) Working not required, so a correct answer scores full marks (unless it comes from obviously incorrect working)	A1	$9.6$ oe, eg $\dfrac{48}{5}$ or $9\dfrac{3}{5}$ . If (a) $x = 9.6$ and (b) $y = 12.5$ then M1A0M1A0.	✓

Full marks: 4/4

Marta, Leo and Freya are paid $\pounds 240$ for clearing a garden.
They share the money in the ratios $3 : 4 : 5$

Leo and Freya each give $\pounds 10$ of their share to Marta.

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

[Total 4 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

Marta, Leo and Freya share

\pounds 240

in the ratios

3 : 4 : 5

, so the money is cut into parts and Marta takes

3

of them, Leo takes

4

and Freya takes

5

Leo hands Marta

\pounds 10

and Freya hands Marta

\pounds 10

, so two separate gifts arrive with Marta and each of the other two makes one.

Find

The ratio

\text{Marta} : \text{Leo} : \text{Freya}

of what the three of them are holding once the gifts have been made, written in its simplest form.

Plan the Solution

Add the three parts of the ratio to see how many equal parts the money is cut into, then find what one part is worth.
Multiply to get the three starting amounts, and add them up: they have to come back to $\pounds 240$ before going any further.
Move the gifts. Marta receives twice, Leo and Freya each give once.
Set the three new amounts side by side as a ratio, then cancel by their highest common factor.

Worked Solution [4 marks]

Rule - Sharing in a ratio:

\text{one part} = \dfrac{\text{total}}{\text{the parts added together}}

, and each person takes their own number of parts. A ratio is in its simplest form once the only whole number that divides every part is

1

Step 1: find what one part of the ratio is worth

3 + 4 + 5 = 12

\dfrac{240}{12} = 20

(Reason: It is the parts added together that the total is divided by, never one part on its own, because

3 : 4 : 5

cuts the money into

12

equal parts. One part is worth

\pounds 20

Step 2: work out what each of them starts with

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

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

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

(Reason: Each of them takes their own number of parts, one part being

\pounds 20

. The three amounts add back to

240

, which is the sign that the sharing has been done correctly.)

Step 3: hand over the two gifts of

\pounds 10

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

\text{Leo: } 80 - 10 = 70

\text{Freya: } 100 - 10 = 90

(Reason: Marta is given

\pounds 10

by Leo and another

\pounds 10

by Freya, so she gains

\pounds 20

altogether, while Leo and Freya are each

\pounds 10

worse off. Nothing leaves the three of them, so the amounts still add to

240

Step 4: write the new amounts as a ratio and cancel

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

. Cancelling by a smaller common factor,

2

5

, does divide all three but leaves a ratio that can still be cancelled, so it is not yet in its simplest form.)

8 : 7 : 9

Verification

Check 1 - the money has only moved: Add the three new amounts. The

\pounds 20

that leaves Leo and Freya arrives with Marta, so the total cannot have changed.

80 + 70 + 90 = 240

Check 2 - undo the two gifts: Give the two gifts back. The three of them should be holding the amounts the ratio the question starts from hands out.

80 - 10 - 10 = 60

70 + 10 = 80

90 + 10 = 100

, and

60 : 80 : 100 = 3 : 4 : 5

Check 3 - test the answer against the total: A ratio of

8 : 7 : 9

has

24

parts in it, so it claims Marta ends with

\dfrac{8}{24}

of the

\pounds 240

, Leo with

\dfrac{7}{24}

and Freya with

\dfrac{9}{24}

. Those three fractions of the total have to give the three amounts back.

\dfrac{8}{24} \times 240 = 80

\dfrac{7}{24} \times 240 = 70

\dfrac{9}{24} \times 240 = 90

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 2 of the people after the two gifts	M1	For the correct values for $2$ of the people after Leo and Freya give Marta $\pounds 10$ . For two of: (Marta) $3 \times [20] + 10 + 10 \; (= 80)$ or $[60] + 10 + 10 \; (= 80)$ ; (Leo) $4 \times [20] - 10 \; (= 70)$ or $[80] - 10 \; (= 70)$ ; (Freya) $5 \times [20] - 10 \; (= 90)$ or $[100] - 10 \; (= 90)$ . A value in square brackets is one the printed scheme puts in quotation marks, so the candidate's own value from the first method mark may be used in its place.	✓
All three 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	$8 : 7 : 9$ . Other orders are acceptable if they are labelled correctly. Working is not required, so a correct answer scores full marks unless it comes from obvious incorrect working.	✓

Full marks: 4/4

Bruno buys a painting for $4\,000$ Swiss francs.
The value of the painting increases by $7\%$ each year.

Work out the value of the painting 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 7 - Exam Solution

Understanding the Question

Given

Price paid for the painting:

4\,000

Swiss francs

Rise in value:

7\%

each year, worked out on the value the year starts with

Time:

3

years

Find

The value of the painting at the end of the

3

years, correct to the nearest Swiss franc

Plan the Solution

Write the $7\%$ rise as a decimal and add it to $1$ to get the multiplier for one year.
Multiply by that number once for each year, always starting from the value the year opened with.
Collect the three multiplications into a single power, $1.07^{3}$ , which is the same calculation in one line.
Round only at the very end, so no part of a franc is lost part way through.

Worked Solution [3 marks]

Rule - Repeated percentage increase: a rise of

p\%

multiplies a value by

1 + \dfrac{p}{100}

, and the same rise repeated over

n

years multiplies it by that number

n

times, so a starting value

P

growing at a rate

r

(as a decimal) is worth

P \times (1 + r)^{n}

Step 1: Turn the

7\%

rise into a one-year multiplier

\dfrac{7}{100} = 0.07

1 + 0.07 = 1.07

(Reason: the painting keeps all of the value it already has and gains

7\%

on top, so one year of growth is a single multiplication by

1.07

. Multiplying by

0.07

on its own gives only the rise,

280

francs, which is what the first method mark accepts and is not the new value)

Step 2: Multiply once for each of the three years

4\,000 \times 1.07 = 4\,280

4\,280 \times 1.07 = 4\,579.6

4\,579.6 \times 1.07 = 4\,900.172

(Reason: each year's

7\%

is worked out on the value that year started with, not on the original

4\,000

, so the three rises are

280

299.6

and

320.572

and they grow as the painting does)

Step 3: Do the same three multiplications in one line

1.07^{3} = 1.225043

4\,000 \times 1.225043 = 4\,900.172

(Reason: multiplying by

1.07

three times is multiplying by

1.07

to the power

3

, so a calculator reaches the same value in a single press. Adding

7\%

of the original three times over instead, which is the multiplier

1 + 0.07 \times 3 = 1.21

, is simple interest and earns nothing here)

Step 4: Round to the nearest Swiss franc

4\,900.172 \approx 4\,900

(Reason: the first digit after the decimal point is

1

, which is below

5

, so the whole number of francs is unchanged and nothing is carried)

4\,900

Swiss francs

Verification

Check 1 - undo the three years: Divide the final value by the multiplier once for each year. If the three years were applied correctly, the price paid comes back exactly.

\dfrac{4\,900.172}{1.07^{3}} = 4\,000

Check 2 - add the three rises separately: Work out each year's rise on its own and add all three to the price. This never forms a multiplier or a power, so it reaches the value a different way.

280 + 299.6 + 320.572 = 900.172

, and

4\,000 + 900.172 = 4\,900.172

Check 3 - measure it against simple interest: Simple interest would add

7\%

4\,000

, that is

280

francs, in each of the

3

years, giving

840

of growth and a value of

4\,840

. Compound growth has to come out above that, and over only

3

years not far above it.

900.172 - 840 = 60.172

of extra growth, so the value is

4\,900.172

, which is above

4\,840

and only a little above it

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$0.07 \times 4\,000 \; (= 280)$	M1	For finding $7\%$ of $4\,000$ or $107\%$ of $4\,000$ : $0.07 \times 4\,000 \; (= 280)$ or $1.07 \times 4\,000 \; (= 4\,280)$ .	✓
$0.07 \times 4\,579.6 \; (= 320.572)$	M1	$4\,000 + 280 \; (= 4\,280)$ oe and $0.07 \times 4\,280 \; (= 299.6)$ and $4\,280 + 299.6 \; (= 4\,579.6)$ and $0.07 \times 4\,579.6 \; (= 320.572)$ , or $280 + 299.6 + 320.572 \; (= 900.172)$ . Every value carried forward in this row is one the printed scheme puts in quotation marks, so the candidate's own earlier value may be used in its place, and the printed scheme writes the last total as $900(.172)$ . The scheme also awards M2 across this row and the one above it for $1.07^{3} \times 4\,000$ or $1.07^{4} \times 4\,000 \; (= 5\,243\ldots)$ .	✓
$4\,900$	A1	Allow answers in the range $4\,900$ to $4\,901$ .	✓
One of the recognised wrong methods, and no other mark awarded	SC B1	If no other mark is awarded, SC B1 for $4\,000 \times 0.07 \times 3 \; (= 840)$ or $4\,000 \times 0.21 \; (= 840)$ or $4\,000 + 4\,000 \times 0.07 \times 3 \; (= 4\,840)$ or $4\,000 \times 1.21 \; (= 4\,840)$ or $0.93 \times 4\,000 \; (= 3\,720)$ or $0.79 \times 4\,000 \; (= 3\,160)$ or $0.93^{3} \times 4\,000 \; (= 3\,217\ldots)$ or $4\,000 \times 1.07^{2} \; (= 4\,579\ldots)$ .	✓
A correct answer with no working shown	Note	Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working. This row earns nothing on its own.	✓

Full marks: 3/3

Find the values of $x$ and $y$ that satisfy both of these equations.

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

You must show clear algebraic working. [3 marks]

x =y =

[Total 3 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

The first equation:

3x + 5y = 8

The second equation:

4x + y = -3.5

Two straight-line equations in the same two unknowns, so a single pair of values fits both at once.

Find

The value of

x

and the value of

y

, each one supported by algebraic working

Plan the Solution

Look for the variable whose coefficients are cheapest to match. The $y$ terms are $5y$ and $y$ , so only the second equation has to be multiplied, and only by $5$ .
Multiply every term of that equation, on both sides, so the new line says exactly what the old one said.
Subtract one equation from the other. The matching $y$ terms cancel and a single equation in $x$ is left.
Put that value of $x$ back into the original equation with the smaller numbers and read off $y$ .
Test the pair in both original equations, because a pair that fits only one of them is not a solution.

Worked Solution [3 marks]

Rule - Elimination: multiply one equation, or both, until the same variable has the same coefficient in each. Then subtract the equations when those matching terms have the same sign, and add them when the signs are opposite. That variable disappears and the other one is left on its own, ready to be found.

Step 1: Number the two equations

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

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

(Reason: numbering them lets every later line say which equation it came from, and working a marker can follow is part of what showing clear algebraic working means. Neither line has been changed:

4x + y = -3.5

is written with the coefficient

1

left unwritten in front of the

y

, exactly as the question prints it)

Step 2: Multiply equation (2) by

5

so that both equations carry

5y

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

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

(Reason: multiplying both sides by the same number leaves the equation saying the same thing, and it turns the lone

y

into the

5y

that equation (1) already has. Every term must be multiplied, the

-3.5

included: multiplying only the left-hand side would give a different equation and the method mark would be lost)

Step 3: Subtract equation (1) from equation (3) to remove the

y

terms

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

17x = -25.5

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

(Reason: both equations carry

+5y

, the same term with the same sign, so subtracting is what removes it; had the signs been opposite the equations would have been added instead. On the right, taking

8

away from

-17.5

moves further below zero, which is why the value of

x

comes out negative)

Step 4: Put

x = -1.5

back into equation (2) 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 one with the smaller numbers, so it is the safer of the two to substitute into. The term that has to cross the equals sign is negative, and moving a negative term across is a subtraction of a negative number, which is the point in this question where the working has to be read most carefully)

Step 5: State the pair the question asks for

x = -1.5

y = 2.5

(Reason: the question asks for two values and the answer line prints a blank for each, so both are written out. A pair counts as a solution only when it fits both of the original equations, which is what the checks below confirm)

x = -1.5

y = 2.5

Verification

Check 1 - test the pair in the first equation: Equation (1) was used to eliminate

y

and never used to work

y

out, so putting both values into it is a genuine test rather than a repeat of the working.

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

, which is the right-hand side of the first equation

Check 2 - test the pair in the second equation: The second equation has to hold as well. A pair that satisfies one equation and not the other is not a solution, and a slip made while substituting would show up here.

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

, which is the right-hand side of the second equation

Check 3 - reach the same pair by eliminating the other variable: Multiply the first equation by

4

and the second by

3

, so that both carry

12x

, then subtract. This route never multiplies by

5

and never substitutes a found value, so it can go wrong in different places from the working above.

12x + 20y = 32

and

12x + 3y = -10.5

, so

17y = 42.5

and

y = 2.5

, and the second equation then gives

x = -1.5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $3x + 5y = 8$ and $20x + 5y = -17.5$ , subtracting: $3x - 20x = 8 - -17.5$ or $-17x = 25.5$ , or $3x + 5(-3.5 - 4x) = 8$ , or $4x + \dfrac{8 - 3x}{5} = -3.5$ , or eg $12x + 20y = 32$ and $12x + 3y = -10.5$ , subtracting: $20y - 3y = 32 - -10.5$ or $17y = 42.5$ , or $3\left(\dfrac{-3.5 - y}{4}\right) + 5y = 8$ , or $4\left(\dfrac{8 - 5y}{3}\right) + y = -3.5$	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). 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.	✓
eg $3 \times (-1.5) + 5y = 8$ 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 $x = \dfrac{8 - 5 \times 2.5}{3}$	M1	Dep 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. The printed scheme puts every $-1.5$ and every $2.5$ in this row inside quotation marks, so a candidate may use their own earlier value in place of the correct one and still earn the mark.	✓
$x = -1.5$ and $y = 2.5$	A1	oe, dep on M1. The printed scheme writes 'Working required' in the working column of this row, so the pair has to be supported by algebraic working rather than stated on its own.	✓

Full marks: 3/3

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

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

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

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

[Total 5 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

(a) The inequality

7 - 3t < 2t + 15

(b) The region

R

on the grid, bounded by a vertical line through

2

on the

x

-axis, a horizontal line through

3

on the

y

-axis, and a sloping line joining

(0, 9)

(9, 0)

All three boundary lines are drawn solid, and the shading runs right up to each of them.

Find

(a) Every value of

t

that makes the inequality true, written as an inequality in

t

(b) Three inequalities which are all true at every point of

R

and not all true anywhere else

Plan the Solution

(a) Treat the inequality like an equation: collect the terms in $t$ on one side and the numbers on the other. Send the $t$ terms to whichever side leaves their coefficient positive, and the sign never has to be reversed.
(a) Divide, then rewrite the statement with $t$ first, because that is the form the answer line asks for.
(b) Take the three boundary lines one at a time and write down the equation of each: two of them are parallel to an axis, and the third needs its gradient and its intercept.
(b) An equation names a line, not a side of it, so fix the direction of each inequality by testing one point that is plainly inside $R$ .
(b) Decide between a strict inequality and one that includes its boundary by looking at how the lines are drawn and how far the shading reaches.

Worked Solution [5 marks]

Rule - Inequalities: every move that is allowed on an equation is allowed here, with one exception. Multiplying or dividing both sides by a NEGATIVE number reverses the inequality sign, and so does writing the whole statement the other way round. Rule - A region on a grid: each boundary line gives one inequality, the equation of the line fixes the

=

part of it, and the side the region lies on fixes which way it points.

Step 1: Collect the terms in

t

on one side and the numbers on the other

7 - 3t < 2t + 15

7 - 15 < 2t + 3t

-8 < 5t

(Reason: adding

3t

to both sides and taking

15

from both sides sends the letters one way and the numbers the other. The terms in

t

are sent to the right-hand side on purpose, because

2t + 3t = 5t

is positive there, and a positive coefficient is what lets the next step divide without touching the inequality sign)

Step 2: Divide both sides by

5

-\dfrac{8}{5} < t

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

(Reason:

5

is positive, so dividing by it leaves the inequality sign exactly as it was. As a decimal

-\dfrac{8}{5}

-1.6

, and the mark scheme accepts either form)

Step 3: Turn the statement round so that

t

is written first

t > -1.6

(Reason: the answer line asks for

t

, so the statement is rewritten with

t

on the left. Reading

-1.6 < t

from right to left says that

t

is the greater of the two, so the symbol turns round with the statement. The accuracy mark is for the symbol pointing the correct way: an answer line reading

t = -1.6

scores the method mark and nothing more)

Step 4: Read off the two boundaries that run parallel to the axes

x = 2

y = 3

(Reason: one boundary is vertical, and every point on a vertical line has the same

x

-coordinate. It meets the

x

-axis at

2

, so its equation is

x = 2

. The other is horizontal, so every point on it has the same

y

-coordinate, and it meets the

y

-axis at

3

)

Step 5: Find the equation of the sloping boundary

\dfrac{0 - 9}{9 - 0} = -1

y = 9 - x

x + y = 9

(Reason: the sloping line passes through

(0, 9)

and

(9, 0)

, so it drops one square for every square it moves across and its gradient is

-1

. It cuts the

y

-axis at

9

, which gives

y = 9 - x

. Adding

x

to both sides puts it in the form the mark scheme uses, and that form says something worth remembering: the two coordinates of every point on this line add up to

9

)

Step 6: Test one point inside

R

to fix the direction of each inequality

3 \geq 2

4 \geq 3

3 + 4 = 7

(Reason: a line cuts the grid in two, and its equation says nothing about which half is wanted. The point

(3, 4)

is clearly inside the shaded region, so all three inequalities have to be true there. Its

x

-coordinate is at least

2

, its

y

-coordinate is at least

3

, and its coordinates add to

7

, which is not more than

9

. One interior point settles all three directions at once)

Step 7: Write down the three inequalities

x \geq 2

y \geq 3

x + y \leq 9

(Reason: each boundary line contributes one inequality and step 6 has fixed which way each one points. The lines are drawn solid and the shading reaches them, so points on the boundaries belong to

R

and each inequality is written with the bar underneath. The printed mark scheme also accepts the strict forms

x > 2

y > 3

and

x + y < 9

, so a student who reads the boundaries as excluded is not penalised)

(a)

t > -1.6

(b)

x \geq 2

y \geq 3

x + y \leq 9

Verification

Check 1 - test a value from each side of the boundary: Take one value from inside the answer and one from outside it, and put each into

7 - 3t

and

2t + 15

separately. This tests the answer against the inequality as the question prints it, rather than against any line of the working.

7 - 3 \times 0 = 7

and

2 \times 0 + 15 = 15

, and

7

is less than

15

, so

t = 0

belongs.

7 - 3 \times (-2) = 13

and

2 \times (-2) + 15 = 11

, and

13

is not less than

11

, so

t = -2

does not

Check 2 - the boundary value itself: Put

-1.6

into both sides. If it really is the end of the solution then the two sides meet there, and meeting is also the reason the value itself is left out.

7 - 3 \times (-1.6) = 11.8

and

2 \times (-1.6) + 15 = 11.8

, which are equal, so the two sides swap over at

-1.6

and the inequality there is strict

Check 3 - the three corners of the region: The corners of

R

are where the boundary lines cross:

(2, 7)

(2, 3)

and

(6, 3)

. Every corner must satisfy all three inequalities, and each should sit exactly on two of the three lines.

2 + 7 = 9

2 + 3 = 5

and

6 + 3 = 9

, so no corner has coordinates adding to more than

9

, and every corner has

x

at least

2

and

y

at least

3

Check 4 - step outside across one boundary at a time: Each of these points lies just outside

R

across a different edge, so each should break exactly one of the three inequalities. A point that broke none would mean the region is larger than the shading, and a point that broke two would mean one inequality is doing another one's work.

(1, 4)

fails only

x \geq 2

;

(3, 2)

fails only

y \geq 3

; and

(5, 5)

gives

5 + 5 = 10

, which fails only

x + y \leq 9

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) eg $-3t - 2t < 15 - 7$ 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) Working not required, so a correct answer scores full marks (unless from obvious incorrect working)	A1	$t > -1.6$ , oe eg $-1.6 < t$ or $t > -\dfrac{8}{5}$ or $-\dfrac{8}{5} < t$ oe. Must have correct inequality symbol on answer line. NB Sight of 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) All of $x \leq 2$ , $y \leq 3$ , $x + y \geq 9$ oe, or all of $x < 2$ , $y < 3$ , $x + y > 9$	SC B2	for all three	✓
(b) All of $x = 2$ , $y = 3$ , $x + y = 9$ oe	SC B1	for all three	✓

Full marks: 5/5

$ABD$ and $ABC$ are right-angled triangles.

$AB = 16 \text{ cm} \qquad BC = 12 \text{ cm} \qquad AD = 1.5 \times AC$

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

[Total 5 marks]

Show solution & mark schemeHide solution & mark scheme

Question 10 - Exam Solution

Understanding the Question

Given

ABD

and

ABC

are right-angled triangles, and in both of them the right angle is at

B

AB = 16 \text{ cm}

BC = 12 \text{ cm}

AD = 1.5 \times AC

C

lies on

BD

, so

BC

and

CD

together make

BD

Find

The length of

CD

, correct to

3

significant figures

Plan the Solution

Use Pythagoras in triangle $ABC$ to find $AC$ . Both short sides are known, so $AC$ is the hypotenuse.
Scale that answer by $1.5$ to get $AD$ .
Use Pythagoras again in triangle $ABD$ to find $BD$ . This time the hypotenuse $AD$ is known, so the squares are subtracted.
Take $BC$ away from $BD$ , because $C$ sits on $BD$ .
Round once, at the very end. Rounding $BD$ first would push the final answer out.

Worked Solution [5 marks]

Pythagoras: in a right-angled triangle the square on the hypotenuse equals the sum of the squares on the other two sides,

a^2 + b^2 = c^2

. Rearranged for a shorter side,

a^2 = c^2 - b^2

. Which form to use is decided by which side is the hypotenuse, never by which numbers are given.

Step 1: find

AC

from triangle

ABC

AC^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 side facing it and is the hypotenuse of triangle

ABC

. The two known squares are added.)

Step 2: scale up to get

AD

AD = 1.5 \times 20 = 30 \text{ cm}

(Reason: The question states that

AD

1.5

times

AC

, and

AC

has just been found.)

Step 3: find

BD

from triangle

ABD

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

BD = \sqrt{644} = 25.3771\ldots \text{ cm}

(Reason: In triangle

ABD

the right angle is again at

B

, so

AD

is the hypotenuse and

BD

is one of the shorter sides. A shorter side is found by subtracting, not by adding.)

Step 4: subtract

BC

to reach

CD

CD = BD - BC = 25.3771\ldots - 12 = 13.3771\ldots

(Reason:

C

lies on

BD

, so

BC

and

CD

make up the whole of

BD

. Keep the unrounded value until the last line.)

Step 5: round to

3

significant figures

CD = 13.4 \text{ cm}

(Reason: The unrounded length is

13.3771\ldots

and the fourth significant figure is

7

, so the third rounds up from

3

4

CD = 13.4 \text{ cm}

(correct to

3

significant figures)

Verification

Check 1: Rebuild

BD

from the answer and test triangle

ABD

the other way round.

BD = 12 + 13.3771\ldots = 25.3771\ldots

, and squaring that gives

644

644 + 256 = 900 = 30^2

, which is

AD^2

, so the rebuilt triangle really is right-angled at

B

Check 2: Redo the middle of the question with trigonometry instead of Pythagoras. In triangle

ABD

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

, so angle

BAD = 57.7690\ldots^\circ

30 \times \sin(57.7690\ldots^\circ) = 25.3771\ldots

, and taking

12

away leaves

13.3771\ldots

, the same length as before.

Check 3: Check the size is sensible.

CD

is only part of

BD

, and

BD

is a shorter side of a right-angled triangle whose hypotenuse

AD

measures

30

cm, so

CD

must be well under

30

cm.

0 < 13.4 < 25.4 < 30

, so

CD

is shorter than

BD

, which is shorter than

AD

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A method in triangle $ABC$	M1	For a correct method using triangle $ABC$ : $(AC^2 =)\,12^2 + 16^2\,(= 144 + 256 = 400)$ or $(BAC =) \tan^{-1}\left(\dfrac{12}{16}\right)(= 36.8(698\ldots))$ or $36.9$ or $(BCA =) \tan^{-1}\left(\dfrac{16}{12}\right)(= 53.1(301\ldots))$ .	✓
Find $AC$	M1	For a correct method to find $AC$ : $(AC =) \sqrt{12^2 + 16^2}\,(= \sqrt{144 + 256} = \sqrt{400} = 20)$ 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 $(AC =) \dfrac{12}{\cos 53.1^\circ}(= 20)$ . The printed scheme puts $36.8$ and $53.1$ inside quotation marks, which means an angle found earlier in the answer may be used in place of the correct one.	✓
A method in triangle $ABD$	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 $CD$ : $(BD^2 =) (1.5 \times 20)^2 - 16^2\,(= 644)$ 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 $(BCA =) \sin^{-1}\left(\dfrac{16}{20}\right)(= 53.1(301\ldots))$ together with $\dfrac{CD}{\sin(180 - 126.9 - 32.2)} = \dfrac{30}{\sin(180 - 53.1)}$ , or equivalent. The printed scheme puts $20$ and $30$ inside quotation marks, so a value found earlier in the answer may be used in place of the correct one.	✓
Find $BD$ or $CD$	M1	For a correct method to find $BD$ or $CD$ : $(BD =) \sqrt{(1.5 \times 20)^2 - 16^2}\,(= 25.3(771)\ldots)$ or $(BD =) \sqrt{30^2 - 16^2}\,(= \sqrt{900 - 256} = \sqrt{644} = 2\sqrt{161} = 25.3(771\ldots))$ or $(BD =) 16 \times \tan 57.7^\circ\,(= 25.3(771\ldots))$ or $(BD =) 30 \times \sin 57.7^\circ\,(= 25.3(771\ldots))$ or $(BD =) \sqrt{16^2 + 30^2 - 2 \times 16 \times 30 \times \cos 57.7^\circ}\,(= 25.3(771\ldots))$ or $(BD =) 30 \times \cos 32.2^\circ\,(= 25.3(771\ldots))$ 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. Here too the printed scheme quotes $20$ and $30$ , so an earlier value may be used.	✓
The answer	A1	awrt $13.4$ . Working is not required, so a correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 5/5

Continue to questions 11 to 19

The remaining 9 questions, with the same full worked solutions and mark schemes

Frequently asked questions

There are 26 questions worth 100 marks in total, sat over 2 hours. It is Higher tier and a calculator is allowed throughout, unlike UK GCSE Maths, where one paper is non-calculator.

Higher tier targets grades 4 to 9, so the lower grades 1 to 3 are only reachable on the tier below. 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 lowest grades on this Higher paper are the ones the two tiers share.

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 Higher tier formulae sheet is printed in the paper. It gives the area of a trapezium, the volume of a prism, the volume and curved surface area of a cylinder, the volume and curved surface area of a cone, the volume and surface area of a sphere, the area of a triangle from two sides and the included angle, the sine rule, the cosine rule, the sum of an arithmetic series and the quadratic formula. Other results, such as Pythagoras theorem and the trigonometric ratios for right-angled triangles, still have to be recalled. 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 4MA1/2H, Wednesday 4 June 2025: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 10 of 26

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

Frequently asked questions

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