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Edexcel IGCSE 4MA1 Paper 1F, November 2024: Worked Solutions, Questions 15 to 26

Sir Faraz Hassan

31 Jul 2026

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Edexcel International GCSE Mathematics A (4MA1)Paper 1F - Foundation Tier - November 2024100 marks · 2 hours · Calculator allowed

Back to questions 1 to 14

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

Original worked solutions for Edexcel International GCSE Mathematics A (4MA1), Paper 1F (Foundation Tier), November 2024 – 100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

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

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

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All 26 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 15 to 26 of 26

The diagram shows the shape $ABCDE$ .
It is formed by joining a right-angled triangle $ABE$ to a square $BCDE$

$ABC$ is a straight line.

$AB = 15$ cm and $AE = 17$ cm

Triangle $ABE$ has a perimeter of $40$ cm

Work out the area of shape $ABCDE$ [4 marks]

cm²

[Total 4 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

Shape

ABCDE

is a right-angled triangle

ABE

joined to a square

BCDE

along the side

BE

ABC

is a straight line, so the right angle of the triangle is at

B

AB = 15

cm and

AE = 17

cm.

The perimeter of triangle

ABE

40

cm.

Find

The area of the whole shape

ABCDE

, in cm². Two shapes, so expect two areas and one addition at the end.

Plan the Solution

Only two of the triangle's three sides are given, and the missing one is $BE$ . The perimeter supplies it in a single subtraction.
$BE$ is also a side of the square, so finding it unlocks both areas at once. This is why the question gives a perimeter rather than the side itself.
Work out the area of the square, then the area of the triangle, then add them.
Nothing needs to be taken away: the triangle and the square meet along $BE$ and do not overlap.

Worked Solution [4 marks]

Rule - Composite shape: split it into parts, work out each area, then add. Area of a triangle =

\dfrac{1}{2} \times \text{base} \times \text{height}

, and area of a square =

\text{side} \times \text{side}

Step 1: Use the perimeter to find

BE

\text{Perimeter of } ABE = AB + AE + BE

40 = 15 + 17 + BE

BE = 40 - 32 = 8 \text{ cm}

(Reason: The perimeter is the total of all three sides, so the unknown side is whatever is left once the two known sides are taken off:

15 + 17 = 32

, and

40 - 32 = 8

Step 2: Work out the area of the square

BCDE

\text{Area of } BCDE = BE \times BE

= 8 \times 8 = 64 \text{ cm}^2

(Reason: All four sides of a square are equal, so

BE = 8

cm makes

BC

CD

and

DE

each

8

cm as well.)

Step 3: Work out the area of the triangle

ABE

\text{Area of } ABE = \dfrac{1}{2} \times AB \times BE

= \dfrac{1}{2} \times 15 \times 8 = 60 \text{ cm}^2

(Reason: The right angle is at

B

, so the two sides that meet there,

AB

and

BE

, are the base and the height. The sloping side

17

cm is the hypotenuse and is never the height.)

Step 4: Add the two areas

\text{Area of } ABCDE = 60 + 64

= 124 \text{ cm}^2

(Reason: The shape is exactly the triangle and the square with nothing counted twice. The join

BE

is a side of each piece, not a piece of area, so the two totals simply add.)

124 \text{ cm}^2

Verification

Check 1: Test

BE

a completely different way. Triangle

ABE

is right-angled at

B

, so Pythagoras must hold with

AE

as the hypotenuse.

15^2 + 8^2 = 225 + 64 = 289 = 17^2

, so

BE = 8

cm is right, and the perimeter really is

15 + 17 + 8 = 40

cm.

Check 2: Treat the shape as one trapezium instead of two pieces.

AC = AB + BC = 15 + 8 = 23

cm, and

ACDE

has parallel sides

AC

and

ED

with height

BE

\dfrac{1}{2} \times (23 + 8) \times 8 = 124

, the same total by a route that never splits the shape up.

Check 3: A size check. The whole shape sits inside a rectangle

23

cm by

8

cm, of area

184

cm². What that rectangle has and the shape does not is the triangle above

AE

, whose legs are

15

cm and

8

cm.

184 - \dfrac{1}{2} \times 15 \times 8 = 184 - 60 = 124

, and

124

is comfortably less than

184

, as it must be.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$40 - (17 + 15) = 8$ or $\sqrt{17^2 - 15^2} = \sqrt{289 - 225} = \sqrt{64} = 8$	M1	For a correct method to find $BE$ . May be seen on the diagram.	✓
$8 \times 8 = 64$	M1	For the area of the square. The scheme writes the side in quotes, so a candidate's own value for $BE$ is used. May be seen on the diagram.	✓
$\dfrac{15 \times 8}{2} = 60$ oe	M1	For the area of the triangle. Again the scheme quotes the side, so a candidate's own value for $BE$ is used. May be seen on the diagram.	✓
$60 + 64 = 124$	A1	Working not required, so a correct answer scores full marks (unless it comes from obviously incorrect working).	✓
Alternative to the two area marks: $\dfrac{1}{2} \times (15 + 8 + 8) \times 8$	M2	Use of the trapezium formula on $ACDE$ earns both area marks at once. It replaces the square and triangle rows above rather than adding to them, so the question is still out of $4$ .	✓

Full marks: 4/4

Here are the first four terms of an arithmetic sequence.

$1 \quad 4 \quad 7 \quad 10$

(a) Work out an expression, in terms of $n$ , for the $n$ th term of this sequence. [2 marks]

A different arithmetic sequence has $n$ th term $5n + 17$

(b) Work out the $12$ th term of this second sequence. [1 mark]

(a)(b)

[Total 3 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

The first four terms of an arithmetic sequence:

1, \; 4, \; 7, \; 10

A second, different arithmetic sequence whose

n

th term is

5n + 17

Find

(a) An expression, in terms of

n

, for the

n

th term of the first sequence. (b) The

12

th term of the second sequence.

Plan the Solution

Subtract each term from the one after it. Equal gaps confirm the sequence is arithmetic and give the common difference $d$ .
The $n$ th term of an arithmetic sequence always starts with $dn$ . So write it as $3n + k$ and pin down $k$ using the first term.
Part (b) needs no listing at all. The $12$ th term is just the value of $5n + 17$ when $n = 12$ .

Worked Solution [3 marks]

Rule - nth term of an arithmetic sequence:

n\text{th term} = a + (n - 1)d

, where

a

is the first term and

d

is the common difference. Multiplied out, this is always

dn + (a - d)

Step 1: check that the sequence really is arithmetic

4 - 1 = 3

7 - 4 = 3

10 - 7 = 3

(Reason: All three gaps are equal, so the sequence is arithmetic with common difference

d = 3

Step 2: start the expression with

3n

n\text{th term} = 3n + k

(Reason: Going up in

3

s means the terms track the

3

times table

3n

, shifted by some fixed number

k

. Reaching this line alone earns the method mark.)

Step 3: find

k

from the first term

3 \times 1 + k = 1

3 + k = 1

k = 1 - 3 = -2

(Reason: Putting

n = 1

into the expression must give the first term,

1

. Rearranging leaves

k

on its own.)

Step 4: write down the

n

th term

n\text{th term} = 3n - 2

(Reason: Test it on the fourth term:

3 \times 4 - 2 = 10

, which is the term printed in the question.)

Step 5: substitute

n = 12

into

5n + 17

5 \times 12 + 17

(Reason: The

12

th term is the value of the expression at

n = 12

, so there is no need to write out all twelve terms.)

Step 6: work out the value

= 60 + 17

= 77

(Reason: Multiplication is done before addition.)

(a)

n\text{th term} = 3n - 2

(b)

77

Verification

Check 1: Put

n = 1

and

n = 4

into

3n - 2

and compare with the terms printed in the question.

3 \times 1 - 2 = 1

and

3 \times 4 - 2 = 10

, the first and fourth terms.

Check 2: Reach part (a) a different way: use the standard formula

a + (n - 1)d

with

a = 1

and

d = 3

, then multiply out.

1 + (n - 1) \times 3 = 3n - 3 + 1 = 3n - 2

Check 3: Reach part (b) without substituting. The second sequence has first term

5 \times 1 + 17 = 22

and rises in

5

s, so its

12

th term is

11

steps further on.

22 + 11 \times 5 = 22 + 55 = 77

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) An expression of the form $3n + k$ , with $k \neq -2$	M1	Method mark for using the common difference as the coefficient of $n$ . Accept $3 \times n + k$ or $n \times 3 + k$ , and $k$ may be zero or absent.	✓
(a) $3n - 2$	A1	Accept any equivalent form, eg $1 + (n - 1)3$ , $3 \times n - 2$ or $n \times 3 - 2$ , and allow $x = 3n - 2$ , $3x - 2$ or $T_n = 3n - 2$ . Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working. An answer written as $n = 3n - 2$ scores M1 only.	✓
(b) $77$	B1	Correct answer only, from $5 \times 12 + 17$ . No method is required for the mark.	✓

Full marks: 3/3

$450$ employees at a city-centre office were asked how they travelled to work on Monday.
Each employee walked or travelled by bus or travelled by car or travelled by bicycle.
Each employee used just one method of travel.

One of these employees is chosen at random.
The table shows information about the probability of each method of travel.

$\begin{array}{|l|c|c|c|c|}\hline \textbf{Method of travel} & \text{walk} & \text{bus} & \text{car} & \text{bicycle} \\ \hline \textbf{Probability} & 0.20 & x & 2x & 0.26 \\ \hline \end{array}$

Work out how many of the $450$ employees travelled by car. [4 marks]

[Total 4 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

450

employees, each of whom used exactly one of four methods of travel

The probability table: walk

0.20

, bus

x

, car

2x

, bicycle

0.26

The car probability is written as

2x

, so it is twice the bus probability

Find

How many of the

450

employees travelled by car.

Plan the Solution

Every employee used one method and no employee used two, so the four probabilities describe every possibility once and must add to $1$ .
Add the two probabilities that are given as numbers, then subtract that total from $1$ to see how much probability is left for the bus and the car together.
That leftover is $x + 2x$ , which is $3x$ , so divide by $3$ to find $x$ and then double it for the car.
Finally turn the car probability into a number of employees by multiplying by $450$ .

Worked Solution [4 marks]

Rule - Probabilities of a complete set of outcomes: when every outcome is listed and no two of them can happen together, the probabilities add to

1

. And the expected number in a group is

\text{probability} \times \text{total}

Step 1: write down that the four probabilities add to 1

0.20 + x + 2x + 0.26 = 1

(Reason: Each employee used just one method, and the four methods are the only ones offered, so the four outcomes cover everything and cannot overlap. Reaching this line, in any equivalent form, earns the first method mark.)

Step 2: collect the two probabilities that are already numbers

0.20 + 0.26 = 0.46

1 - 0.46 = 0.54

3x = 0.54

(Reason: The walkers and the cyclists account for

0.46

of the probability, so

0.54

is left for the bus and the car. On the left,

x + 2x

collects to

3x

Step 3: find

x

, the probability for the bus

x = \dfrac{0.54}{3} = 0.18

(Reason: The leftover probability is shared out in equal lots of

x

, so dividing tells you what one lot is worth.)

Step 4: double it to get the probability for the car

2x = 2 \times 0.18 = 0.36

(Reason: The car cell of the table reads

2x

, which is twice the bus probability, so the car probability is

0.36

Step 5: turn the probability into a number of employees

0.36 \times 450 = 162

(Reason: A probability of

0.36

means

0.36

of the group, and the group is

450

employees.)

162

employees travelled by car

Verification

Check 1: Work out all four group sizes and add them up. They must come back to

450

: walkers

0.20 \times 450 = 90

, bus

0.18 \times 450 = 81

, car

162

, cyclists

0.26 \times 450 = 117

90 + 81 + 162 + 117 = 450

, the number of employees the question states.

Check 2: Reach the answer without ever finding

x

. Count the walkers and the cyclists first, take them off the

450

, then split what is left so that the car group is twice the bus group, which makes the car group two thirds of it.

90 + 117 = 207

, then

450 - 207 = 243

, and

\dfrac{2}{3} \times 243 = 162

Check 3: Put the value of

x

back into the table and add the four probabilities.

0.20 + 0.18 + 0.36 + 0.26 = 1

, so the completed table is a proper probability distribution.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$1 - (0.20 + 0.26) = 0.54$ oe, or $x + 2x + 0.26 + 0.20 = 1$ oe, or $x + 2x = 0.54$ oe	M1	Method mark for showing clear understanding that the total of the probabilities is $1$ . If the probabilities are given as percentages then the $\%$ sign must be seen.	✓
$\dfrac{0.54}{3} = 0.18$ , or $\dfrac{2}{3} \times 0.54 = 0.36$ oe, or $0.54 \times 450 = 243$	M1	For a correct method to find $x$ or $2x$ . The mark scheme writes the $0.54$ in quotation marks, which means a candidate's own earlier value may be used here in its place.	✓
$(2 \times) \, 0.18 \times 450$ oe, or $81$ , or $0.36 \times 450$ oe	M1	Or for $\dfrac{81}{450}$ or $\dfrac{162}{450}$ . Again the $0.18$ may be the candidate's own value.	✓
$162$	A1	Working is not required, so a correct answer scores full marks unless it comes from obviously incorrect working.	✓
Alternative full method, worth the same four marks	note	$0.2 \times 450$ and $0.26 \times 450$ , or $90 + 117 = 207$ , or $0.46 \times 450 = 207$ , earns the first method mark; $450 - 207 = 243$ earns the second; $\dfrac{1}{3} \times 243 = 81$ or $\dfrac{2}{3} \times 243 = 162$ earns the third; and the answer is still $162$ .	✓

Full marks: 4/4

Work out the highest common factor (HCF) of $72$ and $108$
You must show your working clearly. [2 marks]

[Total 2 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

The two numbers

72

and

108

The working has to be shown, so a method must appear on the page and not just the answer.

Find

The highest common factor (HCF) of

72

and

108

- the largest number that divides into both of them exactly.

Plan the Solution

Break each number down into a product of prime factors.
Pick out the primes that appear in both lists, and take the lower power of each one.
Multiply those shared prime factors together to get the HCF.
Check by dividing $72$ and $108$ by the answer - both divisions must come out exactly.

Worked Solution [2 marks]

Rule - Highest common factor: write each number as a product of prime factors, then multiply together only the primes the two numbers share, taking the lower power of each. Taking the higher power gives the lowest common multiple instead, which is a different number.

Step 1: write

72

as a product of prime factors

72 = 2 \times 36 = 2 \times 2 \times 18 = 2 \times 2 \times 2 \times 9

72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^{3} \times 3^{2}

(Reason: (Reason: halving as far as it will go strips out every factor of

2

, and what is left,

9

, is

3 \times 3

.))

Step 2: write

108

as a product of prime factors

108 = 2 \times 54 = 2 \times 2 \times 27

108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^{2} \times 3^{3}

(Reason: (Reason:

27

is odd, so the halving stops after two steps, and

27 = 3 \times 3 \times 3

.))

Step 3: multiply the prime factors the two numbers share

\text{HCF} = 2^{2} \times 3^{2} = 4 \times 9 = 36

(Reason: (Reason:

72

brings three

2

s and two

3

s, while

108

brings two

2

s and three

3

s. Only what BOTH lists can supply may be used, so the lower power of each shared prime is the most that can divide into both numbers.))

36

Verification

Check 1: Divide each of the two numbers by

36

and see whether the division is exact.

\dfrac{72}{36} = 2

and

\dfrac{108}{36} = 3

, both whole numbers, so

36

really is a common factor.

Check 2: Look at the two quotients

2

and

3

. If a larger common factor existed, these two would still share a factor above

1

2

and

3

share no factor above

1

, so nothing bigger than

36

divides both numbers - it is the HIGHEST common factor, not just a common one.

Check 3: List every factor of each number and compare the two lists.

72

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

108

1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108

. The common ones are

1, 2, 3, 4, 6, 9, 12, 18, 36

, and the largest of them is

36

Check 4: Use the fact that the HCF multiplied by the lowest common multiple equals the product of the two numbers. Taking the HIGHER power of each shared prime gives the LCM,

2^{3} \times 3^{3} = 216

36 \times 216 = 7776

and

72 \times 108 = 7776

, so the pair

36

and

216

is consistent -

36

is the HCF and

216

is the LCM.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Any correct valid method with no errors, for example: starting to list at least four different factors of each number; or both prime factorisations seen, $2 \times 2 \times 2 \times 3 \times 3$ and $2 \times 2 \times 3 \times 3 \times 3$ (a factor tree or a ladder diagram counts, and $1$ is ignored); or a fully correct Venn diagram; or another clear method such as a division table.	M1	Method mark for a correct approach with no errors. Partial factorisations are accepted in the same spirit, for example $4 \times 2 \times 3 \times 3$ and $4 \times 3 \times 3 \times 3$ , or $2 \times 36$ and $3 \times 36$ .	✓
$36$	A1	Dependent on the method mark. Working is required. Accept $2^{2} \times 3^{2}$ or equivalent as the final answer.	✓

Full marks: 2/2

Elena is a driving instructor.
She keeps a record of the number of kilometres her car travels each month.

In April, the car travelled $943$ kilometres.
This is $15\%$ more than the number of kilometres the car travelled in March.

Work out the number of kilometres the car travelled in March. [3 marks]

kilometres

[Total 3 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

The car travelled

943

kilometres in April.

The April distance is

15\%

more than the March distance.

Find

The number of kilometres the car travelled in March. The March distance is the amount that was increased, so it is the

100\%

amount.

Plan the Solution

March is the starting amount, so March counts as $100\%$ .
Adding $15\%$ makes April $115\%$ of March, which is a multiplier of $1.15$ .
March was multiplied by $1.15$ to give $943$ , so divide by $1.15$ to get back to March.
Do not take $15\%$ off $943$ - that would be a percentage of the wrong amount.

Worked Solution [3 marks]

Reversing a percentage increase:

\text{new} = \text{original} \times \text{multiplier}

, so

\text{original} = \dfrac{\text{new}}{\text{multiplier}}

Step 1: Turn the increase into a multiplier

100\% + 15\% = 115\%

115\% = \dfrac{115}{100} = 1.15

(Reason: March is the amount being increased, so March is

100\%

. April is

15\%

more, so April is

115\%

of March.)

Step 2: Write the relationship as an equation

m \times 1.15 = 943

(Reason: Let

m

be the number of kilometres travelled in March. Multiplying

m

1.15

gives the April distance,

943

Step 3: Divide by the multiplier

m = \dfrac{943}{1.15} = 820

(Reason: Dividing undoes the multiplication, so the April distance divided by

1.15

gives the March distance.)

820

kilometres

Verification

Check 1: Increase the answer by

15\%

again and see whether April comes back.

820 \times 1.15 = 943

, the April distance.

Check 2: Work out

15\%

820

on its own and add it on.

0.15 \times 820 = 123

and

820 + 123 = 943

Check 3: Use the unitary method instead:

115\%

943

kilometres, so find

1\%

and multiply by

100

\dfrac{943}{115} = 8.2

and

8.2 \times 100 = 820

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$1 + 0.15 = 1.15$ , or $100\% + 15\% = 115\%$ , or $m + 0.15m = 943$ , or $\dfrac{943}{115} = 8.2$ oe	M1	A correct first step: the multiplier, the total percentage, an equation in $m$ , or the value of $1\%$ .	✓
$\dfrac{943}{1.15}$ , or $\dfrac{943}{115} \times 100$ , or $943 \times \dfrac{100}{115}$ , or $8.2 \times 100$	M1 dep	A complete method for the March distance, using their own multiplier or their own $1\%$ value. Dependent on the first method mark.	✓
$820$	A1	Working is not required, so a correct answer scores full marks unless it clearly comes from incorrect working.	✓

Full marks: 3/3

In the diagram, $ABCDE$ is a regular pentagon, and $E$ is joined to a point $F$ that lies outside the pentagon.
Not drawn accurately.

Angle $AEF = 96^{\circ}$
Work out the size of the obtuse angle $FED$
You must show your working clearly. [4 marks]

[Total 4 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

ABCDE

is a regular pentagon.

The point

F

lies outside the pentagon, and

E

is joined to

F

Angle

AEF = 96^{\circ}

Find

The size of the obtuse angle

FED

Plan the Solution

A regular pentagon has $5$ equal interior angles, so work one of them out from the angle sum $(n - 2) \times 180^{\circ}$ .
That interior angle is the pentagon's own angle at $E$ , which is angle $AED$ .
Angles $AEF$ , $AED$ and $FED$ all meet at $E$ and fill one full turn, so take the two known angles away from $360^{\circ}$ .

Worked Solution [4 marks]

Rule - Angles at a point add to

360^{\circ}

. One interior angle of a regular

n

-sided polygon is

\dfrac{(n - 2) \times 180^{\circ}}{n}

Step 1: the angle sum of a pentagon

(5 - 2) \times 180^{\circ} = 540^{\circ}

(Reason: A pentagon can be cut into

3

triangles from one vertex, and each triangle contributes

180^{\circ}

Step 2: one interior angle of the regular pentagon

\dfrac{540^{\circ}}{5} = 108^{\circ}

\angle AED = 108^{\circ}

(Reason: All

5

interior angles of a regular pentagon are equal, so each one is a fifth of

540^{\circ}

. The pentagon's own angle at

E

is the angle between

EA

and

ED

Step 3: the three angles at E fill one full turn

\angle AEF + \angle AED + \angle FED = 360^{\circ}

\angle FED = 360^{\circ} - 96^{\circ} - 108^{\circ} = 156^{\circ}

(Reason: The three angles round the point

E

leave no gap and do not overlap, so together they make one complete turn.)

Step 4: check it is the obtuse angle that has been found

90^{\circ} < 156^{\circ} < 180^{\circ}

(Reason: Turning the other way round from

EF

ED

gives the reflex angle

360^{\circ} - 156^{\circ} = 204^{\circ}

, so

156^{\circ}

is the obtuse one the question asks for.)

\angle FED = 156^{\circ}

Verification

Check 1: Produce

AE

past

E

to a point

G

. Angles on the straight line

AEG

give

\angle FEG = 180^{\circ} - 96^{\circ} = 84^{\circ}

, and the exterior angle of the pentagon at

E

\dfrac{360^{\circ}}{5} = 72^{\circ}

EG

lies inside angle

FED

, so the two pieces add.

84^{\circ} + 72^{\circ} = 156^{\circ}

Check 2: Add the three angles that meet at

E

: the given

96^{\circ}

, the pentagon's own

108^{\circ}

and the answer. They must come to one full turn.

96^{\circ} + 108^{\circ} + 156^{\circ} = 360^{\circ}

Check 3: Is the size sensible? The five interior angles come to

5 \times 108^{\circ} = 540^{\circ}

, which matches the angle sum, and the answer is bigger than a right angle but smaller than a straight line.

90^{\circ} < 156^{\circ} < 180^{\circ}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Start on the pentagon's angles	M1	$(5 - 2) \times 180^{\circ} = 540^{\circ}$ or $\dfrac{360^{\circ}}{5} = 72^{\circ}$	✓
One angle of the pentagon, or the piece on the straight line	M1	$\dfrac{540^{\circ}}{5} = 108^{\circ}$ or $180^{\circ} - 72^{\circ} = 108^{\circ}$ or $180^{\circ} - 96^{\circ} = 84^{\circ}$	✓
A complete method for angle FED	M1	$84^{\circ} + 72^{\circ}$ or $360^{\circ} - (96^{\circ} + 108^{\circ})$ or $180^{\circ} - (108^{\circ} - 84^{\circ})$	✓
The answer	A1	$156$ , and the working must be shown	✓
Note	note	Angles written on the diagram earn these marks only if they come from correct working and are correctly assigned.	✓

Full marks: 4/4

(a) Expand and simplify $(m + 5)(m - 8)$ [2 marks]

(b) Solve $3n - 4 = \dfrac{5n + 6}{3}$
You must show clear algebraic working. [3 marks]

(a)(b) n =

[Total 5 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

Part (a): the product of two brackets,

(m + 5)(m - 8)

Part (b): the equation

3n - 4 = \dfrac{5n + 6}{3}

- a linear equation with

n

on both sides, and the right-hand side written over a denominator of

3

Find

(a) The expansion of

(m + 5)(m - 8)

, simplified to a quadratic in

m

with the like terms collected. (b) The value of

n

. The paper asks for clear algebraic working, so the answer alone earns nothing here.

Plan the Solution

(a) Multiply each of the two terms in the first bracket by each of the two terms in the second. That gives four terms, two of which are terms in $m$ .
(a) Collect those two terms in $m$ . The number term and the term in $m^2$ have nothing to collect with.
(b) The right-hand side is divided by $3$ , so multiply BOTH sides by $3$ to clear the fraction. The whole of the left-hand side is multiplied, not just its first term.
(b) Then gather the $n$ terms on one side and the number terms on the other, and divide to finish.

Worked Solution [5 marks]

Rule - Expand then collect:

(a + b)(c + d) = ac + ad + bc + bd

, and only like terms may be added. Rule - Clear the fraction first: multiplying both sides of an equation by the denominator removes it, and every term on each side is multiplied.

Step 1: multiply each term in the first bracket by each term in the second

(m + 5)(m - 8) = m \times m + m \times (-8) + 5 \times m + 5 \times (-8)

(m + 5)(m - 8) = m^2 - 8m + 5m - 40

(Reason: Each of the two terms in the first bracket has to meet both terms in the second, so there are four products. Keeping the sign with the

8

is what makes the third and fourth products come out as

-8m

and

-40

Step 2: collect the two terms in m

-8m + 5m = -3m

(m + 5)(m - 8) = m^2 - 3m - 40

(Reason: Only

-8m

and

5m

are like terms. The

m^2

term and the

-40

have nothing to pair with, so they are already simplified.)

Step 3: clear the fraction in part (b) by multiplying both sides by 3

3 \times (3n - 4) = 3 \times \dfrac{5n + 6}{3}

9n - 12 = 5n + 6

(Reason: Multiplying by

3

undoes the division by

3

on the right, so the fraction disappears. On the left the

3

multiplies the whole bracket, which is exactly what the mark scheme means by removal of the fraction AND multiplying out the left-hand side.)

Step 4: gather the n terms on one side and the numbers on the other

9n - 5n = 6 + 12

4n = 18

(Reason: Subtracting

5n

from both sides clears the

n

from the right; adding

12

to both sides clears the number from the left. Doing both leaves a single term in

n

equal to a single number.)

Step 5: divide both sides by 4

n = \dfrac{18}{4}

n = \dfrac{9}{2} = 4.5

(Reason: Dividing by the coefficient of

n

leaves

n

on its own. The fraction

\dfrac{18}{4}

cancels by

2

\dfrac{9}{2}

, and the mark scheme accepts either form, or

4.5

(a)

m^2 - 3m - 40

(b)

n = \dfrac{9}{2} = 4.5

Verification

Check 1: Part (a), by substitution. Put

m = 2

into the original brackets and into the expansion. If the expansion is right, the two must give the same number.

(2 + 5)(2 - 8) = 7 \times (-6) = -42

and

2^2 - 3 \times 2 - 40 = 4 - 6 - 40 = -42

Check 2: Part (a), by the zeros. The brackets are zero when

m = -5

and when

m = 8

, so the expansion must be zero at both of those values too.

(-5)^2 - 3 \times (-5) - 40 = 25 + 15 - 40 = 0

and

8^2 - 3 \times 8 - 40 = 64 - 24 - 40 = 0

Check 3: Part (b), by substitution into the ORIGINAL equation, fraction and all. Put

n = 4.5

into each side separately. Left-hand side

3 \times 4.5 - 4 = 13.5 - 4 = 9.5

; right-hand side

\dfrac{5 \times 4.5 + 6}{3} = \dfrac{28.5}{3} = 9.5

. The two sides agree.

Check 4: Part (b), by trapping the solution. Both sides are straight lines, so they cross once. Test

n = 4

and

n = 5

and see the left-hand side overtake the right. At

n = 4

the sides are

8

and

\dfrac{26}{3} \approx 8.67

; at

n = 5

they are

11

and

\dfrac{31}{3} \approx 10.33

. The crossing lies between

4

and

5

, as

4.5

does.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $m^2 - 8m + 5m - 40$	M1	For any $3$ correct terms out of the $4$ , or for all $4$ terms correct ignoring signs, or for $m^2 - 3m$ with anything following, or for the $- 3m - 40$ part alone.	✓
(a) $m^2 - 3m - 40$	A1	Working is not required in part (a), so a correct answer on its own scores both marks, unless it follows obviously incorrect working.	✓
(b) $9n - 12 = 5n + 6$ or equivalent	M1	For removal of the fraction AND multiplying out the left-hand side, or for separating the fraction on the right-hand side within an equation, eg $3n - 4 = \dfrac{5}{3}n + \dfrac{6}{3}$ .	✓
(b) $9n - 5n = 6 + 12$ or $4n = 18$ or equivalent	M1ft	Dependent on a $4$ term equation. For correctly rearranging their $4$ term equation so that the terms in $n$ are on one side and the number terms on the other. Follow through their own equation, eg $-12 - 6 = 5n - 9n$ or $-4n = -18$ scores this mark too.	✓
(b) $n = \dfrac{9}{2}$	A1	Dependent on M2. Working is required in part (b). Or equivalent, eg $\dfrac{18}{4}$ or $4.5$ or $4\dfrac{1}{2}$ .	✓

Full marks: 5/5

$\mathcal{E} = \{23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34\}$
$A = \{\text{even numbers}\}$
$B = \{23, 29, 31\}$
$C = \{\text{multiples of } 3\}$

(a) Write down all the members of the set
(i) $B \cup C$
[1 mark]
(ii) $A' \cap C$ [1 mark]

(b) Is it true that $B \cap C = \varnothing$ ?
Write Yes or No, and give a reason for your answer. [1 mark]

The set $D$ has $4$ members and is such that $D \cap (A \cup C) = \varnothing$

(a)(i)(a)(ii)(b)(c)

[Total 5 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

\mathcal{E} = \{23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34\}

, the universal set, with

12

members

A = \{\text{even numbers}\}

B = \{23, 29, 31\}

C = \{\text{multiples of } 3\}

D

has exactly

4

members, and

D \cap (A \cup C) = \varnothing

Find

The members of

B \cup C

The members of

A' \cap C

Whether

B \cap C = \varnothing

is true, and why The members of

D

Plan the Solution

$A$ and $C$ are described in words, so write both out in full first. Every part after that is read straight off the lists.
Union $\cup$ collects, intersection $\cap$ filters: build $B \cup C$ by listing $B$ and adding what $C$ brings that is new.
For $A' \cap C$ , write $A'$ (everything in $\mathcal{E}$ that is not even) and keep only what also appears in $C$ .
For $D$ , read the condition backwards: sharing nothing with $A \cup C$ means $D$ can only use the numbers left in $\mathcal{E}$ once $A \cup C$ is removed.

Worked Solution [5 marks]

Rule - Set notation:

\cup

keeps everything that is in either set,

\cap

keeps only what is in both,

A'

is everything in

\mathcal{E}

that is not in

A

, and

\varnothing

is the set with no members at all.

Step 1: Write out A and C in full

A = \{24, 26, 28, 30, 32, 34\}

C = \{24, 27, 30, 33\}

(Reason: The even members of

\mathcal{E}

are

24, 26, 28, 30, 32, 34

, and the multiples of

3

between

23

and

34

are

24, 27, 30, 33

. All four parts are answered from these two lists, so it is worth writing them once rather than sifting the numbers again each time.)

Step 2: (a)(i) Build the union B u C

B \cup C = \{23, 29, 31\} \cup \{24, 27, 30, 33\}

B \cup C = \{23, 24, 27, 29, 30, 31, 33\}

(Reason: The union keeps every number that is in

B

or in

C

or in both, and each member is written once only. Nothing appears in both sets here, so all

3

members of

B

and all

4

members of

C

survive, giving

7

numbers.)

Step 3: (a)(ii) Find A' first, then intersect with C

A' = \{23, 25, 27, 29, 31, 33\}

A' \cap C = \{23, 25, 27, 29, 31, 33\} \cap \{24, 27, 30, 33\}

A' \cap C = \{27, 33\}

(Reason:

A'

is everything in

\mathcal{E}

that is not even, which is the odd numbers. The intersection then keeps only what is in both lists, so it is asking for the multiples of

3

that are odd:

24

and

30

are multiples of

3

but they are even, so they are ruled out.)

Step 4: (b) Test whether B and C share anything

B \cap C = \{23, 29, 31\} \cap \{24, 27, 30, 33\}

B \cap C = \varnothing

(Reason: Checking the members of

B

one at a time:

23

29

and

31

are none of them multiples of

3

, so no number lies in both sets. An intersection with no members is the empty set, so the statement is true and the answer is Yes.)

Step 5: (c) Remove A u C from the universal set

A \cup C = \{24, 26, 27, 28, 30, 32, 33, 34\}

n(A \cup C) = 6 + 4 - 2 = 8

D = \{23, 25, 29, 31\}

(Reason:

D \cap (A \cup C) = \varnothing

says

D

shares no member with

A \cup C

, so every member of

D

has to come from what is left of

\mathcal{E}

once

A \cup C

is taken out. Only

23, 25, 29, 31

are left, and that is exactly the

4

members

D

is stated to have, so there is no choice about it.)

(a)(i)

B \cup C = \{23, 24, 27, 29, 30, 31, 33\}

(a)(ii)

A' \cap C = \{27, 33\}

(b) Yes, because there are no multiples of

3

in set

B

(c)

D = \{23, 25, 29, 31\}

Verification

Check 1: Count the union a different way.

n(B \cup C) = n(B) + n(C) - n(B \cap C)

, and

B \cap C

is empty, so nothing is double counted.

3 + 4 - 0 = 7

, and the list in (a)(i) has

7

members

Check 2: Test the two answers to (a)(ii) against both conditions, and test the numbers that were rejected. A member of

A' \cap C

must be odd and a multiple of

3

27 = 3 \times 9

and

33 = 3 \times 11

are both odd; the only other multiples of

3

\mathcal{E}

are

24

and

30

, which are even and so belong to

A

, not

A'

Check 3: Count part (c) instead of listing it.

\mathcal{E}

has

12

members and

A \cup C

has

8

, so count how many numbers are left over and compare with the

4

members

D

must have.

12 - 8 = 4

numbers left over, matching the

4

members of

D

, and none of

23, 25, 29, 31

is even or a multiple of

3

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) $B \cup C = \{23, 24, 27, 29, 30, 31, 33\}$	B1	All seven values $23, 24, 27, 29, 30, 31, 33$ , in any order, with no repeats.	✓
(a)(ii) $A' \cap C = \{27, 33\}$	B1	Both values $27, 33$ , in any order, with no repeats.	✓
(b) Yes, with a correct reason	B1	Yes together with a statement showing correct meanings of intersection and empty set, for example there are no multiples of $3$ in set $B$ , or the two sets have no members in common, or $23, 29, 31$ are not in $C$ . This is not an exhaustive list; allow element or value for member. If Yes is not written on the answer line, it must be stated in the reason.	✓
(c) $D = \{23, 25, 29, 31\}$	B2	B2 for the four correct numbers $23, 25, 29, 31$ and no additions. B1 for three correct values with no more than one incorrect, or for four correct values with no more than one incorrect.	✓

Full marks: 5/5

A solid metal cylinder is standing on a workbench.

The volume of the cylinder is $1575$ cm³
The force exerted by the cylinder on the workbench is $84$ newtons.

$\text{pressure} = \dfrac{\text{force}}{\text{area}}$

Work out the pressure on the workbench due to the cylinder. [3 marks]

newtons/cm²

[Total 3 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

Volume of the cylinder:

1575

cm³

Height of the cylinder:

21

cm, marked on the figure

Force on the workbench:

84

newtons

The formula is given in the question:

\text{pressure} = \dfrac{\text{force}}{\text{area}}

The area the cylinder presses on is not given.

Find

The pressure on the workbench, in newtons/cm² The formula needs an area, and no area is given, so this is a two-step question.

Plan the Solution

The area in the formula is the area actually being pressed on: the flat circular base of the cylinder.
A cylinder is a prism, so its volume is the area of that base multiplied by the height. Dividing the volume by the height therefore undoes the multiplication and leaves the base area.
Then put the force over that area.
The radius is never needed. Working it out first reaches the same base area by a longer road, which is why the mark scheme allows either.

Worked Solution [3 marks]

Rule - Pressure and prisms:

\text{pressure} = \dfrac{\text{force}}{\text{area}}

, and for any prism

\text{volume} = \text{cross-sectional area} \times \text{height}

Step 1: turn the volume and the height into the area of the base

\dfrac{1575}{21} = 75 \text{ cm}^2

(Reason: the cylinder is a prism, so

1575

is the base area multiplied by

21

. Dividing by the height undoes that multiplication and leaves the area of the circular base.)

Step 2: divide the force by that area

\dfrac{84}{75} = 1.12 \text{ newtons/cm}^2

(Reason: pressure is the force spread over the area it presses on, so the

84

newtons goes over the base area worked out in Step 1, not over any other length in the question.)

1.12

newtons/cm²

Verification

Check 1: Multiply the base area back by the height. The volume the question gives should come back.

75 \times 21 = 1575

cm³

Check 2: Multiply the pressure by the area. The force the question gives should come back.

1.12 \times 75 = 84

newtons

Check 3: Take a different road entirely. Substituting

\text{area} = \dfrac{\text{volume}}{\text{height}}

into the pressure formula gives force times height over volume, with no area worked out at all.

\dfrac{84 \times 21}{1575} = 1.12

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Base area from the volume and the height: $\dfrac{1575}{21} = 75$	M1	for a correct method to find the cross-sectional area - either $\dfrac{1575}{21}$ , or finding $r^2$ or $r$ from $1575 = \pi r^2 \times 21$ . Note that $r^2$ and $r$ may be rounded or truncated.	✓
Force over that area: $\dfrac{84}{75}$ or equivalent	M1	for $\dfrac{84}{\text{their area of the circle}}$ . Follow through on the area found in the first step, including one reached through the radius.	✓
$1.12$	A1	for $1.12$ newtons/cm². Accept anything from $1.06$ to $1.121$ , which covers a rounded or truncated radius. Working is not required, so a correct answer scores full marks unless it follows obviously incorrect working.	✓

Full marks: 3/3

The table gives the amount of wheat harvested in each of two farming regions in $2020$

(a) Write $3.5 \times 10^7$ as an ordinary number. [1 mark]

In $2020$ , the Highmarsh region harvested $6\,780\,000$ more tonnes of wheat than the Fenwold region.

(b) Work out the amount of wheat harvested in the Highmarsh region in $2020$
Give your answer in standard form. [2 marks]

(a)(b) tonnes

[Total 3 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

The table gives two amounts, both in standard form: Amberdown

3.5 \times 10^7

tonnes and Fenwold

8.2 \times 10^5

tonnes

Highmarsh harvested

6\,780\,000

more tonnes of wheat than Fenwold

The Highmarsh amount is not in the table. It is described in words, and only in terms of the Fenwold amount.

Find

(a)

3.5 \times 10^7

written as an ordinary number (b) the amount Highmarsh harvested, given in standard form and not as an ordinary number

Plan the Solution

(a) The index says how far the digits move. Multiplying by $10^7$ makes every digit worth ten million times as much, so the point in $3.5$ travels seven places and zeros fill the gaps.
(b) A standard form number and an ordinary number cannot be added as they stand, so turn $8.2 \times 10^5$ into an ordinary number first.
Then add the extra tonnes, and put the total back into standard form.
That last conversion is the step that is easy to lose: an ordinary number is the right amount of wheat but the wrong form of answer, and the question asks for the form.

Worked Solution [3 marks]

Rule - Standard form: a number is written

A \times 10^n

, where

A

is at least

1

and less than

10

, and the index

n

counts the places every digit moves.

Step 1: (a) move the digits seven places

3.5 \times 10^7 = 35\,000\,000

(Reason: the index

7

says multiply by

10

seven times, so the point in

3.5

moves seven places to the right and the empty places are filled with zeros. The result has

8

digits, one more than the index.)

Step 2: (b) write the Fenwold amount as an ordinary number

8.2 \times 10^5 = 820\,000

(Reason: the two amounts that have to be added are written in different forms, and only numbers in the same form can be added. The index

5

moves the point in

8.2

five places to the right.)

Step 3: add the extra tonnes

820\,000 + 6\,780\,000 = 7\,600\,000

(Reason: harvesting

6\,780\,000

more tonnes means that many tonnes on top of the Fenwold amount, so the two are added.)

Step 4: put the total back into standard form

7\,600\,000 = 7.6 \times 10^6

(Reason: standard form needs a number between

1

and

10

in front, so the digits are written as

7.6

. Counting from

7.6

up to

7\,600\,000

is six places, so the index is

6

(a)

35\,000\,000

(b)

7.6 \times 10^6

tonnes

Verification

Check 1: Turn the part (b) answer back into an ordinary number and take the extra tonnes away again. The Fenwold amount must come back.

7\,600\,000 - 6\,780\,000 = 820\,000

Check 2: Work part (b) again without ever leaving standard form. Counted in multiples of

10^6

, the Fenwold amount is

0.82

and the extra is

6.78

, so the two mantissas add directly.

0.82 \times 10^6 + 6.78 \times 10^6 = 7.6 \times 10^6

Check 3: Undo part (a) instead of redoing it: divide the ordinary number by

10^7

. The figure printed in the table must come back.

\dfrac{35\,000\,000}{10^7} = 3.5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $35\,000\,000$	B1	for $35\,000\,000$ . Working is not required, so the number on its own earns the mark, written with spaces or with no separator at all.	✓
(b) $820\,000 + 6\,780\,000$ or equivalent	M1	for a correct method: $8.2 \times 10^5 + 6\,780\,000$ or $820\,000 + 6\,780\,000$ . A correct mixture of ordinary numbers and standard form numbers is allowed. Also award it for the digits reached but not converted, $7\,600\,000$ or $76 \times 10^5$ , and for $7.6 \times 10^n$ with any index other than $6$ , which is the right digits with the places miscounted.	✓
(b) $7.6 \times 10^6$	A1	for $7.6 \times 10^6$ . Working is not required, so a correct answer scores full marks unless it follows obviously incorrect working. An ordinary number left as the final answer scores the method mark only, because the question asks for standard form.	✓

Full marks: 3/3

(a) Simplify $(2p)^{0}$ given that $p > 0$ [1 mark]

$y^{9} \times y^{-3} = y^{n}$

(b) Work out the value of $n$ [1 mark]

(a)(b) n =(c)

[Total 4 marks]

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Question 25 - Exam Solution

Understanding the Question

Given

(a)

(2p)^{0}

, with

p > 0

(b)

y^{9} \times y^{-3} = y^{n}

(c)

(5a^{4}c^{2})^{3}

Three separate index laws, one per part, and no calculator work is needed.

Find

(a) the value of

(2p)^{0}

(b) the value of

n

(c)

(5a^{4}c^{2})^{3}

written as a single product of a number, a power of

a

and a power of

c

Plan the Solution

Part (a): the whole bracket carries the power $0$ , and the condition $p > 0$ is there to guarantee the base is not zero.
Part (b): the two powers share the base $y$ , so multiplying them adds the indices; then match indices on both sides.
Part (c): the power $3$ is outside a bracket holding three factors, so every factor is cubed, and a power of a power multiplies the indices.
Finish each part by checking the answer against a numerical substitution.

Worked Solution [4 marks]

Rule - Index laws:

x^{0} = 1

for any non-zero

x

;

x^{m} \times x^{n} = x^{m+n}

; and

(x^{m})^{n} = x^{mn}

, where every factor inside a bracket is raised to the outside power.

Part (a): use the zero-index law

(2p)^{0} = \dfrac{(2p)^{1}}{(2p)^{1}}

(2p)^{0} = 1

(Reason: (Reason:

p > 0

, so the base

2p

is never zero, and any non-zero quantity divided by itself is

1

. The

2

is inside the bracket, so it carries the power

0

as well and does not survive.))

Part (b): multiply the powers by adding the indices

y^{9} \times y^{-3} = y^{9 + (-3)}

9 + (-3) = 6

y^{9} \times y^{-3} = y^{6}

(Reason: (Reason: the base is

y

on both factors, so the indices add. The negative index means a division, so of the nine

y

factors on the top, three are cancelled and

6

are left.))

Part (b): compare the two sides

y^{6} = y^{n}

n = 6

(Reason: (Reason: the two sides are powers of the same base, so their indices must be equal.))

Part (c): raise every factor inside the bracket to the power 3

(5a^{4}c^{2})^{3} = 5^{3} \times (a^{4})^{3} \times (c^{2})^{3}

(Reason: (Reason: the bracket holds three factors multiplied together, and the outside power applies to each of them. The common slip is to cube the letters and leave the

5

untouched.))

Part (c): work out each factor

5^{3} = 5 \times 5 \times 5 = 125

(a^{4})^{3} = a^{4 \times 3} = a^{12}

(c^{2})^{3} = c^{2 \times 3} = c^{6}

(Reason: (Reason: a power raised to a power multiplies the two indices, because

a^{4}

is written down

3

times.))

Part (c): put the three factors back together

(5a^{4}c^{2})^{3} = 125a^{12}c^{6}

(Reason: (Reason: the coefficient and the two powers are multiplied, so the simplified form is one product with no bracket left.))

(a)

1

(b)

n = 6

(c)

125a^{12}c^{6}

Verification

Check 1: Part (a): put

p = 3

into the bracket first, then apply the power.

(2 \times 3)^{0} = 6^{0} = 1

Check 2: Part (b): put

y = 2

into both sides and work them out as numbers.

512 \times \dfrac{1}{8} = 64

and

2^{6} = 64

Check 3: Part (c): put

a = 2

and

c = 3

into the bracket and into the answer.

(5 \times 16 \times 9)^{3} = 720^{3} = 373\,248\,000

and

125 \times 4096 \times 729 = 373\,248\,000

Check 4: Part (c): count the letters instead of using a law. Writing

5a^{4}c^{2}

down three times gives

5 \times 5 \times 5

at the front,

4 + 4 + 4

factors of

a

and

2 + 2 + 2

factors of

c

125a^{12}c^{6}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Simplify $(2p)^{0}$	B1	Answer $1$ , cao. The whole bracket carries the power $0$ , so $2p$ is not left behind.	✓
(b) Value of $n$ in $y^{9} \times y^{-3} = y^{n}$	B1	Answer $6$ , cao, from $9 + (-3)$ .	✓
(c) Simplify $(5a^{4}c^{2})^{3}$ fully	B2	Answer $125a^{12}c^{6}$ . Multiplication signs between the terms are accepted, and $125a^{12}$ , $125c^{6}$ or $a^{12}c^{6}$ is allowed as long as it is not added to any other term.	✓
(c) Partial credit	B1	A product in the form $ka^{p}c^{q}$ where $2$ from $k$ , $p$ or $q$ are correct, for example $5a^{12}c^{6}$ or $125a^{12}3c^{6}$ .	✓
Note	note	The named error behind the partial credit row is a power of a power being ADDED instead of multiplied: $(a^{4})^{3}$ read as $a^{7}$ rather than $a^{12}$ . This row carries no mark of its own.	✓

Full marks: 4/4

The diagram shows the timber frame for the end of a garden shelter.

The frame is made from four lengths of timber, $AB$ , $AC$ , $BC$ and $MC$

$AC = BC = 9$ m and $AB = 12$ m
angle $AMC = 90^\circ$

Nathan is going to buy lengths of timber to make the frame.

The timber costs $21.50$ euros per metre.
Each length of timber he buys has to be a whole number of metres.

Work out the total cost of the timber Nathan needs to buy.
Show your working clearly. [4 marks]

euros

[Total 4 marks]

Show solution & mark schemeHide solution & mark scheme

Question 26 - Exam Solution

Understanding the Question

Given

Triangle

ABC

with

AC = BC = 9

m and

AB = 12

m, so the triangle is isosceles

M

lies on

AB

with angle

AMC = 90^\circ

, so

MC

is perpendicular to the base

The frame uses four lengths:

AB

AC

BC

and

MC

Timber costs

21.50

euros per metre and is bought in whole metres only

Find

The total cost, in euros, of the four lengths of timber Nathan buys

Plan the Solution

Because $AC = BC$ , the perpendicular $MC$ meets $AB$ at its midpoint, so $AM = 6$ m
Use Pythagoras in the right-angled triangle $AMC$ to find $MC$
Round $MC$ UP to a whole number of metres, because a shorter length would not reach
Add the four whole-metre lengths, then multiply the total by $21.50$

Worked Solution [4 marks]

Rule - Pythagoras in a right-angled triangle:

a^{2} + b^{2} = c^{2}

, so a shorter side is

\sqrt{c^{2} - b^{2}}

. Then, when timber is sold in whole metres, every length is rounded UP before it is priced.

Step 1: find

AM

, half of the base

AM = MB = \dfrac{12}{2} = 6 \text{ m}

(Reason:

AC = BC

, so triangle

ABC

is isosceles and the perpendicular from

C

lands on the midpoint of

AB

)

Step 2: use Pythagoras in triangle

AMC

MC^{2} = 9^{2} - 6^{2} = 81 - 36 = 45

MC = \sqrt{45} = 3\sqrt{5} \approx 6.71 \text{ m}

(Reason: angle

AMC = 90^\circ

, so

AC

is the hypotenuse of triangle

AMC

and the shorter side

MC

comes from subtracting the squares)

Step 3: round

MC

up to a whole number of metres

MC \approx 6.71 \implies \text{buy a } 7 \text{ m length}

(Reason: each length bought has to be a whole number of metres, and a

6

m length would be too short, so the next whole metre up is the one Nathan can use)

Step 4: add the four lengths that are bought

7 + 9 + 9 + 12 = 37 \text{ m}

(Reason: the frame needs

MC

AC

BC

and

AB

, and it is the

7

m length from Step 3 that is actually bought)

Step 5: multiply the total length by the price per metre

37 \times 21.50 = 795.50 \text{ euros}

(Reason: every metre bought costs

21.50

euros, so the cost is the total length multiplied by the price per metre)

795.50

euros

Verification

Check 1: Rebuild the right angle at

M

: the two shorter sides squared are

45

and

6^{2} = 36

, and

45 + 36 = 81

81 = 9^{2}

, which is

AC^{2}

, so triangle

AMC

closes exactly and

MC = \sqrt{45}

is right.

Check 2: Find

MC

by trigonometry instead:

\cos CAM = \dfrac{6}{9}

gives angle

CAM = 48.19^\circ

, and

MC = 6 \times \tan 48.19^\circ

. This gives

MC \approx 6.71

m again, so the same

7

m length is bought and the total is unchanged.

Check 3: Price the lengths one at a time rather than as one total:

7 \times 21.50 = 150.50

9 \times 21.50 = 193.50

twice, and

12 \times 21.50 = 258.00

150.50 + 193.50 + 193.50 + 258.00 = 795.50

euros, which agrees with the boxed answer.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$(MC)^{2} + \left(\dfrac{12}{2}\right)^{2} = 9^{2}$ oe, or $9^{2} - \left(\dfrac{12}{2}\right)^{2} \ (= 81 - 36 = 45)$	M1	A correct Pythagoras statement for $MC$ , using half of $AB$ as the base of the right-angled triangle.	✓
$\sqrt{9^{2} - \left(\dfrac{12}{2}\right)^{2}}$ oe $\left(= \sqrt{45} = 3\sqrt{5} = 6.7(08\ldots)\right)$	M1	The square root taken, so $MC$ is reached as a length rather than left as $45$ .	✓
$(7 + 9 + 9 + 12) \times 21.5(0)$ oe	M1	Their $MC$ rounded up to a whole number of metres, added to the other three lengths, and the total multiplied by the price per metre.	✓
$795.5(0)$	A1	Working required, so a bare correct answer with no method shown does not earn this mark.	✓
Alternative, by trigonometry, with $\dfrac{12}{2} = 6$ as half of $AB$ : $\cos(CAM) = \dfrac{6}{9} \implies CAM = 48.1(896\ldots)^\circ$ and $MC = 6 \times \tan 48.1\ldots^\circ$ or $MC = 9 \times \sin 48.1\ldots^\circ \ (= 6.7\ldots)$	M2	The angle and the length together earn the first two method marks; this replaces the two Pythagoras rows above and is not additional to them.	✓
Answer of awrt $789$ from using $6.7\ldots$ metres of timber for $MC$	SC	The named error: pricing the exact length $6.7\ldots$ m instead of the $7$ m length that has to be bought, giving $36.7 \times 21.50 = 789.05$ .	✓

Full marks: 4/4

Keep revising

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

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Edexcel IGCSE 4MA1 Paper 1F, November 2024: Worked Solutions, Questions 15 to 26

Worked solutions, questions 15 to 26 of 26

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