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Edexcel IGCSE 4MA1/1F, Thursday 15 May 2025: Worked Solutions, Questions 14 to 26

Sir Faraz Hassan

22 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)4MA1/1F - Foundation Tier - Thursday 15 May 2025100 marks · 2 hours · Calculator allowed

Back to questions 1 to 13

This is the rest of the paper. Questions 1 to 13, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/1F (Foundation Tier), June 2025 series, sat Thursday 15 May 2025 –100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

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

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

Download printable PDF

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

Worked solutions, questions 14 to 26 of 26

(a) Multiply out $x(x - 7)$ [1 mark]

(b) Factorise $8y - 10$ [1 mark]

Here is a formula.
$m = 5a + 7b$

(a)(b)(c) a =

[Total 5 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

(a) the product

x(x - 7)

, a single term multiplying a bracket

(b) the expression

8y - 10

, two terms that share a factor

m = 5a + 7b

, with

m = 48

and

b = 3

Find

(a) the expanded form, with no brackets left (b)

8y - 10

written as a number outside a bracket (c) the value of

a

, which need not be a whole number

Plan the Solution

(a) Multiply the $x$ outside the bracket by each term inside it.
(b) Find the highest common factor of $8$ and $10$ , put it outside the bracket, and divide each term by it.
(c) Put the two known values into the formula, then undo the $+ 21$ and the $\times 5$ in that order to leave $a$ on its own.

Worked Solution [5 marks]

Rule - Expanding: multiply the term outside the bracket by every term inside it. Factorising: take the highest common factor outside the bracket, and divide each term by it. Substituting: replace the letters with their values, then undo each operation in the reverse order to leave the unknown by itself.

Step 1: multiply the outside term by each term inside the bracket

x(x - 7) = x \times x - x \times 7 = x^2 - 7x

(Reason: Both terms inside the bracket are multiplied by the

x

outside it, and

x \times x

x^2

Step 2: take the highest common factor out of the two terms

8 = 2 \times 4

10 = 2 \times 5

8y - 10 = 2(4y - 5)

(Reason: The largest number that divides both

8

and

10

2

, so

2

goes outside the bracket and what is left of each term goes inside.)

Step 3: substitute the two given values into the formula

m = 5a + 7b

48 = 5a + 7 \times 3

48 = 5a + 21

(Reason: Replacing

m

with

48

and

b

with

3

turns the formula into an equation with only

a

unknown, and

7 \times 3 = 21

Step 4: undo the

+ 21

to leave

5a

on its own

5a = 48 - 21 = 27

(Reason: The opposite of adding is subtracting, so doing that to both sides removes the constant term and leaves

5a

by itself.)

Step 5: undo the

\times 5

to find

a

a = \dfrac{27}{5}

a = 5.4

(Reason: Dividing both sides by

5

leaves

a

on its own, and

\dfrac{27}{5} = 5.4

exactly, so no rounding is needed.)

(a)

x^2 - 7x

(b)

2(4y - 5)

(c)

a = 5.4

Verification

Check 1: Put

x = 3

into the bracketed form and into the expanded form of part (a). Two forms of the same expression must give the same value for every

x

3(3 - 7) = -12

and

3^2 - 7 \times 3 = -12

Check 2: Multiply out the answer to part (b) and see whether the original expression comes back.

2(4y - 5) = 8y - 10

Check 3: Put

a = 5.4

and

b = 3

into

m = 5a + 7b

and check that

m

comes back out as

48

5 \times 5.4 + 7 \times 3 = 27 + 21 = 48

Check 4: Work part (c) the other way round: rearrange the formula first, and substitute afterwards.

a = \dfrac{48 - 7 \times 3}{5} = \dfrac{27}{5} = 5.4

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $x^2 - 7x$	B1	Correct answer only. Allow the terms the other way round, $-7x + x^2$ .	✓
(b) $2(4y - 5)$	B1	Correct answer only. Allow the terms inside the bracket the other way round, $2(-5 + 4y)$ .	✓
(c) Substituting into the formula, eg $48 = 5a + 7 \times 3$ or $48 = 5a + 21$ or $(5a =) 48 - 7 \times 3 \; (= 27)$ or $(5a =) 48 - 21 \; (= 27)$ , or making $a$ the subject, $a = \dfrac{m - 7b}{5}$	M1	for substituting into the equation, or for making $a$ the subject of the equation. Or equivalent.	✓
(c) $(a =) \dfrac{48 - 7 \times 3}{5}$ or $(a =) \dfrac{48 - 21}{5}$	M1	for a complete method. Or equivalent.	✓
(c) $a = 5.4$	A1	or equivalent, eg $\dfrac{27}{5}$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
(c) Special case	SCB1	for answer of $261$	✓
(c) The special case	Note	$5 \times 48 + 7 \times 3 = 261$ is the value of $m$ when $a = 48$ , so the two letters have been used the wrong way round: $m$ has been worked out from $a$ instead of $a$ from $m$ .	✓

Full marks: 5/5

Marcus sells lavender candles and vanilla candles on a market stall.

The ratio
number of lavender candles he sells : number of vanilla candles he sells $= 5 : 7$

He sells $40$ lavender candles.

Marcus sells each lavender candle for $\$8.50$
He sells each vanilla candle for $\$12.75$

Work out the total amount of money Marcus gets for selling the candles. [4 marks]

[Total 4 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

The ratio of lavender candles sold to vanilla candles sold is

5 : 7

40

lavender candles are sold

One lavender candle sells for

\$8.50

and one vanilla candle sells for

\$12.75

Find

The total amount of money Marcus gets for selling all of the candles

Plan the Solution

Read the ratio carefully. The $40$ is the LAVENDER part of the ratio, not the total number of candles, so $5$ parts are worth $40$ candles.
Divide to find what one part of the ratio is worth, then multiply by $7$ to get the number of vanilla candles.
Work out the money taken on each kind of candle separately, then add the two amounts together.

Worked Solution [4 marks]

Rule - Sharing in a ratio: divide the known quantity by the number of parts it stands for to get one part, then multiply one part by the other number of parts. The money taken on a kind of candle is

\text{price} \times \text{number sold}

Step 1: Find what one part of the ratio stands for

\dfrac{40}{5} = 8

(Reason: The

40

lavender candles are the

5

parts of the ratio, so one part stands for

8

candles.)

Step 2: Work out how many vanilla candles were sold

7 \times 8 = 56

(Reason: The vanilla candles are

7

parts of the ratio, and each part stands for

8

candles.)

Step 3: Work out the money taken on the lavender candles

40 \times 8.50 = 340

(Reason: Every lavender candle brings in

\$8.50

, and

40

of them were sold.)

Step 4: Work out the money taken on the vanilla candles

56 \times 12.75 = 714

(Reason: Multiply the number of vanilla candles by the price of one vanilla candle.)

Step 5: Add the two amounts

340 + 714 = 1\,054

(Reason: The total taken is the money from the lavender candles plus the money from the vanilla candles.)

\$1\,054

Verification

Check 1: The two candle counts must cancel back to the ratio the question gives. Divide each of them by

8

\dfrac{40}{56} = \dfrac{5}{7}

Check 2: Work backwards from the total. Take the lavender money away from it, then divide what is left by the price of one vanilla candle. That must give the number of vanilla candles back.

1\,054 - 340 = 714

, and

\dfrac{714}{12.75} = 56

Check 3: A size check. Altogether

96

candles were sold, at prices between

\$8.50

and

\$12.75

, so the total should be near

96

lots of about

\$10

96 \times 10 = 960

, which is the right size for the answer.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{40}{5} = 8$ or $40 \times 8.50 = 340$	M1	for starting to work with the ratio, or for working out the total money taken on the lavender candles	✓
$7 \times 8 = 56$	M1	for a method to find the number of vanilla candles	✓
$340 + 56 \times 12.75$ or $340 + 714$	M1	for a complete method	✓
$1\,054$	A1	cao. A correct answer scores full marks unless it comes from obviously incorrect working.	✓
If no other marks are awarded	SC B2	for answer of $439(.166\ldots)$	✓
If no other marks are awarded	SC B1	for $141(.66\ldots)$ or $297(.5(0))$	✓
The two special cases	Note	Both come from sharing the $40$ candles in the ratio $5 : 7$ , instead of reading the $40$ as the $5$ parts. $141(.66\ldots)$ is the lavender half of that wrong share and $297(.5(0))$ the vanilla half, and $439(.166\ldots)$ is the two together. The bracketed digits are optional, so a truncated $141$ or $297$ or $439$ is accepted.	✓

Full marks: 4/4

Some members of a music club were asked whether they play the guitar ( $G$ ) or the piano ( $P$ )
The Venn diagram gives information about the results.
The values shown represent numbers of members.

(a) How many of these members play only the guitar or only the piano? [1 mark]

One of these members is chosen at random.

(b) Find the probability that this member plays both the guitar and the piano. [2 marks]

(a)(b)

[Total 3 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

A Venn diagram for

G

(plays the guitar) and

P

(plays the piano)

9

G

only,

6

in the overlap,

8

P

only

2

members inside the box but outside both circles

Find

(a) The number of members who play only the guitar or only the piano (b) The probability that a member chosen at random plays both instruments

Plan the Solution

Part (a) wants the members inside one circle only, so add the two regions that are not in the overlap.
Part (b) needs the size of the whole group first, so add all four regions - the members outside both circles were asked too.
Then put the number in the overlap over that total and see whether the fraction cancels.

Worked Solution [3 marks]

Rule - Probability:

\dfrac{\text{members in the region you want}}{\text{members altogether}}

. The four regions of a Venn diagram do not overlap, so the total is the sum of all four values.

Step 1: add the two regions that lie inside one circle only

9 + 8 = 17

(Reason: The

9

is inside

G

but outside the overlap, so those members play the guitar and not the piano, and the

8

is inside

P

but outside the overlap. The

6

in the middle plays both instruments, so it is not counted here.)

Step 2: work out how many members were asked altogether

9 + 6 + 8 + 2 = 25

(Reason: Every member who was asked appears in exactly one region of the diagram, and that includes the ones sitting inside the box but outside both circles, so the size of the whole group is the sum of all four values.)

Step 3: write the probability of landing in the overlap

\dfrac{6}{25}

\dfrac{6}{25} = 0.24

(Reason: Only the

6

in the overlap play both instruments, and there are

25

members to choose from.

6

and

25

share no common factor, so the fraction is already in its simplest form. The decimal is worth writing down too, because the mark scheme accepts it.)

(a)

17

members(b)

\dfrac{6}{25}

Verification

Check 1: Add the answer to part (a) to the members it leaves out - the

6

who play both and the

2

who play neither. It must come back to the whole group.

17 + 6 + 2 = 25

Check 2: Read the probability as a decimal and take that fraction of the group. It must come back to the number in the overlap.

0.24 \times 25 = 6

members

Check 3: The four regions cover everyone and share nobody, so the four probabilities must add to

1

\dfrac{9}{25} + \dfrac{6}{25} + \dfrac{8}{25} + \dfrac{2}{25} = \dfrac{25}{25} = 1

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $17$	B1	cao - the correct answer only. No working has to be shown for this mark.	✓
(b) $\dfrac{6}{m}$ where $m > 6$ , or $\dfrac{n}{25}$ where $n < 25$ , or $6 : 25$ , or $6$ out of $25$	M1	For a probability with the right numerator or the right denominator. Writing $6$ over anything bigger than $6$ shows the overlap has been picked out; writing anything smaller than $25$ over $25$ shows the whole group has been counted.	✓
(b) $\dfrac{6}{25}$	A1	oe eg $0.24$ , $24\%$	✓
A correct answer scores full marks unless it comes from obviously incorrect working.	Note	Printed against part (b) on the official scheme. A ratio such as $6 : 25$ earns the method mark but not the accuracy mark, because a ratio is not a probability.	✓

Full marks: 3/3

The diagram shows the plan of a wooden deck.
The deck is made up of a rectangle and an isosceles triangle.

Haruka is going to cover the deck with wood sealer.
Enough tins of sealer must be bought to give the whole deck one coat.

Each tin of sealer covers an area of $4 \text{ m}^2$

Work out the smallest number of tins of sealer that Haruka needs to buy.
Show your working clearly. [4 marks]

[Total 4 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

The plan is a rectangle

8

m by

6

m with an isosceles triangle joined along one

6

m edge

The triangle stands

3

m out from that edge, so

3

m is its height

One tin of sealer covers

4 \text{ m}^2

Find

The smallest number of whole tins that will give the deck one coat

Plan the Solution

Split the plan along the dashed line into the rectangle and the triangle, and work out each area on its own.
Add the two areas to get the area of the whole deck.
Divide that total by $4$ to see how many tins' worth of sealer the deck takes.
Round up. Tins are sold whole, so any sealer needed beyond a whole number of tins means buying one more.

Worked Solution [4 marks]

Rule - Composite area: split the shape into parts whose areas you know, then add. A rectangle has area

\text{length} \times \text{width}

and a triangle has area

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

Step 1: the area of the rectangle

8 \times 6 = 48 \text{ m}^2

(Reason: The rectangle is

8

m along the bottom and

6

m up the side, so its area is length times width.)

Step 2: the area of the triangle

\dfrac{1}{2} \times 6 \times 3 = 9 \text{ m}^2

(Reason: The triangle is joined along the

6

m edge, so that edge is its base, and the arrow gives its height as

3

m.)

Step 3: the area of the whole deck

48 + 9 = 57 \text{ m}^2

(Reason: The rectangle and the triangle meet along the dashed line and do not overlap, so the two areas simply add.)

Step 4: how many tins' worth of sealer that is

\dfrac{57}{4} = 14.25

(Reason: Each tin covers

4 \text{ m}^2

, so the total area divided by

4

gives the number of tins the deck takes.)

Step 5: round up to whole tins

14 \times 4 = 56 \text{ m}^2

15 \times 4 = 60 \text{ m}^2

(Reason: Sealer is only sold in whole tins.

14

tins cover

56 \text{ m}^2

, which leaves part of the deck bare, so the

14.25

has to be rounded up rather than down.)

15

tins

Verification

Check 1: Use the symmetry instead of the split. Cutting along the line through the point leaves a trapezium with parallel sides

8

m and

11

m and width

3

m, and the other half is identical.

\dfrac{1}{2} \times (8 + 11) \times 3 = 28.5

and

28.5 \times 2 = 57

, the same total area

Check 2: Count the tins part by part rather than from the total:

\dfrac{48}{4}

for the rectangle and

\dfrac{9}{4}

for the triangle.

12 + 2.25 = 14.25

, which rounds up to the same

15

tins

Check 3: Test the answer against the area it has to cover, from both sides.

15 \times 4 = 60

, enough for

57 \text{ m}^2

with

3 \text{ m}^2

spare, while

14 \times 4 = 56

leaves

1 \text{ m}^2

uncovered

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct method for one relevant area	M1	Any one correct area method, eg $6 \times 8 = 48$ or $\dfrac{1}{2} \times 6 \times 3 = 9$ or $\dfrac{1}{2} \times \dfrac{6}{2} \times 3 = 4.5$ (half the triangle) or $\dfrac{1}{2} \times (11 + 8) \times 3 = 28.5$ (half the plan)	✓
A complete method for the total area, or the tins for one part	M1	From this mark on, follow either the area route or the tins route. Area route - a complete method for the total area, eg $48 + 9 = 57$ or $48 + 4.5 + 4.5 = 57$ or $28.5 \times 2 = 57$ . Tins route - a correct method for the tins needed for part of the deck, eg $\dfrac{48}{4} \; (= 12)$ or $\dfrac{9}{4} \; (= 2.25 \text{ or } 3)$ or $\dfrac{4.5}{4} \; (= 1.125)$ or $\dfrac{28.5}{4} \; (= 7.125)$ . The official scheme prints the $9$ in quotation marks, which means the candidate's own earlier area may be used in its place.	✓
A method for the number of tins	M1 indep	Area route - independent, and follow through from any area that came from a calculation using at least two of the given dimensions, eg $\dfrac{57}{4} = 14.25$ or their area divided by $4$ or $15 \times 4 = 60$ . Tins route - adding the tins found for the parts, eg $12 + 2.25 = 14.25$ or $2 \times 7.125 = 14.25$	✓
The answer	A1 dep on M2	$15$ , from correct working. Working is required, so an unsupported answer earns nothing here.	✓

Full marks: 4/4

A large supermarket asked each of its $100$ members of staff how far they travel to work.
The distance, $d$ km, was recorded for every member of staff.
The table gives information about these distances.

$\begin{array}{|c|c|}\hline \textbf{Distance}\ (d\ \textbf{km}) & \textbf{Frequency} \\ \hline 0 < d \leq 5 & 26 \\ \hline 5 < d \leq 10 & 40 \\ \hline 10 < d \leq 15 & 16 \\ \hline 15 < d \leq 20 & 10 \\ \hline 20 < d \leq 25 & 8 \\ \hline \end{array}$

(a) Write down the modal class. [1 mark]

(b) Work out an estimate for the mean distance. [4 marks]

(a)(b) km

[Total 5 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

The distance,

d

km, travelled to work by each of

100

members of staff, grouped into five classes

The five frequencies

26

40

16

10

and

8

No individual distance is given, only the class that each one falls in

Find

(a) The modal class (b) An estimate for the mean distance, in km

Plan the Solution

For part (a), read the answer straight off the table: the modal class is the class with the largest frequency, and the answer is that class, not the frequency itself.
For part (b), the individual distances are unknown, so treat everybody in a class as if they travelled the midpoint of that class.
Multiply each midpoint by its frequency, add the five products to estimate the total distance travelled, then divide that total by $100$ .

Worked Solution [5 marks]

Rule - Estimated mean of grouped data:

\text{estimated mean} = \dfrac{\sum fx}{\sum f}

, where

x

is the midpoint of a class and

f

is the frequency of that class.

Step 1: Pick out the largest frequency (part (a))

26,\ 40,\ 16,\ 10,\ 8

5 < d \leq 10

(Reason: The largest frequency is

40

, and those

40

members of staff are the ones whose distance lies in the class

5 < d \leq 10

. The modal class is that class. Writing

40

as the answer names how many, not how far.)

Step 2: Find the midpoint of each class

\dfrac{0 + 5}{2} = 2.5

\dfrac{5 + 10}{2} = 7.5

\dfrac{10 + 15}{2} = 12.5

\dfrac{15 + 20}{2} = 17.5

\dfrac{20 + 25}{2} = 22.5

(Reason: Add the two ends of the class and halve. The midpoint is the single value that best represents everybody in that class, which is why the answer to part (b) can only ever be an estimate.)

Step 3: Multiply each midpoint by its frequency

2.5 \times 26 = 65

7.5 \times 40 = 300

12.5 \times 16 = 200

17.5 \times 10 = 175

22.5 \times 8 = 180

(Reason: Each product estimates the total distance travelled by the staff in one class:

26

people at about

2.5

km each accounts for about

65

km.)

Step 4: Add the five products

65 + 300 + 200 + 175 + 180 = 920

(Reason: This is the estimated total distance, in km, travelled to work by all

100

members of staff.)

Step 5: Divide by the total frequency

\dfrac{920}{100} = 9.2

(Reason: Sharing the estimated total distance equally between the

100

members of staff gives the estimated mean. Divide by the total frequency, never by the number of classes.)

(a)

5 < d \leq 10

(b)

9.2

Verification

Check 1: Add the frequency column:

26 + 40 + 16 + 10 + 8 = 100

. Every member of staff has been counted exactly once, so

100

is the right number to divide by.

Check 2: Work the mean out a second way, from an assumed mean of

7.5

. The midpoints differ from

7.5

-5

0

5

10

and

15

, so

26 \times (-5) + 40 \times 0 + 16 \times 5 + 10 \times 10 + 8 \times 15 = 170

7.5 + \dfrac{170}{100} = 9.2

, which is the same estimate reached without using the five products.

Check 3: Bracket the answer. Replace every distance by the smallest value its class allows, then by the largest:

\dfrac{670}{100} = 6.7

and

\dfrac{1170}{100} = 11.7

. The mean has to sit between

6.7

and

11.7

, and

9.2

does. It also sits inside the modal class, which is where most of the staff are.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Write down the class with the greatest frequency	B1	$5 < d \leq 10$ . Allow $5 - 10$ or $5$ to $10$ , or $5 < d < 10$ , or $5 \leq d \leq 10$ , or $5 \leq d < 10$ .	✓
(b) At least $4$ correct products of midpoint and frequency, added	M2	$2.5 \times 26 + 7.5 \times 40 + 12.5 \times 16 + 17.5 \times 10 + 22.5 \times 8 = 920$ , or $65 + 300 + 200 + 175 + 180 = 920$ . The products need not be evaluated, so the first form on its own earns this.	✓
If M2 is not earned: at least $4$ products added, using a value taken consistently from inside each interval (an end point is allowed), or correct midpoints used for at least $4$ products but not added	M1	For example $5 \times 26 + 10 \times 40 + \ldots$ Again the products need not be evaluated.	✓
Divide their total by their total frequency: $\dfrac{920}{100}$	M1	Dependent on at least M1. Division by their own $\sum f$ is allowed, provided the addition, or a total written under the frequency column, is seen.	✓
(b) The estimate of the mean	A1	$9.2$ , or an equivalent value such as $9\dfrac{1}{5}$ or $\dfrac{46}{5}$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
Special case, when no other marks in part (b) are earned: an answer of $6.7$ , $9.7$ or $11.7$	SC B2	Each one names a particular slip. $6.7$ uses the lower end of every class instead of its midpoint, $11.7$ uses the upper end, and $9.7$ uses $3$ , $8$ , $13$ , $18$ and $23$ , each a half kilometre above the true midpoint.	✓

Full marks: 5/5

Lena makes candles.
Each candle costs $6$ Swiss francs to make.

Lena packs the candles into boxes to sell.
Each box contains $4$ candles.

Lena sells $80$ boxes of candles for a total of $2160$ Swiss francs.

(a) Work out the percentage profit Lena makes.
Show your working clearly. [4 marks]

The height of each candle is $9$ cm, correct to the nearest cm

(b) Write down the lower bound of the height. [1 mark]

The weight of each candle is $120$ g, correct to the nearest $10$ g

(a) %(b) cm(c) g

[Total 6 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

Each candle costs

6

Swiss francs to make.

Each box holds

4

candles, and

80

boxes are sold.

The

80

boxes bring in

2160

Swiss francs altogether.

A height of

9

cm, correct to the nearest cm.

A weight of

120

g, correct to the nearest

10

Find

(a) The percentage profit. (b) The lower bound of the height. (c) The upper bound of the weight.

Plan the Solution

Count the candles first: $4$ in each of $80$ boxes.
Cost them all, then take that away from the money taken to leave the profit.
Percentage profit compares the profit with what the candles COST, not with the money taken.
For each bound, halve the accuracy the measurement is given to, then step that far below or above the value stated.

Worked Solution [6 marks]

Rule - Percentage profit:

\text{percentage profit} = \dfrac{\text{profit}}{\text{cost}} \times 100

. Rule - Bounds: a measurement given to the nearest

u

lies within

\dfrac{u}{2}

of the value stated, so the bounds sit half a unit either side of it.

Step 1: Count the candles that are sold

4 \times 80 = 320

(Reason: (Reason: each of the

80

boxes holds

4

candles, so the number of candles is one multiplication.))

Step 2: Work out what those candles cost to make

320 \times 6 = 1920

(Reason: (Reason: every candle costs the same

6

Swiss francs, so the total cost is the number of candles times the cost of one.))

Step 3: Take the cost away from the money taken

2160 - 1920 = 240

(Reason: (Reason: profit is what is left of the

2160

Swiss francs once the making cost has been paid for.))

Step 4: Write the profit as a percentage of the cost

\dfrac{240}{1920} \times 100 = 12.5

(Reason: (Reason: percentage profit measures the profit against what the candles cost to make, so it is the profit that is divided by the cost.))

Step 5: (b) The lower bound of the height

9 - 0.5 = 8.5

(Reason: (Reason: to the nearest cm, half a unit is

0.5

cm, so every height from

8.5

cm up to

9.5

cm rounds to

9

cm, and

8.5

cm is the smallest of them.))

Step 6: (c) The upper bound of the weight

120 + 5 = 125

(Reason: (Reason: to the nearest

10

g, half a unit is

5

g, so the weights that round to

120

g run from

115

g up to

125

g, and

125

g is the upper bound.))

(a)

12.5

%(b)

8.5

cm(c)

125

Verification

Check 1: Work with one box instead of the whole order. One box brings in

\dfrac{2160}{80} = 27

Swiss francs and costs

4 \times 6 = 24

Swiss francs to make.

\dfrac{27 - 24}{24} \times 100 = 12.5

Check 2: Scale the cost back up. A profit of

12.5

per cent means the money taken should be

1.125

times the making cost.

1920 \times 1.125 = 2160

Check 3: Test the two bounds by looking at the interval each one ends. The heights that round to

9

cm run from

8.5

cm to

9.5

cm; the weights that round to

120

g run from

115

g to

125

g. Each interval is centred on the measurement given,

\dfrac{8.5 + 9.5}{2} = 9

and

\dfrac{115 + 125}{2} = 120

, so the bounds are the right distance out.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A method for the total making cost, or for one box, or for the money taken for one box or for one candle: $4 \times 6 \times 80 = 1920$ , or $4 \times 6 = 24$ , or $\dfrac{2160}{80} = 27$ , or $\dfrac{2160}{80 \times 4} = 6.75$	M1	Any one of these four starting methods earns the mark.	✓
Work out the profit, or divide the money taken by the cost: $2160 - 1920 = 240$ , or $27 - 24 = 3$ , or $6.75 - 6 = 0.75$ , or $\dfrac{2160}{1920} = 1.125$	M1	The official scheme puts these figures in quotation marks, which means the candidate's own earlier value may be used here.	✓
A method that reaches one step from the answer: $\dfrac{240}{1920} \times 100$ , or $0.125 \times 100$ , or $\left(\dfrac{2160}{1920} - 1\right) \times 100$ , or $1.125 \times 100 = 112.5$	M1	Getting as far as $\dfrac{1}{8}$ or equivalent, or $0.125$ , or $112.5$ . A candidate who stops at $112.5$ has not taken off the $100$ per cent that is the original cost.	✓
(a) $12.5$	A1	Working is required, and this mark is not awarded unless a method mark has been earned.	✓
(b) $8.5$	B1	Correct answer only.	✓
(c) $125$	B1	Allow $124.\dot{9}$ or $124.99\ldots$ .	✓

Full marks: 6/6

The diagram shows two sides $AB$ and $BC$ of a regular polygon with $n$ sides.
A third straight line is drawn from $B$ , and the two angles it makes are marked on the diagram.

Work out the value of $n$ .
You must show your working clearly. [4 marks]

n =

[Total 4 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

AB

and

BC

are two sides of a regular polygon with

n

sides

B

a third line makes an angle of

50^\circ

with

BC

and an angle of

148^\circ

with

BA

Find

The value of

n

, the number of sides of the polygon

Plan the Solution

The three angles at $B$ make one full turn, so taking the two marked angles away from $360^\circ$ leaves the interior angle $\angle ABC$ of the polygon.
An interior angle and its exterior angle sit on a straight line, so the exterior angle is what is left from $180^\circ$ .
Every exterior angle of a regular polygon is the same size and they add up to $360^\circ$ , so dividing gives the number of sides.

Worked Solution [4 marks]

Rule - Exterior angles: angles at a point add up to

360^\circ

; an interior angle and its exterior angle add up to

180^\circ

; and the exterior angles of any polygon add up to

360^\circ

Step 1: Add the two marked angles at

B

148 + 50 = 198

(Reason: The two marked angles sit side by side at

B

, so between them they take up

198^\circ

of the turn at that point.)

Step 2: Take that away from a full turn

360 - 198 = 162

(Reason: Angles at a point add up to

360^\circ

, so the angle left over is the polygon's interior angle

\angle ABC

, which is

162^\circ

Step 3: Turn the interior angle into the exterior angle

180 - 162 = 18

(Reason: The interior angle and the exterior angle at the same corner lie on a straight line, and angles on a straight line add up to

180^\circ

Step 4: Divide a full turn by the exterior angle

\dfrac{360}{18} = 20

(Reason: A regular polygon has

n

equal exterior angles adding up to

360^\circ

, so the number of them of size

18^\circ

is what

n

must be.)

n = 20

Verification

Check 1: Work backwards from the answer. The interior angles of a polygon with

20

sides add up to

180 \times 18 = 3240

, shared equally between

20

corners.

\dfrac{3240}{20} = 162

, which is the interior angle the diagram gives.

Check 2: Put the interior angle back on the diagram and add all three angles at

B

162 + 148 + 50 = 360

, one complete turn.

Check 3: Walk right round the polygon, turning through one exterior angle at each of the

20

corners.

20 \times 18 = 360

, so the walk faces its starting direction again and the shape closes.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Method for the interior angle at $B$	M1	For $360 - (148 + 50) \; (= 162)$ , or for a start on the exterior angle such as $180 - 50 \; (= 130)$ or $180 - 148 \; (= 32)$ .	✓
Method for the exterior angle	M1	For $180 - 162 \; (= 18)$ , or $148 - 130 \; (= 18)$ , or $50 - 32 \; (= 18)$ ; or for setting up the interior angle sum as $180(n - 2) = 162n$ or $\dfrac{180(n - 2)}{n} = 162$ . The official scheme prints the $162$ , the $130$ and the $32$ in quotation marks, which means the candidate's own interior angle may be used in place of each.	✓
A complete method	M1	For $\dfrac{360}{18}$ on their exterior angle, or for $n = \dfrac{360}{180 - 162}$ written in one line.	✓
The value of $n$	A1	For $20$ . Working is required, so the mark is not given for a bare answer: it depends on a method mark being earned.	✓

Full marks: 4/4

$x^5 \times x^7 = x^m$

(a) Work out the value of $m$ . [1 mark]

$\dfrac{y^8}{y^3} = y^n$

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

(a) m =(b) n =(c)

[Total 4 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

x^5 \times x^7 = x^m

\dfrac{y^8}{y^3} = y^n

The expression

(5a^4r^2)^3

, to be written as one simplified term

Find

The value of

m

The value of

n

A single term for

(5a^4r^2)^3

with no bracket left in it

Plan the Solution

Parts (a) and (b) are the two basic index laws. Powers of the same base multiplied together have their indices added; divided, their indices are taken away.
Part (c) has a bracket, so the index $3$ outside it reaches every factor inside: the $5$ , the $a^4$ and the $r^2$ each get cubed.
Deal with those three factors one at a time. The number is cubed like everything else, and a power raised to a power has its indices multiplied.

Worked Solution [4 marks]

Rule - Index laws:

x^a \times x^b = x^{a+b}

\dfrac{x^a}{x^b} = x^{a-b}

and

(x^a)^b = x^{ab}

. A power written outside a bracket applies to every factor inside it.

Step 1: Part (a) - two powers of the same base multiplied

x^5 \times x^7 = x^{5 + 7} = x^{12}

m = 12

(Reason: The base is

x

on both sides of the multiplication sign, so five

x

's standing beside seven

x

's make twelve

x

's in one product. Comparing that with

x^m

gives the value of

m

Step 2: Part (b) - two powers of the same base divided

\dfrac{y^8}{y^3} = y^{8 - 3} = y^{5}

n = 5

(Reason: Three of the eight

y

's on the top cancel with the three underneath, and five are left. Comparing with

y^n

gives the value of

n

Step 3: Part (c) - give every factor inside the bracket the index

3

(5a^4r^2)^3 = 5^3 \times (a^4)^3 \times (r^2)^3

(Reason: Cubing the bracket means writing its contents out three times and multiplying, so each of the three factors inside is cubed. Missing the

5

here is the commonest way marks are lost in this part.)

Step 4: Part (c) - work out the three factors separately

5^3 = 5 \times 5 \times 5 = 125

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

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

(Reason: The number is cubed exactly as the letters are. For a power of a power the indices multiply, because

a^4

taken three times over is

a

taken

4 \times 3

times.)

Step 5: Part (c) - put the three factors back together

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

(Reason: The three factors are multiplied, so the number goes in front and each letter keeps its new index. Nothing can be collected any further, because

a

and

r

are different letters.)

(a)

m = 12

(b)

n = 5

(c)

125a^{12}r^{6}

Verification

Check 1 - part (a): Put

x = 2

into the left-hand side, working the two powers out separately:

2^5 = 32

and

2^7 = 128

32 \times 128 = 4096

and

2^{12} = 4096

, so the two sides agree and

m = 12

Check 2 - part (b): Put

y = 2

into the fraction: the top is

2^8 = 256

and the bottom is

2^3 = 8

\dfrac{256}{8} = 32

and

2^5 = 32

, so the two sides agree and

n = 5

Check 3 - part (c): Put

a = 2

and

r = 3

into the bracket first, then cube what comes out:

5 \times 16 \times 9 = 720

720^3 = 373\,248\,000

and the answer gives

125 \times 2^{12} \times 3^6 = 373\,248\,000

, the same number.

Check 4 - part (c) without any index law: Write the bracket out three times,

(5a^4r^2)(5a^4r^2)(5a^4r^2)

, and collect the numbers and each letter by hand. The numbers give

5 \times 5 \times 5 = 5^3 = 125

, the

a

's give

4 + 4 + 4 = 12

and the

r

's give

2 + 2 + 2 = 6

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Part (a): the value of $m$	B1	For $12$ . Accept the answer written as $x^{12}$ .	✓
Part (b): the value of $n$	B1	For $5$ . Accept the answer written as $y^5$ .	✓
Part (c): fully simplified	B2	For $125a^{12}r^6$ , with no bracket left and nothing further to collect.	✓
Part (c): partly simplified	B1	Awarded instead of the two marks, for a product in the form $ka^pr^q$ in which two of $k$ , $p$ and $q$ are correct, for example $5a^{12}r^6$ - the candidate who cubes both letters but leaves the $5$ alone. Allow $125a^{12}$ , $125r^6$ or $a^{12}r^6$ so long as they are not added to any other terms.	✓

Full marks: 4/4

A sports shop is holding a clearance sale.
In the sale, all normal prices are reduced by $28\%$
The sale price of a rucksack is $198$ euros.

Work out the normal price of the rucksack. [3 marks]

euros

[Total 3 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

In the sale, every normal price is reduced by

28\%

The sale price of the rucksack is

198

euros.

The reduction is taken off the normal price, and the normal price is the unknown.

Find

The normal price of the rucksack, in euros, before the reduction.

Plan the Solution

A reduction of $28\%$ leaves $72\%$ of the normal price, so work out what is left before anything else.
Write that percentage as a decimal multiplier, so that the sale price is the multiplier times the normal price.
Divide to undo the multiplication, then check by taking the reduction off the answer.

Worked Solution [3 marks]

Rule - Reverse percentage: sale price

= \text{multiplier} \times \text{normal price}

, so normal price

= \dfrac{\text{sale price}}{\text{multiplier}}

Step 1: Work out the percentage that is left

100\% - 28\% = 72\%

(Reason: The normal price is the whole thing,

100\%

. Taking

28\%

away leaves

72\%

of it, and that is what the

198

euros on the sale ticket buys.)

Step 2: Turn that percentage into a multiplier

\dfrac{72}{100} = 0.72

0.72 \times n = 198

(Reason: Per cent means out of a hundred, so

72\%

is the decimal

0.72

. Writing

n

for the normal price in euros, paying

72\%

of it is the same as multiplying it by

0.72

Step 3: Divide to reverse the multiplication

n = \dfrac{198}{0.72}

n = 275

(Reason: Division undoes multiplication, so the normal price is the sale price divided by the multiplier found in Step 2. The answer comes out as a whole number of euros, which is what a price tag should be.)

275

euros

Verification

Check 1: Take

28\%

of the answer and subtract it. The sale price must come back.

\dfrac{28 \times 275}{100} = 77

and

275 - 77 = 198

Check 2: Use the unitary method instead. Divide the sale price by

72

to get

1\%

of the normal price, then take

100

of those.

\dfrac{198}{72} = 2.75

and

2.75 \times 100 = 275

Check 3: Compare the two prices as a fraction. The sale price should be

0.72

of the normal price.

\dfrac{198}{275} = 0.72

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $1 - 0.28 = 0.72$ oe, or $100\% - 28\% = 72\%$ , or $0.72x = 198$ , or $\dfrac{198}{72} = 2.75$ oe	M1	for a correct first step: the multiplier, or the percentage that is left, or an equation for the normal price	✓
eg $(x =) \; \dfrac{198}{0.72}$ oe, or $\dfrac{198}{72} \times 100$ oe, or $2.75 \times 100$	M1	for a complete method: dividing the sale price by the multiplier, or the equivalent unitary method. The official scheme prints the $0.72$ , the $72$ and the $2.75$ in quotation marks, which means the candidate's own value from the first mark may be used in place of each.	✓
$275$	A1	cao, the normal price in euros	✓
Correct answer scores full marks, unless it comes from obvious incorrect working.	Note	guidance for the marker; this row carries no mark of its own	✓

Full marks: 3/3

(a) Solve $x - 4 = \dfrac{3 + 2x}{6}$
You must show clear algebraic working. [3 marks]

(b) (i) Factorise $y^2 - 11y + 30$ [2 marks]

(ii) Hence solve $y^2 - 11y + 30 = 0$ [1 mark]

(a) x =(b)(i)(b)(ii)

[Total 6 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

The equation

x - 4 = \dfrac{3 + 2x}{6}

, which has the unknown on both sides and a fraction on the right.

The quadratic expression

y^2 - 11y + 30

, which is in the form

y^2 + by + c

with

b = -11

and

c = 30

Find

(a) The value of

x

, with the algebra shown. (b)(i)

y^2 - 11y + 30

written as a product of two brackets. (b)(ii) The values of

y

that make that product zero. A quadratic like this one has two of them.

Plan the Solution

(a) Multiply both sides by $6$ so the fraction disappears, expand the bracket, then gather the $x$ terms on one side and the numbers on the other.
(b)(i) Look for the pair of numbers whose product is $30$ and whose sum is $-11$ . A positive product with a negative sum means both numbers are negative.
(b)(ii) The word hence is an instruction: use the brackets from part (b)(i) rather than starting again. A product of two things is zero only when one of them is zero.

Worked Solution [6 marks]

Rule - Clearing a fraction: multiply every term on BOTH sides by the denominator, then solve as usual. Rule - Factorising

y^2 + by + c

: find the pair of numbers whose product is

c

and whose sum is

b

; each one goes into its own bracket.

(a) Multiply both sides by

6

x - 4 = \dfrac{3 + 2x}{6}

6(x - 4) = 3 + 2x

(Reason: The fraction is the only thing stopping ordinary algebra. Multiplying by

6

cancels the denominator on the right, and the left-hand side must be multiplied by

6

too, which is why it goes inside a bracket.)

(a) Expand the bracket

6x - 24 = 3 + 2x

(Reason: The

6

outside multiplies BOTH terms inside the bracket, the

x

and the

4

. This line is where the first method mark is earned, and it is the line most often lost.)

(a) Gather the

x

terms on one side

6x - 2x = 3 + 24

4x = 27

(Reason: Take

2x

from both sides and add

24

to both sides. The four terms then sit where they belong: terms in

x

on the left, numbers on the right.)

(a) Divide both sides by

4

x = \dfrac{27}{4} = 6.75

(Reason:

27

does not divide by

4

exactly, so the fraction is the exact answer. The mark scheme accepts

\dfrac{27}{4}

6.75

and

6\dfrac{3}{4}

, because they are one value written three ways.)

(b)(i) Find the pair of numbers

(-6) \times (-5) = 30

(-6) + (-5) = -11

(Reason: The product

30

is positive and the sum

-11

is negative, so both numbers must be negative. Of the pairs that multiply to

30

, only

-6

and

-5

also add to

-11

(b)(i) Write the expression as two brackets

y^2 - 11y + 30 = (y - 6)(y - 5)

(Reason: Each number of the pair goes into its own bracket. Multiplying out gives

y^2 - 5y - 6y + 30

, and the two middle terms collect back to

-11y

(b)(ii) Set each bracket equal to zero

(y - 6)(y - 5) = 0

y - 6 = 0 \text{ or } y - 5 = 0

y = 6 \text{ or } y = 5

(Reason: Two things multiply to give zero only when one of them IS zero, so each bracket is tried in turn. Because part (b)(i) has already been done, no further algebra is needed here, which is why this part carries only one mark.)

(a)

x = \dfrac{27}{4} = 6.75

(b)(i)

(y - 6)(y - 5)

(b)(ii)

y = 6

y = 5

Verification

Check 1: Put the answer back into the equation the question printed and work the two sides out separately. Left-hand side:

6.75 - 4

. Right-hand side:

\dfrac{3 + 2 \times 6.75}{6}

. Both sides come to

2.75

, so

x = 6.75

balances the equation.

Check 2: Multiply the brackets back out, without looking at the original expression:

(y - 6)(y - 5) = y^2 - 5y - 6y + 30

. That collects to

y^2 - 11y + 30

, which is the expression the question gave.

Check 3: Substitute each solution into the quadratic itself, not into the brackets:

6^2 - 11 \times 6 + 30

and

5^2 - 11 \times 5 + 30

. The first gives

36 - 66 + 30 = 0

and the second gives

25 - 55 + 30 = 0

, so both values really are solutions.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$6x - 24 = 3 + 2x$ or $x - 4 = \dfrac{3}{6} + \dfrac{2}{6}x$	M1	for correct removal of the fraction and expansion of the bracket in a correct equation, or for separating the fraction on the right-hand side in an equation	✓
$6x - 2x = 3 + 24$ or $4x = 27$ or $-24 - 3 = 2x - 6x$ or $-27 = -4x$ or $x - \dfrac{2}{6}x = \dfrac{3}{6} + 4$ or $-4 - \dfrac{3}{6} = \dfrac{2}{6}x - x$	M1ft	Dependent on a $4$ term equation. For correctly rearranging their $4$ term equation so that the terms in $x$ are on one side of the equation and the number terms on the other.	✓
$x = \dfrac{27}{4}$	A1	oe, eg $6.75$ or $6\dfrac{3}{4}$ , dependent on the first M1. Working is required, so a correct answer written down with no algebra scores nothing here.	✓
$(y \pm 6)(y \pm 5)$ or $(6 \pm y)(5 \pm y)$ or $y(y - 6) - 5(y - 6)$ or $y(y - 5) - 6(y - 5)$	M1	for $(y \pm 6)(y \pm 5)$ or $(6 \pm y)(5 \pm y)$ , or for $(y + a)(y + b)$ where $ab = 30$ or $a + b = -11$ ; or for $y(y + a) + b(y + a)$ or $y(y + b) + a(y + b)$ , again where $ab = 30$ or $a + b = -11$	✓
$(y - 6)(y - 5)$	A1	oe, and any letter is allowed in place of $y$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
$y = 6$ , $y = 5$	B1	must follow through from their answer to part (b)(i), where their factors are in the form $(y + a)(y + b)$	✓

Full marks: 6/6

Here is a solid candle in the shape of a cylinder.

The radius of the candle is $8$ cm.
The height of the candle is $h$ cm.

The volume of the candle is $3892$ cm³.

Work out the value of $h$ .
Give your answer correct to one decimal place. [3 marks]

h =

[Total 3 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

A solid candle in the shape of a cylinder

Radius

r = 8

Height

h

cm - the unknown

Volume

V = 3892

cm³

Find

The value of

h

, correct to one decimal place

Plan the Solution

The volume of a cylinder is the area of its circular face multiplied by its height.
Work out the area of the circular face first, from the radius $8$ cm.
Then divide the volume by that area, which leaves $h$ on its own.
Keep the full calculator value all the way through and round only on the last line.

Worked Solution [3 marks]

Rule - Volume of a cylinder:

V = \pi r^{2} h

, where

r

is the radius of the circular face and

h

is the height.

Step 1: put what you know into the formula

V = \pi r^{2} h

3892 = \pi \times 8^{2} \times h

(Reason: The volume and the radius are both given, so

h

is the only letter left in the equation.)

Step 2: work out the area of the circular face

\pi \times 8^{2} = \pi \times 64 = 201.0619\ldots

(Reason: Square the radius first, then multiply by

\pi

- this is the area of the flat circular face of the candle, in square centimetres.)

Step 3: divide to find

h

h = \dfrac{3892}{\pi \times 8^{2}}

h = \dfrac{3892}{201.0619\ldots} = 19.3572\ldots

(Reason:

h

is multiplied by the area, so dividing the volume by the area undoes that and leaves the height.)

Step 4: round to one decimal place

h = 19.4

(Reason: The digit after the first decimal place is

5

, so

19.3572\ldots

rounds up to

19.4

h = 19.4

Verification

Check 1: Reverse the division - multiply the area of the circular face by the height and see whether the volume comes back.

201.0619\ldots \times 19.3572\ldots = 3892

Check 2: Work it out again with

3.14

in place of

\pi

, and then with

\dfrac{22}{7}

\dfrac{3892}{3.14 \times 64} = 19.367\ldots

, still

19.4

to one decimal place, while

\dfrac{22}{7}

gives

19.349\ldots

, which rounds to

19.3

. That pair is exactly why the mark scheme accepts anything from

19.3

19.4

Check 3: Is the size sensible? The smallest box that would hold the candle is

16

cm by

16

cm by the height, and a cylinder fills a little under four fifths of its box.

16 \times 16 \times 19.4 = 4966.4

, and

3892

is about

78

per cent of that - just what a cylinder inside its box should be.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Substitute into $V = \pi r^{2} h$ to reach $3892 = \pi \times 8^{2} \times h$ , or work out $\pi \times 8^{2}$ on its own	M1	A method mark for a correct volume statement. Allow $3.14$ or $\dfrac{22}{7}$ in place of $\pi$ , and allow the area left as $64\pi$ or written as $201\ldots$ .	✓
Divide: $h = \dfrac{3892}{\pi \times 8^{2}}$ , or in two stages, $\dfrac{3892}{64} = 60.8\ldots$ then $\dfrac{60.8\ldots}{\pi}$	M1	A second method mark for the division, in either order. Again allow $3.14$ or $\dfrac{22}{7}$ for $\pi$ .	✓
Answer: $h = 19.4$	A1	Allow anything from $19.3$ to $19.4$ , which covers the approximations for $\pi$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

(a) Express $520$ million in standard form. [1 mark]

(b) Express $8.79 \times 10^{-5}$ as an ordinary number. [1 mark]

(a)(b)(c)

[Total 4 marks]

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Question 25 - Exam Solution

Understanding the Question

Given

The number

520

million.

The number

8.79 \times 10^{-5}

, already written in standard form.

The product

(5 \times 10^{42}) \times (7 \times 10^{-180})

Find

(a) That number written in standard form. (b) The same number written out in full, with no power of ten in it. (c) The product, written in standard form.

Plan the Solution

(a) Write $520$ million out in full, then put the decimal point just after the first digit and count how many places it moved.
(b) A negative index means dividing by that power of ten, so move every digit $5$ places to the right.
(c) Multiply the front numbers, add the indices, then bring the front number back into range and lift the index to match.

Worked Solution [4 marks]

Standard form writes a number as a front number times a power of ten,

A \times 10^{n}

, where the front number is from

1

up to (but never reaching)

10

, and the index is a whole number. To multiply two numbers in standard form, multiply the front numbers and add the indices.

(a) Write the number in full, then count the places

520\,000\,000 = 5.2 \times 100\,000\,000

100\,000\,000 = 10^{8}

520\,000\,000 = 5.2 \times 10^{8}

(Reason: One million is

1\,000\,000

, so

520

million is

520\,000\,000

. The front number has to be from

1

up to

10

, which makes it

5.2

, and the decimal point has travelled

8

places to the left. That count is the index.)

(b) A negative index moves the digits the other way

8.79 \times 10^{-5} = \dfrac{8.79}{10^{5}}

\dfrac{8.79}{100\,000} = 0.0000879

(Reason: Multiplying by

10^{-5}

is the same as dividing by

10^{5}

, so every digit moves

5

places to the right. The

8

lands in the fifth place after the decimal point, with the

7

and the

9

following it.)

5 \times 7 = 35

10^{42} \times 10^{-180} = 10^{42 + (-180)} = 10^{-138}

(5 \times 10^{42}) \times (7 \times 10^{-180}) = 35 \times 10^{-138}

(Reason: A product may be taken in any order, so collect the two front numbers together and the two powers of ten together. Adding the indices gives

42 + (-180) = -138

, and adding a negative index is a subtraction, so the index falls well below zero.)

35 = 3.5 \times 10

35 \times 10^{-138} = 3.5 \times 10^{-137}

(Reason: The front number must be from

1

up to

10

, and

35

is too big for that. Writing

35

3.5 \times 10

hands one more power of ten to the index, so the index rises by one:

-138 + 1 = -137

(a)

5.2 \times 10^{8}

(b)

0.0000879

(c)

3.5 \times 10^{-137}

Verification

Check 1: Multiply the standard-form answer to part (a) back out. It must return the number the question gave.

5.2 \times 10^{8} = 520\,000\,000

Check 2: Send part (b) back the other way. Multiplying the ordinary number by

10^{5}

must return the front number.

0.0000879 \times 10^{5} = 8.79

Check 3: Divide the answer to part (c) by one of the two numbers being multiplied. The other one must come back.

\dfrac{3.5 \times 10^{-137}}{7 \times 10^{-180}} = 5 \times 10^{42}

Check 4: Test both standard-form answers against the definition itself: a front number from one up to ten, and a whole-number index. The front numbers are

5.2

and

3.5

, both in range, and the indices

8

and

-137

are whole numbers, so each answer really is in standard form.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$5.2 \times 10^{8}$ , and nothing else.	✓
(b)	B1	$0.0000879$ , and nothing else. The digits may be written grouped in threes.	✓
(c)	M1	For $35 \times 10^{-138}$ , or $3.5 \times 10 \times 10^{-138}$ , or $3.5$ times a power of ten whose index is anything other than $-137$ . This is the method mark: the multiplying and the adding of indices have been done, but the answer has not been put back into standard form.	✓
(c)	A1	$3.5 \times 10^{-137}$ . A correct answer scores both marks, unless it follows obviously incorrect working.	✓

Full marks: 4/4

The diagram shows a trapezium $ABCD$ that has exactly one line of symmetry.

angle $ADC = 60^{\circ}$
$AD = 12$ cm
$DC = 47$ cm

Calculate the area of the trapezium.
Give your answer correct to $3$ significant figures.
You must show your working clearly. [5 marks]

cm²

[Total 5 marks]

Show solution & mark schemeHide solution & mark scheme

Question 26 - Exam Solution

Understanding the Question

Given

Trapezium

ABCD

, with exactly one line of symmetry

angle

ADC = 60^{\circ}

AD = 12

cm, the sloping side

DC = 47

cm, the long parallel side

Find

The area of the trapezium, correct to

3

significant figures

Plan the Solution

The parallel sides are $AB$ and $DC$ . Only one of them is given, so the area formula needs two missing pieces: the length of $AB$ and the perpendicular height.
Drop a perpendicular from $A$ onto $DC$ . That makes a right-angled triangle with hypotenuse $12$ cm and an angle of $60^{\circ}$ at $D$ .
In that triangle the height is opposite the $60^{\circ}$ , so it comes from the sine. The piece it cuts off along the bottom is next to the $60^{\circ}$ , so it comes from the cosine.
One line of symmetry means an identical triangle sits at the other end, so that same piece comes off $DC$ at both ends.
Then put $AB$ , $DC$ and the height into the trapezium formula and round at the very end.

Worked Solution [5 marks]

Rule - Trapezium:

\text{Area} = \dfrac{1}{2}(a + b)h

, where

a

and

b

are the two parallel sides and

h

is the perpendicular distance between them. In a right-angled triangle the side opposite an angle is

\text{hypotenuse} \times \sin \theta

and the side next to it is

\text{hypotenuse} \times \cos \theta

Step 1: The perpendicular height

h = 12 \times \sin 60^{\circ}

h = 10.3923\ldots \text{ cm}

(Reason: The perpendicular from

A

DC

makes a right-angled triangle in which

AD = 12

cm is the hypotenuse and the height is opposite the

60^{\circ}

angle. The exact height is

6\sqrt{3}

cm, so leave it on the calculator instead of rounding it here.)

Step 2: The piece cut off at each end

12 \times \cos 60^{\circ} = 6 \text{ cm}

(Reason: In the same triangle this piece of

DC

lies next to the

60^{\circ}

angle, so it uses the cosine. Pythagoras agrees:

\sqrt{12^2 - (6\sqrt{3})^2} = 6

Step 3: The short parallel side

AB = 47 - 6 - 6 = 35 \text{ cm}

(Reason: The one line of symmetry makes the two end triangles identical, so the

6

cm from Step 2 comes off

DC

at the

D

end and again at the

C

end.)

Step 4: Into the trapezium formula

\dfrac{1}{2} \times (47 + 35) \times 10.3923 = 426.0843

(Reason: The parallel sides are

47

cm and

35

cm, and

10.3923

is the height to

4

decimal places. Carrying the height unrounded instead gives

426.0845

, which rounds to the same answer - one more reason to round only at the end.)

Step 5: Round to 3 significant figures

426.0843 \approx 426 \text{ cm}^2

(Reason: The first three significant figures are

4

2

and

6

. The next digit is

0

, so the

6

is left alone. The units are

\text{cm}^2

because this is an area.)

426 \text{ cm}^2

Verification

Check 1 - cut the shape a different way: A rectangle

35

cm by

10.3923

cm sits between two triangles, each of base

6

cm and height

10.3923

cm.

35 \times 10.3923 + 2 \times \dfrac{1}{2} \times 6 \times 10.3923 = 426.0843

Check 2 - split along the diagonal: Triangle

ADC

has sides

12

cm and

47

cm with

60^{\circ}

between them, and triangle

ABC

has sides

35

cm and

12

cm with

120^{\circ}

between them, because

AB

is parallel to

DC

. Each area is

\dfrac{1}{2}ab \sin C

, giving

244.2

and

181.9

1

decimal place.

244.2 + 181.9 = 426.1

, which is

426

3

significant figures

Check 3 - is the size sensible? The trapezium contains a rectangle

35

cm by

10.3923

cm and fits inside a rectangle

47

cm by

10.3923

cm, so its area must lie between the areas of those two rectangles.

363.7

and

488.4

are the two bounds, and

426

lies between them

Mark Scheme Breakdown

Step	Mark	Description	Got it?
For example $12 \sin 60^{\circ}$ $(= 6\sqrt{3} = 10.39\ldots)$ or $\sqrt{12^2 - 6^2}$ , or the area of triangle $ADC$ as $\dfrac{1}{2} \times 12 \times 47 \times \sin 60^{\circ}$ $(= 244.2\ldots)$	M1	For a method to find the perpendicular height of the trapezium, or the area of triangle $ADC$ . The first two method marks may be earned in either order.	✓
For example $12 \cos 60^{\circ} = 6$ or $\sqrt{12^2 - (6\sqrt{3})^2} = 6$	M1	For a method to find the base of the end triangle. Condone missing brackets around $6\sqrt{3}$ . Awarded whatever happened earlier in the question.	✓
$AB = 47 - 6 - 6 = 35$	M1	For a method to find the length of $AB$ , taking the end piece off both ends of $DC$ . Taking it off once, giving $41$ , earns nothing here.	✓
For example $\dfrac{1}{2} \times (47 + 35) \times 10.39\ldots$ or $35 \times 10.39\ldots + 2 \times \dfrac{1}{2} \times 6 \times 10.39\ldots$	M1	For a complete method that would give the correct area. Other complete methods score this mark.	✓
$426$	A1	Working must be shown. Accept anything from $420$ to $427$ from correct working.	✓

Full marks: 5/5

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 4MA1/1F, Thursday 15 May 2025: Worked Solutions, Questions 14 to 26

Worked solutions, questions 14 to 26 of 26

Question 14, Calculator allowed

Question 14 - Exam Solution

Question 15, Calculator allowed

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Question 16, Calculator allowed

Question 16 - Exam Solution

Question 17, Calculator allowed

Question 17 - Exam Solution

Question 18, Calculator allowed

Question 18 - Exam Solution

Question 19, Calculator allowed

Question 19 - Exam Solution

Question 20, Calculator allowed

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Question 21, Calculator allowed

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Question 22, Calculator allowed

Question 22 - Exam Solution

Question 23, Calculator allowed

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Question 24, Calculator allowed

Question 24 - Exam Solution

Question 25, Calculator allowed

Question 25 - Exam Solution

Question 26, Calculator allowed

Question 26 - Exam Solution

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