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Edexcel IGCSE 4MA1/1F, Thursday 15 May 2025: Worked Solutions and Mark Schemes

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

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.

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

Worked solutions, questions 1 to 13 of 26

Write a number in each box to make each calculation correct.

(a) $475 + \boxed{\phantom{000}} = 1200$ [1 mark]

(b) $180 \times \boxed{\phantom{000}} = 5760$ [1 mark]

Here are four counters.
Each counter has a number on it.

Using each counter once,

(c)(i) write down the smallest number that can be made, [1 mark]

(ii) write down the largest even number that can be made. [1 mark]

(c)(i)(c)(ii)

[Total 4 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

(a)

475

and the number in the box add up to

1200

(b)

180

multiplied by the number in the box gives

5760

7

4

8

and

5

, each one used once

Find

(a) and (b): the number that belongs in each box. (c)(i): the smallest number the four counters can make. (c)(ii): the largest even number the four counters can make.

Plan the Solution

Parts (a) and (b) are missing-number calculations, so undo the operation on show: subtract in (a), divide in (b).
Every counter is used in part (c), so each number made there has four digits.
For the smallest, start at the thousands column and use the smallest counter left each time. For the largest even, settle the units column first, because the units digit is what decides whether the number is even, and only then fill the front with what is left.

Worked Solution [4 marks]

Rule - Undo, then order: a missing number is found with the inverse operation, subtraction undoing addition and division undoing multiplication; and a number built from given digits is ordered by place value, thousands column first. A number is even exactly when its units digit is even.

Step 1: the box in part (a)

1200 - 475 = 725

(Reason: Adding and subtracting undo each other, so the missing part is the total

1200

less the part already there,

475

. Counting on gives the same number:

25

to reach

500

, then

700

more to reach

1200

Step 2: the box in part (b)

\dfrac{5760}{180} = 32

\dfrac{576}{18} = 32

(Reason: Multiplying and dividing undo each other, so divide

5760

180

. Cancelling one zero from each number first keeps the division the same and makes it easier:

576

shared into

18

groups.)

Step 3: the smallest number, part (c)(i)

4, \; 5, \; 7, \; 8

4578

(Reason: All four counters are used, so the number has four digits, and the thousands column is worth the most. Put the smallest counter,

4

, there, then

5

in the hundreds,

7

in the tens and

8

in the units.)

Step 4: the largest even number, part (c)(ii)

8, \; 7, \; 5, \; 4

8754

8754 > 7548

(Reason: An even number ends in an even digit, so the units counter is

4

8

. Spending the

8

on the units column leaves only

7

5

and

4

for the front and gives

7548

. Using the

4

there instead keeps the

8

for the thousands column and gives

8754

, which is the bigger of the two.)

(a)

725

(b)

32

(c)(i)

4578

(c)(ii)

8754

Verification

Check 1: Put

725

back into part (a) and add.

475 + 725 = 1200

Check 2: Put

32

back into part (b), multiplying the other way round.

32 \times 180 = 5760

Check 3: Hunt for an arrangement smaller than

4578

. Any other one has a bigger counter in the thousands column, or the same

4

there and a bigger counter in the hundreds. The next smallest arrangement is

4587

, which is bigger.

Check 4: Hunt for an even arrangement bigger than

8754

8754

is the biggest arrangement of all, and it already ends in

4

, so nothing even can beat it.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $725$ written in the box	B1	The number that completes the addition: $1200 - 475 = 725$	✓
(b) $32$ written in the box	B1	The number that completes the multiplication: $\dfrac{5760}{180} = 32$	✓
(c)(i) $4578$	B1	The four counters in ascending order, smallest counter in the thousands column	✓
(c)(ii) $8754$	B1	The largest arrangement of the four counters. It ends in $4$ , so it is already even.	✓

Full marks: 4/4

The pictogram shows some information about the number of tins of paint sold by a hardware shop each month from January to April.

(a) How many tins of paint were sold in January? [1 mark]

(b) Work out the total number of tins of paint sold in the four months from January to April. [2 marks]

In May, $22$ tins of paint were sold.

(a)(b)

[Total 4 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

A pictogram of the tins of paint sold each month from January to April, read against a key in which one whole symbol represents

8

tins

January

3

whole symbols, February

4

whole symbols, March

5

whole symbols and a quarter symbol, April

3

whole symbols and a half symbol

The May row is empty, and

22

tins were sold in May

Find

(a) the number of tins of paint sold in January. (b) the total for the four months from January to April. (c) what to draw in the May row of the pictogram.

Plan the Solution

Start at the key. It fixes what one whole symbol is worth, and every reading on the pictogram is built from it.
Value the two part symbols before reading any row that carries one: a quarter of a symbol is a quarter of $8$ , and a half of a symbol is half of $8$ .
Read the four months one at a time, then add the four readings for part (b).
Part (c) runs the key backwards: take $22$ apart into whole eights and a remainder, then turn the remainder into a fraction of a symbol.

Worked Solution [4 marks]

Rule - The key sets the scale: a reading is the number one whole symbol stands for, multiplied by the count of symbols in that row, and a part symbol is worth the same fraction of that number as the fraction of the symbol that is drawn.

Step 1: read the key, then January

3 \times 8 = 24

(Reason: The key fixes the scale: one whole symbol stands for

8

tins. The January row holds

3

whole symbols and no part symbol, so the reading is three lots of

8

Step 2: what a part symbol is worth

\dfrac{1}{4} \times 8 = 2

\dfrac{1}{2} \times 8 = 4

(Reason: A part symbol is worth the same fraction of

8

as the fraction of the symbol that is drawn. March carries a quarter symbol and April carries a half symbol, so both of these are needed before either row can be read.)

Step 3: February and April

4 \times 8 = 32

3 \times 8 + 4 = 28

(Reason: February is

4

whole symbols and nothing else. April is

3

whole symbols, worth

24

, and then the half symbol beside them adds the

4

from Step 2.)

Step 4: March

5 \times 8 + 2 = 42

(Reason: March is the longest row and the one with the smallest part symbol. The

5

whole symbols are worth

40

between them, and the quarter symbol beside them adds the

2

from Step 2.)

Step 5: add the four months

24 + 32 + 42 + 28 = 126

(Reason: Part (b) asks for the four months together, so add the four readings. The

24

and the

32

make

56

, the

42

takes that to

98

, and the

28

finishes it.)

Step 6: showing May on the pictogram

22 = 2 \times 8 + 6

\dfrac{6}{8} = \dfrac{3}{4}

(Reason: This step runs the key backwards. Two whole symbols account for

16

tins and leave

6

over. Six out of the

8

that a whole symbol is worth is three quarters of one, so the May row is drawn as

2

whole symbols followed by a three-quarter symbol.)

(a)

24

tins(b)

126

tins(c) the May row drawn as

2

whole symbols and a

\dfrac{3}{4}

symbol

Verification

Check 1: Total the four months a different way round: count every whole symbol first, then add the two part symbols on their own. January to April hold

15

whole symbols between them, one half symbol and one quarter symbol.

15 \times 8 + 4 + 2 = 126

Check 2: Change the unit and count in quarter symbols, so no row needs a part symbol of its own. A quarter symbol is worth

2

tins, and the four rows hold

12 + 16 + 21 + 14 = 63

quarter symbols.

63 \times 2 = 126

Check 3: Read the completed May row straight back off the pictogram:

2

whole symbols and three more quarter symbols.

2 \times 8 + 3 \times 2 = 22

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $24$	B1	The January row holds $3$ whole symbols, and the key gives $3 \times 8 = 24$ .	✓
(b) $24 + 32 + 42 + 28$ or $15 \times 8 + 4 + 2$	M1	The four monthly readings added, or every whole symbol counted together with the half and the quarter added on. Follow through on the candidate's own answer to part (a), and allow one error or omission in adding the values.	✓
(b) $126$	A1	The total for the four months from January to April.	✓
(c) $2$ whole symbols and a $\dfrac{3}{4}$ symbol drawn in the May row	B1	Or equivalent, so $2$ whole symbols with a $\dfrac{1}{2}$ symbol and a $\dfrac{1}{4}$ symbol beside them scores as well. The mark is for the drawing, so there is no answer line for this part.	✓

Full marks: 4/4

The solid drawn below is a prism.

(a) Write down the number of faces the prism has. [1 mark]

(b) Write down the number of vertices the prism has. [1 mark]

The points $A$ and $B$ both lie on a circle with centre $O$

(d) Write down the mathematical name of the straight line $CD$ [1 mark]

(a)(b)(c)(d)

[Total 4 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

A prism. Its end face is a four-sided shape with one pair of parallel sides, and the three edges that cannot be seen are drawn dashed.

A circle with centre

O

The straight line

AB

, with

A

and

B

on the circle and the line running through

O

The straight line

CD

, which reaches the circle at one point and stays outside it everywhere else.

Find

(a) the number of faces of the prism (b) the number of vertices of the prism (c) the mathematical name of the line

AB

(d) the mathematical name of the line

CD

Plan the Solution

Start from the end face. A prism is that one shape swept straight backwards, so every side of the end face carries one flat face, and every corner of the end face turns up twice, once at each end.
Count the sides of the end face first, then feed that one number through both counting rules and check the pair against Euler's formula.
Ask what $AB$ does to the circle: it joins two points of the circle, and it runs through the centre.
Ask what $CD$ does to the circle: it reaches the circle at a single point and never crosses inside it.

Worked Solution [4 marks]

For a prism whose end face has

n

sides: faces

= n + 2

, vertices

= 2n

and edges

= 3n

. On a circle, a straight line joining two points of the circle is a chord, a chord that passes through the centre is a diameter, and a straight line that meets the circle at exactly one point is a tangent.

(a) Count the faces

4 + 2 = 6

(Reason: The end face is a four-sided shape, so it has

4

sides. One flat face runs back along each of those

4

sides, and the two end faces themselves make

2

more.)

(b) Count the vertices

2 \times 4 = 8

(Reason: Each end face is a four-sided shape, so it has

4

corners, and the prism has

2

end faces. No other corners are made by sweeping the shape backwards, so every vertex of the solid is a corner of one end face.)

AB

OA = OB = r

AB = OA + OB = 2r

(Reason:

A

and

B

are both on the circle, so

AB

is a chord. This chord also passes through the centre

O

, which makes it the longest chord the circle has, twice the radius. A chord through the centre is called a diameter.)

(d) Name the line

CD

OT = r

OT \perp CD

(Reason: Call

T

the one point where

CD

reaches the circle. The radius drawn to that point is at right angles to the line, and every other point of

CD

lies further than

r

from

O

, so the line never crosses inside. A straight line that meets a circle at exactly one point is called a tangent.)

(a)

6

(b)

8

Verification

Check 1: Count the edges instead. There are

4

round the front end face,

4

round the back one and

4

joining the two, so

E = 12

. Euler's formula for a solid like this one is

F + V - E = 2

6 + 8 - 12 = 2

Check 2: Count off the drawing rather than from a rule: the flat top, the long sloping face, the base and the back face are four, and the two end faces make the rest.

4 + 2 = 6

faces, and the

4

corners of the front end face together with the

4

sitting behind them give

8

vertices

Check 3: Count how many points each line has in common with the circle.

AB

has both of its ends on the circle, so it meets the circle twice.

CD

meets it once and stays outside it everywhere else. two points in common for a chord, and a chord through the centre is a diameter; one point in common for a tangent

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$6$	✓
(b)	B1	$8$	✓
(c)	B1	diameter	✓
(d)	B1	tangent	✓

Full marks: 4/4

(a) Write $4 \times 5e$ in its simplest form. [1 mark]

(b) Find the value of $f$ in the equation $f + 14 = 29$ [1 mark]

(a)(b) f =

[Total 2 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

(a) The expression

4 \times 5e

(b) The equation

f + 14 = 29

Find

(a) the expression

4 \times 5e

written in its simplest form (b) the value of

f

Plan the Solution

Part (a) is a multiplication and nothing else. The $5e$ means $5 \times e$ , so the expression is a string of multiplications; gather the two numbers together, multiply them, and write the letter after the result.
Part (b) is an equation, so ask what has been done to $f$ . A number has been added to it, so undo that with the inverse operation, and do it to both sides so the two sides stay equal.
Put each answer back into what the question gave, and check that the two sides still agree.

Worked Solution [2 marks]

Rule - Multiplying a term by a number: multiply the numbers and keep the letter, so

a \times bx = (ab)x

. Rule - Solving a one-step equation: do the inverse operation to both sides, so

x + c = d

gives

x = d - c

(a) Multiply the numbers and keep the letter

4 \times 5e = 4 \times 5 \times e

4 \times 5 = 20

4 \times 5e = 20e

(Reason: The term

5e

already means

5 \times e

, so the whole expression is

4 \times 5 \times e

. Multiplication can be carried out in any order, so the two numbers are multiplied first and the letter is written straight after the

20

. The multiplication sign between a number and a letter is never written down.)

(b) Take 14 off both sides

f + 14 = 29

f = 29 - 14 = 15

(Reason: The

14

has been added to

f

, and the inverse of adding is subtracting. Taking

14

off the left-hand side leaves

f

on its own, and taking the same

14

off the right-hand side keeps the equation balanced.)

(a)

20e

(b)

f = 15

Verification

Check 1: Part (a) again as repeated addition.

4 \times 5e

means

5e

added four times over, so add the four coefficients instead of multiplying.

5 + 5 + 5 + 5 = 20

, so the expression is

20e

Check 2: Put a number in place of the letter. With

e = 3

, work out the question's expression and the answer separately and compare the two values.

4 \times 5 \times 3 = 60

and

20 \times 3 = 60

Check 3: Put

f = 15

back into the equation the question gave and work out the left-hand side.

15 + 14 = 29

, which is the right-hand side

Check 4: Count on instead of subtracting. From

14

up to

20

6

, and from

20

up to

29

9

6 + 9 = 15

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$20e$	✓
(b)	B1	$15$	✓

Full marks: 2/2

(a) Change $0.6$ into a percentage. [1 mark]

(b) Change $\dfrac{19}{100}$ into a decimal. [1 mark]

Here is a rectangle divided into equal parts.
Some of the parts are shaded.

(a) %(b)(c)

[Total 4 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

(a) The decimal

0.6

(b) The fraction

\dfrac{19}{100}

Find

(a)

0.6

written as a percentage (b)

\dfrac{19}{100}

written as a decimal (c) the fraction of the rectangle that is shaded, in its simplest form

Plan the Solution

Per cent means out of $100$ , so part (a) is a change of units and not a change of value. Write the decimal as a number of hundredths, which is the same as multiplying it by $100$ .
Part (b) is the same idea running backwards. The denominator is already $100$ , so the fraction is a number of hundredths, and hundredths are the second column after the decimal point.
Part (c) is a counting question before it is a fraction question. Count how many equal parts the whole rectangle is divided into, count how many of them are shaded, write the shaded count over the total count, then cancel that fraction down.
Nothing in this question needs a calculator, so check each answer by turning it back into what the question gave.

Worked Solution [4 marks]

Rule - Per cent means out of

100

: a decimal becomes a percentage when it is multiplied by

100

, and a percentage becomes a decimal when it is divided by

100

. Rule - Simplest form: a fraction is in its simplest form when the numerator and the denominator share no factor except

1

, which is reached by dividing both of them by their highest common factor.

(a) Write the decimal as a number of hundredths

0.6 = \dfrac{6}{10} = \dfrac{60}{100}

0.6 \times 100 = 60

(Reason: Six tenths and sixty hundredths are the same amount, because the top and the bottom of

\dfrac{6}{10}

have both been multiplied by

10

. A percentage is a number of hundredths, so

0.6

60

parts out of

100

. Multiplying by

100

does the same work in one step, and the per cent sign is what records that the

100

is there.)

(b) Read the fraction off the decimal columns

\dfrac{19}{100} = \dfrac{10}{100} + \dfrac{9}{100}

\dfrac{19}{100} = 0.19

(Reason: Nineteen hundredths splits into ten hundredths, which is one tenth, and nine hundredths. The first column after the decimal point counts tenths and the second counts hundredths, so the

1

goes in the tenths column and the

9

in the hundredths column. A denominator of

100

always gives a decimal with two figures after the point, which is why

0.19

and not

0.019

is right.)

5 \times 4 = 20

(Reason: The rectangle is ruled into

5

parts across and

4

parts down, and every part is the same size, which is what lets them be counted rather than measured. Multiplying the two counts gives the number of parts in the whole rectangle, and that number is the denominator of the fraction.)

1 + 5 + 5 + 5 = 16

(Reason: Take the rows one at a time. The top row has

1

shaded part and

4

white ones, so it contributes

1

. Each of the three rows underneath it is shaded the whole way across, so each of those contributes

5

. Adding the four rows gives the number of shaded parts, and that number is the numerator.)

\dfrac{16}{20} = \dfrac{4 \times 4}{5 \times 4} = \dfrac{4}{5}

(Reason: Shaded parts over total parts is

\dfrac{16}{20}

, which is a correct fraction but not yet the simplest one. Both

16

and

20

are multiples of

4

, and

4

is the highest number that divides into both, so taking that factor out of the top and the bottom leaves

\dfrac{4}{5}

. Now

4

and

5

share no factor except

1

, so the cancelling has gone as far as it can. The question asks for a fraction, so stop here and do not turn it into

0.8

(a)

60\%

(b)

0.19

(c)

\dfrac{4}{5}

Verification

Check 1: Part (a) backwards. A percentage is a number of hundredths, so put the answer over

100

again and see whether the decimal the question gave comes back.

\dfrac{60}{100} = 0.6

, which is the decimal in the question

Check 2: Part (b) backwards, by place value rather than by division. Read

0.19

off the columns:

1

tenth and

9

hundredths, then add the two fractions.

\dfrac{1}{10} + \dfrac{9}{100} = \dfrac{19}{100}

, which is the fraction in the question

Check 3: Part (c) counted the other way round. Count the white parts instead: there are

4

of them out of

20

, which is

\dfrac{1}{5}

of the rectangle. The shaded part and the white part must make one whole rectangle between them.

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

, so the two fractions account for the whole rectangle

Check 4: Part (c) tested against the count. If

\dfrac{4}{5}

of the rectangle is shaded, then taking

\dfrac{4}{5}

of the

20

parts must give back the number of parts that were counted as shaded.

\dfrac{4}{5} \times 20 = 16

, which is the shaded count

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$60$	✓
(b)	B1	$0.19$	✓
(c)	M1	for a correct unsimplified fraction, $\dfrac{16}{20}$ or equivalent such as $\dfrac{8}{10}$ , or for an answer of $\dfrac{1}{5}$ or $0.8$ or $80\%$ , or for any fraction that will cancel written in its simplest form, such as $\dfrac{15}{20} = \dfrac{3}{4}$ or $\dfrac{4}{16} = \dfrac{1}{4}$	✓
(c)	A1	$\dfrac{4}{5}$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
(c)	Note	Do not ignore subsequent working here: $\dfrac{4}{5}$ followed by an answer of $0.8$ or $80\%$ scores M1A0, because the question asks for a fraction.	✓
(c)	Note	The $\dfrac{1}{5}$ that earns the method mark is the error worth naming: it is $\dfrac{4}{20}$ , the fraction left white. The counting and the cancelling are both right and the wrong region has been counted.	✓

Full marks: 4/4

The first four terms of a number sequence are shown below.

$7 \qquad 11 \qquad 15 \qquad 19$

(a) Write down the next term of this sequence.
[1 mark]
(b) Explain how you worked out your answer to part (a).
[1 mark]
(c) Work out the $12\text{th}$ term of this sequence. [1 mark]

(a)(b)(c)

[Total 3 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

The first four terms of the sequence:

7 \qquad 11 \qquad 15 \qquad 19

The terms climb by the same amount every time, so this is an arithmetic sequence.

Find

(a) the term that comes straight after

19

(b) a sentence saying how that term was worked out (c) the

12\text{th}

term

Plan the Solution

Subtract each term from the one after it. If every gap is the same, the sequence has a common difference.
Add that difference on to $19$ for part (a). Part (b) is then just that sentence written down: the mark is for saying what was ADDED, not for saying what the gap is.
Writing out all twelve terms would work for part (c), but it is slow and easy to miscount, so build the $n\text{th}$ term rule and substitute $n = 12$ .

Worked Solution [3 marks]

Rule - Arithmetic sequence: the terms go up in equal steps of

d

, so each term is the one before it plus

d

, and the

n\text{th}

term is

dn + (\text{first term} - d)

Step 1: Find the gap between the terms

11 - 7 = 4

15 - 11 = 4

19 - 15 = 4

(Reason: Every consecutive pair is

4

apart, so the sequence rises in equal steps and the common difference is

d = 4

Step 2: Add one more step on to the last term shown

19 + 4 = 23

(Reason: The next term follows

19

, so it sits one step of

4

further on. This answers part (a), and stating that

4

was added on to the previous term is exactly what part (b) asks for. Writing only that the difference is

4

does not answer part (b), because it never says what was done with it.)

Step 3: Build a rule for the

n\text{th}

term

7 - 4 = 3

n\text{th term} = 4n + 3

(Reason: Each step adds

4

, so the sequence runs alongside the

4

times table

4, 8, 12, 16, \ldots

and every term sits

3

above the matching multiple. That

3

is the first term less one step,

7 - 4

Step 4: Substitute

n = 12

into the rule

4 \times 12 + 3 = 51

(Reason: The rule reaches any term in one line, so the

12\text{th}

term needs no listing and no counting.)

(a)

23

(b) Add

4

to the previous term(c)

51

Verification

Check 1: Reach the

12\text{th}

term without the rule. From the

4\text{th}

term to the

12\text{th}

term is

8

more steps of

4

19 + 8 \times 4 = 51

Check 2: Test the rule on terms that are already printed. Put

n = 4

and

n = 5

into

4n + 3

and compare with the paper and with part (a).

4 \times 4 + 3 = 19

and

4 \times 5 + 3 = 23

Check 3: Sanity check the parity.

4n

is always even, so

4n + 3

is always odd, and the printed terms

7, 11, 15, 19

are all odd.

51

is odd, so it belongs to this sequence

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The next term	B1	cao. The only answer accepted is $23$ .	✓
(b) The explanation	B1	Acceptable: add $4$ , $+4$ , $19 + 4 \; (= 23)$ , (rule is) $4n + 3$ , goes up by $4$ , $4 \times 5 + 3 \; (= 23)$ , $(n) + 4$ , $(3n) + 4$ . Not acceptable: $11 - 7 = 4$ , difference is $4$ , we subtract the first number and the next number and it gives us the answer.	✓
(b)	Note	The three rejected answers all find the gap between the terms but never say it was added on to $19$ . $(3n) + 4$ is accepted even though it is not a correct rule for this sequence: the mark is for saying what was DONE to reach the next term, not for a correct $n$ th term.	✓
(c) The $12\text{th}$ term	B1	cao. The only answer accepted is $51$ . No working is needed for the mark, so a correct answer from listing all twelve terms scores it in full.	✓

Full marks: 3/3

The normal price of a desk lamp is £ $26.80$

Claire has a student discount card that takes $\dfrac{1}{4}$ off the normal price.

Claire buys $2$ of these desk lamps.
She pays with a £ $50$ note.

Work out how much change Claire should receive. [4 marks]

[Total 4 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

Normal price of one desk lamp: £

26.80

The student discount card takes

\dfrac{1}{4}

off that normal price.

Claire buys

2

lamps and hands over a £

50

note.

Find

The change Claire should receive, in pounds.

Plan the Solution

Take a quarter off the normal price, so you know what one lamp actually costs her.
Double that reduced price, because she buys two identical lamps.
Subtract the total cost from the amount she hands over.

Worked Solution [4 marks]

Rule - Discount, then change: taking

\dfrac{1}{4}

off leaves

\dfrac{3}{4}

of the normal price, and the change is the amount handed over minus the total cost.

Step 1: Work out the discount on one lamp

\dfrac{1}{4} \times 26.80 = 6.70

(Reason: the card takes a quarter off, so a quarter of £

26.80

is the money Claire saves on each lamp)

Step 2: Work out the reduced price of one lamp

26.80 - 6.70 = 20.10

(Reason: taking the discount off the normal price leaves what Claire actually pays for one lamp)

Step 3: Work out the cost of two lamps

2 \times 20.10 = 40.20

(Reason: both lamps are the same, so double the price she pays for one of them)

Step 4: Work out the change

50 - 40.20 = 9.80

(Reason: the change is whatever is left of the £

50

note once the total cost has been taken off)

9.80

Verification

Check 1: Add the change back on to the total cost. It has to rebuild the £

50

note exactly.

40.20 + 9.80 = 50

Check 2: Discount the whole bill instead of each lamp. Two lamps at the normal price come to

2 \times 26.80 = 53.60

, and the card takes a quarter off that.

53.60 - \dfrac{1}{4} \times 53.60 = 40.20

Check 3: Redo the whole calculation in pence, where no decimal point can slip.

5000 - 4020 = 980

pence, which is £

9.80

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{1}{4} \times 26.80 \; (= 6.70)$ or $\dfrac{3}{4} \times 26.80 \; (= 20.10)$ or $2 \times 26.80 \; (= 53.60)$	M1	For a method to find the discount on one lamp ( $\dfrac{1}{4} \times 26.80 = 6.70$ ), or the reduced price of one lamp ( $\dfrac{3}{4} \times 26.80 = 20.10$ ), or the cost of two lamps at the normal price ( $2 \times 26.80 = 53.60$ ). Any one of the three earns it, or an equivalent.	✓
$2 \times 20.10 \; (= 40.20)$ or $(26.80 - 6.70) \times 2 \; (= 40.20)$ or $\dfrac{3}{4} \times 53.60 \; (= 40.20)$ or $6.70 \times 3 \times 2 \; (= 40.20)$	M1	For a method to find the reduced price of the two lamps, reaching $40.20$ . Awarded on the candidate's own value from the first mark, so $53.60 - \dfrac{1}{4} \times 53.60 = 40.20$ earns it just as $2 \times 20.10 = 40.20$ does.	✓
$50 - 40.20$ (using their cost of two lamps)	M1	For a complete method: the amount handed over minus the total cost, $50 - 40.20 = 9.80$ .	✓
Correct answer £ $9.80$	A1	For $9.8(0)$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
Special case	SC B2	for an answer of $29.9(0)$ or $36.6(0)$	✓
Special case	SC B1	for an answer of $23.2(0)$ or $\pm 3.1(0)$	✓
The four special-case answers	Note	$29.9(0)$ comes from $50 - 20.10 = 29.90$ , the price of one lamp taken off instead of two. $36.6(0)$ comes from $50 - 2 \times 6.70 = 36.60$ , the money saved doubled instead of the money paid. $23.2(0)$ comes from $50 - 26.80 = 23.20$ , one lamp at the normal price with the discount forgotten. $3.1(0)$ comes from $50 - (53.60 - 6.70) = 3.10$ , one discount taken off a bill for two lamps.	✓

Full marks: 4/4

(a) Fill in the missing values in the table below for $y = 2x - 1$
$\begin{array}{|c|c|c|c|c|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline y & & & -1 & & & 5 \\ \hline\end{array}$
[2 marks]
(b) Draw the graph of $y = 2x - 1$ on the grid below, for values of $x$ from $-2$ to $3$ [2 marks]

[Total 4 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

The straight-line rule

y = 2x - 1

A table whose

x

row runs

-2

-1

0

1

2

3

Two

y

cells already printed:

-1

under

x = 0

, and

5

under

x = 3

A grid reaching from

-6

7

up the

y

-axis, so every value of the table fits on it

Find

(a) The four missing

y

values (b) The graph of

y = 2x - 1

, drawn between

x = -2

and

x = 3

Plan the Solution

Take the columns one at a time: double the $x$ value, then subtract $1$ .
Use the two printed cells as a check - the same rule has to produce them.
Read each column as a coordinate pair and plot the six points.
Join them with one ruled straight line, stopping at $x = -2$ and $x = 3$ .

Worked Solution [4 marks]

Rule - a table of values: substitute each

x

into the equation in turn. Because

y = 2x - 1

is linear, the points all lie on one straight line, and the

y

values climb in equal steps of

2

Step 1: Substitute

x = -2

2 \times (-2) - 1 = -4 - 1 = -5

(Reason: (Reason: doubling

-2

gives

-4

, and the rule then takes

1

away from it.))

Step 2: Substitute

x = -1

and

x = 1

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

2 \times 1 - 1 = 2 - 1 = 1

(Reason: (Reason: every column uses the same two operations, in the same order - double, then subtract

1

.))

Step 3: Substitute

x = 2

2 \times 2 - 1 = 4 - 1 = 3

(Reason: (Reason: that is the last blank cell. The paper had already printed

-1

under

x = 0

and

5

under

x = 3

.))

Step 4: Write the completed table

\begin{array}{|c|c|c|c|c|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline y & -5 & -3 & -1 & 1 & 3 & 5 \\ \hline\end{array}

(Reason: (Reason: read along the

y

row and the values rise by

2

every column. A linear rule must do that, so an uneven gap would mean a cell was worked out wrongly.))

Step 5: Plot the six points

(-2, -5)

(-1, -3)

(0, -1)

(1, 1)

(2, 3)

(3, 5)

(Reason: (Reason: each column of the table is one point - the

x

value across, the

y

value up or down from the

x

-axis.))

Step 6: Rule one line from

x = -2

x = 3

\dfrac{5 - (-5)}{3 - (-2)} = \dfrac{10}{5} = 2

(Reason: (Reason: the six points are in a straight line of gradient

2

, so one ruled line through all of them is the graph. Stop it at the ends of the range the question gives.))

(a)

y

row:

-5

-3

-1

1

3

5

(b) A straight line through

(-2, -5)

and

(3, 5)

, drawn on the grid

Verification

Check 1: Put the two cells the paper had already filled in back through the rule. If

x = 0

and

x = 3

give the printed numbers, the rule has been read off the question correctly.

2 \times 0 - 1 = -1

and

2 \times 3 - 1 = 5

Check 2: Read along the completed

y

row. Each column steps

1

across, so each

y

value should step

2

up.

-5 \to -3 \to -1 \to 1 \to 3 \to 5

- every gap is

2

Check 3: Read the drawn line back. Its gradient from end to end must be the

2

y = 2x - 1

, and it must cross the

y

-axis at

-1

\dfrac{5 - (-5)}{3 - (-2)} = 2

, and the line passes through

(0, -1)

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Every value in the $y$ row correct: $-5$ , $-3$ , $-1$ , $1$ , $3$ , $5$	B2	B2 for all correct values in the table. The official scheme brackets $-1$ and $5$ because the paper prints those two cells already.	✓
(a) Only two or three of the values correct	(B1)	(B1 for 2 or 3 correct values)	✓
(b) At least five points plotted correctly, or a correct line segment through at least three of $(-2, -5)$ , $(-1, -3)$ , $(0, -1)$ , $(1, 1)$ , $(2, 3)$ , $(3, 5)$	M1ft	For at least five points plotted correctly (within the circles on the overlay), followed through on the candidate's own incorrect table provided B1 was scored in part (a); or for a correct line segment through at least three of the six points; or for a straight line of gradient $2$ ; or for a straight line through $(0, -1)$ with a positive gradient.	✓
(b) One correct straight line drawn between $x = -2$ and $x = 3$	A1	For a correct line between $x = -2$ and $x = 3$ , with clear intention to go through all the points, and it must be a line - freehand is allowed. Anything drawn to the left of $x = -2$ or to the right of $x = 3$ is ignored.	✓
Guidance from the official scheme: a correct graph with an empty table	Note	If a fully correct graph is shown but the table in part (a) is left blank, award $2$ marks for part (a) and $2$ marks for part (b).	✓

Full marks: 4/4

Rosalind asked some people how they prefer their potatoes cooked.
They could choose mashed or roasted or chipped or baked.

The pie chart shows some information about their answers.

(a) Use the pie chart to complete the table.

$\begin{array}{|l|c|c|}\hline \textbf{Potato} & \textbf{Frequency} & \textbf{Angle of sector} \\ \hline \text{mashed} & 6 & 72^\circ \\ \hline \text{roasted} & \text{......} & 132^\circ \\ \hline \text{chipped} & \text{......} & \text{......} \\ \hline \text{baked} & \text{......} & 48^\circ \\ \hline \end{array}$ [3 marks]

One of these people is selected at random.

(b) Write down the probability that this person chose mashed. [1 mark]

(b)

[Total 4 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

A pie chart in four sectors: mashed

72^\circ

, roasted

132^\circ

, baked

48^\circ

, and chipped left unmarked.

The table gives the frequency for mashed as

6

The four sector angles add to

360^\circ

Find

(a) The three missing frequencies and the one missing sector angle. (b) The probability that a person picked at random chose mashed.

Plan the Solution

Every person takes up the same angle, so start from the one row that is complete: $6$ people fill $72^\circ$ .
Divide to find the angle that stands for one person.
Divide each of the two other given angles by that to get its frequency.
Take the three given angles off a whole turn to find the angle that is missing, then divide that by the angle for one person.
For (b), the probability is the mashed frequency out of the total frequency, which is the same as the mashed angle out of the whole turn.

Worked Solution [4 marks]

Rule - Pie chart: the sector angles add to

360^\circ

, and every person takes up the same angle, so frequency and angle are in proportion.

Step 1: find the angle that stands for one person

\dfrac{72}{6} = 12

(Reason: The mashed row is the only complete one:

6

people fill

72^\circ

, so one person is worth

12^\circ

Step 2: turn the two other given angles into frequencies

\dfrac{132}{12} = 11

\dfrac{48}{12} = 4

(Reason: A sector holds as many people as there are lots of

12^\circ

inside its angle.)

Step 3: find the missing angle

360 - (72 + 132 + 48) = 108

(Reason: The four sectors fill a whole turn, so take all three of the given angles off a whole turn.)

Step 4: turn that angle into a frequency

\dfrac{108}{12} = 9

(Reason: The chipped sector is

108^\circ

and one person is worth

12^\circ

, so it stands for

9

people.)

Step 5: write down the probability for part (b)

\dfrac{72}{360} = \dfrac{1}{5}

(Reason: The mashed sector is

72^\circ

out of the whole

360^\circ

, which is the same as

6

people out of

30

(a) roasted

11

, chipped

9

, baked

4

, chipped angle

108^\circ

(b)

\dfrac{72}{360} = \dfrac{1}{5}

Verification

Check 1: Add all four sector angles. A pie chart must close, so they have to make one whole turn.

72 + 132 + 108 + 48 = 360

Check 2: Add the four frequencies, then multiply the total by the angle for one person.

6 + 11 + 9 + 4 = 30

and

30 \times 12 = 360

Check 3: Work out part (b) the other way round, from the completed frequency column instead of from the angles.

\dfrac{6}{30} = \dfrac{1}{5}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) eg $\dfrac{72}{6} = 12$ or $\dfrac{132}{72} = \dfrac{11}{6}$ or $\dfrac{48}{72} = \dfrac{2}{3}$ or $360 - (48 + 72 + 132) = 108$	M1	For a correct first step, or for one correct value. Values may be seen on the pie chart.	✓
eg (frequency roasted =) $\dfrac{132}{12} \; (= 11)$ or $6 \times \dfrac{11}{6} \; (= 11)$ OR (frequency chipped =) $\dfrac{108}{12} \; (= 9)$ OR (angle chipped =) $9 \times 12 \; (= 108)$ or $360 - (48 + 72 + 132) \; (= 108)$ OR (frequency baked =) $\dfrac{48}{12} \; (= 4)$ or $6 \times \dfrac{2}{3} \; (= 4)$	M1	For a method to work out three of the values. The official scheme prints the $12$ , the $\dfrac{11}{6}$ , the $108$ , the $9$ and the $\dfrac{2}{3}$ in quotation marks, which means the candidate's own earlier value may be used in place of each. That is what makes this mark a follow-through.	✓
$11$ , $9$ , $4$ , $108^\circ$	A1	cao. A correct answer scores full marks unless it comes from obviously incorrect working.	✓
$\dfrac{72}{360}$	B1ft	(b) oe, eg $\dfrac{1}{5}$ , $\dfrac{6}{30}$ , $20\%$ , $0.2$ . Allow $6$ over their total frequency, dependent on a complete frequency column.	✓

Full marks: 4/4

Packets of tea are loaded into a shipping carton.

Each packet is a cuboid, $16$ cm by $4$ cm by $6$ cm
The carton is a cuboid, $80$ cm by $12$ cm by $42$ cm

Work out the greatest number of packets that can be put into the carton. [3 marks]

[Total 3 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

Each packet is a cuboid,

16

cm by

4

cm by

6

cm.

The carton is a cuboid,

80

cm by

12

cm by

42

cm.

The packets are all the same size, so they can be set out in rows and stacked in layers.

Find

The greatest number of packets that can be put into the carton.

Plan the Solution

Lay one edge of the packet along each edge of the carton: $16$ cm along the $80$ cm side, $4$ cm along the $12$ cm side, $6$ cm up the $42$ cm side.
Divide each carton edge by the packet edge lying along it to see how many fit in that direction.
Every one of those divisions comes out exact, so no space is wasted and no packet has to be cut.
Multiply the three counts together, then check the total against the two volumes.

Worked Solution [3 marks]

Rule - Packing a cuboid: when each edge of the small cuboid divides exactly into the carton edge it lies along, the number that fit is the three counts multiplied together, and that is also

\dfrac{\text{carton volume}}{\text{packet volume}}

Step 1: how many fit along the 80 cm side

\dfrac{80}{16} = 5

(Reason: The

16

cm edge of the packet lies along the

80

cm side of the carton, and it goes in exactly

5

times with nothing left over.)

Step 2: how many fit across the 12 cm side

\dfrac{12}{4} = 3

(Reason: The

4

cm edge of the packet lies across the

12

cm side of the carton, so divide the

12

by that edge.)

Step 3: how many layers fit up the 42 cm side

\dfrac{42}{6} = 7

(Reason: The

6

cm edge stands upright, and

6

divides into

42

exactly, so the packets stack in

7

layers.)

Step 4: multiply the three counts

5 \times 3 \times 7 = 105

(Reason: Each of the

5

places along the carton pairs with each of the

3

places across it, and that whole layer is repeated

7

times, so the three counts multiply.)

105

packets

Verification

Check 1: Work with volumes instead. The packets leave no gaps, so the carton's volume divided by one packet's volume must give the same count.

80 \times 12 \times 42 = 40\,320

and

16 \times 4 \times 6 = 384

, so

\dfrac{40\,320}{384} = 105

Check 2: Count one layer on the floor of the carton first, then count the layers.

\dfrac{80 \times 12}{16 \times 4} = 15

packets in a layer, and

15 \times 7 = 105

Check 3: Turn the packets the other way round to confirm this really is the greatest: put the

6

cm edge across the

12

cm side, so only

2

fit across, and stand the

4

cm edge upright.

\dfrac{42}{4} = 10.5

, so only

10

layers fit, giving

5 \times 2 \times 10 = 100

packets, which is fewer.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $80 \times 12 \times 42 = 40\,320$ or $16 \times 4 \times 6 = 384$ or $\dfrac{42}{6} = 7$ or $\dfrac{12}{4} = 3$ or $\dfrac{80}{16} = 5$ or $\dfrac{80 \times 12}{16 \times 4} = 15$ or $\dfrac{80 \times 42}{16 \times 6} = 35$ or $\dfrac{42 \times 12}{6 \times 4} = 21$	M1	For a method to find the volume of the carton or the volume of one packet, or a method to find the number of packets along one dimension of the carton, or for dividing the areas of corresponding faces.	✓
eg $\dfrac{40\,320}{384}$ or $7 \times 3 \times 5$ or $15 \times 7$ or $35 \times 3$ or $21 \times 5$	M1	For a complete method to find the number of packets. The values used may be the candidate's own from the first mark.	✓
$105$	A1	cao. A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 3/3

A market stall sells trays of dates.
The total cost of $12$ of these trays is $708$ dirhams.
Work out the cost of $5$ of these trays. [2 marks]

dirhams

[Total 2 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

12

trays cost

708

dirhams altogether.

Every tray costs the same, so cost and number of trays are in direct proportion.

Find

The cost of

5

of these trays, in dirhams.

Plan the Solution

Divide the total by $12$ to get the cost of one tray.
Multiply that one-tray cost by $5$ to get the cost of five trays.
Confirm it a second way, by scaling $708$ straight down with the fraction $\dfrac{5}{12}$ .

Worked Solution [2 marks]

Rule - Unitary method: find the cost of ONE item first, then multiply by how many are wanted.

\text{cost of one} = \dfrac{\text{total cost}}{\text{number of trays}}

Step 1: Find the cost of one tray

\dfrac{708}{12} = 59

(Reason:

12

equal trays cost

708

dirhams, so one tray costs a twelfth of that. Dividing by

12

strips the total back to a single tray, which is the value every later step is built from.)

Step 2: Multiply the one-tray cost by the number of trays wanted

5 \times 59 = 295

(Reason: Each tray costs the same

59

dirhams, so several trays cost

59

multiplied by how many there are. The answer is in dirhams because a number of trays multiplied by a cost per tray is a cost.)

295

dirhams

Verification

Check 1: Multiply back. If one tray really costs

59

dirhams, then

12

trays must rebuild the total the question printed.

12 \times 59 = 708

Check 2: A different route to the same answer: scale the total straight to

5

trays with a fraction, without ever finding the cost of one tray.

\dfrac{5}{12} \times 708 = 295

Check 3: Size check.

5

trays is fewer than half of

12

trays, so the cost must come out under half of

708

dirhams, which is

354

dirhams.

295 < 354

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Start to work with proportion, eg the cost of one tray $\dfrac{708}{12} = 59$	M1	Any correct start to the proportion scores this mark, eg $\dfrac{708}{6} = 118$ for $2$ trays, $\dfrac{708}{4} = 177$ for $3$ trays, $\dfrac{708}{3} = 236$ for $4$ trays, $\dfrac{708}{2} = 354$ for $6$ trays, or $\dfrac{12}{5} = 2.4$ , or $\dfrac{708}{2.4}$ , or $708 \times \dfrac{5}{12}$ , or $708 \times 5 = 3540$ .	✓
The cost of $5$ trays	A1	cao $295$ . A correct answer scores full marks unless it comes from obviously incorrect working, so the two marks are earned by $295$ on the answer line even with no working shown.	✓

Full marks: 2/2

Callum went into a bistro at $18\,15$
He left the bistro at $21\,40$

(a) Work out the length of time that Callum spent in the bistro.
Give your answer in hours and minutes. [2 marks]

On Saturday, the bistro sold $120$ meals.
$48$ of these meals were vegetarian meals.

(b) What percentage of the meals sold on Saturday were vegetarian? [2 marks]

(a) hours and minutes(b) %

[Total 4 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

Callum went into the bistro at

18\,15

and left at

21\,40

On Saturday the bistro sold

120

meals

48

of those meals were vegetarian

Find

(a) how long Callum was inside, written in hours and minutes (b) the vegetarian meals as a percentage of all

120

meals

Plan the Solution

(a) An hour on the clock is $60$ minutes, not $100$ , so change both readings into minutes after midnight, subtract, then turn the answer back into hours and minutes.
(b) Write the vegetarian meals as a fraction of the total, then multiply by $100$ .
A calculator is allowed, so check each answer by working backwards from it.

Worked Solution [4 marks]

Rule - Elapsed time and percentage: an hour is

60

minutes, so a clock reading is

\text{hours} \times 60 + \text{minutes}

minutes after midnight; and a part written as a percentage is

\dfrac{\text{part}}{\text{whole}} \times 100

Step 1: (a) Change each clock reading into minutes after midnight

18 \times 60 + 15 = 1095

21 \times 60 + 40 = 1300

(Reason:

18\,15

18

whole hours and

15

minutes after midnight, and every hour is

60

minutes.)

Step 2: (a) Subtract to find how many minutes he was inside

1300 - 1095 = 205

(Reason: The gap between the two readings is the time Callum spent in the bistro, measured in minutes.)

Step 3: (a) Write

205

minutes as hours and minutes

3 \times 60 = 180

205 - 180 = 25

(Reason: Three whole hours use up

180

minutes and leave

25

minutes over. A fourth hour will not fit, because

4 \times 60 = 240

is more than

205

Step 4: (b) Write the vegetarian meals as a fraction of all the meals

\dfrac{48}{120} = \dfrac{2}{5}

(Reason:

48

of the

120

meals were vegetarian, so the fraction is

48

out of

120

. Dividing top and bottom by

24

simplifies it.)

Step 5: (b) Turn the fraction into a percentage

\dfrac{48}{120} \times 100 = 40

(Reason: A percentage is a number out of

100

, so multiplying the fraction by

100

gives the percentage.)

(a)

3

hours

25

minutes(b)

40\%

Verification

Check 1 - part (a): Count on from

18\,15

. Add

3

whole hours, then add

25

minutes.

18\,15

goes to

21\,15

, and

21\,15

goes to

21\,40

, which is when Callum left.

Check 2 - part (a): Turn the answer back into minutes and add it on to the starting reading.

3 \times 60 + 25 = 205

and

1095 + 205 = 1300

, which is

21\,40

Check 3 - part (b): Take the answer as a percentage of the whole and see whether the vegetarian meals come back.

40\%

0.4

as a decimal.

0.4 \times 120 = 48

, the number of vegetarian meals sold.

Check 4 - part (b): The vegetarian and the non-vegetarian percentages must add up to

100

120 - 48 = 72

meals were not vegetarian and

\dfrac{72}{120} = 0.6

, so

60\%

were not vegetarian.

40 + 60 = 100

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Both numbers correct: $3$ hours and $25$ minutes	B2	for $3$ (hours) and $25$ (minutes)	✓
(a) Only one of the two numbers correct	B1	for $3$ (hours) or $25$ (minutes)	✓
(b) A correct first step, for example $\dfrac{48}{120}$ or $\dfrac{48}{120} \times 100$	M1	for a correct first step. Also allow $\dfrac{48}{120} = 0.4$ , the simplified fraction $\dfrac{2}{5}$ , or the complement $\dfrac{120 - 48}{120} \times 100$ which gives $60$ , or equivalent	✓
(b) $40$	A1	cao. A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 4/4

Show that $\dfrac{3}{7} + \dfrac{1}{4} = \dfrac{19}{28}$
You must show all your working. [2 marks]

[Total 2 marks]

Show solution & mark schemeHide solution & mark scheme

Question 13 - Exam Solution

Understanding the Question

Given

The two fractions

\dfrac{3}{7}

and

\dfrac{1}{4}

Their denominators,

7

and

4

, are different, so the fractions count parts of different sizes and cannot be added as they stand.

Find

Show that these two fractions add to exactly

\dfrac{19}{28}

. The answer is already printed in the question, so it is the working that earns both marks - a bare restatement of it scores nothing.

Plan the Solution

Find a denominator that both $7$ and $4$ divide into.
Rewrite each fraction as an equivalent fraction over that denominator.
Add the numerators, keep the denominator, and compare the result with $\dfrac{19}{28}$ .

Worked Solution [2 marks]

Rule - Adding fractions: rewrite both fractions over a common denominator, then add the numerators and leave the denominator alone.

\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{ad}{bd} + \dfrac{cb}{bd} = \dfrac{ad + cb}{bd}

Step 1: Choose a common denominator

7 \times 4 = 28

(Reason:

7

and

4

share no factor other than

1

, so the smallest number both of them divide into is simply their product,

28

. Twenty-eighths are the common unit both fractions can be counted in.)

Step 2: Write each fraction over 28

\dfrac{3}{7} = \dfrac{3 \times 4}{7 \times 4} = \dfrac{12}{28}

\dfrac{1}{4} = \dfrac{1 \times 7}{4 \times 7} = \dfrac{7}{28}

(Reason: Multiplying the top and the bottom of a fraction by the same number does not change its value, only the size of the parts it is counted in. Each denominator needs its own multiplier:

7

has to be scaled by

4

to reach

28

, while

4

has to be scaled by

7

Step 3: Add the numerators

\dfrac{12}{28} + \dfrac{7}{28} = \dfrac{12 + 7}{28} = \dfrac{19}{28}

(Reason: Both amounts are now counted in twenty-eighths, so

12

of them plus

7

of them is

19

of them. The denominator records how big each part is, not how many there are, so it does not change when the parts are added.)

Step 4: Compare with what the question printed

\dfrac{3}{7} + \dfrac{1}{4} = \dfrac{19}{28}

(Reason: The sum has come out as exactly the fraction the question asked for.

19

is prime and is not a factor of

28

, so nothing cancels and the fraction is already in its simplest form. That completes the demonstration.)

\dfrac{3}{7} + \dfrac{1}{4} = \dfrac{12}{28} + \dfrac{7}{28} = \dfrac{19}{28}

as required

Verification

Check 1: Leave fractions behind and work in decimals instead, rounding to

4

decimal places:

\dfrac{3}{7} \approx 0.4286

\dfrac{1}{4} = 0.25

and

\dfrac{19}{28} \approx 0.6786

0.4286 + 0.25 = 0.6786

Check 2: Run the statement backwards. If the sum really is

\dfrac{19}{28}

, then taking

\dfrac{1}{4}

, which is

\dfrac{7}{28}

, back off it must return the first fraction.

\dfrac{19}{28} - \dfrac{7}{28} = \dfrac{12}{28} = \dfrac{3}{7}

Check 3: A different route to the same place: add in a single line with the general rule, without writing the two equivalent fractions down separately.

\dfrac{3 \times 4 + 1 \times 7}{7 \times 4} = \dfrac{19}{28}

Check 4: Size check.

\dfrac{3}{7}

is a little under a half, and a quarter is added to it, so the total must land between a half and three quarters. Written in twenty-eighths those two boundaries are

\dfrac{14}{28}

and

\dfrac{21}{28}

\dfrac{14}{28} < \dfrac{19}{28} < \dfrac{21}{28}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Find a common denominator, with at least one fraction correct, eg $\dfrac{12}{28}$ and $\dfrac{7}{28}$	M1	Any correct start to the common denominator scores this mark: $\dfrac{12}{28}$ and $\dfrac{7}{28}$ , or the unsimplified $\dfrac{12n}{28n}$ and $\dfrac{7n}{28n}$ , or $4 \times 3 = 12$ and $1 \times 7 = 7$ and $4 \times 7 = 28$ , or $\dfrac{3 \times 4}{4 \times 7}$ and $\dfrac{1 \times 7}{4 \times 7}$ . Calculations that would lead to $12$ and $7$ and $28$ also score it, even if they are not yet written as fractions.	✓
A complete correct method leading to $\dfrac{19}{28}$ , eg $\dfrac{12}{28} + \dfrac{7}{28} = \dfrac{19}{28}$	A1	Dependent on the M1. Also accepted: $\dfrac{12n}{28n} + \dfrac{7n}{28n} = \dfrac{19n}{28n} = \dfrac{19}{28}$ , or $12 + 7 = 19$ together with $\dfrac{19}{28}$ , or $4 \times 3 + (1 \times) 7 \; (= 19)$ and $4 \times 7 \; (= 28)$ and $\dfrac{19}{28}$ , or $\dfrac{3 \times 4}{4 \times 7} + \dfrac{(1 \times) 7}{4 \times 7} = \dfrac{19}{28}$ , or the one-line $\dfrac{3 \times 4 + 1 \times 7}{4 \times 7} = \dfrac{19}{28}$ .	✓
Working	Note	A correct final line with no common denominator anywhere earns nothing. $\dfrac{19}{28}$ is printed in the question, so copying it out is not a method.	✓

Full marks: 2/2

Continue to questions 14 to 26

The remaining 13 questions, with the same full worked solutions and mark schemes

Frequently asked questions

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

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

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

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

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

Keep revising

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

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Edexcel IGCSE 4MA1/1F, Thursday 15 May 2025: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 13 of 26

Question 1, Calculator allowed

Question 1 - Exam Solution

Question 2, Calculator allowed

Question 2 - Exam Solution

Question 3, Calculator allowed

Question 3 - Exam Solution

Question 4, Calculator allowed

Question 4 - Exam Solution

Question 5, Calculator allowed

Question 5 - Exam Solution

Question 6, Calculator allowed

Question 6 - Exam Solution

Question 7, Calculator allowed

Question 7 - Exam Solution

Question 8, Calculator allowed

Question 8 - Exam Solution

Question 9, Calculator allowed

Question 9 - Exam Solution

Question 10, Calculator allowed

Question 10 - Exam Solution

Question 11, Calculator allowed

Question 11 - Exam Solution

Question 12, Calculator allowed

Question 12 - Exam Solution

Question 13, Calculator allowed

Question 13 - Exam Solution

Frequently asked questions

Keep revising

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