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

Sir Faraz Hassan

24 Aug 2026

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

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/1FR (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 14 of 26

An online atlas gives the land area, in $\text{km}^2$ , of six states in the USA.

$\begin{array}{|l|c|}\hline \textbf{State} & \textbf{Land area (km}^2\textbf{)} \\ \hline \text{Arizona} & 294\,207 \\ \hline \text{Florida} & 138\,887 \\ \hline \text{Montana} & 376\,962 \\ \hline \text{New Mexico} & 314\,161 \\ \hline \text{Oregon} & 248\,608 \\ \hline \text{Washington} & 172\,119 \\ \hline \end{array}$

(a) Write down the name of the state with the greatest land area. [1 mark]

(b) Round the number $314\,161$ to the nearest hundred. [1 mark]

(d) Work out the total land area of Oregon and Washington. [1 mark]

The land area of Vermont is $23\,871 \text{ km}^2$

(e) Write the number $23\,871$ in words. [1 mark]

(a)(b)(c)(d) km²(e)

[Total 5 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

Six land areas, in

\text{km}^2

: Arizona

294\,207

, Florida

138\,887

, Montana

376\,962

, New Mexico

314\,161

, Oregon

248\,608

, Washington

172\,119

Vermont has a land area of

23\,871 \text{ km}^2

Every land area in the table is a six-digit whole number, so every comparison is a place-value comparison.

Find

(a) The state with the greatest land area. (b)

314\,161

to the nearest hundred. (c) What the digit

4

is worth in

294\,207

. (d) The two land areas added together. (e)

23\,871

written out in words.

Plan the Solution

Line the six land areas up by place value and read from the left, so the largest is found by comparing digits and nothing is added.
For the nearest hundred, look only at the tens digit: $5$ or more sends the hundreds up, and anything less leaves them alone.
For place value, name the column the digit is sitting in, then multiply the digit by that column.
For the total, add in columns from the right, carrying into the next column whenever a column reaches ten.
For the words, split the number at the thousands and write out each block in turn.

Worked Solution [5 marks]

Rule - Place value: in a whole number the columns, reading from the right, are units, tens, hundreds, thousands, ten thousands and hundred thousands. A digit is worth the digit multiplied by its own column, and comparing, rounding, adding and writing in words all read off those same columns.

Step 1: (a) Compare the six land areas

376\,962 > 314\,161 > 294\,207 > 248\,608 > 172\,119 > 138\,887

(Reason: every land area has six digits, so the hundred thousands column decides first. Montana and New Mexico both have

3

there, and Montana then wins on the ten thousands,

7

against

1

)

Step 2: (b) Round

314\,161

to the nearest hundred

314\,100 < 314\,161 < 314\,200

314\,161 - 314\,100 = 61

314\,200 - 314\,161 = 39

314\,161 \approx 314\,200

(Reason: the tens digit is

6

, which is

5

or more, so the hundreds go up. The two gaps say the same thing a second way:

39

is smaller than

61

, so the higher hundred is the nearer one)

Step 3: (c) Read the column the

4

is sitting in

294\,207 = 200\,000 + 90\,000 + 4000 + 200 + 7

4 \times 1000 = 4000

(Reason: counting from the right the columns are units, tens, hundreds, thousands, ten thousands and hundred thousands, so this

4

is in the thousands column and is worth

4

thousands. The digit is a

4

wherever it sits; the column is what gives it its value)

Step 4: (d) Add the two land areas in columns

248\,608 + 172\,119 = 420\,727

(Reason: add from the right. The units give

8 + 9 = 17

, so write

7

and carry

1

into the tens. The thousands then give

8 + 2 = 10

, so write

0

and carry again into the ten thousands)

Step 5: (e) Split

23\,871

into thousands and the rest

23\,871 = 23 \times 1000 + 871

(Reason: say the two blocks in turn:

23

thousand, then

871

as eight hundred and seventy one. The comma in the written answer sits exactly where the number splits)

(a) Montana(b)

314\,200

(c)

4

thousands, or

4000

(d)

420\,727 \text{ km}^2

(e) Twenty three thousand, eight hundred and seventy one

Verification

Check 1: Part (a): take the next largest land area, New Mexico's, away from Montana's. A positive difference means Montana really is ahead.

376\,962 - 314\,161 = 62\,801

, which is positive.

Check 2: Part (b): measure the distance from

314\,161

to each of the two hundreds either side of it.

39

up against

61

down, so

314\,200

is the nearer hundred.

Check 3: Part (c): replace the

4

with a zero and subtract. Whatever is left is what the

4

was worth.

294\,207 - 290\,207 = 4000

Check 4: Part (d): undo the addition by taking Washington's land area back off the total.

420\,727 - 172\,119 = 248\,608

, which is Oregon's land area.

Check 5: Part (d) again, roughly: round both land areas to the nearest thousand and add those instead.

249\,000 + 172\,000 = 421\,000

, only

273

above the exact total.

Check 6: Part (e): read the written answer back as a number.

23\,000 + 800 + 71 = 23\,871

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	Montana	✓
(b)	B1	$314\,200$	✓
(c)	B1	$4$ thousands. Accept $4000$ , thousands	✓
(d)	B1	$420\,727$	✓
(e)	B1	Twenty three thousand, eight hundred (and) seventy one	✓

Full marks: 5/5

Astrid has four tiles.
There is a number on each tile.

Astrid is going to pick at random one of these tiles.

(a) Circle the word in the box below that best describes the likelihood that Astrid will pick a tile with the number $5$ on it.
$\begin{array}{|ccccc|}\hline \text{impossible} & \text{unlikely} & \text{evens} & \text{likely} & \text{certain} \\ \hline \end{array}$ [1 mark]

(b) On the probability scale below, mark with a cross (×) the probability that Astrid will pick a tile with a number less than $6$ on it. [1 mark]

Meera has six tiles each with a number on it.
Four of these numbers are shown below.

When she picks at random one of the six tiles, the probability that she picks a tile with an even number on it is $\dfrac{1}{2}$

[Total 3 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

Astrid's four tiles show

1

2

4

and

5

, and one tile is picked at random.

Meera has six tiles: four of them show

3

3

5

and

6

, and two are blank.

For Meera's tiles,

P(\text{even}) = \dfrac{1}{2}

Find

(a) The word that best describes the likelihood of picking the tile showing

5

. (b) Where the cross goes on the probability scale for a number less than

6

. (c) A number for each of the two blank tiles.

Plan the Solution

Count the tiles that succeed, then write the probability as a fraction of the total number of tiles.
Measure that fraction against $0$ , $\dfrac{1}{2}$ and $1$ to choose the word in (a) and the place on the scale in (b).
For (c), turn the probability into a number of tiles first, then count the even numbers already on show.

Worked Solution [3 marks]

Rule - Probability of an event:

P(\text{event}) = \dfrac{\text{successful outcomes}}{\text{all outcomes}}

, which runs from

0

(impossible) to

1

(certain).

Step 1: (a) count the tiles that show

5

P(5) = \dfrac{1}{4} = 0.25

(Reason: Only one of the four tiles has

5

on it, so one outcome out of four is successful.)

Step 2: (a) measure

0.25

against a half

0 < 0.25 < 0.5

(Reason: The chance is bigger than

0

, so it is not impossible, and smaller than

\dfrac{1}{2}

, so it is less than evens. That band is the word unlikely.)

Step 3: (b) count the tiles showing a number less than

6

P(\text{less than } 6) = \dfrac{4}{4} = 1

(Reason: Every one of

1

2

4

and

5

is smaller than

6

, so no tile can fail. An event that cannot fail is certain, and certain is the right-hand end of the scale.)

Step 4: (c) turn the probability into a number of tiles

\dfrac{1}{2} \times 6 = 3

(Reason: A probability of

\dfrac{1}{2}

out of six tiles means

3

of the six tiles must have an even number on them.)

Step 5: (c) count the even numbers already there

3 - 1 = 2

(Reason: Of

3

3

5

and

6

only the

6

is even, so two more even numbers are needed - and there are exactly two blank tiles. Writing

2

and

4

on them is one possible set.)

(a) unlikely(b) a cross at

1

, the right-hand end of the scale(c)

2

and

4

(any two even numbers)

Verification

Check 1: (a) from the other side. Three of the four tiles are not the

5

, and the two probabilities have to add to

1

. The smaller of the two shares is the unlikely one.

0.25 + 0.75 = 1

Check 2: (b) by listing. The numbers smaller than

6

are

1

2

4

and

5

- all four of the tiles - so no tile is left out.

\dfrac{4}{4} = 1

Check 3: (c) by counting the finished set. Meera's six tiles would read

3

3

5

6

2

4

, and three of those numbers are even.

\dfrac{3}{6} = \dfrac{1}{2}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	unlikely	✓
(b)	B1	× at $1$	✓
(c)	B1	$2$ numbers which are even	✓

Full marks: 3/3

The diagram shows a polygon with $6$ sides.

(a) Measure the length of the side $AB$
Write down the units of your answer. [2 marks]

(b) Measure the size of the angle marked $x$ [1 mark]

(a)(a)(b) degrees(c)

[Total 4 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

A polygon with

6

sides, drawn accurately, with two of its vertices labelled

A

and

B

One angle of the polygon is marked

x

, with a small arc drawn inside it.

No length and no angle is printed on the figure, so both readings have to be taken off the drawing itself.

Find

(a) the length of

AB

, and the units it is measured in (b) the size of the angle marked

x

6

sides

Plan the Solution

Nothing here is worked out. Parts (a) and (b) are read off the page with a ruler and a protractor, and part (c) is recall.
For (a), lay the ruler along $AB$ with its zero mark exactly on $A$ , and read the mark that $B$ falls on. Read to the nearest millimetre.
Then write the units beside the number. The number on its own is only half of that answer, which is why part (a) carries two marks and not one.
For (b), put the centre of the protractor on the vertex where $x$ is marked, lay the zero line along one arm of the angle, and read the scale where the other arm crosses it.
A protractor carries two scales, running in opposite directions. The angle marked $x$ is obtuse - it opens wider than a right angle - so its reading must be more than $90$ , and that is which of the two scales to take.
For (c), count the sides round the outline and name the polygon from that count.

Worked Solution [4 marks]

Ruler:

10

mm make

1

cm, so a reading of

5

cm and

3

mm is

5.3

cm, or

53

mm. Protractor: an acute angle reads under

90

and an obtuse angle reads over it, and the two scales at any one position always add to

180

. Polygons are named by their number of sides:

3

triangle,

4

quadrilateral,

5

pentagon,

6

hexagon,

7

heptagon.

(a) Measure

AB

with a ruler

AB = 5 \text{ cm } 3 \text{ mm}

(Reason: With the zero mark on

A

, the vertex

B

falls between the

5

cm and

6

cm marks, three of the small millimetre marks past the

5

. Read the small marks rather than guessing at the gap.)

(a) Write the reading with its units

5 \text{ cm} = 50 \text{ mm}

50 + 3 = 53

(Reason: The two marks here are for two different things: one for the value and one for the units. Centimetres and millimetres are both accepted, so

5.3

cm and

53

mm are worth the same. A bare

5.3

with nothing after it is not.)

(b) Measure the angle marked

x

with a protractor

x = 109^\circ

(Reason: The centre of the protractor goes on the vertex the arc is drawn at, and the zero line goes along one arm. Both scales are printed on the protractor and only one of them is the answer: the arms open wider than a right angle, so the reading to take is the one above

90

, not the one below it.)

6 \text{ sides} \rightarrow \text{hexagon}

(Reason: Count the straight edges round the outline: there are

6

of them. A polygon is named by that count, and the prefix hexa- means six. The corner that turns in towards the middle of the shape does not change the name - a polygon may have a corner like that and still be a hexagon.)

(a)

5.3

cm (or

53

mm)(b)

x = 109^\circ

Verification

Check 1: Measure the same side in millimetres instead of centimetres, then convert. The ruler reads

53

mm, and dividing by

10

turns millimetres into centimetres.

\dfrac{53}{10} = 5.3

cm, which is the reading part (a) gave

Check 2: Read the protractor's other scale at the same position. The two scales run in opposite directions, so the two readings must account for a straight line between them.

109 + 71 = 180

, so the obtuse reading and the acute one fit together, and the obtuse one is the angle marked

Check 3: Name the shape from its angles rather than from its sides. The angles inside the drawn figure add up to

720

, and the angles of a polygon with

n

sides add up to

(n - 2) \times 180

(6 - 2) \times 180 = 720

, so the figure has

6

sides, and the name for

6

sides is hexagon

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B2	$5.3$ cm or $53$ mm or $5$ cm $3$ mm	✓
(a)	B1	for $5.3$ (allow $5.1$ - $5.5$ ) or $53$ (allow $51$ - $55$ ) or cm with a value from $4.8$ - $5.8$ or mm with a value from $48$ - $58$	✓
(b)	B1	$109$	✓
(c)	B1	hexagon	✓

Full marks: 4/4

(a) Write $\dfrac{9}{10}$ as a decimal. [1 mark]

(b) Write $\dfrac{3}{50}$ as a percentage. [1 mark]

Here is a shape made from identical squares.

(d) One of these fractions is not equivalent to $\dfrac{2}{5}$
Which one?
$\dfrac{20}{50}$ $\dfrac{4}{10}$ $\dfrac{2}{500}$ $\dfrac{10}{25}$ $\dfrac{60}{150}$ [1 mark]

A choir has $32$ members.
$8$ of the members sing tenor.

(e) What fraction of the members do not sing tenor?
Give your fraction in its simplest form. [2 marks]

(a)(b) %(d)(e)

[Total 6 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

Two fractions to rewrite:

\dfrac{9}{10}

as a decimal, and

\dfrac{3}{50}

as a percentage.

A shape made from identical squares,

8

across and

3

down.

Five fractions:

\dfrac{20}{50}

\dfrac{4}{10}

\dfrac{2}{500}

\dfrac{10}{25}

and

\dfrac{60}{150}

. Four of them are equivalent to

\dfrac{2}{5}

and one is not.

A choir of

32

members, of whom

8

sing tenor.

Find

(a)

\dfrac{9}{10}

written as a decimal (b)

\dfrac{3}{50}

written as a percentage (c) how much of the shape to shade so that

\dfrac{5}{8}

of it is shaded (d) the one fraction that is not equivalent to

\dfrac{2}{5}

(e) the fraction of the members who do not sing tenor, in its simplest form

Plan the Solution

For (a), read the fraction as a number of tenths. Tenths are the first place after the decimal point, so a fraction with $10$ underneath needs no working at all once that is seen.
For (b), a percentage is a number of hundredths, so turn the fraction into hundredths. $50$ doubles to $100$ , and whatever is done to the bottom is done to the top.
For (c), count the squares first. The denominator $8$ says how many equal parts the shape is cut into, and the numerator $5$ says how many of those parts to shade.
For (d), cancel each fraction down and see which one does not become $\dfrac{2}{5}$ . Cross-multiplying is a second way to test the same thing.
For (e), subtract first and simplify second. Find how many members do not sing tenor, write that over $32$ , then cancel by the largest number that divides both.
Every part here is about the same idea: multiplying or dividing the top and the bottom of a fraction by the same number leaves its value unchanged.

Worked Solution [6 marks]

Equivalent fractions: multiplying or dividing top and bottom by the same number does not change a fraction's value, so

\dfrac{2}{5} = \dfrac{4}{10} = \dfrac{20}{50}

. Decimals: the first place after the point is tenths and the second is hundredths. Percentages: per cent means out of

100

, so a fraction becomes a percentage once its denominator is

100

, or by multiplying it by

100

. A fraction of an amount: divide by the denominator to get one part, then multiply by the numerator. Simplest form: cancel until the only number dividing top and bottom is

1

(a) Read

\dfrac{9}{10}

as tenths

\dfrac{9}{10} = 0.9

(Reason: The first column after the decimal point is the tenths column, so

9

tenths is a

9

written in that column. Said as a division instead: the fraction bar means divide, and dividing by

10

moves the digit one place to the right, out of the units and into the tenths.)

(b) Scale

\dfrac{3}{50}

to a denominator of

100

\dfrac{3}{50} = \dfrac{3 \times 2}{50 \times 2} = \dfrac{6}{100}

(Reason: Per cent means out of

100

, so the useful denominator is

100

50

doubles to

100

, so the

3

on top doubles as well. Doubling both parts leaves the value of the fraction alone.)

(b) Write those hundredths as a percentage

\dfrac{3}{50} \times 100 = 6

(Reason: Six hundredths is

6

per cent, so the answer is read straight off the scaled fraction. Multiplying the original fraction by

100

is the same conversion done in one line, and it gives the same

6

8 \times 3 = 24

(Reason: The shape is a rectangle of identical squares,

8

across and

3

down, so there are

24

of them altogether. Nothing can be shaded until this total is known, because the fraction is a fraction of these squares.)

\dfrac{5}{8}

\dfrac{24}{8} = 3

5 \times 3 = 15

(Reason: The denominator

8

cuts the shape into

8

equal parts, and

24

squares shared into

8

parts is

3

squares per part. The numerator

5

says to take five of those parts. A column of the grid is exactly one of them, so shading five whole columns is the tidiest way to show it, though any

15

squares would earn the mark.)

(d) Cancel each fraction and compare it with

\dfrac{2}{5}

\dfrac{20}{50} = \dfrac{2 \times 10}{5 \times 10} = \dfrac{2}{5}

\dfrac{4}{10} = \dfrac{2 \times 2}{5 \times 2} = \dfrac{2}{5}

\dfrac{10}{25} = \dfrac{2 \times 5}{5 \times 5} = \dfrac{2}{5}

\dfrac{60}{150} = \dfrac{2 \times 30}{5 \times 30} = \dfrac{2}{5}

(Reason: A fraction is equivalent to

\dfrac{2}{5}

when it is

\dfrac{2}{5}

with both parts multiplied by the same number. Four of the five are, with the multiplier

10

2

5

and

30

in turn.)

(d) Name the one that is left

\dfrac{2}{500} = \dfrac{1}{250}

(Reason: In

\dfrac{2}{500}

the denominator has been multiplied by

100

but the numerator has not, so the two parts have not been scaled by the same number and the value has changed. It cancels to

\dfrac{1}{250}

, which is nowhere near

\dfrac{2}{5}

(e) Count the members who do not sing tenor

32 - 8 = 24

(Reason: The question asks about the members who do not sing tenor, so those are counted first: the whole choir less the

8

who do. Answering with the

8

themselves is the one mistake this part is testing for.)

(e) Write that as a fraction of the choir and simplify it

\dfrac{24}{32} = \dfrac{3 \times 8}{4 \times 8} = \dfrac{3}{4}

(Reason: A fraction is the part over the whole, so it is

24

over

32

. Simplest form means cancelling by the largest number that divides both, and that is

8

: it goes into

24

three times and into

32

four times. Cancelling by

4

instead gives

\dfrac{6}{8}

, which is the right value but not yet the simplest form the question asks for.)

(a)

0.9

(b)

6\%

(c)

15

squares shaded, which is

5

of the

8

columns(d)

\dfrac{2}{500}

(e)

\dfrac{3}{4}

Verification

Check 1: Turn part (a)'s decimal back into a fraction. One decimal place is tenths, so multiplying by

10

should give the numerator the question started with.

0.9 \times 10 = 9

, so the decimal is

\dfrac{9}{10}

again

Check 2: Take part (b)'s percentage back the other way.

6

per cent is

6

hundredths, and both parts of

\dfrac{6}{100}

halve.

\dfrac{6}{100} = \dfrac{3 \times 2}{50 \times 2} = \dfrac{3}{50}

, the fraction the question started from

Check 3: Count what is left unshaded in part (c) instead of what is shaded. Three eighths of the shape should be left, and three eighths of

24

9

24 - 15 = 9

squares left, and

\dfrac{9}{24} = \dfrac{3}{8}

, so the part shaded really is five eighths

Check 4: Test part (d) by cross-multiplying rather than by cancelling. For two equivalent fractions the two cross-products match.

20 \times 5 = 100

and

2 \times 50 = 100

, which match, and the same happens for

\dfrac{4}{10}

\dfrac{10}{25}

and

\dfrac{60}{150}

; but

2 \times 5 = 10

against

2 \times 500 = 1000

, which do not

Check 5: Work part (e) backwards. Take three quarters of the choir and see whether it is the number who do not sing tenor.

\dfrac{3}{4} \times 32 = 24

, and

32 - 24 = 8

, which is the number who do sing tenor

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$0.9$	✓
(b)	B1	$6$	✓
(c)	B1	$15$ squares shaded	✓
(d)	B1	$\dfrac{2}{500}$	✓
(e)	M1	eg $\dfrac{32 - 8}{32}$ oe or $1 - \dfrac{8}{32}$ oe or $\dfrac{32}{32} - \dfrac{8}{32}$ oe or $\dfrac{24}{32}$ or $0.75$ or $\dfrac{1}{4}$	✓
(e)	A1	$\dfrac{3}{4}$	✓

Full marks: 6/6

(a) Write $4e \times 7f$ in its simplest form. [1 mark]

(b) Write $d \times d \times d \times d \times d$ in index form. [1 mark]

(d) Solve the equation $x + 6 = 15$ [1 mark]

(e) Solve the equation $2r - 9 = 14$ [2 marks]

(a)(b)(c)(d) x =(e) r =

[Total 7 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

Two products to simplify:

4e \times 7f

and

d \times d \times d \times d \times d

One sum with two different letters in it:

3a + 2k + a - 7k

Two equations to solve:

x + 6 = 15

and

2r - 9 = 14

Find

(a)

4e \times 7f

written in its simplest form (b)

d \times d \times d \times d \times d

written in index form (c)

3a + 2k + a - 7k

written in its simplest form (d) the value of

x

(e) the value of

r

Plan the Solution

Parts (a) and (b) are products, so nothing is added anywhere in them. In (a) the numbers multiply and the letters are written side by side. In (b) the same letter is multiplied by itself, and that is counted with an index rather than a coefficient.
Part (c) is a sum, so here things do add, but only terms carrying the SAME letter may be put together. Deal with the $a$ terms and the $k$ terms separately, and let each term keep the sign printed in front of it.
Parts (d) and (e) are solved by doing the same thing to both sides until the letter stands alone. In (d) one operation has been applied to $x$ ; in (e) two have been applied to $r$ , so they are undone in reverse order.
A calculator is allowed, but every part here is quicker by hand than by keying it in. What the calculator is worth is the check at the end.

Worked Solution [7 marks]

Products: multiply the numbers, then write the letters after them, so

3p \times 5q = 15pq

. Index form: an index COUNTS how many times a letter is multiplied by itself, so

p \times p \times p = p^{3}

. Like terms: only terms with exactly the same letter can be combined, and each term carries the sign written in front of it. Solving: do the same thing to both sides, undoing the operations in the reverse of the order they were applied.

(a) Multiply the numbers, then write the letters together

4e \times 7f = 4 \times 7 \times e \times f

4 \times 7 = 28

4e \times 7f = 28ef

(Reason: Multiplication may be done in any order, so the two numbers can be brought together and the two letters can be brought together.

4

times

7

28

, and

e \times f

is written

ef

with no sign between the letters. Because the letters are different, there is nothing further to collect and

28ef

is as simple as it gets.)

(b) Count how many times

d

is multiplied by itself

d \times d \times d \times d \times d = d^{5}

(Reason: An index is a count, not a total: it records how many copies of the letter are being multiplied. There are five copies of

d

here, so the index is

5

. Read as the index law, each

d

d^{1}

and multiplying powers of the same letter adds the indices, which gives

1 + 1 + 1 + 1 + 1 = 5

again. Answering

5d

is the slip this part is testing for:

5d

means

d + d + d + d + d

, which is addition, not multiplication.)

a

terms and the

k

terms separately

3a + a = 4a

2k - 7k = -5k

3a + 2k + a - 7k = 4a - 5k

(Reason: Only like terms combine.

3a

and

a

are both counts of

a

, and a letter written on its own means one of it, so

3 + 1 = 4

of them. Each term keeps the sign printed in front of it, so the second pair is

2k

take away

7k

, and

2 - 7 = -5

leaves

-5k

. The two totals cannot then be added to each other, because

a

and

k

stand for different things, so the answer is two terms and not one.)

(d) Undo the

+\,6

x + 6 = 15

x = 15 - 6

x = 9

(Reason: The equation says

6

has been added to

x

. Taking

6

off both sides undoes that and leaves

x

by itself on the left, with

15 - 6

on the right. Both sides were treated the same way, so the equation still balances.)

(e) Undo the

-\,9

first, then undo the

\times\,2

2r - 9 = 14

2r = 14 + 9 = 23

r = \dfrac{23}{2} = 11.5

(Reason: Two things have been done to

r

: it was multiplied by

2

and then

9

was taken off. They are undone in the reverse order, so the

9

goes back on first by adding

9

to both sides, and only after that are both sides halved. Halving

23

does not give a whole number, and it does not have to:

11.5

is a perfectly good solution, and the mark scheme accepts it as

11.5

\dfrac{23}{2}

11\dfrac{1}{2}

(a)

28ef

(b)

d^{5}

(c)

4a - 5k

(d)

x = 9

(e)

r = 11.5

Verification

Check 1: Test parts (a) and (b) with numbers. Two expressions that are really the same give the same value whatever is put in, so take

e = 2

f = 5

and

d = 2

4 \times 2 \times 7 \times 5 = 280

and

28 \times 2 \times 5 = 280

;

2 \times 2 \times 2 \times 2 \times 2 = 32

and

2^{5} = 32

, while the wrong answer

5d

would give only

5 \times 2 = 10

Check 2: Test part (c) the same way, with

a = 2

and

k = 3

. A sign error in the

k

terms is the likeliest mistake, and it would show here at once.

3 \times 2 + 2 \times 3 + 2 - 7 \times 3 = -7

and

4 \times 2 - 5 \times 3 = -7

Check 3: Put each solution back into the equation it came from. A solution is correct exactly when it makes the left-hand side equal the right-hand side.

9 + 6 = 15

and

2 \times 11.5 - 9 = 14

Check 4: Solve part (e) a second way, halving every term first instead of moving the

9

. A different order of working should reach the same value.

r - 4.5 = 7

, so

r = 7 + 4.5 = 11.5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$28ef$	✓
(b)	B1	$d^{5}$	✓
(c)	B2	$4a - 5k$	✓
(c)	(B1)	for $4a$ or $-5k$ or $4a + -5k$	✓
(d)	B1	$9$	✓
(e)	M1	$2r = 14 + 9$ or $2r = 23$ oe or $\dfrac{14 + 9}{2}$ or $2 \times 11.5 - 9 = 14$ , for a correct first step or a correct calculation for $r$	✓
(e)	A1	for $11.5$ or $\dfrac{23}{2}$ or $11\dfrac{1}{2}$	✓

Full marks: 7/7

Zubair has some sacks of flour and some tins of oil.
Each sack has the same weight.
Each tin has the same weight.

The total weight of $7$ sacks and $2$ tins is $27.3$ kg
The total weight of $4$ sacks and $2$ tins is $16.8$ kg

Work out the weight of one tin. [4 marks]

[Total 4 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

Two kinds of item, and every item of one kind has the same weight: sacks of flour, and tins of oil.

7

sacks and

2

tins weigh

27.3

kg altogether.

4

sacks and

2

tins weigh

16.8

kg altogether.

Find

The weight of ONE tin, in kilograms.

Plan the Solution

Give each unknown weight a letter, so that each printed total becomes an equation. Let $s$ be the weight of one sack in kilograms and $t$ the weight of one tin in kilograms.
Look at what the two totals have in COMMON. Both carry $2$ tins, so the tins contribute the same amount to each. Taking the smaller total away from the larger therefore removes the tins completely and leaves only sacks.
What is left is the weight of the extra sacks: $7 - 4 = 3$ of them. From that comes the weight of one sack.
Put the weight of one sack back into either printed total. The sacks in it can then be accounted for, and whatever is left over is the weight of the $2$ tins. Halve it for one tin.
A calculator is allowed, so keep the decimals as they are printed. Nothing here needs converting into grams.

Worked Solution [4 marks]

Rule - Elimination: when two totals contain the SAME number of one item, subtracting one total from the other removes that item entirely, and what is left is a statement about the other item alone. Here both totals contain

2

tins, so the subtraction leaves sacks and nothing else.

Step 1: write each printed total as an equation

7s + 2t = 27.3

4s + 2t = 16.8

(Reason: Every sack weighs the same, so

7

of them weigh

7s

; every tin weighs the same, so

2

of them weigh

2t

. Adding those two amounts gives the printed total of

27.3

kg, and the second total is written the same way. Both letters stand for a weight in kilograms, so both equations are in the same units.)

Step 2: subtract, because both totals carry the same

2

tins

(7s + 2t) - (4s + 2t) = 27.3 - 16.8

27.3 - 16.8 = 10.5

3s = 10.5

(Reason: The

2

tins weigh the same in both totals, so subtracting cancels them:

2t - 2t = 0

. The sacks do not cancel, because there are

7

in one total and only

4

in the other, leaving

7 - 4 = 3

of them. So the whole gap between the two printed totals,

10.5

kg, is the weight of

3

sacks.)

Step 3: divide by

3

to reach one sack

3s = 10.5

s = \dfrac{10.5}{3} = 3.5

(Reason: The

10.5

kg is shared equally between the

3

extra sacks, so it is divided by

3

and not by either of the printed counts. Dividing by the

7

of the first total is the slip this step is easiest to make: the

10.5

kg is not the weight of

7

sacks, it is the weight of the

3

that the first load has and the second does not.)

Step 4: take the

4

sacks off the second total

4 \times 3.5 = 14

16.8 - 14 = 2.8

2t = 2.8

(Reason: The second load is

4

sacks and

2

tins. The sacks in it weigh

4 \times 3.5 = 14

kg, so removing them from the printed

16.8

kg leaves only the tins. The first total would do just as well and is used as a check below.)

Step 5: halve

2.8

to reach one tin

t = \dfrac{2.8}{2} = 1.4

(Reason: The

2.8

kg is the weight of

2

tins, and the question asks for one. Every tin has the same weight, so halving is all that is left to do. The answer is a weight, so it is given in kilograms.)

1.4

Verification

Check 1: A pair of weights is only right if it fits BOTH printed totals, not just the one it was worked out from. Put

3.5

kg for a sack and

1.4

kg for a tin into each of them.

7 \times 3.5 + 2 \times 1.4 = 27.3

and

4 \times 3.5 + 2 \times 1.4 = 16.8

Check 2: Work the question a second way, removing the SACKS instead of the tins. Multiplying the first total by

4

and the second by

7

puts

28

sacks in both, so subtracting now cancels the sacks. This route never divides by

3

and never finds the weight of a sack, so a slip made above cannot repeat itself here.

28s + 14t = 117.6

take away

28s + 8t = 109.2

gives

6t = 8.4

, so

t = 1.4

Check 3: Use the other printed total, the one the working did not use. Taking the

7

sacks out of the

27.3

kg load should leave the same

2

tins.

7 \times 3.5 = 24.5

, so

27.3 - 24.5 = 2.8

for

2

tins, giving

1.4

kg each

Check 4: Test the size of the gap between the two loads. The only difference between them is

3

extra sacks, so

3

sacks must account for the whole of it.

3 \times 3.5 = 10.5

and

27.3 - 16.8 = 10.5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Subtract the two totals	M1	$27.3 - 16.8 \; (= 10.5)$ or $7s + 2t = 27.3$ and $4s + 2t = 16.8$ subtracted to give $3s = 10.5$ , for a correct first step to find the weight of $3$ sacks	✓
Weight of one sack	M1	$\dfrac{10.5}{7 - 4} \; (= 3.5)$ or $s = 3.5$ , for a method to find the weight of one sack. The printed scheme puts the $10.5$ in quotation marks, so the candidate's own value from the first step may be used	✓
Weight of $2$ tins	M1	eg $27.3 - 7 \times 3.5 \; (= 2.8)$ or $16.8 - 4 \times 3.5 \; (= 2.8)$ or $2t = 16.8 - 4 \times 3.5 \; (= 2.8)$ , for a method to find the weight of $2$ tins. The printed scheme puts the $3.5$ in quotation marks, so the candidate's own value may be used	✓
The weight of one tin	A1	$1.4$ , or an equivalent such as $\dfrac{7}{5}$	✓

Full marks: 4/4

$AEC$ and $BED$ are straight lines.

Mateo says that the value of $x$ is $58$

(a) Give a reason why Mateo is correct. [1 mark]

In the diagram, $GHIJ$ is a quadrilateral.

Elena says that $FGH$ is a straight line.

(b) Show that Elena is wrong.
Give a reason for each stage of your working. [4 marks]

(a)

[Total 5 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

AEC

and

BED

are straight lines that cross at

E

, with one angle there marked

58^{\circ}

and the angle facing it across the crossing marked

x^{\circ}

GHIJ

is a quadrilateral with

163^{\circ}

H

, a right angle at

J

and

47^{\circ}

I

. The angle at

G

inside the quadrilateral is not given.

F

lies outside the quadrilateral, beyond

G

, and the angle between

GF

and

GJ

110^{\circ}

Find

(a) A reason why

x = 58

is correct. (b) Working, with a reason at every stage, showing that

FGH

is not a straight line.

Plan the Solution

(a) Nothing needs calculating. The two marked angles are made by the same pair of straight lines crossing, so name the angle fact that connects them.
(b) The angle $\angle JGH$ is missing, so find it first from the quadrilateral, whose four angles add up to $360^{\circ}$ .
(b) Then add the two angles that meet at $G$ and compare the total with the $180^{\circ}$ that a straight line needs. If it misses, the line is not straight.

Worked Solution [5 marks]

Rule - Angle facts: vertically opposite angles are equal; angles on a straight line add up to

180^{\circ}

; the four angles of a quadrilateral add up to

360^{\circ}

Step 1: Part (a), name the fact that links the two marked angles

x = 58

(Reason: The straight lines

AEC

and

BED

cross at

E

, and the two marked angles face each other across that crossing point, so they are vertically opposite. Vertically opposite angles are equal, which is why

x

must be

58

and Mateo is right. No calculation is needed, and the mark here is for the reason, not for the number.)

Step 2: Part (b), find the angle at

G

inside the quadrilateral

163 + 90 + 47 = 300

360 - 300 = 60

(Reason: The four angles of quadrilateral

GHIJ

add up to

360^{\circ}

. Taking the three given angles away from

360^{\circ}

leaves the fourth one, so

\angle JGH = 60^{\circ}

. Note that this is the angle inside the quadrilateral, not the whole angle at

G

Step 3: Part (b), add the two angles that meet at

G

\angle FGH = \angle FGJ + \angle JGH

110 + 60 = 170

(Reason: The angle

\angle FGH

is made of two pieces that sit side by side at

G

: the

110^{\circ}

marked outside the quadrilateral and the angle found in Step 2 inside it.)

Step 4: Part (b), compare with a straight line and conclude

170 \neq 180

(Reason: Angles on a straight line add up to

180^{\circ}

. The two angles at

G

come to

170^{\circ}

, which is

10^{\circ}

short of a half turn, so

F

G

and

H

do not lie on one straight line and Elena is wrong.)

(a) Vertically opposite angles are equal, so

x = 58

(b)

\angle FGH = 110^{\circ} + 60^{\circ} = 170^{\circ}

, not

180^{\circ}

, so

FGH

is not a straight line

Verification

Check 1: (a) Reach

x

through the straight lines instead of the opposite-angle rule. On the straight line

BED

\angle BEC = 180^{\circ} - 58^{\circ} = 122^{\circ}

. On the straight line

AEC

, the angle marked

x^{\circ}

is what is left after that same

122^{\circ}

180 - 122 = 58

, so

x = 58

, reached without using the rule it is being checked against

Check 2: (b) Take Elena at her word. If

FGH

really were straight then

\angle JGH = 180^{\circ} - 110^{\circ} = 70^{\circ}

, and the four angles of the quadrilateral would have to add up to

360^{\circ}

70 + 163 + 90 + 47 = 370

, which is

10

too many, so the assumption cannot hold

Check 3: (b) Put the angle found in Step 2 back into the quadrilateral and confirm the four angles close.

60 + 163 + 90 + 47 = 360

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Reason given	B1	vertically opposite angles are equal, or opposite to $58$ (with or without the degree sign). The printed row underlines vertically, opposite, opposite angles and $58$ as the words the answer must carry.	✓
(b) A method for angle $JGH$	M1	for a method to find angle $JGH$ either using the quadrilateral or assuming line $FGH$ is straight: ( $JGH$ =) $360 - 163 - 90 - 47$ (= $60$ ) or ( $JGH$ =) $180 - 110$ (= $70$ )	✓
(b) The figure that shows Elena is wrong	A1	( $110 + 60$ =) $170$ or $110 + 60 \neq 180$ or ( $180 - 60$ =) $120$ or ( $70 + 163 + 90 + 47$ =) $370$ or $70 + 163 + 90 + 47 \neq 360$ or for ( $180 - 110$ =) $70$ and ( $360 - 163 - 90 - 47$ =) $60$	✓
(b) Reason for the straight-line stage	B1	angles on a straight line add to $180^{\circ}$	✓
(b) Reason for the quadrilateral stage	B1	angles in a quad(rilateral) add up to $360$ (Accept a $4$ -sided shape)	✓
(b) Guidance printed with the scheme	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 5/5

The pictogram gives information about the number of loaves of bread a bakery sold on each of five days.

The number of loaves sold on Friday was $16$ more than the number of loaves sold on Thursday.

Work out the number of loaves sold on Monday. [3 marks]

[Total 3 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

The pictogram gives the number of loaves sold on each of five days, and it prints no key

A whole symbol is a rectangle ruled into

4

small squares, and a part symbol draws only some of those four

Friday's row draws

3

whole symbols and one half symbol; Thursday's row draws

1

whole symbol and one half symbol

Monday's row draws

2

whole symbols and one small square

Friday's number of loaves was

16

more than Thursday's

Find

The number of loaves the bakery sold on Monday

Plan the Solution

The pictogram has no key, so the first job is to build one. Count each row in SMALL SQUARES rather than in whole symbols, because every part symbol here is a whole number of small squares.
Friday and Thursday are the only two rows the question ties together, so the gap between their counts must be worth the $16$ loaves it names.
Divide to find what one small square is worth, then count Monday's small squares and multiply.

Worked Solution [3 marks]

Rule - the key of a pictogram is how many items one symbol stands for. When no key is printed, a stated difference between two rows supplies it: divide the difference in items by the difference in squares, then read every row with the value that gives.

Step 1: count the small squares in the Friday row and in the Thursday row

\text{Friday: } 3 \times 4 + 2 = 14

\text{Thursday: } 1 \times 4 + 2 = 6

(Reason: A whole symbol is

4

small squares and a half symbol is

2

. Friday draws

3

whole symbols and one half symbol, and Thursday draws one whole symbol and one half symbol, so count four for each whole one and two for each half.)

Step 2: the difference between the two rows

14 - 6 = 8

(Reason: Friday's row carries

8

small squares more than Thursday's, and the question says that gap is worth

16

loaves.)

Step 3: the value of one small square

\dfrac{16}{8} = 2

(Reason: Sharing the

16

loaves equally between the

8

extra small squares says what one small square stands for. That is the key the pictogram does not print.)

Step 4: count Monday's small squares

\text{Monday: } 2 \times 4 + 1 = 9

(Reason: Monday draws

2

whole symbols, worth

4

small squares each, and then one small square on its own.)

Step 5: turn Monday's small squares into loaves

9 \times 2 = 18

(Reason: Every small square stands for

2

loaves, so multiply Monday's count of small squares by the key.)

18

loaves

Verification

Check 1: Work in whole symbols instead of small squares. A whole symbol is

4

small squares, so it stands for

8

loaves. Friday draws

3.5

symbols, Thursday draws

1.5

, and Monday draws

2.25

3.5 \times 8 - 1.5 \times 8 = 16

and

2.25 \times 8 = 18

Check 2: Work in half symbols, a third unit again. Friday is

7

halves and Thursday is

3

, so the

16

loaves are spread over four halves. Monday is

4.5

halves.

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

and

4.5 \times 4 = 18

Check 3: Read the whole week with the same key and add it up. The rows hold

9

19

4

6

and

14

small squares, each worth

2

loaves, so every day comes out a whole number of loaves and the two totals must agree.

18 + 38 + 8 + 12 + 28 = 104

and

(9 + 19 + 4 + 6 + 14) \times 2 = 104

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $8$ small squares $= 16$ or $2$ large squares $= 16$ or $2$ small squares $= 4$ or [small square] $= \dfrac{16}{8} \; (= 2)$ or [large square] $= \dfrac{16}{2} \; (= 8)$ or [small square] $= \dfrac{4}{2} \; (= 2)$ or [ $2$ small squares] $= \dfrac{16}{4} \; (= 4)$ or Friday $= 28$ (loaves) and Thursday $= 12$ (loaves)	M1	for starting to work with proportion. May be seen in a square on the pictogram or in working or implied by correct working, or for finding the correct number of loaves sold on Thursday and Friday.	✓
eg $9 \times 2$ oe or $2.25 \times 8$ oe or $4.5 \times 4$ oe. The printed scheme puts the $2$ , the $8$ and the $4$ in quotation marks, so the candidate's own value from the first mark may be used here.	M1	for a complete method to find the loaves sold on Monday.	✓
$18$	A1	cao. A correct answer scores full marks, unless it comes from obvious incorrect working.	✓
No key is printed on this pictogram, so the first mark is the one for producing a key at all.	Note	A common wrong answer is $20$ , from reading Monday's third symbol as a half rather than as a single small square: $2.5 \times 8 = 20$ . That earns the first mark and the second, but not the accuracy mark.	✓

Full marks: 3/3

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

$5 \qquad 12 \qquad 19 \qquad 26$

(i) Write down the fifth term of this sequence. [1 mark]

(ii) Explain how you found that term. [1 mark]

(i)(ii)

[Total 2 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

A number sequence whose first four terms are

5

12

19

and

26

The terms are listed in order, so each one follows on from the term before it.

Find

(i) the fifth term of the sequence. (ii) an explanation, in words, of how that term is reached.

Plan the Solution

Look at what happens from one term to the next, and check that the same thing happens every time.
If the sequence climbs by the same amount each time it is a linear sequence, so the fifth term is the fourth term plus one more of those steps.
Part (ii) asks for the rule in words, so say what is done to a term to reach the one after it.

Worked Solution [2 marks]

Rule - Linear sequences: when the difference between one term and the next is the same all the way along, that difference is the common difference, and the next term is the last term given plus one common difference.

Step 1: find what is added from one term to the next

12 - 5 = 7

19 - 12 = 7

26 - 19 = 7

(Reason: All three gaps are the same, so the sequence climbs in equal steps of

7

. That common difference is what generates every later term.)

Step 2: add the common difference to the last term given

26 + 7 = 33

(Reason: The fifth term comes straight after the fourth term that is printed, so one more step of

7

is added on.)

Step 3: put the rule into words for part (ii)

7n - 2

7 \times 5 - 2 = 33

(Reason: The step is the same every time, so the explanation is simply that

7

is added to the term before. The position-to-term rule

7n - 2

says the same thing in symbols, and it gives the same fifth term.)

(i)

33

(ii) Add

7

to the term before

Verification

Check 1: Work backwards. Taking one step of

7

off the fifth term must land on the fourth term that is printed.

33 - 7 = 26

, which is the last term printed.

Check 2: Count on from the first term instead. Getting from the first term to the fifth takes four steps of

7

5 + 4 \times 7 = 33

, the same term reached a different way.

Check 3: Test the position-to-term rule

7n - 2

on the terms that are printed.

7 \times 1 - 2 = 5

7 \times 2 - 2 = 12

7 \times 3 - 2 = 19

7 \times 4 - 2 = 26

, so the rule fits, and

7 \times 5 - 2 = 33

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(i) $33$	B1	For the answer $33$ .	✓
(ii) Added $7$	B1	Accept eg add $7$ , $(n) + 7$ , $7n - 2$ .	✓

Full marks: 2/2

A music shop keeps a record of how many guitars it sells each week.
The table gives information about the number of guitars sold in each of $30$ weeks.

$\begin{array}{|c|c|}\hline \textbf{Number of guitars sold} & \textbf{Frequency} \\ \hline 1 & 4 \\ \hline 2 & 10 \\ \hline 3 & 5 \\ \hline 4 & 7 \\ \hline 5 & 4 \\ \hline \end{array}$

Work out the mean number of guitars sold per week. [3 marks]

[Total 3 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

A frequency table covering

30

weeks at a music shop, with the number of guitars sold in a week running from

1

5

The frequency column counts weeks:

4

weeks sold one guitar,

10

weeks sold two guitars, and so on down the table.

Find

The mean number of guitars sold per week.

Plan the Solution

Add the frequency column first and check that it comes to $30$ , so the table really does account for every week in the record.
Multiply each number of guitars sold by the frequency beside it, because that frequency says how many weeks sold that many.
Add those products together to get the total number of guitars sold across the whole record.
Divide that total by the total frequency, which shares the guitars equally between the weeks.

Worked Solution [3 marks]

Rule - Mean from a frequency table:

\text{mean} = \dfrac{\text{sum of the products}}{\text{total frequency}}

. Multiply every value by its frequency, add the products, then divide by the total frequency. The frequencies are what make the busy weeks count more than the quiet ones.

Step 1: check the frequencies account for every week

4 + 10 + 5 + 7 + 4 = 30

(Reason: The frequency column counts weeks, so adding it gives the length of the record. It comes to

30

, which matches the

30

weeks in the question, so no week is missing and this total is the divisor to use later.)

Step 2: multiply each number of guitars by its frequency

1 \times 4 = 4

2 \times 10 = 20

3 \times 5 = 15

4 \times 7 = 28

5 \times 4 = 20

(Reason: The second row stands for

10

separate weeks that each sold

2

guitars, so between them those weeks account for

2 \times 10 = 20

guitars. Every row is treated the same way.)

Step 3: add the products to get the total sold

4 + 20 + 15 + 28 + 20 = 87

(Reason: Adding the five products gives the number of guitars sold over all

30

weeks together, which is

87

. Adding the left-hand column instead would only count the different sales figures, not the weeks.)

Step 4: divide the total sold by the total frequency

\dfrac{87}{30} = 2.9

(Reason: The mean shares the guitars equally between the weeks, so the total sold is divided by the total frequency found in step 1. A mean does not have to be a whole number, and a decimal is what a division like this normally gives.)

2.9

guitars per week

Verification

Check 1: Reverse the last step. Multiplying the mean by the

30

weeks has to give the total number of guitars back.

2.9 \times 30 = 87

, the total found in step 3.

Check 2: Work it out a different way, from an assumed mean of

3

. Measure every value against

3

, weight those differences by the frequencies, then correct the assumption.

4 \times (-2) + 10 \times (-1) + 5 \times 0 + 7 \times 1 + 4 \times 2 = -3

, so the mean is

3 + \dfrac{-3}{30} = 2.9

, reached without ever forming the total.

Check 3: Sanity check the size of the answer. A mean must lie between the smallest and the largest value in the table, and it should sit near the values that come up most often.

1 < 2.9 < 5

, and the most common number of sales is

2

, with a second cluster at

4

, so a mean a little below

3

is exactly what to expect.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$1 \times 4 + 2 \times 10 + 3 \times 5 + 4 \times 7 + 5 \times 4 \; (= 87)$ or $4 + 20 + 15 + 28 + 20 \; (= 87)$	M1	For at least $4$ correct products and intention to add. Products may be seen by the side of the table.	✓
" $87$ " divided by $30$ oe	M1	Dep on M1. Allow use of their " $30$ " from adding the frequencies from the table.	✓
$2.9$	A1	Correct answer scores full marks (unless from obvious incorrect working). Accept an answer of $3$ if correct working seen, eg $87$ divided by $30$ oe.	✓

Full marks: 3/3

Draw the graph of $y = 2x - 5$ on the grid below, taking values of $x$ from $-1$ to $4$ . [3 marks]

[Total 3 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

The equation

y = 2x - 5

The values of

x

to use: from

-1

4

A printed grid, which is where the answer goes

Find

The graph of

y = 2x - 5

drawn on the grid, from

x = -1

x = 4

Plan the Solution

Work out $y$ for every whole-number value of $x$ in the range, so the graph is built from points rather than guessed.
Plot each point on the grid, counting the squares across first and then up or down.
Rule one straight line through the points, running it right out to $x = -1$ at one end and $x = 4$ at the other.

Worked Solution [3 marks]

Straight-line graph:

y = 2x - 5

is in the form

y = mx + c

, so its graph is a straight line with gradient

m = 2

and

y

-intercept

c = -5

. Two points fix that line and the others check it.

Step 1: Work out

y

for each value of

x

2 \times (-1) - 5 = -7

2 \times 0 - 5 = -5

2 \times 1 - 5 = -3

2 \times 2 - 5 = -1

2 \times 3 - 5 = 1

2 \times 4 - 5 = 3

(Reason: Each whole-number value of

x

from

-1

4

goes into

y = 2x - 5

in turn: double it, then subtract

5

. Keep the bracket round

(-1)

so the minus sign survives the doubling.)

Step 2: Collect the points into a table

\begin{array}{|c|c|c|c|c|c|c|}\hline x & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & -7 & -5 & -3 & -1 & 1 & 3 \\ \hline\end{array}

(Reason: Each column of the table is one point to plot:

(-1, -7)

(0, -5)

(1, -3)

(2, -1)

(3, 1)

and

(4, 3)

. Every one of them fits on the printed grid, which is a sign the table is right.)

Step 3: Plot the points and rule one straight line through them

\dfrac{3 - (-7)}{4 - (-1)} = \dfrac{10}{5} = 2

(Reason: The points climb

2

squares for every

1

square across, so they lie on one straight line and a ruler through them is the graph. Draw it right across the range, from

x = -1

x = 4

, with no gap at either end: the marks are for the whole segment.)

The straight line through

(-1, -7)

and

(4, 3)

, ruled from

x = -1

x = 4

Verification

Check 1: Read the gradient off the drawn line. From

(-1, -7)

(4, 3)

it rises

10

squares for a run of

5

\dfrac{10}{5} = 2

, which is the gradient

y = 2x - 5

asks for

Check 2: Read where the drawn line crosses the

y

-axis, and compare it with the value the equation gives at

x = 0

2 \times 0 - 5 = -5

, and the line crosses at

(0, -5)

Check 3: Test a point that was not used to fix the two ends. Start at

(-1, -7)

and move

4

squares across, which lifts

y

2 \times 4

-7 + 2 \times 4 = 1

, so the ruled line passes through

(3, 1)

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct line, drawn between $x = -1$ and $x = 4$	B3	for a correct line between $x = -1$ and $x = 4$	✓
A correct line that is short of the full range, or the points plotted and left unjoined	B2	for a correct straight line segment through at least $3$ of $(-1, -7)$ $(0, -5)$ $(1, -3)$ $(2, -1)$ $(3, 1)$ $(4, 3)$ or for all of $(-1, -7)$ $(0, -5)$ $(1, -3)$ $(2, -1)$ $(3, 1)$ $(4, 3)$ plotted but not joined	✓
Some correct points, or a line with one of the two right features	B1	for at least $2$ correct points stated (may be in a table) or for a line drawn with a positive gradient through $(0, -5)$ or for a line with a gradient of $2$	✓

Full marks: 3/3

A sports shop in a Swiss ski resort accepts payment in pounds ( $\pounds$ ) or in Swiss francs.

In the shop, a fleece costs $\pounds 35$ or $42$ Swiss francs.
The cost of a rucksack is $\pounds 54$

Claire works out the cost of the rucksack in Swiss francs.
She uses the same exchange rate that was used for the cost of the fleece.

What is the cost of the rucksack in Swiss francs? [3 marks]

Swiss francs

[Total 3 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

A fleece, priced in both currencies:

\pounds 35

42

Swiss francs

A rucksack, priced in pounds only:

\pounds 54

One exchange rate, used for both items

Find

The cost of the rucksack in Swiss francs

Plan the Solution

The fleece is the only item with a price in each currency, so it is the fleece that fixes the exchange rate. Work the rate out from its two prices first.
Then apply that rate to the rucksack's price of $\pounds 54$ .
Decide which way round the rate goes before reaching for the calculator: francs for each pound, or pounds for each franc. The two are reciprocals, and they give very different answers.

Worked Solution [3 marks]

Exchange rate: an item priced in both currencies fixes the rate for everything else in the shop. Dividing its price in francs by its price in pounds gives the number of francs one pound is worth, and every other price in pounds is then multiplied by that same number.

Step 1: Use the fleece to find the exchange rate

\dfrac{42}{35} = 1.2

(Reason: The fleece's two prices are the same amount of money, so

\pounds 35

and

42

Swiss francs are worth exactly the same. Sharing those francs out over the pounds says how many francs a single pound is worth:

1.2

of them.)

Step 2: Apply the rate to the rucksack

54 \times 1.2 = 64.8

(Reason: Each of the

54

pounds the rucksack costs is worth

1.2

Swiss francs, so the price in francs is

54

lots of

1.2

. The rate multiplies a price in pounds. Dividing by it would send the conversion back the other way and turn francs into pounds, which is the commonest slip on this question.)

64.8

Swiss francs

Verification

Check 1: Turn the answer back into pounds at the same rate. If

64.8

Swiss francs really is the rucksack's price, dividing it by the rate must give back

\pounds 54

\dfrac{64.8}{1.2} = 54

, the price the question started from

Check 2: Compare the two items instead of the two currencies. The rucksack costs

\dfrac{54}{35}

times as much as the fleece, and that must hold whichever currency the prices are written in.

42 \times \dfrac{54}{35} = 64.8

Check 3: Cancel the rate to whole numbers and avoid decimals altogether.

\pounds 35

42

francs cancels by

7

\pounds 5

6

francs, and

\pounds 54

\dfrac{54}{5}

lots of

\pounds 5

6 \times \dfrac{54}{5} = 64.8

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $\dfrac{42}{35}$ $(=1.2)$ or $\dfrac{35}{42}$ $(=0.833...)$ or $\dfrac{54}{35}$ $(=1.54...)$ or $\dfrac{35}{54}$ $(=0.648...)$	M1	a method to find a correct ratio	✓
eg $54 \times 1.2$ or $\dfrac{54}{0.833...}$ or $42 \times 1.54...$ or $\dfrac{42}{0.648...}$	M1	for a complete method, with the candidate's own ratio from the first mark used in place of each quoted rate	✓
$64.8$ - correct answer scores full marks (unless from obvious incorrect working)	A1	accept $64.80$	✓

Full marks: 3/3

The radius of a circle is $6.4$ cm.

Work out the area of this circle.
Give your answer correct to $3$ significant figures.
[2 marks]

cm²

[Total 2 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

A circle whose radius is

r = 6.4

cm.

The length given is the radius, not the diameter, so nothing needs halving first.

Find

The area of the circle. The answer is wanted correct to

3

significant figures.

Plan the Solution

Use the area formula for a circle, $A = \pi r^2$ .
Square the radius first, then multiply that by $\pi$ .
Keep the calculator's full value all the way through, and round only at the very end to $3$ significant figures.

Worked Solution [2 marks]

Rule - Area of a circle:

A = \pi r^2

, where

r

is the radius.

Step 1: Square the radius

r^2 = 6.4^2 = 40.96

(Reason: The formula multiplies

\pi

r^2

, and squaring means multiplying the radius by itself.)

Step 2: Multiply the squared radius by

\pi

A = \pi \times 40.96

A = 128.6796351\ldots

(Reason: The calculator's own value for

\pi

is used, so every digit is still there when the rounding happens.)

Step 3: Round to

3

significant figures

A = 129 \text{ cm}^2

(Reason: The third significant figure of

128.6796351\ldots

is the

8

, and the digit after it is

6

, so the

8

rounds up.)

129 \text{ cm}^2

Verification

Check 1: Work the calculation backwards:

\sqrt{\dfrac{A}{\pi}}

must give back the radius the question started with. The radius comes back as

6.4

cm, which is the length given.

Check 2: Redo the calculation with

\dfrac{22}{7}

in place of

\pi

, the approximation the mark scheme allows.

\dfrac{22}{7} \times 40.96 \approx 128.7

, which still rounds to

129

Check 3: Squeeze it. The circle fits inside a square of side

12.8

cm, and it contains the square whose diagonal is

12.8

cm, so the area must lie between those two square areas.

81.92 < 129 < 163.84

, so the answer is the size a circle of this radius has to be.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $\pi \times 6.4^2 \; \left(= \dfrac{1024}{25}\pi\right)$	M1	allow $3.14\ldots$ or $\dfrac{22}{7}$ for $\pi$	✓
$129$ Correct answer scores full marks (unless from obvious incorrect working)	A1	accept $128$ to $129$	✓

Full marks: 2/2

The diagram shows a plan of a paddock made from three identical rectangles.

The length of each rectangle is $6$ metres.
The width of each rectangle is $3.5$ metres.

Martin puts a fence around the perimeter of the paddock.
He charges $7.60$ euros for each $1$ metre of fence.

Work out how much Martin charges in total for the fence. [4 marks]

euros

[Total 4 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

Three identical rectangles, each

6

m long and

3.5

m wide, joined to make the paddock in the diagram

A fence right round the outside of the paddock, charged at

7.60

euros for each

1

metre

Find

The total charge for the fence, in euros

Plan the Solution

Work out the length of every side that goes round the outside of the shape.
The two short steps are not printed on the diagram, so get them from $6 - 3.5$ .
Add the eight outside lengths to get the perimeter of the paddock.
Multiply that perimeter by $7.60$ euros for each metre.

Worked Solution [4 marks]

Rule - Perimeter: the perimeter is the total distance all the way round the outside of a shape. Every outside edge is counted once, and a line drawn inside the shape is not counted at all.

Step 1: Find the short step at each end of the outline

6 - 3.5 = 2.5

(Reason: (Reason: where one rectangle drops below its neighbour, the exposed piece is what is left of a

6

m length once

3.5

m of it is covered by the rectangle beside it.))

Step 2: Find the two long sides of the outline

6 + 3.5 = 9.5

(Reason: (Reason: the top of the shape runs along the

6

m length of the middle rectangle and then the

3.5

m width of the right-hand one. The bottom does the same on the other two rectangles.))

Step 3: Add the eight sides that go round the outside

3.5 + 2.5 + 9.5 + 6 + 3.5 + 2.5 + 9.5 + 6 = 43

(Reason: (Reason: the fence follows the outline only, so each outside edge is counted once and the two dashed lines inside the paddock are not fence at all.))

Step 4: Charge for every metre of that perimeter

43 \times 7.60 = 326.80

(Reason: (Reason: the charge is

7.60

euros for each

1

metre, so the total is the number of metres multiplied by

7.60

.))

326.80

euros

Verification

Check 1 - count the fence in pieces of each size: Going round the outline, four of the sides are

6

m, four are

3.5

m, and the two short steps are

2.5

m each.

4 \times 6 + 4 \times 3.5 + 2 \times 2.5 = 43

m, the same perimeter

Check 2 - build the perimeter from three separate rectangles: One rectangle on its own has perimeter

2 \times (6 + 3.5) = 19

m, so three separate rectangles give

57

m. Each of the two joins hides a

3.5

m width from both of the rectangles that meet there.

57 - 4 \times 3.5 = 43

m, the same perimeter

Check 3 - split the charge into two easier pieces: Charge for the first

40

m, then for the remaining

3

m, and add the two.

40 \times 7.60 = 304

and

3 \times 7.60 = 22.80

, so

304 + 22.80 = 326.80

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$6 - 3.5 = 2.5$ or $4 \times 6 + 4 \times 3.5 = 38$ or $6 \times 6 + 6 \times 3.5 = 57$ or $2 \times 9.5 \times 3 = 57$ or $19 \times 3 = 57$	M1	for a method to find the missing length (may be shown on the diagram) or for a method to find the length of the solid lines excluding the $2.5$ , may include extra sides added, or for a method to find the perimeter of the $3$ rectangles	✓
$4 \times 6 + 4 \times 3.5 + 2 \times 2.5 = 43$ or $6 \times 6 + 6 \times 3.5 - 4 \times 3.5 = 43$ or $2 \times 9.5 \times 3 - 4 \times 3.5 = 43$ or $19 \times 3 - 4 \times 3.5 = 43$	M1	for a complete method to find the perimeter of the shape. The printed scheme writes the $2.5$ in quotation marks, so the candidate's own earlier value may be used here.	✓
eg $43 \times 7.6$	M1	for a method to find the cost, allow use of their $43$ as long as it is from adding at least $4$ correct lengths including a length of $3.5$ and a length of $6$ , eg $57 \times 7.6(0)$ or $38 \times 7.6(0)$	✓
$326.8(0)$	A1	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 4/4

Continue to questions 15 to 26

The remaining 12 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 1FR and Paper 1HR, 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 4MA1/1FR, Thursday 15 May 2025: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 14 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

Question 14, Calculator allowed

Question 14 - Exam Solution

Frequently asked questions

Keep revising

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