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Edexcel IGCSE 4MA1/2H, Monday 3 June 2024: Worked Solutions and Mark Schemes

Sir Faraz Hassan

12 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)4MA1/2H - Higher Tier - Monday 3 June 2024100 marks · 2 hours · Calculator allowed

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/2H (Higher Tier), June 2024 series, sat Monday 3 June 2024 –100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

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

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

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

Worked solutions, questions 1 to 15 of 25

Eight numbers are written below in order of size.

$h \quad 6 \quad 7 \quad 8 \quad j \quad 16 \quad k \quad k$

where $h$ , $j$ and $k$ are integers.

For these eight numbers:
the median is $10$
the mode is $18$
the range is $13$

Find the value of $h$ , the value of $j$ and the value of $k$ . [3 marks]

h =j =k =

[Total 3 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

Eight numbers, in order of size:

h \quad 6 \quad 7 \quad 8 \quad j \quad 16 \quad k \quad k

h

j

and

k

are integers, and the list is already ordered, so

h

is the smallest number and

k

is the largest.

The median is

10

, the mode is

18

and the range is

13

Find

The value of

h

. The value of

j

. The value of

k

Plan the Solution

Take the three averages one at a time, and start with the one that gives a value on its own. Only $k$ is written twice, so the repeated value in the list is $k$ and the mode names it immediately.
Then the median. There are eight numbers, an even amount, so the median sits halfway between the 4th and the 5th, which are $8$ and $j$ .
Finish with the range, which links the two ends of an ordered list, so it ties $h$ to the $k$ already found.
Write the completed list out at the end and read the three averages back off it, because the values must also leave the list in order of size.

Worked Solution [3 marks]

Rule - For a list already written in order of size: the median of an even amount of numbers is the mean of the two middle ones, the mode is the value that occurs most often, and the range is

\text{largest} - \text{smallest}

Step 1: use the mode to find

k

k = 18

(Reason: (Reason: the mode is the value that occurs most often. Every entry in the list is written once except

k

, which is written twice, so the mode has to be

k

itself. The mode is given as

18

.))

Step 2: use the median to find

j

\dfrac{8 + j}{2} = 10

8 + j = 20

j = 12

(Reason: (Reason: with eight numbers in order the median is the mean of the 4th and the 5th, which are

8

and

j

. Multiplying the median by

2

gives

20

for the two of them together, and taking the

8

away leaves

12

.))

Step 3: use the range to find

h

k - h = 13

18 - h = 13

h = 18 - 13 = 5

(Reason: (Reason: the list is in order of size, so the largest number is

k

, now known to be

18

, and the smallest is

h

. The range is the largest take away the smallest, so

h

sits

13

below

18

.))

Step 4: write the completed list out

5 \quad 6 \quad 7 \quad 8 \quad 12 \quad 16 \quad 18 \quad 18

(Reason: (Reason: the question says the eight numbers are in order of size, so that has to still be true once the three values are put back. Reading left to right, the list never goes down, so the answers are consistent with the way the question presents them.))

h = 5

j = 12

k = 18

Verification

Check 1: Put the three values back and find the middle pair of the completed list. The 4th and 5th numbers are

8

and

12

, so take their mean.

\dfrac{8 + 12}{2} = 10

Check 2: The completed list runs from

5

up to

18

, so subtract the smallest from the largest.

18 - 5 = 13

Check 3: Count how often each value occurs in the completed list, then read it from left to right to confirm it never goes down. Only

18

occurs twice, so the mode is

18

, and

5 \quad 6 \quad 7 \quad 8 \quad 12 \quad 16 \quad 18 \quad 18

is in order of size.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
One of the three values found, or a correct statement leading to one of them	M1	For a correct value for $h$ , $j$ or $k$ , or for a correct statement for one of them, e.g. $k = 18$ , or $\dfrac{8 + j}{2} = 10$ , or $8 + j = 2 \times 10$ , or $10 \times 2 - 8$ , or $j = 12$ , or $k - h = 13$ , or $18 - h = 13$ , or $h = 5$ . The scheme quotes the $18$ in that last statement, so the candidate's own value of $k$ may stand in its place.	✓
Two of the three values found, or correct statements for two of them	M1	For $2$ correct values from $h$ , $j$ or $k$ , or for $2$ correct statements for them.	✓
All three values correct	A1	$h = 5$ , $j = 12$ and $k = 18$ . All three are needed for this mark.	✓
A correct answer written down with little or no working	Note	A correct answer scores full marks, unless it clearly comes from obviously incorrect working.	✓

Full marks: 3/3

(a) On the grid below, draw the straight line whose equation is
(i) $y = 2$
(ii) $x = 6$
(iii) $y = x + 1$
Write the equation of each line beside the line you have drawn. [3 marks]

(b) On the same grid, shade the one region that satisfies all three of these inequalities at once
$y \geq 2 \quad\quad x \leq 6 \quad\quad y \leq x + 1$
Write the letter $R$ inside the region you have shaded. [1 mark]

[Total 4 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

A square grid whose

x

-axis and

y

-axis are both numbered from

0

8

Three equations to draw:

y = 2

x = 6

and

y = x + 1

Three inequalities to satisfy together:

y \geq 2

x \leq 6

and

y \leq x + 1

Find

Part (a): each of the three lines drawn on the grid, with its equation written beside it. Part (b): the one region where all three inequalities hold at once, shaded and marked

R

Plan the Solution

Sort the three equations by shape before drawing anything. $y = 2$ has no $x$ in it, so every point on it sits at the same height and the line is horizontal. $x = 6$ has no $y$ in it, so it is vertical. $y = x + 1$ has both letters, so it slopes.
A straight line needs only two points. Take the two edges of the grid for the first two lines, and substitute two values of $x$ for the third.
Each inequality keeps one side of its own line. Settle which side by testing a single point that is not on any of the lines, rather than by reading the direction off the sign.
The region is where all three sides overlap. Work out the three points where the boundaries cross, so the shading can be drawn exactly, then shade it and mark it $R$ .

Worked Solution [4 marks]

Rule - A line

y = a

is horizontal, a line

x = a

is vertical, and a line

y = mx + c

slopes. An inequality keeps one whole side of its boundary line, and which side is settled by testing one point that is not on the line.

Step 1: draw

y = 2

, the horizontal line

y = 2

(0, 2) \quad \text{and} \quad (8, 2)

(Reason: (Reason: the equation fixes the height and says nothing at all about

x

, so every point

2

units above the

x

-axis lies on it. Joining

(0, 2)

(8, 2)

carries the line right across the grid, which is well over the

2

cm the mark scheme asks for.))

Step 2: draw

x = 6

, the vertical line

x = 6

(6, 0) \quad \text{and} \quad (6, 8)

(Reason: (Reason: this time the equation fixes how far across the point is and leaves the height free, so the line runs straight up from

(6, 0)

(6, 8)

. It is the mirror image of Step 1: no

y

in the equation means vertical, no

x

means horizontal.))

Step 3: draw

y = x + 1

, the sloping line

x = 0: \quad y = 0 + 1 = 1

x = 7: \quad y = 7 + 1 = 8

(0, 1) \quad \text{and} \quad (7, 8)

(Reason: (Reason: both letters appear, so pick two values of

x

and work out the matching

y

x = 8

is no use here because it would give

y = 9

, which is off the top of the grid, so

x = 7

is the last value that still lands on the paper.))

Step 4: test the point

(4, 3)

to fix which side of each line to keep

y \geq 2 \text{ gives } 3 \geq 2

x \leq 6 \text{ gives } 4 \leq 6

y \leq x + 1 \text{ gives } 3 \leq 4 + 1

(Reason: (Reason:

(4, 3)

is not on any of the three lines, so it is strictly on one side of each of them. All three inequalities are true there, so the side to keep is above

y = 2

, to the left of

x = 6

and below

y = x + 1

. Testing a point is safer than reading the direction off the sign, because below and above only mean anything once the line is on the paper.))

Step 5: find the corners, then shade and label the region

y = 2 \text{ meets } y = x + 1: \quad 2 = x + 1 \implies x = 1

y = 2 \text{ meets } x = 6: \quad (6, 2)

x = 6 \text{ meets } y = x + 1: \quad y = 6 + 1 = 7

(1, 2), \quad (6, 2), \quad (6, 7)

(Reason: (Reason: two boundaries cross at every corner, so taking the lines in pairs gives the three corners exactly rather than by eye. The first pair gives the corner on the left, the second the corner on the right, and the third the corner at the top. Shade the triangle those three points enclose and write

R

inside it, well clear of the edges.))

(a)(i)

y = 2

drawn from

(0, 2)

(8, 2)

(a)(ii)

x = 6

drawn from

(6, 0)

(6, 8)

(a)(iii)

y = x + 1

drawn from

(0, 1)

(7, 8)

(b)

R

is the triangle with corners

(1, 2)

(6, 2)

and

(6, 7)

Verification

Check 1: Take a point well inside the shaded triangle,

(4, 3)

, and put it into all three inequalities.

3 \geq 2

4 \leq 6

and

3 \leq 4 + 1

, so

(4, 3)

really is in the region.

Check 2: Step just across each boundary in turn and confirm the point leaves the region:

(4, 1)

below

y = 2

(7, 3)

to the right of

x = 6

, and

(2, 5)

above

y = x + 1

1 < 2

7 > 6

and

5 > 3

, so each point breaks exactly one inequality and all three lines are genuine edges of the region.

Check 3: Count the whole-number points of the region column by column. The column at

x

runs from

y = 2

up to

y = x + 1

, which is

x

points, and the columns run from

x = 1

x = 6

1 + 2 + 3 + 4 + 5 + 6 = 21

points, the highest of them at

(6, 7)

, which is the top corner found in Step 5.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) the line $y = 2$ drawn	B1	For the horizontal line $y = 2$ drawn on the grid.	✓
(a)(ii) the line $x = 6$ drawn	B1	For the vertical line $x = 6$ drawn on the grid.	✓
(a)(iii) the line $y = x + 1$ drawn	B1	For the sloping line $y = x + 1$ drawn on the grid, through $(0, 1)$ and $(7, 8)$ .	✓
Guidance on the three lines of part (a)	Note	A line may be solid, dotted or dashed. Each one must be at least $2$ cm long, and the lines need not be labelled to earn the marks in part (a).	✓
(b) the correct region shaded and marked $R$	B1ft	For the correct region indicated. The follow-through is dependent on at least $2$ of the $3$ marks in part (a), and on a vertical line, a horizontal line and a sloping line of positive gradient all having been drawn.	✓
(b) special case: the two letters read the wrong way round	SC B1	For $y = x + 1$ , $y = 6$ and $x = 2$ drawn, with the region those three enclose shaded. That is what comes out when $y \geq 2$ and $x \leq 6$ are read as $x \geq 2$ and $y \leq 6$ , swapping the two letters and leaving the third inequality alone.	✓

Full marks: 4/4

A cargo aircraft flies from Dubai to Tokyo.
The flight takes $9$ hours $36$ minutes.
The average speed of the aircraft for the whole flight is $820$ km/h.
Work out the total distance the aircraft flies. [3 marks]

[Total 3 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

Time for the flight:

9

hours

36

minutes

Average speed for the whole flight:

820

km/h

Find

The total distance the aircraft flies, in kilometres

Plan the Solution

The speed is given in kilometres per hour, so the time has to be in hours before the two can be multiplied together.
Write the $36$ minutes as a fraction of an hour, then add it to the $9$ whole hours.
Multiply the speed by that time in hours.

Worked Solution [3 marks]

Rule - Speed, distance and time:

\text{distance} = \text{speed} \times \text{time}

, with the time measured in hours whenever the speed is in kilometres per hour.

Step 1: Write the

36

minutes as a fraction of an hour

\dfrac{36}{60} = 0.6

(Reason: An hour is

60

minutes, so

36

minutes is

\dfrac{36}{60}

of an hour, and that fraction cancels to

0.6

of an hour.)

Step 2: Write the whole flight time in hours

9 + 0.6 = 9.6

(Reason: The time is now measured in the same unit as the speed. Writing

9.36

here is the classic trap: that reads the

36

as hundredths of an hour rather than as minutes.)

Step 3: Multiply the speed by the time

820 \times 9.6 = 7\,872 \text{ km}

(Reason: The aircraft covers

820

kilometres in each hour of the flight, so multiplying the speed by the number of hours gives the whole distance.)

7\,872

Verification

Check 1 - divide the distance back by the speed: Dividing the distance by the speed has to give the time back. If it does not come out as

9.6

hours, the multiplication was wrong.

\dfrac{7\,872}{820} = 9.6

hours, which is

9

hours

36

minutes

Check 2 - split the flight into the whole hours and the last part hour: In

9

hours the aircraft covers

820 \times 9 = 7\,380

km, and in the final

0.6

of an hour it covers

820 \times 0.6 = 492

km.

7\,380 + 492 = 7\,872

km, the same total reached by a different split

Check 3 - is the size sensible: Round the speed down to

800

km/h. The distance should then come out a little below the answer, and a flight of nearly ten hours should cover a few thousand kilometres.

800 \times 9.6 = 7\,680

km, just below

7\,872

km, so the answer is the right size

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Convert the time into hours (or into minutes)	M1	For a correct conversion of the time, e.g. $9.6$ hours, or $9\dfrac{36}{60}$ hours, or $9\dfrac{3}{5}$ hours oe, or $576$ minutes	✓
Use $\text{speed} \times \text{time}$ with the time in hours	M1	e.g. $820 \times 9.6$ , or $820 \times \dfrac{576}{60}$ , or $576 \times \dfrac{820}{60}$ , or $576 \times \dfrac{41}{3}$ (allow $13.7$ for $\dfrac{41}{3}$ ) oe. Allow the use of $9.36$ for this mark. Award M2 for the split method $820 \times 9 + \dfrac{820}{60} \times 36 = 7\,380 + 492$ , or for $\dfrac{34\,560}{60 \times 60} \times 820$ oe	✓
Answer	A1	$7\,872$ . A correct answer scores full marks, unless it comes from obviously incorrect working	✓
Special case	SC B1	Award one mark for $7\,675.2$ if no other marks are awarded. That value is $820 \times 9.36$ , the distance produced by reading $9$ hours $36$ minutes as $9.36$ hours instead of $9.6$ hours	✓

Full marks: 3/3

Show that $2\dfrac{4}{7} \times 3\dfrac{1}{9} = 8$
You must show all your working. [3 marks]

[Total 3 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

The first mixed number:

2\dfrac{4}{7}

The second mixed number:

3\dfrac{1}{9}

The value the product has to be shown to equal:

8

Find

Working that reaches exactly

8

, not a decimal that rounds to it. On a show-that question the working IS the answer, so every stage has to be written down.

Plan the Solution

Mixed numbers cannot be multiplied a piece at a time. Multiplying the whole numbers together and the fractions together would give only two of the four products that a multiplication of two sums produces, and it comes to a value nowhere near the target.
So turn each mixed number into a single improper fraction first.
Then cancel before multiplying: $18$ and $9$ share a factor of $9$ , and $28$ and $7$ share a factor of $7$ , which leaves a multiplication of two whole numbers.

Worked Solution [3 marks]

Rule - Multiplying mixed numbers: rewrite each mixed number as an improper fraction using

\dfrac{\text{whole} \times \text{denominator} + \text{numerator}}{\text{denominator}}

, cancel any factor common to a numerator and a denominator, then multiply the numerators together and the denominators together.

Step 1: Write each mixed number as an improper fraction

2\dfrac{4}{7} = \dfrac{2 \times 7 + 4}{7} = \dfrac{18}{7}

3\dfrac{1}{9} = \dfrac{3 \times 9 + 1}{9} = \dfrac{28}{9}

(Reason: One whole is

\dfrac{7}{7}

in the first number and

\dfrac{9}{9}

in the second, so the whole number is multiplied by the denominator and the numerator is added to it. The denominator itself never changes.)

Step 2: Cancel the common factors before multiplying

\dfrac{18}{7} \times \dfrac{28}{9} = \dfrac{2}{1} \times \dfrac{4}{1}

(Reason: The two numerators are multiplied together and the two denominators are multiplied together, so any factor above a line may be cancelled with any factor below either line. Here

18

and

9

both divide by

9

, leaving

2

and

1

; and

28

and

7

both divide by

7

, leaving

4

and

1

Step 3: Multiply what is left

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

(Reason: Both denominators are now

1

, so the product is a whole number, and it is the

8

the question asked to be shown.)

2\dfrac{4}{7} \times 3\dfrac{1}{9} = \dfrac{18}{7} \times \dfrac{28}{9} = 8

as required

Verification

Check 1 - multiply straight out, with no cancelling: Cancelling is a shortcut, so the long way must give the same thing. Multiply the numerators and the denominators as they stand:

18 \times 28 = 504

and

7 \times 9 = 63

\dfrac{504}{63} = 8

, because

8 \times 63 = 504

Check 2 - work backwards from the answer: If the product really is

8

, then dividing

8

by the second mixed number must give the first one back. Dividing by

\dfrac{28}{9}

is the same as multiplying by

\dfrac{9}{28}

8 \times \dfrac{9}{28} = \dfrac{72}{28} = \dfrac{18}{7}

, which is

2\dfrac{4}{7}

again

Check 3 - expand instead of converting: Treat each mixed number as a sum and multiply out all four products of

\left(2 + \dfrac{4}{7}\right)\left(3 + \dfrac{1}{9}\right)

. This route never forms an improper fraction at all, so it tests the conversions rather than repeating them.

6 + \dfrac{2}{9} + \dfrac{12}{7} + \dfrac{4}{63} = 8

, and the two middle terms are exactly what is lost by multiplying only the whole numbers and only the fractions

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Write both mixed numbers as improper fractions	M1	For correct improper fractions $\dfrac{18}{7}$ and $\dfrac{28}{9}$	✓
Cancel, or show a clear intention to multiply	M1dep	For cancelling the fractions fully, or for cancelling partially with a clear intention to multiply, or for not cancelling at all with a clear intention to multiply. An arithmetic error in the multiplication is allowed. For example $\dfrac{18}{7} \times \dfrac{28}{9} = \dfrac{504}{63}$ oe, or a partial cancel such as $\dfrac{2 \times 28}{7} = \dfrac{56}{7}$ , or both fractions written over $63$ as $\dfrac{162}{63} \times \dfrac{196}{63} = \dfrac{31\,752}{3\,969}$ oe	✓
Reach the given value from fully correct working	A1	For a correct answer of $8$ from fully correct working. Working is required: the value is printed in the question, so an unsupported $8$ earns nothing	✓
Note - the alternative presentation	Note	A candidate may write $8 = \dfrac{8}{1}$ , possibly under the given $8$ , and then need only show that their fraction comes to $\dfrac{8}{1}$	✓

Full marks: 3/3

Triangle $ABC$ is shown in the diagram below.

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

x =

[Total 3 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

Triangle

ABC

, with the right angle at

B

AC = 6.5

cm, the side facing the right angle

The angle at

A

34^\circ

BC = x

Find

The value of

x

, correct to one decimal place

Plan the Solution

Stand at the $34^\circ$ angle and name the two sides that matter from there: $AC$ is the hypotenuse and $BC$ is the side opposite the angle.
Opposite together with hypotenuse is the sine ratio, so that is the ratio to write down.
Rearrange to make $x$ the subject, evaluate it with the calculator in degree mode, then round to one decimal place.

Worked Solution [3 marks]

Sine ratio in a right-angled triangle:

\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}

, which rearranges to

\text{opposite} = \text{hypotenuse} \times \sin \theta

Step 1: Name the sides from the angle you know

AC = 6.5\text{ cm}

BC = x\text{ cm}

(Reason: The right angle sits at

B

, so the side facing it,

AC

, is the hypotenuse. Standing at

A

, the side

BC

faces the

34^\circ

angle, so it is the opposite side.)

Step 2: Write the sine ratio for the

34^\circ

angle

\sin 34^\circ = \dfrac{x}{6.5}

(Reason: Opposite over hypotenuse is the sine ratio, and here that is

x

over

6.5

. Only the two lengths named in Step 1 appear, so no other side is needed.)

Step 3: Make

x

the subject and evaluate

x = 6.5 \times \sin 34^\circ = 3.63475\ldots

(Reason: Multiplying both sides of the ratio by

6.5

leaves

x

on its own. Keep the full display value for now, with the calculator in degree mode, and round only at the last step.)

Step 4: Round to one decimal place

x = 3.6

(Reason: The digit in the second decimal place is

3

, which is below

5

, so the digit in the first decimal place is left alone.)

x = 3.6

Verification

Check 1: Work backwards. Put the unrounded answer back into the ratio and take the inverse sine: the angle at

A

has to come back as

34^\circ

\sin^{-1}\left(\dfrac{3.63475}{6.5}\right) = 34.0^\circ

Check 2: Test the triangle a different way, with no trigonometry in the test itself. The other short side is

AB = 5.38874\ldots

, so the squares of the two short sides must add to the square of the hypotenuse, and

6.5^2 = 42.25

5.38874^2 + 3.63475^2 = 42.25

Check 3: A size check. The angle is under

45^\circ

, so the side facing it must be the shorter of the two short sides, and both must be shorter than the hypotenuse.

3.6 < 5.4 < 6.5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Write a correct trigonometric statement for $x$	M1	$\sin 34^\circ = \dfrac{x}{6.5}$ , or $\dfrac{x}{\sin 34^\circ} = \dfrac{6.5}{\sin 90^\circ}$ , or $\cos 56^\circ = \dfrac{x}{6.5}$ , or $6.5^2 - (6.5 \times \cos 34^\circ)^2$ , or equivalent	✓
A fully correct method to find $x$	M1	$x = 6.5 \times \sin 34^\circ$ , or $x = \dfrac{6.5 \times \sin 34^\circ}{\sin 90^\circ}$ , or $x = \sqrt{6.5^2 - (6.5 \times \cos 34^\circ)^2}$ , or $x = 6.5 \times \cos 56^\circ$ , or equivalent	✓
The value of $x$ , correct to one decimal place	A1	awrt $3.6$	✓
Correct answer seen with no working	Note	A correct answer scores full marks, unless it comes from working that is obviously incorrect.	✓

Full marks: 3/3

A cable car moves at a steady speed of $w$ metres per second.
Work out the speed of the cable car in kilometres per hour.
Give your answer in terms of $w$ in its simplest form. [3 marks]

kilometres per hour

[Total 3 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

The speed of the cable car is

w

metres per second.

There are

60

seconds in a minute and

60

minutes in an hour.

There are

1000

metres in

1

kilometre.

Find

The same speed written in kilometres per hour, in terms of

w

, in its simplest form.

Plan the Solution

Change the time unit first: multiply by $3600$ , because one hour is $60 \times 60$ seconds.
Change the distance unit next: divide by $1000$ , because one kilometre is $1000$ metres.
Simplify the single multiplier $\dfrac{3600}{1000}$ and write the answer as a multiple of $w$ .

Worked Solution [3 marks]

Rule - Unit conversion: there are

3600

seconds in one hour and

1000

metres in one kilometre, so a speed in metres per second becomes a speed in kilometres per hour by multiplying by

3600

and then dividing by

1000

Step 1: Change the seconds into hours

60 \times 60 = 3600

w \times 3600 = 3600w

(Reason: There are

60

seconds in a minute and

60

minutes in an hour, so one hour is

3600

seconds. In one hour the cable car covers

3600

lots of

w

metres.)

Step 2: Change the metres into kilometres

\dfrac{3600w}{1000} \text{ kilometres per hour}

(Reason: There are

1000

metres in

1

kilometre, so the number of kilometres is the number of metres over

1000

Step 3: Simplify the multiplier

\dfrac{3600}{1000} = \dfrac{18}{5} = 3.6

\dfrac{3600w}{1000} = 3.6w

(Reason: Dividing the top and the bottom by

200

turns

\dfrac{3600}{1000}

into

\dfrac{18}{5}

, which is

3.6

. The

w

is carried through untouched, so the speed is

3.6

times

w

3.6w

kilometres per hour

Verification

Check 1: Try a real speed. Put

w = 10

: in one hour the cable car covers

10 \times 3600 = 36000

metres, and

36000

metres is

36

kilometres.

3.6 \times 10 = 36

Check 2: Convert the answer back the other way. At

3.6w

kilometres per hour the cable car covers

3600w

metres in

3600

seconds.

\dfrac{3600w}{3600} = w

Check 3: Compare with the fraction form the mark scheme also accepts.

\dfrac{18}{5}

and

3.6

are the same number, so

\dfrac{18}{5}w

and

3.6w

are the same answer.

\dfrac{18}{5} = 3.6

Mark Scheme Breakdown

Step	Mark	Description	Got it?
One correct conversion, or the combined multiplier on its own	M1	For one of $\dfrac{w}{1000}$ , $\dfrac{w}{10^{3}}$ , $w \times 10^{-3}$ , $0.001w$ oe, or $(w \times 60 \times 60)$ oe, or $w \times 3600$ or $\dfrac{w}{\dfrac{1}{3600}}$ oe. Also for $\dfrac{3600}{1000}$ or $\dfrac{18}{5}$ or $3.6$ oe, without a link to $w$ .	✓
A fully correct method, including $w$	M1	For $\dfrac{w \times 60 \times 60}{1000}$ oe, for example $w \times \dfrac{3600}{1000}$ .	✓
The simplified answer	A1	For $3.6w$ , or $\dfrac{18}{5}w$ , or $3\dfrac{3}{5}w$ ; allow $3.6 \times w$ . A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 3/3

The diagram shows the six-sided shape $ABCDEF$ .

$AF = 21 \text{ cm}$
$CD = 15 \text{ cm}$
$AB = FE = 13 \text{ cm}$

$CD$ is parallel to $AF$ .
The perpendicular height of the shape is $h$ cm.
The shape has an area of $390 \text{ cm}^2$ .

Find the value of $h$ . [4 marks]

h =

[Total 4 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

AF = 21 \text{ cm}

CD = 15 \text{ cm}

AB = FE = 13 \text{ cm}

CD

is parallel to

AF

, and the figure marks a right angle at

A

and at

F

The area of

ABCDEF

390 \text{ cm}^2

, and its perpendicular height is

h

Find

The value of

h

, the perpendicular distance from

AF

up to

CD

Plan the Solution

The right angles at $A$ and $F$ make $ABEF$ a rectangle, so cut the shape along $BE$ into a rectangle and a trapezium.
Work out the rectangle's area and take it off $390$ . What is left is the trapezium $BCDE$ .
The trapezium's parallel sides are known, so its area formula gives its height.
Add that height to the $13$ cm below $BE$ to reach $h$ .

Worked Solution [4 marks]

Rule - split a composite shape into parts whose areas you can write down. Rectangle: area is length times width. Trapezium: area is

\dfrac{1}{2}(a + b) \times \text{height}

, where

a

and

b

are the two parallel sides.

Step 1: the area of rectangle

ABEF

21 \times 13 = 273

(Reason: The right angles at

A

and

F

make

AB

and

FE

both perpendicular to

AF

, and the question says they are equal, so

ABEF

is a rectangle

21

cm by

13

cm.)

Step 2: the area left for trapezium

BCDE

390 - 273 = 117

(Reason: The rectangle and the trapezium together make the whole shape, so whatever is left of the

390 \text{ cm}^2

belongs to

BCDE

Step 3: the trapezium's parallel sides

\dfrac{1}{2} \times (21 + 15) = 18

(Reason:

BE

is the same length as

AF

, so the two parallel sides of the trapezium are

21

cm and

15

cm. Halving their sum gives the number that multiplies the trapezium's height.)

Step 4: the height of the trapezium

\dfrac{117}{18} = 6.5

(Reason: Dividing the trapezium's area by that

18

leaves its height, in centimetres.)

Step 5: the perpendicular height of the whole shape

13 + 6.5 = 19.5

(Reason:

h

runs the whole way from

AF

up to

CD

, so it is the rectangle's

13

cm plus the trapezium's

6.5

cm.)

h = 19.5

Verification

Check 1: Put the two parts back together: the rectangle is

273

and the trapezium is

18 \times 6.5

273 + 117 = 390

Check 2: Find the area a completely different way. Draw the rectangle that encloses the shape,

21 \times 19.5 = 409.5

, then cut off the two corner triangles: their bases add to

21 - 15 = 6

and each has height

19.5 - 13 = 6.5

409.5 - \dfrac{1}{2} \times 6 \times 6.5 = 390

Check 3: That second method turns the question into the equation

18h + 39 = 390

. Substitute the answer back into it.

18 \times 19.5 + 39 = 390

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct area calculation linked to the shape, for example $13 \times 21 = 273$ or $\dfrac{1}{2}(15 + 21) \times y$	M1	Any one correct area expression for a part of the shape, or for the whole of it. The trapezium's height may appear as $h - 13$ , or as $x$ or $y$ or any other letter, and even as $h$ itself; all are acceptable. Brackets need not be used for this mark.	✓
The areas of all the parts considered, for example $390 - 273 = 117$ , or $21h$ together with $2 \times \dfrac{1}{2} \times 3(h - 13)$	M1	The parts need not be added or subtracted to give the whole shape for this mark. The scheme quotes the $273$ , so the candidate's own value from the first mark may stand there. The same freedom over the letter used for the trapezium's height applies, and correct use of brackets is required.	✓
A correct calculation for a height, for example $\dfrac{117}{18} = 6.5$ , or a correct equation such as $273 + 18(h - 13) = 390$ or $18h = 351$	M1	Award for reaching the height of the trapezium, or the height of the whole shape, or a correct equation in which that height is the unknown. The scheme quotes the $117$ , so the candidate's own value from the previous mark may stand there. The same freedom over the letter used for that height applies, and correct use of brackets is required here too.	✓
$h = 19.5$	A1	Accept any equivalent form, for example $\dfrac{39}{2}$ .	✓
A correct answer with no working scores all four marks, unless it clearly follows incorrect working.	Note	The mark scheme prints $6.5$ once, in the third M1 row above, as one of the ways that mark can be earned; it says nothing about what a script ending there scores.	✓

Full marks: 4/4

Nikhil orders $600$ seedlings for a plant nursery.
He orders marigold seedlings, aster seedlings and zinnia seedlings so that the number of each kind is in the ratio

$\text{marigold} : \text{aster} : \text{zinnia} = 9 : 4 : 2$

$45\%$ of the marigold seedlings are for orange flowers.

$\dfrac{5}{8}$ of the aster seedlings are for orange flowers.

All of the zinnia seedlings are for orange flowers.

Work out the number of seedlings that are for orange flowers. [5 marks]

[Total 5 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

Total order:

600

seedlings

Ratio

\text{marigold} : \text{aster} : \text{zinnia} = 9 : 4 : 2

45\%

of the marigold seedlings are for orange flowers

\dfrac{5}{8}

of the aster seedlings are for orange flowers

Every zinnia seedling is for orange flowers

Find

The number of the

600

seedlings that are for orange flowers

Plan the Solution

Add the ratio numbers, then divide $600$ by that total to find one part.
Multiply one part by each ratio number to get the size of the three groups.
Take $45\%$ of the marigolds, $\dfrac{5}{8}$ of the asters, and every one of the zinnias.
Add the three orange counts. Each group has its own fraction, so the fractions can never simply be added first.

Worked Solution [5 marks]

Rule - Sharing in a ratio: add the ratio numbers, divide the total by that sum to find one part, then multiply one part by each ratio number to get the group sizes.

Step 1: Add the ratio numbers

9 + 4 + 2 = 15

(Reason: the ratio splits the order into

15

equal parts, not into

3

equal groups)

Step 2: Work out one part

\dfrac{600}{15} = 40

(Reason: dividing the whole order by the number of parts gives the number of seedlings in a single part)

Step 3: Work out the size of each group

\text{marigold: } 9 \times 40 = 360

\text{aster: } 4 \times 40 = 160

\text{zinnia: } 2 \times 40 = 80

(Reason: each group is its own ratio number of parts, and the three come back to

360 + 160 + 80 = 600

, which is the first sign the shares are right)

Step 4: Count the orange marigolds

0.45 \times 360 = 162

(Reason:

45\%

0.45

, and the percentage is of the marigold group only, never of the whole order)

Step 5: Count the orange asters

\dfrac{5}{8} \times 160 = 100

(Reason: five of every eight asters are for orange flowers, so multiply the aster group by that fraction and not by the part left over)

Step 6: Add the three orange counts

162 + 100 + 80 = 342

(Reason: all

80

zinnias are for orange flowers, so the whole zinnia group is counted)

342

seedlings

Verification

Check 1 - count in ratio parts instead: Work out how many of the

15

parts are orange:

0.45 \times 9 = 4.05

parts of marigold,

\dfrac{5}{8} \times 4 = 2.5

parts of aster, and both zinnia parts.

4.05 + 2.5 + 2 = 8.55

parts, and

\dfrac{8.55}{15} \times 600 = 342

Check 2 - one fraction of the whole order: Collapse the three shares into a single fraction of the order:

\dfrac{45}{100} \times \dfrac{9}{15} + \dfrac{5}{8} \times \dfrac{4}{15} + \dfrac{2}{15} = \dfrac{57}{100}

\dfrac{57}{100} \times 600 = 342

, so

57\%

of the order is orange

Check 3 - count what is not orange: The seedlings that are not for orange flowers are

0.55 \times 360 = 198

marigolds,

\left(1 - \dfrac{5}{8}\right) \times 160 = 60

asters and no zinnias at all.

600 - 198 - 60 = 342

, which agrees, and

342

sits between

80

and

600

as it must

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{600}{9 + 4 + 2} = 40$	M1	A correct method to find one share. Also allow $0.45 \times 600 = 270$ or $\dfrac{5}{8} \times 600 = 375$ , or the fraction of a share that is orange, $0.45 \times 9 = 4.05$ .	✓
$2 \times 40 = 80$	M1	A correct method to find the number of zinnia seedlings. $\dfrac{2}{15} \times 600 = 80$ also scores it and implies the first M1, as does the fraction of a share that is orange aster, $\dfrac{5}{8} \times \dfrac{4}{15} = \dfrac{1}{6}$ .	✓
$0.45 \times 360 = 162$	M1	A correct method to find the number of orange marigold seedlings, or the total of the orange parts, $4.05 + 2.5 + 2 = 8.55$ , which implies the first two M marks.	✓
$\dfrac{5}{8} \times 160 = 100$	M1	A correct method to find the number of orange aster seedlings, or multiplying the total of the correct orange shares by the order, $\dfrac{8.55}{9 + 4 + 2} \times 600 = 342$ , which implies all the previous M marks.	✓
$162 + 100 + 80 = 342$	A1	cao. $342$ seedlings are for orange flowers.	✓
A correct answer with no working shown	Note	A correct answer scores full marks, unless it comes from obviously incorrect working. This row earns nothing on its own.	✓

Full marks: 5/5

Marek pays $4\,500$ koruna into a savings bond.
The bond runs for $4$ years and pays $2.4\%$ per year compound interest.

Work out how much money Marek will have in the bond at the end of the $4$ years.
Give your answer correct to the nearest koruna. [3 marks]

koruna

[Total 3 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

Amount paid in:

4\,500

koruna

Interest:

2.4\%

per year, compound

Time:

4

years

Find

The value of the bond at the end of the

4

years, to the nearest koruna

Plan the Solution

Write $2.4\%$ as a decimal and add it to $1$ to get the multiplier for one year.
Multiply by that number once for each year, always starting from the balance the year opened with.
Collect the four multiplications into a single power, $1.024^{4}$ , which is the same calculation in one line.
Round at the very end only, so no part of a koruna is lost part way through.

Worked Solution [3 marks]

Rule - Compound interest: the interest for each year is worked out on the balance at the START of that year, so the balance is multiplied by the same number once per year. After

n

years an amount

P

growing at a rate

r

(as a decimal) is worth

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

Step 1: Turn the rate into a one-year multiplier

\dfrac{2.4}{100} = 0.024

1 + 0.024 = 1.024

(Reason: the balance keeps its whole self and gains

2.4\%

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

1.024

rather than by

0.024

)

Step 2: Multiply once for each of the four years

4\,500 \times 1.024 = 4\,608

4\,608 \times 1.024 = 4\,718.592

4\,718.592 \times 1.024 = 4\,831.838208

4\,831.838208 \times 1.024 = 4\,947.802324992

(Reason: each year's interest is worked out on the balance that year started with, not on the original

4\,500

, which is exactly what makes the interest compound rather than simple)

Step 3: Do the same four multiplications in one line

4\,500 \times 1.024^{4} = 4\,947.802324992

(Reason: multiplying by

1.024

four times is multiplying by

1.024

to the power

4

, so a calculator reaches the same balance in a single press)

Step 4: Round to the nearest koruna

4\,947.802324992 \approx 4\,948

(Reason: the first digit after the decimal point is

8

, so the amount rounds up to the next whole koruna)

4\,948

koruna

Verification

Check 1 - undo the four years: Divide the final balance by the multiplier once for each year. If the four years were applied correctly, the amount paid in comes back exactly.

\dfrac{4\,947.802324992}{1.024^{4}} = 4\,500

Check 2 - compare it with simple interest: Simple interest would pay

4\,500 \times 0.024 = 108

koruna every year, so

4

years of it would be

432

koruna and the bond would hold

4\,932

. Compound interest must pay a little more than that, and never a lot more over

4

years.

4\,947.802324992 - 4\,500 = 447.802324992

of interest, which is more than

432

and only a little more

Check 3 - read the growth as a single percentage: Work out the four-year growth factor on its own, then apply the part above

1

to the amount paid in. This never uses the year-by-year balances, so it is an independent route to the interest.

1.024^{4} = 1.099511627776

, and

0.099511627776 \times 4\,500 = 447.802324992

, the same interest as Check 2

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$4\,500 \times 1.024 = 4\,608$	M1	A correct start to a compound method: $4\,500 \times 1.024$ or equivalent, or the first year's interest on its own, $4\,500 \times 0.024 = 108$ .	✓
$4\,831.838208 \times 1.024 = 4\,947.802324992$	M1	Carrying the multiplication through all four years: $4\,608 \times 1.024 = 4\,718.592$ , then $4\,718.592 \times 1.024 = 4\,831.838208$ , then $4\,831.838208 \times 1.024 = 4\,947.802324992$ . The mark scheme prints these figures in quotation marks, which means the candidate's own running totals are followed through. The power form scores M2 on its own, and the mark scheme allows $4\,500 \times 1.024^{4}$ or $4\,500 \times 1.024^{5}$ for it.	✓
$4\,948$	A1	Anything from $4\,947$ to $4\,948$ is accepted, which covers a candidate who truncates the unrounded $4\,947.802324992$ instead of rounding it.	✓
One of the recognised wrong methods, and no other mark earned	SC B1	Simple interest in place of compound: $4\,500 \times 0.024 \times 4 = 432$ or $0.096 \times 4\,500 = 432$ ; the simple total $4\,500 + 4\,500 \times 0.024 \times 4 = 4\,932$ or $4\,500 \times 1.096 = 4\,932$ ; a depreciation multiplier, $0.976 \times 4\,500 = 4\,392$ or $0.904 \times 4\,500 = 4\,068$ or $0.976^{4} \times 4\,500 = 4\,083.304660992$ ; or three years instead of four, $4\,500 \times 1.024^{3} = 4\,831.838208$ .	✓
A correct answer with no working shown	Note	A correct answer scores full marks, unless it comes from obviously incorrect working. This row earns nothing on its own.	✓

Full marks: 3/3

Solve the simultaneous equations

$6x + 4y = 1$
$3x + 5y = 8$

You must show clear algebraic working. [3 marks]

x =y =

[Total 3 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

The first equation:

6x + 4y = 1

The second equation:

3x + 5y = 8

Two linear equations in the same two unknowns, so one pair of values has to satisfy both at once.

Find

The value of

x

and the value of

y

that fit both equations together. Neither equation is quadratic, so expect exactly one solution pair.

Plan the Solution

Compare the $x$ terms first: $6x$ and $3x$ . Doubling the second equation makes them match, so the first equation needs no work at all.
Subtract one equation from the other to cancel $x$ , leaving a single equation in $y$ .
Solve that for $y$ , then substitute the value back into one of the original equations to reach $x$ .
Test the pair in both original equations at the end, because a slip in either value shows up immediately there.

Worked Solution [3 marks]

Rule - Elimination: multiply one or both equations so that one letter has the same number in front of it in each, then add or subtract the equations to remove that letter. What is left is a single equation in one unknown, and putting its value back into either original equation gives the other unknown.

Step 1: Match the number of

x

in the two equations

3x + 5y = 8

6x + 10y = 16

(Reason: the first equation already carries

6x

, so doubling every term of the second one gives it

6x

as well; multiplying a whole equation by

2

keeps it true, because both sides are doubled together)

Step 2: Subtract to remove

x

, and solve for

y

(6x + 10y) - (6x + 4y) = 16 - 1

6y = 15

y = \dfrac{15}{6} = \dfrac{5}{2} = 2.5

(Reason: both equations now carry

6x

, so taking one from the other cancels

x

and leaves an equation in

y

on its own)

Step 3: Put

y = 2.5

back into an original equation

6x + 4 \times 2.5 = 1

6x + 10 = 1

6x = -9

x = \dfrac{-9}{6} = -\dfrac{3}{2} = -1.5

(Reason: with

y

known, the first equation has a single unknown left in it, and choosing an original equation rather than the doubled one means any slip made in the doubling cannot travel into

x

)

x = -1.5

y = 2.5

Verification

Check 1 - the first equation: Put

x = -1.5

and

y = 2.5

into

6x + 4y = 1

. The left-hand side has to come out as the right-hand side.

6 \times (-1.5) + 4 \times 2.5 = -9 + 10 = 1

Check 2 - the second equation: The second equation was the one that got doubled, so it is the one a multiplying slip would spoil. Put the same pair into

3x + 5y = 8

in its original form.

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

Check 3 - solve again, removing the other letter: Multiply the first equation by

5

and the second by

4

, so both carry

20y

, then subtract. This route never uses the value of

y

found above, so it reaches

x

independently.

30x + 20y = 5

and

12x + 20y = 32

, so

18x = -27

and

x = -1.5

again

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$6x + 10y = 16$ with $6x + 4y = 1$ , giving $6y = 15$	M1	A correct method to eliminate one letter: multiplying one or both equations so that $x$ or $y$ carries the same number in each, together with the correct operation to remove it, which can be shown by $2$ of the $3$ terms being correct for the subtraction or the addition. One arithmetic error in the multiplying is allowed. The other route scores the same mark: a correct substitution of one letter into the other equation, for example $6x + 4 \times \dfrac{8 - 3x}{5} = 1$ or $3 \times \dfrac{1 - 4y}{6} + 5y = 8$ . The mark is for the method and not for what the method produces, although a correct result seen earns it.	✓
$6x + 4 \times 2.5 = 1$	M1dep	A correct method to reach the value of the other letter, dependent on the first M1: substituting the letter already found into either equation, which does not then have to be solved, or starting again with a fresh elimination or substitution.	✓
$x = -1.5$ , $y = 2.5$	A1	Both values, and dependent on the first M1. Any equivalent exact form is accepted, so $-\dfrac{3}{2}$ or $-1\dfrac{1}{2}$ is as good as $-1.5$ , and $\dfrac{5}{2}$ or $2\dfrac{1}{2}$ as good as $2.5$ . It must be a vulgar fraction, a mixed number or a decimal, so a fraction left unfinished, such as $y = \dfrac{12.5}{5}$ , is not accepted.	✓
Working required	Note	The mark scheme prints the words working required beside the answer, and the question asks for clear algebraic working, so a correct pair of values written down with no algebra behind it earns nothing at all. This row carries no mark of its own.	✓

Full marks: 3/3

(i) Write $x^2 + 9x - 22$ as a product of two brackets.
[2 marks]
(ii) Hence solve the equation $x^2 + 9x - 22 = 0$ [1 mark]

(i)(ii)

[Total 3 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

x^2 + 9x - 22

, a quadratic expression in which the coefficient of

x^2

1

Part (ii) sets that same expression equal to

0

, and the word "Hence" says to use the brackets from part (i) rather than start again.

Find

(i) the expression written as a product of two brackets. (ii) the values of

x

that satisfy

x^2 + 9x - 22 = 0

. A quadratic, so expect two solutions.

Plan the Solution

Look for two numbers whose product is $-22$ and whose sum is $9$ .
List the factor pairs of $-22$ and test the sum of each one, so the search is complete rather than lucky.
Write the brackets, then expand them to confirm they give $x^2 + 9x - 22$ back.
For (ii), a product is $0$ only when one of the factors is $0$ , so set each bracket equal to $0$ in turn.

Worked Solution [3 marks]

Factorising

x^2 + bx + c

: find two numbers

p

and

q

with

pq = c

and

p + q = b

. Then

x^2 + bx + c = (x + p)(x + q)

Step 1: Name the two conditions the numbers must meet

x^2 + 9x - 22

pq = -22

p + q = 9

(Reason: Expanding

(x + p)(x + q)

gives

x^2 + (p + q)x + pq

, so the constant term is the product of the two numbers and the coefficient of

x

is their sum. The product is negative, so one number is positive and the other is negative.)

Step 2: Test every factor pair of the constant

-22 = 1 \times (-22) = (-1) \times 22 = 2 \times (-11) = (-2) \times 11

1 + (-22) = -21

(-1) + 22 = 21

2 + (-11) = -9

(-2) + 11 = 9

(Reason: Only four pairs of whole numbers multiply to

-22

, and only the last of them adds to

9

. Listing them all is what makes this a search rather than a guess, so

p = -2

and

q = 11

Step 3: Write the brackets, then expand to check

x^2 + 9x - 22 = (x - 2)(x + 11)

(x - 2)(x + 11) = x^2 + 11x - 2x - 22

x^2 + 11x - 2x - 22 = x^2 + 9x - 22

(Reason: Each number goes inside its own bracket with its own sign, so

-2

gives

(x - 2)

and

11

gives

(x + 11)

. Multiplying out returns the original expression, which is the answer to part (i).)

Step 4: Solve by setting each bracket equal to zero

(x - 2)(x + 11) = 0

x - 2 = 0 \implies x = 2

x + 11 = 0 \implies x = -11

(Reason: Two numbers multiply to

0

only if at least one of them is

0

, so each bracket supplies one solution. Notice the sign flips: a bracket that reads

+ 11

gives the negative solution, and a bracket that reads

- 2

gives the positive one. This is why the question says "Hence": part (i) has already done the work.)

(i)

(x - 2)(x + 11)

(ii)

x = 2 \text{ or } x = -11

Verification

Check 1: Multiply the brackets out term by term:

x \times x

, then

x \times 11

, then

-2 \times x

, then

-2 \times 11

(x - 2)(x + 11) = x^2 + 9x - 22

, the expression the question printed.

Check 2: Substitute each solution back into

x^2 + 9x - 22

. A solution must make the whole expression

0

2^2 + 9 \times 2 - 22 = 0

and

(-11)^2 + 9 \times (-11) - 22 = 0

Check 3: Use the sum and product of the roots, which never touches the brackets. For

x^2 + bx + c

the two solutions add to

-b

and multiply to

c

, so here they must add to

-9

and multiply to

-22

2 + (-11) = -9

and

2 \times (-11) = -22

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(i) $(x \pm 2)(x \pm 11)$	M1	Or $(x + a)(x + b)$ where $ab = -22$ or $a + b = 9$ . The method mark is for the right pair of numbers, whatever the signs.	✓
(i) $(x - 2)(x + 11)$	A1	A correct answer scores full marks, unless it comes from obviously incorrect working.	✓
(ii) $2, -11$	B1ft	Follow through from their factors in part (i): the two solutions must be the ones their own brackets give. A candidate who wrote $(x + 2)(x - 11)$ in (i) and then $-2, 11$ here still earns this mark.	✓

Full marks: 3/3

Bilal records the number of bottles a machine fills each day for $7$ days.

For the first $4$ days, the mean number of bottles filled each day is $11\,800$
For the next $3$ days, the mean number of bottles filled each day is $13\,207$

Work out the mean number of bottles filled each day for the $7$ days. [3 marks]

[Total 3 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

The number of bottles filled is recorded on

7

days in a row.

The mean for the first

4

days is

11\,800

bottles a day.

The mean for the next

3

days is

13\,207

bottles a day.

Find

The mean number of bottles filled each day across all

7

days.

Plan the Solution

Two means cannot simply be averaged with each other here, because the two blocks are different sizes: $4$ days against $3$ days.
Turn each mean back into a total instead. Multiply each mean by the number of days it covers.
Add the two totals to get the total for the whole week, then divide that by $7$ .

Worked Solution [3 marks]

Rule - Mean:

\text{mean} = \dfrac{\text{total}}{\text{number of values}}

, so rearranged,

\text{total} = \text{mean} \times \text{number of values}

Step 1: Total for the first

4

days

4 \times 11\,800 = 47\,200

(Reason: Each of those

4

days contributes the mean of

11\,800

to the total, so the block total is

4

lots of

11\,800

Step 2: Total for the next

3

days

3 \times 13\,207 = 39\,621

(Reason: The same idea for the second block: multiply its mean by the number of days that block covers.)

Step 3: Total for all

7

days

47\,200 + 39\,621 = 86\,821

(Reason: The two blocks cover every one of the

7

days once and do not overlap, so the two totals add to the whole week's total.)

Step 4: Divide the total by the number of days

\dfrac{86\,821}{7} = 12\,403

(Reason: The mean for the week is the total for the week divided by the number of days in it, which is

7

12\,403

bottles

Verification

Check 1: Work backwards. If the mean over the

7

days really is

12\,403

, then multiplying it by

7

must rebuild the total found in Step 3.

7 \times 12\,403 = 86\,821

, which is the same total

Check 2: A different route: start every day at the lower mean

11\,800

. Each of the last

3

days is

13\,207 - 11\,800 = 1\,407

above that, and the extra is shared over

7

days.

11\,800 + \dfrac{3 \times 1\,407}{7} = 11\,800 + 603 = 12\,403

Check 3: Sanity check the size. The answer must sit between the two means, and nearer the lower one, because

4

of the

7

days come from the block with the lower mean.

11\,800 < 12\,403 < 13\,207

, and it is

603

above the lower mean but

804

below the upper one

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$4 \times 11\,800 = 47\,200$ or $3 \times 13\,207 = 39\,621$ or $86\,821$	M1	for one correct product, or for the sum of the two products	✓
$\dfrac{47\,200 + 39\,621}{7}$ or $\dfrac{86\,821}{7}$	M1	for a fully correct method to find the mean for the $7$ days, using their own two block totals	✓
$12\,403$	A1	cao. A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 3/3

A hospital manager records the distance, in km, that each of $70$ nurses travels to work.
The table gives information about these distances.

$\begin{array}{|c|c|}\hline \textbf{Distance } (d \textbf{ km}) & \textbf{Frequency} \\ \hline 0 < d \leq 10 & 7 \\ \hline 10 < d \leq 20 & 17 \\ \hline 20 < d \leq 30 & 18 \\ \hline 30 < d \leq 40 & 14 \\ \hline 40 < d \leq 50 & 10 \\ \hline 50 < d \leq 60 & 4 \\ \hline \end{array}$

(a) Complete the cumulative frequency table.

$\begin{array}{|c|c|}\hline \textbf{Distance } (d \textbf{ km}) & \textbf{Cumulative frequency} \\ \hline 0 < d \leq 10 & \\ \hline 0 < d \leq 20 & \\ \hline 0 < d \leq 30 & \\ \hline 0 < d \leq 40 & \\ \hline 0 < d \leq 50 & \\ \hline 0 < d \leq 60 & \\ \hline \end{array}$ [1 mark]

(b) On the grid below, draw a cumulative frequency graph for your table. [2 marks]

(d) Use your graph to find an estimate for the number of nurses who travel more than $46$ km. [2 marks]

[Total 7 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

The distances, in km, travelled to work by

70

nurses, grouped into six classes of width

10

km.

Frequencies

7

17

18

14

10

4

A grid running from

0

60

km across and

0

70

up.

Find

(a) The six cumulative frequencies. (b) The cumulative frequency graph, drawn on the grid. (c) An estimate for the interquartile range of the distances. (d) An estimate for the number of nurses who travel more than

46

km.

Plan the Solution

Build a running total down the frequency column; the last one must be the whole group.
Plot each running total at the upper end of its class, then join the points up.
Read across from $\dfrac{70}{4}$ and from $\dfrac{3 \times 70}{4}$ on the cumulative frequency axis, down to the distance axis, and subtract the two distances.
For part (d), read up from $46$ km to the graph, then take that reading away from $70$ .

Worked Solution [7 marks]

Rule - Cumulative frequency: plot the running total at the upper end of each class and join the points. The lower quartile sits

\dfrac{n}{4}

of the way up the cumulative frequency axis and the upper quartile

\dfrac{3n}{4}

of the way up. The interquartile range is the gap between the two distances they lead to, never the gap between the two heights.

Step 1: Add the frequencies down the table

7

7 + 17 = 24

24 + 18 = 42

42 + 14 = 56

56 + 10 = 66

66 + 4 = 70

(Reason: Each cumulative frequency counts everyone so far, so the previous total is carried forward and the next class added on. The last total has to come to

70

, the whole group of nurses.)

Step 2: Plot each total at the upper end of its class

(10, 7) \quad (20, 24) \quad (30, 42)

(40, 56) \quad (50, 66) \quad (60, 70)

(Reason: A cumulative total counts everyone at or below the top of the class, so

24

belongs at a distance of

20

km and never at the midpoint

15

km. Joining the points with a smooth curve or with straight line segments both earn the marks.)

Step 3: Read the two quartiles from the graph

\dfrac{70}{4} = 17.5

\dfrac{3 \times 70}{4} = 52.5

Q_1 \approx 16.2 \text{ km}

Q_3 \approx 37.5 \text{ km}

(Reason: Go up the cumulative frequency axis to

17.5

for the lower quartile and to

52.5

for the upper quartile, across to the graph, then straight down to the distance axis. The mark scheme accepts a lower quartile anywhere from

16

18

and an upper quartile from

36

38

Step 4: Subtract for the interquartile range

37.5 - 16.2 = 21.3 \text{ km}

(Reason: The interquartile range is the width of the middle half of the data, so it is a distance. Subtract the two readings taken off the distance axis, not the two heights read up the cumulative frequency axis.)

Step 5: Read the graph at

46

56 + \dfrac{6}{10} \times 10 = 62

(Reason: The graph climbs from

56

40

km to

66

50

km, and

46

km is

\dfrac{6}{10}

of the way along that step. So about

62

nurses travel

46

km or less.)

Step 6: Take that reading away from the total

70 - 62 = 8

(Reason: A cumulative frequency graph gives the number at or below a distance, so the number above it is the whole group minus that reading. The answer has to be a whole number of nurses.)

(a)

7, 24, 42, 56, 66, 70

(b) the six points plotted and joined, as shown above(c)

21.3

km(d)

8

nurses

Verification

Check 1: Add every frequency on its own:

7 + 17 + 18 + 14 + 10 + 4

. The running total finishes on

70

, the number of nurses, so nobody has been lost or counted twice.

Check 2: The lower quartile has to land in the class holding the

17.5

th value, and the upper quartile in the class holding the

52.5

th value.

16.2

lies inside

10 < d \leq 20

and

37.5

lies inside

30 < d \leq 40

, so both readings sit in the right class.

Check 3: Count part (d) from the other end instead. From

46

km to

50

km is

\dfrac{4}{10}

of a class holding

10

nurses, and the last class holds

4

more.

4 + 4 = 8

, which matches the number read off the graph.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Cumulative frequency table	B1	All six values correct: $7, 24, 42, 56, 66, 70$ .	✓
(b) Cumulative frequency graph	B2	Fully correct graph: six points plotted at the upper end of each class and joined with a curve or with line segments. Award B1 for five correct points plotted and joined, or for five or six points plotted but not joined, or for five or six points plotted consistently within each interval rather than at its upper end, at their correct heights and joined, for example at $5$ , $15$ , $25$ , $35$ , $45$ and $55$ . Any of the B1 options may follow through from a part (a) table with one error in it, provided its values are ascending. A bar-chart type graph scores zero marks. Ignore any part of the graph before $(10, 7)$ .	✓
(c) Quartile readings	M1ft	A correct method allowing readings to be taken on the distance axis from cumulative frequency $52.5$ (or $53.25$ ) and from $17.5$ (or $17.75$ ), or equivalent. Follow through from their own graph. The readings themselves are $16$ to $18$ and $36$ to $38$ , but for this mark they need not be correct provided correct working is shown, such as lines or marks at those two cumulative frequencies with the matching points indicated on the distance axis.	✓
(c) Interquartile range	A1ft	A single value from $18$ to $22$ , or follow through from their own cumulative frequency graph, unless it comes from obviously incorrect working.	✓
(d) Reading at $46$ km	M1ft	A line up from $46$ to the graph and a reading across, or a reading of $61$ to $64$ which need not be a whole number, from their own graph.	✓
(d) Number of nurses	A1ft	$6$ , $7$ , $8$ or $9$ , follow through from their own graph. It must be a whole number, and again the mark is not awarded where the answer comes from obviously incorrect working.	✓

Full marks: 7/7

(a) Show that the product $3y(2y + 5)(y + 7)$ can be written as $ay^3 + by^2 + cy$ , where $a$ , $b$ and $c$ are integers. [3 marks]

(b) Solve the equation $\dfrac{2x + 3}{5} + \dfrac{6x - 5}{4} = \dfrac{163}{100}$
You must show clear algebraic working. [4 marks]

x =

[Total 7 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

Part (a): the product of three factors,

3y(2y + 5)(y + 7)

, and the target form

ay^3 + by^2 + cy

with

a

b

and

c

integers.

Part (b): the equation

\dfrac{2x + 3}{5} + \dfrac{6x - 5}{4} = \dfrac{163}{100}

, with denominators

5

4

and

100

Find

(a) The expanded and simplified cubic, with every line of the expansion shown. (b) The value of

x

, with clear algebraic working.

Plan the Solution

(a) Multiply two of the three factors first, simplify, then multiply that result by the third factor. Any order gives the same cubic, so choose the pair that keeps the arithmetic smallest.
(a) Collect the two middle terms before the last multiplication, so only three terms have to be multiplied by $3y$ at the end.
(b) Write the left-hand side over the lowest common denominator $20$ , then multiply both sides by $20$ to clear the fractions.
(b) Gather the $x$ terms on one side and the numbers on the other, then divide.

Worked Solution [7 marks]

Rule - Expanding a triple product: multiply any two factors first, simplify, then multiply by the third. Multiplication is associative, so the order cannot change the answer. Rule - Clearing fractions: multiplying every term of an equation by the lowest common denominator leaves an equivalent equation with no fractions in it, so

\dfrac{p}{20} = \dfrac{q}{100}

becomes

p = \dfrac{20q}{100}

Step 1: Expand the pair of brackets

(2y + 5)(y + 7) = 2y^2 + 14y + 5y + 35

(Reason: Every term in the first bracket multiplies every term in the second, so there are four products to write down:

2y \times y

, then

2y \times 7

, then

5 \times y

, then

5 \times 7

. Leaving the

3y

out until later keeps these numbers small.)

Step 2: Collect the two middle terms

14y + 5y = 19y

(2y + 5)(y + 7) = 2y^2 + 19y + 35

(Reason: The two middle products are both multiples of

y

, so they are like terms and add together. The quadratic is now in three terms, which is what the final multiplication needs.)

Step 3: Multiply every term by

3y

3y \times 2y^2 = 6y^3

3y \times 19y = 57y^2

3y \times 35 = 105y

3y(2y^2 + 19y + 35) = 6y^3 + 57y^2 + 105y

(Reason: Multiply the numbers and add the indices:

y \times y^2 = y^3

and

y \times y = y^2

. Every term of the quadratic must be multiplied, not just the first one.)

Step 4: Compare with the form asked for

ay^3 + by^2 + cy = 6y^3 + 57y^2 + 105y

a = 6, \quad b = 57, \quad c = 105

(Reason: Matching term by term gives the three coefficients. All three are whole numbers, and there is no constant term, so the product really does have the form

ay^3 + by^2 + cy

with

a

b

and

c

integers, which is what part (a) asked to be shown.)

Step 5: Put the left-hand side over one denominator

\dfrac{2x + 3}{5} + \dfrac{6x - 5}{4} = \dfrac{4(2x + 3)}{20} + \dfrac{5(6x - 5)}{20}

\dfrac{4(2x + 3) + 5(6x - 5)}{20} = \dfrac{163}{100}

(Reason: The lowest common denominator of

5

and

4

20

. The first fraction is multiplied top and bottom by

4

and the second by

5

, which changes how each fraction is written but not its value.)

Step 6: Expand and simplify the numerator

4(2x + 3) + 5(6x - 5) = 8x + 12 + 30x - 25

8x + 12 + 30x - 25 = 38x - 13

\dfrac{38x - 13}{20} = \dfrac{163}{100}

(Reason: Take the minus sign with the

5

5 \times (-5) = -25

, not

+25

. Then

8x + 30x = 38x

and

12 - 25 = -13

Step 7: Clear the fractions

38x - 13 = \dfrac{163 \times 20}{100}

38x - 13 = 32.6

(Reason: Multiplying both sides by

20

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

20

as well. Since

163 \times 20 = 3260

, the right-hand side is

32.6

Step 8: Collect the terms and divide

38x = 32.6 + 13

38x = 45.6

x = \dfrac{45.6}{38} = 1.2

(Reason: Adding

13

to both sides puts the number terms on the right, leaving one term in

x

on the left. Dividing by

38

gives the solution, and the decimal is exact rather than rounded.)

(a)

3y(2y + 5)(y + 7) = 6y^3 + 57y^2 + 105y

, so

a = 6

b = 57

c = 105

(b)

x = 1.2

Verification

Check 1: Put

y = 2

into the original product and into the cubic. If the two forms are the same expression, they must give the same number.

3 \times 2 \times 9 \times 9 = 486

and

6 \times 8 + 57 \times 4 + 105 \times 2 = 486

, so the two agree.

Check 2: Expand in a different order: multiply

3y

by the first bracket instead, then by the second.

3y(2y + 5) = 6y^2 + 15y

, and

(6y^2 + 15y)(y + 7) = 6y^3 + 42y^2 + 15y^2 + 105y

, which simplifies to

6y^3 + 57y^2 + 105y

Check 3: Substitute

x = 1.2

into the left-hand side of the original equation.

\dfrac{5.4}{5} + \dfrac{2.2}{4} = 1.08 + 0.55 = 1.63

, and

\dfrac{163}{100} = 1.63

, so both sides match.

Check 4: Solve part (b) a second way: multiply every term of the original equation by

100

instead of

20

20(2x + 3) + 25(6x - 5) = 163

gives

190x - 65 = 163

, so

190x = 228

and

\dfrac{228}{190} = 1.2

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) First expansion	M1	An expansion of one pair of the three factors with only one error, for example $(2y + 5)(y + 7) = 2y^2 + 14y + 5y + 35$ , or $3y(2y + 5) = 6y^2 + 15y$ , or $3y(y + 7) = 3y^2 + 21y$ . Do not award this mark for $6y^2 + 15y + 3y^2 + 21y$ .	✓
(a) Second expansion	M1	Follow through, dependent on the first M1, allowing one further error: $(6y^2 + 15y)(y + 7) = 6y^3 + 42y^2 + 15y^2 + 105y$ , or $(3y^2 + 21y)(2y + 5) = 6y^3 + 15y^2 + 42y^2 + 105y$ , or $3y(2y^2 + 19y + 35) = 6y^3 + 57y^2 + 105y$ . Alternatively M2 for $3$ correct terms out of a maximum of $4$ of $6y^3 + 42y^2 + 15y^2 + 105y$ , and M1 for $2$ correct out of a maximum of $4$ .	✓
(a) Simplified cubic	A1	cao, dependent on M1. The terms may be in any order but must be simplified: $6y^3 + 57y^2 + 105y$ . Accept $a = 6$ , $b = 57$ , $c = 105$ . Working is required.	✓
(b) Common denominator	M1	Writing the fractions over a common denominator (two fractions are enough), or a method to remove the denominator by multiplying each term by, for example, $20$ or $100$ . If the numerator is expanded, allow one error. For example $\dfrac{4(2x + 3) + 5(6x - 5)}{20} = 1.63$ , or $20(2x + 3) + 25(6x - 5) = 163$ , which may all be written over $100$ .	✓
(b) Brackets and fractions removed	M1	Removing the brackets and the fractions on the left-hand side, in an equation with no more than one error from expanding the numerator, or an equation with the terms on the numerator simplified with no more than one such error. For example $8x + 12 + 30x - 25 = 32.6$ , or $40x + 60 + 150x - 125 = 163$ , or $\dfrac{38x - 13}{20} = \dfrac{163}{100}$ .	✓
(b) Terms collected	M1	Terms in $x$ on one side and number terms on the other, in a correct equation. For example $8x + 30x = 32.6 - 12 + 25$ , or $38x = 45.6$ , or $190x = 228$ .	✓
(b) Solution	A1	$1.2$ or equivalent, dependent on M1. Working is required.	✓

Full marks: 7/7

(a) Rearrange the formula
$e = \sqrt{\dfrac{7g + 5}{11 + 2g}}$
to make $g$ the subject. [4 marks]

(b) Solve the inequality $3y^{2} + 4y - 32 > 0$
You must show your working clearly. [3 marks]

(a)(b)

[Total 7 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

The formula

e = \sqrt{\dfrac{7g + 5}{11 + 2g}}

, in which

g

sits under a square root and in both parts of a fraction.

The inequality

3y^{2} + 4y - 32 > 0

, a quadratic whose

y^{2}

coefficient is positive.

Find

(a)

g

written in terms of

e

. (b) every value of

y

that makes the quadratic greater than

0

. Expect two separate regions, not one.

Plan the Solution

(a) Square both sides to remove the root, then multiply by the denominator so that nothing is left underneath a fraction.
(a) Two terms will contain $g$ . Collect them on one side, factorise $g$ out and divide by the bracket that is left.
(b) Factorise the quadratic, then set each bracket equal to $0$ to get the two critical values.
(b) A positive quadratic dips below the axis between its roots and sits above it outside them, so the solution is the two outer regions.

Worked Solution [7 marks]

Rule - Changing the subject: clear roots and fractions first, gather every term containing the new subject on one side, then factorise it out and divide by the bracket. For a quadratic inequality: factorise, find the critical values, then choose the regions from the shape of the curve.

Step 1: square both sides

e^{2} = \dfrac{7g + 5}{11 + 2g}

(Reason: Squaring is the inverse of the square root, and it is applied to the whole of each side, so the left becomes

e^{2}

and the right loses its root sign completely.)

Step 2: multiply by the denominator and expand

e^{2}(11 + 2g) = 7g + 5

11e^{2} + 2e^{2}g = 7g + 5

(Reason: While

g

is still inside the denominator nothing can be collected, so the fraction has to go first. Multiplying both sides by

11 + 2g

clears it, and expanding the bracket puts every term on one level.)

Step 3: gather the terms in g on one side

2e^{2}g - 7g = 5 - 11e^{2}

(Reason: There are now two terms containing

g

, one on each side. Both must end up together before

g

can be taken out as a factor, and everything without a

g

in it moves the other way.)

Step 4: factorise out g, then divide

g(2e^{2} - 7) = 5 - 11e^{2}

g = \dfrac{5 - 11e^{2}}{2e^{2} - 7}

(Reason: As far as

g

is concerned,

2e^{2} - 7

is just the number multiplying it, so dividing both sides by that bracket leaves

g

on its own. That bracket is never zero:

2e^{2} = 7

would need

\dfrac{7g + 5}{11 + 2g}

to equal

\dfrac{7}{2}

, and no value of

g

makes that true.)

Step 5: factorise the quadratic in part (b)

3y^{2} + 4y - 32 = 3y^{2} + 12y - 8y - 32

3y^{2} + 12y - 8y - 32 = 3y(y + 4) - 8(y + 4)

3y(y + 4) - 8(y + 4) = (3y - 8)(y + 4)

(Reason: The middle term is split using two numbers whose product is

3 \times (-32) = -96

and whose sum is

4

. Those numbers are

12

and

-8

, and grouping in pairs then gives the common bracket.)

Step 6: find the critical values

3y - 8 = 0 \implies y = \dfrac{8}{3}

y + 4 = 0 \implies y = -4

(Reason: A product is zero exactly when one of its factors is zero, so the quadratic takes the value

0

at these two values of

y

and nowhere else. They are the boundaries between the positive and the negative parts of the curve, and neither is itself a solution, because the inequality is strict.)

Step 7: choose the regions

y < -4 \text{ or } y > \dfrac{8}{3}

(Reason: The coefficient of

y^{2}

3

, which is positive, so the curve is a U shape: below the axis between the critical values and above it outside them. Testing

y = 0

, which lies between them, gives

-32

, confirming that the middle region is not wanted.)

(a)

g = \dfrac{5 - 11e^{2}}{2e^{2} - 7}

(b)

y < -4 \text{ or } y > \dfrac{8}{3}

Verification

Check 1 - put a number through both formulas: Take

g = 3

in the original:

7 \times 3 + 5 = 26

and

11 + 2 \times 3 = 17

, so

e^{2} = \dfrac{26}{17}

. Then

11e^{2} = \dfrac{286}{17}

and

2e^{2} = \dfrac{52}{17}

. The rearranged formula gives

\dfrac{5 - \dfrac{286}{17}}{\dfrac{52}{17} - 7} = \dfrac{-201}{-67} = 3

, the value

g

started at.

Check 2 - the other accepted form: Multiply the numerator and the denominator of

\dfrac{5 - 11e^{2}}{2e^{2} - 7}

-1

. Every sign flips and the value is unchanged, giving

\dfrac{11e^{2} - 5}{7 - 2e^{2}}

, which is the alternative form the mark scheme accepts, so the two answers are one expression written two ways.

Check 3 - expand the factorisation: Multiply

(3y - 8)(y + 4)

out again and collect the middle terms.

3y^{2} + 12y - 8y - 32 = 3y^{2} + 4y - 32

, which is exactly the quadratic the question gives.

Check 4 - one value from each region: Substitute

y = -5

y = 0

and

y = 3

into

3y^{2} + 4y - 32

, one from each side of the critical values and one from between them. They give

23

-32

and

7

, so the expression is positive on the two outer regions only, which is what the answer claims.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$e^{2} = \dfrac{7g + 5}{11 + 2g}$	M1	For removing the square root.	✓
$11e^{2} + 2e^{2}g = 7g + 5$	M1	For multiplying by the denominator and expanding, in a correct equation.	✓
eg $2e^{2}g - 7g = 5 - 11e^{2}$ or $11e^{2} - 5 = 7g - 2e^{2}g$ oe	M1	For gathering the terms in $g$ on one side and the other terms on the other side, in a correct equation.	✓
$g = \dfrac{5 - 11e^{2}}{2e^{2} - 7}$	A1	oe. Accept for example $g = \dfrac{11e^{2} - 5}{7 - 2e^{2}}$ or $g = \dfrac{\dfrac{5}{e^{2}} - 11}{2 - \dfrac{7}{e^{2}}}$ or $g = \dfrac{\dfrac{5 - 11e^{2}}{e^{2} - 3.5}}{2}$ .	✓
Correct answer scores full marks	Note	unless it comes from obviously incorrect working.	✓
$(3y - 8)(y + 4)$	M1	For a correct factorisation, or correct use of the quadratic formula $\dfrac{-4 \pm \sqrt{4^{2} - 4 \times 3 \times (-32)}}{2 \times 3}$ , or as far as $\dfrac{-4 \pm \sqrt{400}}{6}$ .	✓
$\left(y - \dfrac{8}{3}\right)(y + 4)$	Note	is not a valid factorisation of the given quadratic, so it scores no marks unless it is preceded by dividing the quadratic by $3$ .	✓
$y = \dfrac{8}{3}$ , $y = -4$	A1	dep on M1, for both correct critical values. Allow $2.6$ or better, or $2.7$ .	✓
$y < -4$ , $y > \dfrac{8}{3}$	A1	oe dep on M1, and working is required. Allow $x$ in place of $y$ , and accept the interval form $(-\infty, -4)$ with $\left(\dfrac{8}{3}, \infty\right)$ , or the union of the two.	✓

Full marks: 7/7

Continue to questions 16 to 25

The remaining 10 questions, with the same full worked solutions and mark schemes

Frequently asked questions

There are 25 questions worth 100 marks in total, sat over 2 hours. It is Higher tier and a calculator is allowed throughout, unlike UK GCSE Maths, where one paper is non-calculator.

Higher tier targets grades 4 to 9, so the lower grades 1 to 3 are only reachable on the tier below. About 40 per cent of the questions are targeted at grades 4 and 5 and appear on both Paper 2F and Paper 2H, so the lowest grades on this Higher paper are the ones the two tiers share.

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

Yes, a Higher tier formulae sheet is printed in the paper. It gives the area of a trapezium, the volume of a prism, the volume and curved surface area of a cylinder, the volume and curved surface area of a cone, the volume and surface area of a sphere, the area of a triangle from two sides and the included angle, the sine rule, the cosine rule, the sum of an arithmetic series and the quadratic formula. Other results, such as Pythagoras theorem and the trigonometric ratios for right-angled triangles, still have to be recalled. Nothing may be written on the formulae page.

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

Keep revising

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

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Edexcel IGCSE 4MA1/2H, Monday 3 June 2024: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 15 of 25

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

Question 15, Calculator allowed

Question 15 - Exam Solution

Frequently asked questions

Keep revising

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