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Edexcel IGCSE 4MA1/1HR, Thursday 16 May 2024: Worked Solutions, Questions 14 to 23

Sir Faraz Hassan

10 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)4MA1/1HR - Higher Tier - Thursday 16 May 2024100 marks · 2 hours · Calculator allowed

Back to questions 1 to 13

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

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/1HR (Higher Tier), June 2024 series, sat Thursday 16 May 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.

Download printable PDF

All 23 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 14 to 23 of 23

(a) Expand the brackets and simplify
$(3x + 1)(2 - x)(4 + x)$ [3 marks]

(b) Simplify fully
$\left( \dfrac{a^{3}b}{a^{9}b^{5}} \right)^{-\dfrac{1}{2}}$ [3 marks]

(a)(b)

[Total 6 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

(a) a product of three brackets,

(3x + 1)(2 - x)(4 + x)

(b) a fraction in

a

and

b

raised to the power

-\dfrac{1}{2}

\left( \dfrac{a^{3}b}{a^{9}b^{5}} \right)^{-\dfrac{1}{2}}

Two separate pieces of algebra: one expansion, one index question.

Find

(a) the expansion written as a simplified cubic in

x

(b) one simplified term in

a

and

b

, with positive indices

Plan the Solution

(a) Expand two of the three brackets first, then multiply that quadratic by the bracket that is left over.
(a) Collect like terms and write the cubic in descending powers.
(b) Simplify inside the bracket first, using $\dfrac{a^{m}}{a^{n}} = a^{m-n}$ on each letter separately.
(b) Then apply the power outside the bracket, using $\left( a^{m} \right)^{n} = a^{mn}$ .

Worked Solution [6 marks]

Rule - expand a triple product two brackets at a time; for indices,

\dfrac{a^{m}}{a^{n}} = a^{m-n}

and

\left( a^{m} \right)^{n} = a^{mn}

Step 1: expand the first two brackets

(3x + 1)(2 - x) = 6x - 3x^2 + 2 - x

= -3x^2 + 5x + 2

(Reason: Each term of the first bracket multiplies each term of the second, and then

6x - x = 5x

. Any two of the three brackets may be chosen here; the other two pairings give

-x^2 - 2x + 8

and

3x^2 + 13x + 4

, and both finish at the same cubic.)

Step 2: multiply that quadratic by the third bracket

(-3x^2 + 5x + 2)(4 + x) = -12x^2 - 3x^3 + 20x + 5x^2 + 8 + 2x

(Reason: Each of the three terms of

-3x^2 + 5x + 2

multiplies each of the two terms of

4 + x

, so six terms appear before any tidying up.)

Step 3: collect like terms

-3x^3 + (5 - 12)x^2 + (20 + 2)x + 8

= -3x^3 - 7x^2 + 22x + 8

(Reason: The cubic term and the constant stand alone. The

x^2

column is the one that mixes signs, so it is written out in full before it is added, and the

x

column adds two positive terms.)

Step 4: simplify inside the bracket in part (b)

\dfrac{a^{3}b}{a^{9}b^{5}} = a^{3-9} b^{1-5} = a^{-6}b^{-4}

(Reason: Dividing powers of the same letter subtracts the indices, letter by letter:

3 - 9 = -6

and

1 - 5 = -4

. The single

b

on the top counts as

b^{1}

Step 5: apply the power outside the bracket

\left( a^{-6}b^{-4} \right)^{-\dfrac{1}{2}} = a^{-6 \times -\dfrac{1}{2}} \; b^{-4 \times -\dfrac{1}{2}}

= a^{3}b^{2}

(Reason: Raising a power to a power multiplies the indices. Both products are of two negatives, so both indices come out positive and no negative index is left in the answer.)

(a)

-3x^3 - 7x^2 + 22x + 8

(b)

a^{3}b^{2}

Verification

Check 1: Put

x = 1

into the original product and into the expansion. The brackets become

(3 + 1)(2 - 1)(4 + 1)

4 \times 1 \times 5 = 20

and

-3 - 7 + 22 + 8 = 20

Check 2: Put

x = -1

, which tests the signs that

x = 1

cannot. The brackets become

(-3 + 1)(2 + 1)(4 - 1)

-2 \times 3 \times 3 = -18

and

3 - 7 - 22 + 8 = -18

Check 3: Put

a = 2

and

b = 2

into part (b). The fraction becomes

\dfrac{16}{16\,384} = \dfrac{1}{1\,024}

, and a power of

-\dfrac{1}{2}

means the square root of the reciprocal.

\sqrt{1\,024} = 32

and the answer gives

2^{3} \times 2^{2} = 32

Check 4: Do part (b) the other way round: deal with the negative sign in the index first by turning the fraction upside down,

\left( \dfrac{a^{3}b}{a^{9}b^{5}} \right)^{-\dfrac{1}{2}} = \left( \dfrac{a^{9}b^{5}}{a^{3}b} \right)^{\dfrac{1}{2}}

\left( a^{6}b^{4} \right)^{\dfrac{1}{2}} = a^{3}b^{2}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Expand two of the three brackets	M1	for a correct method to expand two brackets, with at least 3 of the 4 terms correct (or 2 of 3 terms correct), eg $6x - 3x^2 + 2 - x$ . Do not award for two separate pair-expansions written side by side, nor for one pair-expansion with the remaining bracket simply added on rather than multiplied.	✓
(a) Multiply that quadratic by the remaining bracket	M1ft	follow through, dependent on the first M1 and on a quadratic having been reached, for a correct method to multiply by the third bracket - allow one further error, eg $-12x^2 - 3x^3 + 20x + 5x^2 + 8 + 2x$	✓
(a) Collect like terms	A1	for $-3x^3 - 7x^2 + 22x + 8$ or equivalent, but it must be simplified, eg $22x - 3x^3 - 7x^2 + 8$ . A correct answer scores full marks unless it follows obviously incorrect working.	✓
(a) Alternative - expand all three brackets in one go	Note	M2 for a complete expansion showing 8 terms of which at least 4 are correct (M1 only, for at least 4 correct terms from any number of terms), then the same A1 for the simplified answer.	✓
(a) No working shown	B2	if no working is shown, award B2 for 3 correct terms out of a maximum of 4.	✓
(b) One correct simplification	M1	for simplifying the $a$ and the $b$ terms in the fraction, or for applying the power $\dfrac{1}{2}$ to at least 3 of the 4 indices in $a^{3}, b, a^{9}, b^{5}$ , or for applying the negative power to at least 3 of those 4, eg $\left( a^{-6}b^{-4} \right)^{-\dfrac{1}{2}}$ or $\left( \dfrac{a^{9}b^{5}}{a^{3}b} \right)^{\dfrac{1}{2}}$	✓
(b) A second correct simplification	M1	for two of: simplifying the $a$ and the $b$ terms in the fraction; applying the power $\dfrac{1}{2}$ to at least 3 of the 4 indices; applying the negative power to at least 3 of the 4 indices, eg $\left( a^{6}b^{4} \right)^{\dfrac{1}{2}}$ or $\left( \dfrac{1}{a^{3}b^{2}} \right)^{-1}$	✓
(b) Final answer	A1	for $a^{3}b^{2}$ , accepting $\dfrac{1}{a^{-3}b^{-2}}$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 6/6

The diagram shows an isosceles triangle $EFG$ .

$EF = GF$
Angle $EFG = 130^\circ$
The area of triangle $EFG$ is $74$ cm²

Calculate the length of $EF$ .
Give your answer correct to $3$ significant figures. [3 marks]

[Total 3 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

Triangle

EFG

with

EF = GF

- the two equal sides meet at

F

Angle

EFG = 130^\circ

, the angle between those two sides

Area of triangle

EFG

74

cm²

Find

The length of

EF

, correct to

3

significant figures

Plan the Solution

The marked angle sits between the two equal sides, which is exactly the arrangement $\dfrac{1}{2}ab\sin C$ needs - no right angle and no perpendicular height required.
Call each equal side $x$ . Both $a$ and $b$ are then $x$ , so the formula gives one equation in $x^2$ .
Rearrange for $x^2$ , square root, and round only at the very end.

Worked Solution [3 marks]

Rule - Area from two sides and the angle between them:

\text{Area} = \dfrac{1}{2}ab\sin C

, where

C

is the angle between the sides

a

and

b

Step 1: call each equal side

x

and build the area equation

EF = GF = x

\dfrac{1}{2} \times x \times x \times \sin 130^\circ = 74

(Reason: Both sides in the formula are the same length here, and

130^\circ

is the angle between them, so the left-hand side is the area of this triangle.)

Step 2: rearrange to find

x^2

x^2 = \dfrac{2 \times 74}{\sin 130^\circ} = 193.200\ldots

(Reason: Multiply both sides by

2

to clear the half, then divide by

\sin 130^\circ

. Keep the whole decimal on the calculator - rounding at this point moves the third significant figure of the answer.)

Step 3: square root, then round

x = \sqrt{193.200\ldots} = 13.8996\ldots

EF = 13.9 \text{ cm}

(Reason: The square root undoes the square. Only now round:

13.8996\ldots

3

significant figures is

13.9

. The units are centimetres because the area was in square centimetres.)

EF = 13.9

cm (correct to

3

significant figures)

Verification

Check 1: Put the unrounded length back into the area formula:

\dfrac{1}{2} \times 13.8996\ldots \times 13.8996\ldots \times \sin 130^\circ

. It gives

74

cm², the area the question states.

Check 2: A different formula. The perpendicular from

F

EG

splits the

130^\circ

into two halves of

65^\circ

, so the base is

2 \times 13.8996\ldots \times \sin 65^\circ = 25.194\ldots

and the height is

13.8996\ldots \times \cos 65^\circ = 5.874\ldots

. Half of

25.194\ldots \times 5.874\ldots

74

again, from working that never uses

\sin C

Check 3: Is the size sensible?

\sin 130^\circ

is less than

1

, so the area has to be less than

\dfrac{1}{2}x^2

, which is

96.6\ldots

. The given

74

is comfortably below that bound, so a side near

13.9

cm is the right order of size.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
e.g. $\dfrac{1}{2} \times EF \times GF \times \sin 130^\circ = 74$ oe, or $EF \times GF \times \sin 130^\circ = 2 \times 74$ oe	M1	for setting up an equation using the area of a triangle formula	✓
$(EF^2 =) \dfrac{2 \times 74}{\sin 130^\circ}$ $(= 193.2\ldots)$ oe, or $(EF =) \sqrt{\dfrac{2 \times 74}{\sin 130^\circ}}$	M1	for a complete method to find $EF^2$ or $EF$	✓
$13.9$	A1	awrt 13.9	✓
A correct answer scores full marks unless it comes from obviously incorrect working.	Note	no mark of its own - it records how the A1 is applied	✓

Full marks: 3/3

The table gives information about the heights, in metres, of the pine trees in a nature reserve.

$\begin{array}{|c|c|}\hline \textbf{Height }(h\textbf{ metres}) & \textbf{Frequency} \\ \hline 0 < h \leq 2 & 5 \\ \hline 2 < h \leq 5 & 12 \\ \hline 5 < h \leq 10 & 18 \\ \hline 10 < h \leq 20 & 14 \\ \hline 20 < h \leq 35 & 9 \\ \hline \end{array}$

On the grid, draw a histogram for this information. [3 marks]

[Total 3 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

A grouped frequency table for

58

pine trees.

The class widths are

2

3

5

10

and

15

metres, so the classes are not equally wide.

A grid with frequency density up the vertical axis.

Find

The histogram: how tall to draw each of the five bars, and how wide. Because the classes are unequal, the height of a bar is the frequency density, not the frequency. Expect five different heights, one of which reaches

4

at the top of the grid.

Plan the Solution

Work out the width of every class first, by subtracting its two boundaries.
Divide each frequency by its own class width to get that class's frequency density.
Draw each bar from its lower boundary to its upper boundary at that height, with no gaps between bars.
Check by reading the areas back: a bar's area must return the frequency it came from.

Worked Solution [3 marks]

Rule - Histogram with unequal class widths:

\text{frequency density} = \dfrac{\text{frequency}}{\text{class width}}

, so the AREA of a bar is its frequency and the HEIGHT is not.

Step 1: Work out each class width

2 - 0 = 2

5 - 2 = 3

10 - 5 = 5

20 - 10 = 10

35 - 20 = 15

(Reason: The width is the upper boundary minus the lower boundary. The five widths come out as

2

3

5

10

and

15

- all different, which is exactly why the frequencies cannot be used as heights.)

Step 2: Divide each frequency by its own class width

\dfrac{5}{2} = 2.5

\dfrac{12}{3} = 4

\dfrac{18}{5} = 3.6

\dfrac{14}{10} = 1.4

\dfrac{9}{15} = 0.6

(Reason: Each division shares the frequency out over the metres its class covers, so a frequency density is trees per metre of height. Take the width from Step 1, not the upper boundary: the

5 < h \leq 10

class is

5

metres wide, not

10

Step 3: Plot the heights and draw the bars

2.5 = 25 \times 0.1

4 = 40 \times 0.1

3.6 = 36 \times 0.1

1.4 = 14 \times 0.1

0.6 = 6 \times 0.1

(Reason: One small square up the frequency density axis is

0.1

, so every height lands exactly on a ruled line and nothing has to be judged by eye. Each bar runs from its lower boundary to its upper boundary, so the five bars fill the axis from

0

35

with no gaps between them.)

0 < h \leq 2

frequency density

2.5

2 < h \leq 5

frequency density

4

5 < h \leq 10

frequency density

3.6

10 < h \leq 20

frequency density

1.4

20 < h \leq 35

frequency density

0.6

Verification

Check 1: Read the areas back. Width times height must return the frequency that bar was drawn from.

2 \times 2.5 = 5

3 \times 4 = 12

5 \times 3.6 = 18

10 \times 1.4 = 14

15 \times 0.6 = 9

Check 2: Add the five areas. The whole histogram has to hold every tree, and the table lists

58

of them.

5 + 12 + 18 + 14 + 9 = 58

Check 3: A histogram drawn at half height everywhere is still in the correct ratio, and halving ours gives

1.25

2

1.8

0.7

and

0.3

- the five heights the mark scheme names in its special case. Doubling those must return our heights.

2 \times 1.25 = 2.5

2 \times 2 = 4

2 \times 1.8 = 3.6

2 \times 0.7 = 1.4

2 \times 0.3 = 0.6

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Any $3$ correct frequency densities, or $3$ correct bars	M1	Three of $\dfrac{5}{2} = 2.5$ , $\dfrac{12}{3} = 4$ , $\dfrac{18}{5} = 3.6$ , $\dfrac{14}{10} = 1.4$ , $\dfrac{9}{15} = 0.6$ , or equivalent. The working alone earns this - the bars need not be drawn yet.	✓
Any $4$ correct frequency densities, or $4$ correct bars	M1	One more than the first method mark: $4$ of the five, by working or by drawing. This is why a single slip, such as dividing by an upper boundary, still leaves $2$ marks available.	✓
A completely correct histogram	A1	All five bars: correct widths, edge to edge from $0$ to $35$ , and correct heights. Marked with the overlay.	✓
All five bars of correct width, with heights in the correct ratio	SC B2	Awarded if no other marks are earned - for example bars drawn at $1.25$ , $2$ , $1.8$ , $0.7$ and $0.3$ . Those are every frequency density halved, which is what happens when the frequency is divided by twice the class width.	✓
Guidance	Note	A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 3/3

(a) $\left(\sqrt[4]{k^{12}}\right)^{5} = k^{n}$
Work out the value of $n$ . [1 mark]

(b) Write $\dfrac{7}{2 - \sqrt{3}}$ in the form $\sqrt{c} + d$ , where $c$ and $d$ are integers.
You must show your working clearly. [3 marks]

n =(b)

[Total 4 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

(a)

\left(\sqrt[4]{k^{12}}\right)^{5} = k^{n}

- a fourth root raised to a power, written as a single power of

k

(b)

\dfrac{7}{2 - \sqrt{3}}

- a fraction whose denominator contains a surd.

Find

(a) the value of

n

. (b) the integers

c

and

d

for which

\dfrac{7}{2 - \sqrt{3}} = \sqrt{c} + d

. Note the order: the surd is written first, so the whole number

d

is the second term.

Plan the Solution

Part (a): rewrite the fourth root as the power $\dfrac{1}{4}$ , then multiply the indices twice - once for the root and once for the outer power $5$ .
Part (b): multiply the top and the bottom by the conjugate $2 + \sqrt{3}$ . The denominator becomes a difference of two squares, so the surd disappears from it.
Finish part (b) by turning the $7\sqrt{3}$ term into a single square root, because the answer has to look like $\sqrt{c} + d$ .

Worked Solution [4 marks]

Rule - Index laws:

\sqrt[n]{x} = x^{\dfrac{1}{n}}

and

\left(x^{a}\right)^{b} = x^{ab}

. Rule - Rationalising a denominator: multiply above and below by the conjugate, and move a coefficient inside a root with

a\sqrt{b} = \sqrt{a^{2}b}

Step 1: write the fourth root of

k^{12}

as a power

\sqrt[4]{k^{12}} = \left(k^{12}\right)^{\dfrac{1}{4}}

\left(k^{12}\right)^{\dfrac{1}{4}} = k^{12 \times \dfrac{1}{4}} = k^{3}

(Reason: An

n

th root is the power

\dfrac{1}{n}

, and a power raised to a power multiplies the two indices. Here

12 \times \dfrac{1}{4} = 3

, so the fourth root of

k^{12}

is simply

k^{3}

Step 2: raise

k^{3}

to the power

5

\left(k^{3}\right)^{5} = k^{3 \times 5} = k^{15}

k^{n} = k^{15} \implies n = 15

(Reason: The same law applies again:

3 \times 5 = 15

. The two sides are powers of the same base, so the indices must be equal and

n = 15

. The whole of part (a) is one

\dfrac{12}{4} \times 5

calculation.)

Step 3: multiply the top and the bottom by the conjugate

2 + \sqrt{3}

\dfrac{7}{2 - \sqrt{3}} = \dfrac{7\left(2 + \sqrt{3}\right)}{\left(2 - \sqrt{3}\right)\left(2 + \sqrt{3}\right)}

(Reason: The conjugate of

2 - \sqrt{3}

2 + \sqrt{3}

- the same two terms with the sign between them reversed. Multiplying top and bottom by the same thing does not change the value of the fraction, only how it is written.)

Step 4: expand the numerator and the denominator

7\left(2 + \sqrt{3}\right) = 14 + 7\sqrt{3}

\left(2 - \sqrt{3}\right)\left(2 + \sqrt{3}\right) = 4 + 2\sqrt{3} - 2\sqrt{3} - 3 = 1

\dfrac{14 + 7\sqrt{3}}{1} = 14 + 7\sqrt{3}

(Reason: The denominator is a difference of two squares,

2^{2} - \left(\sqrt{3}\right)^{2}

: the two middle terms cancel and

4 - 3 = 1

is left. That is the whole point of the conjugate - the surd has gone from the bottom.)

Step 5: write the

7\sqrt{3}

term as a single square root

7\sqrt{3} = \sqrt{7^2 \times 3} = \sqrt{147}

(Reason: A number in front of a square root can be taken inside it, but only after it has been squared:

a\sqrt{b} = \sqrt{a^{2}b}

. The answer must have the shape

\sqrt{c} + d

, so the coefficient has nowhere else to go.)

Step 6: put the answer in the form

\sqrt{c} + d

14 + 7\sqrt{3} = \sqrt{147} + 14

c = 147

d = 14

(Reason: The question asks for the surd first and the integer second, so the two terms are swapped round. Both

147

and

14

are integers, which is what the question demands of

c

and

d

(a)

n = 15

(b)

\sqrt{147} + 14

Verification

Check 1: Part (a) with a number in place of

k

. Put

k = 2

into the original expression, and into

k^{15}

separately.

\left(\sqrt[4]{2^{12}}\right)^{5} = \left(\sqrt[4]{4096}\right)^{5} = 8^{5} = 32768

and

2^{15} = 32768

Check 2: Part (b) backwards. If the answer is right then multiplying it by the original denominator

2 - \sqrt{3}

must give back the numerator

7

\left(14 + 7\sqrt{3}\right)\left(2 - \sqrt{3}\right) = 28 - 14\sqrt{3} + 14\sqrt{3} - 21 = 7

Check 3: Part (b) as decimals, which is an entirely different test from the algebra. Work out the printed fraction on the calculator, then work out

\sqrt{147} + 14

\dfrac{7}{2 - \sqrt{3}} \approx 26.1244

and

\sqrt{147} + 14 \approx 12.1244 + 14 \approx 26.1244

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Give the single index $n$	B1	$15$ . The scheme also accepts the answer written as $k^{15}$ rather than as a bare number.	✓
(b) Multiply the numerator and the denominator by the conjugate	M1	For multiplying the numerator and the denominator of the fraction by $2 + \sqrt{3}$ or by $-2 - \sqrt{3}$ , for example $\dfrac{7\left(2 + \sqrt{3}\right)}{\left(2 - \sqrt{3}\right)\left(2 + \sqrt{3}\right)}$ or $\dfrac{7\left(-2 - \sqrt{3}\right)}{\left(2 - \sqrt{3}\right)\left(-2 - \sqrt{3}\right)}$ .	✓
(b) Expand to reach a fraction with a rational denominator	M1	Dependent on the previous method mark. For example $\dfrac{14 + 7\sqrt{3}}{4 + 2\sqrt{3} - 2\sqrt{3} - 3}$ or $\dfrac{14 + 7\sqrt{3}}{4 - 3}$ or $\dfrac{14 + 7\sqrt{3}}{1}$ , and equally the negative versions $\dfrac{-14 - 7\sqrt{3}}{-4 - 2\sqrt{3} + 2\sqrt{3} + 3}$ or $\dfrac{-14 - 7\sqrt{3}}{-4 + 3}$ or $\dfrac{-14 - 7\sqrt{3}}{-1}$ .	✓
(b) The answer, with the working shown	A1	$\sqrt{147} + 14$ , dependent on both method marks. Working is required, so the answer alone does not earn this mark - which is why the question says to show the working clearly.	✓
(b) Special case - the right answer without the method	SC	SCB1 for $\sqrt{147} + 14$ gained with no method marks awarded, and SCB2 for $\sqrt{147} + 14$ gained with the first method mark awarded. The error being marked is not an arithmetic one: it is answering from a calculator display, or jumping to $14 + 7\sqrt{3}$ with no rationalising shown, on a question whose marks are mostly for the method.	✓

Full marks: 4/4

The diagram shows two mathematically similar glass storage jars, $P$ and $Q$

The height of jar $P$ is $30$ cm
The height of jar $Q$ is $12$ cm

Given that

$\text{surface area of }P - \text{surface area of }Q = 178.5\text{ cm}^2$

find the surface area of jar $P$ [4 marks]

cm²

[Total 4 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

Two mathematically similar jars,

P

and

Q

Height of

P

30

cm, height of

Q

12

Surface area of

P

minus surface area of

Q

178.5

cm²

Find

The surface area of jar

P

, in cm²

Plan the Solution

Divide the two heights to get the length scale factor.
Square it, because areas scale by the square of the length scale factor.
That turns the two surface areas into $25$ parts and $4$ parts of one size, so the given difference is $21$ of those parts.
Divide $178.5$ by $21$ to get one part, then take $25$ of them.

Worked Solution [4 marks]

Rule - Similar solids: if every length is multiplied by

k

, then every area is multiplied by

k^2

and every volume by

k^3

Step 1: Work out the length scale factor

\dfrac{30}{12} = \dfrac{5}{2} = 2.5

(Reason: the jars are mathematically similar, so every length on

P

2.5

times the matching length on

Q

)

Step 2: Square it to get the area scale factor

\dfrac{5^2}{2^2} = \dfrac{25}{4} = 6.25

(Reason: an area is a length times a length, so it is multiplied by

2.5

twice, not once; the surface areas are therefore in the ratio

25

4

)

Step 3: Turn that ratio into parts

25 - 4 = 21

(Reason: the surface areas are

25

parts and

4

parts of one size, which is the same as calling them

25x

and

4x

, so the difference is the number of parts between them)

Step 4: Work out one part

\dfrac{178.5}{21} = 8.5

(Reason: the difference of

178.5

cm² given in the question is exactly those

21

parts)

Step 5: Scale up to jar P

25 \times 8.5 = 212.5

(Reason: jar

P

25

of those parts, so multiply one part by

25

)

Surface area of jar

P

212.5

cm²

Verification

Check 1: Work backwards from the answer. Divide it by the area scale factor

6.25

to get the smaller surface area, then take the difference.

\dfrac{212.5}{6.25} = 34

and

212.5 - 34 = 178.5

, the difference the question gives

Check 2: Reach it from the other end. The difference is

5.25

times the smaller surface area, so divide by

5.25

first and scale up afterwards.

\dfrac{178.5}{5.25} = 34

and

34 \times 6.25 = 212.5

Check 3: Is it sensible? The two surface areas must be in the ratio

25

4

, and the larger one must be bigger than the difference it was built from.

\dfrac{212.5}{34} = 6.25

, and

212.5

is bigger than

178.5

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Length scale factor from the two heights, e.g. $\dfrac{30}{12} = 2.5$ or the ratio $5$ to $2$	M1	for a method to find the ratio for the lengths, or the linear scale factor	✓
An equation in the surface areas, e.g. $5^2x - 2^2x = 178.5$ or $6.25y - y = 178.5$	M1	for setting up an equation using the surface areas	✓
A complete method, e.g. $\dfrac{178.5}{5^2 - 2^2} \times 5^2$ or $\dfrac{178.5}{6.25 - 1} \times 6.25$	M1	for a complete method	✓
The surface area of jar P	A1	for $212.5$ or an equivalent value	✓
A correct answer with no working	Note	a correct answer scores full marks, unless it comes from obviously incorrect working	✓
Using the length ratio instead of the area ratio, giving $1487.5$	Note	this answer comes from dividing by 3 instead of 21, that is from using the ratio 5 to 2 without squaring it. The first method mark is still earned, because the length ratio itself has been found correctly; the second and the third need the area ratio, so the slip scores 1 of the 4 marks	✓

Full marks: 4/4

The curve $C$ has equation $y = x^3 - 8x^2 - 12x + 5$
Curve $C$ has exactly two stationary points, one at the point $P$ and one at the point $Q$ , such that
$x \text{ coordinate of point } P > x \text{ coordinate of point } Q$
Work out the coordinates of the point $P$ .
You must show clear algebraic working. [5 marks]

[Total 5 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

The curve

C

has equation

y = x^3 - 8x^2 - 12x + 5

, a cubic

There are exactly two stationary points,

P

and

Q

The

x

coordinate of

P

is the greater of the two

Find

The coordinates of

P

, so both an

x

value and a

y

value

Plan the Solution

Differentiate to get $\dfrac{\text{d}y}{\text{d}x}$ , which is the gradient function.
Set the gradient function to zero and solve the quadratic; that gives both stationary $x$ values.
Take the greater of the two, because the question says $P$ is the point with the greater $x$ coordinate.
Substitute that $x$ into the equation of $C$ , not into the derivative, to get the $y$ coordinate.

Worked Solution [5 marks]

Stationary points: solve

\dfrac{\text{d}y}{\text{d}x} = 0

for

x

, then put each

x

back into the equation of the curve to find

y

Step 1: differentiate the equation of

C

y = x^3 - 8x^2 - 12x + 5

\dfrac{\text{d}y}{\text{d}x} = 3x^2 - 16x - 12

(Reason: Each term drops one power and is multiplied by the old power:

x^3

gives

3x^2

-8x^2

gives

-16x

-12x

gives

-12

, and the constant

5

disappears.)

Step 2: a stationary point has zero gradient

3x^2 - 16x - 12 = 0

(Reason: At a stationary point the tangent is horizontal, so the gradient function is zero there. This is a quadratic, which is why

C

has two stationary points and not one.)

Step 3: solve the quadratic

(3x + 2)(x - 6) = 0

x = -\dfrac{2}{3} \text{ or } x = 6

(Reason: Expanding

(3x + 2)(x - 6)

gives

3x^2 - 18x + 2x - 12 = 3x^2 - 16x - 12

, so the factorisation is right, and a product is zero only when one bracket is zero. The formula gives the same two values, since

\dfrac{16 \pm \sqrt{256 + 144}}{6} = \dfrac{16 \pm 20}{6}

Step 4: pick the point with the greater

x

coordinate

6 > -\dfrac{2}{3}

(Reason: The question defines

P

as the stationary point whose

x

coordinate is the greater one, so

P

has

x = 6

and

Q

has

x = -\dfrac{2}{3}

. Only

P

is wanted here.)

Step 5: substitute

x = 6

into the equation of the curve

y = 6^3 - 8 \times 6^2 - 12 \times 6 + 5

y = 216 - 288 - 72 + 5 = -139

(Reason: The stationary point sits on

C

, so its

y

coordinate comes from the curve's own equation, never from the derivative. Note that

8 \times 6^2

squares first and then multiplies, giving

288

, not

48^2

P = (6, -139)

Verification

Check 1: Differentiate a second time,

\dfrac{\text{d}^2y}{\text{d}x^2} = 6x - 16

, and test both stationary values. At

x = 6

6 \times 6 - 16 = 20

, which is positive, so

P

is a minimum. At

x = -\dfrac{2}{3}

6 \times \left(-\dfrac{2}{3}\right) - 16 = -20

, which is negative, so

Q

is a maximum. Two stationary points, one of each kind, exactly as the question states.

Check 2: Rebuild the

y

coordinate by nested multiplication,

((x - 8)x - 12)x + 5

, which uses the coefficients in a different order and never squares or cubes anything.

((6 - 8) \times 6 - 12) \times 6 + 5 = -24 \times 6 + 5 = -144 + 5 = -139

Check 3: Test both roots at once against the quadratic's coefficients: for

ax^2 + bx + c = 0

the roots sum to

-\dfrac{b}{a}

and multiply to

\dfrac{c}{a}

6 + \left(-\dfrac{2}{3}\right) = \dfrac{16}{3}

and

6 \times \left(-\dfrac{2}{3}\right) = -4

, matching

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

and

\dfrac{-12}{3} = -4

, so neither root is wrong.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Differentiate: $\dfrac{\text{d}y}{\text{d}x} = 3x^2 - 16x - 12$	M1	for differentiation with at least two terms correct	✓
Set the derivative equal to zero: $3x^2 - 16x - 12 = 0$	M1ft	(dep on previous M1) for their $\dfrac{\text{d}y}{\text{d}x} = 0$	✓
Solve the quadratic, e.g. $(3x + 2)(x - 6) = 0$ , or $x = \dfrac{-(-16) \pm \sqrt{(-16)^2 - 4 \times 3 \times (-12)}}{2 \times 3}$	M1ft	(dep on 1st M1) for the correct $x$ value (of $6$ ), ignore the other $x$ value, or for solving their three term quadratic equation using any correct method. If factorising, allow brackets which expanded give 2 out of 3 terms correct. If using the formula, allow one sign error and some simplification - as far as $\dfrac{16 \pm \sqrt{256 + 144}}{6}$ . If completing the square, then as far as shown on the left-hand side. The award of this mark implies the previous M mark	✓
Substitute into the equation of the curve: $6^3 - 8 \times 6^2 - 12 \times 6 + 5$	M1ft	(dep on 1st M1) for $x = 6$ substituted into the correct equation for curve $C$ , or (dep on 1st M1 and two values for $x$ ) for their greatest $x$ value substituted into the correct equation for curve $C$ , ignoring any attempt to substitute their least $x$ value	✓
Working required. Coordinates of $P$ : $(6, -139)$	A1	(dep on M2) cao	✓

Full marks: 5/5

(a) Write $2x^2 - 11x + 9$ in the form $a(x - b)^2 - c$ , where $a$ , $b$ and $c$ are numbers to be found. [3 marks]

A curve $C$ has equation $y = 2(x - 3)^2 - 11(x - 3) + 9$
The minimum point on $C$ is $P$

(b) Work out the coordinates of $P$ [2 marks]

(a)(b)

[Total 5 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

the quadratic

2x^2 - 11x + 9

, to be written as

a(x - b)^2 - c

the curve

C

with equation

y = 2(x - 3)^2 - 11(x - 3) + 9

P

is the lowest point on

C

Find

the three numbers

a

b

and

c

the coordinates of

P

, which will be a pair of fractions

Plan the Solution

Take the factor $2$ out of the $x^2$ and $x$ terms only, leaving the $+9$ outside the bracket.
Complete the square inside the bracket, then multiply the bracket by $2$ and combine what is left with the $+9$ .
For part (b), notice that every $x$ of part (a) has become $x - 3$ , so the completed square of part (a) can be reused with $x - 3$ written in place of $x$ .
A squared bracket is never negative, so $y$ is smallest when that bracket is zero. Solve that equation for $x$ and read the $y$ value straight off.

Worked Solution [5 marks]

Rule - Completing the square: take the coefficient of

x^2

outside a bracket, halve the coefficient of

x

inside that bracket, then subtract the square of that half. A squared bracket is never negative, so the whole expression is at its smallest when the bracket equals

0

Step 1: (a) Take the factor of

2

out of the

x

terms

2x^2 - 11x + 9 = 2\left(x^2 - \dfrac{11}{2}x\right) + 9

(Reason: Only the

x^2

and

x

terms carry the factor of

2

. Dividing

-11

2

gives the

-\dfrac{11}{2}

inside, and the

+9

is left outside so nothing has been changed.)

Step 2: Complete the square inside the bracket

x^2 - \dfrac{11}{2}x = \left(x - \dfrac{11}{4}\right)^2 - \dfrac{121}{16}

(Reason: Half of

\dfrac{11}{2}

\dfrac{11}{4}

, so the bracket is

\left(x - \dfrac{11}{4}\right)^2

. Squaring

\dfrac{11}{4}

gives

\dfrac{121}{16}

, and that has to be taken away again because the square put it there.)

Step 3: Multiply the bracket back by

2

and tidy the constant

2\left[\left(x - \dfrac{11}{4}\right)^2 - \dfrac{121}{16}\right] + 9

2 \times \dfrac{121}{16} = \dfrac{121}{8}

\dfrac{121}{8} - 9 = \dfrac{49}{8}

2x^2 - 11x + 9 = 2\left(x - \dfrac{11}{4}\right)^2 - \dfrac{49}{8}

(Reason: The

\dfrac{121}{16}

is inside the bracket, so it is doubled as well and becomes

\dfrac{121}{8}

. Taking that away and adding the

9

leaves

\dfrac{49}{8}

to be taken away.)

Step 4: Read off

a

b

and

c

a = 2

b = \dfrac{11}{4}

c = \dfrac{49}{8}

(Reason: Comparing with

a(x - b)^2 - c

term by term. In decimals these are

2

2.75

and

6.125

, which the mark scheme also accepts.)

Step 5: (b) Spot that

C

is part (a) with

x - 3

written in place of

x

y = 2(x - 3)^2 - 11(x - 3) + 9

y = 2\left((x - 3) - \dfrac{11}{4}\right)^2 - \dfrac{49}{8}

(Reason: Part (a) is true for whatever is written in place of

x

, and here that is

x - 3

in all three places. So no fresh completing of the square is needed.)

Step 6: The square is at its smallest when it is zero

(x - 3) - \dfrac{11}{4} = 0

(Reason: A square is never negative, and the number in front of it,

2

, is positive. So

y

is least exactly when the squared bracket is

0

Step 7: Add the

3

back to get the

x

coordinate

x - 3 = \dfrac{11}{4}

x = \dfrac{11}{4} + 3

\dfrac{11}{4} + \dfrac{12}{4} = \dfrac{23}{4}

(Reason: The bracket is zero when

x - 3 = \dfrac{11}{4}

, so add

3

to both sides. Writing

3

\dfrac{12}{4}

keeps both fractions over the same denominator.)

Step 8: Read off the

y

coordinate

y = -\dfrac{49}{8}

(Reason: With the squared bracket equal to

0

, all that is left of

y

is the

-\dfrac{49}{8}

. In decimals

P

(5.75,\, -6.125)

(a)

2\left(x - \dfrac{11}{4}\right)^2 - \dfrac{49}{8}

(b)

P

is at

\left(\dfrac{23}{4},\, -\dfrac{49}{8}\right)

Verification

Check 1: Expand part (a) again. The two constants must recombine to the original

+9

: doubling

\dfrac{121}{16}

gives

\dfrac{121}{8}

, and the answer takes

\dfrac{49}{8}

away from it.

\dfrac{121}{8} - \dfrac{49}{8} = 9

Check 2: Put

x = \dfrac{23}{4}

straight into the equation of

C

. Then

x - 3 = \dfrac{11}{4}

, so the three terms are

2 \times \dfrac{121}{16}

-11 \times \dfrac{11}{4}

and

+9

, written over

8

\dfrac{121}{8} - \dfrac{242}{8} + \dfrac{72}{8} = -\dfrac{49}{8}

Check 3: A different route to the same

x

. The roots of

2x^2 - 11x + 9 = 0

are

x = 1

and

x = 4.5

, and a parabola is symmetric about the midpoint of its roots. That midpoint is the

b

of part (a), and adding the shift of

3

gives the

x

coordinate of

P

\dfrac{1 + 4.5}{2} = 2.75

and

2.75 + 3 = 5.75

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The factor of $2$ taken out of the $x$ terms: $2\left(x^2 - \dfrac{11}{2}x\right)$	M1	for taking out a factor of $2$	✓
(a) The square completed inside the bracket: $2\left[\left(x - \dfrac{11}{4}\right)^2 - \dfrac{121}{16}\right]$	M1	for correctly completing the square	✓
(a) $2\left(x - \dfrac{11}{4}\right)^2 - \dfrac{49}{8}$	A1	or equivalent, for example $2(x - 2.75)^2 - 6.125$ . Allow $a = 2$ , $b = \dfrac{11}{4}$ , $c = \dfrac{49}{8}$ stated separately. A correct answer scores full marks unless it follows obviously incorrect working.	✓
(a) Alternative method: expand $a(x - b)^2 - c$ to $ax^2 - 2abx + ab^2 - c$ and compare coefficients.	Note	The alternative scheme carries the same three marks: M1 for the correct expansion, M1 for setting up $-2ab = -11$ and $ab^2 - c = 9$ , then A1 for the same answer.	✓
(a) Answer left as $2\left(x - \dfrac{11}{4}\right)^2$ with the constant term never dealt with.	SC B1	a special case: if no other marks are awarded, this earns $1$ mark. The error is stopping before the $\dfrac{121}{16}$ has been doubled and combined with the $+9$ .	✓
(b) $\left(\dfrac{23}{4},\, -\dfrac{49}{8}\right)$	B2	or equivalent, for example $(5.75,\, -6.125)$ . Follow through is allowed from the candidate's own $b$ and $c$ in part (a).	✓
(b) Only one coordinate right, for example $x = \dfrac{23}{4}$ with no $y$ value.	(B1)	for one correct coordinate, follow through allowed. Shown in brackets because it is the partial award inside the B2 above, not a mark on top of it.	✓

Full marks: 5/5

There are $25$ beads in a jar such that

$6$ beads are green
$x$ beads are amber, where $x > 9$
the rest of the beads are cream

Idris takes at random two of the beads from the jar.

The probability that Idris takes one amber bead and one cream bead is $\dfrac{22}{75}$

Calculate the probability that Idris takes $2$ cream beads from the jar.
Show clear algebraic working. [5 marks]

[Total 5 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

A jar holds

25

beads:

6

green,

x

amber and the rest cream, with

x > 9

Two beads are taken at random, one after the other and not replaced.

The probability of one amber and one cream bead is

\dfrac{22}{75}

Find

The probability that both beads are cream, with clear algebraic working.

Plan the Solution

Write the number of cream beads in terms of $x$ : the three colours use up all $25$ beads.
Multiply along one branch for amber then cream, then double it, because one of each colour happens in two orders.
Set that equal to $\dfrac{22}{75}$ and clear the fractions to get a quadratic in $x$ .
Factorise, then use $x > 9$ to choose between the two roots.
Finish with two cream beads, remembering that the second denominator drops to $24$ .

Worked Solution [5 marks]

Rule - Taking two objects without replacement: the second fraction has a denominator one smaller, and one of each colour happens in two orders, so

P(\text{one of each}) = 2 \times P(\text{first colour, then second})

Step 1: Count the cream beads in terms of

x

\text{cream} = 25 - 6 - x = 19 - x

(Reason: (Reason: green, amber and cream are the only colours, so the three counts add to

25

and the cream count is whatever is left.))

Step 2: Write the probability of one amber and one cream bead

P(\text{amber, then cream}) = \dfrac{x}{25} \times \dfrac{19-x}{24}

P(\text{cream, then amber}) = \dfrac{19-x}{25} \times \dfrac{x}{24}

P(\text{one of each}) = 2 \times \dfrac{x}{25} \times \dfrac{19-x}{24}

(Reason: (Reason: the first bead is not replaced, so the second denominator is

24

. The two orders give the same product, so one of them is doubled.))

Step 3: Clear the fractions to get a quadratic in

x

2 \times \dfrac{x}{25} \times \dfrac{19-x}{24} = \dfrac{22}{75}

\dfrac{x(19-x)}{300} = \dfrac{22}{75}

x(19-x) = 300 \times \dfrac{22}{75} = 88

x^2 - 19x + 88 = 0

(Reason: (Reason: the factor

2

has already been folded into the left-hand side, so multiplying both sides by the common denominator clears every fraction at once and collecting the terms on one side gives a quadratic.))

Step 4: Solve the quadratic, then use

x > 9

x^2 - 19x + 88 = 0

(x - 8)(x - 11) = 0

x = 8 \text{ or } x = 11

x > 9 \implies x = 11

(Reason: (Reason:

8 \times 11 = 88

and

8 + 11 = 19

, so the brackets are correct. Both roots satisfy the given probability, which is exactly why the question states

x > 9

- it is the condition that decides between them.))

Step 5: Work out the probability of two cream beads

\text{cream} = 19 - 11 = 8

P(\text{2 cream}) = \dfrac{8}{25} \times \dfrac{7}{24}

= \dfrac{56}{600} = \dfrac{7}{75}

(Reason: (Reason: once one cream bead has been taken there are

7

cream beads left out of

24

, and

\dfrac{56}{600}

cancels by

8

.))

\dfrac{7}{75}

Verification

Check 1: Put

x = 11

back into the probability the question gives. The jar then holds

6

green,

11

amber and

8

cream beads, which is

25

altogether.

2 \times \dfrac{11}{25} \times \dfrac{8}{24} = \dfrac{88}{300} = \dfrac{22}{75}

Check 2: Count pairs instead of using order. There are

\dfrac{8 \times 7}{2} = 28

ways to choose

2

cream beads from

8

, and

\dfrac{25 \times 24}{2} = 300

ways to choose any

2

beads from

25

. Order never enters this method, so it is independent of the branch working.

\dfrac{28}{300} = \dfrac{7}{75}

Check 3: Test the rejected root. If

x = 8

there would be

11

cream beads, and the answer would be different, so the condition

x > 9

really is doing the work of choosing the root.

\dfrac{11}{25} \times \dfrac{10}{24} = \dfrac{11}{60}

, which is not

\dfrac{7}{75}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A correct product for one amber and one cream bead in one order: $\dfrac{x}{25} \times \dfrac{25-(x+6)}{24}$ or $\dfrac{x}{25} \times \dfrac{19-x}{24}$	M1	for a correct product for P(amber, cream), oe	✓
Double it and set it equal to the given probability: $2 \times \left( \dfrac{x}{25} \times \dfrac{19-x}{24} \right) = \dfrac{22}{75}$	M1	for setting up a correct equation in $x$ , oe	✓
Clear the fractions: $2x^2 - 38x + 176 (= 0)$ oe eg $x^2 - 19x + 88 (= 0)$	M1	for dealing with the fractions to set up a correct quadratic equation	✓
$x = 11$ or cream $= 25 - 6 - 11 (= 8)$	M1	for $x = 11$ or cream $= 25 - 6 - 11 (= 8)$ .	✓
Working required. Final answer $\dfrac{7}{75}$	A1	oe eg $0.093...$ or $9.3...\%$ . Working is required, so an answer with no algebraic working scores no marks.	✓

Full marks: 5/5

The diagram shows a cuboid $ABCDEFGH$ .
The face $ADEH$ is the horizontal base.

$AB = 8$ cm, $AD = 12$ cm and $DE = 20$ cm

The point $M$ is the centre of the base $ADEH$ and the point $P$ is the midpoint of the edge $CF$

Find the size of angle $BMP$ .
Give your answer correct to one decimal place. [6 marks]

[Total 6 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

ABCDEFGH

is a cuboid whose horizontal base is the rectangle

ADEH

AB = 8

cm,

AD = 12

cm and

DE = 20

cm.

M

is the centre of the base, so it is the midpoint of each diagonal of

ADEH

P

is the midpoint of

CF

, a top edge running from the front of the solid to the back.

Find

The size of angle

BMP

, correct to one decimal place. Triangle

BMP

is a slanting slice through the solid with no right angle in it, so expect the cosine rule rather than ordinary trigonometry.

Plan the Solution

The cosine rule needs all three sides, so find $MP$ , $BM$ and $BP$ first. Each one is the hypotenuse of a right-angled triangle hidden inside the cuboid.
Every horizontal distance measured from $M$ is a half, because $M$ is the centre of the base and not a corner of it: half of $AD$ is $6$ cm and half of $DE$ is $10$ cm.
$B$ and $P$ are both $8$ cm above the base, so $BP$ is purely horizontal - the easiest of the three.
Finish with the cosine rule rearranged for the angle, and round only on the very last line.

Worked Solution [6 marks]

Rule - Cosine rule for an angle:

\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}

, where

a

is the side facing the angle. Every length inside a cuboid comes from Pythagoras.

Step 1: Work out

MP

MP = \sqrt{(\text{half of } AD)^2 + AB^2}

\sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ cm}

(Reason:

M

and

P

are both

10

cm from the front face, so nothing changes in the front-to-back direction. Going from

M

P

you move across by half of

AD

, which is

6

cm, and straight up by

AB = 8

cm.)

Step 2: Work out

BM

AM = \sqrt{(\text{half of } AD)^2 + (\text{half of } DE)^2}

\sqrt{6^2 + 10^2} = \sqrt{36 + 100} = \sqrt{136}

BM = \sqrt{AM^2 + AB^2}

\sqrt{136 + 64} = \sqrt{200} = 14.14 \text{ cm}

(Reason:

AM

is half of the base diagonal

AE

: across

6

cm and back

10

cm.

B

is straight above

A

, so

AM

and

AB

meet at a right angle and Pythagoras applies again. Keep the

200

rather than the decimal, because it goes straight into the cosine rule.)

Step 3: Work out

BP

BP = \sqrt{AD^2 + (\text{half of } DE)^2}

\sqrt{12^2 + 10^2} = \sqrt{144 + 100} = \sqrt{244} = 15.62 \text{ cm}

(Reason:

B

is directly above

A

and

P

is level with it, both

8

cm up, so

BP

has no vertical part at all. Flatten it onto the top face: across the full

12

cm, then back half of

20

cm.)

Step 4: Put the three sides into the cosine rule

\cos BMP = \dfrac{BM^2 + MP^2 - BP^2}{2 \times BM \times MP}

\dfrac{200 + 100 - 244}{2 \times \sqrt{200} \times 10} = 0.19799

(Reason: The side facing angle

BMP

BP

, so

BP^2 = 244

is the one that gets subtracted. The squares

200

and

100

go in exactly as they came out of Pythagoras, with no rounding on the way.)

Step 5: Take the inverse cosine, then round

BMP = \cos^{-1}(0.19799) = 78.58^{\circ}

BMP = 78.6^{\circ} \text{ to 1 decimal place}

(Reason: A positive cosine means the angle is acute, which is what the picture shows. Rounding is left to this last line so that nothing is lost on the way, and the mark scheme accepts anything from

78.5

78.7

Angle

BMP = 78.6^{\circ}

Verification

Check 1: Drop the cuboid onto axes, with

A

at the origin,

D

(12, 0, 0)

H

(0, 20, 0)

and

B

(0, 0, 8)

. Then

M

(6, 10, 0)

and

P

(12, 10, 8)

. The dot product of the two vectors leaving

M

gives the cosine without the cosine rule being used at all. The vectors are

(-6, -10, 8)

and

(6, 0, 8)

, so the dot product is

-36 + 0 + 64 = 28

. Dividing by

10\sqrt{200}

gives

0.19799

once more, and the same

78.6^{\circ}

Check 2: Work out the other two angles of triangle

BMP

with the same cosine rule and add all three. A triangle that closes must give

180^{\circ}

. The angle at

P

62.55^{\circ}

and the angle at

B

38.87^{\circ}

, and

78.58 + 62.55 + 38.87 = 180.00

Check 3: A shape check that needs almost no arithmetic. In any triangle the longest side lies opposite the largest angle, and the three sides here are

MP = 10

cm,

BM = 14.14

cm and

BP = 15.62

cm.

BP

is the longest, and it is the side facing

M

, so the angle at

M

has to be the biggest of the three - and

78.6^{\circ}

beats

62.6^{\circ}

and

38.9^{\circ}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$MP = \sqrt{8^2 + 6^2} \quad (= 10)$	M1	For a method to find $MP$ . It may be seen in later working, for example in the correct place in the cosine rule. A bare $10$ with no correct method, or not identified as $MP$ , scores M0.	✓
$BM = \sqrt{\left(\sqrt{10^2 + 6^2}\right)^2 + 8^2} \quad (= \sqrt{200} = 10\sqrt{2} = 14.1\ldots)$	M1	For a method to find $BM$ .	✓
$BP = \sqrt{12^2 + 10^2} \quad (= \sqrt{244} = 2\sqrt{61} = 15.6\ldots)$	M1	For a method to find $BP$ .	✓
For example $244 = 10^2 + 200 - 2 \times 10 \times \sqrt{200} \cos BMP$ , or $\cos BMP = \dfrac{10^2 + 200 - 244}{2 \times 10 \times \sqrt{200}}$ oe	M1	For correct substitution into the cosine rule. The scheme prints each length in quotation marks, so a candidate's own values for $MP$ , $BM$ and $BP$ follow through.	✓
$BMP = \cos^{-1}\left(\dfrac{10^2 + 200 - 244}{2 \times 10 \times \sqrt{200}}\right)$ oe	M1	For a complete correct method to find angle $BMP$ .	✓
$78.6$	A1	Accept anything from $78.5$ to $78.7$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 6/6

The first three terms of an arithmetic sequence are shown below.

$(4x - 14), \quad (x + 2), \quad (7x - 9)$

Work out, as an integer, the sum of the first $40$ terms of the sequence.
You must show clear algebraic working. [4 marks]

[Total 4 marks]

Show solution & mark schemeHide solution & mark scheme

Question 23 - Exam Solution

Understanding the Question

Given

The first three terms of an arithmetic sequence are

(4x - 14)

(x + 2)

and

(7x - 9)

In an arithmetic sequence consecutive terms differ by the same amount, the common difference

d

Find

The sum of the first

40

terms of the sequence, written as an integer.

Plan the Solution

The gap from the first term to the second equals the gap from the second to the third. Write that as one equation in $x$ and solve it.
Put the value of $x$ back into the three expressions to get the first term $a$ and the common difference $d$ .
Substitute $a$ , $d$ and $n = 40$ into the sum formula for an arithmetic series.

Worked Solution [4 marks]

Rule - Arithmetic series:

S_n = \dfrac{n}{2}\left[2a + (n - 1)d\right]

, where

a

is the first term,

d

the common difference and

n

the number of terms.

Step 1: the two gaps are equal

(x + 2) - (4x - 14) = (7x - 9) - (x + 2)

-3x + 16 = 6x - 11

(Reason: The sequence is arithmetic, so the second term minus the first equals the third minus the second. Removing the brackets is where marks are lost:

-(4x - 14)

gives

-4x + 14

, not

-4x - 14

Step 2: solve for

x

16 + 11 = 6x + 3x

27 = 9x

x = 3

(Reason: Collecting the

x

terms on the side where they stay positive avoids a sign slip. Dividing

27

9

gives

x = 3

Step 3: the first term and the common difference

4 \times 3 - 14 = -2

3 + 2 = 5

7 \times 3 - 9 = 12

5 - (-2) = 7

12 - 5 = 7

(Reason: Substituting

x = 3

turns the three expressions into the numbers

-2

5

and

12

. Both gaps come to

7

, which confirms the sequence really is arithmetic, so

a = -2

and

d = 7

Step 4: sum the first

40

terms

S_{40} = \dfrac{40}{2}\left[2 \times (-2) + 39 \times 7\right]

20 \times (-4 + 273) = 20 \times 269 = 5380

(Reason: The formula uses

(n - 1)

lots of the common difference, not

n

lots, because the first term already sits at the start of the sequence. With

n = 40

that is

39

lots of

7

, giving

-4 + 273 = 269

inside the bracket.)

5380

Verification

Check 1: Put

x = 3

back into the three printed expressions and compare the two gaps. The terms are

-2

5

12

, with gaps

5 - (-2) = 7

and

12 - 5 = 7

, so the sequence is arithmetic.

Check 2: Pair the terms instead of using the formula. The

40

th term is

-2 + 39 \times 7 = 271

, and the

40

terms fall into

20

pairs, each pair worth

-2 + 271

20 \times 269 = 5380

, the same total.

Check 3: Use the mean of the first and last terms. That mean is

\dfrac{-2 + 271}{2} = 134.5

, and there are

40

terms.

134.5 \times 40 = 5380

, so all three methods agree.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Set up an equation in $x$	M1	For $(x + 2) - (4x - 14) = (7x - 9) - (x + 2)$ oe, e.g. $6x - 11 = 16 - 3x$ , or two simultaneous equations in $x$ and $d$ .	✓
Correct values, or correct substitution	M1	For $x = 3$ and $a = -2$ and $d = 7$ , or $x = 3$ with $S_{40}$ expressed in terms of $x$ . Allow $(40 - 1)$ for $39$ .	✓
Substitute into the sum formula	M1	For $\dfrac{40}{2}\left(2 \times (-2) + 39 \times 7\right)$ oe. Allow their own $a$ and their own $d$ , or their own $x$ as long as it is clearly stated. Allow $(40 - 1)$ for $39$ .	✓
The integer sum	A1	For $5380$ , dependent on at least one method mark.	✓
Working required	Note	The paper asks for clear algebraic working and the accuracy mark is dependent, so a correct answer written down with no algebraic working earns nothing. The common slip is taking $40$ lots of the common difference instead of $39$ , which gives $5520$ . The third method mark needs the 39 itself - the scheme allows only (40 - 1) as another way of writing it - so that slip scores 2 of the 4 marks.	✓

Full marks: 4/4

Keep revising

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

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Edexcel IGCSE 4MA1/1HR, Thursday 16 May 2024: Worked Solutions, Questions 14 to 23

Worked solutions, questions 14 to 23 of 23

Question 14, Calculator allowed

Question 14 - Exam Solution

Question 15, Calculator allowed

Question 15 - Exam Solution

Question 16, Calculator allowed

Question 16 - Exam Solution

Question 17, Calculator allowed

Question 17 - Exam Solution

Question 18, Calculator allowed

Question 18 - Exam Solution

Question 19, Calculator allowed

Question 19 - Exam Solution

Question 20, Calculator allowed

Question 20 - Exam Solution

Question 21, Calculator allowed

Question 21 - Exam Solution

Question 22, Calculator allowed

Question 22 - Exam Solution

Question 23, Calculator allowed

Question 23 - Exam Solution

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