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Edexcel IGCSE 4MA1/2H, Monday 3 June 2024: Worked Solutions, Questions 16 to 25

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

Back to questions 1 to 15

This is the rest of the paper. Questions 1 to 15, the paper's overview and the frequently asked questions are on the first page.

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/2H (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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All 25 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 16 to 25 of 25

$60$ members of a community centre were asked if they would like to join clubs for chess ( $C$ ), for gardening ( $G$ ) or for baking ( $B$ )

Of these members
$9$ chose chess, gardening and baking
$17$ chose chess and gardening
$16$ chose gardening and baking
$20$ chose chess and baking
$28$ chose gardening
$39$ chose baking
$2$ chose none of the clubs

(a) Using this information, complete the Venn diagram to show the numbers of members in each subset. [3 marks]

One of the members is chosen at random.
Given that this member chose gardening,

(b) find the probability that this member also chose chess. [2 marks]

(d) Find $\text{n}([G \cup B] \cap C)$ [1 mark]

(b)(c)(d)

[Total 7 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

60

members were asked about three clubs: chess

C

, gardening

G

and baking

B

\text{n}(C \cap G \cap B) = 9

\text{n}(C \cap G) = 17

\text{n}(G \cap B) = 16

\text{n}(C \cap B) = 20

\text{n}(G) = 28

and

\text{n}(B) = 39

, and

2

members joined no club at all

Find

(a) the number of members in each of the eight regions of the Venn diagram (b) the probability that a member chose chess, given that the member chose gardening (c)

\text{n}(G \cap C')

(d)

\text{n}([G \cup B] \cap C)

Plan the Solution

Start in the middle. The $9$ who joined all three clubs go in the centre region, and every other total in the question already includes them.
Work outwards. Take the centre off each pair total to get the members in exactly two clubs, then take the filled regions off $28$ and $39$ to get the members in exactly one.
Finish with the $60$ . The last region is whatever is left once the $2$ outside the circles are set aside.
Then read parts (b), (c) and (d) straight off the completed diagram - no new working is needed.

Worked Solution [7 marks]

Rule - overlapping sets: a pair total counts everyone in both circles, including everyone in all three. So

\text{n}(C \cap G) = 17

leaves

17 - 9 = 8

members in chess and gardening but not baking.

Step 1: put the

9

who joined all three in the centre

\text{n}(C \cap G \cap B) = 9

(Reason: The centre region lies inside all three circles, so it holds exactly the members who chose every club. Filling it first is what makes every later subtraction possible.)

Step 2: take the centre off each pair total

17 - 9 = 8

16 - 9 = 7

20 - 9 = 11

(Reason: Each pair total already contains the

9

in the centre, so removing them leaves the members in exactly two of the clubs:

8

in chess and gardening only,

7

in gardening and baking only, and

11

in chess and baking only.)

Step 3: fill the gardening-only and baking-only regions

28 - 8 - 9 - 7 = 4

39 - 11 - 9 - 7 = 12

(Reason: The

28

who chose gardening are spread over four regions, so take away the three that are already filled. The

39

who chose baking work the same way.)

Step 4: use the

60

to finish the chess-only region

8 + 9 + 11 + 4 + 7 + 12 = 51

60 - 2 - 51 = 7

(Reason: Every region except chess-only is now known. The

2

who joined nothing sit outside the circles, so they come off the total before the rest is shared out. The diagram is complete.)

Step 5, part (b): restrict to the gardening circle

4 + 8 + 9 + 7 = 28

8 + 9 = 17

\text{P}(C \mid G) = \dfrac{17}{28}

(Reason: Being told the member chose gardening narrows the choice from

60

members to the

28

inside that circle. Of those, the

8

and the

9

also chose chess.)

Step 6, part (c):

\text{n}(G \cap C')

4 + 7 = 11

(Reason:

C'

is everything outside the chess circle, so keep only the parts of the gardening circle that lie outside

C

: gardening only and gardening with baking.)

Step 7, part (d):

\text{n}([G \cup B] \cap C)

8 + 9 + 11 = 28

(Reason:

G \cup B

is everything in the gardening or baking circles, and intersecting with

C

keeps the parts of the chess circle that overlap them - so every region of

C

except chess only.)

(a) chess only

7

; chess and gardening only

8

; gardening only

4

; chess and baking only

11

; all three

9

; gardening and baking only

7

; baking only

12

; outside

2

(b)

\dfrac{17}{28}

(c)

11

(d)

28

Verification

Check 1 - every member is somewhere: Add all eight regions of the completed diagram, including the

2

outside the circles. The result must be the

60

members who were asked.

7 + 8 + 4 + 11 + 9 + 7 + 12 + 2 = 60

Check 2 - the three pair totals come back: For each pair of clubs, add the exactly-two region to the centre and compare with the

17

16

and

20

the question gives.

8 + 9 = 17

7 + 9 = 16

11 + 9 = 20

Check 3 - part (d) a second way: Everyone in the chess circle who is not in chess alone must have chosen gardening or baking, so part (d) is the whole chess circle less the chess-only region.

7 + 8 + 9 + 11 = 35

and

35 - 7 = 28

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) the chess-only region	B1	For $7$ in chess only.	✓
(a) the other seven regions	B2	For all $7$ of the other regions correct: $8$ , $4$ , $11$ , $9$ , $7$ , $12$ and the $2$ outside the circles.	✓
(a) most of the other regions	(B1)	For $4$ , $5$ or $6$ of those other seven regions correct.	✓
(b) the conditional probability	B2	$\dfrac{17}{28}$ or equivalent: $0.61$ or $61\%$ or $0.607$ or $60.7\%$ or better. Follow through from the candidate's own Venn diagram, or work from the values given in the question text.	✓
(b) a partly correct fraction	(B1)	For $17$ as the numerator, or $28$ as the denominator, of a fraction between $0$ and $1$ .	✓
(c) the gardening members outside chess	B1ft	$11$ , follow through from the candidate's own Venn diagram, or from the values given in the question text.	✓
(d) the chess members who also chose gardening or baking	B1ft	$28$ , follow through from the candidate's own Venn diagram, or from the values given in the question text.	✓
Follow through in (b), (c) and (d)	Note	Follow through is only available where the candidate has actually written numbers into the regions of the Venn diagram.	✓

Full marks: 7/7

$Q$ is directly proportional to the square root of $d$
When $d = 324$ , $Q = 4.5$
Work out a formula for $Q$ in terms of $d$ [3 marks]

[Total 3 marks]

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Question 17 - Exam Solution

Understanding the Question

Given

Q \propto \sqrt{d}

- a direct proportion to the square root, not to

d

itself

One pair of matching values:

Q = 4.5

when

d = 324

Find

A formula for

Q

in terms of

d

The answer must be written as

Q = \ldots

, so the constant has to be worked out first

Plan the Solution

Replace the proportion sign by a constant of proportionality $k$ to get an equation.
Substitute the pair $Q = 4.5$ , $d = 324$ into that equation. $324$ is a square number, so the root is exact.
Solve for $k$ , then put its value back into the equation and leave $d$ as a letter.

Worked Solution [3 marks]

Rule - Direct proportion:

Q \propto \sqrt{d}

means

Q = k\sqrt{d}

for one fixed constant

k

. A single pair of values is enough to fix

k

, and the finished formula must show

d

as a letter.

Step 1: write the proportion as an equation

Q = k\sqrt{d}

(Reason: Directly proportional means one quantity is a fixed multiple of the other, so a single constant

k

links them. The multiplier is applied to

\sqrt{d}

, not to

d

, because that is what the question says is proportional to

Q

Step 2: substitute the pair of values

4.5 = k\sqrt{324}

\sqrt{324} = 18

4.5 = 18k

(Reason: The given pair obeys the same rule, so putting it in turns the equation into one with only

k

unknown. Here

18^2 = 324

, so the square root is exact and nothing is rounded.)

Step 3: solve for the constant

k = \dfrac{4.5}{18} = 0.25

(Reason: Divide both sides by the number that is multiplying

k

. The constant is a pure number: it carries no

Q

and no

d

Step 4: write the formula

Q = 0.25\sqrt{d}

(Reason: Putting the value of

k

back into

Q = k\sqrt{d}

gives what was asked for. The equivalent forms

Q = \dfrac{\sqrt{d}}{4}

and

Q = \sqrt{\dfrac{d}{16}}

are also accepted, but an answer that does not start

Q =

is not.)

Q = 0.25\sqrt{d}

Verification

Check 1: Put

d = 324

back into the formula. The square root of

324

18

0.25 \times 18 = 4.5

, which is the value of

Q

the question gives.

Check 2: Test one of the equivalent forms the mark scheme accepts,

Q = \sqrt{\dfrac{d}{16}}

, at the same value of

d

. It reaches the answer without dividing by a root at all.

\sqrt{\dfrac{324}{16}} = \sqrt{20.25} = 4.5

, the same value again.

Check 3: A structure check rather than an arithmetic one. Proportion to

\sqrt{d}

means

d

must be multiplied by

4

for

Q

to double. Take

d = 4 \times 324 = 1296

0.25\sqrt{1296} = 0.25 \times 36 = 9

, which is exactly twice

4.5

. A formula using

d

instead of

\sqrt{d}

would have given four times as much.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Writes the proportion as an equation, eg $Q = k\sqrt{d}$ or $kQ = \sqrt{d}$ or $Q = \sqrt{kd}$	M1	Any correct algebraic form with a constant of proportionality, or equivalent, provided $k \neq 1$ . Use of $\propto$ in place of $=$ is condoned for the method marks.	✓
Substitutes the given pair, eg $4.5 = k \times \sqrt{324}$ , or states $k = 0.25$	M1	Or equivalent value of the constant, eg $k = \dfrac{1}{4}$ . Award both method marks for this line if $Q = k\sqrt{d}$ is never written down.	✓
Gives the formula $Q = 0.25\sqrt{d}$	A1	Or equivalent, but it must be in the form $Q = \ldots$ eg $Q = \dfrac{\sqrt{d}}{4}$ or $Q = \sqrt{\dfrac{d}{16}}$ . A correct answer scores full marks unless it comes from obviously incorrect working. A value of $k$ alone, with no formula, does not earn this mark.	✓

Full marks: 3/3

The straight line $P$ has equation $5y + 2x = 7$

The straight line $Q$ is perpendicular to $P$
Work out the gradient of $Q$ [2 marks]

[Total 2 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

The straight line

P

has equation

5y + 2x = 7

- written with the

x

and

y

terms on the same side, not as

y = mx + c

The straight line

Q

is perpendicular to

P

Find

The gradient of

Q

A gradient is a number, so the answer carries no unit. Nothing is asked about where either line sits, so the

7

will not appear in the answer

Plan the Solution

Rearrange $5y + 2x = 7$ into the form $y = mx + c$ , because the gradient is only readable once $y$ is the subject.
Read the gradient of $P$ as the number multiplying $x$ , sign included.
Apply the perpendicular rule to that gradient to get the gradient of $Q$ . Then check the two gradients multiply to $-1$ .

Worked Solution [2 marks]

Rule - Perpendicular gradients: if two lines are perpendicular then

m_{1} \times m_{2} = -1

, so each gradient is the negative reciprocal of the other - turn the fraction upside down and change its sign. A horizontal line and a vertical line are the one exception, because a vertical line has no gradient.

Step 1: make

y

the subject

5y + 2x = 7

5y = 7 - 2x

y = \dfrac{7 - 2x}{5}

y = \dfrac{7}{5} - \dfrac{2}{5}x

(Reason: Subtracting

2x

from both sides and then dividing every term by

5

leaves

y

on its own. Splitting the single fraction into two is what makes the coefficient of

x

visible; the mark scheme also accepts the unsplit form

y = \dfrac{7 - 2x}{5}

for the method mark.)

Step 2: read the gradient of

P

m_{1} = -\dfrac{2}{5} = -0.4

(Reason: Comparing

y = \dfrac{7}{5} - \dfrac{2}{5}x

with

y = mx + c

gives

c = \dfrac{7}{5}

and a gradient of

-\dfrac{2}{5}

. The minus sign is part of the gradient, not a stray subtraction:

P

falls from left to right.)

Step 3: turn that gradient into the gradient of

Q

m_{1} \times m_{2} = -1

-\dfrac{2}{5} \times m_{2} = -1

m_{2} = \dfrac{5}{2} = 2.5

(Reason: Dividing both sides by

-\dfrac{2}{5}

is the same as multiplying by its reciprocal, so the gradient of

Q

is the gradient of

P

turned upside down with its sign changed. One line falls and the other rises, so the two gradients must have opposite signs.)

\dfrac{5}{2} = 2.5

Verification

Check 1: Multiply the two gradients together. Perpendicular lines must give

-1

, and this is the definition the answer has to satisfy.

-\dfrac{2}{5} \times \dfrac{5}{2} = -\dfrac{10}{10} = -1

, so the two gradients really are perpendicular. Keeping the sign would have given

+1

instead.

Check 2: Get the gradient of

P

from two points instead of from

y = mx + c

, so the rearranging is never used. Both

(1, 1)

and

(6, -1)

satisfy

5y + 2x = 7

\dfrac{-1 - 1}{6 - 1} = -\dfrac{2}{5}

, the same gradient for

P

that Step 2 read off, so Step 1 was rearranged correctly.

Check 3: A geometry check, using no gradient rule at all. From

(1, 1)

, a step of

5

right and

2

down stays on

P

and lands on

(6, -1)

. Turning that step a quarter turn gives

2

right and

5

up, landing on

(3, 6)

. The three squared lengths are

5^2 + 2^2 = 29

2^2 + 5^2 = 29

and

3^2 + 7^2 = 58

29 + 29 = 58

, so by the converse of Pythagoras the corner at

(1, 1)

is a right angle. The turned step rises

5

for every

2

across, a gradient of

\dfrac{5}{2}

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Rearranges $5y + 2x = 7$ so the gradient can be seen, eg $y = \dfrac{7 - 2x}{5}$ or $y = -0.4x + \ldots$ , or writes an equation whose gradient is $2.5$ oe, eg $y = \dfrac{5}{2}x + \ldots$	M1	Oe. Writing the gradient of $P$ as $-\dfrac{2}{5}$ on its own earns this mark. Otherwise the rearrangement need not be simplified, provided the gradient of $P$ can be read from it. The term $\dfrac{5}{2}x$ written on its own also earns this mark, so a candidate who goes straight to the perpendicular line's equation is not penalised for skipping the gradient of $P$ .	✓
States the gradient of the perpendicular line, $\dfrac{5}{2}$	A1	Oe, eg $2.5$ . The gradient must be stated, but isw if it is stated and then used inside an equation. A correct answer scores full marks unless it comes from obviously incorrect working, so no working is needed for the $2$ marks.	✓
Special case: an equation of a line with gradient $\dfrac{5}{2}$ is given, but $\dfrac{5}{2}$ is never written on its own	SC B1	Awarded only if no other mark is earned. The named error is answering with a whole equation when the question asked for a number: the perpendicular gradient has in fact been found, so one mark is recovered, but the answer as written does not state it.	✓

Full marks: 2/2

$G = \dfrac{c}{2f - 3h}$

Each of the three values below has been rounded.

$c = 8$ correct to the nearest whole number
$f = 6.62$ correct to $2$ decimal places
$h = 1.2$ correct to $1$ decimal place

Find the lower bound for the value of $G$
Give your answer correct to $3$ decimal places.
You must show your working clearly. [3 marks]

[Total 3 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

G = \dfrac{c}{2f - 3h}

c = 8

, correct to the nearest whole number

f = 6.62

, correct to

2

decimal places

h = 1.2

, correct to

1

decimal place

Find

The lower bound for the value of

G

, correct to

3

decimal places

Plan the Solution

Turn each rounded value into a pair of bounds by going half a rounding unit either side.
Decide which bound of each letter makes $G$ as small as it can be.
Substitute those bounds, work the denominator out on its own, then divide.
Round the result to $3$ decimal places.

Worked Solution [3 marks]

Rule - Bounds: a rounded value lies within half a rounding unit of the value stated, so

8

to the nearest whole number means

7.5 \leq c < 8.5

. A fraction is smallest when its numerator is smallest and its denominator is largest, so for

G = \dfrac{c}{2f - 3h}

take the lower bound of

c

, the upper bound of

f

and the lower bound of

h

Step 1: Write down the bounds for each letter

7.5 \leq c < 8.5

6.615 \leq f < 6.625

1.15 \leq h < 1.25

(Reason: Each value can be out by half of the unit it was rounded to. Half of

1

0.5

, half of

0.01

0.005

, and half of

0.1

0.05

Step 2: Choose the bound of each letter that makes G smallest

\dfrac{7.5}{2 \times 6.625 - 3 \times 1.15}

(Reason: The numerator must be as small as possible, so

c

takes its lower bound

7.5

. The denominator must be as large as possible, so

f

takes its upper bound

6.625

while

h

takes its lower bound

1.15

, because

3h

is taken away. This is the trap in the question: the lower bound of

G

needs the upper bound of one letter and the lower bound of the other.)

Step 3: Work out the largest possible denominator

2 \times 6.625 = 13.25

3 \times 1.15 = 3.45

13.25 - 3.45 = 9.8

(Reason: Deal with the denominator on its own before dividing. Doubling the upper bound of

f

and then subtracting three lots of the lower bound of

h

gives the largest value the denominator can take.)

Step 4: Divide, then round to 3 decimal places

\dfrac{7.5}{9.8} = \dfrac{75}{98} = 0.7653061224

0.7653061224 \approx 0.765

(Reason: Multiplying top and bottom by

10

turns

\dfrac{7.5}{9.8}

into

\dfrac{75}{98}

, which is a tidier thing to key in. The fourth decimal digit is

3

, so the third digit stays as it is.)

Lower bound

= 0.765

Verification

Check 1: Multiply the answer back by the largest denominator. It should return the smallest numerator,

7.5

0.7653061224 \times 9.8 = 7.5

Check 2: Swap each chosen bound for the other one, one at a time. If the right three bounds were picked, every swap must make

G

bigger.

\dfrac{8.5}{9.8} = 0.8673469388

\dfrac{7.5}{9.78} = 0.7668711656

\dfrac{7.5}{9.5} = 0.7894736842

, all bigger than

0.765

Check 3: Confirm the denominator can never turn negative, otherwise the largest denominator would not give the smallest

G

. The smallest it can be pairs the lower bound of

f

with the upper bound of

h

2 \times 6.615 - 3 \times 1.25 = 13.23 - 3.75 = 9.48

, which is positive

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Write down any one correct bound, for example $7.5 \leq c < 8.5$	B1	For a correct upper or lower bound: $7.5$ , $8.5$ , $6.615$ , $6.625$ , $1.15$ , $1.25$ . Allow $8.4\dot{9}$ for $8.5$ , $6.624\dot{9}$ for $6.5$ and $1.24\dot{9}$ for $1.25$ . That $6.5$ appears to be a slip for $6.625$ , which is what $6.624\dot{9}$ equals.	✓
$\dfrac{7.5}{2 \times 6.625 - 3 \times 1.15}$	M1	For the lower bound of $c$ over $2$ times the upper bound of $f$ minus $3$ times the lower bound of $h$ , where $7.5 \leq \text{LB}_c < 8$ , $6.62 < \text{UB}_f \leq 6.625$ and $1.15 \leq \text{LB}_h < 1.2$ .	✓
A candidate who drops the $2$ and the $3$ from the denominator and works out $\dfrac{7.5}{6.625 - 1.15} = 1.369863$	SC B1	A special case, awarded in addition to the first B1. The error is reading $2f - 3h$ as $f - h$ : the bounds themselves are chosen correctly, but the coefficients are dropped.	✓
$0.765$	A1	awrt $0.765$ (the full value is $0.7653061224$ ). Working is required. Dependent on completely correct bounds, and dependent on the M1.	✓

Full marks: 3/3

Given that $k = x - y$ and $x = \dfrac{1}{4y}$
write $\dfrac{5k}{x + 2}$ in the form $\dfrac{a - by^2}{c + dy}$ , where $a$ , $b$ , $c$ and $d$ are integers. [3 marks]

[Total 3 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

k = x - y

x = \dfrac{1}{4y}

Two facts linking

k

x

and

y

. The required form contains only

y

, so

k

and

x

must both disappear.

Find

\dfrac{5k}{x + 2}

written as

\dfrac{a - by^2}{c + dy}

. The four integers

a

b

c

and

d

Plan the Solution

Replace $x$ by $\dfrac{1}{4y}$ everywhere, so that $y$ is the only letter left.
Write the top, $5k$ , as one fraction over $4y$ .
Write the bottom, $x + 2$ , as one fraction over the same $4y$ .
Divide the two fractions. The $4y$ cancels, which is what leaves integers behind.

Worked Solution [3 marks]

Rule - Divide by a fraction by multiplying by its reciprocal:

\dfrac{P}{Q}

divided by

\dfrac{R}{S}

\dfrac{P}{Q} \times \dfrac{S}{R}

. Here

Q

and

S

are both

4y

, so they cancel each other.

Step 1: write

k

as a single fraction over

4y

k = x - y = \dfrac{1}{4y} - y

y = \dfrac{4y^2}{4y}

k = \dfrac{1}{4y} - \dfrac{4y^2}{4y} = \dfrac{1 - 4y^2}{4y}

(Reason: Two fractions can only be subtracted over a common denominator. Multiplying

y

top and bottom by

4y

turns it into

\dfrac{4y^2}{4y}

, which changes how it is written but not what it is worth.)

Step 2: write

x + 2

over the same denominator

x + 2 = \dfrac{1}{4y} + 2

2 = \dfrac{8y}{4y}

x + 2 = \dfrac{1}{4y} + \dfrac{8y}{4y} = \dfrac{1 + 8y}{4y}

(Reason: Since

2 \times 4y = 8y

, the whole number

2

\dfrac{8y}{4y}

. This numerator,

1 + 8y

, is the denominator the final answer keeps.)

Step 3: multiply the top by

5

5k = 5 \times \dfrac{1 - 4y^2}{4y} = \dfrac{5 - 20y^2}{4y}

(Reason: The

5

multiplies the numerator as a whole, so every term inside the bracket is multiplied by it. The denominator is untouched.)

Step 4: divide, and cancel the

4y

\dfrac{5k}{x + 2} = \dfrac{5 - 20y^2}{4y} \times \dfrac{4y}{1 + 8y}

= \dfrac{5 - 20y^2}{1 + 8y}

(Reason: Dividing by

\dfrac{1 + 8y}{4y}

is the same as multiplying by

\dfrac{4y}{1 + 8y}

. The

4y

now appears once above and once below, so it cancels and no fraction is left inside a fraction.)

Step 5: read off the four integers

\dfrac{a - by^2}{c + dy} = \dfrac{5 - 20y^2}{1 + 8y}

a = 5,\; b = 20,\; c = 1,\; d = 8

(Reason: Comparing with the required form,

a

is the constant on top,

b

is the number subtracted from it, and

c

and

d

are the two numbers underneath. All four are integers, as the question demands.)

\dfrac{5k}{x + 2} = \dfrac{5 - 20y^2}{1 + 8y}

a = 5

b = 20

c = 1

d = 8

Verification

Check 1: Put

y = 1

into the original. Then

x = \dfrac{1}{4}

k = \dfrac{1}{4} - 1 = -\dfrac{3}{4}

and

x + 2 = \dfrac{9}{4}

, so the original is

-\dfrac{15}{4} \times \dfrac{4}{9}

. The original gives

-\dfrac{5}{3}

, and the answer gives

\dfrac{5 - 20}{1 + 8} = -\dfrac{15}{9} = -\dfrac{5}{3}

. They agree.

Check 2: Now a negative value, which tests the signs. Put

y = -1

. Then

x = -\dfrac{1}{4}

k = -\dfrac{1}{4} + 1 = \dfrac{3}{4}

and

x + 2 = \dfrac{7}{4}

. The original gives

\dfrac{15}{4} \times \dfrac{4}{7} = \dfrac{15}{7}

, and the answer gives

\dfrac{5 - 20}{1 - 8} = \dfrac{15}{7}

. They agree again, and this time both are positive.

Check 3: Check the FORM, not just the value. The question asks for

\dfrac{a - by^2}{c + dy}

: a constant minus a multiple of

y^2

on top, and a constant plus a multiple of

y

underneath, with all four numbers integers.

5 - 20y^2

over

1 + 8y

matches that pattern exactly, and neither

k

nor

x

survives anywhere in it.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Substitute $x = \dfrac{1}{4y}$ into $\dfrac{5k}{x + 2}$ , giving $5\left(\dfrac{1}{4y} - y\right)$ over $\dfrac{1}{4y} + 2$ .	M1	A correct substitution leaving only values of $y$ ; or an expression containing $xy$ (not just $x$ ); or a correct denominator of $1 + 8y$ .	✓
Multiply every term by $4y$ , or write top and bottom over $4y$ : $\dfrac{5 - 20y^2}{4y}$ over $\dfrac{1 + 8y}{4y}$ .	M1	Multiplying every term by $4y$ or a multiple of $4y$ ; or writing numerator and denominator over $4y$ or a multiple of it; or a correct expansion containing an $xy$ term ( $xy$ may be replaced by $0.25$ ); or three of $a$ , $b$ , $c$ , $d$ correct in the required form.	✓
$\dfrac{5 - 20y^2}{1 + 8y}$	A1	Or equivalent, e.g. $\dfrac{-5 + 20y^2}{-1 - 8y}$ or $\dfrac{20 - 80y^2}{4 + 32y}$ , i.e. $a = 5$ , $b = 20$ , $c = 1$ , $d = 8$ or any common multiple of them.	✓
A correct answer written down with no working	Note	A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 3/3

The diagram shows a circle drawn inside a square $ABCD$ .
Each side of the square is a tangent to the circle.

The shaded regions have a total area of $80$ cm².

Work out the length of $AC$ .
Give your answer correct to $3$ significant figures. [5 marks]

[Total 5 marks]

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Question 21 - Exam Solution

Understanding the Question

Given

A square

ABCD

with a circle inside it, and every side of the square is a tangent to that circle.

The shaded regions together have an area of

80

cm².

The figure is not drawn accurately, so nothing may be measured off it.

Find

The length of

AC

, which is a diagonal of the square, correct to

3

significant figures.

Plan the Solution

Turn the word tangent into a length. Each side touches the circle once, so the circle is squeezed between opposite sides and its diameter is the side of the square. Call the radius $r$ ; the side is then $2r$ .
Shaded means the square with the circle taken out of it, so $80$ becomes one equation in $r$ alone.
$4 - \pi$ is just a number, so $r^2$ comes out by dividing. Nothing here needs $r$ to be a tidy value.
$AC$ is the diagonal of a square, so Pythagoras across two sides finishes it. Keep the full accuracy and round once, at the end.

Worked Solution [5 marks]

Rule - Circle inscribed in a square: the side of the square is

2r

, the shaded area is

(2r)^2 - \pi r^2

, and the diagonal follows from Pythagoras,

AC^2 = x^2 + x^2

, where

x

is the side.

Step 1: turn the tangents into a length

x = 2r

(Reason: A tangent meets a circle exactly once, so the circle is held tight between each pair of opposite sides. The distance across the circle is therefore the whole side of the square: if the radius is

r

, the side

x

2r

. This is the only place the word tangent is used, and it is what makes one letter enough.)

Step 2: write the shaded area in that one letter

(2r)^2 - \pi r^2 = 80

4r^2 - \pi r^2 = 80

(Reason: The shaded parts are what is left when the circle is cut out of the square, so their total area is the area of the square minus the area of the circle. The square has side

2r

and the circle has radius

r

, so both areas are written with the same letter. Note that

(2r)^2

expands to

4r^2

, not to

2r^2

- the

2

is squared as well.)

Step 3: factorise, then divide

r^2(4 - \pi) = 80

r^2 = \dfrac{80}{4 - \pi} = 93.1958\ldots

(Reason: Both terms carry

r^2

, so it comes out as a common factor and leaves the bracket

4 - \pi

, which is only a number, about

0.8584

. Dividing

80

by it gives

r^2

in one step, and there is no need to find

r

itself.)

Step 4: square the side of the square

x^2 = (2r)^2 = 4r^2

4 \times 93.1958\ldots = 372.7833\ldots

(Reason: The diagonal is built from the side, not from the radius. Squaring

2r

multiplies

r^2

4

, so the square of the side is four times the

93.1958\ldots

found above. Keeping

x^2

rather than

x

saves a root that Pythagoras would only square again.)

Step 5: Pythagoras across the square

AC^2 = x^2 + x^2 = 2x^2

2 \times 372.7833\ldots = 745.5667\ldots

AC = \sqrt{745.5667\ldots} = 27.3050\ldots

(Reason: The sides

AB

and

BC

meet at a right angle at

B

, and

AC

is the hypotenuse of that right-angled triangle. Both sides are the same length, so the square of the diagonal is simply twice the square of the side, and the square root of it is the length wanted.)

Step 6: round to 3 significant figures

AC = 27.3050\ldots

AC = 27.3 \text{ cm}

(Reason: The first three significant figures of

27.3050\ldots

are

2

7

and

3

, and the digit after them is

0

, so nothing rounds up. The answer is a length, so it is measured in centimetres.)

AC = 27.3

Verification

Check 1: Rebuild both areas from

r^2 = 93.1958\ldots

and see whether the shading comes back. The square has area

4r^2

and the circle has area

\pi r^2

, so subtracting them must return the area the question gives.

372.7833 - 292.7833 = 80

, which is the

80

cm² of shading the question started from

Check 2: A test that does not depend on the size at all. For a circle inscribed in any square the shaded share is fixed at

1 - \dfrac{\pi}{4}

, so dividing the given

80

by the area of the square found above should reproduce it.

\dfrac{80}{372.7833} = 0.2146

and

1 - \dfrac{\pi}{4} = 0.2146

, so the square has come out the right size

Check 3: Finally test the diagonal itself. In every square, whatever its size, the diagonal divided by the side is

\sqrt{2}

. The side itself is the square root of the

372.7833\ldots

found in step 4, which is

19.3075\ldots

, so the ratio can be tested against

1.4142\ldots

\dfrac{27.3050}{19.3075} = 1.4142

, which is

\sqrt{2}

to four decimal places

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Write the total shaded area in one variable, e.g. $(2r)^2 - \pi r^2$	M1	A correct expression for the area of the shaded parts in one variable only. For this mark only, accept it without the brackets, e.g. $2r^2 - \pi r^2$ , and accept the same thing written in the side of the square, $x^2 - \pi \times (0.5x)^2$ . Any letter may be used.	✓
$4r^2 - \pi r^2 = 80$	M1	A correct equation in one variable with the brackets expanded, or equivalent, e.g. $r^2 - 0.25\pi r^2 = 20$ , $x^2 - 0.25\pi x^2 = 80$ or $4x^2 - \pi x^2 = 320$ . It may be seen later in the working.	✓
$r^2 = \dfrac{80}{4 - \pi} = 93.19\ldots$	M1	A correct expression for the radius squared or the radius, or for the side of the square squared or the side of the square, e.g. $r = \sqrt{\dfrac{80}{4 - \pi}} = 9.65\ldots$ , $x^2 = \dfrac{80}{1 - 0.25\pi} = 372.78\ldots$ or $x = \sqrt{\dfrac{320}{4 - \pi}} = 19.307\ldots$ .	✓
A correct calculation for $AC$ , e.g. $\sqrt{(2 \times 9.65)^2 + (2 \times 9.65)^2}$	M1	For a correct calculation to find the length of $AC$ from the candidate's own radius or side, e.g. $2 \times \sqrt{9.65^2 + 9.65^2}$ , $\sqrt{19.307^2 + 19.307^2}$ , $\sqrt{8 \times \dfrac{80}{4 - \pi}}$ or $\dfrac{2 \times 9.65}{\sin 45}$ .	✓
The length of $AC$	A1	$27.3$ . Accept any value from $27.3$ to $27.5$ , which covers the different points at which a candidate may round.	✓
A correct answer written down with no working	Note	A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 5/5

The straight line $L$ has equation $x + y = 5$
The curve $C$ has equation $2x^2 + 3y^2 = 210$

Work out the coordinates of the points of intersection of $L$ and $C$ .
You must show clear algebraic working. [5 marks]

[Total 5 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

The straight line

L

x + y = 5

The curve

C

2x^2 + 3y^2 = 210

One equation is linear; the other is quadratic in both variables.

Find

The coordinates of every point that lies on

L

and on

C

at the same time. A quadratic appears on the way, so expect two solution pairs.

Plan the Solution

Make $y$ the subject of the linear equation.
Substitute that expression into the curve's equation, so one equation in $x$ alone is left.
Expand, collect like terms and write the result as $ax^2 + bx + c = 0$ .
Divide through by the common factor, then factorise to get both values of $x$ .
Put each value of $x$ back into $y = 5 - x$ to find its partner.

Worked Solution [5 marks]

Rule - Substitution (line into curve): make one variable the subject of the linear equation, substitute it into the quadratic equation, solve the quadratic that results, then substitute each root back into the linear equation to find the other coordinate.

Step 1: Make

y

the subject of the line's equation

x + y = 5

y = 5 - x

(Reason: The curve's equation contains

x^2

and

y^2

, so it is the straight line that gets substituted into the curve, never the other way round. Rearranging the curve instead would bring in a square root.)

Step 2: Substitute into the curve's equation

2x^2 + 3y^2 = 210

2x^2 + 3(5 - x)^2 = 210

(Reason: Every

y

in the curve's equation is replaced by

5 - x

, which leaves one equation in

x

alone. Keep the bracket:

(5 - x)^2

is not

25 + x^2

Step 3: Expand and collect into a three-term quadratic

(5 - x)^2 = 25 - 10x + x^2

2x^2 + 3(25 - 10x + x^2) = 210

2x^2 + 75 - 30x + 3x^2 = 210

5x^2 - 30x + 75 = 210

5x^2 - 30x - 135 = 0

(Reason: Square the bracket first, then multiply every term inside it by

3

. The two

x^2

terms collect to

5x^2

, and taking

210

from both sides sets the quadratic equal to zero.)

Step 4: Solve the quadratic

x^2 - 6x - 27 = 0

(x - 9)(x + 3) = 0

x = 9 \text{ or } x = -3

(Reason: Every term has a factor of

5

, so dividing through by

5

gives a simpler quadratic with exactly the same roots. Two numbers with product

-27

and sum

-6

are

-9

and

3

Step 5: Find the

y

that goes with each

x

y = 5 - 9 = -4

y = 5 - (-3) = 8

(Reason: Both points lie on the line, so each

x

takes its own

y

from

y = 5 - x

. Pairing them the wrong way round gives two points that are on neither curve, and it is the commonest way to lose the final mark here.)

(9, -4)

(-3, 8)

Verification

Check 1 - the point $(9, -4)$ : Put

x = 9

and

y = -4

into both original equations. A point of intersection must satisfy each of them.

9 + (-4) = 5

and

2 \times 9^2 + 3 \times (-4)^2 = 162 + 48 = 210

Check 2 - the point $(-3, 8)$ : Put

x = -3

and

y = 8

into both original equations, squaring the negative carefully.

-3 + 8 = 5

and

2 \times (-3)^2 + 3 \times 8^2 = 18 + 192 = 210

Check 3 - the two roots against the quadratic: Add and multiply the roots without re-solving. For

x^2 - 6x - 27 = 0

the sum of the roots must be

6

and their product must be

-27

9 + (-3) = 6

and

9 \times (-3) = -27

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Substitute $y = 5 - x$ into the curve's equation	M1	Substitution of $x = \pm 5 \pm y$ or $y = \pm 5 \pm x$ into $2x^2 + 3y^2 = 210$ , or a correct equation formed using $x = \pm 5 \pm y$ or $y = \pm 5 \pm x$ , to obtain an equation in $x$ only or in $y$ only. For example $2x^2 + 3(5 - x)^2 = 210$ .	✓
Multiply out and collect terms	M1	For multiplying out and collecting terms, forming a three-term quadratic in any form $ax^2 + bx + c = 0$ where at least $2$ of the coefficients are correct. For example $5x^2 - 30x - 135 = 0$ or $x^2 - 6x - 27 = 0$ , or $5y^2 - 20y - 160 = 0$ or $y^2 - 4y - 32 = 0$ .	✓
Complete method to solve their three-term quadratic	M1	For a complete method to solve their three-term quadratic: correct factorisation, for example $(x - 9)(x + 3) = 0$ ; or substitution into the formula, allowing one sign error and some simplification, as far as $\dfrac{6 \pm \sqrt{36 + 108}}{2}$ or $\dfrac{4 \pm \sqrt{16 + 128}}{2}$ ; or completing the square, for example $(x - 3)^2 - 3^2 - 27 = 0$ ; or for seeing $x = 9$ , $x = -3$ or $y = 8$ , $y = -4$ . Incorrect labelling of $x$ and $y$ is allowed at this stage.	✓
Substitute the two found values back to get the other variable	M1ft	For substituting their $2$ found values of $x$ (or of $y$ ) into a suitable equation, for example $y = 5 - 9$ and $y = 5 - (-3)$ , with use of the quadratic equation allowed; or fully correct values for the other variable. Substitution must be seen for incorrect $x$ or $y$ values, and the labels for $x$ and $y$ must be correct here.	✓
Both pairs of coordinates	A1	$(9, -4)$ and $(-3, 8)$ . Working is required, and the mark is dependent on both of the first two method marks, so an answer written down with no algebra behind it does not earn it.	✓

Full marks: 5/5

Simplify
$\dfrac{30 \times 25^{2x+7}}{\sqrt{180} \times \left(\sqrt{5}\right)^{4x+9}}$
Write your answer in the form $5^{w}$ , where $w$ is an expression in terms of $x$ .
You must show each stage of your working clearly. [3 marks]

[Total 3 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

The expression

\dfrac{30 \times 25^{2x+7}}{\sqrt{180} \times \left(\sqrt{5}\right)^{4x+9}}

Every base in it is built from

5

25 = 5^{2}

30 = 6 \times 5

and

180 = 36 \times 5

Find

The whole expression as one power of

5

, written

5^{w}

w

as an expression in terms of

x

, so the

x

terms must not cancel away.

Plan the Solution

Take the square factor out of $\sqrt{180}$ , and split $30$ into a whole number times $5$ .
Rewrite both powers in base $5$ , using $\sqrt{5} = 5^{\dfrac{1}{2}}$ .
Add the indices on the top, and add the indices on the bottom.
Cancel the whole-number factor, then subtract the indices to divide.

Worked Solution [3 marks]

Rule - Index laws in a single base:

(a^{m})^{n} = a^{mn}

a^{m} \times a^{n} = a^{m+n}

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

and

\sqrt{a} = a^{\dfrac{1}{2}}

Step 1: Write

30

and

\sqrt{180}

in terms of

5

30 = 6 \times 5

\sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} = 6 \times 5^{\dfrac{1}{2}}

(Reason: Pulling the square factor

36

out of

180

leaves

\sqrt{5}

behind, and it puts the same factor of

6

on the top and on the bottom.)

Step 2: Rewrite both powers in base

5

25^{2x+7} = (5^{2})^{2x+7} = 5^{4x+14}

\left(\sqrt{5}\right)^{4x+9} = \left(5^{\dfrac{1}{2}}\right)^{4x+9} = 5^{\dfrac{4x+9}{2}}

(Reason: Raising a power to a power multiplies the two indices, and the multiplier acts on the whole of the index inside the bracket, never on part of it.)

Step 3: Collect the numerator and the denominator

30 \times 25^{2x+7} = 6 \times 5 \times 5^{4x+14} = 6 \times 5^{4x+15}

\sqrt{180} \times \left(\sqrt{5}\right)^{4x+9} = 6 \times 5^{\dfrac{1}{2}} \times 5^{\dfrac{4x+9}{2}} = 6 \times 5^{2x+5}

(Reason: A lone

5

counts as

5^{1}

, so the top index is

1 + (4x+14) = 4x+15

. Underneath,

\dfrac{1}{2} + \dfrac{4x+9}{2} = \dfrac{4x+10}{2} = 2x+5

Step 4: Divide by subtracting the indices

\dfrac{6 \times 5^{4x+15}}{6 \times 5^{2x+5}} = 5^{(4x+15)-(2x+5)}

5^{(4x+15)-(2x+5)} = 5^{2x+10}

(Reason: The factor of

6

cancels, and subtracting the indices gives

4x - 2x = 2x

and

15 - 5 = 10

5^{2x+10}

, so

w = 2x + 10

Verification

Check 1: Put

x = 0

into the original expression. The top is

30 \times 25^{7} = 183\,105\,468\,750

and the bottom is

6\sqrt{5} \times \left(\sqrt{5}\right)^{9} = 6 \times 5^{5} = 18\,750

\dfrac{183\,105\,468\,750}{18\,750} = 9\,765\,625 = 5^{10}

, and the answer gives

2(0) + 10 = 10

Check 2: Group the plain numbers first:

\dfrac{30}{\sqrt{180}} = \dfrac{30}{6\sqrt{5}} = \dfrac{5}{\sqrt{5}} = \sqrt{5}

. The two powers then leave

5^{(4x+14)-\dfrac{4x+9}{2}} = 5^{\dfrac{4x+19}{2}}

5^{\dfrac{1}{2}} \times 5^{\dfrac{4x+19}{2}} = 5^{\dfrac{4x+20}{2}} = 5^{2x+10}

, the same answer from a completely different grouping.

Check 3: Count the

x

terms and the constants separately. The top index is

4x+15

and the bottom index is

2x+5

4x - 2x = 2x

and

15 - 5 = 10

, so

w = 2x + 10

still contains

x

, as the question requires.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Any two of the conversions into base $5$ : $30 = 6 \times 5$ , $\sqrt{180} = 6\sqrt{5}$ , $25^{2x+7} = (5^{2})^{2x+7}$ , $\left(\sqrt{5}\right)^{4x+9} = \left(5^{\dfrac{1}{2}}\right)^{4x+9}$	M1	Or equivalent: $30 = 5 \times 2 \times 3$ , $\sqrt{180} = 2 \times 3 \times \sqrt{5}$ , $5^{2(2x+7)}$ , $5^{\dfrac{1}{2}(4x+9)}$ , $\dfrac{30}{\sqrt{180}} = \sqrt{5}$ . Also award for $5^{4x+15}$ as the numerator, or $5^{2x+5}$ as the denominator.	✓
One correct combined expression in terms of $6$ (or $2$ and $3$ ) and powers of $5$ , for example $\dfrac{6 \times 5 \times 5^{4x+14}}{6 \times 5^{0.5} \times 5^{2x+4.5}}$	M1	Some cancelling may already have taken place, for example $\dfrac{5^{4x+15}}{5^{2x+5}}$ . Also award for a fully correct expression in terms of $\sqrt{5}$ and powers, $\dfrac{\left(\sqrt{5}\right)^{8x+30}}{\left(\sqrt{5}\right)^{4x+10}}$ , or in terms of $25$ and powers, $\dfrac{25^{2x+7.5}}{25^{x+2.5}}$ .	✓
$5^{2x+10}$	A1	Dependent on both method marks. Allow $w = 2x + 10$ . Working must be shown.	✓

Full marks: 3/3

The diagram shows the quadrilateral $OACB$ .
In this quadrilateral $\overrightarrow{OA} = 4\mathbf{a}$ , $\overrightarrow{OB} = 3\mathbf{b}$ and $\overrightarrow{BC} = 2\mathbf{a} + \mathbf{b}$

(a) Work out $\overrightarrow{AC}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .
Write your answer as simply as possible. [2 marks]

The point $P$ lies on $AC$ so that $AP:PC = 3:2$
The point $Q$ is placed so that $OPQ$ is a straight line and $BCQ$ is a straight line.

(b) Use a vector method to find $\overrightarrow{OQ}$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .
Write your answer as simply as possible.
You must show your working clearly. [4 marks]

(a) AC =(b) OQ =

[Total 6 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

A quadrilateral

OACB

with

\overrightarrow{OA} = 4\mathbf{a}

\overrightarrow{OB} = 3\mathbf{b}

and

\overrightarrow{BC} = 2\mathbf{a} + \mathbf{b}

P

lies on

AC

with

AP:PC = 3:2

OPQ

and

BCQ

are both straight lines

\mathbf{a}

and

\mathbf{b}

are not parallel, so an

\mathbf{a}

part can never be turned into a

\mathbf{b}

part

Find

(a)

\overrightarrow{AC}

in terms of

\mathbf{a}

and

\mathbf{b}

, simplified (b)

\overrightarrow{OQ}

in terms of

\mathbf{a}

and

\mathbf{b}

, with the vector working shown

Plan the Solution

For (a), travel from $A$ to $C$ along sides whose vectors are given: back along $OA$ , then up $OB$ , then along $BC$ .
For (b), find $\overrightarrow{OP}$ first, because $AP:PC = 3:2$ fixes $\overrightarrow{AP}$ as a fraction of the $\overrightarrow{AC}$ just found.
Write $\overrightarrow{OQ}$ twice, once along each straight line, each time with its own unknown multiplier.
Match the $\mathbf{a}$ parts and match the $\mathbf{b}$ parts. That is two equations in two unknowns, so both multipliers can be found.

Worked Solution [6 marks]

Rule - Vector journeys and straight lines: every route between two points gives the same vector, and travelling against an arrow reverses the sign. If

X

Y

and

Z

lie on one straight line then

\overrightarrow{XZ} = k\,\overrightarrow{XY}

for some number

k

Step 1: travel from

A

C

the long way round

\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{BC}

\overrightarrow{AC} = -4\mathbf{a} + 3\mathbf{b} + (2\mathbf{a} + \mathbf{b})

\overrightarrow{AC} = 4\mathbf{b} - 2\mathbf{a}

(Reason: There is no vector given straight from

A

C

, so go round three sides that are given. Going from

A

O

is against the arrow, so

4\mathbf{a}

becomes

-4\mathbf{a}

. Then

-4\mathbf{a} + 2\mathbf{a} = -2\mathbf{a}

and

3\mathbf{b} + \mathbf{b} = 4\mathbf{b}

Step 2: find

\overrightarrow{OP}

AP:PC = 3:2 \implies \overrightarrow{AP} = \dfrac{3}{5}\overrightarrow{AC}

\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP}

\overrightarrow{OP} = 4\mathbf{a} + \dfrac{3}{5}(4\mathbf{b} - 2\mathbf{a}) = \dfrac{14}{5}\mathbf{a} + \dfrac{12}{5}\mathbf{b}

(Reason: The ratio

3:2

cuts

AC

into

5

equal parts, and

AP

3

of those parts, measured from

A

Step 3: use the straight line

OPQ

\overrightarrow{OQ} = \lambda\,\overrightarrow{OP}

\overrightarrow{OQ} = \lambda\left(\dfrac{14}{5}\mathbf{a} + \dfrac{12}{5}\mathbf{b}\right)

(Reason:

O

P

and

Q

lie on one straight line, so

\overrightarrow{OQ}

is parallel to

\overrightarrow{OP}

, and parallel vectors are multiples of one another. The multiplier

\lambda

is not known yet.)

Step 4: use the straight line

BCQ

\overrightarrow{OQ} = \overrightarrow{OB} + \mu\,\overrightarrow{BC}

\overrightarrow{OQ} = 3\mathbf{b} + \mu(2\mathbf{a} + \mathbf{b})

(Reason:

B

C

and

Q

lie on one straight line, so

\overrightarrow{BQ}

is a multiple of

\overrightarrow{BC}

. This is a second, completely separate way of reaching

Q

Step 5: match the two expressions

\dfrac{14}{5}\lambda = 2\mu

\dfrac{12}{5}\lambda = 3 + \mu

(Reason: Both expressions are the same vector

\overrightarrow{OQ}

. Since

\mathbf{a}

and

\mathbf{b}

are not parallel, they can only be equal if the

\mathbf{a}

parts agree and the

\mathbf{b}

parts agree.)

Step 6: solve for

\lambda

\mu = \dfrac{7}{5}\lambda

\dfrac{12}{5}\lambda = 3 + \dfrac{7}{5}\lambda

\lambda = 3

(Reason: Halving the first equation gives

\mu

in terms of

\lambda

. Substituting it into the second removes

\mu

, and

\dfrac{12}{5} - \dfrac{7}{5} = 1

, which leaves

\lambda = 3

directly.)

Step 7: put

\lambda = 3

back in

\overrightarrow{OQ} = 3\left(\dfrac{14}{5}\mathbf{a} + \dfrac{12}{5}\mathbf{b}\right)

\overrightarrow{OQ} = \dfrac{42}{5}\mathbf{a} + \dfrac{36}{5}\mathbf{b}

(Reason: Multiply each part by

3

. Written as decimals this is

8.4\mathbf{a} + 7.2\mathbf{b}

, which is the same answer in a different form.)

(a)

\overrightarrow{AC} = 4\mathbf{b} - 2\mathbf{a}

(b)

\overrightarrow{OQ} = \dfrac{42}{5}\mathbf{a} + \dfrac{36}{5}\mathbf{b}

Verification

Check 1: Test part (a) on the other route round the quadrilateral. If

\overrightarrow{AC}

is right then

\overrightarrow{OA} + \overrightarrow{AC}

must equal

\overrightarrow{OB} + \overrightarrow{BC}

, because both are

\overrightarrow{OC}

4\mathbf{a} + (4\mathbf{b} - 2\mathbf{a}) = 2\mathbf{a} + 4\mathbf{b}

and

3\mathbf{b} + (2\mathbf{a} + \mathbf{b}) = 2\mathbf{a} + 4\mathbf{b}

Check 2: Test part (b) on the line that was not used to find

\lambda

. From

\mu = \dfrac{7}{5}\lambda

with

\lambda = 3

comes

\mu = \dfrac{21}{5}

. Put that into the

BCQ

expression from Step 4.

3\mathbf{b} + \dfrac{21}{5}(2\mathbf{a} + \mathbf{b}) = \dfrac{42}{5}\mathbf{a} + \dfrac{36}{5}\mathbf{b}

Check 3: Replace the letters with numbers and do the whole thing again as coordinates. Take

\mathbf{a} = (3, 1)

and

\mathbf{b} = (-1, 2)

, which are not parallel. Then

A = (12, 4)

B = (-3, 6)

C = (2, 10)

and

P = (6, 7.6)

. Find where the line through

O

and

P

crosses the line through

B

and

C

. The two lines cross at

(18, 22.8)

, and the answer rebuilds that point exactly:

8.4 \times 3 + 7.2 \times (-1) = 18

and

8.4 \times 1 + 7.2 \times 2 = 22.8

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) A complete vector route from A to C	M1	eg $\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OB} + \overrightarrow{BC}$ or eg $-4\mathbf{a} + 3\mathbf{b} + 2\mathbf{a} + \mathbf{b}$	✓
(a) The simplified answer	A1	$4\mathbf{b} - 2\mathbf{a}$ or equivalent, but it must be simplified, eg $-2\mathbf{a} + 4\mathbf{b}$ or $2(2\mathbf{b} - \mathbf{a})$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
(b) An expression for $\overrightarrow{OP}$	M1ft	Follow through their $\overrightarrow{AC}$ . eg $\overrightarrow{OP} = 4\mathbf{a} + \dfrac{3}{5}(4\mathbf{b} - 2\mathbf{a})$ or eg $\overrightarrow{OP} = 3\mathbf{b} + 2\mathbf{a} + \mathbf{b} - \dfrac{2}{5}(4\mathbf{b} - 2\mathbf{a})$ , either giving $\dfrac{14}{5}\mathbf{a} + \dfrac{12}{5}\mathbf{b}$ or $2.8\mathbf{a} + 2.4\mathbf{b}$ . It may appear as part of another vector equation.	✓
(b) One correct expression for $\overrightarrow{OQ}$ or $\overrightarrow{PQ}$	M1ft	Follow through their $\overrightarrow{AC}$ , or equivalent. eg $\overrightarrow{OQ} = \lambda\left(\dfrac{14}{5}\mathbf{a} + \dfrac{12}{5}\mathbf{b}\right)$ or $\overrightarrow{OQ} = 3\mathbf{b} + \mu(2\mathbf{a} + \mathbf{b})$ or $\overrightarrow{OQ} = 4\mathbf{b} + 2\mathbf{a} + \omega(2\mathbf{a} + \mathbf{b})$ or $\overrightarrow{PQ} = k(2.8\mathbf{a} + 2.4\mathbf{b})$ . This mark can be awarded even if the previous mark was not.	✓
(b) Two correct expressions for $\overrightarrow{OQ}$ or $\overrightarrow{PQ}$	M1ft	Two of the expressions above, or equivalent, follow through their $\overrightarrow{AC}$ . Two are needed because one alone still has an unknown multiplier in it.	✓
(b) The final vector	A1	$\dfrac{42}{5}\mathbf{a} + \dfrac{36}{5}\mathbf{b}$ or equivalent, eg $8.4\mathbf{a} + 7.2\mathbf{b}$ . Working is required, and this mark depends on both method marks in part (b).	✓

Full marks: 6/6

The sketch shows the curve with equation $y = 2\sin(x + 60)^{\circ}$
The points $A$ and $B$ are marked on the curve.

(i) Find the coordinates of the point $A$ [1 mark]

(ii) Find the coordinates of the point $B$ [1 mark]

(i)(ii)

[Total 2 marks]

Show solution & mark schemeHide solution & mark scheme

Question 25 - Exam Solution

Understanding the Question

Given

The curve

y = 2\sin(x + 60)^{\circ}

, sketched on a grid with no numbers on either axis

The point

A

at the top of the first hill, and the point

B

where the curve crosses the

x

-axis after the lowest point

Find

The coordinates of

A

The coordinates of

B

Plan the Solution

Treat the whole bracket $x + 60$ as the angle. Whatever a plain sine does at some angle, this curve does when the bracket reaches that angle.
A sine never rises above $1$ , so the greatest height here is $2 \times 1$ . Find the $x$ that puts the bracket at $90$ .
For $B$ , put $y = 0$ , list every angle in view whose sine is $0$ , and pick the crossing the sketch marks.

Worked Solution [2 marks]

For

y = a\sin(x + c)^{\circ}

the greatest value is

a

, reached when

x + c = 90

. The curve meets the

x

-axis when

x + c = 0, \; 180, \; 360, \ldots

Step 1: Read what the sketch is telling you

y = 2\sin(x + 60)^{\circ}

(Reason: The dot at

A

sits at the top of the curve, so

A

is where

y

is as large as it gets. The dot at

B

sits on the

x

-axis, so at

B

the value of

y

0

Step 2: The

x

-coordinate of

A

\sin(x + 60)^{\circ} = 1

x + 60 = 90

x = 90 - 60 = 30

(Reason: A sine reaches its largest value,

1

, when its own angle is

90

. Here the angle is the whole bracket, so set the bracket equal to

90

and move the

60

across.)

Step 3: The

y

-coordinate of

A

y = 2 \times 1 = 2

A = (30, 2)

(Reason: At the top of the hill the sine is exactly

1

, so the height is just the

2

in front of it. That

2

is how far the curve reaches above and below the

x

-axis.)

Step 4: The coordinates of

B

2\sin(x + 60)^{\circ} = 0

x + 60 = 0, \; 180, \; 360

x = -60, \; 120, \; 300

B = (300, 0)

(Reason: A sine is

0

0

180

and

360

, so those three angles give the three crossings in view. The sketch marks

B

on the way back up, after the lowest point, which is the third of them.)

(i)

A = (30, 2)

(ii)

B = (300, 0)

Verification

Check 1: Put

x = 30

back into the equation. The bracket becomes

30 + 60 = 90

, and the sine of

90

1

2 \times 1 = 2

, which is the height the sketch draws

A

at.

Check 2: Put

x = 300

back in. The bracket becomes

300 + 60 = 360

, and the sine of

360

0

2 \times 0 = 0

, so

B

really does sit on the

x

-axis.

Check 3: Read the curve as

y = 2\sin x^{\circ}

slid

60

to the left. A plain

2\sin x^{\circ}

peaks at

90

and rises through zero at

360

90 - 60 = 30

and

360 - 60 = 300

, which is both answers over again from a different direction.

Check 4: One full wave is

360

, and a peak sits three quarters of a wave before the next rising crossing, which is

270

300 - 30 = 270

, so the two answers are the right distance apart.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(i) The coordinates of $A$	B1	cao for $(30, 2)$ . Both coordinates are needed, and in that order.	✓
(ii) The coordinates of $B$	B1	cao for $(300, 0)$ . Correct answer only, so $300$ on its own does not score.	✓

Full marks: 2/2

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 4MA1/2H, Monday 3 June 2024: Worked Solutions, Questions 16 to 25

Worked solutions, questions 16 to 25 of 25

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

Question 24, Calculator allowed

Question 24 - Exam Solution

Question 25, Calculator allowed

Question 25 - Exam Solution

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