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

Q: How many questions are on Edexcel 4MA1/1HR, sat Thursday 15 May 2025, and how long is the exam?

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

Q: What grades can you get from the Higher paper?

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

Q: Do you have to show your working?

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

Q: Is a formulae sheet given in the exam?

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

Q: Where can I get the official question paper and mark scheme?

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

Sir Faraz Hassan

31 Aug 2026

Table of Contents▾

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

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/1HR (Higher Tier), June 2025 series, sat Thursday 15 May 2025 –100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

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

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

Download printable PDF

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

Worked solutions, questions 1 to 11 of 26

Work out the lowest common multiple (LCM) of $45$ and $70$ [2 marks]

[Total 2 marks]

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Question 1 - Exam Solution

Understanding the Question

Given

45

and

70

Two whole numbers, and neither one divides into the other.

Find

The lowest common multiple of

45

and

70

That is the smallest positive whole number that both

45

and

70

divide into exactly.

Plan the Solution

Split $45$ and $70$ into products of prime factors.
Collect every prime that appears in either list.
For each of those primes, keep the higher of its two powers.
Multiply the kept factors together.
Check by dividing the answer by each of the two numbers.

Worked Solution [2 marks]

Rule - LCM from prime factors: write each number as a product of primes, then multiply together the highest power of every prime that appears in either number.

Step 1: write

45

as a product of primes

45 = 3 \times 15

15 = 3 \times 5

45 = 3^{2} \times 5

(Reason: Keep dividing by the smallest prime that goes in. The factor

3

is used twice, so it is written as

3^{2}

Step 2: write

70

as a product of primes

70 = 2 \times 35

35 = 5 \times 7

70 = 2 \times 5 \times 7

(Reason:

70

is even, so it starts with

2

. Here every prime is used only once, so there are no powers to record.)

Step 3: take the highest power of every prime

\text{primes used: } 2,\ 3,\ 5,\ 7

\text{LCM} = 2 \times 3^{2} \times 5 \times 7

(Reason: The prime

3

appears twice in

45

and not at all in

70

, so

3^{2}

is kept, not

3

. Each of

2

5

and

7

appears once at most, so one of each is kept.)

Step 4: multiply the kept factors together

2 \times 3^{2} \times 5 \times 7 = 630

(Reason: Multiply the kept factors in any order. The product is the smallest number that both

45

and

70

divide into exactly.)

630

Verification

Check 1: A common multiple must divide exactly by both numbers, leaving no remainder.

\dfrac{630}{45} = 14

and

\dfrac{630}{70} = 9

Check 2: Count up in

70

s and stop at the first number that

45

divides into. Stopping at the first one is what makes it the LOWEST common multiple. The ninth multiple of

70

630

, and none of the eight before it is a multiple of

45

Check 3: The only prime shared by

45

and

70

5

, so their HCF is

5

. For any two numbers, the LCM multiplied by the HCF equals the product of the numbers.

\dfrac{45 \times 70}{5} = 630

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$45, 90, 135, 180 \ldots$ and $70, 140, 210, 280 \ldots$ or $2, 5, 7$ and $3, 3, 5$ or a Venn diagram with $2$ and $7$ in one circle, $3$ and $3$ in the other, and $5$ in the overlap or $\dfrac{45 \times 70}{5}$ or $2, 3, 3, 5, 7$ oe or a table with $5$ against $45$ and $70$ , giving $9$ and $14$ or $5, 9, 14$ oe	M1	For any correct valid method, eg for starting to list at least four multiples of each number, or $2, 5, 7$ and $3, 3, 5$ seen (may be in a factor tree, ignore $1$ ), or a fully correct Venn diagram, or $5, 9, 14$ oe (could be in a table).	✓
Correct answer scores full marks (unless from obvious incorrect working)	A1	$630$ . Allow $2 \times 3^{2} \times 5 \times 7$ oe, eg $5 \times 9 \times 14$ .	✓

Full marks: 2/2

The length of a footbridge is $142.8$ m, correct to $1$ decimal place.

(i) Write down the lower bound of the length of the footbridge. [1 mark]

(ii) Write down the upper bound of the length of the footbridge. [1 mark]

(i)(ii)

[Total 2 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

The length of a footbridge is

142.8

m, correct to

1

decimal place.

That is a rounded measurement, so the true length is not exactly

142.8

Find

The lower bound of the length of the footbridge The upper bound of the length of the footbridge Each part asks for one value in metres, and neither asks for any working.

Plan the Solution

Write down the values either side of $142.8$ on a scale marked in tenths of a metre: $142.7$ and $142.9$ .
The gap between them is the unit the length was rounded to, $0.1$ m.
Halve that unit. The true length can be at most $0.05$ m away from $142.8$ m.
Subtract $0.05$ for the lower bound, then add $0.05$ for the upper bound.
Check by rounding each bound back, and by measuring the width of the interval between them.

Worked Solution [2 marks]

Rule - Bounds of a rounded measurement: halve the unit the measurement was rounded to, then subtract that half for the lower bound and add it for the upper bound.

Step 1: find the unit the length was rounded to

142.7 \qquad 142.8 \qquad 142.9

142.9 - 142.8 = 0.1

(Reason: Correct to

1

decimal place means the length was rounded to the nearest tenth of a metre, so the values it could have been rounded to sit

0.1

m apart.)

Step 2: halve the rounding unit

\dfrac{0.1}{2} = 0.05

(Reason: Every length between the two halfway points rounds to

142.8

, and those halfway points sit

0.05

m either side of it.)

Step 3: subtract the half-unit for the lower bound

142.8 - 0.05 = 142.75

(Reason: The shortest length that still rounds up to

142.8

is the one exactly

0.05

m below it, so take the half-unit off and not the whole

0.1

m.)

Step 4: add the half-unit for the upper bound

142.8 + 0.05 = 142.85

(Reason: Going the other way, a length just under

142.85

m still rounds down to

142.8

m, so

142.85

is where that stops.)

Step 5: state the interval the true length lies in

142.75 \leq \text{length} < 142.85

(Reason: The lower bound is included, because

142.75

itself rounds up to

142.8

. The upper bound is not, because

142.85

rounds up to

142.9

. That is why the mark scheme also accepts

142.8499\ldots

for part (ii).)

(i)

142.75

m(ii)

142.85

Verification

Check 1: Round the values at each edge back to

1

decimal place and see where the answer changes.

142.75

rounds to

142.8

, while

142.749

rounds to

142.7

. At the top,

142.8499

still rounds to

142.8

Check 2: Each bound must be the same distance from the measurement that was written down.

142.8 - 142.75 = 0.05

and

142.85 - 142.8 = 0.05

Check 3: The interval must be exactly one rounding unit wide, with the given length at its midpoint.

142.85 - 142.75 = 0.1

and

\dfrac{142.75 + 142.85}{2} = 142.8

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(i) lower bound	B1	$142.75$	✓
(ii) upper bound	B1	$142.85$ . Accept $142.8499\ldots$ or $142.84\dot{9}$ .	✓

Full marks: 2/2

Show that the product of $2\dfrac{1}{4}$ and $1\dfrac{5}{7}$ is $3\dfrac{6}{7}$ . [3 marks]

[Total 3 marks]

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Question 3 - Exam Solution

Understanding the Question

Given

The two mixed numbers

2\dfrac{1}{4}

and

1\dfrac{5}{7}

The value their product is claimed to have,

3\dfrac{6}{7}

The answer is already printed, so the marks are for the working that reaches it, not for the value.

Find

Working that takes

2\dfrac{1}{4} \times 1\dfrac{5}{7}

3\dfrac{6}{7}

, with every line shown.

Plan the Solution

Write each mixed number as an improper fraction, because a whole number and a fraction cannot be multiplied separately and then joined back together.
Multiply the numerators together and the denominators together, which gives $\dfrac{108}{28}$ .
Cancel that down to $\dfrac{27}{7}$ , then turn it back into a mixed number.
Finish on the value the question prints, so the last line of the working is the statement being shown. A calculator is allowed on this paper, but a decimal answer earns nothing here.

Worked Solution [3 marks]

Rule - Multiplying mixed numbers: write each one as an improper fraction, then

\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}

, and change the result back to a mixed number.

Step 1: write each mixed number as an improper fraction

2\dfrac{1}{4} = \dfrac{4 \times 2 + 1}{4} = \dfrac{9}{4}

1\dfrac{5}{7} = \dfrac{7 \times 1 + 5}{7} = \dfrac{12}{7}

(Reason: A mixed number is a whole number plus a fraction, so multiply the whole number by the denominator and add the numerator on. Four quarters make one whole, so

2\dfrac{1}{4}

is nine quarters, and seven sevenths make one whole, so

1\dfrac{5}{7}

is twelve sevenths. The denominator never changes.)

Step 2: multiply the numerators, and multiply the denominators

\dfrac{9}{4} \times \dfrac{12}{7} = \dfrac{9 \times 12}{4 \times 7} = \dfrac{108}{28}

(Reason: Two fractions are multiplied by multiplying the tops together and the bottoms together. This is the line the second mark is for, and the mark scheme asks to see either

\dfrac{108}{28}

or the cancelling that replaces it.)

Step 3: cancel the fraction down

\dfrac{108}{28} = \dfrac{27 \times 4}{7 \times 4} = \dfrac{27}{7}

(Reason: The highest common factor of

108

and

28

4

, and taking that factor out of the top and the bottom leaves

\dfrac{27}{7}

Step 4: write the improper fraction back as a mixed number

\dfrac{27}{7} = \dfrac{21}{7} + \dfrac{6}{7} = 3 + \dfrac{6}{7}

3 + \dfrac{6}{7} = 3\dfrac{6}{7}

(Reason: Seven goes into

27

three times with

6

left over, so twenty-seven sevenths is three whole ones and

\dfrac{6}{7}

of another. That is the value the question asks for, so the statement is shown.)

2\dfrac{1}{4} \times 1\dfrac{5}{7} = \dfrac{108}{28} = \dfrac{27}{7} = 3\dfrac{6}{7}

as required

Verification

Check 1: Cancel the

4

into the

12

before multiplying instead of afterwards, which is the other order the mark scheme allows.

\dfrac{9}{4} \times \dfrac{12}{7} = \dfrac{9 \times 3}{1 \times 7} = \dfrac{27}{7}

, the same simplified fraction, reached without ever writing

\dfrac{108}{28}

Check 2: Split each mixed number into its two parts and multiply out

\left(2 + \dfrac{1}{4}\right)\left(1 + \dfrac{5}{7}\right)

, a route that forms no improper fraction at all.

2 + \dfrac{10}{7} + \dfrac{1}{4} + \dfrac{5}{28} = \dfrac{108}{28}

, which is the same product as Step 2 reached.

Check 3: Work backwards: turn the printed answer

3\dfrac{6}{7}

into an improper fraction and compare it with the product.

\dfrac{7 \times 3 + 6}{7} = \dfrac{27}{7} = \dfrac{108}{28}

, so the two ends of the working meet.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Both mixed numbers written as improper fractions	M1	For $2\dfrac{1}{4}$ and $1\dfrac{5}{7}$ expressed as improper fractions, eg $\dfrac{9}{4}$ and $\dfrac{12}{7}$ .	✓
Correct cancelling, or the multiplication carried out without cancelling	M1	Correct cancelling, or multiplication of numerators and denominators without cancelling, eg $\dfrac{9}{4} \times \dfrac{12}{7} = \dfrac{108}{28}$ or equivalent, eg $\dfrac{63}{28} \times \dfrac{48}{28} = \dfrac{3024}{784}$ . Cancelling the $4$ into the $12$ before multiplying scores this mark just as the multiplication does.	✓
Conclusion reached from correct working	A1	Dependent on M2, for a conclusion to $3\dfrac{6}{7}$ from correct working - either sight of the result of the multiplication, eg $\dfrac{108}{28}$ or equivalent, must be seen, or correct cancelling prior to the multiplication to $\dfrac{27}{7}$ . Working is required.	✓
Decimals	Note	Use of decimals scores no marks unless as a check.	✓

Full marks: 3/3

A stall at a school summer fair uses the biased $5$ -sided spinner shown below.
When the spinner is spun, it can land on orange or on pink or on brown or on black or on white.

The table gives information about the probability of the spinner landing on each colour.

$\begin{array}{|c|c|c|c|c|c|}\hline \textbf{Colour} & \text{orange} & \text{pink} & \text{brown} & \text{black} & \text{white} \\ \hline \textbf{Probability} & 0.12 & 0.20 & 0.38 & 4x & x \\ \hline \end{array}$

Chloe spins the spinner once.

(a) Work out the probability that the spinner lands on orange or on pink or on brown. [1 mark]

Oliver spins the spinner $350$ times.

(b) Work out an estimate for the number of times the spinner lands on black. [4 marks]

(a)(b)

[Total 5 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

Orange

0.12

, pink

0.20

, brown

0.38

, black

4x

, white

x

One spin lands on exactly one colour, so these five outcomes are all of them.

The spinner is spun once in part (a), and

350

times in part (b).

Find

(a) The probability of orange or pink or brown (b) An estimate of how many of the

350

spins land on black

Plan the Solution

Part (a) needs no algebra. The three probabilities are printed and the outcomes cannot happen together, so add them.
For part (b), all five probabilities total $1$ , so $4x$ and $x$ share whatever the first three leave over.
That leftover is $5$ equal shares of $x$ . Work out one share, then take $4$ of them.
Multiply the probability of black by $350$ to estimate the number of spins.

Worked Solution [5 marks]

Rule - for one spin the probabilities of all the possible outcomes add up to

1

, and an estimate for the number of times an outcome happens is

\text{probability} \times \text{number of trials}

Step 1: add the three probabilities that are given

0.12 + 0.20 + 0.38 = 0.7

(Reason: One spin cannot land on two colours, so orange, pink and brown cannot happen together. For outcomes that cannot happen together, the probability of one or another of them is the sum.)

Step 2: find how much probability is left for black and white

1 - 0.7 = 0.3

(Reason: The five probabilities must total

1

, so whatever the first three do not use belongs to

4x

and

x

together.)

Step 3: work out the value of

x

5x = 0.3

\dfrac{0.3}{5} = 0.06

(Reason: Black is

4x

and white is

x

, which is

5

equal shares of

x

, so the

0.3

is divided by

5

Step 4: work out the probability of black

4 \times 0.06 = 0.24

(Reason: Black is

4x

, so it is worth four of those shares.)

Step 5: estimate the number of spins that land on black

0.24 \times 350 = 84

(Reason: An estimate for the number of times an outcome happens is its probability multiplied by the number of trials. It is an estimate, not a promise:

350

real spins would rarely give exactly this.)

(a)

0.7

(b)

84

Verification

Check 1: Put both answers back into the table and add all five probabilities. With

4x = 0.24

and

x = 0.06

the total must be

1

0.12 + 0.20 + 0.38 + 0.24 + 0.06 = 1

Check 2: Count the spins instead. Work out the expected number for each of the three printed colours, subtract them all from

350

, then split what is left in the ratio

4 : 1

350 - 42 - 70 - 133 = 105

and

\dfrac{105}{5} \times 4 = 84

Check 3: Turn the answer back into a probability. If

84

of the

350

spins are black, the probability of black should come back.

\dfrac{84}{350} = 0.24

, which is

4x

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	for $0.7$ oe, eg $\dfrac{7}{10}$ oe or $70\%$ or $\dfrac{0.7}{1}$ . If probabilities are given as percentages then the % sign must be seen.	✓
(b)	M1ft	for $1 - (0.12 + 0.20 + 0.38) = 0.3$ oe, or $1 - 0.7 = 0.3$ oe using their $0.7$ , or $0.12 + 0.20 + 0.38 + 4x + x = 1$ oe, or their $0.7 \times 350 = 245$ oe, or $0.12 \times 350 = 42$ , or $0.38 \times 350 = 133$ . Follow through their $0.7$ from part (a). If probabilities are given as percentages then the % sign must be seen.	✓
(b)	M1	for their $\dfrac{0.3}{5} = 0.06$ or their $\dfrac{0.3}{5} \times 4 = 0.24$ or $0.24$ , or $x = 0.06$ or $4x = 0.24$ , or their $0.3 \times 350 = 105$ oe, or $350 - 245 = 105$ oe using their $245$ , or $350 - 42 - 0.20 \times 350 - 133 = 105$ oe.	✓
(b)	M1	for their $0.06 \times 350 = 21$ oe, or their $\dfrac{105}{5} = 21$ oe, or their $0.06 \times 4 \times 350$ oe, or their $0.24 \times 350$ . Or for $\dfrac{21}{350}$ or $\dfrac{84}{350}$ .	✓
(b)	A1	for $84$ cao. A correct answer scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 5/5

$\mathscr{E} = \{1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 10,\ 11,\ 12\}$
$A = \{2,\ 4,\ 6,\ 8,\ 10,\ 12\}$
$B = \{3,\ 6,\ 9,\ 12\}$
$C = \{1,\ 3,\ 5,\ 7,\ 9,\ 11\}$

(a) Write down the members of the set
(i) $A \cup B$
(ii) $B'$ [2 marks]

$\mathscr{E} = \{1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 10,\ 11,\ 12\}$
$A = \{2,\ 4,\ 6,\ 8,\ 10,\ 12\}$
$B = \{3,\ 6,\ 9,\ 12\}$
$C = \{1,\ 3,\ 5,\ 7,\ 9,\ 11\}$

$\begin{array}{|cccccc|}\hline \quad \mathscr{E} \quad & \quad \cap \quad & \quad \cup \quad & \quad \varnothing \quad & \quad \in \quad & \quad \notin \quad \\ \hline \end{array}$

(b) Choose a symbol from the box and write it on each dotted line so that each statement below is true.
(i) $A \cap C =$ ...........................
(ii) $13$ ........................... $\mathscr{E}$ [2 marks]

(a)(i)(a)(ii)

[Total 4 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

\mathscr{E} = \{1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 10,\ 11,\ 12\}

, the universal set, with

12

members

A = \{2,\ 4,\ 6,\ 8,\ 10,\ 12\}

, the even members of

\mathscr{E}

B = \{3,\ 6,\ 9,\ 12\}

, the multiples of

3

\mathscr{E}

C = \{1,\ 3,\ 5,\ 7,\ 9,\ 11\}

, the odd members of

\mathscr{E}

The box offers six symbols:

\mathscr{E}

\cap

\cup

\varnothing

\in

and

\notin

Find

(a) the members of

A \cup B

and the members of

B'

; (b) the symbol that makes

A \cap C = \ldots

true, and the symbol that makes

13 \ldots \mathscr{E}

true.

Plan the Solution

$\cup$ means union, so for (a)(i) take everything that is in $A$ , in $B$ , or in both, and write each member once, in order.
The dash in $B'$ means complement, so for (a)(ii) keep everything in $\mathscr{E}$ that is not in $B$ .
$\cap$ means intersection, so for (b)(i) look for numbers that appear in $A$ and in $C$ .
For part (b), first ask what each blank sits between. $\varnothing$ and $\mathscr{E}$ are sets, $\cap$ and $\cup$ join two sets, and $\in$ and $\notin$ go between a number and a set.

Worked Solution [4 marks]

Rule - Set notation:

A \cup B

is everything in either set,

A \cap B

is only what is in both,

B'

is everything in

\mathscr{E}

that is not in

B

, and

\varnothing

is the set with no members.

Step 1: List

A \cup B

A = \{2,\ 4,\ 6,\ 8,\ 10,\ 12\}

B = \{3,\ 6,\ 9,\ 12\}

A \cup B = \{2,\ 3,\ 4,\ 6,\ 8,\ 9,\ 10,\ 12\}

(Reason: Union collects every member of

A

together with every member of

B

. Only

3

and

9

are new, because

6

and

12

are already in

A

, and a member is never written twice.)

Step 2: List

B'

\mathscr{E} = \{1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 10,\ 11,\ 12\}

B' = \{1,\ 2,\ 4,\ 5,\ 7,\ 8,\ 10,\ 11\}

(Reason: The complement is everything left in

\mathscr{E}

once the members of

B

are crossed out, so

3

6

9

and

12

go, and the eight numbers that are left stay.)

Step 3: Work out

A \cap C

A = \{2,\ 4,\ 6,\ 8,\ 10,\ 12\}

C = \{1,\ 3,\ 5,\ 7,\ 9,\ 11\}

A \cap C = \varnothing

(Reason: Every member of

A

is even and every member of

C

is odd, so no number can belong to both. An intersection with no members is the empty set, written

\varnothing

Step 4: Compare

13

with

\mathscr{E}

\mathscr{E} = \{1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9,\ 10,\ 11,\ 12\}

13 \notin \mathscr{E}

(Reason: This blank sits between a number and a set, so it takes

\in

\notin

. The largest member of

\mathscr{E}

12

, so

13

is not one of its members and the statement is made true by

\notin

(a)(i)

A \cup B = \{2,\ 3,\ 4,\ 6,\ 8,\ 9,\ 10,\ 12\}

(a)(ii)

B' = \{1,\ 2,\ 4,\ 5,\ 7,\ 8,\ 10,\ 11\}

(b)(i)

A \cap C = \varnothing

(b)(ii)

13 \notin \mathscr{E}

Verification

Check 1 - count $A \cup B$ a different way:

A

has

6

members and

B

has

4

, and the two share

6

and

12

, so the union should be

2

members shorter than

6 + 4

6 + 4 - 2 = 8

, and the listed union has

8

members.

Check 2 - $B$ and $B'$ must fill $\mathscr{E}$ between them: Every member of

\mathscr{E}

is either in

B

or in

B'

and never in both, so the two sizes must add to

12

and the two lists must share nothing.

4 + 8 = 12

, and

\{3,\ 6,\ 9,\ 12\}

and

\{1,\ 2,\ 4,\ 5,\ 7,\ 8,\ 10,\ 11\}

have no member in common.

Check 3 - rule out the other symbols in the box: In (b)(i) the blank has to name a set, and only

\varnothing

and

\mathscr{E}

are sets, so

\cap

\cup

\in

and

\notin

cannot go there. In (b)(ii) the blank joins a number to a set, so only

\in

and

\notin

can go there.

\mathscr{E}

has

12

members while

A \cap C

has none, which leaves

\varnothing

; and

13

is greater than

12

, so

\in

would be false and

\notin

is left.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)(i) $A \cup B$	B1	$2,\ 3,\ 4,\ 6,\ 8,\ 9,\ 10,\ 12$	✓
(a)(ii) $B'$	B1	$1,\ 2,\ 4,\ 5,\ 7,\ 8,\ 10,\ 11$	✓
(b)(i) $A \cap C = \ldots$	B1	$\varnothing$	✓
(b)(ii) $13 \ldots \mathscr{E}$	B1	$\notin$	✓

Full marks: 4/4

(a) Write down the inequality that the number line above shows. [2 marks]

(b) Solve the inequality $7a - 5 \leq 3a + 28$
You must show clear algebraic working. [2 marks]

(a)(b)

[Total 4 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

A number line carrying an open circle above

-2

and a solid circle above

1

, with a thick line joining them.

The inequality

7a - 5 \leq 3a + 28

Find

(a) the inequality the number line shows (b) the solution of

7a - 5 \leq 3a + 28

, with the algebra shown

Plan the Solution

Read the two ends off the number line first: the shading starts at one tick and stops at another.
Then let each circle choose its own sign. An open circle leaves its value out, so it gives $<$ or $>$ . A solid circle takes its value in, so it gives $\leq$ or $\geq$ .
For part (b), work exactly as you would with an equation: gather the $a$ terms on one side, the numbers on the other, then divide.
Watch the one thing that is different from an equation. Multiplying or dividing by a negative number turns the sign round. Here you divide by $4$ , which is positive, so nothing turns.

Worked Solution [4 marks]

Rule - an inequality is solved like an equation, except that multiplying or dividing both sides by a negative number reverses the sign. On a number line an open circle leaves its value out and a solid circle takes it in.

(a) Step 1 - read the two ends off the line

\text{open circle above } -2

\text{solid circle above } 1

(Reason: The thick line runs from the circle above

-2

to the circle above

1

, so those two numbers are the ends of the region.)

(a) Step 2 - let each circle choose its sign

x > -2

x \leq 1

(Reason: The left circle is hollow, so

-2

itself is not included and the sign is strict. The right circle is filled, so

1

itself is included.)

(a) Step 3 - write the two conditions as one inequality

-2 < x \leq 1

(Reason: Both conditions hold at once, so

x

sits between the two ends, with only the right-hand end taken in.)

(b) Step 1 - collect the

a

terms on one side

7a - 5 \leq 3a + 28

7a - 3a \leq 28 + 5

(Reason: Subtract

3a

from both sides and add

5

to both sides. Adding and subtracting never turn the sign round, so it stays as it is.)

(b) Step 2 - simplify each side

4a \leq 33

(Reason: On the left

7a - 3a

leaves

4a

, and on the right

28 + 5

makes

33

(b) Step 3 - divide both sides by

4

a \leq \dfrac{33}{4}

a \leq 8.25

(Reason:

4

is positive, so dividing by it leaves the sign pointing the same way. The paper allows a calculator, but the working still has to be shown.)

(a)

-2 < x \leq 1

(b)

a \leq 8.25

Verification

Check 1 - test each end of the number line, and a point between them: Take

x = 0

, which sits under the thick line:

-2 < 0

and

0 \leq 1

are both true. Take

x = -2

, where the circle is hollow:

-2 < -2

is false, so

-2

is left out, as the open circle asks. Take

x = 1

, where the circle is filled:

1 \leq 1

is true, so

1

is taken in. The inequality agrees with the line at both ends and in between.

Check 2 - put the boundary value back into both sides: At the boundary the two sides must be equal, because

8.25

is exactly where the inequality turns into an equation.

7 \times 8.25 - 5 = 52.75

and

3 \times 8.25 + 28 = 52.75

Check 3 - test a value on each side of the boundary: At

a = 8

7 \times 8 - 5 = 51

and

3 \times 8 + 28 = 52

, so the inequality holds. At

a = 9

7 \times 9 - 5 = 58

and

3 \times 9 + 28 = 55

, so it fails. Values below the boundary work and values above it do not, so the sign points the right way.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $-2 < x \leq 1$	B2	accept $1 \geq x > -2$ or $x > -2, x \leq 1$	✓
(a) $-2 < x$ or $x \leq 1$ or $-2 \leq x < 1$ or $-2 \leq x \leq 1$ or $-2 < x < 1$	B1	if not B2 then B1 for any one of these	✓
(a) a letter other than $x$	Note	Condone use of a variable other than $x$ but not $0$ .	✓
$7a - 3a \leq 28 + 5$ or $4a \leq 33$ or $-5 - 28 \leq 3a - 7a$ or $-33 \leq -4a$	M1	for $a$ terms on one side and numbers on the other. Condone $=$ rather than $\leq$ or any other sign for this mark.	✓
$a \leq 8.25$ (working required)	A1	(dep on M1) oe eg $a \leq \dfrac{33}{4}$ or $a \leq 8\dfrac{1}{4}$ or $8.25 \geq a$ . Must have correct sign on answer line.	✓
(b) the correct answer seen in the working space only	Note	Sight of the correct answer in the working space and just $8.25$ on the answer line gains M1 only.	✓

Full marks: 4/4

Convert a speed of $50x$ metres per second into kilometres per hour. [3 marks]

km/h

[Total 3 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

A speed of

50x

metres per second.

Two facts about the units:

1

hour is

3600

seconds, and

1

kilometre is

1000

metres.

Find

The same speed written in kilometres per hour. No value is given for

x

, so it stays in the answer.

Plan the Solution

Change the time first: a speed given per second covers $3600$ times as far in one hour, so multiply by $3600$ .
Change the distance next: $1000$ metres make $1$ kilometre, so divide by $1000$ .
Carry the $x$ through every line. It is part of the speed, not something to be solved for.
Finish by doing both moves at once, as a single multiplier, and check the two agree.

Worked Solution [3 marks]

Rule - metres per second to kilometres per hour: multiply by the seconds in an hour,

60 \times 60 = 3600

, then divide by the metres in a kilometre,

1000

. The two together are one multiplier,

\dfrac{3600}{1000} = 3.6

Step 1: change the seconds into hours

50x \times 3600 = 180\,000x

(Reason: An hour is

3600

seconds, so whatever the speed covers in one second it covers

3600

times over in one hour. The distance is still in metres, so this is a speed in metres per hour.)

Step 2: change the metres into kilometres

\dfrac{180\,000x}{1000} = 180x

(Reason: There are

1000

metres in a kilometre, so dividing by

1000

turns metres per hour into kilometres per hour. Both units have now been changed, so the conversion is complete.)

Step 3: the same conversion as one multiplier

\dfrac{3600}{1000} = 3.6

50x \times 3.6 = 180x

(Reason: Multiplying by

3600

and dividing by

1000

can be done in one move, because every speed in metres per second is

3.6

times as many kilometres per hour. It reaches the same answer, which is the point of doing it.)

180x

km/h

Verification

Check 1: Test it on a number. Take

x = 2

, so the speed is

100

metres per second. In one hour it covers

100 \times 3600 = 360\,000

metres, and that is

360

kilometres. The answer gives

180 \times 2 = 360

, the same speed.

Check 2: Run the conversion backwards.

180x

kilometres per hour is

180\,000x

metres per hour, and an hour is

3600

seconds.

\dfrac{180\,000x}{3600} = 50x

metres per second, which is the speed the question gave.

Check 3: Use the other standard fact instead:

1

kilometre per hour is

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

metres per second, so divide the speed by

\dfrac{5}{18}

50x \times \dfrac{18}{5} = 180x

, the same answer along a different route.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{50x}{1000} \; (= 0.05x)$ oe or $50x \times 60 \times 60 \; (= 180\,000x)$ oe or $\dfrac{50x}{\dfrac{1}{3600}} \; (= 180\,000x)$ oe or $\dfrac{50x}{1000} \times 60 \; (= 3x)$ or $\dfrac{3600}{1000}$ or $\dfrac{18}{5}$ or $3.6$ or $\dfrac{1000}{3600}$ or $\dfrac{5}{18}$ or $0.277(77\ldots)$	M1	Condone omission of $x$ for this mark	✓
eg $\dfrac{50x \times 60 \times 60}{1000}$ oe or $50x \times 3.6$ oe or $\dfrac{50x}{\dfrac{1000}{3600}}$ oe or $180$	M1	For a complete method including $x$ or for an answer of $180$	✓
$180x$	A1	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 3/3

(a) Simplify $a^{6} \times a^{10}$ [1 mark]

(b) Simplify $\dfrac{c^{30}}{c^{12}}$ [1 mark]

(a)(b)(c)(i)(c)(ii)

[Total 5 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

Part (a): the product

a^{6} \times a^{10}

, two powers of the same letter multiplied together.

Part (b): the quotient

\dfrac{c^{30}}{c^{12}}

, two powers of the same letter divided.

Part (c): the quadratic expression

y^{2} - 10y + 21

, whose

y^{2}

term has coefficient

1

Find

Part (a): the product written as a single power of

a

. Part (b): the quotient written as a single power of

c

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

y

that satisfy the equation. It is a quadratic, so expect two solutions.

Plan the Solution

Parts (a) and (b) are the index laws. Multiplying powers of one letter adds the indices; dividing them subtracts.
Part (c)(i): the coefficient of $y^{2}$ is $1$ , so look for two numbers whose product is $21$ and whose sum is $-10$ .
The product is positive while the sum is negative, so both numbers are negative. That leaves very few pairs to test.
Part (c)(ii): the word Hence means carry the brackets down from part (c)(i). A product is zero only when one of its brackets is zero.

Worked Solution [5 marks]

Rule - Index laws and factorising:

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

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

, and

y^{2} + by + c = (y + p)(y + q)

when

pq = c

and

p + q = b

Step 1 - Part (a): combine the two powers

a^{6} \times a^{10} = a^{6+10} = a^{16}

(Reason: A power of

a

is a run of

a

s multiplied together, so a product of two powers is one longer run:

6

of them followed by

10

more.)

Step 2 - Part (b): cancel the common powers

\dfrac{c^{30}}{c^{12}} = c^{30-12} = c^{18}

(Reason: Each

c

on the bottom cancels one

c

on the top.

12

of the

30

are cancelled and

18

are left.)

Step 3 - Part (c)(i): find the pair of numbers

p \times q = 21 \quad \text{and} \quad p + q = -10

(-3) \times (-7) = 21

(-3) + (-7) = -10

(Reason: The factor pairs of

21

are

1

with

21

, and

3

with

7

. Both numbers must be negative for the sum to be negative, and only

-3

with

-7

adds to

-10

Step 4 - Part (c)(i): write down the two brackets

y^{2} - 10y + 21 = (y - 3)(y - 7)

(Reason:

p = -3

gives the bracket

(y - 3)

, and

q = -7

gives

(y - 7)

Step 5 - Part (c)(ii): set each bracket to zero

(y - 3)(y - 7) = 0

y - 3 = 0 \quad \text{or} \quad y - 7 = 0

y = 3 \quad \text{or} \quad y = 7

(Reason: Two numbers multiply to give zero only when at least one of them is zero, so each bracket is taken to be zero in turn. This is why part (c)(ii) says Hence: the brackets are already there from part (c)(i).)

(a)

a^{16}

(b)

c^{18}

(c)(i)

(y - 3)(y - 7)

(c)(ii)

y = 3

y = 7

Verification

Check 1 - part (a): Substitute

a = 2

and work each side out as an ordinary number.

2^{6} \times 2^{10} = 2^{16}

, and both sides come to

65\,536

Check 2 - part (b): A division is checked by multiplying back: the answer times the divisor must return the top of the fraction.

c^{18} \times c^{12} = c^{18+12} = c^{30}

Check 3 - part (c)(i): Expand the two brackets again and collect the

y

terms.

(y - 3)(y - 7) = y^{2} - 7y - 3y + 21 = y^{2} - 10y + 21

Check 4 - part (c)(ii): Substitute each solution into the left-hand side of the equation; both must give zero.

3^{2} - 10 \times 3 + 21 = 0

and

7^{2} - 10 \times 7 + 21 = 0

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $a^{16}$	B1	for $a^{16}$	✓
(b) $c^{18}$	B1	for $c^{18}$	✓
(c)(i) $(y \pm 3)(y \pm 7)$	M1	for $(y \pm 3)(y \pm 7)$ or for $(y \pm a)(y \pm b)$ with $ab = 21$ or $a + b = -10$	✓
(c)(i) $(y - 3)(y - 7)$	A1	for correct factors	✓
(c)(i)	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓
(c)(ii) $3$ , $7$	B1ft	ft dep on factorising in the form $(y \pm p)(y \pm q)$	✓

Full marks: 5/5

The diagram shows triangle $PQR$ , in which angle $PQR$ is a right angle.

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

[Total 3 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

Triangle

PQR

, with the right angle at

Q

The side

PQ

6.5

cm long.

The angle at

R

, between

RQ

and

RP

, is

24^\circ

Find

The length of

QR

, the side that joins the right angle to the

24^\circ

angle. The answer is wanted correct to

3

significant figures, so the rounding is part of what is being marked.

Plan the Solution

Stand at $R$ and name the two sides that matter: $PQ$ is opposite the angle and $QR$ is adjacent to it.
Opposite and adjacent together are the tangent ratio, so the hypotenuse $PR$ is not needed here. A route through the hypotenuse does exist, and it is two steps longer; it is Check 3 below.
The unknown starts underneath the fraction, so rearrange first and only then reach for the calculator: $QR$ ends up as $6.5$ divided by the tangent.
Keep the calculator's full value to the end and round once, to $3$ significant figures.

Worked Solution [3 marks]

Rule - Tangent ratio in a right-angled triangle:

\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}

, so a known angle and a known opposite side give the adjacent side.

Step 1 - Name the sides as the angle at

R

sees them

\tan 24^\circ = \dfrac{PQ}{QR}

\tan 24^\circ = \dfrac{6.5}{QR}

(Reason: Seen from

R

, the side

PQ

is opposite the angle and

QR

runs from the angle to the right angle, so

QR

is the adjacent side. Opposite over adjacent is the tangent, and the hypotenuse

PR

does not appear in it.)

Step 2 - Rearrange to make

QR

the subject

QR \times \tan 24^\circ = 6.5

QR = \dfrac{6.5}{\tan 24^\circ}

(Reason:

QR

is underneath the fraction, so multiply both sides by it first and then divide both sides by

\tan 24^\circ

. Doing the rearranging before any button is pressed is what keeps the two operations the right way round.)

Step 3 - Work the value out on the calculator

\tan 24^\circ = 0.4452287

\dfrac{6.5}{0.4452287} = 14.5992

(Reason: The calculator must be in degree mode - in radian mode the same keystrokes read

24

as radians and return a different number altogether. The size of the answer is the check on the operation:

\tan 24^\circ

is less than

1

, and a small angle at

R

has to leave

QR

longer than

6.5

cm, not shorter.)

Step 4 - Round to

3

significant figures

14.5992\ldots \to 14.6

QR = 14.6\text{ cm}

(Reason: The first three significant figures are the

1

, the

4

and the

5

. The digit after them is a

9

, so the

5

rounds up to a

6

QR = 14.6

Verification

Check 1 - multiply back: A division is checked by multiplying back. If the length is right, the length times

\tan 24^\circ

must return the side

PQ

0.4452287 \times 14.5992 = 6.5

1

decimal place

Check 2 - work from the third angle instead: The three angles add to

180

degrees, so the angle at

P

180 - 90 - 24 = 66

degrees. Seen from

P

the side

QR

is the opposite one, so this time it is

6.5

multiplied by the tangent rather than divided by it.

6.5 \times 2.246037 = 14.5992

, the same length from a different ratio

Check 3 - Pythagoras through the hypotenuse: The sine ratio gives the hypotenuse

PR

from the same two given values,

\dfrac{6.5}{0.4067366} = 15.9809

, and the three sides of a right-angled triangle must then satisfy Pythagoras.

15.9809^{2} - 6.5^{2} = 213.14

and

14.5992^{2} = 213.14

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $\tan 24 = \dfrac{6.5}{QR}$ or $\dfrac{6.5}{\sin 24} = \dfrac{QR}{\sin(180 - 90 - 24)}$ oe or $\tan(180 - 90 - 24) = \dfrac{QR}{6.5}$ or $(PR =) \dfrac{6.5}{\sin 24} = 15.9\ldots$ and $6.5^{2} + QR^{2} = 15.9^{2}$ , where the $15.9$ may be the candidate's own value for $PR$	M1	for setting up a trig equation in $QR$ or for a complete method to find $PR$ and then setting up Pythagoras or a trig equation for $QR$	✓
eg $(QR =) \dfrac{6.5}{\tan 24}$ or $(QR =) \dfrac{6.5}{\sin 24} \times \sin 66$ or $(QR =) 6.5 \tan 66$ where $66 = 180 - 90 - 24$ or $(QR =) \sqrt{15.9^{2} - 6.5^{2}}$ , again with the candidate's own value for $PR$	M1	for a complete method	✓
$14.6$	A1	accept $14.5$ to $14.61$	✓
Answer	Note	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 3/3

The diagram shows two water containers at a garden centre.
One is a cuboid trough and the other is a cylindrical drum.

The trough measures $35$ cm by $28$ cm by $20$ cm
The surface of the water in the trough is $9$ cm above the base of the trough.

The drum has a radius of $10$ cm and a height of $33$ cm
The drum is completely full of water.

Freya is going to pour all the water from the drum into the trough.

Show that the trough will not be completely full of water. [3 marks]

[Total 3 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

A cuboid trough measuring

35

cm by

28

cm by

20

cm, holding water to a depth of

9

cm.

A cylindrical drum of radius

10

cm and height

33

cm, completely full of water.

All the water in the drum is poured into the trough, and none of it is spilt.

Find

A comparison rather than a single value: the room the trough still has against the water the drum holds. The demand is show that, so every figure has to be written down. A conclusion with no working earns nothing here, and there is no answer line to fill in.

Plan the Solution

Work out how much empty space the trough still has. The base is the same rectangle at every height, so it is the base area times the height that is still empty.
Work out the volume of the drum with $V = \pi r^{2} h$ . The drum is completely full, so that volume is exactly the water that is poured in.
Compare the two. If the water poured in is smaller than the space waiting for it, the trough cannot fill up.
Keep $\pi$ on the calculator until the last line. Rounding it early is what pushes a volume outside the range the mark scheme accepts.

Worked Solution [3 marks]

Rule - Volume of a cuboid:

V = l \times w \times h

. Volume of a cylinder:

V = \pi r^{2} h

. A container is full when the water in it has the same volume as the container itself.

Step 1 - Work out the space still empty in the trough

35 \times 28 = 980

(20 - 9) \times 980 = 10\,780

(Reason: Every horizontal slice of the trough is the same rectangle, so one base area serves every depth. The water already stands

9

cm up a

20

cm side, which leaves

20 - 9 = 11

cm of height empty above it, and that empty part is a cuboid of its own.)

Step 2 - Work out the volume of water in the drum

V = \pi r^{2} h

V = \pi \times 10^{2} \times 33 = 3300\pi = 10\,367.25\ldots

(Reason: The radius is

10

cm and the height is

33

cm, so squaring the radius gives

100

and

100 \times 33 = 3300

. The drum is completely full, so this volume is exactly the water that is poured out of it.)

Step 3 - Compare the water poured in with the space available

10\,367.25\ldots < 10\,780

10\,780 - 10\,367.25\ldots = 412.74\ldots

(Reason: The drum holds less water than the trough has room for, so all of it fits and none of it spills. The gap between the two figures is the volume that never gets filled: about

413

cm³ of the trough is still empty once the pouring is finished, so the trough is not completely full of water.)

Water poured in

3300\pi = 10\,367.25\ldots

cm³, space waiting for it

10\,780

cm³The trough is left

412.74\ldots

cm³ short of full, so it is not completely full of water

Verification

Check 1 - compare the total volumes instead: Add the two lots of water and compare with the trough's own volume. The water already there is

9 \times 980 = 8820

cm³ and the trough itself holds

20 \times 980 = 19\,600

cm³.

8820 + 10\,367.25\ldots = 19\,187.25\ldots

, which is less than

19\,600

Check 2 - work out the depth the water would stand at: Divide the total volume of water by the area of the base. If the trough were completely full the depth would read

20

cm exactly.

\dfrac{19\,187.25\ldots}{980} = 19.57\ldots

cm, which is below the

20

cm rim

Check 3 - reach the shortfall by two different subtractions: The volume left empty can be taken from the space above the water or from the trough's full capacity. Two different subtractions have to land on the same figure.

10\,780 - 10\,367.25\ldots = 412.74\ldots

and

19\,600 - 19\,187.25\ldots = 412.74\ldots

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(volume of water =) $9 \times 35 \times 28 \; (= 8820)$ or (total volume of cuboid =) $20 \times 35 \times 28 \; (= 19\,600)$ or (volume of space =) $(20 - 9) \times 35 \times 28 \; (= 10\,780)$	M1	for a method to find a relevant volume for the cuboid	✓
$\pi \times 10^{2} \times 33 \; (= 3300\pi \text{ or } 10\,367(.25\ldots))$ oe	M1	(indep) for a method to find the volume of the cylinder, accept a volume in the range $10\,362$ to $10\,368.6$ . Allow $3.14\ldots$ or $\dfrac{22}{7}$ for $\pi$	✓
(total volume of water =) $8820 + 10\,367(.25\ldots) \; (= 19\,187(.25\ldots))$ (difference between volumes of both solids =) $19\,600 - 10\,367(.25\ldots) \; (= 9232(.74\ldots))$ (volume not filled =) $19\,600 - 8820 - 10\,367(.25\ldots) \; (= 412(.74\ldots))$ Answer column: Shown	A1	correct workings with accurate figures, eg $\begin{array}{\|c\|c\|}\hline \textbf{Value 1} & \textbf{Value 2} \\ \hline 10\,780 & 10\,367(.25\ldots) \; \text{accept } 10\,362 \text{ to } 10\,372 \\ \hline 19\,600 & 19\,187(.25\ldots) \; \text{accept } 19\,182 \text{ to } 19\,192 \\ \hline 8820 & 9232(.74\ldots) \; \text{accept } 9228 \text{ to } 9238 \\ \hline 412(.74\ldots) \text{ or } 413 \; \text{accept } 408 \text{ to } 418 & \text{none needed} \\ \hline \end{array}$	✓
Working required	Note	printed in italic in the working column of the A1 row: the marks here are for the figures on the page, so a bare statement that the trough will not fill scores nothing	✓

Full marks: 3/3

Jian invests money for $2$ years.
The amount invested with Meridian Bank is $\$2500$ and the amount invested with Halewood Bank is $\$3000$ .

$\begin{array}{|c|}\hline \textbf{Meridian Bank} \\ \text{Invests } \$2500 \\ \text{amount of money invested} : \text{total interest after 2 years} = 20 : 3 \\ \hline\end{array}$

$\begin{array}{|c|}\hline \textbf{Halewood Bank} \\ \text{Invests } \$3000 \\ 4\%\text{ per year compound interest for 2 years} \\ \hline\end{array}$

Jian receives more interest from Meridian Bank than from Halewood Bank.
How much more? [5 marks]

[Total 5 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

Meridian Bank:

\$2500

invested, with amount invested to total interest after

2

years in the ratio

20 : 3

Halewood Bank:

\$3000

invested at

4\%

per year compound interest for

2

years

One panel states its interest as a ratio, the other as a compound percentage - two different pieces of work before anything can be compared.

Find

How much more interest Meridian Bank pays than Halewood Bank, in dollars. Both interest amounts have to be found first; neither panel states one directly.

Plan the Solution

Read the Meridian panel as a ratio: $20$ parts is the amount invested and $3$ parts is the interest. Divide $\$2500$ by $20$ to get one part, then take $3$ of them.
Read the Halewood panel as compound interest: multiply $\$3000$ by $1.04$ once for each year, which grows the balance to a total.
That total still contains the money invested, so take the amount invested off it to leave the interest on its own.
Subtract the smaller interest from the larger one. Compare interest with interest - never the two totals, because the amounts invested are different.

Worked Solution [5 marks]

Rule - Compound interest: the total after

n

years is

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

, and the interest is that total minus the amount invested

P

. Rule - Ratio: one part is the whole divided by the number of parts.

Step 1: Find one part of the Meridian ratio

\dfrac{2500}{20} = 125

(Reason: the ratio

20 : 3

sets the amount invested against

20

parts, and

\$2500

was invested, so one part is worth

\$125

Step 2: Work out the interest from Meridian Bank

125 \times 3 = 375

(Reason: the second number in the ratio is the interest, and it is

3

parts, so the interest is

3

lots of

\$125

Step 3: Grow the Halewood investment for two years

3000 \times 1.04^{2} = 3244.80

(Reason: adding

4\%

multiplies a balance by

1.04

, and that happens once for each of the

2

years, so the multiplier is

1.04^{2}

Step 4: Take off the amount invested to leave the interest

3244.80 - 3000 = 244.80

(Reason: the

\$3244.80

is the whole balance, and

\$3000

of it is the money that was put in, so the rest is what the account earned.)

Step 5: Compare the two amounts of interest

375 - 244.80 = 130.20

(Reason: both amounts of interest are now known, so subtracting the smaller from the larger says how much more the first bank pays.)

\$130.20

Verification

Check 1 - rebuild the Meridian account a different way: The ratio

20 : 3

means the account returns

23

parts altogether. One part is

\$125

, so the balance after

2

years can be built first and the amount invested taken off afterwards.

23 \times 125 = 2875

, and

2875 - 2500 = 375

, which is the interest found in Step 2.

Check 2 - add the Halewood interest year by year: Year

1

earns

4\%

\$3000

. That interest is added on, so year

2

earns

4\%

of the larger balance. The two years together should give the amount found in Step 4.

0.04 \times 3000 = 120

, then

0.04 \times 3120 = 124.8

, and

120 + 124.8 = 244.8

Check 3 - is the size sensible? Meridian pays

\dfrac{3}{20}

\$2500

, and

\dfrac{3}{20}

15\%

. Halewood pays

4\%

twice on a growing balance, so a little over

8\%

\$3000

. The two amounts of interest should therefore be close to

\$375

and

\$245

, leaving a gap of roughly

\$130

. The difference found,

\$130.20

, is the size that estimate predicts, and it is Meridian that is ahead.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Interest from Meridian Bank	M1	For a method to find the interest for Meridian Bank: $\dfrac{2500}{20} \times 3 \; (= 375)$ oe, or $125 \times 3 \; (= 375)$ , or $\dfrac{7500}{20} \; (= 375)$ . The same method on the other amount also earns it: $\dfrac{3000}{20} \times 3 \; (= 450)$ oe, or $150 \times 3 \; (= 450)$ , or $\dfrac{9000}{20} \; (= 450)$ . An answer of $2875$ or $3450$ implies this method mark.	✓
Four per cent, or one hundred and four per cent, of an amount	M1	For finding $4\%$ or $104\%$ of $3000$ or $2500$ : $0.04 \times 3000 \; (= 120)$ oe, or $0.04 \times 2500 \; (= 100)$ , or $1.04 \times 3000 \; (= 3120)$ oe, or $1.04 \times 2500 \; (= 2600)$ . Or award M2 here for $3000 \times 1.04^{2} \; (= 3244.8)$ or $2500 \times 1.04^{2} \; (= 2704)$ .	✓
Total in the Halewood account	M1	For completing the method to find the total amount for Halewood Bank: $1.04 \times 3120 \; (= 3244.8)$ oe, or $1.04 \times 2600 \; (= 2704)$ , where the candidate's own value from the row above may be used in place of $3120$ or $2600$ . Or $3000 \times 1.04^{3} \; (= 3374.59)$ or $2500 \times 1.04^{3} \; (= 2812.16)$ .	✓
Interest from Halewood Bank	M1	For a complete method to find the interest for Halewood Bank, eg $3244.8 - 3000 \; (= 244.8)$ or $2704 - 2500 \; (= 204)$ , where the candidate's own total may be used in place of $3244.8$ or $2704$ .	✓
How much more	A1	$130.2$ , with or without the trailing zero ( $130.20$ ). A correct answer scores full marks unless it comes from obviously incorrect working.	✓
Special case	SC	If neither the second nor the third method mark is gained, award one special-case method mark for $0.08 \times 3000 \; (= 240)$ oe, or $240$ , or $1.08 \times 3000 \; (= 3240)$ , or $0.08 \times 2500 \; (= 200)$ oe, or $200$ , or $1.08 \times 2500 \; (= 2700)$ , or $3000 \times (1 - 0.04)^{2} \; (= 2764.8)$ , or $2500 \times (1 - 0.04)^{2} \; (= 2304)$ . These are the values of two named slips: simple interest over the two years, and a $4\%$ decrease each year in place of an increase.	✓
Notation	Note	Accept $(1 + 0.04)$ or $\left(1 + \dfrac{4}{100}\right)$ as equivalent to $1.04$ throughout.	✓

Full marks: 5/5

Continue to questions 12 to 19

The remaining 8 questions, with the same full worked solutions and mark schemes

Frequently asked questions

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

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

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

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

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

Keep revising

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

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

Worked solutions, questions 1 to 11 of 26

Question 1, Calculator allowed

Question 1 - Exam Solution

Question 2, Calculator allowed

Question 2 - Exam Solution

Question 3, Calculator allowed

Question 3 - Exam Solution

Question 4, Calculator allowed

Question 4 - Exam Solution

Question 5, Calculator allowed

Question 5 - Exam Solution

Question 6, Calculator allowed

Question 6 - Exam Solution

Question 7, Calculator allowed

Question 7 - Exam Solution

Question 8, Calculator allowed

Question 8 - Exam Solution

Question 9, Calculator allowed

Question 9 - Exam Solution

Question 10, Calculator allowed

Question 10 - Exam Solution

Question 11, Calculator allowed

Question 11 - Exam Solution

Frequently asked questions

Keep revising

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