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Edexcel IGCSE 4MA1/1FR, Thursday 15 May 2025: Worked Solutions, Questions 15 to 26

Sir Faraz Hassan

24 Aug 2026

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

Back to questions 1 to 14

This is the rest of the paper. Questions 1 to 14, the paper's overview and the frequently asked questions are on the first page.

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

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

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

Download printable PDF

All 26 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 15 to 26 of 26

Fiona took the written test to qualify as a football referee.
Fiona scored $119$ out of $140$ marks on the test.

Work out Fiona's score as a percentage. [2 marks]

[Total 2 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

A written test marked out of

140

marks

Fiona's score on that test:

119

marks

Find

Fiona's score written as a percentage

Plan the Solution

Write the score as a fraction of the total number of marks on the test.
Cancel that fraction down: $119$ and $140$ share a factor of $7$ .
Multiply the cancelled fraction by $100$ , because a percentage counts parts per $100$ .

Worked Solution [2 marks]

Rule - A score as a percentage: put the score over the total, then multiply by

100

. The fraction says what part of the whole test was earned, and multiplying by

100

re-counts that part in hundredths.

Step 1: Write the score as a fraction of the whole test

\dfrac{119}{140}

(Reason: (Reason: the whole test is

140

marks, so the total goes underneath and the

119

marks that were scored go on top.))

Step 2: Cancel the fraction to its simplest form

119 = 7 \times 17

140 = 7 \times 20

\dfrac{119}{140} = \dfrac{17}{20}

(Reason: (Reason:

7

is the highest common factor of the two numbers, and cancelling it leaves

20

underneath, which scales up to

100

in one step.))

Step 3: Scale the fraction up to a percentage

\dfrac{17}{20} \times 100 = 17 \times 5 = 85

(Reason: (Reason:

20

goes into

100

exactly

5

times, so each twentieth of the test is worth

5

per cent.))

85

Verification

Check 1 - turn the percentage back into marks: Take

85

per cent of the

140

marks the test is out of.

0.85 \times 140 = 119

, the marks Fiona scored

Check 2 - count the marks that were dropped: Fiona missed

140 - 119 = 21

marks, and

\dfrac{21}{140} = \dfrac{3}{20}

, which is

15

per cent of the test.

100 - 15 = 85

, the same percentage

Check 3 - build the score from blocks of ten per cent: A tenth of

140

14

marks, so

80

per cent is

112

marks and

5

per cent is

7

marks.

112 + 7 = 119

marks, which is

80 + 5 = 85

per cent

Mark Scheme Breakdown

Step	Mark	Description	Got it?
eg $\dfrac{119}{140} \times 100$ or $0.85 \times 100$ or $\dfrac{17}{20} \times 100$ oe	M1	for a correct method to write the score as a percentage	✓
$85$	A1	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 2/2

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

[Total 2 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

The two numbers

45

and

70

Both are whole numbers, so each one can be written as a product of prime factors

Find

The lowest common multiple (LCM) of

45

and

70

- the smallest number that both of them divide into exactly

Plan the Solution

Write each number as a product of prime factors, using a factor tree or repeated division by $2$ , $3$ , $5$ , $7$ .
Build the LCM by multiplying together the highest power of every prime that appears in either list.
Check the result divides exactly by both numbers, and cross-check it against the product of the two numbers divided by their highest common factor.

Worked Solution [2 marks]

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 list.

Step 1: Write

45

as a product of prime factors

45 = 3 \times 3 \times 5 = 3^{2} \times 5

(Reason:

45

divides by

3

to give

15

, and

15 = 3 \times 5

. Both

3

and

5

are prime, so the tree stops there.)

Step 2: Write

70

as a product of prime factors

70 = 2 \times 35 = 2 \times 5 \times 7

(Reason:

70

is even, so take out

2

first, leaving

35

. Then

35 = 5 \times 7

, and both of those are prime as well.)

Step 3: Multiply the highest power of every prime

2^{1}, \quad 3^{2}, \quad 5^{1}, \quad 7^{1}

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

(Reason: The LCM has to contain

45 = 3^{2} \times 5

, so it needs two

3

s, and it has to contain

70 = 2 \times 5 \times 7

, so it needs a

2

and a

7

. One

5

is enough, because neither number uses more than one.)

630

Verification

Check 1: Divide the answer by each of the two numbers. A common multiple must give a whole number both times.

\dfrac{630}{45} = 14

and

\dfrac{630}{70} = 9

, both whole numbers

Check 2: Use the other standard route: the product of the two numbers divided by their highest common factor. The only prime

45

and

70

share is

5

, so the highest common factor is

5

\dfrac{45 \times 70}{5} = \dfrac{3150}{5} = 630

Check 3: Confirm nothing smaller works: list every multiple of

70

below

630

and test each one for divisibility by

45

70

140

210

280

350

420

490

560

- not one of them is a multiple of

45

, so

630

really is the lowest

Mark Scheme Breakdown

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

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 17 - Exam Solution

Understanding the Question

Given

A footbridge whose length is

142.8

m, correct to

1

decimal place.

That length has been rounded, so the true length is not exactly

142.8

Find

The lower bound of the length. The upper bound of the length.

Plan the Solution

Find the rounding unit. Correct to $1$ decimal place means rounded to the nearest $0.1$ m.
Halve that unit, because a rounded value can be at most half a unit away from the true length.
Take half a unit off for the lower bound, and put half a unit on for the upper bound.

Worked Solution [2 marks]

Rule - Bounds: a measurement written as

x

correct to the nearest

u

has lower bound

x - \dfrac{u}{2}

and upper bound

x + \dfrac{u}{2}

Step 1: Find half of one rounding unit

\text{rounding unit} = 0.1

\dfrac{0.1}{2} = 0.05

(Reason: Correct to

1

decimal place means rounded to the nearest

0.1

m, so the true length is less than

0.05

m away from the value written down.)

Step 2: Take half a unit off for the lower bound

142.8 - 0.05 = 142.75

(Reason: The lower bound is the smallest length that still rounds to the stated length, so half a rounding unit comes off.)

Step 3: Put half a unit on for the upper bound

142.8 + 0.05 = 142.85

(Reason: Any length below

142.85

m rounds down to

142.8

m, while

142.85

m itself rounds up to

142.9

m, so it is the value the length stops short of.)

Step 4: Write the error interval

142.75 \leq L < 142.85

(Reason: Writing

L

for the true length in metres, the lower bound is included because it does round to

142.8

m, and the upper bound is not.)

(i)

142.75

m(ii)

142.85

Verification

Check 1: Round the lower bound back to

1

decimal place. A halfway value rounds up, so it should come back to the stated length.

142.75

rounds to

142.8

, so the lower bound is a length the footbridge could actually have.

Check 2: The two bounds must be one whole rounding unit apart, with the stated length exactly halfway between them.

142.85 - 142.75 = 0.1

and

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

Check 3: Test a length just below the upper bound, and then the upper bound itself.

142.8499

rounds to

142.8

, while

142.85

rounds to

142.9

, which is why the upper bound is a value the length never reaches.

Mark Scheme Breakdown

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

Full marks: 2/2

Show that $2\dfrac{1}{4} \times 1\dfrac{5}{7} = 3\dfrac{6}{7}$
You must show all your working. [3 marks]

[Total 3 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

The product of two mixed numbers,

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

The value it is claimed to come to,

3\dfrac{6}{7}

Find

Working that shows the product really is

3\dfrac{6}{7}

. The answer is printed in the question, so every mark here is for the working, not for the number.

Plan the Solution

Write each mixed number as an improper fraction. Mixed numbers cannot be multiplied whole part by whole part and fraction part by fraction part, so this has to come first.
Multiply the numerators together and the denominators together. Cancelling a common factor before multiplying is allowed and keeps the numbers small - the mark scheme accepts either order.
Simplify the result, then turn it back into a mixed number so that it matches the form the question states.

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 finally write the result back as a mixed number.

Step 1: Write each mixed number as an improper fraction

2 \times 4 + 1 = 9

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

1 \times 7 + 5 = 12

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

(Reason: A mixed number is a whole number added to a fraction, so

2\dfrac{1}{4}

means

2 + \dfrac{1}{4}

. To collect that into a single fraction, multiply the whole number by the denominator and add the numerator, keeping the denominator unchanged. The same rule turns

1\dfrac{5}{7}

into sevenths. This first stage carries a mark of its own because the whole method depends on it.)

Step 2: Multiply the numerators, and multiply the denominators

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

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

(Reason: Two fractions are multiplied straight across: numerator times numerator over denominator times denominator. A common denominator is not needed here - that is only for adding and subtracting - so

9 \times 12 = 108

goes on top and

4 \times 7 = 28

underneath.)

Step 3: Simplify the fraction

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

(Reason: The highest common factor of

108

and

28

4

, so each is written as

4

times something and the shared factor is removed. Cancelling that

4

into the

12

before multiplying gives the same

\dfrac{27}{7}

with much smaller numbers, and the mark scheme accepts either order.)

Step 4: Write the improper fraction as a mixed number

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

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

(Reason: Seven goes into

27

three whole times, using up

21

and leaving

6

sevenths over. That is the form the question prints, so the working has now shown exactly what it was asked to show.)

2\dfrac{1}{4} \times 1\dfrac{5}{7} = \dfrac{9}{4} \times \dfrac{12}{7} = \dfrac{108}{28} = \dfrac{27}{7}

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

, as required

Verification

Check 1: Clear both denominators. Multiplying by

4 \times 7 = 28

must leave a whole number on each side, and the two whole numbers must agree.

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

and

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

Check 2: Run it backwards. Dividing the product by

1\dfrac{5}{7}

means multiplying by its reciprocal

\dfrac{7}{12}

, which must return the first factor.

\dfrac{27}{7} \times \dfrac{7}{12} = \dfrac{27}{12} = \dfrac{9}{4} = 2\dfrac{1}{4}

Check 3: Reach the same total without improper fractions at all, by expanding

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

as four separate products and putting them over

28

2 + \dfrac{10}{7} + \dfrac{1}{4} + \dfrac{5}{28} = \dfrac{56 + 40 + 7 + 5}{28} = \dfrac{108}{28}

Check 4: A size check.

1\dfrac{5}{7}

lies between

1

and

2

, so the product must lie between

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

and

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

2\dfrac{1}{4} < 3\dfrac{6}{7} < 4\dfrac{1}{2}

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 it	M1	correct cancelling or multiplication of numerators and denominators without cancelling, eg $\dfrac{9}{4} \times \dfrac{12}{7}$ with the $4$ cancelled to $1$ and the $12$ cancelled to $3$ , or $\dfrac{9}{4} \times \dfrac{12}{7} = \dfrac{108}{28}$ oe eg $\dfrac{63}{28} \times \dfrac{48}{28} = \dfrac{3024}{784}$	✓
Conclusion reached from correct working	A1	dep on M2, for conclusion to $3\dfrac{6}{7}$ from correct working - either sight of the result of the multiplication, eg $\dfrac{108}{28}$ oe must be seen, or correct cancelling prior to the multiplication to $\dfrac{27}{7}$	✓
Guidance printed with the scheme	Note	NB: use of decimals scores no marks unless as a check. Working required, and the answer column reads shown.	✓

Full marks: 3/3

Here is a biased $5$ -sided spinner.
When the spinner is spun, it can land on a star or on a moon or on a sun or on a leaf or on a bell.

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

$\begin{array}{|c|c|c|c|c|c|}\hline \textbf{Symbol} & \text{star} & \text{moon} & \text{sun} & \text{leaf} & \text{bell} \\ \hline \textbf{Probability} & 0.12 & 0.20 & 0.38 & 4x & x \\ \hline \end{array}$

Hannah spins the spinner once.

(a) Work out the probability that the spinner lands on a star or on a moon or on a sun. [1 mark]

Oliver spins the spinner $350$ times.

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

(a)(b)

[Total 5 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

The spinner is biased and has

5

sections: a star, a moon, a sun, a leaf and a bell.

P(\text{star}) = 0.12

P(\text{moon}) = 0.20

and

P(\text{sun}) = 0.38

P(\text{leaf}) = 4x

and

P(\text{bell}) = x

, so the last two sections are in the ratio

4 : 1

In part (b) the spinner is spun

350

times.

Find

(a) The probability that one spin lands on a star or on a moon or on a sun. (b) An estimate of how many of the

350

spins land on a leaf.

Plan the Solution

One spin cannot land on two sections at once, so for part (a) the three probabilities simply add.
Every spin lands on one of the five sections, so all five probabilities add to $1$ . Taking the part (a) total off $1$ leaves the probability of a leaf or a bell.
That leftover is $4x + x = 5x$ , so divide it by $5$ to get $x$ , then take four of those to get $4x$ .
An estimate of how often something happens is its probability multiplied by the number of spins.

Worked Solution [5 marks]

Rule - Mutually exclusive outcomes: their probabilities add, and the probabilities of all the possible outcomes add to

1

. An expected frequency is the probability multiplied by the number of trials.

Step 1: Add the three probabilities the table gives

0.12 + 0.20 + 0.38 = 0.7

(Reason: One spin lands on a single section, so a star, a moon and a sun cannot happen together. For outcomes like that the probabilities add, and this is the answer to part (a).)

Step 2: Find what is left for the leaf and the bell

4x + x = 1 - 0.7

5x = 0.3

(Reason: The five sections are all that can happen, so their probabilities add to

1

. Whatever is not a star, a moon or a sun must be a leaf or a bell.)

Step 3: Solve for

x

x = \dfrac{0.3}{5} = 0.06

(Reason: The leftover

0.3

is shared as

4x + x

, which is

5

equal parts, so divide by the number of parts.)

Step 4: Write down the probability of a leaf

P(\text{leaf}) = 4x = 4 \times 0.06 = 0.24

(Reason: The table gives the leaf

4x

, which is four of those equal parts, and the bell keeps the remaining one at

0.06

Step 5: Turn the probability into an estimate for

350

spins

0.24 \times 350 = 84

(Reason: The spinner is biased, but the probability still says what share of the spins to expect, so multiplying by

350

estimates how many of them land on a leaf.)

(a)

0.7

(b)

84

times

Verification

Check 1: Put the two found probabilities back into the table and add all five. They must come to

1

0.12 + 0.20 + 0.38 + 0.24 + 0.06 = 1

Check 2: Estimate the number of spins for every section and add them. They must come to

350

42 + 70 + 133 + 84 + 21 = 350

Check 3: Reach part (b) without

x

at all. A leaf or a bell has probability

0.3

, so about

0.3 \times 350 = 105

spins land on one of the two, and the leaf takes

4

of every

5

of them.

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

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) The probability of a star or a moon or a sun.	B1	$0.7$ 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) A correct first step.	M1ft	eg $1 - (0.12 + 0.2 + 0.38) = 0.3$ oe or $1 - 0.7 = 0.3$ oe or $0.12 + 0.20 + 0.38 + 4x + x = 1$ oe or $0.7 \times 350 = 245$ 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) A correct second step.	M1	eg $\dfrac{0.3}{5} = 0.06$ or $\dfrac{0.3}{5} \times 4 = 0.24$ or $0.24$ or $x = 0.06$ or $4x = 0.24$ or $0.3 \times 350 = 105$ oe or $350 - 245 = 105$ oe or $350 - 42 - 0.20 \times 350 - 133 = 105$ oe, in each case using their $0.3$ , their $245$ , their $42$ and their $133$ .	✓
(b) A correct third step.	M1	eg $0.06 \times 350 = 21$ oe or $\dfrac{105}{5} = 21$ oe or $0.06 \times 4 \times 350$ oe or $0.24 \times 350$ , in each case using their $0.06$ , their $105$ or their $0.24$ . Or for $\dfrac{21}{350}$ or $\dfrac{84}{350}$ .	✓
(b) The estimate.	A1	$84$ cao. A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 5/5

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

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

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

(b) Complete each statement below by writing one symbol from the box on the dotted line, so that the statement is true.
(i) $A \cap C$ = ...............
(ii) $13$ ............... $\mathcal{E}$ [2 marks]

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

[Total 4 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

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

- the universal set, the whole numbers from

1

12

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

- the even members of

\mathcal{E}

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

- the multiples of

3

\mathcal{E}

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

- the odd members of

\mathcal{E}

Find

(a)(i) every member of

A \cup B

(a)(ii) every member of

B'

(b)(i) the symbol from the box that makes the statement about

A \cap C

true (b)(ii) the symbol from the box that correctly links

13

and

\mathcal{E}

Plan the Solution

Read $\cup$ as "in A, or in B, or in both", and write each member once however many sets it belongs to.
Read the dash in $B'$ as "not in B", then work through $\mathcal{E}$ keeping everything B leaves out.
For (b)(i), compare A and C member by member and see what, if anything, they share.
For (b)(ii), look for $13$ among the members listed in $\mathcal{E}$ .

Worked Solution [4 marks]

Rule - Union, intersection and complement:

A \cup B

holds every member of A together with every member of B;

A \cap C

holds only the members A and C share; and

B'

holds every member of

\mathcal{E}

that is not in B.

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: The union collects everything that is in A or in B.

6

and

12

sit in both sets, but a member is written once however many sets it belongs to, so the list holds

8

members rather than

10

. Writing them in order makes a missed member easy to spot.)

Step 2: list

B'

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

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

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

(Reason: The dash means "not in B", so start from

\mathcal{E}

and cross out each of the four members B contains. Everything still standing belongs to

B'

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 be in both. A set with no members at all is the empty set, written

\varnothing

, which is the symbol wanted from the box.)

Step 4: decide where

13

sits

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

13 \notin \mathcal{E}

(Reason: The universal set stops at

12

, so

13

is not one of its members.

\notin

is read as "is not a member of", so it is the symbol that makes the statement true.)

(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 \mathcal{E}

Verification

Check 1: Count the union a second way instead of reading the list again: A has

6

members and B has

4

, and the two members they share,

6

and

12

, have been counted twice.

6 + 4 - 2 = 8

, and the answer to (a)(i) holds

8

members.

Check 2: A set and its complement must share nothing and, between them, account for every member of

\mathcal{E}

exactly once. Count B against

B'

4 + 8 = 12

, the number of members of

\mathcal{E}

, and no number appears in both lists.

Check 3: Test the two symbols against the members themselves: hunt for a single number that is in both A and C, then hunt for

13

in the list for

\mathcal{E}

. Neither hunt finds anything, so

A \cap C = \varnothing

and

13 \notin \mathcal{E}

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)	B1	$\varnothing$	✓
(b)(ii)	B1	$\notin$	✓

Full marks: 4/4

(a) The number line above shows an inequality.
Write down this inequality. [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 21 - Exam Solution

Understanding the Question

Given

(a) A number line marked from

-3

3

, carrying an unfilled circle at

-2

, a filled circle at

1

, and a solid line joining them.

(b) The inequality

7a - 5 \leq 3a + 28

, with the unknown on both sides.

Find

(a) The inequality that the number line shows, written in symbols. (b) Every value of

a

for which the inequality is true.

Plan the Solution

(a) Read the two ends separately. An unfilled circle leaves its own value out; a filled circle keeps it in.
(a) Write the two ends either side of $x$ so that one statement carries both.
(b) Gather the $a$ terms on one side of the sign and the plain numbers on the other.
(b) Divide by the number in front of $a$ . It is positive, so the sign still points the same way.

Worked Solution [4 marks]

Rule - Reading a number line: an unfilled circle means that value is not included, so the sign is

<

>

; a filled circle means it is included, so the sign is

\leq

\geq

. Rule - Solving a linear inequality: add or subtract the same amount on both sides, then divide by the coefficient of the unknown. Dividing by a positive number leaves the sign pointing the same way; dividing by a negative number turns it round.

(a) Step 1: read the left-hand end

-2 < x

(Reason: The circle above

-2

is not filled in, so

-2

itself is not one of the values. Everything the line covers is greater than

-2

(a) Step 2: read the right-hand end

x \leq 1

(Reason: The circle above

1

is filled in, so

1

is one of the values. Everything the line covers is less than or equal to

1

(a) Step 3: put the two ends into one inequality

-2 < x \leq 1

(Reason: The line runs from one end to the other with nothing missing in between, so

x

lies between the two values. Writing them either side of

x

keeps both ends, and both signs, in one statement.)

(b) Step 4: collect the

a

terms on one side and the numbers on the other

7a - 5 \leq 3a + 28

7a - 3a \leq 28 + 5

4a \leq 33

(Reason: Subtract

3a

from both sides and add

5

to both sides. Adding or subtracting the same amount on both sides never changes the direction of the sign, so the

\leq

is untouched.)

(b) Step 5: divide both sides by

4

a \leq \dfrac{33}{4}

\dfrac{33}{4} = 8.25

(Reason:

4

is positive, so dividing by it leaves the sign pointing the same way. The answer line wants a value, and

\dfrac{33}{4}

written as a decimal is

8.25

(a)

-2 < x \leq 1

(b)

a \leq 8.25

Verification

Check 1 - part (a), the two ends: Test each end against its own circle. The unfilled circle at

-2

must be left out by the inequality, and the filled circle at

1

must be kept in.

-2 < -2

is not true, so

-2

is left out, and

1 \leq 1

is true, so

1

is kept in. Both agree with the circles.

Check 2 - part (a), a value on the line and a value off it: Take

0

, which the drawn line passes over, and

2

, which it does not reach.

-2 < 0 \leq 1

is true, and

2

is greater than

1

so it fails the right-hand end. The inequality covers exactly what the picture shades.

Check 3 - part (b), a value inside and a value outside: Put

8

in place of

a

in the original inequality, then

9

. One is below

8.25

and one is above it, so the first should work and the second should not.

7 \times 8 - 5 = 51

and

3 \times 8 + 28 = 52

, and

51 \leq 52

, so

8

works. But

7 \times 9 - 5 = 58

and

3 \times 9 + 28 = 55

, and

58

is greater than

55

, so

9

does not.

Check 4 - part (b), the boundary itself: At the boundary the two sides should come to the same value, since that is where

\leq

changes from true to false. Put

8.25

in place of

a

7 \times 8.25 - 5 = 52.75

and

3 \times 8.25 + 28 = 52.75

, so the two sides meet exactly at

8.25

, which is why it is included in the answer.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $-2 < x \leq 1$	B2	accept $1 \geq x > -2$ or $x > -2$ , $x \leq 1$ if not B2 then B1 for $-2 < x$ or $x \leq 1$ or $-2 \leq x < 1$ or $-2 \leq x \leq 1$ or $-2 < x < 1$ Condone use of a variable other than $x$ but not $0$	✓
(b) $7a - 3a \leq 28 + 5$ 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 an equals sign rather than $\leq$ , or any other sign, for this mark.	✓
(b) Working required. $a \leq 8.25$	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 (sight of correct answer in working space and just $8.25$ on answer line gains M1 only)	✓

Full marks: 4/4

A coach travels a distance of $1262$ km from Lahore to Karachi.
The coach takes $17\dfrac{1}{2}$ hours.

(a) Work out the average speed of the coach.
Give your answer, in km/h, correct to the nearest whole number. [2 marks]

(b) Change a speed of $50x$ metres per second to a speed in kilometres per hour. [3 marks]

(a) km/h(b) km/h

[Total 5 marks]

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Question 22 - Exam Solution

Understanding the Question

Given

(a) A journey of

1262

km that takes

17\dfrac{1}{2}

hours, so a distance in kilometres and a time in hours.

(b) A speed of

50x

metres per second, with the letter

x

standing for a number that is never given.

Find

(a) The average speed in kilometres per hour, rounded to the nearest whole number. (b) The same speed measured in kilometres per hour, still written in terms of

x

Plan the Solution

(a) Turn the mixed number of hours into a decimal, so the division can be typed straight into the calculator.
(a) Divide the distance by the time. Both are already in the units the answer asks for, so no conversion is needed afterwards.
(b) Change the two units one at a time: seconds into hours first, then metres into kilometres.
(b) Finish by noting that the two steps together are one multiplication by $3.6$ , which is worth remembering for any change from metres per second to kilometres per hour.

Worked Solution [5 marks]

Rule - Average speed:

\text{average speed} = \dfrac{\text{distance}}{\text{time}}

, with the distance and the time measured in the units the answer asks for. Rule - Changing the units of a speed: change the distance unit and the time unit separately. Multiplying by

3600

turns a rate per second into a rate per hour, and dividing by

1000

turns metres into kilometres, so metres per second become kilometres per hour on multiplying by

\dfrac{3600}{1000} = 3.6

(a) Step 1: write the time as a decimal number of hours

17\dfrac{1}{2} = 17 + 0.5 = 17.5

(Reason: Half an hour is

0.5

of an hour, so the mixed number becomes

17.5

. The calculator needs a single decimal number of hours before the division can be done.)

(a) Step 2: divide the distance by the time

\text{average speed} = \dfrac{\text{distance in km}}{\text{time in hours}}

\dfrac{1262}{17.5} \approx 72.11

(Reason: Speed is distance divided by time. The distance is already in kilometres and the time is now in hours, so the division gives a speed in kilometres per hour with nothing left to convert.)

(a) Step 3: round to the nearest whole number

\text{average speed} \approx 72 \text{ km/h}

(Reason: The first digit after the decimal point is

1

, which is below

5

, so the whole-number part is left as it stands. The working was already in km/h, which is the unit the answer line asks for.)

(b) Step 4: change the seconds into hours

1 \text{ hour} = 60 \times 60 = 3600 \text{ seconds}

50x \times 3600 = 180\,000x

(Reason: A speed of

50x

metres every second covers

3600

times as far in an hour, because an hour is

3600

seconds. So the coach covers

180\,000x

metres in one hour.)

(b) Step 5: change the metres into kilometres

1 \text{ km} = 1000 \text{ m}

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

(Reason: There are

1000

metres in a kilometre, so dividing the number of metres by

1000

gives the number of kilometres. The

x

is only a multiplier and rides through both steps untouched.)

(b) Step 6: do the same thing in one multiplication

\dfrac{3600}{1000} = 3.6

50x \times 3.6 = 180x

(Reason: Multiplying by

3600

and then dividing by

1000

is one multiplication by

3.6

, and it gives the same

180x

. Either route earns the marks; this one is quicker once the factor is known.)

(a)

72

km/h(b)

180x

km/h

Verification

Check 1 - part (a), put the rounded speed back into the journey: Multiply the answer by the time and see how close the distance comes back. Rounding

72.11

down to

72

should leave the journey slightly short, not long.

72 \times 17.5 = 1260

, which is

2

km short of

1262

. A shortfall that small over

17\dfrac{1}{2}

hours is exactly what rounding down by

0.11

km/h produces.

Check 2 - part (a), work the time back out instead: Turn the division round: divide the distance by the answer and see whether the time given comes back.

\dfrac{1262}{72} \approx 17.53

hours, against the

17.5

hours the question gives. The gap is a few tenths of a minute and comes only from the rounding.

Check 3 - part (b), test it on one value: Put

x = 1

, which makes the speed a plain

50

metres per second, and work out the distance covered in one hour from scratch.

50 \times 3600 = 180\,000

metres in an hour, and

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

kilometres, so the speed is

180

km/h. That is

180x

with

x = 1

Check 4 - part (b), reverse the conversion: Take the answer back the other way. Turning

180x

km/h into metres per second must return the speed the question started with.

180x \times 1000 = 180\,000x

metres in an hour, and

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

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

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $\dfrac{1262}{17.5}$ oe	M1	$\dfrac{1262}{\text{their time}}$ their time may be an incorrect conversion to a decimal time eg $17.3$ or from an attempt at converting to minutes eg $1050$	✓
(a) $72$	A1	accept $72.1$ or $72.11\ldots$	✓
(b) $\dfrac{50x}{1000}\ (= 0.05x)$ 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	✓
(b) $\dfrac{50x \times 60 \times 60}{1000}$ 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 $180$	✓
(b) $180x$	A1	Correct answer scores full marks (unless from obvious incorrect working)	✓

Full marks: 5/5

(a) Multiply out the brackets in $x(x - 3)$ [1 mark]

(b) Rearrange the formula $m = \dfrac{t + 4}{5}$ to make $t$ the subject. [2 marks]

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

(e) (i) Factorise $y^{2} - 10y + 21$ [2 marks]

(ii) Hence solve $y^{2} - 10y + 21 = 0$ [1 mark]

(a)(b)(c)(d)(e)(i)(e)(ii)

[Total 8 marks]

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Question 23 - Exam Solution

Understanding the Question

Given

A bracket to expand:

x(x - 3)

A formula:

m = \dfrac{t + 4}{5}

Two powers to simplify:

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

and

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

A quadratic expression:

y^{2} - 10y + 21

Find

(a) the bracket multiplied out (b)

t

written in terms of

m

a

and of

c

(e)(i) the two brackets, and (ii) the values of

y

that make the expression zero

Plan the Solution

(a) Multiply each term inside the bracket by the $x$ outside it.
(b) Clear the fraction first by multiplying both sides by $5$ , then subtract $4$ from both sides.
(c) and (d) Use the index laws: powers of the same letter are multiplied by adding the indices and divided by subtracting them.
(e) Look for two numbers with product $21$ and sum $-10$ . Part (ii) says "hence", so read the solutions straight off the brackets rather than starting again.

Worked Solution [8 marks]

Rule - the four tools this question needs: expand by multiplying every term inside the bracket,

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: (a) multiply the bracket out

x(x - 3) = x \times x - x \times 3

= x^{2} - 3x

(Reason: Both terms inside the bracket are multiplied by the

x

outside, and

x \times x

x^{2}

, not

2x

Step 2: (b) clear the fraction

m = \dfrac{t + 4}{5}

5m = t + 4

(Reason: The whole of

t + 4

is being divided by

5

, so multiplying both sides by

5

undoes it in one move and leaves

t

on top.)

Step 3: (b) make

t

the subject

5m - 4 = t

t = 5m - 4

(Reason: Taking

4

from both sides leaves

t

on its own. Writing it with

t

on the left is what "make

t

the subject" asks for.)

Step 4: (c) multiply the powers of

a

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

(Reason: The letter stays the same and the two indices are combined into a single power in one step.)

Step 5: (d) divide the powers of

c

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

(Reason: Twelve of the

c

's on the top cancel with the twelve on the bottom, so the indices are subtracted.)

Step 6: (e)(i) find the pair of numbers, then factorise

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

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

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

(Reason: The constant

21

is positive and the

y

term is negative, so both numbers must be negative.

-3

and

-7

are the only pair with product

21

and sum

-10

Step 7: (e)(ii) solve using the brackets

(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 one of them is zero, so each bracket is set to zero in turn. This is why part (i) was asked first.)

(a)

x^{2} - 3x

(b)

t = 5m - 4

(c)

a^{16}

(d)

c^{18}

(e)(i)

(y - 3)(y - 7)

(e)(ii)

y = 3

y = 7

Verification

Check 1 - part (a): Put

x = 5

into both forms. The bracket gives

5 \times 2

and the expanded expression gives

25 - 15

10 = 10

Check 2 - part (b): Put the rearranged

t

back into the original formula. The

-4

and the

+4

cancel, leaving

5m

over

5

\dfrac{(5m - 4) + 4}{5} = m

Check 3 - parts (c) and (d): Count the letters instead of using the laws.

a^{6}

is six

a

's and

a^{10}

is ten more, which is sixteen

a

's altogether. For (d), reverse it: multiplying the answer by

c^{12}

must return

c^{30}

a^{16}

and

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

Check 4 - part (e): Multiply the brackets back out:

y^{2} - 7y - 3y + 21

. Then substitute each solution into the original expression:

9 - 30 + 21

and

49 - 70 + 21

y^{2} - 10y + 21

, and both solutions give

0

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a)	B1	$x^{2} - 3x$	✓
(b)	M1	for a correct first step, eg $5m = t + 4$ or $m = \dfrac{t}{5} + \dfrac{4}{5}$	✓
(b)	A1	$t = 5m - 4$ , oe eg $t = 5\left(m - \dfrac{4}{5}\right)$ or $t = -4 + 5m$ . $5m - 4$ only on the answer line scores M1 unless $t = 5m - 4$ is seen in the working, then score M1A1. Correct answer scores full marks (unless from obvious incorrect working).	✓
(c)	B1	$a^{16}$	✓
(d)	B1	$c^{18}$	✓
(e)(i)	M1	for $(y \pm 3)(y \pm 7)$ or for $(y \pm a)(y \pm b)$ with $ab = 21$ or $a + b = -10$	✓
(e)(i)	A1	for correct factors $(y - 3)(y - 7)$ . Correct answer scores full marks (unless from obvious incorrect working).	✓
(e)(ii)	B1	$3$ , $7$ ft dep on factorising in the form $(y \pm p)(y \pm q)$	✓

Full marks: 8/8

Triangle $PQR$ is shown in the diagram.
The right angle is at $Q$ .

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

[Total 3 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

Triangle

PQR

has its right angle at

Q

The diagram labels the vertical side:

PQ = 6.5

cm.

The angle marked at

R

24^\circ

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

Find

The length of

QR

, correct to

3

significant figures.

Plan the Solution

Stand at the $24^\circ$ angle and name the two sides the question involves. $PQ$ is across the triangle from it, so it is the opposite side; $QR$ runs from it to the right angle, so it is the adjacent side.
Opposite with adjacent is the tangent ratio, so the hypotenuse $PR$ is never needed and Pythagoras is not the tool here.
The unknown lands underneath the fraction, so rearrange first and press the calculator once.
Keep the whole calculator display, and round to $3$ significant figures only at the very end.

Worked Solution [3 marks]

Rule - Tangent: in a right-angled triangle

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

, where the two sides are named from the angle

\theta

itself. Set the calculator to degrees before pressing anything.

Step 1: Choose the ratio that uses the two sides in the question

\tan R = \dfrac{PQ}{QR}

(Reason: Seen from

R

, the side

PQ

is opposite and the side

QR

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

PR

does not appear in it.)

Step 2: Put in the angle and the length the diagram gives

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

(Reason: The angle at

R

24^\circ

, and the side opposite it is

PQ

, which is

6.5

cm. Everything in the equation is now a number except

QR

Step 3: Rearrange to make

QR

the subject

QR \times \tan 24^\circ = 6.5

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

(Reason: The unknown is underneath, so multiply both sides by

QR

to lift it up, then divide both sides by

\tan 24^\circ

to leave it alone.)

Step 4: Work it out, with the calculator in degrees

QR = 14.5992\ldots

(Reason: Type the division in one go and keep the whole display. Rounding here and then carrying the rounded number on is how an answer drifts outside the range the mark scheme accepts.)

Step 5: Round to

3

significant figures

QR = 14.6 \text{ cm}

(Reason: The first three significant figures are

1

4

and

5

. The digit after them is

9

, so the

5

rounds up to

6

. The question asks for a length, so the unit goes on the answer line.)

QR = 14.6

Verification

Check 1: Put the answer back where it came from. Multiplying

14.5992\ldots

\tan 24^\circ

has to give back the

6.5

cm the diagram prints.

14.5992\ldots \times \tan 24^\circ = 6.5

Check 2: Come at it from the other acute angle. The angles of a triangle add to

180^\circ

, so angle

QPR

180 - 90 - 24 = 66

degrees, and from

P

it is

QR

that is opposite and

6.5

cm that is adjacent.

QR = 6.5 \times \tan 66^\circ = 14.5992\ldots

Check 3: Take a route with no tangent in it at all: the sine ratio gives the hypotenuse

PR

, and Pythagoras then gives

QR

. The answer must also come out longer than

6.5

cm, because

24^\circ

is less than

45^\circ

PR = \dfrac{6.5}{\sin 24^\circ} = 15.9808\ldots

QR = \sqrt{PR^2 - 6.5^2} = 14.5992\ldots

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 $6.5^2 + QR^2 = 15.9^2$	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 $QR$ . The printed scheme puts $15.9$ in quotation marks, which means the candidate's own value for $PR$ may be used there.	✓
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 $(QR =) \sqrt{15.9^2 - 6.5^2}$	M1	for a complete method	✓
$14.6$ Correct answer scores full marks (unless from obvious incorrect working)	A1	accept $14.5$ to $14.61$	✓

Full marks: 3/3

The diagram shows two rainwater tanks at a plant nursery.
One tank is a cuboid and the other tank is a cylinder.

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

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

Rowena is going to pour all the water from the cylindrical tank into the cuboid tank.

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

[Total 3 marks]

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Question 25 - Exam Solution

Understanding the Question

Given

Cuboid tank:

35

cm by

28

cm by

20

Water already standing in the cuboid tank:

9

cm deep

Cylindrical tank: radius

10

cm, height

33

cm, completely full of water

Find

Show that pouring all the water from the cylindrical tank into the cuboid tank does not fill the cuboid tank. A show-that question, so the working is the answer: two volumes and a comparison between them.

Plan the Solution

Find the depth of empty space above the water in the cuboid tank, then the volume of that space.
Find the volume of water in the cylindrical tank.
Compare the two. If the water poured in is less than the space waiting for it, the tank cannot end up full.

Worked Solution [3 marks]

Volume of a cuboid: multiply the three dimensions together. Volume of a cylinder:

\pi r^{2} h

Step 1: the depth of empty space above the water

20 - 9 = 11

(Reason: The tank is

20

cm deep and the water reaches

9

cm up it, so

11

cm of depth is still empty.)

Step 2: the volume of that empty space

11 \times 35 \times 28 = 10\,780

(Reason: The empty part of the tank is a cuboid in its own right: the same

35

cm by

28

cm base, with the empty depth from Step 1 as its height.)

Step 3: the volume of water in the cylindrical tank

10^{2} \times 33 = 3300

\pi \times 3300 = 10\,367.3

(Reason: Volume of a cylinder is

\pi r^{2} h

, with

r = 10

and

h = 33

. Holding the

\pi

back until the last line keeps the rounding out of the working; to one decimal place the volume is

10\,367.3

cm³.)

Step 4: compare the water with the space

10\,367.3 < 10\,780

10\,780 - 10\,367.3 = 412.7

(Reason: There is less water in the cylinder than there is space waiting for it, so all of it fits and

412.7

cm³ of the cuboid tank is still empty afterwards.)

10\,367.3

cm³ of water is poured into

10\,780

cm³ of empty space, leaving about

412.7

cm³ unfilled, so the cuboid tank will not be completely full of water.

Verification

Check 1: Work with totals rather than with the gap. The water already in the cuboid tank is

35 \times 28 \times 9 = 8\,820

cm³, and the tank holds

35 \times 28 \times 20 = 19\,600

cm³ altogether. After the pour there is

8\,820 + 10\,367.3 = 19\,187.3

cm³ of water, and

19\,600 - 19\,187.3 = 412.7

cm³ of the tank is still empty. The same shortfall, from the other end.

Check 2: A bound, with no rounding in it at all. Since

\pi

is smaller than

3.15

, the cylinder must hold less than

3300 \times 3.15 = 10\,395

cm³. Under

10\,395

cm³ of water going into

10\,780

cm³ of space, so the tank cannot fill however the rounding is done.

Check 3: The empty depth is

11

of the tank's

20

cm, so the empty space should be

\dfrac{11}{20}

of the whole tank:

\dfrac{11}{20} \times 19\,600 = 10\,780

cm³. The same

10\,780

cm³ as Step 2, reached from the tank's capacity instead of from its base area.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(volume of water =) $9 \times 35 \times 28 \; (= 8\,820)$ 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	(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 =) " $8\,820$ " $+$ " $10\,367(.25\ldots)$ " $(= 19\,187(.25\ldots))$ or (difference between the volumes of both solids =) " $19\,600$ " $-$ " $10\,367(.25\ldots)$ " $(= 9\,232(.74\ldots))$ or (volume not filled =) " $19\,600$ " $-$ " $8\,820$ " $-$ " $10\,367(.25\ldots)$ " $(= 412(.74\ldots))$ Shown	A1	correct workings with accurate figures, eg $10\,780$ with $10\,367(.25\ldots)$ (accept $10\,362$ to $10\,372$ ), or $19\,600$ with $19\,187(.25\ldots)$ (accept $19\,182$ to $19\,192$ ), or $8\,820$ with $9\,232(.74\ldots)$ (accept $9\,228$ to $9\,238$ ), or $412(.74\ldots)$ or $413$ (accept $408$ to $418$ ) with no second value needed. The values in quotation marks in the step column may be the candidate's own earlier values.	✓
Working required	Note	A show-that question: the conclusion on its own earns nothing. Both volumes and the comparison between them have to be on the page. The printed scheme puts Shown in the answer column and Working required beneath the working.	✓

Full marks: 3/3

Wei puts money into two savings plans. Each plan runs for $2$ years.
He puts $\$2500$ into the Harbour plan and $\$3000$ into the Meridian plan.

$\begin{array}{|c|}\hline \textbf{Harbour plan} \\ \text{Invests } \$2500 \\ \text{amount invested} : \text{interest after 2 years} = 20 : 3 \\ \hline\end{array}$

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

Wei receives more interest from the Harbour plan than from the Meridian plan.

How much more? [5 marks]

[Total 5 marks]

Show solution & mark schemeHide solution & mark scheme

Question 26 - Exam Solution

Understanding the Question

Given

Harbour plan:

\$2500

invested, with amount invested : interest after

2

years in the ratio

20 : 3

Meridian plan:

\$3000

invested at

4\%

per year compound interest, for

2

years

Find

The interest the Harbour plan pays The interest the Meridian plan pays How much more the Harbour plan pays than the Meridian plan

Plan the Solution

The Harbour ratio compares the amount invested with the interest, so the $\$2500$ invested is $20$ equal shares. Divide by $20$ for one share, then take $3$ shares for the interest.
Compound interest is not the same amount every year: each year's $4\%$ is worked out on the new total, so multiply by $1.04$ once for each year.
That gives the Meridian TOTAL, not its interest, so subtract the $\$3000$ invested.
Only then subtract: interest against interest, never a total against an interest.

Worked Solution [5 marks]

Rule - Ratio share: one share is the amount divided by the number of shares in it. Rule - Compound growth: multiply by

1 + r

once for each year, where

r

is the yearly rate written as a decimal, then subtract the amount invested to leave the interest.

Step 1: The interest the Harbour plan pays

\dfrac{2500}{20} = 125

3 \times 125 = 375

(Reason: the ratio

20 : 3

compares the amount invested with the interest, so the

\$2500

invested is

20

equal shares. One share is worth

\$125

, and the interest is

3

of those shares)

Step 2: The Meridian total after both years

1 + \dfrac{4}{100} = 1.04

3000 \times 1.04^2 = 3244.8

(Reason: adding

4\%

means keeping the

100\%

already there and adding

4

parts more, which is one multiplication by

1.04

. The second year's

4\%

is charged on the new total, not on the original

\$3000

, so the multiplier is used twice)

Step 3: The interest the Meridian plan pays

3244.8 - 3000 = 244.8

(Reason: the

\$3244.80

is everything in the plan, so it still contains the

\$3000

Wei put in. Taking that out leaves the interest on its own)

Step 4: How much more the Harbour plan pays

375 - 244.8 = 130.2

(Reason: both figures are interest, so taking one from the other compares like with like; subtracting a total from an interest would compare two different things)

\$130.20

Verification

Check 1 - rebuild the Harbour ratio: Divide both parts of

2500 : 375

125

and see whether the ratio really is

20 : 3

\dfrac{2500}{125} = 20

and

\dfrac{375}{125} = 3

, so the ratio is

20 : 3

Check 2 - the Meridian plan one year at a time: Instead of squaring, add the interest twice over:

3000 \times 1.04 = 3120

at the end of the first year.

3120 \times 1.04 = 3244.8

, the same total, so the interest is

3244.8 - 3000 = 244.8

Check 3 - put the difference back: Add the smaller interest on to the difference. It must give the larger interest.

244.8 + 130.2 = 375

, the Harbour interest

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Interest for the Harbour plan	M1	for a method to find the interest for the Harbour plan: $\dfrac{2500}{20} \times 3 = 375$ oe, or $125 \times 3 = 375$ , or $\dfrac{7500}{20} = 375$ oe; or $\dfrac{3000}{20} \times 3 = 450$ oe, or $150 \times 3 = 450$ , or $\dfrac{9000}{20} = 450$ oe. An answer of $2875$ or $3450$ implies this method mark.	✓
One year's growth on the Meridian plan	M1	for finding $4\%$ or $104\%$ of $3000$ or of $2500$ : $0.04 \times 3000 = 120$ oe, or $0.04 \times 2500 = 100$ oe, or $1.04 \times 3000 = 3120$ oe, or $1.04 \times 2500 = 2600$ oe.	✓
The Meridian total in one step	M2	as an alternative to that method mark and the next one, for $3000 \times 1.04^2 = 3244.8$ or $2500 \times 1.04^2 = 2704$ .	✓
The Meridian total after both years	M1	for completing the method to find the total amount for the Meridian plan: $1.04 \times 3120 = 3244.8$ oe, or $1.04 \times 2600 = 2704$ oe, where the candidate's own value from the previous mark may stand in place of $3120$ or $2600$ ; or $3000 \times 1.04^3 = 3374.59$ or $2500 \times 1.04^3 = 2812.16$ .	✓
Interest for the Meridian plan	M1	for a complete method to find the interest for the Meridian plan, for example $3244.8 - 3000 = 244.8$ or $2704 - 2500 = 204$ , where the candidate's own total may stand in place of $3244.8$ or $2704$ .	✓
The answer	A1	for $130.2$ or $130.20$ . A correct answer scores full marks unless it comes from obviously incorrect working.	✓
Special case	SC	if none of the second or third method marks is gained, award M1 for $0.08 \times 3000 = 240$ oe, or $1.08 \times 3000 = 3240$ , or $0.08 \times 2500 = 200$ oe, or $1.08 \times 2500 = 2700$ , or $3000 \times (1 - 0.04)^2 = 2764.8$ , or $2500 \times (1 - 0.04)^2 = 2304$ . Any of $240$ , $3240$ , $200$ or $2700$ seen on its own is enough. This is the student who treats the yearly interest as the same amount every year, so the second year's interest is taken on the original amount again.	✓
Note	Note	accept $(1 + 0.04)$ or $\left(1 + \dfrac{4}{100}\right)$ as equivalent to $1.04$ throughout.	✓

Full marks: 5/5

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 4MA1/1FR, Thursday 15 May 2025: Worked Solutions, Questions 15 to 26

Worked solutions, questions 15 to 26 of 26

Question 15, Calculator allowed

Question 15 - Exam Solution

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Question 20 - Exam Solution

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Question 25 - Exam Solution

Question 26, Calculator allowed

Question 26 - Exam Solution

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