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Edexcel IGCSE 4MA1/1HR, Thursday 16 May 2024: Worked Solutions and Mark Schemes

Q: How many questions are on Edexcel 4MA1/1HR, sat Thursday 16 May 2024, and how long is the exam?

There are 23 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

10 Aug 2026

Table of Contents▾

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

Original worked solutions for Edexcel International GCSE Mathematics A, Paper 4MA1/1HR (Higher Tier), June 2024 series, sat Thursday 16 May 2024 –100 marks, 2 hours, calculator allowed. The questions have been reworded; all numerical values match the original paper. The official question paper and mark scheme are published by Pearson Edexcel. This resource reproduces neither the exam paper nor the official mark scheme.

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

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

Download printable PDF

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

Worked solutions, questions 1 to 13 of 23

Here are six number tiles from a puzzle set.
Five of the tiles already have a number painted on them. The sixth tile is still blank.

Work out the number that must be painted on the blank tile so that the six numbers have a mean of $11$ [3 marks]

[Total 3 marks]

Show solution & mark schemeHide solution & mark scheme

Question 1 - Exam Solution

Understanding the Question

Given

The five painted numbers are

16

15

3

2

and

9

There are

6

tiles altogether, so six numbers go into the mean.

The mean of those six numbers must come out at

11

Find

The number painted on the blank tile.

Plan the Solution

A mean is a total shared out equally, so work backwards: the mean tells you what the total has to be.
Multiply the mean by $6$ to get the total the six numbers must have.
Add the five painted numbers, then take that sum away from the total. What is left is the missing number.

Worked Solution [3 marks]

Rule - Mean: the mean is the total of the numbers divided by how many numbers there are, so the total is

\text{mean} \times \text{how many}

Step 1: work out the total the six numbers must have

11 \times 6 = 66

(Reason: A mean of

11

means the six numbers share out to

11

each, so their total is the mean multiplied by how many tiles there are. Every tile counts, blank or not.)

Step 2: add the five numbers that are already painted

16 + 15 + 3 + 2 + 9 = 45

(Reason: These five cannot change, so this much of the total is already accounted for.)

Step 3: subtract to find the missing number

66 - 45 = 21

(Reason: If the blank tile carries

x

, then

45 + x = 66

, so

x

is what is left when

45

is taken from

66

21

Verification

Check 1: Paint

21

on the blank tile and work the mean out forwards, the usual way round.

\dfrac{16 + 15 + 3 + 2 + 9 + 21}{6} = 11

Check 2: Measure each painted number from the mean of

11

instead. The five sit

+5

+4

-8

-9

and

-2

away from it, and for the mean to be right the amounts above and below must cancel out.

5 + 4 - 8 - 9 - 2 = -10

so the sixth number has to sit

10

above the mean, giving

11 + 10 = 21

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The total six numbers need for a mean of $11$	M1	A correct calculation for the total, $11 \times 6 = 66$ , or a correct equation for the last tile using $x$ .	✓
The five painted numbers totalled, then subtracted	M1	A correct equation for $x$ with no fraction in it, or a correct calculation for the number on the last tile, $66 - 45$ , where $16 + 15 + 3 + 2 + 9 = 45$ .	✓
The number on the last tile	A1	cao $21$ . A correct answer scores full marks unless it clearly comes from incorrect working, and if the answer line is left blank, check the tile itself.	✓

Full marks: 3/3

The five-sided spinner shown below is biased.

The table gives the probability that the spinner lands on each number when it is spun once.

$\begin{array}{|c|c|c|c|c|c|}\hline \textbf{Number} & 1 & 2 & 3 & 4 & 5 \\ \hline \textbf{Probability} & 2x & 0.27 & 0.04 & x & 0.12 \\ \hline \end{array}$

Bethany is going to spin the spinner $400$ times.

Work out an estimate for the number of times the spinner will land on an odd number. [4 marks]

[Total 4 marks]

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Question 2 - Exam Solution

Understanding the Question

Given

The spinner is biased and can land on

1

2

3

4

5

Reading the table in order, the probabilities are

2x

0.27

0.04

x

and

0.12

The spinner is spun

400

times.

Find

An estimate for the number of spins that land on an odd number, that is on

1

3

5

Plan the Solution

The spinner must land on something, so the five probabilities total $1$ . That one equation is enough to find $x$ .
Watch the two rows that hold the unknown: $1$ carries $2x$ and $4$ carries $x$ , so there are three lots of $x$ altogether, not two.
Add the probabilities of the three odd numbers, then multiply by $400$ to turn that probability into an expected number of spins.

Worked Solution [4 marks]

Rule - the probabilities of all the possible outcomes add up to

1

, and an estimate for a frequency is

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

Step 1: write down the total probability

2x + 0.27 + 0.04 + x + 0.12 = 1

3x + 0.43 = 1

(Reason: the spinner has to land on one of the five numbers, so the five probabilities account for every spin and must total

1

; collecting the

2x

with the

x

gives

3x

)

Step 2: solve for x

3x = 1 - 0.43 = 0.57

x = \dfrac{0.57}{3} = 0.19

(Reason: take the three known probabilities off

1

, then split what is left equally between the three lots of

x

)

Step 3: add the probabilities of the odd numbers

P(\text{odd}) = P(1) + P(3) + P(5)

P(\text{odd}) = 2 \times 0.19 + 0.04 + 0.12 = 0.54

(Reason: the odd numbers on this spinner are

1

3

and

5

, and the probability of landing on one of them is the sum of the three separate probabilities because the spinner cannot land on two numbers at once)

Step 4: scale the probability up to 400 spins

400 \times 0.54 = 216

(Reason: an estimate for how often an event happens is its probability multiplied by the number of trials, and here there are

400

spins)

An estimate of

216

spins landing on an odd number

Verification

Check 1: Put

x = 0.19

back into the table and add all five probabilities, including the

2x

row as

0.38

0.38 + 0.27 + 0.04 + 0.19 + 0.12 = 1

Check 2: Work in spins instead of probabilities, never finding

x

at all:

0.27 \times 400 = 108

0.04 \times 400 = 16

and

0.12 \times 400 = 48

, which leaves

400 - 108 - 16 - 48 = 228

spins to share between

1

and

4

in the ratio

2 : 1

, so

152

and

76

152 + 16 + 48 = 216

Check 3: Estimate the even numbers instead and take them off the

400

spins. The even numbers are

2

and

4

, with probability

0.27 + 0.19 = 0.46

, so

0.46 \times 400 = 184

400 - 184 = 216

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Uses the fact that the probabilities total $1$ : $2x + 0.27 + 0.04 + x + 0.12 = 1$ or $1 - (0.27 + 0.04 + 0.12) = 0.57$	M1	For a clear understanding that the total of the probabilities is $1$ . Also earned by finding estimates for the other three numbers: $0.27 \times 400 = 108$ , $0.04 \times 400 = 16$ and $0.12 \times 400 = 48$	✓
$x = \dfrac{0.57}{3} = 0.19$ or $2x = \dfrac{0.57}{3} \times 2 = 0.38$	M1	For a method to find the value of $x$ or of $2x$ . By the frequency route, for $\dfrac{400 - 108 - 16 - 48}{3} = 76$ or $76 \times 2 = 152$	✓
$(2 \times 0.19 + 0.04 + 0.12) \times 400$ or $2 \times 76 + 16 + 48$	M1	For a complete method: the three odd probabilities added and then scaled to the number of spins, or the three odd estimates added directly	✓
The estimate: $216$	A1	For an answer of $216$	✓
Guidance on this question	Note	A correct answer scores full marks unless it comes from obviously incorrect working. An answer left as $\dfrac{216}{400}$ or equivalent is a probability rather than a number of spins and scores M3A0	✓

Full marks: 4/4

Matteo sells plain plant pots and painted plant pots on his market stall.

He sells a total of $200$ pots such that

$\text{the number of plain pots sold} : \text{the number of painted pots sold} = 3 : 2$

Matteo sells the plain pots for $\pounds 1.50$ each.
He sells the painted pots for $\pounds 1.75$ each.

$40\%$ of the price of a plain pot is profit.
$60\%$ of the price of a painted pot is profit.

Work out the total profit Matteo makes when he sells all $200$ pots. [5 marks]

[Total 5 marks]

Show solution & mark schemeHide solution & mark scheme

Question 3 - Exam Solution

Understanding the Question

Given

Matteo sells

200

pots altogether, some plain and some painted.

The plain pots and the painted pots are sold in the ratio

3 : 2

A plain pot sells for

\pounds 1.50

and a painted pot sells for

\pounds 1.75

40\%

of the price of a plain pot is profit, and

60\%

of the price of a painted pot is profit.

Find

The total profit Matteo makes once all

200

pots are sold.

Plan the Solution

Share the $200$ pots in the ratio $3 : 2$ to find how many of each kind are sold.
Turn each number of pots into the money taken for that kind, by multiplying by the price of one pot.
The two profit rates are different, so they can never be combined into a single percentage of the total takings. Apply each rate to the money taken for its own kind of pot.
Add the two profits together.

Worked Solution [5 marks]

Rule - to share an amount in a given ratio, divide by the total number of shares to find one share; and a percentage of an amount is that percentage, written as a decimal, multiplied by the amount.

Step 1: find one share of the ratio

3 + 2 = 5

\dfrac{200}{5} = 40

(Reason: the ratio

3 : 2

splits the pots into

5

equal shares, so dividing the

200

pots by

5

gives the size of one share)

Step 2: find how many pots of each kind are sold

3 \times 40 = 120

2 \times 40 = 80

(Reason: the plain pots take

3

of the shares and the painted pots take

2

, and the two counts add back to

200

, which is the check that the sharing is right)

Step 3: work out the money taken for each kind

120 \times 1.50 = 180

80 \times 1.75 = 140

(Reason: multiply the number of pots by the price of one pot, so the plain pots take

\pounds 180

and the painted pots take

\pounds 140

)

Step 4: take the profit rate of each kind out of its own takings

0.4 \times 180 = 72

0.6 \times 140 = 84

(Reason: each kind of pot has its own profit rate, so the plain rate is applied to the

\pounds 180

and the painted rate is applied to the

\pounds 140

; a percentage is turned into a decimal before multiplying)

Step 5: add the two profits

72 + 84 = 156

(Reason: the total profit is the profit made on the plain pots plus the profit made on the painted pots)

A total profit of

\pounds 156

Verification

Check 1: Work out the profit on a single pot first, then multiply by how many of that kind are sold:

0.4 \times 1.50 = 0.60

on a plain pot and

0.6 \times 1.75 = 1.05

on a painted pot.

120 \times 0.60 + 80 \times 1.05 = 156

Check 2: Come at it from the other side. The part of each price that is not profit is

60\%

of a plain pot and

40\%

of a painted pot, so the total cost is

0.6 \times 180 + 0.4 \times 140 = 164

, and the total takings are

180 + 140 = 320

320 - 164 = 156

Check 3: Count the pots whose whole price is profit instead:

0.4 \times 120 = 48

plain pots and

0.6 \times 80 = 48

painted pots.

48 \times 1.50 + 48 \times 1.75 = 156

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Divides by the total number of shares: $\dfrac{200}{3 + 2}$ which gives $40$	M1	For a method to find one share of the ratio, $\dfrac{200}{3 + 2} = 40$	✓
$3 \times 40 = 120$ and $2 \times 40 = 80$	M1	For a method to find the number of plain pots and the number of painted pots. Both are needed: $3 \times 40 = 120$ and $2 \times 40 = 80$	✓
$120 \times 1.50 = 180$ and $80 \times 1.75 = 140$	M1	For a method to find the money taken from the plain pots and the money taken from the painted pots, or the number of pots that are entirely profit, or the profit on a single pot of either kind. Examples of the last two routes: $0.4 \times 120 = 48$ and $0.6 \times 80 = 48$ ; $0.4 \times 1.50 = 0.60$ and $0.6 \times 1.75 = 1.05$ . The printed scheme disagrees with itself on that last route - its condition allows the profit on a single pot of one kind or the other, while its worked example shows both - and it is the condition that carries the mark, so one is enough	✓
$0.4 \times 180 = 72$ and $0.6 \times 140 = 84$	M1	For a complete method to find the total profit on the plain pots and the total profit on the painted pots: $0.4 \times 180 = 72$ and $0.6 \times 140 = 84$ . By the other two routes, $48 \times 1.50 = 72$ and $48 \times 1.75 = 84$ , or $120 \times 0.60 = 72$ and $80 \times 1.05 = 84$	✓
The total profit: $156$	A1	cao. For an answer of $156$ . A correct answer scores full marks unless it comes from obviously incorrect working	✓
An answer of $164$ or $174$	SC	Award SCB4 for either. $164$ is the two profit rates swapped, $0.6 \times 180 + 0.4 \times 140 = 164$ - which is why it is not a near miss but the total cost of the pots instead of the profit. $174$ is the ratio taken the other way round, giving $80$ plain and $120$ painted, so $0.4 \times 120 + 0.6 \times 210 = 174$	✓

Full marks: 5/5

Show that
$\dfrac{2\dfrac{1}{3}}{5\dfrac{1}{4}} = \dfrac{4}{9}$
You must show every stage of your working. [3 marks]

[Total 3 marks]

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Question 4 - Exam Solution

Understanding the Question

Given

One mixed number divided by another,

\dfrac{2\dfrac{1}{3}}{5\dfrac{1}{4}}

The fraction it is claimed to come to,

\dfrac{4}{9}

, is printed for you.

Find

A complete piece of working that starts at

\dfrac{2\dfrac{1}{3}}{5\dfrac{1}{4}}

and finishes at

\dfrac{4}{9}

. Because the answer is already printed, the working is the whole of the mark. An unsupported answer earns nothing here.

Plan the Solution

Turn each mixed number into a top-heavy (improper) fraction. The division rule only works on those, so $2\dfrac{1}{3}$ and $5\dfrac{1}{4}$ cannot be used as they stand.
Turn the dividing fraction upside down and multiply by it instead.
Multiply the numerators together and the denominators together.
Cancel the result down to its lowest terms, and check that what is left is $\dfrac{4}{9}$ .

Worked Solution [3 marks]

Rule - Dividing by a fraction: turn the second fraction upside down and multiply. In symbols,

\dfrac{a}{b}

divided by

\dfrac{c}{d}

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

, and the first fraction is left exactly as it is.

Step 1: write

2\dfrac{1}{3}

and

5\dfrac{1}{4}

as improper fractions

2 + \dfrac{1}{3} = \dfrac{6}{3} + \dfrac{1}{3} = \dfrac{7}{3}

5 + \dfrac{1}{4} = \dfrac{20}{4} + \dfrac{1}{4} = \dfrac{21}{4}

(Reason: A mixed number is a whole part plus a fraction part, so

2\dfrac{1}{3}

means

2 + \dfrac{1}{3}

. One whole is

\dfrac{3}{3}

, so two wholes are

\dfrac{6}{3}

. The same count of quarters turns

5\dfrac{1}{4}

into

\dfrac{21}{4}

. The quicker version of the same thing is to multiply the whole number by the denominator and add the numerator.)

Step 2: replace the division with a multiplication

\dfrac{\dfrac{7}{3}}{\dfrac{21}{4}} = \dfrac{7}{3} \times \dfrac{4}{21}

(Reason: Dividing by

\dfrac{21}{4}

asks how many lots of

\dfrac{21}{4}

will fit, and that is the same as taking

\dfrac{4}{21}

of the amount. Only the fraction being divided by is turned over; the first one is copied down unchanged.)

Step 3: multiply along the top and along the bottom

\dfrac{7}{3} \times \dfrac{4}{21} = \dfrac{7 \times 4}{3 \times 21} = \dfrac{28}{63}

(Reason: Multiplying fractions needs no common denominator: the two numerators multiply to give the new numerator, and the two denominators multiply to give the new denominator.)

Step 4: cancel down to lowest terms

\dfrac{28}{63} = \dfrac{4 \times 7}{9 \times 7} = \dfrac{4}{9}

(Reason: Both

28

and

63

are multiples of

7

, and

7

is the highest number that divides into both, so cancelling it leaves the fraction in its lowest terms. Nothing above

1

divides into both

4

and

9

, so this is as far as it goes, and it is the fraction the question asked us to reach.)

\dfrac{2\dfrac{1}{3}}{5\dfrac{1}{4}} = \dfrac{4}{9}

as required

Verification

Check 1: Do the division the other way the mark scheme allows, with no reciprocal at all. Put both mixed numbers over the common denominator

12

2\dfrac{1}{3}

\dfrac{28}{12}

and

5\dfrac{1}{4}

\dfrac{63}{12}

. Twelfths divided by twelfths leaves a plain ratio of the two counts.

\dfrac{\dfrac{28}{12}}{\dfrac{63}{12}} = \dfrac{28}{63} = \dfrac{4}{9}

Check 2: Run the division backwards. If the answer is right, multiplying it by the divisor

5\dfrac{1}{4}

must return the dividend

2\dfrac{1}{3}

\dfrac{4}{9} \times \dfrac{21}{4} = \dfrac{84}{36} = \dfrac{7}{3}

, and

\dfrac{7}{3}

2\dfrac{1}{3}

Check 3: Test the cancelling on its own, without repeating it. Two fractions are equal exactly when their cross products match, so compare

\dfrac{28}{63}

with

\dfrac{4}{9}

that way.

28 \times 9 = 252 = 63 \times 4

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Both mixed numbers written as improper fractions	M1	For $\dfrac{7}{3}$ and $\dfrac{21}{4}$ . A candidate who has already inverted the second fraction and written $\dfrac{4}{21}$ rather than $\dfrac{21}{4}$ still earns this mark.	✓
The division turned into a multiplication, or both fractions put over one denominator	M1	For the intention to multiply the correct improper fraction by the inverted fraction, $\dfrac{7}{3} \times \dfrac{4}{21}$ or equivalent, for example $\dfrac{49}{21} \times \dfrac{4}{21}$ . Also earned for writing the two fractions over the same common denominator, $\dfrac{28}{12}$ and $\dfrac{63}{12}$ .	✓
The working completed to the printed fraction	A1	For correctly completing to reach the required answer, for example $\dfrac{7}{3} \times \dfrac{4}{21} = \dfrac{28}{63} = \dfrac{4}{9}$ , or $\dfrac{49}{21} \times \dfrac{4}{21} = \dfrac{196}{441} = \dfrac{4}{9}$ , or the common-denominator route finishing at $\dfrac{28}{63} = \dfrac{4}{9}$ . Cancelling the sevens inside the multiplication before multiplying is equally acceptable.	✓
Guidance carried on the mark scheme	Note	Working is required. The mark scheme header names question 4 as one of the questions where a correct answer is not taken to imply a correct method, so the printed fraction copied out on its own scores nothing. Decimals written only as a check are ignored, neither credited nor penalised, provided the fraction working is there.	✓

Full marks: 3/3

Radomir pays $5200$ euros into a savings bond.
The bond pays $2.5\%$ per year compound interest.
Radomir leaves his money in the bond for $4$ years.

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

euros

[Total 3 marks]

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Question 5 - Exam Solution

Understanding the Question

Given

Paid into the bond:

5200

euros

Compound interest:

2.5\%

per year

Time left in the bond:

4

years

Find

The amount in the bond after

4

years, correct to the nearest euro

Plan the Solution

Turn the rate into a multiplier for one year: $100\% + 2.5\% = 102.5\%$ , which is $1.025$ .
Compound interest means each year's interest is worked out on the new balance, so that same multiplier is used once for every year.
Four years is therefore four multiplications by $1.025$ , which is the same as multiplying by $1.025^{4}$ .
Keep the full decimal all the way through and round only at the very end, or the rounding error grows with every year.

Worked Solution [3 marks]

Rule - Compound interest:

\text{final amount} = P \times \left(1 + \dfrac{r}{100}\right)^{n}

, where

P

is the amount paid in,

r

is the percentage rate per year and

n

is the number of years.

Step 1: Write the yearly rate as a multiplier

1 + \dfrac{2.5}{100} = 1.025

(Reason: The money is not replaced by the interest, it is added to it, so the bond is worth

102.5\%

of what it was worth a year earlier. Writing that percentage as the decimal

1.025

turns a year in the bond into a single multiplication.)

Step 2: Multiply once for each year, to see what compound interest is doing

5200 \times 1.025 = 5330

5330 \times 1.025 = 5463.25

5463.25 \times 1.025 = 5599.83125

5599.83125 \times 1.025 = 5739.82703125

(Reason: Each line starts from the balance the line above finished with, which is exactly what makes the interest compound: the second year earns

133.25

euros where the first earned only

130

euros, because the second year's

2.5\%

is taken on a bigger amount.)

Step 3: Do the same thing in one line with a power

5200 \times 1.025^{4}

1.025^{4} = 1.103812890625

5200 \times 1.103812890625 = 5739.82703125

(Reason: Multiplying by

1.025

four times in a row is multiplying by

1.025^{4}

once, so the whole of Step 2 collapses into one calculation. In the exam this is the line to write, and the power key on the calculator does the rest.)

Step 4: Round to the nearest euro

5739.82703125 \text{ euros} \implies 5740 \text{ euros}

(Reason: The question asks for the nearest euro. The first digit after the decimal point is

8

, so the amount rounds up. Rounding here, at the end, and not earlier, is what keeps the answer inside the accepted

5739

5740

5740

euros

Verification

Check 1: Undo the four years. Dividing the final amount by

1.025^{4}

must give back exactly the amount that was paid in.

\dfrac{5739.82703125}{1.103812890625} = 5200

, the original amount

Check 2: Look at the interest on its own. Four years of

2.5\%

simple interest would be

4 \times 130 = 520

euros, so compound interest must come to a little more than that, and only a little.

5739.82703125 - 5200 = 539.82703125

, which beats

520

by about

20

euros

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$5200 \times 1.025 = 5330$ or $5200 \times 0.025 = 130$	M1	For a method to find $2.5\%$ of $5200$ , or $102.5\%$ of $5200$ . Any equivalent working earns this.	✓
$5330 \times 1.025 = 5463.25$ and $5463.25 \times 1.025 = 5599.83\ldots$ and $5599.83\ldots \times 1.025 = 5739.8\ldots$	M1	For a complete method: the multiplier used once for each of the $4$ years. The mark scheme quotes the candidate's own values, so this follows through from a wrong balance in an earlier year.	✓
$5200 \times 1.025^{4}$	M2	The power method written in one line earns both method marks at once. The mark scheme also allows M2 for $5200 \times 1.025^{5} = 5883\ldots$ , which is a complete method spoiled only by using $5$ years instead of $4$ .	✓
The value written on the answer line	A1	Accept anything from $5739$ to $5740$ , which covers a candidate who truncated rather than rounded. A correct answer scores full marks unless it comes from obviously incorrect working.	✓
If no other mark has been earned	SC B1	For any one of $5200 \times 0.1 (= 520)$ oe (reading the rate as $10\%$ ), $5200 \times 1.1 (= 5720)$ oe (one year at $10\%$ ), $5200 \times 0.9 (= 4680)$ oe (a $10\%$ decrease), $5200 \times 0.975 (= 5070)$ oe (one year of decrease instead of increase) or $5200 \times 0.975^{4} (= 4699\ldots)$ oe (four years of decrease). Accept (1 + 0.025) as equivalent to 1.025 throughout, but not (1 + 2.5%).	✓
Equivalent forms of the multiplier	Note	Accept $(1 + 0.025)$ as equivalent to $1.025$ throughout, but do not accept $(1 + 2.5\%)$ .	✓

Full marks: 3/3

The diagram shows a solid cylinder cut from a length of hardwood.

The cylinder has radius $8$ cm and height $h$ cm.
The volume of the cylinder is $1208$ cm³

(a) Work out the value of $h$ .
Give your answer correct to the nearest whole number. [2 marks]

The density of the hardwood is $1.25$ g/cm³

(b) Work out the mass of the cylinder.
Give your answer in kilograms. [2 marks]

(a) h =(b) kilograms

[Total 4 marks]

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Question 6 - Exam Solution

Understanding the Question

Given

A solid hardwood cylinder of radius

8

cm and height

h

cm.

Its volume is

1208

cm³, so on this solid the volume is known and the height is not.

The hardwood has density

1.25

g/cm³, so each cm³ of it has mass

1.25

Find

(a) The height

h

, correct to the nearest whole number. (b) The mass of the whole cylinder, in kilograms.

Plan the Solution

A cylinder is a circular face swept through its height, so $V = \pi r^2 h$ . Putting the radius and the volume in leaves $h$ as the only unknown.
Work out the area of the circular face first, then divide the volume by it. Keep the calculator's own $\pi$ all the way through and round once, at the very end of part (a).
Density is the mass of one cm³, so $\text{mass} = \text{density} \times \text{volume}$ gives the mass in grams. The question wants kilograms, and $1000$ g make $1$ kg.
Part (b) does not need the answer to part (a). The volume is already given as $1208$ cm³, so use that rather than one rebuilt from a rounded height.

Worked Solution [4 marks]

Rule - Cylinder and density:

V = \pi r^2 h

, and

\text{mass} = \text{density} \times \text{volume}

Step 1: Put the radius and the volume into the cylinder formula

\pi \times 8^2 \times h = 1208

(Reason: The circular face has radius

8

cm, so its area is

\pi \times 8^2

cm², and the volume of a cylinder is that area multiplied by the height. Writing the given volume on the other side turns the formula into an equation in

h

Step 2: Work out the area of the circular face

8^2 = 64

\pi \times 64 = 201.0619\ldots

(Reason: Squaring the radius first keeps the working in one place: the face is

64\pi

cm², a little over

201

cm². Squaring the diameter here instead of the radius is the slip this question catches most often.)

Step 3: Make

h

the subject

h = \dfrac{1208}{\pi \times 8^2}

(Reason: The face area multiplies the height, so dividing both sides by that area leaves the height on its own.)

Step 4: Work the height out on the calculator

h = \dfrac{1208}{201.0619\ldots} = 6.0081\ldots

(Reason: Use the

\pi

key rather than a rounded value. Rounding the

201.0619\ldots

before dividing moves the last figures of the height.)

Step 5: Round to the nearest whole number

6.0081\ldots \approx 6

(Reason: The first figure after the decimal point is

0

, which is below

5

, so the whole-number part stays at

6

Step 6: Work out the mass of the cylinder in grams

1.25 \times 1208 = 1510

(Reason: Each cm³ of the wood has mass

1.25

g and the solid holds

1208

cm³ of it, so the mass is

1.25

g taken

1208

times. The volume used is the one the question states, not one rebuilt from the rounded height.)

Step 7: Change the grams into kilograms

\dfrac{1510}{1000} = 1.51

(Reason: There are

1000

g in

1

kg, so the number of kilograms is the number of grams divided by

1000

. The question asks for kilograms, so an answer left in grams is not finished.)

(a)

h = 6

(b)

1.51

kilograms

Verification

Check 1: Put the unrounded height back into the volume formula. The volume the question states must come back.

\pi \times 8^2 \times 6.0081\ldots = 1208

cm³

Check 2: Is the size sensible, with no calculator? The face is a little over

200

cm², and a height of

6

cm on a face of

200

cm² falls just short of the stated volume.

200 \times 6 = 1200

, just under

1208

, so the true height sits a little above

6

and still rounds to

6

Check 3: Divide the mass in grams by the volume. Density is the mass of one cm³, so the density given in the question must return.

\dfrac{1510}{1208} = 1.25

g/cm³

Check 4: Turn the kilograms back into grams, which undoes the last step.

1.51 \times 1000 = 1510

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) An equation in $h$ built from the volume of the cylinder, $\pi \times 8^2 \times h = 1208$ , or a correct calculation for $h$ , $\dfrac{1208}{\pi \times 8^2}$	M1	The method may be seen in stages: $\pi \times 8^2 = 201.06\ldots$ followed by a division by that figure earns the mark exactly as the single line does. Squaring the diameter, $16$ , in place of the radius does not.	✓
$6$	A1	Accept anything from $6$ to $6.02$ . The band is stated with no reason on the paper, and this is the reason: a candidate who types a rounded $\pi$ lands a little high, at $\dfrac{1208}{3.14 \times 64} = 6.0111\ldots$ , and the band is exactly wide enough for that and no wider. A correct answer scores both marks unless it comes from obviously incorrect working.	✓
(b) An equation built from density as mass per unit volume, $\dfrac{m}{1208} = 1.25$ , or a calculation for the mass, $1208 \times 1.25 \ (= 1510)$	M1	Any correct method for the mass earns this, including changing the density to $0.00125$ kg/cm³ first and multiplying by $1208$ , which reaches kilograms in a single line.	✓
$1.51$	A1	The answer must be in kilograms. $1510$ left in grams shows the method but not the change of unit, so it scores the method mark alone. A correct answer scores both marks unless it comes from obviously incorrect working.	✓
Part (b) is marked from the volume the question states, $1208$ cm³, not from a volume rebuilt out of the rounded height.	Note	Rebuilding it from $h = 6$ gives $1206.37\ldots$ cm³ and a mass of $1507.96\ldots$ g. That still rounds to $1.51$ kg, so it is not penalised here, but the given volume is the one the mark scheme's own calculation uses.	✓

Full marks: 4/4

(a) Simplify $\dfrac{g^{9}}{g^{2}}$ [1 mark]

(b) Expand $5k^{2}(k^{3} + 4)$ [2 marks]

(d) Solve the inequality $7 - 2y < 3y - 12$ [3 marks]

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

[Total 9 marks]

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Question 7 - Exam Solution

Understanding the Question

Given

\dfrac{g^{9}}{g^{2}}

- one power of

g

divided by another.

5k^{2}(k^{3} + 4)

- a single term multiplying a bracket.

x^{2} - 2x - 63

- a quadratic whose

x^{2}

coefficient is

1

7 - 2y < 3y - 12

- a linear inequality with

y

on both sides.

Find

(a) the quotient written as a single power of

g

. (b) the bracket expanded, one term at a time. (c)(i) the quadratic as a product of two brackets, then (ii) the two values of

x

that follow from them. (d) the range of values of

y

that satisfy the inequality.

Plan the Solution

(a) Dividing powers of the same base subtracts the indices, so this is $9 - 2$ .
(b) Multiply each of the two terms inside the bracket by $5k^{2}$ , adding indices as you go.
(c)(i) Look for two integers whose product is $-63$ and whose sum is $-2$ .
(c)(ii) A product is zero only when one of its factors is zero, so read one solution off each bracket. This is why the question says "Hence" - the factors are already done.
(d) Rearrange it exactly as you would an equation, but reverse the inequality sign if you divide by a negative number.

Worked Solution [9 marks]

Rule - Dividing powers:

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

. Expanding: multiply every term inside the bracket by the term outside. Factorising

x^{2} + bx + c

: find two integers with product

c

and sum

b

. Inequalities: solve as an equation, but reverse the sign if you multiply or divide by a negative number.

(a) Subtract the indices

\dfrac{g^{9}}{g^{2}} = g^{9-2} = g^{7}

(Reason: The numerator is

9

copies of

g

and the denominator is

2

of them, so

2

cancel and

7

are left.)

(b) Multiply each term in the bracket by

5k^{2}

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

5k^{2} \times 4 = 20k^{2}

(Reason: Multiplying powers of the same base adds the indices, so

k^{2} \times k^{3} = k^{2+3} = k^{5}

. The second term has no

k

in the bracket, so its power of

k

is unchanged.)

(b) Write the two terms as one expression

5k^{2}(k^{3} + 4) = 5k^{5} + 20k^{2}

(Reason: The powers

k^{5}

and

k^{2}

are different, so the two terms will not collect together. This is the final answer.)

(c)(i) Find two integers with product

-63

and sum

-2

7 \times (-9) = -63

7 + (-9) = -2

(Reason: The product is negative, so one integer is positive and the other is negative. The pairs that multiply to

-63

are

1

and

-63

3

and

-21

7

and

-9

and their opposites, and only

7

and

-9

add to

-2

(c)(i) Write the quadratic as two brackets

x^{2} - 2x - 63 = (x + 7)(x - 9)

(Reason: The two integers go straight into the brackets. Expanding gives

x^{2} - 9x + 7x - 63

, and the middle terms collect to

-2x

(c)(ii) Set each bracket equal to zero

(x + 7)(x - 9) = 0

x + 7 = 0 \implies x = -7

x - 9 = 0 \implies x = 9

(Reason: A product of two numbers is zero only when at least one of them is zero, so each bracket gives one solution.)

(d) Collect the

y

terms on the left and the numbers on the right

7 - 2y < 3y - 12

-2y - 3y < -12 - 7

(Reason: Subtract

3y

from both sides and subtract

7

from both sides. Adding or subtracting never changes the direction of an inequality.)

(d) Simplify each side

-5y < -19

(Reason:

-2y - 3y = -5y

and

-12 - 7 = -19

(d) Divide both sides by

-5

and reverse the inequality

y > \dfrac{19}{5}

(Reason: Dividing by a negative number reverses the direction of an inequality, so the "less than" sign turns into a "greater than" sign. As a decimal the boundary is

3.8

, and the boundary itself is not included.)

(a)

g^{7}

(b)

5k^{5} + 20k^{2}

(c)(i)

(x + 7)(x - 9)

(c)(ii)

x = -7

and

x = 9

(d)

y > \dfrac{19}{5}

Verification

Check 1: Part (a) with

g = 2

. The numerator is

2^{9} = 512

and the denominator is

2^{2} = 4

\dfrac{512}{4} = 128

and

2^{7} = 128

Check 2: Part (b) with

k = 2

. The bracket is

2^{3} + 4 = 12

and the term outside is

5 \times 4 = 20

20 \times 12 = 240

and

5 \times 32 + 20 \times 4 = 240

Check 3: Expand the brackets from part (c)(i) and compare with the quadratic the question printed.

(x + 7)(x - 9) = x^{2} - 9x + 7x - 63 = x^{2} - 2x - 63

Check 4: Put both solutions from part (c)(ii) back into

x^{2} - 2x - 63

81 - 18 - 63 = 0

and

49 + 14 - 63 = 0

Check 5: Test part (d) either side of

3.8

, using

y = 4

and then

y = 3

. At

y = 4

the two sides are

-1

and

0

, and

-1

is less than

0

, so it works. At

y = 3

the two sides are

1

and

-3

, and

1

is not less than

-3

, so it does not.

Check 6: The boundary itself. Put

y = \dfrac{19}{5}

into both sides. Both sides come to

-\dfrac{3}{5}

, so at the boundary the two sides are equal, which confirms the inequality is strict.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $\dfrac{g^{9}}{g^{2}} = g^{7}$	B1	For $g^{7}$ .	✓
(b) $5k^{5} + 20k^{2}$	B2	For $5k^{5} + 20k^{2}$ . Award $1$ mark only, B1, for either $5k^{5}$ or $20k^{2}$ alone.	✓
(c)(i) $(x \pm 7)(x \pm 9)$	M1	For $(x \pm 7)(x \pm 9)$ , or for $(x + a)(x + b)$ where $ab = -63$ or $a + b = -2$ , with $a$ and $b$ integers.	✓
(c)(i) $(x + 7)(x - 9)$	A1	For the correct factors. A correct answer scores full marks unless it comes from obviously incorrect working.	✓
(c)(ii) $-7, 9$	B1	Must follow through from part (c)(i), and is dependent on factorising in the form $(x + p)(x + q)$ where $p$ and $q$ are integers.	✓
(d) $-2y - 3y < -12 - 7$ or $7 + 12 < 3y + 2y$	M1	For a rearrangement with the $y$ terms on one side and the numerical terms on the other in a correct inequality, or for the correct simplification of the $y$ terms or of the numbers on one side in a correct inequality. The sign may be $=$ or the incorrect inequality sign.	✓
(d) $-5y < -19$ or $19 < 5y$	M1	For the correct simplification of the $y$ terms on one side and the numbers on the other side in a correct inequality, or for a correct inequality with the wrong sign. The sign may be $=$ or the incorrect inequality sign.	✓
(d) $y > \dfrac{19}{5}$	A1	Or equivalent, for example $y > 3.8$ or $3.8 < y$ . It must be given as the correct inequality on the answer line. A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 9/9

$ABCD$ is a trapezium.

Angle $DAB$ and angle $ADC$ are right angles.
$AD = 15$ cm
$DC = 14$ cm
The area of the trapezium is $360$ cm²
Work out the perimeter of the trapezium. [6 marks]

[Total 6 marks]

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Question 8 - Exam Solution

Understanding the Question

Given

ABCD

is a trapezium, with angle

DAB

and angle

ADC

both right angles

AD = 15

cm and

DC = 14

Area of

ABCD

360

cm²

Find

The perimeter of

ABCD

, in cm - so all four sides are needed, and two of them are not given

Plan the Solution

Both $AB$ and $DC$ are perpendicular to $AD$ , so they are the parallel sides and $AD$ is the distance between them.
Put the given area into the trapezium formula and solve for the missing parallel side $AB$ .
Drop a perpendicular from $C$ to $AB$ , meeting it at $M$ . That cuts the trapezium into rectangle $AMCD$ and right-angled triangle $MBC$ .
Use Pythagoras in triangle $MBC$ to find the slant side $CB$ , then add the four sides.

Worked Solution [6 marks]

Rule - Trapezium:

\text{Area} = \dfrac{a + b}{2} \times h

, where

a

and

b

are the parallel sides and

h

is the perpendicular distance between them.

Step 1: name the parallel sides and the height

a = DC = 14

h = AD = 15

(Reason: Angle

DAB

and angle

ADC

are both right angles, so

AB

and

DC

are both perpendicular to

AD

. That makes them the parallel pair, with

AD

as the distance between them and

AB

as the unknown side

b

Step 2: put the area into the formula

\dfrac{14 + AB}{2} \times 15 = 360

(Reason: Everything in the formula is now known except

AB

, so the area gives one equation in one unknown.)

Step 3: solve for

AB

14 + AB = \dfrac{2 \times 360}{15}

14 + AB = 48

AB = 48 - 14 = 34

(Reason: Multiply both sides by

2

and divide by

15

to undo the formula. That leaves the two parallel sides adding to

48

, so take the known one,

DC = 14

, off the total.)

Step 4: split the trapezium at

M

AM = DC = 14

MC = AD = 15

MB = 34 - 14 = 20

(Reason:

M

is the point on

AB

with

MC

perpendicular to

AB

. Then

AMCD

has four right angles, so it is a rectangle and its opposite sides are equal. What is left of

AB

is the base of the triangle.)

Step 5: Pythagoras in triangle

MBC

CB^2 = 15^2 + 20^2

CB^2 = 225 + 400 = 625

CB = \sqrt{625} = 25

(Reason: Angle

BMC

is a right angle, so the square on the slant side

CB

is the sum of the squares on

MC

and

MB

Step 6: add the four sides

\text{Perimeter} = AB + BC + CD + DA

\text{Perimeter} = 34 + 25 + 14 + 15 = 88

(Reason: The perimeter is the distance all the way round the outside, so each of the four sides is counted exactly once. The dashed line

MC

is a construction line and is not part of it.)

88

Verification

Check 1: Put

AB = 34

back into the trapezium formula and see whether the given area returns.

\dfrac{34 + 14}{2} \times 15 = 24 \times 15 = 360

Check 2: Work the area out a second way, as rectangle

AMCD

plus triangle

MBC

, which never uses the trapezium formula at all.

14 \times 15 + \dfrac{20 \times 15}{2} = 210 + 150 = 360

Check 3: Test

CB

without Pythagoras. The legs

15

and

20

are

5

times

3

and

4

, so triangle

MBC

is a scaled

3

4

5

triangle and its hypotenuse must be

5

times

5

5 \times 5 = 25

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{14 + AB}{2} \times 15 = 360$ oe, or $360 - 14 \times 15 (=150)$ oe	M1	for setting up an equation using the area of the trapezium, or for a method to find the area of the triangle	✓
$AB = 34$ or $MB = 20$	A1	could be seen on the diagram, where $M$ is the point on $AB$ with $MC$ perpendicular to $AB$	✓
$(CB^2 =) 15^2 + 20^2 (= 625)$ or $(CB^2 =) 15^2 + MB^2$	M1	allow use of their $MB$	✓
$(CB =) \sqrt{15^2 + 20^2} (= 25)$ or $(CB =) \sqrt{15^2 + MB^2}$	M1	allow use of their $MB$	✓
$14 + 15 + 34 + 25$ oe, or $14 + 15 + 14 + MB + CB$ oe	M1ft	dep on the previous two M marks, for a method to find the perimeter of the trapezium; allow use of their $MB$ and their $CB$	✓
$88$	A1	cao. A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 6/6

The straight line $L$ is drawn on the grid below.

Work out an equation of $L$ .
Give your answer in the form $y = mx + c$ [3 marks]

[Total 3 marks]

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Question 9 - Exam Solution

Understanding the Question

Given

A straight line

L

drawn on a grid, with both axes numbered.

One square on the grid stands for

1

unit across and

1

unit up.

The line runs right across the grid, from the left edge to the right edge.

Find

An equation of

L

, written in the form

y = mx + c

. That is two numbers: the gradient

m

, and the value

c

where the line crosses the

y

-axis.

Plan the Solution

Pick two points where the line passes exactly through a crossing of the grid lines, so both coordinates can be read with nothing estimated.
Divide the change in $y$ by the change in $x$ between those two points. That is the gradient.
Read the value of $y$ where the line cuts the $y$ -axis. That is $c$ .
Put both numbers into $y = mx + c$ , then test the equation on a third point of the line.

Worked Solution [3 marks]

Rule - Straight line:

y = mx + c

, where

m

is the gradient

\dfrac{\text{change in } y}{\text{change in } x}

and

c

is the value of

y

where the line crosses the

y

-axis.

Step 1: read two exact points off the line

(-4, 3) \text{ and } (4, -1)

(Reason: The line meets each edge of the grid at a crossing of the grid lines, so both readings are exact. A point taken from the middle of a square would only be an estimate.)

Step 2: work out the gradient

m

\dfrac{-1 - 3}{4 - (-4)} = \dfrac{-4}{8} = -\dfrac{1}{2}

m = -\dfrac{1}{2}

(Reason: Going from

(-4, 3)

(4, -1)

, the

y

value falls by

4

while the

x

value rises by

8

. A line that falls from left to right has a negative gradient.)

Step 3: read the value of

c

c = 1

(Reason: The line crosses the

y

-axis one square above the origin, at

(0, 1)

, and that value of

y

c

Step 4: write the equation in the form the question asks for

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

(Reason: The gradient goes in front of the

x

and the intercept goes on the end. Writing

y = -0.5x + 1

says exactly the same thing and is equally acceptable.)

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

Verification

Check 1: Put

x = 4

into the equation and compare it with the point the line reaches at the right-hand edge of the grid.

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

, and the grid shows the line at

(4, -1)

Check 2: Test a point that was not used to build the equation: the line cuts the

x

-axis at

(2, 0)

-\dfrac{1}{2} \times 2 + 1 = 0

, so

(2, 0)

lies on the line, as the grid shows.

Check 3: Check the shape of the answer rather than a number: a gradient of

-\dfrac{1}{2}

means the line should drop one square for every two squares it moves to the right. Across the whole grid the line moves

8

squares right and drops

4

squares, which is one down for every two across.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
One correct piece of the equation: the gradient $-\dfrac{1}{2}$ , or $-\dfrac{1}{2}x + c$ , or $mx + 1$ with $m \neq 0$	B1	The gradient alone scores, however it is written - $-0.5$ , $\dfrac{-1}{2}$ , or the working $\dfrac{3 - (-1)}{-4 - 4}$ . The commonest loss is dividing the wrong way round, change in $x$ over change in $y$ , which gives $-2$ and scores nothing.	✓
Both parts correct, but not yet written as $y = mx + c$	B2	Award for $-\dfrac{1}{2}x + 1$ with no $y =$ in front, or for a correct equation in another form such as $2y + x = 2$ . Also award for $y = -\dfrac{1}{2}x + c$ or for $y = mx + 1$ with $m \neq 0$ .	✓
The equation in the form asked for: $y = -\dfrac{1}{2}x + 1$	B3	Accept any equivalent, for example $y = -0.5x + 1$ , provided it is in the form $y = mx + c$ . A correct equation left as $2y + x = 2$ scores $2$ of the $3$ marks, because the question names the form it wants.	✓

Full marks: 3/3

Here are the numbers of loaves of bread left unsold by a bakery at the end of each of its last $11$ days.

$0 \quad 1 \quad 2 \quad 2 \quad 3 \quad 4 \quad 4 \quad 6 \quad 7 \quad 9 \quad 11$

Work out the interquartile range of the numbers of loaves. [2 marks]

[Total 2 marks]

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Question 10 - Exam Solution

Understanding the Question

Given

The numbers of unsold loaves on

11

days, already written in ascending order:

0, 1, 2, 2, 3, 4, 4, 6, 7, 9, 11

A list, not a table, so every value counts once - including the repeated

2

and the repeated

4

Find

The interquartile range,

IQR = Q_3 - Q_1

, where

Q_1

is the lower quartile and

Q_3

is the upper quartile. One number, and it is a spread, not an average - expect it to be smaller than the range.

Plan the Solution

Check the list really is in order, then count how many values there are: $n = 11$ .
Use the position rule. With $n = 11$ every quartile position comes out a whole number, so no averaging of two values is needed.
Find $Q_1$ and $Q_3$ by counting along the list to those positions - the mark scheme gives the method mark for these two values alone.
Subtract: $Q_3 - Q_1$ .

Worked Solution [2 marks]

Rule - Quartiles by position: for

n

values written in order,

Q_1

is the

\dfrac{n+1}{4}

th value, the median is the

\dfrac{n+1}{2}

th value,

Q_3

is the

\dfrac{3(n+1)}{4}

th value, and

IQR = Q_3 - Q_1

Step 1: Put the values in order and count them

0, 1, 2, 2, 3, 4, 4, 6, 7, 9, 11

n = 11

(Reason: (Reason: quartiles are positions in an ORDERED list. This list is already ascending, so nothing has to be moved - but the count

n = 11

is what every position below is worked out from.))

Step 2: Find the middle value

\dfrac{11+1}{2} = 6

\text{median} = \text{6th value} = 4

(Reason: (Reason: the median splits the list into

5

values below it and

5

above it, which is what makes the quartile positions land on whole numbers. The mark scheme accepts the median

4

as evidence of the method.))

Step 3: Find the lower quartile

\dfrac{11+1}{4} = 3

Q_1 = \text{3rd value} = 2

(Reason: (Reason:

\dfrac{n+1}{4}

gives a POSITION, not a value. Counting along the list to that position is what turns it into the lower quartile.))

Step 4: Find the upper quartile

\dfrac{3(11+1)}{4} = 9

Q_3 = \text{9th value} = 7

(Reason: (Reason: three quarters of the way along, so the

9

th of the

11

values. Counting from the other end it is the

3

rd from the top, which is a quick way to check it.))

Step 5: Subtract the quartiles

IQR = Q_3 - Q_1 = 7 - 2 = 5

(Reason: (Reason: the interquartile range is the width of the middle half of the data, so it is always the upper quartile MINUS the lower quartile - never the two added, and never the median subtracted from anything.))

Interquartile range =

5

loaves

Verification

Check 1: Use the other standard method: delete the median

4

and take the middle of each half. Lower half

0, 1, 2, 2, 3

has middle

2

; upper half

4, 6, 7, 9, 11

has middle

7

7 - 2 = 5

Check 2: Count in from both ends independently, so neither quartile is found from the other. The

3

rd value up from the bottom is

2

, and the

3

rd value down from the top is

7

7 - 2 = 5

Check 3: Sanity check the size. The range is

11 - 0 = 11

, and the interquartile range measures only the middle half, so it must come out smaller than the range.

5 < 11

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Both quartiles correctly identified: $2$ and $7$	M1	Method mark for reading off both quartiles. Working that identifies the median $(4)$ may also be seen and is accepted as part of the method.	✓
$7 - 2 = 5$	A1	cao. A correct answer of $5$ scores full marks, unless it comes from obviously incorrect working.	✓

Full marks: 2/2

Two numbers, $A$ and $B$ , are written below as products of their prime factors.

$A = 2^{5} \times 5 \times 7^{2}$
$B = 2^{3} \times 5^{3} \times 7^{4}$

(a) Write down the highest common factor (HCF) of $5A$ and $2B$
Write your answer as a product of prime factors. [2 marks]

$A = 2^{5} \times 5 \times 7^{2}$
$B = 2^{3} \times 5^{3} \times 7^{4}$

(b) Work out the value of $(AB)^{2}$
Write your answer as a product of prime factors. [2 marks]

(a)(b)

[Total 4 marks]

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Question 11 - Exam Solution

Understanding the Question

Given

A = 2^{5} \times 5 \times 7^{2}

B = 2^{3} \times 5^{3} \times 7^{4}

Both numbers are already broken into their prime factors, so no factor tree is needed.

Find

(a) The highest common factor of

5A

and

2B

, written as a product of prime factors (b) The value of

(AB)^{2}

, written as a product of prime factors

Plan the Solution

Multiply each given number by its extra factor first: the $5$ in $5A$ raises the index of $5$ by one, and the $2$ in $2B$ raises the index of $2$ by one.
For the HCF, compare the two lists prime by prime and keep the lower power each time.
For part (b), build $AB$ by adding the indices of each prime, then square the result by doubling every index.
Never multiply the numbers out: the answer is wanted as a product of prime factors, and the indices carry all the information.

Worked Solution [4 marks]

Rule - Index laws on prime factors: to multiply two numbers, add the indices of each prime; to square a number, double every index; and for the HCF, keep the LOWEST power of each prime that appears in both numbers.

Step 1: Write

5A

and

2B

as products of prime factors

5A = 5 \times 2^{5} \times 5 \times 7^{2} = 2^{5} \times 5^{2} \times 7^{2}

2B = 2 \times 2^{3} \times 5^{3} \times 7^{4} = 2^{4} \times 5^{3} \times 7^{4}

(Reason: The extra

5

joins the one

5

already in

A

, giving

5^{2}

; the extra

2

joins the

2^{3}

B

, giving

2^{4}

Step 2: Keep the lowest power of each prime

\text{lower of } 2^{5} \text{ and } 2^{4} \text{ is } 2^{4}

\text{lower of } 5^{2} \text{ and } 5^{3} \text{ is } 5^{2}

\text{lower of } 7^{2} \text{ and } 7^{4} \text{ is } 7^{2}

2^{4} \times 5^{2} \times 7^{2} = 19\,600

(Reason: A common factor cannot use more copies of a prime than the smaller supply allows, so

2^{4}

is as far as the twos go. The product is the HCF; the numerical value

19\,600

is shown only as a check, since the answer is wanted in index form.)

Step 3: Write

AB

as a product of prime factors

AB = 2^{5+3} \times 5^{1+3} \times 7^{2+4} = 2^{8} \times 5^{4} \times 7^{6}

(Reason: Every prime in

A

meets the same prime in

B

, and when two powers of one prime are multiplied together the indices are combined by ADDING them, never by multiplying them.)

Step 4: Square

AB

by doubling every index

(AB)^{2} = 2^{8 \times 2} \times 5^{4 \times 2} \times 7^{6 \times 2} = 2^{16} \times 5^{8} \times 7^{12}

(Reason: Squaring is multiplying the number by itself, so each index is added to a copy of itself, which doubles it:

8

becomes

16

4

becomes

8

and

6

becomes

12

(a)

2^{4} \times 5^{2} \times 7^{2}

(b)

2^{16} \times 5^{8} \times 7^{12}

Verification

Check 1: Divide both numbers by the HCF and see whether anything is left in common.

5A = 39\,200

and

2B = 4\,802\,000

\dfrac{39\,200}{19\,600} = 2

and

\dfrac{4\,802\,000}{19\,600} = 245

, and

2

and

245

share no factor, so nothing more can be taken out.

Check 2: Use the HCF and LCM together. The LCM takes the HIGHEST power of each prime, giving

2^{5} \times 5^{3} \times 7^{4} = 9\,604\,000

, and HCF times LCM must equal the product of the two numbers.

19\,600 \times 9\,604\,000 = 39\,200 \times 4\,802\,000

, so the pair of answers is consistent.

Check 3: Square each number first instead of multiplying first:

A^{2} = 2^{10} \times 5^{2} \times 7^{4}

and

B^{2} = 2^{6} \times 5^{6} \times 7^{8}

2^{10} \times 5^{2} \times 7^{4} \times 2^{6} \times 5^{6} \times 7^{8} = 2^{16} \times 5^{8} \times 7^{12}

, the same answer by a different order of working.

Check 4: A square number has an even index on every prime, because each index has been doubled.

16

8

and

12

are all even, so the answer really is a perfect square.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) Correct HCF as a product of prime factors	B2	$2^{4} \times 5^{2} \times 7^{2}$ or equivalent, for example $2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 7 \times 7$	✓
(a) Partially correct HCF	B1	For $2^{m} \times 5^{n} \times 7^{p}$ with two of $m = 4$ , $n = 2$ , $p = 2$ , or for $19\,600$ without sight of the correct factorisation, or a fully correct Venn diagram for $5A$ and $2B$ , or an answer of $2^{3} \times 5 \times 7^{2}$ or equivalent	✓
(b) A correct product of prime factors on the way to the answer	M1	For any correct product of prime factors, for example $AB = 2^{8} \times 5^{4} \times 7^{6}$ , or $A^{2} = 2^{10} \times 5^{2} \times 7^{4}$ , or $B^{2} = 2^{6} \times 5^{6} \times 7^{8}$ , or for $(AB)^{2}$ evaluated as $3.5 \ldots \times 10^{20}$ , oe	✓
(b) Correct answer as a product of prime factors	A1	$2^{16} \times 5^{8} \times 7^{12}$ or equivalent. A correct answer scores full marks unless it comes from obviously incorrect working.	✓
(b) Special case, if no other marks are earned	SC B1	For $2^{c} \times 5^{d} \times 7^{f}$ with two of $c = 16$ , $d = 8$ , $f = 12$ . The error behind it is doubling only some of the indices.	✓

Full marks: 4/4

Solve the simultaneous equations
$4x + 3y = 9.6$
$6x + 5y = 16.8$
You must show clear algebraic working. [4 marks]

x =y =

[Total 4 marks]

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Question 12 - Exam Solution

Understanding the Question

Given

4x + 3y = 9.6

6x + 5y = 16.8

Two linear equations in the same two unknowns, so one solution pair is expected.

Find

The value of

x

and the value of

y

that satisfy both equations at once.

Plan the Solution

Both equations are linear, so use elimination: scale them until one unknown has the same coefficient in each.
The $x$ terms are $4x$ and $6x$ , and the lowest common multiple of $4$ and $6$ is $12$ , so multiply the first equation by $3$ and the second by $2$ .
Both scaled equations then have $+12x$ , so subtracting removes $x$ and leaves $y$ on its own.
Substitute that value back into an original equation to get $x$ , then test the pair in both original equations.

Worked Solution [4 marks]

Rule - Elimination: multiply each equation so that one unknown has the same coefficient in both, subtract to remove that unknown, then substitute the value found back into an original equation.

Step 1: Match the

x

terms

4x + 3y = 9.6 \implies 12x + 9y = 28.8

6x + 5y = 16.8 \implies 12x + 10y = 33.6

(Reason: (Reason: multiplying every term of the first equation by

3

and every term of the second by

2

makes both

x

coefficients

12

. Multiplying an equation through by a number does not change its solutions.))

Step 2: Subtract to remove

x

(12x + 10y) - (12x + 9y) = 33.6 - 28.8

y = 4.8

(Reason: (Reason: the

12x

terms are identical and cancel, and

10y - 9y = y

, so one subtraction leaves

y

alone.))

Step 3: Substitute

y = 4.8

into the first equation

4x + 3 \times 4.8 = 9.6

4x + 14.4 = 9.6

4x = 9.6 - 14.4 = -4.8

x = \dfrac{-4.8}{4} = -1.2

(Reason: (Reason: replacing

y

with

4.8

leaves an equation in

x

alone. The

14.4

moves across the equals sign, so it changes sign, and dividing by

4

finishes it.))

x = -1.2

y = 4.8

Verification

Check 1: Put

x = -1.2

and

y = 4.8

back into the first equation.

4 \times (-1.2) + 3 \times 4.8 = -4.8 + 14.4 = 9.6

Check 2: The same pair must also satisfy the second equation, which was never used to find

y

6 \times (-1.2) + 5 \times 4.8 = -7.2 + 24 = 16.8

Check 3: Eliminate the other way round: multiply the first equation by

5

and the second by

3

to remove

y

instead, giving

20x + 15y = 48

and

18x + 15y = 50.4

2x = 48 - 50.4 = -2.4

, so

x = -1.2

as before.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Multiply one or both equations so that one unknown has matching coefficients, with the correct operation selected, for example $12x + 9y = 28.8$ with $12x + 10y = 33.6$ , or $20x + 15y = 48$ with $18x + 15y = 50.4$ .	M1	For multiplication of one or both equations with the correct operation selected (allow one arithmetic error). If the operation is not shown, assume it is the one that at least $2$ of the $3$ terms have been calculated for. A correct rearrangement of one equation with substitution into the other earns this mark instead.	✓
First unknown found: $x = -1.2$ or $y = 4.8$ .	A1	Either value, or equivalent. dep on M1	✓
Substitute the value found into one of the original equations, or repeat the scaling and subtraction for the second unknown.	dM1	For substitution of the found variable, or for repeating the steps of the first M1 for the second variable. dep on M1	✓
Both values stated, with the algebraic working shown: $x = -1.2$ and $y = 4.8$ .	A1	Both values, or equivalent. Working required. dep on M1	✓

Full marks: 4/4

$A$ , $B$ , $C$ and $D$ are four points on a circle whose centre is $O$ .
Angle $BCD$ is $128^{\circ}$ .

Work out the size of angle $OBD$ .
Give a reason for each stage of your working. [5 marks]

angle OBD =

[Total 5 marks]

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Question 13 - Exam Solution

Understanding the Question

Given

A

B

C

and

D

lie on a circle with centre

O

Angle

BCD

128^{\circ}

, marked at

C

on the diagram

OB

and

OD

are radii of the same circle, so they are equal in length

Find

The size of angle

OBD

, in degrees A reason for every stage of the working, not just the arithmetic

Plan the Solution

The four vertices of $ABCD$ all sit on the circle, so it is a cyclic quadrilateral. Its opposite angles add up to $180$ degrees, and that turns the marked angle at $C$ into angle $BAD$ .
Angle $BAD$ and angle $BOD$ stand on the same arc $BD$ , one at the circumference and one at the centre, so the angle at the centre is double the other one.
Triangle $OBD$ has two sides that are radii, so it is isosceles. Sharing what is left of $180$ degrees between its two equal base angles gives the answer.

Worked Solution [5 marks]

Rule - three circle facts in a chain: opposite angles of a cyclic quadrilateral add to

180

degrees; the angle at the centre is twice the angle at the circumference standing on the same arc; and the base angles of an isosceles triangle are equal.

Step 1: find angle

BAD

180 - 128 = 52

(Reason:

ABCD

is a cyclic quadrilateral, so angle

BAD

and angle

BCD

are opposite angles and add up to

180

degrees)

Step 2: find the obtuse angle

BOD

at the centre

2 \times 52 = 104

(Reason: angle

BAD

and angle

BOD

both stand on arc

BD

, one at the circumference and one at the centre, and the angle at the centre is twice the angle at the circumference)

Step 3: share the rest of triangle

OBD

\dfrac{180 - 104}{2} = 38

(Reason:

OB

and

OD

are radii, so triangle

OBD

is isosceles; angle

OBD

and angle

ODB

are equal base angles, so they share what is left of

180

degrees equally)

angle

OBD

38^{\circ}

Verification

Check 1 - do the three angles of triangle $OBD$ add up: The triangle has the apex angle at

O

and the two equal base angles at

B

and

D

. Add all three.

104 + 38 + 38 = 180

Check 2 - go round by the reflex angle instead: Double angle

BCD

to get the reflex angle at the centre, take that from

360

to get the obtuse angle

BOD

, then halve what is left of the triangle.

2 \times 128 = 256

, then

360 - 256 = 104

, and

\dfrac{180 - 104}{2} = 38

Check 3 - run the chain backwards from the answer: Start at

38

degrees, rebuild the angle at the centre, halve it for the angle at the circumference, and see whether the marked angle at

C

comes back.

180 - 2 \times 38 = 104

, then

\dfrac{104}{2} = 52

, and

180 - 52 = 128

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Angle $BAD$ : $180 - 128 = 52$	M1	Award for $180 - 128$ , or for the reflex angle route $2 \times 128 = 256$ . Angles must be identified either by notation or by being correctly positioned on the diagram.	✓
Obtuse angle $BOD$ : $2 \times 52 (= 104)$	M1 dep	Dependent on the first M1. Award for $2 \times 52$ , or for $360 - 256 (= 104)$ on the reflex route. The scheme prints both of those values in quotation marks, so a candidate's own earlier angle follows through. Angles must again be identified by notation or by position on the diagram.	✓
Angle $OBD$ = $38$	A1	A correct answer of $38$ scores the first three marks on its own, unless it comes from obviously incorrect working.	✓
All three reasons stated correctly for the method used	B2	Dependent on a fully correct method for finding angle $OBD$ . The reasons are: opposite angles of a cyclic quadrilateral sum to $180$ degrees; the angle at the centre is twice the angle at the circumference, or equally the angle at the circumference is half the angle at the centre; the base angles of an isosceles triangle are equal and the angles in a triangle add to $180$ degrees. On the reflex route, angles around a point adding to $360$ degrees replaces the cyclic quadrilateral reason.	✓
Only one correct circle theorem given for their method	(B1)	Dependent on the first M1. One correct circle theorem for their own method earns one of the two reasoning marks.	✓
The reflex route is worth full marks	Note	A candidate who writes reflex $BOD$ first is on the mark scheme's second route: $2 \times 128 = 256$ then $360 - 256 = 104$ , which reaches the same obtuse angle at the centre.	✓

Full marks: 5/5

Continue to questions 14 to 23

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

Frequently asked questions

There are 23 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 16 May 2024: Worked Solutions and Mark Schemes

Worked solutions, questions 1 to 13 of 23

Question 1, Calculator allowed

Question 1 - Exam Solution

Question 2, Calculator allowed

Question 2 - Exam Solution

Question 3, Calculator allowed

Question 3 - Exam Solution

Question 4, Calculator allowed

Question 4 - Exam Solution

Question 5, Calculator allowed

Question 5 - Exam Solution

Question 6, Calculator allowed

Question 6 - Exam Solution

Question 7, Calculator allowed

Question 7 - Exam Solution

Question 8, Calculator allowed

Question 8 - Exam Solution

Question 9, Calculator allowed

Question 9 - Exam Solution

Question 10, Calculator allowed

Question 10 - Exam Solution

Question 11, Calculator allowed

Question 11 - Exam Solution

Question 12, Calculator allowed

Question 12 - Exam Solution

Question 13, Calculator allowed

Question 13 - Exam Solution

Frequently asked questions

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