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Edexcel IGCSE 4MA1 Paper 1FR, November 2024: Worked Solutions, Questions 14 to 24

Sir Faraz Hassan

1 Aug 2026

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Edexcel International GCSE Mathematics A (4MA1)Paper 1FR - Foundation Tier - November 2024100 marks · 2 hours · Calculator allowed

Back to questions 1 to 13

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

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

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

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

Download printable PDF

All 24 questions with a full worked solution and mark scheme - free PDF

Worked solutions, questions 14 to 24 of 24

Use your calculator to work out the value of

$\dfrac{\sqrt{17.8 \times 19.2}}{3.4^{2} \times 0.23}$

Write down all the figures on your calculator display. [2 marks]

[Total 2 marks]

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Question 14 - Exam Solution

Understanding the Question

Given

The calculation

\dfrac{\sqrt{17.8 \times 19.2}}{3.4^{2} \times 0.23}

, to be done on a calculator.

A square root on the top of the fraction, and a square followed by a multiplication on the bottom.

Find

The value of the expression. Every figure the calculator display shows, so the answer is not rounded.

Plan the Solution

Work out the top and the bottom separately, then divide one by the other.
Top: multiply $17.8$ by $19.2$ first, then take the square root of the product.
Bottom: square $3.4$ first, then multiply by $0.23$ .
Keep the unrounded value on the display all the way through, and copy the final display in full.

Worked Solution [2 marks]

Rule - a fraction bar is a division that acts on the whole of the top and the whole of the bottom, so each one is completed before dividing; and inside the bottom,

3.4^{2}

is squared before it is multiplied by

0.23

Step 1: multiply the two numbers under the square root

17.8 \times 19.2 = 341.76

(Reason: The root sign covers the whole product, so the multiplication is done first and the root is taken of the single number it gives.)

Step 2: take the square root to get the numerator

\sqrt{341.76} = 18.48675201\ldots

(Reason: Leave this on the display or store it in the calculator's memory. Rounding here would change the later figures, and the question asks for all of them.)

Step 3: work out the denominator

3.4^{2} = 11.56

11.56 \times 0.23 = 2.6588

(Reason: The square is worked out before the multiplication, following the order of operations. Squaring the whole product instead would give a different bottom line.)

Step 4: divide the numerator by the denominator

\dfrac{18.48675201\ldots}{2.6588} = 6.953043483\ldots

(Reason: The division is done last, using the unrounded numerator from Step 2.)

Step 5: write down every figure on the display

6.953043483

(Reason: The question asks for all the figures shown, so nothing is rounded off. A display with room for more figures shows

6.95304348325

, and any answer that starts

6.95304

and continues with the display's own figures is the same answer.)

6.953043483

Verification

Check 1: Reverse the division. Multiplying the answer by the denominator must give the numerator back.

6.953043483 \times 2.6588 = 18.48675201\ldots

, which is the numerator from Step 2

Check 2: Estimate with easy numbers:

\sqrt{18 \times 19} = \sqrt{342} \approx 18.5

on the top, and

3.4^{2} \times 0.23 \approx 11.6 \times 0.23 \approx 2.7

on the bottom.

\dfrac{18.5}{2.7} \approx 6.9

, which is close to

6.95

, so the answer is the right size

Check 3: Undo both operations at once. Squaring the answer and multiplying by

2.6588^{2}

must return the number that was under the root.

6.953043483^{2} \times 2.6588^{2} \approx 341.76

, which is

17.8 \times 19.2

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$18.4(867\ldots)$ or $2.65(88)$ or $6.95$ or $6.953$ or $6.9530$	M1	For a correct numerator $\sqrt{17.8 \times 19.2} = 18.4(867\ldots)$ , or a correct denominator $3.4^{2} \times 0.23 = 2.65(88)$ , or the correct answer rounded or truncated to $3$ , $4$ or $5$ significant figures. The figures in brackets need not be shown.	✓
$6.95304(3483\ldots)$	A1	Correct answer scores full marks (unless from obvious incorrect working). The bracketed figures are the ones a display may or may not show, so $6.953043483$ and any longer reading of the same display are accepted.	✓

Full marks: 2/2

The diagram shows a wooden block and a storage trunk with a lid.

The block is a cube of side $5$ cm.
Nathan has a large supply of these blocks.

The inside of the trunk is a cuboid measuring $27$ cm by $35$ cm by $40$ cm.

Nathan packs as many blocks as possible into the trunk so that the lid will still shut.

Work out the volume of the space inside the trunk that is not filled with blocks. [4 marks]

cm³

[Total 4 marks]

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Question 15 - Exam Solution

Understanding the Question

Given

A block that is a cube of side

5

cm, and a supply of identical blocks.

A trunk whose inside is a cuboid measuring

27

cm by

35

cm by

40

cm.

As many blocks as possible are packed in, and the lid must still shut, so no block may stick out.

Find

The volume of the space inside the trunk that no block fills. The answer is a volume, so it is measured in

\text{cm}^3

Plan the Solution

Divide each inside length by the length of a block to see how many blocks fit along it, and round each answer DOWN, because part of a block cannot be packed.
Multiply the three counts together for the number of blocks that fit.
Work out the volume of one block, then the volume of all of them.
Take the volume of the blocks away from the volume of the trunk.

Worked Solution [4 marks]

Rule - blocks only fit a whole number of times along an edge, so every count is rounded down; then the volume of space left is the volume of the trunk take away the total volume of the blocks.

Step 1: how many blocks fit along each inside length

\dfrac{27}{5} = 5.4

\dfrac{35}{5} = 7

\dfrac{40}{5} = 8

(Reason: Only whole blocks can be packed, so each count is rounded down:

5

blocks along the

27

cm side,

7

along the

35

cm side and

8

along the

40

cm side.)

Step 2: the number of blocks that fit in the trunk

5 \times 7 \times 8 = 280

(Reason: The blocks fill complete rows, complete columns and complete layers, so the three counts multiply together.)

Step 3: the volume the blocks take up

5 \times 5 \times 5 = 125 \text{ cm}^3

280 \times 125 = 35\,000 \text{ cm}^3

(Reason: Each block is a cube, so its volume is

5

cubed, and every packed block has that same volume.)

Step 4: the volume inside the trunk

27 \times 35 \times 40 = 37\,800 \text{ cm}^3

(Reason: The INSIDE lengths are the ones to use, because it is the space inside the trunk that is being filled.)

Step 5: take the blocks away from the trunk

37\,800 - 35\,000 = 2800 \text{ cm}^3

(Reason: What is left is the space no block reaches: a gap of

2

cm right across the top, because five layers of blocks stand only

25

cm tall.)

2800 \text{ cm}^3

Verification

Check 1: Look at the gap at the top instead. Five layers of blocks stand

25

cm tall, so

27 - 25 = 2

cm of height is empty right across the floor area of the trunk.

2 \times 35 \times 40 = 2800 \text{ cm}^3

, the same answer from a completely different starting point

Check 2: Count in block-sized spaces. The trunk has room for

\dfrac{37\,800}{125} = 302.4

block-sized spaces, but only

280

of them actually hold a block.

302.4 - 280 = 22.4

spaces are empty, and

22.4 \times 125 = 2800

Check 3: Put the two volumes back together. The blocks and the empty space must fill the inside of the trunk exactly.

35\,000 + 2800 = 37\,800 \text{ cm}^3

, which is the volume worked out in Step 4

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{35}{5} = 7$ or $\dfrac{27}{5} = 5.4$ or $\dfrac{40}{5} = 8$ or $5 \times 5 \times 5 = 125$ or $27 \times 35 \times 40 = 37\,800$ or $25 \times 35 \times 40 = 35\,000$	M1	For finding the number of blocks that fit along one inside length of the trunk, or the volume of one block, or the volume taken up by the blocks.	✓
$5 \times 7 \times 8 = 280$ and $5 \times 5 \times 5 = 125$ , or $27 - 5 \times 5 = 2$ , or $27 \times 35 \times 40 = 37\,800$ and $25 \times 35 \times 40 = 35\,000$ , or $\dfrac{37\,800}{125} - 5 \times 7 \times 8 = 22.4$	M1	For the total number of blocks that will fit together with the volume of one block; or the number of centimetres of height left with no block in it; or the volume of the trunk together with the total volume of the blocks; or the space left at the top written as a number of blocks. Note that a count of $6$ layers along the $27$ cm side is not a count of blocks that fit: those layers would stand $30$ cm tall and the lid would not shut.	✓
$37\,800 - 280 \times 125$ or equivalent, e.g. $37\,800 - 35\,000$ or $2 \times 35 \times 40$ or $22.4 \times 125$	M1	For a fully correct method to find the volume of the space left. The official scheme prints the values of this row in quotation marks, which is its way of saying that the candidate's own earlier values may be used, provided the method itself is complete.	✓
$2800$	A1	cao. A correct answer scores full marks unless it comes from obviously incorrect working. The unit is given on the answer line, so it need not be repeated.	✓

Full marks: 4/4

Here are six tiles.
Five of the tiles have a number printed on them.

Work out the number that should be printed on the last tile so that the mean of the six numbers will be $11$ [3 marks]

[Total 3 marks]

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Question 16 - Exam Solution

Understanding the Question

Given

Six tiles in a row. Five of them are printed with

16

15

3

2

and

9

The mean of all six numbers has to be

11

The sixth tile is blank, so its number is the only thing that can be chosen.

Find

The number that belongs on the blank tile.

Plan the Solution

A mean of $11$ spread over $6$ numbers fixes what those numbers must add up to, so start with that total.
Add the five numbers that are already printed.
Whatever is left over from the total is the number for the blank tile.

Worked Solution [3 marks]

Mean:

\text{total} = \text{mean} \times \text{number of values}

, so a known mean can always be turned back into a total.

Step 1: turn the mean into the total the six numbers must make

11 \times 6 = 66

(Reason: The mean is the total shared equally between the tiles, so multiplying the mean by how many numbers there are gives that total back.)

Step 2: add the five numbers that are already printed

16 + 15 + 3 + 2 + 9 = 45

(Reason: These five cannot be changed, so their sum is the part of the

66

that is already on the tiles.)

Step 3: take the five away from the total

45 + x = 66

66 - 45 = 21

(Reason: The blank tile has to make up the difference between what the five printed numbers give and the total the mean demands.)

21

Verification

Check 1: Print

21

on the last tile, add all six numbers, then divide by

6

16 + 15 + 3 + 2 + 9 + 21 = 66

and

\dfrac{66}{6} = 11

, which is the mean asked for.

Check 2: Measure each printed number against

11

instead of totalling:

16

5

above,

15

4

above,

3

8

below,

2

9

below and

9

2

below. Those five sit

10

below the mean altogether, so the last number must sit

10

above it:

11 + 10 = 21

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Work out the total the six numbers need, or write the mean as an equation	M1	$11 \times 6 = 66$ or $\dfrac{16 + 15 + 3 + 2 + 9 + x}{6} = 11$ - a correct calculation for the total, or a correct equation for the last tile using $x$ .	✓
Reach an equation with no fraction in it, or the subtraction that gives the last number	M1	$16 + 15 + 3 + 2 + 9 + x = 66$ oe, eg $45 + x = 66$ or $66 - (16 + 15 + 3 + 2 + 9)$ .	✓
The number printed on the last tile	A1	$21$ - a correct answer scores full marks unless it comes from obviously incorrect working. If the answer line is blank, check the tile.	✓

Full marks: 3/3

A five-sided spinner is used in a board game.
The spinner is biased.

The table gives information about the probability that, when the spinner is spun once, it will land on each number.

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

Rosalind 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 17 - Exam Solution

Understanding the Question

Given

The spinner is biased, so the five numbers are not equally likely.

The probabilities of landing on

1

2

3

4

and

5

are

2x

0.27

0.04

x

and

0.12

The spinner is to be spun

400

times.

Find

An estimate for how many of the

400

spins land on an odd number. The odd numbers on this spinner are

1

3

and

5

Plan the Solution

Use the fact that the five probabilities have a total of $1$ to form an equation in $x$ .
Solve that equation, then double the answer, because the probability of landing on $1$ is $2x$ .
Add the three odd probabilities, then multiply that total by $400$ .

Worked Solution [4 marks]

Rule - Expected frequency: the probabilities of all the possible outcomes have a total of

1

, and an estimate for the number of times an event happens is its probability multiplied by the number of trials.

Step 1: Write down what the five probabilities must total

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

(Reason: Every spin lands on exactly one of the five numbers, so the five probabilities have a total of

1

Step 2: Collect the like terms

0.27 + 0.04 + 0.12 = 0.43

3x + 0.43 = 1

(Reason: The three known probabilities come to

0.43

, and

2x

and

x

combine to give

3x

Step 3: Solve for

x

3x = 1 - 0.43 = 0.57

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

2x = 2 \times 0.19 = 0.38

(Reason: Take

0.43

from both sides, then divide by

3

. The table gives the probability of landing on

1

2x

, which is double this.)

Step 4: Add the probabilities of the odd numbers

0.38 + 0.04 + 0.12 = 0.54

(Reason: The odd numbers are

1

3

and

5

, so the three probabilities to add are

2x

0.04

and

0.12

Step 5: Multiply the probability by the number of spins

0.54 \times 400 = 216

(Reason: An estimate for how often an event happens is its probability multiplied by the number of trials, and the spinner is spun

400

times.)

216

Verification

Check 1 - do the five probabilities total 1: With

x = 0.19

the two unknown entries in the table are

0.38

and

0.19

, so the even probabilities come to

0.27 + 0.19 = 0.46

. Odd and even together give

0.54 + 0.46 = 1

, so nothing has been left out and nothing counted twice.

Check 2 - count the spins instead of the probabilities: Estimate the even outcomes directly:

0.27 \times 400 = 108

for the

2

and

0.19 \times 400 = 76

for the

4

. The even spins come to

108 + 76 = 184

, so

400 - 184 = 216

spins are left for the odd numbers.

Check 3 - is the size sensible? A little over half of the probability sits on the odd numbers, so a little over half of the

400

spins should land on one of them. Half of

400

200

, and

216

is a little more than that.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
Use the total probability: $1 - (0.27 + 0.04 + 0.12) = 0.57$ or $2x + 0.27 + 0.04 + x + 0.12 = 1$	M1	for showing a clear understanding that the total of the probabilities is $1$ , or for finding estimates for the number of times the spinner will land on $2$ , $3$ and $5$	✓
Find the unknown entry: $\dfrac{0.57}{3} = 0.19$ , or $2x = 0.38$ , or $\dfrac{400 - 108 - 16 - 48}{3} = 76$	M1	for a method to find the value of $x$ or of $2x$ , or an estimate for the number of times the spinner will land on $4$ or on $1$	✓
A complete method: $(2 \times 0.19 + 0.04 + 0.12) \times 400$ or $2 \times 76 + 16 + 48$	M1	for a complete method	✓
The answer $216$	A1	for an answer of $216$ . An answer of $\dfrac{216}{400}$ scores M3A0.	✓
Note	note	A correct answer scores full marks unless it comes from obviously incorrect working.	✓

Full marks: 4/4

Alessandro sells plain bagels and sesame bagels on a market stall.

He sells a total of $200$ bagels such that

the number of plain bagels sold : the number of sesame bagels sold = $3 : 2$

Alessandro sells the plain bagels for £ $1.50$ each.
He sells the sesame bagels for £ $1.75$ each.

$40\%$ of the price of a plain bagel is profit.
$60\%$ of the price of a sesame bagel is profit.

Work out Alessandro's total profit when he sells all $200$ bagels. [5 marks]

[Total 5 marks]

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Question 18 - Exam Solution

Understanding the Question

Given

A total of

200

bagels are sold

plain : sesame =

3 : 2

A plain bagel costs £

1.50

and a sesame bagel costs £

1.75

Profit is

40\%

of a plain bagel's price, and

60\%

of a sesame bagel's price

Find

The total profit, in pounds, on all

200

bagels

Plan the Solution

Share the $200$ bagels in the ratio $3 : 2$ to find how many of each kind are sold.
Multiply each count by its own price, to get the money taken on the plain bagels and the money taken on the sesame bagels.
Take $40\%$ of the plain money and $60\%$ of the sesame money. The two percentages are different, so the two kinds must be kept apart.
Add the two profits. Taking one single percentage of the whole £ $320$ taken would not work, because neither $40\%$ nor $60\%$ applies to all of it.

Worked Solution [5 marks]

Rule - Ratio then percentage: divide by the total number of parts to get one share, multiply up to get each quantity, then take each percentage of its own amount and add.

Step 1: Find one share of the ratio

3 + 2 = 5

\dfrac{200}{5} = 40

(Reason: The ratio

3 : 2

splits the

200

bagels into

5

equal shares, so one share is

40

bagels.)

Step 2: Find how many of each kind are sold

3 \times 40 = 120 \text{ plain}

2 \times 40 = 80 \text{ sesame}

(Reason: Plain takes

3

of the shares and sesame takes

2

. Check that they rebuild the total:

120 + 80 = 200

Step 3: Work out the money taken on each kind

120 \times 1.50 = 180

80 \times 1.75 = 140

(Reason: The two kinds are priced differently, so their takings are worked out separately: £

180

from the plain bagels and £

140

from the sesame bagels.)

Step 4: Take each percentage of its own amount

0.4 \times 180 = 72

0.6 \times 140 = 84

(Reason: Write each percentage as a decimal first:

40\% = 0.4

and

60\% = 0.6

. The

40\%

belongs to the plain takings only, and the

60\%

to the sesame takings only.)

Step 5: Add the two profits

72 + 84 = 156

(Reason: The profit on the plain bagels and the profit on the sesame bagels together give Alessandro's total profit on all

200

bagels.)

156

Verification

Check 1: Work out the profit on ONE bagel of each kind first. A plain bagel gives

0.4 \times 1.50 = 0.60

and a sesame bagel gives

0.6 \times 1.75 = 1.05

. Then multiply by how many are sold:

120 \times 0.60 = 72

and

80 \times 1.05 = 84

72 + 84 = 156

, the same total from a completely different order of working

Check 2: Count how many bagels' worth of money is profit.

40\%

120

plain bagels is

48

bagels' worth, and

60\%

80

sesame bagels is also

48

bagels' worth. Now price those:

48 \times 1.50 = 72

and

48 \times 1.75 = 84

156

again, reached without ever finding the takings

Check 3: A size check. Altogether Alessandro takes

180 + 140 = 320

pounds. Part of that earns

40\%

profit and part earns

60\%

, so the answer must sit strictly between those two percentages of £

320

\dfrac{156}{320} = 0.4875

, which lies between

0.4

and

0.6

as it must

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{200}{3 + 2} = 40$	M1	for a method to find one 'share' of the ratio	✓
$3 \times 40 = 120$ and $2 \times 40 = 80$	M1	for a method to find the number of plain bagels and the number of sesame bagels	✓
$120 \times 1.50 = 180$ and $80 \times 1.75 = 140$	M1	for a method to find the money taken from the plain bagels and from the sesame bagels, or the number of bagels that are entirely profit ( $0.4 \times 120 = 48$ and $0.6 \times 80 = 48$ ), or the profit on a single plain bagel or a single sesame bagel ( $0.4 \times 1.50 = 0.60$ or $0.6 \times 1.75 = 1.05$ )	✓
$0.4 \times 180 = 72$ and $0.6 \times 140 = 84$	M1	for a complete method to find the total profit on the plain bagels and the total profit on the sesame bagels	✓
$72 + 84 = 156$	A1	cao. A correct answer scores full marks unless it follows obviously incorrect working. The answer is £ $156$ .	✓
Special case	SC	award SC B4 for an answer of $164$ or $174$ . An answer of $164$ comes from swapping the two percentages over, $0.6 \times 180 + 0.4 \times 140$ ; an answer of $174$ comes from reading the ratio the wrong way round, so $80$ plain bagels and $120$ sesame ones. This row carries no mark of its own.	✓

Full marks: 5/5

Show that $2\dfrac{1}{3}$ divided by $5\dfrac{1}{4}$ gives $\dfrac{4}{9}$ .
You must show your working. [3 marks]

[Total 3 marks]

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Question 19 - Exam Solution

Understanding the Question

Given

Two mixed numbers,

2\dfrac{1}{3}

and

5\dfrac{1}{4}

, and the first is to be divided by the second.

The result that has to be reached:

\dfrac{4}{9}

Find

A complete line of working that starts at

2\dfrac{1}{3}

and

5\dfrac{1}{4}

and finishes at

\dfrac{4}{9}

. This is a show that question, so the working is the answer. Quoting the result on its own, or reaching it only in decimals, earns nothing.

Plan the Solution

Write each mixed number as an improper fraction first. A mixed number cannot be divided as it stands, because the whole number and the fraction would have to be handled separately.
Replace the division by a multiplication. Dividing by a fraction is the same as multiplying by its reciprocal, so $\dfrac{21}{4}$ is turned upside down to give $\dfrac{4}{21}$ , and the first fraction is left alone.
Multiply the numerators together and the denominators together, then cancel the result down to its lowest terms and confirm it is $\dfrac{4}{9}$ .

Worked Solution [3 marks]

Dividing by a fraction means multiplying by its reciprocal, so

\dfrac{a}{b}

divided by

\dfrac{c}{d}

becomes

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

. Only the second fraction is turned upside down.

Step 1: Write each mixed number as an improper fraction

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

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

(Reason: Multiply the whole number by the denominator and add the numerator, keeping the same denominator. There are

2 \times 3 + 1 = 7

thirds in

2\dfrac{1}{3}

, and

5 \times 4 + 1 = 21

quarters in

5\dfrac{1}{4}

Step 2: Turn the division into a multiplication

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

(Reason: The reciprocal of

\dfrac{21}{4}

\dfrac{4}{21}

, so dividing by

\dfrac{21}{4}

is the same as multiplying by

\dfrac{4}{21}

. Only the second fraction is inverted; the first one stays exactly as it is.)

Step 3: Multiply the numerators, and multiply the denominators

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

(Reason: When two fractions are multiplied, the two numerators give the new numerator and the two denominators give the new denominator. There is no need for a common denominator here, because this is a multiplication and not an addition.)

Step 4: Cancel down to lowest terms

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

(Reason: Both numbers are multiples of

7

, since

28 = 7 \times 4

and

63 = 7 \times 9

. Dividing the top and the bottom by

7

leaves

\dfrac{4}{9}

, which is the result the question asked for.)

\dfrac{7}{3} \times \dfrac{4}{21} = \dfrac{28}{63} = \dfrac{4}{9}

, as required

Verification

Check 1: Multiplying undoes dividing, so multiplying the result by

5\dfrac{1}{4}

must give back

2\dfrac{1}{3}

\dfrac{4}{9} \times \dfrac{21}{4} = \dfrac{84}{36} = \dfrac{7}{3} = 2\dfrac{1}{3}

Check 2: The second route in the mark scheme: put both improper fractions over the common denominator

12

, after which the division is just one numerator over the other.

\dfrac{7}{3} = \dfrac{28}{12}

and

\dfrac{21}{4} = \dfrac{63}{12}

, giving

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

again.

Check 3: A calculator check in decimals:

2\dfrac{1}{3}

2.3333

and

5\dfrac{1}{4}

5.25

, each to 4 decimal places. The division gives

0.4444

, and

\dfrac{4}{9}

0.4444

to 4 decimal places, so the two agree. The mark scheme allows decimals as a check only, never in place of the fraction working.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
$\dfrac{7}{3}$ and $\dfrac{21}{4}$	M1	For both mixed numbers written as improper fractions. A candidate who writes $\dfrac{4}{21}$ rather than $\dfrac{21}{4}$ has still earned this mark, because inverting the second fraction is the next step anyway.	✓
$\dfrac{7}{3} \times \dfrac{4}{21}$ or equivalent, for example $\dfrac{49}{21} \times \dfrac{4}{21}$ , or $\dfrac{28}{12}$ and $\dfrac{63}{12}$	M1	For the intention to multiply the correct improper fraction by the inverted fraction, or for writing the two fractions over the same common denominator.	✓
$\dfrac{7}{3} \times \dfrac{4}{21} = \dfrac{28}{63} = \dfrac{4}{9}$ or equivalent, correctly shown	A1	For completing the working correctly to reach the required answer. Working is required, so the printed result copied out on its own scores nothing.	✓
Decimal working	Note	Ignore any decimals used as checking. A decimal answer may sit alongside the fraction working, but it cannot replace it.	✓

Full marks: 3/3

Miroslav puts $5200$ euros into a savings bond at his local credit union.
The bond runs for $4$ years and it pays $2.5\%$ per year compound interest.

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

euros

[Total 3 marks]

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Question 20 - Exam Solution

Understanding the Question

Given

Amount put in at the start:

5200

euros

Time in the bond:

4

years

Compound interest:

2.5\%

per year, so each year's interest is worked out on the balance at the start of that year

Find

The amount in the bond after

4

years, correct to the nearest euro

Plan the Solution

Turn the $2.5\%$ increase into a decimal multiplier.
Compound interest applies that multiplier once for every year, so raise it to the power $4$ rather than multiplying the interest by $4$ .
Round to the nearest euro only at the very end, so no accuracy is lost partway through.

Worked Solution [3 marks]

Rule - Compound growth:

\text{final amount} = \text{starting amount} \times (\text{yearly multiplier})^{n}

, where

n

is the number of years.

Step 1: Write the

2.5\%

increase as a multiplier

100\% + 2.5\% = 102.5\%

102.5\% = \dfrac{102.5}{100} = 1.025

(Reason: Adding

2.5\%

to a balance leaves

102.5\%

of it, and

102.5\%

as a decimal is

1.025

, so one year of interest is one multiplication.)

Step 2: Apply the multiplier once for each of the

4

years

\text{amount} = 5200 \times 1.025 \times 1.025 \times 1.025 \times 1.025

\text{amount} = 5200 \times 1.025^{4}

(Reason: With compound interest each year's interest is paid on the new balance, so the multiplier acts on the previous year's amount. Four years means four multiplications, which is written

1.025^{4}

Step 3: Work out the value

5200 \times 1.025^{4} = 5200 \times 1.103812890625 = 5739.82703125

(Reason: Keep the full figure the calculator gives here. Rounding partway through a compound interest calculation moves the final amount, because every later year is worked out from it.)

Step 4: Round to the nearest euro

5739.82703125 \approx 5740

(Reason: The first digit after the decimal point is

8

, which is

5

or more, so the whole-euro part rounds up from

5739

5740

5740

euros

Verification

Check 1: Build the balance one year at a time instead of using a power, multiplying by

1.025

four times.

5200 \to 5330 \to 5463.25 \to 5599.83125 \to 5739.82703125

Check 2: Add the four separate interest payments to the

5200

euros put in. They grow a little each year because the balance grows.

130 + 133.25 + 136.58125 + 139.99578125 = 539.82703125

, and

5200 + 539.82703125 = 5739.82703125

Check 3: Compare with simple interest, where the same

130

euros would be paid every year. Compound interest must come out a little higher, but not wildly higher.

5200 + 4 \times 130 = 5720

, and

5720 < 5739.83

, so the compound amount is above it by under

20

euros.

Check 4: Work backwards: undo the four years of growth and the starting amount should reappear.

\dfrac{5739.82703125}{1.025^{4}} = 5200

Mark Scheme Breakdown

Step	Mark	Description	Got it?
A method to find $2.5\%$ of $5200$ , or $102.5\%$ of $5200$ : $5200 \times 1.025 \ (= 5330)$ or $5200 \times 0.025 \ (= 130)$	M1	One year's growth, by either the multiplier or the interest itself. $(1 + 0.025)$ counts as $1.025$ .	✓
A complete method for all four years: $5330 \times 1.025 \ (= 5463.25)$ and $5463.25 \times 1.025 \ (= 5599.83\ldots)$ and $5599.83\ldots \times 1.025 \ (= 5739.8\ldots)$	M1	The scheme quotes each yearly amount, so this follows through on the candidate's own earlier figures provided the method is a complete four-year one.	✓
Alternative to the two method marks above, in one line: $5200 \times 1.025^{4}$ , or $5200 \times 1.025^{5} \ (= 5883\ldots)$	M2	The single-power method earns both method marks at once. The $1.025^{5}$ version still shows compound growth, so it earns the method marks even though counting $5$ years loses the accuracy mark.	✓
$5740$	A1	Accept anything from $5739$ to $5740$ , which covers a candidate who truncates $5739.82\ldots$ instead of rounding it. A correct answer scores full marks unless it comes from obviously incorrect working.	✓
Special case, if no other marks are awarded: $5200 \times 0.1 \ (= 520)$ , $5200 \times 1.1 \ (= 5720)$ , $5200 \times 0.9 \ (= 4680)$ , $5200 \times 0.975 \ (= 5070)$ or $5200 \times 0.975^{4} \ (= 4699\ldots)$	B1	Each names one specific error. The $0.1$ and $1.1$ lines come from treating $4$ years at $2.5\%$ as a single $10\%$ ; the $0.9$ , $0.975$ and $0.975^{4}$ lines come from taking the interest off instead of adding it on.	✓
Accept $(1 + 0.025)$ as equivalent to $1.025$ throughout, but do not accept $(1 + 2.5\%)$ .	note	$(1 + 2.5\%)$ mixes a number with a percentage, so it is not a multiplier and earns nothing on its own.	✓

Full marks: 3/3

The diagram shows a solid cylinder turned from a single block of beech wood.

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 beech wood 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 21 - Exam Solution

Understanding the Question

Given

A solid cylinder of radius

8

cm and height

h

cm.

Its volume is

1208

cm³, so the volume is known and the height is not.

The wood has density

1.25

g/cm³, which means every cm³ of it has mass

1.25

Find

(a) The height

h

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

Plan the Solution

Put the radius and the volume into $V = \pi r^2 h$ and solve the equation for $h$ .
Keep the unrounded height on the calculator and round only at the very end of part (a), so no accuracy is lost partway through.
For part (b) use $\text{mass} = \text{density} \times \text{volume}$ . The density is in grams per cm³, so the mass arrives in grams and is changed to kilograms last.
Part (b) does not need the answer to part (a). The volume is already given as $1208$ cm³, so use that rather than a volume 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 radius is

8

cm, so the circular face has area

\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 has area

64\pi

cm², which is about

201

cm².)

Step 3: Rearrange to make

h

the subject

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

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

Step 4: Work out the height

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

(Reason: Use the calculator's own

\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 digit after the decimal point is

0

, which is less than

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: Every cm³ of the wood has mass

1.25

g and there are

1208

cm³ of it, so the mass is

1.25

g taken

1208

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

Step 7: Change the mass from grams to 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 the grams answer on its own 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 gives must reappear.

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

cm³

Check 2: Divide the mass in grams by the volume. Density is mass per unit volume, so the density given in the question must come back.

\dfrac{1510}{1208} = 1.25

g/cm³

Check 3: Convert the kilograms back to grams, which undoes the last step.

1.51 \times 1000 = 1510

Check 4: Is the size sensible? A height of exactly

6

cm on a face of about

201

cm² would give a volume a little below the one stated, so the true height must be a little above

6

6 \times 201 = 1206

, just under

1208

, so a height slightly above

6

is right and it still rounds to

6

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 it just as the single line does. Using the diameter, $16$ , instead of the radius does not.	✓
$6$	A1	Accept anything from $6$ to $6.02$ . The band exists because a candidate working with a rounded $\pi$ lands a little above the true height: $\dfrac{1208}{3.14 \times 64} = 6.011\ldots$ . 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 converting the density to $0.00125$ kg/cm³ first and multiplying by $1208$ , which reaches the kilograms in one line.	✓
$1.51$	A1	The answer must be in kilograms. $1510$ left in grams shows the method but not the conversion, 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 gives, $1208$ cm³, not from a volume rebuilt out of a 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 here, so it is not penalised, but the given volume is the one the mark scheme's own calculation uses.	✓

Full marks: 4/4

(a) Write $\dfrac{g^{9}}{g^{2}}$ as a single power of $g$ . [1 mark]

(b) Multiply out the brackets $5k^{2}(k^{3} + 4)$ . [2 marks]

(ii) Hence solve the equation $x^{2} - 2x - 63 = 0$ . [1 mark]

(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 22 - Exam Solution

Understanding the Question

Given

(a)

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

- one power of

g

divided by another power of the same letter.

(b)

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

- a single term standing outside a bracket.

(c)

x^{2} - 2x - 63

- a quadratic whose

x^{2}

coefficient is

1

(d)

7 - 2y < 3y - 12

- a linear inequality with

y

on both sides.

Find

(a) The quotient written as one power of

g

. (b) The expansion, as two terms. (c) (i) The two brackets, and (ii) the solutions that follow from them. A quadratic has two, so expect two values of

x

. (d) The values of

y

that satisfy the inequality, given as an inequality.

Plan the Solution

(a) Divide powers of one base by subtracting the indices.
(b) Multiply each term inside the bracket by $5k^{2}$ , adding indices wherever two powers of $k$ meet.
(c) (i) Hunt for two integers with product $-63$ and sum $-2$ . (ii) A product is zero only when one factor is zero, so read a solution off each bracket.
(d) Move the $y$ terms to the side that leaves a positive coefficient, then divide. Doing it that way means the inequality sign never has to be reversed.

Worked Solution [9 marks]

Index laws:

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

and

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

. For

x^{2} + bx + c

, two integers with product

c

and sum

b

give the brackets. An inequality is solved exactly like an equation, except that multiplying or dividing both sides by a negative number reverses the sign.

Step 1: subtract the indices in

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

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

= g^{7}

(Reason: Dividing powers of the same base subtracts the indices,

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

, because

2

of the

9

factors of

g

cancel and

7

survive.)

Step 2: multiply

k^{3}

5k^{2}

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

(Reason: Multiplying powers of the same base combines the two indices

2

and

3

into a single index, and the

5

is simply carried along.)

Step 3: multiply

4

5k^{2}

, then write both terms

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

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

(Reason: The

4

carries no

k

, so only the numbers multiply,

5 \times 4 = 20

, and the

k^{2}

is unchanged. Each term inside the bracket gets multiplied, so there are two terms in the answer.)

Step 4: find two integers with the right product and sum

7 \times (-9) = -63

7 + (-9) = -2

(Reason: For

x^{2} + bx + c

with

b = -2

and

c = -63

, the pair must multiply to

c

and add to

b

. A negative product means one integer is negative, and the negative one is the larger in size because the sum is negative.)

Step 5: write the factorised form

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

(Reason: The coefficient of

x^{2}

1

, so each bracket starts with

x

, and one integer of the pair goes into each bracket.)

Step 6: set each bracket equal to zero

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

x + 7 = 0 \quad \text{or} \quad x - 9 = 0

x = -7 \quad \text{or} \quad 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. This is why part (i) makes part (ii) short.)

Step 7: collect the

y

terms on the right

7 - 2y < 3y - 12

7 + 12 < 3y + 2y

19 < 5y

(Reason: Add

2y

to both sides and add

12

to both sides. Adding never changes the direction of the sign, and gathering the

y

terms on the right leaves

5y

, a positive coefficient.)

Step 8: divide both sides by

5

\dfrac{19}{5} < y

y > \dfrac{19}{5} = 3.8

(Reason: Dividing by the positive number

5

leaves the sign as it is. Reading the statement from the other end turns

\dfrac{19}{5} < y

into

y > \dfrac{19}{5}

, which is the form the answer line asks for.)

(a)

g^{7}

(b)

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

(x + 7)(x - 9)

x = -7

x = 9

(d)

y > \dfrac{19}{5}

, that is

y > 3.8

Verification

Check 1: Part (a) with a number in place of the letter. Put

g = 2

, so that

g^{9} = 512

and

g^{2} = 4

\dfrac{512}{4} = 128

and

2^{7} = 128

Check 2: Part (b) evaluated both ways. Put

k = 2

into the bracketed form and into the expansion.

5 \times 4 \times (8 + 4) = 240

and

5 \times 32 + 20 \times 4 = 240

Check 3: Part (c) both ways round. Multiply the brackets back out, then put each solution into the original quadratic.

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

, and

9^{2} - 2(9) - 63 = 0

(-7)^{2} - 2(-7) - 63 = 0

Check 4: Part (d) tested on either side of

3.8

. Substitute

y = 4

and then

y = 3

into the original inequality. At

y = 4

it reads

-1 < 0

, which is true. At

y = 3

it reads

1 < -3

, which is false.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
(a) $g^{7}$	B1	The quotient written as one power. No working is needed.	✓
(b) $5k^{5} + 20k^{2}$	B2	Both terms correct. Award B1 only for $5k^{5}$ or for $20k^{2}$ 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	Follow through from (c)(i), dependent on factorising in the form $(x + p)(x + q)$ where $p$ and $q$ are integers.	✓
(d) $7 + 12 < 3y + 2y$	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) $19 < 5y$ or $-5y < -19$	M1	For the correct simplification of the $y$ terms on one side and the numbers on the other in a correct inequality, or a correct inequality with the wrong sign. The sign may be $=$ or the incorrect inequality sign. Accept $y = \dfrac{19}{5}$ or equivalent here.	✓
(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

The diagram shows a trapezium $ABCD$ .

Angle $DAB$ and angle $ADC$ are right angles.
$AD = 15$ cm and $DC = 14$ 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 23 - Exam Solution

Understanding the Question

Given

Trapezium

ABCD

, with

AB

parallel to

DC

AD = 15

cm and

DC = 14

cm, with right angles at

A

and at

D

The area of the trapezium is

360

cm².

Find

The perimeter of the trapezium. The top side

AB

and the sloping side

CB

are both unknown, so the area has to give one before the other can be found.

Plan the Solution

Only $AB$ is missing from the area formula, so put the area of $360$ into it and solve for $AB$ .
Drop a perpendicular from $C$ onto $AB$ , meeting it at $M$ . That splits the trapezium into a rectangle and a right-angled triangle.
The triangle has short sides $15$ and $MB$ , so Pythagoras' theorem gives the sloping side $CB$ .
Add the four sides for the perimeter.

Worked Solution [6 marks]

Area of a trapezium: half the sum of the two parallel sides, times the distance between them,

\dfrac{a + b}{2} \times h

. Pythagoras' theorem in a right-angled triangle:

c^2 = a^2 + b^2

, where

c

is the hypotenuse.

Step 1: use the area to find

AB

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

AB + 14 = \dfrac{2 \times 360}{15} = 48

AB = 48 - 14 = 34

(Reason: The parallel sides are

AB

and

DC

, and

AD

is the distance between them because it is perpendicular to both. Multiplying by

2

and dividing by

15

undoes the formula.)

Step 2: drop a perpendicular from

C

onto

AB

AM = DC = 14

CM = AD = 15

MB = AB - AM

34 - 14 = 20

(Reason:

AMCD

has four right angles, so it is a rectangle and its opposite sides are equal. What is left of the top side is the base of the right-angled triangle

MCB

Step 3: Pythagoras' theorem in triangle

MCB

CB^2 = CM^2 + MB^2

15^2 + 20^2 = 225 + 400 = 625

CB = \sqrt{625} = 25

(Reason: Angle

CMB

is a right angle, so

CB

is the hypotenuse. Squaring the two short sides and adding gives

625

, and the square root of that is the length itself.)

Step 4: add the four sides

AD + DC + CB + BA

15 + 14 + 25 + 34 = 88

(Reason: The perimeter is the distance all the way round: the two lengths given in the question, the sloping side from Step 3, and the whole top side from Step 1.)

88

Verification

Check 1: Put

AB = 34

back into the area formula: half of

34 + 14

, times the height

15

\dfrac{48}{2} \times 15 = 360

, which is the area the question gives.

Check 2: The right-angled triangle has short sides

15

and

20

, and those are

5

times

3

and

5

times

4

, so it is a scaled up

3

4

5

triangle.

5 \times 5 = 25

, the same sloping side that Pythagoras' theorem gave.

Check 3: Rebuild the area a different way: rectangle

AMCD

15 \times 14

and triangle

MCB

is half of

15 \times 20

210 + 150 = 360

, so the split of the top side into

14

and

20

is right.

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 $MB = 20$ , where $M$ is the point on $AB$ for which $MC$ is perpendicular to $AB$	A1	Could be seen on the diagram.	✓
$(CB^2 =) \; 15^2 + 20^2 \; (= 625)$ , 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	Dependent on the previous two method marks. For a method to find the perimeter of the trapezium, allowing 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 $L$ has been drawn on the grid below.

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

[Total 3 marks]

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Question 24 - Exam Solution

Understanding the Question

Given

A straight line

L

drawn on a square grid

The grid runs from

-4

4

across and from

-2

4

up, one square to each unit

Find

An equation of

L

written in the form

y = mx + c

Plan the Solution

Pick two points where $L$ passes exactly through the corner of a grid square, so both coordinates can be read without estimating.
Work out the gradient $m$ from those two points.
Read $c$ off the graph where the line crosses the $y$ -axis.
Put $m$ and $c$ into $y = mx + c$ .

Worked Solution [3 marks]

Gradient and intercept: for a straight line,

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

, and

c

is the

y

-value where the line crosses the

y

-axis.

Step 1: read two points off the line

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

(Reason: The line passes exactly through the corner of a grid square at each of these two points, so both coordinates can be read off the grid instead of estimated. Any two exact points on the line will do.)

Step 2: work out the gradient

m

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

(Reason: Going from

(-4, 3)

(4, -1)

the line moves

8

squares to the right and

4

squares down, so the change in

y

goes on top of the change in

x

and the gradient is negative.)

Step 3: read the value of

c

c = 1

(Reason: The line crosses the

y

-axis at

(0, 1)

, and in

y = mx + c

the number

c

is that crossing value.)

Step 4: write the equation in the form

y = mx + c

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

(Reason: Putting

m = -\dfrac{1}{2}

and

c = 1

into

y = mx + c

gives the equation. The question asks for this exact form, so leave it written this way.)

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

Verification

Check 1: Put

x = 4

into the answer. It should give back the second point that was read off the grid,

(4, -1)

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

Check 2: A different point again: the line crosses the

x

-axis at

(2, 0)

, so putting

x = 2

in must give zero.

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

Check 3: Count squares on the grid. A gradient of

-\dfrac{1}{2}

means the line falls one square for every two squares it travels to the right. From

(0, 1)

going

2

right and

1

down reaches

(2, 0)

, which is on the line.

Mark Scheme Breakdown

Step	Mark	Description	Got it?
The gradient on its own, or one of the two numbers in the equation	B1	for $m = -\dfrac{1}{2}$ or gradient $= -\dfrac{1}{2}$ oe, eg $\dfrac{3 - (-1)}{-4 - 4}$ , or for $-\dfrac{1}{2}x + c$ , or for $mx + 1$ where $m$ is not zero	✓
Both numbers right, or a correct equation written the wrong way round	B2	for $-\dfrac{1}{2}x + 1$ without the $y =$ , or for $y = -\dfrac{1}{2}x + c$ , or for $y = mx + 1$ where $m$ is not zero, or for a correct equation in the wrong form, eg $2y + x = 2$	✓
The equation of $L$ in the form the question asks for	B3	for $y = -\dfrac{1}{2}x + 1$ oe, eg $y = -0.5x + 1$	✓
Note on the form of the answer	note	A correct equation written any other way, such as $2y + x = 2$ , earns B2 and not B3, because the question asks for the $y = mx + c$ form. Reading the gradient upside down, as $-2$ instead of $-\dfrac{1}{2}$ , scores no marks for the gradient.	✓

Full marks: 3/3

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.

past-papersedexcel4ma1igcseworked-solutionsmark-scheme

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Edexcel IGCSE 4MA1 Paper 1FR, November 2024: Worked Solutions, Questions 14 to 24

Worked solutions, questions 14 to 24 of 24

Question 14, Calculator allowed

Question 14 - Exam Solution

Question 15, Calculator allowed

Question 15 - Exam Solution

Question 16, Calculator allowed

Question 16 - Exam Solution

Question 17, Calculator allowed

Question 17 - Exam Solution

Question 18, Calculator allowed

Question 18 - Exam Solution

Question 19, Calculator allowed

Question 19 - Exam Solution

Question 20, Calculator allowed

Question 20 - Exam Solution

Question 21, Calculator allowed

Question 21 - Exam Solution

Question 22, Calculator allowed

Question 22 - Exam Solution

Question 23, Calculator allowed

Question 23 - Exam Solution

Question 24, Calculator allowed

Question 24 - Exam Solution

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