Maths Edexcel SKE

Kath Hipkiss

13/11/24

Aims of the session

• Identify key topics that students lost marks on in the GCSE 2024 exam at grade 4,5, 7
• Hot topics that always appear on the GCSE exam for grade 4/5 students and grade 7 students
• Approaches to maths interventions to ensure maximum outcomes
• Approaches to problem solving questions for low and middle attainers
• What more they can do to ensure their outcomes are above national average.

Hot topics that always appear on the GCSE exam for grade 4/5 students and grade 7 students

• Both tiers of papers had a mixed reception; students found Papers 1 and 3 particularly challenging, but had an easier time with Paper 2
• Foundation tier had a higher proportion of procedural questions and a lower proportion of multi-step unfamiliar problems compared to the “average” paper, whereas the Higher paper was much closer to average distributions
• As usual, some topics appeared twice or more across the series, and some topics weren’t assessed in depth. These include trigonometry and circle mensuration in Foundation, and systems of non-linear equations in Higher.

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Paper 1F

Summary

• Improve numeracy skills, including multiplication and division skills when working with integers, decimals, and fractions (including mixed numbers)
• The interpretation of problems involving ratio, particularly combined ratios
• Considering the context of the problem they are solving and would their answer be feasible
• Knowing, stating and applying angle facts
• Describing a single transformation
• Solving equations and inequalities and working with number machines and inverse number machines
• Processes for estimation (with emphasis of rounding a value to 1 significant figure) and identification of whether the chosen rounding leads to an overestimate or an underestimate of the answer
• Finding the equation of a given line
• The understanding of set notation
• Identification of different types of percentage question – including when reversed percentage change is required and how this should be performed
• Encouraging students to set out their working using logical steps
• Encouraging students to read the question carefully so that they understand what they are required to do and what format the answer should be stated in
• Interpretation of questions involving context and multiple step processes.

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Paper 2F

Summary

• Read questions carefully, including checking whether the magnitude of an answer is sensible, units are appropriate, and the level of accuracy required is shown
• Practise questions involving ratio and forming algebraic expressions
• Practise using equipment such as rulers, protractors and compasses
• Practise using calculator functions and negative numbers
• Use formal methods when working with fractions or percentages

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Paper 3F

Summary

• Learn the names of the parts of the circle
• Practise drawing pie charts from a frequency table
• Check their answers, in particular when completing two-way tables
• Practise drawing graphs of linear functions
• Understand the geometrical rules for angles and practise using these to answer longer, multi-step problems
• Practise using their calculators to deal with questions involving substitution
• Use tables and diagrams to extract information from a question to enable them to navigate through a problem
• Understand the difference between simple and compound interest
• Come prepared with the necessary equipment, particularly a calculator of which they are familiar
• Show their workings out especially when calculating percentages rather than just writing 15% of ... or 15% × ...
• Understand that the sum of the probabilities of all outcomes for an event is 1 and cannot be greater than 1
• Become more familiar with including appropriate units to answers
• Practise plans and elevations
• Practise articulating their mathematics competently, using the correct mathematical language. A regular practice of mathematical discussions would help in the writing of explanations
• Ensure incorrect working is properly crossed out and it is clear which response they wish the examiner to consider as their final workings/answer

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Paper 1H

• Communicate clearly when a decision and a reason are required

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• Develop an understanding of the purpose of rounding so that they can choose appropriate. rounded values when working out an estimate

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• Become more familiar with the set notation for the complement of a set

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• Practise applying the rules of indices in problems, including the use of negative and fractional indices

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• Practise solving problems involving similar triangles

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• Ensure that axes are correctly labelled on graphs and charts

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• Practise drawing and interpreting histograms

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Paper 2H

• Ensure they know how and when to use a calculator efficiently, but not to forget manual methods. Use of brackets and edit function will help with substitution on a calculator
• Practise articulating their mathematics in verbal discussions to aid them in written responses
• Use tables and diagrams to extract information from a question so they can better navigate through the problem
• Know the relationships between volume, area and length scale factors
• Provide unambiguous explanation whenever required by the question
• Cross out any work they do not wish the examiners to mark, leaving one clear solution
• Whenever possible use accurate figures without prematurely rounding or truncating in presenting work

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Paper 3H

• Write down all steps in your working, ensuring work is structured and easy to follow to have access to as many marks as possible. This can partly be done by annotating given diagrams eg labelling sides and angles

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• Check all your working for any errors and to make sure you have not miscopied numbers or algebraic expressions either given in the question or in a previous line of your working. Along with this check that you have answered the question fully eg giving the answer in the required form

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• Practice factorising quadratics, particularly those where a ≠ 1 or b = 0 and check this by expanding the brackets

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• Take care with algebraic notation and ensure that brackets are used when necessary

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• Learn and know how to use all the necessary formulae eg the formula for the volume of a pyramid, the Sine Rule and the formula for the area of a triangle

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• Only round your answer at the final stage to avoid losing out on the final accuracy mark

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• Be concise when answering questions that require an explanation and don’t add in unnecessary calculations that risk the addition of an incorrect statement and the mark(s) being deducted

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Foundation Review

Number

• 73% of marks within Number were for Structure and Calculation.
• 50% of these marks were for Calculations using four operations with integers and fractions.

Algebra

• 40% of the marks within Algebra were in Notation, Vocabulary and Manipulation, including factorising expressions, substitution and working with algebraic expressions.
• 35% of the marks in Algebra were in Graphs (significantly more than the average).

Ratio and proportion

• Many of the marks within Ratio & Proportion were for ratio calculations and work with units of measurement, including unit conversions and compound measures.
• Fewer marks than average for this strand.

Geometry

• An unusual skew towards Properties and Calculations within Geometry.
• Perimeter and area focused solely on rectilinear shapes.
• Very limited assessment of Pythagoras.
• Trigonometry was not assessed.

Probability and Statistics

• More Statistics questions than Probability.
• More Probability on Paper 3 and more Statistics on Paper 2.
• No assessment on sampling or scatter graphs in Statistics.

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Higher Review

• Algebra and Geometry were the main focus in the Higher series.
• Many Geometry marks came from Mensuration and Calculations.
• More work on Structure and Calculations.
• Fractions, Decimals and Percentages sub-strand attracted fewer marks than average.
• Sequences sub-strand of Algebra was more prevalent than in some previous series.
• Graphs accounted for 10% of the total marks for the whole series, significantly higher than usual.
• Statistics and Probability had gaps within both strands.
• Charts and Graphs attracted far more marks than Averages in Statistics.
• A large proportion of Probability work focused on Combined Events.

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Return of more complex problems to the Higher tier

Unlike Foundation, where we saw a higher proportion of standard procedural questions and a lower proportion of multi-step unfamiliar problems, the complexity distribution for the Higher papers is much closer to the average for all series.

We’ve seen a marked uptick in the marks available for multi-step questions compared to last year’s June papers. 43% in 2024 compared to 34% in 2023. This may explain why some students found these papers challenging.

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Hot topics to look at with students/staff from both tiers

Areas to focus to maximise results F tier

• Foundation

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Areas to focus to maximise results H tier

• Higher

Approaches to problem solving questions for low and middle attainers

Planning for the year ahead

• The primary goal for intervention for Year 11s for this term must be to hit the ground running with their GCSE intervention strategies already in place. Likely, your core group of pupils has already been identified during face-to-face teaching sessions in the autumn term. Organising something for them from the word ‘go’ will set the pace for the year ahead.
• For maximum impact, these sessions should link with what’s taught during class time. However, if this isn’t feasible, I’d suggest some intensive work on calculation, proportion, numeracy and problem-solving skills, particularly for Foundation candidates.
• We must also be particularly proactive in identifying pupils with topic-specific GCSE maths revision gaps; they may be able to plug these gaps themselves with support and direction to appropriate resources, or they may benefit from a very short burst of additional tuition.
• Finally, although I’ve focused almost exclusively on Year 11, almost all of what I’ve suggested is also relevant for Years 7-10. Where resources and funding allow, similar programmes for the other year groups could help to mitigate some of the lost learning over the last eighteen months. 

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What more they can do to ensure their outcomes are above national average.

Here are some teaching strategies to integrate the insights from this year’s exam papers into your teaching for the upcoming academic year:

• Incorporate challenging or unusual problems from this year’s exams into teaching materials for the next cohort of Year 11s.
• Repeated or scaffolded variations of challenging problems can be especially beneficial.
• Include the less-standard topic crossovers referenced more frequently in your lessons.
• Use student feedback from this year’s cohort to identify which questions they found difficult on the exams and determine if current learning materials need adjustments to address these gaps.
• Consider how you will use question-level analysis for your cohort. Are there specific teaching gaps within your department not reflected nationwide? This may indicate a need for departmental CPD or sharing of best practices on key topics.

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Using a calculator

• Share these clips with your students

GCSE Maths Revision Checklist - Foundation

GCSE Maths Revision Checklist - Foundation

GCSE Maths Revision Checklist - Higher

GCSE Maths Revision Checklist - Higher

Summer 2024 Exemplar

1MA1 1F 2F 3F

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Senior Examiner’s feedback on student responses

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GCSE (9-1)

Mathematics (1MA1)

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Contents

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© Pearson Education 2024

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Foundation Paper 1

Paper 1F�Question 9(b)

9 Here is a number machine.

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(b) Work out the input when the output is 28

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(2)

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Mark Scheme

Handout

Paper 1F�Question 9(b) – Examiner comments

Part (b) was well answered. Where candidates did not correctly find the input which would give an output of 28, they were often able to gain partial credit for a start to a method to use inverse operations, often by working out 28 + 10.

The most common incorrect answers were 46 or 9 which came from the result of using an input of 28 rather than using this as the output or from subtracting 10 from 28 and then dividing by 2. Where candidates knew to add 10 and then divide by 2, dividing 38 by 2 proved problematic. There were many instances of students performing the inverse operations correctly, getting the answer 19 then writing a different number on the answer line. Other candidates obtained an incorrect answer of 36 by calculating (28 – 10) × 2, therefore only partially using inverse operations. Many candidates found it useful using the number machine to write their workings, helping them to answer the questions step by step.

Paper 1F�Question 9(b) – Performance

M1 for start of a method to find input using inverse operations. 28+10 is sufficient for this mark to be awarded.

A0

A0 as the answer is incorrect.

1/2

M1

M0 Candidate has a partial inverse calculation. �Starting with 28 + 10 would lead to a correct final answer but starting with 28 - 10 would not.

A0

A0 as the answer is incorrect. A0 will always follow M0 as an accuracy mark cannot be awarded following an incorrect method.

Note: The mark scheme states “starts method” so the subsequent division by 2 is not sufficient for the mark.

0/2

M0

M1 The candidate has used an algebraic approach and worked with inverse operations.

If the correct answer had not been obtained, then the M mark could be awarded for 28+10 as part of the method to solve the equation.

A1 as the answer is correct.

2/2

A1

M1

Paper 1F�Question 9(c)

9 Here is a number machine.

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(c) Show that there is a number for which the output is the same as the input.

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(2)

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Mark Scheme

Handout

Paper 1F�Question 9(c) – Examiner comments

In this question candidates were asked to show that there is a number for which the output of the number machine was the same as the input. Many candidates were able to correctly identify 10 as this number and the majority showed that this was the case either by indicating that 10 × 2 = 20 and 20 − 10 = 10 or by showing the number machine with an input of 10 and an output of 10. Where candidates did not show that an input of 10 gave an output of 10, they were often able to gain partial credit for identifying that 10 was the value which met the requirements of the question without showing that this input gave a matching output or for a correctly evaluated trial. Partial credit was also available for candidates who had made a correct trial of a number different to 10. The most common incorrect answers seen were where candidates changed the number machine used in order to make their input and output match, often by changing the amount subtracted.

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Paper 1F�Question 9(c) – Performance

M1 for a correctly evaluated trial.

C0

C0 Candidate has not shown that an input of 10 gives an output of 10.

Note: If there is more than one trial, then we only need one to be correctly evaluated to award M1.

1/2

M1

M0 Whilst the trial can be for any value, it must be evaluated correctly to score. There is an arithmetic error here.

C0

C0 Candidate has not shown that an input of 10 gives an output of 10.

Note: If the function machine in the working space had an input of 13 or an output of 28 correctly evaluated, it would be acceptable for M1, as it is seen in their working.

However, if an input of 13 or an output of 28 has been shown by the original function machine (diagram) we cannot award M1 for the trial as it is not clear that it is the working for (c) as stated in the additional guidance.

0/2

M0

M1 for a correctly evaluated trial.

C1 for showing that an input of 10 gives an output of 10.

Note: One trial here (input of 10) is sufficient for both marks. We do not need to see two separate trials.

2/2

C1

M1

Paper 1F�Question 12

12 The diagram shows a triangle and a rectangle.

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The perimeter of the rectangle is a quarter of the perimeter of the triangle.

Work out the length of the rectangle.

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....................................................... cm

(Total for Question 12 is 4 marks)

Mark Scheme

Handout

Paper 1F�Question 12 – Examiner comments

A minority of candidates were able to correctly work with the perimeter of the triangle, find a quarter of this and complete the process to find the missing side length of the rectangle. Some candidates had a correct complete process, but mistakenly gave the perimeter or the rectangle (20cm) as their answer. There were also a number of candidates who correctly reached the perimeter of the rectangle but did not know how to find the missing length in the rectangle often dividing by 4 (working with area) rather than subtracting the two 4cm sides and dividing by 2. Some found the perimeter of the triangle but could not progress further. Others subtracted 4cm and did not consider the other 4cm on the rectangle or divided 20cm by 4cm and gave an answer of 5cm. They then divided 16cm by 2 to get an answer of 8cm. Common incorrect approaches worked with area or tried to use the quarter as a scale factor dividing a single side length of the triangle by 4.

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Paper 1F�Question 12 – Performance

P1 for process to find the perimeter of the triangle 36 + 30 + 14 which is incorrectly evaluated as 110.

A0

A0 as the answer is incorrect.

Note: "80" must come from a correct process but may be incorrectly evaluated.

3/4

P1

P1

P1

P1 for "80" ÷ 4, this is seen as 110 ÷ 2 = 60 (incorrectly evaluated) and 60 ÷ 2 = 30.

P1 for ("20" - 4 - 4) ÷ 2, this is seen as 30 - 4 - 4 = 22

and 22 ÷ 2 = 11.

P1 for a process to find perimeter of the triangle, 36 + 14 + 30.

A1

A1 for answer 6.

Note: We condone the incorrect use of = in the calculation shown i.e. 36 + 14 + 30 is not equal to 6.

4/4

P1

P1

P1

P1 for "80" ÷ 4.

P1 for ("20" - 4 - 4) ÷ 2.

P0 as there is no process to find the perimeter of the triangle.

A0

A0 as the answer is incorrect.

Note: If the candidate had divided all 3 sides of the triangle by 4 AND then added them together, we could have awarded P1P1. See Additional Guidance in the mark scheme.

0/4

P0

P0

P0

P0 as there is no calculation of "80" ÷ 4. Although we see values divided by 4 these are not from a correct process to find the

perimeter of the triangle.

P0 as there is no complete process.

Paper 1F�Question 22(a)

22 The diagram shows a plan of a floor.

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Petra is going to cover the floor with paint.

Petra has 3 tins of paint.

There are 2.5 litres of paint in each tin.

Petra thinks 1 litre of paint will cover 10 m² of floor.

(a) Assuming Petra is correct, does she have enough paint to cover the floor?

You must show all your working.

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(4)

Mark Scheme

Handout

Paper 1F�Question 22(b)

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Actually, 1 litre of paint will cover 11 m² of floor.

(b) Does this affect your answer to part (a)?

You must give a reason for your answer.

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(1)

(Total for Question 22 is 5 marks)

Mark Scheme

Paper 1F�Question 22 – Examiner comments

This two-part question was attempted by the vast majority of candidates. It was well answered by slightly less than half of the candidates with many of these being able to give a fully correct answer to both parts. �In part (a) candidates who were successful generally calculated the area of the floor by splitting it into two rectangles (5m by 10m and 3m by 6m or 8m by 6m and 4m and 5m). Some were able to go on to find the area that the paint would cover (75m²) or find the volume of paint required to cover the floor together with the volume of paint available and give a correct conclusion for full marks. Some candidates did not show full working to support their conclusion, for example, finding the area as 68 and the amount of paint as 7.5 litres but then not finding the comparable figures (6.8 or 75). A few candidates did not answer the question as yes/no, instead saying that Petra was correct. Where only partially correct answers were seen these often involved finding the total area of the floor without making further meaningful progress, finding the area of one shape, finding the area covered by the 3 tins available or finding the number of litres required for their area. Common incorrect approaches included working with the perimeter or multiplying together three or more of the side lengths. When considering the floor area that could be covered it was common to see candidates incorrectly working with 100 after misinterpreting the 10m² coverage. Some candidates did not show their calculations and made arithmetic errors and therefore could not score the corresponding process marks: obtaining 70.5 instead of 75 was a common example of this. �In part (b) candidates were told that the paint actually covered a greater area per litre than had been suggested and were asked if this affected their answer to (a). Where candidates had concluded that there was sufficient paint in (a) they were often able to correctly identify that there would be no effect as less paint would be required. Some candidates gave a correct reason – “she would still have enough” but paired incorrectly with the answer yes. Where candidates had incorrect answers to (a) they were often still able to identify the effect that the change would have made for their figures. Common errors included giving the incorrect decision on the effect for the figures that they had found in (a) and including incorrect figures in the statement which then prevented the award of the C mark.

Paper 1F�Question 22 – Performance

Part (a)

P0 No attempt to calculate an area.

A0

A0 No decision made.

1/5

P0

C0

P1

P0 No attempt to calculate the total area.

Part (b)

C0 No decision made.

P1 Complete process to find the area covered by 3 tins of paint.

P0

Note: We don't need to see 7.5 x 10 explicitly. Finding the area covered by 3 tins of paint can be implied by 75.

Part (a)

P1 for a process to calculate an area.

A1

A1 for a correct decision 'Yes' supported by correct figures.

5/5

P1

C1

P1

P1 for a complete process to find the total area.

Part (b)

C1 for correct decision and reason.

P1 Complete process to find the area covered by 3 tins of paint.

P1

Note: Here we see a fully correct response with the correct figures as in the mark scheme.

However, other combinations of figures may be seen, for example if this response had shown the 68 m² and said that there were 7m² of paint left over then this implies that they have the 68m² and 75m².

You might also see the correct process leading to 6.8 litres and

reference to 0.7 litres more which together with 'yes' would be a fully correct response.

Note: When awarding the C1 mark in part (b), any figures included in the statement must be correct for their area in part (a).

Part (a)

P1 for a process to find the area of one shape. Here this can be awarded for 8 x 10 or for 4 x 3.

A0

A0 Although they say 'yes’, which is the correct decision, they do not have the required supporting figures.

3/5

P0

C1

P1

P1 for a complete process to find the total area.

Part (b)

C1 for a correct decision with supporting reason.

P0 as there is no process to work out the number of tins needed, the number of litres needed, or the area covered by 3 tins.

P1

Note: Part (b) is a strict follow through and requires that we have seen a decision in part (a).

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If we see 'yes’ in part (a) then we can easily check part (b) - the acceptable examples or similar responses would all be correct.

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If we see 'no' in part (a) then in part (b) the required answer will depend on [area] in part (a). This means that they need to have an [area] in part (a).

Paper 1F�Question 26

26 Kasim has some small jars, some medium jars and some large jars.

He has a total of 400 jars.

3/8 of the 400 jars are empty.

For the empty jars,

number of small jars : number of medium jars = 3 : 4

number of medium jars : number of large jars = 1 : 2

Work out the percentage of Kasim’s jars that are empty small jars.

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.......................................................%

(Total for Question 26 is 5 marks)

First method on the Mark Scheme

Handout

Paper 1F�Question 26

26 Kasim has some small jars, some medium jars and some large jars.

He has a total of 400 jars.

3/8 of the 400 jars are empty.

For the empty jars,

number of small jars : number of medium jars = 3 : 4

number of medium jars : number of large jars = 1 : 2

Work out the percentage of Kasim’s jars that are empty small jars.

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.......................................................%

(Total for Question 26 is 5 marks)

Second method on the Mark Scheme

Paper 1F�Question 26 – Examiner comments

Paper 1F�Question 26 – Performance

First method on the mark scheme�P1 for 150 empty jars identified next to the question wording.

P0

P0 There is no attempt to work out the percentage. The answer is just the number of empty small jars.

3/5

P1

A0

P1

P1 for the process to deal with the ratios. i.e. 3 : 4 : 8.

A0 for an incorrect answer.

P1 for the process to find the number of empty small jars. �i.e. 30 : 40 : 80.

P1

Note: Here the 30 (empty small jars) is seen in the working. However, if we had a response where they did not have 30 in the

working and 30 appeared only on the answer line, they would not be awarded the third P mark as this is 30% not 30 empty small jars.

Note: In the first method, for the 1st P1 we do not need to see the answer of 150, just the complete process to find it e.g. 400/8 x 3 oe. Whilst we would accept 3/8 x 400 for correct process, we will not accept 3/8 of 400 for evidence of correct process without seeing 150.

First method on the mark scheme�P0 No attempt has been made to find the number of empty jars.

P0

P0 Their process to find percentage is not correct.

1/5

P0

A0

P1

P1 for the process to deal with the ratios. Either 3 : 4 AND 4 : 8 (note that 3 : 4 is in the question so they don't have to write it again) or 3 : 4 : 8 would gain this mark.

A0 Incorrect answer.

P0 No process is seen to find the number of empty small jars. Their 30 comes from 3 multiplied by 10 then divided by 100 (30%) which is an incorrect process.

P0

Note: For the 4^th P mark, if this candidate had written 30/400 alone without x 100, this mark would not be awarded as the process for calculating a percentage is not complete.

This response could be using the first or second method on the mark scheme

First method on the mark scheme�P1 for the process to find the number of empty jars. Here we see

400 divided by 8 (= 50) and 50 x 3 (=150).

P1

P1 for the process to find percentage shown. Here 30/400 x 100.

4/5

P1

A0

P1

P1 for the start of a process to work with ratios. Here we could award the mark for seeing 3 : 4 and 4 : 8 or for the next step 3 : 4 : 8.

A0 Incorrect answer given.

P1 for the process to find the number of empty SMALL jars. �30 : 40 : 80 is enough for this mark.

P1

Note: Processes may be seen in stages and may be equivalent to those exemplified in the mark scheme, for example for the 3rd P mark working out 150/15 then 10 x 3.

Note: You can see here that the candidate has used a highlighter on parts of the question. Whilst the use of highlighters on the question is absolutely fine to assist when answering it, please ensure your students know not to use them on their working or answers. As you can see here, the highlighter has made the writing beneath almost impossible to read (and in other cases it is impossible to read) when the script is scanned.

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Foundation Paper 2

Paper 2F�Question 14(b)

14 On Monday, Lizzie cycled 36 kilometres in 3 hours.

(a) Work out Lizzie’s average speed.

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...................................................... kilometres per hour

(2)

On Tuesday, Lizzie cycled 36 kilometres at an average speed of 16 kilometres per hour.

Lizzie says that the total time she cycled on Monday and Tuesday was less than

5 hours 20 minutes.

(b) Is Lizzie correct?

You must show how you get your answer.

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(3)

(Total for Question 14 is 5 marks)

Mark Scheme

Handout

Paper 2F�Question 14(b) – Examiner comments

Part (b) was answered less well than part (a), with only a small proportion giving a fully correct answer. Partial marks were often awarded, with candidates often gaining one mark for 2.25 and most then continuing to find 5.25 to gain the second mark. The third mark was often lost for an inability to correctly convert the decimal to minutes, with 5 hours and 25 minutes along with the wrong decision being made due to this. Some successfully converted fully to minutes and then compared but this approach was rarely seen. Converting time from decimal form to hours and minutes continues to be an area of weakness for students on this tier. Ensuring students are confident with converting to and from time base 60 to decimal base 10 and developing understanding of compound measure skills involving speed distance and time with a calculator would be beneficial at all grades.

Paper 2F�Question 14(b) – Performance

P1 This candidate has worked in minutes to find speed by first finding how many minutes for 1 km then multiplying by 36 to find the full time to cycle 36 km.

Note: Final comparable figures need to be given in consistent

units, eg 315 mins and 320 mins or just 5 hours and 15 mins as they have done here.

3/3

P1

C1

P1

P1 A complete process as they continue in minutes and correctly add 180 minutes instead of adding 3 hours.

C1 The decision with correct supporting figures is next to the question and units stated are correct.

P1 A correct process shown to find time (36 divided by 16).

Note: Units are not needed for process marks and can be ignored.

2/3

P1

C0

P1

P1 Full process to find figures to compare.�5 hours 25 mins is acceptable for this mark. Misinterpretation of 5.25 is only acceptable for P marks.

C0 The C mark can only be awarded if we see “yes” and correct figures. If units are written, they must be correct for C mark.

P1 for 2h15 (minutes).

Note: Supporting figures are considered to be the minimal amount of working needed for the process marks.

2/3

P1

C0

P1

P1 Full process using correct units is implied by 5 hours 15 mins, with working to find time for Tuesday shown in a triangle formula.

C0 No decision is made.

Paper 2F�Question 18

18 Mrs Simpson organised a school trip for 66 children.

The total cost of the trip was £1800

The school paid 56% of the total cost.

The rest of the total cost was divided equally between the 66 children.

Work out how much money each child paid.

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£......................................................

(Total for Question 18 is 3 marks)

Mark Scheme

Handout

Paper 2F�Question 18 – Examiner comments

Whilst many candidates gained all 3 marks available in this small problem, many students struggled to find 56% of 1800 using a valid method as the first step. Calculations attempted unsuccessfully were often a non-calculator method such as a build-up or partitioning method, with the working shown often being insufficient for students to be given credit. Students should be encouraged to use a calculator appropriately and be confident in using decimal multipliers. Of the candidates who could calculate 56%, many of those then misunderstood the question and did not subtract this from 1800 and some also did not divide by 66. The vast majority of candidates did, however, realise and therefore gained credit for using [44%] ÷ 66.

As with the majority of questions requiring the use of percentages, many students still believe writing 56% of a number or 56% = … is sufficient evidence of a correct process but such statements gain no credit unless a correct value is given. Use of the percentage button on the calculator may account for this but there is evidence to support the conclusion that many students do not know how to use this function correctly.

Paper 2F�Question 18 – Performance

P1 The learner has stated 56% = 905 which is not a process nor is it accurate. However, we can award P1 for 895 / 66.

895 is their clear intended value for 44%.

1/3

P0

A0

P1

P0 No correct complete process.

A0 Incorrect answer.

P1 An incorrect build up method to find 56% of 1800 (= 930). However, we do see 870 / 66 which is [44%]/66 so we can award P1 (1800 - 930 = 870 not seen but implied).

1/3

P0

A0

P1

P0 No correct complete process.

A0 Incorrect answer.

P1 1008 is an accurate figure from 56% x 1800, this divided by 66 (= 15.272..)

1/3

P0

A0

P1

P0 No correct complete process.

A0 Incorrect answer.

Note: 56% x 1800 is not considered a process and does not get any marks on its own.

Paper 2F�Question 23

23 There are only red discs, blue discs and yellow discs in a bag.

There are 24 yellow discs in the bag.

Mel is going to take at random a disc from the bag.

The probability that the disc will be yellow is 0.16

the number of red discs : the number of blue discs = 5 : 4

Work out the number of red discs in the bag.

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(Total for Question 23 is 4 marks)

Mark Scheme

Handout

Paper 2F�Question 23 – Examiner comments

This question was a good discriminator for this cohort, with the full range of different possible scores seen. Almost all candidates attempted the question and very few blank responses were seen. Many were able to gain the first mark, usually by working out 1 − 0.16 and thereby showing some understanding of probability and some gained the first mark for working correctly with ratio to find the total number of discs in the bag, but this was seen less frequently. A score of 2 marks was also often seen, usually where students got as far as a probability of 0.46 but then did not know how to go further, or for working with probability and ratio separately but not knowing how to combine both skills correctly to make further progress through the problem.

Of the candidates who understood how to approach the problem and showed a complete process, many rounded their numbers prematurely, often reaching answers like 69, 70.5 or 71 and scored all but the accuracy mark. Centres should remind students to work with accurate values fully until the end of the question and avoid premature rounding or truncating.

Paper 2F�Question 23 – Performance

P1 for 1 - 0.16 (=0.84).

3/4

P1

A0

P1

P1 for 0.465. This figure comes from rounding of 0.09333... to 2sf, but the process is correct.

A0 Incorrect answer.

P1 for 69.75, which is 0.465 x 24 / 0.16.

P1

Note: This candidate gets confused and writes their number of blue discs on the answer line. �If they had written 69.75 (seen at bottom-left) this would still be A0 as it is an inaccurate answer.

P1 for 1 - 0.16.

2/4

P1

A0

P1

P1 for [probability] / 9 x 5 (seen twice on the page - either is acceptable for this mark).

A0 No answer.

P0 No complete correct process.

P0

P1 for 0.84 for working correctly with probability

2/4

P1

A0

P1

P1 for [probability] / 9 x 5 (use of ratio correctly with probability)

A0 Incorrect answer.

P0 No complete correct process.

P0

Note: 0.46 or better or 0.47 may imply P2.

If followed by an answer of 69 or 71 (or 70.5) with or without showing multiplication by "150", the third mark is awarded as the correct process using rounded or truncated values is implied.

Paper 2F�Question 25

Mark Scheme

Handout

Paper 2F�Question 25 – Examiner comments

Paper 2F�Question 25 – Performance

Note: Students are allowed to choose a number for the total sweets they may have. Hopefully, it is a multiple of 21. If not, then you can allow decimal answers, rounded or truncated to 2 decimal places.

P1 Clearly shown process of 100/21 to find one share but there is an arithmetic error, it should be 4.76. Then correctly finds 3/7 of 63 = 27 as number given to Andy. The value 63 is Tina's 14 x

4.5).

3/4

P1

C0

P1

P1 Subtracts this value from Tina's share and finds 12.5% of the remainder and adds to Luke.

C0 Not supported by correct figures due to the initial arithmetic error.

P1 Finds all the final amounts for the three people.

P1

P1 Correct value for sweets given to Andy. (Always look at the whole page).

2/4

P1

C0

P0

P1 for the values for Andy AND Tina after first exchange. May be implied by ratio 7:6:8.

P0C0 No further process.

P1

P1 for process to find the number of sweets Tina gives to Andy, implied by the −6 on the first row of working.

Note: This is the minimal working required for this question. 7:7:7 alone without working scores zero marks.

4/4

P1

C1

P1

P1 for the values for Andy AND Tina after first exchange. May be implied by ratio 7:6:8.

C1 for “yes” supported by correct figures.

P1 for process to find final amounts, implied by 7 : 7 : 7 ratio.

P1

Foundation Paper 3

Paper 3F�Question 9

9 Lucia is going on a skiing holiday.

The cost of ski hire is £26 per day.

The cost of a lift pass is £45 per day.

The cost of ski lessons is £23.50 per hour.

Lucia will pay for

ski hire for 5 days

a lift pass for 4 days

ski lessons for 8 hours.

Lucia has £500

Show that Lucia has enough money to pay for the total cost of ski hire, the lift pass and the ski lessons.

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(Total for Question 9 is 3 marks)

Mark Scheme

Handout

Paper 3F�Question 9 – Examiner comments

This question was answered very well with almost the entire cohort scoring full marks. The most common process seen was to calculate the total for 5 days of ski hire, 4 days of lift passes and 8 hours of ski lessons and find the sum of the 3 totals, leading to an answer of 498. Of those that did not gain full marks, the most common errors were usually related to digits, e.g. writing down 23.5 × 8 = 180 instead of 188, presumably copying down incorrectly from the calculator. Some candidates opted to use a repeated addition method e.g. 26 + 26 + 26 + 26 + 26 = 130; this was a successful, although not efficient, process for some candidates; but led to errors for the majority who used it.

Paper 3F�Question 9 – Performance

P1 for a start to the process of finding the total cost 130 or 180 or 188. As values are correct, we do not need to see the process of getting them.

C1

C1 Correct figures and complete process.

Note: If a candidate reaches 498 and then says “1 left”, we will award C1. Condone further incorrect calculations as long as the statement doesn’t contradict.

3/3

P1

P1

P1 for a complete process of adding their values.

P1 for a start to the process of finding the total cost.

C0

C0 Incorrect figure(s).

Note: A statement or decision is not needed but if a statement is given for the correct figure, it must not contradict ie

'she does not have enough’ with a total of 498 would be C0.

2/3

P1

P1

P1 for a complete process. They have shown addition of their amounts and even though the 184 is incorrect we can see it has come from a correct process (4 x 45) in their workings.

P1 for a start to the process of finding the money left after paying costs 500 − 26 (= 474) or 500 − 23.50 (= 476.5).

C0

C0 No complete process with correct figures shown.

Note: This could be looked at as choice. Mark in accordance with the general marking guidance in the mark scheme (note 4). As both approaches score 1 mark this candidate gains P1.

1/3

P0

P1

P0 as there is no complete process to get to the answer.

Note: This could also be awarded for 26 + 45 + 23.50 (= 94.5(0)).

Paper 3F�Question 15

15 Which is greater

15% of 88 or 20% of 62?

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You must show all your working.

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(Total for Question 15 is 3 marks)

Mark Scheme

Handout

Paper 3F�Question 15 – Examiner comments

The majority of candidates scored 3 marks on this question for reaching two comparable figures and making a correct decision. A small number of candidates gained 0 marks and it was evident that they found working with percentages challenging or they showed an incorrect process such as, for example, finding 115% of 88 and 120% of 62 or 85% of 88 and 80% of 62. It should be noted that 15% × 88 is not considered a mathematical process and therefore will not gain credit unless it is accompanied by the correct figure (e.g. 15% × 88 = 13.2 would gain the mark but without the 13.2 it would not).

Many candidates used non-calculator processes that resulted in errors in arithmetic, or the full process was not shown and therefore credit could not be awarded.

Paper 3F�Question 15 – Performance

P1 for first step towards finding comparable figures �eg 6.2 + 6.2 (= 12.4).

C0

C0 despite identifying 15% of 88, this mark cannot be awarded if either of the values are incorrect. (12.8 should be 13.2).

1/3

P1

P0

P1 for first step towards finding comparable figures.

C0

C0 for not having a correct decision from correct values.

2/3

P1

P1

P1 for process to find two comparable figures. The process to find 20% of 62 is correct (62 ÷ 10 = 6.2 and then 6.2 x 2) but they have given an incorrect value.

P1P1 for process to find two comparable figures.

C1

C1 for the decision 15% of 88 from correct values of 13.2 and 12.4.

Note: Circling or underlining etc the correct answer or crossing

out the incorrect answer is enough for the decision as long as the correct figures are seen.

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13.2 > 12.4 is acceptable as a decision.

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You can ignore an incorrect difference after a correct decision from correct figures unless it contradicts eg if a student reaches 13.2 and 12.4, says 15% of 88 is greater, and then does �13.2 − 12.4 = 0.6 and says 13.2 is 0.6 bigger, we will ignore this as it does not contradict the decision, and you can award 3 marks.

3/3

P1

P1

Paper 3F�Question 18(a)

18

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(a) On the grid above, rotate the shaded shape 180° about (0, 0)

(2)

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Mark Scheme

Handout

Paper 3F�Question 18(a) – Examiner comments

For part (a), a minority of the cohort were able to produce a correct rotation. It was common to see the shape rotated by 180° with an incorrect centre; this was awarded 1 mark. A similar number of candidates were unable to gain any marks, common incorrect answers included reflections and translations.

Paper 3F�Question 18(a) – Performance

B1 for the shape rotated 180° about any centre (not (0,0)).

Note: A reflection in the x-axis would also score B1, as it would get a mark for rotation of the shape by 180° about any centre.

1/2

B1

B0 Shape has not been rotated 180° - it appears to have been translated.

Note: Remember to check if 3 out of 4 vertices are correct

0/2

B0

B2 Correct shape drawn at correct vertices.

2/2

B2

Paper 3F�Question 18(b)

18

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Mike was asked to

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‘Reflect shape A in the line with equation x = 3’

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Mike’s answer is shown on the grid.

His answer is wrong.

(b) Explain why.

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..............................................................................................................

.............................................................................................................. (1)

(Total for Question 18 is 3 marks)

Mark Scheme

Handout

Paper 3F�Question 18(b) – Examiner comments

Part (b) was answered less well with a small number of candidates gaining 1 mark for a correct explanation. For those who did gain the mark, explaining that he has reflected in y = 3 was the most common correct answer seen. A whole variety of incorrect answers were seen, the most common of which are highlighted in the mark scheme. Some candidates drew the correct reflection, but this was only awarded the mark if accompanied by an explanation such as ‘it should be here’. It was clear that many candidates were unable to describe lines with the correct equation.

Paper 3F�Question 18(b) – Performance

Note: The candidate is just repeating the words given in the question.

0/1

C0

C0 An incorrect response.

Note: If the student draws the correct reflection in line x = 3 shown on diagram with supporting comment eg it 'should be here', award C1. Without the supporting comment, eg the correct reflection drawn and nothing else, it is C0.

0/1

C0

Note: It is quite common for candidates to use 'equation' for

'line', because of the wording in the question and we are condoning this.

1/1

C1

Paper 3F�Question 20

20 ABC and BCD are isosceles triangles.

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AB = BC = CD

Angle CAB = 81°

Angle BCD = 4 × angle ABC

Find

the size of angle ABC : the size of angle CBD

Give your answer in the form 1 : n

You must show all your working.

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(Total for Question 20 is 5 marks)

Mark Scheme

Handout

Paper 3F�Question 20 – Examiner comments

This question was not well attempted by candidates. Many homed in on the statement ‘Angle BCD = 4 × angle ABC’ and multiplied 81 by 4 and went from there, gaining 0 marks. It is advisable for future cohorts to approach angles questions by taking in all the information given, including the diagram, and then working through systematically using the geometrical rules they know. For the other candidates, 1, 2, 3, 4 and 5 marks were all awarded, with 1 and 5 being most common; those that scored 1 mark achieved ABC = 18 but were then unable to interpret the other information correctly and made no further progress. A few candidates thought that angle ABD was a right angle and hence labelled angle CBD as 72 from 90 – 18. Correct 3-letter notation was rarely seen and a significant number of students lost marks as a result. Some candidates misinterpreted the isosceles triangle and thought that angle BAC = angle ABC. Marks were lost by students writing the angles in the incorrect positions on the diagram thus creating a contradiction where no 3-letter notation were used for the angles in the working, with the most common being 72 being placed as angle CBD on the diagram.

Paper 3F�Question 20 – Performance

M1 for method to find angle ABC = 18°.

A1

A1 for correct answer of 1 : 3.

5/5

M1

M1

M1 for method to find angle BCD = 72°.

M1

M1

M1 for writing ABC : CBD in ratio form eg 18 : 54 - this is dependent on previous M3 marks.

M1 for method to find angle CBD = 54°.

Note: These first 3 M1 marks can be awarded for them being seen unambiguously on the diagram.

M1 for method to find angle ABC = 18°.

A0

Note: You can ignore any geometric reasons given, whether they are correct or incorrect.

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If the student has not used 3-letter notation or labelled angles on the diagram, you can award method marks as long as it is clear which angle they are working out and is not contradicted, eg if a student works out 18 but then puts it in the wrong place for example at ACB, this is M0.

1/5

M0

M1

M0 for incorrect method to find BCD. Although we see 72

this has not been labelled using 3 letter notation or on the diagram at BCD and instead labelled as CBD.

M0

M0

M0M0A0 No further marks can be gained as the 3rd M needs to have "72" coming from previous correct working and the 4th M & A mark are dep on M3.

M1 for method to find angle ABC = 18°.

A0

A0 for incorrect answer.

4/5

M1

M1

M1 for method to find angle BCD = 72°.

M1

M1

M1 for writing ABC : CBD in ratio form eg 18:54 - this is dependent on previous M3.

M1 for method to find angle CBD = 54°.

Paper 3F�Question 26

26 Habib has two identical tins.

He puts 600 grams of flour into one of the tins.

The flour fills the tin completely.

The density of the flour is 0.6 g/cm³

Habib puts 600 grams of salt into the other tin.

The salt does not fill the tin completely.

The volume of the space in the tin that is not filled with salt is 700 cm³

Work out the density of the salt.

You must show all your working.

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(Total for Question 26 is 4 marks)

Mark Scheme