Cambridge IGCSE Mathematics - International 0607 — 2024 May/June Paper 6 · Variant 1

This document has 16 pages. Any blank pages are indicated. [Turn over * 4 8 3 6 8 7 4 7 4 0 * Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 Investigation and Modelling (Extended) May/June 2024 1 hour 40 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer both part A (Questions 1 to 7) and part B (Questions 8 to 13). ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● You should use a graphic display calculator where appropriate. ● You may use tracing paper. ● You must show all necessary working clearly, including sketches, to gain full marks for correct methods. ● In this paper you will be awarded marks for providing full reasons, examples and steps in your working to communicate your mathematics clearly and precisely. INFORMATION ● The total mark for this paper is 60. ● The number of marks for each question or part question is shown in brackets [ ]. DC (DE/SW) 332294/2 © UCLES 2024

2 0607/61/M/J/24 © UCLES 2024 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 7) SUMS OF POWERS (30 marks) You are advised to spend no more than 50 minutes on this part. This investigation looks at connections between the sum of the positive integers, 1 2 3 + + + … , the sum of their squares, the sum of their cubes and the sum of their 5th powers. Example 1 2 3 4 5 6 7 8 9 10 11 12 78 + + + + + + + + + + + = The numbers can also be added using this method. Step 1 Write down the first half of the numbers in a row. 1 2 3 … 6 Step 2 Write down the second half of the numbers underneath the first half but in reverse order. 12 11 10 … 7 Step 3 Add each column of two numbers to make a third row. 13 13 13 … 13 Step 4 Find the total of the numbers in the third row by writing the calculation as a multiplication. 6 13 78 # = 1 Use the method to complete the sum of the first 60 positive integers. That is 1 2 3 + + + … 60 1830 + = . 1 2 3 … … … … … … … … … … … … … # … = 1830 [3]

3 0607/61/M/J/24 © UCLES 2024 [Turn over 2 Use the method to calculate the sum of the first 128 positive integers. That is 1 2 3 + + + … 128 + . … [3] 3 Complete the table. Use Question 1, Question 2 and any patterns you notice. Number of positive integers, starting at 1 Multiplication Sum 12 6 13 # 78 26 13 27 # 351 Question 1 60 1830 Question 2 128 204 20 910 [1] 4 1 2 3 + + + … n + has n positive integers and its sum is T. Find a formula for T in terms of n. … [2]

4 0607/61/M/J/24 © UCLES 2024 5 (a) Complete the table. Sum of first n positive integers Sum of first n square numbers T S written as a fraction with denominator 3 n Calculation Sum (T ) Calculation Sum (S ) 1 1 1 12 1 3 3 2 1 2 + 3 1 2 2 2 + 5 3 5 3 1 2 3 + + 6 1 2 3 2 2 2 + + 14 3 4 1 2 3 4 + + + 10 1 2 3 4 2 2 2 2 + + + 5 1 2 3 4 5 + + + + 15 1 2 3 4 5 2 2 2 2 2 + + + + 55 [3] (b) Find an expression for T S in terms of n. … [3] (c) The sum of the first 60 positive integers is 1830. Find the sum of the first 60 square numbers. … [2]

5 0607/61/M/J/24 © UCLES 2024 [Turn over (d) Use Question 4 and Question 5(b) to help you find a formula for S in terms of n. Write your answer as a single fraction. … [1] 6 (a) Complete the table. Sum of first n positive integers Sum of first n cube numbers Calculation Sum (T ) Calculation Sum (C ) 1 2 + 3 1 2 3 3 + 9 1 2 3 + + 6 1 2 3 3 3 3 + + 36 1 2 3 4 + + + 10 1 2 3 4 3 3 3 3 + + + 100 1 2 3 4 5 + + + + 15 1 2 3 4 5 3 3 3 3 3 + + + + 225 1 2 3 4 5 6 + + + + + 1 2 3 4 5 6 3 3 3 3 3 3 + + + + + [1] (b) Write a formula for C in terms of T. … [1] (c) 1 2 3 + + + … 60 1830 + = Calculate 1 2 3 3 3 3 + + + … 603 + . … [2]

6 0607/61/M/J/24 © UCLES 2024 7 (a) Complete the table. Use Question 6(a) to help you. Sum of first n cube numbers Sum of first n 5th powers C F as a fraction with denominator 3 n Calculation Sum (C ) Calculation Sum (F ) 1 13 1 15 1 1 3 3 = 2 1 2 3 3 + 9 1 2 5 5 + 33 9 33 3 11 = 3 1 2 3 3 3 3 + + 36 1 2 3 5 5 5 + + 276 3 36 276 23 = 4 1 2 3 4 3 3 3 3 + + + 100 1 2 3 4 5 5 5 5 + + + 1300 5 1 2 3 4 5 3 3 3 3 3 + + + + 225 1 2 3 4 5 5 5 5 5 5 + + + + 4425 3 225 4425 59 = 6 1 2 3 4 5 6 3 3 3 3 3 3 + + + + + 1 2 3 4 5 6 5 5 5 5 5 5 + + + + + [2]

7 0607/61/M/J/24 © UCLES 2024 [Turn over (b) The fraction C F is written with a denominator of 3. Find an expression for the numerator in terms of n. … [4] (c) 1 2 3 + + + … 60 1830 + = Calculate 1 2 3 5 5 5 + + + … 605 + . Write down all the numbers on your calculator display. … [2]

8 0607/61/M/J/24 © UCLES 2024 The modelling starts on the next page.

9 0607/61/M/J/24 © UCLES 2024 [Turn over B MODELLING (QUESTIONS 8 to 13) INCOME INEQUALITY (30 marks) You are advised to spend no more than 50 minutes on this part. This task looks at a model to measure the spread of income within the population of a country. In this task, x is the decimal fraction of the population of a country and y is the decimal fraction of the total income for the country. Examples When x 1 = this is the total population of a country. When . x 0 5 = this is half of the population of a country. When y 1 = this is the total income for the country. When . y 0 5 = this is half of the total income for the country. The graph shows how the total income of a country is shared among the population. 00 0.1 0.2 0.3 0.4 0.5 Fraction of population (x) Fraction of total income (y) 0.6 0.7 0.8 0.9 1.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 U T The point T shows that 0.3 of the population of a country earn 0.3 of the total income. The point U shows that 0.6 of the population of a country earn 0.6 of the total income. 8 (a) Mark a point on the line and label it Z. Complete this statement for your point. The point Z shows that … of the population of a country earn … of the total income. [1] (b) For this graph there is perfect equality because the income is shared equally among the population of a country. The graph shows the line of perfect equality. Write down the equation of the line. … [1]

10 0607/61/M/J/24 © UCLES 2024 9 In reality, the total income of a country is not shared equally, so there is income inequality. In 1905 the American economist Max Lorenz invented the Lorenz curve to show income inequality. A Lorenz curve is always on or below the line of perfect equality. The graph shows the Lorenz curve for one country. 00 0.1 0.2 0.3 0.4 0.5 Fraction of population (x) Fraction of total income (y) 0.6 0.7 0.8 0.9 1.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 V Lorenz curve Line of perfect equality The point V on this Lorenz curve shows that the poorest 0.7 of the population only earn 0.4 of the total income. (a) Use another point on the Lorenz curve to make a similar statement. … … [1] (b) From the statement for point V we can also say that the richest 0.3 of the population earn 0.6 of the total income. (i) Write calculations to show why the values in this statement are correct. … … [1] (ii) Write a similar statement for the point that you chose in part (a). … … [1]

11 0607/61/M/J/24 © UCLES 2024 [Turn over 10 In 1912 the Italian statistician Corrado Gini invented the Gini coefficient. The Gini coefficient measures how much income inequality there is in a country. 00 Fraction of population (x) Fraction of total income (y) 1 1 Lorenz curve Line of perfect equality The Gini coefficient is two times the shaded area between the line of perfect equality and the Lorenz curve. This means that the greater the shaded area, the greater the income inequality. (a) When there is perfect equality of income write down the Gini coefficient. … [1] (b) When the income inequality is a maximum, the shaded area will be as large as possible. Find the Gini coefficient when there is maximum inequality of income. … [2]

12 0607/61/M/J/24 © UCLES 2024 11 Mei uses these steps to model the Gini coefficient for the curve in Question 9. Step 1: Plot point V from Question 9 on the Lorenz curve. Step 2: Approximate the area below the curve with a rectangle and two triangles. Step 3: Calculate the total area of the rectangle and the two triangles. This is T. Step 4: Calculate the shaded area by subtracting T from the area of the large triangle with vertices (0, 0), (1, 1) and (1, 0). Step 5: Multiply the result by 2. 00 0.1 0.2 0.3 0.4 0.5 Fraction of population (x) Fraction of total income (y) 0.6 0.7 0.8 0.9 1.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 V Line of perfect equality Steps 1 and 2 in Mei’s model have been done for you on the graph. Do Steps 3, 4 and 5 to calculate her approximation of the Gini coefficient. … [6]

13 0607/61/M/J/24 © UCLES 2024 [Turn over 12 Mei decides to use a point P(x, y) in Step 1 of her model. 0 0 Fraction of population x y P(x, y) Fraction of total income 1 1 (a) Find the area in Step 3 of her model as an expression without brackets in terms of x and y. … [5] (b) Do steps 4 and 5 in her model to show that her approximation, G, for the Gini coefficient is x y - . [2] (c) Give a reason why G will be smaller than the actual Gini coefficient. … [1]

14 0607/61/M/J/24 © UCLES 2024 13 Mei knows that her approximation, G x y = - , is always smaller than the actual Gini coefficient. So, to make her model as accurate as possible, she takes the maximum value of G x y = - . (a) For Country A, the equation of the Lorenz curve is y x2 = . Find Mei’s most accurate estimate for the Gini coefficient by finding the maximum of G x x2 = - . … [2]

15 0607/61/M/J/24 © UCLES 2024 (b) (i) Write down the coordinates of the two points that must be on every Lorenz curve. … [1] (ii) For Country B, the equation of the Lorenz curve is y a x 1 = - - where a is a constant. Use part (i) to find the value of a. … [1] (iii) Sketch the graph of G x y = - for Country B. G 0.5 0 x 1 0 [2] (c) Use the most accurate estimates in Mei’s model of the Gini coefficient to compare income inequality in Country A and Country B. … [2]

16 0607/61/M/J/24 © UCLES 2024 Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. BLANK PAGE

This document consists of 7 printed pages. © Cambridge University Press & Assessment 2024 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 (Extended) May/June 2024 MARK SCHEME Maximum Mark: 60 Published This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers. Cambridge International will not enter into discussions about these mark schemes. Cambridge International is publishing the mark schemes for the May/June 2024 series for most Cambridge IGCSE, Cambridge International A and AS Level and Cambridge Pre-U components, and some Cambridge O Level components.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 2 of 7 Generic Marking Principles These general marking principles must be applied by all examiners when marking candidate answers. They should be applied alongside the specific content of the mark scheme or generic level descriptions for a question. Each question paper and mark scheme will also comply with these marking principles. GENERIC MARKING PRINCIPLE 1: Marks must be awarded in line with:  the specific content of the mark scheme or the generic level descriptors for the question  the specific skills defined in the mark scheme or in the generic level descriptors for the question  the standard of response required by a candidate as exemplified by the standardisation scripts. GENERIC MARKING PRINCIPLE 2: Marks awarded are always whole marks (not half marks, or other fractions). GENERIC MARKING PRINCIPLE 3: Marks must be awarded positively:  marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate  marks are awarded when candidates clearly demonstrate what they know and can do  marks are not deducted for errors  marks are not deducted for omissions  answers should only be judged on the quality of spelling, punctuation and grammar when these features are specifically assessed by the question as indicated by the mark scheme. The meaning, however, should be unambiguous. GENERIC MARKING PRINCIPLE 4: Rules must be applied consistently, e.g. in situations where candidates have not followed instructions or in the application of generic level descriptors. GENERIC MARKING PRINCIPLE 5: Marks should be awarded using the full range of marks defined in the mark scheme for the question (however; the use of the full mark range may be limited according to the quality of the candidate responses seen). GENERIC MARKING PRINCIPLE 6: Marks awarded are based solely on the requirements as defined in the mark scheme. Marks should not be awarded with grade thresholds or grade descriptors in mind.

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 3 of 7 Mathematics-Specific Marking Principles 1 Unless a particular method has been specified in the question, full marks may be awarded for any correct method. However, if a calculation is required then no marks will be awarded for a scale drawing. 2 Unless specified in the question, non-integer answers may be given as fractions, decimals or in standard form. Ignore superfluous zeros, provided that the degree of accuracy is not affected. 3 Allow alternative conventions for notation if used consistently throughout the paper, e.g. commas being used as decimal points. 4 Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). 5 Where a candidate has misread a number or sign in the question and used that value consistently throughout, provided that number does not alter the difficulty or the method required, award all marks earned and deduct just 1 A or B mark for the misread. 6 Recovery within working is allowed, e.g. a notation error in the working where the following line of working makes the candidate’s intent clear. MARK SCHEME NOTES The following notes are intended to aid interpretation of mark schemes in general, but individual mark schemes may include marks awarded for specific reasons outside the scope of these notes. Types of mark M Method marks, awarded for a valid method applied to the problem. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. For accuracy marks to be given, the associated Method mark must be earned or implied. B Mark for a correct result or statement independent of Method marks. When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. The notation ‘dep’ is used to indicate that a particular M or B mark is dependent on an earlier mark in the scheme. Abbreviations awrt answers which round to cao correct answer only dep dependent FT follow through after error isw ignore subsequent working nfww not from wrong working oe or equivalent rot rounded or truncated SC Special Case soi seen or implied

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 4 of 7 Question Answer Marks Partial Marks 1 30 60 59 58 … 31 61 61 61 … 61 30 × 61 3 B1 for the first two rows B1 for 61 seen in the third row B1 for 30  61 2 129 seen as the sum of 1 and 128 and the sum of one other pair of values C1 64  their 129 C1 8256 1 3 102 × 205 in last row of the middle column 1 4 T = 2 n (n + 1) oe 2 Mark final answer for 2 or B1 B1 for an incorrectly expressed product e.g. 2 n × n + 1 5(a) 14 7 3 30 9 3 55 11 3 2 B1 for one correct in the final column 14 6 or 30 10 or 3 1 or 3 or 55 15 in the correct cell or linked to a correct fraction with denominator 3 C1 5(b) Three first differences of 2 or 2 3 seen C1 2 1 3 n  oe, isw 2 B1 for 2 3 n seen or implied or B1 for a final answer of 2n + 1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 5 of 7 Question Answer Marks Partial Marks 5(c) Substitution of T = 1830 and n = 60 to form a calculation that will generate the answer without further rearrangement e.g. 1830 × 2 60 1 3   C1 FT their 5(b) 73 810 1 FT their 5(b) providing it is of the form 3 an b  where a and b are non- zero constants 5(d) ( 1)(2 1) 6 n n n S    oe seen, isw 1 FT their 2 n (n + 1)  3 an b  providing their 2 n (n + 1) is a polynomial with at least 2 terms in terms of n and a and b are non-zero constants 6(a) 21 441 1 6(b) C = T 2 1 Mark final answer 6(c) 18302 C1 3 348 900 1 7(a) 39 3 83 3 in last column 2 B1 for each or B1 for 12201 441 and 1300 100 or 13 7(b) Three second differences of 4 or 4 3 seen C1 2n2 + 2n – 1 oe 3 mark final answer for 3 marks or B2 or B1 B2 for 2n2 + 2n [ +c] or B1 for 2n2 [+ bn + c] If C1 B0 or C0 B0 scored, SC1 for a correct answer seen and then spoiled 7(c) Substitution of C = 18302 and n = 60 to form a calculation that will generate the answer without further rearrangement e.g. 2 2 60 2 60 1 3348900 3    C1 FT their 3 348 900 and 2 (2 – 1) 3 2 their n n  if the numerator is quadratic with at least two terms 8 170 199 700 1 8(a) Correct values for their Z 1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 6 of 7 Question Answer Marks Partial Marks 8(b) y = x 1 9(a) Correct statement for a different point e.g. [the poorest] 0.8 of [the] population [only] earn 0.5 of [the total] income 1 9(b)(i) 1 – 0.7 = 0.3 and 1 – 0.4 = 0.6 oe 1 9(b)(ii) Correct statement for their point from (a) e.g. [the richest] 0.2 of [the] population earn 0.5 of [the total] income 1 FT their (a) 10(a) 0 1 10(b) Correct shaded triangle or calculation indicating the area of triangle = 1 2 , times 2 e.g.   1 2 1 1 2   or correct sketch C1 May be seen on the graph 1 1 11 1 0.7 0.4 2   + 0.3 × 0.4 + 1 0.3 0.6 2   C3 C2 for two correct area calculations or C1 for two from 0.14, 0.12, 0.09 0.35 seen 1 1[ 1 1] 2  – their 0.35 C1 0.3 or correct evaluation of 2(0.5 – their 0.35) 1 FT their 0.35 providing the result is positive and less than 1 12(a) 1 2 x y  + (1 – x)×y + 1 (1 ) (1 ) 2 x y     C3 C2 for two correct area terms or C1 for one correct area term C1 for horizontal line drawn and (1 – x) or (1 – y) seen on diagram or used appropriately in at least one calculation 1 1 1 1 1 2 2 2 2 2 xy y xy x y xy       oe isw 2 B1 for one term out of seven missing or incorrect or for (1 – x)(1 – y) = 1 – x – y + xy seen in working

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2024 © Cambridge University Press & Assessment 2024 Page 7 of 7 Question Answer Marks Partial Marks 12(b) 1 1 1 1 2 2 2 2 y x          or 1 1 1 1 2 2 2 2 y x    oe M1 FT 1 2 their 12(a) providing their 12(a) is an expression with at least one x term and at least one y term Multiplication by 2 leading correctly to given answer x – y A1 dep on a correct expression for T in (a) 12(c) Valid reason suggesting that the area found in the approximate method is less than the shaded area: A triangle takes up part of the shaded area oe 1 13(a) Sketch of G = x – x2 with x = 0.5 indicated on the sketch or 0.5 – 0.52 C1 0.25 1 13(b)(i) (0, 0) and (1, 1) 1 13(b)(ii) 1 1 13(b)(iii) Correct sketch 2 B1 for curve with one turning point and passing through (0, 0) and (1, 0) 13(c) [Maximum = ] 0.25 C1 Same [income inequality] 1 dep on correct maximum point for country B and the correct answer in 13(a)