Cambridge IGCSE Mathematics - International 0607 — 2022 May/June Paper 6 · Variant 1

This document has 16 pages. Any blank pages are indicated. [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 Investigation and Modelling (Extended) May/June 2022 1 hour 40 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer both part A (Questions 1 to 4) and part B (Questions 5 to 7). ● 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 (LK/SG) 303084/2 © UCLES 2022 * 0 0 7 4 3 0 8 0 3 6 *

2 0607/61/M/J/22 © UCLES 2022 Answer both parts A and B. A INVESTIGATION (QUESTIONS 1 to 4) STORAGE BINS (30 marks) You are advised to spend no more than 50 minutes on this part. This investigation looks at different methods to store items in storage bins. Amara wants to use the smallest number of storage bins possible. Each bin can hold a maximum total mass. 1 Amara uses this method. Method 1 Put each item in the first bin that can hold its mass. Example These are the masses, in kg, of four items. 6 7 4 2 The maximum total mass that each bin can hold is 10 kg. The tables show how Amara puts these items into bins. Amara puts the first item in bin 1. 4 kg of storage is unused in this bin. Bin Mass of items in bin Unused mass in bin 1 6 4 2 3 The second item will not go in bin 1 because it is more than 4 kg. Amara puts the second item in bin 2. Bin Mass of items in bin Unused mass in bin 1 6 4 2 7 3 3 The third item is 4 kg. Amara puts this in bin 1. Bin 1 is now full. The fourth item will go in bin 2. Bin 3 is not used. Bin Mass of items in bin Unused mass in bin 1 6, 4 4 0 2 7, 2 3 1 3 Amara needs two bins which can hold a total of 20 kg. 1 kg out of the total of 20 kg of storage is unused.

3 0607/61/M/J/22 © UCLES 2022 [Turn over (a) These are the masses, in kg, of ten items. 38 6 21 50 32 7 15 9 27 25 The maximum total mass that each bin can hold is 60 kg. Amara uses Method 1 to put these ten items into bins. The table shows how she puts the first 6 items into bins. Bin Mass of items in bin Unused mass in bin 1 38, 6, 7 22 16 9 2 21, 32 39 7 3 50 10 4 5 Complete Amara’s table to show that she needs 5 bins. [4] (b) These are the masses, in kg, of six items. 8 16 13 10 5 3 The maximum total mass that each bin can hold is 20 kg. Bin Mass of items in bin Unused mass in bin 1 8 12 2 3 4 5 Use Method 1 to complete the table for all six items. The first item has been put in for you. You may not need all the bins. [2]

4 0607/61/M/J/22 © UCLES 2022 2 Amara wants to see if she can use fewer bins. She puts her items in order of mass before she puts them in bins. She uses this method. Method 2 Put the masses in order, largest first. Then use Method 1. These are the masses, in kg, of the ten items from Question 1(a). 38 6 21 50 32 7 15 9 27 25 (a) Write these ten masses in order, largest first. … , … , … , … , … , … , … , … , … , … [1] (b) The maximum total mass that each bin can hold is 60 kg. Complete the table using Method 2. Bin Mass of items in bin Unused mass in bin 1 2 3 4 5 [2]

5 0607/61/M/J/22 © UCLES 2022 [Turn over (c) Find and compare the percentage of unused storage for these ten items when using Method 1 and Method 2. Use your answers from Question 1(a) and Question 2(b). [4] (d) A best solution uses the smallest possible number of bins. A set of items with a total mass of 270 kg is put into 4 bins. The maximum total mass that each bin can hold is 80 kg. Show that this is a best solution. [2]

6 0607/61/M/J/22 © UCLES 2022 (e) Amara tries another way to improve Method 1. Method 3 Look for items that combine to make as many full bins as possible and place these first. For the remaining items, use Method 2. These are the masses, in kg, of eight items. 21 10 30 19 13 7 28 4 The maximum total mass that each bin can hold is 40 kg. Does Method 3 give a best solution for these items? Show how you decide. Bin Mass of items in bin Unused mass in bin 1 2 3 4 5 [4]

7 0607/61/M/J/22 © UCLES 2022 [Turn over 3 Amara has six items. The mass, in kg, of each item is an integer. These are the masses of her six items. 9 4 7 15 x y The maximum total mass that each bin can hold is 20 kg. Amara finds that Method 1 needs fewer bins than Method 2. Find the value of x and the value of y. x = … y = … [4]

8 0607/61/M/J/22 © UCLES 2022 4 Amara now also uses the dimensions of each bin and of each item to find the maximum number of items that will go into a bin. Each item is a cuboid and is the same height as the bin. Amara looks at the rectangular top of each bin to work out the maximum number of items that will go into it. Storage bin View of top of storage bin NOT TO SCALE A bin is now full either because it contains the maximum total mass or because it contains the maximum number of items that will go into it. She uses this method. Method 4 Work out the maximum number of items for each bin using dimensions. Then use Method 2. (a) The rectangular top of Amara’s bin measures 90 cm by 90 cm. The rectangular top of each item measures 60 cm by 30 cm. Find the maximum number of items that will go into a bin using these dimensions. … [2]

9 0607/61/M/J/22 © UCLES 2022 [Turn over (b) These are the masses, in kg, of eleven items with tops measuring 60 cm by 30 cm. 41 33 22 18 16 14 8 7 6 5 4 The maximum total mass that each bin can hold is 90 kg. The top of each bin measures 90 cm by 90 cm. Each bin costs $13.50 . Use Method 4 to put these items into bins. Show that this is a best solution and find the cost of this solution. Bin Mass of items in bin Unused mass in bin Number of items 1 2 3 4 5 $ … [5]

10 0607/61/M/J/22 © UCLES 2022 B MODELLING (QUESTIONS 5 to 7) HALF-LIVES (30 marks) You are advised to spend no more than 50 minutes on this part. Some chemicals decay naturally. This means that their mass reduces. This task is about the decreasing mass of these chemicals. The half-life is the time it takes for the mass of a chemical to become half of its mass. Lee uses different models to look at some of these chemicals. 5 (a) The half-life of chemical A is 100 minutes. Lee has 400 grams of this chemical. Lee uses this model for the mass, in grams, of chemical A which remains at time t minutes. ( ) A t 400 2 1 t 100 = b l At the end of 1 half-life, t = 100 and half of the mass of chemical A, 200 g, remains. At the end of 2 half-lives, t = 200 and 100 g of chemical A remains. (i) Find the mass, in grams, of chemical A which remains at the end of 4 half-lives. … [2] (ii) On the axes below, sketch the graph of ( ) A y t = for t 0 600 G G . t y 0 600 Time (minutes) Mass (grams) [2] (iii) Find the decrease in mass from t = 50 to t = 100. … [2]

11 0607/61/M/J/22 © UCLES 2022 [Turn over (iv) Find the number of half-lives when the mass remaining is 10 grams. Give your answer correct to 2 decimal places. … [3] (b) The half-life of chemical B is 30 minutes. Lee has 200 grams of this chemical. (i) Change the model from part (a) to find the mass, in grams, of chemical B which remains at time t minutes. ( ) B t 2 1 # ff kk = e o [1] (ii) Find the fraction of the mass of chemical B which remains at the end of 300 minutes. … [2] (c) The quarter-life is the time it takes for the mass of a chemical to become a quarter of its mass. At the end of 1 quarter-life, one quarter of the mass remains. The quarter-life of chemical C is 48 minutes. Lee has 240 grams of this chemical. He changes the model from part (a) so that it finds the mass, in grams, of chemical C which remains at time t minutes. His model is valid for 6 quarter-lives. Complete Lee’s model. ( ) C t = … for … t G G … [3]

12 0607/61/M/J/22 © UCLES 2022 6 Lee now uses this model in his calculations for the mass of a chemical which remains at time t minutes. ( ) N N t 3kt 0 # = where k is a constant (a) (i) Explain why N0 is the mass of the chemical Lee starts with. … [1] (ii) Write an expression in terms of N0 for the mass remaining at the end of 1 half-life. … [1] (iii) Lee thinks 3 2 1 kH = , for any starting mass N0 and any half-life H minutes. Show that Lee is correct. [1] (b) The half-life of chemical D is 10 minutes. (i) Use logarithms to show that the value of k for chemical D is .0 063 - , correct to 3 decimal places. [2]

13 0607/61/M/J/22 © UCLES 2022 [Turn over (ii) Lee has 40 grams of chemical D. Use N(t) to find the time it takes for the mass of chemical D to become 10% of its mass. … [4]

14 0607/61/M/J/22 © UCLES 2022 7 Lee has 60 grams of chemical E. This decreases by 11 grams in 1.6 minutes. (a) Use ( ) N N t 3kt 0 # = , where k is a constant, to find the half-life of chemical E in minutes. … [5] (b) Change the model from Question 5(a) to find the mass, in grams, of chemical E which remains at time t minutes and use it to check your answer to Question 7(a). [1]

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16 0607/61/M/J/22 © UCLES 2022 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 10 printed pages. © UCLES 2022 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/61 Paper 6 (Extended) May/June 2022 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 2022 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 2022 © UCLES 2022 Page 2 of 10 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 descriptors 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 2022 © UCLES 2022 Page 3 of 10 Maths-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, 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 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 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 2022 © UCLES 2022 Page 4 of 10 Question Answer Marks Partial Marks 1(a) Bin Mass of items in bin Unused mass in bin 1 38, 6, 7, 9 22 16 9 [0] 2 21, 32 39 7 3 50 10 4 15, 27 45 18 5 25 35 4 B1 for 15 as first entry in bin 4 B1 for 60 – their 15 unused in bin 4 B1 for 9 as the only extra entry in bin 1 1(b) Bin Mass of items in bin Unused mass in bin 1 8, 10 12 2 2 16, 3 [4] 1 3 13, 5 [7] 2 4 5 2 B1 for one correct row or for all 5 items correctly placed or for finding the correct unused masses for their positions of all 5 items or for rows 2 and 3 reversed 2(a) 50, 38, 32, 27, 25, 21, 15, 9, 7, 6 1 2(b) Bin Mass of items in bin Unused mass in bin 1 50, 9 [10] 1 2 38, 21 [22] 1 3 32, 27 [28] 1 4 25, 15, 7, 6 [35] [20] [13] 7 5 2 B1 for 2 correct rows or for their first 5 items correctly placed

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 5 of 10 Question Answer Marks Partial Marks 2(c) Correct calculation to find either percentage e.g. Method 1 [ ] 70 100 300 × or [ ] 5 60 230 100 5 60 × − × × Method 2 [ ] 10 100 240 × or [ ] 4 60 230 100 4 60 × − × × C1 Method 1 23.3[3...%] Method 2 4.17 or 4.166 to 4.167[%] Valid comparison e.g. Method 2 gives approximately 19% less waste 3 B1 for each percentage or SC1 for Method 1 70 100 30.4[3...%] 230   × =     or Method 2 10 100 4.35 230   × =     or 4.347 to 4.348[%] or SC1 for the correct percentages incorrectly labelled B1 FT for a valid comparison, dependent on their two percentages 2(d) 270 ÷ 80 = 3.375 oe or 3 × 80 + 30 [=] 270 oe or 4 × 80 = 320 and 3 × 80 = 240 oe or 270 ÷ 3 = 90 and 270 ÷ 4 = 67.5 oe 2 M1 for 270 ÷ 80 or 3.375 or 4 × 80 = 320 or 3 × 80 = 240 oe or 270 ÷ 3 = 90 or 270 ÷ 4 = 67.5 oe

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 6 of 10 Question Answer Marks Partial Marks 2(e) 28, 13, 7, 4 or 30, 28, 21, 13 C1 oe or oe Bin Mass of items in bin Unused mass in bin 1 30, 10 [10] [0] 2 21, 19 [19] [0] 3 28, 7, 4 [12] [5] [1] 4 13 [27] 5 Bin Mass of items in bin Unused mass in bin 1 10, 19, 7, 4 [30] [11] [4] [0] 2 30 [10] 3 28 [12] 4 21, 13 [19] [6] 5 2 B1 for the first 2 rows correct or B1 for the last 2 rows correct or B1 for the first 2 rows correct or B1 for the last 2 rows correct 3.3 or [3 × 40 =] 120 and 132 seen or 132 ÷ 3 = 44 seen or an argument such as Bin 1 30 10 full Bin 2 21 19 full Bin 3 has to have 28 and 13 to improve on 4 bins, but 28 + 13 = 41 > 40 So Bin 4 is required. so [4 bins] is a best solution 1 FT their number of bins used

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 7 of 10 Question Answer Marks Partial Marks 3 Bin Mass of items in bin 1 9, 4, 7 2 15, C1 Bin Mass of items in bin 1 15, 4 2 9, 7 C1 x + y ⩽ 5 oe or Bin Mass of items in bin 1 9, 4, 7 2 15, 3, 2 Bin Mass of items in bin 1 15, 4 2 9, 7, 3 3 2 C1 x = 2 and y = 3 or x = 3 and y = 2 1 4(a) Diagram showing 3 or 4 items correctly or reasoning using dimensions e.g. ‘3 items uses 90 by 60 and that leaves 90 by 30, so one more will fit’ C1 4 1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 8 of 10 Question Answer Marks Partial Marks 4(b) Bin Mass of items Unused mass in bin Items 1 41, 33, 16 [49] [16] [0] [3] 2 22, 18, 14, 8 [68] [50] [36] [28] [4] 3 7, 6, 5, 4 [83] [77] [72] [68] [4] 2 B1 for any two correct rows Evidence of 3 bins as best solution for items: 11 ÷ 4 = 2.75 or 2 × 4 = 8 and 3 × 4 = 12 oe 2 B1 for evidence of best solution for mass e.g. Mass 174 ÷ 90 = 1.9[33...] or 1 bin = 90 and 2 × 90 = 180 oe and 174 seen OR partial evidence of best solution for items e.g. 11 ÷ 4 or 2.75 or 12 ÷ 4 or 2 × 4 = 8 or 3 × 4 = 12 40.5[0] 1 dep on correctly packing the 3 bins, ignoring unused masses/number of items or evidence of the 3 bins as best solution 5(a)(i) 100 ÷ 2 = 50, 50 ÷ 2= 25 oe or 400 100 1 400 2       or 4 1 400 2       C1 25 1 5(a)(ii) Correct sketch 600 t y 0 1 y-intercept at 400 C1 5(a)(iii) 50 100 100 100 1 1 400 400 2 2       −           or 282.8... – 200 oe seen C1 82.8 1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 9 of 10 Question Answer Marks Partial Marks 5(a)(iv) A sketch of y = 10 on the diagram in (ii) or on a new sketch in this part or 100 1 10 400 2   =     t C1 532[.19...÷ 100] or log(0.025) log(0.5) [× 100] C1 5.32 1 Alternative method 1 10 400 2   =     T (C1) Sketch of relevant functions or log(0.025) log(0.5) (C1) 5.32 (1) 5(b)(i) 30 1 B( ) 2 200  =     t t 1 5(b)(ii) 300 30 1 200 2       or 10 1 200 2      or 300 30 1 2       or 10 1 2       oe C1 1 1024 1 5(c) [ ] 48 1 C( ) 240 4   =     t t 2 B1 for two of 240, 1 4 or 48 t oe correctly used in the model for 0 ⩽ t ⩽ 288 1 6(a)(i) Valid explanation e.g. N(0) = N0 × 30 or N(0) = N0 × 1 or When t = 0, 3kt = 1 1 6(a)(ii) 0 N 2 oe 1 6(a)(iii) 0 0 N N 3 2 = × kH leading to 1 3 2 = kH 1

0607/61 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 10 of 10 Question Answer Marks Partial Marks 6(b)(i) 310k = 0.5 soi M1 1 log 2 10 log3 = k or [ ] 1 log 2 10log3 = k oe leading to k = −0.0630[9...] or – 0.0631 A1 6(b)(ii) relevant sketch of intersecting graphs or 0.063 4 40 3− = × t or 0.063 1 3 10 − = t oe C1 y = 4 or y = 0.1 indicated on relevant sketch of intersecting graphs or 1 log10 0.063 log3 = − t C1 33.2 to 33.3 1 Correct units stated e.g. Minutes C1 7(a) relevant sketch of intersecting graphs or 1.6 49 60 3 = × k or 1.6 49 3 60 = k C1 y = 49 or y = 49 60 indicated on relevant sketch of intersecting graphs or evidence of a calculation for the value of k e.g. 49 log 60 1.6 log3 = k C1 k = −0.115[2...] 1 relevant sketch of intersecting graphs to solve 30 = 60 0.1152... 3− × H or Evidence of a calculation for the value of the half-life H e.g. 1 log 2 0.115 log3 = − H C1 FT their value of k awrt 5.48 or 5.49 1 7(b) Valid verification e.g. 1.6 5.476... 1 60 2       = 49 or 0.292… 1 60 2       = 49 1