Cambridge IGCSE Mathematics - International 0607 — 2022 May/June Paper 5 · Variant 1

This document has 8 pages. [Turn over Cambridge IGCSE™ * 1 3 0 6 2 6 3 9 7 2 * DC (PQ) 303072/2 © UCLES 2022 CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/51 Paper 5 Investigation (Core) May/June 2022 1 hour 10 minutes You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer all questions. ● 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 36. ● The number of marks for each question or part question is shown in brackets [ ].

2 0607/51/M/J/22 © UCLES 2022 Answer all the questions. INVESTIGATION STORAGE BINS 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. Bin Mass of items in bin Unused mass in bin 1 6 4 2 3 4 kg of storage is unused in this bin. The second item will not go in bin 1 Bin Mass of items in bin Unused mass in bin 1 6 4 2 7 3 3 because it is more than 4 kg. Amara puts the second item in bin 2. The third item is 4 kg. Bin Mass of items in bin Unused mass in bin 1 6, 4 4 0 2 7, 2 3 1 3 Amara puts this in bin 1. Bin 1 is now full. The fourth item will go in bin 2. Bin 3 is not used. 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 0607/51/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 (i) Complete Amara’s table to show that she needs 5 bins. [4] (ii) Work out the total unused mass in the 5 bins. … [2] (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 0607/51/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 [3] (c) Work out the difference in the total unused mass when using Method 1 and Method 2. Use your answers from Question 1(a)(ii) and Question 2(b). … [2]

5 0607/51/M/J/22 © UCLES 2022 [Turn over 3 A best solution uses the smallest possible number of bins. (a) (i) 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] (ii) Show that the solution in Question 1(b) is a best solution. [2] (b) Amara knows that for a particular set of items a best solution is 6 bins. The maximum total mass that each bin can hold is 5 kg. The total mass of the items is 27.5 kg. Work out the amount of unused storage for a best solution for these items. … [2]

6 0607/51/M/J/22 © UCLES 2022 4 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. (a) 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 [6]

7 0607/51/M/J/22 © UCLES 2022 [Turn over (b) Amara puts nine items into bins using Method 3. The maximum total mass that each bin can hold is 40 kg. Bin Mass of items in bin Unused mass in bin 1 18, 22 0 2 32, 5, 3 0 3 32 8 4 19, 15 21 6 5 12 28 Amara only wants to use 4 bins. She removes the last item she packed and divides it into two smaller items with the same total mass. She puts each of these two items into a bin that can hold its mass. Work out how much the percentage of unused storage changes when Amara uses 4 bins instead of 5 bins. … [5] Question 5 is printed on the next page.

8 0607/51/M/J/22 © UCLES 2022 5 These are the masses, in kg, of eight items. 31 10 39 20 29 47 50 12 The maximum total mass that each bin can hold is 60 kg. Each bin Amara uses costs $13.50 . Use Method 2 or Method 3 to put these items into bins to give a best solution. Find the cost of this solution. Bin Mass of items in bin Unused mass in bin 1 2 3 4 5 $ … [5] 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.

This document consists of 7 printed pages. © UCLES 2022 [Turn over Cambridge IGCSE™ CAMBRIDGE INTERNATIONAL MATHEMATICS 0607/51 Paper 5 (Core) May/June 2022 MARK SCHEME Maximum Mark: 36 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.

0607/51 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 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 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.

0607/51 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 3 of 7 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

0607/51 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 4 of 7 Question Answer Marks Partial Marks 1(a)(i) 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 9 as the only extra entry in row 1 B1 for 15 as first entry in row 4 B1 for first unused = 60 – their first entry in row 4 1(a)(ii) [0 +] 7 + 10 + their 18 + their 35 or 5  60 – 230 C1 FT their table 70 1 FT their table 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 B1FT for finding the correct unused masses for their positions of all 5 items 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 3 B2 for first 5 items correctly placed or for 2 correct rows or B1 for first 3 items correctly placed or for 1 correct row For B2 or B1 FT their 2(a)

0607/51 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 5 of 7 Question Answer Marks Partial Marks 2(c) their 1 + 1 + 1 + 7 or 60  their 4 – 230 oe or their 70 – their 10 C1 FT their 2(b) table and their 1(a)(ii) 60 1 3(a)(i) 270 ÷ 80 oe = 3.375 or 3  80 + 30 [=] 270 oe soi or 4  80 = 320 and 3  80 = 240 oe or 270 ÷ 3 = 90 and 270 ÷ 4 = 67.5 oe 2 B1 for 270 ÷ 80 or 3.375 or 3  80 + 30 oe soi or 4  80 = 320 or 3  80 = 240 oe, or 270 ÷ 3 = 90 or 270 ÷ 4 = 67.5 oe, 3(a)(ii) 2.75 or 3  20 = 60 and 2  20 = 40 and 55 seen 2 B1 for 55 ÷ 20 or 3  20 = 60 and 55 seen 3(b) 6  5 – 27.5 or 30 – 27.5 or (5 – 27.5 ÷ 6) × 6 C1 2.5 1

0607/51 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 6 of 7 Question Answer Marks Partial Marks 4(a) 28, 13, 7, 4 or 30, 28, 21, 13 C1 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 or 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 3 B2 for first two rows correct or B1 for one of the first two rows correct Yes oe AND 3.3 or [3  40 =] 120 and 132 seen or 132 ÷ 3 = 44 seen 2 B1 for 132 ÷ 40 or 3.3 or either [3  40 =] 120 or 132 seen 4(b) Evidence of 12 removed from bin 5 and 2 items totalling 12 correctly placed in bins 3 and 4 C1   42 100 200  or   8 6 28 100 5 40     oe or   2 100 160  or   2 100 4 40   oe C1 FT their 2 from the table [decreases by] 19.75[%] or 19.8[%] nfww 3 B1 for 21[%] or 1.25[%] seen M1 for their 21  their 1.25 or 42 100 200   2 100 160  seen

0607/51 Cambridge IGCSE – Mark Scheme PUBLISHED May/June 2022 © UCLES 2022 Page 7 of 7 Question Answer Marks Partial Marks 5 3.96... so 4 bins best solution [therefore min cost] C1 Method 2: 50, 47, 39, 31, 29, 20, 12, 10 or Method 3: 47, 39, 20, 12 C1 Method 2 Bin Mass of items Unused mass in bin 1 50, 10 [10] 0 2 47, 12 13 1 3 39, 20 21 1 4 31, 29 [29] 0 2 B1 for first 6 items correctly placed or 2 correct rows or Method 3 Bin Mass of items Unused mass in bin 1 31, 29 [29] 0 2 50, 10 [10] 0 3 47, 12 13 1 4 39, 20 21 1 [$] 54 1