GCE Further Mathematics A Y534/01: Discrete Mathematics Advanced Subsidiary GCE Mark Scheme for Autumn 2021

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Oxford Cambridge and RSA Examinations GCE Further Mathematics A Y534/01: Discrete Mathematics Advanced Subsidiary GCE Mark Scheme for Autumn 2021Oxford Cambridge and RSA Examinations OCR (Oxford ... Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. © OCR 2021Y534/01 Mark Scheme October 2021 Annotations and abbreviations Annotation in RM assessor Meaning  and  BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded0,1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0,1 SC Special case ^ Omission sign MR Misread BP Blank Page Seen Highlighting Other abbreviations in mark scheme Meaning dep* Mark dependent on a previous mark,indicated by *.The * may be omitted if only one previous M mark cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working AG Answer given awrt Anything which roundsto BC By Calculator DR This question included the instruction: In this question you mustshow detailedreasoning.Y534/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 1 (a) 5 partitions into a set of size 1 and a set of size 4 B1 1.1 5 where smaller set has size 1 or 5C1 = 5 {X | X, X, X, X} 10 partitions into a set of size 2 and a set of size 3 B1 2.5 10 where smaller set has size 2, with an explanation of why it is 10 {X, X | X, X, X} because there are 5C2 choices for the set of size 2 (note the totalof 15 is given in the question) e.g. 5C2 = 10 or (5×4) ÷ 2 = 10 or 4 + 3 + 2 + 1 = 10 Alternative solution {A}, {B, C, D, E} {B}, {A, C, D, E} B1 List (or any equivalent) that has exactly 5 distinct cases where {C}, {A, B, D, E} {E}, {A, B, C, D} {D}, {A, B, C, E} smaller set has size 1 May just list one set, e.g. {A}, {B}, {C}, {D}, {E} {A, B}, {C, D, E} {A, C}, {B, D, E} List (or any equivalent) that has 10 distinct cases sets where smaller {A, D}, {B. C. E} {B, C}, {A, D, E} {B, E}, {A, C, D} {C, E}, {A, B, D} {A, E}, {B, C, D} {B, D}, {A, C, E} {C, D}, {A, B, E} {D, E}, {A, B, C} B1 set has size 2 May just list one set, e.g. {A, B}, {A, C} {A, D}, {A, E}, {B, C}, {B, D}, {B, E}, {C, D}, {C, E}, {D, E} [2] 1 (b) Partitions into sets of sizes 1, 1 and 3 M1 1.1 Considering cases where set sizes are 1, 1, 3 5 × 4 ÷ 2 = 10 partitions of this type A1 2.1 Explanationof why there are 10 of these e.g. 5C3 = 10 or 5 × 4 ÷ 2 = 10 or a list of the cases Partitions into sets of sizes 1, 2 and 2 M1 1.1 Considering cases where set sizes are 1, 2, 2 5 × (4C2 ÷ 2) = 5 × 3 = 15 partitions of this type A1 2.1 Explainingwhy there are 15 of these e.g a relevant calculation or list of cases [4] 1 (c) 10 partitions into sets of sizes 1, 1, 1, 2 M1 2.1 Trying to dealwith the cases when there are more than 3 subsets 1 partition into sets of sizes 1, 1, 1, 1, 1 May be implied from answer 51 15 + 25 + 10 + 1 = 51 A1 1.1 51 [2] 1 (d) Number line is split into 6 pieces B1 2.1 6 pieces But there are 8 numbers B1 2.2a Using pigeonhole, or explainingwhy there must be at least one piece Hence result by the pigeonhole principle with two or more numbers [2]Y534/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 2 (a) (i) Next-fit method M1 A1 [2] 1.1 1.1 Bins 1 and 2 correct All correct 2 (a) (ii) First-fit method M1 A1 [2] 1.1 1.1 Bins 1 and 2 correct All correct 2 (a) (iii) 23 18 15 12 8 7 5 First-fit decreasingmethod M1 A1 [2] 1.1 1.1 Ordered list may be seen Bins 1 and 2 correct All correct 2 (b) With ‘online’ lists the items are presented one at a time and the whole list is not known untilthe end. With next-fit and first-fit the items are placed in the order they appear in the list, so these methods can be used ‘offline’ or ‘online’. However, for first-fit decreasingthe whole list needs to be known before it can be sorted, so first-fit decreasing can only be used for an ‘offline’ list. B1 B1 [2] 1.2 2.3 Evidence of understandingwhat ‘online’ means Evidence of realising that ffd cannot be used with an online list (or implied from an appropriate statement aboutnext-fit andfirst-fit) Bin 1 12 Bin 2 23 Bin 3 15 Bin 4 18 8 Bin 5 7 5 Bin 1 12 15 Bin 2 23 7 Bin 3 18 8 Bin 4 5 Bin 5 Bin 1 23 7 Bin 2 18 12 Bin 3 15 8 5 Bin 4 Bin 5Y534/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 2 (c) 88 ÷ 4 = 22, so M is at least 22 But it is not possible to fill 4 bins of capacity22 Since 22 – 18 = 4 which is less than 5 So the 23 would have to be split as 4 and 19 And then there is no 3 to go with the 19 M = 23 is possible e.g. 23 – x and x, 18 + 5, 15 + 8, 12 + 7 Hence, least M is 23 M1 A1 B1 [3] 1.1 2.4 2.2a Identifyingthat M must be at least 22 Showing that M = 22 is not possible Fully correct explanation Showing that M = 23 is possible 3 (a) ABXE B1 [1] 1.1 3 (b) M1 A1 M1 A1 [4] 1.1 1.1 1.1 1.1 AB = 0.6, AC = 1.1 AD = 2.7, AE = 2.8 DF = 0.6, EF = 0.7 BF = 2.7, CF = 2.5 AF = 3.3 or ft from other values 3 (c) AB = 0.6 AC = 1.1 B D CE = 1.8 EF = 0.7 A X Y • F DF = 0.6 4.8 C E M1 A1 B1 ft [3] 3.1b 3.2 a 1.1 A graph that connects {A, B, C, D, E, F} with or without X and/or Y Correct tree drawn or arcs listed, including CX and XE 4.8 (km) or totalfor their tree 3 (d) Adaptingthe answer to part (c) B – A – C – X – Y – E – F – D M1 A1 3.1b 1.1 Any walk or cycle that starts at B anduses every vertex atleast once, including X and Y cao [2] A 0.6 B 1.1 1.7 C 2.7 2.1 2.5 D 2.8 2.2 1.8 1.2 E 3.3 2.7 2.5 0.6 0.7 FY534/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 4 (a) (i) Points won Mia by Mia X Y Z B1 1.1 15, 9 and 6 allcorrect 4 (a) (ii) Points won Mia by Li X Y Z [1] B1 1.1 Subtract 4.5 from each value in the table showingpoints won by Li All correct Or any positive multiple of this table Alternative solution Points won by Li minus points won by Mia B1 All correct (not follow through from (a)) 4 (b) 4 (c) Row Z dominates row Y Column Ydominates column Z Points won by Li Mia X Y Li X 0.5 -10.5 Z -5.5 -0.5 Rowminima are –10.5 and –5.5 Row maximin = –5.5 Li’s play-safe is strategy Z Mia should play strategy X [1] B1 B1 [2] M1 A1 ft B1 ft [3] 1.1 1.1 1.1 1.1 1.1 Row Y removed, seen in resulting 2×2 table Column Z removed, seen in resulting 2×2 table Follow through their table from part (a)(ii) or using original table Follow through their table from part (b) or original tables Working may be seen on table from part (b) Seen or implied from rowminima, or identifyingplay-safe for Li Or indicated in table, or implied from their row maximin value stated Correct choice for their table from part (b) e.g. X Y X 5 -6 Z -1 4 e.g. X Y X 1 -21 Z -11 -1 X 4 15 9 Li Y 11 6 5 Z 10 5 1 X 0.5 -10.5 -4.5 Li Y -6.5 -1.5 -0.5 Z -5.5 -0.5 3.5 Points won Mia by Li X Y Z X 1 -21 -9 Li Y -13 -3 -1 Z -11 -1 7Y534/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 5 (a) (i) Ignore any extra lines (e.g. profit lines) or working for parts (b), (c) M1 1.1 Line 2x + 3y = 12 through (6, 0) and (0, 4) M1 1.1 Line x + y = 10 through at least two of (10, 0), (2, 8), (4, 6), (6, 4), (8, 2) and (0, 10) M1 1.1 Line 5x + 2y = 30 through at least two of (6, 0), (4, 5), (2, 10) and (0, 15) A1 1.1 Feasible region identified and correct [4] 5 (a) (ii) x y P = 4x – y M1 3.1a ‘Determine’ so method must be seen, not implied Checkingat least two of their vertices or sliding a profit line (a line of gradient 4 anywhere on graph or indicating the vertex (6, 0)) 0 4 –4 0 10 –10 3.33 6.67 6.67 6 0 24 Maximum P = 24 A1 1.1 24 [2] 5 (b) FR has boundaries x = 0, x + y = k, 2x + 3y = 12 Not graphical x + y = k and 2x + 3y = 12 or 4x – y = 3 M1 3.4 Vertex where 2x + 3y = 12 and x + y = k or profit on line x + y = k Profit line 4x – y = 3 cuts 2x + 3y = 12 at (1.5, 3) M1 3.1a Calculate where profit = 3 on boundary 2x + 3y = 12 or (1.5, 3) k = 4.5 A1 2.2a 4.5 Alternative solution 4x – (k – x) = 3 ⇒ 3x – k = 3 M1 Use x + y = k to substitute for y (or x) in 4x – y = 3 and 2x + 3(k – x) = 1 ⇒ 3k – x = 12 M1 Form a second simultaneous equationin the sameunknowns k = 4.5 A1 4.5 [3]Y534/01 Mark Scheme October 2021 Question Answer Marks AO Guidance 5 (c) Profit line 4x – y = 3 cuts 5x + 2y = 30 at 3 , 1 13 13 k = 141 13 M1 A1 [2] 3.1a 2.2a Not graphical Or 2 1 , 8 1 or (2.7 to 2.8, 8.0 to 8.2) 13 13 Or 10 11 or 10.8 to 10.9 13 6 (a) (i) 4 4 B(1) 6 7 A(4) D(1) 0 0 5 5 E(2) C(3) 8 8 F(1) H(1) G(1) 9 9 M1 A1 M1 M1 3.3 1.1 3.4 3.4 Activity network with A, B and C correct D, E, F, G, H and dummycorrect (accept directions missing) Forward pass attempted, or implied from min duration correct Backward pass attempted, or implied from criticalactivities correct 7 7 Minimum time = 9 hours A1 1.1 9 A, B, E, G, H have no float A1 1.1 A, B, E, G, H (in any order) and no others [6] 6 (a) (ii) Assuming that there are enough workers for each activity Resourcingmay restrict howmany activities can happen together B1 [1] 3.5b A reason why it may not always be possible to do allthe activities that are needed at the same time NOT an assumptionabout the durations or immediate predecessors or that would delay the start time of an activity (e.g. weather or delays in arrivalof materials) 6 (b) Earliest time that E can start is 5 hours from start B1 1.1 5 (all the activities that must be done before E havemin completion time 5) If there are not enough workers then A, B, C may need to be done one after another, taking8 hours. And E could also be delayed untilafter Dand F, giving a latest start time for E of 10 hours M1 A1 3.5a 2.2b Recognising that tasks may be done sequentially (or implied from answer 8, 9 or 10) 10 (all the activities that can be done before E have totalduration10 – startingE after 10 would be an unecessary delay) [3] 6 (c) Extend the durationof Dto 3 hours B1 [1] 3.5c Or add an activity immediately after Dof duration2 hoursOCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre [Show More]

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