OCR GCE Further Mathematics A Y544/01: Discrete Mathematics Advanced GCE Mark Scheme for Autumn 2021

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Question Answer Mark AO Guidance 2 (a) (i) 1 because the graph is connected B1 1.1 ‘1’ and explaining why 0 is not possible [1] (ii) 4 because if there is a vertex of degree 5 then the other th... ree vertices must all be even and have degree sum 16 – (3+4+5) = 4, which is not possible B1 1.1 ‘4’ and explaining why 5 is not possible [1] 2 (b) 1, 2, 2, 3, 4, 4 2, 2, 2, 3, 3, 4 M1 A1 1.1 1.2 Values in either list correct, given in any order Both lists correct, in this form, each in increasing order (allow decreasing order) and no others [2] 2 (c) J K L M N Indegree 2 2 4 2 2 Outdegree 2 2 2 4 2 M1 A1 1.1 1.2 Indegree = column total, outdegree = row total Both rows correct, although rows may be interchanged Completely correct [2] 2 (d) (i) Not Eulerian, for a digraph to be Eulerian the indegree must equal the outdegree for each vertex B1 1.1 Not Eulerian, indegree  outdegree for L (or M) Discussion of odd/even vertex degrees is wrong here, need to use indegrees and outdegrees for a digraph Saying ‘Eulerian since all even’ is also wrong [1] (ii) Connected but not simple, there are two arcs from M to L B1 1.1 Not simple with a valid reason Alternative method Not simple, there is a (directed) loop at M [1]Y544/01 Mark Scheme October 2021 8 Question Answer Mark AO Guidance 3 (a) 150 cards shared equally between 7 (= 6 players and stack) = 21.43 So at least one has 22 or more cards B1 2.2a Or 217 = 147 < 150 (conclusion may be implied, since given in question) [1] 3 (b) One-digit numbers: 1, 4, 6, 7, 8, 9 are not red = 6 cards Two-digit numbers: tens digit must be 1, 4, 6, 7, 8, 9 and units digit can be any of 0, 1, 4, 6, 7, 8, 9 = 67 = 42 cards Three-digit numbers: Hundreds digit is 1, tens digit must be 0, 1, 4 and units digit can be any of 0, 1, 4, 6, 7, 8, 9 = 37 = 21 cards 6 + 42 + 21 = 69 cards with no red digits M1 A1 1.1 1.1 Listing/describing 6 one-digit numbers with no red digits and making a good attempt at counting the number of two-digit numbers with no red digits Or appropriate sight of any two of 6, 42, 21 Or implied from final answer 69 Final answer 69 Alternative method One-digit numbers: 2, 3, 5 are red = 3 cards Two-digit numbers: 9 + 21 + 18 = 48 RR = 3×3 = 9; RN = 3×7 = 21; NR = 6×3 = 18 Three-digit numbers: 6 + 15 + 9 = 30 1RR = 2×3 = 6; 1RN = 2×7 + 1 = 15; 1NR = 3×3 = 9 3 + 48 + 30 = 81 cards with at least one red digit 150 – 81 = 69 cards with no red digits Listing/describing 3 one-digit numbers with a red digit and making a good attempt at counting the number of two-digit numbers with at least one red digit Or appropriate sight of any two of 3, 48, 30 Or implied from 81 or from final answer 69 Final answer 69 [2] 3 (c) Cards in set A are multiples of 2, so units digit is 0, 2, 4, 6 or 8 One-digit numbers: 4, 6, 8 are not red = 3 cards Two-digit numbers: tens digit can be any of 1, 4, 6, 7, 8, 9 and units digit can be any of 0, 4, 6, 8 = 64 = 24 cards Three-digit numbers: tens digit must be 0, 1, 4 = 34 = 12 cards 3+24+12 = 39 cards in A with no red digits M1 A1 3.1a 1.1 4, 6, 8 or 0, 4, 6, 8 as units digit Trying to count even cards (or odd cards) with no red digits (or with red digits) [Show More]

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