Centre Number Candidate Number For Examiner s Use Surname Other Names Candidate Signature Examiner s Initials Mathematics Unit Decision 1 Wednesday 9 June 2010 General Certificate of Education Advanced Subsidiary Examination June 2010 1.30 pm to 3.00 pm For this paper you must have: the blue AQA booklet of formulae and statistical tables. You may use a graphics calculator. Time allowed 1 hour 30 minutes MD01 Instructions Use black ink or black ball-point pen. Pencil or coloured pencil should only be used for drawing. Fill in the boxes at the top of this page. Answer all questions. Write the question part reference (eg (a), (b)(i) etc) in the left-hand margin. You must answer the questions in the spaces provided. Do not write outside the box around each page. Show all necessary working; otherwise marks for method may be lost. Do all rough work in this book. Cross through any work that you do not want to be marked. The final answer to questions requiring the use of calculators should be given to three significant figures, unless stated otherwise. Information The marks for questions are shown in brackets. The maximum mark for this paper is 75. Question 1 2 3 4 5 6 7 8 TOTAL Mark (JUN10MD0101) P28322/Jun10/MD01 6/6/6/ MD01
2 Do not write outside the box Answer all questions in the spaces provided. 1 (a) Draw a bipartite graph representing the following adjacency matrix. 1 2 3 4 5 A 1 0 0 0 1 B 0 1 1 1 0 C 0 1 1 1 0 D 1 0 0 0 1 E 1 0 0 0 1 (2 marks) (b) If A, B, C, D and E represent five people and 1, 2, 3, 4 and 5 represent five tasks to which they are to be assigned, explain why a complete matching is impossible. (2 marks) QUESTION PART REFERENCE (02) P28322/Jun10/MD01
4 Do not write outside the box 2 (a) (i) Use a bubble sort to rearrange the following numbers into ascending order, showing the list of numbers after each pass. 6 2 3 5 4 (3 marks) (ii) Write down the number of comparisons on the first pass. (1 mark) (b) (i) Use a shuttle sort to rearrange the following numbers into ascending order, showing the list of numbers after each pass. 6 2 3 5 4 (4 marks) (ii) Write down the number of comparisons on the first pass. (1 mark) QUESTION PART REFERENCE (04) P28322/Jun10/MD01
6 Do not write outside the box 3 The network shows 10 towns. The times, in minutes, to travel between pairs of towns are indicated on the edges. A 12 B 27 C 21 22 17 16 21 14 D E F 8 18 22 G 20 20 19 11 H 6 I 25 26 10 9 J (a) Use Kruskal s algorithm, showing the order in which you select the edges, to find a minimum spanning tree for the 10 towns. (6 marks) (b) State the length of your minimum spanning tree. (1 mark) (c) Draw your minimum spanning tree. (3 marks) (d) If Prim s algorithm, starting at B, had been used to find the minimum spanning tree, state which edge would have been the final edge to complete the minimum spanning tree. (1 mark) QUESTION PART REFERENCE (06) P28322/Jun10/MD01
8 Do not write outside the box 4 The network below shows 13 towns. The times, in minutes, to travel between pairs of towns are indicated on the edges. The total of all the times is 384 minutes. (a) (b) (i) (ii) Use Dijkstra s algorithm on the network below, starting from M, to find the minimum time to travel from M to each of the other towns. (7 marks) Find the travelling time of an optimum Chinese postman route around the network, starting and finishing at M. (6 marks) State the number of times that the vertex F would appear in a corresponding route. (1 mark) QUESTION PART REFERENCE (a) B 12 A 13 D 11 C 7 19 6 19 10 E F 20 20 20 11 13 20 G 40 H 15 9 22 8 11 I 40 J 5 K 6 L 6 12 9 M (08) P28322/Jun10/MD01
10 Do not write outside the box 5 Phil, a squash coach, wishes to buy some equipment for his club. In a town centre, there are six shops, G, I, N, R, S and T, that sell the equipment. The time, in seconds, to walk between each pair of shops is shown in the table. Phil intends to check prices by visiting each of the six shops before returning to his starting point. G I N R S T G 81 82 86 72 76 I 81 80 82 68 73 N 82 80 84 70 74 R 86 82 84 74 70 S 72 68 70 74 64 T 76 73 74 70 64 (a) (b) Use the nearest neighbour algorithm starting from S to find an upper bound for Phil s minimum walking time. (4 marks) Write down a tour starting from N which has a total walking time equal to your answer to part (a). (1 mark) (c) By deleting S, find a lower bound for Phil s minimum walking time. (5 marks) QUESTION PART REFERENCE (10) P28322/Jun10/MD01
12 Do not write outside the box 6 Phil is to buy some squash balls for his club. There are three different types of ball that he can buy: slow, medium and fast. He must buy at least 190 slow balls, at least 50 medium balls and at least 50 fast balls. He must buy at least 300 balls in total. Each slow ball costs 2.50, each medium ball costs 2.00 and each fast ball costs 2.00. He must spend no more than 1000 in total. At least 60% of the balls that he buys must be slow balls. Phil buys x slow balls, y medium balls and z fast balls. (a) Find six inequalities that model Phil s situation. (4 marks) (b) Phil decides to buy the same number of medium balls as fast balls. (i) Show that the inequalities found in part (a) simplify to give x 5 190, y 5 50, x þ 2y 5 300, 5x þ 8y 4 2000, y 4 1 3 x (2 marks) (ii) Phil sells all the balls that he buys to members of the club. He sells each slow ball for 3.00, each medium ball for 2.25 and each fast ball for 2.25. He wishes to maximise his profit. On Figure 1 on page 14, draw a diagram to enable this problem to be solved graphically, indicating the feasible region and the direction of an objective line. (7 marks) (iii) Find Phil s maximum possible profit and state the number of each type of ball that he must buy to obtain this maximum profit. (4 marks) QUESTION PART REFERENCE (12) P28322/Jun10/MD01
14 Do not write outside the box QUESTION PART REFERENCE (b)(ii) Figure 1 ~ x x 5 190, y 5 50, x þ 2y 5 300, 5x þ 8y 4 2000, y 4 1 3 x 300 350 400 450 50 100 150 200 250 ~ y 300 250 200 150 100 50 0 0 (14) P28322/Jun10/MD01
16 Do not write outside the box 7 A student is testing a numerical method for finding an approximation for p. The algorithm that the student is using is as follows. Line 10 Input A, B, C, D, E Line 20 Let A ¼ A þ 2 Line 30 Let B ¼ B Line 40 Let C ¼ B A Line 50 Let D ¼ D þ C Line 60 Let E ¼ðD 3:14Þ 2 Line 70 If E < 0:05 then go to Line 90 Line 80 Go to Line 20 Line 90 Print p is approximately, D Line 100 End Trace the algorithm in the case where the input values are A ¼ 1, B ¼ 4, C ¼ 0, D ¼ 4, E ¼ 0 (6 marks) QUESTION PART REFERENCE (16) P28322/Jun10/MD01
18 Do not write outside the box 8 A simple connected graph has six vertices. (a) One vertex has degree x. State the greatest and least possible values of x. (2 marks) (b) The six vertices have degrees x 2, x 2, x,2x 4, 2x 4, 4x 12 Find the value of x, justifying your answer. (2 marks) QUESTION PART REFERENCE (18) P28322/Jun10/MD01
MD01 - AQA GCE Mark Scheme 2010 June series Key to mark scheme and abbreviations used in marking M m or dm A B E mark is for method mark is dependent on one or more M marks and is for method mark is dependent on M or m marks and is for accuracy mark is independent of M or m marks and is for method and accuracy mark is for explanation or ft or F follow through from previous incorrect result MC mis-copy CAO correct answer only MR mis-read CSO correct solution only RA required accuracy AWFW anything which falls within FW further work AWRT anything which rounds to ISW ignore subsequent work ACF any correct form FIW from incorrect work AG answer given BOD given benefit of doubt SC special case WR work replaced by candidate OE or equivalent FB formulae book A2,1 2 or 1 (or 0) accuracy marks NOS not on scheme x EE deduct x marks for each error G graph NMS no method shown c candidate PI possibly implied sf significant figure(s) SCA substantially correct approach dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be provided on the mark scheme. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded. 3
MD01 - AQA GCE Mark Scheme 2010 June series MD01 Q Solution Marks Total Comments 1(a) M1 Bipartite graph, 2 sets of (some) vertices, labelled, 6+ edges. 2 All correct (b) 3 letters matched to 2 numbers impossible or E1 OE; PI by subsequent E1 2 letters matched to 3 numbers impossible A, D, E matched to 1, 5 impossible or E1 2 OE B, C matched to 2, 3, 4 impossible Total 4 4
MD01 - AQA GCE Mark Scheme 2010 June series MD01 (cont) Q Solution Marks Total Comments 2(a)(i) (6 2 3 5 4) 2 3 5 4 6 M1 Bubble, condone 1 slip but must have 6 at end of first pass 1st pass correct 2 3 4 5 6 2 3 4 5 6 3 All correct, these 3 lines only Or reverse: (6 2 3 5 4) 2 6 3 4 5 M1 Bubble, condone 1 slip but must have 2 at start of 1st pass 1st pass correct 2 3 6 4 5 2 3 4 6 5 2 3 4 5 6 All correct these 4 lines only (ii) 4 B1 1 NOTE (6 2 3 5 4) 2 3 5 4 6 2 3 5 4 6 2 3 5 4 6 2 3 4 5 6 scores M0 (b)(i) (6 2 3 5 4) 2 6 3 5 4 M1 Shuttle swap 2 and 6 only on 1st pass 2 3 6 5 4 2nd pass 2 3 5 6 4 3rd pass 2 3 4 5 6 4 All correct (ii) 1 B1 1 Total 9 5
MD01 - AQA GCE Mark Scheme 2010 June series MD01 (cont) Q Solution Marks Total Comments 3(a) M1 Kruskal s, 6 + edges stated, not just HI 6 lengths, (no cycles) must be in ascending DE 8 order (condone 1 slip only) IJ 9 IG CG BE 11 14 17 AB BF FI 12 16 19 B1 9 edges IJ 3rd AB 5th BF 7th 6 All correct (b) 112 B1 1 (c) M1 tree 7+ edges 9 edges 3 All correct, including labelling (d) CG B1 1 Total 11 6
MD01 - AQA GCE Mark Scheme 2010 June series MD01 (cont) Q Solution Marks Total Comments 4(a) M1 SCA, cancelling at 2+ vertices Correct values at K, condone no box at 11 m1 3 values at F m1 2 values at E or G m1 2 values at A or C All correct including final values at vertices boxed B1 7 49 at B (b)(i) Odd vertices A, B, C, M E1 PI, CAO AB + CM = 25 + 48 or 73 M1 3 correct sets of lettered pairs of AC + BM = 24 + 49 or 73 candidate s vertices AM + BC = 47 + 23 or 70 A2,1 3 correct, 2 correct Min = 384 + 70 F PI, 384 plus their shortest = 454 B1 6 SC 454 with no working, or 454 with route 2/6 Route without 454 0/6 (ii) 4 B1 1 Total 14 7
MD01 - AQA GCE Mark Scheme 2010 June series MD01 (cont) Q Solution Marks Total Comments 5(a) S T R I N G S M1 Tour starting from any vertex 64 70 82 80 82 72 m1 Visits all other vertices only once Correct order = 450 B1 4 Note: If solution on a matrix then order of selection of vertices must be clearly shown (b) N G S T R I N B1F 1 Must have scored M2 in part (a) Or reverse (c) Delete S M1 Clear method: spanning tree (edges or diagram, not just numbers) with one vertex deleted AND adding 2 edges from deleted vertex (condone double shortest edge from deleted vertex) B1 Spanning tree with 4 edges (may include S) Correct MST + F 2 shortest from candidate s deleted vertex (not shortest edge doubled) = 425 5 SC 425 without earning first M1: 2/5 Total 10 8
MD01 - AQA GCE Mark Scheme 2010 June series MD01 (cont) Q Solution Marks Total Comments 6(a) x 190, y 50, z 50 oe B1 x+ y + z 300 oe B1 2.5x+ 2y+ 2z 1000 oe B1 Strict inequalities: penalise first two ( 5x 4y 4z 2000) + + instances only 60 x ( x+ y+ z) oe B1 4 100 ( 2x 3y+ 3z) (b)(i) y = z x 190, y 50 (ii) x+ 2y 300 oe M1 5x+ 8y 2000 2x 6y y 1 3 x x+ y+ y 300 or 5x+ 4y+ 4y 2000 or 2x 3y+ 3y ie at least one clear line of working showing substitution of y = z oe 2 AG All correct (3 or become and ) For all lines must be correct to 1 2 square horizontal or vertical B1 x = 190, y = 50 B1 through (0,150) and (300,0) B1 through (0,250) and (400,0) M1 y= mx through (0,0) through (300,100) B1 Region must have all lines correct and labelled region (condone lack of shading) B1 7 A correct objective line 9
MD01 - AQA GCE Mark Scheme 2010 June series MD01 (cont) Q Solution Marks Total Comments 6 (b)(iii) 1 1 1 1 1 P = x+ y+ z or x+ y 2 4 4 2 2 M1 PI Max at ( 320,50 ) B1 Profit (160 + 25) = 185 Note: (with no working) 185 3/4 Buys 320 slow, 50 medium, 50 fast B1 4 320 slow, 50 medium, 50 fast 2/4 320 slow, 50 medium, 50 fast and 185 4/4 Total 17 10
MD01 - AQA GCE Mark Scheme 2010 June series MD01 (cont) Q Solution Marks Total Comments 7 A B C D E (1 4 0 4 0) 3 4 5 4 7 4 4 3 (awrt 1.33) 4 5 8 3 (awrt 2.67) 52 15 (awrt 3.5) 4 304 7 105 (awrt (awrt 0.571) 2.9) 0.22404 444 (awrt 0.22) 0.10671 111 (awrt 0.11) 0.0599 (awrt 0.06) M1 M1 M1 1st pass to candidate s 8 3 1st pass all correct to E = 0.22 2nd pass to candidate s 52 15 2nd pass correct to E = 0.11 3rd pass to candidate s 304 105 9 4 4 9 (awrt 0.444) π is approximately 3.34 1052 315 (awrt 3.34) 0.03987 (awrt 0.04) 6 Total 6 All correct and no extra line Final answer 1052 or awrt 3.34 315 11
MD01 - AQA GCE Mark Scheme 2010 June series MD01 (cont) Q Solution Marks Total Comments 8(a) Max 5 B1 Min 1 B1 2 Do not allow 1 or 5 (b) 4x 12 1 (or >0) 13 x 4 Or 4 x 12 5 (or <6) 17 M1 Any one of these inequalities x 4 Or OR 2x 4 5 (or < 6) Exhaustive check of all values from 1 to 5 inclusive, condone one omission. x 9 2 x = 4 2 First inequality and one of the other two, or completely correct exhaustive check, and x = 4 Alternative solution Sum of degrees = 11x 24 must be even x is even x 2 > 0 x > 2 x 5 Hence x = 4 M1 Total 4 TOTAL 75 12
Scaled mark unit grade boundaries - June 2010 exams A-level Max. Scaled Mark Grade Boundaries and A Conversion Points Code Title Scaled Mark A A B C D E HBIO4 GCE HUMAN BIOLOGY UNIT 4 90 67 62 57 52 47 43 HBIO5 GCE HUMAN BIOLOGY UNIT 5 90 70 65 60 55 50 46 HBI6T GCE HUMAN BIOLOGY UNIT 6T 50 42 39 36 33 30 27 HBI6X GCE HUMAN BIOLOGY UNIT 6X 50 37 34 31 29 27 25 INFO1 GCE INFO AND COMM TECH UNIT 1 80-49 43 37 31 26 INFO2 GCE INFO AND COMM TECH UNIT 2 80-49 44 39 34 29 INFO3 GCE INFO AND COMM TECH UNIT 3 100 77 69 61 53 46 39 INFO4 GCE INFO AND COMM TECH UNIT 4 70 63 56 49 42 35 29 LAW01 GCE LAW UNIT 1 96-69 62 56 50 44 LAW02 GCE LAW UNIT 2 94-64 55 46 38 30 LAW03 GCE LAW UNIT 3 80 66 59 52 46 40 34 LAW04 GCE LAW UNIT 4 85 69 62 55 48 41 34 XMCAS GCE MATHEMATICS UNIT XMCAS 125-105 93 81 70 59 MD01 GCE MATHEMATICS UNIT D01 75-61 54 48 42 36 MFP1 GCE MATHEMATICS UNIT FP1 75-68 61 54 47 40 MM1A GCE MATHEMATICS UNIT M1A 100-81 71 61 52 43 MM1A/W GCE MATHEMATICS UNIT M1A - WRITTEN 75 61 33 MM1A/C GCE MATHEMATICS UNIT M1A - COURSEWORK 25 20 10 MM1B GCE MATHEMATICS UNIT M1B 75-64 56 48 40 32 MPC1 GCE MATHEMATICS UNIT PC1 75-63 55 47 40 33 MS1A GCE MATHEMATICS UNIT S1A 100-75 67 59 51 43 MS/SS1A/W GCE MATHEMATICS UNIT S1A - WRITTEN 75 55 33 MS/SS1A/C GCE MATHEMATICS UNIT S1A - COURSEWORK 25 20 10