1. The graph is neither Eulerian or Semi – Eulerian as it has 4 ODD vertices.

2. The COMBINATIONS are AB + CI = 100 + 440 = 540 AC + BI = 150 + 450 = 600 AI + BC = 380 + 120 = 500 REPEAT the SHORTEST DISTANCE AI + BC = 380 + 120 = 500 TOTAL DISTANCE 2090 + 500 = 2590 The ODD VERTICES are noted down A, B, C, I

3. The Examiner is looking for the following points The ODD VERTICES are noted down A, B, C, I The COMBINATIONS with DISTANCES AB + CI = 100 + 440 = 540 AC + BI = 150 + 450 = 600 AI + BC = 380 + 120 = 500 REPEAT the SHORTEST DISTANCE AI + BC = 380 + 120 = 500 TOTAL DISTANCE 2090 + 500 = 2590

4. TOTAL NUMBER OF STATUES SEEN IS = B C D E F G H I J 2 + 2 + 3 + 2 + 2 + 3 + 1 + 2 + 1 = 18 Statues As the number of statues seen is found by dividing the edges (INCLUDING THE REPEATED EDGES) at each vertex by 2

5. The ODD VERTICES are noted down A, C, D, F The COMBINATIONS with DISTANCES AC + DF = 18 + 22 = 40 AD + CF = 32 + 30 = 62 AF + CD = 12 + 30 = 42 REPEAT the SHORTEST DISTANCE AC + DF = 18 + 22 = 40 TOTAL DISTANCE 164 + 40 = 204

6. b) Start / Finish A or C Repeat DF = 22 TOTAL DISTANCE 164 + 22 = 186 ci) You need to repeat the shortest pair = AF Repeat AF = 12 TOTAL DISTANCE 164 + 12 = 176 cii) Start / Finish at C / D

7. The ODD VERTICES are noted down E, H, J, K The COMBINATIONS with DISTANCES EH + JK = 69 + 131 = 200 EJ + HK = 93 + 106 = 199 EK + HJ = 129 + 142 = 271 REPEAT the SHORTEST DISTANCE EJ + HK = 93 + 106 = 199 TOTAL DISTANCE 1135 + 199 = 1334

8. See next slide ai) aii) 4b) ODD Vertices at A, D, K, H AD (27) + KH (30) = 57 AH (20) + DK (20) = 40 AK (46) + DH (40) = 86 Repeat AH (20) + DK (20) = 40 Optimum Route 308 + 40 = 348 minutes

9. June 08 The ODD VERTICES are noted down A, B, C, D The COMBINATIONS with DISTANCES AB + CD = 270 + 270 = 540 AC + BD = 290 + 290 = 580 AD + BC = 260 + 270 = 530 REPEAT the SHORTEST DISTANCE AD + BC = 260 + 270 = 530 TOTAL DISTANCE 1920 + 530 = 2450

10. bi) Repeats BC (Length = 1920 + 270 =) 2190 (metres) ci) She needs to repeat the shortest repeated distance which is AD = 260 (Length = 1920 + 260 =) 2180 (metres) cii) B or C

11. AC + DG AD + CG AG + CD AC (19) +DG (15) = 34 AD (24) +CG (10) =34 AG (19) +CD (6) = 25 Repeat AG + CD = 25 ODD vertices at A, C, D, G Total route is 167mins + (AG + CD) 25 =192 minutes D to A = 24

12. The ODD VERTICES are noted down B, C, F, H The COMBINATIONS with DISTANCES BC + FH = 160 + 320 = 480 BF + CH = 280 + 520 = 800 BH + CF = 360 + 210 = 570 REPEAT the SHORTEST DISTANCE BC + FH = 160 + 320 = 480 TOTAL DISTANCE 2410 + 480 = 2890

13. ODD vertices at B, C, D, E BC + DE BD + CE BE + CD BC (22) +DE (18) = 40 BD (38) +CE (27) =65 BE (22) +CD (16) = 38 Repeat BE + CD = 38 Total route is 307m + (BE + CD) 38 =345 metres

14. June 2010 Odd vertices A, B, C, M AB + CM = 25 + 48 or 73 AC + BM = 24 + 49 or 73 AM + BC = 47 + 23 or 70 Repeat AM + BC = 70 Min = 384 + 70 = 454 bii) 8 edges at F so F will be visited 8 ÷ 2 = 4

15. The graph is neither Eulerian or semi Eulerian as it has 4 odd vertices

16. These are the ODD vertices AB + GH AG + BH AH + BG AB (180) +GH (165) = 345 AG (90) +BH (210) =300 AH (150) +BG (210) = 360 Repeat AG + BH = 300 Length of route = 1215 + 300 = 1515 metres

17. Add the repeated edges on to your graph AG (90) + BH (210) = 300 Repeated edges F now has 6 edges so will be visited 6 ÷ 2 times = 3 H now has 4 edges so will be visited 4 ÷ 2 times = 2

20. ODD vertices at A, C, D, F AC + DF AF + CD AD + CF AC (14) + 32 AF (10) +36 AD (26) + 50 Repeat AC + DF = 32 DF (18) = CD (26) = CF (24) = Total route is 150 + (BE + CD) 32 = 182 km

21. As he is starting at an ODD vertex and completing his journey at an ODD vertex (A and C) he only needs to repeat FD Total = 150 + 18 = 168 km

22. ci) She needs to start at an ODD vertex and finish at an ODD vertex and REPEAT the SMALLEST value of the pairs of ODD VERTICES Smallest distance is AF (10) Distance therefore is 150 + 10 = = 160 km cii) If she repeats AF she must start / finish at C and D

23. These are the ODD vertices

24. You need to work out lowest value for all combinations CE + KH = 35 + 24 = 59 CK + EH =25 + 40 = 65 CH + EK =25 +30 = 55 Repeat CH + EK = 55 Total = 224 (all edges) + 55 (CH + EK) = 279 Odd vertices at C, E, H, K

25. Show the repeated edges on the matrix for CH and EK. EK CH As J has SIX edges, it will be visited 6 ÷ 2 = 3 TIMES

26. BD + FH = 210 + 210 = 420 BF + DH = 200 + 180 = 380 BH + DF =260 + 340 =600 Repeat BF + DH = 380 Total = 2430 (all edges) + 380 (BF + DH) = 2810 Odd vertices at B, D, F, H 5b) Repeats D, F 2430 +340 = 2770 5b) Repeat the shortest D, H = 180 2430 + 180 = 2610 Start/finish at B/F

27. Odd vertices A, C, L, G AC + LG = 12 + 15 = 27 AL + CG = 11 + 15 = 26 AG + CL = 21 + 9 = 30 Repeat AL + CG = 26 Min = 134 + 26 = 160 bii) 8 edges at B so B will be visited 8 ÷ 2 = 4