AS Maths Decision Paper January 2006 Model Answers

It is important students have a copy of the questions as you go through the model answers.

A U B V C W D X E Y F Z

D – X – B – V – C – Z then Final Match F – Y – E – W – A – U Initial Match A U B V C W D X E Y F Z A U B V C W D X E Y F Z D – X – B / – V – C / – Z then Final Match F – Y – E / – W – A / – U AU, BV, CZ, DX, EW, FY

Initial Set 18 23 12 7 26 19 16 24 After 1st Pass 12 7 16 18 23 26 19 24 After 2nd Pass 7 12 16 18 19 23 26 24 After 3rd Pass 7 12 16 18 19 23 24 26 All correct Indicates pivots Indicates used pivots

3ai) Edges in a spanning tree are n – 1 vertices, so 10 – 1 means there would be 9 edges 3aii) Edges in a spanning tree are n – 1 vertices

Minimum Spanning Tree = 89 miles GI = 5 AB = 6 EI = 7 BD = 8 IJ = 10 JH = 11 AF = 13 DE = 14 CG = 15 Total = 89 Each added edge must be in ASCENDING order Minimum Spanning Tree = 89 miles See next slide

A B C D E F G H I J

x ≥ 20 y ≥ 10 x + y ≤ 100 2x + y ≤ 160 y ≤ x + 40 (y – x ≤ 40) To find the maximum and / or minimum values contained within the feasible region, test each vertex Maximum at (2x + 3y) = (30, 70) = 270 Maximum at (3x + 2y) = (60, 40) = 260 x x Minimum at (-2x + y) = (75, 10) = -140 The 5 inequalities that define the region are x x ≥ 20 y ≥ 10 x x x + y ≤ 100 2x + y ≤ 160 y ≤ x + 40 (y – x ≤ 40)

Corresponding Route is A , C, E, D, F, G, H, I, J 20 25 14 13 22 21 26 38 37 50 51 8 14 32 31 44 43 Corresponding Route is A , C, E, D, F, G, H, I, J

P R T I A M Print Line 10 400 Line 20 5 3 Line 30 60 Line 40 460 Line 50 Line 60 12.8 Line 70 12.8

P R T A K I M Print Line 10 400 Line 20 5 3 Line 30 400 Line 40 Line 50 Line 60 1 Line 70 20 Line 80 420 Line 60 2 Line 70 21 Line 80 441 Line 60 3 Line 70 22.05 463.05 Line 80 12.9 Line 100 12.9 Line 110

a) As there are 4 odd vertices at A, B, C and I the graph is neither eulerian or semi – eulerian.

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 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

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

Route L,N,O,L L → N → O → L = Total = 70 35 + 20 + 15 Route L,O,N,L L → O → N → L = Total = 95 30 + 40 + 25

Any cycle that visits each vertex once only and returns to the start vertex. An example being L → N → O → P → R → S → L

= 145 minutes ci) S → P → O → L → N → R → S 20 + 25 + 15 + 35 + 25 + 25 = cii) It is a TOUR that can be IMPROVED upon

= 138 minutes ciii) S → R → O → L → N → P → S 30 + 17 + 15 + 35 + 21 + 20 =

Type A 5x + 4y + 3z ≤ 180 (3 hours is 180 minutes) Type B 12x + 8y + 10z ≤ 240 (4 hours is 240 minutes) Type C 24x + 12y + 18z ≤ 540 (9 hours is 540 minutes) Simplified becomes Type A 5x + 4y + 3z ≤ 180 (No Common Factors) Type B 6x + 4y + 5z ≤ 120 (Divide by 2) Type C 4x + 2y + 3z ≤ 90 (Divide by 6)

x > y y > z x ≥ 0.4 (x + y + z) 5x ≥ 2 (x + y + z) 5x ≥ 2x + 2y + 2z) 3x ≥ 2y + 2z