1. AS Maths
Decision Paper
January 2006
Model Answers

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

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

5. 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

6. Initial Set
After 1st Pass
After 2nd Pass
After 3rd Pass
All correct
Indicates pivots
Indicates used pivots

7. 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

8. 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

9. A B C D E F G H I J

10. 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)

12. Corresponding Route is A , C, E, D, F, G, H, I, J20 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

14. 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

15. 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

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

18. The COMBINATIONS are
AB + CI = = 540
AC + BI = = 600
AI + BC = = 500
REPEAT the SHORTEST DISTANCE
AI + BC = = 500
TOTAL DISTANCE
= 2590

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

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

22. Route L,N,O,L
L → N → O → L =
Total = 70
Route L,O,N,L
L → O → N → L =
Total = 95

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

24. = 145 minutes ci) S → P → O → L → N → R → S=
cii) It is a TOUR that can be IMPROVED upon

25. = 138 minutes ciii) S → R → O → L → N → P → S=

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

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