1. AS Maths
Decision Paper
January 2010
Model Answers

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

3. 1a)

4. A M
B N
C P
D R
E S
F V

5. D – R – B – N – C – V F – R – D – S – E – P – A – M A M B N C P D R
INITIAL MATCH
A M
B N
C P
D R
E S
F V
D – R – B /
– N – C /
– V
F – R – D /
– S – E /
– P – A /
– M
FINAL MATCH
AM, BN, CV, DS, EP, FR

6. Initial List
1c 0s
1c 1s
1c 1s
1c 1s
1c 1s
1c 1s
1c 1s
After First Pass
1c 1s
1c 1s
1c 1s
1c 1s
1c 1s
1c 1s
After Second Pass
1c 0s
1c 1s
1c 0s
1c 1s
1c 1s
After Third Pass
After Fourth Pass
After Fifth Pass
After Sixth Pass
Comparisons
Swaps
1st Pass
2nd Pass
3rd Pass 7 6 6 6 5 3

7. x ≥ 0, y ≥ 0
Write the inequalities as equations
x + 4y ≤ 36
4x + y ≤ 68
Plot these on the graph and shade the UNWANTED region
y ≤ 2x
y ≥ ¼x

8. 4x + y = 68
x + 4y = 36
y = 2x
FEASIBLE
REGION
y = ¼x

9. Maximum value for P = x + 5y is

10. AC = 13
AE = 14
EI = 15
CD = 16
CH = 20
EF = 21
FB = 19
BG = 19
Total Spanning tree = 137

11. G B C A F H E D I

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

13. B E C D A B =
12.0 or 12
B D A C E B =
13.5

14. The best upper bound is the lowest value, so 12.0
From To
The best upper bound is the lowest value, so 12.0
B A D E C B =
12.1

16. Line 10 A B N T D H E 1 5 2
Line 20
Line 30 1
Line 40 2
Line 50 1
Line 60
126
Line 70 3
Line 90
180
Line 70 5
Line 110
Print Area (T x E) = 180

17. Line 10 A B N T D H E 1 5 4
Line 20
Line 30 1
Line 40 1
Line 50
0.5
Line 60
126
Line 70 2
Line 90
142
Line 70 3
Line 90
196
Line 70 4
Line 90
324
Line 70 5
Line 110
Print Area (T x E) = 162

19. 25 24 5 15 27
38 + x + y 10 9 20
18 + x + y 50 8 19
28 + 3x + y 18
As all three routes are the same weight then 16
3x + y = 22 6 28
x + y = 12
2x = 10
28 + 3x + y = 50, so 3x + y = 22
x = 5
5 + y = 12
y = 7
and 38 + x + y = 50, so x + y = 12 26 25
So x = 5 and y = 7

21. Type A x + 3y + 4z
≤ 360
Type B x + y + 5z
≤ 300
Type C x + 3y + 2z
≤ 400
Type A > Type B x + 3y + 4z >
3x + y + 5z
You must leave at least ONE value on the left, as y would be the only value to leave a positive value, bring the y’s over from the right and the x and z’s to the left.
2y > x + z

22. 5x + 4y + 9z ≥ 4x + 3y + 2z
So x + y + 7z ≥ 0
4x + 3y + 2z ≥ 0.4 (9x + 7y + 11z)
20x + 15y + 10z ≥ 2 (9x + 7y + 11z)
20x + 15y + 10z ≥ 18x + 14y + 22z
You must leave at least ONE value on the left, as x and y would leave positive values bring the x and y’s over from the right
2x + y ≥ 12z