1. AS Maths
Decision Paper
June 2011
Model Answers

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

3. Grade Boundaries
Grade A B C D E
Marks 59 52 45 39 33

5. A 1
B 2
C 3
D 4
E 5
F 6

6. Final Match – 1 then E – 5 – D – 2 – B A1, B6, C4, D2, E3, F5
Initial Match
A 1
B 2
C 3
D 4
E 5
F 6
A 1
B 2
C 3
D 4
E 5
F 6
One solution is
Final Match
E – 5 – D
– 2 – B
– 1 then / /
A1, B6, C4, D2, E3, F5
F – 5 – E
– 3 – A
– 1 – B
– 6 / / /

8. x < 6 x < 4 Initial List 6 4 x 2 11 After 1st pass 4 x 2 6 11
6 swap with 4,
6 swap with x,
6 swap with 2,
x < 6
After 2nd pass x
x < 4
4 swap with x,
4 swap with 2,

9. x < 11 x > 2 x = 3 Initial List 11 x 2 4 6
After 1st pass x
x < 11
11 swap with x
After 2nd pass 2 x
x > 2
11 swap with 2
x swap with 2
c) So x > 2, but < 4 (part 1) then
x = 3

11. AC = 10 CH = 9 FH = 8 CE = 10 ED = 11 Could use CD GH = 13 DB = 14 1 82 6 5 4 7 3
CH = 9
FH = 8
CE = 10
ED = 11
Could use CD
GH = 13
DB = 14

12. 75p AC = 10 CH = 9 FH = 8 CE = 10 ED = 11 GH = 13 DB = 14
Total Spanning Tree = 13 14 G B
75p

13. 70p Delete ALL edges connected to H Total Spanning Tree =10 C 10 E 11 9 11 8 F H D 14 13 14
Total Spanning Tree =
70p G B
From the matrix TWO more edges need to be added,
(AF 11 and any one of GC, GA, GD and GF all 14 are the lowest)

15. 7 18 14 13 17 29 10 22 23 9 8 6 14 13 18 16

16. Route O, F, S, T, E, D
16 minutes
Route O, F, S, R, B

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

20. 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 =
= 168 km

21. 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 =
= 160 km
cii) If she repeats AF she must start / finish at C and D

23. Line A B C D E 10 6 20 7 30
300 40
6.5 50
25.375
100
6.5 40
6.75 50
-7.547 80
6.75 40
6.625 50
9.92
100
6.625 40
6.6875 50
0.92

24. First reason – there is no output
Second reason – Need to know an interval within which the cube root lies at the outset

26. 200x y = → x + 5y ≥ 25
200x y = → 2x + 15y ≥ 60
100x y = → x + 25y ≥ 40
(C =) 2.5x +15y

27. Feasible Region x + 5y ≥ 25 2x + 15y ≥ 60 x + 25y ≥ 40 Objective Line
(C =) 2.5x +15y
2.5x + 15y = 37.5
x = 15, y = 2.5
Objective Line

28. Feasible Region 15 DIY, 2 Trade Cost (C =) 2.5x +15y
2.5 x x 2 =
= £67.50
Objective Line

30. P → U → S → R
This is shorter than any other route

31. 40 24 26 14 40 16 28 26 24 16 12 10 26 28 12 14 26 10

32. ci) Q U S T P R Q
ci) Q → U → S → T → P → R → Q =
119 miles

33. ci) Q U S T P R Q
Q → U
U → S
S → T
T → U → P
P → U → S
S → R
R → Q
cii) Q → U → S → T → U → P → U → S → R → Q

34. Lower Bound 52 + 31 = 83 minutes Remove Q Q 20 11 P U T S R 16 10 1412 12
Either UT or ST as both12
Total = 52
Add lowest 2 from Q
QU = 11 + QR = 20 = 31 Q R 20 U 11
Lower Bound =
83 minutes
This is not an optimal route as at least one edge has to be repeated