1. AS Maths
Decision Paper
January 2012
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
Mark 62 56 50 44 39

4. 4 Sub Groups 2 Sub Groups Arrange each Sub Group 36 37 24 25 13 16
2 Sub Groups
Arrange each
Sub Group

6. Then A and C can only be matched to 1
Ques 2a
Ques 2b A B C D E F 1 2 3 4 5 6
F must match with 6
∴ E must match with 5
∴ B must match with 2
Then A and C can only be matched to 1
This makes it impossible as 2 people can’t be allocated to the same task.

7. ED = 6
You could have used CD = 10
AC = 8
AD = 10
FG = 11
BE = 12
CF = 16
Total = 63
If you use AD you can’t use CD as it forms a loop
Length of spanning tree = 63

8. The two minimum spanning trees areB E C A D F G B E C A D F G
Note – You must label each vertex with its appropriate letter and put highlight the vertices

9. These are the ODD vertices

10. 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

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

12. Change all the inequalities to equations
y = 20
x + y = 25
5x + 2y = 100
y = 4x
y = 2x
Plot the lines on the graph
Shade the region that doesn’t represents the inequality

13. y = 4x
y = 2x
Feasible
Region
y = 20
x + y = 25
5x + 2y = 100

14. You can use co-ordinates or an objective line
If you use co-ordinates they have to be integers
Minimum for P = x + 2y is x = 5 and y = 2 is P = 45
Minimum for P = -x + y is x = 10 and y = 20 is P = 10

15. See next slide

16. 28 48 47 39 37 83 94 93 92 91
106
101
137
122
121
135
149
145

17. Write the corresponding route
A B E F G H I J
Or the reverse
AJ was 145 miles. It has been reduced by 10 miles so is now
135 miles
The new road connects B to G. You know AB is 28 and GJ is 44 miles. Therefore the new road must be 135 (total distance) – 72 (unaffected distances AB + GJ)
63 miles

19. 8 7 13 4 8 7 13 4 10 19 10 19

20. bi) B A D E F G C B
= 80 miles
bii) B A D E F G E C A B
You have to visit E from G before going to C and
A from C before going to B
biii) The best upper bound is the LOWEST value
= 76 miles

21. AB = 2 3 5 6 1 2 4
BD = 3
CA = 6
FD = 12
GF = 20
Total = 43

22. 7cii
E has been deleted 8 7
Use the table not the diagram to find the two nearest neighbours to E, which are 13 4 8 7 13 4 10 19 10 19
ED = 4 + EB = 7
AB = 2
BD = 3
7 + 4 =11
= 54
CA = 6
FD = 12
GF = 20
Total = 43

23. 7ciii) The best lower bound is the HIGHEST value =64
7d) 64 ≤ T ≤ 76

25. 4x – 13 > 0 (as it has to be a positive integer)
Therefore 4x > 13
Then x > 13 4
So x > 3.25
So x ≥ 4 if it is a positive integer

26. So 2x + 3 > 3x - 5 3x - 5 > x + 1 x + 1 > 4x - 13
After the first pass (2x + 3) is the largest value
After the second pass
(3x -5) is the larger than
After the third pass
(x + 1) is the larger than
So 2x + 3 > 3x - 5
3x - 5 > x + 1
x + 1 > 4x - 13
2x + 3 > x + 1
3x - 5 > 4x - 13
2x + 3 > 4x - 13

27. x + 1 > 4x - 13
So 1 > 3x - 13
So 14 > 3x
Then 3x < 14
So x < 14 3
From part 1 we also know that x > 13 4
Therefore < x < 14 3
As x is an integer it must = 4
x = 4