AS Maths Decision Paper January 2012 Model Answers

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

Grade Boundaries Grade A B C D E Mark 62 56 50 44 39

4 Sub Groups 2 Sub Groups Arrange each Sub Group 36 37 24 25 13 16 36 37 24 25 13 16 11 12 2 Sub Groups 36 24 13 11 37 25 16 12 13 16 36 37 11 12 24 25 Arrange each Sub Group 13 11 16 12 36 24 37 25 11 12 13 16 24 25 36 37

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.

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

The two minimum spanning trees are B 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

These are the ODD vertices

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

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.

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

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

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

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

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

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

bi) B A D E F G C B 2 4 4 10 20 32 8 = 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

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

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 43 + 11 = 54 CA = 6 FD = 12 GF = 20 Total = 43

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

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

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

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 13 4 < x < 14 3 As x is an integer it must = 4 x = 4