AS Maths Decision Paper January 2010 Model Answers

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

1a)

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 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

Initial List 13 16 10 11 4 12 6 7 1c 0s 1c 1s 1c 1s 1c 1s 1c 1s 1c 1s 1c 1s After First Pass 13 10 11 4 12 6 7 16 1c 1s 1c 1s 1c 1s 1c 1s 1c 1s 1c 1s After Second Pass 10 11 4 12 6 7 13 16 1c 0s 1c 1s 1c 0s 1c 1s 1c 1s After Third Pass 10 4 11 6 7 12 13 16 After Fourth Pass 4 10 6 7 11 12 13 16 After Fifth Pass 4 6 7 10 11 12 13 16 After Sixth Pass 4 6 7 10 11 12 13 16 Comparisons Swaps 1st Pass 2nd Pass 3rd Pass 7 6 6 6 5 3

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

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

Maximum value for P = x + 5y is

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

G B C A F H E D I

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

B E C D A B 3.7 + 1.9 + 2.7 + 2.0 + 1.7 = 12.0 or 12 B D A C E B 1.8 + 2.0 + 1.9 + 4.2 + 3.6 = 13.5

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 1.7 + 2.0 + 1.7 + 4.2 + 2.5 = 12.1

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

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

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

Type A 2x + 3y + 4z ≤ 360 Type B 3x + y + 5z ≤ 300 Type C 4x + 3y + 2z ≤ 400 Type A > Type B 2x + 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

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