AS Maths Decision Paper June 2011 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 Marks 59 52 45 39 33

A 1 B 2 C 3 D 4 E 5 F 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 / / /

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 2 4 6 11 x < 4 4 swap with x, 4 swap with 2,

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

AC = 10 CH = 9 FH = 8 CE = 10 ED = 11 Could use CD GH = 13 DB = 14 1 8 2 6 5 4 7 3 CH = 9 FH = 8 CE = 10 ED = 11 Could use CD GH = 13 DB = 14

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

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)

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

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

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 150 + (BE + CD) 32 = 182 km

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 = 150 + 18 = 168 km

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 150 + 10 = = 160 km cii) If she repeats AF she must start / finish at C and D

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

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

200x + 1000y = 5000 → x + 5y ≥ 25 200x + 1500y = 6000 → 2x + 15y ≥ 60 100x + 2500y = 4000 → x + 25y ≥ 40 (C =) 2.5x +15y

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

Feasible Region 15 DIY, 2 Trade Cost (C =) 2.5x +15y 2.5 x 15 + 15 x 2 = 37.5 + 30 = £67.50 Objective Line

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

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

ci) Q U S T P R Q ci) Q → U → S → T → P → R → Q 11 + 10 + 12 + 26 + 40 + 20 = 119 miles

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

Lower Bound 52 + 31 = 83 minutes Remove Q Q 20 11 P U T S R 16 10 14 12 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 52 + 31 = 83 minutes This is not an optimal route as at least one edge has to be repeated