AS Maths Decision Paper January 2011 Model Answers

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

GradeABCDE Marks Grade Boundaries

1a) A B C D E F This is an adjacency matrix

Initial Matchings A1B2C3D4E5F6A1B2C3D4E5F6 A - 5/ B - 6 then C - 4/ E - 2 / D - 6 / B - 3/ F - 1 You must show your Final Matchings Match A5, B3, C4, D6, E2, F1 There are alternate ways to find the final matchings One potential matching is

a) The pivots used after the first pass and therefore for the second pass are 7 and 22 b) First Pass 7 Second Pass 5 Third Pass 3 c) No, because 16 and 19 have not been compared

EB = 5 EH = 7 AB = 8 HI = 9 AD = 10 DG = 4 EF = 12 FC = 6 Length of spanning tree = =61

A H F B E C I D G A H F B E C I D G Original Spanning tree = 61 AB and BE dropped = -13 GE added= 11 New Spanning tree = (61 – ) = 59 aiii) b)

A – D – F – I – J

The minimum time from A to G is 12 minutes If you use the subway (CG) this time is reduced. You must use AC = the subway Therefore the subway (CG) must be < 4.5 (12 – 7.5) The minimum time from A to J is not reduced. The time taken from A to J is 18 minutes (last question) If you use the subway, then to get to J you must use AC and GJ = 16.5 minutes ( ) As the minimum time is not reduced then CG ≥ 1.5 The time taken for CG must be 1.5 ≤ x < 4.5

The graph is neither Eulerian or semi Eulerian as it has 4 odd vertices

These are the ODD vertices AB + GH AG + BH AH + BG AB (180) +GH (165) = 345 AG (90) +BH (210) =300 AH (150) +BG (210) = 360 Repeat AG + BH = 300 Length of route = = 1515 metres

Add the repeated edges on to your graph AG (90) + BH (210) = 300 Repeated edges F now has 6 edges so will be visited 6 ÷ 2 times = 3 H now has 4 edges so will be visited 4 ÷ 2 times = 2

A D C B E A complete graph for k is shown 5 Starting at A there are 4 unused edges From B there are 3 unused edges From C there are 2 unused edges From D there is 1 unused edge From E there are 0 unused edges Total number of edges is = 10 Edges

A D C B E aii) The number of edges in a minimum spanning tree for graph k is n – 1 vertices = 5 5 – 1 = 4 aiii) As a HAMILTONIAN cycle visits each vertex once, returning to the start vertex, then 5 Edges (An example is shown)

b) A simple graph has six vertices. It also has 9 edges an is Eulerian which means every vertex will have an even number of edges leaving them A B C D E F A has 4 edges, B has 2 edges, C has 4 edges, D has 2 edges, E has 4 edges and F has 2 edges

BACDEBBACDEB = 33 BAEDCBBAEDCB = 41

You are left with A, B, D, E the nearest neighbour produces a spanning tree like this A B 3 E 4 D 10 = 17 Now add the two shortest neighbours to C which are CD = 5 andCA = 11 C 11 D 5 A Total is = 33

A B 3 E 4 D 10 C 11 5 It is an optimal journey

KEY INFORMATION

X Line 10 0 Line 20 A B 20 8 Line 6010 Line Line 60 5 Line Line Line 60 Print 160 Line 602 Line Line Line Line 90 Ques 8

The algorithm defines the MULTIPLICATION of A and B If Line 50 read If A = 1, then go to Line 80 it would create A CONTINUOUS LOOP as it will never reach LINE 90

Tins 6x + 9y + 9z ≤ 600Simplified 2x + 3y + 3z ≤ 200 Packets 9x + 6y + 9z ≤ 600 Simplified 3x + 2y + 3z ≤ 200 Bottles 6x + 12y + 18z ≥ 480Simplified x + 2y + 3z ≥ 80

So z = y 2x + 3y + 3z ≤ 200 → 3x + 2y + 3z ≤ 200 → x + 2y + 3z ≥ 80 → As z = y, then put the z and y values together as y’s 2x + 6y ≤ 200 → 3x + 5y ≤ 200 x + 3y ≤ 100 x + 5y ≤ 80

Write each inequality as an equation 3x + 5y ≤ 200 becomes x + 3y ≤ 100 becomes x + 5y ≤ 80 becomes x + 3y = 100 3x + 5y = 200 x + 5y = 80 Also remember that x, y and z are also ≥ 0 Now plot the graph See next slide

x + 5y = 80 x + 3y = 100 3x + 5y = 200 x ≤ 0 y ≤ 0 Feasible Region iii) Maximum is when x = 25 and y = 25 Remember x + 2y as there are 3 variables, x, y and z and z = y So x + 2y = = 75 iv) 25 basic 25 standard 25 luxury ii)