general certificate of education june 2008 advanced ... · june 2008 advanced subsidiary...

General Certificate of Education

June 2008

Advanced Subsidiary Examination

MATHEMATICS MD01Unit Decision 1

Friday 6 June 2008 1.30 pm to 3.00 pm

For this paper you must have:* an 8-page answer book

* the blue AQA booklet of formulae and statistical tables

* an insert for use in Questions 6 and 7 (enclosed).

You may use a graphics calculator.

Time allowed: 1 hour 30 minutes

Instructions* Use black ink or black ball-point pen. Pencil or coloured pencil should only be used for

drawing.* Write the information required on the front of your answer book. The Examining Body for this

paper is AQA. The Paper Reference is MD01.* Answer all questions.* Show all necessary working; otherwise marks for method may be lost.* The final answer to questions requiring the use of calculators should be given to three significant

figures, unless stated otherwise.* Fill in the boxes at the top of the insert.

Information* The maximum mark for this paper is 75.* The marks for questions are shown in brackets.

P5653/Jun08/MD01 6/6/6/6/ MD01

2

P5653/Jun08/MD01

Answer all questions.

1 Six people, A, B, C, D, E and F, are to be matched to six tasks, 1, 2, 3, 4, 5 and 6.

The following adjacency matrix shows the possible matching of people to tasks.

Task 1 Task 2 Task 3 Task 4 Task 5 Task 6

A 0 0 1 0 1 1

B 0 1 0 1 0 0

C 0 1 0 0 0 1

D 0 0 0 1 0 0

E 1 0 1 0 1 0

F 0 0 0 1 1 0

(a) Show this information on a bipartite graph. (2 marks)

(b) Initially, A is matched to task 3, B to task 4, C to task 2 and E to task 5. From this

initial matching, use the maximum matching algorithm to obtain a complete matching.

List your complete matching. (5 marks)

2 (a) Use a quick sort to rearrange the following letters into alphabetical order. You must

indicate the pivot that you use at each pass.

P B M N J K R D (5 marks)

(b) (i) Find the maximum number of swaps needed to rearrange a list of 8 numbers into

ascending order when using a bubble sort. (1 mark)

(ii) A list of 8 numbers was rearranged into ascending order using a bubble sort. The

maximum number of swaps was needed. What can be deduced about the original

list of numbers? (1 mark)

3

P5653/Jun08/MD01

3 (a) (i) State the number of edges in a minimum spanning tree of a network with

11 vertices. (1 mark)

(ii) State the number of edges in a minimum spanning tree of a network with

n vertices. (1 mark)

(b) The following network has 11 vertices, A, B, ..., K. The number on each edge

represents the distance, in miles, between a pair of vertices.

(i) Use Prim’s algorithm, starting from A, to find a minimum spanning tree for the

network. (5 marks)

(ii) Find the length of your minimum spanning tree. (1 mark)

(iii) Draw your minimum spanning tree. (2 marks)

G

D

B

A

C

F

J

K

I

E

18

15

17

12

16

18

19

17

H

18

18

21

13

21 14

18

17

19

20

16

17

Turn over

s

4

P5653/Jun08/MD01

4 David, a tourist, wishes to visit five places in Rome: Basilica (B), Coliseum (C),

Pantheon (P), Trevi Fountain (T) and Vatican (V ). He is to start his tour at one of the

places, visit each of the other places, before returning to his starting place.

The table shows the times, in minutes, to travel between these places. David wishes to keep

his travelling time to a minimum.

B C P T V

B – 43 57 52 18

C 43 – 18 13 56

P 57 18 – 8 48

T 52 13 8 – 51

V 18 56 48 51 –

(a) (i) Find the total travelling time for the tour TPVBCT . (1 mark)

(ii) Find the total travelling time for David’s tour using the nearest neighbour

algorithm starting from T . (4 marks)

(iii) Explain why your answer to part (a)(ii) is an upper bound for David’s minimum

total travelling time. (2 marks)

(b) (i) By deleting B, find a lower bound for the total travelling time for the minimum

tour. (5 marks)

(ii) Explain why your answer to part (b)(i) is a lower bound for David’s minimum

total travelling time. (2 marks)

(c) Sketch a network showing the edges that give the lower bound found in part (b)(i) and

comment on its significance. (2 marks)

5

P5653/Jun08/MD01

5 The diagram shows a network of sixteen roads on a housing estate. The number on each

edge is the length, in metres, of the road. The total length of the sixteen roads is

1920 metres.

(a) Chris, an ice-cream salesman, travels along each road at least once, starting and

finishing at the point A. Find the length of an optimal ‘Chinese postman’ route for

Chris. (6 marks)

(b) Pascal, a paperboy, starts at A and walks along each road at least once before finishing

at D. Find the length of an optimal route for Pascal. (2 marks)

(c) Millie is to walk along all the roads at least once delivering leaflets. She can start her

journey at any point and she can finish her journey at any point.

(i) Find the length of an optimal route for Millie. (2 marks)

(ii) State the points from which Millie could start in order to achieve this optimal

route. (1 mark)

300

A B

E F30120 120

40 40

I

40 40

30H G120120

20 30 300280

D

290

C

Total Length = 1920 metres

Turn over

s

6

P5653/Jun08/MD01

6 [Figure 1, printed on the insert, is provided for use in this question.]

A factory makes two types of lock, standard and large, on a particular day.

On that day:

the maximum number of standard locks that the factory can make is 100;

the maximum number of large locks that the factory can make is 80;

the factory must make at least 60 locks in total;

the factory must make more large locks than standard locks.

Each standard lock requires 2 screws and each large lock requires 8 screws, and on that day

the factory must use at least 320 screws.

On that day, the factory makes x standard locks and y large locks.

Each standard lock costs £1.50 to make and each large lock costs £3 to make.

The manager of the factory wishes to minimise the cost of making the locks.

(a) Formulate the manager’s situation as a linear programming problem. (5 marks)

(b) On Figure 1, draw a suitable diagram to enable the problem to be solved graphically,

indicating the feasible region and the direction of the objective line. (6 marks)

(c) Find the values of x and y that correspond to the minimum cost. Hence find this

minimum cost. (4 marks)

7

P5653/Jun08/MD01

7 [Figure 2, printed on the insert, is provided for use in this question.]

The following network has eight vertices, A, B, ..., H , and edges connecting some pairs of

vertices. The number on each edge is its weight. The weights on the edges EH and GH are

functions of x and y.

Given that there are three routes from A to H with the same minimum weight, use Dijkstra’s

algorithm on Figure 2 to find:

(a) this minimum weight; (6 marks)

(b) the values of x and y. (3 marks)

END OF QUESTIONS

B E

AD F

H

C G14

103

9

14

85

1012

15

22

12

9

2xþ y

3x� 2y

2

P5653/Jun08/MD01

Figure 1 (for use in Question 6)

~y

100 –

80 –

60 –

40 –

20 –

0 – ––––––

0 60 80 1004020

~

x

3

P5653/Jun08/MD01

BE

AD

FH

CG

14

10

39

14

85

10

12

15

22

12

9

2xþy

3x�2y

Figure

2(foruse

inQuestion7)

MD01 - AQA GCE Mark Scheme 2008 June series

3

Key to mark scheme and abbreviations used in marking M mark is for method m or dM mark is dependent on one or more M marks and is for method A mark is dependent on M or m marks and is for accuracy B mark is independent of M or m marks and is for method and accuracy E mark is for explanation

or ft or F follow through from previous incorrect result

MC

mis-copy

CAO correct answer only MR mis-read CSO correct solution only RA required accuracy AWFW anything which falls within FW further work AWRT anything which rounds to ISW ignore subsequent work ACF any correct form FIW from incorrect work AG answer given BOD given benefit of doubt SC special case WR work replaced by candidate OE or equivalent FB formulae book A2,1 2 or 1 (or 0) accuracy marks NOS not on scheme –x EE deduct x marks for each error G graph NMS no method shown c candidate PI possibly implied sf significant figure(s) SCA substantially correct approach dp decimal place(s) No Method Shown Where the question specifically requires a particular method to be used, we must usually see evidence of use of this method for any marks to be awarded. However, there are situations in some units where part marks would be appropriate, particularly when similar techniques are involved. Your Principal Examiner will alert you to these and details will be provided on the mark scheme. Where the answer can be reasonably obtained without showing working and it is very unlikely that the correct answer can be obtained by using an incorrect method, we must award full marks. However, the obvious penalty to candidates showing no working is that incorrect answers, however close, earn no marks. Where a question asks the candidate to state or write down a result, no method need be shown for full marks. Where the permitted calculator has functions which reasonably allow the solution of the question directly, the correct answer without working earns full marks, unless it is given to less than the degree of accuracy accepted in the mark scheme, when it gains no marks. Otherwise we require evidence of a correct method for any marks to be awarded.


4

MD01 Q Solution Marks Total Comments

1(a)

M1

A1

2

Bipartite graph: 2 sets of vertices with at least one edge All correct

(b) A3, B4, C2, E5 Initial match

Start from D, F or 1, 6 M1 M1

1st path must go beyond 2nd letter/number2nd path

⎫⎬⎭

eg 4 ( ) /D B F− −/

If working is only on diagram, the path(s) must be clear, and only 1 path per diagram can be credited. If 2 paths shown on one diagram, max mark M1A1

Accept paths in reverse order 4 ( ) 2 ( ) 6D B C− − − − −/ / A1 1st correct path 5 ( ) 1F E− − −/ A1 2nd correct path or or 4( ) 2( ) 6F B C− − − − −/ / 5( ) 3( ) 6F E A− − − − −/ / 4( ) 5( ) 1D F E− − − − −/ / 4( ) 2( ) 6( ) 3( ) 1D B C A E− − − − − − − − −/ / / / Match: A3, B2, C6, D4, E1, F5 B1 5 Must be clearly stated or indicated Total 7

2(a)

P B M N J K R D

B M N J K D P R

B M N J K D P R

B J K D M N P R

B D J

K M N

P R

M1

A1

A1

A1

B1

5

Using quick sort First pass (based on their pivot) A correct third pass All passes correct

Consistent pivots clearly labelled (at least three passes)

(b)(i) 28 B1 1

(ii) In reverse order B1 1 Allow descending Total 7


5

MD01 (cont) Q Solution Marks Total Comments 3(a)(i) 10 B1 1

(ii) n – 1 B1 1

(b) Condone candidates attempting all of part

(b) together / in different order

(i)

AB BC BD CF DG or FJ GK JK KJ GK KH or KI KI IE EI KH

M1

A1 A1

A1 B1

5

Using Prim’s BD 3rd CF 4th All correct 10 edges

(ii) (Length =) 155 B1 1

(iii)

M1

A1

2

Spanning tree with at least 8 edges Any cycle scores M0 Correct and labelled Alternative: FJ instead of DG:

Total 10


6

MD01 (cont) Q Solution Marks Total Comments

4(a)(i) 130 B1 1 T P V B C T 8 48 18 43 13

(ii) T P C B V T

8 18 43 18 51 M1

M1

Tour (vertices or edges) starting from T (Letters not numbers) Visits all vertices starting from T

A1 Correct order = 138 B1 4

(iii) A possible solution, eg tour E1 OE

May be improved on E1 2 Allow ‘can’ in this case as (i) < (ii) OE

(b)(i) PT, CT, PV

M1

A1

Spanning tree with 3 edges Correct

+ 2 shortest from B

m1

A1

2 edges from B Correct

(Lower bound =) 130 A1 5 CSO

(ii) May not exist E1 OE Cannot be lowered E1 2 OE

(c)

B1

Tour or optimum or same as (a)(i) E1 2 Lower bound = Upper bound Total 16


7


5(a) Odds A, B, C, D M1 PI (but A, B, C, D must be mentioned) m1

Considering 3 sets of pairings of odd

vertices, eg AB with CD etc + = 270 + 270 = 540

+ = 290 +290 = 580+ = 260 +270 = 530

AB CDAC BDAD BC

⎫⎪⎬⎪⎭

A2,1,0 A1 for 2 correct, A2 for all correct

Repeat AD, BC A1F

Follow through their shortest pairing PI by adding 530 to 1920 Or AEHD or DHEA and BFGC or CGFB listed in any route

(Length = 1920 + 530 =) 2450 (metres) B1 6

(b) Repeats BC E1

PI by BFGC or CGFB listed in a complete route or adding 270 / subtracting 260

(Length = 1920 + 270 =) 2190 (metres) B1 2 2450 – 260 = 2190 (2190 with no evidence scores E0B1)

(c)(i) Min. repeat AD E1 PI by AEHD or DHEA listed in a complete route or adding 260 / subtracting 270

(Length = 1920 + 260 =) 2180 (metres) B1 2 2450 – 270 = 2180 (2180 with no evidence scores E0B1)

(ii) B, C B1 1 Condone start at B, finish at C (or reverse) Total 11


8


6(a) All inequalities must be as below 100, 80x y B1 Both x y+ 60 B1 x y< B1 2 8 320x y+ B1 OE (minimise C =) 1.5x + 3y B1 5

B1

B1 × 3

100, 80 1within square2

from (0,0) to (80,80)Other lines

x y= = ⎫⎪⎬⎪⎭

B1 Feasible Region CAO (must have scored

B4 for drawing lines) (condone x = y as solid line)

B1 6 An Objective Line with gradient –0.5

(b)

(c) Considering an extreme point in their region

M1

Min at intersect of 60x y+ = 4 160x y+ = A1

PI by indication on diagram or 2 126 333 3

x y= =

Considering a pair of integer values where

26 28, 32 34x y M1

(C =) £141 at (26, 34)

or £141 at (28, 33) A1 4

Total 15


9


7(a)

M1 SCA; cancelling at 2 (or more) vertices A1 Correct at D M1 2 values at E M1 2 values at G A1

All correct (condone 0 missing at A and missing expressions in x and y at H)

(Min =) 43 B1 6 Accept 43 at H

(b) 2x + y = p 3x – 2y = q

M1

Obtaining a pair of equations in this form or (22) + 2x + y = (43) and (22) + 3x – 2y = (43)

2 21x y+ = and 3 – 2 21x y = 9x = A1 CAO 3y = A1 3 CAO NMS: both correct M1A2

one/none correct M0A0 Total 9 TOTAL 75

Scaled mark component grade boundaries - June 2008 exams

Component MaximumCode Component Title Scaled Mark A B C D E

Scaled Mark Grade Boundaries

GCE

klm

ICT4 GCE INFO AND COMM TECH UNIT 4 90 61 55 49 44 39ICT5 GCE INFO AND COMM TECH UNIT 5 90 69 63 57 52 47ICT6 GCE INFO AND COMM TECH UNIT 6 90 59 51 43 36 29

LAW1 GCE LAW UNIT 1 65 50 45 41 37 33LAW2 GCE LAW UNIT 2 65 46 40 35 30 25LAW3 GCE LAW UNIT 3 65 45 40 35 30 26LAW4 GCE LAW UNIT 4 85 58 53 48 43 39LAW5 GCE LAW UNIT 5 85 57 53 49 45 41LAW6 GCE LAW UNIT 6 70 48 43 39 35 31

MD01 GCE MATHEMATICS UNIT D01 75 60 52 45 38 31MD02 GCE MATHEMATICS UNIT D02 75 58 50 43 36 29MFP1 GCE MATHEMATICS UNIT FP1 75 63 55 48 41 34MFP2 GCE MATHEMATICS UNIT FP2 75 58 51 44 37 30MFP3 GCE MATHEMATICS UNIT FP3 75 63 55 47 39 31MFP4 GCE MATHEMATICS UNIT FP4 75 66 58 51 44 37MM03 GCE MATHEMATICS UNIT M03 75 56 48 40 33 26MM04 GCE MATHEMATICS UNIT M04 75 54 46 39 32 25MM05 GCE MATHEMATICS UNIT M05 75 60 52 44 36 29

MM1A/C GCE MATHEMATICS UNIT M1A - COURSEWORK 25 20 18 15 13 10MM1A/W GCE MATHEMATICS UNIT M1A - WRITTEN 75 60 51 43 35 28

MM1B GCE MATHEMATICS UNIT M1B 75 61 52 43 34 25MM2A/C GCE MATHEMATICS UNIT M2A - COURSEWORK 25 20 18 15 13 10MM2A/W GCE MATHEMATICS UNIT M2A - WRITTEN 75 55 48 40 34 28

MM2B GCE MATHEMATICS UNIT M2B 75 53 46 39 33 27MPC1 GCE MATHEMATICS UNIT PC1 75 59 51 43 35 28MPC2 GCE MATHEMATICS UNIT PC2 75 60 52 44 37 30

D.Douis

Highlight

general certificate of education june 2008 advanced ... · june 2008 advanced subsidiary...

Documents