General Certificate of Education

January 2009

Advanced Subsidiary Examination

MATHEMATICS MD01Unit Decision 1

Wednesday 21 January 2009 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 3 and 4 (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.

P12641/Jan09/MD01 6/6/6/ MD01

2

P12641/Jan09/MD01

Answer all questions.

1 The following network shows the lengths, in miles, of roads connecting 11 villages,

A, B, ..., K.

(a) Starting from G and showing your working at each stage, use Prim’s algorithm to find a

minimum spanning tree for the network. (6 marks)

(b) State the length of your minimum spanning tree. (1 mark)

(c) Draw your minimum spanning tree. (3 marks)

B E I10 16

17 11

D

14 7 14 23

A22 12 5 22

K

6 15 23 21

G H

8 24

C 20 F 19 J

3

P12641/Jan09/MD01

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

following bipartite graph shows the tasks that each of the people is able to undertake.

(a) Represent this information in an adjacency matrix. (2 marks)

(b) Initially, B is assigned to task 1, C to task 2, D to task 4, and E to task 5.

Demonstrate, by using an algorithm from this initial matching, how each person can be

allocated to a task. (5 marks)

Turn over for the next question

A

B

C

D

E

F

1

2

3

4

5

6

Turn over

s

4

P12641/Jan09/MD01

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

The diagram shows roads connecting some places of interest in Berlin. The numbers

represent the times taken, in minutes, to walk along the roads.

The total of all walking times is 167 minutes.

(a) Mia is staying at D and is to visit H .

(i) Use Dijkstra’s algorithm on Figure 1 to find the minimum time to walk from

D to H . (6 marks)

(ii) Write down the corresponding route. (1 mark)

(b) Each day, Leon has to deliver leaflets along all of the roads. He must start and finish

at A.

(i) Use your answer to part (a) to write down the shortest walking time from D to A.

(1 mark)

(ii) Find the walking time of an optimum Chinese Postman route for Leon. (6 marks)

B

D 7 E

83 11

23

CF A

H

5

G

20

12 3

9

306

5 15 10

5

P12641/Jan09/MD01

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

Each year, farmer Giles buys some goats, pigs and sheep.

He must buy at least 110 animals.

He must buy at least as many pigs as goats.

The total of the number of pigs and the number of sheep that he buys must not be

greater than 150.

Each goat costs £16, each pig costs £8 and each sheep costs £24.

He has £3120 to spend on the animals.

At the end of the year, Giles sells all of the animals. He makes a profit of £70 on each goat,

£30 on each pig and £50 on each sheep. Giles wishes to maximize his total profit, £P.

Each year, Giles buys x goats, y pigs and z sheep.

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

(b) One year, Giles buys 30 sheep.

(i) Show that the constraints for Giles’s situation for this year can be modelled by

y5 x , 2xþ y4 300 , xþ y5 80 , y4 120 (2 marks)

(ii) On Figure 2, draw a suitable diagram to enable the problem to be solved

graphically, indicating the feasible region and the direction of the objective line.

(8 marks)

(iii) Find Giles’s maximum profit for this year and the number of each animal that he

must buy to obtain this maximum profit. (3 marks)

Turn over for the next question

Turn over

s

6

P12641/Jan09/MD01

5 A student is using the algorithm below to find an approximate value offfiffiffi

2p

.

Line 10 Let A ¼ 1 , B ¼ 3 , C ¼ 0

Line 20 Let D ¼ 1 , E ¼ 2 , F ¼ 0

Line 30 Let G ¼ B=E

Line 40 Let H ¼ G2

Line 50 If ðH � 2Þ2 < 0:0001 then go to Line 130

Line 60 Let C ¼ 2Bþ A

Line 70 Let A ¼ B

Line 80 Let B ¼ C

Line 90 Let F ¼ 2E þ D

Line 100 Let D ¼ E

Line 110 Let E ¼ F

Line 120 Go to Line 30

Line 130 Print ‘ffiffiffi

2p

is approximately’, B=E

Line 140 Stop

Trace the algorithm. (6 marks)

6 A connected graph G has five vertices and has eight edges with lengths 8, 10, 10, 11, 13, 17,

17 and 18.

(a) Find the minimum length of a minimum spanning tree for G. (2 marks)

(b) Find the maximum length of a minimum spanning tree for G. (2 marks)

(c) Draw a sketch to show a possible graph G when the length of the minimum spanning

tree is 53. (3 marks)

7

P12641/Jan09/MD01

7 Liam is taking part in a treasure hunt. There are five clues to be solved and they are at the

points A, B, C, D and E. The table shows the distances between pairs of points. All of the

distances are functions of x, where x is an integer.

Liam must travel to all five points, starting and finishing at A.

A B C D E

A – xþ 6 2x� 4 3x� 7 4x� 14

B xþ 6 – 3x� 7 3x� 9 xþ 9

C 2x� 4 3x� 7 – 2x� 1 xþ 8

D 3x� 7 3x� 9 2x� 1 – 2x� 2

E 4x� 14 xþ 9 xþ 8 2x� 2 –

(a) The nearest point to A is C.

(i) By considering AC and AB , show that x < 10 . (2 marks)

(ii) Find two other inequalities in x. (2 marks)

(b) The nearest neighbour algorithm, starting from A, gives a unique minimum tour

ACDEBA .

(i) By considering the fact that Liam’s tour visits D immediately after C, find two

further inequalities in x. (3 marks)

(ii) Find the value of the integer x. (4 marks)

(iii) Hence find the total distance travelled by Liam if he uses this tour. (2 marks)

END OF QUESTIONS

2

P12641/Jan09/MD01

Figure 1 (for use in Question 3)

B

D 7 E

83 11

23

CF A

H

5

G

20

12 3

9

306

5 15 10

3

P12641/Jan09/MD01

Figure 2 (for use in Question 4)

~y

140 –

120 –

100 –

80 –

60 –

40 –

20 –

0 –

~–––––––

0 40 80 120 x20 60 100

MD01 - AQA GCE Mark Scheme 2009 January 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.

MD01 - AQA GCE Mark Scheme 2009 January series

4

MD01 Q Solution Marks Total Comments

1(a) GH

(5)

GE (7) HJ (8) BE (10) BD (11) IH (14) DC (15) AC (6) FJ (19) HK (22)

M1

B1 A1 A1

A1

A1

6

SCA allow Prim’s from any vertex but not Kruskal or path – min of 8 edges 10 edges HJ 3rd BE 4th AC 8th All correct

(b) 117 B1 1

(c)

(Possibly shown in part (a))

M1

A1

A1

3

MST (8+ edges) 10 edges All correct (+ vertices labelled)

Total 10 2(a) Labelled 6 × 6 matrix with ‘1’s M1 Must have ‘1’s not ‘ ’s

1 2 3 4 5 6

A 0 1 0 0 0 0 B 1 1 1 0 0 0 C 1 1 0 0 0 0 D 0 0 0 1 0 1 E 0 0 0 1 1 0 F 0 0 0 0 1 0

Or A B C D E F

1 0 1 1 0 0 0 2 1 1 1 0 0 0 3 0 1 0 0 0 0 4 0 0 0 1 1 0 5 0 0 0 0 1 1 6 0 0 0 1 0 0

OE A1 2 Must have ‘0’s not ‘–’s or blank

(b) M1 A – 2 −/ C or 3 – B −/ 1 2 1 3A C B− − − − −/ / A1 M1 F – 5 −/ E or 6 – D −/ 4 5 4 6F E D− − − − −/ / A1 Match: A2, C1, B3, F5, E4, D6 B1 5 If working on diagram: Only one path on each half M1A1M1A1 as above – start point must

be shown, otherwise M0 Total 7

MD01 - AQA GCE Mark Scheme 2009 January series

5

MD01 (cont) Q Solution Marks Total Comments

3(a)(i)

M1 Cancelling at at least 2 vertices A1 Correct at F m1 2 different values at B A1 Correct at G – depends only on M1 m1 4 different values at H A1 6 All correct – no extra values Alternative if working from H: 0 , 10 , 23H A B 21 , 25F 24 , 29 ,

36

C

D 35( ) 34 , 20 , 30G E 29 27

(M1) (A1) (m1)

SCA Correct at B 2 values at F

(A1) Correct at E (m1) 2 or 3 values at D (A1) All correct

(ii) Route: DEFBAH B1 1 Or reverse

(b)(i) 24 B1 1

(ii) (Odds) A, C, D, G only E1 PI AC + DG = 19 +15 or 34 M1 3 sets of pairs AD + CG = 24 +10 or 34 A2,1,0 AG + CD = 19 + 6 or 25 (Repeat AG + CD) Length = 25 + 167 A1F 167 + their shortest pairing = 192 B1 6 Total 14

( )

MD01 - AQA GCE Mark Scheme 2009 January series

6

MD01 (cont) Q Solution Marks Total Comments

4(a) 110x y z+ + ≥

B1

–1 for strict inequalities (max) –1 for using g, p, s instead of x, y, z (max)

y x≥ B1 150y z+ ≤ B1

16 8 24 3120x y z+ + ≤ ISW B1 ( )2 3 390x y z+ + ≤

( ) 70 30 50P x y z= + + B1 5

(b)(i) 30z = M1 Justify by correctly substituting into at least one of their inequalities

80x y+ ≥ (or 30 110x y+ + ≥ ) ( )y x≥ 120y ≤ (or 30 150y + ≤ )

2 300x y+ ≤ (or 2 90 390x y+ + ≤ OE) A1 2 Correctly substituting into all 3 inequalities

( )70 30 1500P x y= + + AG

(ii) B1

B1

B1

M1

A1

B1

M1

A1

8

120y =

80x y+ = y x= , correct at (40, 40) and (100, 100)

2 300x y+ = , –ve gradient with one correct point in the interval 80 120x≤ ≤ Correct at (100, 100) and (90, 120) Correct region labelled

OL: gradient of 73

− or 37

−

Gradient = 73

−

(iii) Considering (90, 120) and/or (100, 100) M1 Ignore other points being considered

(£) 11500 A1 100 goats, 100 pigs, 30 sheep A1 3 CAO Total 18

MD01 - AQA GCE Mark Scheme 2009 January series

7

MD01 (cont) Q Solution Marks Total Comments

5

A B C D E F G H 1 3 0 1 2 0 1.5 2.25 7

3 7 5 2 5 1.4 1.96 17

7 17 12 5 12 1.416 2.007

172 is approximately12

⎛ ⎞⎜ ⎟⎝ ⎠

M1 A1 M1

A1 M1

A1

6

Condone equivalent fractions

SCA – finding a value for G 1st pass G, H correct 2nd pass – finding a new value for C All correct on pass 3rd pass C = 17 or their (2B+A) AWRT 1.417 All correct (allow 2.005 to 2.008) and no further passes

Total 6 6(a) Min MST M1 4 edges

= 8 + 10 +10 + 11 = 39 A1 2

(b) Max MST = 8 + 17 +17 +18 M1 8 + 18 + 2 others = 60 A1 2

(c)

M1

A1

A1

3

Connected graph with 5 vertices (all edges numbered, from G) MST = 53 8, 11, 17, 17 or 8, 10, 17, 18 other edges OE (other possibilities not shown) (all edges numbered, from G)

Total 7

MD01 - AQA GCE Mark Scheme 2009 January series

8

MD01 (cont) Q Solution Marks Total Comments 7(a)(i) 2 4 6x x− < + M1 2 4x − <

10x∴ < CSO A1 2 AG

(ii) 2 4 3 7 OEx x− < − 2 4 4 14 OEx x− < −

B1 B1

2

Allow any expression in matrix > 0 Allow any expression in matrix > 0

35

xx

= >⎛ ⎞⎜ ⎟>⎝ ⎠

(b)(i) 2 1x − < M1 Condone ≤ for method mark only

2 1 3 7x x− < − A1 2 1 8x x− < + A1 3

(ii) ( 6)x⇒ > 9x < B1 Possibly earned in (b)(i) 2 2 3 9x x− < − M1 Condone ≤ for method mark only 7x > A1 8x = B1 4

(iii) A C D E B A 12 15 14 17 14 M1

8 8x + with their integer x

= 72 A1 2 CAO (unsupported 72 scores M0A0) Total 13 TOTAL 75

Scaled mark component grade boundaries - January 2009 exams

Component MaximumCode Component Title Scaled Mark A B C D E

Scaled Mark Grade Boundaries

GCE (Legacy)

klm

LAW5 GCE LAW UNIT 5 85 57 53 49 45 41

MD01 GCE MATHEMATICS UNIT D01 75 60 52 45 38 31MD02 GCE MATHEMATICS UNIT D02 75 61 53 46 39 32MFP1 GCE MATHEMATICS UNIT FP1 75 58 51 44 37 31MFP2 GCE MATHEMATICS UNIT FP2 75 60 52 44 36 29MFP3 GCE MATHEMATICS UNIT FP3 75 59 51 44 37 30MFP4 GCE MATHEMATICS UNIT FP4 75 56 49 42 35 28

MM1A/C GCE MATHEMATICS UNIT M1A - COURSEWORK 25 20 17 14 12 10MM1A/W GCE MATHEMATICS UNIT M1A - WRITTEN 75 64 56 48 40 33

MM1B GCE MATHEMATICS UNIT M1B 75 64 56 49 42 35MM2B GCE MATHEMATICS UNIT M2B 75 61 53 45 37 30MPC1 GCE MATHEMATICS UNIT PC1 75 62 54 46 39 32MPC2 GCE MATHEMATICS UNIT PC2 75 65 57 49 41 34MPC3 GCE MATHEMATICS UNIT PC3 75 59 51 43 36 29MPC4 GCE MATHEMATICS UNIT PC4 75 60 52 44 36 29

MS/SS1A/W GCE MATHS/STATISTICS UNIT 1A - WRITTEN 75 60 52 44 37 30MS/SS1A/C GCE MATHS/STATISTICS UNIT 1A - COURSEWORK 25 20 17 14 12 10

MS1B GCE MATHEMATICS UNIT S1B 75 59 51 43 35 28MS2B GCE MATHEMATICS UNIT S2B 75 56 49 42 36 30

XMCAS GCE MATHEMATICS UNIT XMCAS 125 101 89 77 66 55XMCA2 GCE MATHEMATICS UNIT XMCA2 125 100 88 76 64 53

MED1 GCE MEDIA STUDIES UNIT 1 60 38 33 28 23 19MED2 GCE MEDIA STUDIES UNIT 2 60 41 35 30 25 20MED3 GCE MEDIA STUDIES UNIT 3 100 72 64 56 48 40MED4 GCE MEDIA STUDIES UNIT 4 60 42 38 34 30 26