4771 mark scheme june 2005 mark scheme 4771 …mei.org.uk/files/papers/all packs/ms_mei_d1.pdf4771...

4771 Mark Scheme June 2005

Mark Scheme 4771June 2005


1. (i) Any connected tree. 12 connections (ii) 14 connections (iii) e.g. He might be able to save cable by using it. e.g. To avoid overloading. (iv) Yes. A minimum connector is a tree. This gives the min number of arcs (n–1). This gives the minimum no of connections (2(n–1)).

M1 A1 B1 B1 B1 B1 B1 B1

2.

(i) Janet John (ii) Yes Janet’s route traces west and south walls plus

"attachments". John’s route traces north and east walls plus

"attachments". − or equivalent (Any “islands” are irrelevant.) (iii) Yes (iv) Yes All avenues covered by forward and backward pass (i.e.

by John's original route + Janet's route).

M1 A1 A1 M1 A1 B1 B1 B1


3. (i) (ii) Critical − A, D and C (iii) Total float for B = 2 Independent float for B = 1 Total float for E = 1 Independent float for E = 0

M1

A1 B1 B1 B1 B1 both total floats A1 B's independent A1 E's independent

5

D A 3

0 3 5 0 5

3

4

C 1 2

23 E B


4. (i) P Q R S T U V 45 14 12 15 25 31 49 P T S C V U S C (ii) PV ST CR RT UV Length = 80 TU QR (iii) CP reduced to 26 CV reduced to 34 (iv) UV replaced by PQ New length = 74 (v) Q Semi-Eulerian. (Order of P changed from 3 to 4, but

order of Q changed from 2 to 3 − so still 2 odd vertices.) or Cross the bridge and proceed as before or A valid route

B1 starting at C M1 Dijkstra A1 labels A1 order of labelling A1 working values B1 B1 B1 M1 A1 first 5 A1 last 2 B1 length B1 (both and no more) B1 M1 A1

P

U

S R

C

Q

14

V

20

22 818

8

16

16

10 T

14

10 12

15

15 0 1

14 3

26 12 2

12

14

31

45 7 45

25 5 25

31

15 4

49 49 6 8


5. (i) eg. 00–19 → 0 20–49 → 1 50–69 → 2 70–84 → 3 85–99 → 4 (ii) 1, 0, 2, 3, 1, 3, 4, 3, 0, 0 (iii) eg. 00–15 → 0 16–39 → 1 40–63 → 2 64–95 → 3 96–99 → ignore (iv) 1, 0, 1, 0, 1, 1, 3, 3, 2, 2 (v) Day 0 1 2 3 4 5 6 7 8 9 10 Stock 3 3 3 2 0 0 0 0 0 2 4 Disptd 0 0 0 0 1 0 2 1 0 0 0 (vi) Day 0 1 2 3 4 5 6 7 8 9 10 Stock 3 3 3 2 0 1 0 0 1 3 5 Disptd 0 0 0 0 0 0 0 1 0 0 0 Only 1 disappointed under new policy against 4 under

old policy. Not definitely, but pretty convincingly.

M1 sca at proportions A1 M1 A1 M1 missing some A1 times B1 one ignored B1 rest M1 A1 A1 M1 using both ret dists A1 A1 B1 B1


6. (i) Let f be the number of litres of Flowerbase produced B1

M1 A1

Let g be the number of litres of Growmuch produced Max 9f + 20g s.t. 0.75f + 0.5g ≤ 12000 M1 A1

A1

f + 2g ≤ 25000 (ii) B1 labels + scales

B1 B1 lines B1 shading

M1 A1

Max profit = £2500 by producing 12500 litres of

Growmuch (iii) No effect

B1 M1 A1

(iv) No effect The profit on Flowerbase will be reduced by more than

that suffered by Growmuch, since it uses more fibre. The objective gradient will thus increase from −9/20, making it even less attractive to produce any Flowerbase.

(v) £3000

B1

f

2500

24000

12500

16000 1440

25000

g

(11500, 6750)2385

Mark Scheme 4771January 2006

1.

2. (ii) Merges ordered lists to give an ordered list (iii) 7 (iv) Max = x + y – 1 Min = min (x, y)

M1 sca A1 to first step 3 inc. A1 to second step 3 A1 rest B1 B1 B1 B1

(i) & (ii) Critical: A, E (iii) A, E and D 6 days

B1 C OK B1 D OK B1 E OK M1 early and late A1 times B1 critical B1 B1

0 0

5 5

5 5

5 5

10 10B 3

A 5

C 3

D 4

E 5

(i)

Step number

List 1 List 2 A B List 3

1 3 4 4 3 4 3 3

2, 34, 35, 56 34, 35, 56 35, 56 35, 56 35, 56 56 56

13, 22, 34, 81, 90, 92 22, 34, 81, 90, 92 22, 34, 81, 90, 92 34, 81, 90, 92 81, 90, 92 81, 90, 92 90, 92 90, 92 90, 92

2 34 34 34 35 35 56 56

13 13 22 34 34 81 81 81

2 2, 13 2, 13, 22 2, 13, 22, 34 2, 13, 22, 34, 34 2, 13, 22, 34, 34, 35 2, 13, 22, 34, 34, 35, 56, 81, 90, 92

3. (i) Ins and outs One more out than in at D. Vice-versa at A. Start at D and end at A (ii) Existence – A B D C A Uniqueness – Only alternative is A B C …!!! Extra arc – New possibility A D C B … !!! (iii) B D C A B

M1 A1 B1 B1 M1 A1 A1 B1

4. (i) 12.5 kg 250 g (of butter) 10 kg 3 kg (of sugar) (ii) Identification of variables e.g. Let x = kg of toffee made Let y = kg of fudge made Max x + y st 100x + 150y ≤ 1500 800x + 700y ≤ 10000 Make 9 kg toffee and 4 kg fudge (iii) 12.5 kg of toffee and no fudge – either by comparing

68.75 with 67.50 with 45, or by a gradient argument Toffee price must decrease by £0.36, or to £5.14.

B1 B1 B1 B1 B1 B1 B1 B1 axes labelled and

scaled B1 butter line B1 sugar line B1 shading B1 max x+y + solution M1 A1 B1 B1

x

y

10

2714

12.5 15

(9,4)

5. (i) Total length = 57 miles Might be used to determine where to lay pipes or cables to

connect the towns. (ii) Shortest route: AGE Length: 24 (iii) Shortens mst to 53 miles (√ by 4) New shortest route ABGE – 23 miles (√ by 1)

M1 A1 selections A1 order of selecting A1 deletions B1 B1 B1 M1 sca Dijkstra A1 labels A1 order of labelling A1 working values B1 B1 B1 B1 B1

1 2 5 6 4 7 3 A B C D E F G

A – 10 – – – 12 15 B 10 – 15 20 – – 8 C – 15 – 7 – – 11 D – 20 7 – 20 – 13 E – – – 20 – 17 9 F 12 – – – 17 – 13 G 15 8 11 13 9 13 –

A

B C

D

E F

G

A

B C

D

E F

G

10

15

7

20

17

12

15

8 11

20

9 13 13

5 24

29 24

3 12

12

1 0

30 28

25

2 10

10

4 15

15

6. (i) e.g. 0 – 6 petrol 7 – 9 other (ii) e.g. 0 – 2 1 min 3 – 6 1.5 mins 7 – 8 2 mins 9 2.5 mins (iii) e.g. 00 – 13 1 min 14 – 41 1.5 mins 42 – 69 2 mins 70 – 83 2.5 mins 84 – 97 3 mins 98, 99 reject Two digits – fewer rejects (v) 24.5/10 = 2.45 mins

B1 M1 A1 M1 some rejected A1 2 rejected A1 B1 B1 arrival times M1 types M1 service start M1 service duration M1 service end M1 time in shop A1 M1 A1

(iv) Customer number

Inter-arrival time

Arrival time

Type of customer

Arrival at till

Time at till

Departure time

Queuing + paying

1 1 1 F 1 1 2 1 2 0.5 1.5 N 2 2 4 2.5 3 3.5 5 N 5 1.5 6.5 1.5 4 3 8 F 8 1.5 9.5 1.5 5 1 9 F 9.5 1 10.5 1.5 6 0.5 9.5 F 10.5 1 11.5 2 7 1.5 11 F 11.5 2.5 14 3 8 2 13 N 14 2.5 16.5 3.5 9 2 15 F 16.5 2 18.5 3.5

10 0 5 15 5 F 18 5 1 5 20 4 5


(i) Least weight route: A F B G E D Weight = 10 (ii) 11 From working value. Can't be bettered since new least

weight must be bigger than 10.

M1 sca Dijkstra A1 labels A1 order of labelling A1 working values B1 B1 B1 B1

2.

(i) e.g. a tree (ii) 13 (iii) 14 (iv) e.g.

M1 A1 B1 B1 B1 M1 A1 A1

G F

E D

C

B A

2

5

3

6 12

1

6

1

8 3

2

7

1 0

3 35 3

2 22

4 55

6 912 10

7 10

11 10

5 67 6

×

× × × × ×

×

×

× ×

× × ×

×⊗ ⊗


3. (i) M = 1 f(M) = –1 L = 1 M = 1.5 f(M) = 0.25 R = 1.5 (ii) Solves equations (Allow "Finds root 2".) (iii) A termination condition

B1 B1 B1 B1 B1 B1 B1 B1

4.

(i) & (ii) Critical activities: A, C, E (iii) (iv) 2 hours (resource smoothing on A/B, but extra time needed for

D/E). (v) P – Q – R – S Q, R T Q, R U R V S, T, U W U

M1 sca activity-on-arc A1 A, B, C A1 D A1 E B1 forward pass (1.25 at end of B/dummy) B1 backward pass (1.25 at start of dummy/D) B1 M1 A1 M1 A1 B1 B1 B1 B1 B1

0 0

1.25 1.25

1 1

1.75 1.75

1.25 1.25

B 0.5

A 1

C 0.25

D 0.25

E 0.5

1

0.5 hours

people

A

B

C

D

E E A


5. (i) Let x be the number of hours spent at badminton Let y be the number of hours spent at squash 3x + 4y ≤ 11 1.5x + 1.75y ≤ 5 (ii) (iii) x + 2y (iv) 22/4 > 5 > 10/3, so 5.5 at (0, 11/4) (v) Squash courts sold in whole hours 1 hour badminton and 2 hours squash per week (vi) 3 hours of badminton and no squash

B1 B1 B1 B1 axes labelled and

scaled B1 line B1 line B1 shading B1 intercepts B1 (1, 2) B1 M1 A1 B1 B1 B1 B1

x

y

11/3 10/3

11/4 20/7

(1, 2)


6. (i) year 1: 00 – 09 failure, otherwise no failure year 2: 00 – 04 year 3: 00 – 01 year 4: 00 – 19 year 5: 00 – 19 year 6: 00 – 29 (ii)(A) (B) 0.6 (iii) (A) if no failure then continue after year 3 – but using rules

for yrs 1 to 3 (B) (C) 0.3 (iv) more repetitions

M1 A1 A1 M1 ticks and crosses A1 run 1 A1 runs 2–4 A1 runs 5–7 B1 runs 8–10 B1 B1 B1 M1 A1 runs 1–5 A1 runs 6–10 B1 B1

Run 1

Run 2

Run 3

Run 4

Run 5

Run 6

Run 7

Run 8

Run 9

Run 10

year 1

√ √ √ √ x √ √ x √ √

year 2

√ √ √ √ √ √ √ √

year 3

√ √ √ √ √ √ √ √

year 4

√ √ √ x √ √ x √

√ √ √ √ √

Run 1

Run 2

Run 3

Run 4

Run 5

Run 6

Run 7

Run 8

Run 9

Run 10

year 1

√ √ √ √ x √ √ x √ √

year 2

√ √ √ √ √ √ √ √

year 3

√ √ √ √ √ √ √ √

year 4

√ √ √ √ √ √ x √

√ √ √ √ √ √ √

D1 June ‘06 6(ii) (A) Run 1 Run 2 Run 3 Run 4 Run 5 Run 6 Run 7 Run 8 Run 9 Run 10 Year 1 Year 2 Year 3 Year 4 Year 5 Year 6 6(iii) (B) Run 1 Run 2 Run 3 Run 4 Run 5 Run 6 Run 7 Run 8 Run 9 Run 10 Year 1 Year 2 Year 3 Year 4 Year 5 Year 6

83

Mark Scheme 4771January 2007

4771 Mark Scheme Jan 2007

84

1. (i) (ii) Any two of 1 or 2 or 3 or 5 or 7 (iii)

(iv)

(v) A tree

B1 B1 B1 M1 branching tree A1 M1 branching tree A1 B1

2. (i) 109; 32; 3; 523; 58 32; 3; 109; 58; 523 4 comparisons and 3 swaps 3; 32; 58; 109; 523 3 and 2 3; 32; 58; 109; 523 2 and 0 3; 32; 58; 109; 523 1 and 0 10 and 5 in total (ii) 523; 109; 58; 32; 3 10 swaps (iii) 1.5 × 1002 = 15000 seconds = 4 hrs 10 mins

M1 A1 only if all iterations

completed B1 B1 B1 B1 M1 A1 hours and minutes

3. (i) e.g. 0, 1 → A 2, 3 → B 4, 5 → C 6, 7 → D 8, 9 → E (ii) e.g: 3, 4, 4, 4, 1 (iii) In the above simulation mean = 3.2 (Correct expectation is 2.5 – geometric rand variable) (iv) More repetitions

M1 A1 proportions OK B1 efficient M1 A1 M1 A1 B1


85

4. (i) (ii) See above Critical activities: A; B; D; F; G; I; K Duration = 46 (iii) E: total float = 1; independent float = 1 H: 1 and 0 J: 14 and 13 C: 2 and 2 (iv) Tiler (I) – 2 days – £500 Electrician (D) – 1 day – £300 Bricklayer (B) – 1 day – £350

M1 activity-on-arc A1 single start and

end A1 dummy 1 A1 dummy 2 A1 rest M1 A1 forward pass M1 A1 backward pass B1 critical activities B1 duration B1 total floats B1 independent floats B1 tiler B1 electrician B1 bricklayer

10 10 17 17

0 0

22 22

41 41

30 31

31 31

28 28 22 22 A 10

B 7

C 10

D 5 E 4

F 6

H 2

G 3

I 10

J 2 46 46

K 5


86

5. (i) Let x be the number of m2 of lawn. Let y be the number of m2 of flower beds. x + y ≥ 1000 0.80x + 0.40y ≤ 500, i.e. 2x + y ≤ 1250 y ≥ 2x x ≥ 200 Minimise 0.15x + 0.25y (ii) & (iii) Lay 250 m2 of lawn and 750 m2 of flower beds. Annual maintenance = £225. (iv) Intersection of y ≥ 2x & area constraint is at

(333.33,666.67) so max useful capital is £533.33. So £33.33.

B1 B1 B1 B1 B1 B1 B1 B1 axes labelled +

scaled B4 lines B1 shading M1 A1 B1 (allow £533.33)

x 1000 625

1250

1000

200

y

(200,800)

(200,850)

(250,750)

230

242.5

225


87

6. (i) DtoE; BtoD; CtoE; DtoF; AtoB Total length = 20 (iii) e.g. Total length = 37 (iii) Lengths are 27 and 28. Shorter and more nearly equal.

M1 A1 no BC nor BE A1 B1 B1 B1 reduced table M1 delete/select/delete A1 first 2 rows A1 rest of table A1 order B1 B1 B1 B1 B1 B1

1 3 4 6 2 5 A B C D E F A – – – – 12 – B – – 5 – 6 6 C – 5 – 8 – 7 D – – 8 – – – E 12 6 – – – 7 F – 6 7 – 7 –

F

E D

C

B A

F

E D

C

B A


94

4771 Mark Scheme June 2007 1. (i) (ii) No. Two arcs AC. (iii)

(iv) No. ABDC(train)A is a cycle.

M1 4 nodes and 5 arcs A1 M1 A1 M1 5 nodes and 5 arcs A1 M1 A1

D C

B A

D C(train)

B A

C(bus)

2. (i) Rucksack 1: 14; 6 Rucksack 2: 11; 9 final item will not fit. (ii) Order: 14, 11, 9, 6, 6 Rucksack 1: 14; 11 Rucksack 2: 9; 6; 6 (iii) Rucksack 1: 14; 9 Rucksack 2: 11; 6; 6 e.g. weights.

M1 6 must be in R1 A1 B1 B1 ordering M1 11 in R1 A1 B1 B1

3. Optimum of 15.4 at x = 6.8 and y = 0.6.

B1 axes scaled & used M1 lines A1 B1 shading M1 two intersection A1 points M1 solution A1 (or by using the objective gradient to identify the optimal point)

2.8 4

7 8 14

(4, 2)

(6.8, 0.6) 8.4

14

14

15.4

x

y

95

4771 Mark Scheme June 2007 4. (i)

Activity Duration (minutes)

Immediate predecessors

A Rig foresail 3 – B Lower sprayhood 2 – C Start engine 3 – D Pump out bilges 4 C E Rig mainsail 1 B F Cast off mooring ropes 1 A, C, E G Motor out of harbour 10 D, F H Raise foresail 3 A I Raise mainsail 4 E J Stop engine and start sailing 1 G, H, I

(ii)

Critical activities: C; D; G; J Project duration: 18 minutes

(iii) H and I (iv) 25 mins Must do A, B, E, C, F, D (in appropriate order) then H and I with G,

then J. (v) 18 mins

e.g. Colin does C, D Crew does A, B, E, F Thence G et al

B1 A, B, C, D, E, H & I B1 F B1 G and J M1 A1 forward pass M1 A1 backward pass

B1 B1 B1 B1 B1 B1 B1 B1 B1

B 2

A

F 1 3 6

2 5 0 0

3 6

3 3 7 7

17 173 6 C 3

G 10

D 4

H 3

18 18

J 1 I 4

E 1

96

4771 Mark Scheme June 2007 5. (i) & (ii) Route: G A F C D Weight: 17 (iii) Route: G B C F E D or G B A E D Weight: 6 Any capacitated route application. (iv) Compute min(label, arc) and update working value if result is

larger than current working value. Label unlabelled vertex with largest working value.

M1 A1 arcs A1 arc weights M1 Dijkstra A1 labels A1 order of labelling A2 working values

B1 B1 B1 B1 B1 B1 B1 B1

F

E D

C

B

A

G 11

10 3 8 5

2

7

6

6

5

14

2 55

1 0 4 1414

6 15 15

7 17 5 14 15 14

3 8 8

or 4

or 5

19 17 if E is fourth

20

97

4771 Mark Scheme June 2007 6. (i)(a) e.g. Dry: 00 – 39 Wet: 40 – 69 Snowy: 70 – 99 (b) e.g. Dry: 00 – 19 Wet: 20 – 69 Snowy: 70 – 99 (c) e.g. Dry: 00 – 27 Wet: 28 – 55 Snowy: 56 – 97 Reject: 98 & 99 (ii) D (today) → D → S → S → W → S → D → D (iii) 3/7 (or 4/8) (iv) a (much) longer simulation run, with a "settling in" period ignored.

M1 proportions A1 efficient M1 proportions A1 efficient M1 reject some A1 proportions A1 reject 2 M1 applying their rules

sometimes A1 dry rules A1 wet rules A1 snowy rules B1

(v) Defining days as dry, wet or snowy is problematical. Assuming that the transition probabilities remain constant. Weather depends on more than just previous day's weather

B1 B1 B1 B1

98

4771 Mark Scheme January 2008

58

4771 Decision Mathematics 1

1 (i) 6 routes M→A→I→T→Pi→C M→A→I→T→Pi→R→C M→A→I→T→Pi→H→R→C M→V→I→T→Pi→C M→V→I→T→Pi→R→C M→V→I→T→Pi→H→R→C (ii) 6 routes M→A→I→Pa→Pi→C M→A→I→Pa→Pi→R→C M→A→I→Pa→Pi→H→R→C M→V→I→Pa→Pi→C M→V→I→Pa→Pi→R→C M→V→I→Pa→Pi→H→R→C (iii) M→V→I→Pa→Pi→H→R→Pi→C A R→H (iv) e.g. P→T→I→V→M→A→I→Pa→P→H→R→C→P→R

B1 B1 B1 B1 B1 M1 ends at R A2 (–1 each

error/omission)


59

2.

(i) lines shading (3.6, 0.6) 25.8 at (3.6, 0.6) versus 21 and 24 (or profit line) (ii) 25 at (3, 1)

B2 B1 B1 graph or sim. eqns M1 A1 B1 B1

y

x


60

3.

y = 2008 c = 2008/100 = 20 n = 2008 – 19 x (2008/19) = 2008 – 19 x (105) = 13 k = 3/25 = 0 i = 20 – 5 – 20 / 3 + 19 × 13 + 15 = 271 i = 1 i = 1 – 0 = 1 j = 2008 + 502 + 1 + 2 – 20 + 5 = 2498 j = 6 p = – 5 m = 3 d = 23 So 23rd March

B1 B1 B1 B1 B1 B1 B1 B1


61

4. (i) e.g. 0–3→brown 4–7→blue 8–9→green (ii) e.g. 0–1→brown 2–5→blue 6–7→green 8–9→reject (iii) e.g. Eye colours

Parent 1 brown

brown

brown blue

Parent 2 brown blue brow

n blue

Offspring brown

brown

brown

brown

brown

green blue gree

n brown

brown

brown blue brow

n green

brown green

brown blue brow

n green

brown blue

M1 A1 proportions OK A1 efficient M1 some rejected A2 proportions OK (–1 each error) A1 efficient B1 br/br→br (4 times) B1 br/gr→bl B1 gr/gr→gr M1 br/bl rule A1 application A1 application B1 bl/bl application M1 gr/bl rule A1 application


62

5. (i)&(ii) e.g. time − 55 weeks critical − A; B; F; G; J (iii) 50 weeks (49 weeks if G crashed rather than H) (iv) E − 1 week F − 3 weeks J − 2 weeks (G − 1 week, if crashed) (v) £115000 (£121000)

M1 sca (activity on arc) A1 dummy activities + E

and F A1 A, B, C, D A1 G, H, I, J M1 forward pass A1 M1 backward pass A1 B1 cao B1 cao B1 M1 A3 A1

D

B 22

J 5

C 10

F 11

G 9 I 3

H 10 0

4

8 30

30

41

50 55

55 55

55 50

41

30

30 8

30

A 8

E 9


63

6. (i) e.g. Total length = 2.2 km (ii) Prim: connect in nearest to connected set Kruskal: Shortest arc s.t. no cycles (iii) Arcs used: AD, DE, EF, FG, DI, IH, AB or DB, FC or BC Total length = 2.7 km (AB&FC) or 2.9 km (AB&BC) or 2.4 km

(DB&FC) or 2.6 km (DB&BC)

M1 connecting tree A1 DE A1 FC, FG A1 AD, DI, FH A1 2 of length 0.4 M1 A1 M1 name A1 description M1 Dijkstra A1 working values (see vertex G) A1 order of

labelling A1 labels M1 arcs counted A1 only once A1

0.2

0.2

0.2

0.2

0.1

0.1

0.1 0.1 0.3

A

B C

D E

F G

H I

0 1

0.7

0.3

2 0.3

0.6

0.4

3 0.4

0.9

4 0.6

1.0

1.5

5 0.7

1.1

6 0.9 1.1

7 1.0

8 1.1

9 1.1

0.2

0.2

0.2

0.2

0.1

0.1

0.1 0.1 0.3

A

B C

D E

F G

H I


83

4771 Decision Mathematics 1 Solutions 1. Objective has maximum value of 24 at (20,4)

M1 A1 third line B1 shading B1 (0,20) and (22,0) B1 (9,14) B1 (20,4) M1 A1 solution or M1 A1 B1, B1 scale (implied OK), B1 profit line, B1 (20,4) M1 A1 (20,4) A1 (24)

x

y

(20,4)

(22,0)

(9,14)

(0,20) or by profit line


84

2. (i)

X Y 5, 14, 153, 6, 24, 2, 14, 15 5, 14, 153 5, 2 5, 14, 6, 24,14, 15 5, 14, 24 5 14, 6, 14, 15, 14, 15 14, 6 14, 14

Answer = 14 Comparisons = 30 (ii)

X Y 5, 14, 153, 6, 24, 2, 14 5, 14, 153 5, 2 5, 14, 6, 24,14 5, 14, 24 5 14, 6, 14 14 14, 6 14

Answer = 14 Comparisons = 24 (iii) Median (iv) Time taken approximately proportional to square of length

of list (or twice length takes four times the time, or equivalent).

M1 A1 A1 M1 A1 A1 B1 B1

3. (i) T1→T2 T1→T3→T2

T1→T3 T1→T2→T3 T1→T2→T3→T4 T1→T3→T4 (ii) T4→T3→T2→T1 T4→T3→T1

T4→T3→T1→T2 T4→T3→T2 T4→T3 (iii) 22 (iv) 11

M1 A1 M1 A1 M1 allow for 23 A1 M1 halving (not 11.5) A1


85

4. (i) e.g. 00–09→1 10–39→2 40–79→3 80–89→4 90–99→5 (ii) e.g. 00–15→1 16–47→2 48–55→3 56–79→4 80–87→5 88–95→6 96, 97, 98, 99 reject (iii) & (iv) Sim. no.

Cars arriving after Joe – time interval│number of passengers

Time to 15 passengers (minutes)

1 3│2 2│1 1│2 2│2 3│1 6 2 3│1 2│2 1│4 1│2 5│1 6 3 5│1 2│2 2│1 3│4 2│2 12 4 4│6 3│2 4│1 1│2 2│3 4 5 5│1 4│1 3│2 5│4 2│2 17 6 4│4 4│2 5│3 1│4 1│4 8 7 4│1 4│2 3│1 5│4 1│3 16 8 2│2 2│2 2│4 3│5 1│2 6 9 1│1 1│1 1│1 1│1 1│2 5

10 2│4 3│2 2│6 2│5 2│1 5 (v) 0.8 more runs

M1 A1 proportions OK A1 efficient M1 some rejected A2 proportions OK (–1 each error) A1 efficient M1 A2 (–1 each error) M1 simulation A1 time intervals A1 passengers A1 time to wait

B1 B1


86

5. (a)(i) Activity D. Depends on A and B in project 1, but on A, B and C in

project 2. (ii) Project 1: Duration is 5 for x<3, thence x+2. Project 2: Duration is 5 for x<2, thence x+3 (b) (i) & (ii) Project duration – 7 Critical activities – T, X

M1 A1 A1 B1 "5" B1 B1 beyond 5 M1 activity-on-arc A1 single start and single end A2 precedences (–1 each error) M1 A1 forward pass M1 A1 backward pass B1 B1

R 2

Y 3

X 2

S 1

T 5

W 3

Z 1

0 0

2

2

5

5

7 7

6

5

3

3


87

6. (i)

Order of inclusion

1 3 6 4 5 2

A B C D E F

A – 10 7 – 9 5

B 10 – – 1 – 4

C 7 – – – 3 –

D – 1 – – 2 –

E 9 – 3 2 – –

F 5 4 – – – –

Arcs: AF, FB, BD, DE, EC Length: 15 (ii) & (iii) Arcs: AF, FB, BD, AC, AE Length: 26 (iv) Cubic n applications of Dijkstra, which is quadratic

M1 A1 select A1 delete A1 order B1 B1 B1 arcs B1 lengths M1 Dijkstra A1 working values A1 order of labelling A1 labels M1 A1 B1 B1

F

E D

C

B A 10

7

9

5

1 4

3

2

1 0

10

7

9

5

2 5

9

3 7

4(5) 9

(11)10

5(4) 9 6 10


62


1. (i) (ii) Can't both have someone shaking hands with everyone and

someone not shaking hands at all. (iii) n arcs leaving By (ii) only n–1 destinations

M1 bipartite A1 one arc from each

letter A1 David A1 rest B1 0 ⇒ ~3 B1 3 ⇒ ~0 B1 B1

2. (i)

n i j k 5 1 3 3 2 2 8 3 1 13 4 0 16

k = 16 (ii) f(5) = 125/6 – 35/6 + 1 = 90/6 + 1 = 16

(Need to see 125 or 20.83•

for A1) (iii) cubic complexity

B1 B1 B1 B1 B1 M1 substituting A1 B1

B

C

D

A

1

2

3

0


63

3. (i)

Cheapest: £11 [start (£2 starter)] → A (£3 main) → E (£3 main) → B (£1 main) → F (£2 sweet) → [end] (ii) repeated mains ! directed network

M1 Dijkstra A1 order A1 labels A1 working values B1 £11 B1 route B1 B1

start

1042

13 10 6

sweet £2

7 27

end

1 0

2 2 2

3 4 4

5 8 10 8

4 5 5

6 9

10 9 7 10 10

8 11 12 11

12

63

A B C

D E F

G


64

4. (i) e.g. 00–47→90 48–79→80 80–95→40 96, 97,98, 99 ignore (ii) smaller proportion rejected (iii) e.g. 90, 90, 90, 80 350 (iv) e.g. 90, 80, 90, 80 340 80, 90, 80, 80 330 90, 40, 80, 90 300 40, 90, 90, 90 310 90, 90, 90, 90 360 80, 80, 40, 90 290 80, 80, 80, 90 330 90, 80, 90, 90 350 90, 40, 40, 80 250 prob (load>325) = 0.6 (v) e.g. family groups

M1 some rejected A3 correct proportions (–

1 each error) A1 efficient B1 M1 A1 A1√ M1 A3 (–1 each error) √ M1 A1 B1

5. (i)&(ii) e.g. time − 60 minutes critical − A; C; E; F; G; H (iii) A and B at £300 A; C; G; H B; E; F

M1 sca (activity on arc) A1 single start & end A1 dummy A1 rest M1 forward pass A1 M1 backward pass A1 B1 √ B1 cao B1 2 out of A, B, E B1 A B1 B B1 300 from A and B B1 B1

D 20

A 30

C 15

G 10

B 25

E 25

F 5

H 50 0

20

30 45

30

55

55

60 60

55

55

30

45 30

55


65

6. (i) xi represents the number of tonnes produced in month i x2 ≤ x3

x1 + x2 ≤ 12 (ii) Substitute x3 = 20 – x1 – x2 x2 ≤ x3 → x1 + 2x2 ≤ 20 Min 2000x1 + 2200x2 + 2500x3 → Max 500x1 + 300x2 (iii) Production plan: 6 tonnes in month 1 6 tonnes in month 2 8 tonnes in month 3 Cost = £45200

M1 quantities A1 tonnes B1 B1 M1 A1 A1 M1 sca A3 lines A1 shading M1 >1 evaluated

point or profit line

A1 (6, 6) or 4800 M1 √ all 3 A1 cao

x1

x2

(4, 8) 4400

(0, 10) 3000

(6, 6) 4800

12

12


84


Question 1 (i) 1 and 6, 2 and 5, 3 and 4 (ii) (iii)

B1 M1 10 to 14 edges A4 (–1 each edge error) B1 identification B1 sketch

Question 2. (i) A's c takes 2, leaving 3. You have to take 1. A's c takes one and you lose. (ii) A's c takes 3 leaving 3. Then as above. (iii) A's c takes 3 leaving 4. You can then take 1, leading to a win.

M1 A1 A1 M1 A1 M1 A1 A1

Vertex 1

Vertex 5

Vertex 4 Vertex 3

Vertex 2

Vertex 6


85

Question 3. (i) 61 at (11, 7) (ii) Intersection of 2x+5y=60 and x+y=18 is at (10,8) 10 + 2×8 = 26

M1 optimisation A1 M1 A1

x

(5, 10) 55

(11, 7) 61

(18, 0) 54

(0, 12) 48

B3 lines B1 shading


86

Question 4. (i) e.g. 0–4 exit 5–9 other vertex (ii) e.g. 0.5, 0.5, 1.9 (Theoretical answers: 2/3, 1/3, 2) (Gambler's ruin) (iii) e.g. 0–2 exit 3–5 next vertex in cycle 6–8 other vertex 9–ignore and re-draw (iv) e.g. 0.7, 0.1, 0.2 (Theoretical probs are 0.5, 0.25, 0.25) (Markov chain)

B1 B1 M1 process with exits A1 B1 probabilities M1 duration A1 M1 ignore DM1 conditionality A1 equal prob A1 efficient M1 A2 M1 A1

1 A ExA 2 A B A B A B ExB3 A ExA 4 A B A B A ExA 5 A B ExB 6 A B A B A B ExB7 A B A B ExB 8 A ExA 9 A B ExB 10 A ExA

1 A B A B A ExA 2 A C A ExA 3 A ExA 4 A B C B C ExC 5 A ExA 6 A C A B ExB 7 A ExA 8 A B C ExC 9 A ExA

10 A ExA


87

Question 5.

(i) e.g. (ii) Order: AE; AD; DB; DC or AD; AE; DB; DC or AD; DB; AE; DC Length: 65 km OR Order: AD; DB; DE; DC Length: 66 km (iii) Length: 53 km Advice: Close BC, AE and BD (iv) facility (e.g. anglers) distances (e.g. B to C)

M1 A1 connectivity A1 lengths M1 connected tree A2 (–1 each error) A1 B1 M1 connected tree A2 (–1 each error)

B1 B3 B1

E

B

D C

A

B

E

D C

A E

B

D C

A

20

18

15

16

17

15

15 and/or 20

25

B

E

D C

A

X


88

Question 6. (i)&(ii) time − 230 minutes critical − A; B; E; F; H (iii) e.g. Least time = 240 mins Minimum project completion times assumes no resource

constraints.

M1 sca (activity on arc) A1 single start & end A1 dummy A1 rest M1 forward pass A1 M1 backward pass A1 B1 B1 cao M1 cascade A2 B1 Joan/Keith B1 B1

E 16D 16

G 16F 16

B 16A 16

H 16

C 16

D 60

A 100

B 60

E 40

C 150

H 10

F 20 0

60

100 160 200

230

220 220

230

200

G 30

160 190

100

200

160

4771 Mark Scheme

59


1 (i) & (ii)

Critical activities: A and D

M1 activity-on-arc

A1 C and E OK A1 D OK M1 forward A1 pass M1 backward A1 pass B1

2 (i)

M1 subgraph A1 M1 Changing

colours A1 top right A1 bottom left A1 not singletons

B1

(ii) The rule does not specify a well-defined and terminating set of actions.

B1

A 3

B 2

C 3

D 5

E 1

0 0

3 3

3 3

6 7

8 8

Red

Blue

Red

Red

Red Blue

Blue

Red

Blue

Blue

Red

Red

Blue Blue

Red

Blue

Subgraph

Swap colours on connected vertices and complete

4771 Mark Scheme

60

3 (i) No repeated arcs. No loops B1 B1

(ii) Two disconnected sets, {A,B,D,F} and {C,E,G,H} M1 A1

(iii)

6 arcs added

M1 A1 B1

(iv)

4×4 = 16 or 8

12 28 12 162

− = − =

B1

A

B

C

D

E

F

H

G

4771 Mark Scheme

61

4

(i) e.g. Let x be the number of adult seats sold. Let y be the number of child seats sold. x + y ≤ 120 x + y ≥ 100 x ≥ y

M1 A1 B1 B1 B1

(ii)

B3 lines (scale must be

clear) B1 shading (axes must be

clear) B1 point +

amount M1 point A1 amount M1 point A1 amount

(vi) 6000+60c > 10000 => c ≥ 67

M1 A1

y

x 120

120

100

100

(iii) £12000

(iv) £9000

(v) £7500

4771 Mark Scheme

62

5 (i)

& (ii)

shortest route: A E C F distance: 26 miles

M1 network A1 arcs A1 lengths M1 Dijkstra A1 working

values B1 order of

labelling B1 labels B1 B1

(iii) CE CD AE CF AD BF AB EF total length of connector = 45

M1 5 arc connector

A1 AD not included

A1 all OK, inc order

B1

B1

(iv) A 3 miles (or length = 9)

B 2 miles (or length = 10)

B1 B1

A B

C

D E

F

22

13

12

6

11

5

10 26

1

12

22

10

10

0

2

15 36

12 3

15 4

26

5 22

6 26

A B

C

D E

F

13

6

11

5

10

4771 Mark Scheme

63

6 (i) e.g. 0, 1, 2 fall

3, 4, 5, 6, 7, 8 not fall 9 redraw

M1 ignore at least 1A1 proportions

correct A1 efficient

(ii) apple r n fall?

1 1 yes 2 3 no 3 8 no 4 0 yes 5 2 yes 6 7 no Three apples fall in this simulation.

M1 A2 –1 each error B1√

(iii) apple r n fall?

2 0 yes 3 1 yes 6 4 no apple r n fall? 6 4 no apple r n fall? 6 8 no apple r n fall? 6 0 yes 5 days before all have fallen

M1 A2 –1 each error A1√

(iv) apple r n fall? 1 picked 2 1 yes 3 3 no 4 8 no 5 0 yes 6 2 yes apple r n fall? 3 picked 4 7 no apple r n fall? 4 picked 3 days before none left

M1 A2 –1 each error B1√

(v) more simulations

B1


1

1. (i)

AB 12 AB AC 13 AC AD 29 ABD AE 35 ABDE AF 22 ACF

(ii) 5

M1 Dijkstra A1 working values B1 order of labelling B1 labels B1 AB and AC B1 AD and AF B1 AE B1

2. (i) 3 8 6 4 12 2 24 1 24 (ii) 26 42 52 21 104 10 208 5 416 2 832 1 1092 (iii) multiplication

M1 doubling and halving M1 deleting and summing A1 cao M1 doubling and halving M1 deleting DM summing A1 cao B1

A B

C

D E

F

0 1

12

13

45

12 2

29

24

13 3

37

22

22 4

29 5

35

35 6

12

17

45

6

9

13 12

24


2

3. (i) (ii) (iii) The graphs represent traffic flows within the junctions.

They do not take account of flows approaching or leaving the junctions.

(Graphs are not planar if these flows are added, so traffic flows have to cross.)

B1 B1 B1 B1 12 arcs B1 connectvity B1 3 out of each in vertex B1 3 into each out vertex B1


3

4. (i) Each small tile has area 100 cm2 so 1000x Similarly 900y So 1000x + 900y ≥ 400×300 = 120000 (ii) y ≤ 100 10x ≤ 9y (iii) e.g. minimise 1.5x + 2y Integer solution required, so x=60, y=67, cost = 224 (iv) wastage or design

M1 areas A1 tile areas A1 B1 B1 B1 B1 B3 lines B1 shading M1 solving A1 x = 59-61 y = 66-68 A1 220-228 B2

(30, 100) 245

(60, 2366 )

223.33

(90, 100) y

x

100


4

5. (i) e.g. 0 to 4 –> stagger left 5 to 9 –> stagger right + accumulation (ii) probably one of: (iii) repeat relative frequency (iv) e.g. 0 to 2 –> stagger left 3 to 8 –> stagger right 9 reject and redraw

M1 A1 B1 M1 A1 B1 B1 M1 reject some A1 proportions A1 efficient

(v) e.g. run 1 R L R L L Rrun 2 R L R R L Rrun 3 R R L L L Lrun 4 L L R L R Rrun 5 R R R * run 6 L R R R R *run 7 R R L R R *run 8 R R L R R *run 9 R R R *run 10 L R R L R R

Probability estimate = 0.5 (Theoretical = 0.73 + 5×0.74×0.3 = 0.70315)

M1 A2 (–1 each wrong row) B1 falling in M1 probability A1


5

6. (i) & (ii) Duration = 24 months Critical : A; F; J; G (iii) Crash F by 1 month and G by 1 month at a cost of £6m. (iv) Crash G by 2 months at a cost of £8m.

M1 activity-on-arc A1 D, E, H and K A1 F A1 I and J A1 G M1 forward pass A1 M1 backward pass A1 B1 cao B1 cao B1 F by 1 month B1 G by 1 month B1 £6m M1 G only A1 £8m

A 4

C 7

B 2

K 12

G 6

0 0

4 4

4 4

18 18

24 24

7 8

11 11

H 3

F 7

I 12

D 12

E 5

J 7

4771 Mark Scheme January 2011 1. (i) (ii) 6 (iii) e.g. 4 arcs and (e.g.) {A}, {B, C, D, E} (iv) Reference to parts (i) and (ii), in reverse − or similar

M1 A1 M1 attempt at complete

connectivity A1 B1 B1 B1 B1

M1 any and only 3 of the 4 A1 all M1 5, 6 or 7 A1 6 4 ... set of 1 ... disjoint set of 4 SC M1 6 arcs A1 appropriate sets ... disjoint of size 2 and 3

A B

DE

C

1


2. (i)

B1

B1

B1 (ii)

B1

B1 B1 B1 1,2 3 4

B1 5,6

7

8

cao cao ... allow extra second line of 5678 D, but with ‒1 cao cao

cao cao award the last two B1s only for contiguous blocks of 3 tests from line 3 allow extraneous lines but ‒1 once only, and only from the last two B1s

Test number

Sample drawn from flagons numbered

Result (D = dead, A = alive)

A 1 1, 2, 3, 4 2 5, 6 A 3 7 D 4 8 A

Test number

Sample drawn from flagons numbered

Result (D = dead, A = alive)

D 1 1, 2, 3, 4 2 5, 6, 7, 8 D 3 1, 2 A 4 3 D 5 4 A 6 5, 6 A 7 7 D 8 8 A

2

4771 Mark Scheme January 2011 3. (i) Shortest distance = 27 Shortest route … ABCEF (ii) Because F was the final vertex labelled. (iii) Because if there were to be a shorter route than BCEF

from B to F, then A to B followed by it would give a shorter route from A to F.

or “B is en route”

M1 Dijkstra A1 working values B1 order of labelling B1 labels B1 B1 B1 B1

cao

A

F

E D

C

B 5

7

10

5

9 6

22

25

37

28

19

05

19

28

2 5

27 21

37

1

42 3 12

22

1227 6

225

27

214

3

4771 Mark Scheme January 2011 4. (i) (ii)&(iii) (iv) critical activities: E; F; J duration: 7 minutes task: A B C D E F G H I J float: 4.5 4.5 4 4 0 0 3.5 4 4 0

B1 A, C, E and G B1 B, D and F B1 H, I and J M1 activity-on-arc A1 A, G, C ,E, B, D, F A1 H, I, J

M1 A1 forward pass M1 A1 backward pass B1 B1 B1

no follow through no multiple starts no multiple ends but no follow of activity-on-node ditto

cao cao cao blank=0

Task Description Duration Immediate predecessor(s) (mins) – A Fill kettle and switch on 0.5

B Boil kettle 1.5 A – C Cut bread and put in toaster 0.5

D Toast bread 2 C – E Put eggs in pan of water and

light gas 1

F Boil eggs 5 E – G Put tablecloth, cutlery and

crockery on table 2.5

H Make tea and put on table 0.5 B; G I Collect toast and put on table 0.5 D; G J Put eggs in cups and put on

table 1 F; G

B 0.5 5 6.5 2.5 1.5

A 0.5 H 0.5

G 0 6 7 2.5 7 0 2.5 I C

0.5 0.5 D 0.5 4.5 2.5 6.5 E 2 J 1 1

F 1 6 1 6 5

4

4771 Mark Scheme January 2011 (v) e.g.

M1 cascade or

condensed cascade A1 activities other than

B, D and F non-overlapping

A1 correctly finish in 7

need to have 9 or 10 activities E C A G H I J cao

J E F D C

G H I A B

5

4771 Mark Scheme January 2011 5. (i) e.g. 00–04 6 05–29 7 30–79 8 80–99 9 (ii) e.g. 00–09 goal 10–99 no goal (iii) e.g. 8 0 1 0 0 0 0 0 0 so 1 goal (iv) e.g. 00–31 5 32–63 6 64–79 7 80–95 8 96–99 reject and redraw (v) e.g. 6 0 0 1 0 0 0 so 1 goal (vi) Each scored 10 goals. Nothing to choose between

them. (vii) More repetitions

M1 rule using 2-digit nos

A1 correct proportions A1 efficient B1 B1 B1 B1 M1 2 or more rejected A1 correct proportions A1 efficient B1 M1 A1 M1 A1 B1

complete rule required rule (i) need to see which are converted ... their 8 and rule (ii) their 8 and rule (ii) ... ignore previous line allow part (iv) if seen elsewhere 3 or 4 rejected in part (v) below expect either 00‒11 or 88‒99 for goal any other rule must be declared to score marks rule (iv) their 6 ... need to see which are converted goals scored one, the other or indifferent, depending on goals scored “greater number of random numbers” 0 “more accurate data” 0 Also no “or”s! 3-digit RNs 0

6


7

6. (i) Thousands of litres of A in stock = 2 b −4 (ii) 5(a+2) + 6(b+4) ≥ 61 (a+2) + (b+4) ≤ 12 giving a + b ≤ 6 (iii) (iv) Increase stock levels of A by 9000 litres. Reduce stock levels of B by 3000. (v) New stock levels are 11000 of A and 1000 of B. 511000 + 61000 = 61000 11000 + 1000 = 12000

B1 B1 M1 A1 M1 A1 B4 lines B1 shading B1 B1 B1 B1 B1

cao watch for fluke their negative gradient stock line shape = or

Give the marks for 9000, ‒3000, or equivalent 200 litres on both (iv) SC correct answer from nowhere OK Allow comment only for the “fully stocked” B1.

a

(9, −3)

6 5.4

b

6

4.5


4771, June 2011, Markscheme 1.

(i) (ii) 14 (iii) 47 (iv) (0, 0) and (1, 0) (v) Explanation should recognise that a line is a set of

points − not appropriate in this context.

B1 3 to 4 deleted B1 1 to 4 deleted B1 4 to 4 added B1 M1 A1 cao B1 B1

-1 for each arc in error Award method mark if answer correct, or if wrong but with a sum of products shown. Award only if correct points are specified in some way. e.g. “Intermediate points have no meaning.” e.g. “Can’t have one and a half pairs of shoes.” (sic)

number of people

num

ber o

f pai

rs o

f sho

es

0

1

2

3

4

5

0

1

2

3

4

5

5


2. (i) X = min(25, 8.5) = 8.5 or equivalent Y = min(5, 42.5) = 5 oe X* = (85–10)/10 = 7.5 oe Y* = (25–8.5)/5 = 3.3 oe (ii) Avoids tiny feasible regions.

B1 cao B1 cao B1 cao B1 cao B1 allow ft B1 cao B1 cao B1

OK if only seen once or more on graph OK if only seen once or more on graph OK if only seen on graph OK if only seen on graph sensibly scaled for their X and Y e.g. disallow if either of the lines in the question could intersect both axes. lines - can extend to beyond segment condone minor errors in plotting (e.g. 8.5 plotted at 9) need comment on size of region

(8.5, 3.3)

(0, 5)

(8.5, 0)

(7.5, 5)

6


3. (i) e.g. 1, 2, 3 1 4 2 5, 6 3 (ii) e.g. 1, 2 1 3 2 4 3 (5, 6 reject and throw again) (iii) non uniform allows 100

M1 A1 A1 M1 reject some A1 reject two A1 rest

B1 B1

function with domain {1,2,3,4,5,6} and range {1,2,3} (special cases are possible – if correct!) proportions 3:2:1 all OK (Special cases are possible – if correct! e.g. allow throwing die twice and allocating correct proportions of 36.)

“101 values” OK no credit for, e.g. “3 is not a two-digit number”

7


4. (i) e.g. x = number of large houses y = number of standard houses

land: 200x + 120y <= 120000 oe cash: 60x + 50y <= 42400 oe market: x <= 0.5y oe (ii) (iii) intersection of y=2x and 6x+5y=4240, (265, 530) 2650

(iv) their 60x + 50y <= 45000 or line from their (0, 900) to (750, 0)

Best point is at the intersection of the land constraint and the new cash constraint, and not on y=2x

(214, 643) 2785

M1 A1

B1 B1 B1 B1 line 1, allow ft B1 line 2, allow ft B1 line 3, allow ft B1 feasible region M1 correct point, cao A1

B1 ft

M1 comparison of two (or more) points

A1

M1 correct point, cao A1

M1 for variables for large and for standard A1 for “number”

use “isw” for incorrect simplifications -1 once only for any “ < ” for instance, if x <= 2y in part (i), then allow correct graph of x <= 0.5y or ft graph of x <= 2y plotting tolerance on axis intersection points – within correct small square must consider 3 lines ft if region includes y-axis interval from origin upwards allow any clear indication of feasible region ignore any indication(s) of boundary lines included or excluded identification only - coordinates not required here their 4x+3y from (260-280, 520-540)

can be implied from final M1 working

not just ringing points their identified best point is not on y=2x or an axis

identification, coordinates not required here bedrooms - their 4x+3y from (200-220, 620-660)

706.67 x

y 1000

(265, 530) 2650

600

848

8


9

5. (i) (ii) (iii) critical activities: A; Pl; Fo; W; R; E; Deco project duration = 41 days act A Pl Dm Fo W Pb R Fl E WD Dc

M1 Fl correct A1 rest

M1 at least one correct nontrivial join

A1 forward pass M1 at least one correct

nontrivial burst A1 backward pass

B1 cao B1 cao B1 A, Pl, Dm, Fo, W B1 rest B1 B1 one of R/WD

excluding start node

cao cao – most see zeros, dashes or empty spaces won’t do SC1 for a convincing but not specific answer, e.g. “A dummy is needed to cater for both joint and separate precedences”.

float 0 0 21 0 0 2 0 1 0 4 0 (iv) Fl has both W and Pb as immediate predecessors. R and WD have only W as immediate predecessor.

10 24 31

32 36

41 41

36

31 24 10

Pl 14

Demo3

A10 W3

Pb2 Fl 2

R3 Fo4

WD1

E2 Deco5

0 28

34 34 31

28 0

Activity Immediate predecessors A – Pl A Demo – Fo Pl; Demo W Fo Pb Fo R W Fl Pb; W E R; Fl WD W Deco WD; E


(v) (vi) new duration = 42 days critical activities: A; Pl; Fo; W; C; R; E; Deco

M1 C between W and R A1 Fl + dummy OK A1 WD OK B1 cao

both needed

WD Pl

Demo

A W

Pb Fl

R Fo C E Deco

10

4771 me June 2011

11

Mark Sche

6. (i)

1 7 9 8 2 10 3 6 11 5 4 P S F Ln Br Nr Bm Ld Nc Lv M

P − 150 − 240 125 − − − − − − S 150 − 150 80 105 − 135 − − − − F − 150 − 80 − − − − − − −

Ln 240 80 80 − 120 115 120 − − − − Br 125 105 − 120 − 230 90 − − − − Nr − − − 115 230 − 160 175 255 − − Bm − 135 − 120 90 160 − 120 − − 90 Ld − − − − − 175 120 − 210 100 90 Nc − − − − − 255 − 210 − 175 − Lv − − − − − − − 100 175 − 35 M − − − − − − 90 90 − 35 −

Length = 985 miles

M1 tabular

Prim A2 choosings A1 crossings B1 cao B1 cao

125 in P column and 90 in Br column ringed, with both rows crossed all circles in correct place; -1 each error (watch for one error making two changes to a row) all rows crossed out except, possibly, Nc row. accept convincing transpose

Nr

F

Nc

Ln

Ld

S

Bm

Br

Lv

M

P


(ii) Advantage: shortest length of track Disadvantage: tree, no redundancy fragility (breakdown et al) Disadvantage: some journeys are not shortest paths (iii) Route: P S Ln Nr Distance: 345 miles

(iv) Distance by min connector = 425 miles

B1 cao B1 B1 M1 Dijkstra A1 working

values B1 labels B1 order of

labelling B1 cao B1 cao

B1 ft their mc

allow cost minimisation could say “no cycles” disallow comments relating to direct connectivity, or relating to more stops “longer journeys” or “takes longer” allowed allow “min connector arcs may be more expensive” oe don’t allow two marks for the same point described differently. e.g. longer journeys/more time/more upkeep correct working values (no extras) at Ln and Nr, and working values only superseded at Ln and Nr (ignore Nc for this M) (need to check Nc here)

Nc

Br

P S

Bm

F

Ln

Nr

Ld M

Lv

150

175

35 100

90

90

210

255

175120

160

90

230

120

135 115

120

240 125

150

80 80105

9 340 340 545 515

8 335 7 305

335 305

10 345 355 345

4 215 215

2 125 230 5 125 240 230

0 1

150 3 6 300 150 300

12


Question Answer Marks Guidance 1 (i) & (ii) B1 connectivity B1 lengths B1 Dijkstra Award if wv’s OK at C. B1 working values allow legitimate later and other than at C larger wv’s which are listed, but not used. Disregard F. B1 order of labelling SC ... If possible follow for B1 labels these two marks. following

errors in network

B1 B1

Route: AECG Distance: 8

[8]

3 3 3

C

G

0 1

A

F

E D

B

3

5

1

2

6

4

1

5

4

8 3

8 4

5

2 3 9 6 8

8

1

6 6

7

3

9

5 7 5 7

5


Question Answer Marks Guidance 2 (i) A L R B f(L) f(R)

3 3.382 3.618 4 2.146 1.910

B1 B1

R and L -1 once only for incorrect f(R) and f(L) accuracy, but condone 1.91. Surds OK, but lose the accuracy mark. (Q says 3dp.)

3.382 3.618 3.764 4 1.910 1.875

B1 B1 B1

A L and R f(L) and F(R)

3.618 B1 A [6] 2 (ii) Saves a function evaluation B1 Has to be a comment about

function values.

[1] 2 (iii) eg

Setting the control on a gas fire to achieve a room temperature of 20C. Function could be (temp–20)2. (This example shows that optimising can be used to “achieve”.) Note that the domain cannot be time based ... i.e finding when something occurred. One cannot go back in time to take a reading!

B1

Optimisation with need to “Deepest point in seabed” sample at discrete intervals. example seen. This is

acceptable, assuming that depth soundings are taken at points, and ignoring the fact that the domain is two dimensional rather than one dimensional.

[1]

6


3 (i) M1 directed graph on 3 vertices A1 all correct M1 undirected on 3 vertices Arcs must either have an

arrow at each end. or no arrows.

A1 all correct

“shares at least one element with” “is a subset of”

[4] 3 (ii) M1 R subset of Q Allow area split in two, with A1 no other subsets third area. eg B1

B1 PQ PQ’

If P and R shown intersecting

then can score M1 A1 B0 B0.

eg

[4]

X Y

Z

X Y

Z

Q R P Q R P

7


Question Answer Marks Guidance 4 (i) Let x be the number of type X motors produced. M1 adequate definition Strict inequalities are Let y be the number of type Y motors produced. A1 “number of” equally OK 10x + 12y 200 B1 x 5 and y 5 B1 0.5x + 0.3y 7 B1 [5] 4 (ii) B1 inclined line B1 inclined line B1 x=5 and y=5 B1 shading follow line errors if shape is

The guidance level of

accuracy throughout this question is 0.25 on the x coordinate and 0.25 on the y coordinate.

(Look at (8,10) first.)

Inaccurate sketch with axis intercepts given is OK.

[4]

y

23.3

x

(16, 8)

16.7

14 5

5

(8,10) 1500

(5,12.5) 1375

(11,5) 1450

15

(iv) (10,6.7)

18

8


Question Answer Marks Guidance 4 (iii) Profit = 100X + 70Y

(5,12.5) or (5,12) 1375 or 1340 (8,10) 1500 (11,5) 1450 £1500 profit.

B1

M1

A1

optimisation either profit line or

evaluating and comparing at their 3 appropriate points

(OK if on graph) 1500 seen cao

SC B1 for 1500 without the preceding M mark

[3] 4 (iv)

Solution in range 41

32

41 6,10 =

691.6641.6,25,1075.9

Identification of one of (9,7), (10,6) and (11,5). Evaluation at all three of (9,7) (10,6) (11,5) 1390 1420 1450 So 11 of X and 5 of Y

B1

B1

M1

A1

cao looking for 326,10

cao cao

[4]

9


Question Answer Marks Guidance 5 (i) eg 07 double M1 reject Rejection can be implied. 8 single A1 correct proportions 9 reject and re-draw [2] 5 (ii) eg 05 double M1 reject Rejection can be implied. 6,7 single A1 correct proportions Ignore rule for (4,0). 8,9 reject and re-draw [2] 5 (iii) e.g. day doubles singles random number

selection 1 5 0 2 4 1 5 double 3 3 2 9, 4 double 4 2 3 0 double 5 1 4 Probability of drawing a single bag on day 5 is now 4/6.

M1 A1

M1 A1

M1 A1

M1 A1

[8]

For the simulation M1’s you need to see a random number being used with their rules

allow 5 shown as used on RN list. selection must show RN(s) explicitly Follow a candidate who new scenario seen explicitly, manages correctly to go from not implied by day 4 rule (4,1) to (4,0). It will then gain

M1 if it correctly goes to (3,1) on day 4, with A1 if shows no simulation needed.

a correct day 4 rule rule must be seen selection and new scenario needs RN explicit. Allow new

scenario if seen in subsequent probability calculation.

denominator = 6 Can be implied by 2/3 or 1/3 if numerator correct for their simulation.

10


Question Answer Marks Guidance 5 (iv) 4 simulations, each ending with 6 bags

all scenarios correct

M1

A1

[2]

Condone one slip. Condone simulating at (4,0) if correctly done. 6 bags can be implied by probs of thirds or sixths.

5 (v) Either averaging correct probabilities or sum of singles/30 M1 Correct computation, but allow 1 slip or omission. A1 Correct answer for their simulations.

[2]

11


12

Question Answer Marks Guidance 6 (i)

&

M1 activity on arc

(ii) A1 at least 1 dummy for E and F A1 precedences for D A1 precedences for G A1 rest eg. penalise multiple starts M1 forward pass A1 M1

A1

B1

backward pass If OK at start of dummy. If there is no dummy then these two marks are not available.

B1

Minimum completion time = 14 mins Critical activities ... A, B, D, G, H

[11] 6 (iii) 2 people

B1

[1]

6 (iv) 1 person ... 15.5 mins B1

[1]

6 (v) B1 network B1 time with small oven

time = 35.5 minutes [2]

6 (vi) revised time = 26.5 minutes

B1

[1]

time with large oven

A 0.5

F 0 .5

14 14 H 104 4 G

0.53.5 3.5 D 2 E

0.5

B 1 C

0.5

0.5 0.5

0.5 3.5

1.5 1.5

O2

O3

O1

0 0

P2

P3

P1

Oxford Cambridge and RSA Examinations

GCE

Mathematics (MEI) Advanced Subsidiary GCE

Unit 4771: Decision Mathematics 1

Mark Scheme for January 2013


5

Question Answer Marks Guidance 1 (i) M1 Dijkstra (if working values

correct at D) A1 working values B1 order of labelling B1 labels Route ... ABDCF Time ... 51 minutes B1 route and time [5] (ii) B1 methodology indicated B1 correct min connector

Time ... 52 minutes B1 cao [3]

B D

F

C

A

5

E

18

19

10

15

B D

F

C

A

41

5

17

2

17

32 31 30

10

4

56 51

51 6

3 15

1 0

E

18

21

19 32 21

10

15

15 10

33

5 33 30


6

Question Answer Marks Guidance 2 (i) bipartite B1 cao [1] (ii) 100 M1

A1 [2]

allow for 200 cao

(iii) B1 Darcy correct B1 Elizabeth correct B1 Panto characters correct [3] (iv) 58 18 + (8 5) M1 allow for 98 A1 cao [2]

A

B

Charming

Darcy

E

F

G

H

I

J

V W Cinderella Ugly sister 1 Ugly sister 2 Ugly sister 3 Elizabeth X Y Z


7

Question Answer Marks Guidance 3 (i) Step 1 x = 0.44

Step 2 oldr = 1 B1 cao

Step 345 i = 1, j = 0.5, k = 0.5 Step 6 change = 0.22

Step 7 newr = 1.22 Step 9 oldr = 1.22

B1 set-up (i.e. as far as 1.22)

Steps 10 11 12 i = 2, j = –0.5, k = –0.125 B1 3 steps correct Step 13 change = –0.0242 Step 14 newr = 1.1958 B1 new estimate (1.1958) Step 15 |change| = 0.0242 Step 9 oldr = 1.1958 Steps 10 11 12 i = 3, j = –1.5, k = 0.0625 Step 13 change = 0.005324 Step 14 newr = 1.201124 Step 15 |change| = 0.005324 Step 9 oldr = 1.201124 B1 iteration (to 1.201124) Steps 10 11 12 i = 4, j = –2.5, k = –0.03906 Step 13 change = –0.0014641 Step 14 newr = 1.1996599 Step 15 |change| = 0.0014641 Step 17 1.1996599 B1 iteration and end [6] (ii) 1 – 0.22 – 0.0242 – 0.005324 – 0.0014641 = 0.7490119 M1

A1 use of –0.44 as shown SC1 (cao) for algorithm repeated or answer only

[2]


8

Question Answer Marks Guidance 4 (i) M1 activity on arc & A1 single start and end (ii) A1 A, B, C OK A1 J, K, L OK A1 rest OK [5] M1

A1 forward pass (must have at least one join correct

M1

A1 backward pass (must have at least one burst correct)

Minimum completion time = 155 minutes B1 cao Critical activities are C, D, E, F, G, J, K and M B1 cao [6] 4 (iii) eg

Kate C C C D D D E1 F1 F1 F1 F1 F1 F1 H1 H1 H1 H1 H1 H1Pete A B E2 F2 F2 F2 F2 F2 F2 H2 H2 H2 H2 H2 H2

B1 ABCD cont. G1 G1 G1 G1 G1 G1 I1 I1 I1 J1 J1 J1 J1 K1 K1 K1 K1 K1 K1

cont G2 G2 G2 G2 G2 G2 I2 I2 I2 J2 J2 J2 J2 K2 K2 K2 K2 K2 K2B1 rest ... watch for M’s after

K’s and L’s cont L1 L1 L1 M1 M1

cont L2 L2 L2 M2 M2

215 minutes (3 hours and 35 minutes) B1 cao [3] 4 (iv) Two more people would be needed, so that the H’s and I’s could be done at the same time as the F’s and

G’s, and so that the two L’s could be done at the same time as the two K’s B1 B1

cao reasoning

[2]

C 15

A 5

15 15 0 0

65 65

35 35 30 30 H 30

D 15

E 5

M 10

F 30

J 20

B 5

95 95

65 80

5 10

I 15

G 30

K 30

L 15

115 130

115 115

145 145 155 155


9

Question Answer Marks Guidance 5 (i) e.g. 0 0 M1 either 0.2 for 1 or 0.3 for

2 1, 2 1 A1 all proportions correct 3, 4, 5 2 6, 7 3 8, 9 4 [2] 5 (ii) random number 5 3 0 2 4 7 9 1 1 8 M1 8 outcomes correct number of occupants 2 2 0 1 2 3 4 1 1 4 A1 all correct [2] 5 (iii) e.g. 0, 1 child B1 must use all 10 digits 2 – 9 adult cao [1] 5 (iv) chair 1 2 3 4 5 6 7 8 9 10

occ1 6 A 0 C 9 6 A 2 A 9 A 1 C 5 A 6 A 2 A occ2 2 A 6 A 5 2 1 C 1 C 4 A 8 1 9 A occ3 3 7 2 1 3 6 A 6 A 5 3 5 A occ4 3 1 1 2 8 0 6 A 0 5 1 C

M1 8 chairs OK

number of children = 5 A1 all OK number of adults = 15 [2] 5 (v) 40 children and 120 adults B1 FT... by 8 [1] 5 (vi) e.g. 00 – 06 0 M1 ignore some 07 – 13 1 A1 proportions correct 14 – 34 2 A1 efficient 35 – 55 3 56 – 90 4 91 – 99 ignore and “redraw” [3]

random number

child (C) or adult (A)


10

Question Answer Marks Guidance 5 (vii) random number 23 65 07 99 37 45 M1 3 OK number of occupants 2 4 1 – 3 3 A1 all correct FT [2] 5 (viii) chair 1 2 3 4 5

occ1 1 C 9 A 6 A 8 A 1 C occ2 2 A 2 A 8 0 C 8 A occ3 6 3 A 2 2 A 1 C occ4 4 6 A 1 9 4

number of children = 4 number of adults = 9 B1 FT ... all correct 64 children and 144 adults B1 FT ... by 16 [2] 5 (ix) greater reliability or more representative B1 [1]


11

Question Answer Marks Guidance 6 (i) e.g. Let x be the number of hats which Jean knits B1 must say “number of” Let y be the number of scarves which Jean knits B1 or vice-versa of course 1.5x + 3y ≤ 75, i.e. x + 2y ≤ 50 B1 simplification not 4x + 2.5y ≤ 100, i.e. 8x + 5y ≤ 200 B1 required x ≤ 20 and y ≤ 20 B1 both

B1 B1 B1 B1 lines (cao) B1 shading ... follow any set

of two horizontal, two vertical and two negatively inclined lines which give a hexagon in the bottom left corner.

[10]

50 25 hats - x

scarves - y

25

40

20

20

(20, 8) 220

(13.64, 18.18) 277.27

(10, 20) 270


12

Question Answer Marks Guidance 6 (ii) Objective = 7x + 10y B1 objective Best non-integer point

Solution ... (12, 19) 274, (13, 18) 271 or (14, 17) 268

M1 A1

considering profits at their three points as indicated cao

So 12 hats and 19 scarves B1 cao [4] 6 (iii) 10 hats and 20 scarves B1 cao £34 B1 FT ... their answer – 240 [2]


5

Question Answer Marks Guidance 1 (i)

M1

A1

simple and connected but not complete. (Ignore directions) cao

e.g.

B1

planar - cao

[3] 1 (ii) e.g.

M1 A1

exactly 3 vertices cao

[2] 1 (iii) B1 complete graph on 4

letters M1 4 regions A1 cao (planar OK) [3]

A B

C D

A B

C D

1

2 3

A B

C D

1 2

3 4

A B

C D

2 1

3 4

3

A B

C D

2 1


6

Question Answer Marks Guidance

2 (i) comps swaps i=1 9 7 3 11 5 13 5 3 i=2 7 3 9 5 11 13 4 3 i=3 3 7 5 9 11 13 3 2 i=4 3 5 7 9 11 13 2 1 i=5 3 5 7 9 11 13 1 0

B1 B1 B1

B1 B1

i=2 row OK i=3 row OK FT i=4 and 5 rows OK cao comparisons swaps

[5] 2 (ii) comparisons 6 B1 cao (OK if in 2 parts) swaps 3 B1 cao (OK if in 2 parts) [2] 2 (iii) further swaps 6 B1 cao [1]


7


3 (i) B1 Dijkstra – C correct B1 other working values B1 order of labelling B1 labels Note that D and G could

be labelled in the reverse order.

AB 13 ABC 26 ABCD 39 ABE 44 ABCF 35 ABCG 39 ABCDH 52 ABCGI 46

B1 B1

first 4 pairs second 4 pairs

[6] 3 (ii) Turn distances to times throughout the network.

Add 10 mins to every arc incident upon C. (or do Dijkstra twice, once with C deleted, and compare with the adjusted time through C)

E1

E1

Explanations needed, not answers any correct logic

[2]

B

D G

C

A

E

F

I H 25

17

7

9 13

17 19

13

11 9

23 13

51 27 13 31

13 1 0

13

51 27

2 13

44 26

3 26

35 39

39

5 39

6 39 4 35

54 52 52 46

7 44

8 46 9 52


8


4 (i) & (ii)

M1 activity on arc A1 single start and end A1 A, B, C OK A1 D, F, I OK A1 rest OK [5] M1 forward pass (must have at

least one join correct A1 FT M1 backward pass (must have

at least one burst correct) A1 FT Minimum completion time = 100 minutes

Critical activities are A, C, D, I, J and L B1 B1

cao cao

[6] 4 (iii) e.g. Critical activities (100 mins) + others.

e.g. B has to be done whilst A is underway. B1 Needs a comparison of

times, possibly implied. [1] 4 (iv) (If L omitted in (i) ignore omission here.)

e.g.

M1

A1

A1 A1

diagram like this or attempted cascade ... no more than 1 omitted activity nowhere needing more than 2 people precedences correct fully correct, inc who does what

[4]

B 10

A 30

100 100 30 30

30 30

15 80

0 0 85 85 60 60 50 50

H 10

C 20

D 10

G 5

F 5

E 15

I 25

J 10

L 5

K 15

95 95

G C A D I J L Simon

Friend B E K H F 10 25

30

40 50 60

85 95

100


9


5 (i) e.g. Let x be the number of snowboards B1 Let y be the number of (pairs of) skis B1 or vice-versa of course x + y ≤ 600 B1 x ≤ 250 and y ≤ 500 B1 both 1.1x ≤ y B1

B1 FT horizontal line B1 FT vertical line B1 FT positive slope line B1 x+y = 600 Note ... error tolerance of

+/- half a small square within feasible region.

B1 shading ... follow any

pentagon bounded by the y-axis, a horizontal line, a vertical line, a negatively inclined line and a positively inclined line

[10]

600 250 boards - x

skis - y

500

600

(100,500) 29000

(250,350) 27500

(285.7, 314.3)


10

Question Answer Marks Guidance 5 (ii) Objective = 40x + 50y B1 objective

M1 considering profits at the

two indicated points of their pentagon (or using a profit line)

29000 at (100,500) 27500 at (250,350) Solution ... 100 snowboards and 500 pairs of skis

A1 cao www

[3] 5 (iii) €10 or more B1 cao (allow €51 etc) [1] 5 (iv) 35 snowboards M1 moving to appropriate

new feasible point on their negatively inclined line

A1 cao... integer! (allowing 30 to 40 for graphical inaccuracy)

[2]


11

Question Answer Marks Guidance 6 (i) e.g. 0, 1, 2 1

3, 4, 5, 6, 7 2 M1 either 3 numbers for 1 or

5 numbers for 2 8 3 A1 all proportions correct 9 4 [2] 6 (ii) random number 5 3 2 4 7 9 1 1 8

time interval (mins) 2 2 1 2 2 4 1 1 3 arrival times 0 2 4 5 7 9 13 14 15 18

M1 all outcomes achieved with first 2 correct for their rule

A1 all correct FT B1 accumulation [3] 6 (iii) e.g. 00 ‒ 13 0.1 M1 ignore some 14 ‒ 41 0.25 A1 proportions correct 42 ‒ 83 1 A1 efficient (fewer than 7

rejected) 84 ‒ 97 2 98, 99 ignore and “redraw” [3] 6 (iv) random number 23 15 01 32 45 47 86 71 17 83

processing time 0.25 0.25 0.1 0.25 1 1 2 1 0.25 1 M1 first 4 customers correct

for their rule A1 all correct FT [2] 6 (v) e.g. 0 ‒ 5 1 B1 6 ‒ 9 0.25 [1] 6 (vi) random number 8 3 0 1 4 0 2 5 7 6 B1 FT payment time 0.25 1 1 1 1 1 1 1 0.25 0.25 [1] 6 (vii) arrival 0 2 4 5 7 9 13 14 15 18 M1 deals with a wait correctly departure 0.5 3.25 5.1 6.35 9 11 16 18 18.5 19.75 A1 all correct FT [2] 6 (viii) arrival 0 2 4 5 7 9 13 14 15 18 M1 deals with last 3 correctly departure 0.5 3.25 5.1 6.35 9 11 16 18 15.5 19.25 A1 all correct FT [2]


1

1. Annotations and abbreviations Annotation in scoris Meaning

Blank Page – this annotation must be used on all blank pages within an answer booklet (structured or unstructured) and on each page of an additional object where there is no candidate response.

and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in mark scheme

Meaning

E1 Mark for explaining U1 Mark for correct units G1 Mark for a correct feature on a graph M1 dep* Method mark dependent on a previous mark, indicated by * cao Correct answer only oe Or equivalent rot Rounded or truncated soi Seen or implied www Without wrong working


2

2. Subject-specific Marking Instructions for GCE Mathematics (MEI) Decision strand a Annotations should be used whenever appropriate during your marking.

The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.

b An element of professional judgement is required in the marking of any written paper. Remember that the mark scheme is designed to assist in marking incorrect solutions. Correct solutions leading to correct answers are awarded full marks but work must not be judged on the answer alone, and answers that are given in the question, especially, must be validly obtained; key steps in the working must always be looked at and anything unfamiliar must be investigated thoroughly. Correct but unfamiliar or unexpected methods are often signalled by a correct result following an apparently incorrect method. Such work must be carefully assessed. When a candidate adopts a method which does not correspond to the mark scheme, award marks according to the spirit of the basic scheme; if you are in any doubt whatsoever (especially if several marks or candidates are involved) you should contact your Team Leader.

c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks.


3

E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.

d When a part of a question has two or more ‘method’ steps, the M marks are in principle independent unless the scheme specifically says otherwise; and similarly where there are several B marks allocated. (The notation ‘dep *’ is used to indicate that a particular mark is dependent on an earlier, asterisked, mark in the scheme.) Of course, in practice it may happen that when a candidate has once gone wrong in a part of a question, the work from there on is worthless so that no more marks can sensibly be given. On the other hand, when two or more steps are successfully run together by the candidate, the earlier marks are implied and full credit must be given.

e The abbreviation ft implies that the A or B mark indicated is allowed for work correctly following on from previously incorrect results. Otherwise, A and B marks are given for correct work only — differences in notation are of course permitted. A (accuracy) marks are not given for answers obtained from incorrect working. When A or B marks are awarded for work at an intermediate stage of a solution, there may be various alternatives that are equally acceptable. In such cases, exactly what is acceptable will be detailed in the mark scheme rationale. If this is not the case please consult your Team Leader. Sometimes the answer to one part of a question is used in a later part of the same question. In this case, A marks will often be ‘follow through’. In such cases you must ensure that you refer back to the answer of the previous part question even if this is not shown within the image zone. You may find it easier to mark follow through questions candidate-by-candidate rather than question-by-question.

f Wrong or missing units in an answer should not lead to the loss of a mark unless the scheme specifically indicates otherwise. Candidates are expected to give numerical answers to an appropriate degree of accuracy, with 3 significant figures often being the norm. Small variations in the degree of accuracy to which an answer is given (e.g. 2 or 4 significant figures where 3 is expected) should not normally be penalised, while answers which are grossly over- or under-specified should normally result in the loss of a mark. The situation regarding any particular cases where the accuracy of the answer may be a marking issue should be detailed in the mark scheme rationale. If in doubt, contact your Team Leader.

g Rules for replaced work If a candidate attempts a question more than once, and indicates which attempt he/she wishes to be marked, then examiners should do as the candidate requests.


4

If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.

h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.


1


1 (i)

M1 12 vertices

A1 connectivity

(all 18 arcs and no extras)

[2]

1 (ii) 4 ( or “>2” or “multiple” ... not “some”) odd nodes ... top steps, pool, front steps, olive ...

so neither Eulerian nor semi-Eulerian., but not just “not Eulerian”. (This terminology not required.)

B1

[1]

1 (iii) start/end at pool/top steps, or vice versa M1

e.g. po–pd-fd-po-pa-pd-bd-fd-fs-gat-ol-fs-ol-gar-bd-pa-ts-fi-or-ts (20 nodes, 19 arcs) A1

path from front steps to the olive tree B1 must be stated

[3]

fig

top steps

back door

patio

pool door pool

front door

front steps

gate olive

garage

orange


2


1 (iv) Possible answer:

No repetition of any arc needed

M1

Start/stop are front steps/olive A1

Alternative answer:

By repeating fs/ol or ol/fs ...

(M1)

can start and stop at same point, e.g. front door. (A1)

[2]

2 (i) e.g.

0,1,2 coffee

3,4,5,6,7,8 tea

(9 reject and redraw)

M1

reject

A1

proportions + efficient,

ie using 9 digits (so allow

00, 01, ..., 09)

[2]

2 (ii) Ten simulated coffees or teas, corresponding to their rule and the given random digits.

e.g. T C C T C T T C T C

e.g. C T T T T C T T C T

B1

[1]

2 (iii) e.g.

Coffee at breakfast

00-54 coffee

55-99 tea

B1

Breakfast drink must be

specified.

Tea at breakfast

00-14 tea

15-99 coffee

B1

Breakfast drink must be

specified.

[2]


3


2 (iv) Ten simulated coffees or teas, using answers to part (ii) to define which rule to use.

e.g. C C T C C C C C T C

e.g. C C T C C T C C C C

e.g. C C C C T T C C C T

M1

A1

first 4, ref part (ii)

ft errors in (ii)

[2]

2 (v) Accumulating and computing the proportion.

e.g.

C - 65%

B1 ft

[1]


4


3 (i) ACD is 7+ 2 = 9 (< 12)

or AFD is 3+ 8 =11 (< 12)

B1

needs numerical

justification

AD could by via some point of interest, or over difficult terrain, or ... The triangle inequality applies to

triangles!

B1

[2]

3 (ii)

A 1 B 3 C 4 D5 E (6) F 2

A 6 7 12 3

B 6 10 8

C 7 10 2

D 12 2 9 8

E 8 9

F 3 8

26 (miles)

M1

M1

M1

A1

B1

B1

[6]

starting at and crossing

row A (i.e. no selection in

row)

selecting FA and BA

(or first two arcs following

wrong start)

numbering columns A, F

and B (similarly)

all correct

(dependent on 3 Ms)

(can cross all rows)

cao (weights not needed)

cao

A

B

C

D

E

F

6

7

3

8

2

or transpose


5


4 (i)

&

(ii)

e.g.

minimum completion time = 7.5 hours

critical activities – A, B, E, F, G

(or ABEG + ABEF)

M1 Activity on arc

A1 Single start and end

A1 A, B, C, D (precedences)

A1 E (precedences)

A1 F and G (all correct)

M1 A1

forward pass

M1 A1

backward pass

B1 time (cao)

B1 critical activities (cao)

[11]

4 (iii) e.g.

B1

not ft

B1

must be labelled or to

scale (e.g. on the squares

provided)

Can be written out instead.

[2]

4 (iv) 8.0 hours or delay 0.5 hours B1 cao ISW if needed

A, C, D B1 cao

8.5 hours or delay of 1 hour B1 cao ISW if needed

[3]

B

5

A

0.5 0.5 0.5

2.5 5.5

5.5 5.5

0 0

7.5 7.5

7.5 7.5 6.5 6.5

C

2 D

1.5

G

1

F

1 E

1

A B

F C D

0.5 2.5 4.0

5.5

7.5

7.5

6.5

G E

Needs to be clear what is done by whom.

This doesn’t necessarily require people

being labelled ... but might.


6


5 (a)

(i) 6 3 10 5 16 8 4 2 1 4 2 1 ... (can stop at second “4”)

M1 6 3 10

A1

[2]

5 (a) (ii) 256 128 64 32 16 8 4 2 1 4 2 1 ...

(as above, or can note repetition from “16”)

M1 256 128 64

A1

[2]

5 (a) (iii) e.g. Step 25 If n is 1 then stop. (Any step number between 21 and 29, or indicated in some other way.)

B1 ISW, but “Step 35” is

wrong.

[1]

5 (a) (iv) Need to know that all chosen numbers lead to 1. B1

[1]

5 (b) (i) Box 1: 2 1 6 A B C

Box 2: 3 3 D E

Box 3: 5 F

B1

3 boxes B1

[2]

5 (b) (ii) 1 2 3 3 5 6 B A D E F C B A E D F C

Box 1: 1 2 3 3 B A D E

Box 2: 5 F

Box 3: 6 C

B1

sorted increasing

B1

[2]

5 (b) (iii) (6 5 3 3 2 1) (C F D E A B) (C F E D A B)

Box 1: 6 3 1 C D B

Box 2: 5 3 2 F E A

M1

placing a “3” or D or E

into box 1

A1

[2]


7


5 (b) (iv) e.g. (for fitting into boxes of size 10)

6 3 3 2 2 2 2

Reducing order/first fit:

Box 1: 6 3

Box 2: 3 2 2 2

Box 3: 2

Optimal:

Box 1: 6 2 2

Box 2: 3 3 2 2

M1

valid example

A1

correctly doing it

[2]

5 (b) (v) 30 (60/6)2 = 3000 secs ... 50 minutes M1

multiplying 30 by a

squared value

A1

[2]


8


6 (i) Let x be the number of (10s of) litres of stew and y the number of (10s of) litres of soup that Ian makes.

B1

“number of”, referring to

soup & stew

B1 identification of soup and

stew variables

Carrots: 0.15x + 0.1y < 100, i.e. 3x + 2y < 2000 B1 -1 each scaling or

Beans: 0.1x + 0.075y < 70, i.e, 4x + 3y < 2800 B1 systematic error, e.g.

Tomatoes: 0.15x + 0.15y < 110, i.e. 3x + 3y < 2200 B1 equalities

[5]

6 (ii) Intercepts are (666.7,0) and (0,1000)

(700,0) and (0,933.3)

(733.3,0) and (0,733.3)

B1

axes consistently labelled

and scaled

B1 line 1

B1 line 2

B1

line 3 all subject to

negative gradients

B1 shading giving feasible

quadrilateral bounded by

axes ... or identified by

vertices

[5]

1000

733.3

y

x

933.3

733.3

700 666.7

bod shading here

Ignore “soup” and “stew” labelling on axes unless no variable

labelling.

-1 if variables swapped in error.

-1 if systematic scaling error (following inequalities in 6(i).

broken axis scores 0 for 6(ii)


9


6 (iii)

Line 2 irrelevant. Comparing at (0, 733.3), (533.3 10, 200 10) and (666.7, 0)

(accuracy quoted is for graphical solutions).

Max profit at intersection of lines 1 and 3 (533.33,200) with profit £3466.67 (accuracy from 3375 to 3560)

(cf £3333.33 and £2933.33)

M1

comparing 3 vertices (not

origin) or profit line with

approximately correct

gradient (–5/4)

So make 533.33 litres of stew and 200 litres of soup, A1 stew and soup (cao)

giving a profit of £3466.67 (3375 – 3560). A1 profit (cao)

[3]

6 (iv)

Best solution now at (0, 933.3) ... profit £3733.33 (£373.33)

M1

So best new solution uses 30 kg extra tomatoes (140 kg total)

A1

30kg (allow 140 new total)

cao

Extra profit is £(3733.33 – 3466.67 – 30*2.5) = £191.67

A1

(allow £3658.33 new total)

cao

[3]


GCE

Mathematics (MEI)

Unit 4771: Decision Mathematics 1

Advanced Subsidiary GCE

Mark Scheme for June 2015

OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2015


3

Annotations and abbreviations

Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in mark scheme

Meaning



4

Subject-specific Marking Instructions for GCE Mathematics (MEI) Decision strand a Annotations should be used whenever appropriate during your marking.

The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.


c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks.


5

E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result. Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.





If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last


6

(complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.

h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.


7


1 (i)

M1

At least two directed arcs,

each from the top of a lift

to the bottom

A1 all 4 correct

[2]

1 (ii) (Angus has to repeat all of the chairlifts.)

He has to repeat A either because two ski runs deliver skiers to it, or because it serves two ski runs. B1

He has to repeat B and C …

… either because two ski runs deliver skiers to them, or because they serve two ski runs or because of ski

run 4.

M1

A1

[3]

1 (iii) Angus has to repeat ski run 3 because he has to repeat chairlifts B and/or C (or runs 4 and 5). M1 run 3

A1 for explanation

[2]

1 (iv) This would have to be represented by an arc from chairlift C to chairlift D, but in a bipartite graph an arc

can only connect two elements which are not in the same set. In this case the sets are chairlifts and ski

runs.

B1 needs to be contextualised

[1]

C

1

3

4

5

B

A


8


2 (i) i 1 2 3

m1 2

c1 8

m2 2

c2 5

m3 4

c3 3

j 1 2 3

a 2 3 4 1

b 3 4 1 5 2

M1

A1

j 1

a 2

b 3

as and bs

(4’s and 5’s not essential)

d1 2

x1 1

y1 7 B1 for 1 and 7

d2 -2

x2 2.5

y2 13

B1 for 2.5 and 13

d3 0

x3

y3

B1 for 0

Print area M1 use of print area

(1, 7)

(2.5, 13)

parallel A1 3 copied, inc “parallel”

[7]

2

(ii)

Finds the line intersections

B1

[1]


9


3 (i) At least 50% coffee (allow more than)

(so number of coffee filters ≥ number of tea bags,

so number tea bags ≤ number of coffee filters.)

B1

referral to sales info to

get ≤ (allow <)

At most 75% coffee (allow less than)

so number of coffee filters ≤ 3 number of tea bags,

so number of tea bags ≥ 1/3 number of coffee filters.

B1 referral to sales info +

explanation of 1/3 to get

≥ (allow >)

[2]

3 (ii) Let x be the number of coffee filters.

Let y be the number of tea bags ... or vice versa.

B1

“number of” essential

B1

“500” line

B1

£50 line

B1

lines from (i)

B1cao

shading

[5]

3 (iii) Coffee – 75% of 500. Tea – 50% of 500. B1cao

[1]

x

y

1000

500

500


10


4 (a)

B1

Dijkstra

award only if correct at E

B1

other working values

B1

order of labelling

B1 labels

AB 3

AC 2

AD 7

AFE 5

AF 3

B1

routes

B1 lengths

[6]

A B

C F 8

3

2

1

3

0

2 3

2 2

3/4 3

3/4 3 7

6

5

3

E D

4

5

7

5 6 7

2

8 5


11


4 (b) (i)

M1

tree or attempt at Prim

A1

Length = 15 B1

[3]

4 (b) (ii) Removes AE, AD, CE then BC M1 AE, AD, CE (in order)

A1 BC only

[2]

4 (b) (iii) It will remain connected. B1

There will be no cycles left. B1

Removing a largest possible arc at each stage guarantees a minimum spanning tree. B1

[3]

4 (b) (iv) (n2-3n+2)/2 (or equivalent) arcs for Jill to remove. B1 algebraic simplification

not needed

More than Prim if n is 5 or more B1

[2]

A B

C F

3

2

5

3

E D

2


12


5 (i)&(ii)

M1

activity on arc

A1 F & I

A1 J

A1 K

A1 rest

[5]

M1A1 forward pass

M1A1 backward pass

minimum completion time = 55 minutes B1cao time

critical activities – A, E, F, G, H, J, K B1cao critical activities

[6]

5 (iii)

e.g. (each cell represents 5 minutes)

1st person A E E F G J K

2nd person B D I I I I I I I

other activities C C C C

H H H H

M1

A, E, F, G allocated OK

A1

B, D, I, J, K OK

B1 C and H correctly timed

[3]

5 (iv) e.g.

1st person A D I I I I I I J K

2nd person B E E F G I

B1

a correct schedule for two

people

50 minutes

B1

50 minutes seen

[2]

B

5

E

10 F

5

D

5

H

20

0 0

5 5

10 30

G

5

I

30

C

20

J

5

10 15

15 15 20 20 25 25

45 45

50 50

55 55

A

5

K

5


13


6 (i) e.g.

French 0, 1, 2, 3, 4, 5, 6 Greek

7, 8, 9 French

Greek 0, 1, 2, 3, 4, 5 French

6, 7, 8, 9 Greek

B1

French

M1

A1

proportions

efficient

[3]

6 (ii) Using Greek rule

Using French rule

e.g. F G G G F G F G G G

M1 Greek

M1 French

A1

Computing observed probabilities

e.g. P(F)=0.3 and P(G)=0.7

(Long run probabilities are 6/13 French and 7/13 Greek.)

B1

[4]

6 (iii) e.g.

French 0, 1 French

2, 3, 4, 5, 6, 7 Greek

8, 9 Hungarian

B1

Greek 0, 1, 2, 3, 4 French

5, 6, 7 Greek

8, 9 Hungarian

B1

Hungarian 0, 1, 2 French

3, 4, 5 Greek

6, 7, 8 Hungarian

9 reject and redraw

M1 reject one (or more)

A1 proportions

A1 efficient

[5]


14

6 (iv) Greek rule applied in correct circumstances and correctly B1

French rule applied in correct circumstances and correctly B1

Hungarian rule applied in correct circumstances and correctly

e.g. F F H F G H F G F F

B1

so P(F)=0.6, P(G)=0.2, P(H)=0.2

(Long run proportions are 56/169, 74/169 and 39/169.) B1

[4]

Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee Registered in England Registered Office; 1 Hills Road, Cambridge, CB1 2EU Registered Company Number: 3484466 OCR is an exempt Charity OCR (Oxford Cambridge and RSA Examinations) Head office Telephone: 01223 552552 Facsimile: 01223 552553 © OCR 2015

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mailto:[email protected]

http://www.ocr.org.uk/


GCE

Mathematics (MEI)

Unit 4771: Decision Mathematics 1 Advanced Subsidiary GCE

Mark Scheme for June 2016

OCR (Oxford Cambridge and RSA) is a leading UK awarding body, providing a wide range of qualifications to meet the needs of candidates of all ages and abilities. OCR qualifications include AS/A Levels, Diplomas, GCSEs, Cambridge Nationals, Cambridge Technicals, Functional Skills, Key Skills, Entry Level qualifications, NVQs and vocational qualifications in areas such as IT, business, languages, teaching/training, administration and secretarial skills. It is also responsible for developing new specifications to meet national requirements and the needs of students and teachers. OCR is a not-for-profit organisation; any surplus made is invested back into the establishment to help towards the development of qualifications and support, which keep pace with the changing needs of today’s society. This mark scheme is published as an aid to teachers and students, to indicate the requirements of the examination. It shows the basis on which marks were awarded by examiners. It does not indicate the details of the discussions which took place at an examiners’ meeting before marking commenced. All examiners are instructed that alternative correct answers and unexpected approaches in candidates’ scripts must be given marks that fairly reflect the relevant knowledge and skills demonstrated. Mark schemes should be read in conjunction with the published question papers and the report on the examination. OCR will not enter into any discussion or correspondence in connection with this mark scheme. © OCR 2016


3

1. Annotations and abbreviations Annotation in scoris Meaning and BOD Benefit of doubt FT Follow through ISW Ignore subsequent working M0, M1 Method mark awarded 0, 1 A0, A1 Accuracy mark awarded 0, 1 B0, B1 Independent mark awarded 0, 1 SC Special case ^ Omission sign MR Misread Highlighting Other abbreviations in mark scheme

Meaning



4

2. Subject-specific Marking Instructions for GCE Mathematics (MEI) Decision strand

a Annotations should be used whenever appropriate during your marking. The A, M and B annotations must be used on your standardisation scripts for responses that are not awarded either 0 or full marks. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. For subsequent marking you must make it clear how you have arrived at the mark you have awarded.


c The following types of marks are available. M A suitable method has been selected and applied in a manner which shows that the method is essentially understood. Method marks are not usually lost for numerical errors, algebraic slips or errors in units. However, it is not usually sufficient for a candidate just to indicate an intention of using some method or just to quote a formula; the formula or idea must be applied to the specific problem in hand, eg by substituting the relevant quantities into the formula. In some cases the nature of the errors allowed for the award of an M mark may be specified. A Accuracy mark, awarded for a correct answer or intermediate step correctly obtained. Accuracy marks cannot be given unless the associated Method mark is earned (or implied). Therefore M0 A1 cannot ever be awarded. B Mark for a correct result or statement independent of Method marks. E A given result is to be established or a result has to be explained. This usually requires more working or explanation than the establishment of an unknown result.


5

Unless otherwise indicated, marks once gained cannot subsequently be lost, eg wrong working following a correct form of answer is ignored. Sometimes this is reinforced in the mark scheme by the abbreviation isw. However, this would not apply to a case where a candidate passes through the correct answer as part of a wrong argument.





If there are two or more attempts at a question which have not been crossed out, examiners should mark what appears to be the last (complete) attempt and ignore the others. NB Follow these maths-specific instructions rather than those in the assessor handbook.


6

h For a genuine misreading (of numbers or symbols) which is such that the object and the difficulty of the question remain

unaltered, mark according to the scheme but following through from the candidate’s data. A penalty is then applied; 1 mark is generally appropriate, though this may differ for some units. This is achieved by withholding one A mark in the question. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error.


7


1 (i) e.g. 0, 1, 2, 3 win

4, 5, 6, 7, 8, 9 lose

or 0, 1, 2, 3, 4, 5 lose

6, 7, 8, 9 win

M1

A1

correct rule

efficient rule

disallow 1, 8, 3, 5 win

disallow 6, 7, 1, 3 win

(ii) In the worst case Pierre will suffer 3 consecutive losses, of £100, £200 and £400 respectively. He will then

be unable to fund the next bet of £800.

B1 100, 200, 400 may by

implied, eg by 700 lost or

300 left

(iii) e.g. or

1, 6, 4 win 1, 6, 4 win

8, 6, 4 lose 8, 6, 4 win

8, 7, 4 lose 8, 7, 4 win

3, 1, 1 win 3, 1, 1 lose

5, 3, 2 win 5, 3, 2 lose

M1

A1

or

M1

A1

correct identification of

first win and of first loss

rest

numbers L/W for 1 6 4

and 8 6 4

rest + interpretation

(iv) 100 x no of wins + 700 x no of losses

eg1 … -£1100 eg2 … £3900

eg1 … on average a loss of £220 per application of the strategy eg2 … on average £780 left

M1

A1

A1

weighted sum or 5

monetary outcomes –

implied OK

(Their “700” OK here)

correct sum following

their simulation but not

their 700.

their sum/5


8


2 (i) n 1 2 3 4 5 6 7

p 1 0.962 0.888 0.785 0.664 0.537 0.413

M1

M1

A1

A1

n=2 … awrt 0.96

n=3 … awrt 0.88 or 0.89

n=4 … awrt 0.79

stopping at n=7 with p<0.5

(ii) Need to select 7 cards for the probability of repetition on the list to exceed 0.5. B1

B1

their “7”

P(repetition) exceeds 0.5

(iii) Step 1 Set n = 1.

Step 2 Set p = 1.

Step 3 Set n = n + 1.

Step 4 Set p = p (366-n)/365.

Step 5 If p < 0.5 then stop.

Step 6 Go to Step 3.

B1

both changes (step 4) and

no others

(iv) Because they do not have the same frequency of occurrence (probability) as other birthdays. B1


9


3 (i)

B1

B1

B1

adjacency graph all correct

cao

complement graph correct

cao

three colours

(ii)

A

B Two disjoint, complementary and complete subgraphs can be identified (in several ways)

M1

A1

A1

M1

A1

top front adjacency OK

adjacency graph all correct

complement graph correct

cao

two subgraphs

complete

base

front back

left right

base

front back

left right

So 3 colours

are needed

top front top back

top left top right

base front base back

base left base right

top front top back

top left top right

base front base back

base left base right


10


4 (i)

M1

A1

3 widths + waste

4 (ii) 0.06x + 0.21y + 0.04z (m2)

M1

A1

3 areas

any units OK ... ignore

scaling

4 (iii) 2x + y > 20 B1

4 (iv) y + 3z > 24

B1

4 (v) Use of z = 20 – x – y

Minimise 0.02x + 0.17y (constant of 0.8 not needed but OK if there)

st 2x + y > 20

-3x – 2y > -36 or 3x + 2y < 36

M1

A1

A1

“minimise” not needed -

given

z m

4cm waste

32cm

32cm

32cm


11


4

(vi)

B1

B1

B1

M1

M1

A1

A1

line (cao)

line (cao)

shading – follow two neg

grad lines making a

triangle with base on x-

axis

objective valued at (10,0)

and at (4,12)

z =10

both

waste

Cut 10m according to plan x and

10m according to plan z.

Gives ...

20m of material with width 47cm

30m of material with width 32cm

1m2 of waste.

y

x 10 12

18

20

(4, 12)

20 or 100 or 0.2 or 1

212 or 292 or 2.12 or 2.92

Min waste at (10, 0)


12


5 (i)

Activity Duration (weeks) Immediate predecessors

A 1 ‒

B 1 ‒

C 1 ‒

D 8 A

E 6 B

F 6 C, I

G 6 ‒

H 1 D, E, F,G

I 2 ‒

J 4 A, B, C

K 2 H, J

B1

B3

A, B, C, G and I

-1 for 1 or 2 wrong rows



5 (ii)+

(iii)

Critical activities ... A, D, H, K

M1

A1

A1

A1

A1

M1

A1

M1

A1

B1cao

activity on arc

immediate predecessors

A, B, C, I, G


D (A), E(B), F(C,I,)


J(A,B,C), H(D,E,F,G)


K(J,H) + rest

forward pass

backward pass

1 6

0 0

1 1

1 3

2 3

1 3

6 9

9 9

12 12 10 10

A

1

C

1

B

1

I

2

G

6

F

6

D 8

E

6

H

1

J

4 K

2


13


5 (iv) eg

Week 1 2 3 4 5 6 7 8 9 10 11 12

CPA start time 0 1 2 3 4 5 6 7 8 9 10 11

Room 1 A D D D D D D D D H K K

Room 2 B E E E E E E F F

Room 3 C F F F F

G G G G G G

I I

J J J J

B1cao

B1cao

F (ES2, LF9)

rest


14


6 (a) (i) e.g.

Route – A D C Z U W X or A B D C Z U W X Distance ‒ 48 km

B1

B1

B1

B1

B1 B1

Dijkstra – E correct

other working values

order of labelling

labels

6 (a) (ii) No difference, but allow “one fewer”, as 15 (CZ) does not need to be added on to determine the route.

Part (i) is effectively using Dijkstra on the left network then Dijkstra on right network, but starting at 30

instead of 0 on the right network.

M1

A1

Need comment re just

one arc connecting the

two networks.

4

2 4

16 15

6 15

1 0

8

3 8

15 14

5 14

9

4 9

30

7 30

41

9 41

49

49 48

11 48

39

8 39

46 45

10 45 A

B

C

D

E F

12

13 10

5

9

8

6

5

7

4

Z

U V

W

X

Y

11

9

9

10

7

8

4 7

3

15


15


6 (b) (i) Order of choice ... AB, BD DE, DC, EF

M1

A1

B1

Kruskal (first 2 arcs

identified – OK if by

length)

min connector

6 (b) (ii) Order of inclusion ... U, W, X, V Y, Z M1

A1

Prim (first 2 vertices

identified - OK to say

UW, WX, etc, but in that

order. Not OK to identify

by lengths)

B1

min connector

6 (b) (iii) Total length = 72 km. B1cao

B1

27 + 30

+ 15 for the pass

A

B

C

D

E F

5

6

5

7

4

Z

U V

W

X

Y

9

7

4 7

3

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