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Wednesday 15 June 2016 – MorningAS GCE MATHEMATICS (MEI)4771/01 Decision Mathematics 1

QUESTION PAPER

*6382480783*

INSTRUCTIONS TO CANDIDATESThese instructions are the same on the Printed Answer Book and the Question Paper.• The Question Paper will be found inside the Printed Answer Book.• Write your name, centre number and candidate number in the spaces provided on the

Printed Answer Book. Please write clearly and in capital letters.• Write your answer to each question in the space provided in the Printed Answer

Book. Additional paper may be used if necessary but you must clearly show your candidate number, centre number and question number(s).

• Use black ink. HB pencil may be used for graphs and diagrams only.• Read each question carefully. Make sure you know what you have to do before starting

your answer.• Answer all the questions.• Do not write in the bar codes.• You are permitted to use a scientific or graphical calculator in this paper.• Final answers should be given to a degree of accuracy appropriate to the context.

INFORMATION FOR CANDIDATESThis information is the same on the Printed Answer Book and the Question Paper.• The number of marks is given in brackets [ ] at the end of each question or part question

on the Question Paper.• You are advised that an answer may receive no marks unless you show sufficient detail

of the working to indicate that a correct method is being used.• The total number of marks for this paper is 72.• The Printed Answer Book consists of 16 pages. The Question Paper consists of 8 pages.

Any blank pages are indicated.

INSTRUCTIONS TO EXAMS OFFICER/INVIGILATOR• Do not send this Question Paper for marking; it should be retained in the centre or

recycled. Please contact OCR Copyright should you wish to re-use this document.

OCR is an exempt CharityTurn over

© OCR 2016 [K/102/2651]DC (NH/SG) 126744/2

Candidates answer on the Printed Answer Book.

OCR supplied materials:• Printed Answer Book 4771/01• MEI Examination Formulae and Tables (MF2)

Other materials required:• Scientific or graphical calculator

Duration: 1 hour 30 minutes

Oxford Cambridge and RSA

2

4771/01 Jun16© OCR 2016

Section A (24 marks)

1 Pierre knows that, if he gambles, he will lose money in the long run. Nicolas tries to convince him that this is not the case.

Pierre stakes a sum of money in a casino game. If he wins then he gets back his stake plus the same amount again. If he loses then he loses his stake.

Nicolas says that Pierre can guarantee to win by repeatedly playing the game, even though the probability of winning an individual game is less than 0.5.

His idea is that Pierre should bet in the first game with a stake of £100. If he wins then he stops, as he will have won £100.

If he loses then he plays again with a stake of £200. If he wins then he has lost £100 and won £200. This gives a total gain of £100, and he stops.

If he loses then he plays again with a stake of £400. If he wins this time he has lost £100 and £200 and won £400. This gives a total gain of £100, and he stops.

Nicolas’s advice is that Pierre simply has to continue in this way, doubling his stake every time that he loses, until he eventually wins. Nicolas says that this guarantees that Pierre will win £100.

You are to simulate what might happen if Pierre tries this strategy in a casino game in which the probability of him winning an individual game is 0.4, and in which he has £1000 available.

(i) Give an efficient rule for using 1-digit random numbers to simulate the outcomes of individual games, given that the probability of Pierre winning an individual game is 0.4. [2]

(ii) Explain why at most three random digits are needed for one simulation of Nicolas’s strategy, given that Pierre is starting with £1000. [1]

(iii) Simulate five applications of Nicolas’s strategy, using the five sets of three 1-digit random numbers in your answer book. [2]

(iv) Summarise the results of your simulations, giving your mean result. [3]

3

4771/01 Jun16 Turn over© OCR 2016

2 A bag contains 26 cards. A different letter of the alphabet is written on each one. A card is chosen at random and its letter is written down. The card is returned to the bag. The bag is shaken and the process is repeated several times.

Tania wants to investigate the probability of a letter appearing twice. She wants to know how many cards need to be chosen for this probability to exceed 0.5. Tania uses the following algorithm.

Step 1 Set 1n = Step 2 Set 1p = Step 3 Set 1n n= +

Step 4 Set 2627p p n

#=-c m

Step 5 If 0.5p 1 then stop Step 6 Go to Step 3

(i) Run the algorithm. [4]

(ii) Interpret your results. [2]

A well-known problem asks how many randomly-chosen people need to be assembled in a room before the probability of at least two of them sharing a birthday exceeds 0.5 (ignoring anyone born on 29 February).

(iii) Modify Tania’s algorithm to answer the birthday problem. (Do not attempt to run your modified algorithm.) [1]

(iv) Why have 29 February birthdays been excluded? [1]

4

4771/01 Jun16© OCR 2016

3 The adjacency graph of a cube is shown.

Vertices on the graph represent faces of the cube. Two vertices are connected by an arc if the corresponding faces of the cube share an edge.

The second graph is the complement of the adjacency graph, i.e. the graph that consists of the same vertices together with the arcs that are not in the adjacency graph.

top

base

front

left right

back

top

base

front

left right

back

Throughout this question we wish to colour solids so that two faces that share an edge have different colours. The second graph shows that the minimum number of colours required for a cube is three, one colour for the top and base, one for the front and back, and one for the left and right.

(i) Draw the adjacency graph for a square-based pyramid, and draw its complement. Hence find the minimum number of colours needed to colour a square-based pyramid. [3]

(ii) (A) Draw the adjacency graph for an octahedron, and draw its complement. [3]

(B) An octahedron can be coloured using just two colours. Explain how this relates to the complement of the adjacency graph. [2]

5

4771/01 Jun16 Turn over© OCR 2016

Section B (48 marks)

4 Two products are to be made from material that is supplied in a single roll, 20 m long and 1 m wide. The two products require widths of 47 cm and 32 cm respectively.

Two ways of cutting lengths of material are shown in the plans below.

waste

length x metres

47 cm

47 cm

6 cm

waste

length y metres

47 cm

32 cm

21 cm

(i) Given that there should be no unnecessary waste, draw one other cutting plan that might be used for a cut of length z metres. [2]

(ii) Write down an expression for the total area that is wasted in terms of x, y and z. [2]

All of the roll is to be cut, so x + y + z = 20.

There needs to be a total length of at least 20 metres of the material for the first product, the one requiring width 47 cm.

(iii) Write this as a linear constraint on the variables. [1]

There needs to be a total length of at least 24 metres of the material for the second product, the one requiring width 32 cm.

(iv) Write this as a linear constraint on the variables. [1]

(v) Formulate an LP in terms of x and y to minimise the area that is wasted. You will need to use the relationship x + y + z = 20, together with your answers to parts (ii), (iii) and (iv). [3]

(vi) Solve your LP graphically, and interpret the solution. [7]

6

4771/01 Jun16© OCR 2016

5 A village amateur dramatic society is planning its annual pantomime. Three rooms in the village hall have been booked for one evening per week for 12 weeks. The following activities must take place. Their durations are shown.

Activity Duration (weeks)

A choose lead actors 1

B choose rest of actors 1

C choose dancers 1

D rehearse lead actors 8

E rehearse rest of actors 6

F rehearse dancers 6

G prepare scenery 6

H install scenery 1

I prepare music 2

J make costumes 4

K dress rehearsals 2

Each activity needs a room except for activities G, I and J.

Choosing actors and dancers can be done in the same week. Rehearsals can begin after this.

Rehearsing the dancers cannot begin until the music has been prepared.

The scenery must be installed after rehearsals, but before dress rehearsals.

Making the costumes cannot start until after the actors and dancers have been chosen.

Everything must be ready for the dress rehearsals in the final two weeks of the 12-week preparation period.

(i) Complete the table in your answer book by showing the immediate predecessors for each activity. [4]

(ii) Draw an activity on arc network for these activities. [5]

(iii) Mark on your network the early time and the late time for each event. Give the critical activities. [5]

It is discovered that there is a double booking and that one of the rooms will not be available after week 6.

(iv) Using the space provided, produce a schedule showing how the pantomime can be ready in time for its first performance. [2]

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4771/01 Jun16© OCR 2016

6 A mountain ridge separates two populated areas. Networks representing roads connecting the villages in each area are shown below. The numbers on the arcs represent distances in kilometres.

B

A

F

D

E

C12

56

105

7

13159

4

8

9

Z

11

U 8 V

4

W

79

X10

Y

73

There is also a mountain road of length 15 kilometres connecting C to Z.

(a) A national bus company needs a route from A to X.

(i) Use Dijkstra’s algorithm on the complete network, including CZ, to find the shortest route from A to X. Give the route and its corresponding distance. [6]

(ii) Would it need fewer computations to use Dijkstra’s algorithm on the network for villages A to F to find the shortest route from A to C, and then use Dijkstra’s algorithm on the network for villages U to Z to find the shortest route from Z to X? Give a brief justification for your answer. [2]

(b) The local council needs to discover which roads it should keep clear of snow during the winter to keep all the villages connected, and the corresponding total length of road.

(i) Use Kruskal’s algorithm on the network for villages A to F to find a minimum connector for , B, C, D, E, FA" ,. Show your use of the algorithm. Draw your minimum connector. [3]

(ii) Use Prim’s algorithm on the network for villages U to Z to find a minimum connector for ,U V, W, X, Y, Z" ,, starting at U. Show your use of the algorithm. Draw your minimum

connector. [3]

(iii) What is the total length of road that the council must keep clear of snow? [2]

END OF QUESTION PAPER

8

4771/01 Jun16© OCR 2016


Copyright Information

OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download from our public website (www.ocr.org.uk) after the live examination series.

If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.

For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.

OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

Wednesday 15 June 2016 – MorningAS GCE MATHEMATICS (MEI)4771/01 Decision Mathematics 1

PRINTED ANSWER BOOK

INSTRUCTIONS TO CANDIDATESThese instructions are the same on the Printed Answer Book and the Question Paper.• The Question Paper will be found inside the Printed Answer Book.• Write your name, centre number and candidate number in the spaces provided on the

Printed Answer Book. Please write clearly and in capital letters.• Write your answer to each question in the space provided in the Printed Answer

Book. Additional paper may be used if necessary but you must clearly show your candidate number, centre number and question number(s).

• Use black ink. HB pencil may be used for graphs and diagrams only.• Read each question carefully. Make sure you know what you have to do before starting

your answer.• Answer all the questions.• Do not write in the bar codes.• You are permitted to use a scientific or graphical calculator in this paper.• Final answers should be given to a degree of accuracy appropriate to the context.

INFORMATION FOR CANDIDATESThis information is the same on the Printed Answer Book and the Question Paper.• The number of marks is given in brackets [ ] at the end of each question or part

question on the Question Paper.• You are advised that an answer may receive no marks unless you show sufficient detail

of the working to indicate that a correct method is being used.• The total number of marks for this paper is 72.• The Printed Answer Book consists of 16 pages. The Question Paper consists of 8 pages.

Any blank pages are indicated.

* 4 7 7 1 0 1 *

OCR is an exempt CharityTurn over

© OCR 2016 [K/102/2651]DC (NH/SG) 126745/3

Candidates answer on this Printed Answer Book.

OCR supplied materials:• Question Paper 4771/01 (inserted)• MEI Examination Formulae and Tables (MF2)

Other materials required:• Scientific or graphical calculator

*6382625658*

Duration: 1 hour 30 minutes


2

© OCR 2016

1 (i)

1 (ii)

1 (iii) Random digits

1 6 4

8 6 4

8 7 4

3 1 1

5 3 2

1 (iv)

3

Turn over© OCR 2016

2 (i)

2 (ii)

2 (iii)

2 (iv)

4

© OCR 2016

3 (i)

front

left

Adjacency graph

base

right

back front

left

Complement

base

right

back

Minimum number of colours ..................

3 (ii)(A)

top front

top left

base front

base left

Adjacency graph

base back

base right

top right

top back top front

top left

base front

base left

base back

base right

top right

top backComplement

3 (ii)(B)

5


Spare templates for 3(i) and 3(ii) (A)

front

left

Adjacency graph

base

right

back front

left

Complement

base

right

back

top front

top left

base front

base left

Adjacency graph

base back

base right

top right

top back top front

top left

base front

base left

base back

base right

top right

top backComplement

6

© OCR 2016

4 (i) z metres

waste

4 (ii)

4 (iii)

4 (iv)

4 (v)

4 (vi)

(answer space continued on next page)

7


4 (vi) (continued) A spare copy of this graph can be found on page 12.

8

© OCR 2016

5 (i)Activity Duration (weeks) Immediate predecessors

A 1

B 1

C 1

D 8

E 6

F 6

G 6

H 1

I 2

J 4

K 2

5 (ii) &(iii)

9


5 (iv) Week 1 2 3 4 5 6 7 8 9 10 11 12Room 1Room 2Room 3

Spare for 5 (iv)

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

Spare space for 5 (ii) & (iii)

10

© OCR 2016

6(a)(i)

A

F

8

4

B

5

6

U

11 4

9

12

7

D

C

15

Z

9

8

W

5

13 10

E Y10

73

7

V

9

X

Route …

Distance …

6(a)(ii)

6(b)(i)

C

D

EF

A

B

11

© OCR 2016

6(b)(ii)

V

W

X

Y

Z

U

6(b)(iii)

Turn over

12

© OCR 2016

4 (vi) Spare copy of graph for question 4(vi).

13

© OCR 2016

BLANK PAGE

PLEASE DO NOT WRITE ON THIS PAGE

14

© OCR 2016

BLANK PAGE


15

© OCR 2016

BLANK PAGE


16

© OCR 2016



Copyright Information

OCR is committed to seeking permission to reproduce all third-party content that it uses in its assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download from our public website (www.ocr.org.uk) after the live examination series.

If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.

For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.

OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.

Oxford Cambridge and RSA Examinations

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

4771 Mark Scheme June 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

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


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.

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.

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.

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.

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

Question Answer Marks Guidance

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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OCR Report to Centres – June 2016

47

4771 Decision Mathematics 1

General Comments: As always, many of the candidates for this paper exhibited poor communication skills. Communication skills go hand-in-glove with thinking skills. Comments on Individual Questions: Question No.1 The first question carried an obvious moral tale about gambling, which will perhaps be well used in future revision classes. Regrettably, the message will have eluded the many candidates who failed with part (iv). Parts (i), (ii) and (iii) were answered well, which makes the repeated failures in part (iv) all the more surprising. Having considered what happens, and then simulated what happens, it should have been clear that the mean required in part (iv) was the mean monetary outcome. This shows that the expectation is for a substantial loss. Instead many candidates computed simulated probabilities of winning/losing. This misses the point when wins were always £100, but when losses in this scenario were £700 a time. Question No. 2

This question was also well answered. The most common mistake in part (i) was to give only the updating probabilities, instead of using them in the updating calculation. The other parts were answered well, with many candidates referring to unequal probabilities in their answers to (iv).

Question No. 3 Parts (i) and (ii)(A) were answered well, but very few marks were scored in (ii)(B). It had been hoped that the solution for the cube, described in the question, would have pointed the way. Inevitably there were candidates who described a colouring, which does not answer the question.

Question No. 4

This question proved to be too difficult for nearly all candidates. Very few decent graphs were seen. Most could do parts (i) and (ii), but thereafter confusion reigned. Candidates could not get to grips with the scenario, and it was common to see x > (or ≥) 20 for (iii) and y > (or ≥) 24 for (iv). The instruction seemed clear enough in (v), but there were very few attempts to use z = 20 – x – y. Given all that, few could make any coherent sense of where they had got to in terms of drawing the feasible region.

OCR Report to Centres – June 2016

48

Question No. 5 This question started by requiring candidates to model written precedence expressions by a list of immediate predecessors. This made it more difficult than usual to produce the activity-on-arc network in part (ii). Creditable attempts were seen, but many candidates lost marks in their use of dummy activities. For instance, many candidates did not realise that the following logic is OK for F (immediate predecessors of C and I) but not for J (A, B and C), since it also has I as one of its immediate predecessors. Question No. 6

Part (a) was answered well. In part (b) examiners needed to be convinced that Kruskal was being used in (i) and Prim in (ii). The safest way to do this is to show the order of including arcs for Kruskal, and the order of including vertices for Prim.

0 0

A J

C

B F

I

Published: 17 August 2016 Version 1.0 1

GCE Mathematics (MEI)

Max Mark a b c d e u 4751 01 C1 – MEI Introduction to advanced mathematics (AS) Raw 72 63 57 52 47 42 0 UMS 100 80 70 60 50 40 0 4752 01 C2 – MEI Concepts for advanced mathematics (AS) Raw 72 56 49 42 35 29 0 UMS 100 80 70 60 50 40 0

4753 01 (C3) MEI Methods for Advanced Mathematics with Coursework: Written Paper

Raw

72

58

52

47

42

36

0 4753 02 (C3) MEI Methods for Advanced Mathematics with

Coursework: Coursework

Raw

18

15

13

11

9

8

0 4753 82 (C3) MEI Methods for Advanced Mathematics with

Coursework: Carried Forward Coursework Mark

Raw

18

15

13

11

9

8

0 UMS 100 80 70 60 50 40 0 4754 01 C4 – MEI Applications of advanced mathematics (A2) Raw 90 64 57 51 45 39 0 UMS 100 80 70 60 50 40 0

4755 01 FP1 – MEI Further concepts for advanced mathematics (AS)

Raw

72

59

53

48

43

38

0 UMS 100 80 70 60 50 40 0

4756 01 FP2 – MEI Further methods for advanced mathematics (A2)

Raw

72

60

54

48

43

38

0 UMS 100 80 70 60 50 40 0

4757 01 FP3 – MEI Further applications of advanced mathematics (A2)

Raw

72

60

54

49

44

39

0 UMS 100 80 70 60 50 40 0

4758 01 (DE) MEI Differential Equations with Coursework: Written Paper

Raw

72

67

61

55

49

43

0 4758 02 (DE) MEI Differential Equations with Coursework:

Coursework

Raw

18

15

13

11

9

8

0 4758 82 (DE) MEI Differential Equations with Coursework: Carried

Forward Coursework Mark

Raw

18

15

13

11

9

8

0 UMS 100 80 70 60 50 40 0 4761 01 M1 – MEI Mechanics 1 (AS) Raw 72 58 50 43 36 29 0 UMS 100 80 70 60 50 40 0 4762 01 M2 – MEI Mechanics 2 (A2) Raw 72 59 53 47 41 36 0 UMS 100 80 70 60 50 40 0 4763 01 M3 – MEI Mechanics 3 (A2) Raw 72 60 53 46 40 34 0 UMS 100 80 70 60 50 40 0 4764 01 M4 – MEI Mechanics 4 (A2) Raw 72 55 48 41 34 27 0 UMS 100 80 70 60 50 40 0 4766 01 S1 – MEI Statistics 1 (AS) Raw 72 59 52 46 40 34 0 UMS 100 80 70 60 50 40 0 4767 01 S2 – MEI Statistics 2 (A2) Raw 72 60 55 50 45 40 0 UMS 100 80 70 60 50 40 0 4768 01 S3 – MEI Statistics 3 (A2) Raw 72 60 54 48 42 37 0 UMS 100 80 70 60 50 40 0 4769 01 S4 – MEI Statistics 4 (A2) Raw 72 56 49 42 35 28 0 UMS 100 80 70 60 50 40 0 4771 01 D1 – MEI Decision mathematics 1 (AS) Raw 72 48 43 38 34 30 0 UMS 100 80 70 60 50 40 0 4772 01 D2 – MEI Decision mathematics 2 (A2) Raw 72 55 50 45 40 36 0 UMS 100 80 70 60 50 40 0 4773 01 DC – MEI Decision mathematics computation (A2) Raw 72 46 40 34 29 24 0 UMS 100 80 70 60 50 40 0

4776 01 (NM) MEI Numerical Methods with Coursework: Written Paper

Raw

72

55

49

44

39

33

0 4776 02 (NM) MEI Numerical Methods with Coursework:

Coursework

Raw

18

14

12

10

8

7

0 4776 82 (NM) MEI Numerical Methods with Coursework: Carried

Forward Coursework Mark

Raw

18

14

12

10

8

7

0 UMS 100 80 70 60 50 40 0 4777 01 NC – MEI Numerical computation (A2) Raw 72 55 47 39 32 25 0 UMS 100 80 70 60 50 40 0 4798 01 FPT - Further pure mathematics with technology (A2) Raw 72 57 49 41 33 26 0


UMS 100 80 70 60 50 40 0

GCE Statistics (MEI)

Max Mark a b c d e u G241 01 Statistics 1 MEI (Z1) Raw 72 59 52 46 40 34 0 UMS 100 80 70 60 50 40 0 G242 01 Statistics 2 MEI (Z2) Raw 72 55 48 41 34 27 0 UMS 100 80 70 60 50 40 0 G243 01 Statistics 3 MEI (Z3) Raw 72 56 48 41 34 27 0 UMS 100 80 70 60 50 40 0 GCE Quantitative Methods (MEI) Max Mark a b c d e u G244 01 Introduction to Quantitative Methods MEI

Raw 72 58 50 43 36 28 0

G244 02 Introduction to Quantitative Methods MEI Raw 18 14 12 10 8 7 0 UMS 100 80 70 60 50 40 0 G245 01 Statistics 1 MEI Raw 72 59 52 46 40 34 0 UMS 100 80 70 60 50 40 0 G246 01 Decision 1 MEI Raw 72 48 43 38 34 30 0 UMS 100 80 70 60 50 40 0 Level 3 Certificate and FSMQ raw mark grade boundaries June 2016 series

For more information about results and grade calculations, see www.ocr.org.uk/ocr-for/learners-and-parents/getting-your-results

Level 3 Certificate Mathematics for Engineering

H860 01 Mathematics for Engineering H860 02 Mathematics for Engineering

Level 3 Certificate Mathematical Techniques and Applications for Engineers

Max Mark a* a b c d e u

This unit has no entries in June 2016

Max Mark a* a b c d e u H865 01 Component 1 Raw 60 48 42 36 30 24 18 0 Level 3 Certificate Mathematics - Quantitative Reasoning (MEI) (GQ Reform)

Max Mark a b c d e u H866 01 Introduction to quantitative reasoning Raw 72 55 47 39 31 23 0 H866 02 Critical maths Raw 60 47 41 35 29 23 0 Overall 132 111 96 81 66 51 0 Level 3 Certificate Mathematics - Quantitive Problem Solving (MEI) (GQ Reform) Max Mark a b c d e u H867 01 Introduction to quantitative reasoning Raw 72 55 47 39 31 23 0 H867 02 Statistical problem solving Raw 60 40 34 28 23 18 0 Overall 132 103 88 73 59 45 0 Advanced Free Standing Mathematics Qualification (FSMQ) Max Mark a b c d e u 6993 01 Additional Mathematics Raw 100 59 51 44 37 30 0 Intermediate Free Standing Mathematics Qualification (FSMQ) Max Mark a b c d e u 6989 01 Foundations of Advanced Mathematics (MEI) Raw 40 35 30 25 20 16 0

http://www.ocr.org.uk/ocr-for/learners-and-parents/getting-your-results


Version Details of change 1.1 Correction to Overall grade boundaries for H866

Correction to Overall grade boundaries for H867

Documents

Oxford Cambridge and RSA Wednesday 15 June 2016 – Morningmei.org.uk/files/papers/D1_2016_June.pdf · Tania uses the following algorithm. ... Mark on your network the early time