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Monday 6 June 2016 – AfternoonFSMQ ADVANCED LEVEL

6993/01 Additional Mathematics

QUESTION PAPER

*6361255741*

INSTRUCTIONS TO CANDIDATES

These 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. If additional space is required, you should use the lined page(s) at the end of this booklet. The question number(s) must be clearly shown.

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

your answer.• 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 correct to three significant figures where appropriate.

INFORMATION FOR CANDIDATES

This 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 100.• The Printed Answer Book consists of 20 pages. The Question Paper consists of 8 pages.

Any blank pages are indicated.

INSTRUCTION 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 [100/2548/0]DC (ST/FD) 126097/2

Candidates answer on the Printed Answer Book.

OCR supplied materials:• Printed Answer Book 6993/01

Other materials required:• Scientific or graphical calculator

Duration: 2 hours

Oxford Cambridge and RSA

2

6993/01 Jun16© OCR 2016

Formulae Sheet: 6993 Additional Mathematics

In any triangle ABC

Cosine rule a2=b2+c2–2bccosA

C

b a

c BA

Binomial expansion

Whennisapositiveinteger

(a+b)n=an+(n1)an–1b+(n2)an–2b2+…+(nr )an–rbr+…+bn

where

(nr )=nCr= n!––––––––r!(n–r)!

3

6993/01 Jun16 Turn over© OCR 2016

Section A

Answerallthequestions.

1 Solvetheinequality ( )x x1 2 3 42- - . [3]

2 Thegradientfunctionofacurveisgivenby xy

x x3 4 2dd 2= - + .

Findtheequationofthecurve,giventhatitpassesthroughthepoint(1,3). [4]

3 Findallthevaluesofxintherange x0 3601 1c cthatsatisfy3sinx=4cosx. [4]

4 Youaregiventhat ( )f x x x x 623= - + - .

Showthat

(i) (x-2)isafactoroff(x), [1]

(ii) theequationf(x)=0hasonlyonerealroot. [4]

5 JohndrawsatriangleABCwithsidesAB=12cm,BC=16cmandAC=20cm.However,hecanonlymeasurethesidestothenearestcentimetre.

(i) StatethesmallestpossiblelengthofABinJohn’sdrawing. [1]

(ii) HencecalculatethelargestpossiblevalueoftheangleBinJohn’sdrawing. [3]

6 Twocarsareinitiallyatrestfacinginthesamedirectiononastraightroad.CarAis100maheadofcarB.Thetwocarsstartfromrestatthesamemoment.

CarAmoveswithconstantaccelerationof1.5ms-2andcarBmoveswithconstantaccelerationof2ms-2.

Find

(i) thedistancethatcarBtravelsbeforeitovertakescarA, [4]

(ii) thespeedofcarBatthemomentwhenitovertakescarA. [2]

4

6993/01 Jun16© OCR 2016

7 Anextensiontotheroofofahouseisshowninthediagrambelow.

C

A

P

Q

D

B

M

2 m

1.5 m

Theridge,CD,andthelinesABandPQarehorizontal.PQisperpendiculartoCD. MisthemidpointofAB.ThelinePMisvertical.

APBisanisoscelestrianglewithheight2metresandbaselength3metres. AnglePQMis45°.

Find

(i) thelengthofPQ, [1]

(ii) theanglePBQ. [4]

8 (i) Writedownthebinomialexpansionof(1+δ)3. [2]

(ii) Henceexplainwhy,ifδissmall,(1+δ)3≈1+3δ. [≈means‘isapproximatelyequalto’] [1]

Youaregiventhattheequationx3-0.9x-0.206=0hasarootveryclosetox=1.

(iii) Substitutex=1+δintotheequationandusetheapproximationinpart(ii) tofindanestimateofthisroot,correctto3significantfigures.Showallyourworking. [4]

9 Acurvehasequationy=x3-3x2-3x+4. PointsPandQlieonthecurve.ThecoordinatesofPare(3,-5).

(i) FindtheequationofthetangenttothecurveatP. [4]

ThetangenttothecurveatQisparalleltothetangenttothecurveatP.

(ii) FindthecoordinatesofQ. [3]

5


10 (i) OntheaxesgiveninthePrintedAnswerBook,indicatetheregionforwhichthefollowinginequalitieshold.Youshouldshadetheregionthatisnotsatisfiedbytheinequalities.

x y4 3 30G+ y x2H x 1H [5]

(ii) Findthemaximumvalueof7x+4ysubjecttotheseconditions. [2]

6

6993/01 Jun16© OCR 2016

Section B

11 Arailwaytrackrunsdueeast-westandiscrossedatObyaroadrunningduesouth-north,asshownbelow.Thecrossinghasnobarriers.

NO

A

P

QB400 m

100 m

InitiallyatrainisatpointB,400mfromO,andacarisatpointA,100mfromO.Thetrainistravellingataconstantspeedof25ms-1towardsOandthecaristravellingataconstantspeedof20ms-1towardsO.

AttimetsecondsthetrainisatpointQandthecarisatpointP.

(i) FindexpressionsforthedistancesOPandOQasfunctionsoft. [2]

(ii) Thedistancebetweenthecarandthetrainattimetsisxm.Findaformulaforx2intermsoft.Giveyourformulaintheformx a bt ct2 2= + + wherea,bandcaretobedetermined. [3]

(iii) Differentiatethisformulawithrespecttotandfindthetimeatwhichx2isaminimum.Hencefindtheshortestdistancebetweenthecarandthetrain. [6]

(iv) ShowthatthecarpassespointObeforethetrain. [1]

12 ThelineL1hasequation3x-y=1andthepointPhascoordinates(8,3).

(i) FindtheequationofthelineL2whichpassesthroughPandisperpendiculartolineL1. [3]

(ii) FindthecoordinatesofthepointQwhereL1andL2intersect. [3]

(iii) FindthelengthPQ. [2]

(iv) WritedowntheequationofthecirclethathascentrePandlineL1asatangent. [1]

(v) FindtheequationoftheotherlinethatisatangenttothecircleandisparalleltolineL1. [3]

7


13 The cost of a packet of buns in a local supermarket is x pence and the cost of a loaf of bread isx+75pence.

(i) Writeanexpressionforthenumberofpacketsofbunsthatcanbeboughtfor£5.40andanexpressionforthenumberofloavesthatcanbeboughtfor£5.40. [2]

Thenumberofpacketsofbunsthatcanbeboughtfor£5.40is5morethanthenumberofloavesthatcanbeboughtfor£5.40.

(ii) Usingthisinformationandyouranswertopart(i),deriveanequationinxandshowthatitsimplifiestox2+75x-8100=0. [5]

(iii) Solvethisequationtofindthecostofapacketofbunsandthecostofaloafofbread. [5]

Question 14 is printed overleaf

8

6993/01 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.

14 Theequationofacurveisgivenbyy=x3+ax2+bx+1.ThepointsP(-3,7)andQ(1,3)lieonthecurve.

(i) Formtwoequationsinaandb.Solvetheseequationstoshowthata=3andb=-2. [4]

(ii) Findthemidpoint,R,ofthelinePQandshowthatRliesonthecurve. [2]

ThediagrambelowshowsthecurveandthelinePRQ.

y

x

PA1

R

A2

Q

O

Theareabetween thecurveand the line segmentPR isA1and theareabetween thecurveand the linesegmentRQisA2.

(iii) ShowthatA1=A2. [6]END OF QUESTION PAPER

Monday 6 June 2016 – AfternoonFSMQ ADVANCED LEVEL

6993/01 Additional Mathematics

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. If additional space is required, you should use the lined page(s) at the end of this booklet. The question number(s) must be clearly shown.

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

your answer.• 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 correct to three significant figures where appropriate.

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 100.• The Printed Answer Book consists of 20 pages. The Question Paper consists of 8 pages.

Any blank pages are indicated.

* 6 9 9 3 0 1 *

OCR is an exempt CharityTurn over

© OCR 2016 [100/2548/0]DC (ST/FD) 126098/1

Candidates answer on the Printed Answer Book.

OCR supplied materials:• Question Paper 6993/01 (inserted)

Other materials required:• Scientific or graphical calculator

*6361296134*

Duration: 2 hours


2

© OCR 2016

Section A

1

2

3

Turn over© OCR 2016

3

4 (i)

4 (ii)

4

© OCR 2016

5 (i)

5 (ii)

5


6 (i)

6 (ii)

6

© OCR 2016

7 (i)

7 (ii)

7


8 (i)

8 (ii)

8 (iii)

8

© OCR 2016

9 (i)

9 (ii)

9


10 (i)

1O

1

2

3

4

5

6

7

8

9

10

2 3 4 5 6 7 8 9 10

There is a spare copy of this graph on page 18. If you wish to offer a second attempt, then you must cross through the attempt on this page.

10 (ii)

10

© OCR 2016

Section B

11 (i)

11 (ii)

11


11 (iii)

11 (iv)

12

© OCR 2016

12 (i)

12 (ii)

13


12 (iii)

12 (iv)

12 (v)

14

© OCR 2016

13 (i)

13 (ii)

15


13 (iii)

16

© OCR 2016

14 (i)

14 (ii)

17


14 (iii)

18

© OCR 2016

This is a spare copy of the graph for Question 10(i).Only write on this page if you want to offer a second attempt at the graph.If you do so, then you must cross through the first attempt on page 9.

1O

1

2

3

4

5

6

7

8

9

10

2 3 4 5 6 7 8 9 10

19

© OCR 2016

ADDITIONAL ANSWER SPACE

If additional space is required, you should use the following lined page(s). The question number(s) must be clearly shown in the margin(s).

20

© 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.

FSMQ

Additional Mathematics Unit 6993: Additional Mathematics Free Standing Mathematics Qualification

Mark Scheme for June 2016

Oxford Cambridge and RSA Examinations

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

6993 Mark Scheme June 2016

Annotations and abbreviations

Annotation 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

AG Answer given M1 dep Method mark dependent on a previous method mark(s) cao Correct answer only oe Or equivalent soi Seen or implied www Without wrong working

3


Subject-specific Marking Instructions for Additional Mathematics

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 can never be awarded.

BMark for a correct result or statement independent of Method marks.

4


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.

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.

5


Section A

Question Answer Marks Guidance 1

( )

1 2( 3) 4or

Either 7 or 6 in either of above7 or 1.17 or 1.166

x xa bx a bx

a b

x

− − >⇒ > − < −

= =

⇒ <

M1

A1

A1

Expand and collect soi www isw

Do not allow = anywhere even if final answer correct

3


( )

2

3 2

3 2

d 3 4 2d

2 2 ( ) Satisfied by (1, 3) 3 1 2 2

2

2 2 2

y x xx

y x x x cc

c

y x x x

= − +

⇒ = − + +⇒ = − + +

⇒ =

⇒ = − + +

oe

M1

A1 M1dep A1

Int: At least 1 power increased by 1: Beware mult by x Three terms ignoring c Substitution Complete simplified equation

4

6



43sin 4cos tan3

53.1(3)and 180 53.13 233(.13)

x x x

xx

= ⇒ =

⇒ == + =

Alternative: Square, use Pythagoras M1

cos 0.6 or sin 0.8x x⇒ = ± = ± A1(must include ±) Gives 53.1 A1 Or 233 B1 only if no extra values in range

M1 A1

A1 B1

For tanx

For 43

One angle (53 not acceptable) ft Other angle B0 any extra values in range, ignore any outside range

1 14 3Allow tan or tan3 4

k k− − = =

4

Question Answer Marks Guidance 4 (i) 8 − 4 + 2 − 6 = 0

Alternative: Demonstration that 2f ( ) ( 2)( 3)x x x x= − + +

B1

must be seen i.e. powers evaluated

1 (ii)

( )( )

2

2

2

f ( ) ( 2)( 3)4ac 1 4 3 ( 11) 0

or 0.5 2.75 0 so only one root or no other roots

x x x xD b

x

= − + +

= − = − × = − <

+ + ≠

M1

A1 M1

A1

Factorise: Any 2 correct terms of 3 term quadratic seen. For long division: first two terms For quad factor Numerical evidence must be seen on correct quadratic. Last statement must be seen. Condone reference to (x − 2) being the root.

If quad factor is found in (i) then give credit in (ii) if seen in (ii) e.g. 11 won't work−

4

7


Question Answer Marks Guidance

5 (i) 11.5 B1 One number only seen or AB clearly identified

1 (ii) Use 11.5, 15.5 and 20.5

( )2 2 2

0

11.5 15.5 20.5cos 0.1339...2 11.5 15.5

97.7

B

B

+ −= = −

× ×

⇒ =

B1

M1

A1

Correct use of cosine rule on correct angle using values rounding to given values Answers rounding to 97.7

i.e. range [11.5,12.5]. [15.5,16.5],[19.5,20.5] Values must be consistent.

3

Question Answer Marks Guidance 6 (i) ( )

( )

( )

2

2

2 2

3Distance for A: 4

Distance for B: 3 3 3100 or 100 s or 1004 4 4300 or 400 or 20

B travels 400 m

t

t

s s s t t

s t

⇒ = ± ± = ± =

⇒ = =⇒

B1

B1

M1

A1

soi; ignore 100 soi; ignore 100 Equating distances leading to one of the 6 forms www

SC4 www for trial and error giving correct answer.

4 (ii) 2 2

2

1

Using 22.2.400 1600

40 m s

v u asvv −

= +

⇒ = =

⇒ =

M1

A1

And using a = 2 and their s from (i) www

Or complete and equivalent method. Allow missing u

2

8



7 (i) 2

B1

1 (ii) For PB:

2 2PB 2 1.5 2.52tan 0.8

2.5Angle PBQ 38.7

= + =

⇒ = =

⇒ = a

PBQ

M1 A1

M1

A1

Using their PQ and PB Alternatively for the last two marks: Attempt to find QB and use it with sin, cos or sine rule or cosine rule

If PB is found in (i) then give credit in (ii) only if seen in (ii). n.b. 10.25QB =

4

Question Answer Marks Guidance 8 (i)

( )3 3 2 2 3

2 3

1 1 3.1 3.1

1 3 3

δ δ δ δ

δ δ δ

+ = + + +

= + + +

B1

B1

Unsimplified expansion soi Can be by expansion

2 (ii) Because, if δ is small, then (terms in) (3)δ2 and

δ3 are very small and can be ignored B1 "ignored" or similar must be

seen e.g. neglected, eliminated

1 (iii) ( ) ( )

( )

31 0.9(1 ) 0.206 0

1 3 0.9(1 ) 0.206 02.1 0.106

0.05.....1.05...x

δ δ

δ δδ

δ

+ − + − =

⇒ + − + − =

⇒ =⇒ =⇒ =

M1

M1dep

A1 A1

Sub Using result of (ii) 3sf or better

4

9



9 (i)

2d 3 6 3d

dWhen 3, 6d

Equation of tangent is5 6( 3)

6 23

y x xx

yxx

y xy x

= − −

= =

⇒+ = −

⇒ = −oeoe

M1

A1

M1dep

A1

Diffn. At least one power reduced by 1. Beware division by x Any valid form using their gradient and (3, −5). oe only 3 terms

Ignore +c

4 (ii)

( )( )

2

2

d 3 6 3 6d

2 3 03 1 0

Q is where 1, 3

y x xx

x xx x

x y

= − − =

⇒ − − =

⇒ − + =

⇒ = − =

M1

A1

A1

Equating their gradient fn and their 6 Correct factorisation www cao www

Ignore (3, −5) as a possible answer.

2dSC3 if 2 1 g 2 in (i)d

and Q is correct.

y x xx

= − − ⇒ =

3

10



10 (i)

B1 B1 B1

B1 B1

One line 2nd line 3rd line Shading x ≤ 1 Other shading. Allow ft if gradients of lines are the same sign as the correct lines.

Allow one small square tolerance on each axis

5 (ii) Max value at intersection which is (3, 6)

=45 B1 B1

soi e.g. 45 gets 2

2

11


Section B

Question Answer Marks Guidance 11 (i) (OP =) 100 – 20t

(OQ =) 400 – 25t

B1 B1

isw isw

Ignore labels

2 (ii)

( ) ( ) ( )2 22

2 2

100 20 400 25

170000 24000 1025

x t t

x t t

= − + −

= − +

M1

A1 A1

Use of Pythagoras on their expressions. Soi ignore lack of x2

Final answer must include x2

Condone use of 20t − 100 etc for full marks

3 (iii) ( )

( )

2

2

d 24000 2050dt

0 24000 480when 11.7 2050 41

Then 29512

172(m)

x t

t

x

x

= − +

=

= = =

=

⇒ =

oe

M1

A1

M1dep

A1

M1dep

A1

Diffn of their fn : reduction of power in at least one term Correct numerical expression isw Set = 0 and attempt to solve Allow correct answer even if premature division in (i) Substitute their t(providing t > 0). Dep on both M

Ignore incorrect constant from (ii) Ignore notation on lhs SC 1 for b + 2ct

6 (iv) Car takes 5 secs to reach O

Train takes 16 secs B1 Numerical evidence for both required Accept other valid explanations

1

12



12 (i) x + 3y = k or 13

y x c= − + or 13

y bx a

−= −

−

substitute (8, 3) gives x + 3y = 17 oe

M1

M1dep

A1

3 term equation isw

1717 or 3

1 173 3

k c

y x

= =

= − +

3 (ii) Solve their L2 with y = 3x − 1 simultaneously:

x = 2, y = 5

M1 A1 A1

Must lead to a value for x or y

SC3 Checking points and finding that (2, 5) lies on both

3 (iii) ( ) ( ) ( )

( )2 22 8 2 3 5 40

40 2 10 6.32

d

d

= − + − =

⇒ = = =

M1

A1

Application of Pythagoras

2 (iv) ( ) ( )2 28 3 40− + − =x y B1 FT from (iii) Allow 6.322 oe

1 (v) The point is on the other end of the diameter:

(2, 5) to (8, 3) is 62

−

(14,1)⇒ 3x – y = c satisfied by (14, 1) 3 41x y⇒ − = oe

M1

A1

A1

Using (8, 3) and their Q from (ii). Only 3 terms

Alternatively: 2 58, 3 M1

2 214, 1 A1

x y

x y

+ += =

⇒ = =

3

13



13 (i) 540 540,75x x +

oe

B1 B1

Condone 5.4 5.40 5.40, or 0.75 75x x x+ +

Ignore any labels.

Allow 540nx

≤ etc

2 (ii)

( )( )

2

540 540 5 75

540 75 540 5 ( 75)

540 75 5 ( 75)

75 8100 0

x x

x x x x

x x

x x

⇒ = ++

⇒ + = + +

⇒ × = +

⇒ + − =

oe

oe

M1

A1 M1

A1

A1

For forming 3 term eqn using their terms from (i) Condone −5 Correct eqn Clear both fractions. Eqn must have 3 terms with x and x ±75 involved in denominator for 2 terms AG. At least 1 intermediate step must be seen

May start again Any wrong algebra gets final A0

5 (iii)

( )( )

2 75 8100 0

60 135 06075( 135) or 60 75

x x

x xxx

+ − =

⇒ − + =

⇒ =⇒ + = +

Buns 60p, loaf of bread 135p oe

M1

A1

A1 A1 A1

Solving given quadratic by factorisation that would lead to 2 terms correct when expanded soi Or correct formula soi Correct factorisation or correct substitution soi by final answer cao www - units must be given

Ignore −135 Correct answer only full marks

5

14


Question Answer Marks Guidance 14 (i)

( )

3 27 ( 3) ( 3) 3 1and 3 1 1

9 3 33 and 1

3, 2

a ba b

a b a b

a b

⇒ = − + − − += + + +

⇒ − = + =

⇒ = = −

oeoe

B1 B1

M1

A1

1st equation 2nd equation Solve their eqns leading to either a or b Both AG

Need not be simplified for either Need to see at least one intermediate step

4 (ii) Midpoint is (−1, 5)

Show (−1, 5) lies on curve. B1 B1

Must see −1 + 3 + 2 + 1 = 5

i.e. powers must be evaluated

2 (iii) ( )

( )1

2

A Area under curve area under line

or A Area under line area under curve

= ± −

= ± −

( ) ( )4

3 2 3 2Area under curve = 3 2 1 d4xx x x x x x x c+ − + = + − + +∫

11 811 1 1 27 9 3 124 4

11 75 12 16 12 44 4

A = − − − − − − − − = − − − − = − =

21 18 1 1 1 1 1 1 8 4 44 4

A = − + − + − − − − = − =

B1

M1 A1

M1dep

A1 A1

For sight of one attempt to find a difference of areas For integration, ignore limits Integration correct Correct limits for one curve integral (For A1, −3 to −1 and for A2, −1 to 1) For A1 www For A2 www

At least 3 powers increased by 1. Watch for multiplication by x Could be wrong way round but must be subtracted n.b. an answer of −4 should be explained for credit of A1

6

15


Question Answer Marks Guidance Alternative 1: if subtraction is before integration.

( ) ( ) ( )4 2

3 2 31A = 3 2 1 (4 ) d 3

4 2

7 9 44 4

x xx x x x x x x c+ − + − − = + − − +

= − − =

∫

( ) ( ) ( )4 2

3 2 32 A = (4 ) ( 3 2 1) d 3

4 2

9 7 44 4

x xx x x x x x x c− − + − + = − − + + +

= − − =

∫

B1

M1

A1 M1dep

A1

A1

For subtracting their y = 4 – x from curve For either integration, ignore limits Either integration correct Correct limits for one curve integral (For A1, −3 to −1 and for A2, −1 to 1) For A1 www For A2 www

Could be subtracted in either order Could be wrong way round but must be subtracted. n.b. an answer of −4 should be explained for credit of A1

Alternative 2

( )3 2 3 2

3

3 2 1 (4 ) 3 3

( 1) 4( 1)

y x x x x x x x

y x x

= + − + − − = + − −

= + − +

This is an odd function relative to x = −1. The function therefore has1800 rotational symmetry about (−1, 0) So A1 = A2

B1

M1 A1 M1 A1 A1

For subtracting their y = 4 – x from curve Writing as a function of (x + 1) Understanding of odd function Rotational symmetry Conclusion

16

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Oxford Cambridge and RSA Examinations

FSMQ

Additional Mathematics

Unit 6993: Paper 1 Free Standing Mathematics Qualification

OCR Report to Centres 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 report on the examination provides information on the performance of candidates which it is hoped will be useful to teachers in their preparation of candidates for future examinations. It is intended to be constructive and informative and to promote better understanding of the specification content, of the operation of the scheme of assessment and of the application of assessment criteria. Reports should be read in conjunction with the published question papers and mark schemes for the examination. OCR will not enter into any discussion or correspondence in connection with this report. © OCR 2016

CONTENTS

Additional Mathematics FSMQ (6993)

OCR REPORT TO CENTRES

Content Page Additional Mathematics – 6993 4

OCR Report to Centres – June 2016

4

Additional Mathematics – 6993

General

Candidates found one or two questions tricky and the mean score was a little down on last year.

Centres will be aware that the assessment cannot test all the specification content in one

examination; the requirement is to cover the whole specification within a few years. There is no

requirement within the specification for specific proportions of the assessment per topic.

Consequently, the percentage of the topics will vary from year to year. Candidates who are

thoroughly prepared for the examination will find no difficulty with this, but those who are

prepared for specific topics could struggle.

Question 1

This was an easy start to the paper and most candidates obtained the correct answer. The most

common error was incorrect processing of signs leading to 1 2 6 4x x and a final answer of

5

6x . Much less common, though more disturbing at this level, was to write

1 2( 3) 1( 3)x x . Very few had to be penalised for using treating the question as an

equation.

Question 2

There were very few incorrect answers to this question. A very small percentage of candidates

differentiated and it was rare to see the coordinates substituted the wrong way round. Just a few

were penalised for not writing the equation properly as ‘y =.....’.

Question 3

The most common error was to obtain 3

tan4

x rather than 4

tan3

x . The second most common

error was in not obtaining the 2nd

value correctly.

The alternative method was to square both sides and use Pythagoras.

However, there are problems with this approach because the result 2 216 9sin or cos

25 25x x

will yield 4 values for x, 2 of which satisfy the equation 4

tan3

x . There is a fair amount of

extra work to obtain the correct answer by this method as the values that do not satisfy the given

equation need to be rejected. This is a typical case where an alternative method is perfectly

acceptable but takes very much more time.

Question 4

(i) Most substituted the correct value but there were a few who substituted x = 2. This was

a “show that” question and so all the calculations need to be shown.

An alternative method was, instead of using the factor theorem, to divide f(x) by (x 2) to show

that there was no remainder. This was a lot of work for 1 mark, but fortunately on this occasion

what was done could be used in part (ii).

(ii) The derivation of the quadratic factor was required here but credited if it was seen in (i).

Clearly it was not sufficient to say that the quadratic factor did not factorise. It was necessary to

set it equal to 0 and to attempt to solve, usually by showing that the discriminant was negative.

Some substituted factors of 6 into the cubic in an attempt to show 2 was the only root which was

also clearly insufficient as it depends on any roots being integers.


5

It was disappointing to see candidates who had been successful so far writing “therefore (x 2) is

the only root”.

Question 5

(i) This was successfully completed by the vast majority with just a few adding together the

smallest lengths for all three sides.

(ii) Most candidates correctly used the cosine rule and gained the method mark without

necessarily finding the correct angle. A number made several attempts with different

combinations of lengths without appreciating the geometry of the problem. Some candidates

appeared to be wary of using 20.5 as an upper value and opted for 20.4. Some even used

inconsistent values such as 11.5 and 12.5 in the numerator and denominator of the cosine rule. It

was a good discriminator as candidates were required to think to select the correct combination of

lengths.

Question 6

(i) The majority of candidates found this a very challenging question with the 100 m

difference creating a conceptual difficulty in applying the equations of motion. A significant

number wrote down all the equations of motion that they could remember, made a faltering start

and moved on. A number did successfully obtain the correct answer by trial and error but others

amazingly got 400 m by rather more dubious means. Often the values of 1.5 and 2 were used as

velocities rather than accelerations and a common result was 2 × 200 s. This question clearly

differentiated between those who could apply their knowledge to a problem and those who could

not.

(ii) This part was not attempted by many who had failed in part (i) but those who had found a

distance in part (i) often obtained the method mark in this part.

Question 7

The three dimensional nature of this problem caused difficulty to many candidates who could not

appreciate the nature of the isosceles and right-angled triangles.

(i) Many candidates used unnecessary calculations and the result was not stated often. Many

found PB in this part and did not mention it in part (ii).

(ii) Strong candidates realised that triangle PQB was right-angled at P and proceeded to get 4

marks in a couple of lines. Others, however, found BQ by very long-winded ways including the

application of the sine or cosine rule in the right-angled triangle. Some did not appreciate where

the right angle was and used the incorrect ratio. This was another question that tested the ability

of candidates to apply their knowledge in unfamiliar circumstances.

Question 8

(i) While most candidates were successful with this question a surprising number struggled

with the expansion. Even those who expanded (1+δ)(1+ δ) then multiplied the result by (1+δ)

often made careless numerical errors. The most successful candidates used Pascal’s triangle,

although there were still careless errors, leaving the answer as 13+3δ+3δ

2+δ

3, forgetting the

coefficients or giving every term a coefficient of 3.

(ii) Few students were clear or confident in their explanations. Most recognised the squared

and cubed terms would be smaller, and have less significance or make less difference but few

proceeded to say they were so small to be negligible, or approximated to zero.

(iii) Most correctly substituted and a number used the guidance provided in part (ii). However,

too many candidates expanded 0.9(1+δ) incorrectly, obtaining 0.9+0.9δ and thus an incorrect

answer. Even those who did this correctly and found δ, often did not go on to find x. This shows a

lack of attention to detail that is concerning at this level of mathematics.


6

Question 9

(i) This part was generally well answered. The vast majority of candidates differentiated

correctly. Some then set this equal to zero and attempted to solve it but most substituted

x = 3. There was again some careless numerical work resulting in 12 as the gradient rather than

6, however generally the gradient was used with the correct coordinates to find the equation of

the line. There was some carelessness arithmetic. For instance, )3(65 xy often

became 136 xy . A number of students used the normal gradient instead, a lack of

thoroughness in reading the question.

(ii) This part was less well answered. Candidates recognised that the gradient of 6 was

relevant and needed to be used but many were unsure how to use it. A number tried to use the

equation of a line with 6 and obviously had no direction with which to continue the question.

Many candidates did correctly equate d

d

y

xand ‘their’ 6 and then solve correctly to obtain the

correct x value. From this point there was again a disappointing number of careless errors when

substituting to find y; this lack of scrupulousness cost marks here and elsewhere in the paper.

Question 10

(i) This question, in common with previous years, was very well answered, the vast majority

of candidates achieving full marks. There were the usual and expected mistakes of shading the

wrong sides, not plotting accurately enough, or confusing the axes so the lines were plotted

incorrectly, but on the whole candidates were very successful.

(ii) This part was less successful than part (i) although candidates were generally accurate. A

few just identified the relevant point and did not find the maximum amount; others attempted

decimal points or points outside the feasible region resulting in incorrect answers.

Question 11

A large number of candidates made little or no attempt at this question, possibly because they

thought it was simply a mechanics question. Very few scored full marks.

(i) The most common error was to say OP = 20t and similarly for OQ.

(ii) Of those who made it successfully to this part, a large number failed to get the final A1

mark through faulty expansion – not least because of the number of 0s involved!

(iii) A fair number made some headway this part, though by no means all equated their

derivative to zero. Some candidates seemed to be uncomfortable with finding d

d

y

x when “y” was

not a function of x but of x2.

(iv) This part was generally very well answered even by those who had made little headway in

the other parts.

Question 12

This was an easily accessible question, but despite a high level of competence in algebraic

manipulation, better training in setting out solutions more systematically would lead to fewer

errors.

(i) Many candidates understood the concepts behind this question and were able to access

this part. Very few were unable to produce the correct "negative reciprocal". Where there were

errors it was often in the collecting together of the constants after the values (8, 3) had been

substituted into the correct equation.

(ii) It was a rare candidate who failed to recognise that this was a question on simultaneous

equations. Most chose the sensible method of substitution to solve them and did so successfully,


7

although elimination was also seen. Testing which points lay on both lines was rarely seen. A

small minority made mistakes in basic arithmetic and were unaware that they had done so;

centres should encourage students to check their answers. Rearrangement of the equation

3x y = 1 to y = 1 3x was an error seen not infrequently.

(iii) Many candidates knew how to find the length of a line. Those who had obtained the

coordinates of Q successfully in (ii) went on to get this part correct on the whole. Many

candidates converted to decimal form, but centres would do well to encourage surd form,

especially as decimal form usually encourages writing numbers to, say, 3 significant figures.

(iv) Nearly all students knew the basic formula for the equation of a circle and most knew that

the radius was their answer to part (iii). One fairly common error was to halve the answer

obtained in part (iii). Another was to give the radius and not the radius squared. Here, candidates

with an approximate answer in (iii), obtained only an approximate answer here, and this was

penalised.

(v) This was, as expected, very discriminatory part with only a small minority of candidates

knowing how to find the point on the line, although many knew that parallel lines have equal

gradients. Candidates should be encouraged to draw their own rough diagrams for geometry

questions to help them with problem solving strategies. With a diagram, the vector can clearly be

seen and the point found. Candidates who embarked on finding the points of intersection of the

circle with a general line of the form y = 3x + c, presumably to find the values of c that gave

coincident points, could not work it through to completion. This would be another example of an

alternative, long-winded, method that would require much more work than the marks available

would justify.

Question 13

Although this question was of a familiar type, and expressed in terms of money, which usually

makes it easier for candidates to realise what is going on, it still proved challenging to many.

(i) The majority of candidates who were making a reasonable attempt at the paper were able

to gain both marks, albeit some of them having mixed units.

(ii) When there were two fractions from part (i) to be combined into an equation, the expected

error of 5 being added to the wrong term was quite common. A surprisingly common error was

to multiply one of the equations by 5, in effect, saying that five times as many buns could be

bought rather than five more.

Assuming there was an equation of the correct form, fractions were usually cleared successfully.

A significant number of candidates who had the wrong units in their initial equation then realised

part way through that there was a problem and tried to adjust, not always successfully.

Consequently, there was plenty of 'fudging' evident. This also included those who had added 5 to

the wrong term in their initial equation and obtained the wrong sign in the printed answer.

(iii) A good number of candidates showed sound examination technique by picking up the

question at this point, even when the first two parts had defeated them. Ironically, those who

didn’t and left this part blank, had already demonstrated success on similar work elsewhere in the

paper!

The formula was the usual method of solution, and the correct answers found and usually

expressed in context. Some candidates continue to misquote the quadratic formula or make

careless slips when substituting in values.

Those who chose to factorise were normally successful. Completing the square, as expected, was

the least common and least successful method of solution.

Some candidates should have been alerted by their impractical answers: loaves for 7p, for

example!


8

Question 14

(i) Many candidates were able to answer this question correctly. The type of mistakes that

were seen when forming the two equations included candidates not dealing with the negative sign

correctly when squaring or to form a quadratic by having a2 in the equation.

Most candidates were able to go on correctly to solve the simultaneous equations with sufficient

working shown. Some candidates did not read the question and showed that the values worked

rather than solving the equations.

Candidates who had more success in solving the two equations usually attempted to do so with

the equations in their simplified form.

(ii) Most candidates were able to find the midpoint but not all went on to show that this point

was on the curve. Some attempted to but did not show sufficient evidence by evaluating the

indices.

(iii) Many candidates appreciated the need for integration for this question. Many were able to

integrate the curve correctly and the majority recognising that this was not the complete answer.

Finding the equation of the line was good but the combination of subtracting two areas with

negative limits proved too much for many. This was compounded by the fact that most candidates

used integration to find the area under the line, rather than find the area of a trapezium.

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 2016

OCR (Oxford Cambridge and RSA Examinations) 1 Hills Road Cambridge CB1 2EU OCR Customer Contact Centre

Education and Learning Telephone: 01223 553998 Facsimile: 01223 552627 Email: [email protected] www.ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored

mailto:[email protected]

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

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

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