Oxford Cambridge and RSA Tuesday 19 June 2018 – Afternoon · 2018. 8. 28. · Tuesday 19 June 2018 – Afternoon FSMQ ADVANCED LEVEL 6993/01 Additional Mathematics QUESTION PAPER

Tuesday 19 June 2018 – AfternoonFSMQ ADVANCED LEVEL6993/01 Additional Mathematics

QUESTION PAPER

*4396589493*

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.• 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 barcodes.• 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.

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

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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 Jun18© OCR 2018

Formulae Sheet: 6993 Additional Mathematics

In any triangle ABC

Cosine rule a2 = b2 + c2 – 2bc cos A

C

b a

c BA

Binomial expansion

When n is a positive integer

(a + b)n = an + (n1) an – 1b + (n

2) an – 2b2 + … + (nr ) an – rbr + … + bn

where

(nr ) = nCr =

n!––––––––r!(n – r)!

3

6993/01 Jun18 Turn over© OCR 2018

Answer all the questions.

Section A

1 Solve the inequality ( )x x2 1 3 21- + - . [3]

2 The gradient function of a curve is given by xy

dd

x x2 2 3 2= + - .

Find the equation of the curve given that it passes through the point (2, 3). [4]

3 The graph of y x x2 32= + - is given on the grid in the Printed Answer Book.

(i) Write down the solution to the equation x x2 3 02 + - = . [1]

(ii) By plotting the appropriate straight line on the grid, find the solution to the equation x x 062 - =- . [3]

4 You are given that the acute angle i is such that sin 51

i = .

Find the exact value of each of the following.

• cos i

• tan i [4]

4

6993/01 Jun18© OCR 2018

5 Fig. 5 shows part of the graph of the curve with equation y x x6 22 3= - .

21

43

65

87

9

–1 1 2 3 4

y

x–1

0

Fig. 5

Find the area of the shaded region enclosed by the curve and the x-axis. [4]

6 (i) Solve these simultaneous equations. x y3 4 18+ = x y7 3 5- = [4]

(ii) Draw a rough sketch of the lines to demonstrate graphically the solution to part (i). [2]

7 (i) Find the coordinates of the points where the line y x7 9= - cuts the curve y x x2 52= + - . [4]

(ii) Determine whether the line is a normal to the curve at either of the points of intersection. [3]

8 (i) Simplify the equation xx a x

42 0+

+-= , leaving your answer in the form ( )x p q2+ =

where p is an integer and q is given in terms of the constant a. [3]

(ii) Hence write down the range of values of a for which the equation has real roots. [2]

(iii) Using your answer to part (i), solve the equation when a 1=- , giving your answers exactly. [2]

5


9 The proportion of people who are left-handed is 20%.

(a) For a group of 10 students chosen at random, use the binomial distribution to find the probability that

(i) no student is left-handed, [2] (ii) exactly 4 students are left-handed. [3]

(b) State the conditions necessary for the binomial distribution to be valid. [2]

10 Fig. 10 shows an “up and over” garage door, XY, that is 200 cm long. There is a small wheel at the point G on the door. The wheel runs freely up a groove in a fixed vertical door frame, AB. A metal rod AP is fixed to the top of the door frame, A, and is also fixed to the point P on the door. The rod is hinged at both ends.

GP = PX = AP = 60 cm and YG = 80 cm. When the door is closed, Y is at B and X is at A. When the door is fully open, G is at A and the door is

horizontal, 200 cm above the horizontal ground.

X

Y

B

A

P

Not to scaleG

Fig. 10

(i) Explain why P is the centre of the circle through A, G and X. [1]

(ii) Hence show that AX is horizontal whatever the position of the garage door. [1] (iii) Find the height of Y above the ground when angle AGP = 40°. [4]

6

6993/01 Jun18© OCR 2018

Section B

11 A circle has centre (0, 3) and radius 3.

(i) Show that the equation of the circle is x y ky 02 2+ - = where k is to be determined. [2]

The line y mx 2= - passes through the point P (0, –2) and is a tangent to the circle.

(ii) Find the two possible values of m. [6]

The two tangents from P meet the circle at the points A and B respectively.

(iii) Find the lengths PA and PB. [4]

12 The shape shown in Fig. 12 is made of metal rods. ABCD is a rectangle. AB = CD = y cm and BC = DA = 6x cm. AED is an isosceles triangle with height 4x cm and AE = ED.

4x6x

y A

D

Not to scale

C

B

E

Fig 12

(i) Show that the perimeter, p cm, can be written as p = 16x + 2y. [3]

You are given that p = 96.

(ii) Show that the area of the shape, A cm2, can be written as A x x288 36 2= - . [3]

(iii) Find the maximum area of the shape as x and y vary and find the values of x and y for this area. [6]

7


13 Jessie walks at 3 km per hour in a straight line from a point B to a point C, a distance of 5 km. C is on a bearing 050° from B, as shown in Fig 13.1.

Brandon sets out at the same time as Jessie. He starts from a point A which is 2 km due East of B. He walks at 2 km per hour directly to C.

N Not to scale

C

AB 2 km

5 km

50°

Fig. 13.1

(i) Calculate the distance AC, correct to 3 significant figures. [4] (ii) Show that Brandon arrives at C approximately 11 minutes after Jessie arrives. [3] Charlie also sets out at the same time as Jessie. He walks in a straight line from A at 2 km per hour to meet

Jessie at a point X on BC, as shown in Fig. 13.2. He arrives at the point X at the same time as Jessie.

N Not to scale

C

AB 2 km

X

50°

Fig. 13.2

(iii) Show that there are two possible positions for X and find the bearing on which Charlie must walk in each case. [5]

8

6993/01 Jun18© OCR 2018


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 Two cars, P and Q, accelerate from rest from a point O at the same time.

(a) P accelerates uniformly at 2 m s–2.

(i) Write down the formula for the displacement, s metres, of P at time t seconds after leaving O. [1]

(ii) Using appropriate units, find the time taken for P to reach a speed of 90 km h–1. [3]

(b) Q accelerates from rest with variable acceleration a m s–2 where, at time t seconds, a = 1 + kt , where k is a positive constant. Q passes P when t = 10.

(i) Find the value of k. [5]

(ii) Show that at the time when P reaches 90 km h–1, Q is travelling at a speed just less than 130 km h–1. [3]

END OF QUESTION PAPER

Tuesday 19 June 2018 – AfternoonFSMQ ADVANCED LEVEL6993/01 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. 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.• 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 barcodes.• 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 *

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© OCR 2018 [100/2548/0]DC (LK/KN) 154339/6

Candidates answer on the Printed Answer Book.

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

Other materials required:• Scientific or graphical calculator

*4396776393*

Duration: 2 hours


2

© OCR 2018

Section A

1

2

3

Turn over© OCR 2018

3

3 (i)

3 (ii)

y

x

–5

0 1 2 3 4–1–2–3–4–5

5

10

15

4

© OCR 2018

4

5


5

6

© OCR 2018

6 (i)

7


6 (ii)

8

© OCR 2018

7 (i)

7 (ii)

9


8 (i)

8 (ii)

8 (iii)

10

© OCR 2018

9 (a) (i)

9 (a) (ii)

9 (b)

11


10 (i)

10 (ii)

10 (iii)

12

© OCR 2018

Section B

11 (i)

11 (ii)

13


11 (iii)

14

© OCR 2018

12 (i)

12 (ii)

15


12 (iii)

16

© OCR 2018

13 (i)

13 (ii)

17


13 (iii)

18

© OCR 2018

14 (a)(i)

14 (a)(ii)

19


14 (b)(i)

20

© OCR 2018

14 (b)(ii)


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

FSMQ

Additional Mathematics

Unit 6993: Additional Mathematics Free Standing Mathematics Qualification

Mark Scheme for June 2018

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 2018

6993/01 Mark Scheme June 2018

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

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

Marking Instructions 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.


5

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;

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 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.

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.

6993 Mark Scheme June 2018

6

Section A

Question Answer Marks Guidance

1 2 1 3( 2)

2 4 5

4 7

7

4

x x

x

x

x

B1

B1

B1

Remove brackets giving rhs 1 + 3x 6

or better

Ft Result in the form ax b oe

[3]


2

2

2 3

2 3

d2 2 3

d

2

Through ( 2, 3)

3 4 4 8

3

2 3

yx x

x

y x x x c

c

c

y x x x

M1

A1

M1

A1

Attempt to integrate – at least two

powers increased by 1

Coefficients unsimplified

Dep. Substitute in their function

Equation must be given

Beware of mult by x

[4]


7


3 (i) (x =) 3, 1

B1 Both, by any means Allow (-3, 0) and (1, 0)

[1]

(ii) The line is (y =) 3x + 3

The solution is 3, 2

B1

B1

B1

Equation (soi by plot)

Correctly plotted

Dep on previous 2

Algebraic soln not acceptable.

Allow (3, 12) and (2,3)

[3]


4 2 2 1

cos 1 sin 125

24 2cos or 6

25 5

M1

A1

Use of Pythagoras

Isw

allow 0.96

sin 1 5tan

cos 5 2 6

1 1 1 or or 6

1224 2 6

M1

A1

Use of tan ratio

Isw

1allow

5 0.96

M0 if approximate values used

[4]

Alternatively:

Find third side of triangle

24

24giving cos

25

1and tan

24

M1

A1isw

A1 isw

A1 isw


8


5

3

2 3

0

43

34

3

0

6 2 d

22

22

154 40 0

2

113

2

x x x

xx

xA x

M1

A1

M1

A1

Integration – ignore limits

Dep. Substitution of x = 3 (and x =

0 soi). Any other limits M0

i.e. both powers increased by 1

Allow unsimplified

SC Answer only or www seen B4

[4]


6 (i) (i) 3: 9 12 54

(ii) 4: 28 12 20

Add: 37 74

2

3

x y

x y

x

x

y

M1

M1

A1

A1

Making a coefficient the same

Elimination

Alternatively soln by substitution

SC Answer only or www seen B4

[4]

(ii) Sketch to show two lines, one +ve gradient

and one –ve,

intersecting at their point from (i)

B1

B1

Two lines

Dep. Their intersection

[2]


9


7 (i)

2

2

7 9 2 5

5 4 0

4 1 0

1,4

2, 19

1, 2 , 4,19

x x x

x x

x x

x

y

M1

A1

A1

A1

Equate

Correct quadratic

Both x-values (or both y-values or one

pair)

Both coordinates

Alt: Make x subject and substitute

to give 2 17 38 0y y

Allow x = 1, y = 2

and x = 4, y = 19

SC. Answer (i.e. both pairs) only

or www B4

[4]

(ii) d2 2

d

dAt (1, 2), 2 2 4

d

dAt (4,19), 8 2 10

d

1 1Grad normals = and 7 so no.

4 10

yx

x

y

x

y

x

oe

M1

A1

A1

Diffn

Both values

Correct comparison

Alternative:

Diffn M1

One value seen and correct

numeric comparison A1

[3]


10


8 (i)

2

2

2

2

20

4

4 4 2 0

2 4

2 1 1 4

1 1 4

x a x

x

x a x x

x x a

x x a

x a

M1

M1

A1

Clear fractions on lhs

Collection of terms to a 3 term

quadratic and attempt to complete the

square

Correct final form

"Attempt" means make lhs include 2 22x px p

[3]

(ii) Roots if 0

1

4

their q

a

≥

≤

M1

A1

Soi. Allow use of >

ft their q . correct inequality.

Allow = here only if ans is correct.

Allow expansion of quadratic and

use of discriminant

[2]

(iii) 2

1 5

1 5

x

x

M1

A1

Substitute to obtain quadratic in form

2

x p n

Both required isw

Allow use of formula

[2]


11


9 (a) (i) 104

0.107(4)5

M1

A1

Correct power and p

Awrt isw

One term only

[2]

(ii) 4 610 1 4

210 0.000419434 5 5

0.088(08)

M1

A1

A1

Includes correct powers and a

coefficient

210 soi

Awrt isw

[3]

(b) Fixed number of trials

Each trial has two outcomes

Fixed probability for success

Independent trials

B1

B1

Any one correct

Another correct

Ignore incorrect answers or other

answers

[2]


10 (i) Because AP = XP = GP (= radius) B1

[1]

(ii) Angle GAX is angle in semicircle and since

BA is vertical, XA must be horizontal. B1 Accept any valid method

[1]

(iii) Finding AG = 120cos40 ( = 91.93) oe

Finding depth of Y below G =80cos40

(=61.28)

Ht of Y above ground = 200 AG depth

of Y below G

=46.8 cm

M1

M1

M1

A1

Need not be numeric

Distance of Y below AX =

200cos40 M1(correct triangle) M1

(correct ratio)

= 153.2 cm

So height above ground = 200 –

153.2 M1

= 46.8 cm A1

[4]


12

Section B


11 (i)

22

2 2

2 2

3 9

6 9 9

6 0

6

x y

x y y

x y y

k

M1

A1

Isw

[2]

(ii)

22

2 2

2 2

2

They meet when 2 6 2 0

1 10 16 0

Tangent if coincident roots

10 4.16 1

36 64

4

3

x mx mx

m x mx

m m

m

m

oe

oe

M1

A1

M1

A1

A1

A1

Substitute line into their curve

ft Allow bracket expanded

Dep. Attempt to find coincident roots

using "b24ac"

Or substitute for x to give

quadratic in y:

2 2 2

4 2

1 4 6 4 0

36 64 0

y m y m

m m

[6]

(iii) In triangle PCA, PC = 5

CA = 3

By Pythagoras:

PA = PB = 4

B1

B1

M1

A1

soi

soi

Both

[4]


13


12 (i) p = 2y + 6x + AE + DE

AE = DE = 5x

Giving p = 2y + 6x + 5x + 5x

p = 2y + 16x

M1

B1

A1

Adding

www soi

www AG

algebra must be correct

N./B. 2 23 4 25 5x x x x

or 2 2 23 4 25 5x x x x

could earn M1 B0 A0

[3]

(ii) 2

2

2

6 12

3 (96 16 ) 12

288 36

A xy x

x x x

x x

M1

M1

A1

Calculate the area

Substitute correct

expression for y

AG

[3]

(iii) 2

2

288 36

d288 72

d

2880 when 4

72

288 4 36 16 576cm

96 6416

2

A x x

Ax

x

x

A

y

Alternatively:

2 2 2

2

288 36 36 8 36 16 16 8

36 16 4

which has its greatest value when 4

36 16 576

48 8 16

A x x x x x x

x

x

A

y x

M1 A1

M1 A1

A1

A1

M1

A1

M1

A1

A1

A1

Diffn - reduce each

power by 1

Set = 0

x

area

y

SC Graph of fn goes through

(0,0) and (8,0) so being

quadratic means max value at x

= 4 B4

Area B1

y value B1

Other symmetrical points may

be used.

[6]


14


13 (i) 2AC 25 4 2.2.5cos 40

13.68

AC = 3.70 km

M1

A1

A1

A1

Cos formula

Correct subs soi

AC2 soi

cao

[4]

(ii)

5 hrs 100 mins

3

3.7hrs 111 mins

2

111 100 11 mins

B1

B1

B1

www AG

J = 100 mins

B = 111 mins

[3]

(iii)

0

0

0

0

sin sin 40

3 2

3sin sin 40 0.9642

2

74.6

Bearing 345

or 105.4

Bearing 015

oe

M1

A1

A1

A1

A1

Sin rule with denominators in

proportion 3:2

Soi

One angle awrt

Correct bearing awrt

2nd angle plus bearing A1awrt

(Allow 15º) awrt

Solution using cosine rule

acceptable.

Alternatively 2nd angle

Then both bearings

[5]


15


14 (a) (i) 2 21

22

s t t B1

[1]

(ii) 1 1 2 290 km h 25 m s or 2ms = 25920 km hr

2 25 2

12.5 secs

v t t

t

B1

M1

A1

Units must be given - others are

possible

Application of v = u + at with

consistent units

Units must be given

Beware mixing of units which

could give 12.5

[3]

(b) (i) When t = 10 sP = 100

For Q:

2

2 3

1

2

2 6

1000At 10 50 100

6

3000.3

1000

Q

Q

a kt

ktv t

t kts

kt s

k

B1

M1

A1

A1

A1

Seen anywhere

Integrating wrt t - both powers

increased by 1

Must not include c.

Must not include ct + d.

[5]

(ii) When t = 12.5 2 2

1

0.3 12.512.5

2 2

35.94 m s

60 6035.94 129.4 km h

1000

Q

ktv t

M1

A1

A1

Inserting their t and their k into

velocity eqn

AG

[3]

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Published: 15 August 2018 Version 1.0 1

Unit level raw mark and UMS grade boundaries June 2018 series

For more information about results and grade calculations, see https://www.ocr.org.uk/students/getting-your-results/

AS GCE / Advanced GCE / AS GCE Double Award / Advanced GCE Double Award

AS & Advanced GCE MathematicsMax Mark a b c d e u

4721 01 C1 Core mathematics 1 (AS) Raw 72 61 55 50 45 40 0UMS 100 80 70 60 50 40 0

4722 01 C2 Core mathematics 2 (AS) Raw 72 55 49 43 37 31 0UMS 100 80 70 60 50 40 0

4723 01 C3 Core mathematics 3 (A2) Raw 72 55 48 41 34 28 0UMS 100 80 70 60 50 40 0

4724 01 C4 Core mathematics 4 (A2) Raw 72 54 47 40 34 28 0UMS 100 80 70 60 50 40 0

4725 01 FP1 Further pure mathematics 1 (AS) Raw 72 56 50 45 40 35 0UMS 100 80 70 60 50 40 0

4726 01 FP2 Further pure mathematics 2 (A2) Raw 72 59 53 47 41 35 0UMS 100 80 70 60 50 40 0

4727 01 FP3 Further pure mathematics 3 (A2) Raw 72 47 41 36 31 26 0UMS 100 80 70 60 50 40 0

4728 01 M1 Mechanics 1 (AS) Raw 72 60 51 42 34 26 0UMS 100 80 70 60 50 40 0

4729 01 M2 Mechanics 2 (A2) Raw 72 53 46 39 32 26 0UMS 100 80 70 60 50 40 0



4732 01 S1 – Probability and statistics 1 (AS) Raw 72 57 50 43 36 29 0UMS 100 80 70 60 50 40 0

4733 01 S2 – Probability and statistics 2 (A2) Raw 72 56 49 42 35 28 0UMS 100 80 70 60 50 40 0



4736 01 D1 – Decision mathematics 1 (AS) Raw 72 55 48 42 36 30 0UMS 100 80 70 60 50 40 0

4737 01 D2 – Decision mathematics 2 (A2) Raw 72 58 53 48 44 40 0UMS 100 80 70 60 50 40 0

Version 1.0 11

AS & Advanced GCE Mathematics (MEI)Max Mark a b c d e u

4751 01 C1 – Introduction to advanced mathematics (AS) Raw 72 60 55 50 45 40 0UMS 100 80 70 60 50 40 0

4752 01 C2 – Concepts for advanced mathematics (AS) Raw 72 53 47 41 36 31 0UMS 100 80 70 60 50 40 0

4753 01 (C3) Methods for Advanced Mathematics (A2): Written Paper Raw 72 61 56 51 46 40 04753 02 (C3) Methods for Advanced Mathematics (A2): Coursework Raw 18 15 13 11 9 8 0

4753 82 (C3) Methods for Advanced Mathematics (A2): Carried ForwardCoursework Mark Raw 18 15 13 11 9 8 0

UMS 100 80 70 60 50 40 04754 01 C4 – Applications of advanced mathematics (A2) Raw 90 63 56 49 43 37 0

UMS 100 80 70 60 50 40 04755 01 FP1 – Further concepts for advanced mathematics (AS) Raw 72 55 51 47 43 40 0

UMS 100 80 70 60 50 40 04756 01 FP2 – Further methods for advanced mathematics (A2) Raw 72 48 42 36 31 26 0

UMS 100 80 70 60 50 40 04757 01 FP3 – Further applications of advanced mathematics (A2) Raw 72 63 56 49 42 35 0

UMS 100 80 70 60 50 40 04758 01 (DE) Differential Equations (A2): Written Paper Raw 72 61 54 48 42 35 04758 02 (DE) Differential Equations (A2): Coursework Raw 18 15 13 11 9 8 04758 82 (DE) Differential Equations (A2): Carried Forward Coursework Mark Raw 18 15 13 11 9 8 0

UMS 100 80 70 60 50 40 04761 01 M1 – Mechanics 1 (AS) Raw 72 51 44 37 31 25 0

UMS 100 80 70 60 50 40 04762 01 M2 – Mechanics 2 (A2) Raw 72 59 53 47 41 35 0



UMS 100 80 70 60 50 40 04766 01 S1 – Statistics 1 (AS) Raw 72 59 53 47 42 37 0

UMS 100 80 70 60 50 40 04767 01 S2 – Statistics 2 (A2) Raw 72 54 47 41 35 29 0



UMS 100 80 70 60 50 40 04771 01 D1 – Decision mathematics 1 (AS) Raw 72 50 44 38 32 26 0

UMS 100 80 70 60 50 40 04772 01 D2 – Decision mathematics 2 (A2) Raw 72 55 51 47 43 39 0

UMS 100 80 70 60 50 40 04773 01 DC – Decision mathematics computation (A2) Raw 72 46 40 34 29 24 0

UMS 100 80 70 60 50 40 04776 01 (NM) Numerical Methods (AS): Written Paper Raw 72 57 52 48 44 39 04776 02 (NM) Numerical Methods (AS): Coursework Raw 18 14 12 10 8 7 0

4776 82 (NM) Numerical Methods (AS): Carried Forward Coursework Mark Raw 18 14 12 10 8 7 0

UMS 100 80 70 60 50 40 04777 01 NC – Numerical computation (A2) Raw 72 55 47 39 32 25 0

UMS 100 80 70 60 50 40 04798 01 FPT - Further pure mathematics with technology (A2) Raw 72 57 49 41 33 26 0

UMS 100 80 70 60 50 40 0

AS GCE Statistics (MEI)Max Mark a b c d e u

G241 01 Statistics 1 MEI Raw 72UMS 100 80 70 60 50 40 0



No entry in June 2018



AS GCE Quantitative Methods (MEI)Max Mark a b c d e u

G244 Raw 72 58 50 43 36 28 0G244

01 Introduction to Quantitative Methods (Written Paper)02 Introduction to Quantitative Methods (Coursework) Raw 18 14 12 10 8 7 0

UMS 100 80 70 60 50 40 0G245 01 Statistics 1 Raw 72 61 55 49 43 37 0

UMS 100 80 70 60 50 40 0G246 01 Decision Mathematics 1 Raw 72 50 44 38 32 26 0

UMS 100 80 70 60 50 40 0

Published: 15 August 2018 Version 1.0 1

Level 3 Certificate, Level 3 Extended Project and FSMQ raw mark grade boundaries June 2018 series

Level 3 Certificate Mathematics - Quantitative Methods (MEI)Max Mark a b c d e u

G244 A 01 Introduction to Quantitative Methods with Coursework (WrittenPaper) Raw 72 58 50 43 36 28 0

G244 A 02 Introduction to Quantitative Methods with Coursework(Coursework) Raw 18 14 12 10 8 7 0

UMS 100 80 70 60 50 40 0Overall 90 72 62 53 44 35 0

Level 3 Certificate Mathematics - Quantitative Reasoning (MEI)Max Mark a b c d e u

H866 01 Introduction to quantitative reasoning Raw 72 56 49 42 35 28 0H866 02 Critical maths Raw 60 44 39 34 29 24 0

*To create the overall boundaries, component 02 is weighted to give marks out of 72 Overall 144 109 96 83 70 57 0

Level 3 Certificate Mathematics - Quantitative Problem Solving (MEI)Max Mark a b c d e u

H867 01 Introduction to quantitative reasoning Raw 72 56 49 42 35 28 0H867 02 Statistical problem solving Raw 60 40 36 32 28 24 0

*To create the overall boundaries, component 02 is weighted to give marks out of 72 Overall 144 104 92 80 69 57 0

Advanced Free Standing Mathematics Qualification (FSMQ)Max Mark a b c d e u

6993 01 Additional Mathematics Raw 100 56 50 44 38 33 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

For more information about results and grade calculations, see https://www.ocr.org.uk/students/getting-your-results/

Documents

Oxford Cambridge and RSA Tuesday 19 June 2018 – Afternoon · 2018. 8. 28. · Tuesday 19 June 2018 – Afternoon FSMQ ADVANCED LEVEL 6993/01 Additional Mathematics QUESTION PAPER