Y420/01 Mark Scheme October 2021 3 2. Subject-specific Marking Instructions for AS/A Level Further Mathematics B (MEI) a a Annotations must be used during your marking. For a response awarded zero (or full) marks a single appropriate annotation (cross, tick, M0 or ^) is sufficient, but not required. For responses that are not awarded either 0 or full marks, you must make it clear how you have arrived at the mark you have awarded and all responses must have enough annotation for a reviewer to decide if the mark awarded is correct without having to mark it independently. It is vital that you annotate standardisation scripts fully to show how the marks have been awarded. Award NR (No Response) - if there is nothing written at all in the answer space and no attempt elsewhere in the script - OR if there is a comment which does not in any way relate to the question (e.g. ‘can’t do’, ‘don’t know’) - OR if there is a mark (e.g. a dash, a question mark, a picture) which isn’t an attempt at the question. Note: Award 0 marks only for an attempt that earns no credit (including copying out the question). If a candidate uses the answer space for one question to answer another, for example using the space for 8(b) to answer 8(a), then give benefit of doubt unless it is ambiguous for which part it is intended. 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 always 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, escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. If you are in any doubt whatsoever you should contact your Team Leader. Y420/01 Mark Scheme October 2021 4 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, e.g. 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 method mark may usually be implied by a correct answer unless the question includes the DR statement, the command words “Determine” or “Show that”, or some other indication that the method must be given explicitly. 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. Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. 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, what is acceptable will be detailed in the mark scheme. If this is not the case, please escalate the question to your Team Leader who will decide on a course of action with the Principal Examiner. 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. Y420/01 Mark Scheme October 2021 5 f Unless units are specifically requested, there is no penalty for wrong or missing units as long as the answer is numerically correct and expressed either in SI or in the units of the question. (e.g. lengths will be assumed to be in metres unless in a particular question all the lengths are in km, when this would be assumed to be the unspecified unit.) We are usually quite flexible about the accuracy to which the final answer is expressed; over-specification is usually only penalised where the scheme explicitly says so.  When a value is given in the paper only accept an answer correct to at least as many significant figures as the given value.  When a value is not given in the paper accept any answer that agrees with the correct value to 2 s.f. unless a different level of accuracy has been asked for in the question, or the mark scheme specifies an acceptable range. NB for Specification A the rubric specifies 3 s.f. as standard, so this statement reads “3 s.f” Follow through should be used so that only one mark in any question is lost for each distinct accuracy error. Candidates using a value of 9.80, 9.81 or 10 for g should usually be penalised for any final accuracy marks which do not agree to the value found with 9.8 which is given in the rubric. g Rules for replaced work and multiple attempts:  If one attempt is clearly indicated as the one to mark, or only one is left uncrossed out, then mark that attempt and ignore the others.  If more than one attempt is left not crossed out, then mark the last attempt unless it only repeats part of the first attempt or is substantially less complete.  if a candidate crosses out all of their attempts, the assessor should attempt to mark the crossed out answer(s) as above and award marks appropriately. 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 or B mark in the question. Marks designated as cao may be awarded as long as there are no other errors. If a candidate corrects the misread in a later part, do not continue to follow through. E marks are lost unless, by chance, the given results are established by equivalent working. Note that a miscopy of the candidate’s own working is not a misread but an accuracy error. i If a calculator is used, some answers may be obtained with little or no working visible. Allow full marks for correct answers provided that there is nothing in the wording of the question specifying that analytical methods are required such as the bold “In this question you must show detailed reasoning”, or the command words “Show” and “Determine. Where an answer is wrong but there is some evidence of method, allow appropriate method marks. Wrong answers with no supporting method score zero. If in doubt, consult your Team Leader. j If in any case the scheme operates with considerable unfairness consult your Team Leader. Y420/01 Mark Scheme October 2021 6 Question Answer Marks AOs Guidance 1 (a) 1 (2 1)(2 1) 2 1 2 1 A B r r r r = + − + − + 1 (2 1) (2 1) A r B r = + + − 1 1 2 2 r A =  = 1 1 2 2 r B = −  = − M1 A1 A1 [3] 1.1a 1.1 1.1 or cover-up method 1 (b) ( ) 1 1 1 1 1 1 (2 1)(2 1) 2 2 1 2 1 n n r r r r r r = = = − − + − +   ( ) 1 1 1 1 1 1 1 1 1 ... 2 3 3 5 2 3 2 1 2 1 2 1 n n n n = − + − + + − + − − − − + ( ) 1 1 1 2 2 1 n = − + 2 1 n n = + M1* M1de p* A1 A1 [4] 1.1 1.2 1.1 1.1 showing cancellation clearly SC B3 For fully correct summation with A = - ½ and B = ½ Y420/01 Mark Scheme October 2021 7 Question Answer Marks AOs Guidance 2 DR 2 d 1 6 2 d 1 4 y x x =   − 1 4 3 4 d 1 when , 6 2 8 3 d y x x = =   = B1 M1 A1 A1 [4] 1.1 1.1 1.1 1.1 𝑘 √1−4𝑥2 Chain rule (x 2) Fully correct 𝑑𝑦 𝑑𝑥 Alternative solution DR sin 𝑦 6 = 2𝑥 𝑑𝑥 𝑑𝑦= 1 12 cos 𝑦 6 𝑑𝑦 𝑑𝑥= 12 cos𝑦 6 and 𝑥= 1 4 , 𝑦= 𝜋 →𝑑𝑦 𝑑𝑥= 8√3 M1 M1 A1 A1 For 𝑑𝑥 𝑑𝑦= 𝑘cos 𝜋 6 Using chain rule (× 1 6) Or 2 d 1 6 2 d 1 4 y x x =   − Y420/01 Mark Scheme October 2021 8 Question Answer Marks AOs Guidance 3 (a) DR 1 8 z = 1 3 arg( ) 4 z  = B1 E1 [2] 1.1 1.1 Must see some reasoning for arg(z) 3 (b) DR 1 2 8 2 2 z z = = ( ) 1 2 3 7 arg 4 6 12 z z    = − = so ( ) 1 2 7 7 2 cos isin 12 12 z z   = + B1 M1 A1 B1 [4] 1.1 1.1 1.1 1.1 ( ) 1 1 2 2 arg arg( ) arg( ) z z z z = − used 4 DR 1 2 1 1 1 mean d 1 ( 1) 1 4 x x − = −− +   1 2 1 1 4 1 1 d 8 x x − = +   1 1 1 2arctan 2 8 x −   =   1 arctan 2 0.554 2 =  B1 M1 A1 A1 [4] 1.1 1.1 1.1 1.1 or 2 2 2 1 1 1 2 d 2 2 1 u x u u − =  +  2 2 1 arctan 4 u −   =   M1 for rearranging denominator correctly into appropriate form or for karctan2x Y420/01 Mark Scheme October 2021 9 Question Answer Marks AOs Guidance 5 (a) 2 2 1 ln(1 2 ) 2 (2 ) 2 2 2 x x x x x +  − = − B1 [1] 1.1 5 (b) ln(1.2) 0.18  % error = 0.18 ln(1.2) 100 ln(1.2) −  = ()1.27% B1ft M1 A1 [3] 1.1 1.1 1.1 5 (c) 1 1 2 2 1 2 1 x x − −   the Maclaurin series is not convergent for this approximation when x = 1 M1 A1 [2] 1.1 2.3 substitute x = 1 in quadratic 6 ( )( ) ( ) 2 1 2 2 2 2 2 x x y y x y + = − − substituting y mx = and 2 2 ( 2 ) x y m x y − = + 2 2 ( 2 ) x mx m x mx − = + 2 2 3 2 0 m m  + − = 1 2 2, m  = − M1 M1 A1 A1 [4] 1.1 2.1 1.1 2.2a Could see mx instead of y in the initial matrix multiplication for this mark Y420/01 Mark Scheme October 2021 10 Question Answer Marks AOs Guidance 7 1 1 0 1 1 2 1 4 2 2 r r r − = + = = −  so true for n = 1 Assume true for n = k so 1 1 1 2 4 2 2 k r k r r k − − = + = −  1 1 1 1 2 1 4 2 2 2 k r k k r r k k + − − = + + = − +  2 4 1 4 2k k k + − − = − 1 2 4 2k k + + = − so true for n = k + 1 So true for n = 1 and if true for n = k then true for n = k + 1  true for all n B1 M1 M1 M1 A1 A1 [6] 2.1 2.1 2.1 1.1 2.2a 2.2a Dependent on all previous marks awarded Y420/01 Mark Scheme October 2021 11 Question Answer Marks AOs Guidance 8 (a) 1 9 ( ) 4     −   = − 𝛼= (±) 3 2 Sum of roots 1 1   = + = 2 1 0    − + = 1 i 3 1 1 i 3 , 2 2     = = so roots are 3 1 i 3 and 2 2   M1 A1 M1 A1 A1 2.1 2.2a 2.1 2.1 2.2a Alternative solution ( ) 1 ( )( )( ) x x x x     − + − − ( ) ( ) 4 3 2 2 2 2 1 1 (1 ) x x x x        = + + + − − + − 2 9 3 4 2    =  =  2 1 1 1 0     + =  − + = 1 i 3 1 1 i 3 , 2 2     = = so roots are 3 1 i 3 and 2 2   M1 A1 A1 M1 A1 [5] Y420/01 Mark Scheme October 2021 12 Question Answer Marks AOs Guidance 8 (b) Sum of pairs ( ) 1         = − + + − − + 2 5 1 5 4 p  = − = −  = − Sum of triples ( ) 2 1 9 9 1 4 4    = − + = − = − so 9 q = M1 A1 M1 A1 2.1 2.2a 2.1 2.2a Unsimplified simplified Alternative solution ( )( )( )( ) 3 3 1 i 3 1 i 3 2 2 2 2 x x x x + − − + − − ( ) 2 2 9 ( 1) 4 x x x = − − + 4 3 2 5 9 9 4 4 4 x x x x = − − + − so p = 5 and q = 9 M1 A1 A1 A1 Award for either 𝑥2 𝑜𝑟 𝑥 coefficient correct [4] Y420/01 Mark Scheme October 2021 13 Question Answer Marks AOs Guidance 9 (a) M1 A1 [2] 1.1 1.1 ( )( ) ( ) 1 0 0 1 1 0 0 1 1 0 2 1 0 0 1 1 0 2 1 1 − − − = − − − A′ (1, 2), B′ (1, 1), C′ (0, 1) plotted correctly SC B1 For unlabelled diagram with no working 9 (b) (i) det 1 1 0 ( 2) 1 = −−− = − M B1 [1] 1.1 9 (b) (ii) area is preserved orientation is reversed B1 B1 [2] 1.1 1.1 9 (c) (i) Reflection in y axis then shear invariant line y-axis, mapping (1, 0) to (1, 2) B1 M1 A1 3.1a 3.1a 1.1 oe, e.g. mapping (1, 0) to (1, 2) Alternatively Shear invariant line y-axis, mapping (1, 0) to (1, 2) then reflection in y-axis M1 A1 B1 [3] 9 (c) (ii) Reflection: ( ) 1 0 0 1 − Shear: ( ) 1 0 2 1 ( )( ) ( ) 1 0 1 0 1 0 2 1 0 1 2 1 − − = − B1 B1 B1 [3] 1.1 1.1 2.2a Or shear ( ) 1 0 2 1 − first then reflection 1 0 0 1 −       ( )( ) ( ) 1 0 1 0 1 0 0 1 2 1 2 1 − − = − − Must match the order described in ci for final mark Y420/01 Mark Scheme October 2021 14 Question Answer Marks AOs Guidance 10 (a) B1 B1 [2] 1.1 1.1 z = 1 other two roots forming correct equilateral triangle 10 (b) (i) 1 3i 2 + = ( ) arg 1 3i 3  + = i 9 3 2e z  = 7i 9 3 2e  5i 9 3 2e  − B1 B1 B1ft B1ft B1ft [5] 1.1 1.1 2.5 1.1 1.1 if roots correct but in cis form, withhold one mark only 10 (b) (ii) Rotated through 20 (oe) Enlarged by 3 2 B1 B1 [2] 3.1a 3.1a 10 (b) (iii) ( ) ( ) 5i 2i 9 3 5i i 7i 9 9 9 2i 3 3 3 3 3 3 2 e 1 e 2 e 2 e 2 e 1 e       − − − + + = − ( ) 5i 9 2i 3 2i 3 2 e 1 e 0 1 e    − − = = − M1 A1 [2] 3.1a 2.2a sum of GP formula May use cos 𝜃+ 𝑖sin 𝜃 but must take out factor for M1 10 (b) (iv) Imaginary part of sum of roots is zero ( ) 7 5 sin sin sin 0 9 9 9     + + − = sin20 sin140 sin( 100 ) sin100  + = − − =  M1 A1 [2] 2.1 2.2a AG 1 Y420/01 Mark Scheme October 2021 15 Question Answer Marks AOs Guidance 11 (a) (i) 1 1 2 ( 3) ( 2)  = + + − − u .v 8  = + B1 [1] 1.1 11 (a) (ii) 1 4 1 2 2 3 3 2 2 1               =  = −      − − −      u v oe i, j, k form B1 B1 [2] 1.1a 1.1 For one of i,j,k correct For all i,j,k correct 11 (b) (i) angle between 2 3 + − i j k and 2 2 + − i j k taking 2, = 10 cos 14 9 =  = 27.0 or 0.472 rad M1 M1 A1 [3] 3.1a 1.1 1.1 11 (b) (ii) direction vectors are 3 3 + − i j k and 2 2 , + − i j k so take 3 = with 3, 4 3 5 =  = + + u v i j k 4 3 0 1 20 distance 3 0 4 2 2 50 5 2 5 2 ( 2) −     = − = =   −−   . M1 M1 A1 [3] 3.1a 1.1 1.1 Y420/01 Mark Scheme October 2021 16 Question Answer Marks AOs Guidance 12 Let DOA , COA    =  = z w + is represented by C arg( ) z w   + = + BOC =  (diagonal of rhombus bisects BOA) arg arg ( 2 ) z w    + = + + 2( )   = + so 1 2 arg( ) (arg arg ) z w z w + = + B1 M1 M1 A1 [4] 3.1a 2.1 2.1 2.2a finds arg arg z w + in terms of ,  AG 13 2 2 d d 2 3 2e d d x y y y x x + − = AE: 2 2 3 0   + −= 3, 1   = − CF: 3 e e x x y A B − = + PI: ex y Cx = d (e e ) d x x y C x x = + 2 2 d (2e e ) d x x y C x x = + (2e e ) 2 (e e ) 3 e 2e x x x x x x C x C x Cx  + + + − = 1 2 4 2 C C  =  = GS: 3 1 2 e e e x x x y A B x − = + + M1 A1 M1 A1 A1 M1 A1 [7] 2.1 2.1 2.1 1.1 1.1 2.1 2.2a Y420/01 Mark Scheme October 2021 17 Question Answer Marks AOs Guidance 14 (a) Let cos 2sin cos( ) R     + = − cos cos sin sin R R     = + cos 1, sin 2 R R   = = 5 R = tan 2 = 1.107 = so 5 cos( 1.107) r a  = − r is maximum when cos( 1.107) 1 − = 1.107 rad   = polar coordinates are 5 , 1.107 a     M1A1 M1 A1 M1 A1 A1 3.1a 1.1 1.1 1.1 3.1a 1.1 1.1 or arctan(2) or arctan(2) Alternative solution d ( sin 2cos ) d r a   = − + r is maximum when d 0 d r = 2cos sin 0   − + = tan 2 1.107    =  = when 1.107, 5 or 2.236 r a a = = polar coordinates are [2.236a, 1.107] M1A1 M1 M1 A1B1 B1 2.24 or better [7] 14 (b) (i) 2 cos 2 sin r ar ar   = + 2 2 2 x y ax ay  + = + 2 2 2 5 1 2 4 ( ) ( ) x a y a a  − + − = This is the cartesian equation of a circle radius 1 2 5a M1 M1 A1 M1 A1 A1 [6] 3.1a 3.1a 1.1 2.1 2.2a 3.2a attempt to find cartesian eqn multiplying by r completing the square Y420/01 Mark Scheme October 2021 18 Question Answer Marks AOs Guidance 14 (b) (ii) centre 1 2 5 , 1.107 a     B1 [1] 1.1 Y420/01 Mark Scheme October 2021 19 Question Answer Marks AOs Guidance 15 4 7 1 2 5 4 ( 17) ( 9) 7 7 2 3 1 k k − − = −− − − +  117 9k = + so det = 0 when k = −13 4 13 7 4 (1) 2 5 (2) 2 3 2 (3) x y z x y z l x y z − − + = − + = + + = (1) 2 (3): 7 9 8 (3) 2 (2): 7 9 2 2 y z y z l +  − + = − − = − so 2 2 8 5 l l −= = M1 M1 A1 M1 M1 A1 [6] 3.1a 1.1 1.1 1.1 1.1 1.1 finding det of matrix of coeffs setting det = 0 k = 13 finding eqn in 2 variables finding 2nd eqn in 2 variables l = 5 or B2 for use of linear dependency to find k or M2 for 2(2)  3(3) Y420/01 Mark Scheme October 2021 20 Question Answer Marks AOs Guidance Alternative method 16 (a) 2 2 e e LHS 2 u u − + = ( ) 2 e e RHS 1 2 2 u u − − = + ( ) 2 2 e 2 e 1 2 4 u u − −+ = + ( ) 2 2 2 2 e 2 e e e 1 2 2 u u u u − − − + + = + = = LHS B1 B1 B1 B1 [4] 2.1 2.1 2.1 2.2a 16 (b) Let d 2sinh 2cosh d x x u u u =  = 2 arsinh1 2 2 2 2 0 0 4sinh d 2cosh d 4 4 4sinh x u x u u x u = + +       arsinh1 2 0 4sinh 2cosh d 2cosh u u u u =  arsinh1 2 0 4sinh d u u =  M1 A1 M1 A1 3.1a 1.1 2.1 2.1 2 2 1 sinh cosh u u + = used arsinh1 0 (2cosh2 2) d u u = −    arsinh1 0 sinh2 2 u u = − ( ) arsinh1 ln 1 2 = + ( ) ( ) 2ln(1 2) 2ln(1 2) 1 2 e e 2ln 1 2 + − +  − − + ( ) ( ) ( ) 2 2 ln(1 2) ln(1 2) 1 2 e e 2ln 1 2 − + +   = − − +   ( ) ( ) ( ) 2 2 1 2 1 2 1 2 2ln 1 2 −   = + − + − +   M1 A1 B1 M1 M1 2.1 2.1 2.1 2.1 2.1 2 cosh2 1 2sinh u u = + used soi or sinh2 2sinh cosh u u u = M1 2 2sinh 1 sinh u u = + M1 2 2 = Y420/01 Mark Scheme October 2021 21 Question Answer Marks AOs Guidance ( ) ( )( ) 1 1 1 2 1 2 2 1 1 2 1 2 1 2 − − + = = = − + + − giving ( ) ( )   ( ) 1 2 3 2 2 3 2 2 2ln 1 2 = + − − − + ( ) 2 2 2ln 1 2 = − + A1 2.2a AG Alternative for last 6 marks arsinh1 2 0 (e e ) d u u u − = −  arsinh1 2 2 0 (e 2 e ) d u u u − = − +  arsinh1 2 2 0 1 1 2 2 e 2 e u u u −   = − −   ( ) arsinh1 ln 1 2 = + ( ) ( ) 2 2ln 1 2 e 1 2 + = + ( ) ( ) ( ) 2 2 2ln 1 2 e 1 2 2 1 − − + = + = − giving ( ) 3 2 2 3 2 2 2ln 1 2 2 2 + − − − + ( ) 2 2 2ln 1 2 = − + M1 A1 B1 M1 M1 A1 1 2 sinh (e e ) u u u − = − used AG [10] Y420/01 Mark Scheme October 2021 22 Question Answer Marks AOs Guidance 17 (a) (i) at litres of chemical in, bt litres of mixture out amount of liquid in container = 1 ( ) a b t + − litres  proportion of chemical = 1 ( ) x a b t + − M1 A1 [2] 3.1b 2.4 AG 17 (a) (ii) rate of chemical in = a litres/hr rate of chemical out = 1 ( ) bx a b t + − litres/hr d d 1 ( ) x bx a t a b t  = −+ − d d 1 ( ) x bx a t a b t  + = + − M1 A1 [2] 3.3 3.3 AG 17 (b) (i) d d x ax a t + = 1 d d 1 x a t x  = −   ln(1 )x at c − − = + when 0, 0 0 t x c = =  = 1 e at x − − = 1 e at x − = − M1 A1 B1 A1 [4] 1.1 1.1 3.3 3.4 separating variables or IF eat e e at at x c = + c = 1 17 (b) (ii) 1 2 1 e a − = − a = ln 2= 0.693 rate of inflow = 0.693 l/min M1 A1 [2] 3.3 3.4 Y420/01 Mark Scheme October 2021 23 Question Answer Marks AOs Guidance 17 (c) (i) when 1 t a = the container has no liquid left B1 [1] 3.5b oe e.g. volume negative when 1 t a  17 (c) (ii) d 2 d 1 x ax a t at + = − 2 d 1 IF e a t at −  = 2ln(1 ) 2 e (1 ) at at − − − = = − ( ) 2 2 d (1 ) (1 ) d x at a at t − −  − = − 2 1 (1 ) (1 ) x at at c − −  − = − + when t = 0, x = 0  c = 1 2 (1 ) (1 ) (1 ) x at at at at  = − − − = − 2 d 1 2 0 when d 2 x a a t t t a = − = = 1 1 1 2 4 4 x = − = so maximum amount of chemical is 0.25 l B1 M1 A1 M1 A1 M1 A1 M1 A1 [9] 1.1 3.1a 1.1 1.1 1.1 3.4 1.1 3.4 3.2a or completing the square OCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA OCR Customer Contact Centre Education and Learning Telephone: 01223 553998 Facsimile: 01223 552627 Email: general.qualifications@ocr.org.uk www.ocr.org.uk For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored