H240/02 Mark Scheme October 2021 3 2. Subject-specific Marking Instructions for A Level Mathematics 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. H240/02 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. 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. H240/02 Mark Scheme October 2021 5 f 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 3 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 B (MEI) the rubric is not specific about the level of accuracy required, so this statement reads “2 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. 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” or “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. H240/02 Mark Scheme October 2021 6 Question Answer Mark Guidance 1 (a) –4e–4x oe B2 B1 for e–4x seen as part of answer, not in denominator [2] 1 (b) By quotient rule 2 2 ( 1) 2 1 ( 1) x x x x + × − × + B1 B1 B1: correct denominator & 1 correct term in numerator, any form ISW Alternative method – chain rule or –(x + 1)–2 × x2 + (x + 1)–1 × 2x B1 B1 B1 for either term correct ISW (= 2 ( 2) ( 1) x x x + + or 2 2 2 2 1 x x x x + + + ) [2] 2 f(x) is positive on both sides of the 1st root oe B2 B1 for "The graph touches the x-axis" or “repeated root” Curve does not cross the x-axis (near root) or “It is a stationary point” Sign does not change (near the root) No negative value (near the root) Ignore all else, eg “inflection” [2] 3 a + 14d = 88 (i) M1 M1 for one error, eg a + 9d = 88 A1 10 2 (2a + 9d) = 310 (ii) M1 M1 for 15 2 (2a + 14d) = 310 or for one error, eg 5(2a + 18d) = 310 A1 Substitute from (i) into (ii) M1 Attempt substitution, elimination or to verify solutions found BC. Condone one arithmetical error 10 2 (2(88 – 14d) + 9d) = 310 SC for last 2 marks: Correct answers, substitution not seen: B1B0, dep M1M1 176 – 19d = 62 d = 6, a = 4 A1 cao [6] H240/02 Mark Scheme October 2021 7 Question Answer Mark Guidance 4 (a) 6000 B1 [1] 4 (b) 2000 B1f ft their (a) − 4000 [1] 4 (c) Oscillates or Goes up and down. oe B1 Ignore all else Fluctuates. Moves in a cycle NOT “Increases for 1st 6 months then decreases” [1] 4 (d) 30t = 360 M1 May be implied by answer Time to return to initial size = 12 months A1 Allow t = 12, or t = 12 months, or just 12 [2] 4 (e) 4500 = 5000 – 1000cos(30t)o M1 Substitute P = 4500 May be implied by next line cos(30t) o = 0.5 A1 Correct rearrangement 30t = 60 or 300 (both) M1 Attempt 30t = cos-1(their 0.5), giving α and 360 − α. Condone 30t = 5 3 3 , π π 2nd time P = 4500 is when t = 10 A1 or after 10 months. Allow t = 10 months, or just 10 SC. (If not gained 1st M1A1)Correct answer with no or inadequate working and/or T&I: t = 10 stated: B2; t = 10 embedded: B1B0 Alternative methods for 2nd M1A1 30t = 60 or −60 (both) (t = 2 or −2) M1 30t = 60 (t = 2) 2nd time P = 4500 is when t = −2 + 12 = 10 A1 (end of 1st cycle at t = 12) 2nd time P = 4500 is when t = 12 − 2 = 10 30t = 60 (t = 2) M1 6 − 2 = 4; t = 6 + 4 = 10 A1 [4] 4 (f) eg P = 5000 – 1000e–tcos (30t)o P = 5000 – 1000e–ktcos (30t)o (k > 0) B1 or other good answers Answers in words must be equivalent to one of these eg P = 5000 − (1000cos(30t)o)1/t P = 5000 – 1000 t cos (30t)o. (t > 0) [1] H240/02 Mark Scheme October 2021 8 Question Answer Mark Guidance 5 (a) Midpoint AB is (3.5, 5.5); Gradient AB = 1 7 − B1 Both. Allow midpoint = ( 0 7 6 5 2 2 , + + ) ISW Gradient of perpendicular bisector −1/( 1 7 − ) M1 (= 7) y – 5.5 = 7(x – 3.5) oe ISW A1 cao. Correct answer, no working or inadequate working: SC B2 Midpt AB is (3.5, 5.5); Gradient AB = 1 7 − B1 Both (y = 7x + c) 5.5 = 7× 3.5 + c M1 ft their midpt and gradient, NOT 1 7 − y = 7x – 19 A1 cao. Any correct form x2 + (y – 6)2 = (x – 7)2 + (y – 5)2 M1 M1 Attempt expansion –12y + 36 = –14x –10y +49 + 25 ISW A1 cao. Any correct form eg y = 7x – 19 [3] 5 (b) Perpendicular bisector of BC is x + 7y – 17=0 OR of CA is 4y = 3x – 1 B1 Any correct form for another perp bisector Example method, perp bisectors of AB & BC: x + 7(7x – 19) – 17 = 0 (⇒x = 3) M1 Attempt solve simultaneously equations of two perpendicular bisectors. Can be implied Alternative method for 1st two marks Grad BC is 7 so BC & AB perpendicular M1 Hence AC is a diameter B1 Centre is (3, 2) B1 cao. NB, if centre = (3, 2) without clear working, B0M0B1 eg Radius2 = 32 + (6 – 2)2 = 25 M1 Correct method for r2 or r using their centre & A or B or C Equn of circle is (x – 3)2 + (y – 2)2 = 25 or x2 – 6x + y2 – 4y = 12 oe A1ft ISW. ft their centre & radius, dep both M1 marks [5] H240/02 Mark Scheme October 2021 9 Question Answer Mark Guidance 6 (a) 0.1(1 + 1.12 + 1.22 + 1.32 + 1.42) NB. Check working 0.1(1.12 + 1.22 + 1.32 + 1.42 + 1.52 ) B1 Both seen oe [1] 6 (b) 0.79 B1 Not 0.7915 About half way between the last two bounds B1 condone "The mean of the last two bounds" or other sensible or (0.784 + 0.799) ÷ 2 = 0.79 Allow UB and LB are converging towards 0.79 oe Ignore all else The two B1 marks are independent [2] 6 (c) 1.5 δ 0 1 lim δ x x y x → = ∑ B1 for δ 0 lim δ x y x → ∑ . Allow x2 instead of y B1 for limits, dep using Σ not integral. 1.5 δ 0 1 lim δ x y x → ∑ B1B0 7 cos( δ ) cos x x x + − Allow h or other letter for δx throughout = cos cosδ sin sin δ cos x x x x x − − B1 δ 0 lim x→ cos cosδ sin sinδ cos δ x x x x x x − − M1 or δ 0 lim x→ cos( δ ) cos δ x x x x + − or may be seen later. Must include δ 0 lim x→ . as δx → 0: cos δx → 1 or 1 – 2 (δ ) 2 x Allow cos δx = 1 for small δx (or 1 – 2 (δ ) 2 x ) and sinδ δ x x → 1 or sin δx → δx M1 Allow sin δx = δx for small δx Both must be explicitly stated for M1 ( δ 0 lim x→ cos sin δ cos δ x x x x x − − ) If not stated but implied, M0, but can still possibly gain final A1 = –sin x A1 Dep on at least B1M1 gained, and approximations either seen explicitly or seen substituted. and nothing incorrect seen NB. cos x − sin x − cos x = − sin x is incorrect and scores A0 H240/02 Mark Scheme October 2021 10 Question Answer Mark Guidance [4] 8 (a) b = –1, c = 1 B1 or (n + 1)(n2 – n + 1) [1] 8 (b) n + 1 (or n2–n+1) is a factor of K B1 Stated. Allow x instead of n NOT n 3+1 can be expressed as (n+1)(n2−n+1) n > 2 so n + 1 > 1 or n + 1 > 3 or n + 1 ≠ 1 B1 Must see n > 2 (n2–n+1 is a factor of K and n2–n+1>1 or ≠ 1) Allow omission of this step Assume these factors are equal Let n2 – n + 1 = n + 1 M1 ⇒n2 – 2n = 0 n = 0 or 2. A1 n > 2 so both invalid; hence 2 distinct factors A1 Conclusion stated, from correct working seen. Dep at least B1M1 and correct reasoning given Ignore attempted proofs that either factor ≠ K SC: (n+1) > 1 or n + 1 > 3 (or n2–n+1 > 1) B1 (n+1) & (n2−n+1) are factors of K B1 [5] 9 (a) Summary method: OM  = 1 2 (b + c) or b + 1 2 (−b + c) oe B1 Can be implied AM  or MA  attempted in terms of a, b and c M1 May be included in working, eg AX  = 2 3 ( 1 2 (b + c) – a) (= ±( 1 2 (b + c) – a) oe) Not necessarily correct OX  = a + 2 3 AM  or OM  + 1 3 MA  oe attempted in terms of a, b and c M1 Not necessarily correct OX  = 1 3 (a + b + c) A1 or equivalent simplified form H240/02 Mark Scheme October 2021 11 Question Answer Mark Guidance 9 (a) Examples of methods using the above Other correct methods may be seen Allow inadequate notation ctd OM  = 1 2 (b + c) B1 AX  = 2 3 AM  = 2 3 ( 1 2 (b + c) – a) M1 for AM  = ( 1 2 (b + c) – a) implied OX  = a + 2 3 AM  = a + 2 3 ( 1 2 (b + c) – a) M1 = 1 3 (a + b + c) A1 BM  = 1 2 (−b + c) AM  = AO  + OB  + BM  = −a + 1 2 b + 1 2 c B1 M1 Implied OX  = a + 2 3 (−a + 1 2 b + 1 2 c) M1 = 1 3 (a + b + c) A1 OM  = 1 2 (b + c) B1 XM  = 1 3 AM  = 1 3 ( 1 2 (b + c) – a) M1 for AM  = ( 1 2 (b + c) – a) implied OX  = OM  − 1 3 AM  equivalent to OM  + 1 3 MA  = 1 2 (b + c) − 1 3 ( 1 2 (b + c) – a) M1 = 1 3 (a + b + c) A1 H240/02 Mark Scheme October 2021 12 Question Answer Mark Guidance [4] 9 (b) OF  = a + b + c B1* Hence X lies on OF, so AM and OF intersect B1dep Both statements needed. NB dep on B1 [2] 10 (a) Very likely weight will increase with time oe B1 Or eg "Expect weight to increase with time" oe or He is only looking for positive correlation "Foetuses grow" oe [1] Ignore all else 10 (b) H0: ρ = 0 Allow other letters B1 B1B0 for 1 error, eg undefined ρ or 2-tail H1: ρ > 0 where ρ is the correlation B1 coefficient for the population For hypotheses in words, not using parameter: or where ρ is the correlation coefficient H0: There is no correlation between time and weight between time and weight H1: There is positive correlation between time and weight B1B0 But omission of “positive”: B0B0 Comp 0.722 with 0.6851 M1 Reject H0. Condone Accept H1 M1 May be implied by conclusion There is evidence of (positive linear) correlation between time from conception to birth and weight of new-born babies A1 Allow without "positive” and without “linear" In context, not definite Or eg It appears that birth weight increases with time (from conception to birth) [5] 11 (a) Population large oe B1 or, eg Would take too long to contact all students. H240/02 Mark Scheme October 2021 13 Question Answer Mark Guidance [1] NOT “Easier” 11 (b) eg: Includes students from all years (or ages) Numbers in years in correct proportions B1 or: Different years may have different numbers of students Different years might like different music NOT “It’s more representative” or “Takes all students into account” “You get a range of people” “It avoids bias” [1] 11 (c) 21 + 2×4.2 = 29.4 B1 Allow “30 is more than 2 sds away from the mean” 22.9 + 1.5(22.9 – 18.0) = 30.25 B1 Allow “30 is less than 1.5×IQR from UQ” Unclear whether 30 is an outlier B1 or eg "It depends which definition you use." Any comment implying uncertainty Ignore comments about mean ± 3 sds or mean ± 1 sd [3] 11 (d) NB Allow 2 sf throughout H0: μ = 20 B1 Allow other letters, not X unless defined. Not X H1: μ > 20 where μ = pop mean time spent oe B1 B1B0 for 1 error eg 2-tail or: 2-tail B1B0 μ = sample mean implied B1B0 undefined μ B1B0 Not include value 20 B0B0 not in terms of parameter B1B0 eg H0 = 20 etc: B0B0 X ~ N(20, 2 4.2 60 ) and X = 21 M1 Correct distribution and value of X . stated or implied eg by 0.0326 or 0.967 or 20.9 or 1.84 or0.000335 even if within incorrect statement eg P(X = 21) = 0.0326 Condone 2 2 2 4.2 4.2 60 60 or or 4.2 60 P( X > 21) = 0.0326 A1 BC Allow 2 sf, ie 0.033 Compare 0.05 A1 Dep 0.0326 or 1.84 or 0.9674 or P(X > 21) or P(X > 21) soi Must compare like with like, H240/02 Mark Scheme October 2021 14 Question Answer Mark Guidance eg NOT prob cf z-value or large prob cf small prob or CV cf wrong end of acceptance region 11 (d) Alternative methods for M1A1A1 ctd or 20 4 2 60 a− ÷ . = 1.645 (a = 20.9) M1 Condone 2 2 2 4.2 4.2 60 60 or or 4.2 60 CV = 20.9 A1 21 > 20.9 or 21 not in acceptance region A1 or 21 20 4 2 60 − ÷ . (= 1.84) M1 Condone 2 2 2 4.2 4.2 60 60 or or 4.2 60 zcalc = 1.84 A1 Compare 1.645 A1 X ~ N(20, 2 4.2 60 ) and X = 1260 60 or 21 M1 Condone 2 2 2 4.2 4.2 60 60 or or 4.2 60 P( X < 21) = 0.9674 A1 BC Compare 0.95 A1 Reject H0 Condon Accept H1 M1 Dependent on clearly valid comparison of like with like. Dep 0.0485 or 0.951 or 20.9 or 1.84 or P(X > 21) or P(X > 21) soi May be implied by conclusion, eg “There is evidence that mean time is > 20 hours” M1A1 There is evidence that (mean) time spent is > 20 hours A1f In context, not definite; eg “Mean time is > 20 hours": A0 or eg there is evidence to support Zac’s belief But “There is evidence to reject H0 and that mean time is > 20 h” H240/02 Mark Scheme October 2021 15 Question Answer Mark Guidance M1A1 [7] Allow opposite conclusion, ft their values, if above conditions met NB Use of 4.22/60 as sd gives p = 0.000335 and loses 2nd A1. But potentially can score all the other 6 marks 12 (a) 0 B1 Allow 0% [1] 12 (b) B1 B1 B1 [3] Ignore extra branches if no probabilities or p = 0 B2: 8 correct branches and probs OR names, no extra branches B2: 7 correct branches, probs and names, no extra branches B1: 8 correct branches without probs & names. No extra branches B1: 6 correct branches, probs and names. Ignore extra branches Ignore products at ends 12 (c) 0.3×0.3 + 0.7×0.3×0.3 + 0.3×0.7×0.3 M1 All correct M2 ft their diagram or 0.09 + 0.063 + 0.063 oe M1 Two products correct M1 ft their diagram = 27 125 or 0.216 A1 SC Correct answer with no working: B2 [3] 12 (d) 0.7 × 0.3 + 0.3 × 0.7 or 0.21 + 0.21 M1 or 1 – (0.72 + 0.32) or 1 − (0.49 + 0.09) Condone missing brackets or 0.7×0.3×0.7 + 0.7×0.3×0.3 + 0.3×0.7×0.7 + 0.3×0.7×0.3 oe or 2 × 0.147 + 2× 0.063 oe Wholly correct method ft their diagram = 21 50 or 0.42 A1 SC Correct answer with no working: B1 [2] 12 (e) P(B wins and 3 points) ft their diagram for M-marks = 0.7×0.3×0.3 + 0.3×0.7×0.3 or 2 × 0.063 M2 soi M1 for one correct product or 0.063 (= 0.126 oe) P(B wins & 3 points) P(B wins) = '0.126' '0.216' M1 Must attempt P(B wins & 3 points) Their (c) NOT (d) or 0.3 0.3+0.7 0.3 0.3+0.3 0.7 0.3 × × × × × A 0.7 A (A) 0.7 0.3 B (0.7) 0.3 B A 0.7 (0.3) A 0.7 0.3 (B) B 0.3 B H240/02 Mark Scheme October 2021 16 Question Answer Mark Guidance = 7 12 A1 Allow 0.583 (3 sf) SC no working 7 12 B2 But minimal working is OK, eg 2 0.063 0.216 × = 7 12 M2M1A1 [4] 13 (a) A: High private and low public B2 Allow B1 for A or B or C with one correct factor only. Ignore else OR B: Low bicycle and low public OR B: Low bicycle and high private All answers can be implied OR C: Low bicycle and high private Ignore all else [2] 13 (b) Pie charts allow comparison of proportions B1 NOT: Bar charts don’t show proportions unless also state pies do. Pie charts show proportions oe Assume “They” means pie charts. NOT: It’s easier to compare data Ignore all else Pie charts don’t easily show which is greatest Allow percentages instead of proportions [1] 13 (c) (i) C and D B1 Both, no others Larger (or high or most) proportion public transport B1 Dep applied to at least one of C and D, even if other LA mentioned Can be implied. Ignore all else B-marks are independent [2] 13 (c) (ii) LDS says “The method of travel used is for the longest part, by distance, of the usual journey to work” Correct answers therefore should assume users of P&R will report only private, not public Method of travel is the type used for the longest part of the journey B1 "Disagree" may be implied Most people using Park-and-Ride would still H240/02 Mark Scheme October 2021 17 Question Answer Mark Guidance say they were using private transport. B1 independent Ignore all else People will still have to travel by car, so the results won’t be very different B2 13 (c) (ii) ctd Alternatives, assuming (incorrectly) users of P&R will report both methods or just public: Public transport increase or private decrease B1 “Agree” may be implied B0 Ignore all else P&R will result in more people using private B1 Not just “Increase in private” without justification transport, so this will show an increase B0 Users of P&R will use both public and private, so not clear B1 B0 No, because other changes might have been made that affect the proportions. B2 If P&R users report both, or just public, then yes, public will show increase B2 Recognition of issue as to whether P&R users should report private or public B1 Other sensible answers may be seen not covered in this MS Hence unclear whether change will show B1 [2] 14 (a) k(1 + 1 1 1 2 3 4 + + ) = 1 M1 k × 25 12 = 1 or eg 25 3 k = 4 or 25k = 12. Correct equation involving multiple of k hence k = 12 25 AG A1 Must see previous line and answer H240/02 Mark Scheme October 2021 18 Question Answer Mark Guidance or verification: 12 25 + 6 25 + 4 25 + 3 25 = 1 M1 A1 [2] 14 (b) 1 2 3 4 12 25 6 25 4 25 3 25 B1 or equivalent exact values [1] 14 (c) (3, 1, 1) (4, 1, 1) (4, 2, 1) (4, 1, 2) M1 At least three of these seen or implied. No extras or repeats. 4 25 ×( 12 25 )2 + 3 25 ×( 12 25 )2 + 3 25 × 6 25 × 12 25 + 3 25 × 12 25 × 6 25 oe M1 At least two correct terms, no incorrect coefficients; ft their table. Allow in terms of k = 288 3125 or 0.09216 A1 Allow 0.0922 (3 sf) [3] 14 (d) (1, 1, 1, 1, 3) (1, 1, 1, 2, 2) B1 B1 B1B1 for both sets in any order, without extras. Both soi. B1 for both sets in any order, with extras. ( 12 25 )4 × 4 25 × 5 + ( 12 25 )3 × ( 6 25 )2 × 5C2 oe M1 ( 12 25 )4 × 4 25 or ( 12 25 )3 × ( 6 25 )2 oe seen. Ignore coeffs. ft their table A1 For either ( 12 25 )4 × 4 25 × 5 or ( 12 25 )3 × ( 6 25 )2 × 5C2 oe ft their table Alternative method for M1A1 or ( 12 25 )4× 4 25 ×(4+1)+( 12 25 )3×( 6 25 )2×(4C2+4) M1 oe A1 For either ( 12 25 )4× 4 25 ×(4+1) or ( 12 25 )3×( 6 25 )2×(4C2+4) oe = 41472 390625 or 0.10616832 A1 Allow 0.106 (3 sf) [5] 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