GCSE (9–1) Mathematics J560/06: Paper 6 (Higher tier) General Certificate of Secondary Education Mark Scheme

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GCSE (9–1) Mathematics J560/06: Paper 6 (Higher tier) General Certificate of Secondary Education Mark Scheme Oxford Cambridge and RSA Examinations H GCSE (9–1) Mathematics J560/06: Pap... er 6 (Higher tier) General Certificate of Secondary Education Mark Scheme for DRAFT November 2021Oxford 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 2021 DRAFTH J560/06 GCSE MATHEMATICS Paper 6 (Higher tier) Monday 8 November 2021 – Morning 1 hour 30 minutes MARK SCHEME MAXIMUM MARK 100 Post-standardisation Version: Final Last updated: 21/11/21 (FOR OFFICE USE ONLY) This document consists of 23 pages DRAFTJ560/06 Mark Scheme November 2021 2 Annotations available in RM Assessor. These must be used whenever appropriate during your marking. Annotation Meaning Correct Incorrect Benefit of doubt Follow through Ignore subsequent working (after correct answer obtained), provided method has been completed Method mark awarded 0 Method mark awarded 1 Method mark awarded 2 Accuracy mark awarded 1 Independent mark awarded 1 Independent mark awarded 2 Misread Special case Omission sign Blank page Seen DRAFTJ560/06 Mark Scheme November 2021 3 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. Subject-Specific Marking Instructions 1. M marks are for using a correct method and are not lost for purely numerical errors. A marks are for an accurate answer and depend on preceding M (method) marks. Therefore M0 A1 cannot be awarded. B marks are independent of M (method) marks and are for a correct final answer, a partially correct answer, or a correct intermediate stage. SC marks are for special cases that are worthy of some credit. 2. The following abbreviations are commonly found in GCSE Mathematics mark schemes. - figs 237, for example, means any answer with only these digits. You should ignore leading or trailing zeros and any decimal point e.g. 237000, 2.37, 2.370, 0.00237 would be acceptable but 23070 or 2374 would not. - isw means ignore subsequent working after correct answer obtained and applies as a default. - nfww means not from wrong working. - oe means or equivalent. - rot means rounded or truncated. - soi means seen or implied. - dep means that the marks are dependent on the marks indicated. You must check that the candidate has met all the criteria specified for the mark to be awarded. - with correct working means that full marks must not be awarded without some working. The required minimum amount of working will be defined in the guidance column and SC marks given for unsupported answers. 3. Anything in the mark scheme which is in square brackets […] is not required for the mark to be earned, but if present it must be correct. 4. Unless the command word requires that working is shown and the working required is stated in the mark scheme, then if the correct answer is clearly given and is not from wrong working full marks should be awarded. Do not award the marks if the answer was obtained from an incorrect method, i.e. incorrect working is seen and the correct answer clearly follows from it. DRAFTJ560/06 Mark Scheme November 2021 4 5. Where follow through (FT) is indicated in the mark scheme, marks can be awarded where the candidate’s work follows correctly from a previous answer whether or not it was correct. For questions with FT available you must ensure that you refer back to the relevant previous answer. You may find it easier to mark these questions candidate by candidate rather than question by question. Figures or expressions that are being followed through are sometimes encompassed by single quotation marks after the word their for clarity, e.g. FT 180 × (their ‘37’ + 16), or FT 300 – (their ‘52 + 72’). Answers to part questions which are being followed through are indicated by e.g. FT 3 × their (a). 6. In questions with no final answer line, make no deductions for wrong work after an acceptable answer (i.e. isw) unless the mark scheme says otherwise, indicated by the instruction ‘mark final answer’. 7. In questions with a final answer line and incorrect answer given: (i) If the correct answer is seen in the body of working and the answer given on the answer line is a clear transcription error allow full marks unless the mark scheme says ‘mark final answer’. Place the annotation ✓ next to the correct answer. (ii) If the correct answer is seen in the body of working but the answer line is blank, allow full marks. Place the annotation ✓ next to the correct answer. (iii) If the correct answer is seen in the body of working but a completely different answer is seen on the answer line, then accuracy marks for the answer are lost. Method marks could still be awarded if there is no other method leading to the incorrect answer. Use the M0, M1, M2 annotations as appropriate and place the annotation  next to the wrong answer. 8. In questions with a final answer line: (i) If one answer is provided on the answer line, mark the method that leads to that answer. A correct step, value or statement that is not part of the method that leads to the given answer should be awarded M0 and/or B0. (ii) If more than one answer is provided on the answer line and there is a single method provided, award method marks only. (iii) If more than one answer is provided on the answer line and there is more than one method provided, award marks for the poorer response unless the candidate has clearly indicated which method is to be marked. 9. In questions with no final answer line: (i) If a single response is provided, mark as usual. DRAFTJ560/06 Mark Scheme November 2021 5 (ii) If more than one response is provided, award marks for the poorer response unless the candidate has clearly indicated which response is to be marked. 10. When the data of a question is consistently misread in such a way as not to alter the nature or difficulty of the question, please follow the candidate’s work and allow follow through for A and B marks. Deduct 1 mark from any A or B marks earned and record this by using the MR annotation. M marks are not deducted for misreads. If a candidate corrects the misread in a later part, do not continue to follow through, but award A and B marks for the correct answer only. 11. Unless the question asks for an answer to a specific degree of accuracy, always mark at the greatest number of significant figures even if this is rounded or truncated on the answer line. For example, an answer in the mark scheme is 15.75, which is seen in the working. The candidate then rounds or truncates this to 15.8, 15 or 16 on the answer line. Allow full marks for the 15.75. 12. Ranges of answers given in the mark scheme are always inclusive. 13. For methods not provided for in the mark scheme give as far as possible equivalent marks for equivalent work. If in doubt, consult your Team Leader. 14. If in any case the mark scheme operates with considerable unfairness consult your Team Leader. DRAFTJ560/06 Mark Scheme November 2021 6 Question Answer Marks Part marks and guidance 1 (a) 5120 1 (b) Topozero, Tana, Mweru, Ladoga, Victoria or 986, 3200, 5120, 18 100, 68 900 oe in standard form 2 B1 for Topozero as smallest or Victoria as largest or all in correct reverse order 9.86  102, 3.20  103, 5.12  103, 1.81  104, 6.89  104 condoning superfluous zeros and slip in index (c) 1.5 × 104 nfww isw 4 B3 for 15000 oe or 1.49[0..] × 104 or B2 for 14900 oe or M1 for figs 181 – figs 32 If 0 scored SC1 for their value correctly rounded to 2 significant figures eg 15000 may be 15 × 103 Subtraction may be implied by figs 15 or figs 149 Their unrounded value must be seen 2 (a) 285 2 M1 for 760 ÷ (2 + 3 +3) soi by 95 (b) 24 2 M1 for 2 36 3  oe Allow (0.66 or 0.7)  36 for M1 only 3 2x + 7 as final answer 2 B1 for each part or M1 for 3x + 6 or -x + 1 4 (a) Triangle at (-8, 6), (-8, 2), (0, 6) 2 B1 for reflection in x = k or in y = 0 Mark intention, condoning freehand DRAFTJ560/06 Mark Scheme November 2021 7 Question Answer Marks Part marks and guidance (b) Enlargement 1 4 or 0.25 (0, -6) 3 B1 for each element Marks spoilt if extra transformations Condone omission of brackets Accept centre as a vector 5 (a) 0.14, 0.09, (0.19), 0.2[0], 0.13, 0.25 2 B1 for three or four correct relative frequencies in the correct place Accept fractions (b) (i) [Unbiased dice] would have each [rf=] 0.16-0.17 or [Unbiased dice] would have each [f=] 50 or comment about very unequal [relative] frequencies and implied comparison 1 Accept “about 0.16” Accept “about 50” Not enough to say one number was rolled the most. Must say 6 [and 4] or some numbers are much higher or 2 or 5 or some numbers are much lower (ii) need larger sample oe 1 DRAFTJ560/06 Mark Scheme November 2021 8 Question Answer Marks Part marks and guidance 6 5 : 6 nfww 4 B3 for 5kn : 6kn k>0 or equivalent correct unsimplified ratio seen OR M1 for two ratios with a common number of mints implied by … : 10k and 10k : … seen, k>0 with one correct ratio or for 2.5n : 5 seen A1 for 5kn : 10k : 6kn Accept for all part marks n replaced by a consistent integer Eg 2.5n : 3n or 5n : 6n or 10 : 12 etc May be seen as two separate ratios Eg 5n : 10 and 10 : 6n or 10 : 20 and 20 : 12 etc For M1 the examples just require the common 10 or the common 20 etc 7 (a) Ruled bisector of angle ABC to reach CD with construction arcs 2 B1 for correct ruled bisector at least 2cm long by eye with no construction arcs or correct construction arcs with no/wrong bisector drawn Tolerance ±2˚ Construction arcs on AB and on BC and two intersecting arcs from these (b) Arc, centre C, radius 5 cm, intersecting their line twice or intersecting BC and CD or two points marked on their line that are 5 cm from C Locus of line within arc from C rad indicated 2 1dep B1 for any arc, centre C, intersecting their line at least once or intersecting BC or CD or short arc (at least 1 cm), centre C, radius 5 cm Dep on at least B1 in (a) and B1 in (b) Tolerance 4.8 – 5.2 cm Max B1 for freehand, all within template DRAFTJ560/06 Mark Scheme November 2021 9 Question Answer Marks Part marks and guidance 8 (a) 54 nfww 4 B3 for 90 min and 144 min or for 0.9h or B2 for 90 min or 144 min or for A: 1.5h and B 2.4h or M1 for evidence of dist = speed /time For B3 and B2 accept 1 h 30 min and 2 h 24 min (b) 1000x/3600 oe isw 2 B1 for 1000x [m/h] or x/3600 [km/s] or x1000/3600 oe For 2 marks, final answer must not have any units within the expression isw wrong simplification after correct answer. Accept x/3.6 and (0.277 to 0.28)x For B1 allow x × 1000 or x ÷ 3600 or these clearly implied in a longer calculation 9 (a) 59.5 60.5 2 B1 for either one correct or both correct DRAFT but reversedJ560/06 Mark Scheme November 2021 10 Question Answer Marks Part marks and guidance (b) (i) Accept any correctly matched wall and 6 × cupboard where values quoted satisfy: wall < 6 × cupboard where 362.5 ≤ wall < 363 and 362.5 < 6 × cupboard ≤ 363 OR wall ÷ 6 < cupboard where 362.5 ≤ wall < 363 and 60.416̇ to 60.42 < cupboard ≤ 60.5 OR wall ÷ cupboard < 6 where 362.5 ≤ wall < 363 and 60.416̇ to 60.42 < cupboard ≤ 60.5 3 B1 for 362.5 ≤ wall value < 363 B1 for 362.5 < 6 × cupboard value ≤ 363 or 60.416 to 60.42 < cupboard value ≤ 60.5 eg [lower bound of] wall is 362.5 [upper bound of] cupboard is 60.5 362.5 ÷ 60.5 = 5.9[9…] < 6 (ii) 6.5 cm 3 M2 for 363.5 – 6 × their lower bound of cupboard or 363.5 – 357 or M1 for [upper bound of wall =] 363.5 or for 6 × their lower bound of cupboard or [6 cupboards =] 357 DRAFTJ560/06 Mark Scheme November 2021 11 Question Answer Marks Part marks and guidance 10 (a) Using interior angles: ((10 – 2) × 180) ÷ 10 or 1440 ÷ 10 seen [Int angle of triangle =] 60 in working 360 – (144 + 60) oe [= 156] 1 1 1 Using exterior angles: 360 ÷ 10 seen [Ext angle of triangle =] 120 in working 36 + 120 [= 156] Alternative method: 1 for 360 ÷ 10 seen 1 for [Int angle of triangle =] 60 in working 1 for 180 – (60 – 36) [= 156] If 0 scored SC1 for 24, 36, 60, 120 or 144 shown in correct place on diagram Mark the working. Mark angles on diagram only if 0 scored. Working backwards from 156 to 144 [to 10 sides] scores 0 (b) 15 2 M1 for   360 180 156 n = − or 180 2 ( ) 156 n n − = 11 (a) 3 B2 for 8, 9 or 11 correctly placed or B1 for the total of F = 17 or for the total of G = 20 or for all 3 regions add up to 28 or for 17 – x , x, 20 – x Do not accept a blank region to represent 0 8 9 11 DRAFTJ560/06 Mark Scheme November 2021 12 Question Answer Marks Part marks and guidance (b) 88/435 oe or 0.202(…) with correct working 5 B1 for 8 30 oe soi or 11 30 soi M1 for P(F only, G only) [+] P(G only, F only) M1 for P(F only, G only) = �ℎ��� 8 30 × �ℎ��� 11 29 or 30 or �ℎ��� 11 30 × �ℎ��� 8 29 or 30 A1 for 88/870 or 44/435 or 0.101(…) If 0, 1 or 2 scored, instead award SC3 for answer 88/435 oe or 0.202(…) with no or insufficient working If 0 or 1 scored, instead award SC2 for 88/870 or 44/435 or 0.101(…) with no or insufficient working If 0 scored SC1 for 88/450, 44/225 or 0.195[5..] to 0.196 with no working “Correct working” requires evidence of at least M1M1 eg correct branches identified on tree or implied by their subsequent calculation FT their (a) their 8 and their 11 are FT their (a) Likely incorrect answers with working: B1M1M1 for answer 88/450, 44/225 or 0.195[5..] to 0.196 B1M0M1 for answer 88/900, 44/450, DRAFT22/225 or 0.097[7..] to 0.098J560/06 Mark Scheme November 2021 13 Question Answer Marks Part marks and guidance 12 answer with at least 4 sf rounding to 46.9 with correct working 6 M1 for [vol =] 235 ÷ 7.78 [=30.2…] AND M2 for   =  3 30.2... 3 4 their r oe [=7.2…] or M1 for their 30.2 = 4 3 3  r A1 for r = 1.93… AND M1 for [SA =] 4 1.93...    their 2 If 0 scored SC1 for [r =] 1.93… with no working “Correct working” requires evidence of at least M1 AND M1 AND M1 ie using formulas for density, volume and surface area After their 30.2 = 4 3 3  r , r = 1.93… scores M2A1 Condone working in reverse for a maximum of 2 marks: M1 for 46.9 = 4πr2 A1 for r = 1.93... 13 3 x 4 y = − + oe simplified form 4 B3 formfor correct equation not in required OR B1 for gradient of perp line = -¼ oe soi M1 for y – 1 = their grad × (x – 8) or 1 = their grad × 8 + c M1 for correct simplification to y = mx + c form of their y – 1 = their grad × (x – 8) or using their c = 1 – their grad × 8 their grad may be 4 ie they are finding the equation of the parallel line. Max M1 if their grad is ‘m’ DRAFTJ560/06 Mark Scheme November 2021 14 Question Answer Marks Part marks and guidance 14 (a) 30 y x = oe 3 M1 for y k x oe B1 for [k =] 30 eg condone 36 k y = for M1 (b) 2.25 oe 3 B2 for √� = 3 2 oe or M1 for 20 their 30 x = or 20 5 = √36 √� DRAFTJ560/06 Mark Scheme November 2021 15 Question Answer Marks Part marks and guidance 15 (a) 23 – 5 × 2 – 1 = -3 33 – 5 × 3 – 1 = 11 Sign change so solution between x = 2 and x = 3 3 M2 for 23 – 5 × 2 – 1 = -3 and 33 – 5 × 3 – 1 = 11 or M1 for 23 – 5 × 2 – 1 or 33 – 5 × 3 – 1 soi by -3 or 11 Alternative method After x3 – 5x = 1 seen M2 for 23 – 5 × 2 = -2 and 33 – 5 × 3 = 12 A1 for -2 < 1 and 12 > 1 so solution between x = 2 and x = 3 OR M1 for 23 – 5 × 2 or 33 – 5 × 3 soi by -2 or 12 Alternative method SC3 for using an iterative equation that converges to a value in the range 2.25 to 2.35 and concluding statement that 2 < 2.25 to 2.35 < 3 oe or SC2 for using an iterative equation that converges to a value in the range 2.25 to 2.35 Accept other values of x used between 2 and 3 (see table in part (b)). For full marks, the two values need to produce a sign change. Examples just sufficient for third mark include: change of sign -3 < 0 < 11 x = 2 gives an answer < 0 and x = 3 gives an > 0 Examples insufficient for third mark: so x lies between 2 and 3 If within part (a) candidates refer to their working in part (b), award marks DRAFTfor this final alternative method.J560/06 Mark Scheme November 2021 16 Question Answer Marks Part marks and guidance (b) Two correct evaluations in the range 2.25 to 2.35, one which gives a positive value and the other giving a negative value 2.3 M3 and A1dep M2 for two correct evaluations between 2 and 3, one which gives a positive value and the other giving a negative value or M1 for one correct evaluation between 2 and 3 Dependent on achieving at least M2 Alternative method M1 rearranges to a correct iterative formula (converging or diverging) M1 attempts first iteration (either substitution seen or found to at least 2dp (rot) M1 continues iteration(s) to reach x in the range 2.25 to 2.35 A1 for 2.3 If 0 scored SC1 for answer 2.3 with no worthwhile working Likely values: accept rot to 2+sf x x3 −5x – 1 x x3 −5x – 1 2.1 -2.239 2.25 -0.859 2.2 -1.352 2.26 -0,757 2.25 -0.859 2.27 -0.653 2.3 -0.333 2.28 -0.548 2.4 0.824 2.29 -0.441 2.5 2.125 2.30 -0.333 2.6 3.576 2.31 -0.224 2.7 5.183 2.32 -0.113 2.75 6.047 2.33 -0.001 2.8 6.952 2.34 0.113 2.9 8.889 2.35 0.228 Condone missing subscripts If within part (b) candidates refer to their working in part (a), award up to full marks for part (b). DRAFTJ560/06 Mark Scheme November 2021 17 Question Answer Marks Part marks and guidance 16 (a) Subst into correct formula (may be implied) and partial simplification 25 = 20t – 4t2 seen and correct completion to 4t2 – 20t + 25 = 0 2 1dep B2 for 25 20 8 1 2 2 = −   t t oe or 25 = 20t +(−4)t2 or B1 for subst eg 25 = 20t + 1 2 (-8)t2 Dep on previous 2 marks Only accept 25 = 20t – 4t2 if subst seen For B1, condone ambiguity caused by missing brackets (b) 2.5 oe 3 M2 for (2t – 5) (2t – 5) or M1 for any two factors that give two correct terms when expanded or for partial factorisation 2t(2t – 5) – 5(2t – 5) OR M2 for  20 400 400 8 t =  − or better M1 for   ( ) ( ) 20 20 4 4 25 2 2 4 t − −  − −   =  with at most one error eg a sign error, short fraction line, short root but condone missing brackets DRAFTJ560/06 Mark Scheme November 2021 18 Question Answer Marks Part marks and guidance (c) Shows v = 0 and concludes “stationary” 3 M1 for [v2 =] 202 + 2(-8)25 or [v =] 20 + (-8)×their (b) A1 v = 0 If 0 scored, instead award SC2 for v = 0 and other values substituted into a relevant equation as a correct check or SC1 for v = 0 DRAFTJ560/06 Mark Scheme November 2021 19 Question Answer Marks Part marks and guidance 17 113 to 114 with correct working 5 Check diagram for incorrect triangle. Correct triangle or no triangle indicated: M2 for √6.82 + 2.82 [= 7.35…] or for full alternative method or M1 for 6.82 + 2.82 [= 54.08 or 54.1] AND M2 for 5.6 5.6 4 5.6 7.35...  +    1 2 their or M1 for 1 2 × 5.6 × their 7.35… Incorrect triangle indicated: M1 for complete Pythagoras to find hypotenuse with a maximum of one incorrect dimension If 0 or M1 scored, instead award SC2 for answer 113 to 114 with no or insufficient working If 0 scored SC1 for 7.35 either correctly placed on diagram or with no working “Correct working” requires evidence of at least M2 AND M1 ie correct triangle with Pythagoras and area of a triangle eg Finding AC, AO, AE then the height (AC = 7.919…, AE = 7.868…) eg EB found as √6.82 + 2.82 is using DRAFTan incorrect dimension for OBJ560/06 Mark Scheme November 2021 20 Question Answer Marks Part marks and guidance 18 = 4 − 3 17 y t or � = −4 17−3� 5 B4 for 4 3 17 t − or 2 1.5 8.5 y t = − as final answer OR M2 for 10y + 4 = 3ty – 7y or 5 1.5 3.5 + = − 2 t y or M1 for 2(5� + 2) � = 3� − 7 or ( ) 3 7 5 2 2 y t y − + = or 10y + 4 or 3ty – 7y seen or5 2 3 7 2 t y − + = or 5 2 y 1.5 3.5 t + y = − M1ft for correctly collecting y terms on one side and non-y terms on the other (need not be simplified at this stage) M1ft for factorising their 2 or 3 terms To award full marks, solution must be correct eg 4 = 3ty – 7y – 10y or 2 1.5 3.5 5 t y = − − ft for formulae of equal difficulty (eg must include a ty term oe) eg 4 = y (3t – 17) or = 1 − 2 1.5 8.5 y t DRAFTOCR (Oxford Cambridge and RSA Examinations) The Triangle Building Shaftesbury Road Cambridge CB2 8EA [Show More]

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