Mark Scheme (Results) January 2019 Pearson Edexcel .

Mark Scheme (Results)January 2019Pearson Edexcel International GCSEIn Mathematics A (4MA1) Higher TierPaper 2HR

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General Marking Guidance All candidates must receive the same treatment. Examiners mustmark the first candidate in exactly the same way as they mark thelast. Mark schemes should be applied positively. Candidates must berewarded for what they have shown they can do rather thanpenalised for omissions. Examiners should mark according to the mark scheme not accordingto their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark schemeshould be used appropriately. All the marks on the mark scheme are designed to be awarded.Examiners should always award full marks if deserved, i.e. if theanswer matches the mark scheme.Examiners should also be prepared to award zero marks if thecandidate’s response is not worthy of credit according to the markscheme. Where some judgement is required, mark schemes will provide theprinciples by which marks will be awarded and exemplification maybe limited. When examiners are in doubt regarding the application of the markscheme to a candidate’s response, the team leader must beconsulted. Crossed out work should be marked UNLESS the candidate hasreplaced it with an alternative response.

Types of marko M marks: method markso A marks: accuracy markso B marks: unconditional accuracy marks (independent of Mmarks)Abbreviationso cao – correct answer onlyo ft – follow througho isw – ignore subsequent workingo SC - special caseo oe – or equivalent (and appropriate)o dep – dependento indep – independento eeoo – each error or omission No workingIf no working is shown then correct answers normally score fullmarksIf no working is shown then incorrect (even though nearly correct)answers score no marks. With workingIf there is a wrong answer indicated on the answer line always checkthe working in the body of the script (and on any diagrams), andaward any marks appropriate from the mark scheme.If it is clear from the working that the “correct” answer has beenobtained from incorrect working, award 0 marks.Any case of suspected misread loses A (and B) marks on that part,but can gain the M marks.If working is crossed out and still legible, then it should be given anyappropriate marks, as long as it has not been replaced by alternativework.If there is a choice of methods shown, then no marks should beawarded, unless the answer on the answer line makes clear themethod that has been used.If there is no answer on the answer line then check the working foran obvious answer. Ignoring subsequent workIt is appropriate to ignore subsequent work when the additionalwork does not change the answer in a way that is inappropriate for

the question: eg. Incorrect cancelling of a fraction that wouldotherwise be correct.It is not appropriate to ignore subsequent work when the additionalwork essentially makes the answer incorrect eg algebra.Transcription errors occur when candidates present a correctanswer in working, and write it incorrectly on the answer line; markthe correct answer. Parts of questionsUnless allowed by the mark scheme, the marks allocated to one partof the question CANNOT be awarded in another.

Apart from questions 10b, 11, 12, 21, 22 (where the mark scheme states otherwise) the correct answer, unless clearly obtained from anincorrect method, should be taken to imply a correct method.Question1Working(a)(b)2 16 6 18 10 19 14 27 18 20 32 108 190 378 360( 1068)AnswerMarkNotes12 d 161B110.684M2f d for at least 4 products withcorrect mid- interval values andintention to add.If not M2 then award M1for d used consistently for at least4 products within interval(including end points) andintention to addorfor at least 4 correct productswith correct mid-interval valueswith no intention to add‘1068’ 100M1dep on at least M1 Allow divisionby their f provided additionor total under column seenA1Accept 10, 10.7 and 11

QuestionWorking0.5 6 6 5 ( 90)2319.3Mark NotesM1Correct method for volume of A0.5 3² 5 ( 22.5 70.6858.) or 3² 5 ( 45 141.37166.)M1Correct method for volume of Bor correct volume of cylinder‘90’ – ‘70.6858.’M1Correct method to find thedifference in the volumeA119 – 19.4M1for 6n k (k may be 0 or absent)oeA1oe eg 10 (n – 1)6 or n 6 4M1continuing sequence and writingat least 5 terms of 2nd sequence –allow one error or(a)(b)Answer6n 4.40, 46,.e.g. 22 or 46422 2, 1, 6, 13, 22, 33 46 .for a correct equation ft part (a)6n 4 n² 3 oeA1or other number in bothsequences eg 2

QuestionWorking0.07 10 800 ( 756) oe4AnswerMark11 556310 800 ‘756’NotesM1M1M2 for 1.07 10 800 oeA1(a)5P in correct regionon overlay2M1A16(b) (i)154(ii)332360 8 ( 45 ) or 180 – (360 8) ( 135)or1924540 112 112 84( 116) or2Fully correct position for PB1150 – 158 ft from diagramB1330 – 334 ft from diagramM1Correct method to find theinterior or exterior angle ofoctagonM1Correct method to find a missingangle from pentagonM1Complete method6 180( 135) oe8e.g.Correct bearing ( 2 ) or correctdistance ( 2 mm)232540 308( 116) or( 116)22e.g. ‘135’ – ‘116’ or 180 – ‘116’ – ‘45’A1

QuestionWorkingAnswer47.578Mark Notes4M1tan '35 ' x15or tan'55 ' or15xx15 sin 35 sin 55M1x 15 tan ‘35’ ( 10.5.) or15( 10.5 ) orx tan'55 '15x sin 35 ( 10.5 )sin 5510.5 37M1(a)A13k 3 m2B2Forming a right-angled trianglewith angle 125 – 90 marked or 55markedawrt 47.5B1 for an answer in the formakxmy with 2 correct froma 3, x 3, y 1(b)7 1 4x 17 1 or71 17 x 44 4(7 1) 4 x (17 1) 4 or7 117 1 x 4 44 42 x 4.53M1or one side of the inequalitycorrect, e.g. 2 or 4.5M1A1Accept x 2, x 4.5

Question9Working6000 0.015 ( 90) or6000 1.015 ( 6090)AnswerMarkNotes368.183M1or forM2 for4 1.5 60001006000 1.0154( 360) or 6360(6000 ‘90’) 0.015 ( 91.35)(‘6090’ ‘91.35’) 0.015 ( 92.72)M1for completemethod (4years) for totalvalue or sightof 6368 .A1accept 368 – 368.20(‘6090’ ‘91.35’ ‘92.72’) 0.015( 94.11.)

Question10Working(a)(b)e.g. 2(8 – 2x) – 3(2x – 3) 4 6 orAnswerMark4x4y32B2B1 for 2 correct terms of 3 in aproduct0.13M1For method to deal withfractionsNotes2(8 2 x ) 3(2 x 3) 4 or66eg. finds a commondenominator (6 or a multipleof 6)16 4 x 6 x 9 4 or66or2(8 2 x ) 3(2 x 3) 4 or6multiplies by commonmultiple in a correct equation. 10 x 25 4 oe6Condone one error inexpansione.g. 16 – 4x – 6x 9 24 or 10 x 25 24 oeM1For method to expand bracketsand multiplies by commondenominator in a correctequation.Condone one error inexpansion(c)m2 1ef3f 3m 2e2A1oe dep on M1M1for squaring the mA1oe must have f

Question11Working1x4 xorx or y 2 or y 222gradient of L1 0.5 oee.g. y 2 e.g.AnswerMarkYes, with correctgradients shown tomake 1 whenmultiplied39 7 7 9( 2) or( 2)7 1 1 7NotesM1M1A12 0.5 1 and yesAlternative scheme for 111x4 xorx or y 2 or y 222gradient of L1 0.5 oee.g. y 2 7 2( 1) c or 9 2(7) c (c 5)Yes, with correctequation shown tobe valid by using thegiven points3M1M1dep on M1 for substituting( 1, 7) or (7, 9) into y 2 x cto find value of cA1Uses the other point in y 2 x 5to show it is valid and yes

QuestionWorking0, 4, 6, 9, 17, 21, 32, 42, 51, 69, 102121 0.65 1.22 ( 2.87) or100 65 122 ( 287)13Answer45300Mark Notes33M1for identifying 6 or 51 fromordered list or attempt to find 3rdand 9th seen (from an ordered list)M1for identifying 6 and 51A1for 45M1oeNote: 861 3 287 is M0861 2.87 or(861 287) 100 oeM1A114(a)(i)a²1B1(ii)ab41B1(iii)1b31B1oex 4, y 33M1or x y 7,M12x y 11A1x 4 and y 3(b)177 147 2187 ( 81) or a 812187 81 ( 27) or b 27

Question15Working(a)0.3³ 0.7Answer0.0756Mark Notes34 0.3³ 0.7(b)161 – 0.74 oe0.7599(a)23M1oeM1oeA1189or 0.075 or 0.0762500M1Fully correct methodA10.759 – 0.7560M1for 2 in the middle and one from1 or 3 or 5 in the correct place inthe Venn diagram(b)3281M1for any 4 correct entriesA1for a fully correct answerincluding 62 outside the circlesinside the rectangleB1ft

Question17Working(a)(b)18(a)4.75 0.2552.735 0.035(0, 1), (90, 0), (180, 1), (270, 0), (360, 1)Answer1.2178Curve throughgiven coordinatesMark Notes222M1for 4.75 or 0.255 seenA11.21125M12.735 or 0.035 seenA178.142857.M1for a translation of the curveparallel to the x axisorfor a curve going through 3correct points(b)(180, 4)2A1fully correctM11 coordinate correct x 2 or a sketch of sin A1for (180, 4)0

QuestionWorking19AnswerBD9.3 sin 97 sin 589.3BD sin 97 ( 10.8846.)sin 580.5 ‘10.88.’ 11.2 sin47 ( 44.57.)66.0Mark Notes5M10.5 ‘10.88.’ 9.3 sin25 ( 21.39.) or0.5 ‘4.63458 ’ 9.3 sin97 ( 21.39.)20(a) 3(x – 2)² 57 3 3(x² 4x) 7 or 3 x 2 4 x 37 3 3((x – 2)² 4) 7 or 3 x 2 4 or 2M1M1Complete method to find areaBCDM1Complete method to find areaABDA1Allow 65.9 – 66.1M1or expanding a(x² 2bx b²) cM1 12 2ab or 7 ab² cA1or a 3, b 2, c 5B1ft from (a)3(x – 2)² 12 7(b)x 21

Question21AnswerWorking( 0.05, 0.3)(10x – 3)(x 1) 6xMark NotesM1for a correct equation to findpoints A and B10x² x – 3 ( 0)M1for rearranging equation in theform ax2 bx c ( 0)(5x 3)(2x – 1) ( 0) orM1dep on M1for solving thequadratic equation usingfactorisation or using the formulaor by completing the squarex 0.6 and x 0.5 (y 3.6 and y 3)A1Both x values correct dep on M2' 0.6 ' '0.5 '' 3.6 ' '3'or22M1dep on M1x 1 12 4 10 3 2 106or10 x 0.05 0.025 3 02oeA1

QuestionWorking22 (5r ) 2 r2 Answer4545or (3r ) 2 3603609 18Mark6NotesM1oe4545or (0.6 r ) 2 3603600.64 r² 4545 81 (3r ) 2 or360360 24545 81 r2 (0.6r ) 2 360360 2M1r² (40.5 8) (1 – 0.36) or r² 506.25 oeM1 (5r ) 2 M2 foroe16 r² (r 22.5)8145 or 23608145 2360or 1 share 4.5 or r 22.5 orOA 13.5 or AP 9r² (40.5 8) (25 – 9) or r² 80.25 oe(r 4.5)(AB ) 2 '13.5 ' 45 27 or 360 8 45 45 oe(PQ ) 2 ' 22.5 ' 360 8 ' 27' 45Perimeter ' ' '9 ' '9 '88M1dep on M3or2 '13.5 ' ' 22.5 ' M1dep on M4A1oe45 9 360

Question23WorkingAnswerMark814NotesM1A correct formula involving 10th term20(2a 19 d ) 1290 oe2M1Correct formula for sum of first 20 termsa 93 or d 3A1A correct value for a or dA1dep on M2a 9d 66 oe

Jan 05, 2019 · January 2019 Pearson Edexcel International GCSE In Mathematics A (4MA1) Higher Tier Paper 2HR . Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the world’s leading learning company. We provide a wide range of qualifications including academic,