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Specification B – Practice Paper AUnit 3 HigherMark SchemeGCSEGCSE Mathematics (Modular)Paper: 5MB3H_01Edexcel Limited. Registered in England and Wales No. 4496750Registered Office: One90 High Holborn, London WC1V 7BH
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GCSE MATHEMATICS UNIT 3HIGHER PRACTICE PAPER A MARKSCHEMENOTES ON MARKING PRINCIPLES1 Types of markM marks: method marksA marks: accuracy marksB marks: unconditional accuracy marks (independent of M marks)2 Abbreviationscao – correct answer only ft – follow throughisw – ignore subsequent working SC: special caseoe – or equivalent (and appropriate) dep – dependentindep - independent3 No workingIf no working is shown then correct answers normally score full marksIf no working is shown then incorrect (even though nearly correct) answers score no marks.4 With workingIf there is a wrong answer indicated on the answer line always check the working in the body of the script (and on anydiagrams), and award any marks appropriate from the mark scheme.If working is crossed out and still legible, then it should be given any appropriate marks, as long as it has not beenreplaced by alternative work.If it is clear from the working that the “correct” answer has been obtained from incorrect working, award 0 marks. Sendthe response to review, and discuss each of these situations with your Team Leader.If there is no answer on the answer line then check the working for an obvious answer.Any case of suspected misread loses A (and B) marks on that part, but can gain the M marks. Discuss each of thesesituations with your Team Leader.If there is a choice of methods shown, then no marks should be awarded, unless the answer on the answer line makesclear the method that has been used.Paper: 5MB3H_01 2Session: Practice Paper A
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GCSE MATHEMATICS UNIT 3HIGHER PRACTICE PAPER A MARKSCHEME5 Follow through marksFollow through marks which involve a single stage calculation can be awarded without working since you can check theanswer yourself, but if ambiguous do not award.Follow through marks which involve more than one stage of calculation can only be awarded on sight of the relevantworking, even if it appears obvious that there is only one way you could get the answer given.6 Ignoring subsequent workIt is appropriate to ignore subsequent work when the additional work does not change the answer in a way that isinappropriate for the question: e.g. incorrect canceling of a fraction that would otherwise be correctIt is not appropriate to ignore subsequent work when the additional work essentially makes the answer incorrect e.g.algebra.Transcription errors occur when candidates present a correct answer in working, and write it incorrectly on the answerline; mark the correct answer.7 ProbabilityProbability answers must be given a fractions, percentages or decimals. If a candidate gives a decimal equivalent to aprobability, this should be written to at least 2 decimal places (unless tenths).Incorrect notation should lose the accuracy marks, but be awarded any implied method marks.If a probability answer is given on the answer line using both incorrect and correct notation, award the marks.If a probability fraction is given then cancelled incorrectly, ignore the incorrectly cancelled answer.8 Linear equationsFull marks can be gained if the solution alone is given on the answer line, or otherwise unambiguously indicated inworking (without contradiction elsewhere). Where the correct solution only is shown substituted, but not identified asthe solution, the accuracy mark is lost but any method marks can be awarded.9 Parts of questionsUnless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another.10 Use of ranges for answersIf an answer is within a range this is inclusive, unless otherwise stated.Paper: 5MB3H_01 3Session: Practice Paper A
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GCSE MATHEMATICS UNIT 3HIGHER PRACTICE PAPER A MARKSCHEME5MB3H Practice paper APaper: 5MB3H_01 4Session: Practice Paper A
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GCSE MATHEMATICS UNIT 3HIGHER PRACTICE PAPER A MARKSCHEMEQuestion Working Answer Mark Notes*127065 – 25192 = 18731873 × 11.45 = 21445.85p£214.4585214.46 5M1 for attempt to subtract the two meter readingsA1 for 1873M1 for “1873” × 11.45A1 21445.85C1 for correctly interpreting their final answer as money2(a) 10 1 B1 cao(b)(i) 10 00 1 B1 cao(b)(ii) 30 1 B1 cao(c) 11 20 1 B1 cao(d) 24 2M1 for attempt to relate a correct distance and time e.g.12 km in 30 minutesA1 cao33x + 3x + x + 2 + x + 2 = 448x + 4 = 448x = 40x = 5Area = 15 × 575 5M1 for attempt to form equation in x equating to 44 withsight of 3x or x + 2M1 for establishing 8x + 4 = 44 oeA1 for x = 5M1 for 3 × “5” × “5”A1 cao4(a)Correctrotation2M1 for their rotated triangle in the correct orientationA1 for their rotated triangle in the correct position(b)Correcttranslation2B1 for triangle moved 6 cm rightB1 for triangle moved 1 cm down(c)Reflectionin y = x2B1 for reflectionB1 for y = x oe533+ 2 × 3 = 334³ + 2 × 4 = 723.53+ 2 × 3.5 = 49.8753.63+ 2 × 3.6 = 53.8563.7³ + 2 × 3.7 = 58.0533.8³ + 2 × 3.8 = 62.4723.75³ + 2 × 3.75 = 60.2343753.7 4B2 for a trial between 3.7 x 3.8 inclusive(B1 for a trial between 3 x 4 evaluated)B1 for a different trial 3.7 < x 3.75B1 (dep on at least one previous B1) for 3.7Accept trials correct to 3 d.p. rounded or truncated.NB: no working scores, no marks even if answer is correct.Question Working Answer Mark NotesPaper: 5MB3H_01 5Session: Practice Paper A
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GCSE MATHEMATICS UNIT 3HIGHER PRACTICE PAPER A MARKSCHEME6(a)–4, –3, –2, –1, 0,1, 22B2 for –4, –3, –2, –1, 0, 1, 2B1 –4, –3, –2, –1, 0, 1 with inclusion of –5 or omission of2(b) –2 ≤ x < 4 2B2 for –2 ≤ x < 4B1 for sight of –2 and 4 with one sign correct785 × 2 = 17020170100× = 34Tyres cost £204Fitting costs 18 +2018100× = 18+3.60= 21.60 Total = 204 + 21.60225.60 4M1 for attempt to find 20% of 85 or 170 or 18M1 for adding one 20% to one costM1 for complete methodA1 cao82 m³ = 2 × 100 × 100 × 100= 2 000 000 cm³2 000 000 ÷ 1000 =2000 3M1 for attempt to change 2 m³ into cm³ by multiplyingby 100 × 100 × 100 oeM1 for attempt to change “2 000 000” into litres bydividing by 1000A1 cao98² + 16² = 64 + 256 = 320√320 = 17.8885438217.89 3M1 for 8² + 16²M1 for √(8² + 16²)A1 for 17.89 accept 17.8885(4382)10186 000 × 60 × 60 × 24 × 365.25= 5 869 713 600 0005.87 × 10123M1 for 186 000 × 60 × 60 × 24 × 365.25A1 for 5 869 713 600 000 oeA1 for 5.87 × 101211(a) 9, –3, 3 2B2 for all 3 missing values correct(B1 for one missing value correct)(b) Drawing graph 2B2 for fully correct curved line(B1 for 6 of their points correctly plotted with attemptto join with curves)(c)x = 4 andx = –12M1 for attempt to draw line across at y = 3 or attemptto factoriseA1 for x = 4 and x = –1Question Working Answer Mark NotesPaper: 5MB3H_01 6Session: Practice Paper A
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GCSE MATHEMATICS UNIT 3HIGHER PRACTICE PAPER A MARKSCHEME12(a)Ratio of lengths is 3 : 12 or 1:42 × 4 =8 2M1 for establishing the ratio is 3 : 12 or 1:4 or 2 × 4A1 for 8(b)10 ÷ 4 = 2.52.5 + 1012.5 3M1 for using the correct ratio or 10 ÷ 4M1 for “2.5” + 10A1 cao13Half base = 10 × cos 50°= 6.427876Height = 10 × sin 50°= 7.6604444Area = ½ × 6.427876 × 7.6604444= 24.62019383AltBase of triangle = 2 × 10 × cos 50= 12.85575Area = ½ 10 × 12.85575 × sin 50°24.6 6M1 for attempt to find the height of the triangle orsight of cos 50°or for attempt to find half the base orsight of sin 50°M1 for 10 × cos 50 or 10 × sin 50M1 for 10 × cos 50 and 10 × sin 50°A1 for 6.42 or 7.66 or 12.85575M1 for attempt to find area of triangle either by halfbase × ht or by half a × b × sin 50°A1 cao1470 % = 49001% = 70100% = 70 × 1007000 3M1 for establishing that 4900 is equivalent to 70% oforiginal priceM1 for 4900 ÷ 0.7 or 4900 ÷ 70 × 100A1 for 700015Vol ratio is 27 : 125Length ratio is 3 : 5Area ratio is 9 : 25Large area is 36 ÷ 9 × 25AltArea scale factor is 25 ÷ 9Large area = 36 × 25 ÷ 9 oe100 3M1 for attempt to cube root 27 and 125 or sight of 3 : 5M1 for establishing the area ratio is 9:25 or scale factoris 25÷ 9A1 for 100*16(a)(i)(ii)6.756.652B1 accept 6.749999…B1 cao(b)Upper bound = 26.95 ÷ 6.65 = 4.05263Lower bound = 26.85 ÷ 6.75 = 3.977784.052633.977783M1 for attempt to divide Upper bound of area by lowerbound of lengthM1 for attempt to divide lower bound of area by upperbound of lengthC1 for upper bound is 4.05 and lower bound is 3.98(c)4 with correctreason 2B1 for selecting 4C1 for reason that states that 4.05 and 3.98 are onlythe same number to the nearest whole i.e. 4 cmQuestion Working Answer Mark NotesPaper: 5MB3H_01 7Session: Practice Paper A
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GCSE MATHEMATICS UNIT 3HIGHER PRACTICE PAPER A MARKSCHEME*17 3M2 for attempt to draw a circle centre (0, 0) withradius 4(M1 for attempt to draw a closed curve either withcentre (0, 0) or with radius 4)C1 for reason that states that the point (1,2) lies insidethe region contained by the curve defined byx² + y² = 16 and any straight line or line extendedwould have to cut the curve twice, once on each side ofthe point oe189534332−=−−+ xxxx3(x – 3) – 4(x + 3) = 5x3x – 9 – 4x – 12 = 5x–x – 21 = 5x–21 = 6x–3.5 4M1 for multiplying both sides by (x + 3)(x – 3) to get3(x – 3) – 4(x + 3) = 5x(M1 for attempt to multiply by 3 by (x +3) or 4 by (x – 3)M1 dep for attempt to move variables to one side andconstant terms to the otherA1 for –3.5Paper: 5MB3H_01 8Session: Practice Paper A
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