Mark Scheme (Results) Summer 017 Pearson Edexcel International GCSE In Mathematics B (4MB0) Paper 0
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General Marking Guidance All candidates must receive the same treatment. Examiners must mark the first candidate in exactly the same way as they mark the last. Mark schemes should be applied positively. Candidates must be rewarded for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according to their perception of where the grade boundaries may lie. There is no ceiling on achievement. All marks on the mark scheme should 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 the answer matches the mark scheme. Examiners should also be prepared to award zero marks if the candidate s response is not worthy of credit according to the mark scheme. Where some judgement is required, mark schemes will provide the principles by which marks will be awarded and exemplification may be limited. When examiners are in doubt regarding the application of the mark scheme to a candidate s response, the team leader must be consulted. Crossed out work should be marked UNLESS the candidate has replaced it with an alternative response. Types of mark o M marks: method marks o A marks: accuracy marks o B marks: unconditional accuracy marks (independent of M marks) Abbreviations o cao correct answer only o ft follow through o isw ignore subsequent working o SC - special case o oe or equivalent (and appropriate) o dep dependent o indep independent o eeoo each error or omission
No working If no working is shown then correct answers normally score full marks If no working is shown then incorrect (even though nearly correct) answers score no marks. With working If there is a wrong answer indicated on the answer line always check the working in the body of the script (and on any diagrams), and award any marks appropriate from the mark scheme. If it is clear from the working that the correct answer has been obtained 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 If working is crossed out and still legible, then it should be given any appropriate marks, as long as it has not been replaced by alternative work. If there is a choice of methods shown, then no marks should be awarded, unless the answer on the answer line makes clear the method that has been used. If there is no answer on the answer line then check the working for an obvious answer. Ignoring subsequent work It is appropriate to ignore subsequent work when the additional work does not change the answer in a way that is inappropriate for the question: eg. Incorrect cancelling of a fraction that would otherwise be correct. It is not appropriate to ignore subsequent work when the additional work essentially makes the answer incorrect eg algebra. Transcription errors occur when candidates present a correct answer in working, and write it incorrectly on the answer line; mark the correct answer. Parts of questions Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another
1 (a) B = {, 3, 4, 6, 9, 1} B1 1 C 4,6,8,9,10,1 C,3,5,7,11 (b) B C= {4, 6, 9, 1} B1 ft 1 NB: (1) ft on (a) () Null set scores B0 (c) One of A B = {, 4, 6, 1} or B C = {, 3} B1 ft NB: ft on (a) and A or C seen AB B C = {} (cso) B1 n A B B C = 1 (cso) B1 3 5 NB: Condone missing brackets Total 5 marks
3 (a) y x 3 x x (allow 1 sign slip, oe) dy 6 3 dx x x d dx ) (one term correct but not from 3 (DEP) All terms correct (oe) A1 3 [OR Product Method (Not in Spec): dy 1 x 3 " " 1 "" 3 dx x x (with at least one derivative correct) (M) dy 6 (cao) (oe) (A1) 3 dx x x 3 x 3x x 3 OR Quotient Method (Not in Spec) : y x dy dx 3 x x x x x x x " 6 6 " 3 3 " " x 4 (with at least one derivative correct) (M) dy 6 (cao) (oe) (A1)] 3 dx x x dy 6 (b) ( x ) " " 3 dx (subst. x = in their answer to (a)) dy 13 ( x ), 3.5 (cso) A1 5 dx 4 Total 5 marks
3 Penalise missing monetary units in answers ONCE only (a) 150 $0.60 (oe) 100 $0.90 (accept 90 cents) A1 (b) Total selling price = (00 $0.60) 10 = $144.00 100 To make 0% overall, needs to earn from 80 pineapples = $144 10 "$0.90" (=$36.00) (DEP) Selling price per remaining pineapple = "$36.00" 80 (DEP) 0 100 [OR Total profit = 00 $0.60 = $4 () Allowable loss made on selling remainder $0.30 10 $4 (= $1) ((DEP)) Price per remaining pineapple = $0.60 $1 / 80 ( (DEP)) OR 10150% 80 X % 00 10% (oe) () 0010% 10150% X % (oe) (=75%) ((DEP)) 80 $0.60 75% (oe) ((DEP)) OR $X = selling price of the remaining 80 pineapples X $ 0.910 $ 80 10% $ 00 0.6 (oe) () $X = 10 100 80 00 0.6 0.9 10 (oe) ((DEP)) $X = "$36.00" 80 (ie has correct method for 10%)] ((DEP)) Selling price of remainder = $0.45 (accept 45 cents) A1 4 6
Total 6 marks
4 1y 17, x x 1 x y 17, x y 17 y = 4 (cao) A1 x " y" x 9 4x OR (substitution) " y" " y" " y" z 0 (oe) " x" " x" x = 3 OR z =5 (cao) A1 " y" " y" " y" z 0 (oe) OR " x"" y" " x" z 3 (oe) " x" " x" (substitution) z =5 OR x = 3 (cao) A1 6 NB: (1) x, y and z denote use of their numerical values () Order of entry on marks into epen boxes: A1 for y A1 for x A1 for z Special Case: If the cand. has failed to score any marks then recheck their working and award (only) for a correct element equation and nothing else. Total 6 marks
5 (a) c = 3h (oe) B1 1 (b) 4c + h = 700 (oe) B1 1 c (c) 4 3h + h = 700 OR 4c 700 3 OR Correct elimination of one of c or h One of h = 50 or c = 150 A1 Both h = 50 and c = 150 A1 3 (d) "150" 400 "50" 8 3 ( )40 400 (cao) A1 7 Special Case: (a) 3c = h B0; (b) 4c + h = 700 B1; (c) Correct elimination of one of c or h c = 70 and h = 10 A1 A0 (d) A0 Total 7 marks
1 6 (a) fg = 3 x x fg: x 1, A1 x x (b) (i) y 1x 3x OR x 1y 3y h -1 x : x, 3 x x 3x NB: Penalise incorrect notation in their answers to (a) and (b)(i) ONCE. A1 (ii) 3 OR 3 x OR x = 3 B1ft 3 NB: ft on their inverse function, h -1 (whose denominator must be of the form ax + b, where a and b are constants). x x 3 x 3 x x (removing their denominators, which must both (c) contain x and one of which must be of the form ax + b) (oe) x 3x 4x 6 x (expanding) (DEP) [OR fg 1 x 1 h 3 " " x x x 1 " " x (oe, eg simplified) () 3x 6 x x x 4 = 4x (oe, removing denominators and expanding) ((DEP))] x 6 (cso) A1 3 8 Total 8 marks
7 Penalise lack of labelling once only (a) Triangle A drawn and labelled B1 1 (b) Two correct construction lines through ( (oe) Triangle B = 5 9 9 4 4 drawn A(-1 eeoo) 3 NB: (1) Award A if (1, ) not indicated and no construction lines but B drawn correctly () Award A1 if (1, ) not indicated and no construction lines but out 3 coordinates of B drawn correctly (3) Deduction of the loss of A marks starts from the last box in epen 1 0 (c) 5 9 9 1 4 4 0 NB: Allow at most one coord incorrectly stated in their matrix 5 9 9 = 1 (cao) A1 Triangle C drawn A1 ft 3 NB: (1) ft on their triangle B () Correct triangle C drawn with no working scores full marks but (3) If triangle B is not fully correct, then we must see working for any marks even if triangle C is drawn correctly. (d) Translation 3 3 B1 B1 (3/) B1 ( In epen, score the last two Bs as B1 B0 if not given as a column vector and no explanation. () If more than one transformation stated, score B0 B0 B0
R Translation (B1) 1 1 in the x direction (oe) (B1) 3 in the positive y direction (oe) (B1) 3] 10 Total 10 marks
8 (a) One branch correct Two branches correct B1 B1 All three branches correct B1 3 NB: Penalise incorrect branches from the end box in epen (b) 3 4 x B1 1 (c) 3 1 " x" (oe) 4 8 1 x (oe), awrt 0.167, 0.16, 16.7% (cso) A1 6 1 1 " 1 " (oe) 4 10 (d) kx
1 1 1 1 k " " 4 6 10 (subst. x, oe) (DEP) k = 18 5, 3.6, 3.59 (if using 0.167) (cso) A1 3 (e) 3 1 " 1 " " " 4 6 3 1 1 1 " " + OR 4 6 10 3 5 1 4 6 4 5 3 1 1 18 1 1 " " + 1 " " " " 4 6 4 5 6 (DEP) [OR Complementary approach: 3 1 1 18 1 " " " " " " () 4 6 4 5 6 3 1 1 18 1 1 " " " " " " 4 6 4 5 6 ((DEP))] 9, 0.75, 7.5% (cso) A1 3 1 40 Total 1 marks
9 Penalise not corrected answers ONCE only (a) (i) 7 FE tan30 7 and DE OR [ sin 30 DE 14 FE = 7 3 = 1.1 1.1 (cm) FE 14 7 ()] A1 NB: Accept 7 3 as a correct, but not corrected, answer (ii) 10 7 OR [ AF (oe) AE = FE + AF (= 7 3 51 = 19.6578408 ) (DEP) 7 sin DAF OR cos ADF so that DAF 44.4 OR ADF 45.6 10 ( ADE 105.6 ) () 10sin"105.6" AE ((DEP))] sin30 AE = 19.(from 3sf), 19.3 (cm) (iii) EB 40 " AE" (= 18.89564 ) (oe) EB = 35.0 (from 19.3), 35.1 (cm) A1 7 A1 (b) Triangle CXB Method: " FE" tan XCB 30 XCB.0 " AE" sin ABC 40 ( ABC = 8.7 (from 3sf), 8.8) CXB 180 ( XCB + ABC ) ( = 19., 19.3 (3 sf)) (DEP) NB: Use the above ordering of the s when entering marks into epen CX 30 " EB" sin" ABC " sin" CXB" (DEP)
CX 30 " EB" sin" ABC " sin" CXB" (DEP) OR [ CX = CF + FX Method " FE" tan XCB 30 XCB.0 () " AE" sin ABC 40 ( ABC = 8.7 (from 3sf), 8.8) () XAF 180 90 " ABC " (= 61.3) ((DEP)) NB: Use the above ordering of the s when entering marks into epen 30 cos" XCB " (CF = 3.36) (oe) CF [BUT NB: can be (INDEP) if Pythagoras used for CF, for example, CF 30 " FE" and so this method is independent of the first three M marks.] OR FX " AF " sin" XAF " sin 180 " XAF " 180 90 " XCB " (FX = 8.095) ie correct method for CF (see BUT NB above) OR FX ((DEP)) ( AFX 68, AXF 50.7 ) CX = CF + FX ((DEP)) NB: Must have correct method for both CF and FX for above (DEP) ] OR [ ACX Method " FE" tan XCB XCB.0 (()) 30
" AE" sin ABC ( ABC = 8.7 (from 3sf), 8.8) (()) 40 NB: Use the above ordering of the s when entering marks into epen AC EB EB ABC 40 (30 " ") 40 (30 " ") cos" " (AC = 35.6536) (((DEP)) CAB : sin CAB sin" ABC " 30 " EB" " AC " ( CAB 118.498, NB it s obtuse) (((DEP)) CX: CX " AC " sin" CAB " sin " XCB " " ABC " (((DEP))] CX = 40.3, 40.4, 40.5 (m) A1 6 13 Total 13 marks
10 (a) (i) OC = a + b B1 (ii) BC = b + a + b BC = a + b " " 3 1 " " 3 (iii) AD = b ab OR AD a b a b A1 AD = 4 a b (oe) A1 5 3 3 (b) AE = nb a B1 1 (c) nb a = 1 " 4 " 3 a 3 b (oe) NB: This must be a vector equation. Comp. of a: Comp of b: 1 1 (DEP) 3 (cao) A1 3 4 1 n 3 " " n = (cao) A1 5 NB: n = can be correctly obtained from a geometric argument eg ADC BD 1 s are similar AND BE BDE AD b () then n = (A1) Special Case: Cand. uses AE AD (M0) then () for component of a, () for component of b, A0 A0 (d) OA EC b a b a or AC OE b with cc B1 1 (e) Area of 1 BDE 30 "" OR 1 " " 3 30 " " 3 = 7.5 A1
Total 14 marks
11 (a) -1.9 B1.75 Penalise nc ONCE only (b)curve NB: (1) Plotting accuracy is -1 mark for straight line segments (penalise ONCE only) each point missed each missed segment each point not plotted each point incorrectly plotted tramlines (penalise ONCE only) very poor curve B3ft (-1eeoo) 3 1 small square () ft just on ( ) and (1.5,.75 ) (3) Deduct errors from the last epen box (c) Tangent drawn at (,.67) (line touching curve at (,.67) and not intersecting curve ((It should hopefully going through (1.5, 4.67) and (.5, 0.67) but this is not essential for the.) Tangent is y = x + 10.67) 1 small square y x = "4.67" ".67" ( 4.0) "1.5" "" NB: (1) ft from using the coords of a point on their drawn tangent (DEP) () Use of calculus scores M0 A0 Accept in range -4 ±0.5 A1 3 ( Range: "4.7".67 "4.6".67 4.3 and 3.71 ) "1.55" "1.475"
(Scales: y range ±0.05 and x range ±0.05) NB: coordinates values are 1 0.05 which is small square (d) Rewriting 8 5 7 3 3 8 5 1 3 3 3 x x 7x as 3 x 7x 4 x (ie allow 1 sign slip) Correct ie 5 1 x A1 3 Draw 5 1 y x B1 ft 3 (eg ft on line going through (0, 1 ) and ( 3 10, ) 1 small square ) 5 1 NB: Award B1ft for correct drawing of cand s y x if A1 above not awarded 3 A1 1.1 A1 1.97 A1 6 14 (ft is ± 0.05 (ss = ±0.05)) The A1s are dependent on the 1 st A1 and 5 1 y x drawn correctly. 3 () Use the above ordering of the A1ft when entering marks into epen Total 14 marks TOTAL 100 MARKS
9 (a) Relating area to frequency e.g. by showing: FD of 5 seen or written on top of FD axis oe OR cm cm square = (frequency) 10 OR 1 cm 1 cm square = (frequency).5 OR 10 mm squares = (frequency) 1 OR 1 mm square = (frequency) 0.1 B1 Passengers travelling 0 km: 0 (using FD width of bar) 4 10, 16.5, 40 1, 400 0.1 (i.e. method that follows from previous mark) Number of passengers = 40 A1 3 (b) using FDs: number of passengers = 40 + 50 + 4 15 + 3 5 OR 40 + 50 + 6 10 + 1.5 10 OR 40 + 50 + 4.5 + 6.5 OR 40 + 50 + 60 1 + 15 1 OR 40 + 50 + 600 0.1 + 150 0.1 (i.e. 40 + 50 + 60 + 15) Total number of passengers = 165 A1 (c) 75 15 5,, 165 33 11 awrt 0.455 B1 1 6 Total 6 marks TOTAL 100 MARKS
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