Mark Scheme (Results) January 013 Internatinal GCSE Mathematics B (4MB0) Paper 01
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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 shuld be applied psitively. Candidates must be rewarded fr what they have shwn they can d rather than penalised fr missins. Examiners shuld mark accrding t the mark scheme nt accrding t their perceptin f where the grade bundaries may lie. There is n ceiling n achievement. All marks n the mark scheme shuld be used apprpriately. All the marks n the mark scheme are designed t be awarded. Examiners shuld always award full marks if deserved, i.e. if the answer matches the mark scheme. Examiners shuld als be prepared t award zer marks if the candidate s respnse is nt wrthy f credit accrding t the mark scheme. Where sme judgement is required, mark schemes will prvide the principles by which marks will be awarded and exemplificatin may be limited. When examiners are in dubt regarding the applicatin f the mark scheme t a candidate s respnse, the team leader must be cnsulted. Crssed ut wrk shuld be marked UNLESS the candidate has replaced it with an alternative respnse. Types f mark M marks: methd marks A marks: accuracy marks B marks: uncnditinal accuracy marks (independent f M marks) Abbreviatins N wrking ca crrect answer nly ft fllw thrugh isw ignre subsequent wrking SC - special case e r equivalent (and apprpriate) dep dependent indep independent ee each errr r missin If n wrking is shwn then crrect answers nrmally scre full marks If n wrking is shwn then incrrect (even thugh nearly crrect) answers scre n marks. With wrking If there is a wrng answer indicated n the answer line always check the wrking in the bdy f the script (and n any diagrams), and award any marks apprpriate frm the mark scheme.
If it is clear frm the wrking that the crrect answer has been btained frm incrrect wrking, award 0 marks. Any case f suspected misread lses A (and B) marks n that part, but can gain the M marks. If wrking is crssed ut and still legible, then it shuld be given any apprpriate marks, as lng as it has nt been replaced by alternative wrk. If there is a chice f methds shwn, then n marks shuld be awarded, unless the answer n the answer line makes clear the methd that has been used. If there is n answer n the answer line then check the wrking fr an bvius answer. Fllw thrugh marks Fllw thrugh marks which invlve a single stage calculatin can be awarded withut wrking since yu can check the answer yurself, but if ambiguus d nt award. Fllw thrugh marks which invlve mre than ne stage f calculatin can nly be awarded n sight f the relevant wrking, even if it appears bvius that there is nly ne way yu culd get the answer given. Ignring subsequent wrk It is apprpriate t ignre subsequent wrk when the additinal wrk des nt change the answer in a way that is inapprpriate fr the questin: eg. Incrrect cancelling f a fractin that wuld therwise be crrect. It is nt apprpriate t ignre subsequent wrk when the additinal wrk essentially makes the answer incrrect eg algebra. Transcriptin errrs ccur when candidates present a crrect answer in wrking, and write it incrrectly n the answer line; mark the crrect answer. Parts f questins Unless allwed by the mark scheme, the marks allcated t ne part f the questin CANNOT be awarded in anther.
4MB0 January 013 Paper 1 Mark Scheme 1. 3, 9, 31, 37 B1, B1 Ttal marks. 6(3) p () 4 (subst.) p = 7 A1 Ttal marks 3. 6 3, 15 3 5, 7=3 3 3 At least tw crrect decmpsitins 70 A1 Ttal marks 4. (a) 3 B1 1 (b) 3 B1 1 Ttal marks 5. 3,.5, B1, B1 Ttal marks 6. 360 11 OR 180 68 (.e) 48 r S68 W A1 Ttal marks 7. A crrect decmpsitin f 500 int r mre numbers leading t 10 and n 10 5 A1 Ttal marks
8. 30 B1 150 B1 Ttal marks 9. (a) 30, 4 B1,B1 (b) Accept differences increase by B1 1 3 differences are,4,6,8,10,1,... sequence is ging up plus then plus 4 then plus 6 ( ) n u n n Ttal 3 marks 10. Sight f -1 (but nt fr x 1 ) Accept f(0) 1 f -1 r [-1, ) r x 1 1 A1 Ttal marks 11. 3 d ( v u) t (remving denminatr) 3d vt ut (.e: islating term invlving u) dep OR: vt d ut () 3 3 vt ut d (dep) 3 3 OR: v u 3 d t 3d u v t () (dep) u 3d vt vt 3d t t (.e u v 3d t ) A1 3 3 Ttal 3 marks
1. circumference = 6378000m = 6378000 km 1000 (cnverting metres t km) 1 756 OR 1 800 (ca) A1 3 3 Ttal 3 marks 13. (a) 3.95 (.e. 19 3 0 ) B1 1 (b).395 10 1 OR.395 10 B1 ft 1 (c) 4.0 B1 ft 1 3 Ttal 3 marks 14. (a) f, g B1 1 (b) a, b, c, f, g, h B1 1 (c) h B1 1 3 Ttal 3 marks 15. (a) 3 1 1 1 B1 1 10 7 (b) B(-1ee) 3 30 1 Ttal 3 marks 16. (a) 4 (cm ) B1 1 (b) perp dist 8 = 4 r perp dist = 7 sin ABC () perp dist = 5.5 cm (awrt) A1 3 Ttal 3 marks
17. 4x 4 6x 3 (.e.) Crrectly remving denminatrs ind 7 8 r 14 16 r -0.875 A1 3 3 Ttal 3 marks 18. 11 1 1 b 1 1 1 3 1 1 11 b (1 1 ) dep 3 1 17 1 b (.e.) dep 6 35 b 34 35 (ca) A1 4 4 Ttal 4 marks 19. (a) 6 = AD AD OR sin r cs 45 6 18.4 cm A1 (b) Sectr ADCP = 1 4 "18.3847..." Shaded ABC = "18.3487..." - 1 4 "18.3847..." dep 7.4 cm A1 3 5 Ttal 5 marks
0. crrect attempt at balancing equatins crrect decisin t add/subtract dep OR crrect attempt t make x/y the subject Substituting fr x (r y) () (dep) x 19 A1 y 14 A1 4 4 Ttal 4 marks 1. (a) 18 x, 16 x B1 3 B1 (b) "(18 x)" x "(16 x )" "3" 3 (.e) x = 5 A1 4 Ttal 4 marks. (a) c = 1 1 5 4 4 (subst) 15 0 OR 5 3 4 A1 (b) c = "( 15)" "( 0)" c = 5 A1ft 4 Ttal 4 marks
3. (a) 5000 100 80 650 A1 (b) 5000 4300 100 5000 14% A1 4 Ttal 4 marks 4. 7 3x (.e.) x 5 (.e.) x OR x A1 1 1 3, 1, 0, 1, A1 4 4 Ttal 4 marks Ntes: Bth Ms must be cllected in rder t cllect final A mark. Ignre any weak inequalities (prvided right way rund) 5. (a) Prb = 1 4 5 (.e) 13 0 (0.65, 65%) A1 (b) 5 40 40 5 r 100 99 100 99 5 40 40 5 + 100 99 100 99 (.e.) dep 0 (0.0, 0.%) 99 A1 3 5 5 40 SC: With replacement ( ) 100 100, M0, A0 Ttal 5 marks
6. (a) 117 3 108 118 10 1 13 14 10 117.4 (cm) A1 (b) 117 (cm) B1 1 117 118 (c) r placing data in rder 117.5 (cm) A1 5 Ttal 5 marks 7. (a) AOC 80 (e) B1 OAC 50 B1 Suitable valid reasn (nly ONE required) eg at centre f circle (= at circum) r base angle f iss. B1 3 (b) OCD 90 (e) B1 ADC 100 (Iss. ) B1 Suitable valid reasn (nly ONE required) eg betw. tangent and radius Or sum f quadrilateral B1 3 6 Ttal 6 marks
8. (a) perpendicular bisectr cnstructin arcs Lcus drawn accurately A1 (b) C labelled and ABC drawn B1 1 (c) Arc, rad. 4cm, centre C drawn at least inside ABC Arc, rad. 6 cm, centre B drawn at least inside ABC Area shaded crrectly A1ft 3 6 Ttal 6 marks 15 9. (a) tan BAC = (.e) 0 BAC = 36.9 A1 (b) AB = 0 15 AB = 5 cm A1 (c) AD = 0 x cs 30 ( = 17.3) 1 ABD = "17.3" "5" sin(30 "36.9") dep ABD = 199 cm A1 3 7 Ttal 7 marks TOTAL FOR PAPER : 100 MARKS END
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