Mark Scheme (Results) Summer International GCSE Mathematics (4MA0) Paper 3H. Level 1 / Level 2 Certificate in Mathematics (KMA0) Paper 3H

Transcription

1 Mark Scheme (Results) Summer 01 International GCSE Mathematics (4MA0) Paper H Level 1 / Level Certificate in Mathematics (KMA0) Paper H

2 Edexcel and BTEC Qualifications Edexcel and BTEC qualifications come from Pearson, the wld s leading learning company. We provide a wide range of qualifications including academic, vocational, occupational and specific programmes f employers. F further infmation, please visit our website at Our website subject pages hold useful resources, suppt material and live feeds from our subject adviss giving you access to a ptal of infmation. If you have any subject specific questions about this specification that require the help of a subject specialist, you may find our Ask The Expert service helpful. Pearson: helping people progress, everywhere Our aim is to help everyone progress in their lives through education. We believe in every kind of learning, f all kinds of people, wherever they are in the wld. We ve been involved in education f over 150 years, and by wking across 70 countries, in 100 languages, we have built an international reputation f our commitment to high standards and raising achievement through innovation in education. Find out me about how we can help you and your students at: Summer 01 Publications Code UG067 All the material in this publication is copyright Pearson Education Ltd 01

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4 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 f what they have shown they can do rather than penalised f omissions. Examiners should mark accding to the mark scheme not accding 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 wthy of credit accding 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 wk 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 crect answer only o ft follow through o isw igne subsequent wking o SC - special case o oe equivalent (and appropriate)

5 o dep dependent o indep independent o eeoo each err omission No wking If no wking is shown then crect answers nmally sce full marks the mark scheme will make it clear when this does not apply. If no wking is shown then increct (even though nearly crect) answers sce no marks. With wking If there is a wrong answer indicated on the answer line always check the wking 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 wking that the crect answer has been obtained from increct wking, award 0 marks. Any case of suspected misread loses A (and B) marks on that part, but can gain the M marks. If wking is crossed out and still legible, then it should be given any appropriate marks, as long as it has not been replaced by alternative wk. If there is a choice of methods shown, then the lower mark should be awarded, unless it is clear which method the candidate has chosen. If there is no answer on the answer line then check the wking f an answer. Igning subsequent wk It is appropriate to igne subsequent wk when the additional wk does not change the answer in a way that is inappropriate f the question: eg. Increct cancelling of a fraction that would otherwise be crect. It is not appropriate to igne subsequent wk when the additional wk essentially makes the answer increct eg algebra. Transcription errs occur when candidates present a crect answer in wking, and write it increctly on the answer line; mark the crect answer. Parts of questions Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another.

6 Apart from Questions 9, 1, 18, 0 and 1 (where the mark scheme states otherwise) the crect answer, unless clearly obtained by an increct method, should be taken to imply a crect method. 1. (a) M1 f oe 5 6 oe (b) ratio 6 : 5400 oe inc 0.6 : 54, 6 cm : 5400 cm, 0.6 m : 54 m (condone omission of units from one side) fraction m oe inc (condone omission of m units from either numerat denominat) 45 A1 cao M M1 f ratio fraction with no units : oe oe eg 6 : 54,, 6 : 540,, 60 : 54, , 1 : 1.5, 54 6, 1 : A1 cao Do not award A1 f 150 cm, 150n etc Total 5 marks

7 . (a) ( ) + 4 ( ) (b) M1 f substitution f crect evaluation of either A1 cao M1 f crect substitution rearrangement 8 = 4 + 4k A x ( k) = oe x 4k = 8 4k = 6 M1 f crect rearrangement of crect substitution 1.5 oe A1 SC M f 8 4 Total 5 marks. (a) 9 1 B1 cao (b) 5 1 B1 cao (c) n 4 6 n 10 5 = 5 oe 5 = 5 oe n 4 6 = oe n 10 = oe M1 n 10 5 = 5 5 n + 10 oe 5 = 5 n 1 5 = 5 1 A1 cao SC If M0, award B1 f an answer of 1 5 Total 4 marks

8 M1 f squaring and adding M1 (dep) f square root 6.71 A1 f answer rounding to 6.71 Total marks B f 1 8 in any der B1 f three positive whole numbers with either a sum of 1 a range of 7 SC Award B1 f Total marks 6. Lines x = 5 and y = drawn B1 Lines may be full broken Line y = x drawn B1 Igne additional lines R shown B1 Condone omission of label Accept shading in shading out, if consistent Award full marks f crect region labelled R even if no shading Total marks

9 7. 9 height = 6 4 M1 height = 4 A M "4" "4" 46 A1 cao Total 4 marks 8. (a) (b) M1 f A1 f ans rounding to oe M f 440 oe M1 f 17.6% = = = 1.76x x 5 seen 500 A1 cao Total 5 marks

10 9. (a) 6x = 6 x 1 = M1 f crect expansion (6x seen) crect division of both sides by (x 1 = ) May be implied by second M1 6x = 6 + 6x = 9 6x 9 = 0 x = + 1 x = x = oe M1 A1 f crect rearrangement Also award f 6x = x = 7 6x 7 = 0 if preceded by 6x 1 = 6 Award marks if answer is crect and at least one method mark sced

11 9. (b) 4(y + 1) = (y ) 4 M1 f clear intention to multiply both sides by 1 by a multiple of 1 eg 4(y + 1) = (y ) y = y y + 1 y 1 = 1 4 May be implied by second M1 by 8y + 1 = y 8y + 4 = y 8y + 1 = y 6 Also award this mark f 4(y + 1) ( y ) = 1 1 8y + 4 = y 6 M1 f crect expansion of brackets crect rearrangement of crect terms 8y + 4 y 6 eg 8y y = 6 4, = 5y = 6 4 8y y = 10 5y = 10 5y = y 8y = 10 5y = 10 5y + 10 = 0 M1 1 1 f crect rearrangement with y terms on one side and numbers on the other AND collection of terms on at least one side f 5y y + 10 = 0 oe f = 0 oe 1 oe A1 Award 4 marks if answer is crect and at least one method mark sced Total 7 marks

12 9. (b) Alternative method = y y = 4 5 = 6 y 4 M1 f crect expansion y M1 y M1 f crect rearrangement of crect terms f crect collection of crect terms on both sides oe A1 Award 4 marks if answer is crect and at least one method mark sced Total 7 marks 10. (a) (b) M1 f finding at least 4 crect products and summing them 78 5 M1 (dep) f division by 5 Accept division by their 5, if addition shown..1 oe inc, M A1 1 oe A1 5 Also accept.1 if both method marks sced

13 10. (c)(i) (ii) oe 5 M1 f oe oe A1 f oe inc M1 M1 f one crect product f sum of all crect products oe A1 f oe inc Note f (c)(ii): sample space method award marks f crect answer; otherwise no marks SC M1 f M1 f SC Sample space method award marks 66 ; otherwise no marks. f 65 Total 10 marks

14 11. (a) (b) M f A1 cao 15 1 scale fact = 5 M f M1 Also award f May be implied by A1 cao

15 11. (c) 5 5 M1 f squaring their scale fact (must be one of 5, 4, 5 1, 4 1 ) 19 f oe.8 f complete crect method of finding vert ht (h cm) of ABC and vert ht (H cm) of PQR 1 eg ".8" h = 4 h = ( ) ".8" 4 H = "5" ( ) ".8" 50 A1 f 50 f answer which rounds to 50.0 ft only from their scale fact of 4 ie if M1 sced f 4 16, award A1 f an answer of Total 7 marks

16 1. (a) l = 15 indicated on graph 70-7 inc stated (b) 0 and and 4 60 indicated on 4 cumulative frequency axis stated and stated M1 9 A1 Accept 8-10 inc M inc A1 An answer in the range 5-6 inc with no indication of method sces marks BUT do not award A1 if an answer in the range 5-6 inc has clearly been obtained by finding the difference between two values, one both of which are outside the ranges and F example, if wking is do not award A1. Total 4 marks

17 1. finds int angle of finds ext angle of 5 M1 ( 5 ) 180 pentagon pentagon f 5 ( 5 ) A1 f Award M1A1 f int angle of pentagon shown as 108 ext angle shown as 7 on printed diagram on candidate s own diagram If there is clear evidence the candidate thinks the interi angle is 7 the exteri angle is 108, do not award the above two marks. int angle of polygon = 144 B1 ext angle of polygon = ( n ) = 144 oe 6 n f int angle of polygon = 144 ext angle of polygon = 6 Award B1 f int angle of polygon shown as 144 ext angle shown as 6 on printed diagram candidate s own diagram M ( n ) f = 144 oe 6 n 10 A1 f 10 cao Award no marks f an answer of 10 with no wking Award 5 marks f an answer of 10 if at least the first M1A1 are awarded Total 5 marks

18 14. (a) y = x 6 y = 6 x M1 May be implied by second M1 by y = x oe 6 y = x oe M1 y x + c even if value of c is increct = finds codinates of points on the line eg (, 0), (0, ), table, sketch showing line cutting x-axis at and y-axis at f crect rearrangement of y = x 6 with y as subject vert difference f clear attempt to use hiz difference f their two points on L oe A1 f oe inc decimal equivalent rounded truncated to at least dp Do not award A1 f x

19 14. (b) M1 f crect 9 = " " 6 + c substitution into " y = " x + c using their answer to (a) oe y = x + 5 A1 f y = x + 5 oe inc x y = 15 ft from their answer to (a) SC If M0 A0, award B1 f answer with y = omitted which would otherwise sce M1 A1 eg x + 5, x if ans to (a) is SC Award B if y 9 = " "( x 6) seen; then isw SC Award B1 f x y = k where k 15 and k 6 with no wking SC If M0 A0, award B1 f " y = " x + c where c 5 c 0 (ie do not award this mark f y = " " x + 5 y = " " x ) does not ft from (a) Total 5 marks

20 15. (OB =) 8sin M1 (BD =) 4 8 M1 A complete crect method eg (BC =) " 8"cos 6 M1.6 A1 f ans rounding to.6 (.619 ) Total 4 marks oe M M1 f A1 Also award f 40 if M sced Total marks 17. (a) a b B B1 f 81 B1 f a 8 b 4 (b) c 4 B B1 f B1 f c 4 Total 4 marks

21 NB The mark scheme f Q18 covers the majity of methods but there are other possible approaches. If you encounter a mathematically crect method which is not covered and (i) the answer is crect award full marks (ii) the answer is not crect send the response, appropriately annotated, to Review. 18. COE = x 6 B OCD = x 69 x 4 + x ODC = x x 69 x Accept x + y = y x = 4 (where OCD = ODC = y) B1 B1 May be stated, marked on diagram part of an equation B1 f each crect expression f an angle up to a max of COD =180 4x 111 x x = 69 M M1f a crect unsimplified equation in x eg x + x = = x + x 69 x = x x + x = x + x + x = x = x Award all B marks if M1 M sced. A1 cao Award 6 marks f an answer of if M1 M sced Total 6 marks

22 19. eg 7 1 π sin M1 M1 M1 7 f oe inc 5 60 f π 5.4 value which rounds to 91.6 seen f completely crect method of finding the 1 area of triangle OAB eg 5.4 sin sin cos ( 18.1 ) A1 f either area crectly evaluated may be rounded truncated to 1 dp A1 f answer rounding to 4.46 (π ) f answer rounding to 4.45 ( ) If all M1s sced, award 5 marks f an answer which rounds to Total 5 marks

23 seen 4 B1 Also accept &, throughout B1 Also award f.5 if first B1 sced ie if seen 6.5 M1 dep on both B1s 7.5 A1 cao Award 4 marks if answer is crect and both B marks sced Total 4 marks

24 1. x = 0 x May be implied by second M1 5 M1 0 y y = May be implied by second M1 x + x 0( = 0) M1 y 89y + 800( = 0) ( x 5)( x + 4)( = 0) x ( x + 4) 5( x + 4)( = 0) x ( x 5) + 4(x 5)( = 0) M1 ( y 5)( y )( = 0) y ( y ) 5( y )( = 0) y ( y 5) (y 5)( = 0) ± 4 ( 0) ± ± 169 ± x =, x = 4 x x 5 5 =, y = = 4, y = A1 A1 89 ± ( 89) ± ± ± y =, y = dep on all method marks 5 5 x =, y = x = 4, y = dep on all preceding marks Accept answers given as codinates Total 5 marks

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26 Further copies of this publication are available from Edexcel Publications, Adamsway, Mansfield, Notts, NG18 4FN Telephone Fax Order Code UG067 Summer 01 F me infmation on Edexcel qualifications, please visit our website Pearson Education Limited. Registered company number 8788 with its registered office at Edinburgh Gate, Harlow, Essex CM0 JE