2017 Mathematics Paper 1 (Non-calculator) Finalised Marking Instructions

Transcription

1 National Qualifications Mathematics Paper 1 (Non-calculator) N5 Finalised Marking Instructions Scottish Qualifications Authority 017 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is reproduced, SQA should be clearly acknowledged as the source. If it is to be used for any other purpose, written permission must be obtained from permissions@sqa.org.uk. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the necessary copyright clearance. SQA s NQ Assessment team may be able to direct you to the secondary sources. These ing instructions have been prepared by examination teams for use by SQA appointed ers when ing external course assessments. This publication must not be reproduced for commercial or trade purposes. page 10

2 General ing principles for National 5 Mathematics This information is provided to help you understand the general principles you must apply when ing candidate responses to questions in this paper. These principles must be read in conjunction with the detailed ing instructions, which identify the key features required in candidate responses. For each question the ing instructions are generally in two sections, namely Illustrative Scheme and Generic Scheme. The illustrative scheme covers methods which are commonly seen throughout the ing. The generic scheme indicates the rationale for which each is awarded. In general, ers should use the illustrative scheme and only use the generic scheme where a candidate has used a method not covered in the illustrative scheme. (a) (b) (c) (d) (e) (f) (g) (h) (i) Marks for each candidate response must always be assigned in line with these general ing principles and the detailed ing instructions for this assessment. Marking should always be positive. This means that, for each candidate response, s are accumulated for the demonstration of relevant skills, knowledge and understanding: they are not deducted from a maximum on the basis of errors or omissions. If a specific candidate response does not seem to be covered by either the principles or detailed ing instructions, and you are uncertain how to assess it, you must seek guidance from your Team Leader. Credit must be assigned in accordance with the specific assessment guidelines. One is available for each. There are no half s. Working subsequent to an error must be followed through, with possible credit for the subsequent working, provided that the level of difficulty involved is approximately similar. Where, subsequent to an error, the working for a follow through has been eased, the follow through cannot be awarded. As indicated on the front of the question paper, full credit should only be given where the solution contains appropriate working. Unless specifically mentioned in the ing instructions, a correct answer with no working receives no credit. Candidates may use any mathematically correct method to answer questions except in cases where a particular method is specified or excluded. As a consequence of an error perceived to be trivial, casual or insignificant, eg 66 1 candidates lose the opportunity of gaining a. However, note the second example in comment (j). page 0

3 (j) Where a transcription error (paper to script or within script) occurs, the candidate should normally lose the opportunity to be awarded the next process, eg This is a transcription error and so the is not awarded. Eased as no longer a solution of a quadratic equation so is not awarded. Exceptionally this error is not treated as a transcription error as the candidate deals with the intended quadratic equation. The candidate has been given the benefit of the doubt and all s awarded. x x x x x 0 x 1 x x x x x 0 ( x )( x1) 0 x 1 or (k) Horizontal/vertical ing Where a question results in two pairs of solutions, this technique should be applied, but only if indicated in the detailed ing instructions for the question. Example: x = x = y = 5 y = 7 Horizontal: 5 x = and x = Vertical: 5 x = and y = 5 6 y = 5 and y = 7 6 x = and y = 7 Markers should choose whichever method benefits the candidate, but not a combination of both. (l) In final answers, unless specifically mentioned in the detailed ing instructions, numerical values should be simplified as far as possible, eg: 15 1 must be simplified to 5 or must be simplified to must be simplified to 8* 5 must be simplified to must be simplified to 15 *The square root of perfect squares up to and including 100 must be known. (m) Commonly Observed Responses (COR) are shown in the ing instructions to help common and/or non-routine solutions. CORs may also be used as a guide when ing similar non-routine candidate responses. page 0

4 (n) Unless specifically mentioned in the ing instructions, the following should not be penalised: Working subsequent to a correct answer Correct working in the wrong part of a question Legitimate variations in numerical answers/algebraic expressions, eg angles in degrees rounded to nearest degree Omission of units Bad form (bad form only becomes bad form if subsequent working is correct), eg ( x x x )(x 1) written as ( x x x ) x 1 x x 6x x x x x credit written as x 5x 8x 7x gains full Repeated error within a question, but not between questions or papers (o) (p) (q) (r) In any Show that question, where the candidate has to arrive at a required result, the last of that part is not available as a follow-through from a previous error unless specified in the detailed ing instructions. All working should be carefully checked, even where a fundamental misunderstanding is apparent early in the candidate's response. Marks may still be available later in the question so reference must be made continually to the ing instructions. The appearance of the correct answer does not necessarily indicate that the candidate has gained all the available s. Scored-out working which has not been replaced should be ed where still legible. However, if the scored out working has been replaced, only the work which has not been scored out should be ed. Where a candidate has made multiple attempts using the same strategy and not identified their final answer, all attempts and award the lowest. Where a candidate has tried different valid strategies, apply the above ruling to attempts within each strategy and then award the highest resultant. For example: Strategy 1 attempt 1 is worth s. Strategy 1 attempt is worth s. From the attempts using strategy 1, the resultant would be. Strategy attempt 1 is worth 1. Strategy attempt is worth 5 s. From the attempts using strategy, the resultant would be 1. In this case, award s. page 0

5 Detailed ing instructions for each question. Question Generic scheme Illustrative scheme 1. Ans: 10 1 substitute into x x evaluate x x Correct answer without working award 0/. Accept 5 5 for 1. For subsequent incorrect working, is not available 1. (a) For. For. For (b) For 5 ( 5) ( 5) 5 10 award / 5 ( 5) ( 5) 5 10 x 15 award 1/ award 0/ 5 ( 5) 10 award 0/. Ans: 16 1 find quartiles calculate semi-interquartile range 1 18, Correct answer without working award 0/. Accept quartiles indicated in the list or on a diagram for 1 1. For award 0/ page 05

6 . Ans: 9 1 start simplification and know how to divide fractions consistent answer or 9 1. Correct answer without working award 0/. Do not penalise incorrect conversion of to a mixed number award 1/ award 1/ award 1/ page 06

7 . Ans: x 5x 10x 1 start to expand complete expansion collect like terms which must include a term in x and a negative coefficient 1 evidence of any correct terms eg x 8x x x 8x x x 1x x 5x 10x 1. Correct answer with no working award /. For subsequent incorrect working, the final is not available 1. For eg x 8x x x 1x x 5x 1x award /. For eg x x 1x x 10x award /. For x 8x x x 1x x 11x 1x award 1/ page 07

8 5. Ans: B0,6,6, C,,9 1 Coordinate B Coordinate C 1 (0,6,6) (,,9) 1. The maximum available is 1/ where (a) brackets are omitted (b) answers are given in component form. For (6,6,0) and (9,,) [repeated error] award 1/ 1. For 0,6,6 and,,9 award 1/ 0. For 6 and 6 9 award 1/ 0. For eg 6 and 0 9 award 0/ page 08

9 6. Ans: y x Method 1: y b mx a 1 find gradient substitute gradient and a point y b m x a into state equation in simplest form 1 8 or equivalent eg y x y x or equivalent 8 Method : y mx c 1 find gradient substitute gradient and a point into y mx c state equation in simplest form 8 ¹ 8 ² eg c ³ y x or equivalent 1. Correct answer without working award /. BEWARE 1 is not available for 6 8 ( 1) 6 ( ) 8 or 1 1. For a final answer of y x 1 award /. y x 8 8 [ m 1, 6] award /. y x 8 8 [ m, ] award /. m 1 y 6 1( x ( 1)) y 1 x 7 award / page 09

10 7. Ans: cm 1 correct substitution into area of triangle formula calculate area cm 1. Correct answer without working award 1/ 1. For 118sin award 1/. For 118 sin award 0/. For award 0/. For (a) or award / (b) or award 1/ (c) or award 0/ 8. Ans: x 5 1 expand bracket collect like terms solve for x 1 x 6 x 10 or 10 x x5 or 5 x 1. Correct answer without valid working award 0/ Treat guess and check as invalid working Commonly Observed Responses 1. For 19 x 15 x 6 x 10 x 5 award 1/. For 19 x 15 x x 6 x award / 55. For 19 x 18( x ) 19 x 18x x x 17 award /. For (a) 19 x 15 x 6 x 10 x 5 x 5 award / (b) 19 x 15 x 6 x 10 x 5 award / page 10

11 9. Ans: 6 Method 1 1 calculate size of angle OBD calculate size of angle ODB (ODB = OBD) calculate size of angle CAB 1 OBD = ODB = CAB = 6 Method 1 calculate size of angle ABC calculate size of angle OCB (OCB = 90 ABC) calculate the size of angle CAB 1 ABC = OCB = 58 CAB = 6 1. Check both methods and award the higher.. Full s may be awarded for information ed on the diagram.. Where information is not ed on the diagram then working must clearly attach calculations to named angles.. For an answer of 6 with no relevant working award 0/ 5. Where candidate uses triangle ABO, is available for ABO = 90 and answer to CAB = 90 AOB eg OBD = ; AOB = ; ABO = 90 and CAB = 58 award / page 11

12 10. Ans: b Fc t or equivalent ¹ multiply by c ² subtract t ¹ ² Fc t b b Fc t divide by t b Fc 1. Correct answer without working / 1. For. For. For c f t b t Fc b Fc t b award / award / award / page 1

13 11. a Ans: a 1 valid common denominator answer in simplest form 1 a or a or a a a a 1. Correct answer without working award /. For subsequent incorrect working, the final is not available a 1 eg a a a 1. For a a a award 1/ award 0/ 1. For. For a a a a a a a award 1/ award 1/ page 1

14 1. Ans: a, b Method 1 1 find x find x x substitute into formula and start to evaluate find values of a and b 1 x 9, 0,,1, 18 a, b or Method 1 find x and x x0 and x 98 1 substitute into formula start to evaluate find values of a and b 18 a, b or 1. Correct answer without working award 0/. For a, b with valid working award /. m is only available for simplifying where m is not a perfect square n page 1

15 1. Ans: ( 5, 5 5 ) 1 evidence of scaling (match x or y coefficients) 1 eg 9x y = 6 x + y = 19 follow a valid strategy through to produce values for x and y state correct x and y coordinates of P values for x and y x 5, y Correct answer without working award 0/. For a solution obtained by guess and check award 0/ 1. For x 5, y 55 (55, 5) with valid working award / page 15

16 1. (a) Ans: a = state value of a Evidence may appear on the graph. Accept...( x 5).... Where no answer appears in (a), check (b) for evidence of a 5 eg 8 = ( + 5) + b (b) Ans: b = ¹ substitute (, 8) into equation ² state value of b ¹ 8 = ( + 5) + b ² 1. Correct answer without working award / 1. Evidence may appear on the graph. An incorrect answer in (a) must be followed through (working must be shown) with the possibility of awarding /. 1. For (a) a and (b) 8 b with or without working award (a) 0/1 and (b) 0/ page 16

17 15. Ans: 6 5 Method 1 1 find scale factor form equation find x Method 1 form equation start to solve find x or x x or x x x x 1 6 or equivalent 5 7 7x = 5(x +.6) or equivalent 65 Method 1 state ratio start to solve find x 1 5: x: 6 stated or implied by Method 1 state ratio start to solve find x 1 PR 6 7 PR 7 6 ( 91) Correct answer without working award 0/ 1. 5 x x award 1/ [END OF MARKING INSTRUCTIONS] page 17

18 National Qualifications Mathematics Paper National 5 Finalised Marking Instructions Scottish Qualifications Authority 017 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is reproduced, SQA should be clearly acknowledged as the source. If it is to be used for any other purpose, written permission must be obtained from permissions@sqa.org.uk. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the necessary copyright clearance. SQA s NQ Assessment team may be able to direct you to the secondary sources. These ing instructions have been prepared by examination teams for use by SQA appointed ers when ing external course assessments. This publication must not be reproduced for commercial or trade purposes.

19 General ing principles for National 5 Mathematics This information is provided to help you understand the general principles you must apply when ing candidate responses to questions in this Paper. These principles must be read in conjunction with the detailed ing instructions, which identify the key features required in candidate responses. For each question the ing instructions are generally in two sections, namely Illustrative Scheme and Generic Scheme. The illustrative scheme covers methods which are commonly seen throughout the ing. The generic scheme indicates the rationale for which each is awarded. In general, ers should use the illustrative scheme and only use the generic scheme where a candidate has used a method not covered in the illustrative scheme. (a) (b) (c) (d) (e) (f) (g) (h) (i) Marks for each candidate response must always be assigned in line with these general ing principles and the detailed ing instructions for this assessment. Marking should always be positive. This means that, for each candidate response, s are accumulated for the demonstration of relevant skills, knowledge and understanding: they are not deducted from a maximum on the basis of errors or omissions. If a specific candidate response does not seem to be covered by either the principles or detailed ing instructions, and you are uncertain how to assess it, you must seek guidance from your Team Leader. Credit must be assigned in accordance with the specific assessment guidelines. One is available for each. There are no half s. Working subsequent to an error must be followed through, with possible credit for the subsequent working, provided that the level of difficulty involved is approximately similar. Where, subsequent to an error, the working for a follow through has been eased, the follow through cannot be awarded. As indicated on the front of the question paper, full credit should only be given where the solution contains appropriate working. Unless specifically mentioned in the ing instructions, a correct answer with no working receives no credit. Candidates may use any mathematically correct method to answer questions except in cases where a particular method is specified or excluded. As a consequence of an error perceived to be trivial, casual or insignificant, eg 66 1 candidates lose the opportunity of gaining a. However, note the second example in comment (j). page

20 (j) Where a transcription error (paper to script or within script) occurs, the candidate should normally lose the opportunity to be awarded the next process, eg This is a transcription error and so the is not awarded. Eased as no longer a solution of a quadratic equation so is not awarded. Exceptionally this error is not treated as a transcription error as the candidate deals with the intended quadratic equation. The candidate has been given the benefit of the doubt and all s awarded. x x x x x 0 x 1 x x x x x 0 ( x )( x1) 0 x 1 or (k) Horizontal/vertical ing Where a question results in two pairs of solutions, this technique should be applied, but only if indicated in the detailed ing instructions for the question. Example: x = x = y = 5 y = 7 Horizontal: 5 x = and x = Vertical: 5 x = and y = 5 6 y = 5 and y = 7 6 x = and y = 7 Markers should choose whichever method benefits the candidate, but not a combination of both. (l) In final answers, unless specifically mentioned in the detailed ing instructions, numerical values should be simplified as far as possible, eg: 15 1 must be simplified to 5 or must be simplified to must be simplified to 8* 5 must be simplified to must be simplified to 15 *The square root of perfect squares up to and including 100 must be known. (m) Commonly Observed Responses (COR) are shown in the ing instructions to help common and/or non-routine solutions. CORs may also be used as a guide when ing similar non-routine candidate responses. page

21 (n) Unless specifically mentioned in the ing instructions, the following should not be penalised: Working subsequent to a correct answer Correct working in the wrong part of a question Legitimate variations in numerical answers/algebraic expressions, eg angles in degrees rounded to nearest degree Omission of units Bad form (bad form only becomes bad form if subsequent working is correct), eg ( x x x )(x 1) written as ( x x x ) x 1 x x 6x x x x x written as credit x 5x 8x 7x gains full Repeated error within a question, but not between questions or papers (o) (p) (q) (r) In any Show that question, where the candidate has to arrive at a required result, the last of that part is not available as a follow-through from a previous error unless specified in the detailed ing instructions. All working should be carefully checked, even where a fundamental misunderstanding is apparent early in the candidate's response. Marks may still be available later in the question so reference must be made continually to the ing instructions. The appearance of the correct answer does not necessarily indicate that the candidate has gained all the available s. Scored-out working which has not been replaced should be ed where still legible. However, if the scored out working has been replaced, only the work which has not been scored out should be ed. Where a candidate has made multiple attempts using the same strategy and not identified their final answer, all attempts and award the lowest. Where a candidate has tried different valid strategies, apply the above ruling to attempts within each strategy and then award the highest resultant. For example: Strategy 1 attempt 1 is worth s. Strategy 1 attempt is worth s. From the attempts using strategy 1, the resultant would be. Strategy attempt 1 is worth 1. Strategy attempt is worth 5 s. From the attempts using strategy, the resultant would be 1. In this case, award s. page

22 Detailed ing instructions for each question Question Generic scheme Illustrative scheme 1. Ans: 1 start process solution Correct answer without working award / No working necessary: award 1/ eg award 1/. 17 award 0/ eg award 0/ page 5

23 . Ans: know how to increase by 5 % know how to calculate value after three years evaluate to nearest Correct answer without working award /. Where an incorrect percentage is used, the working must be followed through to give the possibility of awarding /, eg for = 658, with working award /. Where division is used, (a) along with1 05, 1 is not available eg =105 award / (b) along with an incorrect percentage, 1 and are not available eg =178 award 1/ 1. No working necessary: (a) award / (b) 170 or or 169 award /. Working must be shown: (a) = 105 award / (b) = =16 award 1/ (c) =15 award 1/ (d) = 76 award 1/ (e) =16 award 0/ page 6

24 . Ans: 1m 1 correct substitution into cosine rule cos17 evaluate QR calculate QR m 1. Correct answer without working award 0/. Accept 1 metres with working award /. Where sine rule is used award 0/. Disregard errors due to premature rounding provided there is evidence (a) (-0 8)= award / (b) (-0 8)= award / 5. (a) 07 or 08 (RAD) award / (b) 9 (GRAD) award / Inappropriate use of RAD or GRAD should only be penalised once in either Q, 10 or 15. Working must be shown: = 08( 05...) award 1/. (a) cos17 = award / (b) cos17 10 award / = ( 5...) award / page 7

25 . Ans: x 1, x= 06 1 substitute correctly into quadratic formula evaluate discriminant calculate both values of x correct to one decimal place (stated or implied by ) 1, Correct answer without working award 0/. The final is only available if b ac 0; see CORs - 5. The final is only available when answer requires rounding b -ac award 1/ , 0 6 award 1/ (Beware: candidate may get 7 then change it to 7 ) , 0 6 award / , 06 award 1/ (Beware: candidate may get 7 then change it to 7 ) , 06 award 1/ page 8

26 5. Ans: 00 1 know that 115% 80 begin valid strategy complete calculation within valid strategy 1 115% 80 1% 80 or equivalent For 00 with or without working award /. For 105 or % of 80 or 555 or % of 80 (i) and evidence of 1 award 1/ (ii) otherwise award 0/ = 00 award / %= award /. 15%= award / page 9

27 6. Ans: 180mm 5 1 know to find difference of two volumes substitute correctly into formula for volume of large sphere substitute correctly into formula for volume of small sphere carry out all calculations correctly (must involve difference or sum of two volume calculations and include a fraction) 5 round final answer to significant figures and correct units 1 evidence of difference in two volumes mm 1. Correct answer without working award 0/5. Accept variations in π eg mm. In awarding 5 (a) Intermediate calculations need not be shown eg mm award 5/5 (b) Where intermediate calculations are shown, they must involve at least four significant figures eg mm award /5. Volume of second sphere may be calculated using volume scale factor eg accept 1 for the award of page 10

28 Working must be shown: 1. (a) ( ) 90mm award /5 (b) mm award / mm award / mm award / mm award / mm award / mm award / mm award 1/ mm award / π mm award /5 page 11

29 7. Ans: No, with valid reason Method 1 1 valid strategy (Converse of Pythagoras Theorem in correct triangle with correct combination of sides) and evaluation , 8 comparison and state conclusion Method 1 valid strategy (Pythagoras Theorem in correct triangle with correct combination of sides) 8 19 ; No evaluation comparison and state conclusion length of longest side ; No Method 1 valid strategy (correct substitution into cosine rule to find largest angle in correct triangle) evaluation find angle and state conclusion Method 1 valid strategy (correct substitutions into cosine rule to find angle opposite 6 in triangle A and angle opposite 16 in triangle B) 1 o cos x = 8 19 cos x 0 19 x 101 ; No cos x and cos y evaluation of both cos values cos x and cos y find sum of angles and state conclusion (sum =) 101 ; No page 1

30 1. In Method 1 is not available when evaluations at have not been carried out eg = , = 8 ; ; No award 1/ = , = 8 ; ; No award /. Where the wrong triangle is chosen, is only available for consistent application of Pythagoras or cosine rule; see CORs and = = 5, = 8 ; < ; No award / = 05,19 = 61; ; No award / = 10,16 = 56 ; ; No award 1/ 8 + = 58,19 = 61; ; No award / 5. (a) (b) = 5, = 8 ; The square of the hypotenuse is not equal to the sum of the squares of the other two sides; No award / = 5, = 8 ; The hypotenuse is not equal to the sum of the squares of the other two sides; No award / page 1

31 8. (a) Ans: d - c 1 1 answer 1 d - c or equivalent 1. Accept -c d or d -c. Accept D- C as bad form (b) Ans: d 1 c 1 valid pathway 1 1 TP PR or 1 TQ QR RP correct simplified expression 1 d c or equivalent 1. Correct answer without working award /. Accept D 1 C. TP PV or TQ QR RV alone is not enough for the award of 1. For the award of 1 1 (a) accept d PR but not d PV 1 (b) accept d c RP but not d c RV 1 (c) accept PV ( dc ) but not 1 ( d c ) alone 1 (d) accept RV ( cd ) but not 1 ( c d ) alone 1 1. d c award / page 1

32 9. (a) Ans: x5x factorise 1 x5x 5 (b) Ans: x 5 x 1 start to factorise complete factorising simplify 1 x 5x x5x x 5 x 1. Correct answer without working award /. For x 10 x 1 or x x 5etc award 1/. For subsequent incorrect working, the final is not available eg x 5 7 x award /. is only available when both the numerator and denominator have at least two factors page 15

33 10. Ans: 99kilometres 1 calculate size of angles DEF and DFE correct substitution into sine rule 1 0 and 10 DF 15 sin 0 sin10 rearrange formula 15sin 0 sin10 calculate DF Correct answer without working award 0/. Accept a final answer of 10, with working award /. 1 may be awarded for sizes of angles DEF and DFE ed on the diagram. Where incorrect sizes are used for angles DEF and DFE (a) with prior evidence of angle sizes (ed on diagram or clearly attached to named angles), s, and are available (b) without prior evidence of angle sizes, only s and are available DF (a) with prior evidence of DEF = 0 and DFE = 76 award / (b) without prior evidence of sizes of angles DEF and DFE award / 5. BEWARE sin sin 6. Disregard errors due to premature rounding provided there is evidence 7. Inappropriate use of RAD or GRAD should only be penalised once in either Q, 10 or 15 (a) 7... (RAD) (b) (GRAD) DF sin 6 sin 90 (a) with prior evidence of sizes of angles DEF and DFE s (b) without prior evidence of sizes of angles DEF and DFE award / award / DF award / sin 0 sin 16 DF award 1/ 0 10 page 16

34 11. Ans: 5 or isolate term in y or divide throughout by 5 state gradient explicitly 1 5y x... or x... 5y or or x 5 y or Correct answer without working award /. Do not accept x or y for the award of 5 5. Where gradient formula is used with two points which (a) lie on the line x5y10 0, award 1 for correct substitution into gradient formula award for correct calculation of gradient (b) do not lie on the line x5y10 0, award 0/ 1. 5 x or 06x (with working) award 1/ page 17

35 1. Ans: 1 x 1 apply m n m n x x x stated or implied by apply 1 n x n x 1 x 1. Correct answer without working award / 1. Accept x for 1. Where a number or letter (excluding n ) other than x is used 1 eg a 1 or 8 award 1/ 1 n award 0/ 1. 1 n award / 1. x award 1/. x award 1/ page 18

36 1. Ans: centimetres ¹ marshal facts and recognise right-angled triangle ¹ consistent Pythagoras statement calculation of x find height of the logo x Correct answer without working award 0/. The final is for doubling the result of a Pythagoras (or trig.) calculation and then adding 8. In the absence of a diagram accept 1 1 x as evidence for the award of 1 and. BEWARE Where a diagram is shown, working must be consistent with the diagram. is not available for an incorrect diagram leading to x Disregard errors due to premature rounding provided there is evidence 1. For x 1 1 x18 height or 6 9 (a) working inconsistent with correct diagram award / (b) working consistent with candidate s diagram award / (cosine rule may be used to calculate x ) (c) no diagram award /. For x 1 x height = or 67 (a) working consistent with candidate s diagram award / (b) no diagram or working not consistent with candidate s diagram award /. For x 1 x height = or 86 (a) working consistent with candidate s diagram award / (cosine rule may be used to calculate x ) (b) no diagram or working not consistent with candidate s diagram award / page 19

37 1. Ans: 8 Method 1 ¹ expression for arc length angle ¹ ² know how to find angle ² calculate angle 8 Method ¹ arc length: circumference ratio 15 ¹ = know how to find angle calculate angle 8 1. Correct answer without working award 0/. Accept variations in. Premature rounding of 15 must be to at least decimal places 1 8. For the award of, the calculation must involve a division by a product. The calculation must include 1 5,, 60 and the candidate s chosen diameter or radius 5. For subsequent incorrect working, the final is not available eg award / 1. For award / For For award / award 0/ page 0

38 15. (a) Ans: 515 metres m ark 1 1 calculate height Inappropriate use of RAD or GRAD should only be penalised once in either Q, 10 or 15 (a) (RAD) (b) (GRAD) ,08 5 award 0/1 (b) Ans: 17 metres 1 calculate minimum height Inappropriate use of RAD or GRAD should only be penalised once in either Q, 10 or 15 (a) 6... (RAD) (b) (GRAD) (c) Ans: 1 and substitute 61 correctly into equation calculate cos x calculate value of x calculate nd value of x cos x 1 cos x page 1

39 m ark 1. Correct answers (a) without working award 1/ (b) by repeated substitution award 1/. Accept and 6 with valid working. Disregard errors due to premature rounding provided there is evidence. Do not penalise omission of degree sign throughout the question 5. Inappropriate use of RAD or GRAD should only be penalised once in either Q, 10 or 15 (a) , (RAD) (b) ,... (GRAD) = 0 + cos x 61= 6 cos x cos x = x =1 5, award /. - cos x= x= 91 9, award / [END OF MARKING INSTRUCTIONS] page