GCSE Mathematics 93702H Applications of Mathematics Unit 2: Higher Tier Mark scheme 93702H November 2015 Version 1.0 Final
Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students responses to questions and that every associate understands and applies it in the same crect way. As preparation f standardisation each associate analyses a number of students scripts. Alternative answers not already covered by the mark scheme are discussed and legislated f. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a wking document, in many cases further developed and expanded on the basis of students reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Further copies of this mark scheme are available from aqa.g.uk Copyright 2015 AQA and its licenss. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges f AQA are permitted to copy material from this booklet f their own internal use, with the following imptant exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even f internal use within the centre.
MARK SCHEME GENERAL CERTIFICATE OF SECONDARY EDUCATION 93702F November 2015 Glossary f Mark Schemes GCSE examinations are marked in such a way as to award positive achievement wherever possible. Thus, f GCSE Mathematics papers, marks are awarded under various categies. M A B ft SC M dep B dep Method marks are awarded f a crect method which could lead to a crect answer. Accuracy marks are awarded when following on from a crect method. It is not necessary to always see the method. This can be implied. Marks awarded independent of method. Follow through marks. Marks awarded f crect wking following a mistake in an earlier step. Special case. Marks awarded within the scheme f a common misinterpretation which has some mathematical wth. A method mark dependent on a previous method mark being awarded. A mark that can only be awarded if a previous independent mark has been awarded. Or equivalent. Accept answers that are equivalent. 1 eg, accept 0.5 as well as 2 [a, b] Accept values between a and b inclusive. 3.14 Allow answers which begin 3.14 eg 3.14, 3.142, 3.149. Use of brackets It is not necessary to see the bracketed wk to award the marks. 3of 23
Examiners should consistently apply the following principles Diagrams Diagrams that have wking on them should be treated like nmal responses. If a diagram has been written on but the crect response is within the answer space, the wk within the answer space should be marked. Wking on diagrams that contradicts wk within the answer space is not to be considered as choice but as wking, and is not, therefe, penalised. Responses which appear to come from increct methods Whenever there is doubt as to whether a candidate has used an increct method to obtain an answer, as a general principle, the benefit of doubt must be given to the candidate. In cases where there is no doubt that the answer has come from increct wking then the candidate should be penalised. Questions which ask candidates to show wking Instructions on marking will be given but usually marks are not awarded to candidates who show no wking. Questions which do not ask candidates to show wking As a general principle, a crect response is awarded full marks. Misread miscopy Candidates often copy values from a question increctly. If the examiner thinks that the candidate has made a genuine misread, then only the accuracy marks (A B marks), up to a maximum of 2 marks are penalised. The method marks can still be awarded. Further wk Once the crect answer has been seen, further wking may be igned unless it gs on to contradict the crect answer. Choice When a choice of answers and/ methods is given, mark each attempt. If both methods are valid then M marks can be awarded but any increct answer method would result in marks being lost. Wk not replaced Erased crossed out wk that is still legible should be marked. Wk replaced Erased crossed out wk that has been replaced is not awarded marks. Premature approximation Rounding off too early can lead to inaccuracy in the final answer. This should be penalised by 1 mark unless instructed otherwise. 4of 23
3.4 2 3 2 2.56 their 2.56 dep 1(a) 1.6 A1 (3.4 + 3 + their 1.6) 3.2 2.5 2 (hours) 30 (minutes) A1ft ft their 1.6 rounded truncated to nearest minute 1(b) ft from 3.4 2 + 3 2 in 1(a) (3.4 + 3 + 4.5 3.2 = 4.5 3 (hours) 24 (minutes) A1ft 40 cm B1 2(a) 5of 23
2 π their 40 80π [251.2, 251.4] 251 252 A1ft their 40 from (a) ft their 40 from (a) Do not allow 251 252 if value outside [251.2, 251.4] seen 2(b) their (a) is 44 [276.3, 276.5] 276 277 Do not allow 276 277 if value outside [276.3, 276.5] seen their (a) is 48 [301.4, 301.632] 301 302 Do not allow 301 302 if value outside [301.4, 301.632] seen their (a) is 80 [502.4, 502.72] 502 503 Do not allow 502 503 if value outside [502.4, 502.72] seen 6of 23
Alternative method 1 3x + 18 = 52 eg x + x + x + 2 9 = 52 3x = 52 18 3x = 34 Isolates term in x f their equation of the fm ax + b =... 11 3 1 11.3(3 ) A1ft ft from M0 M0 Do not allow if their equation is of fm (1)x + b =... 3(a) Sets up and solves a linear equation Q1ft ft their equation Allow one err in the solution of their equation Do not allow if their equation is of fm (1)x + b =... Alternative method 2 52 18 34 their 34 3 11 3 1 11.3(3 ) A1ft Q0 ft from M0 M0 7of 23
Examples 3(a) 3x + 18 = 52 3x = 70 x = 26.7 3x + 18 = 52 3x = 34 x = 102 M0 A1ft Q1ft A0ft Q1ft 2x + 18 = 52 2x = 34 x = 17 x + 18 = 52 x = 34 M0 A1ft Q1ft M0 A0ft Q0ft 3x + 9 = 52 3x = 61 x = 20.33 M0 M0 A0ft Q1ft 52 + 18 = 70 M0 70 3 26.7 A1ft Q0 M0 A1ft Q0 Identifies height of trapezium parallelogram as 8 B1 3(b) 1 (9 + 5) their 8 56 2 (9 + 5) their 8 112 1 (23 + 19) their 8 168 2 224 A1 8of 23
Alternative method 1 75 100 5000 1.5(%) A1 Machine Q makes lower proption of damaged parts Alternative method 2 Q1ft Comparison using their 1.5 Must have gained 0.02 5000 100 A1 4 Machine Q makes lower proption of damaged parts Alternative method 3 Compares f the same number of parts eg f 1000 0.02 1000 20 and 75 5 15 Wks out both calculations crectly eg f 1000 20 and 15 Machine Q makes lower proption of damaged parts Q1ft A1 Q1ft Comparison using their 100 Must have gained Comparison using their values Must have gained 9of 23
18 (red) 6 (blue) B1 Necklace A 35 (3 + 2) 7 Necklace B 5 their 7 3 21 (red) their 7 2 14 (blue) 39 (red) 20 (blue) 19 A1ft ft B0 M2 10 of 23
6(a) 1 45.1 B1 2 30.4 B1 Smooth decreasing curve passing through (0, 50), (0.5, 48.8), (1, their 45.1), (1.5, 39.0), (2, their 30.4), (2.5, 19.4), (3, 5.9), (3.5, -10.0) ft decreasing curve only f B2 B1ft 4 points plotted, ± 2 1 square ft their points ± 2 1 square B2ft 6(b) 6(c) 3.2 B1ft ft their graph ± 2 1 square 11 of 23
7(a) (B) C E A D (B) E A C D (B) A E C D (B) C A E D B2 Mark diagram if answer line blank B1 Arrangement where first three tiles (including B) fit, eg (B) C D Repeated tile can sce B1 max angle PAQ = 55 angle QPB = 136 angle BAQ = 110 angle QPC = 44 angle BAP = 55 This mark implies first 7(b) angle BPA = 55 angle BPA = 55 statement and angle BAP = 55 A1 and Two equal angles Angles may be seen on diagram f M marks Must be a clear statement f A1 12 of 23
Alternative method 1 11 30 330 Allow [10.8, 11.2] 30 [324, 336] 5.5 30 165 12 30 360 [5.3, 5.7] 30 [324, 336] [11.8, 12.2] 30 [354, 366] 6 30 180 [5.8, 6.2] 30 [174, 186] 8(a) their 165 2 their 165 must be a radius eg 180 2 [85 486.5, 85 541] their [85 486.5, 85 541] 40 Units must be compatible [3 419 460, 3 421 640] their [3 419 460, 3 421 640] 1000 1000 [3.41946, 3.42164] [3.41946, 3.42164] and 3.42 A1 [3.41946, 3.42164] must have > 3 sf 13 of 23
Alternative method 2 11 30 330 5.5 30 165 12 30 360 6 30 180 Allow [10.8, 11.2] 30 [324, 336] [5.3, 5.7] 30 [324, 336] [11.8, 12.2] 30 [354, 366] [5.8, 6.2] 30 [174, 186] 8(a) their 330 100 3.3(0) their 165 100 1.65 their 360 100 3.6(0) their 180 100 1.8(0) their 1.65 2 [8.54865, 8.5541] their [8.54865, 8.5541] 40 100 [3.41946, 3.42164] their 1.65 must be a radius Units must be compatible [3.41946, 3.42164] and 3.42 A1 [3.41946, 3.42164] must have > 3 sf 14 of 23
Alternative method 3 5.5 2 [94.98, 95.05] Allow [5.3, 5.7] 2 [88.2, 102.1] their [94.98, 95.05] 40 [3799.2, 3802] 8(a) their [3799.2, 3802] 30 2 [3 419 280, 3 421 800] their [3 419 280, 3 421 800] 1000 1000 [3.41928, 3.4218] [3.41928, 3.4218] and [3.4, 3.422] A1 [3.41928, 3.4218] must have > 3sf 3.42 1000 750 750 4 3000 4.56 4.6 and 750 5 3750 and 8(b) 3.42 1000 3420 5 A1 Answer 5 with no increct wking A1 15 of 23
9(a) 8000 1.25 0 = 8000 1.25 0 = 1 B1 8000 1 = 8000 B0 9(b) Crect curve passing through (0, 8000) (1, 10 000) (2, 12 500) (3, 15 625) (4, 19 531.25) 1 All points ± sq 2 B3 B2 Increasing graph passing through any 4 of (0, 8000) (1, 10 000) (2, 12 500) (3, 15 625) (4, 19 531.25) 1 All points ± sq 2 B1 Any two of (0, 8000) (1, 10 000) (2, 12 500) (3, 15 625) (4, 19 513.25) Seen plotted ± 2 1 sq 16 of 23
9(c) 14 400 seen implied and 2014 B2ft ft line on their increasing graph from 14 400 B1 Marking on graph at 14 400 14 400 seen B1ft value of t given f V = 14 400 on their graph 2014 with no valid wking B0 Alternative method 1 756 36 21 their 21 48 36 28 and dep Do not allow 1344 with no wking from 756 + 588 their 28 48 1344 1344 756 = 588 A1 Do not allow 1344 with no wking from 756 + 588 Alternative method 2 10(a) 36 48 0.75 48 36 1.33(3.) 756 (their 0.75) 2 756 their 1.33(3.) 2 1344 dep Do not allow 1344 with no wking from 756 + 588 1344 756 = 588 A1 Do not allow 1344 with no wking from 756 + 588 17 of 23
588 4 2352 their 2352 0.0105 their 2352 may be 3024 5376 10(b) [24.696, 24.7] A1 Accept 25 if method seen SC2 [31.75, 31.8] [56.4, 56.45] 5 cm B1 11(a) 17 12 7 5 2.5 2 2 1 2 5 A1 11(b) cm/s cm s -1 B1ft eg centimetres per second SC1 5 cm in 2 seconds Allow other units if value also crect 0.025 m/s A1 B1 AE = AD radii (of circle centre A) B1 AE = AD radii (of circle centre A) and AD = ED radii (of circle centre D) B2 AD = ED radii (of circle centre D) 12(a) and AE = AD = ED 18 of 23
1.8 2 3.6 [11.3, 11.3112] 60 their [11.3, 11.3112] 360 May multiply by 2 at this stage which leads to [3.76, 3.78] [1.88, 1.89] 12(b) 2 their [1.88, 1.89] + 3 1.8 dep dep on eg [3.76, 3.78] + 3 1.8 [9.16, 9.1704] 9.2 A1 Any crect attempt at an area during the first 40 seconds May be seen on the diagram eg1 eg2 eg3 1 15 26 2 1 25 26 2 1 40 26 2 13(a) 195 325 (325 195 =) 130 195 and 325 and Yes 130 and Yes A1 A1 19 of 23
Draws tangent at 65 seconds B1 13(b) difference in velocities difference in times f their tangent with at least one component crect [0.5, 0.9] A1ft Must have drawn a tangent ft B0 with a tangent drawn and both components crect 20 of 23
Alternative method 1 sin a = 5.4 5.5 a = sin -1 (5.4 5.5) dep [79, 79.1] A1 [79, 79.1] and No Alternative method 2 Q1ft ft decision f their angle with M2 sced SC2 [56, 56.2] and No SC1 [56, 56.2] sin a = 5.4 x 74 a 76 x = 5.4 sin a [5.56, 5.57] ft their 74 a 76 14 [5.6, 5.62] A1ft Use of a = 75 [5.59, 5.6] [5.56, 5.57] and No [5.6, 5.62] and [5.56, 5.57] and No Q1 Use of a = 76 only Use of a = 74 and 76 SC2 [56, 56.2] and No SC1 [56, 56.2] An angle of 74 gives the longest possible length. As this is too long they need to then try 76 to see if the shtest length is OK As an angle of 76 gives the shtest possible length and this is too long, they don t need to go on to try 74 Candidates who wk throughout with 76 can sce 4 marks Candidates who use 75 can sce a maximum of M2 A1 Q0 74 a 76 means they can use any angle in this range f the first 3 marks. 21 of 23
2 π r 3 3 B1 Must use crect fmula 4 Allow π r 3 2 3 1 π r 2 25 B1 3 Must use crect fmula their 3 2 π r 3 = their 3 1 π r 2 25 Must equate two volumes and have used h = 25 Do not allow if an increct value f r has been substituted eg Allow 4 π r 3 = π r 2 25 3 15 Simplification to 2r 3 = 25r 2 r 3 = 12.5r 2 Must have an expression in r 3 and r 2 r 3 = their 3 1 π r 2 25 2 their π 3 r = 12.5 37.5 A1 F B marks condone substitution of a value f r 22 of 23
BC 2 + 7.5 2 = 12.5 2 12.5 2 7.5 2 12.5 2 7.5 2 10 dep 16 tan (ACB =) 8.3 their BC dep [39.69, 39.7] A1 Accept 40 with crect wking SC2 [29.6, 29.7] 23 of 23