GCSE Mathematics 4365H Paper Mark scheme 4365H November 016 Version/Stage: 1.0 Final
MARK SCHEME GCSE MATHEMATICS 4365H NOVEMBER 016 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 016 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 GCSE MATHEMATICS 4365H NOVEMBER 016 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. If a student uses a method which is not explicitly covered by the mark scheme the same principles of marking should be applied. Credit should be given to any valid methods. Examiners should seek advice from their seni examiner if in any doubt. 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 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. e.g. accept 0.5 as well as 1 [a, b] [a, b) Accept values between a and b inclusive. Accept values a value < b 3.14 Accept answers which begin 3.14 e.g. 3.14, 3.14, 3.1416 Q Use of brackets Marks awarded f quality of written communication It is not necessary to see the bracketed wk to award the marks.
MARK SCHEME GCSE MATHEMATICS 4365H NOVEMBER 016 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 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. Continental notation Accept a comma used instead of a decimal point (f example, in measurements currency), provided that it is clear to the examiner that the candidate intended it to be a decimal point. 4
MARK SCHEME GCSE MATHEMATICS 4365H NOVEMBER 016 Paper Higher Tier B3 B f rotation of parallelogram 90 anticlockwise about P crect four vertices plotted but not joined 1 B1 f any rotation of parallelogram 90 crect four vertices plotted but not joined f rotation of parallelogram 90 anticlockwise about P 60 4 9 7 100 4 4 + 58 (= 100) 58 (100 4) 9 9 9 0 9 7 dep dep on nd dep on both M marks (a) Fully crect table 4 9 7 60 18 0 40 4 9 9 100 Allow use of a letter in the table with the letter wked out in the wking If there are two tables mark their best attempt 58 can be implied by total part time and total not wking 5 of 9
Alternative method 1 4 60 4 60 0.4 18 40 18 40 0.45 eg 40(%) 45(%) 5 9 0 40(%) and 45(%) fmat so comparison can be made 0.4 and 0.45 8 0 and 9 0 40(%) and 45(%) and women 0.4 and 0.45 and women 8 0 and 9 0 Alternative method and women Q1 eg 4 10 and 45. 10 Strand (iii) Crect conclusion with all wking crect (b) 60 4.5 40 18. 7 out of 60 (women) 16 out of 40 (men) 9 out of 0 (women) 8 out of 0 (men).5 and. 4 and 7 16 and 18 8 and 9.5 and. and women 4 and 7 and women 16 and 18 and women Q1 8 and 9 and women Strand (iii) Crect conclusion with all wking crect Allow common numerats f comparison Beware of 40 as there are 40 women (40% are women) 6 of 9
Alternative method 1 180 15 8 (360 15 ) their 8 (360 15 ) ( ) dep 15 90 6 180 their 6 (180 90 their 6) 56 Alternative method 70 (used f the hexagon) 540 used f a pentagon 3 (70 4 15) 11 dep 540 15 15 90 90 56 Angles may be on the diagram but must be in the crect place 8 must be f a crect angle If diagram wking shows that 8 is f an increct angle then the method is increct, eg y = 8 (on diagram in the wrong place) Answer 8 degrees M0 M0 50 5 4 00 50 5 50 00 and 50 4(a) Sand 50 and Cement 00 A0 50 5 = 50, 50 4 = 6.5, Sand 6.5, Cement 50 A0 Allow transcription err if clear in the wking 7 of 9
Alternative method 1 5 3 75 5 4 100 5 5 15 5 3 4 300 75 4 300 5 4 3 100 3 300 75 5 5 5 3 15 3 dep Total cement Sand Mix Total sand Total mix 375 Alternative method (uses part (a)) 5 + 50 75 Total cement 4(b) 00 100 (00 + 50) 15 Sand Mix 100 + 00 300 Total sand 5 + 50 + 100 + 00 dep Total mix 15 + 50 Total mix 375 Alternative method 3 (uses part (a)) Scale fact 1.5 seen implied, eg 75 50 50 1.5 75 00 1.5 300 50 1.5 dep Total sand Total mix 375 8 of 9
5(a) 1 5 4 B B1 f one two crect in the crect place 6 7 of their points plotted crectly tolerance ± ½ square 5(b) Fully crect smooth curve tolerance ± ½ square Curve must be U-shaped and must not curve back in have vertical lines 5(c) [.,.3] and [.3,.] their two values read off from the graph B1 tolerance ± ½ square Do not accept codinates 15 100 0 3 1 100 10 1. 0 15 + 10 1 40 6(a) 10 100 10 1 3 + 1. 4. 3 + 1 dep their 40 100 4 Q1 Strand (i) Rounding down 9 of 9
(85 + 88) 86.5 (0.85 + 0.88) 6(b) 0.865 173 00 86.5% Allow 0.87 method shown 87 100 87% if crect Beware of 6 30 leading to 86.6( )% M0A0 0.87 on its own M0A0 π 6 π 36 7(a) [113, 113.] 36π π36 A0 0 50 1000 their 1000 their [113, 113.] dep 7(b) [886.8, 887] 1000 36π ft ft their part (a) Do not igne increct further wking f the A mark, eg 1000 36π = 964π A0 10 of 9
Alternative method 1 53 46 7 53 million 46 million 7 million 8 Alt 1 of 3 Alt of 3 7 46 ( 100) 0.15( ) dep Accept 0.15 if crect method shown 15.( ) (%) Accept 15(%) if crect method shown Alternative method 53 ( 100) 1.15 46 115.( ) 1.15 1 0.15( ) 115.( ) 100 dep Accept 1.15 if crect method shown Accept 115 if crect method shown Accept 0.15 if crect method shown 15.( ) (%) Accept 15(%) if crect method shown 11 of 9
Alternative method 3 eg Any crectly evaluated percentage of 46 (million) 1(%) is 0.46 (million) 5(%) is.3 (million) 10(%) is 4.6 (million) 8 cont Alt 3 of 3 15(%) (increase) is 5.9 (million) 15.1(%) (increase) is 5.946 (million) 15.(%) (increase) is 5.99 (million) 15.3(%) (increase) is 53.038 (million) 15.4(%) (increase) is 53.084 (million) 15.5(%) (increase) is 53.13 (million) dep 15(%) is 6.9 (million) 15.1(%) is 6.946 (million) 15.(%) is 6.99 (million) 15.3(%) is 7.038 (million) 15.4(%) is 7.084 (million) 15.5(%) is 7.13 (million) and 7 (million) Accept 15(%) with two of the trials listed above ( better), one with an answer 15.( ) (%) below 53 million ( 7 million), the other with an answer above 53 million ( 7 million) Increct number of zeros used f millions cannot sce A mark 15(%) sces at least unless clearly from increct wking 1 of 9
8 x 16x 1 6 (4x + ) 3(4x + ) 6(x + 1) 1x + 6 B1 9 8 x = 1 6 (4x + ) 8 x = 3(4x + ) 8 x = 6(x + 1) 16x = 1x + 6 dep Sets up a crect equation Simplified and bracket expanded 1.5 1 1 3 x = 6 4 B1A0 Trial and improvement is 0 4 13 of 9
sin 1 8 31 = 14.(9 ) 15 and tan (14.(9 )) = 8 h 31 and 8 seen 961 and 64 897 sin 1 8 31 = 14.(9 ) 15 h and cos (14.(9 )) = 31 10 cos 1 8 31 = 75.(0 ) 75 h and tan (75.(0 )) = 8 cos 1 8 31 = 75.(0 ) 75 h and sin (75.(0 )) = 31 31 8 961 64 897 9.9 30 dep 8 tan (14.(9...)) 31 cos (14.(9 )) 8 tan (75.(0 )) 31 sin (75.(0 )) [5, 5.1] B1ft ft their 30 if first sced Note using 31 + 8 gives 105 3 leading to answer 3 M0A0B1 14 of 9
0.3 3 10 and B1 1st pair of branches fully crect 0.7 7 10 11(a) 0.8 8 10 4 5 nd and 3rd pairs of branches fully crect and B 0. 10 1 5 B1 f nd 3rd pairs of branches fully crect 0.3 0. 3 10 10 3 10 1 5 May be seen in part (a) but must be chosen 3 6 and 10 10 100 11(b) 0.06 6 100 3 50 6% ft ft their diagram May be seen in part (a) but must be chosen 15 of 9
Draws a right-angled triangle to wk out gradient using grid lines 8 ( 0) c = seen implied m = 6 1(a) Gradient = 3 seen implied m = 3 dep y = 3x + 3x + y = 3x y = ax + where a 3 A0 A0 Two crect points plotted calculated 1(b) Fully crect straight ruled line Mark intention F the A mark the line must extend from (0, 9) to (9, 0) 16 of 9
Indication of point of intersection of their lines 9 x = 1 x x + 1 x = 9 Eliminates a variable 1(c) y = 1 (9 y) x = 6 and y = 3 (6, 3) ft ft their graph 30x 3 y 7 B B1 f two crect terms Do not igne fw f B 30 x 3 y 7 B1 13(a) 30 x 3 y 7 B1 x 3 y 7 30 7x 3 4y 7 Do not allow addition sign, eg 10x 3 + 3y 7 B1 B1 B0 17 of 9
x 3x + 7x 1 Allow one err x + 4x 1 13(b) Do not igne fw unless attempting to solve the equation x 3x 1 x + 7x 1 (one err) A0 x 1 (two errs) M0A0 x 4x 1 with no other wking (two errs) M0A0 13(c) 8 and x = 8 and x = B1 Any der 13(d) xy (4x + 3y) B B1 f a crect partial factisation ie x (8xy + 6y ) y (8x + 6xy) (4x y + 3xy ) x (4xy + 3y ) y (4x + 3xy) xy (8x + 6y) 18 of 9
Alternative method 1 90 is 75% 90 75 100 dep 10 1 10 40 3 10 40 = 80 10 3 = 80 Alternative method 14 80 is two-thirds 80 is 66.6( )(%) 80 3 dep 10 5 100 10 30 75% 75 100 10 30 90 and 90 10 = 80 75 100 10 19 of 9
10 4 40 5.8 14 30 1 30 40 + 14 + 30 dep Allow one err 15(a) 84 Beware of 30 from an increct method, eg 10 4 =.5, 5.8 = 1.78( ), 30 1 = 30, 30 from wrong wking 6 5 = 30 (first bar) M0 M0 15 < t 5 B1 15(b) 16 1 3 and 5 7 B1 f crect and 1 increct B f 1 crect and 1 increct f 1 crect 0 of 9
S πr = πrh S = πr (h + r) S r = h + r r S r = h + r 17(a) h = S r r h = S r r S r r S r r implies A0 S r = πrh S = π (rh + r ) (not enough) M0 1 of 9
Alternative method 1 (uses part (a)) (h =) 95 r r (h =) S 5. 3 5. 3 5. 3 Crectly substitutes at least one value into their equation (h =) 95 5. 3 5. 3 5. 3 dep Any unsimplified version of the answer 3.66 3.7 B1ft Accept 4 if wking shown ft their value rounded to 1 sf sf Alternative method (uses the iginal equation) 17(b) 95π = πh 5.3 + π 5.3 5.3 Crectly substitutes both values into the iginal equation (h =) 95 5. 3 5. 3 5. 3 dep Any unsimplified version of the answer 3.66 3.7 B1ft Accept 4 if wking shown ft their value rounded to 1 sf sf It a student is following through from an increct part (a) they can sce the first and the B1ft only M0A0B1ft Some useful values 5.3 5.3 = 8.09 5.3 5.3 = 176.49 95 = 98.45 95 5.3 5.3 = 11.95 5.3 = 33.30 of 9
1 y α x k y = x k 0 = (k =) 0 (k =) 80 dep 18(a) 1 = k 1 80 80 y = x k y α x 18(b) 80 5 = x x = 16 ft their equation from part (a) 4 Condone 4 and 4 x sin 19 8 sin 13 x 8 0.35... 0.838... 19(a) 8 sin 19 sin 13 dep 8 0.35... 0.838... 3.1 Accept 3 with wking shown F the method marks accept rounded truncated values 3 of 9
sin 13 = sin 57 B1 f crect and 1 increct 19(b) and B f 1 crect and 1 increct cos 13 = cos 57 f 1 crect and 0 increct 3.1 B1ft ft their answer to part (a) 19(c) 4 of 9
Alternative Method 1 Tube A Tube B Radius Diameter Radius Diameter 0 = πr 0 = πd 10 = πr 10 = πd (r =) 0 π (r =) 10 (r =) [3.18, 3.] 0 d = d = [6.36, 6.4] (r =) 10 π (r =) 5 (r =) [1.59, 1.6] 10 d = d = [3.18, 3.] their 10 [317, 3] π 10 their 5 π 0 0 Alt 1 of 1000 dep [158, 161] 500 [317, 3] and [158, 161] Tube A and [317, 3] and [158, 161] 1000 500 Q1ft Strand (ii) ft conclusion from their volumes provided awarded Tube A and 1000 500 and 5 of 9
Alternative Method radius A = r and radius B = 1 r V A = πr (10) V A = πr h V B = π( 1 r) (0) dep V B = π( 1 r) (h) 0 Alt of V A = πr (10) and V B = π( 1 r) (0) V A = πr h and V B = π( 1 r) (h) Tube A and 10πr and 5πr Tube A and πr h and 1 πr h Q1ft Strand (ii) ft conclusion from their volumes provided awarded 6 of 9
3x = 4x + Equation must be crect 3x 4x (= 0) 4 ( 4 ) 4 3 3 4 16 4 6 Allow one err 4 40 6 1 4 ( 4 ) 4 3 3 4 16 4 6 ft Fully crect f their equation 4 40 6 x = 1.7 and x = 0.4 ft ft their equation One crect answer with no wking, eg x = 1.7 implies 3 marks 7 of 9
Alternative method 1 10 = 1 + 15 1 15 cos A 1 15 10 1 15 0.74(7 ) 0.75 (A =) [41.4, 4] h sin (their 41.64) = 1 dep dep sin [41.4, 4] [0.66, 0.67] Alt 1 of 4 Alt of 4 [7.9, 8] ft ft their angle A Alternative method 1 = 10 + 15 10 15 cos B 10 15 1 10 15 0.60 (B =) [5.8, 53.] h sin (their 5.89) = 10 dep dep sin [5.8, 53.] [0.79, 0.8] [7.9, 8] ft ft their angle B 8 of 9
Alternative method 3 1 x = 10 (15 x) h = 1 x and h = 10 (15 x) 144 x = 100 (5 15x 15x + x ) dep 30x = 5 + 144 100 30x = 69 (x =) 69 30 (x =) 8.97 9 dep cont Alt 3 of 4 Alt 4 of 4 [7.9, 8] ft ft their x, dependent on Alternative method 4 10 y = 1 (15 y) h = 10 y and h = 1 (15 y) 100 y = 144 (5 15y 15y + y ) dep 30y = 5 + 100 144 30y = 181 (y =) 181 30 (y =) 6.03 6 dep [7.9, 8] ft ft their y, dependent on 9 of 9