AQA Qualifications GCSE Mathematics Unit : Higher 4360H Mark scheme 4360H June 016 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 016 AQA and its licenss. All rights reserved. AQA retains the6copyright 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 4360H JUNE 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 M dep A B B dep ft SC oe [a, b] Method marks are awarded f a crect method which could lead to a crect answer. A method mark dependent on a previous method mark being awarded. 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. A mark that can only be awarded if a previous independent mark has been awarded. Follow through marks. Marks awarded following a mistake in an earlier step. Special case. Marks awarded within the scheme f a common misinterpretation which has some mathematical wth. Or equivalent. Accept answers that are equivalent. eg, accept 0.5 as well as Accept values between a and b inclusive. 1 3.14 Accept answers which begin 3.14 eg 3.14, 3.14, 3.1416 3 of 18
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 goes 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 of 18
MARK SCHEME GENERAL CERTIFICATE OF SECONDARY EDUCATION 4360H JUNE 016 7x 3x 4x 3 + 9 3 oe 3x 7x 4x 9 3 3 1 7x 3x = 3 + 9 4x = 3 oe 3x 7x = 9 3 4x = 3 8 ft ft M0 with one rearrangement arithmetic err (third term = ) 4a (fourth term = ) 8a 7a ( = 63) 15a a = 63 7 a = 9 seen implied 8 9 15 9 135 a = 9 is implied by second term 18 third term 36 fourth term 7, not from an increct sequence 3(a) Straight line from (0900, 0) to (1100, 10) B1 3(b) Straight line from (1030, 0) to (100, 10) B1ft ft (100, their 10) from their distance at 1100 in part (a) 5 of 18
80 B1ft ft speed from their distance-time graph f Train B 3(c) If their distance-time graph f train B goes from (1030, 0) to (100, 10) the answer f (c) must be 80 F ft their distance-time graph f Train B must be a straight line f at least 90 minutes 0.8 1550 140 oe 1950 3 1300 oe their 1300 0.05 their 1300 0.95 their 1300 135 their 1300 can be their 140 if greater than 150 140 and 135 as final values 4 (Car) B Q1ft Strand (iii) ft f crect decision based on their values, with at least M sced and one crect final value SC 1368 and 100 116.5(0) and 10.5(0) SC1 1368 116.5(0) 10.5(0) Car A = 140 and Car B = 1300 with crect decision of Car A M0A0Q1ft 6 of 18
Alternative method 1 3 5 (+) 1 1 1 4 6 (+) 4 4 10 4 oe common denominat 5 7 1 their 1 1 1 their 5 10 14 1 their 4 4 4 their 10 14 (blue discs) oe 5 1 1 their must be from 1 4 6 7 Alternative method 5 Multiple of 1 f total number of discs Number of red discs and white discs in ratio 3 : Implied by LCM of 1 eg 6R, 4W Numbers of discs in ratio 3 : : 7 eg 6R, 4W, 14B 7 7 out of 1 on answer line 7 A0 on answer line 1 3 (red) (white) 7 (blue) without 7 on answer line A0 1 4 1 6 1 10 10 M0A0 7 of 18
x(3x 7) B B1 (3x 7x) x(6x 14) SC1 x(3x + 7) 6 Allow multiplication signs f B B1 eg x (3x 7) Condone missing final bracket eg x(3x 7 Accept (x + 0)(3x 7) B B B 7(a) 3,, 1, 0, 1 B1 Any der 3,, 1, 0, 1, B0 8 of 18
B1 10 10 B and 10 7(b) 10 Intention must be clear to indicate x > with minimum of a line drawn to the right of hollow circle positioned at B1 Intention must be clear to indicate x 10 with minimum of a line drawn to the left of filled circle positioned at 10 B1 10 9 of 18
Alternative method 1 150 (5 ) 150 3 50 their 50 7 their 50 5 50 and their 50 100 dep 350 SC1 10 8 Alternative method 5 x 150 x oe 5x = (x + 150) (x =) 100 and (x + 150 =) 50 350 SC1 10 50 and 100 is at least 9(a) 5 B1 9(b) 1 B1 3 7 3 9(c) 1 1 7 7 49 1 3 49 10 of 18
x + y = 7 B1 oe allow = any inequality sign x y > 1 B1 oe 10 x and y > 1 and x + y < 7 Q1 oe Strand (i) crect use of notation SC x and y < 1 and x + y > 7 x > and y 1 and x + y 7 Alternative method 1 x x 0 80 1 4 oe 4(x 0) = x + 80 11 4x 80 = x + 80 x x 0 70 4 4x x = 80 + 80 oe crect expansion of their brackets division sces M 3x = 360 x x = 70 + 0 4 collecting their four terms sces M3 3x 90 4 x = 10 SC3 380 11 of 18
Alternative method x + 80 (x 0) (= 3 parts) 300 (= 3 parts) and 100 (= 1 part) sces M x 0 = 100 x + 80 = 400 sces M3 x = 10 SC3 380 Alternative method 3 11 cont x 0 + x + 80 = 5(x 0) x + 60 = 5x 100 sces M 3x = 360 sces M3 x = 10 SC3 380 x 0 = 4(x + 80) x 0 = 4x + 110 1140 = 3x M0 1(a) 5.83 10-4 B1 941 600 B1 1(b) Accept 941,600 941 600.0( ) B1 1 of 18
7 00 000 000 300 700 10 6 300 7. 10 9 300 4 million oe 1(c) 4 000 000 4 10 6 0.04 10 9 oe.4 10 7 ft ft and their 4 000 000 written in standard fm 15 and 13 B1 13(a) Do not accept increct der eg 13 and 15 B0 n + 5n + n + 6 n + 7n + 6 13(b) (n + a)(n + b) where ab = 6 a + b = 7 (n + 1)(n + 6) with full wking seen igne further wk (n + 1) (n + 6) with full wking seen 13 of 18
Alternative method 1 (Elimination) 3a 5 = b 3a + 3b = 69 a + b = 3 oe b 5 = a 3 64 5 = a + b 3 fms one crect equation 14 3a b = 5 and 3a + 3b = 69 3a b = 5 and a + b = 3 oe fms two crect equations with equal coefficients f one unknown 4a = 8 a = 7 4b = 64 b = 16 oe crectly eliminates one unknown strand (ii) complete and crect algebra a = 7 and b = 16 Q1 SC f a = 7 and b = 16 with one crect equation SC1 f a = 7 and b = 16 without wking using trial and improvement 14 of 18
Alternative method (Substitution) 3a 5 = b 3(a + b) 5 = 64 oe b 5 = a 3 3a + 3b = 69 a + b = 3 64 5 = a + b 3 fms one crect equation 14 cont b = 3a 5 and 3a + 3b = 69 a = 3 b and 3a b = 5 oe 3a + 3(3a 5) = 69 3(3 b) b = 5 fms two crect equations with an unknown as the subject of one of them 1a = 84 a = 7 4b = 64 b = 16 oe crectly substitutes f one unknown and simplifies strand (ii) complete and crect algebra a = 7 and b = 16 Q1 SC f a = 7 and b = 16 with one crect equation SC1 f a = 7 and b = 16 without wking using trial and improvement Be careful f that the equations are not equivalent eg a = 3 b and 3a + 3b = 69 M0 15 (x + 1)(x 1) (x + 5)(x + 1) x 1 x 5 Do not allow further wk 15 of 18
Alternative method 1 x 3x 3x x 6x b = 9 a a = 3 oe a = 6 b = 15 16(a) Alternative method Substitutes a value f x into the identity and obtains a crect equation in a and b a = 6 b = 15 x = 0 gives b = 9 a x = 1 gives 1 + a + b = 4 a x = gives 4 + a + b = 1 a x = 3 gives 9 + 3a + b = a 16(b) B1 16 of 18
17(a) 16 16 4 16 oe 56 18 64 8 do not igne further wk 17(b) 5 5 3 5 3 3 5 10 3 3 allow one err in four terms 8 10 3 a = 8 and b = 10 17 of 18
6(x + 3) ( )(x ) 6x + 18 x 4 x + 4 (x )(x + 3) 6x + 18 x + 4 4x + x x + 3x 6 x + x 6 x 3x 8 = 0 allow three crect terms after expansion igne RHS and denominat allow three crect terms after expansion as denominat RHS 18 crect method to solve their quadratic equation by (x 7)(x + 4) (= 0) crect substitution into the quadratic fmula crect completion of the square crect factisation (x =) 7 and (x =) 4 SC (x =) 7 (x =) 4 Crect substitution into quadratic fmula 3 x ( 3) 1 4 1 8 18 of 18