Mark Scheme (Results) January 018 Pearson Edexcel International GCSE Mathematics A (4MB0) Paper 01R
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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 for what they have shown they can do rather than penalised for omissions. Examiners should mark according to the mark scheme not according 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 worthy of credit according 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. Crossed out work 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 correct answer only o ft follow through o isw ignore subsequent working o SC - special case o oe or equivalent (and appropriate) o dep dependent o indep independent o eeoo each error or omission
No working If no working is shown then correct answers normally score full marks If no working is shown then incorrect (even though nearly correct) answers score no marks. With working If there is a wrong answer indicated on the answer line always check the working 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 working that the correct answer has been obtained from incorrect working, award 0 marks. Any case of suspected misread loses A (and B) marks on that part, but can gain the M marks. If working is crossed out and still legible, then it should be given any appropriate marks, as long as it has not been replaced by alternative work. If there is a choice of methods shown, then no marks should be awarded, unless the answer on the answer line makes clear the method that has been used. If there is no answer on the answer line then check the working for an obvious answer. Ignoring subsequent work It is appropriate to ignore subsequent work when the additional work does not change the answer in a way that is inappropriate for the question: eg. Incorrect cancelling of a fraction that would otherwise be correct. It is not appropriate to ignore subsequent work when the additional work essentially makes the answer incorrect eg algebra. Transcription errors occur when candidates present a correct answer in working, and write it incorrectly on the answer line; mark the correct answer. Parts of questions Unless allowed by the mark scheme, the marks allocated to one part of the question CANNOT be awarded in another.
1 x x 3z z x 3 z OR x x z 3z x z x zx 3z 16 3 7 61 3 17 OR Factor tree 16 61 3 63 306 3 1 10 7 34 (prime factors of 16 or 61) HCF = 18 3 960 km 4.55 litres 91 litres (oe) 48 (km per gallon) 4 (a) 0 1 B1 (b) 1 B1 5 8 7 8 748 35 (m) (oe) for either of the other two lengths of the triangle 6 Shaded area = 4 4 1 4 (oe, can be implied) awrt 3.43 (cm ) 7 360 OR 180 4 4 4 90 4 15 8 1 100 100 8 1 8 100 100 40 (Accept ) OR (oe)
9 BC 1010 1 8 8 (oe) BC = 6 (cm) 10 (1,5), (5,1), (, 4), (4,), (3, 3) OR a 6 x 6 table with 5 possible double rolls identified OR at least 3 correct probability products added All 5 correct probability products added, 1 5 6 5, awrt 0.139, 13.9% 36 11 4 7 3 OY 6 4 OY "3" " 4" Second M mark for the modulus of their OY but not for OX or YX NB: Accept working for YO 5 (obtained from correct working) A0 max if OY 3 4 used 1 BCD 45 or DBC 45 and ACE=60 OR Join AB ABC 60 (Alt. Seg Thm) RBA=75 3 B1 B1 B1 ACB 75 o s on straight line an d Tangents to a circle have the same length. OR s on straight line and Isosceles s OR (From 1 st B1: Alternate Segment Thm twice)) NB: Accept angles on diagram
13 4x 45 9x Condone lack of brackets for only 45 = 5x (oe) x = 3 Accept x 3 14 8 1 or 3 (Pythagoras) (3) 1 8 8 1 8 1 8 1 NB: No working shown scores M0 M0 A0 even if correct answer given. 15 (a) 3, 6, 9, 1, 15 only 1 B1 (b) C = {6, 1, 18,, 48} First B mark in (b) can be implied by A B C {3, 9, 15} (so C ={3, 9,, 45}) 3 B1 B1 B1 A B C {3, 9, 15} NB: ft on (a) n A B C 3(cao) 16 (a) 9 5 4 6 (b) 5 1 1 0 14 8 B (-1eeoo) B (-1eeoo)
17 x1.05 New R or 1.05 4 (oe) y 0.75 0.75 7 = or 1.4 or 140 5 R (oe) 7 " " 1R 5 increase = 100 (oe) R 3 rd M mark for correct method to convert to required % 140 100, 1.4 1.0 100, etc. e.g. 40% w 5y x x 3y 5y x 18 (removing denominators correctly) 5yw xw x 6y 10y 4x (expanding) (OR w5y x 8y x (removing denominators) 5yw xw 16y x (expanding) ) 5yw 16y x xw (collecting terms in y) x w 1, xw y x (oe) 5w16 5w16 19 (a) 579 69 467 679 (oe, no errors) 4 ( ) ( ) (b) 55 85, 378 16 NB: No working seen scores M0 A0 0.67460.675 NB: ft on 4 figure accuracy of their (a) giving their 3 sf answer 1 B1 ft (c) 6.75 10-1 (or better) 1 B1 ft
0 (a) 4 1 (one term correct) t 4 "1 " 0 (equating their t f(t) to 0) t = + (cao) 4 4 "" 8 (oe) "" 8 4.5 (metres) 1 (a) x + y = 550 1 B1 (b) x y 1 50 1 5 50 8600 (oe) 1 B1 (c) x x " 1" 550 50 " 1 5 50 8600 " (oe but complete method to solve SEs for x and y with no errors) NB: c s SEs in (a) and (b) must be linear SEs in x and y with (a) having unit coefs. x = 5 y = 35 (a) 1 4 x OR 3x 6 (oe) 3 x x NB: 3 x scores A (b) B1 ft B1 ft Open circle at x=3 and closed circle at x= One single line joining the two circles
3 (a) One term correct 9x 30x (b) " 9x 30 x" 5 4 9x 30x 5 0 3x 5 (Attempt to factorise c s quadratic) x 4 (a) 6 10 sin ABC sin 50 5 OR 1 OR 1.67 3 3 oe 1 6 sin 50 ABC sin 10 (b) ABC =7.3637.4 awrt AB 10 sin 180 50 " ABC" sin50 10sin 180 50 " ABC " AB sin50 () ( ) (OR AB 6 10 610cos 180 50 " ABC " 6 10 6 10 cos 180 50 " ABC " ) AB AB = 1.741.7 (cm) awrt
5 (a) 1 1 1 of the tank filled in 1 hour, 4 48 4x so 1 1 1 (oe) seen 4 48 4x 3x 48x seen) or 1 1 (isw after correct answer 16 4x (b) ) The 3 taps fill 1 1 15 16 4x 3x " " 15 OR 48x So to fill in tank we must have 3x 15 " " 1 (tank) x 4 tanks of water in 15 hours 4 45x30 48x (removing denominators) 1 1 1 (OR 16 4x 15 () 4x = 40 ()) x = 10
6 (a) B (1 each incorrect pair) Probability pairs (0.3, 0.7), (0.9, 0.1), (0.4, 0.6) (b) 0.3 0.1 0.03, 3% (c) "0.30.9" OR "0.7 0.4" 0.30.9 0.7 0.4 11, 0 0.55, 55%
7 (a) ABC 90 o AC 30 15 AC= 33.54 awrt 33.5 (km) (b) Point X is st BX is perpendicular to CX (see diagram) o BCX 0 15 tan BCA BCA 6.565 30 Bearing of A from C = 70 " BCA" 0 o 4 () () () (OR 30 tan BAC BAC 63.435 15 o bearing of C from A is 00 "63.435" 136.565 bearing of A from C is 360 180 "136.565" (oe) ) o 316.565 awrt 317