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Edexcel Maths Paper 2 MS

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Guidance on the use of codes within this mark scheme M1 – method mark. This mark is generally given for an appropriate method in the context of the question. This mark is given for showing your w ... orking and may be awarded even if working is incorrect. P1 – process mark. This mark is generally given for setting up an appropriate process to find a solution in the context of the question. A1 – accuracy mark. This mark is generally given for a correct answer following correct working. B1 – working mark. This mark is usually given when working and the answer cannot easily be separated. C1 – communication mark. This mark is given for explaining your answer or giving a conclusion in context supported by your working. Some questions require all working to be shown; in such questions, no marks will be given for an answer with no working (even if it is a correct answer). Part Working an or answer examiner might expect to see Mark Notes (a) 2  2  3  7 M1 This mark is given for a 2, 2 3 and 7 seen A1 This mark is given for the correct answer only (b) 60, 120, 180, 240, 300, 360, 420 … M1 This mark is given for a method to find 84, 168, 252, 336, 420 … the LCM or 84 = 2  2  3  7 60 = 2 × 2 × 3 × 5 LCM = 2  2  3  5  7 420 A1 This mark is given for the correct answer only Question 2 (Total 5 marks) Part Working or answer an examiner might expect to see Mark Notes (a) M1 This mark is given for 2 and 10 correctly placed in the intersection 4 6 8 2 10 1 5 M1 This mark is given for 4, 6 and 8 placed in A only or 3 7 9 1 and 5 placed in B only or 3, 7 and 9 placed in (A  B) C1 This mark is given for all numbers correctly placed in the Venn diagram (b) n(A  B) = 2 M1 This mark is given for a method to identify the number of elements in A  B 2 A1 This mark is given for the correct answer 10 only Part Working or answer an examiner might expect to see Mark Notes 3000  5 = 600 P1 This mark is given for a start to the process to solve the problem 1200 : 1800 P1 This mark is given for a process to find the ratio of the number of tins in small boxes to the number of tins in large boxes 1200 : 1800 = 200 : 90 P1 This mark is given for a process to find 6 20 the ratio of the number of small boxes to the number of large boxes 90 = 0.3103448…  31% 290 P1 This mark is given for a process to find to find the percentage of tins in large boxes Carlo is not correct; 31% of the boxes C1 This mark is given for a valid conclusion filled with tins are large boxes supported by correct working ..................................................CONTINUED.................................... [Show More]

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