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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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