Mark Scheme (Results) January 2022 Pearson Edexcel International GCSE In Mathematics B (4MB1) Paper 01. 100% approved pass rate.

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Question Working Answer Mark Notes 1 180 2 2 3 3 5 =     198 2 3 3 11 =    or 2 180 198 3 90 99 3 30 33 2 10 11 5 5 11 11 1 11 1 1 2 M1 A correct method to enable the L ... CM to be found. Correct factorisation of 180 and 198 allow on factor trees. For the table they must divide both numbers by those shaded (or other combination with a product of 18) in grey at some point. All other lines are optional but must be correct if seen (2 and 5 could be combined to give 10). Division could be in a different order. Allow list of multiples – at least 5 correct for each number, excluding 180 and 198 eg. 360,540,720,900,1080 and 396, 594, 792, 990,1188 1980 A1 Allow 2 3 5 11 2 2    Total 2 marks 2 3 2 5 p p q r q ( 2 5 3 − ) or 3 2 5 q p q r p 3 3 2 ( − ) or pq p q r 3 2 2 (6 15 − ) or 3 2 5 pq p q r q ( 2 4 2 − ) or 3 2 5 pq p q r q 2 2 3 ( − ) or 3 a 2 term algebraic expression pq3 ( ) 2 M1 3 2 5 pq p q r 3 2 2 ( − ) A1 cao Total 2 marks 3 11 2 ( 2)2 3 −  − − oe 2 M1 Allow 11 2 (2)2 3 −  − or 11 2 4 3 −  − or 11 2( 2)2 3 − − − – 1 A1 cao Total 2 marks https://britishstudentroom-b430a.web.app/Question Working Answer Mark Notes 4 8 cos 15  = ABC 2 M1 Correct expression for trigonometric ratio of ABC Allow use of Pythagoras with tan or sine. Allow use of sine rule or cosine rule. Eg. AC = − =  15 8 12.7 2 2   and sin sin90 "12.7" 15 ABC = 57.8 A1 awrt 57.8 Total 2 marks 5 20 16 9 6   8 8 6 15     − −     −     − − 2 M1 Multiplying by scalar. Allow 1 numerical or sign error. There is no need to show an intention to subtract. 11 10 14 23   −     − A1 cao Total 2 marks 6 (a) −2, 1 1 B1 cao (b) 3 5 103 m − = oe M1 Forms an equation or a method to find m eg.  (103 2 ( )) 1 3 m − − = + allow use of alternative letters 36 2 A1 cao Total 3 marks https://britishstudentroom-b430a.web.app/Question Working Answer Mark Notes 7 Gradient = 8 6 ( ) 3.5 4 0 − − = − oe 3 M1 An expression to find m or equation involving m only eg. 8 = 4m – 6 y x = "3.5" ... or y x = − ... 6 M1 Accept an equation with a gradient of "3.5" or y-intercept of –6 in any form eg. y – 8 = "3.5"(x – 4) If ft gradient previous M mark must be scored. y x = − 3.5 6 A1 oe accept any simplified form with a unitary coefficient of y eg y x + = 6 3.5 Allow any equivalent fraction for 3.5. Allow 2 7 12 y x = − if y x = − 3.5 6 or y x + = 6 3.5 seen in working, otherwise do not isw Total 3 marks 8 55 143 2 6.5 6.239... 360 72        = =     3 M1 Correct method to find the length of arc AB "6.239..." 2 6.5 +  M1 dep add on the 2 straight edges 19.2 A1 awrt 19.2 accept 143 13 72  + Total 3 marks https://britishstudentroom-b430a.web.app/Question Working Answer Mark Notes 9 125x y 6 3 3 M1 Two terms correct as part of a product. Expect to see as the numerator of a fraction but allow separately. Allow 53 for 125 kx y a b M1 dep Must gain the previous M mark if not fully correct. Must be seen as a single expression involving a product of 3 terms. 2 of k = 5, a = "6"− 3 and b = 3 correct, seen k may be a fraction 5x y 3 3 A1 Total 3 marks 10 AD = BC and  =  BAD ABC are given 3 B1 Accept AD = BC and  =  BAD ABC stated just marking on the diagram is not sufficient AB is a common side M1 Accept AB = AB [ABC and BAD are congruent] by SAS AC = BD A1 Correct conclusion with no incorrect statements. Allow a fully correct statement of the cosine rule for each triangle and a conclusion as equivalent to SAS. To award the mark in this instance both previous marks need to be awarded. Total 3 mar [Show More]

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