Mark Scheme (Results) January 2022 Pearson Edexcel International GCSE In Mathematics B (4MB1) Paper 02

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Question Working Answer Mark Notes 1(a) 2.48 10  8 1 B1 cao (b) 0.000256 1 B1 cao (c) Numerator of 2.37 10  60 or 23.7 10  107 oe or an answer in the form 1.58 10  n or m10108 3 M1 ... Allow eg 23.7 10  59 or (25 1.3 10 −  ) 59 Implied by a correct single value in any form 15.8 10  107 or 158 10  106 oe A1 A correct single value seen in their working but need not be in standard form. 1.58 10  108 A1 Total 5 marks Question Working Answer Mark Notes 2(a) 9 8 7 6 10 7 8 7 + + + + + + a = or 47 8 7 a + = 2 M1 Correct method to find the mean no errors. 9 A1 (b) 6, 7, 7, 8, 9, “9”, 10 2 M1 For ordering allow correct ordering using their value for a. This is often seen in the text of the question 8 A1ft Answer must follow their value for part (a) If a 8 the median is 8 If a 7 the median is 7 If there is no value of a found in part(a) the median is 7.5 Check that a median of 8 does not come from calculating the mean (c) 34 49 11 72 1666 792 874  −  = − =   3 M1 A correct method to find the total age of the passengers who are not pensioners. "874" "874" or 34 11 23 − M1dep Dep on previous M being awarded. A correct method to find the mean age of the passengers who are not pensioners. 38 A1 Total 7 marks https://britishstudentroom-b430a.web.app/Question Working Answer Mark Notes 3 (a) y x = − drawn or 2 points correct 2 M1 Correct triangle A1 (–1, –4) (–5, –2) (–7, –8) (b) 2 points correct or 3 x coordinates correct or 3 y coordinates correct 2 M1 Correct triangle A1 (–7, 3) (–5, –1) (–1, 5) (c) enlargement 3 B1 SF – 0.5 B1 Centre (6, –1) B1 Total 7 marks B C https://britishstudentroom-b430a.web.app/Que Working Ans Mark Notes 4 2 2 40 3 64 4 x x   − + =     or 2 40 4 2 64 3 y y   −   + =   oe 40 3 2 64 4 x x   −   = −   or       40 4 −3 y = − 64 y2 6 M1 For substituting a correct expression for x or y into the quadratic equation to form an (un-simplified) quadratic equation in either x or y. Implied by the 2nd M1 2 2 1600 240 9 64 16 16 16 x x x   + − + =     or 1600 320 16 2 2 64 9 9 9 y y y     − + + =   oe M1 For a correct method to expand 2 40 3 4   − x     or 2 40 4 3   − y     resulting with 3 or 4 terms. Condone 1 error in total (numerical or sign). NB −3x2 and −4y2 will count as 1 error. This must then be subst into the correct equation. 25 2 15 36 0 16 x x − + = or 25 240 576 0 x x 2 − + = or 25 320 1024 2 0 9 9 9 y y − + = or 25 320 1024 0 y y 2 − + = oe M1 For correct 3 term quadratic dep on M1 (one of the 2 above) being awarded. 5 5 6 6 4 4 x x       − −    or (5 24 5 24 x x − − )( ) or (5 32 5 32 3 3 3 3 y y − − )( ) or (5 32 5 32 y y − − )( ) or ( 320) 4(25 1024) − −  2 or ( 240) 4(25 576) − −  2 M1 Solving their 3 term quadratic equation using any correct method - if factorising, allow brackets which expanded give 2 out of 3 terms correct. If using formula or completing the square allow one sign error. Working must be seen. By completing the square must see eg   2 25 24 ... 0 16 5 x     −  =   Allow calculation of the discriminant for their quadratic x y = 4.8 or = 6.4 or using discriminant ( 320) 4(25 1024) 0 − −  = 2 or(− −  = 320 25 1024 9 9 9 )2 4 0 ( ) ( 240) 4(25 576) 0 − −  = 2 or ( 15) 4 36 0 − −  = 2 (2516 ) A1 correct single value for x or y. Award A0 if more than one value of x or more than one value of y is given For discriminant the calculation must be correct and = 0 Only one solution. [therefore line intersects the C only once.] A1dep Dep on all previous method marks awarded and only 1 value of x and/or y given. For a correct conclusion stating only one solution [Show More]

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