0580 MATHEMATICS. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Report for Teachers.
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1 CAMBRIDGE INTERNATIONAL EXAMINATIONS Cambridge International General Certificate of Secondary Education MARK SCHEME f the May/June 205 series 0580 MATHEMATICS 0580/2 Paper 2 (Extended), maximum raw mark 70 This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners meeting befe marking began, which would have considered the acceptability of alternative answers. Mark schemes should be read in conjunction with the question paper and the Principal Examiner Rept f Teachers. Cambridge will not enter into discussions about these mark schemes. Cambridge is publishing the mark schemes f the May/June 205 series f most Cambridge IGCSE, Cambridge International A and AS Level components and some Cambridge O Level components. IGCSE is the registered trademark of Cambridge International Examinations.
2 Page 2 Mark Scheme Syllabus Paper Abbreviations cao crect answer only dep dependent FT follow through after err isw igne subsequent wking oe equivalent SC Special Case nfww not from wrong wking soi seen implied x 2 + 8x 5 final answer 2 B f 2 crect terms in final answer f 2x 2 + x 5x 5 5 Sammy and crect reason with 25.7% oe shown 2 B f 25.7% seen conversion of 2% to fraction and common denominat 44 2 B f seen u w final answer 2 B f 2 crect elements in final answer M2 f ( 4 7) 2 + ( ( 2) ) 2 oe f ( 4 7) oe ( ( 2) ) oe B 5 their cao their equivalent division with 5 5 fractions with common denominats M2 f 2 ( + ) 2 oe f area crect If zero sced B f top speed = 720 m per min total time = 0 sec
3 Page Mark Scheme Syllabus Paper (a) 4n oe final answer (b) n oe final answer 2 f a quadratic expression as final answer n oe in wking 2 8 M2 f 2(2 + 4) 2 = p( 2 + 4) 2 oe f f k = 72 k p = ( q + 4) M2 f f wking out distance speed 4 (a) a + 2b a a ( a 2b) oe e.g. figs figs 280 their speed f wking out km/h to m/s conversion 000 e.g. 4 oe their oe (b) Parallelogram PM equal and parallel to QR PM PS parallel to QR and MR found = a so 2 pairs of parallel sides SC f answer trapezium with reason PM parallel to QR 5 y < 8 y [ x oe and y [ x + 2 oe B2 f either y [ x oe y [ x + 2 oe SC2 f y = x oe and y = x + 2 oe SC f y K x y = x y K x + 2 y = x + 2
4 Page 4 Mark Scheme Syllabus Paper 597 cao 4 B f [9...] B2 f 597.[9...] B f If B sced B0 sced and an attempt at compound interest is shown SC f their 597[...] 5000 evaluated crectly provided answer positive and SC f their final answer rounded crectly to nearest $ from their me accurate answer 7 (a) B f 2,, 5 as prime facts (b) 90 2 B f 90k f listing multiples of each up to Crectly equating one set of coefficients Crect method to eliminate one variable x = 0.8 y = Dependent on the coefficients being the same f one of the variables Crect consistent use of addition subtraction using their equations If zero sced SC f 2 values satisfying one of the iginal equations if no wking shown, but 2 crect answers given 9 (a) f [ 0 ] oe 8 (b) 2 cao 2 f 8 9 oe 0 9 their (a) 20 (a) ( p + t)( y + 2x) final answer 2 B f y ( p t) + x( p + t) p( y + 2 x) + t( y + 2x) + 2 ( ) (b) 7( h + k)( h + k ) final answer 2 2 B f 7 ( h + k) ( h + k) ( h + k) ( 7( h + k) 2)
5 Page 5 Mark Scheme Syllabus Paper cao 4 2 f π 4 9, 48π f 2 4 π 4, 28π 272π f to 284.9, If A0 then B f their final answer rounded crectly to nearest whole number from their me accurate answer dependent on at least 22 (a) f a 2 2 matrix with 2 crect elements (b) a b 2 f 2 c d 4 k soi 5 det = 2 soi 2 (a) (b) x 5 ( + 2) x (c) 9x 0 cao 2 f 5 ( 5 x) (d) 5 x final answer oe 2 f crect first step e.g. y 5 y + x = 5 = x better f interchanging x and y, e.g. not need to be the first step y 5 = x x = 5 y, this does
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