4024 MATHEMATICS (SYLLABUS D)
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1 UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS GCE Ordinary Level MARK SCHEME for the May/June 00 question paper for the guidance of teachers 404 MATHEMATICS (SYLLABUS D) 404/ Paper, maximum raw mark 00 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 before marking began, which would have considered the acceptability of alternative answers. Mark schemes must be read in conjunction with the question papers and the report on the examination. CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the May/June 00 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.
2 Page Mark Scheme: Teachers version Syllabus Paper GCE O LEVEL May/June Section A Qu Answers Mark Comments (a) f(7) = as final answer (b) t 5 = t t = Forms an equation in t and attempts to solve (a) (c) Attempt to make x the subject f (x) = 5x ( 00) 37.5% (b) 30% oe soi o.e [5] SC for (x =)5y + ($)5 (c) (i) $88 (ii) $79.0 $.8(0) cao 3 (a) Rectangle 3 cm by 8 cm ft [8] Accept.8 (b) (i) Constructs perpendicular bisector of ZY Arc of circle radius 9 centre X (ii) Labels the correct region (c) (i) P and Q correctly positioned (ii) (a) 4 ± m cao (b) 07 (± ) cao ft [7] to cross rectangle across rectangle No need to shade but must be correct Dep on correct P and Q UCLES 00
3 Page 3 Mark Scheme: Teachers version Syllabus Paper GCE O LEVEL May/June (a) 4(x 3( x + 3) ( x + 3)(x Single fraction. Brackets not essential. Multiplies the first fraction by (x ) and the second fraction by (x + 3) 8x 4 3x 9 ( x + 3)(x Multiplies out the numerator with at least pair of terms correct 5x 3 ( x + 3)(x oe as final answer (b) Squares both sides of the equation k 3n m = as final answer l (c) For num p ± r q p = 4 and r = 6 s.o.i. or used q = 08 or q = x = 3.07, x =.74 Final answers 5 (a) (i) p = 0.5, q = 0. r = 0.3 (ii) (a) 0.5 (b) (i) 7 (b) seen 0. (ii) soi x = 8 6 (a) Either 36 or 44 correct Other one correct [9] [8] B ft SC for both 3.0 to 3. and.7 to.74 seen Can be implied by x + x + 54 =78 After B0, allow SC for A EC ) = 68 or for sum = 80. A C ) O =, A B ) C = 68, (b) A B ) C = 68, B A ) C B CA ) = 68 Isosceles triangle = 44 and [5] Dep UCLES 00
4 Page 4 Mark Scheme: Teachers version Syllabus Paper GCE O LEVEL May/June (a) Mid value used o.e. Sum of (value frequency) / (hours) (b) 73, 78 (c) Correct scale, points correct and smooth curve (d) (i) 3.3 (hours) (ii) Upper quartile and lower quartile used.5 (hours) S P C ft ft [0] Minus each error P for 5 plots which could form ogive C reasonable curve Read at 40 ft within 0. Upper quartile ft within 0. UCLES 00
5 Page 5 Mark Scheme: Teachers version Syllabus Paper GCE O LEVEL May/June Section B Qu Answers Mark Comments 8 (a) p =.6 stated (b) Scales Five points plotted ft Smooth curve (c) x =.55 to.65 (d) (i) y = x (e) 4 (ii) Line drawn and attempt to read at intersect x =.4(0) to.5(0) S Pft C X L G Lost for ruled lines, incomplete, very thick (f) (i) Correct line drawn T Tangent of gradient part (e) (ii) (0, ) Yft ft from their attempted tangent (iii) y = 4x + Eft [] ft from their gradient and their intercept 90 9 (a) (i) π 6 Correct formula and 90 used 360 (ii) to 8.6(0) cm [Their Indep. Attempt to add radius π 8 Area of cross-section = 800 soi π 8 ] h h = 5.9(0) to 5.9 cm Indep. Forms equation (b) (i) (a) MN = x (b) Area of triangle = their (x x) Area of sector = 6π and Subtraction (ii) 0(6π x ) = 800 x = 0... to 0.3 x = 3.(0) to 3. cm [] Expect justification and a subtraction Forms equation Correct method of solution UCLES 00
6 Page 6 Mark Scheme: Teachers version Syllabus Paper GCE O LEVEL May/June (a) (i) 40 o (ii) or or oe 30 o Correct method leading to solution (b) (i) tan 40 o CT = oe 3 CT = 9.9 to 9.3(0) cm (ii) or Accept 0 + their CT for (their CT ) or ( their CT ) to 650 cm (iii) 0560 to 0600 ft 4 their (b)(ii) (iv) (a) 46 cm 79 cm ft 40 + their (b)(i) rounded up (b) 930 to 980 cm cao [] (a) (i) 6 5 Accept 6 5 but not 6, 5 or (6, 5) (ii) Enlargement Scale factor and not lost if transformation stated, when SC SC scored Centre (4, ) (iii) Shear (iv) y = x (+ c) y = x + Knowing the equation has gradient (b) (i) x-coordinate q y-coordinate p SC for q p (ii) x-coordinate q y-coordinate p SC for q p (iii) W = 0 0 [] UCLES 00
7 Page 7 Mark Scheme: Teachers version Syllabus Paper GCE O LEVEL May/June (a) (i) p q (ii) (p q) + p Correct method 4 3 p q cao 4 (b) (i) (a) 4 7 sin 55 o 67 to 67.5cm (b) Attempt at cosine rule XY = cos to 9.93 (cm) A Correct formula and sign and correct algebra soi SC for 396 to 397 seen (ii) (a) VZ = 5 6 VZ = 3.7 to 3.75 cm (b) 766 cm 3 (Accept ) ft [] Value of 6 and correct use of Pythagoras ft 3 their (b)(i)(a) their (b)(ii)(a) UCLES 00
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