Mark Scheme (Results) November 010 IGCSE IGCSE Mathematics (4400) Paper 4H Higher Tier Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holbn, London WC1V 7BH
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November 010 IGCSE Mathematics (4400) Mark Scheme Paper 4H Apart from Questions 1, 0 and 1(b)(ii) (where the mark scheme states otherwise), the crect answer, unless clearly obtained by an increct method, should be taken to imply a crect method. Question Wking Answer Mark Notes 1. a 10.73 M1 f 10.73.045 + 1.4 =.045... + 1.4 5.3 1.6014 3.44530 A1 f at least first 5 figures b 3.4 1 B1 ft from (a) if non-trivial Total 3 marks. 4 1.5 oe 3 M M1 f 4 1.15 5. 4 75 1 600 310 A1 cao Total 3 marks 3. a (7, 6) B B1 f 7 B1 f 6 b C (3, 10) D (11, ) C (11, ) D (3, 10) B B1 f (3, 10) B1 f (11,) Total 4 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes 4 a 1 (0.3 + 0.1) M1 0.6 A1 cao b 0.1 + 0.6 1 0.3 M1 do not award if ans to (a) > 1 0.7 A1 ft from (a) if ans to (b) < 1 c 0.3 160 M1 4 f 0.3 160 0.3 00 60 4 A1 cao Total 6 marks 5. 50 0.7 1 M1 f 0.7 1 7956 A1 cao Total marks 6. a.6 1.5 M1 f crect substitution 3 3.9 A1 cao b 35 = h.5 M1 f crect substitution 3 crect rearrangement 35 (h =) oe.5 3.4 A1 cao c 3V M1 y = 3V f y = oe h h 3 V A1 3V 3V f ± oe h h h Total 6 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes 7. a Q crect Vertices (6, 10) (9, 10) (6, 16) 3 B3 B f translation of crect shape crect vertices B1 f right-angled triangle with base 3 height 6 in the same b R crect Vertices (10, ) (13, ) (10, ) ientation as P B f R crect ft their Q B1 f translation of 4 to the right down ft their Q c Enlargement with scale fact 3 and centre (1, ) B B1 f Enlargement 3 B1 f (1, ) Award no marks if answer is not a single transfn Total 7 marks. 19.6 50000 3 M1 f 19.6 50000 90 000 100 1000 number with digits 9 50000 ½ km 100 1000 M1 f completing calculation "90000" 19.6 ½ 100 1000 9. A1 cao Total 3 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes 9. x > 1 3 B1 f x > 1 x > 1 oe y > B1 f y > y > oe x + y < oe B1 f x + y < x + y < oe SC B1 if all inequalities reversed Total 3 marks 10. ACO = 1 COB = 4 ACB = 90 OCP = 90 CBP = 111 BCP = 1 10 1 (90 + 1) 10 4 90 10 1 111 4 B1 B1 M1 Angles may be stated marked on diagram 4 A1 Award 4 marks f an answer of 4, unless obtained by a clearly increct method. Total 4 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes 11. a 1350 169 1 3 M1 1 1 M1 1 M1 f 100 100 f 1350 169 1350 169 1 1350 0.94 169 0.06 94 0.063 M1 f 1 0.94 100 94 b 9519 100 9519 oe 1.14 114 M1 f 1350 169 1.06 106. M1 f 1.06 1 106 10 0 Award both method marks f an answer of 6.4, 6.3 better. 6 A1 cao Do not award this mark if a denominat of 169 used. 3 M 9519 100 M f 9519 oe 1.14 114 9519 M1 f, 3.5 seen, 114 9519 114% = 9519, = 1. 14, x 9519 = 1.14x 350 A1 cao Total 6 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes 1. a 5 1 3 M1 f clear attempt oe to use vert difference hiz difference m = A1 f m = y = x + 5 oe B1 ft from their m SC If M0A0, award B1 f y = mx + 5 b y = x + c M1 c 5 y = x + 6 oe A1 ft from (a) SC If M0A0, award B f linear expression in which the coefficient of x is f L = linear expression in which the coefficient of x is oe inc L+x = k SC If M0, award B1 f x + 6 L = x + 6 ft Total 5 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes 13. 11x + x = 10 1x = 10 10 360 10( n ) 4 M1 May be implied by 15 f 1 n n (exteri angle =) 15 A1 360 10( n ) 11 = oe n n 360 360 10 = 11 n n 360 M1 simplified crect equation "15" in which n appears only once eg 360 11 = 10(n ) 360 11 = 10n 360 360 1 = 10 n 4 A1 cao Award 4 marks f an answer of 4 unless clearly obtained by an increct method. Total 4 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes 14. a 3 3 B3 3 Red B1 and crect on LH 9 9 4 9 3 9 Red White 4 3 White Blue Red White branches B All RH branches crect (B1 one RH branch crect ie 3 probabilities) Blue 9 Blue 4 3 Red White b 4 4 + oe 3 M1 9 9 1 Blue M1 16 oe A1 7 9 4 f " " 9 4 " " " " oe 9 f sum of both products 16 f oe 7 9 Award f crect use of probabilities (must be < 1) from their tree diagram. Total 6 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes 15. a 3.6 10 15 1 B1 cao bi Crect expression f xy stated 5 M1 clearly implied with 7 5 evaluated eg 35 10 m + n 3.5 10 (1) 10 m 10 n States clearly implies that xy = 3.5 10 m + n + 1 oe 3.5 10 (1) 10 m + n oe m + n + 1* A1 SC If A1 not sced, award B1 f 35 10 11 seen. *dep on (3.5 ) 10 (1) 10 m 10 n = (3.5 ) 10 1 bii m n = 7 oe B1 f m n = 7 oe inc m = n + 7 m = 3 n = 16 M1 Adding subtracting m + n = 11 and m n = 7 m = 19 n = A1 f both values crect Award 3 marks f both values crect, unless clearly obtained by an increct method. Total 6 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes 16. a k 3 M1 k 1 P = f P = but not f P = V V V Also award f a crect equation in P, V and a constant 1 P = some numerical value V k M1 k 1 = f 1 = f crect 4 4 substitution into an equation which sces first method mark (may be implied by crect evaluation of the constant) 43 A1 P = V Award 3 marks if answer is P = but k is evaluated as 43 in any part b 3V = 43 3V V = 43 M1 f 3V = 43 3V V = 43 V = 144 1 A1 Also accept ± 1 k V Total 5 marks 17. a 1 1 B1 cao b (.5-4) bar height 19 little squares B1 Allow + ½ sq (4-6) bar height 6 little squares B1 Allow + ½ sq Total 3 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes 1. 3 M1 f crect substitution ± 4 3 f this 3 expression with one me of, 4 3 3 crectly evaluated obtains 40 64 4 10 M1 (independent)f crect simplification of discriminant 6.3 0.79,.39 A1 dep on both method marks f values rounding to 0.79 and.39 ( 0.794,.374 ) Total 3 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes 19. a AE 4 = 16 5 M1 0 A1 cao bi 1 5 B1 cao bii 5 + 1 5 + OE "1" M M1 f (cos x = ) 5 OE 5 1 = 5 + 5 cos x 16 + 1 (cos OEC =) "1" = 5 + OE OE 5 cos x 16 + OE "1" 16 OE 16, using the midpoint of CD, cos OEC = 5. 5 OE complete, crect method of finding sin OEC tan OEC 5.5 1 = 16 + 16 cos OEC 1 = 16 + OE 16 OE cos OEC 133.4 A f answer rounding to 133.4 (133.435 ) 55 A1 f 0 oe 0.675 If OEC is used, award A1 f 176 oe 0.675 value 56 rounding to 46.6 seen. If midpoint of CD is used, 5.5 award A1 f oe 0.675 value rounding to 46.6 seen. Total 7 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes 0. x = 7x 10 5 M1 + 10 (may be implied by nd M1) y = 7 x 7x + 10 (= 0) oe M1 y 9y +100 (= 0) oe (x 5)(x ) (= 0) oe M1 (y 4)(y 5) (= 0) 7 ± 9 9 ± 441 7 ± 49 40 9 ± 41 400 7 ± 3 9 ± 1 x =, x = 5 A1 y = 4, y = 5 x =, y = 4 x = 5, y = 5 A1 dep on all method marks dep on all method marks (may be implied by nd M1) Total 5 marks 1. ai a + b 3 B1 aii 3a b B1 aiii ¾ a + ¾ b b + ¼(3a b) 3a B1 ¾(3a b) oe bi collinear, in a (straight) line oe B1 bii ¾ B1 dep on B1 in both (a)(i) and (a)(iii) Total 5 marks IGCSE Mathematics (4400) Paper 4H November 010
Question Wking Answer Mark Notes. ( x + 3)( x ) 4 B1 f crect factisation f 1+ ( x + 4)( x ) crect single fraction, even if unsimplified ( x + 4)( x ) + x + x 6 ( x + 4)( x ) ( x + 4)( x ) + x + x 6 x + x x + 3 x + 3x 14 B1 1+ x + 4 ( x + 4)( x ) x + 3x 14 x + x ( x )[( x + 4) + ( x + 3)] ( x + 4)( x ) x + 4 + x + 3 x + 4 x + 3 B1 + x + 4 x + 4 x + 4 (x + 7)( x ) ( x + 4)( x ) x + 7 B1 x + 4 Total 4 marks TOTAL FOR PAPER: 100 MARKS IGCSE Mathematics (4400) Paper 4H November 010
IGCSE Mathematics (4400) Paper 4H November 010
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