DSE16_Compulsory_P2solE_set1.doc - OXFORD UNIVERSITY PRESS MOCK 16(I COMPULSORY PART PAPER 2 SOLUTION Compulsory Part Paper 2 Question No Key 1 B 2 D 3

# DSE16_Compulsory_P2solE_set1.doc - OXFORD UNIVERSITY PRESS...

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OXFORD UNIVERSITY PRESS MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION Compulsory Part Paper 2 Question No. Key Question No. Key 1. B 31. A 2. D 32. C 3. A 33. C 4. D 34. B 5. C 35. B 6. A 36. D 7. C 37. C 8. B 38. B 9. C 39. D 10. A 40. A 11. D 41. D 12. A 42. C 13. A 43. C 14. D 44. D 15. B 45. A 16. A 17. B 18. B 19. D 20. B 21. D 22. B 23. B 24. A 25. A 26. C 27. B 28. C 29. C 30. B © Oxford University Press 2016 P.1
OXFORD UNIVERSITY PRESS MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION Solutions to Paper 2 1. B 162 648 16 6 = 162 4 648 ) 2 ( ) 3 2 ( = 648 648 648 2 3 2 = 3 648 Alternative method: 162 648 16 6 = 162 4 648 ) 2 ( 6 = 648 648 2 6 = 648 2 6 = 3 648 2. D b b c a 3 = 2 b c a 3 1 = 2 c a 3 = b 3 c c a 3 = b 3 b = c a c 3 3 3. A a 2 + 8 ab + 16 b 2 9 a 2 b 2 = ( a + 4 b ) 2 (3 ab ) 2 = [( a + 4 b ) + 3 ab ][( a + 4 b ) 3 ab ] = ( a + 3 ab + 4 b )( a  3 ab + 4 b ) 4. D 5. C Rewrite the given equation as: ) 2 ( ... .......... .......... .......... 1 3 4 ) 1 ( ...... .......... .......... 1 22 5 3 s r s r (1) 4: 12 r 20 s + 88 = 4 12 r 20 s = 84 ...... (3) (2) 3: 12 r + 9 s = 3 .................... (4) (4) (3): 29 s = 87 s = 3 6. A a < 0 The graph opens downward. y -intercept of the graph = b < 0 The answer is A. 7. C Rewrite the compound inequality as 2 x 5 5 x + 1 and 5 x + 1 < 16. Solving 2 x 5 5 x + 1: 3 x 6 x 2 ................ (1) Solving 5 x + 1 < 16: 5 x < 15 x < 3 .................. (2) x must satisfy (1) and (2). The solutions are  2  x < 3. 8. B f ( k ) = 0 ( k 2 + k )( k + 1) = 0 k ( k + 1)( k + 1) = 0 k = 0 ( rejected ) or 1 f ( x ) = ( x 2 1)( x + 1) Remainder = f ( 2) = [( 2) 2 1]( 2 + 1) =  3 9. C Let x and y be the numbers of boys and girls in the class respectively. x 75% + y 50% = ( x + y ) 65% 0.75 x + 0.5 y = 0.65 x + 0.65 y 0.1 x = 0.15 y y x = 2 3 The ratio of the number of boys to the number of girls is 3 : 2. 10. A Cost = \$[70(1 40%) + 8] = \$50 Profit = \$(70 50) = \$20 © Oxford University Press 2016 P.2
OXFORD UNIVERSITY PRESS MOCK 16(I) COMPULSORY PART PAPER 2 SOLUTION 11. D a : b = 3 : 1 b = a 3 1 b : c = 2 : 3 c = b 2 3 = a 3 1 2 3 = a 2 1 a + 2 b 3 c = 3 a + 2 a 3 1 3 a 2 1 = 3 a 6 1 = 3 a = 18 12. A From the question, f ( x ) = k 1 x + 2 2 x k , where k 1 and k 2 are non-zero constants. f (2) = 5 k 1 (2) + 2 2 2 k = 5 8 k 1 + k 2 = 20 .................. (1) f (3) = 2 k 1 (3) + 2 2 3 k = 2 27 k 1 + k 2 = 18 .................... (2) (2) (1): 19 k 1 = 38 k 1 = 2 Substitute k 1 = 2 into (1). 8(2) + k 2 = 20 k 2 = 36 f ( x ) = 2 x 2 36 x f (1) = 2(1) 2 1 36 =  34 13. A Maximum error of the measurement = 2 1 1 cm = 0.5 cm Least possible volume of the sphere = 3 4 (5 0.5) 3 cm 3 = 3 cm π 2 243 14. D Let T ( n ) be the number of dots in the n th pattern. T (1) = 11 T (2) = T (1) + 6 = 11 + 6 = 17 T (3) = T (2) + 6 = 17 + 6 = 23 T (4) = T (3) + 6 = 23 + 6 = 29 T (5) = T (4) + 6 = 29 + 6 = 35 T (6) = T (5) + 6 = 35 + 6 = 41 T (7) = T (6) + 6 = 41 + 6 = 47 T (8) = T (7) + 6 = 47 + 6 = 53 The number of dots in the 8th pattern is 53. 15. B In ABC , by Pythagoras’ theorem, AC 2 = AB 2 + BC 2 AC = 2 2 30 16 cm = 34 cm ABC ~ CDE ( AAA ) CD AB = CE AC cm 2 1 cm 16 = CE cm 4 3 CE = 25.5 cm Perimeter of ACEF = 2( AC + CE ) = 2(34 + 25.5) cm = 119 cm 16. A Join AE .

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