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Module 5 Paper 2 Higher Tier June 2008

Module 5 Paper 2 Higher Tier June 2008. Area: ½ x 10.4 x 5.5. = 28.6 cm 2. 2 Marks. 5x - 3x = 7 + 4. 2x = 11. x = 5.5. 13 - 5y = 12. 13 – 12 = 5y. 5y = 1. 13 - 5y = 4 x 3. 6 Marks. y = ⅕ or 0.2. Y= 2. 2 Marks. 1 Mark. 1 Mark.

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Module 5 Paper 2 Higher Tier June 2008

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  1. Module 5 Paper 2 Higher Tier June 2008

  2. Area: ½ x 10.4 x 5.5 = 28.6 cm2 2 Marks

  3. 5x - 3x = 7 + 4 2x = 11 x = 5.5 13 - 5y = 12 13 – 12 = 5y 5y = 1 13 - 5y = 4 x 3 6 Marks y = ⅕ or 0.2

  4. Y= 2 2 Marks

  5. 1 Mark

  6. 1 Mark Sum: 25 + 33 + 34 + 35 = 127

  7. n + 8 n + 9 n + 10 n + n + 8 + n + 9 + n + 10 = 4n + 27 4n is always even. 27 is an odd number Even + Odd = Odd 6 marks

  8. Exterior angle of regular Octagon = 360 ÷ 8 = 45⁰ Exterior angle of regular Pentagon = 360 ÷ 5 = 72⁰ Angle a = 45 + 72 = 117⁰ 4 Marks

  9. 5d = c - 2 2 Marks

  10. Need to extend line so it goes from x =-1 to x=5 X X X 5 Marks x = 3.3 y = 1.7

  11. A B Area of A is 3 x 2 = 6cm2 Area of B is 9 x 2 = 18cm2 Area of cross section is 18 + 6 = 24cm2 Volume of block is area of cross section x length: 24 x 65 = 1560 cm3 5 Marks

  12. Equidistant from P and Q 3 marks

  13. Answer : x4 Answer: 6y7z5 Answer: 8p9r6 5 Marks

  14. OPP Use TAN Adj A = Tan-1 0.6875 = 34.5⁰ 3 Marks

  15. Adj Hyp Use Cosine DF = 24 x cos 64 DF = 10.5cm DF = 10.5(209.....) 4 Marks

  16. 30 x 1.20 = 36 15 X 1.10 = 16.5 3 Marks New area = 36 x 16.5 = 594cm2

  17. Answer: ( 5x + 1 ) ( x + 7 ) Answer: 3 ( y2 - 4z2 ) = 3 ( y - 2z ) ( y + 2z) 5 Marks

  18. Use Sine Rule YZ = 14.65cm (14.7cm) 3 Marks

  19. MC = OA (=a) –equal and parallel Hence AC = OM (and parallel) If opposite sides are equal and parallel then OACM is a parallelogram 2 Marks

  20. Volume of cone: ⅓ πr2h = 2400 r2 = 81.85 r = 9.04cm (9.05cm) 3 Marks

  21. Multiply every term by x ( x – 2 ) 4 ( x – 2 ) + 3x = x ( x – 2 ) 4x - 8 + 3x = x2 - 2x Collect terms together on one side and simplify x2 - 9x + 8 = 0 x = 1 or x = 8 Factorise 5 Marks ( x - 1 ) ( x – 8 ) = 0

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