1 / 23

Module 5 Higher June 2005 Paper 2

Module 5 Higher June 2005 Paper 2. 1a. Calculate the area of the circular flower bed. State the units of your answer. Area of a circle is π r 2 (NOT π D). Area is π x 1.7 2 (or π x 1.7 x 1.7) = 9.07 or 9.08 Units mark: m 2. (3). 2.

roden
Download Presentation

Module 5 Higher June 2005 Paper 2

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Module 5 Higher June 2005 Paper 2

  2. 1a Calculate the area of the circular flower bed. State the units of your answer. Area of a circle is πr2 (NOT πD) Area is π x 1.72 (or π x 1.7 x 1.7) = 9.07 or 9.08 Units mark: m2 (3)

  3. 2 On the grid below, draw the graph of y = 7 – x for values of x from 0 to 7. (3)

  4. 3a Rotate the shaded flag 90° anticlockwise about the origin. Label this new flag with the letter A. (3)

  5. 3a Rotate the shaded flag 90° anticlockwise about the origin. Label this new flag with the letter A. (3)

  6. 3b Reflect the shaded flag in the line y = 1. Label this new flag with the letter B. (2)

  7. 4 Use the formula v = u + at to find the value of v when u = – 10, a = 1.8 and t = 3.7 v = u + at = -10 + 1.8 x 3.7 = -10 + 6.66 = -3.34 (2)

  8. 5 Solve the equation 7t – 3 = 6 + t 7t - 3 = 6 + t 6t - 3 = 6 6t = 9 t = 1.5 (3)

  9. 6 Match three of these equations with the graphs shown below.

  10. 7 (a) ABC is a right-angled triangle. AC = 19cm and AB = 9cm. Using Pythagoras 192 = BC2 + 92 Calculate the length of BC. Answer ........................................................cm (3 marks) 192 – 92 = 280 So BC = √280 = 16.7 to 1d.p.

  11. (b) PQR is a right-angled triangle. PQ = 11cm and QR = 24cm. We must find angle PRQ. We have the opposite and the adjacent sides. Calculate the size of angle PRQ. Answer ...............................................degrees (3 marks) Tan PRQ=11/24 So PRQ = tan-10.458… = 24.6° (to 1 d.p.)

  12. 8 Solve the equation 5x – 1 = 3(x + 2) Answer x = ....................................................... (3 marks) Expand the brackets first. 5x – 1 = 3x + 6 • 3x from both sides: • 2x – 1 = 6 7 – 1 = 6 So 2x = 7 x = 3.5

  13. 9. A solution to the equation x3– 8x = 110 lies between x = 5 and x = 6. Use trial and improvement to find this solution. Give your answer to one decimal place. To find out whether 5.3 or 5.4 is the best estimate for X we must try 5.35. Substituting 5.35 gives an answer of 110.33……, slightly too large, so X = 5.3 3 marks

  14. 10 The diagram shows a cylindrical can of beans. The height is 5.7cm. The radius of the base is 3.7cm. Calculate the total surface area of the can. Answer ............................................cm2 (5 marks) Area of side: the circumference x height of can 2π x 3.7 x 5.7 = 132.5….. 132.5 + 86 = 218.5 cm2 (answer should be 219 or 220 since they have only 2 sigfigs in the question Area of top and bottom: Π x 3.72 x 2 = 43.0…. x 2 = 86.0…

  15. 11 (a) Expand and simplify 4(2x – 1) + 3(x + 6)Answer ................................................................. (2 marks) (b) Expand x2 (4 – 2x) Answer ............................................2 marks) 8x -4 + 3x +18 = 11x +14 4x2 – 2x3

  16. Expand and simplify (x + 1)(x – 3)Answer ........................................................ (2 marks) x2 +x -3x - 3 x2 – 2x - 3 • Simplify 2x3y5 x 4x4y (2 marks) 8x7y6

  17. 12 (a) Which of these statements are correct? P all isosceles triangles are similar Q all squares are similar R all parallelograms are similar S all regular pentagons are similarAnswer ................................................................. (2 marks)(b)These two rectangles are similar.   56 ÷ 42 x 27 = 36 So x = 36 Calculate the value of x. (3 marks)

  18. 13 (a) On the grid below,draw the graph ofx2 + y2 = 9 y = 3 (b)Write down the equation of the tangent to the curve at the point (0,3). Answer ......................................................................... (1 mark)

  19. 14 The hour hand of a clock is 4.5cm long. The minute hand is 6.2cm long. 5/12 x 360 = 1500 Calculate the distance between the tips of the hands at 7 o’clock. Answer ................................cm (4 marks) Now we need to use the cosine rule. 4.52 + 6.22 – 2 x 4.5 x 6.2 x cos 1500 = 107 So the distance is √107 = 10.3 (1 d.p.)

  20. 15 A child’s toy is in the shape of a cone on top of a hemisphere. The diameter of the hemisphere is 15cm and the overall height of the toy is 26cm. Volume of a sphere is: 4/3πr3 Volume of a cone is: 1/3πr2h These are both given on the question paper. Calculate the volume of this toy. 4/3 x π x 7.53 = 1767.146 (3 d.p.) So the hemisphere is 883.6cm3 (4 sf) 5 marks Volume of toy = 883.6 + 1089.7 = 1973cm3 (4sf) Height of cone = 26 – 7.5 = 18.5 So 1/3 x π x 7.52 x 18.5 = 1089.7 (5sf)

  21. 16 The perimeter of a rectangle is 25cm. The length of the rectangle is x cm. Width = 12.5 cm - x (a) Write down an expression for the width of the rectangle in terms of x. (1 mark) x(12.5 – x) = 38 12.5x – x2 =38 Multiply by -2 -25x + 2x2 = -76 2x2 -25x +76 = 0 (b) The area of the rectangle is 38cm. Show that 2x2 – 25x +79 = 0 (2 marks)

  22. Solve the equation given in part (b) to find the value of x. Give your answer to 2 decimal places. (3 marks) Major clue that you have to use the formula (given at the front) and that you can’t factorise!! 2x2 – 25x +76 = 0 Use the formula: x = 25 ±√((-25)2 – 4 x 2 x 76) 2 x 2 so x = 7.28 or 5.22 or both

  23. T T 2.5 23.9 6° P Q B Q 3.8 17 The diagram shows a door-wedge with a rectangular horizontal base PQRS. The sloping face PQTU is also rectangular. PQ = 3.8cm and angle TQR = 6o • Plan: • Calc QT using trig • Then calc PT using Pythag The height TR is 2.5cm.  Calculate the length of the diagonal PT. Using sin, sin 6 = 2.5/QT QT = 23.916… Using Pythagoras, 23.92 + 3.82 = 586… PT = √586 = 24.2 cm (5)

More Related