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Comprehensive Solutions to Intermediate 2 Past Paper Questions from 2000

This presentation offers a complete set of Intermediate 2 past paper questions from the year 2000, meticulously organized by topics aligned with the course's specific outcomes. Each question in the main grid can be accessed by clicking its number, while solutions are readily available by pressing the space bar. A formula sheet is also available for reference. This resource is designed to facilitate better understanding and preparation for students by providing clear solutions and structured content.

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Comprehensive Solutions to Intermediate 2 Past Paper Questions from 2000

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  1. PRESS F5 TO START This presentation contains Intermediate 2 past paper questions complete with solutions for the year 2000. The questions are sorted into topics based on the specific outcomes for the Intermediate 2 course. To access a particular question from the main grid click on the question number. To get the solution for a question press the space bar. To access the formula sheet press the button To begin click on Main Grid button. F Main Grid

  2. Formulae List START Page

  3. This is the formula that we use

  4. 2000 Paper 1 Main Grid Solution

  5. Intercept = 10 c = 10 10 5 Equation y= -2x + 10 Main Grid Solution

  6. Main Grid Solution

  7. Triangle in semi circle is right angled 25° Angles in triangle sum to 180° 65° Tangent to circle = 90° 25° BAC = 25° Solution Main Grid

  8. Main Grid Solution

  9. (a) 58 59 59 60 60 61 61 62 62 64 65 Q2 Q3 Q1 Position of median Q2= (11 + 1)/2 = 6th number Lower Quartile is 59 Upper Quartile is 62 (b) The median is 61 which is slightly above the average contents of 60. IQR is 3 (c) The median of 58 is well below the average contents of 60. IQR is 4 so much more varied amounts could be expected in the boxes too Main Grid

  10. (x,y) for (5, 50) y = ax² 50 = a x 5² a = 50 ÷ 25 a = 2 Main Grid Solution

  11. Period = 360 ÷ 2 = 180 So b = 2 Main Grid Solution

  12. Solution Main Grid

  13. 2000 P2 Main Grid Solution

  14. Main Grid Solution

  15. 19.4 Width = radius of table + 19.4 = 40 +19.4 = 59.4 cm Pythagoras x² = 40² - 35² x =√375 = 19.4 cm 35 x 40 Main Grid Solution

  16. Amount due end of yr =£1000 x 1.185 =£1185 Cost of loan = £185 For 3 mths = £185 ÷4 =£46.25 £1000 x 1.015³ = £1045.67 Cost of 3 mth loan =£45.67 Advantage is cheaper by £0.58 Main Grid Solution

  17. Main Grid Solution

  18. (a) 3x + 50y = 88.50 Eq 1 x4 (b) 4x + 60y = 113.00 Eq 2 x3 12x + 200y = 354 12x + 180y = 339 Subtract200y – 180y = 354 – 339 20y = 15 y = 15 ÷20 = 0.75 Put into Eq 1 3x + 50 x 0.75 = 88.50 3x = 88.50 – 37.5 3x = 51 x = 17 Check with Eq 2 4 x 17 + 50 x 0.75 = 113 Car hire is £17 a day Fuel costs £0.75 per litre Main Grid

  19. Main Grid Solution

  20. Area of cross section = (20 x 10) + ¼ x ∏ x 20² = 200 + 314 = 514 cm² Volume = Ah = 514 x 42 = 21 588 =21 600 cm³ (3 sig figs) Volumes are the same, so the area of cross sections are equal since height is the same. Both have same rectangle 20 x 10. So area of area of triangle = ¼circle = 314 cm² So ½ x base x 20 = 314 Base x 20 = 628 Base = 628 ÷ 20 = 31.4 x = 31.4cm Main Grid

  21. Main Grid Solution

  22. D Adj 35m SOH CAH TOA 10° 35.5m 10° A Hyp 34m d B Use Cosine rule Main Grid

  23. Main Grid Solution

  24. Arc AC = arc BC = Perimeter AB = One circumference =∏ x D = 3.14 x 8 = 25.12 cm Total perimeter = 25.12 + 2 x 16.75 = 58.62 cm Main Grid

  25. Main Grid Solution

  26. B9 Main Grid Solution

  27. 3 3 x = 3 x = 3 B(6, 11) (3, 20) Main Grid Solution

  28. √ √ S A T C Main Grid Solution

  29. Main Grid Solution

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