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Intermediate 2 Past Paper Questions with Solutions (2002)

This presentation contains Intermediate 2 past paper questions from 2002, sorted into topics with complete solutions. Access questions and solutions by clicking on the corresponding numbers on the grid.

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Intermediate 2 Past Paper Questions with Solutions (2002)

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  1. PRESS F5 TO START This presentation contains Intermediate 2 past paper questions complete with solutions for the year 2002. 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. 2002 Paper 1 (b) Prob (<72) = Solution Main Grid

  5. m = 5/2 c = 5 Equation: y = 5/2x + 5 Main Grid Solution

  6. 2nd Quadrant: (180° – 60°) = 120° 3rd Quadrant: (180° + 60°) = 240° Main Grid Solution

  7. Main Grid Solution

  8. 4 5 4 1 4 3 2 2 4 6 2 3 4 4 1 3 1 2 3 1 1 Main Grid Solution

  9. Position of median = (21 + 1) ÷ 2 = 11th No. Q1 = (1 + 2)÷2 = 1.5 Q2 (median) = 3 Q3 = 4 (c) The median for football matches is greater (5 > 3) so on average more matches attended than going to cinema. IQR is greater for football matches (6 > 2.5) so more variation in attendance. No. of Cinema Visits (b) No. of football matches 5 (a) 1 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 4 4 5 6 Main Grid

  10. x = 1 • (1, -16) • x = 1 Graph crosses x axis when y = 0 (x – 1)2 – 16 = 0 x2 – 2x + 1 – 16 = 0 x2 – 2x – 15 = 0 (x + 3)(x - 5) = 0 x = -3 and x = 5 So AB = 5 + 3 = 8 units Main Grid Solution

  11. (a) (b) Main Grid Solution

  12. 2002 P2 Main Grid Solution

  13. 3x – 2y = 11 (eq1) x2 2x + 5y = 1 (eq2) x3 6x – 4y = 22 6x + 15y = 3 subtract -4y – 15y = 22 – 3 -19y = 19 y = -1 Check with eq2 2x + 5y = 1 2 x 3 + 5 x -1 = 6 – 5 = 1 √ Solution (3, -1) Sub into equ 1 3x – (2 x -1) = 11 3x + 2 = 11 3x = 11 – 2 3x = 9 x = 3 Main Grid Solution

  14. Main Grid Solution

  15. Mean price is the same so on average milk prices in • local stores and supermarkets are similar. • Standard deviation of the local stores is much higher than the supermarkets, 17.7 > 10.5, so there is more variation in their prices. Main Grid

  16. Main Grid Solution

  17. Circumference of complete ‘circle’ = ∏ x D = 3.14 x 40 = 125.6 cm Fraction of circle pendulum swings thru = 28.6 ÷ 125.6 = 0.227707 Angle pendulum swings thru = 0.227707 x 360° = 81.97° Main Grid

  18. = 3y(y – 2) = (y + 3)(y – 2) Main Grid Solution

  19. Main Grid Solution

  20. Watch for radius! Vol = (⅓ x π x 82 x 32) - (⅓ x π x 52 x 20) = 2143.57 – 523.333 = 1620.24 = 2000 cm³ (1 sig fig) Vol required = vol of whole cone – vol of bottom cone Main Grid

  21. Main Grid Solution

  22. Main Grid Solution

  23. T 122° a 33° B A 80m SOH CAH TOA 51.38m Opp 33° B 1.6m Total height of pole = 21.7 + 1.6 = 23.3m Main Grid

  24. 2.5m a 2.5m 1.5m d 2.5m Pythagoras a² = c² – b² =2.5² – 1.5² a = √4 = 2m d = 2.5 – 2 = 0.5m Solution Main Grid

  25. Newtown’s Population In 2yrs = 50 000 x 1.052 = 55 125 In 3yrs = 50 000 x 1.05³ = 57 881 Coaltown’s Population In 2yrs = 108 000 x 0.82 = 69 120 In 3yrs = 108 000 x 0.8³ = 55 296 Solution In 3 yrs time Newtown’s population will be greater. i.e. 57 881 > 55 296 Main Grid

  26. Swop sides Main Grid Solution

  27. Main Grid Solution

  28. (a) H = 10 + 5Sin10° = 10.87m • H = 12.5m • 10 + 5Sint° = 12.5 • 5Sint° = 2.5 • Sint° = 2.5 ÷ 5 = 0.5 • Sine positive in quadrant 1 and 2 • Acute angle: t = sin-1(0.5) = 30 • 2nd quad: t = 180 – 30 = 150 • t is in seconds, so height is 12.5m • after 30s and 150s Main Grid

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