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PRESS F5 TO START. This presentation contains Intermediate 2 past paper questions complete with solutions for the year 2001. The questions are sorted into topics based on the specific outcomes for the Intermediate 2 course.

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Main Grid

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  1. PRESS F5 TO START This presentation contains Intermediate 2 past paper questions complete with solutions for the year 2001. 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. 2001 Paper 1 (x + 5)(x – 3) Main Grid Solution

  5. Vert = 30 Intercept = 5 Hor = 3 Gradient = vert ÷ horiz = 30 ÷ 3 = 10 Equation is y = 10x + 5 Main Grid Solution

  6. 2x + y = 5 eq 1 x – 3y = 6 eq 2 x2 gives 2x + y = 5 2x – 6y = 12 Subtract y – (-6y) = 5 – 12 7y = -7 y = -1 Sub y = -1 into eq 1 2x + -1 = 5 2x = 6 x = 3 Check with eq 2 x – 3y = 6 3 – 3(-1) = 3 + 3 = 6 √ Pt of intersection (3, -1) Main Grid Solution

  7. Swop sides Main Grid Solution

  8. Lower Q1 = (38+41)/2 = 39 Median = 52 Upper Q3 = (61+62)/2 = 61.5 Position = (33+1)/2 = 17th Prob(>80) = SIQR = (Q3-Q1)/2 = (61.5 – 39)/2 = 11.25 Solution Main Grid

  9. Period = 120° So b = 360 ÷120 = 3 Main Grid Solution

  10. Perimeter of square = 4(2x + 2) = 8x + 8 Perimeter of rectangle = 2(x + 3) + 2L Perimeters equal so 2(x + 3) + 2L = 8x + 8 2x + 6 + 2L = 8x + 8 2L = 8x – 2x + 8 – 6 2L = 6x + 2 L = 3x + 1 Main Grid Solution

  11. Main Grid Solution

  12. 2001 P2 528 000 x 1.0244 = 580 542 = 581 000 nearest 1000 Main Grid Solution

  13. Main Grid Solution

  14. (b) Group B has the same mean of 60, so the average marks of both groups are the same. The standard deviation of Group B is higher than Group A’s (29.8 > 11.03) so Group B’s marks are much more varied. Main Grid

  15. 44 • Symmetrical distribution • (c) Average of 45 Main Grid Solution

  16. Gordon’s distance = 4.4 x 2 = 8.8 km Brian’s distance = 4.8 x 2 = 9.6 km Angle in triangle = 100 – 45 = 55° 8.8 km h 45° 55° 9.6 km h2 = g2 + b2 – 2gbCosH = 9.62 + 8.82 – (2x9.6x8.8xCos55°) = 72.688 h = 8.53 km apart Main Grid Solution

  17. 3 3 x = 2 (c) By symmetry B (5, 0) • When x = 0 • y = (0 – 2)2 – 9 • = 4 – 9 = -5 • Coords C (0, -5) • (2, -9) Main Grid Solution

  18. Main Grid Solution

  19. (a) Vol of cylinder = ∏r²h = 3.14 x 20² x 50 = 62 800 = 63 000 (2s.f.) • Vol of 1 Cone = ⅓ πr²h • Vol of 800 cones = ⅓ πr²h x 800 • (x3) 3V = 800 x πr²h • h = 3V = 3 x 63 000 = 8.36 • πr²x 800 3.14 x 3x3x 800 • Height of one cone = 8.4cm (2 s.f.) Main Grid

  20. = 2x³ + 3x² – x + 8x² + 12x – 4 = 2x³ + 11x² + 11x - 4 Main Grid Solution

  21. Angle at centre = 360 ÷ 5 = 72° 72° 10cm Area of triangle AOB = ½abSinO° = ½ x 10 x 10 x Sin72° = 47.55 cm² Area of pentagon = 47.55 x 5 = 237.8 cm² Main Grid Solution

  22. Main Grid Solution

  23. 9. Main Grid Solution

  24. Adj Opp AB = BC SOH CAH TOA Opp = Adj x Tan70° = 50 x Tan70° AB = 137.37cm (also = BC) 50 50 70 Main Grid More Solution

  25. Curved part = arc AC Perimeter = 191.88 + 137.37 x 2 = 466.6 cm Main Grid

  26. √ S A T C √ Main Grid Solution

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