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Intermediate 2 Past Paper Questions for 2001 with Solutions

Access complete Intermediate 2 past paper questions with solutions from 2001, sorted by topics. Use the formula sheet and follow the provided instructions to begin. Solve various math problems including equations, geometric calculations, and statistical analyses.

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Intermediate 2 Past Paper Questions for 2001 with Solutions

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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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