1. Main Grid This presentation contains Intermediate 2 past paper questions complete with solutions for the year 2003. 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. PRESS F5 TO START F

2. Topic Units 1, 2 & 3 2003 III Significant Figs Scientific Notation % Calculations Volumes of Solids 36 Linear Relationships 4 Multiplying out Factorising 1 Circles: arcs, sectors, symmetry, chords 71 8 Trigonometry Sine Cosine Rules Area of triangle 10 Simultaneous Equations 3 Graphs, Charts Tables Cumulative Freq Dotplot Boxplot 5 fig summary 42 Statistics: Standard Deviation Cumulative Freq Diag Line of Best Fit Probability 25 Algebraic Fractions Change of Subject Surds & Indices 677 1111 Quadratic Functions Graphs, Formula 89 Trigonometry Graphs Equations, Identity 512 Formulae List START Page

3. This is the formula that we use

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6. Main Grid 12345 RedR,1R,2R,3R,4R,5 YellowY,1Y,2Y,3Y,4Y,5 BlueB,1B,2B,3B,4B,5 GreenG,1G,2G,3G,4G,5 2 (a) (b)

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10. Solution a = 3(amplitude 3x ‘normal’ height) b = 2( period = 360 ÷ 2 = 180

11. Main Grid Solution Cancelling by (x + 1)

12. Main Grid Solution 6cm 8cm

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14. 8 (a)7 + 6x – x 2 factorises to (7 – x)(1 + x) (b) Roots are;x = 7 and x = -1 (c Turning Point halfway between –1 and 7. i.e. x = 3 So y = 7 + (6 x 3) – (3 2 ) = 7 + 18 – 9 = 16 Coords of Max T.P.(3,16) Main Grid

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16. Main Grid OLN = 90° (tangent) PLJ = 90° – 47° = 43° PKL = 90° (angle in a semi circle) PLK = 180° – (90° + 31°) = 59° KLJ = PLK + PLJ = 43° + 59° = 102°

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20. Main Grid (a) x + y = 130eq 1 x 50 (b) 30x + 50y = 6000eq 2 50x + 50y = 6500 30x + 50y = 6000 Subtract20x = 500 x = 25 Sub into eq 1 25 + y = 130 y = 105 25 tickets at £30 105 tickets at £50

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23. Main Grid (b)Yes. Mean temperature of 19°C is within range of 15°C to 25°C Standard deviation of 3.65 °C is less than 5 °C so less variation should mean good growth.

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25. Main Grid 6(a) Total area of cross section = area of rectangle + area of circle =(30 x 46) + (3.14 x 15²) =2086.5 cm² Volume of prism = Ah = area of cross section x height = 2086.5 x 25 = 52162.5 cm³ = 52 000 cm³ (2 sig figs) (b)So volume of quarter of a cylinder. Vol = ¼πr 2 h 30 000 = ¼ x 3.14 x r 2 x 20 r 2 = 30 000 x 4 3.14 x 20 r = √1910.8 = 43.7cm

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31. Main Grid 10 (a) (b)

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33. Main Grid Solution S A T C √ √ (a) = 1 (b)