1. Main Grid 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. PRESS F5 TO START F

2. Topic Units 1, 2 & 3 2000 III Significant Figs Scientific Notation % Calculations 4 Volumes of Solids 6 Linear Relationships 2 Multiplying out Factorising 32 Circles: arcs, sectors, symmetry, chords 43 83 8 Trigonometry Sine Cosine Rules Area of triangle 7 Simultaneous Equations 5 Graphs, Charts Tables Cumulative Freq Dotplot Boxplot 5 fig summary 5 Statistics: Standard Deviation Cumulative Freq Diag Line of Best Fit Probability 11 Algebraic Fractions Change of Subject Surds & Indices 89a Quadratic Functions Graphs, Formula 69b 10 Trigonometry Graphs Equations, Identity 711 Formulae List START Page

3. This is the formula that we use

4. Main Grid Solution 2000 Paper 1 MarksFreqCum Freq 522 635 716 828 9210

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

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

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9. Main Grid (a) 58 59 59 60 60 61 61 62 62 64 65 Q1Q1 Q2Q2 Q3Q3 Position of median Q 2 = (11 + 1)/2 = 6 th number Lower Quartile is 59Upper 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

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

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

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13. Main Grid Solution 2000 P2

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

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

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18. Main Grid (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

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20. Main Grid 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

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

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

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26. Main Grid B9 Solution

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

28. Main Grid Solution S A T C √ √

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