Main Grid This presentation contains Intermediate 2 past paper questions complete with solutions for the year 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

Topic Units 1, 2 & III Significant Figs Scientific Notation % Calculations 4 Volumes of Solids 6 Linear Relationships 2 Multiplying out Factorising 32 Circles: arcs, sectors, symmetry, chords 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

This is the formula that we use

Main Grid Solution 2000 Paper 1 MarksFreqCum Freq

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

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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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Main Grid (a) 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

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

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

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Main Grid Solution 2000 P2

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Main Grid Solution x Pythagoras x² = 40² - 35² x =√375 = 19.4 cm Width = radius of table = = 59.4 cm 19.4

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

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Main Grid (a) 3x + 50y = Eq 1 x4 (b) 4x + 60y = Eq 2 x3 12x + 200y = x + 180y = 339 Subtract200y – 180y = 354 – y = 15 y = 15 ÷20 = 0.75 Put into Eq 1 3x + 50 x 0.75 = x = – x = 51 x = 17 Check with Eq 2 4 x x 0.75 = 113 Car hire is £17 a day Fuel costs £0.75 per litre

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Main Grid Area of cross section = (20 x 10) + ¼ x ∏ x 20² = = 514 cm² Volume = Ah = 514 x 42 = = 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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Main Grid 35m 10° Adj Hyp SOH CAH TOA Use Cosine rule 34m 10° D A B 35.5m d

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Main Grid Perimeter AB = One circumference =∏ x D = 3.14 x 8 = cm Arc AC = arc BC = Total perimeter = x = cm

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Main Grid B9 Solution

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

Main Grid Solution S A T C √ √

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