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 10 Volumes of Solids 6 Linear Relationships 2 Multiplying out Factorising 4 Circles: arcs, sectors, symmetry, chords 4 Trigonometry Sine Cosine Rules Area of triangle 1 8 Simultaneous Equations 2 Graphs, Charts Tables Cumulative Freq Dotplot Boxplot 5 fig summary 5 Statistics: Standard Deviation Cumulative Freq Diag Line of Best Fit Probability 13 Algebraic Fractions Change of Subject Surds & Indices Quadratic Functions Graphs, Formula 5 7 Trigonometry Graphs Equations, Identity Formulae List START Page

This is the formula that we use

Main Grid 2002 Paper 1 Solution ScoresFreqCum Freq (b) Prob (<72) =

Main Grid Solution m = 5 / 2 c = 5 Equation:y = 5 / 2 x + 5

Main Grid Solution 2 nd Quadrant: (180° – 60°) = 120° 3 rd Quadrant:(180° + 60°) = 240°

Main Grid Solution

Main Grid Solution

Main Grid Position of median = (21 + 1) ÷ 2 = 11 th No. Q1 = (1 + 2)÷2 = 1.5 Q2 (median) = 3 Q3 = 4 5 (a) (b) No. of Cinema Visits No. of football matches (c) The median for football matches is greater (5 > 3) so on average more matches attended than going to cinema. IQR is greater for football matches (6 > 2.5) so more variation in attendance.

Main Grid Solution (a) (1, -16) (b) x = 1 x = 1 Graph crosses x axis when y = 0 (x – 1) 2 – 16 = 0 x 2 – 2x + 1 – 16 = 0 x 2 – 2x – 15 = 0 (x + 3)(x - 5) = 0 x = -3 and x = 5 So AB = = 8 units

Main Grid Solution (a) (b)

Main Grid Solution 2002 P2

Main Grid Solution Sub into equ 1 3x – (2 x -1) = 11 3x + 2 = 11 3x = 11 – 2 3x = 9 x = 3 3x – 2y = 11(eq1) x2 2x + 5y = 1(eq2) x3 6x – 4y = 22 6x + 15y = 3 subtract -4y – 15y = 22 – 3 -19y = 19 y = -1 Check with eq2 2x + 5y = 1 2 x x -1 = 6 – 5 = 1 √ Solution (3, -1)

Main Grid Solution

Main Grid (b) Mean price is the same so on average milk prices in local stores and supermarkets are similar. Standard deviation of the local stores is much higher than the supermarkets, 17.7 > 10.5, so there is more variation in their prices.

Main Grid Solution

Main Grid Circumference of complete ‘circle’ = ∏ x D = 3.14 x 40 = cm Fraction of circle pendulum swings thru = 28.6 ÷ = Angle pendulum swings thru = x 360° = 81.97°

Main Grid Solution = 3y(y – 2) = (y + 3)(y – 2)

Main Grid Solution

Main Grid Vol required = vol of whole cone – vol of bottom cone Vol = ( ⅓ x π x 8 2 x 32) - ( ⅓ x π x 5 2 x 20) = – = = 2000 cm³ (1 sig fig) Watch for radius!

Main Grid Solution

Main Grid Solution

Main Grid Total height of pole = = 23.3m B 33° 122° T A 80m a B 51.38m Opp 1.6m 33° SOH CAH TOA

Main Grid Solution Pythagoras a² = c² – b² =2.5² – 1.5² a = √4 = 2m 2.5m 1.5m a 2.5m d d = 2.5 – 2 = 0.5m

Main Grid Solution Newtown’s Population In 2yrs = x = In 3yrs = x 1.05³ = Coaltown’s Population In 2yrs = x 0.82 = In 3yrs = x 0.8³ = In 3 yrs time Newtown’s population will be greater. i.e >

Main Grid Solution Swop sides

Main Grid Solution

Main Grid (a) H = Sin10° = 10.87m (b) H = 12.5m Sint° = Sint° = 2.5 Sint° = 2.5 ÷ 5 = 0.5 Sine positive in quadrant 1 and 2 Acute angle: t = sin -1 (0.5) = 30 2 nd quad: t = 180 – 30 = 150 t is in seconds, so height is 12.5m after 30s and 150s