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![Page 1: Main Grid This presentation contains Intermediate 2 past paper questions complete with solutions for the year 2002. The questions are sorted into topics.](https://reader035.fdocuments.net/reader035/viewer/2022062718/56649e6f5503460f94b6c35e/html5/thumbnails/1.jpg)
Main Grid
This presentation contains Intermediate 2 past paper questions complete with solutions for the year 2002.
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
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Topic Units 1, 2 & 32002
I II
Significant FigsScientific Notation
% Calculations 10
Volumes of Solids 6
Linear Relationships 2
Multiplying out Factorising 4
Circles: arcs, sectors, symmetry, chords 4
TrigonometrySine Cosine RulesArea of triangle
1 8
Simultaneous Equations 2
Graphs, Charts TablesCumulative FreqDotplot Boxplot5 fig summary
5
Statistics:Standard Deviation Cumulative Freq Diag Line of Best Fit Probability
1 3
Algebraic Fractions Change of SubjectSurds & Indices
7 9 11
Quadratic FunctionsGraphs, Formula 5 7
Trigonometry GraphsEquations, Identity 3 6 12
For
mul
ae L
ist
ST
AR
T P
age
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This is the formula that we use
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Main Grid
2002 Paper 1
Solution
Scores Freq Cum Freq
70 2 2
71 3 5
72 3 8
73 3 11
74 2 13
75 2 15
76 1 16
(b) Prob (<72) = 16
5
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Main Grid
Solution
m = 5/2 c = 5
Equation: y = 5/2 x + 5
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Main Grid Solution
2nd Quadrant: (180° – 60°) = 120°
3rd Quadrant: (180° + 60°) = 240°
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Main Grid Solution
313
31234
)14)(3(
23
223
2
xxx
xxxxx
xxx
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Main Grid Solution
4 5 4 1 4 3 2 2 4 6 2
3 4 4 1 3 1 2 3 1 1
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Main Grid
1 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4 4 4 5 6
Position of median = (21 + 1) ÷ 2 = 11th 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.
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Main Grid Solution
(a) (1, -16)
(b) x = 1
x = 1
Graph crosses x axis when y = 0
(x – 1)2 – 16 = 0
x2 – 2x + 1 – 16 = 0
x2 – 2x – 15 = 0
(x + 3)(x - 5) = 0
x = -3 and x = 5
So AB = 5 + 3 = 8 units
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Main Grid Solution
5
5253
5259
5245
(a) (b)
2
22
2
1
1
11
x
xx
x
x
xx
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Main Grid Solution
2
0
8.5437
651001202
1
2
1
m
Sin
abSinCArea
2002 P2
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Main Grid Solution
Sub into equ 13x – (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 3 + 5 x -1= 6 – 5= 1 √
Solution (3, -1)
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Main Grid Solution
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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.
49.10 110 5
550
16
3197432524
32524597975897066
319746
191844
6
)438(
)(
736
438 )(
438597975897066
2222222
22
s
x
n
x
n
xxMean
x
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Main Grid Solution
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Main Grid
Circumference of complete ‘circle’ = ∏ x D
= 3.14 x 40
= 125.6 cm
Fraction of circle
pendulum swings thru = 28.6 ÷ 125.6
= 0.227707
Angle pendulum swings thru = 0.227707 x 360°
= 81.97°
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Main Grid Solution
= 3y(y – 2)
= (y + 3)(y – 2)
3) (y
3y
2)– 3)(y (y
2)– 3y(y
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Main Grid Solution
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Main Grid
Vol required = vol of whole cone – vol of bottom cone
Vol = (⅓ x π x 82 x 32) - (⅓ x π x 52 x 20)
= 2143.57 – 523.333
= 1620.24
= 2000 cm³ (1 sig fig)
Watch for radius!
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Main Grid Solution
1d.p. to1.8-or 3.0
1.78-or 28.04
173
4
173
4
)8(93
22
)124()3(3
1 3 2
0132
2
2
or
x
cba
xx
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Main Grid Solution
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Main Grid
m
hOpp
7.21
Sin2538.51
Sin380
0
Total height of pole = 21.7 + 1.6
= 23.3m
mSin
Sin
SinSin
aSinT
t
SinA
a
38.51122
3380a
Sin3308aSin122
multiply) (cross 122
80
33
122)2533(180T Angle 0
B33°
122°
T
A 80m
a
B
51.38mOpp
1.6m33°
SOH CAH TOA
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Main Grid
Solution
Pythagoras
a² = c² – b²
=2.5² – 1.5²
a = √4 = 2m
2.5m
1.5m
a
2.5m
2.5m
d
d = 2.5 – 2
= 0.5m
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Main Grid
Solution
Newtown’s Population
In 2yrs = 50 000 x 1.052 = 55 125
In 3yrs = 50 000 x 1.05³ = 57 881Coaltown’s Population
In 2yrs = 108 000 x 0.82 = 69 120
In 3yrs = 108 000 x 0.8³ = 55 296
In 3 yrs time Newtown’s population will be greater. i.e. 57 881 > 55 296
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Main Grid Solutionx
x
x
xx
3
3
26
26
2
2
2
1
2
3
2
1
2
3
Swop sides
3
2t– r p
2t– r 3p
r 2t 3p
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Main Grid Solution
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Main Grid
(a) H = 10 + 5Sin10° = 10.87m
(b) H = 12.5m
10 + 5Sint° = 12.5
5Sint° = 2.5
Sint° = 2.5 ÷ 5 = 0.5
Sine positive in quadrant 1 and 2
Acute angle: t = sin-1(0.5) = 30
2nd quad: t = 180 – 30 = 150
t is in seconds, so height is 12.5m
after 30s and 150s