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PRESS F5 TO START. 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.

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PRESS F5 TO START

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.

F

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Formulae List

START Page



2003 Paper 1

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Solution


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Solution


2 (a)

(b)

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Solution


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a = 3 (amplitude 3x ‘normal’ height)

b = 2 ( period = 360 ÷ 2 = 180

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Solution


Cancelling by (x + 1)

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Solution


6cm

8cm

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Solution


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Solution


8 (a) 7 + 6x – x2 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) – (32)

= 7 + 18 – 9

= 16

Coords of Max T.P. (3,16)

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Solution


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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  • x + y = 130 eq 1 x 50

  • 30x + 50y = 6000 eq 2

  • 50x + 50y = 6500

  • 30x + 50y = 6000

  • Subtract 20x = 500

  • x = 25

  • Sub into eq 1 25 + y = 130

  • y = 105

  • 25 tickets at £30

  • 105 tickets at £50

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  • 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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(4)

(3)

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Solution


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 = ¼πr2h

30 000 = ¼ x 3.14 x r2 x 20

r2 = 30 000 x 4

3.14 x 20

r = √1910.8 = 43.7cm

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Main Grid of circle

Solution


Main Grid of circle

Solution


Main Grid of circle


Main Grid of circle

Solution


Main Grid of circle

Solution


10 (a) of circle

(b)

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Solution


of circle

(a)

S A

T C

(b)

= 1

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Solution


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