intermediate 2 – additional question bank exit unit 1 please decide which unit you would like to...

INTERMEDIATE 2 – ADDITIONAL QUESTION BANK

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UNIT 1

Please decide which Unit you would like to revise:

Calculations using %Volumes of SolidsLinear RelationshipsAlgebraic OperationsCircles

TrigonometrySimultaneous LinearEquationsGraphs, Charts & TablesStatistics

Algebraic OperationsQuadratic FunctionsFurther Trigonometry

UNIT 2 UNIT 3

Calculations in a SocialContextLogic DiagramsFormulae

UNIT 4


UNIT 1 :

Linear Relationships

Calculations usingPercentages

Circles

AlgebraicOperations

Volumes ofSolids

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UNIT 1 : Calculations usingPercentages

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Please choose a question to attempt from the following:

1 2 3 4

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Calculations using Percentages : Question 1

In 2001, John deposits £650 in the bank at an interest rate of 3.5%.

(a) How much is his deposit worth after 1 year?

(b) In 2002, the interest rate changed to 3.2%, and in 2003 it changed again to 4.1%. Calculate how much interest John will have earned at the end of 2003.

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Calculations using Percentages : Question 1

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Reveal answer only (a) £22.75

(b)£72.75

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Question 1

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(a) 3.5% of 650 = 0.035 x 650 = 22.75

Value after 1 year = 22.75 + 650 = £672.75

OR 650 x 1.035 = £672.75

(b) 3.2% of 672.75 = 0.032 x 672.75 = 21.53

Value at end of 2002 = 672.75 + 21.53 = £694.28

4.1% of 694.28 = 0.041 x 694.28 = 28.47

Value at end of 2003 = 694.28 + 28.47= £722.75

Total interest = 722.75 – 650 = £72.75

OR 672.75 x 1.032 x 1.041 = 722.74

Total Interest = 722.74 – 650 = £72.74

Use the answer from part (a)

(a) 3.5% of 650 = 0.035 x 650 = 22.75

Value after 1 year = 22.75 + 650 = £672.75

OR 650 x 1.035 = £672.75

(b) 3.2% of 672.75 = 0.032 x 672.75 = 21.53

Value at end of 2002 = 672.75 + 21.53 = £694.28

4.1% of 694.28 = 0.041 x 694.28 = 28.47

Value at end of 2003 = 694.28 + 28.47= £722.75

Total interest = 722.75 – 650 = £72.75

OR 672.75 x 1.032 x 1.041 = 722.74

Total Interest = 722.74 – 650 = £72.74Calculations using Percentages Menu

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State calculation for percentage

Clearly add interest and original amount

Interpret interest percentage

Substitute correct value

Clearly calculate total interest

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Calculations using Percentages – Question 2

Joy buys a piano for £350. It depreciates by 15% in the first year and 20% in the second. How much is the piano worth after 2 years?

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£238.00



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15% of 350 = 0.15 x 350 = 52.50

Value after 1 year = 350 - 52.50 = 297.50

20% of 297.50 = 59.50

Value after 2 years = 297.50 – 59.50 = £238.00

OR 350 x 0.85 x 0.8 = £238.00



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15% of 350 = 0.15 x 350 = 52.50

Value after 1 year = 350 - 52.50 = 297.50

20% of 297.50 = 59.50

Value after 2 years = 297.50 – 59.50 = £238.00

OR 350 x 0.85 x 0.8 = £238.00

Know to subtract for depreciation

Use 297.50 rather than 350

Calculate depreciation term

Repeat decrease by a set percentage

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Joe buys a house for £93,000. Three years later it is worth £120,000.

a) Calculate the percentage increase in the value of Joe’s house as a percentage of the original price.

b) Calculate the current value of the house as a percentage of the original price.

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Joe buys a house for £93,000. Three years later it is worth £120,000.

a) Calculate the percentage increase in the value of Joe’s house as a percentage of the original price.

b) Calculate the current value of the house as a percentage of the original price.

(a) 28.3 %(b) 128 %

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(a) Increase = 120 000 – 93 000 = 26 500Joe buys a house for 93,000. Three years later it is worth £120,000.

a)Calculate the percentage increase in the value of Joe’s house as a percentage of the original price.

b)Calculate the current value of the house as a percentage of the original price.

Percentage Increase = %3.28100

93500

26500

(b) %128100

93500

120000


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(a) Increase = 120 000 – 93 000 = 26 500

Percentage Increase = %3.28100

93500

26500

(b) %128100

93500

120000

Calculate actual increase

Express answer as a percentage

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Adam puts money into a bank. It increases by 5% and is now worth £596.40.

How much money did Adam originally put in the bank?

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Adam puts money into a bank. It increases by 5% and is now worth £596.40.


£568.00

Question 4

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105% = 596.40Adam puts money into a bank. It increases by 5% and is now worth £596.40.


1% = 5.68

100% = 568.00

=> £568 invested originally

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105% = 596.40

1% = 5.68

100% = 568.00

=> £568 invested originally

Notice that £596.40 = 105%

Calculate 1%, and similarly 100%

Clearly state original investment


UNIT 1 : Volumes ofSolids



1 2 3

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Volumes of Solids – Question 1

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A spherical football has a radius of 15 cm. Find the volume of the

football., to 3 significant figures.

15cm


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A spherical football has a radius of 15 cm. Find the volume of the

football, to 3 significant figures.

14100 cm³15cm

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V A spherical football has a radius of 15 cm. Find the volume of the football, to 3 significant figures.

3

3

4r

3153

4

33753

4

= 14137.17

= 14100 cm³


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Select appropriate formulaV

3

3

4r

3153

4

33753

4

= 14137.17

= 14100 cm³

Substitute values

Clearly state answer

Round answer to 3 s.f. as requested


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A and B are 2 different shaped candles. A is a cone and B is a cylinder.

Both cost £4.00. Which candle is better value for money? (Justify your

answer.)

12cm

12cm

B

15 cm

18 cm

A


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Both cost £4.00. Which candle is better value for money? (Justify your

answer.)

12cm

12cm

B

15 cm

18 cm

A

Candle B because it has a larger volume and therefore willburn for longer (or similar).

Question 2

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Both cost £4.00. Which candle is better value for money? (Justify your answer.)

hrCone

V 2

3

1

1518183

1

ConeV

34.5089 cm

ConeV

hrCylinder

V 2

121212 Cylinder

V

37.5428 cm

CylinderV

Candle B is better value because it has a larger volume => it will burn longer.(Or similar)


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ConeV 2

3

1

1518183

1

ConeV

34.5089 cm

ConeV

hrCylinder

V 2

121212 Cylinder

V

37.5428 cm

CylinderV


State formula for volume of a cone

Substitute values

State volume in cm³

State formula for volume of a cylinder

Substitute values


Decide which is better value and give a relevant reason


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A watering trough is shown in diagram A. Diagram B gives the

dimensions of the cross-section. Calculate the volume of the trough,

in litres, to 2 significant figures.

Diagram A800 cm

94cm52cm 120 cm

Diagram A


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A watering trough is shown in diagram A. Diagram B gives the

dimensions of the cross-section. Calculate the volume of the trough,

in litres, to 2 significant figures.

Diagram A800 cm

94cm52cm 120 cm

Diagram A

4800 litres

Question 3

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A watering trough is shown in diagram A. Diagram B gives the dimensions of the cross-section. Calculate the volume of the trough, in litres, to 2 significant figures.

hrCone

V 2

3

1

1518183

1

ConeV

34.5089 cm

ConeV

hrCylinder

V 2

121212 Cylinder

V

37.5428 cm

CylinderV



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ConeV 2

3

1

1518183

1

ConeV

34.5089 cm

ConeV

hrCylinder

V 2

121212 Cylinder

V

37.5428 cm

CylinderV


State formula for volume of a cone

Substitute values


State formula for volume of a cylinder

Substitute values


Decide which is better value and give a relevant reason


UNIT 1 : Linear Relationships



1 2 3

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Linear Relationships - Question 1

Draw the graph of 22

1 xy

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Draw the graph of 22

1 xy y

x

2

2

4

4

6

6

8

8

– 2

– 2

2

2

4

4

6

6

8

8

– 2

– 2

Question 1

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Draw the graph of

22

1 xy

X 0 2 4 6Y 2 1 0 -1

y

x

2

2

4

4

6

6

8

8

– 2

– 2

2

2

4

4

6

6

8

8

– 2

– 2


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Marker’s CommentsX 0 2 4 6Y 2 1 0 -1 Clearly mark at least 3 points on

the grid

Extend the line as far as the grid allowsy

x

2

2

4

4

6

6

8

8

– 2

– 2

2

2

4

4

6

6

8

8

– 2

– 2

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Find the equation of the straight line shown in the diagram

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Find the equation of the straight line shown in the diagram

y

x

2

2

4

4

6

6

8

8

– 2

– 2

2

2

4

4

6

6

8

8

– 2

– 2

– 4

– 4

23

2 xy

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Find the equation of the straight line

shown in the diagram

Question 2 Y-intercept = -2

Gradient 12

12

xx

yy

23

2 xy

03

)2(0

3

2

y

x

2

2

4

4

6

6

8

8

– 2

– 2

2

2

4

4

6

6

8

8

– 2

– 2

– 4

– 4


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Marker’s CommentsY-intercept = -2

Gradient

3

203

)2(012

12

xx

yy

23

2 xy

Clearly state y-intercept

Clearly state gradient

Finish with the equation of the straightline of the form y=mx +c

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Does y = 3x -2 pass through the point (5,10)?

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Does y = 3x -2 pass through the point (5,10)?

y = 3x – 2 does not pass through(5, 10)

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Does y = 3x -2 pass through the

point (5,10)?

Question 3

3 x 5 – 2 = 13

Therefore y = 3x – 2 does not pass through(5, 10) as 3 x 5 – 2 ≠ 10

At (5, 10), x = 5 and y = 10


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3 x 5 – 2 = 13

Therefore y = 3x – 2 does not pass through (5, 10) as 3 x 5 – 2 ≠ 10

At (5, 10), x = 5 and y = 10 Substitute values for x and y correctly

State answer clearly


UNIT 1 : AlgebraicOperations



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Algebraic Operations - Question 1

Remove brackets and simplify this expression.

6 – 8(2x -7)

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6 – 8(2x -7)

62 – 16X

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6 – 8(2x -7)

6 – 8(2x -7)

= 6 – 16x + 56

= 62 – 16x

Multiply bracket by -8


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6 – 8(2x -7)

= 6 – 16x + 56

= 62 – 16x

Simplify expression by collecting like terms

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The diagram shows a garden with a rectangular flower bed.

(a) Calculate the area of the whole garden

(b) Calculate the area of the flower bed

(c) Calculate the area of the grassx - 2

x - 1

x + 7

x + 8Flowerbed

Grass

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The diagram shows a garden with a rectangular flower bed.

(a) Calculate the area of the whole garden

(b) Calculate the area of the flower bed

(c) Calculate the area of the grassx - 2

x - 1

x + 7

x + 8Flowerbed

Grass

(a) (x + 7)(x + 8)(b) (x – 2)(x – 1)(c) 18 (x + 3)

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The diagram shows a garden with a rectangular flower bed.(a) Calculate the area of the

whole garden(b) Calculate the area of the

flower bed(c) Calculate the area of the grass

(a) (x + 7)(x + 8)

(b) (x – 2)(x – 1)

(c) Garden = x² + 8x + 7x + 56= x² + 15x + 56

Flower bed = x² - 2x – x + 2= x² - 3x + 2

Grass = Garden – Flower bed= x² + 15x + 56 – (x² - 3x + 2)= x² + 15x + 56 – x² + 3x - 2)= 18x + 54= 18 (x + 3)

Multiply through by -1


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(a) (x + 7)(x + 8)

(b) (x – 2)(x – 1)

(c) Garden = x² + 8x + 7x + 56= x² + 15x + 56

Flower bed = x² - 2x – x + 2= x² - 3x + 2

Simplify the areas for both gardenand flower bed

Clearly state approach for calculatingthe area of grass

Substitute expressions

Grass = Garden – Flower bed= x² + 15x + 56 – (x² - 3x + 2)= x² + 15x + 56 – x² + 3x - 2)= 18x + 54= 18 (x + 3)

Simplify expression

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Factorise

(a) 99t + 198w

(b) 4a² - 36

(c) x ² - 3x -40

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Factorise

(a) 99t + 198w

(b) 4a² - 36

(c) x ² - 3x -40

(a) 99 (t + 2w)(b) (2a – 6)(2a + 6)(c) (x + 5)(x – 8)

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Factorise(a) 99t + 198w(b) 4a² - 36(c) x ² - 3x -40

(a) 99 (t + 2w)

(b) (2a – 6)(2a + 6)

(c) (x + 5) (x – 8)

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(a) 99 (t + 2w)

(b) (2a – 6)(2a + 6)

(c) (x + 5) (x – 8)

Recognise a common factor

Recognise a difference of 2 squares

Recognise factorisation of a quadratic


UNIT 1 :Circles



1 2 3 4

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Circles - Question 1

A pendulum travels along the arc of a circle. It swings from A to B.

The pendulum is 32cm long. Angle AOB = 42º.

Calculate the length of the arc AB. Give your answer to 3 significant

figures.

42 º32cm

A A

O

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figures.

42 º32cm

A A

O

23.5 cm

Question 3

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figures.

= 23.457

= 23.5 cm (to 3 s.f.)

Arc length = 64360

42

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= 23.457

= 23.5 cm (to 3 s.f.)

Find circumference

Find the fraction of the circle

State answer

Round to 3 significant figures

Arc length = 64360

42

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Find the area of the minor sector AOB of the circle with radius = 8cm and angle AOB = 62º. Give your answer to 1 decimal place.

A B

O

62º8cm

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A B

O

62º8cm

34.6 cm

Question 3

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Area sector = 88360

62

06.201360

62

26.34 cm

62.34

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Area sector = 88360

62

26.34

62.34

06.201360

62

cm

Find area of whole circle

Find the fraction of the circle

State answer

Round 1 decimal place

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If the shaded area = 32.6 cm², calculate the length of the arc.

9cm

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9cm

7.2 cm

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Area of sector2r

= Arc length

d

5.565.254

6.32 arc

5.56128.0 Arc

cmArc 2.7

5.56128.0

arc

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Area of sector2r

= Arc length

d

cmArc

Arc

arc

arc

2.7

5.56128.05.56

128.0

5.565.254

6.32

Demonstrate knowledge of fractionof a circle ratio

Substitute values

Begin to solve equation

Solve to find arc length

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Calculate the length AB. O is the centre of the circle with radius 6 m

6m

O

A

B8.4 m

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Calculate the length AB. O is the centre of the circle with radius 6 m

6m

10.3 m

O

A

B8.4 m

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Calculate the length AB. O is the centre of the circle with radius

6 m

O

B

6m

4.2 m

By Pythagoras’ TheoremOB² = 6² - 4.2²

= 36 – 17.64= 18.36

OB = =4.3 m

36.18

If OB =4.3 and OA = radius = 6 m→ AB = 4.3 + 6

= 10.3m

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O

B

6m

4.2 m

By Pythagoras’ TheoremOB² = 6² - 4.2²

= 36 – 17.64= 18.36

OB = =4.3 m

36.18

If OB =4.3 and OA = radius = 6 m→ AB = 4.3 + 6

= 10.3m

Create a a right angled triangle with the hypotenuse = radius

Use Pythagoras’ Theorem to find the missing side of the triangle

State that OA is a radius

Add missing side and OA to calculateOB


UNIT 2 :

Graphs, Charts and Tables

Trigonometry

Statistics

SimultaneousLinear Equations

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UNIT 2 :Trigonometry



1 2 3 4

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Trigonometry : Question 1

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Calculate the exact value of AC,

60º10cmA B

C


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60º10cmA B

C

310

Question 1

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60º10cmA B

C

adj

opp060tan

1060tan 0 AC

310AC

)360(tan 0

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310

)360(tan

1060tan

60tan

0

0

0

AC

AC

adj

oppState trigonometry ratio

Substitute values

Clearly state the exact value of tan 60º

Solve the equation to find an expression for AC


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Calculate the area of ABCD

70º

10cmA B

D C

42º

21cm

18cm20.5cm


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70º

10cmA B

D C

42º

21cm

18cm20.5cm

321.63

Question 1

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70º

10cmA B

D C

42º

21cm

18cm20.5cm

Area of ABD = ½ x 21 x 18 x sin 70º= 177.60cm²

Area of BDC = ½ x 21 x 20.5 x sin 42º= 144.03cm²

Area of ABCD = 177.60 + 144.03= 321.63 cm²

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Split the shape into ABD and BDCArea of ABD = ½ x 21 x 18 x sin 70º= 177.60cm²

Area of BDC = ½ x 21 x 20.5 x sin 42º= 144.03cm²

Area of ABCD = 177.60 + 144.03= 321.63 cm²

Know to use Area of triangle = ½absinC

Add the areas of the 2 shapes togetherto calculate the area of ABCD


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Calculate obtuse angle ABC area of ABCD

18 º46m

26 m

A

B

C


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146.86º

18 º46m

26 m

A

B

C

Question 1

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Angle ABC = 180 – 33.14 = 146.86

18 º46m

26 m

A

B

C

C

c

B

b

A

a

sinsinsin

If ABC is obtuse, then it is in 2nd quadrant

S A

T C14.3355.0sin 1

55.0sin B24

18sin46sin B

Bsin2618sin46

Bsin

46

18sin

26

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If ABC is obtuse, then it is in 2nd quadrant

Angle ABC = 180 – 33.14 = 146.86

14.3355.0sin

55.0sin24

18sin46sin

sin2618sin46sin

46

18sin

26sinsinsin

1

B

B

BB

C

c

B

b

A

a

S A

T C

Know to use Sine Rule to calculate missing angle

Substitute correct values into Sine Rule

Find a value for B

Know that an obtuse angle is greaterthan 90º

Find ABC


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Calculate angle BAC

25m

28m

24m

A

B

C


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Calculate angle BAC

53.49º

25m

28m

24m

A

B

C

Question 1

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Calculate angle BAC bc

acbA

2cos

222

25m

28m

24m

A

B

C

01 49.53595.0cos

595.0cos A

1400

833cos A

1400

5761409cos

A

25282

242528cos

222

A

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01

222

222

49.53595.0cos

595.0cos1400

833cos

1400

5761409cos

25282

242528cos

2cos

A

A

A

A

bc

acbA Use correct approach with Cosine Rule

Substitute correct values into Cosine Rule

Begin to simplify expression

Find an expression for cos A

Find the inverse to calculate a value for A


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Calculate angle BAC

81º18.1 m

x m

A

B

C

16.5m


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Find x.

22.50 m

81º18.1 m

x m

A

B

C

16.5m

Question 5

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Find x.

Abccba cos2222

81º18.1 m

x m

A

B

C

16.5m

02 81cos3.59761.32725.272 x

mx 50.2242.506x

44.9386.5992 x

42.5062 x

0222 81cos1.185.1621.185.16 x

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mx

x

x

x

x

x

Abccba

50.22

42.506

42.506

44.9386.599

81cos3.59761.32725.272

81cos1.185.1621.185.16

cos2

2

2

02

0222

222

Use correct approach with Cosine Rule

Begin to simplify expression

Substitute correct values into Cosine Rule

Find an expression for x²

Find the square root of x² to find x


UNIT 2 : Simultaneous Linear Equations



1 2

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Simultaneous Linear Equations : Question 1

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Solve this system of simultaneous equations graphically;

53

1

22

xy

xy


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x=3, y=4

53

1

22

xy

xy

Question 5

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53

1

22

xy

xy

Point of intersection = (3,4)So x =3, y=4

y

x

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

– 1

– 1

– 2

– 2

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

– 1

– 1

– 2

– 2


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Point of intersection = (3,4)So x =3, y=4

Plot three points on the line 22 xy

Plot three points on the line

53

1 xy

State the point of intersection

Provide a solution for x and y

y

x

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

– 1

– 1

– 2

– 2

1

1

2

2

3

3

4

4

5

5

6

6

7

7

8

8

– 1

– 1

– 2

– 2


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Let £x be the cost of a child’s ticket for a football match. Let £y be the cost of an adult ticket for the same match.

(a) 4 adult’s and 2 children’s tickets cost £115. Write this as an equation in x and y.

(b) 3 adult’s and 3 children’s tickets cost £103.50. Write this as an equation in x and y.

(c) How much for 5 adults and 2 children?


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(a) 4 adult’s and 2 children’s tickets cost £115. Write this as an equation in x and y.

(b) 3 adult’s and 3 children’s tickets cost £103.50. Write this as an equation in x and y.

(c) How much for 5 adults and 2 children?

(a) 2x + 4y = 155(b) 3x + 3y = 103.5(c) £138

Question 5

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a) 4 adult’s and 2 children’s tickets cost £115. Write this as an equation in x and y.

b) 3 adult’s and 3 children’s tickets cost £103.50. Write this as an equation in x and y.

c) How much for 5 adults and 2 children?

(a) 2x + 4y = 115

(b) 3x + 3y = 103.5

(c) 2x + 4y = 115 (1)3x + 3y = 103.5 (2)

(1) x 3 → 6x + 12y = 345 (3)(2) x 2 → 6x + 6y = 207 (4)

(3) - (4) → 6y = 138 y = 23

If y = 23 → 2x + 4 x 23 =115 2x + 92 = 115

2x = 23 x = £11.50

→ 2x + 5y = 2 x 11.5 + 5 x 23 = £138

→ 2 children and 5 adults costs £138


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Marker’s Comments(a) 2x + 4y = 115

(b) 3x + 3y = 103.5

(c) 2x + 4y = 115 (1)3x + 3y = 103.5 (2)

(1) x 3 → 6x + 12y = 345 (3)(2) x 2 → 6x + 6y = 207 (4)

(3) - (4) → 6y = 138 y = 23

If y = 23 → 2x + 4 x 23 =115 2x + 92 = 115

2x = 23 x = £11.50

→ 2x + 5y = 2 x 11.5 + 5 x 23 = £138

→ 2 children and 5 adults costs £138

Attempt to solve system of equationsby elimination

Subtract system of equations to eliminate variable

Substitute solved variable into originalequation, and solve

Provide a solution to original questionby creating new equation and solve.


UNIT 2 : Graphs, Chartsand Tables



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Graphs, Charts and Tables : Question 1

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200 people are asked about their favourite holiday destinations. The table summarises their replies.

Draw a pie chart to illustrate these findings

Destination Europe USA Australia Other

Number of responses

60 80 20 40


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Destination Europe USA Australia Other

Number of responses

60 80 20 40

Europe

USA

Australia

Other

Question 5

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Destination Europe USA Australia OtherNo of 60 80 20

40responses


072360200

40Other

Europe

USA

Australia

Other

036360200

20Australia

0144360200

80USA

0108360200

60Europe


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Check all angles add up to 3600

0

0

0

72360200

40

36360200

20

144360200

80

108360200

60

Other

Australia

USA

Europe

Europe

USA

Australia

Other

Know to divide by 200

Know to multiply by 360 to find anglein circle

Use results to draw a measured piechart


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A bus company gives its journey times in minutes for 1 month.

68 62 93 82 63 67 68 70 75 90

(a)Calculate the median of this set of numbers

(b)Find the upper and lower quartiles

(c)Draw a boxplot showing this information


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68 62 93 82 63 67 68 70 75 90

(a)Calculate the median of this set of numbers

(b)Find the upper and lower quartiles

(c)Draw a boxplot showing this information

(a) Median = 69(b) Q1 = 67

Q3 = 82(c)

62 67 69 82 93

Question 2

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68 62 93 82 63 67 68 70 75 90 Calculate the median of this set of

numbersFind the upper and lower quartilesDraw a boxplot showing this

information

(a) 10 numbers →

62 63 67 68 68 70 75 82 90 93

Median = 70 + 68 = 69 (between 5th and 6th 2 places

22410 R

(b) Q1 at 3rd position = 67 Q3 at 7th position = 82

62 67 69 82 93

(c)


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State the positions of the median, and the quartilesUse results to draw a boxplot with anappropriate scale

(a) 10 numbers →

62 63 67 68 68 70 75 82 90 93

Median = 70 + 68 = 69 (between 5th and 6th 2 places

22410 R

(b) Q1 at 3rd position = 67 Q3 at 7th position = 82

62 67 69 82 93

(c)

Know to split the list into 4 equalsections

Order the list


UNIT 2 :Statistics



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3 4

Statistics : Question 1

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Adam works for a fast food company. The number of burgers he sells each day is logged.

Number of burgers Numbers of days0 15 1810 2815 2620 37

Calculate(i) the median number of

burgers(ii) the quartiles(iii)the semi-interquartile

range


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Number of burgers Numbers of days0 15 1810 2815 2620 37

Calculate(i) the median number of

burgers(ii) the quartiles(iii)the semi-interquartile

range

(i) 15(ii) 10, 20(iii) 5

Question 1

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Number of burgers Numbers of

days0 15 1810 2815 2620 37

CumulativeFrequency

1194773110

Calculate the (i) median number of burgers, (ii) the quartiles (iii) the semi-interquartile range

110 numbers 2274110 r

(i) Q2 (median) between 55th and 56th position=15 burgers

52

10

2

1020

(ii) Q1 at 28th position = 10 burgersQ3 at 83rd place = 20 burgers

(iii) Semi Interquartile range

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State the positions of the median, and the quartiles

Find cumulative frequency

Know to divide by 4

CumulativeFrequency

1194773110

110 numbers 2274110 r

(i) Q2 (median) between 55th and 56th position=15 burgers

(ii) Q1 at 28th position = 10 burgersQ3 at 83rd place = 20 burgers

(iii) Semi Interquartile range5

2

10

2

1020

Know semi-interquartile range is

213 QQ


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The price, in pence/litre, of petrol at 10 city garages is shown below.

84.2 84.4 85.1 83.9 81.0 84.2 85.6 85.2 84.9 84.8

(a)Calculate the mean and the standard deviation of these prices

(b)In 10 rural garages petrol prices had a mean of 88.8 and a standard deviation of 2.4. How do rural prices compare with city prices?


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The price, in pence/litre, of petrol at 10 city garages is shown below.

84.2 84.4 85.1 83.9 81.0 84.2 85.6 85.2 84.9 84.8

(a)Calculate the mean and the standard deviation of these prices

(b)In 10 rural garages petrol prices had a mean of 88.8 and a standard deviation of 2.4. How do rural prices compare with city prices?

(a) 84.3, 1.28(b) The average price in the

city is lower and there is less variation in the prices

Question 1

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The price, in pence/litre, of petrol at 10 city garages is shown below.84.2 84.4 85.1 83.9 81.0 84.2

85.6 85.2 84.9 84.8(a) Calculate the mean and the

standard deviation of these prices

(b) In 10 rural garages petrol prices had a mean of 88.8 and a standard deviation of 2.4. How do rural prices compare with city prices?

31.71130

89.711154

3.843

2

2

x

x

x

910

89.71115431.71130

SD

3.8410

3.843mean

(b) The average price in the city is lower and there is less variation in the prices.

28.1SD

.

7646.1SD

9

821.14SD

9

489.7111531.71130 SD

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Make one valid comparison betweenthe two sets of data.

Know how to apply one of the formula for standard deviation(Note: this is one possible solution)

Know to how to calculate mean 31.71130

89.711154

3.843

2

2

x

x

x

28.1

7646.1

9

821.14

9

489.7111531.71130

910

89.71115431.71130

.

SD

SD

SD

SD

SD

3.8410

3.843mean

(b) The average price in the city is lower and there is less variation in the prices.


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61 cars are parked in a car park. 26 are saloons, 18 are estates and the rest are sportscars.

(a)What is the probability the first car to leave the car park is a sportscar?

(b)The first car to leave is a saloon car. What is the probability the second is an estate?


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61 cars are parked in a car park. 26 are saloons, 18 are estates and the rest are sportscars.

(a)What is the probability the first car to leave the car park is a sportscar?

(b)The first car to leave is a saloon car. What is the probability the second is an estate?

(a) 17/61(b) 3/10

Question 1

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61 cars are parked in a car park. 26 are saloons, 18 are estates and

the rest are sportscars.(a) What is the probability the first

car to leave the car park is a sportscar?

(b) The first car to leave is a saloon car. What is the probability the second is an

estate?

61

17)(

#

#)(

sportscarP

possible

favourablesportscarP

10

3)2(

60

18)2(

estatecarP

estatecarP

nd

nd(b)

(a)

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Know to reduce the total number ofcars by 1

Calculate the number of sportscars

State formula for probability

61

17)(

#

#)(

sportscarP

possible

favourablesportscarP

10

3)2(

60

18)2(

estatecarP

estatecarP

nd

nd(b)

(a)

Know to divide by the total numberof cars

Simplify the fraction as far as possible


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The number of visitors to a car park each day was recorded.

No of visitors No of days

40 11

50 15

60 16

70 13

Calculate the mean number of visitorsto 1 decimal place.


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52.9 visitors


40 11

50 15

60 16

70 13

Calculate the mean number of visitorsto 1 decimal place.

Question 1

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mean = total number of visitorstotal number of days

Calculate the mean number of visitors to 1 decimal place.


40 11

50 15

60 16

70 13

No of visitors x No of days

440

600

960

910

Total = 2910

55

2910

9.52

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Know how to calculate the totalnumber of visitors

Demonstrate how mean is to becalculated

Give answer to 1 decimal place asrequested

mean = total number of visitorstotal number of days

9.52

55

2910

No of visitors x No of days

440

600

960

910

Total = 2910


UNIT 4 :

Formulae

Calculations in aSocial Context

Logic Diagrams

EXIT


UNIT 4 : Calculations in a Social Context



1 2

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3

Calculations in a Social Context : Question 1

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John Stevens works for the sales department of a financial company.

(a) John earns a basic salary of £1800/month plus 9% of his monthly sales. Calculate his gross salary for March when his sales totalled £4632.

(b) John pays 7% of his gross salary into his pension, along with £93.00 for National Insurance contributions and £262.86 for Income Tax. Calculate his net pay for March.


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(b) John pays 7% of his gross salary into his pension, along with £93.00 for National Insurance contributions and £262.86 for Income Tax. Calculate his net pay for March.

(a) £2216.88(b) £1705.84

Question 1

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(b) John pays 7% of his gross salary into his pension, along with £93.00 for National Insurance contributions and £262.86 for Income Tax. Calculate his net pay for

March.

(a) 9% of £4632 = 0.09 x 4632 = £416.88

(b) 7% of £2216.88 = 0.07 x 2216.88 = £155.18

Gross monthly salary = Basic pay + Overtime +

Commission= 1800 + 416.88= £2216.88

Net pay= Gross pay – Total deductions= 2216.88 – 511.04=£1705.84

Total deductions = NI + Tax + Pension= 93.00 + 262.86 + 155.18= £511.04

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Calculate 9% of sales

State how to calculate Gross monthly salary

Use your answer from (a) to calculatepension payment

(a) 9% of £4632 = 0.09 x 4632 = £416.88

Gross monthly salary = Basic pay + Overtime + Commission

= 1800 + 416.88= £2216.88

(b) 7% of £2216.88 = 0.07 x 2216.88 = £155.18

Total deductions = NI + Tax + Pension= 93.00 + 262.86 + 155.18= £511.04

Net pay= Gross pay – Total deductions= 2216.88 – 511.04=£1705.84

Know how to calculate totaldeductions

State how to calculate Net pay


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Colin is a civil servant. His basic annual salary is £35042 per annum. His annual tax allowance is £5125

Calculate his monthly tax bill using the tax rates given in the table below.

Rates of tax on:

First £1920 of taxable income 10%

Next 25558 of taxable income 23%

All remaining taxable income 40%


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£587.16 a month

Rates of tax on:




Question 2

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Taxable Income = 35042 – 5125 = £29917

Annual tax bill = 192 + 5878.34 + 975.60= £7045.94

Rates of tax on:




Lower rate = 10% of 1920 = £192

Basic rate = 23% of 25558 = £5878.34

(1920 + 25558 = 27478)29917 – 27478 = £2439

Higher rate = 40% of 2439 = £975.60

Monthly tax bill = 7045.94/12= £587.16/month

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Find Taxable Income (Gross salary – allowances)

Calculate full allowance at lower rate

Calculate remaining taxable income

Calculate higher rate tax on £2439 oftaxable income

Calculate annual tax bill by addingtax at all bands

Taxable Income = 35042 – 5125 = £29917

Annual tax bill = 192 + 5878.34 + 975.60= £7045.94

Lower rate = 10% of 1920 = £192

Basic rate = 23% of 25558 = £5878.34

(1920 + 25558 = 27478)29917 – 27478 = £2439

Higher rate = 40% of 2439 = £975.60

Monthly tax bill = 7045.94/12= £587.16/month

Calculate full allowance at basic rate

Know to divide by 12 to find monthlytax


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Karen wants to borrow £4000 for 3 years with loan protection.

What is the cost of Karen’s loan?

Loan Amount (£)

With/out Loan Protection

WLP/WOLP12 months 24 months 36 months

1000 WLP

WOLP

£98.79

£92.26

£55.39

£50.51

£41.18

£36.76

2000 WLP

WOLP

£197.42

£190.26

£102.81

£97.53

£85.86

£78.20

3000 WLP

WOLP

£296.38

£276.77

£166.18

£151.54

£123.56

£110.29

4000 WLP

WOLP

£395.46

£387.98

£207.45

£196.72

£193.42

£179.76


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£2963.12

Loan Amount (£)

With/out Loan Protection

WLP/WOLP12 months 24 months 36 months

1000 WLP

WOLP

£98.79

£92.26

£55.39

£50.51

£41.18

£36.76

2000 WLP

WOLP

£197.42

£190.26

£102.81

£97.53

£85.86

£78.20

3000 WLP

WOLP

£296.38

£276.77

£166.18

£151.54

£123.56

£110.29

4000 WLP

WOLP

£395.46

£387.98

£207.45

£196.72

£193.42

£179.76

Question 2

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36 months = 3 years Monthly payment = £193.42

193.42 x 36 = £6963.12

Total cost of loan = 6963.12 – 4000 = £2963.12

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Change 3 years into 36 months

Read correct monthly payment fromloan repayment table

Multiply loan payment by term of loan in months to calculate total repayment

Subtract original loan from totalrepayment to calculate the cost ofthe loan

36 months = 3 years Monthly payment = £193.42

193.42 x 36 = £6963.12

Total cost of loan = 6963.12 – 4000 = £2963.12


UNIT 4 :Logic Diagrams



1 2

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3 4

Logic Diagrams : Question 1

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A fast food company makes different kinds of burgers, beef or chicken, with or without lettuce, with or without onions and with or without ketchup.

Draw a tree diagram to illustrate all the possible combinations of burgers and use it to calculate the probability a customer will want a burger with lettuce.


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P(Lettuce) = 0.5

Question 1

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Beef orChicken?

Beef

Chicken

Lettuce? Onions? Ketchup?

Y

N

Y

N

YN

Y

N

Y

N

Y

N

Y

N

Y

N

Y

N

Y

N

Y

N

Y

N

Y

N

Y

N

BLOK

BLO

BLK

BL

BOK

BO

BK

B

CLOK

CLO

CLK

CLCOK

CO

CK

C

P(Lettuce) = 5.02

1

16

8

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Calculate number of possible outcomes

Beef orChicken?

Beef

Chicken

Lettuce? Onions? Ketchup?

Y

N

Y

N

YN

Y

N

Y

N

Y

N

Y

N

Y

N

Y

N

Y

N

Y

N

Y

N

Y

N

Y

N

BLOK

BLO

BLK

BL

BOK

BO

BK

B

CLOK

CLO

CLK

CLCOK

CO

CK

C

P(Lettuce) = 5.02

1

16

8

Calculate number of outcomes thatinclude lettuceSimplify probability as far as possible


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(a)Draw a network diagram to represent each of the maps below. Label the vertices to represent places and label the arcs to show distances (in miles).

(b)Find a route that covers all 127 miles only once in Map 1.

(c) Is a similar route possible in Map 2? Give a reason for your answer.

Achan Boul

DunkyCarat

Eerie

DweepTull

Lillen

Gore

Bridge Wull

MAP1 MAP2

15

10

31

208

26

17

21

12

181323

26

20


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(a)Draw a network diagram to represent each of the maps below. Label the vertices to represent places and label the arcs to show distances (in miles).

(b)Find a route that covers all 127 miles only once in Map 1.

(c) Is a similar route possible in Map 2? Give a reason for your answer.

(b) B-A-D-E-D-C-B-E(c) No, Map 2 has more than 2 odd vertices

and so is not traversable.

MAP2

Achan Boul

DunkyCarat

Eerie

DweepTull

Lillen

Gore

Bridge Wull

MAP1

15

10

31

208

26

17

21

12

181323

26

20

Question 2

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Draw a network diagram to represent each of the maps below. (a) Label the vertices to represent places and label the arcs to show distances (in miles).

(b) Find a route that covers all 127 miles only once in Map

1.(c) Is a similar route possible in

Map 2? Give a reason for your answer.

(b) B-A-D-E-D-C-B-E

(c) No, Map 2 has more than 2 odd vertices (Dweep, Lillen and Wull) so it is not traversable.

Achan Boul

Dunky Carat

Eerie

Dweep Tull Lillen

Gore

Bridge

Wull1510

31

20

8

26

17

21 12

1813

23 2620

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2 possible network diagrams for thegiven maps

Find the shortest route. Know to begin and end at an odd vertex

Mention that the network is not traversable and support your statement with evidence of more than2 odd vertices

(b) B-A-D-E-D-C-B-E

(c) No, Map 2 has more than 2 odd vertices (Dweep, Lillen and Wull) so it is not traversable.

Achan Boul

Dunky Carat

Eerie

Dweep Tull Lillen

Gore

Bridge

Wull1510

31

20

8

26

17

21 12

1813

23 2620


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(a)How long does it take to boil the potatoes?

(b)Find the critical path for the whole exercise and state the minimum time required to complete the entire job.

A

C

B

D

E

GFStart End

12

21

18

7

A

80

11 60

0

A – Cook fishB – Fry onionsC – Boil potatoesD – Chop fishE – Mash potatoesF – Mix ingrediantsG – Fry cakes


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(a)How long does it take to boil the potatoes?

(b)Find the critical path for the whole exercise and state the minimum time required to complete the entire job.

(a) 21 minutes(b) C-E-F-G (45 minutes)

A

C

B

D

E

GFStart End

12

21

18

7

A

80

11 60

0


Question 3

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(a) How long does it take to boil the potatoes?

(b) Find the critical path for the whole exercise and state the minimum time required to complete the entire job.

(a) Boil potatoes – arc CE = 21 minutes

(b) Longest path = critical path = C-E-F-G

A

C

B

D

E

GFStart End

12

21

18

7

A

80

11 60

0


Minimum time required = longest path = 21 + 7 +11 +6 = 45 minutes

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Recognise Arc CE corresponds to“Boil potatoes”

Know that the longest path is the critical path

Know that the minimum time requiredto complete the job = longest path

(a) Boil potatoes = arc CE = 21 minutes

(b) Longest path = critical path = C-E-F-G

Minimum time required = longest path = 21 + 7 +11 +6 = 45 minutes


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The flowchart shows how to pay for a car. Use the flowchart to find the total paid for a new car costing £12000 if you pay a deposit

Start

Is the car new

?

Pay adeposit

?

deposit of£5000

15% of cost for 6 years

Pay cost

20% of cost for6 years

Stop

Yes

Yes No

No


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£15 800

Start

Is the car new

?

Pay adeposit

?

deposit of£5000

15% of cost for 6 years

Pay cost

20% of cost for6 years

Stop

Yes

Yes No

No

Question 4

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15% of £12000 = £1800 for 1 year

Total cost = 5000 + 10800 = £15800

£1800 x 6 = £10800

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Calculate 15% of cost of car

Calculate repayments over 6 years

Calculate the total cost by adding thedeposit

15% of £12000 = £1800 for 1 year

£1800 x 6 = £10800

Total cost = 5000 + 10800 = £15800


UNIT 4 :Formulae



1 2

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Formulae : Question 1

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An object falls with a velocity v = u + 10t, where t is the time and u is the initial velocity.

• Calculate v when u = 6 m/s and t = 15 secs

• Calculate t when v = 70 m/s and u = 3 m/s


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• Calculate v when u = 6 m/s and t = 15 secs

• Calculate t when v = 70 m/s and u = 3 m/s

(a) v = 156 m/s(b) t = 6.7 m/s

Question 4

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(a) Calculate v when u = 6 m/s and t = 15 secs

(b) Calculate t when v = 70 m/s and u = 3 m/s

(a)

(b)

tuv 10

tuv 10

1506v

15106 v

smv /156

smt /7.610

67t

t10370

t10370

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Substitute values


State solution with units

(a)

(b)

smv

v

v

tuv

/156

1506

15106

10

smt

t

t

t

tuv

/7.6

10

67

10370

10370

10

Substitute values




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It is possible to convert a temperature in degrees Fahrenheit into degrees Centigrade. First subtract 32 from the temperature in Fahrenheit, then multiply by 5 and divide by 9.

What would the temperature be in Centigrade if it was 77 degrees Fahrenheit?


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25ºC

Question 2

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Centigrade (Fahrenheit – 32)

C025

)3277(9

5

)45(9

5

9

5

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State formula

Substitute values



C

Formula

025

)45(9

5

)3277(9

5

9

5

Centigrade (Fahrenheit – 32)


UNIT 3 :

FurtherTrigonometry

More AlgebraicOperations

Quadratic Functions

EXIT


UNIT 3 : AlgebraicOperations



1 2

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3

More Algebraic Operations : Question 1

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Express in its simplest form.72


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Express in its simplest form.72

26

Question 1

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Express in its

simplest form.

729872

98

98

924

322

26


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Demonstrate that

Simplify further

Collect like terms

9872

98

98

924

322

26

baab

8


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Expand

432

12 32 aaaa


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Expand

432

12 32 aaaa

255 3

2a

aa

Question 2

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Expand

432

12 32 aaaa

42322

12 32 aaaaaa

252

5

32 aaa

255 3

2a

aa


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Know

Know

42322

12 32 aaaaaa

252

5

32 aaa

255 3

2a

aa

mnmn aaa

m nm

n

aa


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Express as a fraction with a rational denominator.23

4


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Express as a fraction with a rational denominator.23

4

7

2412

Question 3

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Express as a fraction

with a rational denominator.

23

4

23

23

23

4

)23)(23(

)23(4

223239

2412

7

2412


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Know to multiply by conjugate surd

Multiply out brackets

Simplify expressions

23

23

23

4

)23)(23(

)23(4

223239

2412

7

2412


UNIT 3 : Quadratic Functions



1 2

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Solve x² - x + 12 = 0

Quadratic Functions : Question 1

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Solve x² - x + 12 = 0


x = -3 and x = 4

Question 1

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Solve x² - x + 12 = 0

x² - x + 12 = 0

x + 3 = 0 or x – 4 = 0

(x + 3)(x - 4) = 0

x = - 3 or x = 4


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Factorise equation

Know that either factor could be equal to 0

Solve both equations to find roots

x² - x + 12 = 0

x + 3 = 0 or x – 4 = 0

(x + 3)(x - 4) = 0

x = - 3 or x = 4

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Solve this quadratic equation; 3x² - 14x + 17


x = 3.54 or 1.13

Question 2

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Solve this quadratic equation;

3x² - 14x + 17

a = 3b = -14 c = 12

a

acbbx

2

42

6

5214

6

5214 ORx

32

12341414 2

x

6

5214x

6

79.6

6

21.21ORx

x = 3.54 OR 1.13


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Know to use quadratic formula

Substitute values


Find both solutions for x

a = 3b = -14 c = 12

a

acbbx

2

42

6

5214

6

5214 ORx

32

12341414 2

x

6

5214x

6

79.6

6

21.21ORx

x = 3.54 OR 1.13

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Sketch the graph of the function y = -(x – 4)² - 10


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Sketch the graph of the function y = -(x – 4)² - 10


y

x

2

2

4

4

6

6

– 2

– 2

2

2

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

– 12

– 12

– 14

– 14

– 16

– 16

– 18

– 18

– 20

– 20

– 22

– 22

– 24

– 24

– 26

– 26

– 28

– 28

– 30

– 30

(4, 10)

Question 3

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Sketch the graph of the function

y = -(x – 4)² - 10

Equation of the form y = a(x –b)² + c a = -1b = 4c = -10

Equation of the axis of symmetry; x = 4

Vertex at (4, -10)

At y-intercept, x = 0

y = -(0 – 4) ² - 10 = -( - 4) ² -10 = -16 – 10 = -26

y-intercept at (0, -26)

y

x

2

2

4

4

6

6

– 2

– 2

2

2

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

– 12

– 12

– 14

– 14

– 16

– 16

– 18

– 18

– 20

– 20

– 22

– 22

– 24

– 24

– 26

– 26

– 28

– 28

– 30

– 30

(4, 10)


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Know that equation is of the formy = a(x –b)² + c

Know that x = b is the axis of symmetry

Know the vertex occurs at (b, c)

Let x = 0 for y-intercept

Equation of the form y = a(x –b)² + c a = -1b = 4c = -10

Equation of the axis of symmetry; x = 4

Vertex at (4, -10)

At y-intercept, x = 0

y = -(0 – 4) ² - 10 = -( - 4) ² -10 = -16 – 10 = -26

y-intercept at (0, -26)Solve equation to find y-intercept

Use information to sketch graphwith key points labelled.

y

x

2

2

4

4

6

6

– 2

– 2

2

2

– 2

– 2

– 4

– 4

– 6

– 6

– 8

– 8

– 10

– 10

– 12

– 12

– 14

– 14

– 16

– 16

– 18

– 18

– 20

– 20

– 22

– 22

– 24

– 24

– 26

– 26

– 28

– 28

– 30

– 30

(4, 10)


UNIT 3 : FurtherTrigonometry



1 2

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Solve 3tanxº - 1 = 0 for 0 ≤ x ≤ 360

Further Trigonometry : Question 1

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Solve 3tanxº - 1 = 0 for 0 ≤ x ≤ 360


x = 18.4º and 198.4 º

Question 1

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Solve 3tanxº - 1 = 0 for 0 ≤ x ≤ 360

3 tan xº - 1 = 0

3 tan xº = 1

tan xº =3

1

3

1tan 1x

= 18.4 º

and x = 180 + 18.4 = 198.4 º


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Find x

Know which quadrants the solutionbelongs to, and therefore find othersolution

3 tan xº - 1 = 0

3 tan xº = 1

tan xº =3

1

3

1tan 1x

= 18.4 º

and x = 180 + 18.4 = 198.4 º

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The graph has an equation of the form y = a cos (x – b)º. Write down the values of a and b.


150 330

y

x

1

1

2

2

– 1

– 1

– 2

– 2

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a = 1 and b = 60

150 330

y

x

1

1

2

2

– 1

– 1

– 2

– 2

Question 2

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a = Amplitude = 1

b = Phase angle = 60

150 330

y

x

1

1

2

2

– 1

– 1

– 2

– 2


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Know that a = amplitude and readvalue from grapha = Amplitude = 1

b = Phase angle = 60Know that b = phase angle and calculate from graph

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Sketch the graph for 0 ≤ x ≤ 720


2

1sin2 xy

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Sketch the graph for 0 ≤ x ≤ 720


2

1sin2 xy

y

x

1

1

2

2

– 1

– 1

– 2

– 2

360 720

Question 3

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Sketch the graph

for 0 ≤ x ≤ 720

0

2

1sin2 xy

Amplitude = 2 (y-maximum = 2, y-minimum = -2)

Period = 0.5

y

x

1

1

2

2

– 1

– 1

– 2

– 2

360 720


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Recognise amplitude = 2

State periodicity

Use information to sketch graph,making intercepts with x axis veryclear

Amplitude = 2 (y-maximum = 2, y-minimum = -2)

Period = 0.5

y

x

1

1

2

2

– 1

– 1

– 2

– 2

360 720

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Solve this quadratic equation; 3x² - 14x + 17


intermediate 2 – additional question bank exit unit 1 please decide which unit you would like to...

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