intermediate 2 – additional question bank exit unit 1 please decide which unit you would like to...
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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
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 1 :
Linear Relationships
Calculations usingPercentages
Circles
AlgebraicOperations
Volumes ofSolids
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 1 : Calculations usingPercentages
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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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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.
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.
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
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?
Question 2
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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
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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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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Calculations using Percentages – Question 3
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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Calculations using Percentages – Question 3
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 %
Question 3
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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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Calculations using Percentages – Question 4
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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Calculations using Percentages – Question 4
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?
£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.
How much money did Adam originally put in the bank?
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
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 1 : Volumes ofSolids
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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
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.
14100 cm³15cm
Question 1
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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
Volumes of Solids – Question 2
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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
Volumes of Solids – Question 2
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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
Candle B because it has a larger volume and therefore willburn for longer (or similar).
Question 2
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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.)
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
Candle B is better value because it has a larger volume => it will burn longer.(Or similar)
State formula for volume of a cone
Substitute values
State volume in cm³
State formula for volume of a cylinder
Substitute values
State volume in cm³
Decide which is better value and give a relevant reason
Volumes of Solids – 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.
Diagram A800 cm
94cm52cm 120 cm
Diagram A
Volumes of Solids – 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.
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
Candle B is better value because it has a larger volume => it will burn longer.(Or similar)
Volume of Solids Menu
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Marker’s Commentshr
ConeV 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)
State formula for volume of a cone
Substitute values
State volume in cm³
State formula for volume of a cylinder
Substitute values
State volume in cm³
Decide which is better value and give a relevant reason
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 1 : Linear Relationships
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Linear Relationships - Question 1
Draw the graph of 22
1 xy
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Linear Relationships - Question 1
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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Linear Relationships - Question 2
Find the equation of the straight line shown in the diagram
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Linear Relationships - Question 2
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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Linear Relationships - Question 3
Does y = 3x -2 pass through the point (5,10)?
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Linear Relationships - Question 3
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
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
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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Algebraic Operations - Question 1
Remove brackets and simplify this expression.
6 – 8(2x -7)
62 – 16X
Question 1
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Remove brackets and simplify this expression.
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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Algebraic Operations - Question 2
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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Algebraic Operations - Question 2
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)
Question 2
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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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Algebraic Operations - Question 3
Factorise
(a) 99t + 198w
(b) 4a² - 36
(c) x ² - 3x -40
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Algebraic Operations - Question 3
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)
Question 3
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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
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 1 :Circles
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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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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
23.5 cm
Question 3
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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.
= 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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Circles - Question 2
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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Circles - Question 2
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
34.6 cm
Question 3
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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.
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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Circles - Question 3
If the shaded area = 32.6 cm², calculate the length of the arc.
9cm
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Circles - Question 3
If the shaded area = 32.6 cm², calculate the length of the arc.
9cm
7.2 cm
Question 3
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If the shaded area = 32.6 cm², calculate the length of the arc.
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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Circles - Question 4
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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Circles - Question 4
Calculate the length AB. O is the centre of the circle with radius 6 m
6m
10.3 m
O
A
B8.4 m
Question 3
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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
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 2 :
Graphs, Charts and Tables
Trigonometry
Statistics
SimultaneousLinear Equations
EXIT
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 2 :Trigonometry
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5
Trigonometry : Question 1
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Calculate the exact value of AC,
60º10cmA B
C
Trigonometry : Question 1
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Calculate the exact value of AC,
60º10cmA B
C
310
Question 1
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Calculate the exact value of AC,
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
Trigonometry : Question 2
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Calculate the area of ABCD
70º
10cmA B
D C
42º
21cm
18cm20.5cm
Trigonometry : Question 2
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Calculate the area of ABCD
70º
10cmA B
D C
42º
21cm
18cm20.5cm
321.63
Question 1
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Calculate the area of ABCD
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
Trigonometry : Question 3
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Calculate obtuse angle ABC area of ABCD
18 º46m
26 m
A
B
C
Trigonometry : Question 3
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Calculate obtuse angle ABC area of ABCD
146.86º
18 º46m
26 m
A
B
C
Question 1
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Calculate obtuse angle ABC area of ABCD
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
Trigonometry : Question 4
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Calculate angle BAC
25m
28m
24m
A
B
C
Trigonometry : Question 4
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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
Trigonometry : Question 5
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Calculate angle BAC
81º18.1 m
x m
A
B
C
16.5m
Trigonometry : Question 5
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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
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 2 : Simultaneous Linear Equations
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Solve this system of simultaneous equations graphically;
53
1
22
xy
xy
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Solve this system of simultaneous equations graphically;
x=3, y=4
53
1
22
xy
xy
Question 5
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Solve this system of simultaneous equations graphically;
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
Simultaneous Linear Equations : Question 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?
Simultaneous Linear Equations : Question 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?
(a) 2x + 4y = 155(b) 3x + 3y = 103.5(c) £138
Question 5
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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?
(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.
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
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
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
Europe
USA
Australia
Other
Question 5
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200 people are asked about their favourite holiday destinations. The table summarises their replies.
Destination Europe USA Australia OtherNo of 60 80 20
40responses
Draw a pie chart to illustrate these findings
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
Graphs, Charts and Tables : Question 2
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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
Graphs, Charts and Tables : Question 2
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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
(a) Median = 69(b) Q1 = 67
Q3 = 82(c)
62 67 69 82 93
Question 2
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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 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
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 2 :Statistics
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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
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
(i) 15(ii) 10, 20(iii) 5
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
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
Statistics : Question 2
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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?
Statistics : Question 2
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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.
Statistics : Question 3
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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?
Statistics : Question 3
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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
Statistics : Question 4
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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.
Statistics : Question 4
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The number of visitors to a car park each day was recorded.
52.9 visitors
No of visitors No of days
40 11
50 15
60 16
70 13
Calculate the mean number of visitorsto 1 decimal place.
Question 1
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The number of visitors to a car park each day was recorded.
mean = total number of visitorstotal number of days
Calculate the mean number of visitors to 1 decimal place.
No of visitors No of days
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
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 4 :
Formulae
Calculations in aSocial Context
Logic Diagrams
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INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 4 : Calculations in a Social Context
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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.
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.
(a) £2216.88(b) £1705.84
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.
(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
Calculations in a Social Context : Question 2
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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%
Calculations in a Social Context : Question 2
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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.
£587.16 a month
Rates of tax on:
First £1920 of taxable income 10%
Next 25558 of taxable income 23%
All remaining taxable income 40%
Question 2
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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.
Taxable Income = 35042 – 5125 = £29917
Annual tax bill = 192 + 5878.34 + 975.60= £7045.94
Rates of tax on:
First £1920 of taxable income 10%
Next 25558 of taxable income 23%
All remaining taxable income 40%
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
Calculations in a Social Context : Question 3
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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
Calculations in a Social Context : Question 3
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Karen wants to borrow £4000 for 3 years with loan protection.
What is the cost of Karen’s loan?
£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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Karen wants to borrow £4000 for 3 years with loan protection.
What is the cost of Karen’s loan?
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
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 4 :Logic Diagrams
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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.
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.
P(Lettuce) = 0.5
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.
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
Logic Diagrams : Question 2
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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
Logic Diagrams : Question 2
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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
Logic Diagrams : 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
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
Logic Diagrams : 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) 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
A – Cook fishB – Fry onionsC – Boil potatoesD – Chop fishE – Mash potatoesF – Mix ingrediantsG – Fry cakes
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
A – Cook fishB – Fry onionsC – Boil potatoesD – Chop fishE – Mash potatoesF – Mix ingrediantsG – Fry cakes
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
Logic Diagrams : Question 4
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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
Logic Diagrams : Question 4
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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
£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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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
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
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 4 :Formulae
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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
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
(a) v = 156 m/s(b) t = 6.7 m/s
Question 4
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An object falls with a velocity v = u + 10t, where t is the time and u is the initial velocity.
(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
Begin to solve equation
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
Begin to solve equation
State solution with units
Formulae : Question 2
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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?
Formulae : Question 2
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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?
25ºC
Question 2
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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?
Centigrade (Fahrenheit – 32)
C025
)3277(9
5
)45(9
5
9
5
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State formula
Substitute values
Begin to solve equation
State solution with units
C
Formula
025
)45(9
5
)3277(9
5
9
5
Centigrade (Fahrenheit – 32)
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 3 :
FurtherTrigonometry
More AlgebraicOperations
Quadratic Functions
EXIT
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 3 : AlgebraicOperations
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More Algebraic Operations : Question 1
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Express in its simplest form.72
More Algebraic Operations : Question 1
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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
More Algebraic Operations : Question 2
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Expand
432
12 32 aaaa
More Algebraic Operations : Question 2
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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
More Algebraic Operations : Question 3
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Express as a fraction with a rational denominator.23
4
More Algebraic Operations : Question 3
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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
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 3 : Quadratic Functions
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Solve x² - x + 12 = 0
Quadratic Functions : Question 1
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Solve x² - x + 12 = 0
Quadratic Functions : Question 1
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
Quadratic Functions : Question 2
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
Begin to solve equation
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
Quadratic Functions : Question 3
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Sketch the graph of the function y = -(x – 4)² - 10
Quadratic Functions : Question 3
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)
INTERMEDIATE 2 – ADDITIONAL QUESTION BANK
UNIT 3 : FurtherTrigonometry
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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
Further Trigonometry : Question 1
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.
Further Trigonometry : Question 2
150 330
y
x
1
1
2
2
– 1
– 1
– 2
– 2
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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.
Further Trigonometry : Question 2
a = 1 and b = 60
150 330
y
x
1
1
2
2
– 1
– 1
– 2
– 2
Question 2
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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.
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
Further Trigonometry : Question 30
2
1sin2 xy
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Sketch the graph for 0 ≤ x ≤ 720
Further Trigonometry : Question 30
2
1sin2 xy
y
x
1
1
2
2
– 1
– 1
– 2
– 2
360 720
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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
Quadratic Functions : Question 2