INTERMEDIATE 2 –
ADDITIONAL QUESTION BANK
Please decide which Unit you would like to revise:
UNIT 1
UNIT 2
UNIT 3
Calculations using %
Volumes of Solids
Linear Relationships
Algebraic Operations
Circles
Trigonometry
Simultaneous Linear
Equations
Graphs, Charts &
Tables
Statistics
Algebraic Operations
Quadratic Functions
Further Trigonometry
EXIT
UNIT 4
Calculations in a Social
Context
Logic Diagrams
Formulae
INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
UNIT 1 :
Calculations using
Percentages
Volumes of
Solids
Linear
Relationships
Algebraic
Operations
Circles
EXIT
INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 1 :
Calculations using
Percentages
Please choose a question to attempt from the following:
1
EXIT
2
3
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Unit 1 Menu
4
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.
Reveal answer only
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EXIT
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.
Reveal answer only
Go to full solution
(a) £22.75
(b) £72.75
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EXIT
Question 1
(a) 3.5% of 650 = 0.035 x 650 = 22.75
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?
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
(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.
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
Begin Solution
Total interest = 722.75 – 650 = £72.75
Continue Solution
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Percentages Menu
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OR
672.75 x 1.032 x 1.041 = 722.74
Total Interest = 722.74 – 650 = £72.74
Marker’s Comments
(a) 3.5% of 650 = 0.035 x 650 = 22.75
State calculation for percentage
Value after 1 year = 22.75 + 650 = £672.75
Clearly add interest and original amount
OR
Interpret interest percentage
650 x 1.035 = £672.75
(b) 3.2% of 672.75 = 0.032 x 672.75 = 21.53
Use the answer from part (a)
Value at end of 2002 = 672.75 + 21.53
= £694.28
4.1% of 694.28 = 0.041 x 694.28 = 28.47
Substitute correct value
Value at end of 2003 = 694.28 + 28.47
= £722.75
Clearly calculate total interest
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
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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?
Reveal answer only
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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?
Reveal answer only
£238.00
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EXIT
Question 2
15% of 350 = 0.15 x 350 = 52.50
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?
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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Marker’s Comments
Know to subtract for depreciation
15% of 350 = 0.15 x 350 = 52.50
Value after 1 year = 350 - 52.50 = 297.50
Use 297.50 rather than 350
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
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.
Reveal answer only
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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.
Reveal answer only
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EXIT
(a) 28.3 %
(b) 128 %
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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(a) Increase = 120 000 – 93 000 = 26 500
Percentage Increase
= 26500
93500
(b)
 100  28 .3%
120000
 100  128 %
93500
Marker’s Comments
Calculate actual increase
(a) Increase = 120 000 – 93 000 = 26 500
Percentage Increase
= 26500
93500
(b)
 100  28 .3%
Express answer as a percentage
120000
 100  128 %
93500
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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?
Reveal answer only
£568.00
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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?
Begin Solution
Continue Solution
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105% = 596.40
1% = 5.68
100% = 568.00
=> £568 invested originally
Marker’s Comments
105% = 596.40
1% = 5.68
Notice that £596.40 = 105%
Calculate 1%, and similarly 100%
100% = 568.00
Clearly state original investment
=> £568 invested originally
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 1 :
Volumes of
Solids
Please choose a question to attempt from the following:
1
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2
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Unit 1 Menu
3
Volumes of Solids – Question 1
A spherical football has a radius of 15 cm. Find the volume of the
football., to 3 significant figures.
Reveal answer only
15cm
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Volumes of Solids – Question 1
A spherical football has a radius of 15 cm. Find the volume of the
football, to 3 significant figures.
Reveal answer only
15cm
14100 cm³
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Question 1
V
A spherical football has a
radius of 15 cm. Find the
volume of the football, to 3
significant figures.
4 3
 r
3
4
    15 3
3
4
    3375
3
= 14137.17
= 14100 cm³
Begin Solution
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Marker’s Comments
V
4 3
 r
3
4
    15 3
3
4
    3375
3
Select appropriate formula
Substitute values
Clearly state answer
Round answer to 3 s.f. as requested
= 14137.17
= 14100 cm³
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Volumes of Solids – Question 2
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.)
Reveal answer only
15 cm
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18 cm
A
Go to Volume of Solids Menu
12cm
12cm
B
EXIT
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Volumes of Solids – Question 2
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.)
Reveal answer only
15 cm
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18 cm
A
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12cm
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12cm
B
EXIT
Candle B because it has a larger volume and therefore will
burn for longer (or similar).
Question 2
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.)
Begin Solution
VCo n e
1 2
 r h
3
1
    18  18  15
Co n e
3
V
VCone
 5089.4cm
3
2
VCylinder
 r h
VCylinder
   12  12  12
VCylinder
3
 5428.7 cm
Continue Solution
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Candle B is better value because it has a
larger volume => it will burn longer.
(Or similar)
VCo n e
Marker’s Comments
1 2
 r h
3
State formula for volume of a cone
1
    18  18  15
Co n e
3
V
VCone
 5089.4cm
3
Substitute values
State volume in cm³
State formula for volume of a cylinder
Substitute values
VCylinder
 r h
State volume in cm³
VCylinder
   12  12  12
Decide which is better value and give a
relevant reason
VCylinder
2
3
 5428.7 cm
Candle B is better value because it has a
larger volume => it will burn longer.
(Or similar)
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Solids Menu
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Volumes of Solids – Question 3
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.
800 cm
Diagram A
Reveal answer only
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Diagram A
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94cm
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52cm
120 cm
Volumes of Solids – Question 3
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.
800 cm
Diagram A
Reveal answer only
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Diagram A
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94cm
52cm
120 cm
4800 litres
EXIT
Question 3
VCo n e
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.
Begin Solution
Continue Solution
1 2
 r h
3
1
    18  18  15
Co n e
3
V
VCone
 5089.4cm
3
2
VCylinder
 r h
VCylinder
   12  12  12
VCylinder
3
 5428.7 cm
Markers Comments
Volume of Solids Menu
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Candle B is better value because it has a
larger volume => it will burn longer.
(Or similar)
VCo n e
Marker’s Comments
1 2
 r h
3
State formula for volume of a cone
1
    18  18  15
Co n e
3
V
VCone
 5089.4cm
3
Substitute values
State volume in cm³
State formula for volume of a cylinder
Substitute values
VCylinder
 r h
State volume in cm³
VCylinder
   12  12  12
Decide which is better value and give a
relevant reason
VCylinder
2
3
 5428.7 cm
Next Comment
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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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 1 :
Linear
Relationships
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1
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2
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Unit 1 Menu
3
Linear Relationships - Question 1
1
Draw the graph of y   x  2
2
Reveal answer only
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8– 2
2
4
6
2– 2
4
6
8
Linear Relationships - Question 1
1
Draw the graph of y   x  2
2
y
8
6
Reveal answer only
4
2
Go to full solution
– 2
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EXIT
– 2
2
4
6
8 x
Question 1
Draw the graph of
y
8– 2
2
4
6
2– 2
4
6
8
X
Y
0
2
1
x2
2
2
1
4
0
y
8
6
4
2
– 2
– 2
Begin Solution
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2
4
6
8 x
6
-1
Marker’s Comments
8– 2
2
4
6
2– 2
4
6
8
X
Y
0
2
2
1
4
0
6
-1
Clearly mark at least 3 points on
the grid
y
Extend the line as far as the grid allows
8
6
4
2
– 2
2
4
6
8 x
– 2
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Linear Relationships
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Linear Relationships - Question 2
Find the equation of the straight line shown in the diagram
Reveal answer only
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EXIT
8– 4
2
4
6
2
2– 2
4
6
8
Linear Relationships - Question 2
Find the equation of the straight line shown in the diagram
y
8
6
Reveal answer only
4
2
Go to full solution
– 2
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2
4
– 2
– 4
Go to Linear Relationships Menu
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EXIT
y
2
x2
3
6
8 x
Question 2
Y-intercept = -2
Find the equation of the straight line
Gradient

y 2  y1
x 2  x1

0  ( 2)
30

2
3
shown in the diagram
y
8
6
4
2
– 2
2
4
6
8 x
– 2
– 4
Begin Solution
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2
y  x2
3
Y-intercept = -2
Gradient
y 2  y1

x 2  x1
0  ( 2 )

30
2

3
Marker’s Comments
Clearly state y-intercept
Clearly state gradient
Finish with the equation of the straight
line of the form y=mx +c
2
y  x2
3
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Linear
Relationships Menu
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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)?
Reveal answer only
y = 3x – 2 does not pass through
(5, 10)
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Question 3
Does y = 3x -2 pass through the
At (5, 10), x = 5 and y = 10
point (5,10)?
3 x 5 – 2 = 13
Therefore y = 3x – 2 does not pass through
(5, 10) as 3 x 5 – 2 ≠ 10
Begin Solution
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Marker’s Comments
At (5, 10), x = 5 and y = 10
Substitute values for x and y correctly
3 x 5 – 2 = 13
Therefore y = 3x – 2 does not pass
through (5, 10) as 3 x 5 – 2 ≠ 10
State answer clearly
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 1 :
Algebraic
Operations
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1
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2
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3
Algebraic Operations - Question 1
Remove brackets and simplify this expression.
6 – 8(2x -7)
Reveal answer only
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Algebraic Operations - Question 1
Remove brackets and simplify this expression.
6 – 8(2x -7)
Reveal answer only
62 – 16X
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EXIT
Question 1
Remove brackets and simplify
this expression.
6 – 8(2x -7)
6 – 8(2x -7)
= 6 – 16x + 56
= 62 – 16x
Begin Solution
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Marker’s Comments
6 – 8(2x -7)
= 6 – 16x + 56
Multiply bracket by -8
Simplify expression by collecting
like terms
= 62 – 16x
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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
x+7
(c) Calculate the area of the grass
x-2
Reveal answer only
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EXIT
Flower
bed
Grass
x-1
x+8
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
x+7
(c) Calculate the area of the grass
x-2
Reveal answer only
Go to full solution
Go to Marker’s Comments
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EXIT
Flower
bed
x-1
Grass
(a) (x + 7)(x + 8)
(b) (x – 2)(x – 1)
(c) 18 (x + 3)
x+8
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
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
Begin Solution
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= Garden – Flower bed
= x² + 15x + 56 – (x² - 3x + 2)
= x² + 15x + 56 – x² + 3x - 2)
= 18x + 54
= 18 (x + 3)
Marker’s Comments
(a) (x + 7)(x + 8)
Simplify the areas for both garden
and flower bed
(b) (x – 2)(x – 1)
(c) Garden
= x² + 8x + 7x + 56
= x² + 15x + 56
Clearly state approach for calculating
the area of grass
Flower bed
= x² - 2x – x + 2
= x² - 3x + 2
Substitute expressions
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
Simplify expression
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Algebraic Operations
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Algebraic Operations - Question 3
Factorise
(a) 99t + 198w
(b) 4a² - 36
(c) x ² - 3x -40
Reveal answer only
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Algebraic Operations - Question 3
Factorise
(a) 99t + 198w
(b) 4a² - 36
(c) x ² - 3x -40
Reveal answer only
Go to full solution
(a) 99 (t + 2w)
(b) (2a – 6)(2a + 6)
(c) (x + 5)(x – 8)
Go to Marker’s Comments
Go to Algebraic Operations Menu
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EXIT
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)
Begin Solution
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Marker’s Comments
(a) 99 (t + 2w)
Recognise a common factor
(b) (2a – 6)(2a + 6)
Recognise a difference of 2 squares
(c) (x + 5) (x – 8)
Recognise factorisation of a quadratic
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 1 :
Circles
Please choose a question to attempt from the following:
1
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2
3
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Unit 1 Menu
4
5
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.
O
Reveal answer only
42 º
32cm
Go to full solution
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EXIT
A
A
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.
O
Reveal answer only
42 º
32cm
Go to full solution
Go to Marker’s Comments
A
Go to Circle Menu
23.5 cm
Go to Main Menu
EXIT
A
Question 3
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.
Begin Solution
Continue Solution
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Arc length =
42
   64
360
= 23.457
= 23.5 cm (to 3 s.f.)
Marker’s Comments
Arc length =
42
   64
360
Find circumference
Find the fraction of the circle
= 23.457
State answer
= 23.5 cm (to 3 s.f.)
Round to 3 significant figures
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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.
Reveal answer only
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O
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8cm
62º
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EXIT
A
B
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.
Reveal answer only
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O
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8cm
62º
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A
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34.6 cm
EXIT
B
Question 3
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 =
62
 88
360
62

 201 .06
360
 34.62
 34.6cm2
Begin Solution
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Marker’s Comments
Area sector =
62
 88
360
62
 201.06
360
 34.62

 34.6cm2
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.
Reveal answer only
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9cm
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EXIT
Circles - Question 3
If the shaded area = 32.6 cm², calculate the length of the arc.
Reveal answer only
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9cm
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7.2 cm
EXIT
Question 3
If the shaded area = 32.6 cm²,
calculate the length of the arc.
Area of sector
r 2
= Arc length
d
32.6
arc

254 .5 56.5
arc
0.128 
56.5
Arc  0.128  56.5
Arc  7.2cm
Begin Solution
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Marker’s Comments
Area of sector
r 2
= Arc length
d
32.6
arc

254.5 56.5
arc
0.128 
56.5
Arc  0.128 56.5
Arc  7.2cm
Demonstrate knowledge of fraction
of a circle ratio
Substitute values
Begin to solve equation
Solve to find arc length
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Circle Menu
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Circles - Question 4
Calculate the length AB. O is the centre of the circle with radius 6 m
A
Reveal answer only
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O
6m
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B
8.4 m
Circles - Question 4
Calculate the length AB. O is the centre of the circle with radius 6 m
A
Reveal answer only
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O
6m
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B
8.4 m
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10.3 m
EXIT
Question 3
Calculate the length AB. O is the
centre of the circle with radius
O
6m
6m
4.2 m
B
By Pythagoras’ Theorem
OB²
= 6² - 4.2²
= 36 – 17.64
= 18.36
OB
= 18.36
=4.3 m
Begin Solution
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If OB =4.3 and OA = radius = 6 m
→ AB = 4.3 + 6
= 10.3m
Marker’s Comments
O
6m
4.2 m
Create a a right angled triangle with
the hypotenuse = radius
B
By Pythagoras’ Theorem
OB²
= 6² - 4.2²
= 36 – 17.64
= 18.36
OB
= 18.36
=4.3 m
If OB =4.3 and OA = radius = 6 m
→ AB = 4.3 + 6
= 10.3m
Use Pythagoras’ Theorem to find the
missing side of the triangle
State that OA is a radius
Add missing side and OA to calculate
OB
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
UNIT 2 :
EXIT
Trigonometry
Simultaneous
Linear Equations
Graphs, Charts
and Tables
Statistics
INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 2 :
Trigonometry
Please choose a question to attempt from the following:
1
EXIT
2
3
Back to
Unit 2 Menu
4
5
Trigonometry : Question 1
Calculate the exact value of AC,
A
10cm
60º
B
C
Reveal answer only
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EXIT
Trigonometry : Question 1
Calculate the exact value of AC,
A
10cm
60º
B
C
Reveal answer only
Go to full solution
Go to Marker’s Comments
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Go to Main Menu
EXIT
10 3
Question 1
opp
tan 60 
adj
AC
0
tan 60 
10
0
Calculate the exact value of
AC,
A
10cm
60º
B
(tan600  3)
AC  10 3
C
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Marker’s Comments
opp
t an 60 
adj
AC
0
t an 60 
10
(t an600  3)
0
AC  10 3
State trigonometry ratio
Substitute values
Clearly state the exact value of tan 60º
Solve the equation to find an
expression for AC
Next Comment
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Trigonometry : Question 2
Calculate the area of ABCD
A
18cm
10cm
B
42º
21cm
20.5cm
70º
D
C
Reveal answer only
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EXIT
Trigonometry : Question 2
Calculate the area of ABCD
A
18cm
10cm
B
42º
21cm
20.5cm
70º
D
C
Reveal answer only
321.63
Go to full solution
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EXIT
Question 1
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
Calculate the area of ABCD
A
18cm
10cm
B
42º
21cm
20.5cm
70º
D
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C
Area of ABCD
= 177.60 + 144.03
= 321.63 cm²
Marker’s Comments
Area of ABD
= ½ x 21 x 18 x sin 70º
= 177.60cm²
Split the shape into ABD and BDC
Know to use Area of triangle = ½absinC
Area of BDC
= ½ x 21 x 20.5 x sin 42º
= 144.03cm²
Add the areas of the 2 shapes together
to calculate the area of ABCD
Area of ABCD
= 177.60 + 144.03
= 321.63 cm²
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Trigonometry : Question 3
Calculate obtuse angle ABC area of ABCD
B
26 m
A
18 º
46m
Reveal answer only
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EXIT
C
Trigonometry : Question 3
Calculate obtuse angle ABC area of ABCD
B
26 m
A
18 º
C
46m
Reveal answer only
146.86º
Go to full solution
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EXIT
Question 1
Calculate obtuse angle ABC area
of ABCD
B
26 m
A
18 º
46m
C
a
b
c


sin A sin B sin C
26
46

sin 18 sin B
46 sin 18  26 sin B
46 sin 18
sin B 
24
sin B  0.55
sin 1 0.55  33.14
S
T
A
C
If ABC is obtuse, then it is in 2nd quadrant
Begin Solution
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Angle ABC = 180 – 33.14 = 146.86
Marker’s Comments
a
b
c


sin A sin B sin C
26
46

sin 18 sin B
46sin 18  26sin B
46sin 18
24
sin B  0.55
Know to use Sine Rule to calculate
missing angle
Substitute correct values into Sine Rule
S
A
sin B 
T
C
sin 1 0.55  33.14
If ABC is obtuse, then it is in 2nd quadrant
Find a value for B
Know that an obtuse angle is greater
than 90º
Find ABC
Angle ABC = 180 – 33.14 = 146.86
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Trigonometry : Question 4
Calculate angle BAC
B
24m
25m
A
28m
C
Reveal answer only
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EXIT
Trigonometry : Question 4
Calculate angle BAC
B
24m
25m
A
28m
C
Reveal answer only
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EXIT
53.49º
Question 1
b2  c2  a 2
cos A 
2bc
282  252  242
cos A 
2  28 25
Calculate angle BAC
B
24m
25m
A
28m
C
1409  576
cos A 
1400
833
cos A 
1400
cos A  0.595
cos1 0.595  53.490
Begin Solution
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Marker’s Comments
b2  c2  a2
cos A 
2bc
282  252  242
cos A 
2  28 25
1409 576
cos A 
1400
833
cos A 
1400
cos A  0.595
cos1 0.595  53.490
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
Next Comment
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Trigonometry : Question 5
Calculate angle BAC
B
16.5m
A
xm
81º
18.1 m
C
Reveal answer only
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EXIT
Trigonometry : Question 5
Find x.
B
16.5m
A
xm
81º
18.1 m
Reveal answer only
C
22.50 m
Go to full solution
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EXIT
Question 5
Find x.
B
16.5m
A
xm
81º
18.1 m
C
a 2  b2  c 2  2bc cos A
x 2  16.52  18.12  2 16.5 18.1cos810
x 2  272.25  327.61 597.3 cos810
x 2  599.86  93.44
x 2  506.42
x  506.42
x  22.50 m
Begin Solution
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Marker’s Comments
a 2  b 2  c 2  2bc cos A
x 2  16.52  18.12  2 16.5 18.1cos810
Use correct approach with Cosine Rule
x  272.25  327.61 597.3 cos81
Substitute correct values into Cosine
Rule
x 2  599.86  93.44
Begin to simplify expression
x 2  506.42
Find an expression for x²
2
x  506.42
x  22.50m
0
Find the square root of x² to find x
Next Comment
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 2 :
Simultaneous
Linear Equations
Please choose a question to attempt from the following:
1
EXIT
2
Back to
Unit 2 Menu
Simultaneous Linear Equations : Question 1
Solve this system of simultaneous equations graphically;
y  2 x  2
1
y   x5
3
Reveal answer only
Go to full solution
Go to Marker’s Comments
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EXIT
Simultaneous Linear Equations : Question 1
Solve this system of simultaneous equations graphically;
y  2 x  2
1
y   x5
3
Reveal answer only
x=3, y=4
Go to full solution
Go to Marker’s Comments
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EXIT
Question 5
y
8
Solve this system of simultaneous
equations graphically;
y  2 x  2
1
y   x5
3
7
6
5
4
3
2
1
– 2 – 1
– 1
1
2
3
4
5
6
– 2
Begin Solution
Continue Solution
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Point of intersection = (3,4)
So x =3, y=4
7
8
x
Marker’s Comments
y
8
Plot three points on the line
7
y  2 x  2
6
5
Plot three points on the line
1
y   x5
3
4
3
2
1
– 2 – 1
– 1
1
2
3
4
5
6
7
8
x
State the point of intersection
– 2
Point of intersection = (3,4)
So x =3, y=4
Provide a solution for x and y
Next Comment
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Simultaneous Linear Equations : Question 2
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?
Reveal answer only
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EXIT
Simultaneous Linear Equations : Question 2
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?
Reveal answer only
Go to full solution
(a) 2x + 4y = 155
(b) 3x + 3y = 103.5
(c) £138
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EXIT
Question 5
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?
Begin Solution
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(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
(a) 2x + 4y = 115
Marker’s Comments
(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
Attempt to solve system of equations
by elimination
Subtract system of equations to
eliminate variable
Substitute solved variable into original
equation, and solve
Provide a solution to original question
by creating new equation and solve.
→ 2x + 5y = 2 x 11.5 + 5 x 23 = £138
→ 2 children and 5 adults costs £138
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 2 :
Graphs, Charts
and Tables
Please choose a question to attempt from the following:
1
EXIT
2
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Unit 2 Menu
Graphs, Charts and Tables : Question 1
200 people are asked about their favourite holiday destinations.
The table summarises their replies.
Destination Europe USA
Australia Other
Number of
responses
60
80
20
40
Draw a pie chart to illustrate these findings
Reveal answer only
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Go to Marker’s Comments
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EXIT
Graphs, Charts and Tables : Question 1
200 people are asked about their favourite holiday destinations.
The table summarises their replies.
Destination Europe USA
Australia Other
Number of
responses
60
80
20
40
Draw a pie chart to illustrate these findings
Reveal answer only
Europe
USA
Go to full solution
Go to Marker’s Comments
Go to Graphs, Charts and Tables Menu
Go to Main Menu
EXIT
Australia
Other
Question 5
200 people are asked about their
favourite holiday destinations.
The table summarises their
replies.
Destination Europe USA Australia Other
No of
60
80
20
40
responses
60
 360  108 0
200
80
USA 
 360  1440
200
20
Australia 
 360  360
200
Europe 
Other 
Draw a pie chart to illustrate these
findings
Begin Solution
Continue Solution
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Tables Menu
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40
 360  720
200
Europe
USA
Australia
Other
Marker’s Comments
60
 360  1080
200
80
USA 
 360  1440
200
20
Australia 
 360  360
200
40
Other 
 360  720
200
Europe
Know to divide by 200
Know to multiply by 360 to find angle
in circle
Check all angles add up to 360
Use results to draw a measured pie
chart
Europe
USA
Australia
Other
Next Comment
Graphs, Charts and
Tables Menu
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Graphs, Charts and Tables : Question 2
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
Reveal answer only
Go to full solution
Go to Marker’s Comments
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EXIT
Graphs, Charts and Tables : Question 2
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
Reveal answer only
Go to full solution
(a) Median = 69
(b) Q1 = 67
Q3 = 82
(c)
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EXIT
62
67 69
82
93
Question 2
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
numbers
Find the upper and lower quartiles
Draw a boxplot showing this
information
(a) 10 numbers → 10  4  2 R 2
62 63 67 68 68 70 75 82 90 93
Median = 70 + 68 = 69 (between 5th and 6th
2
places
(b) Q1 at 3rd position = 67
Q3 at 7th position = 82
(c)
62
Begin Solution
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67 69
82
93
Marker’s Comments
(a) 10 numbers → 10  4  2 R 2
Know to split the list into 4 equal
sections
62 63 67 68 68 70 75 82 90 93
Median = 70 + 68 = 69 (between 5th and 6th
2
places
3rd
(b) Q1 at
position = 67
Q3 at 7th position = 82
Order the list
State the positions of the median,
and the quartiles
Use results to draw a boxplot with an
appropriate scale
(c)
62
67 69
82
93
Next Comment
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 2 :
Statistics
Please choose a question to attempt from the following:
1
EXIT
2
3
Back to
Unit 2 Menu
4
Statistics : Question 1
Adam works for a fast food company. The number of burgers he
sells each day is logged.
Number of burgers
Numbers of days
Calculate
0
1
(i) the median number of
5
18
burgers
10
28
(ii) the quartiles
15
26
20
37
(iii)the semi-interquartile
range
Reveal answer only
Go to full solution
Go to Marker’s Comments
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EXIT
Statistics : Question 1
Adam works for a fast food company. The number of burgers he
sells each day is logged.
Number of burgers
Numbers of days
Calculate
0
1
(i) the median number of
5
18
burgers
10
28
(ii) the quartiles
15
26
20
37
(iii)the semi-interquartile
range
Reveal answer only
Go to full solution
Go to Marker’s Comments
Go to Statistics Menu
Go to Main Menu
EXIT
(i) 15
(ii) 10, 20
(iii) 5
Question 1
Adam works for a fast food
company. The number of
burgers he sells each day is
logged.
Number of burgers Numbers
of days
0
1
5
18
10
28
15
26
Calculate
the (i) median number of
20
37
burgers, (ii) the quartiles
(iii) the semi-interquartile range
Begin Solution
Continue Solution
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Cumulative
Frequency
1
19
47
73
110
110 numbers  110  4  27 r 2
(i) Q2 (median) between 55th and 56th position
=15 burgers
(ii) Q1 at 28th position = 10 burgers
Q3 at 83rd place = 20 burgers
(iii) Semi Interquartile range 
20  10 10

5
2
2
Cumulative
Frequency
1
19
47
73
110
110 numbers  110  4  27 r 2
Marker’s Comments
Know to divide by 4
Find cumulative frequency
(i) Q2 (median) between 55th and 56th position State the positions of the median,
and the quartiles
=15 burgers
(ii) Q1 at 28th position = 10 burgers
Know semi-interquartile range is
Q3 at 83rd place = 20 burgers
Q3  Q1

(iii) Semi Interquartile range 20  10 10

2

2
5
2
Next Comment
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Statistics : Question 2
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?
Reveal answer only
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EXIT
Statistics : Question 2
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?
Reveal answer only
Go to full solution
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EXIT
(a) 84.3, 1.28
(b) The average price in the
city is lower and there is
less variation in the prices
Question 1
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?
 x  843.3
 x   711154.89
 x  71130.31
2
mean 
2
SD 
843 .3
 84.3
10
711154.89
10
9
71130.31
SD 
71130.31 71115.489
9
SD 
14.821
9
.
Begin Solution
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Markers Comments
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SD  1.6467
SD  1.28
(b) The average price in the city is lower and
there is less variation in the prices.
 x  843.3
 x   711154.89
 x  71130.31
2
mean 
2
SD 
843 .3
 84.3
10
711154.89
10
9
71130.31
SD 
71130.31 71115.489
9
SD 
14.821
9
.
SD  1.6467
Marker’s Comments
Know to how to calculate mean
Know how to apply one of the
formula for standard deviation
(Note: this is one possible solution)
Make one valid comparison between
the two sets of data.
SD  1.28
(b) The average price in the city is lower and
there is less variation in the prices.
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Statistics : Question 3
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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EXIT
Statistics : Question 3
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?
Reveal answer only
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(a) 17/61
(b) 3/10
Question 1
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?
Begin Solution
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Markers Comments
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(a) P( sportscar) 
P( sportscar) 
(b)
# favourable
# possible
17
61
P(2 nd car  estate) 
18
60
P(2 nd car  estate) 
3
10
# favourable
(a) P( sportscar) 
# possible
Marker’s Comments
State formula for probability
Calculate the number of sportscars
P( sportscar) 
(b)
17
61
Know to divide by the total number
of cars
18
P(2 car  estate) 
60
nd
P(2 nd car  estate) 
3
10
Know to reduce the total number of
cars by 1
Simplify the fraction as far as
possible
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Statistics : Question 4
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 visitors
to 1 decimal place.
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EXIT
Statistics : Question 4
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 visitors
to 1 decimal place.
Reveal answer only
Go to full solution
Go to Marker’s Comments
Go to Statistics Menu
Go to Main Menu
EXIT
52.9 visitors
Question 1
The number of visitors to a car
park each day was recorded.
Calculate the mean number of
visitors to 1 decimal place.
No of visitors x No of days
440
600
960
No of visitors No of days
40
11
50
15
60
16
70
13
910
Total = 2910
mean = total number of visitors
total number of days

Begin Solution
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2910
55
 52 .9
Marker’s Comments
No of visitors x No of days
440
600
960
Know how to calculate the total
number of visitors
Demonstrate how mean is to be
calculated
910
Total = 2910
mean = total number of visitors
total number of days
Give answer to 1 decimal place as
requested
2910

55
 52.9
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
UNIT 4 :
Calculations in a
Social Context
Formulae
EXIT
Logic Diagrams
INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 4 :
Calculations in a
Social Context
Please choose a question to attempt from the following:
1
EXIT
2
3
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Unit 4 Menu
Calculations in a Social Context : Question 1
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.
Reveal answer only
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EXIT
Calculations in a Social Context : Question 1
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.
Reveal answer only
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(a) £2216.88
(b) £1705.84
Question 1
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.
Begin Solution
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Markers Comments
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(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
(a) 9% of £4632 = 0.09 x 4632 = £416.88
Gross monthly salary
= Basic pay + Overtime + Commission
= 1800 + 416.88
= £2216.88
Marker’s Comments
Calculate 9% of sales
State how to calculate Gross
monthly salary
(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
Use your answer from (a) to calculate
pension payment
Know how to calculate total
deductions
State how to calculate Net pay
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Calculations in a Social Context : Question 2
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%
Reveal answer only
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EXIT
Calculations in a Social Context : Question 2
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%
Reveal answer only
Go to full solution
Go to Marker’s Comments
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Go to Main Menu
EXIT
£587.16 a month
Taxable Income = 35042 – 5125 = £29917
Question 2
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.
First £1920 of taxable income
10%
Next 25558 of taxable income
23%
All remaining taxable income
40%
Continue Solution
Markers Comments
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Basic rate = 23% of 25558 = £5878.34
(1920 + 25558 = 27478)
29917 – 27478 = £2439
Higher rate = 40% of 2439 = £975.60
Rates of tax on:
Begin Solution
Lower rate = 10% of 1920 = £192
Annual tax bill
= 192 + 5878.34 + 975.60
= £7045.94
Monthly tax bill = 7045.94/12
= £587.16/month
Taxable Income = 35042 – 5125 = £29917
Marker’s Comments
Lower rate = 10% of 1920 = £192
Find Taxable Income
(Gross salary – allowances)
Basic rate = 23% of 25558 = £5878.34
Calculate full allowance at lower rate
(1920 + 25558 = 27478)
29917 – 27478 = £2439
Calculate full allowance at basic rate
Calculate remaining taxable income
Higher rate = 40% of 2439 = £975.60
Annual tax bill
= 192 + 5878.34 + 975.60
= £7045.94
Monthly tax bill = 7045.94/12
= £587.16/month
Calculate higher rate tax on £2439 of
taxable income
Calculate annual tax bill by adding
tax at all bands
Know to divide by 12 to find monthly
tax
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Calculations in a Social Context : Question 3
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/WOLP
12 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
Reveal answer only
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Calculations in a Social Context : Question 3
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/WOLP
12 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
Reveal answer only
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EXIT
£2963.12
Go to Social Context Menu
Go to Main Menu
Question 2
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
Begin Solution
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Marker’s Comments
36 months = 3 years
Monthly payment = £193.42
Change 3 years into 36 months
193.42 x 36 = £6963.12
Read correct monthly payment from
loan repayment table
Total cost of loan
= 6963.12 – 4000 = £2963.12
Multiply loan payment by term of
loan in months to calculate total
repayment
Subtract original loan from total
repayment to calculate the cost of
the loan
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 4 :
Logic Diagrams
Please choose a question to attempt from the following:
1
EXIT
2
3
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Unit 4 Menu
4
Logic Diagrams : Question 1
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.
Reveal answer only
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Go to Marker’s Comments
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Go to Main Menu
EXIT
Logic Diagrams : Question 1
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.
Reveal answer only
Go to full solution
Go to Marker’s Comments
Go to Logic Diagrams Menu
Go to Main Menu
EXIT
P(Lettuce) = 0.5
Question 1
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 or
Chicken?
Lettuce?
Onions?
Ketchup?
Y
Y
Y
N
Beef
N
Y
N
Y
Chicken
Y
N
N
Y
N
Begin Solution
Continue Solution
Markers Comments
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P(Lettuce) =
8 1
  0 .5
16 2
BLOK
N
BLO
Y
BLK
N
BL
Y
BOK
N
BO
Y
BK
N
B
Y
CLOK
N
CLO
Y
CLK
N
Y
CL
COK
N
CO
Y
CK
N
C
Beef or
Chicken?
Lettuce?
Onions?
Ketchup?
Y
Y
Y
N
Beef
N
Y
N
Y
Chicken
Y
N
N
Y
N
P(Lettuce) =
8 1
  0 .5
16 2
Marker’s Comments
BLOK
N
BLO
Y
BLK
N
BL
Y
BOK
N
BO
Y
BK
N
B
Y
CLOK
N
CLO
Y
CLK
N
Y
CL
COK
N
CO
Y
CK
N
C
Calculate number of possible outcomes
Calculate number of outcomes that
include lettuce
Simplify probability as far as possible
Next Comment
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Logic Diagrams : Question 2
(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.
21
Tull
26
Achan
Reveal answer only
Dweep
Boul
8
Lillen
20 Carat
Go to full solution
10
Go to Marker’s Comments
Go to Logic Diagrams Menu
EXIT
23
Dunky
17
18
13
Gore
20
31
15
Go to Main Menu
12
Eerie
Bridge
MAP1
26
MAP2
Wull
Logic Diagrams : Question 2
(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.
21
Tull
26
Achan
Reveal answer only
Dweep
Boul
8
Lillen
20 Carat
Go to full solution
10
Go to Marker’s Comments
Go to Logic Diagrams Menu
EXIT
23
Dunky
17
18
13
Gore
20
31
15
Go to Main Menu
12
Eerie
Bridge
MAP1
26
Wull
MAP2
(b) B-A-D-E-D-C-B-E
(c) No, Map 2 has more than 2 odd vertices
and so is not traversable.
Question 2
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.
Begin Solution
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Markers Comments
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Achan
Boul
26
Dweep
Tull
12
21
8
Eerie
31
17
15
Dunky
Carat
Wull
20
23
20
18
13
Gore
10
Lillen
26
Bridge
(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
26
Dweep
Tull
12
21
8
Eerie
31
17
15
Carat
Wull
20
23
20
Dunky
18
13
Gore
10
Marker’s Comments
Lillen
26
2 possible network diagrams for the
given maps
Find the shortest route. Know to
begin and end at an odd vertex
Bridge
(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.
Mention that the network is not
traversable and support your
statement with evidence of more than
2 odd vertices
Next Comment
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Logic Diagrams : Question 3
(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
Reveal answer only
0
Start
Go to full solution
Go to Marker’s Comments
Go to Logic Diagrams Menu
Go to Main Menu
EXIT
0
0
D
18
12
8
B
C
F
21
E
11
G
6
End
7
A
A – Cook fish
B – Fry onions
C – Boil potatoes
D – Chop fish
E – Mash potatoes
F – Mix ingrediants
G – Fry cakes
Logic Diagrams : Question 3
(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
Reveal answer only
0
Start
Go to full solution
Go to Marker’s Comments
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Go to Main Menu
(a) 21 minutes
(b) C-E-F-G (45 minutes)
EXIT
0
0
D
18
12
8
B
C
F
21
E
11
G
6
End
7
A
A – Cook fish
B – Fry onions
C – Boil potatoes
D – Chop fish
E – Mash potatoes
F – Mix ingrediants
G – Fry cakes
A – Cook fish
B – Fry onions
C – Boil potatoes
D – Chop fish
E – Mash potatoes
F – Mix ingrediants
G – Fry cakes
Question 3
(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
0
Start
0
0
12
F
21
(b) Longest path = critical path = C-E-F-G
8
B
C
(a) Boil potatoes – arc CE = 21 minutes
D
18
E
11
G
6
End
7
A
Begin Solution
Continue Solution
Markers Comments
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Minimum time required = longest path
= 21 + 7 +11 +6
= 45 minutes
Marker’s Comments
(a) Boil potatoes = arc CE = 21 minutes
Recognise Arc CE corresponds to
“Boil potatoes”
(b) Longest path = critical path = C-E-F-G
Know that the longest path is the
critical path
Minimum time required = longest path
= 21 + 7 +11 +6
= 45 minutes
Know that the minimum time required
to complete the job = longest path
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Logic Diagrams : Question 4
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
Reveal answer only
Yes
Go to full solution
Yes
Go to Marker’s Comments
Go to Logic Diagrams Menu
Pay a
deposit
?
deposit of
£5000
20% of
cost for
6 years
Go to Main Menu
15% of
cost for
6 years
Stop
EXIT
No
Is the
car new
?
No
Pay cost
Logic Diagrams : Question 4
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
Reveal answer only
Yes
Go to full solution
Yes
Go to Marker’s Comments
Go to Logic Diagrams Menu
Pay a
deposit
?
deposit of
£5000
20% of
cost for
6 years
Go to Main Menu
£15 800
15% of
cost for
6 years
Stop
EXIT
No
Is the
car new
?
No
Pay cost
Question 4
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
Begin Solution
Continue Solution
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15% of £12000 = £1800 for 1 year
£1800 x 6 = £10800
Total cost = 5000 + 10800 = £15800
Marker’s Comments
15% of £12000 = £1800 for 1 year
Calculate 15% of cost of car
£1800 x 6 = £10800
Calculate repayments over 6 years
Total cost = 5000 + 10800 = £15800
Calculate the total cost by adding the
deposit
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 4 :
Formulae
Please choose a question to attempt from the following:
1
EXIT
2
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Unit 4 Menu
Formulae : Question 1
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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Formulae : Question 1
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
Reveal answer only
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EXIT
(a) v = 156 m/s
(b) t = 6.7 m/s
Question 4
(a)
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
v  u  10t
v  6  10 15
v  6  150
v  156 m / s
(b)
v  u  10t
70  3  10t
Begin Solution
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70  3  10t
67
t
10
t  6.7m / s
(a) v  u  10t
Marker’s Comments
Substitute values
v  6  10 15
Begin to solve equation
v  6  150
State solution with units
v  156m / s
(b) v  u  10t
70  3  10t
Substitute values
Begin to solve equation
State solution with units
70  3  10t
t
67
10
t  6.7 m / s
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Formulae : Question 2
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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Formulae : Question 2
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?
Reveal answer only
25ºC
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Question 2
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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Continue Solution
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5
(Fahrenheit – 32)
9
5
 (77  32)
9
5
  (45)
9
Centigrade 
 250 C
Marker’s Comments
Form ula
5
Centigrade  (Fahrenheit – 32)
9
5
 (77  32)
9
5
  (45)
9
State formula
Substitute values
Begin to solve equation
State solution with units
 250 C
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
UNIT 3 :
More Algebraic
Operations
Further
Trigonometry
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Quadratic
Functions
INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
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UNIT 3 :
Algebraic
Operations
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1
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2
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3
More Algebraic Operations : Question 1
Express 72 in its simplest form.
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More Algebraic Operations : Question 1
Express 72 in its simplest form.
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6 2
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Question 1
Express 72 in its
simplest form.
72  8  9
 8 9
 8 9
 4 2  9
 2 2 3
6 2
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72  8  9
 8 9
 8 9
ab  a  b
Demonstrate that
Simplify
8
further
Collect like terms
 4 2  9
 2 2 3
6 2
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More Algebraic Operations : Question 2
 12
3
4 


a
a

2
a

3
a
Expand




2
Reveal answer only
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More Algebraic Operations : Question 2
 12
3
4 


a
a

2
a

3
a
Expand




2
Reveal answer only
a 5  2a 5 
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3
a2
Question 2
 12
3
4 


a
a

2
a

3
a
Expand




1
2
 a  a  a 2  2a 3  a 2  3a 4
2
2
5
2
 a  2a 5  3a 2
3
 a  2a  2
a
5
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5
Marker’s Comments
1
2
 a  a  a  2a  a  3a
2
2
5
2
 a  2a 5  3a 2
3
2
4
Know
Know
a n  a m  a n m
n
m
a  m an
3
 a  2a  2
a
5
5
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More Algebraic Operations : Question 3
4
Express
as a fraction with a rational denominator.
3 2
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More Algebraic Operations : Question 3
4
Express
as a fraction with a rational denominator.
3 2
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12 4 2
7
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Question 3
4
Express
as a fraction
3 2
4
3 2

3 2 3 2

with a rational denominator.
4(3  2 )
(3  2 )(3  2 )
12  4 2

93 2 3 2 2

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12 4 2
7
Marker’s Comments
4
3 2

3 2 3 2

4(3  2 )
(3  2 )(3  2 )

12  4 2
93 2 3 2 2

12 4 2
7
Know to multiply by conjugate surd
Multiply out brackets
Simplify expressions
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
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UNIT 3 :
Quadratic
Functions
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1
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2
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3
Quadratic Functions : Question 1
Solve x² - x + 12 = 0
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Quadratic Functions : Question 1
Solve x² - x + 12 = 0
Reveal answer only
x = -3 and x = 4
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Question 1
Solve x² - x + 12 = 0
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x² - x + 12 = 0
(x + 3)(x - 4) = 0
x+3=0
or
x–4=0
x=-3
or
x=4
Marker’s Comments
x² - x + 12 = 0
Factorise equation
(x + 3)(x - 4) = 0
Know that either factor could be
equal to 0
x+3=0
or
x–4=0
x=-3
or
x=4
Solve both equations to find roots
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Quadratic Functions : Question 2
Solve this quadratic equation; 3x² - 14x + 17
Reveal answer only
x = 3.54 or 1.13
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Question 2
Solve this quadratic equation;
3x² - 14x + 17
a=3
b = -14
c = 12
 b  b 2  4ac
x
2a
14  142  4  3 12
x
23
x
14 52
6
14  52
14  52
OR
6
6
21.21
6.79
x
OR
6
6
x
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x = 3.54 OR 1.13
Marker’s Comments
a=3
b = -14
c = 12
Know to use quadratic formula
Substitute values
 b  b 2  4ac
x
2a
Begin to solve equation
14  142  4  3 12
x
23
Find both solutions for x
x
14 52
6
14  52
14  52
OR
6
6
21.21
6.79
x
OR
6
6
x
x = 3.54 OR 1.13
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Quadratic Functions : Question 3
Sketch the graph of the function y = -(x – 4)² - 10
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2– 30
2
4
6
8
10
12
14
16
18
20
22
24
26
28
2– 2
4
6
Quadratic Functions : Question 3
Sketch the graph of the function y = -(x – 4)² - 10
y
2
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– 2 – 2
– 4
– 6
– 8
– 10
– 12
– 14
– 16
– 18
– 20
– 22
– 24
– 26
– 28
– 30
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EXIT
2
4
(4, 10)
6 x
Equation of the form y = a(x –b)² + c
Question 3
Sketch the graph of the function
a = -1
b=4
c = -10
Equation of the axis of symmetry; x = 4
y = -(x – 4)² - 10
Vertex at (4, -10)
At y-intercept, x = 0
2– 30
2
4
6
8
10
12
14
16
18
20
22
24
26
28
2– 2
4
6
y = -(0 – 4) ² - 10
= -( - 4) ² -10
= -16 – 10
= -26
y-intercept at (0, -26)
y
2
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– 2 – 2
– 4
– 6
– 8
– 10
– 12
– 14
– 16
– 18
– 20
– 22
– 24
– 26
– 28
– 30
2
4
(4, 10)
6 x
Equation of the form y = a(x –b)² + c
a = -1
b=4
c = -10
Equation of the axis of symmetry; x = 4
At y-intercept, x = 0
y = -(0 – 4) ² - 10
= -( - 4) ² -10
= -16 – 10
= -26
Know the vertex occurs at (b, c)
y-intercept at (0, -26)
2
4
Let x = 0 for y-intercept
Solve equation to find y-intercept
Use information to sketch graph
with key points labelled.
y
2
Know that equation is of the form
y = a(x –b)² + c
Know that x = b is the axis of
symmetry
Vertex at (4, -10)
– 2 – 2
– 4
– 6
– 8
– 10
– 12
– 14
– 16
– 18
– 20
– 22
– 24
– 26
– 28
– 30
Marker’s Comments
6 x
(4, 10)
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INTERMEDIATE 2 – ADDITIONAL QUESTION
BANK
You have chosen to study:
UNIT 3 :
Further
Trigonometry
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1
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2
3
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Further Trigonometry : Question 1
Solve 3tanxº - 1 = 0 for 0 ≤ x ≤ 360
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Further Trigonometry : Question 1
Solve 3tanxº - 1 = 0 for 0 ≤ x ≤ 360
Reveal answer only
x = 18.4º and 198.4 º
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Question 1
Solve 3tanxº - 1 = 0 for 0 ≤ x ≤ 360
3 tan xº - 1 = 0
3 tan xº = 1
1
tan xº =
3
1
x  tan1  
 3
= 18.4 º
and x = 180 + 18.4 = 198.4 º
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Marker’s Comments
3 tan xº - 1 = 0
3 tan xº = 1
1
tan xº =
3
Find x
Know which quadrants the solution
belongs to, and therefore find other
solution
1
x  tan1  
 3
= 18.4 º
and x = 180 + 18.4 = 198.4 º
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Further Trigonometry : Question 2
The graph has an equation of the form y = a cos (x – b)º. Write
down the values of a and b.
y
2
1
150
– 1
Reveal answer only
– 2
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330
x
Further Trigonometry : Question 2
The graph has an equation of the form y = a cos (x – b)º. Write
down the values of a and b.
y
2
1
150
330
– 1
Reveal answer only
– 2
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a = 1 and b = 60
x
Question 2
a = Amplitude = 1
The graph has an equation of
the form y = a cos (x – b)º.
Write down the values of a and
b.y
b = Phase angle = 60
2
1
150
330
x
– 1
– 2
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a = Amplitude = 1
Know that a = amplitude and read
value from graph
b = Phase angle = 60
Know that b = phase angle and
calculate from graph
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Further Trigonometry : Question 3
1 0
Sketch the graph y  2 sin x for 0 ≤ x ≤ 720
2
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1– 2
2
1
Further Trigonometry : Question 3
1 0
Sketch the graph y  2 sin x for 0 ≤ x ≤ 720
2
y
2
1
Reveal answer only
360
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– 1
– 2
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720 x
Question 3
1– 2
2
1
1 0
Sketch the graph y  2 sin x
2
Amplitude = 2
(y-maximum = 2, y-minimum = -2)
Period = 0.5
for 0 ≤ x ≤ 720
y
2
1
360
– 1
– 2
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720 x
Marker’s Comments
Recognise amplitude = 2
Amplitude = 2
(y-maximum = 2, y-minimum = -2)
State periodicity
Period = 0.5
Use information to sketch graph,
making intercepts with x axis very
clear
y
2
1
360
720 x
– 1
– 2
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Quadratic Functions : Question 2
Solve this quadratic equation; 3x² - 14x + 17
Reveal answer only
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