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GCSE Linear Questions (H)
β€’ Question set 1
β€’ Question set 2
β€’ Question set 3
β€’ Question set 4
β€’ Question set 5
β€’ Question set 6
β€’ Question set 7
β€’ Question set 8
β€’ Question set 9
β€’ Question set 10
β€’ Question set 11
β€’ Question set 12
β€’ Question set 13
β€’ Question set 14
β€’ Question set 15
β€’ Question set 16
β€’ Question set 17
β€’ Question set 18
β€’ Question set 19
β€’ Question set 20
Odd numbers – Non
calculator
Even numbers -
Calculator
Number Algebra
Shape, Space, Measure Handling Data
Solve the following simultaneous
equations
Arrange these in ascending order
25
, 64
1
2, 2βˆ’1
, 80
, 16
1
4
5π‘₯ + 2𝑦 = 16
3π‘₯ βˆ’ 𝑦 = 14
8 9 14 ?
The mean of the numbers of these 4
cards is 9. What is the number on the
fourth card?
4 + 3x
x + 6
The perimeter is equal to 32cm. What
is the value of x?
Number
Arrange these in ascending order
25, 64
1
2, 2βˆ’1, 80, 16
1
4
25 = 2 x 2 x 2 x 2 x 2 = 32
64
1
2 = 64 = 8
2βˆ’1 =
1
21
=
1
2
80 = 1
16
1
4 =
4
16 = 2
2βˆ’1
, 80
, 16
1
4, 64
1
2, 25
Algebra
Solve the following simultaneous
equations 5π‘₯ + 2𝑦 = 16
3π‘₯ βˆ’ 𝑦 = 14
a
b
Multiply equation b by 2,
6π‘₯ βˆ’ 2𝑦 = 28 c
Add equations a and c
11π‘₯ = 44
𝒙 = πŸ’
Substitute x back into one of the
original equations and solve
Shape, Space, Measure 4 + 3x
x + 6
The perimeter is equal to 32cm. What
is the value of x?
Perimeter = 32cm
Perimeter = 4 + 3π‘₯ + π‘₯ + 6 + 4 + 3π‘₯ + π‘₯ + 6
= 8π‘₯ + 20
Therefore
8π‘₯ + 20 = 32
8π‘₯ = 12
π‘₯ = 1.5
Handling Data
8 9 14 ?
The mean of the numbers of these 4
cards is 9. What is the number on the
fourth card?
Total of the four cards is the mean
multiplied by four
9 Γ— 4 = 36
So find the missing card by subtraction
36 βˆ’ 8 βˆ’ 9 βˆ’ 14 = 5
Number Algebra
Shape, Space, Measure Handling Data
Expand and Simplify the followingA bank account gains 6% compound
interest per annum. If Tom puts Β£700
into his account, how much could he
expect after 6 years?
4 π‘₯ βˆ’ 5 βˆ’ 3(2π‘₯ βˆ’ 6)
(π‘₯ + 2)(π‘₯ βˆ’ 7)
𝑑 + 5 2
Calculate the mean number of cars per
household
35Β°
Work out the missing length x
π‘₯
15
No. of cars Frequency
0 4
1 8
2 7
3 2
Number
A bank account gains 6% compound
interest per annum. If Tom puts Β£700
into his account, how much could he
expect after 6 years?
To calculate the money after one year of interest, multiply by 1.06
π‘Œπ‘’π‘Žπ‘Ÿ π‘œπ‘›π‘’ βˆ’ 700 Γ— 1.06
π‘Œπ‘’π‘Žπ‘Ÿ π‘‘π‘€π‘œ βˆ’ 700 Γ— 1.06 Γ— 1.06
π‘Œπ‘’π‘Žπ‘Ÿ π‘‘β„Žπ‘Ÿπ‘’π‘’ βˆ’ 700 Γ— 1.06 Γ— 1.06 Γ— 1.06
π‘Œπ‘’π‘Žπ‘Ÿ 𝑠𝑖π‘₯ βˆ’ 700 Γ— 1.06 Γ— 1.06 Γ— 1.06 Γ— 1.06 Γ— 1.06 Γ— 1.06
Algebra
Expand and Simplify the following
4 π‘₯ βˆ’ 5 βˆ’ 3(2π‘₯ βˆ’ 6)
(π‘₯ + 2)(π‘₯ βˆ’ 7)
𝑑 + 5 2
4 π‘₯ βˆ’ 5 βˆ’ 3 2π‘₯ βˆ’ 6
= 4π‘₯ βˆ’ 20 βˆ’ 6π‘₯ + 18
= βˆ’2π‘₯ βˆ’ 2
π‘₯ + 2 π‘₯ βˆ’ 7 = π‘₯2 βˆ’ 7π‘₯ + 2π‘₯ βˆ’ 14
= π‘₯2
βˆ’ 5π‘₯ βˆ’ 14
𝑑 + 5 2 = 𝑑 + 5 𝑑 + 5
= 𝑑2
+ 5𝑑 + 5𝑑 + 25
= 𝑑2
+ 10𝑑 + 25
Shape, Space, Measure
35Β°
Work out the missing length x
π‘₯
15
35Β°
π‘₯
15 Label your sides (in play)
h
a
Decide your triangle
h
a
c
Cover up what you are looking for and
write down your formula
β„Ž =
π‘Ž
cos πœƒ
β„Ž =
15
cos 35
β„Ž = 18.31161883
β„Ž = 18.3 (3. 𝑠. 𝑓)
Handling Data
Calculate the mean number of cars per
household
No. of cars Frequency
0 4
1 8
2 7
3 2
No. of cars Frequency Mean
0 4 4 x 0 = 0
1 8 8 x 1 =8
2 7 7 x 2 = 14
3 2 3 x 2 = 6
Total 21 28
π‘€π‘’π‘Žπ‘› =
28
21
= 1. 3
Number Algebra
Shape, Space, Measure Handling Data
Write out the nth term for each
sequence. Hence work out what the
10th and 100th term will be.
Approximate the answer to
12.31 Γ— 16.9
0.394 Γ— 0.216
5, 8, 11, 14, …
10, 4, βˆ’2, βˆ’8, βˆ’14
4, 7, 12, 19, 28, …
6, 11, 18, 27, 38, …
Three cards are drawn from a deck and
replaced each time. What is the
probability of drawing 3 hearts?
Iron has a density of 8g/cm3 . What will
be the mass of the above cuboid?
4cm
2cm
5cm
Number
Approximate the answer to
12.31 Γ— 16.9
0.394 Γ— 0.216
Round all numbers to 1 significant figure
10 Γ— 20
0.4 Γ— 0.2
Calculate this sum
200
0.8
Use equivalent fractions to help divide by a decimal
200
0.8
=
2000
8
= 250
Algebra
Write out the nth term for each
sequence. Hence work out what the
10th and 100th term will be.
5, 8, 11, 14, …
10, 4, βˆ’2, βˆ’8, βˆ’14
4, 7, 12, 19, 28, …
6, 11, 18, 27, 38, …
Sequence nth term 10th term 100th term
5, 8, 11, 14, … 3𝑛 + 2 32 302
10, 4, βˆ’2, βˆ’8, βˆ’14, … 16 βˆ’ 6𝑛 -44 -584
4, 7, 12, 19, 28, … 𝑛2 + 3 103 10003
6, 11, 18, 27, 38, … 𝑛2 + 2𝑛 + 3 123 10203
Shape, Space, Measure
Iron has a density of 8g/cm3 . What will
be the mass of the above cuboid?
4cm
2cm
5cm
πΆπ‘Ÿπ‘œπ‘ π‘  π‘ π‘’π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 4π‘π‘š Γ— 5π‘π‘š
= 20π‘π‘š2
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = πΆπ‘Ÿπ‘œπ‘ π‘  π‘ π‘’π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž Γ— π‘‘π‘’π‘π‘‘β„Ž
= 20 Γ— 2
= 40π‘π‘š3
The density is 8𝑔/π‘π‘š3
this means, every π‘π‘š3
weighs 8g. So
the mass will be
π‘€π‘Žπ‘ π‘  = 40 Γ— 8
= 320𝑔
Handling Data
Three cards are drawn from a deck and
replaced each time. What is the
probability of drawing 3 hearts?
π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ π»π‘’π‘Žπ‘Ÿπ‘‘ 𝐴𝑁𝐷 π»π‘’π‘Žπ‘Ÿπ‘‘ 𝐴𝑁𝐷 π»π‘’π‘Žπ‘Ÿπ‘‘ =
1
4
Γ—
1
4
Γ—
1
4
𝑃 𝐻𝐻𝐻 =
1
64
Number Algebra
Shape, Space, Measure Handling Data
Solve the following quadratic equation,
leave your answers to 3 significant
figures.
πΆπ‘Žπ‘™π‘π‘’π‘™π‘Žπ‘‘π‘’
4.2 Γ— 7.3
5.2 βˆ’ 9.3
Write down your full calculator display
Round this number to 3 significant
figures
3π‘₯2 βˆ’ 5π‘₯ = 18
A school of 800 pupils want to do a
survey on school dinners. They
decided to take a stratified sample of
30 pupils. How many of each year
group should they ask?
π‘₯
Work out the size of angle x
18
13
Year 7 Year 8 Year 9 Year 10 Year 11
182 124 128 195 171
Number πΆπ‘Žπ‘™π‘π‘’π‘™π‘Žπ‘‘π‘’
4.2 Γ— 7.3
5.2 βˆ’ 9.3
Write down your full calculator display
Round this number to 3 significant figures
For Casio fx-83GT plus
Type
4 . 2 x 7 .
3 5 . 2 -
9 . 3 = 𝑺 β‰ͺ=≫D
βˆ’7.478048
βˆ’7.48
Algebra Solve the following quadratic equation,
leave your answers to 3 significant
figures.
3π‘₯2 βˆ’ 5π‘₯ = 18
Make it look like a usual quadratic
3π‘₯2 βˆ’ 5π‘₯ = 18
3π‘₯2
βˆ’ 5π‘₯ βˆ’ 18 = 0
Difficult to factorise οƒ  so use the formula.
πΉπ‘œπ‘Ÿ π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0, π‘€β„Žπ‘’π‘Ÿπ‘’ π‘Ž β‰  0
π‘₯ =
βˆ’π‘ Β± 𝑏2 βˆ’ 4π‘Žπ‘
2π‘Ž
π‘Ž = 3
𝑏 = βˆ’5
𝑐 = βˆ’18
π‘₯ =
βˆ’ βˆ’5 Β± βˆ’5 2 βˆ’ 4 Γ— 3 Γ— βˆ’18
2 Γ— 3
π‘₯ =
5 Β± 25 + 216
6
π‘₯ =
5 + 241
6
π‘œπ‘Ÿ π‘₯ =
5 βˆ’ 241
6
π‘₯ = 3.42 π‘œπ‘Ÿ π‘₯ = βˆ’1.75
Shape, Space, Measure
π‘₯
Work out the size of angle x
18
13
Label your sides (in play)
o
a
Decide your triangle
a
o
t
Cover up what you are looking for and
write down your formula
tan π‘₯ =
π‘œ
π‘Ž
tan π‘₯ =
18
13
π‘₯ = tanβˆ’1
18
13
π‘₯ = 54.2Β°
π‘₯
18
13
Handling Data
A school of 800 pupils want to do a
survey on school dinners. They
decided to take a stratified sample of
30 pupils. How many of each year
group should they ask?
Year 7 Year 8 Year 9 Year 10 Year 11
182 124 128 195 171
30 pupils out of 5 year groups –
must mean 6 from each year
group? Wrong.
We take a stratified sample – this
means we take a fair
representation of each year
group. There should be more year
7 than year 8 in the sample.
Year 7 =
182
800
Γ— 30 = 6.825 = 7 𝑝𝑒𝑝𝑖𝑙𝑠
Year 8 =
124
800
Γ— 30 = 4.65 = 5 𝑝𝑒𝑝𝑖𝑙𝑠
Year 9 =
128
800
Γ— 30 = 4.8 = 5 𝑝𝑒𝑝𝑖𝑙𝑠
Year 10 =
195
800
Γ— 30 = 7.3125 = 7 𝑝𝑒𝑝𝑖𝑙𝑠
Year 11 =
171
800
Γ— 30 = 6.4125 = 6 𝑝𝑒𝑝𝑖𝑙𝑠
Check the student numbers add up
to your sample size.
7 + 5 + 5 + 7 + 6 = 30
Each year group gets a fair
representation.
Number Algebra
Shape, Space, Measure Handling Data
Factorise fullySimplify the following
48 =
3 Γ— 12 =
2 5 Γ— 3 10 =
300 βˆ’ 75 =
5π‘₯4 𝑦3 𝑧2 βˆ’ 15π‘₯6 𝑦5 𝑧
π‘₯2
+ 11π‘₯ βˆ’ 12
π‘₯2
βˆ’ 64
9π‘₯2 βˆ’ 25𝑦2
4, 6, 2, 5, 7, 9, 9, 2, 1, 4, 8
Draw a box plot for the following data
Calculate the area of the rectangle
3 cm
5π‘₯ βˆ’ 3 cm
3π‘₯ + 5 cm
Number Simplify the following
48 =
3 Γ— 12 =
2 5 Γ— 3 10 =
300 βˆ’ 75 =
See Surds or the think SQUARE numbers
48 = 16 Γ— 3
= 16 3
= 4 3
What is the largest square number
which is a factor of 48. 16
3 Γ— 12 = 36
= 6
2 5 Γ— 3 10 = 6 50
= 6 25 Γ— 2
= 6 25 2
= 30 2
When adding fractions, the denominator needs to be the same number.
Similarly, when adding Surds, the Surds need to be the same number –
so simplify them
300 = 100 Γ— 3
= 100 3
= 10 3
75 = 25 Γ— 3
= 25 3
= 5 3
300 βˆ’ 75 =
10 3 βˆ’ 5 3 = 5 3
Algebra
5π‘₯4 𝑦3 𝑧2 βˆ’ 15π‘₯6 𝑦5 𝑧
π‘₯2 + 11π‘₯ βˆ’ 12
π‘₯2 βˆ’ 64
9π‘₯2 βˆ’ 25𝑦2
Factorise fully
5π‘₯4
𝑦3
𝑧2
βˆ’ 15π‘₯6
𝑦5
𝑧 = 5π‘₯4
𝑦3
𝑧(𝑧 βˆ’ 3π‘₯2
𝑦2
)
π‘₯2
+ 11π‘₯ βˆ’ 12 = π‘₯ + 12 π‘₯ βˆ’ 11
π‘₯2 βˆ’ 64 = π‘₯ + 8 π‘₯ βˆ’ 8
9π‘₯2 βˆ’ 25𝑦2 = (3π‘₯ + 5𝑦)(3π‘₯ βˆ’ 5𝑦)
Shape, Space, Measure
Calculate the area of the rectangle
3 cm
5π‘₯ βˆ’ 3 cm
3π‘₯ + 5 cm
Remember the features about a rectangle – two pairs of equal sides!
You can set up an equation using this information to work out the
length of the rectangle.
5π‘₯ βˆ’ 3 = 3π‘₯ + 5
2π‘₯ βˆ’ 3 = 5
2π‘₯ = 8
π‘₯ = 4
Therefore the length is 3 x 4 + 5 = 17cm.
π΄π‘Ÿπ‘’π‘Ž = 17 Γ— 3
= 51π‘π‘š2
Handling Data
4, 6, 2, 5, 7, 9, 9, 2, 1, 4, 8
Draw a box plot for the following data
When working with Quartiles, Median and Range – it is always useful to
arrange your data in size order.
1, 2, 2, 4, 4, 5, 6, 7, 8, 9, 9
Need a few pieces of information for a box plot.
Highest value – 9
Lowest value – 1
The Median, since there are 11 numbers, the middle number will be the
11+1
2
= 6th number. So the 6th number is 5
Lower quartile will be the
11+1
4
= 3π‘Ÿπ‘‘ number in our list. So LQ = 2
Upper quartile will be the
3 11+1
4
= 9π‘‘β„Ž number in our list. UQ = 8
Number Algebra
Shape, Space, Measure Handling Data
Using trial and improvement to find a
solution to 1 decimal place
The lengths of a room have been
calculated to the nearest metre.
Calculate the greatest and least area that
the room could be.
π‘₯3 βˆ’ 5π‘₯ = 50
The probability of Man Utd winning a
match under David Moyes is 0.3 and
losing is 0.2.
Man Utd play 3 matches, what is the
probability that out of these, 2 are
won and one is drawn.
Calculate the volume and surface area.
Leave your answer to 3 significant
figures.
8 π‘π‘š
5 π‘π‘š
9
4
Number The lengths of a room have been
calculated to the nearest metre.
Calculate the greatest and least area that
the room could be.
9
4
Lower bound
3.5
8.5
π΄π‘Ÿπ‘’π‘Ž = 3.5 Γ— 8.5
= 29.75 π‘š2 Upper bound
9.5
4.5
π΄π‘Ÿπ‘’π‘Ž = 4.5 Γ— 9.5
= 42.75 π‘š2
Algebra Using trial and improvement to find a
solution to 1 decimal place
π‘₯3 βˆ’ 5π‘₯ = 50
𝒙 𝒙 πŸ‘
βˆ’ πŸ“π’™ Comment
4 43
βˆ’ 5 Γ— 4 = 44 Low
5 53
βˆ’ 5 Γ— 5 = 100 High
4.3 4.33
βˆ’ 5 Γ— 4.3 = 58.007 High
4.2 4.23
βˆ’ 5 Γ— 4.2 = 53.088 High
4.1 4.13
βˆ’ 5 Γ— 4.1 = 48.421 Low
Draw a table – it helps!
4.153 βˆ’ 5 Γ— 4.15 = 50.723375
Since 4.15 is too high, everything about it must be too high as well.
Therefore the solution to 1 decimal place is 4.1
πΆπ‘Ÿπ‘œπ‘ π‘  π‘ π‘’π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘–π‘Ÿπ‘π‘™π‘’
= πœ‹ Γ— π‘Ÿ2
= πœ‹ Γ— πŸ’2
= 16πœ‹ cm2
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = πΆπ‘Ÿπ‘œπ‘ π‘  π‘ π‘’π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž Γ— π‘‘π‘’π‘π‘‘β„Ž
= 16πœ‹ π‘₯ 5
= 80πœ‹
= 251.3274123
= 251 π‘π‘š3
Shape, Space, Measure
Calculate the volume and surface area.
Leave your answer to 3 significant
figures.
8 π‘π‘š
5 π‘π‘š
Handling Data
The probability of Man Utd winning a
match under David Moyes is 0.3 and
losing is 0.2.
Man Utd play 3 matches, what is the
probability that out of these, 2 are
won and one is drawn.
Probability of drawing a match is
1 βˆ’ 0.3 βˆ’ 0.2 = 0.5
List all the possible combinations
Win Win Draw
Win Draw Win
Draw Win Win
𝑃 π‘Š 𝐴𝑁𝐷 π‘Š 𝐴𝑁𝐷 𝐷 = 0.3 Γ— 0.3 Γ— 0.5 = 0.045
𝑃 π‘Š 𝐴𝑁𝐷 𝐷 𝐴𝑁𝐷 π‘Š = 0.3 Γ— 0.5 Γ— 0.3 = 0.045
𝑃 𝐷 𝐴𝑁𝐷 π‘Š 𝐴𝑁𝐷 π‘Š = 0.5 Γ— 0.3 Γ— 0.3 = 0.045
Therefore, the probability of winning two and drawing one
match is 0.045 + 0.045 + 0.045 = 0.135
Number Algebra
Shape, Space, Measure Handling Data
Solve the following
Leave all answers as mixed numbers
1
3
+
4
5
=
3
5
7
βˆ’ 1
2
3
=
4
9
Γ—
3
8
=
4π‘₯ + 6 = π‘₯ βˆ’ 12
7π‘₯ βˆ’ 8 = 20 βˆ’ 3π‘₯
5π‘₯ + 4
6
=
3π‘₯ + 8
5
Draw a cumulative frequency graph for the
following data. Construct a box plot from this
information.
Calculate the distance between the
coordinates (-1, 4) and (11, 9)
Marks Frequency
0 – 10 2
11 – 20 7
21 – 30 13
31 – 40 8
Number Leave all answers as mixed numbers
1
3
+
4
5
=
3
5
7
βˆ’ 1
2
3
=
4
9
Γ—
3
8
=
1
3
+
4
5
=
5
15
+
12
15
=
17
15
= 𝟐
𝟐
πŸπŸ“ 3
5
7
βˆ’ 1
2
3
=
26
5
βˆ’
5
3
=
78
15
βˆ’
25
15
=
53
15
= πŸ‘
πŸ–
πŸπŸ“
4
9
Γ—
3
8
=
12
72
=
𝟏
πŸ”
Algebra Solve the following
4π‘₯ + 6 = π‘₯ βˆ’ 12
7π‘₯ βˆ’ 8 = 20 βˆ’ 3π‘₯
5π‘₯ + 4
6
=
3π‘₯ + 8
5
4π‘₯ + 6 = π‘₯ βˆ’ 12
βˆ’π‘₯ 3π‘₯ + 6 = βˆ’12 βˆ’π‘₯
βˆ’6 3π‘₯ = βˆ’18 βˆ’6
Γ· 3 𝒙 = βˆ’πŸ” Γ· 3
7π‘₯ βˆ’ 8 = 20 βˆ’ 3π‘₯
+3π‘₯ 10π‘₯ βˆ’ 8 = 20 +3π‘₯
+8 10π‘₯ = 28 +8
Γ· 10 𝒙 = 𝟐. πŸ– Γ· 10
5π‘₯ + 4
6
=
3π‘₯ + 8
5
Γ— 5
25π‘₯ + 20
6
= 3π‘₯ + 8 Γ— 5
Γ— 6 25π‘₯ + 20 = 18π‘₯ + 48 Γ— 6
βˆ’18π‘₯ 7π‘₯ + 20 = 48 βˆ’18π‘₯
βˆ’20 7π‘₯ = 28 βˆ’20
Γ· 7 𝒙 = πŸ’ (Γ· 7)
Shape, Space, Measure Calculate the distance between the
coordinates (-1, 4) and (11, 9)
(-1,4)
(11,9)
11 – βˆ’1 = 12
9 βˆ’ 4 = 5
π‘ˆπ‘ π‘’ π‘ƒπ‘¦π‘‘β„Žπ‘Žπ‘”π‘œπ‘Ÿπ‘Žπ‘  π‘‘π‘œ π‘π‘Žπ‘™π‘π‘’π‘™π‘Žπ‘‘π‘’ π‘‘β„Žπ‘’ π‘™π‘’π‘›π‘”π‘‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ 𝑙𝑖𝑛𝑒
π‘Ž2
+ 𝑏2
= 𝑐2
52 + 122 = 𝑐2
25 + 144 = 𝑐2
169 = 𝑐2
πŸπŸ‘ = 𝒄
Handling Data
Draw a cumulative frequency graph for the
following data. Construct a box plot from this
information.
Marks Frequency
0 – 10 2
11 – 20 7
21 – 30 13
31 – 40 8Marks Frequency Cumulative
Frequency
0 – 10 2 2
11 – 20 7 9
21 – 30 13 22
31 – 40 8 30
Plot your graph using the end
points and cumulative
frequency.
Number Algebra
Shape, Space, Measure Handling Data
Rearrange the formula to make x the
subject
Tins of paint are on offer, buy 5 get 1
free. John the painter needs 27 tins of
paint. If a tin of paint costs Β£3.43, how
much will John have to pay?
𝑦 = 4π‘₯ βˆ’ 2
jπ‘₯ + 𝑑 = π‘˜π‘₯ βˆ’ π‘š
Calculate an estimate for the mean
Table on time of goal scored
Calculate the area of the sector and
the arc length. Leave all answers to 1
decimal place.
132Β° 3 π‘π‘š
Time of match Frequency
0 < π‘₯ ≀ 15 8
15 < π‘₯ ≀ 30 4
30 < π‘₯ ≀ 45 5
45 < π‘₯ ≀ 60 7
60 < π‘₯ ≀ 75 7
75 < π‘₯ ≀ 90 10
Number Tins of paint are on offer, buy 5 get 1 free. John
the painter needs 27 tins of paint. If a tin of
paint costs Β£3.43, how much will John have to
pay?
If John buys 5 tins he gets 6. So using this, if he buys 20
tins, he will actually get 24. Therefore he only needs to
purchase another 3 tins to have 27.
John needs to buy 23 tins.
23 Γ— 3.43 = Β£78.89
Algebra Rearrange the formula to make x the
subject
𝑦 = 4π‘₯ βˆ’ 2
jπ‘₯ + 𝑑 = π‘˜π‘₯ βˆ’ π‘šπ‘¦ = 4π‘₯ βˆ’ 2
+2 𝑦 + 2 = 4π‘₯ +2
Γ· 4
𝑦 + 2
4
= π‘₯ Γ· 4
π‘€π‘Žπ‘˜π‘’ π‘₯ π‘‘β„Žπ‘’ 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 𝑏𝑦 𝑝𝑒𝑑𝑑𝑖𝑛𝑔 π‘œπ‘› π‘‘β„Žπ‘’ 𝑙𝑒𝑓𝑑 β„Žπ‘Žπ‘›π‘‘ 𝑠𝑖𝑑𝑒.
π‘₯ =
𝑦 + 2
4
jπ‘₯ + 𝑑 = π‘˜π‘₯ βˆ’ π‘š
βˆ’π‘˜π‘₯ 𝑗π‘₯ βˆ’ π‘˜π‘₯ + 𝑑 = βˆ’π‘š βˆ’π‘˜π‘₯
βˆ’π‘‘ 𝑗π‘₯ βˆ’ π‘˜π‘₯ = βˆ’π‘‘ βˆ’ π‘š βˆ’π‘‘
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’ π‘‘π‘œ π’Šπ’”π’π’π’‚π’•π’† π‘₯
π‘₯ 𝑗 βˆ’ π‘˜ = βˆ’π‘‘ βˆ’ π‘š
Γ· 𝑗 βˆ’ π‘˜ π‘₯ =
βˆ’π‘‘ βˆ’ π‘š
𝑗 βˆ’ π‘˜
Γ· 𝑗 βˆ’ π‘˜
Shape, Space, Measure
Calculate the area of the sector and
the arc length. Leave all answers to 1
decimal place.
132Β° 3 π‘π‘š
3 π‘π‘š
Calculate the area and circumference of
the full circle.
π΄π‘Ÿπ‘’π‘Ž = πœ‹ Γ— π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘ 2
π΄π‘Ÿπ‘’π‘Ž = πœ‹ Γ— 32
π΄π‘Ÿπ‘’π‘Ž = 9πœ‹
πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = πœ‹ Γ— π‘‘π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ
πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = πœ‹ Γ— 6
πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = 6πœ‹
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘†π‘’π‘π‘‘π‘œπ‘Ÿ =
9πœ‹
360
Γ— 132
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘†π‘’π‘π‘‘π‘œπ‘Ÿ = 10.36725576
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘†π‘’π‘π‘‘π‘œπ‘Ÿ = 10.4 π‘π‘š2
π΄π‘Ÿπ‘ πΏπ‘’π‘›π‘”π‘‘β„Ž =
6πœ‹
360
Γ— 132
π΄π‘Ÿπ‘ πΏπ‘’π‘›π‘”π‘‘β„Ž = 6.911503838
π΄π‘Ÿπ‘ πΏπ‘’π‘›π‘”π‘‘β„Ž = 6.9 π‘π‘š
Handling Data Calculate an estimate for the mean
Table on time of goal scored
Time of
match
Frequency Midpoint Midpoint x
Frequency
0 < π‘₯ ≀ 15 8 7.5 60
15 < π‘₯ ≀ 30 4 22.5 90
30 < π‘₯ ≀ 45 5 37.5 187.5
45 < π‘₯ ≀ 60 7 52.5 367.5
60 < π‘₯ ≀ 75 7 67.5 472.5
75 < π‘₯ ≀ 90 10 82.5 825
Total 41 2002.5
Time of match Frequency
0 < π‘₯ ≀ 15 8
15 < π‘₯ ≀ 30 4
30 < π‘₯ ≀ 45 5
45 < π‘₯ ≀ 60 7
60 < π‘₯ ≀ 75 7
75 < π‘₯ ≀ 90 10
Not sure what time the goal was
scored – so we use the mid point
as an estimate.
π‘€π‘’π‘Žπ‘› =
2002.5
41
π‘€π‘’π‘Žπ‘› = 48.8
A goal was scored on average
at 48.8 minutes.
Number Algebra
Shape, Space, Measure Handling Data
Evaluate
80 =
64
1
2 =
7βˆ’2 =
4
81
1
2
=
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’
π‘₯2
+ 7π‘₯ + 12
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’
3π‘₯2 + 14π‘₯ + 15
𝐻𝑒𝑛𝑐𝑒 π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
π‘₯2 + 7π‘₯ + 12
3π‘₯2 + 14π‘₯ + 15
Draw a histogram for the following data.
Work out the value of x in this regular
pentagon
Marks Frequency
0 < x ≀ 5 3
5 < x ≀ 15 14
15 < x ≀ 30 18
30 < x ≀ 40 85x - 12
Number
Evaluate
80 =
64
1
2 =
7βˆ’2 =
4
81
1
2
=
80 = 1
64
1
2 = 64
= 8
7βˆ’2
=
1
72
=
1
49
4
81
1
2
=
4
81
=
4
81
=
2
9
Algebra πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’
π‘₯2 + 7π‘₯ + 12
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’
3π‘₯2
+ 14π‘₯ + 15
𝐻𝑒𝑛𝑐𝑒 π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
π‘₯2
+ 7π‘₯ + 12
3π‘₯2 + 14π‘₯ + 15
π‘₯2 + 7π‘₯ + 12 = π‘₯ + 3 π‘₯ + 4
3π‘₯2 + 14π‘₯ + 15 = 3π‘₯ + 5 π‘₯ + 3
π‘₯2
+ 7π‘₯ + 12
3π‘₯2 + 14π‘₯ + 15
=
π‘₯ + 3 π‘₯ + 4
3π‘₯ + 5 π‘₯ + 3
=
π‘₯ + 4
3π‘₯ + 5
Shape, Space, Measure Work out the value of x in this regular
pentagon
5x - 12
Interior angles of a pentagon add up to 540˚.
Regular pentagon has equal angles, 540 Γ· 5 = 108Β°
Therefore, 5π‘₯ βˆ’ 12 = 108
5π‘₯ = 120
π‘₯ = 24Β°
Handling Data
Draw a histogram for the following data.
Marks Frequency
0 < x ≀ 5 3
5 < x ≀ 15 14
15 < x ≀ 30 18
30 < x ≀ 40 8
Marks Frequency Frequency Density
0 < x ≀ 5 3 3 Γ· 5 = 0.6
5 < x ≀ 15 14 14 Γ· 10 = 1.4
15 < x ≀ 30 18 18 Γ· 15 = 1.2
30 < x ≀ 40 8 8 Γ· 10 = 0.8
Calculate the frequency density by
frequency Γ· class width.
Area of the bar is the frequency
Number Algebra
Shape, Space, Measure Handling Data
Fill in the table of values and complete
the quadratic graph, 𝑦 = π‘₯2 βˆ’ 5π‘₯ + 6
for -4 ≀ x ≀ 4
1) Out of a class of 28, 19 of the pupils
support Barnsley. What percentage
of pupils do not support Barnsley.
2) An antique was bought for Β£210, it
was later sold for Β£400. What
percentage of the price it was sold
for, was profit?
There are 7 green balls and 3 red balls
in a bag. A ball is chosen at random
and not replaced.
What is the probability of picking 3
balls and
a) Them being all the same colour
b) 2 green balls and a redCalculate the area of the Isosceles
triangle
x -4 -3 -2 -1 0 1 2 3 4
y 42 2
42Β°
6 π‘π‘š
Number 1) Out of a class of 28, 19 of the pupils
support Barnsley. What percentage
of pupils do not support Barnsley.
2) An antique was bought for Β£210, it
was later sold for Β£400. What
percentage of the price it was sold
for, was profit?
19 pupils support Barnsley,
this means 9 pupils don’t
support Barnsley.
9
28
= 9 Γ· 28
= 0.32142857
= 32%
Β£190 was profit. So
190
400
=190 Γ· 400
= 0.475
= 47.5%
Algebra Fill in the table of values and complete
the quadratic graph, 𝑦 = π‘₯2 βˆ’ 5π‘₯ + 6
for -4 ≀ x ≀ 4
x -4 -3 -2 -1 0 1 2 3 4
y 42 2
x -4 -3 -2 -1 0 1 2 3 4
y 42 30 20 12 6 2 0 0 2
Shape, Space, Measure
Calculate the area of the Isosceles
triangle
42Β°
6 π‘π‘š
π΄π‘Ÿπ‘’π‘Ž =
π‘π‘Žπ‘ π‘’ Γ— β„Žπ‘’π‘–π‘”β„Žπ‘‘
2
21Β°
3 π‘π‘š
Need to calculate the height of
the triangle. Use Trigonometry
Label your sides (in play)
o
a
Decide your triangle
a
o
t
Cover up what you are looking for and
write down your formula
π‘œ = tan π‘₯ Γ— π‘Ž
π‘œ = tan 42 Γ— 3
π‘œ = 2.701
π΄π‘Ÿπ‘’π‘Ž =
6 Γ— 2.701
2
π΄π‘Ÿπ‘’π‘Ž = 8.1 π‘π‘š2 (3. 𝑠. 𝑓)
Handling Data There are 7 green balls and 3 red balls
in a bag. A ball is chosen at random and
not replaced.
What is the probability of picking 3 balls
and
a) Them being all the same colour
b) 2 green balls and a red
a) List the combinations –
All three green or all three red
𝑃 𝑅 𝐴𝑁𝐷 𝑅 𝐴𝑁𝐷 𝑅 =
3
10
Γ—
2
9
Γ—
1
8
=
6
720
𝑃 𝐺 𝐴𝑁𝐷 𝐺 𝐴𝑁𝐷 𝐺 =
7
10
Γ—
6
9
Γ—
5
8
=
210
720
𝑃 𝐴𝑙𝑙 π‘”π‘Ÿπ‘’π‘’π‘› 𝒐𝒓 𝐴𝑙𝑙 π‘Ÿπ‘’π‘‘ =
6
720
+
210
720
=
πŸπŸπŸ”
πŸ•πŸπŸŽ
=
πŸ‘
𝟏𝟎
b) List the combinations –
GGR or GRG of RGG
𝑃 𝐺𝐺𝑅 =
7
10
Γ—
6
9
Γ—
3
8
=
126
720
𝑃 𝐺𝑅𝐺 =
7
10
Γ—
3
9
Γ—
6
8
=
126
720
𝑃 𝑅𝐺𝐺 =
3
10
Γ—
7
9
Γ—
6
8
=
126
720
𝑃 𝐺𝐺𝑅 𝒐𝒓 𝐺𝑅𝐺 𝒐𝒓 𝑅𝐺𝐺 =
126
720
+
126
720
+
126
720
=
πŸ‘πŸ•πŸ–
πŸ•πŸπŸŽ
=
𝟐𝟏
πŸ’πŸŽ
Number Algebra
Shape, Space, Measure Handling Data
Estimate
387 βˆ’ 43
0.18
𝐸π‘₯π‘π‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
3 2π‘₯ βˆ’ 5 βˆ’ 7 6 βˆ’ π‘₯
3π‘₯3 𝑦2 𝑧5 3
(3π‘₯ βˆ’ 2𝑦)(2π‘₯ + 5𝑦)
Below is a table to show how James spends his
day. Draw a pie chart to represent this data
Calculate the perimeter of this
isosceles triangle
5x - 711 - x
4x + 6
Activity Hours
Sleeping 9 hours
Work 8 hours
Eating/Cleaning/Cooking 3 hours
Reading 2 hours
Commuting 2 hours
Number
Round all numbers to 1 significant figure
400 βˆ’ 40
0.2
Calculate this sum
360
0.2
Use equivalent fractions to help divide by a decimal
360
0.2
=
3600
2
= 1800
Estimate
387 βˆ’ 43
0.18
Algebra
𝐸π‘₯π‘π‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
3 2π‘₯ βˆ’ 5 βˆ’ 7 6 βˆ’ π‘₯
3π‘₯3 𝑦2 𝑧5 3
(3π‘₯ βˆ’ 2𝑦)(2π‘₯ + 5𝑦)
3 2π‘₯ βˆ’ 5 βˆ’ 7 6 βˆ’ π‘₯
6π‘₯ βˆ’ 15 βˆ’ 42 + 7π‘₯
= 13π‘₯ βˆ’ 57
3π‘₯3 𝑦2 𝑧5 3 = 3π‘₯3 𝑦2 𝑧5 Γ— 3π‘₯3 𝑦2 𝑧5 Γ— 3π‘₯3 𝑦2 𝑧5
= 27π‘₯9 𝑦6 𝑧15
3π‘₯ βˆ’ 2𝑦 2π‘₯ + 5𝑦 = 6π‘₯2 + 15π‘₯𝑦 βˆ’ 4π‘₯𝑦 βˆ’ 10𝑦2
= 6π‘₯2
+ 11π‘₯𝑦 βˆ’ 10𝑦2
Shape, Space, Measure
Calculate the perimeter of this
isosceles triangle
5x - 711 - x
4x + 6
Isosceles – two sides are equal.
11 βˆ’ π‘₯ = 5π‘₯ βˆ’ 7
+π‘₯ 11 = 6π‘₯ βˆ’ 7 +π‘₯
+7 18 = 6π‘₯ +7
Γ· 6 3 = π‘₯ (Γ· 6)
Perimeter = 11 βˆ’ π‘₯ + 5π‘₯ βˆ’ 7 + 4π‘₯ + 6
= 8π‘₯ + 10
Substitute π‘₯ = 3, π‘‘π‘œ 𝑔𝑒𝑑 8 Γ— 3 + 10 = πŸ‘πŸ’
Handling Data Below is a table to show how James spends his
day. Draw a pie chart to represent this data
Activity Hours
Sleeping 9 hours
Work 8 hours
Eating/Cleaning/Cooking 3 hours
Reading 2 hours
Commuting 2 hours
Activity Hours Degrees
Sleeping 9 hours 135
Work 8 hours 120
Eating/Cleaning/Cooking 3 hours 45
Reading 2 hours 30
Commuting 2 hours 30
24 hours in a day. Find out what
each hour is worth
360 Γ· 24 = 15Β°
Each hour is worth 15˚ on our pie
chart. So sleeping is 9 x 15˚ = 135˚
Number Algebra
Shape, Space, Measure Handling Data
Solve the following Simultaneous
Equations
5π‘₯ βˆ’ 𝑦 = 9
15π‘₯ βˆ’ 2𝑦 = 24
2π‘₯ + 3𝑦 = 58
1) A gardens perimeter is 34m
(rounded to the nearest m), garden
fence panels are 130cm (rounded to
the nearest 10cm).
a) What is the most number of
panels that may be needed?
b) What is the least numbers of panels
that may be need?
Probability of Eric passing his Maths
exam is 0.7, the probability of passing
his English exam is independent of
this, and is 0.6.
What is the probability of Eric passing
at least one of his exams?
What is the straight line distance
between the coordinates (4, -5)
and ( 10, -3)
Number 1) A gardens perimeter is 34m (rounded to
the nearest m), garden fence panels are
130cm (rounded to the nearest 10cm).
a) What is the most number of panels
that may be needed?
b) What is the least numbers of panels that
may be need?
Change measurements into the same
units. 34m = 3400cm.
a) The most number of fence panels
needed are when you have a
large perimeter and small fence
panels.
Upper bound for perimeter = 3450cm
Lower bound for fence panel = 125cm
How many β€˜small’ fence panels will you need for a β€˜large’ perimeter.
πŸ‘πŸ’πŸ“πŸŽ Γ· πŸπŸπŸ“ = πŸπŸ•. πŸ” You would need 28 panels!
b) The least number of fence panels needed
are when you have a small perimeter and large
fence panels
Lower bound for perimeter = 3350cm
Upper bound for fence panel = 135cm
How many β€˜large’ fence panels will you need for a β€˜small’ perimeter.
πŸ‘πŸ‘πŸ“πŸŽ Γ· πŸπŸ‘πŸ“ = πŸπŸ’. πŸ–πŸ πŸ’ You would need 25 panels!
Algebra Solve the following
Simultaneous Equations
5π‘₯ βˆ’ 𝑦 = 9
15π‘₯ βˆ’ 2𝑦 = 24
5π‘₯ βˆ’ 𝑦 = 9 (π‘Ž)
15π‘₯ βˆ’ 2𝑦 = 24 𝑏
𝑁𝑒𝑒𝑑 π‘‘π‘œ π‘šπ‘Žπ‘˜π‘’ π‘Ž π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘’π‘›π‘‘ π‘’π‘žπ‘’π‘Žπ‘™
π‘Ž Γ— 2 10π‘₯ βˆ’ 2𝑦 = 18 𝑐
𝑏 βˆ’ 𝑐 5π‘₯ = 6
Γ· 5 𝒙 = 𝟏. 𝟐 Γ· 5
𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘π‘Žπ‘π‘˜ π‘–π‘›π‘‘π‘œ π‘Ž
5 1.2 βˆ’ 𝑦 = 9
6 βˆ’ 𝑦 = 9
βˆ’6 βˆ’ 𝑦 = 3 βˆ’6
Γ— βˆ’1 π’š = βˆ’πŸ‘ (Γ— βˆ’1)
2π‘₯ + 3𝑦 = 58 (π‘Ž)
7π‘₯ = 6𝑦 + 5 𝑏
π‘…π‘’π‘Žπ‘Ÿπ‘Ÿπ‘Žπ‘›π‘”π‘’ 𝑏 π‘‘π‘œ 𝑔𝑒𝑑 π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘’π‘›π‘‘π‘  π‘œπ‘› 𝑙𝑒𝑓𝑑 β„Žπ‘Žπ‘›π‘‘ 𝑠𝑖𝑑𝑒
βˆ’6y 7π‘₯ βˆ’ 6𝑦 = 5 βˆ’6𝑦
π‘Ž Γ— 2 4π‘₯ + 6𝑦 = 116 𝑐
𝑏 + 𝑐 11π‘₯ = 121
Γ· 11 𝒙 = 𝟏𝟏 Γ· 11
𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘π‘Žπ‘π‘˜ π‘–π‘›π‘‘π‘œ π‘Ž
2 11 + 3𝑦 = 58
22 + 3𝑦 = 58
βˆ’22 3𝑦 = 36 βˆ’22
Γ· 3 π’š = 𝟏𝟐 (Γ· 3)
Shape, Space, Measure What is the straight
line distance between
the coordinates (4, -5)
and ( 10, -3)
(4,-5)
(10,-3)
βˆ’3 βˆ’ βˆ’5 = 2
10 βˆ’ 4 = 6
π‘ˆπ‘ π‘’ π‘ƒπ‘¦π‘‘β„Žπ‘Žπ‘”π‘œπ‘Ÿπ‘Žπ‘  π‘‘π‘œ π‘π‘Žπ‘™π‘π‘’π‘™π‘Žπ‘‘π‘’ π‘‘β„Žπ‘’ π‘™π‘’π‘›π‘”π‘‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ 𝑙𝑖𝑛𝑒
π‘Ž2
+ 𝑏2
= 𝑐2
62 + 22 = 𝑐2
36 + 4 = 𝑐2
40 = 𝑐2
6.32 (3.s.f)= 𝒄
Handling Data Probability of Eric passing his Maths
exam is 0.7, the probability of passing
his English exam is independent of
this, and is 0.6.
What is the probability of Eric passing
at least one of his exams?
List the combinations –
Pass Maths (0.7), Fail English (0.4)
Pass English (0.6), Fail Maths (0.3)
Pass English (0.6), Pass Maths (0.7)
𝑃 π‘ƒπ‘Žπ‘ π‘ π‘€ 𝐴𝑁𝐷 πΉπ‘Žπ‘–π‘™πΈ = 0.7 Γ— 0.4
= 0.28
𝑃 π‘ƒπ‘Žπ‘ π‘ πΈ π‘Žπ‘›π‘‘ πΉπ‘Žπ‘–π‘™π‘€ = 0.6 Γ— 0.3
= 0.18
𝑃 π‘ƒπ‘Žπ‘ π‘ πΈ π‘Žπ‘›π‘‘ π‘ƒπ‘Žπ‘ π‘ π‘€ = 0.6 Γ— 0.7
= 0.42
𝑃 π‘ƒπ‘Žπ‘ π‘  π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ = 0.28 + 0.18 + 0.42
= 𝟎. πŸ–πŸ–
Alternative Solution.
Probability of passing at least once = 1 – probability of not passing either exam
𝑃 πΉπ‘Žπ‘–π‘™πΈ π‘Žπ‘›π‘‘ πΉπ‘Žπ‘–π‘™π‘€ = 0.4 Γ— 0.3
= 0.12
𝟏 βˆ’ 𝟎. 𝟏𝟐 = 𝟎. πŸ–πŸ–
Number Algebra
Shape, Space, Measure Handling Data
Put the following, in ascending order
π‘Ÿ0, 2βˆ’5, 81
1
2,
1
4
βˆ’2
,
3
2
2
π‘†π‘œπ‘™π‘£π‘’
4π‘₯ βˆ’ 1 = 2 βˆ’ π‘₯
10π‘₯ + 5 = 8π‘₯ βˆ’ 9
π‘₯ βˆ’ 1
2
+
4π‘₯ βˆ’ 3
3
= 4
Construct a frequency polygon for the
following
The area of the rectangle is 60π‘π‘š2
calculate the perimeter
π‘₯ βˆ’ 7 Cost x (Β£) Frequency
0 ≀ x < 10 8
10 ≀ x < 20 11
20 ≀ x < 30 5
30 ≀ x < 40 14
40 ≀ x < 50 6
π‘₯
Number Put the following, in ascending order
π‘Ÿ0, 2βˆ’5, 81
1
2,
1
4
βˆ’2
,
3
2
2π‘Ÿ0 = 1 (anything to the power 0 is
equal to 1.
2βˆ’5
=
1
25 =
1
32
81
1
2 = 81 = 9
1
4
βˆ’2
=
4
1
2
=
42
12
=
16
1
= 16
3
2
2
=
32
22 =
9
4
Therefore,
πŸβˆ’πŸ“, 𝒓 𝟎,
πŸ‘
𝟐
𝟐
, πŸ–πŸ
𝟏
𝟐,
𝟏
πŸ’
βˆ’πŸ
Algebra π‘†π‘œπ‘™π‘£π‘’
4π‘₯ βˆ’ 1 = 2 βˆ’ π‘₯
10π‘₯ + 5 = 8π‘₯ βˆ’ 9
π‘₯ βˆ’ 1
2
+
4π‘₯ βˆ’ 3
3
= 4
4π‘₯ βˆ’ 1 = 2 βˆ’ π‘₯
+π‘₯ 5π‘₯ βˆ’ 1 = 2 +π‘₯
+1 5π‘₯ = 3 +1
Γ· 5 𝒙 =
πŸ‘
πŸ“
= 𝟎. πŸ” (Γ· 5)
10π‘₯ + 5 = 8π‘₯ βˆ’ 9
βˆ’8π‘₯ 2π‘₯ + 5 = βˆ’9 βˆ’8π‘₯
βˆ’5 2π‘₯ = βˆ’14 βˆ’5
Γ· 2 𝒙 = βˆ’πŸ• (Γ· 2)
π‘₯ βˆ’ 1
2
+
4π‘₯ βˆ’ 3
3
= 4
Γ— 2 π‘₯ βˆ’ 1 +
2 4π‘₯ βˆ’ 3
3
= 8 Γ— 2
Γ— 3 3 π‘₯ βˆ’ 1 + 2 4π‘₯ βˆ’ 3 = 24 Γ— 3
3π‘₯ βˆ’ 3 + 8π‘₯ βˆ’ 6 = 24
11π‘₯ βˆ’ 9 = 24
+9 11π‘₯ = 33 +9
Γ· 11 𝒙 = πŸ‘ (Γ· 11)
Shape, Space, Measure The area of the rectangle is 60π‘π‘š2
calculate the perimeter
π‘₯ βˆ’ 7
π‘₯
π΄π‘Ÿπ‘’π‘Ž = 60π‘π‘š2
π΄π‘Ÿπ‘’π‘Ž = π‘₯ π‘₯ βˆ’ 7
= π‘₯2 βˆ’ 7π‘₯
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, π‘₯2
βˆ’ 7π‘₯ = 60
π‘€π‘Žπ‘˜π‘’ 𝑖𝑑 π‘™π‘œπ‘œπ‘˜ π‘™π‘–π‘˜π‘’ π‘Ž π‘’π‘ π‘’π‘Žπ‘™ π‘žπ‘’π‘Žπ‘‘π‘Ÿπ‘Žπ‘‘π‘–π‘
βˆ’60 π‘₯2
βˆ’ 7π‘₯ βˆ’ 60 = 0 βˆ’60
π‘π‘œπ‘€ π‘ π‘œπ‘™π‘£π‘’. 𝐼𝑑 𝑀𝑖𝑙𝑙 π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’, 𝑏𝑒𝑑 π‘¦π‘œπ‘’ π‘π‘Žπ‘› 𝑒𝑠𝑒 π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž
π‘₯ βˆ’ 12 π‘₯ + 5 = 0
π‘₯ βˆ’ 12 = 0 π‘œπ‘Ÿ π‘₯ + 5 = 0
π‘₯ = 12 π‘œπ‘Ÿ π‘₯ = βˆ’5
We are working with lengths, so we will ignore the -5, π‘₯ = 12.
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = π‘₯ βˆ’ 7 + π‘₯ + π‘₯ βˆ’ 7 + π‘₯
= 4π‘₯ βˆ’ 14
𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 𝑖𝑛 π‘₯ = 12
= 4 12 βˆ’ 14
= πŸ‘πŸ’π’„π’Ž
Handling Data Construct a frequency polygon for the
following
Cost x (Β£) Frequency
0 ≀ x < 10 8
10 ≀ x < 20 11
20 ≀ x < 30 5
30 ≀ x < 40 14
40 ≀ x < 50 6
See polygon – think straight lines.
Is this 8 things at Β£0 or Β£9.99? We
can’t be sure so use the midpoint.
Number Algebra
Shape, Space, Measure Handling Data
Calculate the π‘›π‘‘β„Ž term for the
following sequences
3, 7, 11, 15, 19, …
8, 11, 16, 23, 32, …
2, 7, 14, 23, 34, …
Barnsley Football club is organising travel for
an away game. 1300 adults and 500 juniors
want to go. Each coach holds 48 people and
costs Β£320 to hire. Tickets to the match are
Β£18 for adults and Β£10 for juniors.
The club is charging adults Β£26 and juniors
Β£14 for travel and a ticket. How much profit
does the club make out the trip?
Calculate an estimate for the meanThese two shapes are Mathematical Similar.
Calculate the missing volume and surface
area
4cm
12cm
Surface area =
Volume = 72π‘π‘š3 Surface area = 729π‘π‘š2
Volume =
Cost x (Β£) Frequency
0 ≀ x < 10 8
10 ≀ x < 20 11
20 ≀ x < 30 5
30 ≀ x < 40 14
40 ≀ x < 50 6
Number Barnsley Football club is organising travel for
an away game. 1300 adults and 500 juniors
want to go. Each coach holds 48 people and
costs Β£320 to hire. Tickets to the match are
Β£18 for adults and Β£10 for juniors.
The club is charging adults Β£26 and juniors
Β£14 for travel and a ticket. How much profit
does the club make out the trip?
Cost to supporter
1300 Γ— Β£26 = Β£33,800
500 Γ— Β£14 = Β£7,000
π‘‡π‘œπ‘‘π‘Žπ‘™ = Β£40,800
Cost to the club
Ticket costs
1300 Γ— Β£18 = Β£23,400
500 Γ— Β£10 = Β£5,000
π‘‡π‘œπ‘‘π‘Žπ‘™ = Β£28,400
Coach costs
Coaches needed: (1300 +
Profit to the club
Β£40,800 βˆ’ Β£40,560 = Β£240
Algebra Calculate the π‘›π‘‘β„Ž term for the
following sequences
3, 7, 11, 15, 19, …
8, 11, 16, 23, 32, …
2, 7, 14, 23, 34, …
3, 7, 11, 15, 19, …
πΌπ‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘–π‘›π‘” 𝑏𝑦 2 π‘’π‘Žπ‘β„Ž π‘‘π‘–π‘šπ‘’
2𝑛 + 1
8, 11, 16, 23, 32, …
3 5 7 9
2 2 2 𝑛2
1, 4, 9, 16, 25, …
+7
2, 7, 14, 23, 34, …
5 7 9 11
2 2 2
𝑛2
1, 4, 9, 16, 25, …
1, 3, 5, 7, 9
+ 2𝑛 βˆ’ 1
Shape, Space, Measure These two shapes are Mathematical Similar.
Calculate the missing volume and surface
area
4cm
12cm
Surface area =
Volume = 72π‘π‘š3 Surface area = 729π‘π‘š2
Volume =
Mathematical similar – they are
in scale.
π‘†π‘π‘Žπ‘™π‘’ πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ =
12
4
= 3
Shape B has been enlarged by
the linear scale factor of 3.
The surface area will therefore
be 32
as big.
729
32
= 81π‘π‘š2
The volume will be 33
as big.
72 Γ— 33
= 1944π‘π‘š3
A
B
πŸ–πŸπ’„π’Ž 𝟐
πŸπŸ—πŸ’πŸ’π’„π’Ž πŸ‘
Top Tip:
Area is measured in π’„π’Ž 𝟐
, so when
enlarging a shape – don’t forget to
square the scale factor when working
out the enlarged area.
Volume is measured in π’„π’Ž πŸ‘
, so when
enlarging a shape – don’t forget to
cube the scale factor when working
out the enlarged volume.
Calculate an estimate for the mean
Time of
match
Frequency Midpoint Midpoint x
Frequency
0 ≀ x < 10 8 5 40
10 ≀ x < 20 11 15 165
20 ≀ x < 30 5 25 125
30 ≀ x < 40 14 35 490
40 ≀ x < 50 6 45 270
Total 44 1090
Not sure what time the cost is –
so we use the mid point as an
estimate.
π‘€π‘’π‘Žπ‘› =
1090
44
π‘€π‘’π‘Žπ‘› = Β£24.78
Handling Data
Cost x (Β£) Frequency
0 ≀ x < 10 8
10 ≀ x < 20 11
20 ≀ x < 30 5
30 ≀ x < 40 14
40 ≀ x < 50 6
Number Algebra
Shape, Space, Measure Handling Data
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’
4π‘₯2
βˆ’ 18π‘₯
π‘₯2
βˆ’ 12π‘₯ + 32
6π‘₯2 + 13π‘₯ βˆ’ 5
25π‘₯2 βˆ’ 9𝑦2
Give two criticisms of the following
questionnaire
A rectangles width to length ratio is
2:5, if the perimeter is 84cm, what is
the area?
2.1 x 0.48 =
0.021 x 4.8 =
162.9 Γ· 0.09 =
Do you agree that Barnsley FC are by far the
greatest team the world has ever seen?
Strongly agree Agree Not sure
Number 2.1 x 0.48 =
0.021 x 4.8 =
162.9 Γ· 0.09 =
Ignore the decimal places and do your calculations when multiplying
decimals.
𝟐𝟏 Γ— πŸ’πŸ– = πŸπŸŽπŸŽπŸ–
Now add your decimal point into your answer. The question had three
decimal places – so your answer needs three decimal places.
1 008.
Ignore the decimal places and do your calculations when multiplying
decimals.
𝟐𝟏 Γ— πŸ’πŸ– = πŸπŸŽπŸŽπŸ–
Now add your decimal point into your answer. The question had four
decimal places – so your answer needs four decimal places.
1 008
.
Use equivalent fractions when dividing with decimals.
162.9
0.09
=
16290
9
= 1810
Algebra
πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’
4π‘₯2
βˆ’ 18π‘₯
π‘₯2
βˆ’ 12π‘₯ + 32
6π‘₯2 + 13π‘₯ βˆ’ 5
25π‘₯2 βˆ’ 9𝑦2
4π‘₯2 βˆ’ 18π‘₯ = 2π‘₯ 2π‘₯ βˆ’ 9
π‘₯2 βˆ’ 12π‘₯ + 32 = π‘₯ βˆ’ 8 π‘₯ βˆ’ 4
6π‘₯2 + 13π‘₯ βˆ’ 5 = (3π‘₯ βˆ’ 1)(2π‘₯ + 5)
25π‘₯2 βˆ’ 9𝑦2 = (5π‘₯ βˆ’ 3𝑦)(5π‘₯ + 3𝑦)
Shape, Space, Measure
A rectangles width to length ratio is
2:5, if the perimeter is 84cm, what is
the area?
5π‘₯
2π‘₯
π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = 84π‘π‘š
π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = 5π‘₯ + 2π‘₯ + 5π‘₯ + 2π‘₯
= 14π‘₯
π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, 14π‘₯ = 84
Γ· 14 π‘₯ = 6 Γ· 14
π΄π‘Ÿπ‘’π‘Ž = 2π‘₯ Γ— 5π‘₯
= 10π‘₯2
𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘₯ = 6
π΄π‘Ÿπ‘’π‘Ž = 10 62
= πŸ‘πŸ”πŸŽπ’„π’Ž 𝟐
Handling Data Give two criticisms of the following
questionnaire
Do you agree that Barnsley FC are by far the
greatest team the world has ever seen?
Strongly agree Agree Not sure
1. A leading question, do you agree, puts pressure on
the person answering the question.
2. No response boxes for if you disagree (but why
would you?)
Number Algebra
Shape, Space, Measure Handling Data
Fill in the table of values and complete
the cubic graph, 𝑦 = π‘₯3 βˆ’ 2π‘₯ +1 for
-3 ≀ x ≀ 3
2
3
7
βˆ’ 4
1
4
=
5
8
Γ—
4
15
=
4
5
Γ·
6
7
=
A bag contains 3 blue balls, 5 red balls
and 4 green balls. When a ball is taken
out, it is not replaced. What is the
probability of picking three balls that
are all the same colour.
Calculate the area of the following
quadrilateral
x -3 -2 -1 0 1 2 3
y -3 1 22
Number
2
3
7
βˆ’ 4
1
4
=
5
8
Γ—
4
15
=
4
5
Γ·
6
7
=
For Casio fx-83GT plus
Type
shift 32 7
- shift 14 4 =
βˆ’
51
28
5 8 4 15 =x
1
6
4 5 6 7 =Γ·
14
15
Algebra Fill in the table of values and complete
the cubic graph, 𝑦 = π‘₯3 βˆ’ 2π‘₯ +1 for
-3 ≀ x ≀ 3
x -3 -2 -1 0 1 2 3
y -3 1 22
x -3 -2 -1 0 1 2 3
y -20 -3 2 1 0 5 22
Shape, Space, Measure Calculate the area of the following
quadrilateralWill need these
(found at the front
of your exam)
Find the area of the two
triangles and add these
together...
1
2
Sin B
b
=
Sin A
a
𝑆𝑖𝑛 B
3
=
Sin 83
12
𝑆𝑖𝑛 𝐡 =
3𝑆𝑖𝑛 83
12
𝐡 = π‘†π‘–π‘›βˆ’1
3 sin 83
12
𝐡 = 14.367°
1
B
π‘‡β„Žπ‘’ π‘™π‘Žπ‘ π‘‘ π‘’π‘›π‘˜π‘›π‘œπ‘€π‘› π‘Žπ‘›π‘”π‘™π‘’ 𝑖𝑛 π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 1
180 βˆ’ 83 βˆ’ 14.367 = 82.633Β°
2
𝐹𝑖𝑛𝑑 π‘Žπ‘› π‘Žπ‘›π‘”π‘™π‘’ 𝑖𝑛 π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 2 𝑒𝑠𝑖𝑛𝑔 π‘‘β„Žπ‘’ π‘π‘œπ‘ π‘–π‘›π‘’ π‘Ÿπ‘’π‘™π‘’
π‘Žπ‘›π‘¦ π‘Žπ‘›π‘”π‘™π‘’ 𝑖𝑠 π‘Žπ‘π‘π‘™π‘–π‘π‘Žπ‘π‘™π‘’
π‘Ž2
= 𝑏2
+ 𝑐2
βˆ’ 2𝑏𝑐 πΆπ‘œπ‘  𝐴
122
= 102
+ 42
βˆ’ 2 Γ— 10 Γ— 4 πΆπ‘œπ‘  𝐴
144 = 116 βˆ’ 80πΆπ‘œπ‘  𝐴
28 = βˆ’80 πΆπ‘œπ‘  𝐴
βˆ’
28
80
= πΆπ‘œπ‘  𝐴
πΆπ‘œπ‘ βˆ’1
βˆ’
28
80
= 𝐴
𝐴 = 110.487
Shape, Space, Measure Calculate the area of the following
quadrilateralWill need these
(found at the front
of your exam)
Find the area of the two
triangles and add these
together...
1
2
π‘ˆπ‘ π‘’ π΄π‘Ÿπ‘’π‘Ž =
1
2
π‘Žπ‘π‘†π‘–π‘› 𝐢
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 1 =
1
2
Γ— 12 Γ— 3 Γ— 𝑆𝑖𝑛 82.633
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 1 = 17.851π‘π‘š2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 2 =
1
2
Γ— 10 Γ— 4 Γ— 𝑆𝑖𝑛 110.487
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 2 = 18.735π‘π‘š2
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘„π‘’π‘Žπ‘‘π‘Ÿπ‘–π‘™π‘Žπ‘‘π‘’π‘Ÿπ‘Žπ‘™ = 17.851 + 18.735
= πŸ‘πŸ”. πŸ“πŸ–πŸ”
Handling Data A bag contains 3 blue balls, 5 red balls
and 4 green balls. When a ball is taken
out, it is not replaced. What is the
probability of picking three balls that
are all the same colour.
List the combinations –
Blue, Blue, Blue,
Red, Red, Red,
Green, Green, Green
𝑃 𝐡 𝐴𝑁𝐷 𝐡 𝐴𝑁𝐷 𝐡 =
3
12
Γ—
2
11
Γ—
1
10
=
6
1320
𝑃 𝑅 𝐴𝑁𝐷 𝑅 𝐴𝑁𝐷 𝑅 =
5
12
Γ—
4
11
Γ—
3
10
𝑃 𝐺 𝐴𝑁𝐷 𝐺 𝐴𝑁𝐷 𝐺 =
4
12
Γ—
3
11
Γ—
2
10
𝑃 𝐡𝐡𝐡 𝑂𝑅 𝑅𝑅𝑅 𝑂𝑅 𝐺𝐺𝐺 =
6
1320
+
60
1320
+
24
1320
=
90
1320
=
3
44
Number Algebra
Shape, Space, Measure Handling Data
π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
β„Ž0
3π‘Ÿ2
𝑑4
𝑧 3
10π‘₯𝑦2
𝑧4
2π‘₯3 𝑦2 𝑧
π‘₯2
+ 7π‘₯ + 12
π‘₯ + 4
The mean weight of ten footballers is 73.5kg. A
new player comes along and the mean weight
goes down to 72kg. How much does the new
player weigh?
Calculate the area of the isosceles
triangle
Calculate the area and perimeter of
the rectangle. Leave your answer in
Surd form if applicable
18
2
10cm
13cm
Number Calculate the area and perimeter of
the rectangle. Leave your answer in
Surd form if applicable
18
2
π΄π‘Ÿπ‘’π‘Ž = 18 Γ— 2
= 36
= πŸ” π’–π’π’Šπ’•π’” 𝟐
π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = 18 + 2 + 18 + 2
= 9 Γ— 2 + 2 + 9 Γ— 2 + 2
= 9 2 + 2 + 9 2 + 2
= 3 2 + 2 + 3 2 + 2
= πŸ– 𝟐 π’–π’π’Šπ’•π’”
Algebra π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
β„Ž0
3π‘Ÿ2
𝑑4
𝑧 3
10π‘₯𝑦2
𝑧4
2π‘₯3 𝑦2 𝑧
π‘₯2
+ 7π‘₯ + 12
π‘₯ + 4
β„Ž0 = 1
3π‘Ÿ2 𝑑4 𝑧 3 = 3π‘Ÿ2 𝑑4 𝑧 Γ— 3π‘Ÿ2 𝑑4 𝑧 Γ— 3π‘Ÿ2 𝑑4 𝑧
= 27π‘Ÿ6
𝑑12
𝑧3
10π‘₯𝑦2 𝑧4
2π‘₯3 𝑦2 𝑧
= 5π‘₯βˆ’2 𝑧3 =
5𝑧3
π‘₯2
π‘₯2
+ 7π‘₯ + 12
π‘₯ + 4
=
π‘₯ + 3 π‘₯ + 4
(π‘₯ + 4)
= π‘₯ + 3
Shape, Space, Measure Calculate the area of the
isosceles triangle
10cm
13cm
π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘Ž π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ =
π‘π‘Žπ‘ π‘’ Γ— β„Žπ‘’π‘–π‘”β„Žπ‘‘
2
Need to calculate the perpendicular
height of the triangle.
13cm
5cm
h π‘ˆπ‘ π‘’ π‘ƒπ‘¦π‘‘β„Žπ‘Žπ‘”π‘œπ‘Ÿπ‘Žπ‘  π‘‘π‘œ π‘π‘Žπ‘™π‘π‘’π‘™π‘Žπ‘‘π‘’ π‘‘β„Žπ‘’ β„Žπ‘’π‘–π‘”β„Žπ‘‘
π‘Ž2
+ 𝑏2
= 𝑐2
52
+ 𝑏2
= 132
25 + 𝑏2 = 169
𝑏2 = 144
𝑏 = 12
π»π‘’π‘–π‘”β„Žπ‘‘ 𝑖𝑠 12π‘π‘š. π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’,
π΄π‘Ÿπ‘’π‘Ž =
12 Γ— 5
2
π΄π‘Ÿπ‘’π‘Ž = πŸ‘πŸŽπ’„π’Ž 𝟐
Handling Data The mean weight of ten footballers is
73.5kg. A new player comes along and the
mean weight goes down to 72kg. How much
does the new player weigh?
10 footballers in total weigh
10 Γ— 73.5π‘˜π‘” = 735π‘˜π‘”.
11 footballers in total weigh
11 Γ— 72 = 792π‘˜π‘”
The eleventh footballer must have weight
792 βˆ’ 735 = πŸ“πŸ•π’Œπ’ˆ
Number Algebra
Shape, Space, Measure Handling Data
The formula to convert Celsius to
Fahrenheit is
9
5
𝐢 + 32 = 𝐹
Use your calculator to work out the value of
3.92 βˆ’ 1.42
Write down the full display.
Round your answer to 2 significant figures
Complete a histogram for the following
data
Height, h (cm) Frequency
151 ≀ h < 153 64
153 ≀ h < 154 43
154 ≀ h < 155 47
155 ≀ h < 159 96
159 ≀ h < 160 12
What is the temperature in Fahrenheit,
when it 18˚ Celsius?
What is the temperature in Celsius
when it is 53.6˚ Fahrenheit?
An exterior angle of a
regular polygon is 30˚.
Work out the number of
sides of the polygon.
Number Use your calculator to work out the value of
3.92 βˆ’ 1.42
Write down the full display.
Round your answer to 2 significant figures
3
For Casio fx-83GT plus
Type
. 9 - 1 . 4 𝒙 𝟐 =
3.640054945
3.6
𝒙 𝟐
Algebra The formula to convert Celsius to
Fahrenheit is
9
5
𝐢 + 32 = 𝐹
What is the temperature in Fahrenheit,
when it 18˚ Celsius?
What is the temperature in Celsius
when it is 53.6˚ Fahrenheit?
9 Γ— 18
5
+ 32 = ℉
64.4 = ℉
πŸπŸ–β„ƒ = πŸ”πŸ’. πŸ’β„‰
9
5
𝐢 + 32 = 53.6
βˆ’32
9
5
𝐢 = 21.6 βˆ’32
Γ— 5 9𝐢 = 108 Γ— 5
÷ 9 𝐢 = 12 ÷ 9
πŸπŸβ„ƒ = πŸ“πŸ‘. πŸ”β„‰
Shape, Space, Measure
An exterior angle of a
regular polygon is 30˚.
Work out the number of
sides of the polygon.
Exterior angles add up to 360Β°, therefore
360 Γ· 30 = 12
The polygon has 12 sides (Dodecagon)
30Β°
Handling Data
Complete a histogram for
the following data
Height, h (cm) Frequency
151 ≀ h < 153 64
153 ≀ h < 154 43
154 ≀ h < 155 47
155 ≀ h < 159 96
159 ≀ h < 160 12
Marks Frequency Frequency Density
151 ≀ h < 153 64 64 Γ· 2 = 32
153 ≀ h < 154 43 43 Γ· 1 = 43
154 ≀ h < 155 47 47 Γ· 1 = 47
155 ≀ h < 159 96 96 Γ· 4 = 24
159 ≀ h < 160 12 12 Γ· 1 = 12
Calculate the frequency density by
frequency Γ· class width.
Area of the bar is the frequency
Number Algebra
Shape, Space, Measure Handling Data
𝐸π‘₯π‘π‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
3𝑦 4𝑦 βˆ’ 2
4 π‘₯ βˆ’ 2 βˆ’ 5 π‘₯ + 3
π‘₯ + 4 π‘₯ βˆ’ 7
(3π‘Ž + 2𝑏)(2π‘Ž βˆ’ 3𝑏)
What is the probability of rolling a dice
three times and getting two square
numbers and a prime number.
Convert 8π‘π‘š3 to π‘šπ‘š3
Convert 40,000π‘π‘š2 to π‘š2
Four pumps usually empty water
from a tank in 1 hour 36 minutes.
One of the pumps breaks down.
How long will three pumps, working
at the same rate, take to empty the
same tank.
Number
Four pumps usually empty water
from a tank in 1 hour 36 minutes.
One of the pumps breaks down.
How long will three pumps, working
at the same rate, take to empty the
same tank.
1 hour 36 minutes = 96 minutes
If four pumps take 96 minutes to
empty a tank, that must mean one
pump would take four times as long
4 Γ— 96 = 384 π‘šπ‘–π‘›π‘’π‘‘π‘’π‘ 
If three pumps were to do this, they
would do it in a third of the time
384 Γ· 3 = 128 π‘šπ‘–π‘›π‘’π‘‘π‘’π‘ 
128 minutes = 2 hours 8 minutes.
Algebra 𝐸π‘₯π‘π‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦
3𝑦 4𝑦 βˆ’ 2
4 π‘₯ βˆ’ 2 βˆ’ 5 π‘₯ + 3
π‘₯ + 4 π‘₯ βˆ’ 7
(3π‘Ž + 2𝑏)(2π‘Ž βˆ’ 3𝑏)
3𝑦 4𝑦 βˆ’ 2 = 12y2 βˆ’ 6y
4 π‘₯ βˆ’ 2 βˆ’ 5 π‘₯ + 3 = 4π‘₯ βˆ’ 8 βˆ’ 5π‘₯ βˆ’ 15
= βˆ’π‘₯ βˆ’ 23
π‘₯ + 4 π‘₯ βˆ’ 7 = π‘₯2
βˆ’ 7π‘₯ + 4π‘₯ βˆ’ 28
= π‘₯2 βˆ’ 3π‘₯ βˆ’ 28
3π‘Ž + 2𝑏 2π‘Ž βˆ’ 3𝑏 = 6π‘Ž2
βˆ’ 9π‘Žπ‘ + 4π‘Žπ‘ βˆ’ 6𝑏2
= 6π‘Ž2 βˆ’ 5π‘Žπ‘ βˆ’ 6𝑏2
Shape, Space, Measure
Convert 8π‘π‘š3 to π‘šπ‘š3
Convert 40,000π‘π‘š2 to π‘š2
1cm
1cm
1cm
10mm
10mm
10mm
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 1π‘π‘š Γ— 1π‘π‘š Γ— 1π‘π‘š
= 1π‘π‘š3
π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 10π‘šπ‘š Γ— 10π‘šπ‘š Γ— 10π‘šπ‘š
= 1000π‘šπ‘š3
π‘¬π’—π’†π’“π’š πŸπ’„π’Ž πŸ‘
π’Šπ’” π’†π’’π’–π’Šπ’—π’‚π’π’†π’π’• 𝒕𝒐 πŸπŸŽπŸŽπŸŽπ’Žπ’Ž πŸ‘
,
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆, πŸ–π’„π’Ž πŸ‘ = πŸ–πŸŽπŸŽπŸŽπ’Žπ’Ž πŸ‘
1m
1m
100cm
100cm
π΄π‘Ÿπ‘’π‘Ž = 1π‘š Γ— 1π‘š
= 1π‘š2
π΄π‘Ÿπ‘’π‘Ž = 100π‘π‘š Γ— 100π‘π‘š
= 10,000π‘π‘š2
π‘¬π’—π’†π’“π’š πŸπ’Ž 𝟐 π’Šπ’” π’†π’’π’–π’Šπ’—π’‚π’π’†π’π’• 𝒕𝒐 𝟏𝟎, πŸŽπŸŽπŸŽπ’„π’Ž 𝟐,
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆, πŸ’πŸŽ, πŸŽπŸŽπŸŽπ’„π’Ž 𝟐
= πŸ’π’Ž 𝟐
Handling Data What is the probability of rolling a dice
three times and getting two square
numbers and a prime number.
List the combinations –
Square, Square, Prime
Square, Prime, Square
Prime, Square, Square
𝑃 𝑆 𝐴𝑁𝐷 𝑆 𝐴𝑁𝐷 𝑃 =
1
3
Γ—
1
3
Γ—
1
2
=
1
18
𝑃 𝑆 𝐴𝑁𝐷 𝑃 𝐴𝑁𝐷 𝑆 =
1
3
Γ—
1
2
Γ—
1
3
=
1
18
𝑃 𝑃 𝐴𝑁𝐷 𝑆 𝐴𝑁𝐷 𝑆 =
1
2
Γ—
1
3
Γ—
1
3
=
1
18
Square numbers 1, 4
𝑃 π‘†π‘žπ‘’π‘Žπ‘Ÿπ‘’ =
1
3
Prime numbers 2, 3, 5
𝑃 π‘ƒπ‘Ÿπ‘–π‘šπ‘’ =
1
2 𝑃 𝑆𝑆𝑃 𝑂𝑅 𝑆𝑃𝑆 𝑂𝑅 𝑃𝑆𝑆 =
1
18
+
1
18
+
1
18
=
3
18
=
1
6
Number Algebra
Shape, Space, Measure Handling Data
Write down all the integers n that
satisfy
5 ≀ 𝑛 < 9
βˆ’3 < 2𝑛 ≀ 1
βˆ’2 ≀ 4𝑛 + 2 ≀ 0
Β£1 = $1.67
A laptop costs Β£320 in the UK, and $530 in
the USA. Where is the laptop cheaper, and by
how much?
The table shows van rentals for the
company VansRCool. Calculate a 3
point moving average for the months
below
Hayley can run 100m in 13
seconds. What is her average
speed in miles per hour?
Number
Β£1 = $1.67
A laptop costs Β£320 in the UK, and $530 in
the USA. Where is the laptop cheaper, and by
how much?
UK price Β£320
USA price $530 Γ· 1.67 = Β£317.37
USA cheaper by Β£2.63
UK price Β£320 Γ— 1.67 = $534.40
USA price $530
USA cheaper by $4.40
OR
Algebra
𝑛 = 5, 6, 7, π‘œπ‘Ÿ 8
Write down all the integers n that
satisfy
5 ≀ 𝑛 < 9
βˆ’3 < 2𝑛 ≀ 1
βˆ’2 ≀ 4𝑛 + 2 ≀ 0
βˆ’3 < 2𝑛 ≀ 1
Γ· 2 βˆ’ 1.5 < 𝑛 ≀ 1 Γ· 2
𝑛 = βˆ’1, 0 π‘œπ‘Ÿ 1
βˆ’2 ≀ 4𝑛 + 2 ≀ 0
βˆ’2 βˆ’ 4 ≀ 4𝑛 ≀ βˆ’2 βˆ’2
Γ· 4 βˆ’ 1 ≀ 𝑛 ≀ βˆ’0.5 Γ· 4
𝑛 = βˆ’1, 0
Shape, Space, Measure
Key Facts:
60 seconds in a minute
60 minutes in a hour
1600m in a mile.
Hayley can run 100m in 13
seconds. What is her average
speed in miles per hour?
100 π‘šπ‘’π‘‘π‘Ÿπ‘’π‘  𝑖𝑛 13 π‘ π‘’π‘π‘œπ‘›π‘‘π‘ 
100
13
π‘šπ‘’π‘‘π‘Ÿπ‘’π‘  𝑖𝑛 1 π‘ π‘’π‘π‘œπ‘›π‘‘
60 Γ— 100
13
π‘šπ‘’π‘‘π‘Ÿπ‘’π‘  𝑖𝑛 1 π‘šπ‘–π‘›π‘’π‘‘π‘’
60 Γ— 60 Γ— 100
13
π‘šπ‘’π‘‘π‘Ÿπ‘’π‘  𝑖𝑛 1 β„Žπ‘œπ‘’π‘Ÿ
360000
13
π‘šπ‘’π‘‘π‘Ÿπ‘’π‘  𝑖𝑛 1 β„Žπ‘œπ‘’π‘Ÿ
360000
13 Γ— 1600
π‘šπ‘–π‘™π‘’π‘  𝑖𝑛 1 β„Žπ‘œπ‘’π‘Ÿ.
17.3 miles per hour
Handling Data
3 point moving average – work
out the mean in groups of 3
The table shows van rentals for the
company VansRCool. Calculate a 3
point moving average for the months
below
9 + 22 + 37
3
= 22. 6
22 + 37 + 14
3
= 24. 3
37 + 14 + 18
3
= 23
14 + 18 + 24
3
= 18. 6
π‘‡β„Žπ‘’ 3 π‘π‘œπ‘–π‘›π‘‘ π‘šπ‘œπ‘£π‘–π‘›π‘” π‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ 𝑖𝑠 22.6, 24.3, 23, 18.6

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GCSE Linear Starters Higher

  • 1. GCSE Linear Questions (H) β€’ Question set 1 β€’ Question set 2 β€’ Question set 3 β€’ Question set 4 β€’ Question set 5 β€’ Question set 6 β€’ Question set 7 β€’ Question set 8 β€’ Question set 9 β€’ Question set 10 β€’ Question set 11 β€’ Question set 12 β€’ Question set 13 β€’ Question set 14 β€’ Question set 15 β€’ Question set 16 β€’ Question set 17 β€’ Question set 18 β€’ Question set 19 β€’ Question set 20 Odd numbers – Non calculator Even numbers - Calculator
  • 2. Number Algebra Shape, Space, Measure Handling Data Solve the following simultaneous equations Arrange these in ascending order 25 , 64 1 2, 2βˆ’1 , 80 , 16 1 4 5π‘₯ + 2𝑦 = 16 3π‘₯ βˆ’ 𝑦 = 14 8 9 14 ? The mean of the numbers of these 4 cards is 9. What is the number on the fourth card? 4 + 3x x + 6 The perimeter is equal to 32cm. What is the value of x?
  • 3. Number Arrange these in ascending order 25, 64 1 2, 2βˆ’1, 80, 16 1 4 25 = 2 x 2 x 2 x 2 x 2 = 32 64 1 2 = 64 = 8 2βˆ’1 = 1 21 = 1 2 80 = 1 16 1 4 = 4 16 = 2 2βˆ’1 , 80 , 16 1 4, 64 1 2, 25
  • 4. Algebra Solve the following simultaneous equations 5π‘₯ + 2𝑦 = 16 3π‘₯ βˆ’ 𝑦 = 14 a b Multiply equation b by 2, 6π‘₯ βˆ’ 2𝑦 = 28 c Add equations a and c 11π‘₯ = 44 𝒙 = πŸ’ Substitute x back into one of the original equations and solve
  • 5. Shape, Space, Measure 4 + 3x x + 6 The perimeter is equal to 32cm. What is the value of x? Perimeter = 32cm Perimeter = 4 + 3π‘₯ + π‘₯ + 6 + 4 + 3π‘₯ + π‘₯ + 6 = 8π‘₯ + 20 Therefore 8π‘₯ + 20 = 32 8π‘₯ = 12 π‘₯ = 1.5
  • 6. Handling Data 8 9 14 ? The mean of the numbers of these 4 cards is 9. What is the number on the fourth card? Total of the four cards is the mean multiplied by four 9 Γ— 4 = 36 So find the missing card by subtraction 36 βˆ’ 8 βˆ’ 9 βˆ’ 14 = 5
  • 7. Number Algebra Shape, Space, Measure Handling Data Expand and Simplify the followingA bank account gains 6% compound interest per annum. If Tom puts Β£700 into his account, how much could he expect after 6 years? 4 π‘₯ βˆ’ 5 βˆ’ 3(2π‘₯ βˆ’ 6) (π‘₯ + 2)(π‘₯ βˆ’ 7) 𝑑 + 5 2 Calculate the mean number of cars per household 35Β° Work out the missing length x π‘₯ 15 No. of cars Frequency 0 4 1 8 2 7 3 2
  • 8. Number A bank account gains 6% compound interest per annum. If Tom puts Β£700 into his account, how much could he expect after 6 years? To calculate the money after one year of interest, multiply by 1.06 π‘Œπ‘’π‘Žπ‘Ÿ π‘œπ‘›π‘’ βˆ’ 700 Γ— 1.06 π‘Œπ‘’π‘Žπ‘Ÿ π‘‘π‘€π‘œ βˆ’ 700 Γ— 1.06 Γ— 1.06 π‘Œπ‘’π‘Žπ‘Ÿ π‘‘β„Žπ‘Ÿπ‘’π‘’ βˆ’ 700 Γ— 1.06 Γ— 1.06 Γ— 1.06 π‘Œπ‘’π‘Žπ‘Ÿ 𝑠𝑖π‘₯ βˆ’ 700 Γ— 1.06 Γ— 1.06 Γ— 1.06 Γ— 1.06 Γ— 1.06 Γ— 1.06
  • 9. Algebra Expand and Simplify the following 4 π‘₯ βˆ’ 5 βˆ’ 3(2π‘₯ βˆ’ 6) (π‘₯ + 2)(π‘₯ βˆ’ 7) 𝑑 + 5 2 4 π‘₯ βˆ’ 5 βˆ’ 3 2π‘₯ βˆ’ 6 = 4π‘₯ βˆ’ 20 βˆ’ 6π‘₯ + 18 = βˆ’2π‘₯ βˆ’ 2 π‘₯ + 2 π‘₯ βˆ’ 7 = π‘₯2 βˆ’ 7π‘₯ + 2π‘₯ βˆ’ 14 = π‘₯2 βˆ’ 5π‘₯ βˆ’ 14 𝑑 + 5 2 = 𝑑 + 5 𝑑 + 5 = 𝑑2 + 5𝑑 + 5𝑑 + 25 = 𝑑2 + 10𝑑 + 25
  • 10. Shape, Space, Measure 35Β° Work out the missing length x π‘₯ 15 35Β° π‘₯ 15 Label your sides (in play) h a Decide your triangle h a c Cover up what you are looking for and write down your formula β„Ž = π‘Ž cos πœƒ β„Ž = 15 cos 35 β„Ž = 18.31161883 β„Ž = 18.3 (3. 𝑠. 𝑓)
  • 11. Handling Data Calculate the mean number of cars per household No. of cars Frequency 0 4 1 8 2 7 3 2 No. of cars Frequency Mean 0 4 4 x 0 = 0 1 8 8 x 1 =8 2 7 7 x 2 = 14 3 2 3 x 2 = 6 Total 21 28 π‘€π‘’π‘Žπ‘› = 28 21 = 1. 3
  • 12. Number Algebra Shape, Space, Measure Handling Data Write out the nth term for each sequence. Hence work out what the 10th and 100th term will be. Approximate the answer to 12.31 Γ— 16.9 0.394 Γ— 0.216 5, 8, 11, 14, … 10, 4, βˆ’2, βˆ’8, βˆ’14 4, 7, 12, 19, 28, … 6, 11, 18, 27, 38, … Three cards are drawn from a deck and replaced each time. What is the probability of drawing 3 hearts? Iron has a density of 8g/cm3 . What will be the mass of the above cuboid? 4cm 2cm 5cm
  • 13. Number Approximate the answer to 12.31 Γ— 16.9 0.394 Γ— 0.216 Round all numbers to 1 significant figure 10 Γ— 20 0.4 Γ— 0.2 Calculate this sum 200 0.8 Use equivalent fractions to help divide by a decimal 200 0.8 = 2000 8 = 250
  • 14. Algebra Write out the nth term for each sequence. Hence work out what the 10th and 100th term will be. 5, 8, 11, 14, … 10, 4, βˆ’2, βˆ’8, βˆ’14 4, 7, 12, 19, 28, … 6, 11, 18, 27, 38, … Sequence nth term 10th term 100th term 5, 8, 11, 14, … 3𝑛 + 2 32 302 10, 4, βˆ’2, βˆ’8, βˆ’14, … 16 βˆ’ 6𝑛 -44 -584 4, 7, 12, 19, 28, … 𝑛2 + 3 103 10003 6, 11, 18, 27, 38, … 𝑛2 + 2𝑛 + 3 123 10203
  • 15. Shape, Space, Measure Iron has a density of 8g/cm3 . What will be the mass of the above cuboid? 4cm 2cm 5cm πΆπ‘Ÿπ‘œπ‘ π‘  π‘ π‘’π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = 4π‘π‘š Γ— 5π‘π‘š = 20π‘π‘š2 π‘‰π‘œπ‘™π‘’π‘šπ‘’ = πΆπ‘Ÿπ‘œπ‘ π‘  π‘ π‘’π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž Γ— π‘‘π‘’π‘π‘‘β„Ž = 20 Γ— 2 = 40π‘π‘š3 The density is 8𝑔/π‘π‘š3 this means, every π‘π‘š3 weighs 8g. So the mass will be π‘€π‘Žπ‘ π‘  = 40 Γ— 8 = 320𝑔
  • 16. Handling Data Three cards are drawn from a deck and replaced each time. What is the probability of drawing 3 hearts? π‘ƒπ‘Ÿπ‘œπ‘π‘Žπ‘π‘–π‘™π‘–π‘‘π‘¦ π»π‘’π‘Žπ‘Ÿπ‘‘ 𝐴𝑁𝐷 π»π‘’π‘Žπ‘Ÿπ‘‘ 𝐴𝑁𝐷 π»π‘’π‘Žπ‘Ÿπ‘‘ = 1 4 Γ— 1 4 Γ— 1 4 𝑃 𝐻𝐻𝐻 = 1 64
  • 17. Number Algebra Shape, Space, Measure Handling Data Solve the following quadratic equation, leave your answers to 3 significant figures. πΆπ‘Žπ‘™π‘π‘’π‘™π‘Žπ‘‘π‘’ 4.2 Γ— 7.3 5.2 βˆ’ 9.3 Write down your full calculator display Round this number to 3 significant figures 3π‘₯2 βˆ’ 5π‘₯ = 18 A school of 800 pupils want to do a survey on school dinners. They decided to take a stratified sample of 30 pupils. How many of each year group should they ask? π‘₯ Work out the size of angle x 18 13 Year 7 Year 8 Year 9 Year 10 Year 11 182 124 128 195 171
  • 18. Number πΆπ‘Žπ‘™π‘π‘’π‘™π‘Žπ‘‘π‘’ 4.2 Γ— 7.3 5.2 βˆ’ 9.3 Write down your full calculator display Round this number to 3 significant figures For Casio fx-83GT plus Type 4 . 2 x 7 . 3 5 . 2 - 9 . 3 = 𝑺 β‰ͺ=≫D βˆ’7.478048 βˆ’7.48
  • 19. Algebra Solve the following quadratic equation, leave your answers to 3 significant figures. 3π‘₯2 βˆ’ 5π‘₯ = 18 Make it look like a usual quadratic 3π‘₯2 βˆ’ 5π‘₯ = 18 3π‘₯2 βˆ’ 5π‘₯ βˆ’ 18 = 0 Difficult to factorise οƒ  so use the formula. πΉπ‘œπ‘Ÿ π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 = 0, π‘€β„Žπ‘’π‘Ÿπ‘’ π‘Ž β‰  0 π‘₯ = βˆ’π‘ Β± 𝑏2 βˆ’ 4π‘Žπ‘ 2π‘Ž π‘Ž = 3 𝑏 = βˆ’5 𝑐 = βˆ’18 π‘₯ = βˆ’ βˆ’5 Β± βˆ’5 2 βˆ’ 4 Γ— 3 Γ— βˆ’18 2 Γ— 3 π‘₯ = 5 Β± 25 + 216 6 π‘₯ = 5 + 241 6 π‘œπ‘Ÿ π‘₯ = 5 βˆ’ 241 6 π‘₯ = 3.42 π‘œπ‘Ÿ π‘₯ = βˆ’1.75
  • 20. Shape, Space, Measure π‘₯ Work out the size of angle x 18 13 Label your sides (in play) o a Decide your triangle a o t Cover up what you are looking for and write down your formula tan π‘₯ = π‘œ π‘Ž tan π‘₯ = 18 13 π‘₯ = tanβˆ’1 18 13 π‘₯ = 54.2Β° π‘₯ 18 13
  • 21. Handling Data A school of 800 pupils want to do a survey on school dinners. They decided to take a stratified sample of 30 pupils. How many of each year group should they ask? Year 7 Year 8 Year 9 Year 10 Year 11 182 124 128 195 171 30 pupils out of 5 year groups – must mean 6 from each year group? Wrong. We take a stratified sample – this means we take a fair representation of each year group. There should be more year 7 than year 8 in the sample. Year 7 = 182 800 Γ— 30 = 6.825 = 7 𝑝𝑒𝑝𝑖𝑙𝑠 Year 8 = 124 800 Γ— 30 = 4.65 = 5 𝑝𝑒𝑝𝑖𝑙𝑠 Year 9 = 128 800 Γ— 30 = 4.8 = 5 𝑝𝑒𝑝𝑖𝑙𝑠 Year 10 = 195 800 Γ— 30 = 7.3125 = 7 𝑝𝑒𝑝𝑖𝑙𝑠 Year 11 = 171 800 Γ— 30 = 6.4125 = 6 𝑝𝑒𝑝𝑖𝑙𝑠 Check the student numbers add up to your sample size. 7 + 5 + 5 + 7 + 6 = 30 Each year group gets a fair representation.
  • 22. Number Algebra Shape, Space, Measure Handling Data Factorise fullySimplify the following 48 = 3 Γ— 12 = 2 5 Γ— 3 10 = 300 βˆ’ 75 = 5π‘₯4 𝑦3 𝑧2 βˆ’ 15π‘₯6 𝑦5 𝑧 π‘₯2 + 11π‘₯ βˆ’ 12 π‘₯2 βˆ’ 64 9π‘₯2 βˆ’ 25𝑦2 4, 6, 2, 5, 7, 9, 9, 2, 1, 4, 8 Draw a box plot for the following data Calculate the area of the rectangle 3 cm 5π‘₯ βˆ’ 3 cm 3π‘₯ + 5 cm
  • 23. Number Simplify the following 48 = 3 Γ— 12 = 2 5 Γ— 3 10 = 300 βˆ’ 75 = See Surds or the think SQUARE numbers 48 = 16 Γ— 3 = 16 3 = 4 3 What is the largest square number which is a factor of 48. 16 3 Γ— 12 = 36 = 6 2 5 Γ— 3 10 = 6 50 = 6 25 Γ— 2 = 6 25 2 = 30 2 When adding fractions, the denominator needs to be the same number. Similarly, when adding Surds, the Surds need to be the same number – so simplify them 300 = 100 Γ— 3 = 100 3 = 10 3 75 = 25 Γ— 3 = 25 3 = 5 3 300 βˆ’ 75 = 10 3 βˆ’ 5 3 = 5 3
  • 24. Algebra 5π‘₯4 𝑦3 𝑧2 βˆ’ 15π‘₯6 𝑦5 𝑧 π‘₯2 + 11π‘₯ βˆ’ 12 π‘₯2 βˆ’ 64 9π‘₯2 βˆ’ 25𝑦2 Factorise fully 5π‘₯4 𝑦3 𝑧2 βˆ’ 15π‘₯6 𝑦5 𝑧 = 5π‘₯4 𝑦3 𝑧(𝑧 βˆ’ 3π‘₯2 𝑦2 ) π‘₯2 + 11π‘₯ βˆ’ 12 = π‘₯ + 12 π‘₯ βˆ’ 11 π‘₯2 βˆ’ 64 = π‘₯ + 8 π‘₯ βˆ’ 8 9π‘₯2 βˆ’ 25𝑦2 = (3π‘₯ + 5𝑦)(3π‘₯ βˆ’ 5𝑦)
  • 25. Shape, Space, Measure Calculate the area of the rectangle 3 cm 5π‘₯ βˆ’ 3 cm 3π‘₯ + 5 cm Remember the features about a rectangle – two pairs of equal sides! You can set up an equation using this information to work out the length of the rectangle. 5π‘₯ βˆ’ 3 = 3π‘₯ + 5 2π‘₯ βˆ’ 3 = 5 2π‘₯ = 8 π‘₯ = 4 Therefore the length is 3 x 4 + 5 = 17cm. π΄π‘Ÿπ‘’π‘Ž = 17 Γ— 3 = 51π‘π‘š2
  • 26. Handling Data 4, 6, 2, 5, 7, 9, 9, 2, 1, 4, 8 Draw a box plot for the following data When working with Quartiles, Median and Range – it is always useful to arrange your data in size order. 1, 2, 2, 4, 4, 5, 6, 7, 8, 9, 9 Need a few pieces of information for a box plot. Highest value – 9 Lowest value – 1 The Median, since there are 11 numbers, the middle number will be the 11+1 2 = 6th number. So the 6th number is 5 Lower quartile will be the 11+1 4 = 3π‘Ÿπ‘‘ number in our list. So LQ = 2 Upper quartile will be the 3 11+1 4 = 9π‘‘β„Ž number in our list. UQ = 8
  • 27. Number Algebra Shape, Space, Measure Handling Data Using trial and improvement to find a solution to 1 decimal place The lengths of a room have been calculated to the nearest metre. Calculate the greatest and least area that the room could be. π‘₯3 βˆ’ 5π‘₯ = 50 The probability of Man Utd winning a match under David Moyes is 0.3 and losing is 0.2. Man Utd play 3 matches, what is the probability that out of these, 2 are won and one is drawn. Calculate the volume and surface area. Leave your answer to 3 significant figures. 8 π‘π‘š 5 π‘π‘š 9 4
  • 28. Number The lengths of a room have been calculated to the nearest metre. Calculate the greatest and least area that the room could be. 9 4 Lower bound 3.5 8.5 π΄π‘Ÿπ‘’π‘Ž = 3.5 Γ— 8.5 = 29.75 π‘š2 Upper bound 9.5 4.5 π΄π‘Ÿπ‘’π‘Ž = 4.5 Γ— 9.5 = 42.75 π‘š2
  • 29. Algebra Using trial and improvement to find a solution to 1 decimal place π‘₯3 βˆ’ 5π‘₯ = 50 𝒙 𝒙 πŸ‘ βˆ’ πŸ“π’™ Comment 4 43 βˆ’ 5 Γ— 4 = 44 Low 5 53 βˆ’ 5 Γ— 5 = 100 High 4.3 4.33 βˆ’ 5 Γ— 4.3 = 58.007 High 4.2 4.23 βˆ’ 5 Γ— 4.2 = 53.088 High 4.1 4.13 βˆ’ 5 Γ— 4.1 = 48.421 Low Draw a table – it helps! 4.153 βˆ’ 5 Γ— 4.15 = 50.723375 Since 4.15 is too high, everything about it must be too high as well. Therefore the solution to 1 decimal place is 4.1
  • 30. πΆπ‘Ÿπ‘œπ‘ π‘  π‘ π‘’π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž = π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘π‘–π‘Ÿπ‘π‘™π‘’ = πœ‹ Γ— π‘Ÿ2 = πœ‹ Γ— πŸ’2 = 16πœ‹ cm2 π‘‰π‘œπ‘™π‘’π‘šπ‘’ = πΆπ‘Ÿπ‘œπ‘ π‘  π‘ π‘’π‘π‘‘π‘–π‘œπ‘›π‘Žπ‘™ π‘Žπ‘Ÿπ‘’π‘Ž Γ— π‘‘π‘’π‘π‘‘β„Ž = 16πœ‹ π‘₯ 5 = 80πœ‹ = 251.3274123 = 251 π‘π‘š3 Shape, Space, Measure Calculate the volume and surface area. Leave your answer to 3 significant figures. 8 π‘π‘š 5 π‘π‘š
  • 31. Handling Data The probability of Man Utd winning a match under David Moyes is 0.3 and losing is 0.2. Man Utd play 3 matches, what is the probability that out of these, 2 are won and one is drawn. Probability of drawing a match is 1 βˆ’ 0.3 βˆ’ 0.2 = 0.5 List all the possible combinations Win Win Draw Win Draw Win Draw Win Win 𝑃 π‘Š 𝐴𝑁𝐷 π‘Š 𝐴𝑁𝐷 𝐷 = 0.3 Γ— 0.3 Γ— 0.5 = 0.045 𝑃 π‘Š 𝐴𝑁𝐷 𝐷 𝐴𝑁𝐷 π‘Š = 0.3 Γ— 0.5 Γ— 0.3 = 0.045 𝑃 𝐷 𝐴𝑁𝐷 π‘Š 𝐴𝑁𝐷 π‘Š = 0.5 Γ— 0.3 Γ— 0.3 = 0.045 Therefore, the probability of winning two and drawing one match is 0.045 + 0.045 + 0.045 = 0.135
  • 32. Number Algebra Shape, Space, Measure Handling Data Solve the following Leave all answers as mixed numbers 1 3 + 4 5 = 3 5 7 βˆ’ 1 2 3 = 4 9 Γ— 3 8 = 4π‘₯ + 6 = π‘₯ βˆ’ 12 7π‘₯ βˆ’ 8 = 20 βˆ’ 3π‘₯ 5π‘₯ + 4 6 = 3π‘₯ + 8 5 Draw a cumulative frequency graph for the following data. Construct a box plot from this information. Calculate the distance between the coordinates (-1, 4) and (11, 9) Marks Frequency 0 – 10 2 11 – 20 7 21 – 30 13 31 – 40 8
  • 33. Number Leave all answers as mixed numbers 1 3 + 4 5 = 3 5 7 βˆ’ 1 2 3 = 4 9 Γ— 3 8 = 1 3 + 4 5 = 5 15 + 12 15 = 17 15 = 𝟐 𝟐 πŸπŸ“ 3 5 7 βˆ’ 1 2 3 = 26 5 βˆ’ 5 3 = 78 15 βˆ’ 25 15 = 53 15 = πŸ‘ πŸ– πŸπŸ“ 4 9 Γ— 3 8 = 12 72 = 𝟏 πŸ”
  • 34. Algebra Solve the following 4π‘₯ + 6 = π‘₯ βˆ’ 12 7π‘₯ βˆ’ 8 = 20 βˆ’ 3π‘₯ 5π‘₯ + 4 6 = 3π‘₯ + 8 5 4π‘₯ + 6 = π‘₯ βˆ’ 12 βˆ’π‘₯ 3π‘₯ + 6 = βˆ’12 βˆ’π‘₯ βˆ’6 3π‘₯ = βˆ’18 βˆ’6 Γ· 3 𝒙 = βˆ’πŸ” Γ· 3 7π‘₯ βˆ’ 8 = 20 βˆ’ 3π‘₯ +3π‘₯ 10π‘₯ βˆ’ 8 = 20 +3π‘₯ +8 10π‘₯ = 28 +8 Γ· 10 𝒙 = 𝟐. πŸ– Γ· 10 5π‘₯ + 4 6 = 3π‘₯ + 8 5 Γ— 5 25π‘₯ + 20 6 = 3π‘₯ + 8 Γ— 5 Γ— 6 25π‘₯ + 20 = 18π‘₯ + 48 Γ— 6 βˆ’18π‘₯ 7π‘₯ + 20 = 48 βˆ’18π‘₯ βˆ’20 7π‘₯ = 28 βˆ’20 Γ· 7 𝒙 = πŸ’ (Γ· 7)
  • 35. Shape, Space, Measure Calculate the distance between the coordinates (-1, 4) and (11, 9) (-1,4) (11,9) 11 – βˆ’1 = 12 9 βˆ’ 4 = 5 π‘ˆπ‘ π‘’ π‘ƒπ‘¦π‘‘β„Žπ‘Žπ‘”π‘œπ‘Ÿπ‘Žπ‘  π‘‘π‘œ π‘π‘Žπ‘™π‘π‘’π‘™π‘Žπ‘‘π‘’ π‘‘β„Žπ‘’ π‘™π‘’π‘›π‘”π‘‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ 𝑙𝑖𝑛𝑒 π‘Ž2 + 𝑏2 = 𝑐2 52 + 122 = 𝑐2 25 + 144 = 𝑐2 169 = 𝑐2 πŸπŸ‘ = 𝒄
  • 36. Handling Data Draw a cumulative frequency graph for the following data. Construct a box plot from this information. Marks Frequency 0 – 10 2 11 – 20 7 21 – 30 13 31 – 40 8Marks Frequency Cumulative Frequency 0 – 10 2 2 11 – 20 7 9 21 – 30 13 22 31 – 40 8 30 Plot your graph using the end points and cumulative frequency.
  • 37. Number Algebra Shape, Space, Measure Handling Data Rearrange the formula to make x the subject Tins of paint are on offer, buy 5 get 1 free. John the painter needs 27 tins of paint. If a tin of paint costs Β£3.43, how much will John have to pay? 𝑦 = 4π‘₯ βˆ’ 2 jπ‘₯ + 𝑑 = π‘˜π‘₯ βˆ’ π‘š Calculate an estimate for the mean Table on time of goal scored Calculate the area of the sector and the arc length. Leave all answers to 1 decimal place. 132Β° 3 π‘π‘š Time of match Frequency 0 < π‘₯ ≀ 15 8 15 < π‘₯ ≀ 30 4 30 < π‘₯ ≀ 45 5 45 < π‘₯ ≀ 60 7 60 < π‘₯ ≀ 75 7 75 < π‘₯ ≀ 90 10
  • 38. Number Tins of paint are on offer, buy 5 get 1 free. John the painter needs 27 tins of paint. If a tin of paint costs Β£3.43, how much will John have to pay? If John buys 5 tins he gets 6. So using this, if he buys 20 tins, he will actually get 24. Therefore he only needs to purchase another 3 tins to have 27. John needs to buy 23 tins. 23 Γ— 3.43 = Β£78.89
  • 39. Algebra Rearrange the formula to make x the subject 𝑦 = 4π‘₯ βˆ’ 2 jπ‘₯ + 𝑑 = π‘˜π‘₯ βˆ’ π‘šπ‘¦ = 4π‘₯ βˆ’ 2 +2 𝑦 + 2 = 4π‘₯ +2 Γ· 4 𝑦 + 2 4 = π‘₯ Γ· 4 π‘€π‘Žπ‘˜π‘’ π‘₯ π‘‘β„Žπ‘’ 𝑠𝑒𝑏𝑗𝑒𝑐𝑑 𝑏𝑦 𝑝𝑒𝑑𝑑𝑖𝑛𝑔 π‘œπ‘› π‘‘β„Žπ‘’ 𝑙𝑒𝑓𝑑 β„Žπ‘Žπ‘›π‘‘ 𝑠𝑖𝑑𝑒. π‘₯ = 𝑦 + 2 4 jπ‘₯ + 𝑑 = π‘˜π‘₯ βˆ’ π‘š βˆ’π‘˜π‘₯ 𝑗π‘₯ βˆ’ π‘˜π‘₯ + 𝑑 = βˆ’π‘š βˆ’π‘˜π‘₯ βˆ’π‘‘ 𝑗π‘₯ βˆ’ π‘˜π‘₯ = βˆ’π‘‘ βˆ’ π‘š βˆ’π‘‘ πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’ π‘‘π‘œ π’Šπ’”π’π’π’‚π’•π’† π‘₯ π‘₯ 𝑗 βˆ’ π‘˜ = βˆ’π‘‘ βˆ’ π‘š Γ· 𝑗 βˆ’ π‘˜ π‘₯ = βˆ’π‘‘ βˆ’ π‘š 𝑗 βˆ’ π‘˜ Γ· 𝑗 βˆ’ π‘˜
  • 40. Shape, Space, Measure Calculate the area of the sector and the arc length. Leave all answers to 1 decimal place. 132Β° 3 π‘π‘š 3 π‘π‘š Calculate the area and circumference of the full circle. π΄π‘Ÿπ‘’π‘Ž = πœ‹ Γ— π‘Ÿπ‘Žπ‘‘π‘–π‘’π‘ 2 π΄π‘Ÿπ‘’π‘Ž = πœ‹ Γ— 32 π΄π‘Ÿπ‘’π‘Ž = 9πœ‹ πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = πœ‹ Γ— π‘‘π‘–π‘Žπ‘šπ‘’π‘‘π‘’π‘Ÿ πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = πœ‹ Γ— 6 πΆπ‘–π‘Ÿπ‘π‘’π‘šπ‘“π‘’π‘Ÿπ‘’π‘›π‘π‘’ = 6πœ‹ π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘†π‘’π‘π‘‘π‘œπ‘Ÿ = 9πœ‹ 360 Γ— 132 π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘†π‘’π‘π‘‘π‘œπ‘Ÿ = 10.36725576 π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘†π‘’π‘π‘‘π‘œπ‘Ÿ = 10.4 π‘π‘š2 π΄π‘Ÿπ‘ πΏπ‘’π‘›π‘”π‘‘β„Ž = 6πœ‹ 360 Γ— 132 π΄π‘Ÿπ‘ πΏπ‘’π‘›π‘”π‘‘β„Ž = 6.911503838 π΄π‘Ÿπ‘ πΏπ‘’π‘›π‘”π‘‘β„Ž = 6.9 π‘π‘š
  • 41. Handling Data Calculate an estimate for the mean Table on time of goal scored Time of match Frequency Midpoint Midpoint x Frequency 0 < π‘₯ ≀ 15 8 7.5 60 15 < π‘₯ ≀ 30 4 22.5 90 30 < π‘₯ ≀ 45 5 37.5 187.5 45 < π‘₯ ≀ 60 7 52.5 367.5 60 < π‘₯ ≀ 75 7 67.5 472.5 75 < π‘₯ ≀ 90 10 82.5 825 Total 41 2002.5 Time of match Frequency 0 < π‘₯ ≀ 15 8 15 < π‘₯ ≀ 30 4 30 < π‘₯ ≀ 45 5 45 < π‘₯ ≀ 60 7 60 < π‘₯ ≀ 75 7 75 < π‘₯ ≀ 90 10 Not sure what time the goal was scored – so we use the mid point as an estimate. π‘€π‘’π‘Žπ‘› = 2002.5 41 π‘€π‘’π‘Žπ‘› = 48.8 A goal was scored on average at 48.8 minutes.
  • 42. Number Algebra Shape, Space, Measure Handling Data Evaluate 80 = 64 1 2 = 7βˆ’2 = 4 81 1 2 = πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’ π‘₯2 + 7π‘₯ + 12 πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’ 3π‘₯2 + 14π‘₯ + 15 𝐻𝑒𝑛𝑐𝑒 π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ π‘₯2 + 7π‘₯ + 12 3π‘₯2 + 14π‘₯ + 15 Draw a histogram for the following data. Work out the value of x in this regular pentagon Marks Frequency 0 < x ≀ 5 3 5 < x ≀ 15 14 15 < x ≀ 30 18 30 < x ≀ 40 85x - 12
  • 43. Number Evaluate 80 = 64 1 2 = 7βˆ’2 = 4 81 1 2 = 80 = 1 64 1 2 = 64 = 8 7βˆ’2 = 1 72 = 1 49 4 81 1 2 = 4 81 = 4 81 = 2 9
  • 44. Algebra πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’ π‘₯2 + 7π‘₯ + 12 πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’ 3π‘₯2 + 14π‘₯ + 15 𝐻𝑒𝑛𝑐𝑒 π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ π‘₯2 + 7π‘₯ + 12 3π‘₯2 + 14π‘₯ + 15 π‘₯2 + 7π‘₯ + 12 = π‘₯ + 3 π‘₯ + 4 3π‘₯2 + 14π‘₯ + 15 = 3π‘₯ + 5 π‘₯ + 3 π‘₯2 + 7π‘₯ + 12 3π‘₯2 + 14π‘₯ + 15 = π‘₯ + 3 π‘₯ + 4 3π‘₯ + 5 π‘₯ + 3 = π‘₯ + 4 3π‘₯ + 5
  • 45. Shape, Space, Measure Work out the value of x in this regular pentagon 5x - 12 Interior angles of a pentagon add up to 540˚. Regular pentagon has equal angles, 540 Γ· 5 = 108Β° Therefore, 5π‘₯ βˆ’ 12 = 108 5π‘₯ = 120 π‘₯ = 24Β°
  • 46. Handling Data Draw a histogram for the following data. Marks Frequency 0 < x ≀ 5 3 5 < x ≀ 15 14 15 < x ≀ 30 18 30 < x ≀ 40 8 Marks Frequency Frequency Density 0 < x ≀ 5 3 3 Γ· 5 = 0.6 5 < x ≀ 15 14 14 Γ· 10 = 1.4 15 < x ≀ 30 18 18 Γ· 15 = 1.2 30 < x ≀ 40 8 8 Γ· 10 = 0.8 Calculate the frequency density by frequency Γ· class width. Area of the bar is the frequency
  • 47. Number Algebra Shape, Space, Measure Handling Data Fill in the table of values and complete the quadratic graph, 𝑦 = π‘₯2 βˆ’ 5π‘₯ + 6 for -4 ≀ x ≀ 4 1) Out of a class of 28, 19 of the pupils support Barnsley. What percentage of pupils do not support Barnsley. 2) An antique was bought for Β£210, it was later sold for Β£400. What percentage of the price it was sold for, was profit? There are 7 green balls and 3 red balls in a bag. A ball is chosen at random and not replaced. What is the probability of picking 3 balls and a) Them being all the same colour b) 2 green balls and a redCalculate the area of the Isosceles triangle x -4 -3 -2 -1 0 1 2 3 4 y 42 2 42Β° 6 π‘π‘š
  • 48. Number 1) Out of a class of 28, 19 of the pupils support Barnsley. What percentage of pupils do not support Barnsley. 2) An antique was bought for Β£210, it was later sold for Β£400. What percentage of the price it was sold for, was profit? 19 pupils support Barnsley, this means 9 pupils don’t support Barnsley. 9 28 = 9 Γ· 28 = 0.32142857 = 32% Β£190 was profit. So 190 400 =190 Γ· 400 = 0.475 = 47.5%
  • 49. Algebra Fill in the table of values and complete the quadratic graph, 𝑦 = π‘₯2 βˆ’ 5π‘₯ + 6 for -4 ≀ x ≀ 4 x -4 -3 -2 -1 0 1 2 3 4 y 42 2 x -4 -3 -2 -1 0 1 2 3 4 y 42 30 20 12 6 2 0 0 2
  • 50. Shape, Space, Measure Calculate the area of the Isosceles triangle 42Β° 6 π‘π‘š π΄π‘Ÿπ‘’π‘Ž = π‘π‘Žπ‘ π‘’ Γ— β„Žπ‘’π‘–π‘”β„Žπ‘‘ 2 21Β° 3 π‘π‘š Need to calculate the height of the triangle. Use Trigonometry Label your sides (in play) o a Decide your triangle a o t Cover up what you are looking for and write down your formula π‘œ = tan π‘₯ Γ— π‘Ž π‘œ = tan 42 Γ— 3 π‘œ = 2.701 π΄π‘Ÿπ‘’π‘Ž = 6 Γ— 2.701 2 π΄π‘Ÿπ‘’π‘Ž = 8.1 π‘π‘š2 (3. 𝑠. 𝑓)
  • 51. Handling Data There are 7 green balls and 3 red balls in a bag. A ball is chosen at random and not replaced. What is the probability of picking 3 balls and a) Them being all the same colour b) 2 green balls and a red a) List the combinations – All three green or all three red 𝑃 𝑅 𝐴𝑁𝐷 𝑅 𝐴𝑁𝐷 𝑅 = 3 10 Γ— 2 9 Γ— 1 8 = 6 720 𝑃 𝐺 𝐴𝑁𝐷 𝐺 𝐴𝑁𝐷 𝐺 = 7 10 Γ— 6 9 Γ— 5 8 = 210 720 𝑃 𝐴𝑙𝑙 π‘”π‘Ÿπ‘’π‘’π‘› 𝒐𝒓 𝐴𝑙𝑙 π‘Ÿπ‘’π‘‘ = 6 720 + 210 720 = πŸπŸπŸ” πŸ•πŸπŸŽ = πŸ‘ 𝟏𝟎 b) List the combinations – GGR or GRG of RGG 𝑃 𝐺𝐺𝑅 = 7 10 Γ— 6 9 Γ— 3 8 = 126 720 𝑃 𝐺𝑅𝐺 = 7 10 Γ— 3 9 Γ— 6 8 = 126 720 𝑃 𝑅𝐺𝐺 = 3 10 Γ— 7 9 Γ— 6 8 = 126 720 𝑃 𝐺𝐺𝑅 𝒐𝒓 𝐺𝑅𝐺 𝒐𝒓 𝑅𝐺𝐺 = 126 720 + 126 720 + 126 720 = πŸ‘πŸ•πŸ– πŸ•πŸπŸŽ = 𝟐𝟏 πŸ’πŸŽ
  • 52. Number Algebra Shape, Space, Measure Handling Data Estimate 387 βˆ’ 43 0.18 𝐸π‘₯π‘π‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 3 2π‘₯ βˆ’ 5 βˆ’ 7 6 βˆ’ π‘₯ 3π‘₯3 𝑦2 𝑧5 3 (3π‘₯ βˆ’ 2𝑦)(2π‘₯ + 5𝑦) Below is a table to show how James spends his day. Draw a pie chart to represent this data Calculate the perimeter of this isosceles triangle 5x - 711 - x 4x + 6 Activity Hours Sleeping 9 hours Work 8 hours Eating/Cleaning/Cooking 3 hours Reading 2 hours Commuting 2 hours
  • 53. Number Round all numbers to 1 significant figure 400 βˆ’ 40 0.2 Calculate this sum 360 0.2 Use equivalent fractions to help divide by a decimal 360 0.2 = 3600 2 = 1800 Estimate 387 βˆ’ 43 0.18
  • 54. Algebra 𝐸π‘₯π‘π‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 3 2π‘₯ βˆ’ 5 βˆ’ 7 6 βˆ’ π‘₯ 3π‘₯3 𝑦2 𝑧5 3 (3π‘₯ βˆ’ 2𝑦)(2π‘₯ + 5𝑦) 3 2π‘₯ βˆ’ 5 βˆ’ 7 6 βˆ’ π‘₯ 6π‘₯ βˆ’ 15 βˆ’ 42 + 7π‘₯ = 13π‘₯ βˆ’ 57 3π‘₯3 𝑦2 𝑧5 3 = 3π‘₯3 𝑦2 𝑧5 Γ— 3π‘₯3 𝑦2 𝑧5 Γ— 3π‘₯3 𝑦2 𝑧5 = 27π‘₯9 𝑦6 𝑧15 3π‘₯ βˆ’ 2𝑦 2π‘₯ + 5𝑦 = 6π‘₯2 + 15π‘₯𝑦 βˆ’ 4π‘₯𝑦 βˆ’ 10𝑦2 = 6π‘₯2 + 11π‘₯𝑦 βˆ’ 10𝑦2
  • 55. Shape, Space, Measure Calculate the perimeter of this isosceles triangle 5x - 711 - x 4x + 6 Isosceles – two sides are equal. 11 βˆ’ π‘₯ = 5π‘₯ βˆ’ 7 +π‘₯ 11 = 6π‘₯ βˆ’ 7 +π‘₯ +7 18 = 6π‘₯ +7 Γ· 6 3 = π‘₯ (Γ· 6) Perimeter = 11 βˆ’ π‘₯ + 5π‘₯ βˆ’ 7 + 4π‘₯ + 6 = 8π‘₯ + 10 Substitute π‘₯ = 3, π‘‘π‘œ 𝑔𝑒𝑑 8 Γ— 3 + 10 = πŸ‘πŸ’
  • 56. Handling Data Below is a table to show how James spends his day. Draw a pie chart to represent this data Activity Hours Sleeping 9 hours Work 8 hours Eating/Cleaning/Cooking 3 hours Reading 2 hours Commuting 2 hours Activity Hours Degrees Sleeping 9 hours 135 Work 8 hours 120 Eating/Cleaning/Cooking 3 hours 45 Reading 2 hours 30 Commuting 2 hours 30 24 hours in a day. Find out what each hour is worth 360 Γ· 24 = 15Β° Each hour is worth 15˚ on our pie chart. So sleeping is 9 x 15˚ = 135˚
  • 57. Number Algebra Shape, Space, Measure Handling Data Solve the following Simultaneous Equations 5π‘₯ βˆ’ 𝑦 = 9 15π‘₯ βˆ’ 2𝑦 = 24 2π‘₯ + 3𝑦 = 58 1) A gardens perimeter is 34m (rounded to the nearest m), garden fence panels are 130cm (rounded to the nearest 10cm). a) What is the most number of panels that may be needed? b) What is the least numbers of panels that may be need? Probability of Eric passing his Maths exam is 0.7, the probability of passing his English exam is independent of this, and is 0.6. What is the probability of Eric passing at least one of his exams? What is the straight line distance between the coordinates (4, -5) and ( 10, -3)
  • 58. Number 1) A gardens perimeter is 34m (rounded to the nearest m), garden fence panels are 130cm (rounded to the nearest 10cm). a) What is the most number of panels that may be needed? b) What is the least numbers of panels that may be need? Change measurements into the same units. 34m = 3400cm. a) The most number of fence panels needed are when you have a large perimeter and small fence panels. Upper bound for perimeter = 3450cm Lower bound for fence panel = 125cm How many β€˜small’ fence panels will you need for a β€˜large’ perimeter. πŸ‘πŸ’πŸ“πŸŽ Γ· πŸπŸπŸ“ = πŸπŸ•. πŸ” You would need 28 panels! b) The least number of fence panels needed are when you have a small perimeter and large fence panels Lower bound for perimeter = 3350cm Upper bound for fence panel = 135cm How many β€˜large’ fence panels will you need for a β€˜small’ perimeter. πŸ‘πŸ‘πŸ“πŸŽ Γ· πŸπŸ‘πŸ“ = πŸπŸ’. πŸ–πŸ πŸ’ You would need 25 panels!
  • 59. Algebra Solve the following Simultaneous Equations 5π‘₯ βˆ’ 𝑦 = 9 15π‘₯ βˆ’ 2𝑦 = 24 5π‘₯ βˆ’ 𝑦 = 9 (π‘Ž) 15π‘₯ βˆ’ 2𝑦 = 24 𝑏 𝑁𝑒𝑒𝑑 π‘‘π‘œ π‘šπ‘Žπ‘˜π‘’ π‘Ž π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘’π‘›π‘‘ π‘’π‘žπ‘’π‘Žπ‘™ π‘Ž Γ— 2 10π‘₯ βˆ’ 2𝑦 = 18 𝑐 𝑏 βˆ’ 𝑐 5π‘₯ = 6 Γ· 5 𝒙 = 𝟏. 𝟐 Γ· 5 𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘π‘Žπ‘π‘˜ π‘–π‘›π‘‘π‘œ π‘Ž 5 1.2 βˆ’ 𝑦 = 9 6 βˆ’ 𝑦 = 9 βˆ’6 βˆ’ 𝑦 = 3 βˆ’6 Γ— βˆ’1 π’š = βˆ’πŸ‘ (Γ— βˆ’1) 2π‘₯ + 3𝑦 = 58 (π‘Ž) 7π‘₯ = 6𝑦 + 5 𝑏 π‘…π‘’π‘Žπ‘Ÿπ‘Ÿπ‘Žπ‘›π‘”π‘’ 𝑏 π‘‘π‘œ 𝑔𝑒𝑑 π‘π‘œπ‘’π‘“π‘“π‘–π‘π‘’π‘›π‘‘π‘  π‘œπ‘› 𝑙𝑒𝑓𝑑 β„Žπ‘Žπ‘›π‘‘ 𝑠𝑖𝑑𝑒 βˆ’6y 7π‘₯ βˆ’ 6𝑦 = 5 βˆ’6𝑦 π‘Ž Γ— 2 4π‘₯ + 6𝑦 = 116 𝑐 𝑏 + 𝑐 11π‘₯ = 121 Γ· 11 𝒙 = 𝟏𝟏 Γ· 11 𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘π‘Žπ‘π‘˜ π‘–π‘›π‘‘π‘œ π‘Ž 2 11 + 3𝑦 = 58 22 + 3𝑦 = 58 βˆ’22 3𝑦 = 36 βˆ’22 Γ· 3 π’š = 𝟏𝟐 (Γ· 3)
  • 60. Shape, Space, Measure What is the straight line distance between the coordinates (4, -5) and ( 10, -3) (4,-5) (10,-3) βˆ’3 βˆ’ βˆ’5 = 2 10 βˆ’ 4 = 6 π‘ˆπ‘ π‘’ π‘ƒπ‘¦π‘‘β„Žπ‘Žπ‘”π‘œπ‘Ÿπ‘Žπ‘  π‘‘π‘œ π‘π‘Žπ‘™π‘π‘’π‘™π‘Žπ‘‘π‘’ π‘‘β„Žπ‘’ π‘™π‘’π‘›π‘”π‘‘β„Ž π‘œπ‘“ π‘‘β„Žπ‘’ 𝑙𝑖𝑛𝑒 π‘Ž2 + 𝑏2 = 𝑐2 62 + 22 = 𝑐2 36 + 4 = 𝑐2 40 = 𝑐2 6.32 (3.s.f)= 𝒄
  • 61. Handling Data Probability of Eric passing his Maths exam is 0.7, the probability of passing his English exam is independent of this, and is 0.6. What is the probability of Eric passing at least one of his exams? List the combinations – Pass Maths (0.7), Fail English (0.4) Pass English (0.6), Fail Maths (0.3) Pass English (0.6), Pass Maths (0.7) 𝑃 π‘ƒπ‘Žπ‘ π‘ π‘€ 𝐴𝑁𝐷 πΉπ‘Žπ‘–π‘™πΈ = 0.7 Γ— 0.4 = 0.28 𝑃 π‘ƒπ‘Žπ‘ π‘ πΈ π‘Žπ‘›π‘‘ πΉπ‘Žπ‘–π‘™π‘€ = 0.6 Γ— 0.3 = 0.18 𝑃 π‘ƒπ‘Žπ‘ π‘ πΈ π‘Žπ‘›π‘‘ π‘ƒπ‘Žπ‘ π‘ π‘€ = 0.6 Γ— 0.7 = 0.42 𝑃 π‘ƒπ‘Žπ‘ π‘  π‘Žπ‘‘ π‘™π‘’π‘Žπ‘ π‘‘ π‘œπ‘›π‘’ = 0.28 + 0.18 + 0.42 = 𝟎. πŸ–πŸ– Alternative Solution. Probability of passing at least once = 1 – probability of not passing either exam 𝑃 πΉπ‘Žπ‘–π‘™πΈ π‘Žπ‘›π‘‘ πΉπ‘Žπ‘–π‘™π‘€ = 0.4 Γ— 0.3 = 0.12 𝟏 βˆ’ 𝟎. 𝟏𝟐 = 𝟎. πŸ–πŸ–
  • 62. Number Algebra Shape, Space, Measure Handling Data Put the following, in ascending order π‘Ÿ0, 2βˆ’5, 81 1 2, 1 4 βˆ’2 , 3 2 2 π‘†π‘œπ‘™π‘£π‘’ 4π‘₯ βˆ’ 1 = 2 βˆ’ π‘₯ 10π‘₯ + 5 = 8π‘₯ βˆ’ 9 π‘₯ βˆ’ 1 2 + 4π‘₯ βˆ’ 3 3 = 4 Construct a frequency polygon for the following The area of the rectangle is 60π‘π‘š2 calculate the perimeter π‘₯ βˆ’ 7 Cost x (Β£) Frequency 0 ≀ x < 10 8 10 ≀ x < 20 11 20 ≀ x < 30 5 30 ≀ x < 40 14 40 ≀ x < 50 6 π‘₯
  • 63. Number Put the following, in ascending order π‘Ÿ0, 2βˆ’5, 81 1 2, 1 4 βˆ’2 , 3 2 2π‘Ÿ0 = 1 (anything to the power 0 is equal to 1. 2βˆ’5 = 1 25 = 1 32 81 1 2 = 81 = 9 1 4 βˆ’2 = 4 1 2 = 42 12 = 16 1 = 16 3 2 2 = 32 22 = 9 4 Therefore, πŸβˆ’πŸ“, 𝒓 𝟎, πŸ‘ 𝟐 𝟐 , πŸ–πŸ 𝟏 𝟐, 𝟏 πŸ’ βˆ’πŸ
  • 64. Algebra π‘†π‘œπ‘™π‘£π‘’ 4π‘₯ βˆ’ 1 = 2 βˆ’ π‘₯ 10π‘₯ + 5 = 8π‘₯ βˆ’ 9 π‘₯ βˆ’ 1 2 + 4π‘₯ βˆ’ 3 3 = 4 4π‘₯ βˆ’ 1 = 2 βˆ’ π‘₯ +π‘₯ 5π‘₯ βˆ’ 1 = 2 +π‘₯ +1 5π‘₯ = 3 +1 Γ· 5 𝒙 = πŸ‘ πŸ“ = 𝟎. πŸ” (Γ· 5) 10π‘₯ + 5 = 8π‘₯ βˆ’ 9 βˆ’8π‘₯ 2π‘₯ + 5 = βˆ’9 βˆ’8π‘₯ βˆ’5 2π‘₯ = βˆ’14 βˆ’5 Γ· 2 𝒙 = βˆ’πŸ• (Γ· 2) π‘₯ βˆ’ 1 2 + 4π‘₯ βˆ’ 3 3 = 4 Γ— 2 π‘₯ βˆ’ 1 + 2 4π‘₯ βˆ’ 3 3 = 8 Γ— 2 Γ— 3 3 π‘₯ βˆ’ 1 + 2 4π‘₯ βˆ’ 3 = 24 Γ— 3 3π‘₯ βˆ’ 3 + 8π‘₯ βˆ’ 6 = 24 11π‘₯ βˆ’ 9 = 24 +9 11π‘₯ = 33 +9 Γ· 11 𝒙 = πŸ‘ (Γ· 11)
  • 65. Shape, Space, Measure The area of the rectangle is 60π‘π‘š2 calculate the perimeter π‘₯ βˆ’ 7 π‘₯ π΄π‘Ÿπ‘’π‘Ž = 60π‘π‘š2 π΄π‘Ÿπ‘’π‘Ž = π‘₯ π‘₯ βˆ’ 7 = π‘₯2 βˆ’ 7π‘₯ π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, π‘₯2 βˆ’ 7π‘₯ = 60 π‘€π‘Žπ‘˜π‘’ 𝑖𝑑 π‘™π‘œπ‘œπ‘˜ π‘™π‘–π‘˜π‘’ π‘Ž π‘’π‘ π‘’π‘Žπ‘™ π‘žπ‘’π‘Žπ‘‘π‘Ÿπ‘Žπ‘‘π‘–π‘ βˆ’60 π‘₯2 βˆ’ 7π‘₯ βˆ’ 60 = 0 βˆ’60 π‘π‘œπ‘€ π‘ π‘œπ‘™π‘£π‘’. 𝐼𝑑 𝑀𝑖𝑙𝑙 π‘“π‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’, 𝑏𝑒𝑑 π‘¦π‘œπ‘’ π‘π‘Žπ‘› 𝑒𝑠𝑒 π‘‘β„Žπ‘’ π‘“π‘œπ‘Ÿπ‘šπ‘’π‘™π‘Ž π‘₯ βˆ’ 12 π‘₯ + 5 = 0 π‘₯ βˆ’ 12 = 0 π‘œπ‘Ÿ π‘₯ + 5 = 0 π‘₯ = 12 π‘œπ‘Ÿ π‘₯ = βˆ’5 We are working with lengths, so we will ignore the -5, π‘₯ = 12. π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = π‘₯ βˆ’ 7 + π‘₯ + π‘₯ βˆ’ 7 + π‘₯ = 4π‘₯ βˆ’ 14 𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 𝑖𝑛 π‘₯ = 12 = 4 12 βˆ’ 14 = πŸ‘πŸ’π’„π’Ž
  • 66. Handling Data Construct a frequency polygon for the following Cost x (Β£) Frequency 0 ≀ x < 10 8 10 ≀ x < 20 11 20 ≀ x < 30 5 30 ≀ x < 40 14 40 ≀ x < 50 6 See polygon – think straight lines. Is this 8 things at Β£0 or Β£9.99? We can’t be sure so use the midpoint.
  • 67. Number Algebra Shape, Space, Measure Handling Data Calculate the π‘›π‘‘β„Ž term for the following sequences 3, 7, 11, 15, 19, … 8, 11, 16, 23, 32, … 2, 7, 14, 23, 34, … Barnsley Football club is organising travel for an away game. 1300 adults and 500 juniors want to go. Each coach holds 48 people and costs Β£320 to hire. Tickets to the match are Β£18 for adults and Β£10 for juniors. The club is charging adults Β£26 and juniors Β£14 for travel and a ticket. How much profit does the club make out the trip? Calculate an estimate for the meanThese two shapes are Mathematical Similar. Calculate the missing volume and surface area 4cm 12cm Surface area = Volume = 72π‘π‘š3 Surface area = 729π‘π‘š2 Volume = Cost x (Β£) Frequency 0 ≀ x < 10 8 10 ≀ x < 20 11 20 ≀ x < 30 5 30 ≀ x < 40 14 40 ≀ x < 50 6
  • 68. Number Barnsley Football club is organising travel for an away game. 1300 adults and 500 juniors want to go. Each coach holds 48 people and costs Β£320 to hire. Tickets to the match are Β£18 for adults and Β£10 for juniors. The club is charging adults Β£26 and juniors Β£14 for travel and a ticket. How much profit does the club make out the trip? Cost to supporter 1300 Γ— Β£26 = Β£33,800 500 Γ— Β£14 = Β£7,000 π‘‡π‘œπ‘‘π‘Žπ‘™ = Β£40,800 Cost to the club Ticket costs 1300 Γ— Β£18 = Β£23,400 500 Γ— Β£10 = Β£5,000 π‘‡π‘œπ‘‘π‘Žπ‘™ = Β£28,400 Coach costs Coaches needed: (1300 + Profit to the club Β£40,800 βˆ’ Β£40,560 = Β£240
  • 69. Algebra Calculate the π‘›π‘‘β„Ž term for the following sequences 3, 7, 11, 15, 19, … 8, 11, 16, 23, 32, … 2, 7, 14, 23, 34, … 3, 7, 11, 15, 19, … πΌπ‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘–π‘›π‘” 𝑏𝑦 2 π‘’π‘Žπ‘β„Ž π‘‘π‘–π‘šπ‘’ 2𝑛 + 1 8, 11, 16, 23, 32, … 3 5 7 9 2 2 2 𝑛2 1, 4, 9, 16, 25, … +7 2, 7, 14, 23, 34, … 5 7 9 11 2 2 2 𝑛2 1, 4, 9, 16, 25, … 1, 3, 5, 7, 9 + 2𝑛 βˆ’ 1
  • 70. Shape, Space, Measure These two shapes are Mathematical Similar. Calculate the missing volume and surface area 4cm 12cm Surface area = Volume = 72π‘π‘š3 Surface area = 729π‘π‘š2 Volume = Mathematical similar – they are in scale. π‘†π‘π‘Žπ‘™π‘’ πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿ = 12 4 = 3 Shape B has been enlarged by the linear scale factor of 3. The surface area will therefore be 32 as big. 729 32 = 81π‘π‘š2 The volume will be 33 as big. 72 Γ— 33 = 1944π‘π‘š3 A B πŸ–πŸπ’„π’Ž 𝟐 πŸπŸ—πŸ’πŸ’π’„π’Ž πŸ‘ Top Tip: Area is measured in π’„π’Ž 𝟐 , so when enlarging a shape – don’t forget to square the scale factor when working out the enlarged area. Volume is measured in π’„π’Ž πŸ‘ , so when enlarging a shape – don’t forget to cube the scale factor when working out the enlarged volume.
  • 71. Calculate an estimate for the mean Time of match Frequency Midpoint Midpoint x Frequency 0 ≀ x < 10 8 5 40 10 ≀ x < 20 11 15 165 20 ≀ x < 30 5 25 125 30 ≀ x < 40 14 35 490 40 ≀ x < 50 6 45 270 Total 44 1090 Not sure what time the cost is – so we use the mid point as an estimate. π‘€π‘’π‘Žπ‘› = 1090 44 π‘€π‘’π‘Žπ‘› = Β£24.78 Handling Data Cost x (Β£) Frequency 0 ≀ x < 10 8 10 ≀ x < 20 11 20 ≀ x < 30 5 30 ≀ x < 40 14 40 ≀ x < 50 6
  • 72. Number Algebra Shape, Space, Measure Handling Data πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’ 4π‘₯2 βˆ’ 18π‘₯ π‘₯2 βˆ’ 12π‘₯ + 32 6π‘₯2 + 13π‘₯ βˆ’ 5 25π‘₯2 βˆ’ 9𝑦2 Give two criticisms of the following questionnaire A rectangles width to length ratio is 2:5, if the perimeter is 84cm, what is the area? 2.1 x 0.48 = 0.021 x 4.8 = 162.9 Γ· 0.09 = Do you agree that Barnsley FC are by far the greatest team the world has ever seen? Strongly agree Agree Not sure
  • 73. Number 2.1 x 0.48 = 0.021 x 4.8 = 162.9 Γ· 0.09 = Ignore the decimal places and do your calculations when multiplying decimals. 𝟐𝟏 Γ— πŸ’πŸ– = πŸπŸŽπŸŽπŸ– Now add your decimal point into your answer. The question had three decimal places – so your answer needs three decimal places. 1 008. Ignore the decimal places and do your calculations when multiplying decimals. 𝟐𝟏 Γ— πŸ’πŸ– = πŸπŸŽπŸŽπŸ– Now add your decimal point into your answer. The question had four decimal places – so your answer needs four decimal places. 1 008 . Use equivalent fractions when dividing with decimals. 162.9 0.09 = 16290 9 = 1810
  • 74. Algebra πΉπ‘Žπ‘π‘‘π‘œπ‘Ÿπ‘–π‘ π‘’ 4π‘₯2 βˆ’ 18π‘₯ π‘₯2 βˆ’ 12π‘₯ + 32 6π‘₯2 + 13π‘₯ βˆ’ 5 25π‘₯2 βˆ’ 9𝑦2 4π‘₯2 βˆ’ 18π‘₯ = 2π‘₯ 2π‘₯ βˆ’ 9 π‘₯2 βˆ’ 12π‘₯ + 32 = π‘₯ βˆ’ 8 π‘₯ βˆ’ 4 6π‘₯2 + 13π‘₯ βˆ’ 5 = (3π‘₯ βˆ’ 1)(2π‘₯ + 5) 25π‘₯2 βˆ’ 9𝑦2 = (5π‘₯ βˆ’ 3𝑦)(5π‘₯ + 3𝑦)
  • 75. Shape, Space, Measure A rectangles width to length ratio is 2:5, if the perimeter is 84cm, what is the area? 5π‘₯ 2π‘₯ π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = 84π‘π‘š π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = 5π‘₯ + 2π‘₯ + 5π‘₯ + 2π‘₯ = 14π‘₯ π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, 14π‘₯ = 84 Γ· 14 π‘₯ = 6 Γ· 14 π΄π‘Ÿπ‘’π‘Ž = 2π‘₯ Γ— 5π‘₯ = 10π‘₯2 𝑆𝑒𝑏𝑠𝑑𝑖𝑑𝑒𝑑𝑒 π‘₯ = 6 π΄π‘Ÿπ‘’π‘Ž = 10 62 = πŸ‘πŸ”πŸŽπ’„π’Ž 𝟐
  • 76. Handling Data Give two criticisms of the following questionnaire Do you agree that Barnsley FC are by far the greatest team the world has ever seen? Strongly agree Agree Not sure 1. A leading question, do you agree, puts pressure on the person answering the question. 2. No response boxes for if you disagree (but why would you?)
  • 77. Number Algebra Shape, Space, Measure Handling Data Fill in the table of values and complete the cubic graph, 𝑦 = π‘₯3 βˆ’ 2π‘₯ +1 for -3 ≀ x ≀ 3 2 3 7 βˆ’ 4 1 4 = 5 8 Γ— 4 15 = 4 5 Γ· 6 7 = A bag contains 3 blue balls, 5 red balls and 4 green balls. When a ball is taken out, it is not replaced. What is the probability of picking three balls that are all the same colour. Calculate the area of the following quadrilateral x -3 -2 -1 0 1 2 3 y -3 1 22
  • 78. Number 2 3 7 βˆ’ 4 1 4 = 5 8 Γ— 4 15 = 4 5 Γ· 6 7 = For Casio fx-83GT plus Type shift 32 7 - shift 14 4 = βˆ’ 51 28 5 8 4 15 =x 1 6 4 5 6 7 =Γ· 14 15
  • 79. Algebra Fill in the table of values and complete the cubic graph, 𝑦 = π‘₯3 βˆ’ 2π‘₯ +1 for -3 ≀ x ≀ 3 x -3 -2 -1 0 1 2 3 y -3 1 22 x -3 -2 -1 0 1 2 3 y -20 -3 2 1 0 5 22
  • 80. Shape, Space, Measure Calculate the area of the following quadrilateralWill need these (found at the front of your exam) Find the area of the two triangles and add these together... 1 2 Sin B b = Sin A a 𝑆𝑖𝑛 B 3 = Sin 83 12 𝑆𝑖𝑛 𝐡 = 3𝑆𝑖𝑛 83 12 𝐡 = π‘†π‘–π‘›βˆ’1 3 sin 83 12 𝐡 = 14.367Β° 1 B π‘‡β„Žπ‘’ π‘™π‘Žπ‘ π‘‘ π‘’π‘›π‘˜π‘›π‘œπ‘€π‘› π‘Žπ‘›π‘”π‘™π‘’ 𝑖𝑛 π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 1 180 βˆ’ 83 βˆ’ 14.367 = 82.633Β° 2 𝐹𝑖𝑛𝑑 π‘Žπ‘› π‘Žπ‘›π‘”π‘™π‘’ 𝑖𝑛 π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 2 𝑒𝑠𝑖𝑛𝑔 π‘‘β„Žπ‘’ π‘π‘œπ‘ π‘–π‘›π‘’ π‘Ÿπ‘’π‘™π‘’ π‘Žπ‘›π‘¦ π‘Žπ‘›π‘”π‘™π‘’ 𝑖𝑠 π‘Žπ‘π‘π‘™π‘–π‘π‘Žπ‘π‘™π‘’ π‘Ž2 = 𝑏2 + 𝑐2 βˆ’ 2𝑏𝑐 πΆπ‘œπ‘  𝐴 122 = 102 + 42 βˆ’ 2 Γ— 10 Γ— 4 πΆπ‘œπ‘  𝐴 144 = 116 βˆ’ 80πΆπ‘œπ‘  𝐴 28 = βˆ’80 πΆπ‘œπ‘  𝐴 βˆ’ 28 80 = πΆπ‘œπ‘  𝐴 πΆπ‘œπ‘ βˆ’1 βˆ’ 28 80 = 𝐴 𝐴 = 110.487
  • 81. Shape, Space, Measure Calculate the area of the following quadrilateralWill need these (found at the front of your exam) Find the area of the two triangles and add these together... 1 2 π‘ˆπ‘ π‘’ π΄π‘Ÿπ‘’π‘Ž = 1 2 π‘Žπ‘π‘†π‘–π‘› 𝐢 π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 1 = 1 2 Γ— 12 Γ— 3 Γ— 𝑆𝑖𝑛 82.633 π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 1 = 17.851π‘π‘š2 π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 2 = 1 2 Γ— 10 Γ— 4 Γ— 𝑆𝑖𝑛 110.487 π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘‡π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ 2 = 18.735π‘π‘š2 π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘„π‘’π‘Žπ‘‘π‘Ÿπ‘–π‘™π‘Žπ‘‘π‘’π‘Ÿπ‘Žπ‘™ = 17.851 + 18.735 = πŸ‘πŸ”. πŸ“πŸ–πŸ”
  • 82. Handling Data A bag contains 3 blue balls, 5 red balls and 4 green balls. When a ball is taken out, it is not replaced. What is the probability of picking three balls that are all the same colour. List the combinations – Blue, Blue, Blue, Red, Red, Red, Green, Green, Green 𝑃 𝐡 𝐴𝑁𝐷 𝐡 𝐴𝑁𝐷 𝐡 = 3 12 Γ— 2 11 Γ— 1 10 = 6 1320 𝑃 𝑅 𝐴𝑁𝐷 𝑅 𝐴𝑁𝐷 𝑅 = 5 12 Γ— 4 11 Γ— 3 10 𝑃 𝐺 𝐴𝑁𝐷 𝐺 𝐴𝑁𝐷 𝐺 = 4 12 Γ— 3 11 Γ— 2 10 𝑃 𝐡𝐡𝐡 𝑂𝑅 𝑅𝑅𝑅 𝑂𝑅 𝐺𝐺𝐺 = 6 1320 + 60 1320 + 24 1320 = 90 1320 = 3 44
  • 83. Number Algebra Shape, Space, Measure Handling Data π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ β„Ž0 3π‘Ÿ2 𝑑4 𝑧 3 10π‘₯𝑦2 𝑧4 2π‘₯3 𝑦2 𝑧 π‘₯2 + 7π‘₯ + 12 π‘₯ + 4 The mean weight of ten footballers is 73.5kg. A new player comes along and the mean weight goes down to 72kg. How much does the new player weigh? Calculate the area of the isosceles triangle Calculate the area and perimeter of the rectangle. Leave your answer in Surd form if applicable 18 2 10cm 13cm
  • 84. Number Calculate the area and perimeter of the rectangle. Leave your answer in Surd form if applicable 18 2 π΄π‘Ÿπ‘’π‘Ž = 18 Γ— 2 = 36 = πŸ” π’–π’π’Šπ’•π’” 𝟐 π‘ƒπ‘’π‘Ÿπ‘–π‘šπ‘’π‘‘π‘’π‘Ÿ = 18 + 2 + 18 + 2 = 9 Γ— 2 + 2 + 9 Γ— 2 + 2 = 9 2 + 2 + 9 2 + 2 = 3 2 + 2 + 3 2 + 2 = πŸ– 𝟐 π’–π’π’Šπ’•π’”
  • 85. Algebra π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ β„Ž0 3π‘Ÿ2 𝑑4 𝑧 3 10π‘₯𝑦2 𝑧4 2π‘₯3 𝑦2 𝑧 π‘₯2 + 7π‘₯ + 12 π‘₯ + 4 β„Ž0 = 1 3π‘Ÿ2 𝑑4 𝑧 3 = 3π‘Ÿ2 𝑑4 𝑧 Γ— 3π‘Ÿ2 𝑑4 𝑧 Γ— 3π‘Ÿ2 𝑑4 𝑧 = 27π‘Ÿ6 𝑑12 𝑧3 10π‘₯𝑦2 𝑧4 2π‘₯3 𝑦2 𝑧 = 5π‘₯βˆ’2 𝑧3 = 5𝑧3 π‘₯2 π‘₯2 + 7π‘₯ + 12 π‘₯ + 4 = π‘₯ + 3 π‘₯ + 4 (π‘₯ + 4) = π‘₯ + 3
  • 86. Shape, Space, Measure Calculate the area of the isosceles triangle 10cm 13cm π΄π‘Ÿπ‘’π‘Ž π‘œπ‘“ π‘Ž π‘‘π‘Ÿπ‘–π‘Žπ‘›π‘”π‘™π‘’ = π‘π‘Žπ‘ π‘’ Γ— β„Žπ‘’π‘–π‘”β„Žπ‘‘ 2 Need to calculate the perpendicular height of the triangle. 13cm 5cm h π‘ˆπ‘ π‘’ π‘ƒπ‘¦π‘‘β„Žπ‘Žπ‘”π‘œπ‘Ÿπ‘Žπ‘  π‘‘π‘œ π‘π‘Žπ‘™π‘π‘’π‘™π‘Žπ‘‘π‘’ π‘‘β„Žπ‘’ β„Žπ‘’π‘–π‘”β„Žπ‘‘ π‘Ž2 + 𝑏2 = 𝑐2 52 + 𝑏2 = 132 25 + 𝑏2 = 169 𝑏2 = 144 𝑏 = 12 π»π‘’π‘–π‘”β„Žπ‘‘ 𝑖𝑠 12π‘π‘š. π‘‡β„Žπ‘’π‘Ÿπ‘’π‘“π‘œπ‘Ÿπ‘’, π΄π‘Ÿπ‘’π‘Ž = 12 Γ— 5 2 π΄π‘Ÿπ‘’π‘Ž = πŸ‘πŸŽπ’„π’Ž 𝟐
  • 87. Handling Data The mean weight of ten footballers is 73.5kg. A new player comes along and the mean weight goes down to 72kg. How much does the new player weigh? 10 footballers in total weigh 10 Γ— 73.5π‘˜π‘” = 735π‘˜π‘”. 11 footballers in total weigh 11 Γ— 72 = 792π‘˜π‘” The eleventh footballer must have weight 792 βˆ’ 735 = πŸ“πŸ•π’Œπ’ˆ
  • 88. Number Algebra Shape, Space, Measure Handling Data The formula to convert Celsius to Fahrenheit is 9 5 𝐢 + 32 = 𝐹 Use your calculator to work out the value of 3.92 βˆ’ 1.42 Write down the full display. Round your answer to 2 significant figures Complete a histogram for the following data Height, h (cm) Frequency 151 ≀ h < 153 64 153 ≀ h < 154 43 154 ≀ h < 155 47 155 ≀ h < 159 96 159 ≀ h < 160 12 What is the temperature in Fahrenheit, when it 18˚ Celsius? What is the temperature in Celsius when it is 53.6˚ Fahrenheit? An exterior angle of a regular polygon is 30˚. Work out the number of sides of the polygon.
  • 89. Number Use your calculator to work out the value of 3.92 βˆ’ 1.42 Write down the full display. Round your answer to 2 significant figures 3 For Casio fx-83GT plus Type . 9 - 1 . 4 𝒙 𝟐 = 3.640054945 3.6 𝒙 𝟐
  • 90. Algebra The formula to convert Celsius to Fahrenheit is 9 5 𝐢 + 32 = 𝐹 What is the temperature in Fahrenheit, when it 18˚ Celsius? What is the temperature in Celsius when it is 53.6˚ Fahrenheit? 9 Γ— 18 5 + 32 = ℉ 64.4 = ℉ πŸπŸ–β„ƒ = πŸ”πŸ’. πŸ’β„‰ 9 5 𝐢 + 32 = 53.6 βˆ’32 9 5 𝐢 = 21.6 βˆ’32 Γ— 5 9𝐢 = 108 Γ— 5 Γ· 9 𝐢 = 12 Γ· 9 πŸπŸβ„ƒ = πŸ“πŸ‘. πŸ”β„‰
  • 91. Shape, Space, Measure An exterior angle of a regular polygon is 30˚. Work out the number of sides of the polygon. Exterior angles add up to 360Β°, therefore 360 Γ· 30 = 12 The polygon has 12 sides (Dodecagon) 30Β°
  • 92. Handling Data Complete a histogram for the following data Height, h (cm) Frequency 151 ≀ h < 153 64 153 ≀ h < 154 43 154 ≀ h < 155 47 155 ≀ h < 159 96 159 ≀ h < 160 12 Marks Frequency Frequency Density 151 ≀ h < 153 64 64 Γ· 2 = 32 153 ≀ h < 154 43 43 Γ· 1 = 43 154 ≀ h < 155 47 47 Γ· 1 = 47 155 ≀ h < 159 96 96 Γ· 4 = 24 159 ≀ h < 160 12 12 Γ· 1 = 12 Calculate the frequency density by frequency Γ· class width. Area of the bar is the frequency
  • 93. Number Algebra Shape, Space, Measure Handling Data 𝐸π‘₯π‘π‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 3𝑦 4𝑦 βˆ’ 2 4 π‘₯ βˆ’ 2 βˆ’ 5 π‘₯ + 3 π‘₯ + 4 π‘₯ βˆ’ 7 (3π‘Ž + 2𝑏)(2π‘Ž βˆ’ 3𝑏) What is the probability of rolling a dice three times and getting two square numbers and a prime number. Convert 8π‘π‘š3 to π‘šπ‘š3 Convert 40,000π‘π‘š2 to π‘š2 Four pumps usually empty water from a tank in 1 hour 36 minutes. One of the pumps breaks down. How long will three pumps, working at the same rate, take to empty the same tank.
  • 94. Number Four pumps usually empty water from a tank in 1 hour 36 minutes. One of the pumps breaks down. How long will three pumps, working at the same rate, take to empty the same tank. 1 hour 36 minutes = 96 minutes If four pumps take 96 minutes to empty a tank, that must mean one pump would take four times as long 4 Γ— 96 = 384 π‘šπ‘–π‘›π‘’π‘‘π‘’π‘  If three pumps were to do this, they would do it in a third of the time 384 Γ· 3 = 128 π‘šπ‘–π‘›π‘’π‘‘π‘’π‘  128 minutes = 2 hours 8 minutes.
  • 95. Algebra 𝐸π‘₯π‘π‘Žπ‘›π‘‘ π‘Žπ‘›π‘‘ π‘†π‘–π‘šπ‘π‘™π‘–π‘“π‘¦ 3𝑦 4𝑦 βˆ’ 2 4 π‘₯ βˆ’ 2 βˆ’ 5 π‘₯ + 3 π‘₯ + 4 π‘₯ βˆ’ 7 (3π‘Ž + 2𝑏)(2π‘Ž βˆ’ 3𝑏) 3𝑦 4𝑦 βˆ’ 2 = 12y2 βˆ’ 6y 4 π‘₯ βˆ’ 2 βˆ’ 5 π‘₯ + 3 = 4π‘₯ βˆ’ 8 βˆ’ 5π‘₯ βˆ’ 15 = βˆ’π‘₯ βˆ’ 23 π‘₯ + 4 π‘₯ βˆ’ 7 = π‘₯2 βˆ’ 7π‘₯ + 4π‘₯ βˆ’ 28 = π‘₯2 βˆ’ 3π‘₯ βˆ’ 28 3π‘Ž + 2𝑏 2π‘Ž βˆ’ 3𝑏 = 6π‘Ž2 βˆ’ 9π‘Žπ‘ + 4π‘Žπ‘ βˆ’ 6𝑏2 = 6π‘Ž2 βˆ’ 5π‘Žπ‘ βˆ’ 6𝑏2
  • 96. Shape, Space, Measure Convert 8π‘π‘š3 to π‘šπ‘š3 Convert 40,000π‘π‘š2 to π‘š2 1cm 1cm 1cm 10mm 10mm 10mm π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 1π‘π‘š Γ— 1π‘π‘š Γ— 1π‘π‘š = 1π‘π‘š3 π‘‰π‘œπ‘™π‘’π‘šπ‘’ = 10π‘šπ‘š Γ— 10π‘šπ‘š Γ— 10π‘šπ‘š = 1000π‘šπ‘š3 π‘¬π’—π’†π’“π’š πŸπ’„π’Ž πŸ‘ π’Šπ’” π’†π’’π’–π’Šπ’—π’‚π’π’†π’π’• 𝒕𝒐 πŸπŸŽπŸŽπŸŽπ’Žπ’Ž πŸ‘ , 𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆, πŸ–π’„π’Ž πŸ‘ = πŸ–πŸŽπŸŽπŸŽπ’Žπ’Ž πŸ‘ 1m 1m 100cm 100cm π΄π‘Ÿπ‘’π‘Ž = 1π‘š Γ— 1π‘š = 1π‘š2 π΄π‘Ÿπ‘’π‘Ž = 100π‘π‘š Γ— 100π‘π‘š = 10,000π‘π‘š2 π‘¬π’—π’†π’“π’š πŸπ’Ž 𝟐 π’Šπ’” π’†π’’π’–π’Šπ’—π’‚π’π’†π’π’• 𝒕𝒐 𝟏𝟎, πŸŽπŸŽπŸŽπ’„π’Ž 𝟐, 𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆, πŸ’πŸŽ, πŸŽπŸŽπŸŽπ’„π’Ž 𝟐 = πŸ’π’Ž 𝟐
  • 97. Handling Data What is the probability of rolling a dice three times and getting two square numbers and a prime number. List the combinations – Square, Square, Prime Square, Prime, Square Prime, Square, Square 𝑃 𝑆 𝐴𝑁𝐷 𝑆 𝐴𝑁𝐷 𝑃 = 1 3 Γ— 1 3 Γ— 1 2 = 1 18 𝑃 𝑆 𝐴𝑁𝐷 𝑃 𝐴𝑁𝐷 𝑆 = 1 3 Γ— 1 2 Γ— 1 3 = 1 18 𝑃 𝑃 𝐴𝑁𝐷 𝑆 𝐴𝑁𝐷 𝑆 = 1 2 Γ— 1 3 Γ— 1 3 = 1 18 Square numbers 1, 4 𝑃 π‘†π‘žπ‘’π‘Žπ‘Ÿπ‘’ = 1 3 Prime numbers 2, 3, 5 𝑃 π‘ƒπ‘Ÿπ‘–π‘šπ‘’ = 1 2 𝑃 𝑆𝑆𝑃 𝑂𝑅 𝑆𝑃𝑆 𝑂𝑅 𝑃𝑆𝑆 = 1 18 + 1 18 + 1 18 = 3 18 = 1 6
  • 98. Number Algebra Shape, Space, Measure Handling Data Write down all the integers n that satisfy 5 ≀ 𝑛 < 9 βˆ’3 < 2𝑛 ≀ 1 βˆ’2 ≀ 4𝑛 + 2 ≀ 0 Β£1 = $1.67 A laptop costs Β£320 in the UK, and $530 in the USA. Where is the laptop cheaper, and by how much? The table shows van rentals for the company VansRCool. Calculate a 3 point moving average for the months below Hayley can run 100m in 13 seconds. What is her average speed in miles per hour?
  • 99. Number Β£1 = $1.67 A laptop costs Β£320 in the UK, and $530 in the USA. Where is the laptop cheaper, and by how much? UK price Β£320 USA price $530 Γ· 1.67 = Β£317.37 USA cheaper by Β£2.63 UK price Β£320 Γ— 1.67 = $534.40 USA price $530 USA cheaper by $4.40 OR
  • 100. Algebra 𝑛 = 5, 6, 7, π‘œπ‘Ÿ 8 Write down all the integers n that satisfy 5 ≀ 𝑛 < 9 βˆ’3 < 2𝑛 ≀ 1 βˆ’2 ≀ 4𝑛 + 2 ≀ 0 βˆ’3 < 2𝑛 ≀ 1 Γ· 2 βˆ’ 1.5 < 𝑛 ≀ 1 Γ· 2 𝑛 = βˆ’1, 0 π‘œπ‘Ÿ 1 βˆ’2 ≀ 4𝑛 + 2 ≀ 0 βˆ’2 βˆ’ 4 ≀ 4𝑛 ≀ βˆ’2 βˆ’2 Γ· 4 βˆ’ 1 ≀ 𝑛 ≀ βˆ’0.5 Γ· 4 𝑛 = βˆ’1, 0
  • 101. Shape, Space, Measure Key Facts: 60 seconds in a minute 60 minutes in a hour 1600m in a mile. Hayley can run 100m in 13 seconds. What is her average speed in miles per hour? 100 π‘šπ‘’π‘‘π‘Ÿπ‘’π‘  𝑖𝑛 13 π‘ π‘’π‘π‘œπ‘›π‘‘π‘  100 13 π‘šπ‘’π‘‘π‘Ÿπ‘’π‘  𝑖𝑛 1 π‘ π‘’π‘π‘œπ‘›π‘‘ 60 Γ— 100 13 π‘šπ‘’π‘‘π‘Ÿπ‘’π‘  𝑖𝑛 1 π‘šπ‘–π‘›π‘’π‘‘π‘’ 60 Γ— 60 Γ— 100 13 π‘šπ‘’π‘‘π‘Ÿπ‘’π‘  𝑖𝑛 1 β„Žπ‘œπ‘’π‘Ÿ 360000 13 π‘šπ‘’π‘‘π‘Ÿπ‘’π‘  𝑖𝑛 1 β„Žπ‘œπ‘’π‘Ÿ 360000 13 Γ— 1600 π‘šπ‘–π‘™π‘’π‘  𝑖𝑛 1 β„Žπ‘œπ‘’π‘Ÿ. 17.3 miles per hour
  • 102. Handling Data 3 point moving average – work out the mean in groups of 3 The table shows van rentals for the company VansRCool. Calculate a 3 point moving average for the months below 9 + 22 + 37 3 = 22. 6 22 + 37 + 14 3 = 24. 3 37 + 14 + 18 3 = 23 14 + 18 + 24 3 = 18. 6 π‘‡β„Žπ‘’ 3 π‘π‘œπ‘–π‘›π‘‘ π‘šπ‘œπ‘£π‘–π‘›π‘” π‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ 𝑖𝑠 22.6, 24.3, 23, 18.6

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