Prime numbers
Prime numbers only have two factors.
A Prime Number is a
number that has exactly two
factors
The first few prime numbers are:- 2, 3, 5, 7, 11, 13, 17, 19,…….
Examples
• 1 is NOT prime as it only has 1 factor
• 9 is NOT prime because it can also be divided by 3 so it has 3 factors
Highest Common Factor / Lowest Common Multiple
Highest Common Factor
“The largest number that is a Factor of 2 or more others”
🡪 For example the HCF of 15 and 25 is 5
Lowest Common Multiple
“The smallest number that is a multiple of 2 or more numbers”
🡪 For example the LCM of 20 and 30 is 60
Product Of Primes
PRIME NUMBERS
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, …
EXAMPLES
Find the product of prime factors for the following numbers:
(a)
(b)
32
8
4
2
10
4
2
2
60
2
5
6
2
3
2
2
ANSWER = 2 X 2 X 2 X 2 X 2
=
ANSWER = 2 X 2 X 3 X 5
=
X3X5
HCF and LCM
• What is the HCF and LCM of 50 and 80?
50 = 2 x 5 x 5
Numbers common
to both 50 and
80 go in the
middle…
80 = 2 x 2 x 2 x 5 x
2
50
5
80
5
2
HCF – Multiply the middle numbers
🡪 2 x 5 = 10
Use a
Venn
diagram
2
2
2
LCM – Multiply all the numbers
400
🡪 5x5x2x2x2x2=
HCF and LCM
• What is the HCF and LCM of 105 and 42?
105 = 3 x 5 x 7
Numbers common
to both 105 and
42 go in the
middle…
5
42 = 2 x 3 x 7
3
2
Use a
Venn
diagram
7
HCF – Multiply the middle numbers
🡪 3 x 7 = 21
LCM – Multiply all the numbers
🡪 2 x 3 x 5 x 7 = 210
Percentages
of
x
EXAMPLES
(a) 13% of £567
x
(b) 76% of 4972km
x
13
100
x
76
100
x £567
(c) 62.5% of 782 people
x 4972km
62.5
100
x 782 people
÷
÷
÷
= £73.71
= 3778.72km
= 488.75 people
= 489 people
One number as a percentage of
another
There are 35 sweets in a bag. Four of the sweets are
orange flavour.
What percentage of sweets are
orange flavour?
Start by writing the proportion of orange sweets as a fraction.
4 out of 35 =
4
35
Then convert the fraction to a percentage.
4
35
× 100% =
11. 43 %
10%
÷10
1%
÷100
EXAMPLES
Work out the final amount when:
(a) £60 is increased by 20%
10% = £6
x2
x2
20% = £12
(b) 20miles is decreased by 40%
10% = 2miles x4
x4
40% = 8miles
We now need to
increase £60 by £12
We now need to decrease
20miles by 8miles
£60 + £12 = £72
20miles - 8miles = 12miles
(c) 80minutes is decreased by 35%
10% = 8mins
x3
x3
30% = 24mins
5% = 4mins
35% = 28mins
We now need to decrease
80mins by 28mins
80mins - 28mins = 52mins
(d) 120kg is increased by 17.5%
10% = 12kg
÷2
÷2
5% = 6kg
÷2
÷2
2.5% = 3kg
17.5% = 21kg
We now need to increase
120kg by 21kg
120kg + 21kg = 141kg
Reverse percentages – method 1
Original Amount
100% value
EXAMPLES
(a) Mavis paid £43.90 for a skirt that had been reduced by 15% in a sale.
What was the original price?
Decrease of 15%
85% value
÷ 85
x 100
85% = £43.90
1% = £0.5165
100% = £51.65
÷ 85
x 100
Original Price
(b) Junior received a 8% pay rise. His new salary is £19200. What was his
original salary?
Increase of 8%
108% value
÷ 108
x 100
108% = £19200
1% = £177.78
100% = £17778
Original Salary
÷ 108
x 100
Reverse percentages – method 2
Example – Increase
Example – Decrease
The cost of a train ticket from Liverpool to London
has gone up by 25%. It now costs £100.
What was the price before the increase?
Emma bought a pair of trainers in a 20% off sale for
£45.60
Work out the price of the trainers before the sale.
Compound Interest
Worked Example 1
Long Method
£2000 is invested at 6% compound interest for 3 years.
Find: (a) the amount in the account at the end of the period.
and (b) the interest accrued.
Amount after 1 year = 2000 + 6% of 2000 = 2000 + 120 = £2120
Amount after 2 years = 2120 + 6% of 2120 = 2120 + 127.20 = £2247.20
Amount after 3 years = 2247.20 + 6% of 2247.20 = 2247.20 + 134.83 = £2382.03
Interest accrued = £2382.03 – £2000 = £382.03
Long Method
Depreciation – decreases in value
A van is bought for £10,000. It depreciates at
a rate of 20%. Find it’s value after 3 years.
Multiplier
Decrease: 100 - 20= 80%
80 ÷ 100 = 0.8
Year
Initial
Value at end of year
1
2
10,000
8,000
10000 × 0.8 = 8,000
3
6,400
6400 × 0.8 = 5,120
8000 × 0.8 = 6,400
Upper and Lower bound
Easily getting bounds
Value: 250m
To find bounds, just add or subtract half the accuracy!
Correct to: nearest 10m
Lower Bound = 250 – (10/2)
? = 245m
Upper Bound = 250 + (10/2)
? = 255m
Value: 423mm
Correct to: nearest mm
?
?
Mini Exercise
Work out the lower and upper bound for each value given the
accuracy,
7cm
5.3m
200km
8.04mm
3m
830 litres
830 litres
Accuracy
Nearest cm
To 1 dp
Nearest 10km
To 2 dp
Nearest cm
2sf
3sf
Lower Bound
6.5cm
5.25m?
195km?
8.035mm
?
2.995m
?
825 litres
?
829.5 litres
?
Upper Bound
7.5cm
5.35m?
205km?
8.045mm
?
3.005m?
835 litres
?
830.5 litres
?
Upper and Lower bound
A general rule
• To find the upper bound we have to add half of the unit we rounded to for
example:
•
•
•
•
If we rounded to the nearest 10, we add on 5 (half of 10)
If we rounded to the nearest 1 (whole number), we add on 0.5 (half of 1)
If we rounded to the nearest 0.1 (1 dp), we add on 0.05 (half of 0.1)
If we rounded to the nearest 0.01 (2 dp), we add on 0.005 (half of 0.01)
• And if we want the lower bound, we subtract half of the unit
Upper and lower bound
calculations
Maximum (Upper
Bound)
+
x
÷
Minimum (Lower bound)
Upper and Lower bound calculations
Formula
We want
Calculation
If we want a big number, we
choose the biggest for each!
?
?
?
?
?
If we want small, we start
with a small number and
divide by a big.
Graphs – mid point
Graphs – Drawing
Graphs – parallel lines
Graphs –perpendicular lines
Solving
EXAMPLES
Solve the following equations:
(a) 7𝑥 + 9 = 2𝑥 + 19
- 2𝑥
-9
5𝑥
=
÷5
𝑥
+ 4𝑥
- 2𝑥
5𝑥 + 9 =
=
(b) 3𝑥 - 1 = 6 – 4𝑥
+ 4𝑥
19
7𝑥 - 1 = 6
-9
+1
+1
= 7
10
7𝑥
÷5
÷7
÷7
2
𝑥
= 1
Solving fractions
EXAMPLES
Solve each of the following equations:
2
+
(b)
=
2
+
2x3
=
2
+
6
=
2
=
2
3
6
x6
4
x6
5𝑥 - 21 = 12
+ 21
-
7
4x7
28
28
x 28
5
=
5
=
5
=
5
x 28
3𝑥 + 19 = 140
- 19
+ 21
5𝑥
= 33
3𝑥
÷5
÷5
÷3
𝑥
=
𝑥
- 19
= 121
÷3
Factorising Quadratics
s = +8
p = +15
+5
+3
s = -10
p = +24
s = +4
p = - 21
+7
- 3
-6
-4
Factorising Quadratics
•
Factorising Quadratics
•
s = +11
p = 2 x +15
= +30
What two numbers
have a product of +30
and a sum of +11?
Factorising Quadratics
•
s = -33
p = 5 x -14
= -70
What two numbers
have a product of -70
and a sum of -33?
Factorising Quadratics
•
s = -11
p = 6 x +3
= +18
What two numbers
have a product of +18
and a sum of -11?
Simplifying – Algebraic fractions
Simplify
Simplifying – Algebraic fractions
Simplify
Compound Measures: Speed/Distance/Time
1. A car travels 126 miles in 3
hours. Find its average speed.
d
s=
t
126
s=
3
= 42 mph
Average speed 1
d
s=
t
d
t=
s
d = st
Speed – Distance - Time
(when time is NOT a whole number)
Formula Triangle
M
D
V
Examples
1) A piece of metal weighs 40g
has a volume of 4cm3. What
is it’s density?
10 g/cm3
2) What is the volume of a
piece of rock that has a mass
of 72kg and a density of
9kg/m3?
8m3
Pressure
• If my car pushes on the ground with a force 150N
over an area of 2m2, what is it’s pressure?
How do we get he ‘P’ by itself?What we want
to find out
P=F/A
F
PxA
P = 150 / 2
P = 75N/m2
Pressure
•
What we want
to find out
F=PxA
F = 100 x 1.2
F = 120N
CIRCUMFERENCE OF CIRCLES
diameter
EXAMPLES
(a)Find the circumference of the circle:
8cm
(b)Find the circumference of the circle:
11m
radius
Circumference of a circle = 3.14 x 8
Circumference of a circle = 25.1cm (1d.p.)
Circumference of a circle = 3.14 x 22
Circumference of a circle = 69.1m(1d.p.)
Creative
Thinker
Effective
Participator
Independent
Enquirer
Reflective
Learner
PLT SkillsAREA OF CIRCLES AND SEMI-CIRCLES
Self Manager
Which ones are you
using?
radius
EXAMPLES
(a)Find the area of the circle:
(b) Find the area of the circle:
4cm
14m
diameter
Area of circle = 3.14 x 4 x 4
Area of circle = 50.2cm 2 (1d.p.)
Area of circle = 3.14 x 7 x 7
Area of circle = 153.9 m2 (1d.p.)
(c)Find the area of the semi-circle:
Area of semi-circle =
Area of semi-circle =
20cm
2
2
Area of semi-circle = 157cm 2
Team
Worker
Using Pythagoras’ Theorem
We can use Pythagoras’ Theorem to find
the longest side in a right –angled
triangle
Find the Length of side x
1. Square
2.
3.
Add
Square
Root
130
92 = 81
72 = 49
x2 = 130
x= √
x = 11.4cm
Example
1
7cm
x
9cm
Level 7
Using Pythagoras’ Theorem
We can use Pythagoras’ Theorem to find
a Short side in a right –angled triangle
Find the Length of side x
1. Square
122 = 144
72 = 49
2.
Subtract x2 = 95
3.
Square
Root
x = √ 95
x = 9.7cm
Example
3
x
12cm
7cm
Level 7
Creative
Thinker
Effective
Participator
Independent
Enquirer
Reflective
Learner
Self Manager
Team
Worker
Which ones are you
PLT Skills
TRIGONOMETRY
using?
INTRODUCTION – LABELLING SIDES CORRECTLY
This is opposite the
right-angle.
This is opposite
the angle.
This is next to the angle.
Creative
Thinker
Effective
Participator
Independent
Enquirer
Reflective
Learner
Self Manager
Which ones are you
using?
PLT Skills
TRIGONOMETRY
INTRODUCTION
Some Old Hag Cracked All Her Teeth On Asparagus
SOH CAH TOA
o
s
H
Team
Worker
A
o
C H
T A
Creative
Thinker
Effective
Participator
PLT Skills
EXAMPLES
Independent
Enquirer
Reflective
Learner
Which ones are you
using?
TRIGONOMETRY
A
1) Work out the missing angles below:
(a)
H
O
(b)
A
Team
Worker
Self Manager
H
A
(c)
O
H
O
SUBSTITUTE
WRITELABEL
DOWNIN
THE
THE
THE
SIDES
FORMULA
VALUES SUBSTITUTE
WRITELABEL
DOWNIN
THE
THE
THE
SIDES
FORMULA
VALUES SUBSTITUTE
WRITELABEL
DOWNIN
THE
THE
THE
SIDES
FORMULA
VALUES
4
3
10
sin θ =
cos θ =
tan θ =
9
11
6
3
10
-1 4
θ = 74.2°(3s.f) θ = tan -1
θ = 59.0°(3s.f)
θ = sin 9 θ = 26.4°(3s.f) θ = cos-1
11
6
2) Work out the missing lengths below:
(a)
(b)
H
H
O
(c)
O
H
A
A
WRITE
DOWNTHE
THE
FORMULA
LABEL
SIDES
SUBSTITUTE
IN
THE
VALUES
b
sin 64 =
9
O
A
WRITE
DOWNTHE
THE
FORMULA
LABEL
SIDES
SUBSTITUTE
IN
THE
VALUES SUBSTITUTE
WRITE
DOWNTHE
THE
FORMULA
LABEL
SIDES
IN
THE
VALUES
b
cos 33 =
14
4
b = 9 x sin 64 b = 8.09cm(3s.f) b = 14 x cos 33 b = 11.7cm(3s.f) b =
b=
tan 46
Volume and Surface area
Questionnaire design
Questionnaires should:
1.
2.
3.
4.
5.
6.
7.
use simple language;
ask short questions which can be answered precisely;
provide tick boxes;
avoid open-ended questions;
avoid leading questions.
Where applicable provide a timeframe
No overlapping option boxes
STEM AND LEAF DIAGRAMS
EXAMPLES
1)The speeds of vehicles in 30mph zone
are recorded. Below are the results:
39,42, 35, 29,28, 42, 31, 33
32, 35, 35, 27, 37, 23, 29, 27, 45
(a) Draw a stem and leaf diagram to
Smallest = 23
represent this data.
Biggest = 45
2 3778 99
3 12 35 5 5 7 9
4 225
Key: 3I5 means 35
(b) Work out the mode Most common
35mph
(c) Work out the median
33mph
(d) Work out the range
Biggest - Smallest
45 - 23
22
(e) What fraction were over the 30mph limit?
Eleven were more
than 30mph
11
17
Estimated Mean
median