GCSE Mathematics AQA YEAR 10
Transcript of GCSE Mathematics AQA YEAR 10
NAME:_____________________________
TUTOR GROUP:______________________
Knowledge Organiser
MATHEMATICS GCSE Mathematics AQA
MA
THEM
ATI
CS
GC
SE M
ath
em
atic
s A
QA
YEAR 10
(Spring) 2021-2023
Topic 10 2.6 Perimeter, Area and Volume I
Topic/Skill Definition/Tips Example Your turn
1 Metric System
A system of measures based on:
- the metre for length
- the kilogram for mass
- the second for time
Length: mm, cm, m, km
Mass: mg, g, kg
Volume: ml, cl, l
1 𝑘𝑖𝑙𝑜𝑚𝑒𝑡𝑟𝑒 = 1000 𝑚𝑒𝑡𝑟𝑒𝑠 1 𝑚𝑒𝑡𝑟𝑒 = 100 𝑐𝑒𝑛𝑡𝑖𝑚𝑒𝑡𝑟𝑒𝑠
1 𝑐𝑒𝑛𝑡𝑖𝑚𝑒𝑡𝑟𝑒 = 10 𝑚𝑖𝑙𝑙𝑖𝑚𝑒𝑡𝑟𝑒𝑠
1 𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚 = 1000 𝑔𝑟𝑎𝑚𝑠
3 km = _______ m
7.53 km = ____ m
4 m = ______ cm
6.7m = _____ cm
5cm = _____ mm
10.7cm = ___ mm
6kg = _______ g
0.9kg = _____ g
4000m = ____ km
6475m = ____ km
900cm = _____ m
550cm = _____ m
80mm = ____ cm
47mm = ____ cm
8000g = _____ kg
1700g = _____ kg
2 Imperial System
A system of weights and measures originally
developed in England, usually based on human
quantities.
Length: inch, foot, yard, miles
Mass: lb, ounce, stone
Volume: pint, gallon
1𝑙𝑏 = 16 𝑜𝑢𝑛𝑐𝑒𝑠 (oz) 1 𝑓𝑜𝑜𝑡 = 12 𝑖𝑛𝑐ℎ𝑒𝑠 1 𝑔𝑎𝑙𝑙𝑜𝑛 = 8 𝑝𝑖𝑛𝑡𝑠
3lb = _____ oz
1 ½ lb = _____ oz
5lb4oz = ____ oz
3ft = ____ inches
5ft 6 inches
=_____ inches
4gallons =__ pints
80oz = _____ lb
56oz = _____ lb
73oz = __ lb __ oz
60 inches = ___ft
76 inches =
__ft __ inches
48pints=__ gallons
3 Metric and
Imperial Units
Use the unitary method to convert between
metric and imperial units.
See 10 2.8 29.
5 𝑚𝑖𝑙𝑒𝑠 ≈ 8 𝑘𝑖𝑙𝑜𝑚𝑒𝑡𝑟𝑒𝑠 1 𝑔𝑎𝑙𝑙𝑜𝑛 ≈ 4.5 𝑙𝑖𝑡𝑟𝑒𝑠
2.2 𝑝𝑜𝑢𝑛𝑑𝑠 ≈ 1 𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚 1 𝑖𝑛𝑐ℎ = 2.5 𝑐𝑒𝑛𝑡𝑖𝑚𝑒𝑡𝑟𝑒𝑠
10 miles ≈ _____ kilometres
25 miles ≈ _____ kilometres
48 kilometres ≈ _____ miles
68 kilometres ≈ _____ miles
3 gallons ≈ ______ litres
18 litres ≈______ gallons
5 kilograms ≈ _____ pounds
33 pounds ≈ ______ kilograms
4 inches ≈ _____ cm
15 cm ≈ _____ inches
4 Conversion Graph
A line graph to convert one unit to another.
Can be used to convert units (e.g. miles and
kilometres) or currencies ($ and £).
Find the value you know on one axis, read
up/across to the conversion line and read the
equivalent value from the other axis.
8 𝑘𝑚 = 5 𝑚𝑖𝑙𝑒𝑠
Use the conversion graph to complete the
following:
a) 10 miles = ________ km
b) 12 km = ________ miles
c) 130 miles = ______ km
d) 180 km = _____ miles
Draw a conversion graph to convert pounds
sterling (£) into Turkish Lira. Use the exchange
rate £1 = 12 Lira
Use your graph to answer the following:
a) £7 =_____Lira
b) £30 = _______ Lira
c) 66 Lira = £_____
d) 102 Lira = £_____
5 Perimeter
The total distance around the outside of a shape.
Units include: 𝑚𝑚, 𝑐𝑚, 𝑚 etc.
𝑃 = 8 + 5 + 8 + 5 = 26𝑐𝑚
Work out the perimeter of (lengths are all
given in cm):
6 Area
The amount of space inside a shape.
Units include: 𝑚𝑚2, 𝑐𝑚2, 𝑚2
These shapes are drawn on centimeter
squared paper.
Find the area of:
7 Area of a
Rectangle Length x Width
𝐴 = 36𝑐𝑚2
Work out the area of:
8 Area of a
Parallelogram
Base x Perpendicular Height
Not the slant height.
𝐴 = 21𝑐𝑚2
Work out the area of:
9 Area of a Triangle Base x Perpendicular height ÷ 2
𝐴 = 24𝑐𝑚2
Work out the area of:
10 Area of a Kite
Split in to two triangles and use the method
above.
Or
Area of a kite =1
2× diagonal 1 length
× diagonal 2 length 𝐴 =1
2× 8 × 2.2 = 8.8𝑚2
Work out the area of:
11 Area of a
Trapezium
(𝒂 + 𝒃)
𝟐× 𝒉
“Half the sum of the parallel side, times the height
between them. That is how you calculate the area
of a trapezium” 𝐴 = 55𝑐𝑚2
Work out the area of:
12 Compound Shape A shape made up of a combination of other
known shapes put together.
Work out the area of:
13 Area of a Triangle
Use when given the length of two sides and the
included angle.
𝑨𝒓𝒆𝒂 𝒐𝒇 𝒂 𝑻𝒓𝒊𝒂𝒏𝒈𝒍𝒆 =𝟏
𝟐𝒂𝒃 𝐬𝐢𝐧 𝑪
Lengths are all in centimetres.
𝐴 =1
2𝑎𝑏 sin 𝐶
𝐴 =1
2× 7 × 10 × sin 25 = 14.8 cm2
Work out the area of:
All lengths are given in centimetres.
Give your answer rounded to 1 decimal place.
14 Parts of a Circle
Radius – the distance from the centre of a circle
to the edge.
Diameter – the total distance across the width
of a circle through the centre.
Circumference – the total distance around the
outside of a circle.
Chord – a straight line whose end points lie on
a circle.
Tangent – a straight line which touches a circle
at exactly one point.
Arc – a part of the circumference of a circle.
Sector – the region of a circle enclosed by two
radii and their intercepted arc.
Segment – the region bounded by a chord and
the arc created by the chord.
Label the parts of a circle on the diagrams below.
15 Area of a Circle 𝑨 = 𝝅𝒓𝟐 which means ‘pi x radius squared’. If the radius is 5cm, then:
𝐴 = 𝜋 × 52 = 25𝜋 = 78.5𝑐𝑚2
Find:
a) The area of a circle with radius 7cm.
b) The area of a circle with diameter 22cm.
For each one, give your answer:
i) In terms of π.
ii) Correct to 2 decimal places.
16 Circumference of
a Circle
𝑪 = 𝝅𝒅 which means ‘pi x diameter’.
𝑪 = 𝟐𝝅𝒓 which means ‘2 x pi x radius’.
If the diameter is 8cm, then: 𝐶 = 𝜋 × 8 = 8𝜋
= 25.1𝑐𝑚
If the radius is 5cm, then: 𝐶 = 2 × 𝜋 × 5 = 10 𝜋
= 31.4𝑐𝑚
Find:
a) The circumference of a circle with diameter
22cm.
b) The circumference of a circle with radius 7cm.
For each one, give your answer:
i) In terms of π
ii) Correct to 2 decimal places
17 𝜋 (‘pi’)
Pi is the circumference of a circle divided by the
diameter.
𝝅 ≈ 𝟑. 𝟏𝟒
Use your calculator to work out the following.
Write down all of the decimal places shown on your
calculator.
a) 3π
b) 7π + 10
18
H
Arc Length of a
Sector
The arc length is part of the circumference.
Take the angle given as a fraction over 360° and
multiply by the circumference.
Arc Length = 115
360× 𝜋 × 8 = 8.03𝑐𝑚
Find the arc length for:
19
H Area of a Sector
The area of a sector is part of the total area.
Take the angle given as a fraction over 360° and
multiply by the area.
Major sector is the larger sector.
Area = 115
360× 𝜋 × 42 = 16.1𝑐𝑚2
20 Volume
Volume is a measure of the amount of space
inside a solid shape.
Units: 𝑚𝑚3, 𝑐𝑚3, 𝑚3 etc.
For the units below tick to indicate if it is a unit of
length, a unit of area or a unit of volume.
Length Area Volume
cm2
mm
km3
inches
21 Volume of a
Cube/Cuboid
𝑽 = 𝑳𝒆𝒏𝒈𝒕𝒉 × 𝑾𝒊𝒅𝒕𝒉 × 𝑯𝒆𝒊𝒈𝒉𝒕 𝑽 = 𝑳 × 𝑾 × 𝑯
You can also use the Volume of a Prism formula
for a cube/cuboid.
a) Find the volume of a cube with side length 7cm.
b) Find the volume of this cuboid:
22 Volume of a
Prism
𝑽 = 𝑨𝒓𝒆𝒂 𝒐𝒇 𝑪𝒓𝒐𝒔𝒔 𝑺𝒆𝒄𝒕𝒊𝒐𝒏 × 𝑳𝒆𝒏𝒈𝒕𝒉 𝑽 = 𝑨 × 𝑳
Find the volume of the triangular prism:
23 Volume of a
Cylinder 𝑽 = 𝝅𝒓𝟐𝒉
Find the volume of the cylinder:
Topic 10 2.7 Equations and Inequalities
Topic/Skill Definition/Tips Example Your turn
1 Solve
To find the answer/value of something.
Use inverse operations on both sides of the
equation (balancing method) until you find the
value for the letter.
Solve 2𝑥 − 3 = 7
Add 3 on both sides. 2𝑥 = 10
Divide by 2 on both sides. 𝑥 = 5
Solve:
a) 4𝑥 + 3 = 27
b) 6𝑥 − 5 = 37
c) 40 = 4𝑥 − 8
d) 3(2𝑥 + 4) = 42
e) 5𝑥 + 3 = 3𝑥 + 11
f) 7𝑥 − 4 = 2𝑥 + 6
2 Inverse Opposite. The inverse of addition is subtraction.
The inverse of multiplication is division.
The inverse of subtraction is ________
The inverse of division is ____________
The inverse of squaring is ___________
3 Substitution
Replace letters with numbers.
Be careful of 5𝑥2. You need to square first, then
multiply by 5.
𝑎 = 3, 𝑏 = 2 𝑎𝑛𝑑 𝑐 = 5. Find:
1) 2𝑎 = 2 × 3 = 6
2) 3𝑎 − 2𝑏 = 3 × 3 − 2 × 2 = 5
3) 7𝑏2 − 5 = 7 × 22 − 5 = 23
If 𝑎 = 4, 𝑏 = 5 and 𝑐 = −3. Find:
a) 3𝑎
b) 4𝑎 − 3𝑏
c) 2𝑏2 − 7
d) 𝑎 − 2𝑐
e) 3𝑐2 + 5
4 Simultaneous
Equations
A set of two or more equations, each involving
two or more variables (letters).
The solutions to simultaneous equations satisfy
both/all of the equations.
2𝑥 + 𝑦 = 7 3𝑥 − 𝑦 = 8
𝑥 = 3 𝑦 = 1
Solve simultaneously:
a) 4𝑥 + 𝑦 = 14 2𝑥 − 𝑦 = 4
b) 2𝑥 + 3𝑦 = 22 2𝑥 + 𝑦 = 10
c) 5𝑥 + 4𝑦 = 31 5𝑥 − 2𝑦 = 7
5
Solving
Simultaneous
Equations (by
Elimination)
1) Balance the coefficients of one of the variables.
2) Eliminate this variable by adding or subtracting
the equations (Same Sign Subtract, Different Sign
Add).
3) Solve the linear equation you get using the other
variable.
4) Substitute the value you found back into one of
the previous equations.
5) Solve the equation you get.
6) Check that the two values you get satisfy both of
the original equations.
5𝑥 + 2𝑦 = 9 10𝑥 + 3𝑦 = 16
Multiply the first equation by 2. 10𝑥 + 4𝑦 = 18 10𝑥 + 3𝑦 = 16
Same Sign Subtract (+10x on both). 𝑦 = 2
Substitute 𝑦 = 2 into equation. 5𝑥 + 2 × 2 = 9
5𝑥 + 4 = 9 5𝑥 = 5 𝑥 = 1
Solution: 𝑥 = 1, 𝑦 = 2
Solve simultaneously:
a) 2𝑥 + 3𝑦 = 7 𝑥 + 𝑦 = 2
b) 4𝑥 + 𝑦 = 26 3𝑥 + 2𝑦 = 17
c) 3𝑥 − 𝑦 = 11 2𝑥 + 3𝑦 = 0
d) 5𝑥 + 3𝑦 = 49 2𝑥 − 4𝑦 = 4
e) 6𝑥 − 𝑦 = −1 2𝑥 + 3𝑦 = 13
6
Solving
Simultaneous
Equations (by
Substitution)
1) Rearrange one of the equations into the form
𝑦 = ⋯ or 𝑥 = ⋯
2) Substitute the right-hand side of the rearranged
equation into the other equation.
3) Expand and solve this equation.
4) Substitute the value into the 𝑦 = ⋯ or 𝑥 = ⋯
equation.
5) Check that the two values you get satisfy both
of the original equations.
𝑦 − 2𝑥 = 3 3𝑥 + 4𝑦 = 1
Rearrange: 𝑦 − 2𝑥 = 3 → 𝑦 = 2𝑥 + 3
Substitute: 3𝑥 + 4(2𝑥 + 3) = 1
Solve: 3𝑥 + 8𝑥 + 12 = 1 11𝑥 = −11
𝑥 = −1
Substitute: 𝑦 = 2 × −1 + 3 𝑦 = 1
Solution: 𝑥 = −1, 𝑦 = 1
Solve simultaneously:
a) 𝑦 − 3𝑥 = 5 5𝑥 + 2𝑦 = 21
b) 𝑥 + 𝑦 = 4 4𝑥 + 2𝑦 = 18
c) 3𝑥 + 𝑦 = 10 7𝑥 + 4𝑦 = 30
d) 2𝑥 + 𝑦 = 12 6𝑥 − 3𝑦 = 18
7 Not equal to
The symbol ≠ means that two values are not
equal.
𝑎 ≠ 𝑏 means that a is not equal to b.
7 ≠ 3
𝑥 ≠ 0
8 Inequalities and
inequality symbols
𝒙 > 𝟐 means x is greater than 2
𝒙 < 𝟑 means x is less than 3
𝒙 ≥ 𝟏 means x is greater than or equal to 1
𝒙 ≤ 𝟔 means x is less than or equal to 6
State the integers that satisfy −2 < 𝑥 ≤ 4.
-1, 0, 1, 2, 3, 4
State the integers that satisfy:
a) −3 < 𝑥 < 4
b) 0 ≤ 𝑥 < 6
c) −4 ≤ 𝑥 < −1
d) −1 < 𝑥 ≤ 3
Write in words:
e) 𝑥 < 7
f) 𝑥 ≥ −3
g) 2 ≤ 𝑥
9 Inequalities on a
Number Line
Inequalities can be shown on a number line.
Open circles are used for numbers that are less
than or greater than (< 𝑜𝑟 >)
Closed circles are used for numbers that are less
than or equal or greater than or equal (≤ 𝑜𝑟 ≥)
𝑥 ≥ 0
For each of the following inequalities,
i) draw a number line,
ii) represent the inequality on the number
line.
a) 𝑥 > 3
b) −2 ≤ 𝑥 < 5
c) −5 < 𝑥 ≤ 3
d) −4 < 𝑥 < −1
Write down the inequality represented on
each of the number lines below:
𝑥 < 2
−5 ≤ 𝑥 < 4
10
H Graphical
Inequalities
Inequalities can be represented on a coordinate
grid.
If the inequality is strict (𝑥 > 2) then use a dotted
line.
If the inequality is not strict (𝑥 ≤ 6) then use a
solid line.
Shade the region which satisfies all the
inequalities.
Shade the region that satisfies: 𝑦 > 2𝑥, 𝑥 > 1 𝑎𝑛𝑑 𝑦 ≤ 3
Label with an R the region that satisfies:
𝑦 ≥ 3𝑥, 𝑥 > −2 and 𝑦 ≤ 7
Write down the inequalities shown on the
graph below:
Topic 10 2.8 Calculations
Topic/Skill Definition/Tips Example Your turn
1 Numerator
Denominator
The top number of a fraction.
The bottom number of a fraction.
In the fraction 3
5, 3 is the numerator.
5 is the denominator.
In the fraction 4
9
a) What is the numerator?
b) What is the denominator?
c) Write down a fraction with 4 in the
numerator and 7 in the denominator.
2 Unit Fraction
A fraction where the numerator is one and the
denominator is a positive integer.
1
2,
1
3,
1
4 𝑒𝑡𝑐. are examples of unit fractions.
Which of the following are unit fractions? 2
3
1
8
11
12
1
12
1
10
3 Reciprocal
The reciprocal of a number is 1 divided by the
number.
The reciprocal of 𝑥 is 1
𝑥
When we multiply a number by its reciprocal
we get 1. This is called the ‘multiplicative inverse’.
The reciprocal of 5 is 1
5
The reciprocal of 2
3 is
3
2, because
2
3×
3
2= 1
Write down the reciprocal of:
a) 4
b) 7
c) 4
5
d) 8
9
4 Mixed Number A number formed of both an integer part and a
fraction part. 3
2
5 is an example of a mixed number.
Put a ring around all of the mixed numbers
below.
41
5
54
8 6
2
3 3
39
4 5
7
9
5 Simplifying
Fractions
Divide the numerator and denominator by the
highest common factor.
20
45=
4
9
Simplify the fractions below:
a) 12
15
b) 36
48
c) 45
72
6 Equivalent
Fractions Fractions which represent the same value.
2
5=
4
10=
20
50=
60
150 𝑒𝑡𝑐.
Complete the blanks to give equivalent
fractions: 3
4=
6=
40=
12=
12
7 Comparing
Fractions
To compare fractions, they each need to be
rewritten so that they have a common
denominator.
Ascending means smallest to biggest.
Descending means biggest to smallest.
Put in ascending order: 3
4
2
3
5
6
1
2
Common denominator: 9
12
8
12
10
12
6
12
Correct order: 1
2
2
3
3
4
5
6
Put the following fractions into ascending
order: 3
4
5
8
2
3
5
6
7
12
Put the following fractions into descending
order: 4
9
1
3
5
6
1
2
3
4
7
12
8 Fraction of an
Amount Divide by the bottom, times by the top.
Find 2
5 of £60.
60 ÷ 5 = 12 12 × 2 = 24
a) Find 3
4 of 600kg
b) Find 5
9 of $270
c) Find 7
12 of 156m
9 Adding or
Subtracting
Fractions
Find the LCM of the denominators to find a
common denominator.
Use equivalent fractions to change each fraction to
the common denominator.
Then just add or subtract the numerators and
keep the denominator the same.
2
3+
4
5
Multiples of 3: 3, 6, 9, 12, 15..
Multiples of 5: 5, 10, 15..
LCM of 3 and 5 = 15 2
3=
10
15
4
5=
12
15
10
15+
12
15=
22
15= 1
7
15
Work out:
a) 1
2+
3
8
b) 1
3+
2
5
c) 3
7+
1
4
d) 4
5+
3
8 (Give your answer as a mixed number.)
10 Multiplying
Fractions
Multiply the numerators together and multiply
the denominators together.
3
8×
2
9=
6
72=
1
12
Work out each of the following, give your
answer in its simplest form.
a) 1
4×
3
5
b) 2
7×
3
4
11 Dividing Fractions
‘Keep it, Flip it, Change it – KFC’
Keep the first fraction the same
Flip the second fraction upside down
Change the divide to a multiply
Multiply by the reciprocal of the second fraction.
3
4÷
5
6=
3
4×
6
5=
18
20=
9
10
Work out each of the following, give your
answer in its simplest form.
a) 3
5÷
5
8
b) 12
3÷
11
12
c) 23
4÷ 3
1
5
12 Speed, Distance,
Time
Speed = Distance ÷ Time
Distance = Speed x Time
Time = Distance ÷ Speed
Remember the correct units.
Speed = 4mph
Time = 2 hours
Find the Distance.
𝐷 = 𝑆 × 𝑇 = 4 × 2 = 8 𝑚𝑖𝑙𝑒𝑠
Work out the missing values:
Speed Distance Time
1 35mph 280miles
2 50mph 4.5hours
3 210km 3.5hours
4 12m/s 270m
5 50km/h 45 mins
6 4km/h 2400m
13 Density, Mass,
Volume
Density = Mass ÷ Volume
Mass = Density x Volume
Volume = Mass ÷ Density
Remember the correct units.
Density = 80kg/m³
Mass = 2000g
Find the volume.
𝑉 = 𝑀 ÷ 𝐷 = 2 ÷ 80 = 0.025𝑚³
Work out the missing values:
Mass Volume Density
1 500kg 4m3
2 330kg 0.75m3
3 50g 1.25g/cm3
4 190g 50cm3
5 1.5kg 2.5g/cm3
6 0.2m3 2.4g/cm3
14 Pressure, Force,
Area
Pressure = Force ÷ Area
Force = Pressure x Area
Area = Force ÷ Pressure
Remember the correct units.
Pressure = 10 N/m2
Area = 6m²
Find the Force.
𝐹 = 𝑃 × 𝐴 = 10 × 6 = 60 𝑁
Work out the missing values:
Force Pressure Area
1 20 N/m2 7.5 m2
2 80 N 2 m2
3 60 N 40 N/m2
4 50 N/m2 0.3 m2
5 125 N 2.5 m2
6 75 N 50 N/m2
15 Finding 10% To find 10%, divide by 10. 10% of £36 = 36÷10=£3.60
a) Find:
i) 10% of £230
ii) 10% of 34kg
b) 10% of a distance is 45m. What is the whole
distance?
Hint: write as an
improper fraction.
16 Finding 1% To find 1%, divide by 100 1% of £8 = 8÷100 = £0.08
a) Find:
i) 1% of $450
ii) 1% of 25m
b) 1% of an amount is £2.30. What is the total
amount?
17 Percentage
Change
𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒄𝒆
𝑶𝒓𝒊𝒈𝒊𝒏𝒂𝒍× 𝟏𝟎𝟎%
A games console is bought for £200 and sold
for £250.
% change = 50
200× 100 = 25%
a) The cost of a can of drink is increased from
50p to 60p. What percentage increase is this?
b) A dress is reduced in price in a sale from £45
to £36. What percentage reduction is this?
18 Increase or
Decrease by a
Percentage
Non-calculator: Find the percentage and add or
subtract it from the original amount.
Calculator: Find the percentage multiplier and
multiply.
Increase 500 by 20% (Non Calc):
10% of 500 = 50
so 20% of 500 = 100
500 + 100 = 600
Decrease 800 by 17% (Calc):
100%-17%=83%
83% ÷ 100 = 0.83
0.83 x 800 = 664
Non-calculator
a) Increase £360 by 20%
b) Increase 120kg by 60%
c) Decrease $340 by 30%
d) Increase 86kg by 5%
e) Decrease £700 by 35%
Calculator
f) Increase £140 by 23%
g) Increase 84kg by 7.5%
h) Decrease $340 by 24%
i) Decrease 376m by 12.5%
19 Percentage
Multiplier
The number you multiply a quantity by to
increase or decrease it by a percentage.
Increase by 12% 100% + 12%=1.12
Decrease by 12 100% - 12%=0.88
The multiplier for increasing by 12% is 1.12
The multiplier for decreasing by 12% is 0.88
The multiplier for increasing by 100% is 2
Write down the multipliers for:
a) Increasing by 26%
b) Increasing by 7%
c) Increasing by 150%
d) Decreasing by 32%
e) Decreasing by 6%
f) Decreasing by 7.5%
20 Reverse
Percentage
Find the correct percentage given in the
question, then work backwards to find 100%.
Look out for words like ‘before’ or ‘original’.
A jumper was priced at £48.60 after a 10%
reduction. Find its original price.
100% - 10% = 90%
90% = £48.60
1% = £0.54
100% = £54
a) A coat costs £49.50 after a 10% reduction in
its price. Find its original price.
b) Mark buys a car. A year later he sells it for
£3060 which is a 15% loss. Find its original
price.
c) Tina receives a 5% pay rise. Her new pay is
£24,150 per year. What was her previous pay?
d) After a 14% increase in ticket prices, a train
ticket costs £22.23. What was the original ticket
price?
21 Simple Interest Interest calculated as a percentage of the original
amount.
£1000 invested for 3 years at 10% simple
interest.
10% of £1000 = £100
Interest = 3 × £100 = £300
a) Find the interest on £2000 invested for 4
years at 5% simple interest.
b) Find the interest on £1450 invested for 54
months at 4% per year simple interest.
c) Find the total amount of money after £980 is
invested at 6% simple interest for 2 years.
d) Find the total amount of money after a $600
is invested at 2.4% simple interest for 3 years.
22 Compound
Interest
Interest paid on the original amount and the
accumulated interest.
See 19 for help with percentage multipliers.
A bank pays 5% compound interest a year. Bob
invests £3000. How much will he have after 7
years.
3000 × 1.057 = £4221.30
A car depreciates in value by 10% each year.
It’s original value was £18000. Find the value
after 5 years.
18000 × 0.95 = £10628.82
a) A bank pays 2% compound interest per year.
Ben invests £750. How much will he have after
6 years?
b) A bank pays 3.5% compound interest per
year. Mel invests £600. How much will she have
after 4 years?
c) A car depreciates in value by 5% each year.
It’s original value was £9000. Find the value
after 7 years.
d) A lake is decreasing in area by 3.2% per year.
The current area is 9.2km2. What would the
area of the lake be expected to be in 10 years?
23 Ratio
Ratio compares the size of one part to another
part.
Written using the ‘:’ symbol.
Write down the ratio shown below:
Grey : white is _______ : ________
White : grey is _______ : ________
Grey : white is _______ : ________
White : grey is _______ : ________
Grey : white : striped is ____ : ____ : ____
24 Proportion
Proportion compares the size of one part to the
size of the whole.
Usually written as a fraction.
In a class with 13 boys and 9 girls, the
proportion of boys is 13
22 and the proportion of
girls is 9
22 .
a) A class contains 11 boys and 15 girls. Write
down:
i) the proportion of boys
ii) the proportion of girls
b) Green paint is made by mixing blue paint
and yellow paint. Rachel mixes some green
paint using 3
7 blue paint. She uses 210ml of
blue paint. How much yellow paint does she
use?
25 Simplifying Ratios Divide all parts of the ratio by a common factor. 5 : 10 = 1 : 2 (Divide both by 5.)
14 : 21 = 2 : 3 (Divide both by 7.)
Simplify these ratios:
a) 8:10
b) 12:16
c) 20:25
d) 36:42
e) 63:84
f) 6:12:9
g) 8:16:28
h) 21:14:42
26 Ratios in the form 1 ∶ 𝑛
or 𝑛 ∶ 1
Divide both parts of the ratio by one of the
numbers to make one part equal 1.
5 : 7 = 1 : 7
5 in the form 1 : n
5 : 7 = 5
7 : 1 in the form n : 1
Write each of these in
the form 1:n
a) 4:12
b) 3:7
c) 8:3
d) 9:3
Write each of these in
the form n:1
e) 8:4
f) 20:15
g) 7:9
h) 8:15
27 Sharing in a Ratio
1) Add the total parts of the ratio.
2) Divide the amount to be shared by this value to
find the value of one part.
3) Multiply this value by each part of the ratio.
Use only if you know the total.
Share £60 in the ratio 3 : 2 : 1.
3 + 2 + 1 = 6
60 ÷ 6 = 10
3 x 10 = 30, 2 x 10 = 20, 1 x 10 = 10
£30 : £20 : £10
a) Share £240 in the ratio 5:7.
b) Share $800 in the ratio 3:7:6.
c) Orange paint is made by mixing red paint
and yellow paint in the ratio 3:5. Kate wants to
make 4 litres of orange paint.
i) How much red paint should she use?
ii) How much yellow paint should she use?
28 Proportional
Reasoning
Comparing two things using multiplicative
reasoning and applying this to a new situation.
Identify one multiplicative link and use this to find
missing quantities.
a) It takes Claire 20 minutes to type 1 page of
notes.
i) How long would it take her to type 5
pages of notes?
ii)How many pages of notes could she type
in 80 minutes?
b) Fruit squash is made by diluting 30ml of
concentrate with water to make 180ml of drink.
i) How much concentrate is needed to
make 900ml of drink?
ii) How much of the drink can be made
using 105ml of concentrate?
29 Unitary Method
Finding the value of a single unit and then finding
the necessary value by multiplying the single unit
value.
3 cakes require 450g of sugar to make. Find
how much sugar is needed to make 5 cakes.
3 cakes = 450g
So 1 cake = 150g (÷ by 3)
So 5 cakes = 750 g (x by 5)
2 cakes require 500g of flour to make.
a) How much flour would you need to make 5
cakes?
b) 15 litres of petrol costs £18.90. How much
would 23 litres of petrol cost?
30 Currency
conversions
Find the value of units of money from different
countries using an exchange rate.
Foreign currency
= British Currency x Exchange rate
British Currency =
= Foreign Currency ÷ Exchange rate
£1 = €1.18 (Euros)
To convert £150 into Euros multiply 150 by
the exchange rate.
150 × 1.18 = €177
To convert €200 into Pounds divide 200 by
the exchange rate. 200 ÷ 1.18 = £169.49
For the questions below use these exchange
rates:
£1 = $1.38 (US dollars)
£1 = 5.26 zł (Polish Zloty)
£1 = 8.68 kr (Danish Krona)
a) Convert £15 into US dollars.
b) Convert £23 into Polish Zloty.
c) Convert £45 into Danish Krona.
d) Convert $38 into pounds.
e) Convert 120zł into pounds.
f) Convert 230kr into pounds.
g) Which amount is greatest?
Show how you know.
$46.92 £36 199.88zł 277.76kr
31 Ratio already
shared
Find what one part of the ratio is worth using the
unitary method.
Money was shared in the ratio 3:2:5 between
Ann, Bob and Cat. Given that Bob had £16,
found out the total amount of money shared.
£16 = 2 parts
So £8 = 1 part
3 + 2 + 5 = 10 parts, so 8 x 10 = £80
a) Ann, Ben and Carl shared some sweets in
the ratio 4:7:2. Ben received 84 sweets. How
many sweets were there altogether?
b) Dan, Erica and Fran share some money in
the ratio 3:5:6. Erica receives £16 more than
Dan. How much money was shared?
32 Best Buys
Find the unit cost by dividing the price by the
quantity.
The lowest number is the best value.
8 cakes for £1.28 → 16p each (÷by 8)
13 cakes for £2.05 → 15.8p each (÷by 13)
Pack of 13 cakes is best value.
Which pack of lightbulbs is best value?
Which SD card is best value per GB?
33 Rounding
To make a number simpler but keep its value close
to what it was.
If the digit to the right of the rounding digit is less
than 5, round down.
If the digit to the right of the rounding digit is 5 or
more, round up.
74 rounded to the nearest ten is 70, because
74 is closer to 70 than 80.
152,879 rounded to the nearest thousand is
153,000.
Fill in the blanks below with the correct
rounded values.
153 = _______ (nearest 10)
207 = _______ (nearest 10)
195 = _______ (nearest 10)
372 = _______ (nearest 100)
538 = _______ (nearest 100)
3257 = _______ (nearest 100)
2519 = _______ (nearest 1000)
4397 = _______ (nearest 1000)
16.3 = _______ (nearest whole)
23.5 = _______ (nearest whole)
12.07 = _______ (nearest whole)
34 Decimal Place The position of a digit to the right of a decimal
point.
In the number 0.372, the 7 is in the second
decimal place.
0.372 rounded to two decimal places is 0.37,
because the 2 tells us to round down.
Careful with money - don’t write £27.4, instead
write £27.40
For each of the numbers below indicate the
which decimal place the 4 is in.
a) 0.41
b) 32.54
c) 3.034
d) 7.2014
Fill in the blanks below with the correct
rounded values.
e) 2.53 = _______ (1 decimal place)
f) 4.27 = _______ (1 decimal place)
g) 5.328 = _______ (1 decimal place)
h) 7.163 = _______ (2 decimal places)
i) 5.3172 = ______ (2 decimal places)
j) 2.095 = _______ (2 decimal places)
k) 0.4294 = ______ (3 decimal places)
l) 5.9826 = ______ (3 decimal places)
35 Significant Figure
The significant figures of a number are the digits
which carry meaning (i.e. are significant) to the
size of the number.
The first significant figure of a number cannot be
zero.
In a number with a decimal, leading zeros are not
significant.
In the number 0.00821, the first significant
figure is the 8.
In the number 0.00342 the 0’s are not
significant figures – they are there to keep the
correct place value for the other digits.
In the number 2.740, the 0 is a significant
figure. Once we start counting moving from
the left to the right we count every digit.
0.00821 rounded to 2 significant figures is
0.0082.
19357 rounded to 3 significant figures is
19400. We need to include the two zeros at the
end to keep the digits in the same place value
columns.
a) Which digit is the first significant figure in:
i) 32609
ii) 530.27
iii) 2.538
iv) 642.8
b) Which digit is the third significant figure in
each of the numbers in question 1.
c) Round each of the following numbers to the
number of significant figures indicated.
i) 40.53 (2 sf)
ii) 607621 (3sf)
iii) 0.002519 (2sf)
iv) 0.040763 (3sf)
v) 0.059804 (2sf)
36 Truncation
A method of approximating a decimal number by
dropping all decimal places past a certain point
without rounding.
3.14159265… can be truncated to 4 decimal
places to 3.1415 (note that if it had been
rounded to 4 decimal places, it would become
3.1416).
Truncate each of the following to the number
of decimal places indicated.
a) 2.59012 (2 decimal places)
b) 0.298314 (3 decimal places)
c) 12.29023 (1 decimal place)
37 Error Interval
A range of values that a number could have taken
before being rounded or truncated.
An error interval is written using inequalities, with a
lower bound and an upper bound.
Note that the lower bound inequality can be ‘equal
to’, but the upper bound cannot be ‘equal to’.
0.6 has been rounded to 1 decimal place.
The error interval is:
0.55 ≤ 𝑥 < 0.65
The lower bound is 0.55
The upper bound is 0.65
For each of the rounded values below give:
i) the lower bound
ii) the upper bound
iii) the error interval
a) 2400 (nearest 100)
b) 3000 (nearest 100)
c) 180 (nearest 10)
d) 3420 (nearest 10)
e) 12 (nearest whole)
f) 10 (nearest whole)
g) 17.6 (1 decimal place)
h) 21.0 (1 decimal place)
i) 5.72 (2 decimal places)
j) 0.89 (2 decimal places)
k) 9000 (1 significant figure)
l) 1200 (2 significant figures)
m) 9.0 (2 significant figures)
38 Bank Statement
A record of money being paid into a bank account.
Balance is the amount in the
bank account.
Credit is money being paid in –
add this amount to the balance.
Debit is money being paid out –
take this amount away from the
balance.
Complete this bank statement.
Topic 10 2.9 Graphs
Topic/Skill Definition/Tips Example Your turn
1 Function
Machine
Takes an input value, performs some operations
and produces an output value.
2 Function A relationship between two sets of values.
𝑓(𝑥) = 3𝑥2 − 5
‘For any input value, square the term, then
multiply by 3, then subtract 5’.
𝑓(1) = 3 × (1)2 − 5 = −2
It is given that 𝑓(𝑥) = 2𝑥3 − 7
Find:
a) 𝑓(3)
b) 𝑓(−1)
c) 𝑓(0.5)
3 Coordinates
Written in pairs. The first term is the x-
coordinate (movement across). The second
term is the y-coordinate (movement up or
down).
A: (4,7)
B: (-6,-3)
Write down the coordinates of each point
4 Midpoint of a
Line
Method 1: add the x coordinates and divide by
2, add the y coordinates and divide by 2.
Method 2: Sketch the line and find the values
halfway between the two x and two y values.
Find the midpoint between (2,1) and (6,9)
2+6
2= 4 and
1+9
2= 5
So, the midpoint is (4,5)
Find the midpoint of
a) (2,4) and (6,10)
b) (−5,2) and (5, −4)
5 Linear Graph
Straight line graph.
The equation of a linear graph can contain an x-
term, a y-term and a number.
The general equation of a linear graph is 𝒚 = 𝒎𝒙 + 𝒄
Where 𝒎 is the gradient and 𝑐 is the
y-intercept.
Example:
Other examples: 𝑥 = 𝑦 𝑦 = 4 𝑥 = −2 𝑦 = 2𝑥 − 7 2𝑦 −4𝑥 = 12
𝑦 = 𝟐𝑥 − 𝟕 Gradient 2, y-intercept -7
2𝑦 − 4𝑥 = 12
𝑦 = 𝟐𝑥 + 6 Gradient 2, y-intercept +6
a) Which of the following are linear graphs?
i) 𝑦 = 2𝑥 − 3
ii) 𝑦 = 𝑥2 + 1
iii) 𝑦 = 3√𝑥 − 2
iv) 𝑦 = 7
v) 2𝑥 + 5𝑦 = 10
For each of the following, identify:
i) the gradient
ii) the y-intercept
b) 𝑦 = 3𝑥 + 2
c) 𝑦 =1
2𝑥 − 7
d) 2𝑥 + 𝑦 = 6
e) 4𝑥 − 5𝑦 + 10 = 0
6 Plotting Linear
Graphs
Method 1: Table of Values
Construct a table of values to calculate
coordinates.
Method 2: Gradient-Intercept Method (use
when the equation is in the form 𝑦 = 𝑚𝑥 + 𝑐)
1) Plots the y-intercept
2) Using the gradient, plot a second point.
3) Draw a line through the two points plotted.
Method 3: Cover-Up Method (use when the
equation is in the form 𝑎𝑥 + 𝑏𝑦 = 𝑐)
1) Cover the 𝑥 term and solve the resulting
equation. Plot this on the 𝑥 − 𝑎𝑥𝑖𝑠.
2) Cover the 𝑦 term and solve the resulting
equation. Plot this on the 𝑦 − 𝑎𝑥𝑖𝑠.
3) Draw a line through the two points plotted.
a) Complete the table of values
i) 𝑦 = 2𝑥 + 4
ii) 𝑦 = 4𝑥 − 2
b) Use the gradient/intercept method to
plot 𝑦 =1
2𝑥 + 1
c) Use the cover up method to plot
i) 𝑥 + 𝑦 = 3
ii) 2𝑥 + 𝑦 = 4
𝑦 = −2𝑥 − 3
7 Gradient
The gradient of a line is how steep it is.
Gradient = 𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒚
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒙
The gradient can be positive (sloping upwards) or
negative (sloping downwards).
Calculate the gradient of each of the lines
shown below.
a) b)
8
Finding the
Equation of a
Line given a
point and a
gradient
Substitute in the gradient (m) and point (x, y)
into the equation 𝒚 = 𝒎𝒙 + 𝒄 and solve for c.
Find the equation of the line with gradient 4
passing through (2,7).
𝑦 = 𝑚𝑥 + 𝑐
7 = 4 × 2 + 𝑐 𝑐 = −1
𝑦 = 4𝑥 − 1
a) Find the equation of the line with gradient
5 passing through the point (1,3).
b) Find the equation of the line with gradient
-3 passing through the point (2,-5).
c) Find the equation of the line with gradient 1
2
passing through the point (2,1).
9
Finding the
Equation of a
Line given two
points
Use the two points to calculate the gradient.
Then repeat the method above using the
gradient and either of the points.
Find the equation of the line passing through
(6,11) and (2,3)
𝑚 =11 − 3
6 − 2= 2
𝑦 = 𝑚𝑥 + 𝑐
11 = 2 × 6 + 𝑐 𝑐 = −1
𝑦 = 2𝑥 − 1
In each question below give your answer in
the form 𝑦 = 𝑚𝑥 + 𝑐.
a) Find the equation of the line passing
through (2,3) and (5,9).
b) Find the equation of the line passing
through (1,11) and (4,5).
c) Find the equation of the line passing
through (2,3) and (6,5).
d) Find the equation of the line passing
through (-2, 4) and (4,2).
10 Parallel Lines
If two lines are parallel, they will have the same
gradient. The value of m will be the same for
both lines.
Are the lines 𝑦 = 3𝑥 − 1 and 2𝑦 − 6𝑥 + 10 = 0
parallel?
Answer:
Rearrange the second equation into the form 𝑦 = 𝑚𝑥 + 𝑐
2𝑦 − 6𝑥 + 10 = 0 → 𝑦 = 3𝑥 − 5
Since the two gradients are equal (3), the lines
are parallel.
a) Are the lines 𝑦 = 2𝑥 + 3 and 2𝑦 − 4𝑥 = 8
parallel? Show how you know.
b) Are the lines 𝑦 =1
4𝑥 − 1 and 𝑥 − 4𝑦 = 12
parallel? Show how you know.
c) Are the lines 𝑥 + 2𝑦 = 5 and 𝑦 = −2𝑥 + 4
parallel? Show how you know.
d) Find the equation of the line parallel to
𝑦 = 3𝑥 − 5 that passes through the point
(0,7).
e) Find the equation of the line parallel to
𝑦 = −1
2𝑥 + 5 that passes through the point
(2, 3).
11
H
Perpendicular
Lines
If two lines are perpendicular, the product of
their gradients will always equal -1.
The gradient of one line will be the negative
reciprocal of the gradient of the other line.
You may need to rearrange equations of lines to
compare gradients (they need to be in the form
𝑦 = 𝑚𝑥 + 𝑐).
Find the equation of the line perpendicular to
𝑦 = 3𝑥 + 2 which passes through (6,5)
Answer:
As they are perpendicular, the gradient of the
new line will be −1
3 as this is the negative
reciprocal of 3.
𝑦 = 𝑚𝑥 + 𝑐
5 = −1
3× 6 + 𝑐
𝑐 = 7
𝑦 = −1
3𝑥 + 7
Or 3𝑥 + 𝑥 − 7 = 0
a) Find the equation of the line perpendicular
to 𝑦 = 2𝑥 + 5 that passes through the point
(4,9).
b) Find the equation of the line perpendicular
to 𝑦 =1
5𝑥 − 2 that passes through the point
(2,-4).
c) Find the equation of the line perpendicular
to 𝑦 = −2
3𝑥 + 7 that passes through the point
(4,8).
d) Find the equation of the line perpendicular
to 𝑥 + 2𝑦 = 8 that passes through the point
(-2,6).
e) Find the equation of the line perpendicular
to 3𝑥 − 2𝑦 = 12 that passes through the point
(1,-2).
12
Solving
Simultaneous
Equations
(Graphically)
Draw the graphs of the two equations.
The solutions will be where the lines meet.
The solution can be written as a coordinate.
𝑦 = 5 − 𝑥 and 𝑦 = 2𝑥 − 1.
They meet at the point with coordinates (2,3)
so the answer is 𝑥 = 2 and 𝑦 = 3
a) Shown below are the graphs of 𝑦 = 2𝑥 + 2
and 𝑦 = −𝑥 − 4.
Use the graphs to solve the simultaneous
equations 𝑦 = 2𝑥 + 2 and 𝑦 = −𝑥 − 4
b) By drawing the graphs of 𝑦 = 3𝑥 + 1 and
𝑥 + 𝑦 = 7, solve the simultaneous equations
𝑦 = 3𝑥 + 1 and 𝑥 + 𝑦 = 7.
c) By drawing the graphs of 𝑦 = 3𝑥 + 5 and
𝑥 − 2𝑦 − 6 = 0, solve the simultaneous
equations 𝑦 = 3𝑥 + 5 and 𝑥 − 2𝑦 + 6 = 0.
13 Real Life Graphs
Graphs that are supposed to model some real-life
situation.
The actual meaning of the values depends on the
labels and units on each axis.
The gradient might have a contextual meaning.
The y-intercept might have a contextual
meaning.
The area under the graph might have a
contextual meaning.
A graph showing the cost of hiring a ladder for
various numbers of days.
The gradient shows the cost per day. It costs
£3/day to hire the ladder.
The y-intercept shows the additional
cost/deposit/fixed charge (something not
linked to how long the ladder is hired for). The
additional cost is £7.
For the graph shown below,
a) what is the fixed charge?
b) what is the price per mile?
c) A water company charges customers a
fixed standing charge plus an additional cost
for the amount of water, in cubic metres,
used. The graphs shows information about
the cost charged.
i) Write down the fixed standing charge.
ii) Work out the additional cost for each
cubic metre of water used.
14 Distance-Time
Graphs
You can find the speed from the gradient of the
line (Distance ÷ Time)
The steeper the line, the quicker the speed.
A horizontal line means the object is not
moving (stationary).
Calculate the average speeds during the
cycle for Amanda and Brian.
15 Depth of Water
in Containers
Graphs can be used to show how the depth of
water changes as different shaped containers
are filled with water at a constant rate.
Match the containers to the graph showing
the height of water as they fill.
Answers
10 2.6 Perimeter, Area and Volume I
1) 3000m 7530m 400cm 670cm 50mm 107mm 6000g 900g
4km 6.475km 9m 5.5m 8cm 4.7cm 8kg 1.7kg
2) 48oz 24oz 84oz 36 inches 66 inches 32 pints
5lb 3.5 lb 4lb 9oz 5ft 6ft 4inches 6 gallons
3) 16 kilometres 40 kilometres 30 miles 8 ½ miles 13.5 litres 4 gallons 11 pounds 15 kilograms 10cm
4) a) 16km b) 7.5 miles c) 208km d) 112.5 miles
4) Graph Section
a) 84 Lira
b) 360 Lira
c) £5.50
d) £8.50
5) a) 40 cm b) 19 cm 6) a) 6 𝑐𝑚2 b) 9 𝑐𝑚2 7) a) 15 𝑐𝑚2 b) 36 𝑐𝑚2 8) a) 54 𝑐𝑚2 b) 40 𝑐𝑚2 9) a) 20 𝑐𝑚2 b) 35 𝑐𝑚2
10) 10 𝑚2 11) 35 𝑐𝑚2 12) a) 40 𝑐𝑚2 b) 40 𝑐𝑚2 13) 36.1 𝑐𝑚2
14)
Pounds £
Lira
Answers are approximate
(as read from graph).
15) a) i) 49𝜋 𝑐𝑚2 ii) 153.94 𝑐𝑚2 b) i) 121𝜋 𝑐𝑚2 ii) 380.13 𝑐𝑚2 16) a) i) 22𝜋 𝑐𝑚 ii) 69.2 𝑐𝑚 b) i) 14𝜋 𝑐𝑚 ii) 43.98 𝑐𝑚
17) a) 9.424777961 … b) 31.9911485751 … 18) a) 1.95 𝑐𝑚 (2dp) b) 49.74 (2dp) 19) a) 34.21 𝑐𝑚2 b) 30𝜋 𝑐𝑚2
20)
21) a) 343 𝑐𝑚3 b) 1344 𝑐𝑚3 22) 96 𝑐𝑚3 23) 160𝜋 𝑐𝑚3 or 502.7 𝑐𝑚3 (1dp)
10 2.7 Equations and Inequalities
1) a) 𝑥 = 6 b) 𝑥 = 7 c) 𝑥 = 12 d) 𝑥 = 5 e) 𝑥 = 4 f) 𝑥 = 2 2) Addition, multiplication, square rooting. 3) a) 12 b) 1 c) 43 d) 10 e) 32
4) a) 𝑥 = 3, 𝑦 = 2 b) 𝑥 = 2, 𝑦 = 6 c) 𝑥 = 3, 𝑦 = 4 5) a) 𝑥 = −1, 𝑦 = 3 b) 𝑥 = 7, 𝑦 = −2 c) 𝑥 = 3, 𝑦 = −2 d) 𝑥 = 8, 𝑦 = 3 e) 𝑥 = 0.5, 𝑦 = 4
6) a) 𝑥 = 1, 𝑦 = 8 b) 𝑥 = 5, 𝑦 = −1 c) 𝑥 = 2, 𝑦 = 4 d) 𝑥 = 4.5, 𝑦 = 3
8) a) -2, -1, 0, 1, 2, 3 b) 0, 1, 2, 3, 4, 5 c) -4, -3, -2 d) 0, 1, 2, 3 e) x is less than 7 f) x is greater than or equal to -3 g) x is greater than or equal to 2
9)
a)−1 ≤ 𝑥 b) 𝑥 < 2 c) 2 ≤ 𝑥 ≤ 5 d) 𝑥 < 0 e) −2 ≤ 𝑥 < 3 f) 3 < 𝑥 10) a) b) 𝑥 ≥ −1, 𝑦 > −2, 𝑥 + 2𝑦 < 4
Length Area Volume
cm2 √
mm √
km3 √
inches √
R
10 2.8 Calculations
1) a) 4 b) 9 c) 4
7 2)
1
8
1
12
1
10 3) a) 1
4 b)
1
7 c)
5
4 d)
9
8 4) 4 1
5 6
2
3 5
7
9 5) a) 4
5 b) 3
4 c)
5
8
6) 3
4=
6
8=
30
40=
12
16=
9
12 7) a)
7
12
5
8
2
3
3
4
5
6 b) 5
6
3
4
7
12
1
2
4
9
1
3 8) a) 450kg b) $150 c) 91m
9) a) 7
8 b)
11
15 c)
19
28 d) 1 7
40 10) a) 3
20 b)
3
14 11) a) 24
25 b) 20
11 c)
55
64
12) a) 8 hours b) 225 miles c) 60km/h d) 22.5 seconds e) 37.5 km f) 0.6h = 36 min
13) a) 125kg/m3 b) 440 kg/m3 c) 40cm3 d) 3.8 g/cm3 e) 600cm3 = 0.0006 m3 f) 480kg
14) a) 150 N b) 40 N/m2 c) 1.5 m2 d) 15 N e) 50 N/m2 f) 1.5 m2 15) a) i) £23 ii) 3.4kg b) 450m
16) a) i) $4.50 ii) 0.25m = 25cm b) £230 17) a) 20% b) 20%
18) Non-calculator a) £432 b) 192kg c) $238 d) 90.3kg e) £455
Calculator f) £172.20 g) 90.3kg h) $258.40 i) 329m
19) a) 1.26 b) 1.07 c) 2.5 d) 0.68 e) 0.94 f) 0.925 20) a) £55 b) £3600 c) £23,000 d) £19.50 21) a) £400 b) £261 c) £1,097.60 d) $643.20
22) a) £844.62 b) £688.51 c) £6285.04 d) 6.65 km2 (2 dp) 23) a) 3:2, 2:3 b) 5:3, 3:5 c) 3:1:4
24) a) i) 11
26 ii) 15
26 b) 280ml 25) a) 4:5 b) 3:4 c) 4:5 d) 6:7 e) 3:4 f) 2:4:3 g) 2:4:7 h) 3:2:6
26) In the form 1:n a) 1:3 b) 1:7
3 c) 1:
3
8 d) 1:
1
3
In the form n:1 e) 2:1 f) 43:1 g)
7
9:1 h) 8
15: 1
27) a) 100:140 b) 150:350:300 c) i) 1.5 litres ii) 2.5 litres 28) a) i) 100 minutes = 1 hour 40 minutes ii) 4 pages b) i) 150ml ii) 630ml
29) a) 1250g b) £28.98 30) a) $20.70 b) 120.98 zł c) 390.60 kr d) £27.54 e) £22.81 f) £26.50
g) $46.92 = £34, £36, 199.88zł = £38, 277.76kr = £32 so 199.88zł is the greatest amount.
31) a) 156 sweets b) £112
32) a) Pack of 3 is £4 per bulb, pack of 4 is £4.25 per bulb, pack of 8 is £3.75 per bulb. So, the pack of 8 is best value as it has the lowest price per bulb.
b) Type 1 costs £2.395 per SD card which is £0.1496875 per GB.
Type 2 costs £8.996̇ per SD card which is £0.14057292 per GB. Type 2 is the best value per GB, as it has the lowest price per GB.
Type 3 costs £24.745 per SD card which is £0.1933203125 per GB.
33) 150 210 200 400 500 3300 3000 4000 16 24 12
34) a) 1st decimal place b) 2nd decimal place c) 3rd decimal place d) 4th decimal place
e) 2.5 f) 4.3 g) 5.3 h) 7.16 i) 5.32 j) 2.10 (the 0 must be here as this is to 2 decimal places) k) 0.429 l) 5.983
35) a) i) 3 ii) 5 iii) 2 iv) 6 b) i) 6 ii) 0 iii) 3 iv) 2 c) i) 41 ii) 608,000 iii) 0.0025 iv) 0.0408 v) 0.060
36) a) 2.59 b) 0.298 c) 12.2
37) a) i) 2350 ii) 2450 iii) 2350 ≤ 𝑥 < 2450 b) i) 2950 ii) 3050 iii) 2950 ≤ 𝑥 < 3050 c) i) 175 ii) 185 iii) 175 ≤ 𝑥 < 185
d) i) 3415 ii) 3425 iii) 3415 ≤ 𝑥 < 3425 e) i) 11.5 ii) 12.5 iii) 11.5 ≤ 𝑥 < 12.5 f) i) 9.5 ii) 10.5 iii) 9.5 ≤ 𝑥 < 10.5
g) i) 17.55 ii) 17.65 iii) 17.55 ≤ 𝑥 < 17.65 h) i) 20.95 ii) 21.05 iii) 20.95 ≤ 𝑥 < 21.05 i) i) 5.715 ii) 5.725 iii) 5.715 ≤ 𝑥 < 5.725
j) i) 0.885 ii) 0.895 iii) 0.885 ≤ 𝑥 < 0.895 k) i) 8500 ii) 9500 iii) 8500 ≤ 𝑥 < 9500 l) i) 1150 ii) 1250 iii) 1150 ≤ 𝑥 < 1250
m) i) 8.95 ii) 9.05 iii) 8.95 ≤ 𝑥 < 9.05
38)
10 2.9 Graphs
1) a) 5 b) 9.5 c) 72 d) 22 a)7 b) 80 c) 16 d) 7 2) a) 47 b) −9 c) −6.875
3) A = (3,2) B = (-3,1) C = (-2,4) D = (-2,-1) E = (-3,-3) F = (0,-1) G = (1,-3) H = (4,-2)
4) a) (4,7) b) (0,-1)
5) a) i, iv and v b) i) 3 ii) 2 c) i) 1
2 ii) -7 d) i) -2 ii) 6 e) i) 4
5 ii) 2
6) a) i) ii)
b) c) i) c) ii)
7) a) 1 b) −1
2 8) a) 𝑦 = 5𝑥 − 2 b) 𝑦 = −3𝑥 + 1 c) 𝑦 =
1
2𝑥 9) a) 𝑦 = 2𝑥 − 1 b) 𝑦 = −2𝑥 + 13 c) 𝑦 =
1
2𝑥 + 2 d) 𝑦 = −
1
3𝑥 +
10
3
10) a) Second equation can be written as 𝑦 = 2𝑥 + 4. Both lines have gradient 2 parallel.
b) Second equation can be written as 𝑦 =1
4𝑥 − 3. Both lines have gradient
1
4 so parallel.
c) First equation can be written as 𝑦 = −1
2𝑥 +
5
2. Lines have different gradients (-2 and −
1
2) so not parallel.
d) 𝑦 = 3𝑥 + 7 e) 𝑦 = −1
2𝑥 + 4
11) a) 𝑦 = −1
2𝑥 + 11 b) 𝑦 = −5𝑥 + 6 c) 𝑦 =
3
2𝑥 + 2 d) 𝑦 = 2𝑥 + 10 e) 𝑦 = −
2
3𝑥 +
4
3
12) a) 𝑥 = −2 and 𝑦 = −2 b) 𝑥 = 1.5 and 𝑦 = 5.5 c) 𝑥 = −0.8 and 𝑦 = 2.6
13) a) i) $4 ii) $0.5 per mile b) Fixed standing charge = £15, Cost per cubic metre used ≈ £1.125 14) Amanda 20 km/h, Brian 12 km/h 15) A-2 B-3 C-1
Knowledge Organiser
MATHEMATICS GCSE Mathematics AQA