gcse mathematics aqa year 10

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Knowledge Organiser

MATHEMATICS GCSE Mathematics AQA

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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


9 Area of a Triangle Base x Perpendicular height ÷ 2

𝐴 = 24𝑐𝑚2


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


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


12 Compound Shape A shape made up of a combination of other

known shapes put together.


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


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


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


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


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


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


Density = 80kg/m³

Mass = 2000g

Find the volume.

𝑉 = 𝑀 ÷ 𝐷 = 2 ÷ 80 = 0.025𝑚³


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


Pressure = 10 N/m2

Area = 6m²

Find the Force.

𝐹 = 𝑃 × 𝐴 = 10 × 6 = 60 𝑁


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


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


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


c) Find the equation of the line passing


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


c) Are the lines 𝑥 + 2𝑦 = 5 and 𝑦 = −2𝑥 + 4


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.


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

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