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GCSE Maths GRADE BOOSTER HigherRevision Workbook

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2 GCSE MATHS GRADE BOOSTER HIGHER Revision Workbook

Welcome to this exam performance booster workbook for students aiming for a grade 7 or 8 in the new GCSE 9-1 maths exam. The workbook consists of 8 sections which are all designed to not only help reinforce key GCSE maths content, but also refine and sharpen key exam techniques for the three GCSE maths papers as well as building confidence in the essential assessment skills.

GCSE Maths grade booster Higher REVISION WORKbook

Section 1|NUMBERS 1.1 Multiples, Factors, LCM, HCF and Prime Factors Page 4

1.2 Calculations with Fractions Page 6

1.3 Recurring Decimals Page 8

1.4 Powers, Roots and Indices Page 9

1.5 Standard Form Page 10

1.6 Surds Page 11

Section 2|ALGEBRA 2.1 Substitution into Formulae Page 12 2.2 Algebraic Indices Page 13 2.3 Straight Line Graphs Page 15

2.4 Simultaneous Equations – 1 Page 17

2.5 Factorising and Solving Quadratic Equations Page 19 2.6 The Quadratic Formula 1

2.7 Completing The Square Page 22

2.8 Rearranging Formulae Page 23

2.9 Non-Linear Graphs Page 25

2.10 Simultaneous Equations – 2 Page 27

2.11 The Equation of a Circle Page 29

2.12 Algebraic Fractions Page 30

2.13 Transforming Graphs Page 32

Section 3|Ratio, Proportion and Rates of Change 3.1 Percentages – 1 Page 34

3.2 Percentages – 2 Page 36

3.3 Speed, Density and Pressure Page 38

3.4 Direct and Inverse Proportion Page 40

3.5 Real Life Graphs Page 43

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Section 4|Geometry and Measures 4.1 Congruence and Similarity Page 46 4.2 Length, Area and Volume Scale Factors Page 48 4.3 Circle Theorems Page 50 4.4 Pythagoras' Theorem and Trigonometry Page 53 4.5 The Sine & Cosine Rules & Using Area = 1/2ab sin C Page 56 4.6 Vectors Page 58

Section 5|Probability 5.1 Sets and Venn Diagrams Page 60

5.2 Tree Diagrams and Frequency Trees Page 61

5.3 Conditional Probability Page 63

Section 6|Statistics 6.1 Stratified Sampling Page 64

6.2 Frequency Tables – Averages and Range Page 65

6.3 Cumulative Frequency Graphs and Box Plots Page 67

6.4 Using Box Plots Page 69

6.5 Histograms Page 70

Section 7|Problem Solving 7 Problem Solving Page 72

Section 8|Revision and The Exams 8.1 Revision Tips Page 74 8.2 Being Exam Ready Page 75 8.3 In the Exam Hall Page 76 8.4 Formulae and Command Words Page 77


2) a Find the factors and first six multiples of 15 and 18

Factors of 15 =

Multiples of 15 =

Factors of 18 =

Multiples of 18 =

b Use these answers to find the LCM and HCF of 15 and 18

LCM =

HCF =

1|NUMBERS

Questions

Multiples, Factors, LCM, HCF and Prime Factors1.1Key Words CrosswordA

B

3

4 5

1

2

7

6

Across

2 HCF is the acronym for highest _ _ _ _ _ _ factor (6)

3 LCM is the acronym for _ _ _ _ _ _ common multiple (6)

4 In a set diagram, the intersection is where the regions _ _ _ _ _ _ _ (7)

7 In a set diagram, this is the region enclosed by all sets (5)

Down 1 1, 2, 3, 4, 6, 9,12, 18 and 36 are all _ _ _ _ _ _ _ of 36 (7)

5 One kind of diagram used to display sets is a _ _ _ _ diagram (4)

6 A number which has exactly two factors (5)

1) Circle the prime numbers. 3) Complete the blanks on this prime factor tree for 360. Circles represent prime factors.

2 4 7 19 27 37

42 55 61 69 89 90 360

360 = × ×

2

90

15

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Using Set Diagrams to find the LCM and HCFC

1) As a product of its prime factors, 126 can be written as 126 = a × 3b × c

Find the values of a, b and c.

2) Using your answer to question 1, and by finding the prime factors of 105, find the lowest common multiple and highest common factor of 105 and 126. You should use the set diagram started below for this question.

a =

LCM = HCF =

b =

c =

1

2

Exam Questions

(Total for Question 2 is 3 marks)


a) Find the lowest common multiple (LCM) of 12, 14 and 21.

Two buses leave a bus stop at different intervals. The first bus leaves at 24 minute intervals and the second bus leaves at 30 minute intervals.What time will both buses next be at the bus stop together, if they both leave at 11 AM?

(2)b) Find the highest common factor (HCF) of 42 and 72.

(2)

105 126∩ 105 126∩" "

105 126


Mixed Numbers & Improper FractionsB

Calculations with Fractions1.2MethodsA

For his maths homework, Charlie wrote instructions for how to add or subtract, multiply and divide fractions. His dog Felix (who is not fraction friendly) ripped-up these instructions into each of their individual steps.Re-write these instructions in the table below.

Jess has changed these fractions incorrectly. Circle the incorrect answers and write corrections, where appropriate.

Instructions

Change the problem to amultiplication, then calculate

Find the lowest commondenominator

Multiply the numerators

Find the

equivalent fractions

Keep the first fraction the same

Calculate (do not add or subtract the denominators)

Simplify

Simplify

Simplify

Multiply the denominators

Flip the second fraction

To add or subtract fractions: To multiply fractions: To divide fractions:

=3 25

= 283

23

= 2125

25

= 4296

16

165

=2 35

125

=2 34

134

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Multiplying & Dividing FractionsC Multiplying and dividing fractions is a piece of cake (easy)! Work out these calculations. Change any mixed numbers to improper fractions first. Don’t forget to simplify.

25

512

1 × 13

14

2 1× 12

13

2 1÷ 34

58

3 of 2

Work out:

Remember: You only need the lowest common denominator when adding and subtracting.

Adding & Subtracting Fractions D

34

78

+A B C2

358

1 1– 16

2 58

1+

Senior1

Exam QuestionBoston Golf Club has three types of members

Junior Regular SeniorThree-quarters of the club are Regular members, and one-sixth of the club are Senior members.One-third of the Junior members do not play in competitions.

What is the fraction of all the members in the club that are Junior members which play in competitions?


For division: Use KFC; Keep first fraction the same - Flip second fraction - Change ÷ to ×.Hint


Recurring Decimals1.3

Changing Recurring Decimals into FractionsB

Terminating and Recurring Decimals – DO NOT USE A CALCULATORA Identify whether these fractions are terminating or recurring. Circle ‘T’ if terminating and ‘R’ if recurring, and writeeach fraction as a decimal. Do not use a calculator (use the ‘bus-stop’ division method).

Worked exampleWrite 0.125 as a fraction.

QuestionsChange each recurring decimal into a fraction, giving your answer in its simplest form.

Use the space below for working. The first one has been done for you.

433

38

2933

15

511

27

T0.12

T

T

T

T

T

R

R

R

R

R

R

= 33

x = =

x = 0.125 25 100x = 125.25 10x = 1.25 990x = 124

0 . 1 2 1 2 ...4.4 0704070 = 0.1212 ... = 0.12

124990

62495

433

The part that recurs is shown by placing dots over the first and last digits of the recurring pattern.Reminder

a) 0.6 b) 0.75 c) 0.64 d) 0.681

Let x equal the recurring decimalIdentify the recurring digit(s)

Subtract both sides of the two equations

Solve for x, writing x as a fraction (don’t forget to simplify!)

1

Exam Question


Express the recurring decimal 0.206 as a fraction in its simplest form.

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Powers, Roots and Indices1.4

CalculationsA For each of these expressions: (i) Write as a single power; (ii) Evaluate the expression (use the laws of indices in section 2.2 A).

You need to understand what fractional and negative indices mean, and be able to evaluate powers.

1

Exam Question


Evaluate the following, expressing each answer as a fraction in its simplest form.

na ... is n raised to the power of a (n multiplied by itself a times): na = n × n × ... × nis the ath root of n (find the number you raise to the power of a to get n)

n0... any number raised to the power of zero is 1: n0 = 1

a n ...√

a) 3 × 32 b) 7–3 × 7 c) 158 ÷ 156 d) 0.55 ÷ 0.57 d) 36 ÷ 32 × 3–6

Negative IndicesB

Evaluate:

n–a... is the reciprocal of na (flip it - raise it): n–a =

a) 7–1 b) 4–2 c) 5–2 d) 3–3 d) 4–3

1na

Fractional IndicesC

Evaluate:

n ... is the ath root of n (root it): n =

b) 125 c) ( ) d) 64 d) 32

1a

1a

a n√n ... take the bth root of n and raise this answer to the power of a (root it - raise it): n =

ab

a) 6412

13

23

131

27

35

ab (

(b n a√

Negative and Fractional IndicesD

n ... is the reciprocal of n , first write the reciprocal of n then evaluate n (flip it - root it - raise it): 𝑛n = 1

ab

_

ab

_

ab

ab

ab

(

(b n a√Evaluate:

a) 8 b) 16 d) ( ) e) ( )23

_ 18

916

34

_c) ( )4

49

12

_ 53

_ 32

_

a) 1612

b) 3–2

c) ( )278

23

_ (3)

(1)

(1)

Let x equal the recurring decimalIdentify the recurring digit(s)

Subtract both sides of the two equations

Solve for x, writing x as a fraction (don’t forget to simplify!)


Standard Form1.5

Ordinary Numbers into Standard Form and Vice Versa

Calculating with Standard Form

A

B

Examples What is 25,000 in standard form?

Multiplying and Dividing – Multiply or divide the front numbers together and the powers of 10 together and convert the final answer into standard form if not alreadyAdding and Subtracting – Either make sure the powers of 10 are the same, or convert to ordinary numbers

Questions 1 Write the following in standard form

a) 0.014

b) 7600

c) 0.000123

d) 150000

Questions – Give your answers in standard form

a) 1.23 × 102 + 2.7 × 103

d) 8 × 102 × 4 × 10–6

b) 6.23 × 104 – 5.24 × 103

e) 9 × 106 ÷ 3 × 104

c) 4.1 × 10–3 – 2.6 × 10–4

f) 8 × 108 ÷ 4 × 102

2 Write the following as an ordinary number

a) 6.5 × 10–3

b) 2.415 × 103

c) 6.54 × 10–5

d) 2.12 × 10–4

Write 3.75 × 10–3 as an ordinary number?

This number mustbe between 1 and 10

This number indicates the numberof places that the digits move

1

2

Exam QuestionS



There are approximately 5.74 × 109 people in the world today. How much water would be needed if all ofthem decided to have a shower at the same time and used 22.5 litres of water each. Give your answer instandard form.

Given that

find the value of the following, giving your answer in standard form.

A × 10n"

"

25,000 = 2.5 × 104 3.75 × 10–3 = 0.00375

Do not just add in zeros, for example, in the first example don’t write 0.000375Warning

p = 6 × 102, q = 2 × 103, and r = 5 × 102,

8q + 100r p

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Surds1.6

1

2

Exam QuestionS



Express in its simplest form

Rationalise the denominator

Simplifying Surds

Manipulating Surds

A

B

You must be able to simplify surds in calculations, and sometimes as standalone questions. Test yourself, simplify:

Rules Complete the rules for manipulating surds using the expressions opposite:

do nothing!

Questions

a) 48

a) 8 × 2 b) 45 – 20 c) 2 (2 – 2) d) (2 + 3)2

b) 125 c) 50 d) 72√

√ √ √√ √ √ √

√ √ √

a × b =

a + b =

(a + b )2 = (a + b )(a + b ) =

(a + b )(a – b ) = a2 + a b – a b – ( b )2 =

a2 + 2a b + b

a × b

a2 – b

×

=

=

=

b

2

3– 2

√√

√

√

a√√

√

√√√ √

√ √√ √ √

a

b√

a)5

√

b)2

√

a

b± c√

(3 2 – 2)( 3 + 2 2)√ √ √ √

(3)

(1)

a

b± c

±

±√b c

b c√√

×a

b√ b√b√

ab√


Equations of MotionB

2|ALgebra

Substitution into Formulae2.1Substitution into Circle FormulaeA

In the box below write down the circle formulae for area, circumference, arc length and area of a sector. Use these to match up the questions and answers.

The following formulae are equations of motion; v 2 = 𝑢u 2 + 2as, v = u + at, s = ut + at 2. Choose the correct formula and substitute correctly to find the unknown.

Area of a circle, r = 7

Question Answer

47.0 cm

Perimeter of a semi-circle, d = 10 25.7 cm

Area of a sector, r = 14, θ = 60° 153.9 cm2

Perimeter of a sector, r = 11, θ = 130° 102.6 cm2

12

a) Find v when u = 7, a = −4 and t = 6

b) Find s when a = 4, t = 10 and u = 6

c) Find u when v = 12, a = −2 and s = 3

1 This formula works out the tax paid on annual income, T = 0.15(E − 11,000)T is the tax paid in pounds (£), E is the amount earned in pounds (£).Steve pays £2625 in tax.Calculate the amount he earns.

Exam Question


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Laws of IndicesA

It is important that you know the laws of indices and can apply them.

The expressions below have been simplified wrongly. Find the error and work out the correct solution.

Algebraic Indices2.2

Simple EquationsB

Solve each equation. Where appropriate, write each number on the rhs of the equation as a power using the same base as the lhs.

Answer

4

3

1

x 4 = 16

Question

3 x = 27

2x+1 = 32

2 2x = 1024

x 4 = x 5

2

5

x a × x b = x a + b

x a ÷ x b = x a – b

(x a) b = x a × b

x 0 = 1 16a3 ÷ 8a –5 = 2a –5

ab3 × a2b –4 = a3b

(4ab3)2 = 4ab6

7a0 × 8b4 = 0

Example: Solve 2x = 64 2x = 26

x = 6


Exam Question


1 Simplify fully(4a 3b 4) × 2b 3

6a 8b 2

Exam Question


1 Solve the equation 8x × 2x–4 = 256

Harder Equations

Example: b) Solve:

C

You are expected to use positive integer powers and associated real roots (square, cube and higher) and recognise powers of 2, 3, 4 and 5. In these questions, you must write each expression in terms of the lowest common base.

Solve 2x+1 × 4x = 128

2x+1 × (22)x = 27

2x+1 × 22x = 27

x + 1 + 2x = 73x + 1 = 73x = 6x = 2

32x × 3x–1 = 243

a) Solve:

32 × 22x+3 = 14

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Once you have completed each table, plot the graphs on theset of axes for the values betweenx −2 ≤ x ≤ 2

x

y

-2 -1 0 1 2

Complete the tables of values below for the graphs:

Straight Line Graphs2.3Plotting Straight Line GraphsA

y = 3x – 4 and y = – x + 212

y = 3x – 4

y = – x + 212

x

y

-2 -1 0 1

-1

2

1

Use the cover method to plot 3x𝑢− 2y𝑢= 6Challenge

Hence, or otherwise, state the equationof each line in the form:

Equation Line 1:

y =

Gradient Line 2:

m =

Equation Line 2:

y =

Finding the Equation of a Straight LineB

Calculate the gradient, m, of each line,remembering to use the formula:

Gradient Line 1:

m =

y2 – y1

x2 – x1m =

y = mx + c

The diagram shows two straight line graphs.


Parallel and Perpendicular LinesC

The table below shows pairs of straight lines.

State whether the lines are parallel, perpendicular or neither.

y = 2x + 3 and y = –2x + 5

Lines Relationship

y = 3x – 5 and 3y + 6 = –x

y = – x – 5 and x = –2y – 10

2x + 3y = 6 and y = 2 + x

12

32

You may find it helpful to rearrange equations in the form y = mx + c. Remember that parallel lines have the same gradient and that for perpendicular lines: m1 × m2 = –1

Hint

Exam Question


1 Work out the equation of the line which passes through the point (3 , −5) and is:

a) Parallel to the line y = 2x𝑛− 3

b) Perpendicular to the line y = 2x𝑛− 3

y =

y = (2)

(3)

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Each pair of linear simultaneous equations below can be solved using the elimination method, without the need to multiply either equation. Match the equations to the possible first step and the final solution.

Simultaneous Equations – 12.4Basic Simultaneous EquationsA

x = 4, y = 1

x = 4, y = –1

x = 2, y = 5

x = 4

7y = –7

x + y = 7x – 2y = –8

3x + 2y = 103x – 5y = 17

x + y = 52x + y = 9

Equations:

3y = 15

Possible First Step: Final Solution:

Solving Simultaneous Equations GraphicallyB

The graph shows the straight line for the equationy = 2x + 5. By drawing the graph of 2x𝑢+ 3y𝑢= 7 and using a graphical method, solve the following pair simultaneous equations:

𝑦 y = 2x𝑦 + 52x𝑦 + 3y𝑦 = 7

Rearrange the equation to make y the subject.Hint

x = y =


Solving Harder Simultaneous EquationsC

Solve the pairs of simultaneous equations below. Focus initially on equating the coefficients for either values of x or y.

a) b)2x + 3y = 184x – y = 8

4x + 2y = 23y – 12 = 3x

Multiply one or both equations by a multiplier to equate the coefficients of either unknown.

Hint

Exam Question


1 Freddie bought 3 sticky buns and 2 cups of tea for £6.00.Flora bought 2 sticky buns and 4 cups of tea for £6.40.Felix orders a sticky bun and a cup of tea. Work out the cost of Felix’s order.

£

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a) x2 – 49 = 0 b) x2 – 12x + 36 = 4 c) x2 – 5x – 30 = –6

Remember to ensure the expression you are factorising is equated to zero before you start to factorise.Hint

Solve the follow quadratic equations by factorising

All the solutions to the equations below are given, together with some 'extras'. Match the equations to their solutions. Use these solutions to spell an anagram for a mathematical term.

Solving Quadratic Equations

Solving Harder Quadratic Equations

Connect the quadratic expressions below to their factorised form, remembering the rules for factorising.

Use the Add to, multiply to boxes to help you. The first one has been done for you as an example.

Factorising and Solving Quadratic Equations2.5Factorising Quadratic ExpressionsA

B

C

x2 + 3x – 10

Expression

+3 , –10

Add to, multiply to

(x – 2) (x + 5)

Add to, multiply to

Add to +3x

+ 5x – 2xx2 – 10

Multiply to –10

¨ ¨

x2 + 8x + 16

x2 – 16

x2 – 9x + 20

x2 – x – 12

–9 , +20

–1 , –12

0 , –16

+8 , +16

(x + 4) (x – 4)

(x – 4) (x + 3)

(x + 4)2

(x – 4) (x – 5)

x = –2

x = 1.5

x = –3

x = –1.5

2x2 – 3x – 20 = 0

3x2 + 14x + 8 = 4x

4x2 – 4x = 15

Expression Solutions

A

T

N

U

x = –2.5

x = 4

x = 2.5

x = –1

S

D

R

I 13


Exam Question


1 A cuboid has length x cm.The width of the cuboid is 6 cm shorter than its length.The height of the cuboid is half of its length.

cm3

a) The cuboid has a surface area of 112 cm2 Show that the surface area of the cuboid can be expressed as 2x2 − 9x − 56 = 0

(5)

b) Hence, work out the volume of the cuboid.(5)

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Using the discriminant to determine whether the equations have two distinct roots, one repeated root or no real roots.

Jim has incorrectly substituted into the quadratic formula. Can you find and explain his mistakes? (There may be more than one error in each question.)

The Discriminant and Roots

Common Errors

Solve the two quadratic equations, giving your answers to two decimal places.

The Quadratic Formula2.6

Substitution into the Quadratic FormulaA

B

C

a) x2 – 2x + 1 = 0

b2 – 4ac > 0 Two distinct rootsb2 – 4ac = 0 One repeated rootb2 – 4ac < 0 No real roots

x2 – 5x + 4 = 0 3x2 – 3x + 2 = 7

2×1 2×3x =

x =

x =– 5 ± (–5)2 – 4×1×4 – 3 ± –32 – 4×3×5

b) 3x2 – 6x + 5 = 0 c) 3 – 7x – 4x2 = 0

a) x2 – 5x – 8 = 0 b) 2x2 – 4x – 7 = 0

Hint

√ √


x = and

Exam Question1 Solve the equation 5x2 + 11x – 2 = 0, giving your answers correct to 2 decimal places.

The solutions of ax 2 + bx + c = 0, where a ≠ 0, are given by: – b ± √b 2 – 4ac

2a


Completing The Square2.7Writing Expressions in Completed Square Form

Finding the Minimum Point and Sketching

A

B

x2 + 6x – 4

x2 – 6x + 12

x2 + 9x + 4

x2 – 18x + 7

(x – 9)2 – 74

(x – 3)2 + 3

(x + 3)2 – 13

(x + 4.5)2 – 16.25

The expressions on the left have all been written in the form

(x + a)2 + b

Match each expression to its completed square form.


Exam Question1 Given that x2 + ax – 12 = (x + 4)2 + b, where a and b are integers, find the values of a and b.

a) Complete the square on the left hand-side of the equation

x2 − 2x − 2 = 0

b) Find the roots of the equation x2 − 2x − 2 = 0, giving your answers in the form a ± b

c) Find the coordinates of the minimum point on the curve y = x2 − 2x – 2

d) Find the y-intercept of the curve y = x2 − 2x – 2

e) Sketch the curve y = x2 − 2x – 2 on the axes opposite

√

a =

b =

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Worded Problems Involving Equations of MotionB

Rearranging Formulae2.8Multi-Step FormulaeA

Rearrange each of the formulae below to make x the subject. Match the question to the correct answer.

Each question will require multiple steps.

where a = constant acceleration, s = displacement, t = time, u = initial velocity, and v = final velocity

You will be given the following formulae for the equations of motion in the exam:

y = 2x – 4

y = 2x + 4

y = 2x + 4

y = 2x2 + 4

y = (2x + 4)2

Question Answer

√

√

x =y – 4

2√

x =

x =

y2 – 42

y + 42

√x =y – 4

2

x =(y – 4)2

2

v2 = u2 + 2as v = u + at s = ut + at212

a) A hurricane travelling towards Boston Spa accelerates at 0.1 m/s2, for 30 s. If the hurricane is travelling at 5 m/s𝑢when it arrives in Boston Spa, what was its initial velocity?

b) A cheese starting at rest rolls down a hill, accelerating at 1 m/s2, to reach a speed of 15 m/s. Calculate the distance travelled by the cheese.

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