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Key Stage Three Mathematics
KS3 Maths can seem like one
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...but this brilliant CGP book covers everything you’ll need to solve them all!
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The clearest notes and examples by miles...
Of course — that’s what we do best!
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Hundreds of essential questions...
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Complete Study & Practice
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Key Stage Three
Mathematics
Complete
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Everything you need for the whole course!
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Key Stage Three
Mathematics
There’s a lot to learn in KS3 Maths, that’s for sure. Luckily, help is at hand…
This brilliant CGP book is packed with notes and worked examples that make
every topic crystal-clear. We’ve also included plenty of practice questions and
mixed-topic tests (with answers) to make sure you’ve got to grips with it all.
Once you’ve worked through that lot, you should be ready for the practice
exam at the end of the book — it’ll test you on the whole KS3 course!
Complete
Study & Practice
Everything you need for the whole course!
Contents
Section One — Numbers
Section Three — Graphs
Calculating Tips������������������������������������������ 1
Ordering Numbers�������������������������������������� 3
Addition and Subtraction���������������������������� 4
Multiplying by 10, 100, etc.������������������������ 5
Dividing by 10, 100, etc.���������������������������� 6
Multiplying Without a Calculator���������������� 7
Dividing Without a Calculator�������������������� 8
Multiplying and Dividing with Decimals����� 9
Negative Numbers������������������������������������ 10
Practice Questions�������������������������������� 11
Special Types of Number�������������������������� 13
Prime Numbers����������������������������������������� 14
Multiples, Factors and Prime Factors��������� 15
LCM and HCF������������������������������������������� 16
Practice Questions�������������������������������� 17
Fractions, Decimals and Percentages�������� 19
Fractions��������������������������������������������������� 20
Percentage Basics������������������������������������� 23
Practice Questions�������������������������������� 24
Rounding Numbers����������������������������������� 26
Rounding Errors and Estimating���������������� 28
Practice Questions�������������������������������� 29
Powers������������������������������������������������������ 31
Square Roots and Cube Roots������������������� 32
Standard Form������������������������������������������ 33
Practice Questions�������������������������������� 34
Revision Summary for Section One����������� 36
X and Y Coordinates��������������������������������� 57
Straight Line Graphs��������������������������������� 58
Plotting Straight Line Graphs��������������������� 59
Practice Questions�������������������������������� 60
Finding the Gradient��������������������������������� 62
y = mx + c������������������������������������������������ 63
Practice Questions�������������������������������� 65
Reading Off Graphs���������������������������������� 67
Travel and Conversion Graphs������������������ 68
Real-Life Graphs��������������������������������������� 69
Practice Questions�������������������������������� 70
Solving Simultaneous Equations���������������� 72
Quadratic Graphs������������������������������������� 73
Practice Questions�������������������������������� 74
Revision Summary for Section Three��������� 76
Section 2 — Algebra
Algebra — Simplifying Terms��������������������� 37
Algebra — Multiplying Single Brackets������ 39
Algebra — Multiplying Double Brackets���� 40
Algebra — Taking Out Common Factors���� 41
Practice Questions�������������������������������� 42
Solving Equations������������������������������������� 44
Using Formulas����������������������������������������� 46
Making Formulas from Words������������������� 47
Rearranging Formulas������������������������������� 48
Practice Questions�������������������������������� 49
Number Patterns and Sequences��������������� 51
Inequalities����������������������������������������������� 53
Practice Questions�������������������������������� 54
Revision Summary for Section Two����������� 56
Section Four — Ratio, Proportion
and Rates of Change
Ratios������������������������������������������������������� 77
Direct Proportion�������������������������������������� 79
Inverse Proportion������������������������������������ 80
Practice Questions�������������������������������� 81
Percentage Change����������������������������������� 83
Practice Questions�������������������������������� 86
Metric and Imperial Units������������������������� 88
Converting Units��������������������������������������� 89
More Conversions������������������������������������� 90
Maps and Scale Drawings������������������������� 91
Practice Questions�������������������������������� 92
Best Buy��������������������������������������������������� 94
Density and Speed������������������������������������ 95
Practice Questions�������������������������������� 96
Revision Summary for Section Four����������� 98
Section Five —
Geometry and Measures
Section Six —
Probability and Statistics
Symmetry������������������������������������������������� 99
Quadrilaterals����������������������������������������� 100
Triangles and Regular Polygons��������������� 101
Practice Questions������������������������������ 102
Perimeter and Area��������������������������������� 104
Area of Compound Shapes��������������������� 105
Circles���������������������������������������������������� 106
Practice Questions������������������������������ 107
3D Shapes���������������������������������������������� 109
Nets and Surface Area���������������������������� 110
Volume��������������������������������������������������� 112
Practice Questions������������������������������ 113
Angle Basics������������������������������������������� 115
Geometry Rules�������������������������������������� 116
Parallel Lines������������������������������������������� 117
Geometry Problems�������������������������������� 118
Interior and Exterior Angles��������������������� 119
Practice Questions������������������������������ 120
Transformations�������������������������������������� 122
Enlargements������������������������������������������ 124
Congruent Shapes����������������������������������� 125
Similar Shapes���������������������������������������� 126
Practice Questions������������������������������ 127
Triangle Construction������������������������������ 129
Constructions������������������������������������������ 130
Pythagoras’ Theorem������������������������������� 131
Trigonometry������������������������������������������ 132
Practice Questions������������������������������ 134
Revision Summary for Section Five��������� 137
Probability���������������������������������������������� 138
Relative Frequency��������������������������������� 140
Practice Questions������������������������������ 141
Venn Diagrams��������������������������������������� 143
Practice Questions������������������������������ 145
Data������������������������������������������������������� 147
Graphs and Charts���������������������������������� 148
Practice Questions������������������������������ 150
Mean, Mode, Median and Range������������ 152
Frequency Tables and Averages��������������� 153
Grouped Frequency Tables���������������������� 154
Scatter Graphs���������������������������������������� 155
Practice Questions������������������������������ 156
Revision Summary for Section Six����������� 159
Section Seven — Exam Practice
Mixed Practice Tests�������������������������������� 160
Practice Exam:
Practice Paper 1��������������������������������� 170
Practice Paper 2��������������������������������� 180
Answers�������������������������������������������������� 190
Index������������������������������������������������������ 204
Published by CGP
From original material by Richard Parsons.
Editors:
Chloe Anderson, Mary Falkner, Shaun Harrogate, Frances Rooney and Caley Simpson
Contributors:
Angela Duffy, Alan Mason and Garry Rowlands
With thanks to Jane Appleton and Simon Little for the proofreading.
MHS34DK
Clipart from Corel®
Text, design, layout and original illustrations © Coordination Group Publications Ltd. (CGP) 2017.
All rights reserved.
0800 1712 712 • www.cgpbooks.co.uk
Section One — Numbers
1
Calculating Tips
Ah, the glorious maths universe. OK — maybe it’s not all that glorious, but learn it you must.
And there’s plenty of it. Here are some nifty tips and tricks that you’ll need before you get going.
BODMAS
Brackets, Other, Division, Multiplication, Addition, Subtraction
BODMAS tells you the ORDER in which operations should be done:
Work out Brackets first, then Other things like squaring,
then Divide / Multiply groups of numbers before Adding or Subtracting them.
1.
2. Calculate 10 – 23
Work out 4 + 6 ÷ 2
1) Follow BODMAS —
do the division first...
2) ...then the addition:
If you don’t follow the order of
BODMAS, you get:
4 + 6 ÷ 2 = 10 ÷ 2 = 5
=
=
4+6÷2
4+3
7
1) The cube is an ‘other’
so that’s first:
2) Then do the subtraction:
=
=
10 – 23
10 – 8
2
3. Find (8 – 2) × (3 + 4)
1) Start by working out the brackets:
2) And now the multiplication:
=
=
(8 – 2) × (3 + 4)
6×7
42
Don’t Be Scared of Wordy Questions
You’ll come across all sorts of wordy, real-life questions in maths and
you’ll have to work out what the question’s asking you to do. Remember:
1) Read the question carefully. Work out what bit of maths you need to answer it.
2) Underline the information you need to answer the question —
you might not have to use all the numbers they give you.
3) Write out the question in maths and answer it, showing all your working clearly.
Caley’s Clothing is having a 20% off sale on all clothing items.
A shirt originally cost £25. What is the price of the shirt in the sale?
1) The “20% off” tells you this is a percentages question (covered on page 23).
2) You need £25 (the original price) and 20% off (the percentage).
It doesn’t matter what the shop is called or what the item is.
3) You want to take 20% off £25, so:
20% off is the same as 80% EXAM
TIP
80% of £25 = £20
Don’t forget the units
in your final answer
— this is a question
about cost in pounds,
so the units will be £.
Wordy Questions? But this is maths, not English...
It’s really important to check your working on BODMAS questions. You might be certain you
did the calculation right, but it’s surprisingly easy to do some operations in the wrong order.
Section One — Numbers
2
Calculating Tips
Here are a few more handy tips and tricks that’ll help you on the road to maths glory.
Hidden Brackets in Fractions
This is a bit of a funny one — when you have a fraction with calculations on the top or
bottom you have to imagine they’re in brackets and do them first.
×5
Work out 420
+ 2×3 .
1) Imagine the top and bottom are both in brackets. (20 × 5)
(4 + 2 × 3)
= 100
= 100
10 = 10
(4 + 6)
2) Now follow BODMAS to do the calculation.
Calculators
Make sure you know the important features on your calculator and how to use them.
SHIFT (OR 2ND FUNC)
SQUARE, CUBE AND ROOTS
Press this first if you want to use
something written above a button
(e.g. the pi (p) button).
E.g. 4 x2 gives 4 squared = 16.
And 3√ 27 gives the cube root of 27 = 3.
FRACTIONS
E.g. for 41 press 1
4 .
(If you have a button that looks like
a bc instead, use it in the same way.)
For 1 53 press
1
3
3.6
shift
ON
5 (you
might have to press shift first).
To cancel down a fraction,
enter it and press = .
or S¤D button
Pressing the
also switches an answer between
a fraction and a decimal.
MEMORY ( STO, RCL & M+)
840
E.g. for 12 × 8 :
THE ANSWER
S¤D
3
x2
x
(
)
3
STO
RCL
M+
BRACKETS
7
8
9
Del
AC
4
5
6
÷
1
2
.
×
+
–
Ans
=
0
3
x
×10
Before you jot down 3.6,
think about what it means.
E.g. in a money question, it
might mean £3.60.
Press 12 × 8 = and then STO M+ to
The ‘Ans’ button gives the
store the bottom line in the memory.
number you got when you
Then press 840 ÷ RCL M+ = ,
last pressed the ‘=’ button.
and the answer is 8.75.
Calculators use BODMAS
(see above), so if there’s part
of a question you want the
calculator to do FIRST then
put brackets in to tell it so.
PI (p)
(See page 106.)
The calculator stores the number
for pi (= 3.141...). If it’s above
another button as shown here,
press the shift button first.
Calculators are handy but don’t forget to show your working
Your calculator might be a little bit different to the one shown above — no need to panic though.
Just make sure you can do all of the above functions on your own calculator and you’ll be flying.
Section One — Numbers
3
Ordering Numbers
Here’s a nice easy page filled with examples — it’s all about ordering numbers.
Ordering Whole Numbers
Ascending order just means
smallest to largest.
Write these numbers in ascending order:
–53 17 53 1729 754 421 548 –88 2321
1) First put them into groups, the negative ones first:
negative
2-digit 3-digit 4-digit
–53 –88
17 53
754 421 548
1729 2321
2) Then just put each separate group in order of size:
–88 –53
17 53
421 548 754
1729 2321
Ordering Decimals
1) Do the whole number bit first, then the bit after the decimal point.
2) With numbers between 0 and 1, first group them by the number of 0s at the start.
The group with the most 0s at the start comes first.
Write these numbers in order, from smallest to largest:
11.9 13.56 7.143 11.6 7.7 2.6 8.91
1) First order them by the whole number bit from smallest to largest.
2.6
7.143
7.7
8.91
11.9
11.6
13.56
In decimals, like
in whole numbers,
the value of the
digits decreases
from left to right.
2) If two numbers have the same whole number bit,
then order them by the size of the decimal.
2.6
7.143
7.7
8.91
11.6
11.9
tenths
thousandths
hundredths
0.256
13.56
Write these numbers in order, from smallest to largest:
0.1 0.022 0.53 0.0011 0.027 0.023 0.0023
1) These are all between 0 and 1, so group them by the number of 0s at the start:
2 initial 0s 1 initial 0
no initial 0s
0.0011 0.0023
0.022 0.027 0.023
0.1 0.53
2) Once they’re in groups, just order them by comparing the first non-zero digits.
(If the first non-zero digits are the same, look at the next digit along instead.)
0.0011 0.0023
0.022
0.023
0.027
0.1
0.53
Ordering decimals can seem strange at first, just stick with it
Ordering whole numbers is the easy part — it gets a bit more complicated when you come to
ordering decimals. Learn the rules on this page and you’ll be ordering decimals in no time.
Section One — Numbers
4
Addition and Subtraction
You’re going to have to learn how to add and subtract whole numbers and decimals without using
a calculator. Follow the examples on this page carefully to learn how it’s done.
Adding
1) Line up the units
columns of each number.
2) Add up the columns
from right to left.
3) Carry over any spare tens
to the next column.
Add together 342, 231 and 78.
1) 2 + 1 + 8 = 11
Write 1 and
carry the 1
342
23 1
+ 78
1
1
2) 4 + 3 + 7 + 1 = 15
3) 3 + 2 + 1 = 6
Write 5 and
Write 6 and
carry the 1
you’re done.
Remember to
Line up
342
342
add the carried
units
231
231
number too.
+ 78
+ 78
51
651
1 1
1 1
Subtracting
Work out 372 – 324.
1) Line up the units columns of each number.
2) Working right to left, subtract the bottom
number from the top number.
3) If the top number is smaller than the
bottom number, borrow 10 from the left.
6 12
372
– 324
Line up
units
6 12
372
– 324
048
12 – 4 = 8
6–2=4
3–3=0
You can’t do 2 – 4, so
borrow 10 from the left.
And with Decimals...
The method's just the same, but start instead by lining up the decimal points.
1. Work out 0.7 + 32.2 + 1.65.
0.7 0 Decimal points lined up
3 2 .20
+ 1 .6 5
.5 5
1
It often helps to write in
extra zeros to make all the
decimals the same length
7 + 2 + 6 = 15 —
write 5 and carry the 1
0.7 0
3 2 .20
+ 1 .6 5
3 4 .5 5
1
2. Ben has £5 and spends 91p on a pie.
How much does he have left?
Decimal points lined up
£ 5.00
0 is smaller than 1,
– £0. 9 1
so you can’t do 0 – 1.
4 10
£5.00
– £0. 9 1
9
4 10 10
0 + 2 + 1 + carried 1 = 4
£5 .00
– £0. 9 1
£4 .09
Borrow 10...
...then borrow 10 again
10 – 1 = 9
9–9=0
4–0=4
Always line up either the units or the decimal points...
Decimal questions look tough but if you remember to line up the decimals points they are exactly
the same as whole number questions. Don’t forget about the carried or borrowed numbers either.
Section One — Numbers
5
Multiplying by 10, 100, etc.
Learn a few simple rules on this page and you’ll be able to multiply any number by 10, 100, 1000, etc.
1) To Multiply Any Number by 10
Move the decimal point ONE place BIGGER
and if it's needed, ADD A ZERO on the end.
E.g.
1.6 × 10 = 1 6
6213 × 10 = 6 2 1 3 0
672.12 × 10 = 6 7 2 1 . 2
E.g.
3.5 × 100 = 3 5 0
78 × 100 = 7 8 0 0
3.7734 × 100 = 3 7 7 . 3 4
2) To Multiply Any Number by 100
Move the decimal point TWO places
BIGGER and ADD ZEROS if necessary.
3) To Multiply by 1000 or 10 000, the same rule applies:
Move the decimal point so many places
BIGGER and ADD ZEROS if necessary.
E.g.
99.67 × 1000 = 9 9 6 7 0
1.729 × 10 000 = 1 7 2 9 0
You always move the DECIMAL POINT this much:
1 place for 10, 2 places for 100,
3 places for 1000, 4 for 10 000,
etc.
4) To Multiply by Numbers like 20, 300, 8000 etc.
MULTIPLY by 2 or 3 or 8 etc. FIRST,
then move the decimal point so many places BIGGER (
according to how many zeroes there are.
)
Calculate 110 × 500.
1) First multiply by 5...
2) ...then move the decimal point 2 places.
110 × 5 = 550
550 × 100 = 55000
Moving a decimal point bigger means moving it to the right...
Nothing too strenuous on this page. The key thing to remember is that you move the decimal point
so many places bigger according to how many zeroes are in the number you’re multiplying by.
Section One — Numbers
6
Dividing by 10, 100, etc.
This page tells you how to do the opposite of the previous page. Remember — when you multiply by
10, 100, 1000, etc. you move the decimal point so many places bigger... here, you move it smaller.
1) To Divide Any Number by 10
Move the decimal point ONE place
SMALLER and if it’s needed, REMOVE
ZEROS after the decimal point.
E.g.
32.2 ÷ 10 = 3 . 2 2
6541 ÷ 10 = 6 5 4 . 1
4200 ÷ 10 = 4 2 0 . 0 = 4 2 0
2) To Divide Any Number by 100
E.g.
Move the decimal point TWO places SMALLER
and REMOVE ZEROS after the decimal point.
333.8 ÷ 100 = 3 . 3 3 8
160 ÷ 100 = 1 . 6 0 = 1 . 6
1729 ÷ 100 = 1 7 . 2 9
3) To Divide by 1000 or 10 000, the same rule applies:
Move the decimal point so many
places SMALLER and REMOVE
ZEROS after the decimal point.
E.g.
6587 ÷ 1000 = 6 . 5 8 7
978 ÷ 10 000 = 0 . 0 9 7 8
You always move the DECIMAL POINT this much:
1 place for 10, 2 places for 100,
3 places for 1000, 4 for 10 000,
etc.
4) To Divide by Numbers like 40, 300, 7000 etc.
DIVIDE by 4 or 3 or 7 etc. FIRST, then move the decimal
point so many places SMALLER (i.e. to the left
).
Calculate 180 ÷ 200.
1) First divide by 2...
2) ...then move the decimal point 2 places smaller.
180 ÷ 2 = 90
90 ÷ 100 = 0.9
Moving a decimal point smaller means moving it to the left...
Knowing how to divide by multiples of 10, 100, 1000, etc. will make harder topics like percentages
(see page 23) so much easier. So make sure you learn the rules of how it’s done.
Section One — Numbers
7
Multiplying Without a Calculator
Multiplying with a calculator is easy. The real challenge comes when you don’t have a calculator.
There are lots of methods you can use and three popular ones are shown below.
Just make sure you can do it using whichever method you prefer...
The Traditional Method
a) Work out 32 × 18
Split it into separate 3 2
This is 8 × 32
× 1 8
multiplications, then
2 51 6
add up the results in
3 2 0 This is 10 × 32
columns (right to left).
576
This is 256 + 320
b) Work out 4672 × 52
4 6 7 2
×
52
91 31 4 4
2 33 33 61 0 0
2 4 2 9
44
1
This is 4672 × 2
This is 4672 × 50
This is 9344 + 233600
This method’s got lots of
different names — you might
know it as lattice multiplication
or Chinese multiplication.
Other Methods
Here are a couple more methods you can use.
Calculate 48 × 33.
The Grid Method
The ‘Gelosia’ Method:
1) Split up each number into its units and tens
(and hundreds and thousands if it has them).
48 = 40 + 8 and 33 = 30 + 3
2) Draw a grid, with the
‘bits’ of the numbers
round the outside.
40 8
1) Arrange the calculation
as shown and do 4
easy multiplications
to fill up the grid...
30
4 × 3 = 12
3
4 × 3 = 12
40 × 30
3) Multiply the bits
round the edge to
fill each square.
40
30 1200 240
3 120
40 × 3
4) Finally, add up
the numbers in
the squares.
ION
RE VIS
TASK
8 8 × 30
120 0
2 4 0
1 2 0
+
24
1 5 8 4
24
8×3
4
48 × 33
8
3
3
4 8
8 × 3 = 24
1 2 3
2 4
2 3 8 × 3 = 24
1
2 4
2) Then just add up along the diagonals
(going right to left) to get the answer.
4 8
1 2 3
2 4
1
1
5
2
2
4 3
8
4
You might need to carry a
number over when adding up.
All that multiplication and not a calculator in sight...
Work out 25 × 17, 84 × 34 and 19 × 78. Use a different method for each one. (Answers below.)
Section One — Numbers
(Ans: 425, 2856, 1482)
8
Dividing Without a Calculator
OK, time for some dividing without a calculator — ready for another challenge?
Dividing Whole Numbers
There are two common ways to do division — long division and short division.
Here are some examples of both methods at work. Learn the method you find easier.
You might find it helpful to write out the first
few multiples of the number you’re dividing by.
Short Division
What is 848 ÷ 16?
1) Set out the division as shown.
16 8 4 8
2) Look at the first digit under the line.
8 doesn’t divide by 16, so put a zero
above and look at the next digit.
0
16 8 4 8
3) 16 × 5 = 80, so 16 into 84 goes 5 times,
with a remainder of 84 – 80 = 4.
4) 16 into 48 goes 3 times exactly.
05
16 8 448
Multiples of 16:
16 × 1 = 16
16 × 2 = 32
16 × 3 = 48
16 × 4 = 64
16 × 5 = 80
carry the remainder
0 5 3
16 8 4 48
the top line has
the final answer
So 848 ÷ 16 = 53
Long Division
What is 658 ÷ 28?
1) Set out the division as shown.
2) 6 doesn’t divide by 28. Write a zero
above the 6 and look at the next digit.
3) 28 into 65 goes 2 times, so put a 2 above the 5.
4) Take away 2 × 28 = 56 from 65.
Write the answer underneath, and
move the next digit after the 65 down.
5) 28 into 98 goes 3 times, so put a 3 above the 8.
Take away 3 × 28 = 84 from 98. That leaves 14
and there are no more digits to bring down.
6) You’re left with a remainder,
so give that in your final answer.
28 6 5 8
02 3
28 6 5 8
–5 6
98
–8 4
1 4
Multiples of 28:
28 × 1 = 28
28 × 2 = 56
28 × 3 = 84
28 × 4 = 112
658 ÷ 28 = 23 remainder 14
Try out both methods of dividing to see which you prefer...
Most people think that short division is better because it can be quick — but it’s no good getting a
quick answer if it’s not a right answer. Take your time and use whichever method you like the most.
Section One — Numbers
9
Multiplying and Dividing with Decimals
On the last two pages you’ve seen how to multiply and divide whole numbers without a calculator.
Well, decimals are the same if you just ignore the decimal points — worry about them at the end.
Multiplying Decimals
1) Start by ignoring the decimal points. Do the multiplication using whole numbers.
2) Count the total number of digits after the decimal points in the original numbers.
3) Make the answer have the same number of decimal places.
This is worked out on page 7.
Work out 3.2 × 1.8
1) Do the whole-number multiplication:
2) Count the digits after the decimal points:
3) Give the answer the same number of decimal places:
32 × 18 = 576
3.2 × 1.8 has 2 digits after the
decimal points — so will the answer.
3.2 × 1.8 = 5.76
Dividing a Decimal by a Whole Number
For these, you just set the question out like a whole number division
but put the decimal point in the answer right above the one in the question.
What is 49.8 ÷ 6?
1) Put the decimal point in the answer above the one in the question.
0 .
6 4 49 . 8
2) 6 into 4 doesn’t go, so
carry the remainder of 4.
08.
6 4 49 . 1 8
3) 6 goes into 49 8 times, so
carry the remainder of 1.
08.3
6 4 49 . 1 8
4) 6 goes into 18
3 times exactly.
Dividing a Number by a Decimal
Two-for-one here — this works if you’re dividing a whole number (or a decimal) by a decimal.
What is 8.48 ÷ 0.16?
1) The trick here is to write it as a fraction:
2) Get rid of the decimals by multiplying
top and bottom by 100 (see p5):
3) It’s now a decimal-free division
that you know how to solve:
8.48
8.48 ÷ 0.16 = 0.16
848
= 16
848 ÷ 16 = 53
This is worked out on
the previous page.
Multiplying and dividing on one page — you’re so lucky...
You can use any of the methods for multiplying and dividing whole numbers on these questions too.
Make sure you keep an eye on that decimal point and always check your answer makes sense.
Section One — Numbers
10
Negative Numbers
Numbers less than zero are negative. You can add, subtract, multiply and divide with them.
Adding and Subtracting with Negative Numbers
Use the number line for addition and subtraction involving negative numbers:
numbers get lower in this direction
numbers get higher in this direction
–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0
move this way to subtract
1 2 3 4 5 6 7 8 9 10
move this way to add
What is –3 + 5?
Start at –3 and move 5 places in the positive direction:
So –3 + 5 = 2
–4 –3 –2 –1 0 1 2 3
Work out 3 – 6
Start at 3 and move 6 places in the negative direction:
–4 –3 –2 –1 0 1 2 3 4
Find –1 – 5
So 3 – 6 = –3
Start at –1 and move 5 places in the negative direction:
–6 –5 –4 –3 –2 –1 0
So –1 – 5 = –6
Use These Rules for Combining Signs
+ +
+ –
– +
– –
These rules are ONLY TO BE USED WHEN:
1) Multiplying or dividing
makes +
makes –
makes –
makes +
(invisible + sign)
– + makes – so –3 × 5 = –15
Find: a) –3 × 5
– – makes + so –18 ÷ –3 = 6
b) –18 ÷ –3
2) Two signs appear next to each other
ION
RE VIS
TIP
Work out: a) 3 – –9
– – makes + so 3 – –9 = 3 + 9 = 12
– – makes + and + – makes –
so 2 – –8 + –12 = 2 + 8 – 12 = –2
b) 2 – –8 + –12
Are you positive you’ve learnt all about negative numbers?
Always start a calculation by combining signs — if they’re the same it makes + and if they’re
different it makes –. Then it’s just a straightforward add, subtract, multiply or divide question.
Section One — Numbers
11
Warm-Up and Practice Questions
These warm-up questions will test whether you’ve learnt the facts properly. Go back over any bits
you don’t know, then there will be no nasty surprises when it comes to the practice questions.
Warm-up Questions
1) Find the value of:
Try these without
using a calculator.
a) 11 – 5 × 2 b) 3 × 6 + 15 ÷ 5 c) (6 × 3) ÷ (3 × 2)
2) Put these numbers in ascending order:
a) 54 442 –7 304 45 –9 b) 0.222 0.056 0.207 0.0012 0.026 0.0015
3) Work out: a) 113 + 645 + 39 b) 1239 – 387 c) 0.58 + 1.47 + 16.4
4) Carry out the following multiplications and divisions:
a) 4.9 × 100 b) 1729 × 10 c) 3.3 × 2000 d) 700 × 500
e) 3.33 ÷ 10 f) 85.21 ÷ 1000 g) 3900 ÷ 300 h) 56000 ÷ 800
5) Without using a calculator work out:
a) 55 × 18 b) 1502 × 35 c) 128 ÷ 8 d) 550 ÷ 12
6) Carry out the following decimal multiplications and divisions:
a) 1.8 × 34 b) 17.6 × 0.3 c) 36.6 ÷ 0.6 d) 51 ÷ 1.7
7) Work out: a) –6 + 11 b) –5 – 10 c) –3 × –6 d) 21 ÷ –7
Practice Questions
Once you’ve managed to get through the warm-ups, you can have a go at these practice questions.
Here’s a worked example to start you off, and then you’re on your own for some more practice.
1
Jenny has 37 boxes of coloured pencils and each box contains 68 pencils.
a)
How many pencils are there altogether?
3 7
×68
2 95 6
2 24 2 0
25 1 6
This is 8 × 37
This is 60 × 37
1
2516
...................................
[2 marks]
b)
Jenny wants to separate the pencils into boxes of 19 pencils.
How many boxes would she completely fill?
19 × 1 = 19
19 × 2 = 38
19 × 3 = 57
19 × 4 = 76
19 × 5 = 95
Write out the first
few multiples of 19
0 1 3 2 remainder 8
1 9 2 561 46
132
boxes
..........................
[2 marks]
Section One — Numbers
12
Practice Questions
2
Use BODMAS to work out the values of the following calculations.
a)
5–6×4+5
.........................
[2 marks]
55 ×10 ÷ 5
4×5 – 9
b)
.........................
[2 marks]
3
Put each set of numbers in order of size, starting with the smallest.
a)
4.8
5.72
3.09
4.79
5.17
3.5
..................................................................................................................
[2 marks]
b)
0.4
0.303
0.43
0.31
0.44
0.043
..................................................................................................................
[2 marks]
4
A school bought 80 Maths books at a total cost of £1200.
a)
How much does one book cost?
£ .........................
[2 marks]
A book is made up of a front and back cover and 300 pages.
The front and back covers are 1 mm thick each.
The total thickness of the book is 38 mm.
b)
Work out the thickness of one page of the book.
......................... mm
[3 marks]
Section One — Numbers
13
Special Types of Number
There are lots of special number definitions on this page — keep calm and power on through it.
Integers
An integer is another name for a whole number — either a positive or negative number, or zero.
Integers:
–365, 0, 1, 89, 23 567 890
3
2
Not integers: 0.5, 3 , 7 , 13 4 , –6.66, p
You’ll see this word cropping up everywhere,
so make sure you know what it means.
Rational and Irrational Numbers
All numbers fall into one of these two categories:
RATIONAL NUMBERS — numbers that can be written as fractions.
Most numbers you’ll deal with are rational numbers, e.g. integers, fractions and most decimals.
IRRATIONAL NUMBERS — numbers that can’t be written as fractions.
They are special types of decimals that never end and don’t repeat.
Many roots are irrational — like 2 , 3 and 5 . p is irrational too.
Real Numbers
REAL NUMBERS — all the numbers you’ll have come across, e.g. integers,
fractions, decimals, rational and irrational numbers...
1) There are an infinite amount of real numbers — you could never count them all.
(The same goes for integers, rational and irrational numbers.)
2) Wrap your head around this one — between any two real numbers,
there are an infinite amount of real numbers.
Squares and Cubes
Make sure you know what powers are and how they work (see p31).
THE
SQUARES:
THE
CUBES:
12
1
22
4
32 42 52 62 72 82 92 102 112 122 132 142 152
9 16 25 36 49 64 81 100 121 144 169 196 225
(1×1) (2×2) (3×3) (4×4) (5×5) (6×6) (7×7) (8×8) (9×9)(10×10) (11×11) (12×12) (13×13) (14×14) (15×15)
13
23
33
43
53
103
1
8
27
64
125
1000
(1×1×1) (2×2×2) (3×3×3) (4×4×4) (5×5×5) (10×10×10)
Challenge a friend — who can count to infinity quickest...
Lots of different types of number for you to learn here. Remember that all numbers you’ll come
across are real numbers and all real numbers are either rational or irrational numbers.
Section One — Numbers
14
Prime Numbers
There’s one more special type of number you need to know about — the prime numbers...
A PRIME Number Only Divides by 1 and Itself
Prime numbers are all the numbers that only come up in their own times table:
2
3
5
7
11
13
17
19
23
29
31
37 ...
The only way to get ANY PRIME NUMBER is: 1 × ITSELF
E.g.
The only numbers that multiply to give 3 are 1 × 3
The only numbers that multiply to give 19 are 1 × 19
Show that 18 is not a prime number.
Just find another way to make 18 other than 1 × 18:
3 × 6 = 18
18 divides by other numbers apart from 1 and 18, so it isn’t a prime number.
Five Important Facts
1)
2)
3)
4)
5)
1 is NOT a prime number.
2 is the ONLY even prime number.
The first four prime numbers are 2, 3, 5 and 7.
Prime numbers end in 1, 3, 7 or 9 (2 and 5 are the only exceptions to this rule).
But NOT ALL numbers ending in 1, 3, 7 or 9 are primes, as shown here:
(Only the circled ones are primes.)
2 3 5 7
11 13 17 19
21 23 27 29
31 33 37 39
41 43 47 49
51 53 57 59
61 63 67 69
How to FIND Prime Numbers — a very simple method
1) All primes (above 5) end in 1, 3, 7 or 9 — ignore any numbers that don’t end in one of those.
2) To find which of them ACTUALLY ARE primes you only need to divide each one by 3 and 7.
If it doesn’t divide exactly by either 3 or 7 then it’s a prime.
This works for primes up to 120.
Find all the prime numbers in this list:
51, 52, 53, 54, 55, 56, 57, 58, 59
1) Get rid of anything that doesn’t end in 1, 3, 7 or 9:
51, 52, 53, 54, 55, 56, 57, 58, 59
2) Now try dividing 51, 53, 57 and 59 by 3 and 7:
51 ÷ 3 = 17 so 51 is NOT a prime number
53 ÷ 3 = 17.666... and 53 ÷ 7 = 7.571... so 53 is a prime number
57 ÷ 3 = 19 so 57 is NOT a prime number
59 ÷ 3 = 19.666... and 59 ÷ 7 = 8.428... so 59 is a prime number
So the prime numbers in the list are 53 and 59.
ION
RE VIS
TIP
A number is prime if it only divides by 1 and itself...
Remember — 1 is not a prime number, it’s the only exception to the rule.
Section One — Numbers
15
Multiples, Factors and Prime Factors
Make sure you know about prime numbers (p14) before you try to find prime factors.
Multiples and Factors
The MULTIPLES of a number are just the values in its times table.
Find the first 5 multiples of 12.
You just need to find the first 5 numbers in the 12 times table:
12
24
36
48
60
The FACTORS of a number are all the numbers that divide into it exactly. Here’s how to find them:
Find all the factors of 28.
Increasing by
1 each time
1 × 28
2 × 14
3×—
4×7
5×—
6×—
7×4
1) Start off with 1 × the number itself, then try 2 ×,
then 3 × and so on, listing the pairs in rows.
2) Try each one in turn. Cross out the
row if it doesn’t divide exactly.
3) Eventually, when you get a number repeated, stop.
So the factors of 28 are:
1, 2, 4, 7, 14, 28
4) The factors are the numbers you haven’t crossed out.
Finding Prime Factors — The Factor Tree
Any whole number can be written as a string of prime numbers all multiplied together — this is
called a prime factorisation. The easiest way to find a prime factorisation is using a factor tree.
Find the prime factorisation of 280.
1) Start with the number at the top,
and split it into factors as shown.
280
10
28
4
2
2) Every time you get a prime, ring it.
7
2
2
5
So 280 = 2 × 2 × 2 × 5 × 7
= 23 × 5 × 7
You could split 280 into 14 and 20 or 7 and 40
— you’ll always get the same prime factorisation.
3) Keep going until you can’t go further
(i.e. you’re just left with primes),
then write the primes out in order.
4) (Optional) Group numbers that are the
same into powers. (i.e. 2 × 2 × 2 = 23)
The prime factorisation of a number is always the same, no matter how you split it up.
Every number has a unique prime factorisation — no two are the same.
ION
RE VIS
TASK
Any number can be written as a product of prime factors...
If this is all new to you, have another read through it — especially prime factorisation.
Then scribble down a three digit number and try to find the prime factorisation.
Section One — Numbers
16
LCM and HCF
Here are two big fancy names for you — but don’t be put off, they’re both easy.
LCM — ‘Lowest Common Multiple’
‘Lowest Common Multiple’ sounds a bit complicated, but all it means is this:
The SMALLEST number that will DIVIDE BY ALL the numbers in question.
METHOD: 1) LIST the MULTIPLES of ALL the numbers.
2) Find the SMALLEST one that’s in ALL the lists.
3) That’s the LCM.
Find the lowest common multiple (LCM) of:
a) 3 and 4 b) 9 and 15
Multiples of 3:
Multiples of 4:
3, 6, 9, 12, 15, ...
4, 8, 12, 16, ...
Multiples of 9:
Multiples of 15:
So the LCM of 3 and 4 is 12.
9, 18, 27, 36, 45, 54, 63, ...
15, 30, 45, 60, 75, 90, ...
So the LCM of 9 and 15 is 45.
HCF — ‘Highest Common Factor’
‘Highest Common Factor’ — all it means is this:
The BIGGEST number that will DIVIDE INTO ALL the numbers in question.
METHOD: 1) LIST the FACTORS of ALL the numbers.
2) Find the BIGGEST one that’s in ALL the lists.
3) That’s the HCF.
Find the highest common factor (HCF) of 24, 42 and 84.
Factors of 24 are:
Factors of 42 are:
Factors of 84 are:
1, 2, 3, 4, 6, 8, 12, 24
1, 2, 3, 6, 7, 14, 21, 42
1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
So the highest common factor (HCF) of 24, 42 and 84 is 6.
List the multiples or factors when finding LCMs and HCFs
Careful when you’re listing the factors of a number — make sure you use the proper method
(as shown on p15). This way you’re much less likely to make any silly mistakes.
Section One — Numbers
17
Warm-Up and Practice Questions
It’s time to face the music — try these warm-up questions to see how much you’ve learnt.
When you feel like you’re ready you should move on to the tougher practice questions.
Warm-up Questions
Try these without using a calculator.
1)
What’s the difference between
rational numbers and irrational numbers?
2)
Write down the first ten square numbers and the first five cube numbers.
3)
Write down all the prime numbers from this list: 49, 63, 38, 73, 77, 18, 39, 83
4)
Find the first five multiples of: a) 8
5)
Find all the factors of:a) 36b) 56c) 64
6)
Express 300 as a product of its prime factors.
7)
Find the lowest common multiple (LCM) of 9 and 12.
8)
Find the highest common factor (HCF) of 36 and 84.
b) 15
c) 21
Practice Questions
Take a look at the worked question below and see if you can follow the working. Once you’ve
read through it carefully you can have a go at the practice questions on the next page.
1
The factors of a number are all the numbers that divide into it.
a)
Find all the factors of the following numbers.
1, 2, 11, 22
22 ......................................................................................................................................
1, 41
41 ......................................................................................................................................
1, 3, 11, 33
33 ......................................................................................................................................
1, 2, 5, 10, 11, 22, 55, 110
110 ......................................................................................................................................
[4 marks]
b)
Which of the four numbers above is prime? Give a reason for your answer.
41 is the only prime number. It is the only number from the list
...............................................................................................................................................
where the factors are just 1 and itself.
...............................................................................................................................................
[2 marks]
c)
What is the HCF of 22, 33 and 110?
Look at the lists above to find the highest
number which is a factor of 22, 33 and 110.
11
.....................
[1 mark]
Section One — Numbers
18
Practice Questions
2
Find the first two numbers that are both square numbers and cube numbers.
.........................
[2 marks]
3
Factor trees allow you to find the prime factors of a number.
a)
Complete the factor tree.
168
12
.........
.........
14
.........
.........
.........
.........
[2 marks]
b)
Write down the prime factorisation of 168
........................................
[1 mark]
4
Jacob is making some iced cupcakes. One pack of cake mix will make 12 cupcakes and one
tube of icing will cover 15 cupcakes. Jacob doesn’t want any cupcake mix or icing left over.
What is the minimum number of cupcakes that Jacob has to make?
.........................
[3 marks]
Section One — Numbers
19
Fractions, Decimals and Percentages
Fractions, decimals and percentages are three different ways of describing when you’ve got
part of a whole thing. They’re closely related and you can convert between them.
This table shows the really common conversions which you ought to know straight off:
Fractions with a 1 on the
1 1 1
top (e.g. 2 , 3 , 4 , etc.)
are called unit fractions.
Fraction
Decimal
Percentage
1
2
1
4
3
4
1
3
2
3
1
10
2
10
1
5
2
5
0.5
50%
0.25
25%
0.75
75%
0.333333...
1
33 3 %
0.666666...
2
66 3 %
0.1
10%
0.2
20%
0.2
20%
0.4
40%
0.3333... and 0.6666...
are known as ‘recurring’
decimals — the same
pattern of numbers carries
on repeating itself forever.
The more of those conversions you learn, the better — but for those that you don’t know,
you must also learn how to convert between the three types. These are the methods:
Fraction
Fraction
Divide
7
E.g. 20 is 7 ÷ 20
The awkward one
Decimal
× by 100
Percentage
= 0.35
e.g. 0.35 × 100
= 35%
Decimal
÷ by 100
Percentage
Converting decimals to fractions is a bit more awkward.
The digits after the decimal point go on the top, and a power of 10 on the bottom —
with the same number of zeros as there were decimal places.
6
3
7
0.6 = 10 0.3 = 10 0.7 = 10 etc.
12
0.12 = 100 ION
RE VIS
TIP
78
0.78 = 100 5
0.05 = 100 etc.
345
908
0.345 = 1000 0.908 = 1000 24
0.024 = 1000 etc.
These can often
be cancelled down
— see p20.
Fractions, decimals and percentages are all about proportion
Learn the conversions in the table at the top of the page off by heart. Work out any other
conversions using the four methods for converting — decimals to fractions is the tricky one.
Section One — Numbers
20
Fractions
This page tells you how to deal with fractions without using your calculator.
1
4
Equivalent Fractions
4
...is equivalent to... 16
1) Equivalent fractions are equal in size...
2) ...but the numbers on the top and bottom are different.
3) To get from one fraction to an equivalent one — MULTIPLY top and bottom by the SAME NUMBER:
×2
×5
×100
1 =2
2 4
3 = 15
4 20
1 = 100
5 500
×2
×5
×100
Cancelling Down
1) You sometimes need to simplify a fraction by ‘cancelling down’.
2) This means DIVIDING top and bottom by the SAME NUMBER.
3) To get the fraction as simple as possible, you might have to do this more than once:
÷4
4 =1
12 3
÷4
÷10
÷2
20 = 2 = 1
40 4 2
÷10
÷2
÷3
÷2
÷3
72 = 24 = 12 = 4
90 30 15 5
÷3
÷2
÷3
Ordering Fractions
Ordering fractions — first you just have to make the bottom numbers (denominators)
of the fractions equal. Then compare the size of the top numbers (numerators).
Put these fractions in order of size from smallest to largest.
5 3 2
6, 4, 3
EXAM
TIP
1) Look at the denominators of the fractions:
3, 4 and 6
2) They are all factors of 12 so change each fraction
(make equivalent fractions) so the denominator is 12.
5 = 10 3 = 9 2 = 8
6 12 , 4 12 , 3 12
3) Now order the fractions by comparing their numerators.
8 9 10
12 , 12 , 12
4) Put the fractions back into their original form.
2 3 5
3, 4, 6
Always cancel down fractions as far as possible...
A question might ask you to simplify a fraction or give your answer in its simplest form —
don’t be put off, all it means is cancel the fraction down until you can’t cancel any further.
Section One — Numbers
21
Fractions
It’s time for some more adding, subtracting, multiplying and dividing — but this time it’s with fractions.
There are a few more rules to remember so read through this page carefully.
Adding and Subtracting
1) If the bottom numbers are the same, add or subtract the TOP NUMBERS ONLY,
leaving the bottom number as it is.
Adding: 3 + 2 + 1 = 6
Subtracting: 17 – 11 = 6 = 1
7 7 7 7 30 30 30 5
Remember to cancel
your answer down as
far as possible.
2) If the bottom numbers are different, you have to make them
the same using equivalent fractions (see previous page).
5 1
Work out 8 + 3 .
1) Look at the denominators of the fractions:
8 and 3
2) They are both factors of 24 so change each
fraction so the denominator is 24.
5 = 15 1 = 8
8 24 , 3 24
3) Now add the fractions —
simplify your answer if possible.
15 + 8 = 23
24 24 24
Multiplying
1) Multiply the top numbers to find the numerator...
2) ...and multiply the bottom numbers to find the denominator.
E.g. 3 # 4 = 3# 4 = 12
5 7 5#7 35
Dividing
1) Turn the 2nd fraction UPSIDE DOWN...
2) ...and then multiply, as shown above.
E.g. 3 ' 9 = 3 # 5 = 15 = 5
7 5 7 9 63 21
6
7
Fractions turned upside down are called reciprocals — so the reciprocal of 7 is 6 .
1
The reciprocal of a whole number is just 1 divided by it — so the reciprocal of 3 is 3 .
E.g.
7
1= 7
= 7
10 ' 4 10 # 4 40
What fraction of this page have you learnt?
When adding or subtracting fractions, remember that the denominators of the fractions always have
to be equal. When you’re multiplying or dividing, the denominators don’t have to be equal.
Section One — Numbers
22
Fractions
Mixed Numbers
Mixed numbers have an integer part and a fraction part. If you have to do a calculation with
them, just turn them into improper fractions first, then carry on as normal.
In an improper fraction, the
top number is larger than
the bottom number.
1. Write 3 54 as an improper fraction.
1) Think of the mixed number as an addition: 2) Turn the integer part into a fraction:
3 54 = 3 + 54
+
3 + 54 = 155 + 54 = 15 5 4 = 19
5
2. Write 23
3 as a mixed number.
1) Divide the top number by the bottom.
2) The answer gives the whole number part.
3) The remainder goes on top of the fraction.
23 ÷ 3 = 7 remainder 2
23 = 7 2
3
3
You can add, subtract, multiply and divide mixed numbers by turning them into improper fractions
and then using the rules on the previous page.
Adding: 1 5 + 3 1 = 12 + 22 = 34 = 4 6 Subtracting: 5 1 – 1 2 = 16 – 5 = 11 = 3 2
3
3
3 3
3
3
7 7
7
7
7
7
Multiplying: 3 1 × 1 = 7 × 1 = 7 Dividing: 2 ÷ 2 2 = 2 ÷ 8 = 2 × 3 = 6 = 1
3
3 3 3 3 8 24 4
2 4 2 4 8
Finding a Fraction of Something
1) Multiply the ‘something’ by the TOP of the fraction...
2) ...and divide it by the BOTTOM.
E.g. 9 of £360 = £360 × 9 ÷ 20 = £162
20
Finding One Thing as a Fraction of Another
You can write one number as a fraction of another number just by putting the first number over the
second and cancelling down. This works if the first number is bigger than the second number too —
you’ll just end up with a fraction greater than 1.
132 11
144 12
E.g. 132 is 144 = 12 of 144, and 144 is 132 = 11 of 132.
Don’t be put off by mixed number calculations
Mixed number calculations are just the same as normal fraction calculations once you convert to
improper fractions — don’t forget to turn your answer back into a mixed number.
Section One — Numbers
23
Percentage Basics
These simple percentage questions shouldn’t give you much trouble. Especially if you remember:
20 .
1) ‘Per cent’ means ‘out of 100’, so 20% means ‘20 out of 100’ = 100
2) If a question asks you to work out the percentage OF something
you can replace the word OF with a multiplication (×).
Two Different Question Types
Type 1 — “Find x% of y”
Turn the percentage into a decimal, then multiply.
Find 18% of £4.
1) Write 18% as a decimal:
2) Multiply 0.18 by £4:
18% = 18 ÷ 100 = 0.18
0.18 × £4 = £0.72
Find 125% of 600 kg.
1) Write 125% as a decimal:
2) Multiply 1.25 by 600 kg:
125% = 125 ÷ 100 = 1.25
1.25 × 600 kg = 750 kg
Type 2 — “Express x as a percentage of y”
Divide x by y, then multiply by 100.
Give 28p as a percentage of 98p.
Divide 28p by 98p, then multiply by 100:
(28 ÷ 98) × 100 = 28.57% (2 d.p.)
Farmer Littlewood measured the width of his prized pumpkin at the
start and end of the month. At the start of the month it was 84 cm
wide and at the end of the month it was 1.32 m wide. Give the width
at the end of the month as a percentage of the width at the start.
1) Make sure both amounts are in the same units — convert 1.32 m to cm:
1.32 m is 1.32 × 100 = 132 cm
2) Divide 132 cm by 84 cm, then multiply by 100:
(132 ÷ 84) × 100 = 157.14% (2 d.p.)
Make sure you know the percentage basics really well...
There are more percentage questions coming up later in the book (p83-85), so make sure you learn
the basics before you move on. If anything is not clear have another read through these examples.
Section One — Numbers
24
Warm-Up and Practice Questions
It’s about time you had a bit of practice on fractions, decimals and percentages. Have a go at these
warm-up questions first — they’ll be a good indication of how much you’ve learnt.
Warm-up Questions
1)
2)
3)
Try questions 1-5 without using a calculator.
Turn the following decimals into fractions
and reduce them to their simplest form.
a) 0.8b) 0.04c) 0.23d) 0.999
2
3
1
Which is greater: a) 22% or 3 b) 0.6 or 5 c) 40% or 8
20
9
24
Cancel these fractions down as far as possible: a) 28
b) 36
c) 60
5)
1
3
7
Turn these mixed numbers into improper fractions: a) 3 2
b) 6 4
c) 8 8
3 5
2 1
1 9
11 7
1 1
Work out: a) 5 × 6 b) 7 ÷ 8 c) 8 + 8 d) 5 – 5 e) 3 + 4
6)
Find:
a) 39% of 505
7)
Give:
a) 690 m as a percentage of 980 m
4)
b) 123% of 355
b) 1.2 km as a percentage of 740 m
Practice Questions
Have a look at the practice questions below, I’ve written in the answers for you. Don’t be too jealous
though, there’s plenty more to keep you going on the next page. Enjoy.
1
A pet shop has a 30 kilogram sack of bird food. The food needs to be
repacked into bags, each containing three quarters of a kilogram.
How many bags of food can be made?
You need to find how many
3
times 4 goes into 30
Flip the second
fraction upside down
120
4
3
30 ÷ 4 = 30 × 3 = 3 = 40
40
.........................
bags
[2 marks]
2
Approximately 90 000 people attended the last FA Cup Final.
Five eighths of these were male.
How many females were in the crowd?
5
5
3
If 8 of the crowd were male then 1 – 8 = 8 were female.
3
8 of 90 000 = 90 000 × 3 ÷ 8
= 270 000 ÷ 8 = 33 750 females.
33 750
.........................
[2 marks]
Section One — Numbers
25
Practice Questions
3
Place the following numbers in order, beginning with the smallest.
1
13 9
0.12
17%
9%
20
100
50
..........................................................................................
[2 marks]
4
Helen has 21 marbles. 2 of them are blue, 1 are red and the rest are green.
7
3
How many green marbles does Helen have?
.........................
[2 marks]
5
An antique painting originally bought for £4000 is being put up for auction.
a)
There was a bid of 120% of the original price. How much was the bid?
£ .........................
[2 marks]
b)
The painting was sold for £10 000. What is this as a percentage of the original price?
......................... %
[2 marks]
Section One — Numbers
26
Rounding Numbers
You need to be able to use 3 different rounding methods.
We’ll do decimal places first, but there’s the same basic idea behind all three.
If you’re rounding to
2 d.p. the last digit is
the second digit after
the decimal point.
Decimal Places (d.p.)
To round to a given number of decimal places:
1) Identify the position of the ‘last digit’ from the number of decimal places.
2) Then look at the next digit to the right — called the decider.
3) If the decider is 5 or more, then round up the last digit.
If the decider is 4 or less, then leave the last digit as it is.
4) There must be no more digits after the last digit (not even zeros).
What is 21.84 correct to 1 decimal place?
2 1 . 8 4 = 21.8
LAST DIGIT to be written
(1st decimal place because
we’re rounding to 1 d.p.)
DECIDER
The LAST DIGIT stays the same
because the DECIDER is 4 or less.
What is 39.7392739 to 2 decimal places?
39 . 7392739 = 39.74
LAST DIGIT to be written
(2nd decimal place because
we’re rounding to 2 d.p.)
DECIDER
The LAST DIGIT rounds UP
because the DECIDER is 5 or more.
Watch Out for Pesky Nines
If you have to round up a 9 (to 10), replace the 9 with 0, and add 1 to digit on the left.
Round 48.897 to 2.d.p.:
90
48.897
LAST DIGIT
EXAM
TIP
48.89
48.90 to 2 d.p.
The question asks for 2 d.p. so
you must put 48.90 not 48.9.
DECIDER
Always take extra care when the last digit is a 9...
Make sure you always give your answer to the number of decimal places asked for.
Section One — Numbers
27
Rounding Numbers
Here are the other two rounding methods — they’re each slightly different in their own way.
Significant Figures (s.f.)
The 1st significant figure of any number is the first digit which isn’t a zero.
The 2nd, 3rd, etc. significant figures follow straight after the 1st — they’re allowed to be zeros.
0.002309
SIG. FIGS:
1st 2nd 3rd 4th
506.07
1st 2nd 3rd 4th
To round to a given number of significant figures:
1) Find the last digit — if you’re rounding to, say, 3 s.f.,
then the 3rd significant figure is the last digit.
2) Use the digit to the right of it as the decider, just like for d.p.
3) Once you’ve rounded, fill up with zeros, up to but not beyond the decimal point.
Round 1276.7 to 2 significant figures.
LAST DIGIT is the 2nd sig. fig.
Need two zeros to fill
up to decimal point.
1276.7 = 1300
DECIDER is 5 or more
LAST DIGIT rounds UP
To the Nearest Whole Number, Ten, Hundred etc.
You might be asked to round to the nearest whole number, ten, hundred, thousand or million.
1) Identify the last digit, e.g. for the nearest whole number it’s the
units position, and for the ‘nearest ten’ it’s the tens position, etc.
2) Round the last digit and fill in with zeros up to the decimal point.
Round 61729 to the nearest thousand.
LAST DIGIT is in the
Fill in 3 zeros up to decimal point.
‘thousands’ position
6 1 7 2 9 = 62000
DECIDER is 5 or more
LAST DIGIT rounds UP.
Have you figured out how significant this page is?
Lots of rounding for you on this page but it’s basically the same thing every time — identify the last
digit and then use the digit to the right (decider) to decide if you need to round up or leave it.
Section One — Numbers
28
Rounding Errors and Estimating
When you round a number using any of the methods on the previous two pages your
answer won’t be exact — it will have some amount of error.
Errors in Rounding
The error when rounding is given by ROUNDED VALUE – ACTUAL VALUE.
What is the error when 478 is given to 1 significant figure?
1) Round the number to 1 sig. fig.
478 = 500 (1 sig. fig.)
2) Subtract the actual value from the rounded value. 500 – 478 = 22
If you’re given a rounded value and asked to find a range of values that the actual value
could have been, remember:
Whenever a value is rounded to a given unit the actual value
can be up to HALF THE ROUNDING UNIT bigger or smaller.
Between which two values could these rounded values lie?
Half the rounding unit
Smallest value
10 ÷ 2 = 5
70 – 5 = 65
a) 70 to the nearest 10
Biggest value
70 + 5 = 75
b) 1100 to the nearest 100
100 ÷ 2 = 50
1100 – 50 = 1050
1100 + 50 = 1150
c) 9.2 to 1 d.p.
0.1 ÷ 2 = 0.05
9.2 – 0.05 = 9.15
9.2 + 0.05 = 9.25
d) 99 to 2 s.f.
1 ÷ 2 = 0.5
99 – 0.5 = 98.5
99 + 0.5 = 99.5
e) 1.14 to 3 s.f.
0.01 ÷ 2 = 0.005
1.14 – 0.005 = 1.135
1.14 + 0.005 = 1.145
The biggest value doesn’t actually round to the rounded value (it rounds up) — it’s called the
upper limit. You can show this if you give the range of values as an inequality (see p53).
E.g. in part c) above the range of the possible x values would be 9.15 ≤ x < 9.25.
Estimating
When you’re estimating just follow this simple rule:
Usually 1 significant figure is easiest.
Round everything off to nice convenient numbers and then work out the answer.
63.26 × 13.12
.
16.9
Round each number to 1 s.f and do the
63.26 × 13.12 60 × 10 = 600 =
.
30
calculation with the rounded numbers.
16.9
20
20
Estimate the value of
means ‘approximately equal to’.
Errors can be tricky to get your head around at first
The most confusing bit on this page is finding a range of actual values from a rounded value. Make
sure you can find half the rounding unit and then just add and subtract it from the rounded value.
Section One — Numbers
29
Warm-Up and Practice Questions
There’s nothing to be scared of in this section, just a few things to remember. Have a
go at these warm-up questions and find out if you know everything that you should.
Warm-up Questions
1) Round the number 7.8953 to the following number of decimal places:
a) 1 d.p. b) 2 d.p. c) 3 d.p.
2) Round the number 72.567 to the following number of significant figures (s.f.):
a) 1 s.f. b) 2 s.f. c) 3 s.f.
3) Round 2557 to the: a) nearest ten b) nearest hundred c) nearest thousand
4) What is the error when 957 is rounded to the nearest 100?
5) x is 2800 given to 2 significant figures. Give the range of possible x values as an inequality.
6) Estimate the answer to:
67.24 × 102.58
a) 98.6 × 8.1 b) (54.34 + 45.25) ÷ 18.97 c) 10.59 – 3.256
Practice Questions
I’ve taken the liberty of filling in the answer to the practice question below. I just couldn’t resist.
Luckily for you, on the next page there are loads more practice questions.
1
a)
Which of the following is the best estimate of the answer to 61.24 ÷ 5.92?
67891011
60 ÷ 6 = 10
10
.........................
[1 mark]
b)
Which of the following is the best estimate of the answer to 121.6 × 0.49?
66.56065600650
120 × 0.5 = 120 ÷ 2 = 60
Multiplying by a half is
the same as dividing by 2.
60
.........................
[1 mark]
c)
Estimate the answer to 38.4 × 28.2 .
6.17 × 4.02
Round everything to
1 significant figure
40 × 30
1200
6 × 4 = 24
= 50
50
.........................
[2 marks]
Section One — Numbers
30
Practice Questions
2
Harry is organising a school trip for 154 pupils. The cost of the transport is £614.
Harry decides that to cover the cost he must charge pupils £40 each.
Use estimation to check Harry's calculation. Is Harry right?
.............................
[2 marks]
3
Rob is weighing a horse. He says, “the horse is 510 kg to the nearest 10 kg”.
a)
What is the maximum possible error in Rob’s estimation?
......................... kg
[1 mark]
b)
Give the range of possible weights for the horse as an inequality.
........................................................
[2 marks]
4
A groundsman marks out a tennis court.
a)
He makes the court 23 metres long, to the nearest metre.
What is the shortest possible length of the court?
......................... m
[1 mark]
b)
He makes the court 10 metres wide, to the nearest metre.
What is the shortest possible width of the court?
......................... m
[1 mark]
c)
Jordan wants to know how many complete laps of the tennis court she should run to be
sure of running at least 1 km. Use your answer to parts (a) and (b) to find how many
times she should run around the court.
.........................
[2 marks]
Section One — Numbers
31
Powers
You’ve already seen a few powers on page 13 — well, that’s just the tip of the iceberg.
Powers are a very Useful Shorthand
1) Powers are ‘numbers multiplied by themselves so many times’:
2 × 2 × 2 × 2 = 24 (‘two to the power 4’)
4 × 4 × 4 × 4 × 4 × 4 = 46 (‘four to the power 6’)
8 × 8 × 8 × 8 × 8 × 8 × 8 × 8 = 88 (‘eight to the power 8’)
2) The powers of ten are really easy — the power tells you the number of zeros:
101 = 10
102 = 100
103 = 1000
104 = 10 000
x 4 = to get 94 = 6561.
3) Use the x button on your calculator to find powers — press 9
4) Anything to the power 1 is just itself, e.g. 41 = 4, and anything to the power 0 is 1, e.g. 50 = 1.
5) 1 to any power is still 1, e.g. 1457 = 1.
6) When you have a number to a negative power —
turn the number upside down and make the power positive.
4–5 = 15 = 1
1024
4
The Three Power Rules
1) When MULTIPLYING, you ADD the powers.
e.g.
24 × 26 = 24 + 6 = 210
75 × 7–3 = 75 + (–3) = 75 – 3 = 72
These two rules only
work for powers of
the same number.
2) When DIVIDING, you SUBTRACT the powers.
e.g.
411 ÷ 44 = 411 – 4 = 47
y3 ÷ y–4 = y3 – (–4) = y3 + 4 = y7
3) When RAISING one power to another, you MULTIPLY the powers.
e.g.
ION
RE VIS
TIP
(32)4 = 32 × 4 = 38(x4)–3 = x4 × –3 = x–12
Don’t be put off by letters
— they obey the same rules.
Learn all of the rules about powers
The three power rules are important but don’t forget the little things like: anything
to the power 1 is itself, anything to the power 0 is 1 and 1 to any power is still 1.
Section One — Numbers
32
Square Roots and Cube Roots
Square roots and cube roots are the opposites of squaring and cubing.
Square Roots
‘Squared’ means ‘multiplied by itself’: 62 = 6 × 6 = 36
SQUARE ROOT
is the reverse process: 36 = 6
The best way to think of it is:
1. What is
‘Square Root’ means ‘What Number Times by Itself gives...’
2.
81 ?
9 times by
itself gives 81
What is
81 = 9 × 9
So 81 = 9
7.84
3. Find both square roots of 100.
= 2.8
All numbers also have a
NEGATIVE SQUARE ROOT
— it’s just the ‘–’ version of
the normal positive one.
10 × 10 = 100, so positive square root = 10
–10 × –10 = 100, so negative square root = –10
4. Give the exact side length of a square with area 15 cm2.
The side length is found by taking
the square root of the area.
7.84 ?
15 cm
Sometimes a number can only be
given exactly using a
sign.
E.g. 15 can be rounded to
3.873 but 15 is exact.
Cube Roots
‘Cubed’ means ‘multiplied by itself and then by itself again’: 23 = 2 × 2 × 2 = 8
CUBE ROOT 3
is the reverse process: 3 8 = 2
‘Cube Root’ means ‘What Number Times by Itself and then by Itself Again gives...’
1. What is 3 27 ?
3 times by itself and then
by itself again gives 27.
2. What is 3 4913 ?
27 = 3 × 3 × 3
So 3 27 = 3
3
4913
=
17
Unlike square roots there is only one answer.
Work it out in your head or use a calculator.
Higher roots are found in a similar way — 4
‘to the power 4’ and ‘to the power 5’.
and 5
are the reverse processes of
Square roots have two solutions, cube roots only have one
These roots can be a bit tricky but just remember they’re the opposite of powers.
If you’re stuck on a root question, think — what number times by itself (and by itself again) gives...
Section One — Numbers
33
Standard Form
Standard form (or ‘standard index form’) is useful for writing very big or very small numbers in a
more convenient way. A number written in standard form must always be in exactly this form:
This number must always
be between 1 and 10.
(The fancy way of saying this is 1 ≤ A < 10)
This number is just the number of
places the decimal point moves.
A × 10n
Three Rules for Standard Form
1) The front number must always be between 1 and 10.
2) The power of 10, n, is how far the decimal point moves.
3) n is positive for BIG numbers, n is negative for SMALL numbers.
(This is much better than rules based on which way the decimal point moves.)
Three Important Examples
1
Express 259 000 in standard form.
1) Move the decimal point until 259 000 becomes 2.59 (1 £ A < 10).
2) The decimal point has moved 5 places so n = 5, giving: 105.
2. 5 9 0 0 0
3) 259 000 is a big number so n is +5, not –5.
= 2.59 × 105
2 Express 0.00335 in standard form.
1) The decimal point must move 3 places to give 3.35 (1 £ A < 10).
So the power of 10 is 3.
2) Since 0.00335 is a small number it must be 10–3, not 10+3.
3
Write these numbers in order from smallest to largest:
2.25 × 104
7.98 × 10–4
6880
3.12 × 104
6.75 × 103
0. 0 0 3 3 5
= 3.35 × 10–3
0.000134
1) First convert all the numbers into standard form.
6880 = 6.88 × 103 0.000134 = 1.34 × 10–4
2) Now group the numbers with the same power together and order them based on the power.
7.98 × 10–4
1.34 × 10–4
6.88 × 103
6.75 × 103
2.25 × 104
3.12 × 104
3) Finally order each group by the size of the front number — give the numbers in the form
they are given in the question.
0.000134
ION
RE VIS
TASK
7.98 × 10–4
6.75 × 103
6880
2.25 × 104
3.12 × 104
Standard Index form is another name for standard form
Try writing out these numbers in standard form: 72 100 000 and 0.000094. (Answers below.)
Section One — Numbers
(Ans: 7.21 × 107, 9.4 × 10–5)
34
Warm-Up and Practice Questions
There’s really no point just rushing straight on to the next section — practice really
is the best thing you can do to improve your maths skills, so don’t neglect it.
Warm-up Questions
Only use your calculator in question 5.
1)
Work out the following calculations:
a) 23 + 82b) 34 – 52 c) 25 + 33
2)
Use the power rules to simplify:
3)
Use a combination of the power rules to simplify (75 × 713) ÷ 76.
4)
Find both square roots of: a) 9
5)
Use a calculator to find the following to 2 d.p.
6)
Express in standard form:
a) 7 650 000 000
7)
Express as ordinary numbers:
a) 5.6 × 106b) 1.11 × 10–5
a) 45 × 411 b) 79 ÷ 77c) (y3)6
b) 121
a)
c) 169
19 b)
3
643 c)
5
1729
b) 0.000024
Practice Questions
It’s no good learning all the facts if you can’t use your knowledge when it comes to practice
questions. These worked examples will show you how to turn that knowledge into answers.
1
In an online shop, 3 × 104 purchases were made in one day.
The total amount spent on the day was £1.5 million.
a)
Express the total amount spent in standard form.
1 500 000 = 1.5 × 106
1.5 × 10
£ ...................................
6
[1 mark]
b)
Calculate the average amount spent on each purchase.
Divide the total amount spent
by the total number of purchases
2
Evaluate the following expression
Anything to the power 0 is 1
When multiplying you
add the powers
6
1.5 × 10
3 × 10 4
28 × 23 × 1
1× 2 × 26
28+3 × 1
1 × 2 1+6
Split the calculation into
numbers and powers of 10
= 13.5 × 10 4
10
When dividing you
= 0.5 × 106 – 4
subtract
the powers
= 0.5 × 102
= 0.5 × 100 = 50
50
£ ...................................
^2 2h4 × 23 × 1 32
20 × 21 × ^23h2
=
6
[3 marks]
1 to the power anything is 1
When dividing you
subtract the powers
11
= 2 7 = 2 11 – 7 = 2 4 = 16
2
16
...................................
[3 marks]
Section One — Numbers
35
Practice Questions
3
Evaluate the following to two decimal places.
a)
142
...................................
[1 mark]
b)
3
82
...................................
[1 mark]
c)
5
1986
...................................
[1 mark]
4
Look at this table:
a)
91
9
92
81
93
729
94
6561
95
59 049
96
531 441
97
4 782 969
98
43 046 721
Explain how you can use the table to show that 729 × 6561 = 4 782 969.
[2 marks]
b)
Use the table to help you work out the value of 43 046 721.
531 441
...................................
[1 mark]
5
If
1
0.0005, work out 1 + 1 . Give your answer in standard form.
2000 =
2000 20 000
...................................
[3 marks]
Section One — Numbers
36
Revision Summary for Section One
Well, that’s section one done — have a go at these questions to see how much you can remember.
• Try these questions and tick off each one when you get it right.
• When you’ve done all the questions for a topic and are completely happy with it, tick it off.
Ordering Numbers and Arithmetic (p1-10) 
1) Find the value of: a) 4 + 10 ÷ 2
Only use a calculator if a question tells you to.
b) 12 ÷ 3 × 2
c) (8 × 5) ÷ 22 
2) Put these numbers in order of size: 0.02, 54, 11.8, 23.91, 0.09, 0.001, 0.51, 0.9
3) Work out: a) 417 + 194
b) 753 – 157
c) (2.3 + 1.123) – 0.75
4) Find: a) 1.223 × 100
b) 15.12 × 1000
c) 6.75 ÷ 10
d) 1.24 ÷ 200
5) Work out: a) 131 × 19
b) 672 ÷ 14
c) 9.12 × 34
d) 65.65 ÷ 13
6) Work out: a) –8 + 6
b) –4 – 10
c) –7 × –8
d) 81 ÷ –9





Types of Number, Factors and Multiples (p13-16) 
7) Define: a) integers
b) rational numbers
c) irrational numbers
8) Find all the prime numbers between: a) 40 and 50
9) Find: a) the first 5 multiples of 13
d) real numbers
b) 80 and 90
b) all the factors of 84
10) Express the following numbers as the product of prime factors: a) 252
11) Find: a) the LCM of 14 and 8 b) 230
c) 13
b) the HCF of 80 and 48





Fractions, Decimals and Percentages (p19-23) 
12) Write: a) 0.6 as a fraction and a percentage b) 35% as a fraction and a decimal 
4
3
8
5
13) a) Give two fractions equivalent to 5 .
b) Simplify 60 .
c) Which is bigger, 9 or 7 ?

1
7
1 5
2
3
7
5
14) Work out: a) 3 + 9 b) 12 – 2 c) 11 ' 10 d) 10 # 6 
3
2
15) Calculate: a) 9 of 540 b) 7 of 490 
16) Use a calculator to find: a) 15% of 78 b) 154% of £86 c) 99% of £99 
17) Use a calculator to give: a) 7 as a percentage of 187
b) 79p as a percentage of £17.89

Rounding and Estimating (p26-28) 
18) Round: a) 164.353 to 1 d.p.
b) 76 233 to 2 s.f.
c) 765 444 to the nearest ten

19) What is the error when 945 is rounded to the nearest hundred? 
20) Give the range of possible x values when: a) x = 50 to 1 s.f. b) x = 1.73 to 2 d.p. 
22.3 × 54.3
1000.3 × 46.52
21) Estimate the value of a)
b)

0.457
19.5
Powers, Roots and Standard Form (p31-33) 

23) Use the power rules to simplify: a) 62 × 611 b) 39 ÷ 35 c) (93)–6 
24) Use a calculator to find: a) 256 b) 5.56 to 2 d.p. c) 4 256 
25) Express: a) 856 000 000 in standard form b) 5.678 × 10–5 as an ordinary number 
22) Without using a calculator, work out:
Section One — Numbers
a) 106
b) 17291
c) 92 – 25 d) 5–2
Section Two — Algebra
37
Algebra — Simplifying Terms
Algebra really terrifies so many people. But honestly, it’s not that bad. The first step is to make sure
you understand and learn these basic rules for dealing with algebraic expressions.
Terms
Before you can do anything else with algebra, you must understand what a term is:
A TERM IS A COLLECTION OF NUMBERS, LETTERS AND BRACKETS,
ALL MULTIPLIED/DIVIDED TOGETHER
Terms are separated by + and – signs. Every term has a + or – attached to the front of it.
If there’s no sign in front
of the first term, it means
there’s an invisible + sign.
4x2
‘x2’ term
+ 5x
– 2y
‘x’ term ‘y’ term
+ 6y2
+4
‘y2’ term
‘number’ term
Simplifying or ‘Collecting Like Terms’
To simplify an algebraic expression made up of all the same terms, just add or subtract them.
EXAMPLES:
2. Simplify 2s + 3s – s
1. Simplify r + r + r + r
Just add up all the r’s:
r + r + r + r = 4r
‘r’ just
means ‘1r’.
Again, just combine the terms —
don’t forget there’s a ‘–’ before the last s:
2s + 3s – s = 4s
If you have a mixture of different letters, or letters and numbers, it’s a bit more tricky.
To simplify an algebraic expression like this, you combine ‘like terms’
(e.g. all the x terms, all the y terms, all the number terms etc.).
Simplify 7x + 3 – x – 2
number terms
Invisible
+ sign
7x
+3
–x
–2
=
+7x
–x
+3
6x
x-terms
–2
= 6x + 1
+1
1) Put bubbles round each term — be sure you capture the +/– sign in front of each.
2) Then you can move the bubbles into the best order so that like terms are together.
3) Combine like terms.
Simplify 4x + y – x + 3y
y-terms
Invisible
+ sign
4x +y –x
x-terms
+3y =
+4x
–x
3x
+y
+3y
+4y
= 3x + 4y
You can only add or subtract terms if they are ‘like terms’
Be careful you don’t add together different terms — remember x2 and x are different terms.
Section Two — Algebra
38
Algebra — Simplifying Terms
On this page we’ll look at some rules that will help you simplify expressions that have
letters and numbers multiplied together.
Negative Numbers
Negative numbers crop up everywhere so make sure you learn the rules for dealing with them:
+
+
–
–
+
–
+
–
makes +
makes –
makes –
makes +
Use these rules when:
1) Multiplying or dividing.
2) Two signs are together.
e.g. –2 × 3 = –6, –4p × –2 = +8p
e.g. 5 – –4 = 5 + 4 = 9, –x + –y = –x –y
Letters Multiplied Together
Watch out for these combinations of letters in algebra that regularly catch people out:
1) abc means a × b × c and 3a means 3 × a.
The ×’s are often left out to make it clearer.
2) gn2 means g × n × n. Note that only the n is squared, not the g as well.
3) (gn)2 means g × g × n × n. The brackets mean that BOTH letters are squared.
See p31.
4) Powers tell you how many letters are multiplied together — so r6 = r × r × r × r × r × r.
5) –a2 isn’t very clear. It should either be written (–a)2 or –(a2).
EXAMPLES:
es
Careful — h tim 3
,
h
is
itself 3 times
not 3h.
2. Simplify 3r × 2s × 2
1. Simplify h × h × h
You have 3 h’s multiplied together:
h × h × h = h3
For example, (–3)2 = –3 × –3 = 9,
but –(32) = –(3 × 3) = –9
Multiply the numbers together,
then the letters together:
3r × 2s × 2 = 3 × 2 × 2 × r × s = 12rs
Power Rules and Algebra
You can use the power rules from p31 on algebraic expressions too:
1) When multiplying, you add the powers.
2) When dividing, you subtract the powers.
EXAMPLES:
1. Simplify n5 × n2
You’re multiplying,
so add the powers:
n5 × n2 = n5 + 2 = n7
ION
RE VIS
TIP
13
2. Simplify mm 5
You’re dividing, so subtract the powers:
m 13 = 13 - 5 = 8
m
m
m5
The power rules will help you to simplify algebraic expressions
You’ll need to know the power rules from page 31 and the rules for multiplying and dividing
negative numbers for algebra too — if you’re a bit rusty on them, go and take another look.
Section Two — Algebra
39
Algebra — Multiplying Single Brackets
Multiplying out brackets can be a bit tough. No need to panic though, we’ll start with some
simple questions on single brackets — double brackets will come on the next page.
Single Brackets
There are a few key things to remember before you start multiplying out brackets:
1)
2)
3)
4)
The thing outside the brackets multiplies each separate term inside the brackets.
When letters are multiplied together, they are just written next to each other, e.g. pq.
Remember, r × r = r2, and xy2 means x × y × y, but (xy)2 means x × x × y × y.
Remember, a minus outside the bracket REVERSES ALL THE SIGNS when you multiply.
Expand the following:
a) 2(4x + 3) b) 2p(q – 3r)c) –2(5t2 – 6s2)
= (2 × 4x) + (2 × 3)
= 8x + 6
= (2p × q) + (2p × -3r)
= 2pq – 6pr
= (–2 × 5t2) + (–2 × –6s2)
= –10t2 + 12s2
Note: both signs have been
reversed — see point 4.
Collecting Like Terms
You might be given more than one single bracket in the same expression, e.g. 3(x + 5) – 6x(x + 3).
You’ll have to expand each bracket separately and then collect like terms to simplify it (see p37).
EXAMPLES:
1. Expand and simplify 3(2x + 1) + 2x(4 + 3x).
1) Expand each of the brackets separately:
3(2x + 1) + 2x(4 + 3x) = (3 × 2x) + (3 × 1) + (2x × 4) + (2x × 3x)
= 6x + 3 + 8x + 6x2
2) Collect like terms to simplify the expression:
= 6x2 + 14x + 3
2. Expand and simplify 3y(2y + 1) – y(4 + 3y) – 3y(y – 4)
= (3y × 2y) + (3y × 1) + (–y × 4) + (–y × 3y) + (–3y × y) + (–3y × – 4)
= 6y2 + 3y – 4y – 3y2 – 3y2 + 12y
= 6y2 – 3y2 – 3y2 + 3y – 4y + 12y
Careful with the
minus
signs here.
= 11y
Remember to collect like terms after you’ve expanded...
When you’re collecting like terms it’s conventional to write your answer in descending powers.
E.g. the x2 term should come before the x term, and the x term should come before the number term.
Section Two — Algebra
40
Algebra — Multiplying Double Brackets
Double brackets are a bit more tricky than single brackets — this time, you have to
multiply everything in the first bracket by everything in the second bracket.
Double Brackets
DOUBLE BRACKETS — you get 4 terms, and usually 2 of them combine to leave 3 terms.
There’s a handy way to multiply out double brackets — it’s called the FOIL method and works like this:
First — multiply the first term in each bracket together
Outside — multiply the outside terms (i.e. the first term in the
first bracket by the second term in the second bracket)
Inside — multiply the inside terms (i.e. the second term in the
first bracket by the first term in the second bracket)
Last — multiply the second term in each bracket together
Expand and simplify (n ­– 1)(2n + 5)
F
O
I
L
(n ­– 1)(2n + 5) = (n × 2n) + (n × 5) + (–1 × 2n) + (–1 × 5)
=
2n2
+ 5n
=
2n2 + 3n – 5
– 2n
–5
The two n terms
combine together.
Squared Brackets
SQUARED BRACKETS — always write these out as two brackets (to avoid mistakes).
1) Write out the squared brackets as double brackets, e.g. (x + 1)2 = (x + 1)(x + 1).
2) Use the FOIL method above to expand and simplify the expression.
Expand and simplify (4x + 1)2
1) Write out the expression as two brackets:
(4x + 1)2 = (4x + 1)(4x + 1)
2) Expand and simplify using the FOIL method:
F
O
I
L
= (4x × 4x) + (4x × 1) + (1 × 4x) + (1 × 1)
Don’t make the mistake
of thinking that:
(4x + 1)2 = 16x2 + 1.
= 16x2 + 4x + 4x + 1
= 16x2 + 8x + 1
ION
RE VIS
TIP
Always write out squared brackets as double brackets
The FOIL method is a good method to learn for multiplying out double brackets, because if
you do it right, it always gives you 4 terms — if you get fewer, you know you’ve missed a step.
Section Two — Algebra
41
Algebra — Taking Out Common Factors
Now you know how to expand brackets, it’s time to put them back in. This is called factorising.
Factorising — Putting Brackets In
This is the exact reverse of multiplying out brackets. You have to look for common factors
— numbers or letters that go into every term. Here’s the method to follow:
1) Take out the biggest number that goes into all the terms.
2) For each letter in turn, take out the highest power (e.g. x, x2 etc.)
that will go into EVERY term.
3) Open the brackets and fill in all the bits needed to reproduce each term.
4) Check your answer by multiplying out the brackets again.
REMEMBER: The bits taken out and put at the front of the brackets are the common factors.
The bits inside are what get you back to the original terms when you multiply out again.
Taking Out a Number
If both terms of the expression you’re trying to factorise have a number part,
you can look for a common factor of both numbers and take it outside the brackets.
The common factor is the biggest number that the numbers in both terms divide by.
EXAMPLES:
1. Factorise 2x – 6
2 and 6 both
divide by 2.
Decide what you
need to multiply 2 by
to get to 2x and –6.
2(x – 3)
Check: 2(x – 3) = 2x – 6
2. Factorise 10x + 25y
The biggest number that 10
and 25 both divide by is 5.
5(2x + 5y)
Check: 5(2x + 5y) = 10x + 25y
Taking Out a Letter
If the same letter appears in all the terms (but to different powers), you can take out some
power of the letter as a common factor. You might be able to take out a number as well.
EXAMPLES:
1. Factorise y2 – 2y
Highest power of
y in both terms
y(y – 2)
Decide what you
need to multiply y by
to get y2 and –2y.
Check: y(y – 2) = y2 – 2y
EXAM
TIP
2. Factorise 4x2 + 8x
Biggest number that’ll
divide into 4 and 8
4x(x + 2)
Highest power
of x that will go
into both terms
Check: 4x(x + 2) = 4x2 + 8x
The common factor could have a number and letter part
When you’re doing factorisation questions, remember to check your answer at the end by
multiplying out the brackets again. You should end up with the expression you started with.
Section Two — Algebra
42
Warm-Up and Practice Questions
Algebra can seem pretty weird if you’re not used to it, but that’s why you need to learn the
basics really well. Even if you find it easy, it’s practice that really fixes things in your brain.
Warm-up Questions
1)
Simplify the following expressions:
a) a + a + a b) 3d + 7d – 2d c) 8x – y – 2x + 3y d) –3x + 2y – 5x + 7y
2)
Simplify:
3)
Multiply out the brackets:
a) 2(a + 3)b) 3(2u – 5)
4)
5)
a) d × d × d × d × db) 2e × 8fc) g3 × g6d)
Expand and simplify:
a) ( y + 2)( y + 1)
h 11
h4
c) x(3x + 2)d) 2b(2b – 3)
b) (w – 3)2c) (2p + 1)(3p – 1)
Factorise the following expressions fully:
a) 4 – 10x b) 6x + 2y c) –4x – 8d) 10a + 8a2
d) (q + 1)(q – 1)
e) 9b2 – 3bc
Practice Questions
Now that you’ve had some practice of these basic algebra techniques, it’s time to see how you fare
with some exam-style questions. The first one is a worked example to ease you in gently...
1
The rectangle below has an area of 6y2 – 10y.
Area = 6y2 – 10y
a)
Fully factorise 6y2 – 10y.
Biggest number that will
divide into 6 and 10
Highest power of y that
will go into both terms
6y2 – 10y = 2y(3y – 5)
2y(3y – 5)
......................
[2 marks]
b)
If one side of the rectangle has length y, write down the length of the other side.
Take y out of the expression for the
area to find the other side length
6y2 – 10y = y(6y – 10)
(6y – 10)
......................
[1 mark]
Section Two — Algebra
43
Practice Questions
2
Simplify the following expressions.
a)
4x – 2y + 3x + y
.........................
[1 mark]
b)
2x2 – 2x + 6x – 5 – 3x2 + 3x + 7x2
.........................
[1 mark]
3
Expand and simplify the following expressions.
a)
2(5x – 1) – 7x
.........................
[2 marks]
b)
(y – 6)(y + 3)
.........................
[2 marks]
4
Several possible factorisations of 36b + 12 are shown below.
12(3b + 1)
a)
6(6b + 2)
2(18b + 6)
36(b + 1)
4(9b + 3)
One of the expressions is incorrect. Which is it?
.........................
[1 mark]
b)
Which expression is completely factorised?
.........................
[1 mark]
5
Caley’s garden is a square with side length given by the expression 5x + 2.
Find an expression for the area of Caley’s garden, expanding out any brackets.
5x + 2
.................................................
[2 marks]
Section Two — Algebra
44
Solving Equations
Golden Rules
To solve equations, you must find the value of x (or any given letter) that makes the equation true.
To find this value of x, rearrange the equation until you end up with ‘x = ‘ on one side.
Here are a few important points to remember when rearranging.
1) Always do the SAME thing to both sides of the equation.
2) To get rid of something, do the opposite.
The opposite of + is – and the opposite of – is +.
The opposite of × is ÷ and the opposite of ÷ is ×.
3) Keep going until you have a letter on its own.
Solving One-Step Equations
One-step equations are exactly what they say on the tin — one step and you’re done.
EXAMPLES:
1. Solve x + 3 = 7.
This means ‘take
away 3 from
both sides’.
x+3=7
(–3) x + 3 – 3 = 7 – 3
x = 4
3. Solve 2x = 10.
2x = 10
(÷2) 2x ÷ 2 = 10 ÷ 2
x=5
2. Solve x – 2 = 3.
The opposite
of +3 is –3.
2x means 2 × x,
so do the opposite —
divide both sides by 2.
(+2)
x–2=3
x–2+2=3+2
x = 5
4. Solve 2x = 4.
x
2 =4
(×2) 2x × 2 = 4 × 2
x=8
The opposite
of –2 is +2.
x
2 means x ÷ 2,
so do the opposite —
multiply both sides by 2.
Solving Two-Step Equations
If you come across an equation like 8x – 2 = 14 (where there’s an x-term and a number on
the same side), use the methods above to solve it — just do it in two steps:
1) Add or subtract the number first. 2) Multiply or divide to get ‘x = ‘.
Solve the equation 5x + 2 = 12.
5x + 2 = 12
(–2) 5x + 2 – 2 = 12 – 2
5x = 10
5x ÷ 5 = 10 ÷ 5
(÷5)
x = 2
The opposite of +2 is –2, so
subtract 2 from both sides.
The opposite of ×5 is ÷5,
so divide both sides by 5.
To solve equations always take it one step at a time
When you’re solving an equation it’s a good idea to write down what you’re doing at every stage
— put it in brackets next to the equation (like in the examples above).
Section Two — Algebra
45
Solving Equations
Equations with an ‘x’ on Both Sides
For equations like 3x + 1 = x – 7 (where there’s an x-term on each side), you have to:
1) Get all the x’s on one side and all the numbers on the other.
2) Multiply or divide to get ‘x = ‘.
Solve the equation 5x – 7 = 3x + 3.
(–3x)
(+7)
(÷2)
5x – 7 = 3x + 3
5x – 7 – 3x = 3x + 3 – 3x
2x – 7 = 3
2x – 7 + 7 = 3 + 7
2x = 10
2x ÷ 2 = 10 ÷ 2
x=5
Solve the equation
To get the x’s on only one side,
subtract 3x from each side.
Now add 7 to get the
numbers on the other side.
The opposite of ×2 is ÷2,
so divide both sides by 2.
x+3
2 = x – 7.
x+3
Start by multiplying by 2
2 =x–7
to get rid of the fraction.
+
x 3
(×2)
2 × 2 = (x – 7) × 2
To get the x’s on one side,
x + 3 = 2x – 14
subtract x from each side.
x + 3 – x = 2x – 14 – x
(– x)
3 = x – 14
The opposite of –14 is +14,
3 + 14 = x – 14 + 14 so add 14 so both sides.
(+14)
17 = x
Equations with Brackets
If the equation has brackets in, you have to multiply out the brackets before solving it as above.
Solve the equation 3(x – 1) = 2(x + 2).
(–2x)
(+3)
EXAM
TIP
3(x – 1) = 2(x + 2)
3x – 3 = 2x + 4
3x – 3 – 2x = 2x + 4 – 2x
x – 3 = 4
x – 3 + 3 = 4 + 3
x=7
Multiply out the brackets.
To get the x’s on only one side,
subtract 2x from each side.
The opposite of –3 is +3,
so add 3 to each side.
Get all the x’s on one side and all the numbers on the other
You can check your answer to questions like these by putting your x-value back into each side
of the equation. Both sides should give the same number — if they don’t, you’ve gone wrong.
Section Two — Algebra
46
Using Formulas
Formulas come up again and again in maths, so make sure you’re happy with using them.
The first thing you need to be able to do is pretty easy — it’s just putting numbers into them.
Putting Numbers into Formulas
You might be given a formula and asked to work out its value when you put in certain numbers.
All you have to do here is follow this method.
1) Write out the formula.
2) Write it again, directly underneath, but substituting numbers for letters
on the right-hand side (RHS).
3) Work it out in stages. Use BODMAS (see p1) to work things out in the
right order. Write down values for each bit as you go along.
4) DO NOT attempt to do it all in one go on your calculator
— you’re more likely to make mistakes.
L = 3j – 2k. Find the value of L when j = 4 and k = 5.
L = 3j – 2k
L=3×4 – 2×5
L = 12 – 10
L=2
1) Write out the formula.
2) Write it again, substituting numbers for letters on the RHS.
3) Work out the multiplications first, then do the subtraction.
5
The formula for converting from kilometres (k) to miles (m) is m = 8 k.
Use this formula to convert 16 kilometres into miles.
m = 58 k
m = 58 × 16
m = 10 so 16 kilometres = 10 miles
1) Write out the formula.
2) Write it again, substituting numbers for letters.
3) Work out the multiplication.
Wordy Formulas
If you’re given a formula in words rather than letters, don’t panic — you use the same method as above.
To find the age (in years) of a barrowbeetle, Ben uses the formula:
Age = (number of stripes ÷ length of the beetle in cm) – 2.
How old is a barrowbeetle if it has 6 stripes and is 2 cm long?
Age = (stripes ÷ length) – 2
Age = (6 ÷ 2) – 2
Age = 3 – 2
Age = 1 year
1) Write out the formula.
2) Substitute numbers for letters on the RHS.
3) Use BODMAS to work things out in the right order —
do the bit in brackets first, then do the subtraction.
Make sure you’ve put the right numbers into the right places...
That was an easy introduction to formulas... so when you’re ready, let’s move on.
Section Two — Algebra
47
Making Formulas from Words
Before we get started, there are a few definitions you need to know:
EXPRESSION — a collection of terms (see p37). Expressions DON’T have an = sign in them.
EQUATION — an expression with an = sign in it (so you can solve it).
FORMULA — a rule that helps you work something out (it will also have an = sign in it).
Making a Formula from Given Information
Making formulas from words can be a bit confusing as you’re given a lot of information in one go.
You just have to go through it slowly and carefully and extract the maths from it.
Ruby is x years old. Alia is 3 years older than Ruby.
Jeremy is 5 times as old as Ruby.
a) Write an expression for Alia’s age
b) Write an expression for
in terms of x.
Jeremy’s age in terms of x.
Alia is 3 years older,
5 times older
Ruby’s age is x
Ruby’s age is x
so add 3
So Jeremy’s age is 5 × x = 5x
So Alia’s age is x + 3
The cooking time (t minutes) for a turkey is 40 minutes per kilogram
(k), plus an extra 20 minutes. Write a formula for t in terms of k.
A turkey with mass k
will take 40 × k minutes
Don’t forget to add on the
You’re asked for a formula so
t
=
40k
+
20
extra time (20 minutes)
you must include the ‘t = ‘ bit.
Using Your Formula to Solve Equations
Sometimes, you might be asked to write a formula and use it to solve an equation.
A car mechanic charges £100 plus £50 for each part he has to replace (r).
Sally gets a car repair bill of £450. Write a formula for C (the total repair
cost) and use it to find how many parts were replaced on Sally’s car.
Each part (r)
sts £50
co
C = 50r + 100
Don’t forget to add on
the £100 fixed charge.
450 = 50r + 100
Replace C with the value
(–100) 450 – 100 = 50r + 100 – 100
given in the question (£450)
and solve the equation.
350 = 50r
(÷50) 350 ÷ 50 = 50r ÷ 50
7 = r So Sally has 7 parts replaced on her car
EXAM
TIP
Read the question and pick out the information you need
It might help to underline any bits of the question you think you’ll need to use in the formula.
Section Two — Algebra
48
Rearranging Formulas
The subject of a formula is the letter on its own before the = (so x is the subject of x = 2y + 3).
Changing the Subject of a Formula
Golden Rules
Rearranging formulas means making a different letter the subject, e.g. getting ‘y = ’ from
‘x = 3y + 2’ — you have to get the subject on its own. Fortunately, you can use the
same methods that you used for solving equations (see p44-45) — here’s a quick reminder:
1) Always do the SAME thing to both sides of the formula.
2) To get rid of something, do the opposite.
The opposite of + is – and the opposite of – is +.
The opposite of × is ÷ and the opposite of ÷ is ×.
3) Keep going until you have the letter you want on its own.
Rearrange a = 2b – 3 to make b the subject of the formula.
a = 2b – 3
The opposite of –3 is +3,
so add 3 to both sides.
a + 3 = 2b – 3 + 3
(+3)
a + 3 = 2b
The opposite of ×2 is ÷2,
so divide both sides by 2.
(÷2) (a + 3) ÷ 2 = 2b ÷ 2
+
+
a 2 3 = b OR b = a 2 3
Careful here — you
divide the whole side by
2, not just one term.
n
Rearrange m = 3 + 5 to make n the subject of the formula.
m = n3 + 5
The opposite of +5 is –5, so
subtract 5 from both sides.
m – 5 = n3 + 5 – 5
(–5)
m – 5 = n3
The opposite of ÷3 is ×3,
n
so multiply both sides by 3.
(×3) (m – 5) × 3 = 3 × 3
3(m – 5) = n OR n = 3m – 15
Careful here — you
have to square the
whole expression.
Rearrange y =
x – 3 to make x the subject of the formula.
x –3
(+3)
(2)
y=
y+ 3 =
x –3+3
y+3=
x
The opposite of –3 is +3,
so add 3 to both sides.
(y + 3) = ( x )
2
2
The opposite of taking a
square root is squaring.
(y + 3)2 = x OR x = (y + 3)2
ION
RE VIS
TASK
Always do the same thing to both sides of the formula
The only way to get better at rearranging formulas is practice. Look back at the formulas in
the examples on page 46 — have a go at rearranging them to change their subjects.
Section Two — Algebra
49
Warm-Up and Practice Questions
I like algebra, because you can feel clever without actually doing anything that complicated.
Don’t worry if you’re still finding it all a bit of a mystery — it’ll come with practice.
Warm-up Questions
Solve the equations:
2)
3)
Solve the equations: a) 3x + 4 = 7 b) 4x + 6 = 10x
d) 2x2 = 8e) (x + 3)2 = 64
Hint — parts d) and e)
have two solutions.
If a = 3, b = 2, c = 5 and d = –2, find the values of:
2
2
a) adb) d c) –(d )d) a + b + c + d
e) a – b + c – df) ab + bc + cd g) (a + b) ÷ dh) a + b ÷ d
4)
What is the main difference between an equation and an expression?
5)
Hiring a canoe costs £8 per hour plus a deposit of £10. Lily paid £34.
How long did she hire the canoe for?
6)
Rearrange p = 5 + 3w to make w the subject of the formula.
v–5
Rearrange u = 3 to make v the subject of the formula.
7)
a) x + 6 = 9
b) x – 2 = 9 x
d) 3 = 7
c) 5x – 3 = 9 – x
1)
c) 9x = 27
Practice Questions
Now you’re warmed up, have a go at these practice questions. The first two are worked examples.
1
Make d the subject of a = 2b – d
3
Multiply both sides by 3
3a = 2b – d
d + 3a = 2b
Add d to both sides
d = 2b – 3a
Take 3a from both sides
d = 2b – 3a
.........................
[2 marks]
2
The cost, £C, of calling out a technician to repair a computer is C = 25 + 18h + p,
where h is the number of hours worked, and p is the cost of any parts.
a)
Find the cost when the technician needs £30 of parts and works for 2.5 hours.
p = 30, h = 2.5 so C = 25 + (18 × 2.5) + 30
= 25 + 45 + 30
= £100
100
£ .........................
[2 marks]
b)
If the technician didn’t need any parts and charged £79, how long did he work for?
p = 0, C = 79 so 79 = 25 + 18h + 0
(–25)
79 – 25 = 18h
54 = 18h
(÷18) 3=h
3 hours
.........................
[2 marks]
Section Two — Algebra
50
Practice Questions
3
Solve:
a)
7k – 3 = 11
.........................
[1 mark]
b)
3t + 5 = t + 12
.........................
[2 marks]
4
Jan thinks of a positive number. She doubles it, then adds 3. Then she squares the result.
She tells the answer to Sadiq.
a)
Let x stand for the number Jan thinks of.
Write down an expression for the answer she tells Sadiq.
.........................
[2 marks]
b)
The number she tells Sadiq is 81. What number is Jan thinking of?
.........................
[1 mark]
5
The formula V = 1 πr 2 h is used to calculate the volume of a cone.
3
a) Rearrange the formula to make h the subject.
.........................
[2 marks]
b)
Rearrange the formula to make r the subject.
.........................
[2 marks]
Section Two — Algebra
51
Number Patterns and Sequences
Sequences are just patterns of numbers or shapes that follow a rule.
You need to be able to spot what the rule is.
Finding Number Patterns
The trick to finding the rule for number patterns is to write down what you have to do
to get from one number to the next in the gaps between the numbers.
There are 2 main types to look out for:
1) Arithmetic sequences — Add or subtract the same number each time
E.g.
2
5
+3
THE RULE:
8
11
+3
+3
14
+3
...
30
+3
24
–6
‘Add 3 to the previous term’
18
–6
12
...
–6
–6
‘Subtract 6 from the previous term’
2) Geometric sequences — Multiply or divide by the same number each time
E.g.
2
6
×3
18
×3
54
×3
...
40 000
×3
4000
÷10
THE RULE: ‘Multiply the previous term by 3’
400
÷10
40
÷10
...
÷10
‘Divide the previous term by 10’
You might get number patterns that follow a different rule — for example, you might have to
add or subtract a changing number each time, or add together the two previous terms.
You just need to describe the pattern and use your rule to find the next term.
Shape Patterns
If you have a pattern of shapes, you need to be able to continue the pattern. You might also have
to find the rule for the pattern to work out how many shapes there’ll be in a later pattern.
Here are some patterns made of squares.
a) Draw the next pattern in the sequence.
b) Work out how many squares there will be in the 6th pattern.
a) Just continue the pattern — add an
extra square to each of the three legs.
b) Set up a table to find the rule:
Pattern number
Number of squares
1
1
2
4
3
7
4
10
5
13
6
16
The rule is ‘add 3 to the
previous term’. So just keep on
adding 3 to extend the table
until you get to the 6th term.
Always check your rule works for the whole sequence...
When you think you’ve figured out the rule, check it works for all the terms you’ve been given.
If it doesn’t work for one of them then it’s not the correct rule — you’ll have to find a different one.
Section Two — Algebra
52
Number Patterns and Sequences
You might have to “find an expression for the nth term of a sequence” — this is a rule with n in,
like 5n – 3. It gives every term in a sequence when you put in different values for n.
Finding the nth Term of a Sequence
This method works for sequences with a common difference — where you add or subtract the
same number each time (i.e. the difference between each pair of terms is the same).
Find an expression for the nth term of the sequence that starts 2, 8, 14, 20, ...
n:
term:
1
2
2
8
+6
6n:
term:
3
14
+6
1) Find the common difference. It’s 6,
so this tells you ‘6n’ is in the formula.
24
2) List the values of 6n.
+6
6
12
–4
–4
18
2
14
8
4
20
–4
–4
20
So the expression for the nth term is 6n – 4
Check your formula by putting the
first few values of n back in:
n = 1 gives 6n – 4 = 6 – 4 = 2
n = 2 gives 6n – 4 = 12 – 4 = 8
3) Work out what you have to add or
subtract to get from 6n to the term.
So it’s –4.
4) Put ‘6n’ and ‘–4’ together.
Deciding if a Term is in a Sequence
You might be given the nth term and asked if a certain value is in the sequence.
The trick here is to set the expression equal to that value and solve to find n.
If n is a whole number, the value is in the sequence.
A sequence is given by the rule 3n + 4.
a) Find the 6th term in the sequence.
Put n = 6 into the expression:
(3 × 6) + 4 = 18 + 4 = 22
b) Is 29 a term in this sequence?
Set the rule equal to 29 and solve for n —
if n is a whole number then 29 is in the sequence.
3n + 4 = 29
3n = 25
n = 25 ÷ 3 = 8.333...
n is not a whole number, so 29 is not in the sequence 3n + 4.
The nth term is the number in the nth position of the sequence
It might be even easier to decide if a number is in a sequence or not — for example, if the sequence
was all odd numbers, there’s no way that an even number could be in the sequence.
Use your common sense — if all the terms in a sequence end in 3, 44 would not be in the sequence.
Section Two — Algebra
53
Inequalities
Inequalities aren’t too bad — you just have to find the values which make the statement true.
The Inequality Symbols
> means ‘Greater than’
< means ‘Less than’
≥ means ‘Greater than or equal to’
≤ means ‘Less than or equal to’
The one at the BIG end is BIGGEST so x > 4 and 4 < x both say: ‘x is greater than 4’.
Write down all possible values of x in these inequalities.
a) x is a positive integer
such that x ≤ 3.
b) x is a negative
integer such that –5 < x.
x is a positive integer that is
less than or equal to 3:
x is a negative integer that
is greater than –5.
1, 2, 3
–4, –3, –2, –1
Remember — integers
are just whole numbers
(0 isn’t positive or
negative).
Sometimes an inequality will contain more than one symbol. Here’s how you tackle those ones.
x is an integer such that –2 < x ≤ 4. Write down all possible values of x.
Work out what each bit of the inequality is telling you:
–2 < x means that x is greater than –2
x ≤ 4 means that x is less than or equal to 4
Now just write down all the values that x can take:
–1, 0, 1, 2, 3, 4
You Can Show Inequalities on Number Lines
Drawing inequalities on a number line is dead easy — all you have to remember is
that you use an open circle ( ) for > or < and a coloured-in circle ( ) for ≥ or ≤.
Show the inequality –1 < x ≤ 5 on a number line.
Open circle because
–1 isn’t included.
–2
–1
0
Closed circle because
5 is included.
1
2
3
4
5
Don’t miss out or include values that aren’t in the range...
If you have a < or > symbol in your inequality, then you don’t include the number next to it in the
range — if you have a ≤ or ≥ symbol, then you do include the number next to it in the range.
Section Two — Algebra
54
Warm-Up and Practice Questions
Stuff like “find the nth term” will sound like complete gibberish if you haven’t worked through
this section carefully. So try these warm up questions and check they make sense to you.
Warm-up Questions
1)
Find the next two terms in each of the following sequences:
a) 2, 5, 8, ... , ... b) 10, 11, 14, 19, ... , ...
d) 32, 26, 20, ... , ... e) 128, 32, 8, ... , ...
c) 4, 12, 36, ... , ...
2)
A sequence starts 0, 2, 2, 4, 6, 10. Write down the next two terms in the sequence
and give the rule.
3)
Find the nth term of the following sequences: a) 5, 8, 11, 14, ...
4)
Is 38 in the sequence given by the expression 5n + 2?
5)
n is an integer such that –3 ≤ n < 3. Write down all the possible values of n.
6)
Show the inequality –2 ≤ x ≤ 1 on a number line.
b) –3, –13, –23, ...
Practice Questions
Time for some more practice questions. But first, there’s a worked example about patterns and
sequences — read through it carefully before you have a go at the rest of the practice questions.
1
The diagram shows some floor patterns made up of shaded and unshaded square tiles.
a)
Draw the fourth pattern on the blank grid below.
[1 mark]
b)
How many black squares will be in the 5th pattern?
There will be a 5 by 5 square of black tiles in the middle,
and 1 black tile in each corner.
(5 × 5) + 4 = 25 + 4 = 29
29
.........................
[2 marks]
Section Two — Algebra
55
Practice Questions
2
Show the inequality 1 < x ≤ 7 on a number line.
.........................
[2 marks]
3
Find the nth term of the following sequences.
a)
7, 9, 11, 13, …
.........................
[2 marks]
b)
11, 7, 3, –1, …
.........................
[2 marks]
4
Look at these patterns of dots.
a)
Draw the next pattern in the sequence.
.........................
[1 mark]
b)
How many dots would be in the 8th pattern in the sequence?
.........................
[2 marks]
Section Two — Algebra
56
Revision Summary for Section Two
There was a lot of algebra in that section — let’s see how much you remember.
Try these questions and tick off each one when you get it right.
When you’ve done all the questions for a topic and are completely happy with it, tick off the topic.
Algebra (p37-41) 
1) Simplify:a) a + a + a + a + ab) 3b + 8b – 2b 
2) Simplify:a) d + 3e + 5d – 2eb) 9f + 2 – 11f + 7 
3) Simplify:a) g × g × g × gb) m × n × 9 
4) Simplify:a) r7 × r2b) s15 ÷ s9 
5) Expand:a) 3(v + 8)b) –7(2w + 5) 
6) Expand and simplify 4(3y – 4) – 2(4y – 10) 
7) What is factorising? 
8) Factorise:a) 6x + 18b) 14x + 28y 
Solving Equations (p44-45) 
9) Solve:a) x + 12 = 19b) x – 6 = 16 
x
10) Solve:a) 9x = 36b) 4 = 20 
11) Solve:a) 3x + 5 = 14 b) 9x – 6 = x + 10 
12) Solve:a) 2(x + 3) = 18
b) 8(x – 2) = 4x 
Formulas (p46-48) 
13) P = 4m + 7n. Work out the value of P when m = –2 and n = 3. 
9
14) The formula for converting from Celsius (C) to Fahrenheit (F) is F = 5 C + 32.
Use the formula to convert –20 °C into Fahrenheit. 
15) Lucian is organising a camping trip and buys s sleeping bags and t tents.
Sleeping bags cost £8 each and tents cost £15 each. He spends £P in total.
Write a formula for P in terms of s and t. 
16) Rearrange the formula p = 2q – 3 to make q the subject. 
Number Patterns and Sequences (p51-52) 
17) For each of the following sequences, find the next term and write down the rule you used.
a) 2, 8, 14, 20, ... b) 3, 9, 27, 81, ... c) 2, 3, 5, 8, 13, ...

18) Find an expression for the nth term of the sequence that starts 5, 7, 9, 11, ... 
19) Is 55 a term in the sequence given by the expression 9n + 1? 
Inequalities (p53) 
20) Find all the possible positive integer values of x:
a) x < 7
b) 7 < x ≤ 11 
21) Show the inequality –4 < x < 4 on a number line. 
Section Two — Algebra
Section Three — Graphs
57
X and Y Coordinates
The first thing you’ll need to know about graphs is know how to read coordinates from them.
The Four Quadrants
y
5
X and Y
BOTH
POSITIVE
ONE NEGATIVE
4
ONE POSITIVE
A graph has four different quadrants (regions).
3
The top-right region is the easiest because
ALL THE COORDINATES IN IT ARE POSITIVE.
(–5, 3)
You have to be careful in the other regions though,
because the x- and y- coordinates could be negative,
and that makes life much more difficult.
–5 –4 –3 –2 –1 0
–1
2
(4, 2)
1
x
Three important points about coordinates:
1
2
3
4
5
(–4, –2)
–2
X and Y
BOTH
NEGATIVE
(5, –3)
–3
ONE POSITIVE
ONE NEGATIVE
–4
–5
(x,y)
1) The coordinates are always in ALPHABETICAL ORDER, x then y.
2) x is always the flat axis going ACROSS the page.
In other words ’x is a...cross’ Get it — x is a ’×’. (Hilarious isn’t it)
3) Remember it’s always IN THE HOUSE (Æ) and then UP THE STAIRS (­­≠)
so it’s ALONG first and then UP, i.e. x-coordinate first, and then y-coordinate.
The Midpoint of a Line
The ’MIDPOINT OF A LINE SEGMENT’ is the POINT THAT’S BANG IN THE MIDDLE of it.
Finding the coordinates of a midpoint is
pretty easy. Just learn these three steps...
1) Find the average of the x-coordinates.
2) Find the average of the y-coordinates.
3) Put them in brackets.
R and S have coordinates (1, 1) and (5, 6).
Find the midpoint of the line RS.
5
Average of x-coordinates = 1 +
=3
2
6
Average of y-coordinates = 1 +
= 3.5
2
Coordinates of midpoint = (3, 3.5)
6
5
4
3
2
y
S(5, 6)
Midpoint
of RS
1
0
R(1, 1)
1
x
2 3 4 5 6
In the house and up the... oh wait, I’m in a bungalow...
Getting the coordinates the right way round is the first step — find the method that works for you.
Then you can do all sorts of fun things — like working out the midpoint of a straight line.
Section Three — Graphs
58
Straight Line Graphs
Over the next few pages you’ll come across lots of straight lines. Here are the most basic ones.
Horizontal and Vertical Lines: ‘x = a’ and ‘y = a’
x = a is a vertical line
through ‘a’ on the x-axis
6 y
x=05
x = –5
4
3
2
1
x=3
x
1 2 3 4 5 6
-6 -5 -4 -3 -2 -1 0
-1
-2
-3
-4
-5
-6
4
3
2
1
y
y=3
Remember — all the
points on x = 3 have an
y=0
x x-coordinate of 3, and
all the points on y = 3
-6 -5line
-4 -3 -2 -1 0 1 2 3 4 5 6
y = a is a horizontal
-1
have
a y-coordinate of 3.
through ‘a’ on the y-axis
-2
-3
-4
-5
-6
y = –2
The Main Diagonals: ‘y = x’ and ‘y = –x’
5
5
4
3
2
1
y=x
x
-5 -4 -3 -2 -1 0
-1
-2
-3
-4
1 2
3 4
4 y
‘y = –x’ is the main
3
diagonal that goes
2
DOWNHILL from
1
left to right.
0
The x- and y-coordinates-5 -4 -3 -2 -1 0 1 2 3
-1
of each point are
-2
negatives of each other,
-3
y = –x
e.g. (–4, 4).
‘y = x’ is the main
diagonal that goes
UPHILL from
left to right.
y
5
The x- and
y-coordinates of each
point are the same,
e.g. (4, 4).
-4
Other Lines Through the Origin: ‘y = ax’ and ‘y = –ax’
y = ax and y = –ax are the equations for
A SLOPING LINE THROUGH THE ORIGIN.
See p62 for
how to find a
gradient .
The value of ‘a’ (known as the gradient) tells you the steepness of the line.
The bigger ‘a’ is, the steeper the slope. A MINUS SIGN tells you it slopes DOWNHILL.
y
5
y = –2x
1
y=–2x
4
3
2
1
-6 -5 -4 -3 -2 -1 0
-1
-2
y = 3x
1
y= 2x x
1 2
3 4
5 6
Learn the equations for these basic lines before moving on...
These are the easiest straight lines to learn — there are plenty of tougher ones to come.
Take a minute to go back through the page and make sure everything makes sense to you.
Section Three — Graphs
x
4
5
59
Plotting Straight Line Graphs
On this page you get to practise your drawing skills — shame it’s only straight line graphs though.
They can be difficult to get right, but luckily this method will lead you to the correct answer every time:
You might get lucky
and be given a table
in a question.
Don’t worry if it
contains 5 or 6 values.
1) Choose 3 values of x and draw up a table,
2) Work out the corresponding y-values,
3) Plot the coordinates, and draw the line.
Doing the ‘Table of Values’
Draw the graph of y = 2x + 1 for values of x from –3 to 2.
6 y
5
1) Choose 3 easy x-values for your table:
Use x-values from the grid you’re given.
Avoid negative ones if you can.
0
x
1
2
4
3
y
2
1
2) Find the y-values by putting
each x-value into the equation:
When x = 0,
y = 2x + 1
= (2 × 0) + 1 = 1
x
0
1
2
y
1
3
5
x
-6 -5 -4 -3 -2
When x = 2,
y = 2x + 1
= (2 × 2) + 1 = 5
-1 0
-1
1
2
3
-2
-3
-4
-5
-6
-7
Plotting the Points and Drawing the Graph
...continued from above.
3) Plot each pair of x- and y-values from your table.
The table gives the coordinates:
(0, 1), (1, 3) and (2, 5).
4) Now draw a straight line through your points
— remember to extend the line through
all the x-values given in the question.
(2, 5)
(1, 3)
(0, 1)
If one point looks a bit wacky, check 2 things:
– the y-value you worked out in the table
– that you’ve plotted it properly.
Plot three coordinates then draw a straight line through them...
If you have to plot an equation like 3x + y = 5, you’ll have to do a bit of rearranging (see p48)
before you can find the y-values (this equation would become y = 5 – 3x).
Section Three — Graphs
60
Warm-Up and Practice Questions
Now that you’ve had a quick introduction to the world of graphs it’s time for some
practice questions. But first, let’s get you started with a few quick warm-up questions.
Warm-up Questions
1) The coordinates of three corners of a rectangle are (0, 0), (5, 0) and (0, 10).
By drawing these coordinates on a graph, find the coordinates of the fourth corner.
2) A and B have coordinates (3, 2) and (3, 8). Find the midpoint of the line AB.
3) M and N have coordinates (–4, 2) and (2, –1). Find the midpoint of the line MN.
4) From left to right, do these graphs slope uphill or downhill?
a) y = –4xb) –y = –3xc) x + y = 0 d) –x + y = 0
5) Find the y-coordinate of the straight line given by the equation y = –3x + 5 when:
a) x = 0b) x = 10c) x = –50 d) x = 100
6) Draw the graph of y + 2x = 4 for values of x from –2 to 2.
Practice Questions
Now that you’re all warmed up it’s time for some practice questions — I’ve shown you how to do the
first one, but you’re on your own for the whole of the next page.
1
The points A, B and C are three of the vertices of a parallelogram ABCD.
y
A
x
C
B
a)
Write down the coordinates of A, B and C.
(–2, 2)
A = ........................
b)
(–4, –1)
B = ........................
(2, –1)
C = ........................
[3 marks]
Write down the coordinates of the fourth vertex, D, of the parallelogram.
To get from B to A, you go 2 right and 3 up
(4, 2)
— so do the same from C to find vertex D.
.........................
[1 mark]
c)
What is the equation of the line BC?
A horizontal line through –1.
y = –1
.........................
[1 mark]
Section Three — Graphs
61
Practice Questions
2
Look at the rectangle in the diagram below. Give the equations of the lines through:
4 y
a)
AB
.........................
3
A
[1 mark]
B
2
b)
1
BC
.........................
[1 mark]
–4 –3 –2 –1
–1
D
1
2
3
x
4
c)
–2
CD
.........................
C
[1 mark]
–3
d)
–4
DA
.........................
[1 mark]
3
There are some values missing in the table below.
a)
Complete the table of values using the equation y = 2x + 3.
x
–3
–2
–1
1
2
3
5
y
b)
0
[2 marks]
Use the table of values to plot the graph of y = 2x + 3 on the grid below.
y
10
9
8
7
6
5
4
3
2
1
-3 -2 -1 0
-1
-2
-3
-4
1 2 3 x
[2 marks]
c)
On the same grid draw the line y = 2. [1 mark]
d)
Write down the coordinates of the point where these two lines meet.
.........................
[1 mark]
Section Three — Graphs
62
Finding the Gradient
You had a brief introduction to the word gradient on page 58. Well, now you get to find out what it
actually means and how to find it for any straight line — not just ones through the origin.
Finding the Gradient
The gradient of a line is a measure of its slope. The bigger the number, the steeper the line.
Find the gradient of the straight line below.
50
y
A
Two accurate
points
40
30
Change
in y
20
10
-6
-5
-4
-3
-2
-1
0
B
Change in x
1
2
3
4
5
x
6
7
8
-10
-20
Two -30
points that can be read accurately are:
Point-40A: (7, 40)
Point B: (1, 10)
Change in y = 40 – 10 = 30
Change in x = 7 – 1 = 6
9
1) Find two accurate points and
complete the triangle — pick
two points in the top-right
quadrant to make it easier.
2) Find the change in y and the
change in x — make sure you
subtract the y-coordinates
the same way around as you
subtract the x-coordinates.
3) Learn this formula, and use it:
30
Gradient = 6 = 5
As the graph goes uphill,
the gradient is positive.
So the gradient is 5 (not –5).
CHANGE IN Y
GRADIENT =
CHANGE IN X
4) Check the sign is right.
If it slopes uphill from left to right
then it’s positive (
) — if it slopes
downhill then it’s negative (
).
If you subtracted the coordinates the right way round, the sign should be correct.
If it’s not, go back and check your working.
A lot of people forget about step 4 in the method above. It’s easy to divide, get a nice positive number
and give that as the answer. You’ve got to check the sign to make sure it’s not a negative gradient.
It’s all downhill from here...
Learn the four steps and you’ll be able to find the gradient of any straight line.
Take care — you might not be able to pick two points with nice, positive coordinates.
Section Three — Graphs
63
y = mx + c
The title of this page might put you off but don’t let it — you’ll be an expert by the end of the page.
y = mx + c is the Equation of a Straight Line
y = mx + c is the general equation for a straight-line graph, and you need to remember:
’m’ is equal to the GRADIENT of the graph.
’c’ is the value WHERE IT CROSSES THE Y-AXIS and is called the Y-INTERCEPT.
y
y
4
4
3
y = 3x – 1
2
m (gradient) = 3
‘m’ and ‘c’ are always just numbers
— so y = 3x – 1 and y = –x + 2 are
equations of straight lines.
3
c=2
2
m (gradient) = –1
1
1
x
-4
-3
-2
-1
0
1
2
3
x
4
-4
-3
-2
-1
-1
-2
0
1
2
3
4
-1
-2
c = –1
-3
-3
-4
-4
y = –x + 2
You might have to rearrange a straight-line equation to get it into this form:
Straight line:
Rearranged into ‘y = mx +c’
2x + y = 4 →
y = –2x + 4 (m = –2, c = 4)
6 – x = y → y = –x + 6 (m = –1, c = 6)
Finding the Equation of a Straight Line Graph
Find the equation of the line on
the graph in the form y = mx + c.
change in y 40
m = change in x = 20 = 2
1) Find ‘m’ (gradient)
2) Read off ‘c’ (y-intercept)
c = 10
y
60
50
40
Change
in y = 40
30
20
3) Use these to write the
equation in the form
y = mx + c.
ION
RE VIS
TIP
y = 2x + 10
10
Change
in x = 20
0
10
20
30 x
Make sure you can identify values of m and c in an equation
WATCH OUT: people mix up ‘m’ and ‘c’ when they get something like y = 5 + 2x.
Remember, ‘m’ is the number in front of the ‘x’ and ‘c’ is the number on its own.
Section Three — Graphs
64
y = mx + c
You’ve just seen how to write the equation of a straight line. Well the fun doesn’t stop there.
You’ll also need to sketch the lines and show that two lines are parallel from their equations.
Sketching a Straight Line using y = mx + c
Draw the graph of y + 2x = 2.
6
5
4
3
2
1
1) Get the equation into the form y = mx + c.
y + 2x = 2 → y = –2x + 2
2) Put a dot on the y-axis at the value of c.
‘c’ = 2 so put a
dot here-6 -5 -4 -3 -2 -1 0
y
1 along
2 down
x
1 2 3 4 5 6
-1
-2
-3
-4
-5
-6
3) Go along 1 unit and up or down by m. Make another
dot, then repeat this step a few times in both directions.
Go 1 along and 2 down because ‘m’ = –2
4) When you have 4 or 5 dots, draw a straight line through them.
y = –2x + 2
5) Finally check that the gradient looks right.
A gradient of –2 should be downhill left to right — which it is, so it looks OK.
Parallel Lines Have the Same Gradient
4
3
Parallel lines all have the same gradient, which means their
y = mx + c equations all have the same values of m.
So the lines y = 2x + 3, y = 2x and y = 2x – 4 are all parallel.
2
y = 2x + 3
The gradient (m-value) of line P is 3.
y = –3x – 4 has a gradient of –3.
y – 3x = 3 Þ y = 3x + 3 has a gradient of 3.
3x + y = –1 Þ y = –3x – 1 has a gradient of –3.
So y – 3x = 3 is parallel to y = 3x + 2
y = 2x
1
x
-4
Line P is given by the equation y = 3x + 2.
Which of these lines is parallel to line P?
y = –3x – 4
y – 3x = 3
3x + y = –1
y
-3
-2 -1 0
-1
-2
1
2
3
4
y = 2x – 4
-3
-4
1) First find the gradient of line P.
2) Rearrange the other equations
into y = mx + c form and find
their gradients.
3) Find which line has the same
gradient as line P.
Parallel lines have the same gradient and never cross
ION
RE VIS
TASK
Spotting parallel lines is easy once you know how. Check the gradients — if they match, the
lines are parallel. Sketching lines is trickier so look over the above example and understand
1
each step — get some graph
paper and try sketching a straight line equation of your own.
3
Section Three — Graphs
65
Warm-Up and Practice Questions
Some people find straight line graphs easy — they just need to look at an equation and they know
what the graph will look like. If you’re one of them, you’re lucky. If not, don’t worry, you’ll get there.
Warm-up Questions
4 y
1) A square is shown on the right.
3 D
a) Find the gradient of the side: (i) AB (ii) AD
2
b) Write down the equation of the line through points: 1
(i) A and B (ii) A and D
A
C
x
2) Find the gradient of the lines through the points
-4 -3 -2 -1
1 2 3 4
-1
a) (0, 0) and (5, 5) b) (1, 2) and (4, 5)
-2
c) (5, 5) and (10, 15) d) (2, 4) and (6, 2)
-3 B
3) Find values of the gradient (m) and the y-intercept (c)
-4
in each of these straight line equations:
a) y = 3x + 4 b) y – 4x + 3 = 0 c) y – 2x + 3 = 3x – 7 d) 0 = (4/y) – 2
4) Line P goes through points (0, 2) and (3, 8).
Find the equation of Line P in the form y = mx + c — draw a graph if you need to.
5) Draw the graph of y = 2x – 1.
6) Are these lines parallel to the line in question 5? a) y – 2x = 2 b) –y = –2x + 3
Practice Questions
If you do find these questions hard then today might just be your lucky day. I’ve answered a
practice question for you below and there are plenty more for you to do on the next page.
1
John says the line through A and B on the diagram below has a gradient of 2 .
3
y
4
B
3
2
1
0
a)
A
1
2
3
4
5
x
Is he correct? Explain how you know.
Change in y 3 – 0 3
Gradient = Change in x = 3 – 1 = 2
2
3
No he is not correct, the gradient is 2 not 3 .
b)
[2 marks]
Lorna draws a line through the origin (0, 0)
parallel to AB. What is the equation of this line?
Any parallel line will have the same
gradient 23 and the y-intercept is 0.
3
y = 2x
.........................
[1 mark]
Section Three — Graphs
66
Practice Questions
2
a) Plot the points A(1, 3), B(3, 9), C(0, 3) and D(3, 10) on the graph below.
y
10
9
8
7
6
5
4
3
2
1
x
0
1 2 3 4 5 6
b)
[2 marks]
Find the equation of the line joining points A and B
.........................
[2 marks]
c)
Is the line joining points C and D parallel to the line in part b? Explain your answer.
[2 marks]
3
y = mx + c is the general equation for a straight line graph.
a)
On the axes below, draw and label the graph of y = 3x + 1.
15 y
10
5
x
0
1
–5
b)
2
3
4
5
[2 marks]
On the same axes draw a line parallel to y = 3x + 1 and passing through the point (0, –3).
[2 marks]
Section Three — Graphs
67
Reading Off Graphs
Here’s a nice little page for you on reading graphs. Don’t worry if some of the graphs look a little
bit odd — you always use the same method for reading values from them.
Getting Answers from a Graph
You can read values from a
graph using this method:
1) Draw a straight line to the graph from one axis.
2) Then draw a straight line down or across to the other axis.
Straight Line Graphs
a) Find the value of x
when y = 5.
Draw a line across from the
y-axis to the graph at y = 5
and then down to the x-axis.
x=5
b) Find the value of y
when x = 3.
6 y
Draw a line up from the
x-axis to the graph at x = 3
and then across to the y-axis.
4
5
3
2
1
y ≈ 3.2
x
-6 -5 -4 -3 -2 -1 0
-1
Sometimes the answer won’t be a whole
number and you’ll have to estimate —
≈ means ‘is approximately equal to’.
Curvy Graphs
1
2
3
4
5
6
-2
-3
-4
-5
-5 -4 -3 -2 -1 0
-1
-2
-3
-4
-5
a) Find the value of y when x = 2.
Go up from 2 on the x-axis and
then across to the y-axis.
1 2 3 4 5
x
y=2
b) Find the value of x when y = –1.
Go across from –1 on the y-axis
and then up to the x-axis.
x = –4
The graph shows how the number of cows
infected by an alien virus increases over time.
Estimate the number of cows infected after 4 days.
Draw a line up from day 4 to the graph and
then across to find the number of cows infected.
Number of cows infected
This is a reciprocal
graph — the general
equation is y = A .
x
-6
y5
4
3
2
1
600
500
400
300
200
100
Number of cows infected ≈ 260
0
1
2 3 4 5 6
Number of days
This is an exponential graph — the general
equation is y = akx where k is a positive number.
Always draw the lines to make sure you’ve got it right
Reading graphs is easy once you know how — you just use the same method for every graph.
Section Three — Graphs
68
Travel and Conversion Graphs
Here are another two types of graph for you — this time it’s back to straight lines. The axes mean
something different here but you use the same method as on the last page for reading off values.
Travel Graphs
®
1) A TRAVEL GRAPH is always DISTANCE ( ) against TIME (®)
2) FLAT SECTIONS are where it’s STOPPED.
3) The STEEPER the graph the FASTER it’s going.
4) The graph GOING UP means it’s travelling AWAY.
5) The graph COMING DOWN means it’s COMING BACK AGAIN.
a) How far away from home was the car
the 2nd time it stopped?
The car stopped at the flat parts of the graph:
50 km
b) When was the car travelling fastest?
The car was travelling fastest at the steepest part of the graph.
9 am to 9:30 am
60
Distance (kilometres)
This travel graph shows a morning car journey.
50
40
30
20
10
0
8am
9am 10am
Time
11am
Conversion Graphs
These are really easy if you remember the method for reading graphs on p67.
Conversion graphs are used to convert between things like £ and dollars or mph and km/h, etc.
The conversion graph below can be used to convert between miles and kilometres.
a) Convert 25 miles into kilometres.
Draw a line up from 25 on the miles axis till it
hits the line, then go across to the km axis.
40 km
110
100
90
Kilometres
80
b) Convert 80 kilometres into miles.
Draw a line across from 80 on the km axis till it
hits the line, then go down to the miles axis.
50 miles
70
60
50
40
30
20
10
0
10
20
30
40
Miles
50
60
c) Estimate how many kilometres are equal to 35 miles.
Sometimes you won’t be able to find an
exact answer and you’ll have to estimate.
56 km
The gradient of a travel graph shows the speed
You’ll need to be able to read values from a travel graph and interpret what each part of the graph is
telling you. With conversion graphs it’s a simple matter of reading values from the graph.
Section Three — Graphs
69
Real-Life Graphs
Now and then, graphs might reveal some interesting information about the world around you.
Graphs Can Show Billing Structures
Many bills are made up of two charges:1) a fixed charge, and 2) a cost per unit.
E.g. You might pay £20 each month for a TV package, and then be charged £3 for each film you watch.
This graph shows how a monthly mobile phone bill is calculated.
a) How many megabytes (MB) of Internet usage
are included in the fixed monthly cost?
500 MB
b) What is the cost (in pence) for
each additional MB of data?
Monthly Cost (£)
The fixed cost is the horizontal bit of the line.
40
30
£10
20
250 MB
10
The gradient of the sloped line will give
you the cost of each additional MB.
vertical change
Gradient = horizontal change
x2 + y2 = 25
0
10
250 = £0.04 = 4p
0
250
500
750
Internet Usage (MB)
1000
Graphs Can Show Changes with Time
Time
A
C
B
A steeper slope means that the
juice height is changing faster.
Height
Height
Height
Three empty glasses are shown on the right.
Each glass is filled up at a constant rate.
Each graph below shows how the height
of juice in one glass changes.
Match each graph to the correct glass.
Time
Time
Glass C
Glass A
Glass B
Glass C has straight
sides, so the juice
height rises steadily.
Glass A is narrowest at the
bottom, so the juice height
rises fastest at first.
Glass B is narrowest
at the top, so the height
will rise fastest at the
end of the graph.
Look at the axes first to see what the graph is telling you
Questions on billing structures will often be made up of two straight lines showing two different rates
of cost — the steeper the gradient of the line is, the higher the cost per unit.
Section Three — Graphs
70
Warm-Up and Practice Questions
Once you can read off one type of graph you can read them all — so these questions
shouldn’t surprise you. It’s just a matter of working out what the numbers actually mean.
1) The graph on the right shows how the number of bacteria in a
colony changed over time. Use it to answer these questions:
a) how many bacteria were there after 5 days?
b) how long was it before there were 100 000 bacteria?
c) what is the name given to this type of graph?
2) What does the flat section of a travel graph tell you?
Number of bacteria
(in thousands)
Warm-up Questions
3) What does the gradient of a travel graph tell you?
600
500
400
300
200
100
0
4) What is always plotted up the vertical axis of a travel graph?
1
2 3 4 5 6
Number of days
5) Explain how to use a conversion graph to convert between different units.
6) A rental car costs £90 to hire for a week and £10 for each additional day.
Draw a graph to show the cost of a rental car for up to 15 days.
Practice Questions
As promised, there will be no surprises for you in these questions. You should be
used to the drill by now — first, I get to answer a question, then it’s your turn.
1
This graph can be used to convert pounds into US dollars ($).
Use the graph to answer the following questions.
Dollars
($)
Pounds (£)
a)
Estimate how much £50 is in dollars.
20
On the y-axis each square represents 5 = 4 dollars.
$84
.........................
[1 mark]
b)
What is the gradient of the graph?
The graph goes through
(0, 0) and (60, 100).
100 – 0 = 100 = 5
3
60 – 0
60
5
3
.........................
[2 marks]
Section Three — Graphs
71
2
Kara and Kerry did a sponsored walk.
They set off at noon and jogged for the first
quarter of an hour. Then they walked for
the next hour. After a rest, they set off to
walk back. Kerry’s mother drove to meet them
and, as it started to rain, gave them a lift back.
The travel graph on the right shows their journey.
Distance from home (km)
Practice Questions
10
9
8
7
6
5
4
3
2
1
12:00
a)
13:00
14:00
15:00
Time
For how many kilometres did they jog?
......................... km
[1 mark]
b)
How many minutes did they have for the break?
......................... min
[1 mark]
c)
At what time did Kerry’s mother pick them up?
.........................
[1 mark]
d)
How far did they jog and walk altogether?
......................... km
[1 mark]
0
a)
Time
0
Graph B
Time
Distance from home
Graph A
Distance from home
Paul is travelling to his friend's house. Which of these graphs would represent his journey if:
Distance from home
3
0
Graph C
Time
he cycles to the train station, then travels by train and walks the last part of the journey.
.........................
[1 mark]
b)
he drives part of the way, then gets stuck in slow moving traffic and decides to turn
around and come home?
.........................
[1 mark]
Section Three — Graphs
72
Solving Simultaneous Equations
Simultaneous equations — you’ll be given two equations, each with the same two unknown
variables (e.g. x and y) and asked to “solve these equations simultaneously”.
Solving Simultaneous Equations Using Graphs
To solve two equations simultaneously, you need to find values of x and y that give the same answer
when you put them into either equation. Luckily there’s an easy way to find them.
The solution of two simultaneous equations is
simply the x- and y- values where their graphs cross.
So when a question asks you to solve two
simultaneous equations using a graph,
just follow this method and you’ll be flying:
1)
2)
3)
4)
wn for
When the graphs are dra
p 3.
ste
to
you, jump straight
Do a “table of 3 values” for each equation.
Draw the two graphs.
Find the x- and y-values where they cross.
Put the values back into both equations
to make sure they work.
7y
6
Use this graph to solve the equations
y = 3x + 1 and y = 5 – x simultaneously.
y = 3x + 1
5
1) Find the x-value and the y-value where the two lines cross.
4
x = 1 and y = 4
3
2) Check the values — put the x-value into each
equation and check it gives you the correct y-value.
y = (3 × 1) + 1 = 4
y=5–x
2
1
-2 -1 0
-1
-2
y=5–1=4
x
1
2
3
4
5
Draw the graphs of y = x + 1 and y = 6 – 4x and
use the diagram to solve the equations simultaneously.
1) Find 3 values for each equation.
x
y
y=x+1
0
2
1
3
3
4
y
x
y
6
y = 6 – 4x
0
1
2
6
2
–2
5
y=x+1
4
3
2) Draw the two straight line graphs on the same axes.
3) Read off the x- and y-values where the lines cross.
x=1
y=2
4) Check the values work — put x = 1 into each equation.
y=1+1=2
y = 6 – (4 × 1) = 6 – 4 = 2
y = 6 – 4x
2
1
x
-3
-2
-1 0
-1
-2
1
2
The solution is found at the point where the lines cross
Simultaneous equations can seem fairly daunting at first. Just make sure that when you’re given an
equation you can draw the straight line (p59) — then it’s just a matter of finding where they cross.
Section Three — Graphs
3
73
Quadratic Graphs
Quadratic graphs have lovely smooth curves — not a straight line in sight. You’ll still have to be able
to sketch them from the equations though, so take a look below at how it’s done.
Drawing a Quadratic Graph
y
y
5
4
3
2
1
-3
-2
-1
-1
-2
-3
4
y = –2x 2 + 4x
2
y = x2
1
2
3
-1 0
-2
x
1
2
3 x
Quadratic graphs are of the form:
y = anything with x2 (but not higher powers of x).
They all have the same symmetrical bucket shape.
-4
If the x2 bit has a ‘–’ in front of it
then the bucket is upside down.
-6
-8
Complete the table of values for the
equation y = x2 – 1 and then draw the graph.
1) Work out each y-value by substituting the
corresponding x-value into the equation.
2) Plot the points and join them
with a completely smooth curve.
Definitely DON’T use a ruler.
–3 –2 –1 0
8 3 0 –1
x
y
2
3
3
8
x = –2: y = (–2)2 – 1 = 4 – 1 = 3
x = 1:
y = (1)2 – 1 = 1 – 1 = 0
10
x
y
x
8
This point is
obviously wrong
NEVER EVER let one point drag your
line off in some ridiculous direction.
When a graph is generated from an equation,
you never get spikes or lumps — only MISTAKES.
1
0
6
4
x
x
2
-4 -3 -2
x
-1 0x
-2
x
1
x
2
3
4
Solving Quadratic Equations
The graph on the right shows the curve y = x2 – 2x.
Estimate both solutions to the equation x2 – 2x = 3.
x2 – 2x = 3 is what you get when you put y = 3
into the graph’s equation, so:
10
y
5
y=3
1) Draw a line at y = 3.
2) Read the x-values where the curve crosses this line.
The solutions are x = –1 and x = 3.
x
-2
-1 0
x -1
1
2
3
x3
Quadratic equations usually have 2 solutions —
if they aren’t exact then round sensibly (e.g. to 1 d.p.)
ION
RE VIS
TIP
Quadratic graphs should be smooth — never use a ruler
Sketching quadratics can be tough but just remember the method — draw up a table of
values, plot them on a graph and then (the tough bit) join them together with a smooth curve.
Section Three — Graphs
74
Warm-Up and Practice Questions
Simultaneous equations and quadratic graphs are the hardest topics in the graph section. No need to
worry though — there are plenty of practice questions coming up. Try these warm-up questions first.
Warm-up Questions
1) If two simultaneous equations are drawn on a graph, how do you find the solution?
2) a) Draw the graphs of y = 2x – 2 and y = 4 – x on the same axes.
b) Solve the simultaneous equations y = 2x – 2 and y = 4 – x.
3) How can you tell if an equation is a quadratic?
4) a) Complete the table of values below for the equation y = x2 + 3.
x
y
–3
–2
–1
4
0
3
1
2
3
b) Draw the graph of y = x2 + 3 for values of x between –3 and 3.
c) Use your graph to estimate the solutions to x2 + 3 = 6.
Practice Questions
I bet you’re ready for those practice questions now — I know I am, that’s why I’ve done the
first one for you. I just couldn’t help myself. There are some for you to try on the next page.
1
Look at the diagram below. Use it to solve the following simultaneous equations.
y
y = 4x – 2
y=x+4
Read off the
coordinates
where each pair
of lines cross
x
y = 4 – 2x
a)
y=x+4
y = 4x – 2 2
x = ................
6
y = ..................
b)
y=x+4
0
y = 4 – 2xx = ................
4
y = ..................
c)
y = 4x – 2
1
y + 2x = 4x = ................
2
y = ..................
Section Three — Graphs
[1 mark]
[1 mark]
[1 mark]
75
Practice Questions
2
a) On the grid below, draw and label the graph of the equation y = 2x + 1.
[2 marks]
y
7
6
5
4
3
2
1
x
–2 –1 0
–1
1
2
6
3 4 5
7
b)
On the grid, draw and label the graph of the equation x + y = 7.
c)
Use the graphs you have drawn for (a) and (b) to find the values of x and y which satisfy
the simultaneous equations y = 2x + 1 and x + y = 7.
x = ................
3
[2 marks]
y = ..................
[1 mark]
The graph of the equation y = 2x2 – 10 is a curve.
a)
Complete the table of values for the equation y = 2x2 – 10.
x
y
b)
–3
–2
–1
0
1
2
3
[2 marks]
On the grid, plot the points shown in the table and join them up to form a smooth curve.
y
10
5
x
-3
-2
-1
0
1
2
3
-5
-10
[2 marks]
c)
Use your graph to estimate the values of x when y = 0.
..........................................
[1 mark]
Section Three — Graphs
76
Revision Summary for Section Three
Well, that wraps up section three — time to test yourself and find out how much you really know.
• Try these questions and tick off each one when you get it right.
• When you’ve done all the questions for a topic and are completely happy with it, tick off the topic.
Coordinates and Midpoints (p57) 
y
3


2) Find the midpoint of a line segment with endpoints A and C.
–3 –2 –1 0 1 2 3
–1

–2
3) Find the midpoint of a line segment with endpoints B and D. B
2
1) Give the coordinates of points A to D in the diagram on the right.
A
1
D
x
–3
C
Straight-Line Graphs and Their Gradients (p58-62) 
4) What does the equation of a downhill sloping line through the origin look like? 
5) Say whether the graphs of these equations go uphill, downhill or neither (from left to right):
y

a) y = –2
b) y = –3x
c) –y + x = 0
8
R

6) Draw the graph of y = –3x + 2 for the values of x from –2 to 2. 4
6
2

x
S
7) Find the gradient of the lines R and S on the graph to the right.
0
1
2
3
5
4
6
y = mx + c and Parallel Lines (p63-64) 

10
8) What do ‘m’ and ‘c’ represent in y = mx + c?
y
8

9) Find the equation of the graph on the right.
6
10) Draw the graph of y = 3x – 1 using the y = mx + c method. 
4

11) How are the gradients of parallel lines related?
2
x

12) Is 2y + 8x = 5 parallel to y = –4x + 2?
0
1 2 3 4 5
Distance (kilometres)
Reading Off Graphs (p67-69) 
4
13) What does a horizontal line mean on a distance-time graph?

14) The graph on the right shows Bob’s bicycle journey to the shop and back. 2
a) Did he ride faster on his way to the shop or on his way home? 
0
9:00
9:30
10:00 
b) How long did he spend in the shop?
Time
Monthly cost (£)
15) This graph shows the monthly cost of a TV package.
40
a) What is the cost of the fixed monthly rate? 
30
b) How many films does the fixed monthly rate include? 
20

c) Work out the cost per film for additional films.
10

d) Peter watches 7 films one month. What will his bill be?
0
0
2
4
6
8
Number of films watched
Trickier Graphs (p72-73) 

17) What is the shape of a quadratic graph? 
18) a) Make a table of values for the equation y = x2 – 3 for values of x between –3 and 3. 
b) Use your table in part a) to draw the graph of y = x2 – 3. 
16) By drawing the graphs of y = 2x and y = x + 2, solve the equations simultaneously.
Section Three — Graphs
Section Four — Ratio, Proportion and Rates of Change 77
Ratios
Ratios can be a bit nasty, but the examples on the next two pages should cut them down to size.
Reducing Ratios to Their Simplest Form
To reduce a ratio to a simpler form, divide all the numbers in the ratio by the same thing (a bit like
simplifying a fraction). It’s in its simplest form when there’s nothing left you can divide by.
Vicky has 25 pork pies and Jonny has 15 pork pies. Write the number
of Vicky’s pies to Jonny’s as a ratio in its simplest form.
1) First, write the numbers as a ratio — 25 : 15.
2) Simplify the ratio — both numbers have a factor of 5, so divide them by 5.
We can’t reduce this any further, so the simplest form of 25 : 15 is 5 : 3.
25 : 15 ÷5
= 5:3
÷5
A handy trick on your calculator — use the fraction button
If you enter a fraction with the
or a bc button, the calculator
automatically cancels it down when you press = .
8
2
So for the ratio 8 : 12, enter the fraction 12 , and you’ll get the simplified fraction 3 .
Now you just change it back to ratio form, i.e. 2 : 3.
The More Awkward Cases:
1) If the ratio contains decimals or fractions — multiply
Simplify the ratio 1.2 : 1.5 as far as possible.
1) Multiply both sides by 10 to get rid of the decimal parts.
2) Now divide to reduce the ratio to its simplest form.
1.2 : 1.5
×10
= 12 : 15
÷3
÷3
= 4 : 5
×10
2) If the ratio has mixed units — convert to the smaller unit
Reduce the ratio 18 mm : 3.6 cm to its simplest form.
18 mm : 3.6 cm
1) Convert 3.6 cm to millimetres.
= 18 mm : 36 mm
2) Simplify the resulting ratio. Once the units on both
÷18
÷18
sides are the same, get rid of them for the final answer.
=
1:2
3) To get to the form 1 : n or n : 1 — just divide
Reduce 4 : 30 to the form 1 : n.
4 : 30
÷4
÷4
Divide both sides by 4:
= 1 : 7.5
This form is often the most useful,
since it shows the ratio very clearly.
Always give answers to ratio questions in their simplest form
There are lots of different examples on this page, but the method for simplifying them is the same.
Section Four — Ratio, Proportion and Rates of Change
78
Ratios
Another page of ratios for you to work through. You’ll have to know how to scale up a ratio and how
to divide things in a given ratio. It’s not too bad though — take a look below.
Scaling Up Ratios
If you know the ratio between parts and the actual size of one part,
you can scale the ratio up to find the other parts.
Purple paint is made from red paint and blue paint in the ratio 5 : 4.
If 20 pots of red paint are used, how much blue paint is needed?
You need to multiply by 4 to go
from 5 to 20 on the left-hand side
(LHS) — so do that to both sides:
red paint : blue paint
= 5 : 4 ×4
×4
=
20 : 16
So 16 pots of blue paint are needed.
Brenda has a tray of Bourbon biscuits and custard creams in the ratio 6 : 7.
All the biscuits are either Bourbons or custard creams.
She has 18 Bourbons. How many biscuits does Brenda have in total?
Multiply both sides by 3 to go from 6 to 18 on the LHS:
Bourbon : custard cream
= ×3 6 : 7 ×3
So Brenda has 18 Bourbon biscuits and 21 custard creams.
= 18 : 21
So in total she has 18 + 21 = 39 biscuits
Proportional Division
In a proportional division question a TOTAL AMOUNT is split into parts in a certain ratio.
The key word here is PARTS — concentrate on ‘parts’ and it all becomes quite painless:
Kim and Chris share £600 in the ratio 4 : 11. How much does Chris get?
1) ADD UP THE PARTS:
The ratio 4 : 11 means there will be a total of 15 parts:
2) DIVIDE TO FIND ONE “PART”:
Just divide the total amount by the number of parts:
3) Multiply to find the amounts:
We want to know Chris’s share, which is 11 parts:
4 + 11 = 15 parts
£600 ÷ 15 = £40 (= 1 part)
11 parts = 11 × £40 = £440
Learn these simple methods for questions involving ratios
Don’t be put off by complicated terms like “proportional division” — the methods are quite simple.
Make sure you know how to scale up ratios, and the three steps for proportional division.
Section Four — Ratio, Proportion and Rates of Change
79
Direct Proportion
The easiest proportions are direct proportions — the hard stuff is coming on the next page.
Graphing Direct Proportion
Direct proportions tell you how one thing increases (e.g. the amount of flour)
as another increases (e.g. the number of cupcakes you can make).
y
Goes
through
the origin
Two things are in direct proportion if, when you plot them
on a graph, you get a straight line through the origin.
Remember, the general equation for a straight line through the
origin is y = Ax (see p58) where A is a constant (like 2 or 3).
All direct proportions can be written as an equation in this form.
x
Solving Direct Proportion Questions
You can solve all direct proportion questions using the same method.
All you have to do is remember this golden rule:
DIVIDE for ONE, then TIMES for ALL
3 painters can paint 9 rooms per day.
How many rooms per day could 7 painters paint?
1) Start by dividing by 3 to find how many
rooms 1 painter could paint per day.
9 ÷ 3 = 3 rooms per day
2) Then multiply by 7 to find how many
rooms 7 painters could paint per day.
3 × 7 = 21 rooms per day
6 baguettes cost £1.50.
a) How much will 16 baguettes cost?
1) Start by dividing by 6 to find the cost of 1 baguette.
£1.50 ÷ 6 = £0.25
2) Then multiply by 16 to find the cost of 16 baguettes.
£0.25 × 16 = £4.00
b) Let c be the cost in £s and b be the number of baguettes.
Write an equation in the form c = Ab to represent this direct proportion.
1) Find the value of A by putting c = 1.50 and b = 6 into the equation.
2) Put the value of A back into the equation.
ION
RE VIS
TIP
c = 0.25b
1.50 = 6A
0.25 = A
Remember the golden rule for finding direct proportions...
Direct proportions are easy — both things increase together and give a straight line through
the origin when they are plotted on a graph. Don’t forget — divide for one and times for all.
Section Four — Ratio, Proportion and Rates of Change
80
Inverse Proportion
Here’s a trickier type of proportion — but once you’ve learnt this page you’ll be an expert.
Graphing Inverse Proportion
y
When two things are in inverse proportion,
one increases as the other decreases.
This is the graph
of y = Ax .
On the graph you can see that as the value of x increases,
the value of y decreases. E.g. as the number of hours you watch
TV increases, your score on a maths exam will decrease.
A
The general equation for inverse proportion is y = x .
x
Solving Inverse Proportion Questions
On the previous page you saw the ‘divide and times’ method for solving direct proportions.
Well, inverse proportions are the opposite so you have to:
TIMES for ONE, then DIVIDE for ALL
It takes 3 farmers 10 hours to plough a field.
How long would it take 6 farmers?
1) Start by multiplying by 3 to find how long
it would take 1 farmer to plough the field.
10 × 3 = 30 hours for 1 farmer
2) Then divide by 6 to find how long it
would take 6 farmers to plough the field.
30 ÷ 6 = 5 hours for 6 farmers
Note: Another way of looking at this question is that there are twice as many farmers,
so it will take half as long (10 ÷ 2 = 5 hours).
4 bakers can decorate 100 cakes in 5 hours.
a) How long would it take 10 bakers to decorate the same amount of cakes?
Multiply by 4 to find how long it would take 1 baker.
4 × 5 = 20 hours for 1 baker
Divide by 10 to find how long it would take 10 bakers.
20 ÷ 10 = 2 hours for 10 bakers
b) Let b be the number of bakers and h be the time in hours it takes them to decorate
A
100 cakes. Write an equation in the form h = b to represent this inverse proportion.
Find the value of A by putting b = 4 and h = 5 into the equation.
Put the value of A back into the equation.
EXAM
TIP
A
5 = 4 so A = 20
20
h= b
Inverse proportions — times for one and divide for all...
Inverse proportions can be tricky so it’s always a good idea to check your answer to make
sure it makes sense, e.g. if there are more people doing an activity it should take less time.
Section Four — Ratio, Proportion and Rates of Change
81
Warm-Up and Practice Questions
That’s ratios and proportions done and dusted. Check out how much sank in by having a go at these
warm up questions. If you get any wrong, have a look back over the last few pages, then try again.
Warm-up Questions
1)
Simplify the following ratios:
a) 7 : 63b) 2.2 : 5.5c) 8 cm : 48 mm
2)
A fruit punch is made from orange juice and apple juice in the ratio 4 : 7.
If 20 litres of orange juice are used, how much apple juice is used?
3)
Ahmed and Shannon share £500 in the ratio 12 : 13. How much does Ahmed get?
4)
£6400 needs sharing between Dan, Chris and Angela in the ratio 1 : 3 : 4.
How much does each person receive?
5)
a)
b)
6)
5 mathematicians take 20 hours to solve a problem.
How long would it take 25 mathematicians?
4 lumberjacks chop 12 trees in a day. How many trees could 13 lumberjacks
chop in a day?
Write an equation for this proportion in terms of trees chopped (t)
and number of lumberjacks (l).
Practice Questions
By now you should be nicely warmed up — I’d hate for you to pull a mathematical muscle doing these
strenuous practice questions. Have a look at this worked question before you sprint on to the others.
1
A DVD rental shop records the number of rentals of four films in a week.
a)
Film
Number of rentals
Wolverine
11
Les Misérables
24
Real Steel
18
The Prestige
15
Express the number of rentals of Real Steel compared
to The Prestige as a ratio in its simplest form.
Write the numbers as a ratio:
Divide both sides by 3:
÷3
18 : 15
÷3
6:5
6:5
......................
[1 mark]
b)
The following week, there were 30 rentals of Real Steel.
If the ratio of Real Steel rentals to The Prestige rentals remained the same,
how many copies of The Prestige were rented in the following week?
Use the ratio 6 : 5. You need to multiply by
5 to go from 6 to 30 on the left-hand side
— so multiply the right-hand side by 5 too.
×5
6:5
30 : 25
×5
25
......................
[2 marks]
Section Four — Ratio, Proportion and Rates of Change
82
Practice Questions
2
Alice has bought 14 calculators for £35. How much would 22 calculators cost?
£ .........................
[2 marks]
3
Sunita is making a dressing for a salad. She mixes 2 of a bottle of olive oil with 1 of a
5
10
bottle of vinegar. The bottles are the same size.
a)
Express the amount of olive oil to vinegar as a ratio in its simplest form.
.........................
[2 marks]
b)
Sunita uses 800 ml of olive oil. How much vinegar did she use?
......................... ml
[2 marks]
4
Shaun, Ben and David share 50 DVDs in the ratio 5 : 3 : 2. How many DVDs does Shaun get?
.........................
[3 marks]
5
a)
4 people take 5 hours to paint a mural. How long would it take 10 people
to paint the same mural?
......................... hours
[2 marks]
b)
5 people groom 15 dogs in a day. How many dogs could 11 people groom in a day?
.........................
[2 marks]
Section Four — Ratio, Proportion and Rates of Change
83
Percentage Change
Questions about percentage change are a bit trickier than finding basic percentages (see p23).
Never fear though — the next three pages will show you how to handle them.
Find the New Amount After a % Increase or Decrease
There are two different ways of finding the new amount after a percentage increase or decrease:
1) Find the % then Add or Subtract
Turn the percentage into a decimal, then multiply. Add this on to (or subtract from) the original value.
A dress has increased in price by 30%.
It originally cost £40. What is the new price of the dress?
1) Write 30% as a decimal:
30% = 30 ÷ 100 = 0.3
2) Multiply to find 30% of £40:
0.3 × £40 = £12
3) It’s an increase, so add on to the original:
£40 + £12 = £52
2) The Multiplier Method
This time, you first need to find the multiplier — the decimal that represents the percentage change.
E.g.
5% increase is 1.05 (= 1 + 0.05)
26% increase is 1.26 (= 1 + 0.26)
5% decrease is 0.95 (= 1 – 0.05)
26% decrease is 0.74 (= 1 – 0.26)
Then you just multiply the original value by the multiplier.
A % decrease has a multiplier less than 1,
a % increase has a multiplier greater than 1.
A hat is reduced in price by 20% in the sales.
It originally cost £12. What is the new price of the hat?
1) Find the multiplier:
2) Multiply the original value by the multiplier:
20% decrease = 1 – 0.20 = 0.8
£12 × 0.8 = £9.60
A rare painting has increased in value by 75% since it was painted.
It was originally worth £800. What is its new value?
1) Find the multiplier:
2) Multiply the original value by the multiplier:
75% increase = 1 + 0.75 = 1.75
£800 × 1.75 = £1400
Learn both methods then you can pick your favourite...
The multiplier method might be quicker than finding the percentage then adding or subtracting it,
but you should use whichever method you’re most comfortable with.
Section Four — Ratio, Proportion and Rates of Change
84
Percentage Change
This page will show you how to find simple interest and calculate a percentage change.
Calculating simple interest is a useful real-life skill so make sure you learn how to do it.
Simple Interest
Simple interest means a certain percentage of the original amount is paid at regular intervals
(usually once a year). So the amount of interest is the same every time it’s paid.
Elsa invests £1000 in an account which pays 2% simple interest
each year. How much interest will she earn in 5 years?
1) Work out the amount of interest earned in one year:
2% = 2 ÷ 100 = 0.02
2% of £1000 = 0.02 × £1000 = £20
2) Multiply by 5 to get the total interest for 5 years:
5 × £20 = £100
Finding the Percentage Change
This is the formula for giving a change in value as a percentage — LEARN IT, AND USE IT:
Percentage ‘Change’ =
'CHANGE'
× 100
ORIGINAL
Typical questions will ask ‘Find the percentage increase / profit / error’
or ‘Calculate the percentage decrease / loss / discount’, etc.
A house increases in price from £200 000 to £214 000.
Find the percentage increase.
1) Here the ‘change’ is an increase,
so use the formula:
2) Find the actual increase in price:
3) Then use the formula:
increase
percentage increase = original × 100
increase = £214 000 – £200 000 = £14 000
14 000
percentage increase = 200 000 × 100 = 7%
A trader buys cars for £4000 and sells them for £6000.
Find his profit as a percentage.
1) Here the ‘change’ is profit, so use the formula:
profit
percentage profit = original × 100
2) Work out the actual value of the profit:
profit = £6000 – £4000 = £2000
3) Calculate the percentage profit:
2000
percentage profit = 4000 × 100 = 50%
Interest? It’s simple...
There is another type of interest (called compound interest), but don’t worry about that for now.
Section Four — Ratio, Proportion and Rates of Change
85
Percentage Change
The examples on this page are a bit different — this time you have to
find the original value before a percentage change.
Finding the Original Value
This is when you’re given the percentage change and the new value, and have to work out the
original value. There are two different methods you can use here:
1) Divide to find 1% then Multiply to find 100%
1) Write the amount in the question as a percentage of the original value.
2) Divide to find 1% of the original value.
3) Multiply by 100 to give the original value (= 100%).
A car decreases in value by 40% to £7200.
Find what it was worth before the fall.
1) A decrease of 40% means £7200 represents
60% of the original value.
2) Divide by 60 to find 1% of the original value.
3) Then multiply by 100.
Note: The new, not the
original value is given.
£7200 = 60%
÷60
£120 = 1%
×100
£12 000 = 100%
So the original value was £12 000
Always set them out exactly like this example. The trickiest bit is deciding the top % figure
on the right-hand side — the 2nd and 3rd rows are always 1% and 100%.
2) The Multiplier Method
You can also use the multiplier method from p.83 in reverse to find the original value.
1) Find the multiplier.
2) Divide the new value by the multiplier — this will give you the original value.
An antique vase increases in value by 15% to £3450.
Find what it was worth before the rise.
1) Find the multiplier:
2) Divide the new value by the multiplier:
15% increase = 1 + 0.15 = 1.15
3450 ÷ 1.15 = £3000
If it was a decrease of 15%, the multiplier
would be 1 – 0.15 = 0.85 instead.
EXAM
TIP
Think — does the question give the original or the new value?
Again, you can learn both, then pick the method that you prefer. Remember to check your
answer, e.g. if the new value has been increased then the original value should be smaller.
Section Four — Ratio, Proportion and Rates of Change
86
Warm-Up and Practice Questions
Here are some lovely warm-up questions to test what you’ve learned about percentage change.
This is the time to go back over anything you’re not sure about, before it’s too late...
Warm-up Questions
1)
A unicorn costing £4000 is reduced by 15% in the sale. What is its new price?
2)
The attendance at a rugby match was 16 000 last week. This week it was 18 000.
Calculate the percentage increase in attendance.
3)
£200 is invested in an account which pays 3% simple interest each year.
How much interest is earned in 10 years?
4)
A football player decreases in value from £4 million to £2.5 million.
Find the percentage decrease.
5)
A video game console is priced at £66 after a 40% discount. What was its original price?
6)
A sofa is sold for £416, which is 20% less than the original price.
What was its original price?
7)
Molly’s salary increases by 6% to £20 140. What was her salary before the rise?
Practice Questions
OK, before you dive into the practice questions on the next page, read through this worked
practice question — it should give you an idea of how to deal with the next lot.
1
An antique painting originally bought for £4000 is being put up for auction.
a)
The painting sells for £4400.
(i) What profit has been made?
Profit = selling price – original price = 4400 – 4000 = £400
£400
.........................
[1 mark]
(ii) What is the percentage profit?
profit
400
Percentage profit = original × 100 = 4000 × 100 = 10%
10%
.........................
[2 marks]
b)
If the painting had sold for £3500, what would the percentage loss have been?
Loss = original price – selling price = 4000 – 3500 = £500
loss
500
Percentage loss = original × 100 = 4000 × 100 = 12.5%
12.5%
.........................
[2 marks]
Section Four — Ratio, Proportion and Rates of Change
87
Practice Questions
2
A company decides to give all of its part-time employees a pay rise. They are each given
the choice between an extra £10 per week or an extra 10% per week in their wages.
Which of the following employees would be better off choosing 10%?
Mr Patel who earns £102 per week.
Miss Dalton who earns £98.50 per week.
Mrs Ferrar who earns £120 per week.
....................................................................
[4 marks]
3
The following table shows the sale price for a number of items from a bedroom showroom
and the discount applied to each item. Find the original prices of each item.
ITEM
SALE PRICE
DISCOUNT
Single bed
£63.75
15%
ORIGINAL PRICE
.......................................
Double bed
£240
40%
.......................................
Bunk beds
£168
20%
.......................................
[3 marks]
4
At the beginning of Year 7, Jo was 1.25 m tall. At the end of Year 9, she was 1.7 m tall.
a)
Calculate her percentage increase in height.
.........................
[2 marks]
Her percentage increase in height from the end of Year 9 to the end of Year 11
is only half what it was from Year 7 to Year 9.
b)
Calculate Jo’s height at the end of Year 11 to the nearest cm. Give your answer in metres.
......................... m
[2 marks]
Section Four — Ratio, Proportion and Rates of Change
88
Metric and Imperial Units
There’s nothing too bad on this page — just some facts to learn.
Metric Units
1)
2)
3)
4)
5)
Length mm, cm, m, km
Area mm2, cm2, m2, km2,
Volume mm3, cm3, m3, ml, litres
Mass g, kg, tonnes
Speed km/h, m/s
‘Weight’ is often used
instead of ‘mass’ in
everyday language.
MEMORISE THESE KEY FACTS:
1 cm = 10 mm 1 tonne = 1000 kg
1 m = 100 cm
1 litre = 1000 ml
1 km = 1000 m 1 litre = 1000 cm3
1 kg = 1000 g
1 cm3 = 1 ml
Imperial Units
1)
2)
3)
4)
5)
Length
Area
Volume
Mass
Speed
IMPERIAL UNIT CONVERSIONS
1 foot = 12 inches
1 yard = 3 feet
1 gallon = 8 pints
1 stone = 14 pounds (lb)
1 pound = 16 ounces (oz)
inches, feet, yards, miles
square inches, square feet, square miles
cubic inches, cubic feet, gallons, pints
ounces, pounds, stones, tons
mph
Metric-Imperial Conversions
Learn these approximate conversions
to help you change from metric units
to imperial units:
‘≈’ means ‘approximately
equal to’.
APPROXIMATE CONVERSIONS
1 inch ª 2.5 cm
1 kg ª 2.2 pounds (lb)
1 foot ª 30 cm
1 litre ª 1.75 pints
1 gallon ª 4.5 litres
1 mile ª 1.6 km (or 5 miles ª 8 km)
3-Step Method for Converting:
1) Find the conversion factor (always easy).
2) Decide whether to multiply or divide by it
(use the conversion factor to decide if the answer should be bigger or smaller).
3) Work out your answer (and check it).
You’re just going to have to learn all these conversions...
Learn all the units and conversions on this page — you’ll need to know them to use the 3-step method
for converting. Luckily for you, the next page is filled with examples using this method.
Section Four — Ratio, Proportion and Rates of Change
89
Converting Units
Time to try out the 3-step method for converting from the previous page.
Examples:
A zoo has a miniature gorilla called Augustus, who is 30 cm tall.
How tall is he in m?
1
1) Find the conversion factor
1 m = 100 cm, so conversion factor = 100
2) Decide whether to multiply or divide by it
You’d expect more cm than m, so divide
3) Work out the answer
30 ÷ 100 = 0.3 m
Check your answer — 0.3 is less than 30, so this looks sensible.
2
Nick lives 18 miles away from his friend Anna.
How far is this in km?
1 mile ≈ 1.6 km, so conversion factor = 1.6
1) Find the conversion factor
2) Decide whether to multiply or divide by it You’d expect more km than miles, so multiply
3) Work out the answer
18 × 1.6 = 28.8, so 18 miles ≈ 28.8 km
Check your answer — 28.8 is bigger than 18, so this looks sensible.
3
4
Write the following measurements in order of size from smallest to largest:
180 cm3, 1.75 litres, 185 ml
1) First, write all three measurements in the
same unit — I’m going to choose cm3
180 cm3, 1750 cm3, 185 cm3
2) Write them out in order
180 cm3, 185 cm3, 1750 cm3
3) Convert back to the original units
180 cm3, 185 ml, 1.75 litres
Convert 3 hours 15 minutes into minutes.
1) Convert the hours to minutes.
3 hours = 3 × 60 = 180 minutes
2) Add to get the total minutes.
180 + 15 = 195 minutes
Avoid calculators in
time calculations — the
decimal answers they
give are confusing, e.g.
2.5 hours = 2 hours 30 mins,
NOT 2 hours 50 mins.
Learn the method for converting — it’s the same every time
The method for converting between metric and imperial units never changes — only the conversion
factor does. Always remember to check your answer looks sensible.
Section Four — Ratio, Proportion and Rates of Change
90
More Conversions
Converting areas and volumes from one unit to another is a bit treacherous, because 1 m2 definitely
does NOT equal 100 cm2 (whatever you might think). Remember this and read on for why.
Converting Area and Volume Measurements
Be really careful — 1 m = 100 cm DOES NOT mean 1 m2 = 100 cm2 or 1 m3 = 100 cm3.
You won’t slip up if you LEARN THESE RULES:
Volume: units come with a 3, e.g. mm3, cm3,
m3 — use the conversion factor 3 times.
 100 cm 
 100 cm 
Area: units come with a 2, e.g. mm2, cm2, m2
— use the conversion factor 2 times.
1m2
 100 cm 
1.

m
0c
 100 cm  
1 m2 = 100 cm × 100 cm
1 cm2 = 10 mm × 10 mm
EXAMPLES:
1m3
10
1 m3 = 100 cm × 100 cm × 100 cm
1 cm3 = 10 mm × 10 mm × 10 mm
Convert 5 m2 to cm2.
To change area measurements from
m2 to cm2 multiply by 100 twice.
5 × 100 × 100 = 50 000 cm2
2. Convert 80 000 mm3 to cm3.
To change volume measurements from
mm3 to cm3 divide by 10 three times.
80 000 ÷ (10 × 10 × 10) = 80 cm3
You should expect your answers to be really big or really small compared
to the value you’re given — so if they’re not, something’s gone wrong.
Converting Speeds
Don’t panic if you have to convert a speed from, say, miles per hour (mph) to km per hour (km/h)
— if the time part of the units stays the same then it’s just a distance conversion in disguise.
Rhiannon is driving at 30 mph. The speed limit is 50 km/h.
Is she breaking the speed limit?
1) First convert 30 miles into km:
1 mile ≈ 1.6 km, so conversion factor = 1.6
You’d expect more km than miles, so multiply:
30 × 1.6 = 48 km
2) Add in the ‘per hour’ bit
to get the speed:
30 mph = 48 km/h
So Rhiannon isn’t breaking the 50 km/h speed limit.
One final time, in case you forgot: 1 m2 is not equal to 100 cm2
These topics are actually quite simple if you learn the rules for area and volume conversions —
don’t panic if you get really big or really small answers compared to the value you’ve been given.
Section Four — Ratio, Proportion and Rates of Change
91
Maps and Scale Drawings
Scales tell you what a distance on a map or drawing represents in real life. They can be
written in different ways, but they all say something like “1 cm represents 5 km”.
Map Scales
You might come across these kinds of scales on maps:
1 cm = 3 km
— “1 cm represents 3 km”
1 : 2000 — 1 cm on the map means 2000 cm in real life.
Converting to m gives “1 cm represents 20 m”.
0
1
km
Use a ruler — the line’s 2 cm long, so 2 cm means 1 km.
Dividing by 2 gives “1 cm represents 0.5 km”.
See p88 for a
reminder about
conversions.
To convert between maps and real life, learn these rules:
•
•
•
•
Put your scale into the form “1 cm = ...”
To find REAL-LIFE distances, MULTIPLY by the SCALE.
To find MAP distances, DIVIDE by the SCALE.
Always check your answer looks sensible.
Puddleton is 10 km east of Muddleton.
a) How far apart would they be on this map?
Real-life distance = 10 km
Divide for a map distance.
Scale is 1 cm = 5 km
Distance on map = 10 ÷ 5 = 2 cm This looks
sensible.
b) Mark Puddleton on the map.
Measure 2 cm to the east (right) of Muddleton:
N
Muddleton
1 cm = 5 km
Puddleton
Muddleton
1 cm = 5 km
N
Scale Drawings
The rules above work for scale drawings too.
This is a scale drawing of Josephine’s garden.
1 cm represents 2 m. Find the real length
and width of the patio in m.
1) Measure with a ruler.
2) Multiply to get
real-life length.
Length on drawing = 3 cm
Width on drawing = 1.5 cm
Real length = 3 × 2 = 6 m
Real width = 1.5 × 2 = 3 m
Scale drawings
will often be
shown on a grid.
Patio
Real life units are in m.
ION
RE VIS
TASK
Make sure your maps and scale drawings are always accurate
Only one set of rules to learn here, and they work for both maps and scale drawings, so there
are no excuses for not knowing them. Cover the page and make sure you can jot them down.
Section Four — Ratio, Proportion and Rates of Change
92
Warm-Up and Practice Questions
For conversions, just find the conversion factor, then multiply or divide by it. The tricky part is
often choosing whether to multiply or divide, so try these warm-up questions for practice.
Warm-up Questions
1)
Convert:
a) 3.5 cm to mm
b) 2.8 litres to cm3
c) 4.5 gallons to pints.
2)
Convert:
a) 44 lbs to kg
b) 5 feet to cm
c) 20 km to miles.
3)
Write the following measurements in order of size: 44 mm, 0.5 m, 4.2 cm
4)
Convert 5 hours 10 minutes into minutes.
5)
Convert:
6)
The speed limit along a stretch of road is 60 mph. What is this in km/h?
7)
Mako is drawing a scale drawing of a park, using a scale of 1 cm = 50 m. In real life,
a pond measures 25 m by 75 m. What will it measure on the scale drawing?
a) 2.4 cm3 into mm3
b) 400 000 cm2 into m2.
Practice Questions
Have a look at this worked practice question — then try the questions on the next page for yourself.
Once you’ve done them, you’ll have been converted into a conversions expert.
1
Mr Roe is struggling with the metric system.
a)
He knows his car will hold 12 gallons of petrol. How many litres is this?
You’d expect more litres than
gallons, so multiply. Then check
that your answer seems sensible.
1 gallon ≈ 4.5 litres
Conversion factor = 4.5
12 × 4.5 = 54 litres
54
litres
.................
[2 marks]
b)
In a recent medical he weighed 192 lb. What was his weight in kilograms?
You’d expect fewer kg
than pounds, so divide.
1 kg ≈ 2.2 pounds
Conversion factor = 2.2
192 ÷ 2.2 = 87.272... = 87.3 kg (1 d.p.)
87.3
kg
......................
[2 marks]
c)
The doctor also measured his height and found it to be 1.83 m.
Calculate this height in feet and inches to the nearest inch.
1.83 m = 183 cm
Convert 0.1 feet
to inches.
1 foot ≈ 30 cm
You’d expect fewer feet
Conversion factor = 30
than cm, so divide.
183 ÷ 30 = 6.1 feet
1 foot = 12 inches, so 0.1 × 12 = 1.2 inches
6 feet 1.2 inches = 6 feet 1 inch to the nearest inch.
6 feet ..........
1 inch
..........
[3 marks]
Section Four — Ratio, Proportion and Rates of Change
93
Practice Questions
2
Mr Evy attends his local slimming club. Since starting he has lost a total of 33 lb.
a)
How many kilograms is this?
......................... kg
[1 mark]
b)
He needs to lose another 1 stone 8 lb to achieve his target weight.
What will his total weight loss in kilograms be?
......................... kg
[3 marks]
3 Put the following volumes in order of size, starting with the smallest.
2 pints
500 cm3
0.5 gallons
1 litre
........................., ........................., ........................., .........................
[3 marks]
4
Amira works for a company that organises walking holidays.
She is planning a new hike that goes across a moor.
The scale of the map she is using to plan the walk is 1 : 50 000.
a)
Going along the footpath, the distance between the place where the
walkers will stop for lunch and the highest point on the moor is 6 km.
How far apart will these points be on the path on Amira’s map? Give your answer in cm.
......................... cm
[2 marks]
b)
The distance between the start and finish of the hike on the map is 32 cm.
How long is the hike? Give your answer in kilometres.
......................... km
[2 marks]
Section Four — Ratio, Proportion and Rates of Change
94
Best Buy
These questions are all about working out which product is the best value for money.
Best Buy Questions — Find the Amount per Penny
These are similar to the proportion questions you saw on p79, but here you’re comparing the
‘value for money’ of 2 or 3 similar items. For these, follow this GOLDEN RULE...
Divide by the PRICE in pence (to get the amount per penny)
Ralph’s Red Hot Chilli Relish comes in three different sizes, as shown below.
Which of these represents the best value for money?
Ralph’s
Red Hot
Chilli
Relish
Ralph’s
Red Hot
Chilli
Relish
100 g
200 g
100 g 200 g at
at £2.50 £3.20
Ralph’s
Red Hot
Chilli
Relish
400 g
400 g at
£8
The GOLDEN RULE says:
DIVIDE BY THE PRICE IN PENCE to get the amount per penny
In the 100 g jar you get 100 g ÷ 250p = 0.4 g per penny
In the 200 g jar you get 200 g ÷ 320p = 0.625 g per penny
In the 400 g jar you get 400 g ÷ 800p = 0.5 g per penny
The 200 g jar is the best value for money, because you get more relish per penny.
With any question comparing ‘value for money’, DIVIDE BY THE PRICE (in pence) and
it will always be the BIGGEST ANSWER that is the BEST VALUE FOR MONEY.
...or Find the Price per Unit
For some questions, the numbers mean it’s easier to divide by the amount to get the
cost per unit (e.g. per gram, per litre, etc.). In that case, the best buy is the smallest answer
— the lowest cost per unit. Doing the example above in this way, you’d get:
The relish in the 100 g jar costs 250p ÷ 100 g = 2.5p per gram
The relish in the 200 g jar costs 320p ÷ 200 g = 1.6p per gram
The relish in the 400 g jar costs 800p ÷ 400 g = 2p per gram
The 200 g jar is the best value for money, because it’s the cheapest per gram.
Find the amount per penny or find the price per unit...
All you need to know is that the best buy is the biggest amount per penny or the lowest cost per unit.
Section Four — Ratio, Proportion and Rates of Change
95
Density and Speed
Density and speed are both a matter of learning the formulas and putting in the numbers.
Density = Mass ÷ Volume
Density is the mass per unit volume of a substance. It’s usually measured in kg/m3 or g/cm3.
MASS
DENSITY = VOLUME
VOLUME =
MASS
DENSITY
MASS = DENSITY × VOLUME
A formula triangle is a mighty handy tool for remembering formulas.
Here’s the one for density. To remember the order of the letters in the
formula triangle, just think DMV or DiMoV (the Russian agent).
How DO YOU USE Formula Triangles?
1) COVER UP the thing you want to find
and WRITE DOWN what’s left showing.
2) Now PUT IN THE VALUES for the other
two things and WORK IT OUT.
M
D ×V
E.g. to get the formula for density
from the triangle, cover up D
and you’re left with M
V .
Gold has a density of 19.3 g/cm3. A gold ingot has a
volume of 100 cm3. What is the mass of the ingot in g?
1) First write down the formula:
2) Then put in the values and calculate:
mass = density × volume
mass = 19.3 g/cm3 × 100 cm3
= 1930 g
Check your units — If the density is in g/cm3, the
volume must be in cm3 and you’ll get a mass in g.
Speed = Distance ÷ Time
Speed is the distance travelled per unit time — the number of km per hour or metres per second.
SPEED =
DISTANCE
TIME
TIME =
DISTANCE
SPEED
DISTANCE = SPEED × TIME
Here’s the formula triangle for speed — this time, we have the words
SaD Times to help you remember the order of the letters (S D T).
So if it’s a question on speed, distance and time, just say SAD TIMES.
D
S ×T
A car travels 90 miles at 60 miles per hour. How long does this take?
Write down the formula,
put in the values and calculate:
ION
RE VIS
TASK
distance
90 miles
speed = 60 mph = 1.5 hours
= 1 h 30 mins
time =
Don’t be dense — it’s not that hard really...
The best way to remember these formula triangles is to cover the page and scribble them
down. Then try using your formula triangles to reproduce the six formulas above.
Section Four — Ratio, Proportion and Rates of Change
96
Warm-Up and Practice Questions
This is the last set of questions for this section — whizz through these warm-up questions,
practice questions and revision questions, then you can have a well-earned cup of tea.
Warm-up Questions
1)
Which is the better buy: 400 ml of juice for £1.60, or 1 litre of juice for £3?
2)
Which is the best value for money: 100 g of sweets for £1.50, 250 g of sweets for £2.75
or 400 g of sweets for £5?
3)
A lump of metal with a density of 12 g/cm3 weighs 72 g. What is the volume of the lump?
4)
Shaun runs 5 km in 25 minutes. What is his speed in km/min?
5)
The mass, volume and density of various pieces
of metal are recorded in the table on the right.
Fill in the gaps to complete the table.
Mass
Volume
10 kg
100 cm
Density
3
1 cm3
0.02 kg/cm3
500 g
125 g/cm3
15 cm3
1.2 kg/cm3
Practice Questions
Here’s a lovely worked practice question to guide you through the perilous page ahead. Remember
to take your time and write down each step of your working so you don’t make any silly mistakes.
1
Fred walks 5400 m in 72 minutes.
a)
What is his average speed in km/h?
Distance in km = 5400 ÷ 1000 = 5.4 km
Time in hours = 72 ÷ 60 = 1.2 hours
distance
5.4
Speed = time = 1.2 = 4.5 km/h
Convert the distance and time
into the units you need for the
answer — km and hours.
4.5 km/h
.........................
[2 marks]
b)
How far will he travel in 48 minutes at this speed?
Time in hours = 48 ÷ 60 = 0.8 hours
Distance = speed × time = 4.5 × 0.8 = 3.6 km
3.6 km
.........................
[2 marks]
c)
If he kept at the same speed and had travelled 14 km and 400 m by
the end of the day, how long had he been walking, in hours and minutes?
Don’t forget to convert your
14.4
Time = distance = 4.5 = 3.2 hours
answer
into hours and minutes.
speed
3.2 hours = 3 hours and (0.2 × 60) = 12 minutes
3 hours and 12 minutes
.................................................
[3 marks]
Section Four — Ratio, Proportion and Rates of Change
97
Practice Questions
2
a)
A car travels 45 miles at 60 mph. How long will it take in minutes?
......................... mins
[1 mark]
b)
A plane travels 1800 km in 2 hours 30 minutes. What is its average speed?
......................... km/h
[2 marks]
c)
A train travels for 285 minutes at an average speed of 45 km/h.
How far will it travel to the nearest km?
......................... km
[2 marks]
3
Calculate the density of these materials. State the units in your answers.
a)
A mass of 300 kg of gold alloy with a volume of 0.02 m3.
.........................
[2 marks]
b)
A mass of 945 g of aluminium with a volume of 350 cm3.
.........................
[2 marks]
4
Washing up liquid comes in bottles of three different sizes: 500 ml for £2.50,
700 ml for £4.20 or 1.2 litres for £5.40. Which bottle is the best value for money?
.........................
[2 marks]
5
The density of oil in a tank is 800 kg/m3. The mass of the oil is 20 tonnes.
Calculate the volume of oil in the tank (1 tonne = 1000 kg).
.........................
[3 marks]
Section Four — Ratio, Proportion and Rates of Change
98
Revision Summary for Section Four
It’s time for some revision questions to test how much you’ve learnt in section four.
• Try these questions and tick off each one when you get it right.
• When you’ve done all the questions for a topic and are completely happy with it, tick off the topic.
Ratios and Proportion (p77-80) 

2) Simplify the following ratios: a) 2.4 : 3.6 b) 56 cm : 0.49 m 
3) Write the ratio 5 : 18 in the form 1 : n. 
1) Reduce these ratios to their simplest form: a) 14 : 16
b) 27 : 18
c) 65 : 26
4) Ria’s DVD collection is made up of only cartoons and comedies in the ratio 8 : 11.
She has 22 comedies. How many DVDs does she have in total?
5) Isla and Brett share £150 between them in the ratio 7 : 3. How much does Isla get?
6) 18 sweets cost 90p. How much would 25 sweets cost?
7) It takes 4 boys 30 minutes to wash a car. How long would it take 6 boys?
Percentage Change (p83-85) 
8) A shop increases its prices by 12%. How much does a £6 mug cost after the increase?
9) Kamil invests £150 in an account that pays 1% simple interest each year.
How much will there be in the account after 4 years?
10) The cost of a pint of milk increases from 50p to 56p. Find the percentage change.
11) After a 4% decrease, a bike is worth £86.40. What was its original value?








Unit Conversions and Scale Drawings (p88-91) 
12) From memory, write down all the metric unit conversions and all the imperial unit
conversions from page 88. Now do the same for the metric-imperial conversions.

13) Convert the following: a) 80 km to miles b) 2.5 feet to cm c) 10 kg to pounds 
14) Convert 4 hours 35 minutes into minutes. 
15) Convert the following: a) 6000 mm2 into cm2 b) 7800 cm3 to m3 c) 27 mph to km/h 
16) The distance between two towns is 20 miles.
How far apart would they be on a map with a scale of 1 cm : 5 miles?

17) On a scale drawing, the dimensions of a car park are 5 cm by 2.5 cm.
The scale is 1 cm = 10 m. What are the real-life dimensions of the car park? 
Best Buy (p94) 
18) A shop sells three different sizes of cheese: 300 g for £1.50, 450 g for £2
and 750 g for £3. Which is the best buy? 
Density and Speed (p95) 
19) A piece of cake has a volume of 200 cm3 and weighs 150 g. What is its density?
20) Ramin cycles for 3 hours at a speed of 25 km/h. How far does he cycle?
Section Four — Ratio, Proportion and Rates of Change


Section Five — Geometry and Measures
99
Symmetry
There are two types of symmetry you need to know — line symmetry and rotational symmetry.
Line Symmetry
This is where you draw one or more MIRROR LINES across a shape
and both sides will fold exactly together.
2 LINES OF
SYMMETRY
1 LINE OF
SYMMETRY
1 LINE OF
SYMMETRY
3 LINES OF
SYMMETRY
NO LINES OF
SYMMETRY
1 LINE OF
SYMMETRY
Shade two squares on the pattern on the right
to make a pattern with two lines of symmetry,
and draw on the lines of symmetry.
The extra squares
are shown in blue.
These are the lines of symmetry.
Rotational Symmetry
This is where you can rotate the shape into different positions that look exactly the same.
Order 1
Order 2
Order 2
Order 3
Order 4
The ORDER OF ROTATIONAL SYMMETRY is the posh way of saying:
‘how many different positions look the same’.
• You should say the Z-shape above has ‘rotational symmetry of order 2’.
• When a shape has only 1 position (like the T above) you can either say that it has
‘rotational symmetry of order 1’ or that it has ‘NO rotational symmetry’.
ION
RE VIS
TASK
You can use a mirror to find lines of symmetry...
Try standing a mirror on the shapes at the top of the page — when the mirror is on a line of
symmetry, the image of the shape that you can see should look the same as the original shape.
Section Five — Geometry and Measures
100
Quadrilaterals
There are lots of quadrilaterals for you to learn on this page. You’ll need to be able to draw them,
spell the names correctly and learn all of their properties — so read on carefully.
Quadrilaterals (Four-Sided Shapes)
SQUARE
The little square means
it’s a right angle.
RECTANGLE
4 equal angles of 90° (right angles).
4 lines of symmetry,
rotational symmetry of order 4.
Diagonals cross at right angles.
4 equal angles of 90° (right angles).
2 lines of symmetry,
rotational symmetry of order 2.
RHOMBUS (A square pushed over)
PARALLELOGRAM (A rectangle pushed over)
Matching arrows
show parallel sides.
A rhombus is the
same as a diamond.
4 equal sides (opposite sides are parallel).
2 pairs of equal angles.
2 lines of symmetry,
rotational symmetry of order 2.
Diagonals cross at right angles.
TRAPEZIUM
1 pair of parallel sides.
NO lines of symmetry*.
No rotational symmetry.
2 pairs of equal sides (each pair are parallel).
2 pairs of equal angles.
NO lines of symmetry,
rotational symmetry of order 2.
KITE
2 pairs of equal sides.
1 pair of equal angles.
1 line of symmetry.
No rotational symmetry.
Diagonals cross at right angles.
*except for an isosceles trapezium (a trapezium where the non-parallel
sides are the same length), which has 1 line of symmetry.
Trapeziums either have one or no lines of symmetry...
Learn the names of all the shapes and make sure you know how to spell them (parallelogram is a
tricky one). Then learn the properties of each shape, and have a go at drawing them all.
Section Five — Geometry and Measures
101
Triangles and Regular Polygons
There are some more 2D shapes coming up on this page — let’s start off with all
the different types of triangles, then build up to shapes with lots of sides.
Triangles (Three-Sided Shapes)
EQUILATERAL Triangles
3 equal sides and
3 equal angles of 60°.
60°
3 lines of symmetry,
rotational symmetry of order 3.
ISOSCELES Triangles
2 sides the same.
2 angles the same.
1 line of symmetry.
No rotational symmetry.
SCALENE Triangles
60°
60°
These dashes mean
that the two sides are
the same length.
All three sides different.
All three angles different.
No symmetry (pretty obviously).
RIGHT-ANGLED Triangles
1 right angle (90°).
No lines of symmetry*.
*except for a right-angled
isosceles triangle, which
has 1 line of symmetry.
Regular Polygons
A polygon is a many-sided shape. A regular polygon is one where all the sides and angles are the
same. The regular polygons are a never-ending series of shapes with some fancy features.
EQUILATERAL TRIANGLE
3 sides
3 lines of symmetry
Rotational symmetry of order 3
SQUARE
4 sides
4 lines of symmetry
Rotational symmetry of order 4
REGULAR PENTAGON
5 sides
5 lines of symmetry
Rotational symmetry of order 5
REGULAR HEXAGON
6 sides
6 lines of symmetry
Rotational symmetry of order 6
REGULAR HEPTAGON
7 sides
7 lines of symmetry
Rotational symmetry of order 7
REGULAR OCTAGON
8 sides
8 lines of symmetry
Rotational symmetry of order 8
Regular polygons have equal sides and angles
There’s nothing too tricky on this page. Remember, once you know the number of sides of a regular
polygon, you also know the number of lines of symmetry and its order of rotational symmetry.
Section Five — Geometry and Measures
102
Warm-Up and Practice Questions
Same old, same old — race through these warm-up questions first to check you’ve learnt it all.
There’s no point struggling with the practice questions if there’s stuff you don’t know yet.
Warm-up Questions
1)
How many lines of symmetry are there in a ‘T’ shape?
2)
What is the order of rotational symmetry of an ‘H’ shape?
3)
How many lines of symmetry does a rectangle have?
4)
What is the order of rotational symmetry of a regular pentagon?
5)
I am thinking of a shape with four sides. It has two pairs of equal sides
and its diagonals cross at right angles. It has no rotational symmetry.
What is the name of the shape I’m thinking of?
6)
Which quadrilateral has no lines of symmetry but rotational symmetry of order 2?
7)
What type of triangle has 1 line of symmetry?
8)
A regular polygon has 20 sides. How many lines of symmetry does it have?
What is its order of rotational symmetry?
Practice Questions
Right, now you’re warmed up, it’s time to have a go at some slightly harder questions.
There’s a worked practice question to get you started and ease you into the other questions.
1
a)
Draw the lines of symmetry on this square.
You should know that a square
has 4 lines of symmetry.
That way you won’t forget any.
Remember, a line of symmetry is a
line along which the two halves of
a shape will fold exactly together.
[2 marks]
b)
Add a
to make this shape symmetrical. Find the 3 different ways to do it.
(i)(ii)(iii)
Check what you’ve done by adding the line of symmetry. You should be
able to see whether your new shape will fold exactly together along it.
Section Five — Geometry and Measures
[3 marks]
103
Practice Questions
2
Here is a regular polygon.
a)
What is the name of this shape?
.........................................
[1 mark]
b)
How many lines of symmetry does it have?
.........................................
[1 mark]
c)
What is its order of rotational symmetry?
.........................................
[1 mark]
3
Fill in the missing numbers.
a)
A parallelogram has ................. pairs of equal sides (which are parallel).
It has ................. line(s) of symmetry and rotational symmetry of order ................. .
[3 marks]
b)
An equilateral triangle has ................. equal sides and ................. line(s) of symmetry.
[2 marks]
4
Look at the following shapes.
(i)
(ii)
(iii)
(iv)
a)
Draw all the lines of symmetry on each of the shapes.
b)
What is the order of rotational symmetry for each shape?
(i) ..........
(ii) ..........
(iii) ..........
(iv) ..........
(v)
[3 marks]
(v) ..........
[3 marks]
Section Five — Geometry and Measures
104
Perimeter and Area
Perimeter is the distance all the way around the outside of a 2D shape. It’s pretty straightforward if
you use the big blob method. So pay attention — this could be easy marks.
Perimeter — Distance Around the Edge of a Shape
To find a perimeter, you add up the lengths of all the sides.
But the only reliable way to make sure you get all the sides is this:
1) Put a BIG BLOB at one corner and then go around the shape.
2) Write down the LENGTH of every side as you go along.
(Even sides that seem to have NO LENGTH GIVEN — you must work them out.)
3) Keep going until you get back to the BIG BLOB.
Find the perimeter of the shape drawn on the grid below.
Each grid square represents 1 cm2.
Blob
3 cm
3 cm
2 cm
4 cm
1 cm
Perimeter = 4 + 5 + 1 + 2 + 3 + 3
= 18 cm
5 cm
You Must Learn These Four Area Formulas
Area of RECTANGLE = length × width
Remember that area is measured in
square units (e.g. cm2, m2 or km2).
1
Area of TRIANGLE = 2 × base × vertical height
Height
A=l×w
Width
A= 1
2 ×b×h
Base
Length
Note that the height must always be the
vertical height, not the sloping height.
Area of
PARALLELOGRAM
=
base
×
vertical
height
Area of
distance between
average of parallel
=
×
TRAPEZIUM
them (h)
sides (a and b)
a
Height
A=b×h
Base
ION
RE VIS
TIP
h
A= 1
2 × (a + b) × h
b
Four area formulas — you’ll have to learn them all...
These four area formulas are really important, so make sure you’ve memorised them all
before moving on. (Actually, they might even turn out to be useful on the next page too...)
Section Five — Geometry and Measures
105
Area of Compound Shapes
Make sure you know the area formulas from the previous page — you need them again here.
Areas of More Complicated Shapes
You sometimes have to find the area of strange-looking shapes. What you always find with
these questions is that you can break the shape up into simpler ones that you can deal with.
1) SPLIT THEM UP into the basic shapes:
RECTANGLES, TRIANGLES, etc.
2) Work out the area of each bit SEPARATELY.
3) Then ADD THEM ALL TOGETHER.
Basic
Rectangle
Basic
Triangle
Find the area of the shape below.
6 cm
2 cm
Split the shape into a rectangle and triangle as
shown and work out the area of each shape:
1) Area of rectangle = length × width = 6 × 2 = 12 cm2
2) To find the height of the triangle, subtract the height
of the rectangle from the total height of the shape.
Height of triangle = 7 – 2 = 5 cm
7 cm
Area of triangle = 21 × base × height = 21 × 6 × 5 = 15 cm2
3) So the total area of shape = 12 + 15 = 27 cm2
The shape of a school badge is shown on the right.
a) Find the area of the badge.
You need to work out the area of the badge — so split
it into two shapes (a rectangle and a trapezium):
7 cm
3 cm
7 cm
3 cm
3 cm
Area of the rectangle = l × w
= 7 × 3 = 21 cm2
7 cm
3 cm
6 cm
3 cm
Area of the trapezium = ½(a + b) × h
= ½(7 + 3) × 3 = 15 cm2
So the total area of the badge is 21 + 15 = 36 cm2
b) The material needed to make the badge costs 11p per cm2.
Work out the cost of the material needed for each badge.
Just multiply the area by the cost per cm2:
Cost = 36 × 11 = 396p = £3.96
Remember to add up the separate areas at the end...
As long as you know the area formulas, there’s nothing on this page to trip you up — it’s just a case of
splitting up complicated shapes into basic shapes and working out the area of each bit separately.
Section Five — Geometry and Measures
106
Circles
There’s a surprising number of circle terms you need to know — don’t mix them up.
Take your time with this page and learn the definitions of all the terms.
Radius and Diameter
The DIAMETER goes right across the circle, passing through the centre.
The RADIUS goes from the centre of the circle to any point on the edge.
radius
diameter
The DIAMETER IS EXACTLY DOUBLE THE RADIUS
So if the radius is 4 cm, the diameter is 8 cm, and if the diameter is 24 m, the radius is 12 m.
Area, Circumference and p
There are two more important formulas for you to learn — circumference and area of a circle.
The circumference is the distance round the outside of the circle (its perimeter).
1) CIRCUMFERENCE = p × diameter
= p × radius × 2
2) AREA = p × (radius)2
A = p × r2
C = p × D or C = 2 × p × r
p = 3.141592.... = 3.142 (approximately)
The big thing to remember is that p (called “pi”) is just a number
(3.14159...) which is often rounded off to 3.142. You can just use the
p button on your calculator (which is way more accurate).
So a circle with radius 6 cm has a circumference of 2 × p × r = 2 × p × 6 = 37.7 cm (1 d.p.)
and an area of p × r2 = p × 62 = 113.1 cm2 (1 d.p.).
Tangents, Chords, Arcs, Sectors and Segments
Tangent
A TANGENT is a straight line that just touches the outside of a circle.
A CHORD is a line drawn across the inside of a circle.
AN ARC is just part of the circumference of a circle.
Chord
Arc
Arc
A SECTOR is a wedge-shaped area cut right from the centre.
SEGMENTS are the areas you get when you cut a circle with a chord.
Sector
Segment
ION
RE VIS
TIP
Chord
Don’t be scared of using p — it’s just a number...
Learn the names for the parts of a circle, and the formulas for circumference and area.
Don’t get mixed up between sectors and segments — remember a sector is like a slice of cake.
Section Five — Geometry and Measures
107
Warm-Up and Practice Questions
I’m not going to try and pretend that doing warm-up questions is a fun way to pass the time.
They’re just the best way to check that you really do know your stuff.
Warm-up Questions
1)
Find the area of a parallelogram with a base of 8 cm and a height of 5 cm.
2)
A triangle is 5 cm tall and has a base of length 12 cm. Find its area.
3)
A rectangle has an area of 20 cm2 and a length of 5 cm. Find its width.
4)
A square has a perimeter of 24 cm. Find its area.
5)
Find the area of the shape on the right.
6)
If the diameter of a circle is 6 cm, what is its radius?
7)
Find the circumference and area of a circle
with a diameter of 20 cm.
8)
What name is given to a straight line which
touches the circumference of a circle at one point?
6 mm
15 mm
12 mm
Practice Questions
OK, so practice questions might not be your idea of a good time either — but they will help you brush
up your maths skills. So here are some to have a go at. And, as always, a worked one to start you off.
1
The rectangular sheet of metal below has two circles cut out.
25 cm
(not to scale)
8 cm
R
R = 3.0 cm
r
r = 1.5 cm
Find the area of the remaining metal. Give your answer to 1 decimal place.
Area of rectangle = length × width = 25 × 8 = 200 cm2
Area of big circle = pR2 = p × 32 = 28.2743... cm2
Area of small circle = pr2 = p × 1.52 = 7.0685... cm2
So area of the metal is: 200 – 28.2743... – 7.0685... = 164.6570...
= 164.7 cm2 (1 d.p.)
164.7 cm (1 d.p.)
.........................................
2
[3 marks]
Section Five — Geometry and Measures
108
Practice Questions
2
Label the parts of the circle indicated.
a)
b)
o
c)
3
[3 marks]
These shapes each have an area of 12 cm2. Find the missing lengths.
r
q
p
3 cm
8 cm
4 cm
6 cm
p = ............................ ,
4
q = ............................ ,
r = ............................
[3 marks]
The following shape consists of a polygon and a semicircle.
6 cm
10 cm
(not to scale)
6 cm
14 cm
a)
Find the perimeter of the shape. Give your answer to 2 d.p.
......................................... cm
[3 marks]
b)
Find the area of the shape. Give your answer to 2 d.p.
......................................... cm2
[4 marks]
Section Five — Geometry and Measures
109
3D Shapes
Here are some 3D shapes. You’ll definitely have come across some of these before (like cubes and
spheres). Others might be less familiar like regular tetrahedron — you need to know all of them.
Eight Solids to Learn
3D shapes are solid shapes. These are the ones you need to know:
CYLINDER
There’s more
about prisms
on p112.
TRIANGULAR
PRISM
REGULAR
TETRAHEDRON
SQUARE-BASED
PYRAMID
(triangle-based
pyramid)
SPHERE
CUBE
CUBOID
CONE
Different Parts of Solids
Vertex
There are different parts of 3D shapes you need to be able to
spot. These are vertices (corners), faces (the flat bits) and edges.
You might be asked to find the number of vertices, faces and
edges — just count them up, and don’t forget the hidden ones.
Face
Edge
For the cuboid on the right, write down
the number of faces, the number of
edges and the number of vertices.
A cuboid has 6 faces (there’s one on the bottom
and two at the back that you can’t see).
It has 12 edges (again, there are some hidden ones
— the dotted lines in the diagram).
It has 8 vertices (one is hidden).
It’s one vertex but two vertices...
If you’re asked to count the number of faces, edges and vertices of a 3D shape it can be helpful to do
a quick sketch of the shape — you’ll be less likely to miss out a vertex or count the same edge twice.
Section Five — Geometry and Measures
110
Nets and Surface Area
Pencils and rulers at the ready — you might get to do some drawing over the next two pages.
It’s mainly limited to squares, rectangles and triangles, but you might get the odd circle to draw.
Nets and Surface Area
1)
2)
3)
4)
A NET is just a hollow 3D shape folded out flat.
There’s often more than one net that can be drawn for a 3D shape (see the cubes below).
SURFACE AREA only applies to 3D objects — it’s the total area of all the faces added together.
There are two ways to find the surface area:
1) Work out the area of each face and add them all together (don’t forget the hidden faces).
2) Sketch the net, then find the area of the net (this is the method we’ll use on these pages).
Remember — SURFACE AREA OF SOLID = AREA OF NET.
Cubes
These are just some of the nets of a cube
— there are lots more.
Nets of cubes
Cube
A cube has 6 square faces, so its surface area is just 6 × (area of square face).
Cuboids
Find the surface area of this cuboid:
7 cm
2 cm
4 cm
4 cm
Cuboid
4 cm 2 cm
1) Sketch the net of the
shape, and label all
the measurements:
2 cm
7 cm
2 cm
Net of cuboid
2) Then work out the area of each face and add them
up — there are 2 each of 3 different rectangles.
The net of a cuboid is made up of
3 different sized rectangles — there
are 2 of each size.
Surface area = 2(2 × 7) + 2(4 × 7) + 2(4 × 2)
= 28 + 56 + 16 = 100 cm2
To see if you drew the net right, imagine folding it back up...
Even if you’re not asked to draw the net of a shape, it’s often a good idea to do a quick sketch and
label it — it’s really helpful for finding the surface area. Just make sure you get the labels right.
Section Five — Geometry and Measures
111
Nets and Surface Area
Here are some more nets — these ones are a bit tougher because they contain triangles and circles.
Triangular Prisms and Pyramids
Net of triangular prism
Square-Based
Pyramid
Triangular Prism
Net of square-based
pyramid
Find the surface area of the square-based pyramid below.
You can see from the net above that a square-based
pyramid has 1 square face and 4 triangular faces.
8 cm
5 cm
Area of square face = 5 × 5 = 25 cm2
Area of triangular face = ½ × 5 × 8 = 20 cm2
Total surface area = 25 + (4 × 20) = 25 + 80 = 105 cm2
Have a look
back at page
104 for more
on areas.
Cylinders
Cylinders are a bit trickier — it’s probably best to just learn the formula and stick the numbers in.
r
2
2 r
h
Cylinder
h
Net of cylinder
r
Surface area of a CYLINDER = 2prh + 2pr2
Note that the length of the
rectangle is equal to the
circumference of the circular ends.
2
Find the surface area of the cylinder on the right to 1 d.p.
Just put the measurements into the formula:
Surface area of cylinder = 2prh + 2pr2
= (2 × p × 2 × 6) + (2 × p × 22)
= 75.398... + 25.132...
= 100.530... = 100.5 cm2 (1 d.p.)
2 cm
6 cm
The surface area of a solid is equal to the area of its net
You have to be a bit careful when finding the surface area of a triangular prism — the rectangular sides
could be all the same size if the triangle at the end is equilateral, two different sizes if the triangle is
isosceles, or all different sizes if the triangle is scalene. Check before you start and don’t get caught out.
Section Five — Geometry and Measures
112
Volume
Now it’s time to work out the volumes of 3D shapes which means more formulas for you to learn.
Volumes of Cuboids
A cuboid is a rectangular block. Finding its volume is dead easy:
Height
Length
Volume of Cuboid = length × width × height
V=L×W×H
Width
Volumes of Prisms
A PRISM is a solid (3D) object which
is the same shape all the way through
— i.e. it has a CONSTANT AREA
OF CROSS-SECTION.
Constant area
of cross-section
Length
Volume
cross‑section
length of
=
×
of a prism
area
prism
A prism has a cross-section
area of 6 cm2 and a length
of 15 cm. Find its volume.
V = A × L = 6 cm2 × 15 cm = 90 cm3
V=A×L
Cylinder (Circular Prism)
Volume of cylinder = area of circle × height
V = pr2h
h
Constant Area
of Cross-section
r
Rufus has a cylindrical jar of jam with radius 3 cm and
height 10 cm. What is the volume of the jar to 1 d.p.?
Just put the measurements into the formula above:
V = pr2h = p × 32 × 10 = 282.743... cm3
= 282.7 cm3 (1 d.p.)
3 cm
10 cm
Jam
Prisms have the same cross-section all the way through...
A few more formulas to learn on this page, but they’re not too bad really. The prism formula works on
any prism, no matter what the shape of the cross-section — even if it’s something really weird.
Section Five — Geometry and Measures
113
Warm-Up and Practice Questions
You know what’s coming by now — some lovely warm-up questions to test how much you’ve learnt
from this section. Have a go at these little beauties and see how you do.
Warm-up Questions
1)
How many edges does a cube have?
2)
How many vertices does a cuboid have?
3)
Look at the shape that is shown on the right.
a) Write down its name.
b) How many faces, edges
and vertices does this shape have?
4)
Find the surface area of a cube with a side length of 4 cm.
5)
Find the surface area of a cylinder that has a radius of 5 cm and a height of 12 cm.
Give your answer to 1 decimal place.
6)
The cross-sectional area of an octagonal prism is 28 cm2.
The length of the prism is 6 cm.
Find the volume of the prism.
Practice Questions
I’m sure you managed the warm-up questions with no problems at all. Have a read through this
worked practice question, then move on and have a go at the others to really test your skills.
1
The crate below is used to transport car parts all over the world.
It is a cuboid.
0.5 m
0.8 m
2.4 m
Find the volume of the crate. Include the correct units in your answer.
Use the formula for the volume of a cuboid:
V = length × width × height
Don’t forget the units — the
= 2.4 × 0.8 × 0.5
measurements are given in m,
= 0.96 m3
so the volume will be in m3.
0.96 m
....................................
3
[2 marks]
Section Five — Geometry and Measures
114
Practice Questions
2
3
The diagram below shows a triangular prism. State the number of:
a)
vertices
........................
b)
faces
........................
c)
edges
........................
[1 mark]
[1 mark]
[1 mark]
Look at the cuboid and net below.
3 cm
8 cm
5 cm
Use the net to find the total surface area.
.................................... cm2
[3 marks]
4
The cross-section of a tin of baked beans is a circle with a diameter of 7.4 cm.
a)
Find the area of the cross-section. Give your answer to 2 d.p.
.................................... cm2
[2 marks]
b)
The tin is 11 cm high. What is its volume? Give your answer to 2 d.p.
.................................... cm3
[2 marks]
Section Five — Geometry and Measures
115
Angle Basics
A quick basic introduction to angles for you here. You should be able to give the four different types of
angles, their definitions and know how to measure them. It’s all covered below.
Angle Names
Some angles have special names which you need to know.
ACUTE angles
RIGHT angles
Sharp pointy ones
(less than 90º)
Square corners
(exactly 90º)
OBTUSE angles
REFLEX angles
Flatter ones
(between 90º and 180º)
Ones that bend
back on themselves
(more than 180º)
Three-Letter Angle Notation
B
The best way to say which angle you’re talking
about in a diagram is by using THREE letters.
For example in the diagram, angle BAC = 35º.
NOT TO SCALE
25°
20°
35°
A
The middle letter is where the angle is.
The other two letters tell you which
two lines enclose the angle.
30°
D
C
angle ACD = 20º
You might see angles written
in other ways as well —
t C are both
+ABC and AB
the same as angle ABC.
Drawing Angles with a Protractor
Draw a straight horizontal line to be your base line. Put the protractor on the line
so that the middle of the protractor is on one end of the line as shown:
100 1
10
70
12
0 90 80 7
0
1
60
0
0
10
13
1
60
0
0
50
2
1
50
0
3
1
10
20
180 170
160
0
180
170
160
0
10
0
15
20
30
40
0
14
30
40
15
0
14
0
80
Draw a little line or dot next to the angle you’re
drawing (count up in tens from 0° to make sure you
follow the right scale). Here, I’m drawing an angle
of 55º, so I’m using the outside scale.
Be careful — reading from the wrong
scale is a very common error!
Made in England
Then join your base line to the mark you’ve just made with a straight line.
You must join the end of the base line that was in the middle of the protractor.
55º
Angles will always be between 0° and 360°
Learn all of the different angle names and make sure you know how to use a protractor.
Section Five — Geometry and Measures
116
Geometry Rules
The next few pages are full of rules about angles. Here are a few easier ones to get you started.
6 Simple Rules — that’s all
1) Angles in a triangle add up to 180º.
2) Angles on a straight line add up to 180º.
b
a + b + c = 180°
b
a
3) Angles in a quadrilateral add up to 360º.
a
d
b
c
c
a
c
a + b + c = 180°
4) Angles round a point add up to 360º.
Remember that a quadrilateral
is any 4-sided shape.
b
c
d
a
a + b + c + d = 360°
a + b + c + d = 360°
5) Exterior angle of a triangle = sum of opposite interior angles.
Opposite
interior
angles
a+b=d
b
a
c
d
Exterior angle
6) Isosceles triangles have 2 sides the same and 2 angles the same.
In an isosceles triangle, you only need to know one angle to be able to find the other two.
These dashes indicate
two sides the same
length.
These angles
are the same.
Find the size of angle x.
x
75°
The two angles at the bottom
are the same (they’re both 75°), so
75° + 75° + x = 180°
x = 180° – 150°
x = 30°
Angles in a triangle
add up to 180°
No excuses for not learning these six simple rules...
None of the rules here are particularly difficult, but make sure you don’t get them mixed up.
Section Five — Geometry and Measures
117
Parallel Lines
Parallel lines are always exactly the same distance apart (and never meet).
Perpendicular lines are at right angles to each other (they meet at 90°).
Angles Around Parallel Lines
Perpendicular lines
These arrows show that the
lines are parallel.
When a line crosses two parallel lines,
it forms special sets of angles.
1) The two bunches of angles formed at the
points of intersection are the same.
b
a
2) There are only actually two different angles involved
a
b
(labelled a and b here), and they add up to 180º
(from rule 2 on the previous page).
3) Vertically opposite angles (ones opposite each other) are equal
(in the diagram, a and a are vertically opposite, as are b and b).
b
a
a
b
a + b = 180º
Vertically opposite angles
Alternate, Allied and Corresponding Angles
The diagram above has some characteristic shapes to look out for — and each shape contains a
specific pair of angles. The angle pairs are known as alternate, allied and corresponding angles.
You need to spot the characteristic Z, C, U and F shapes:
ALTERNATE angles
ALLIED angles
b
a
a + b = 180º
a
a
a
Alternate angles are the same.
They are found in a Z-shape.
Allied angles add up to 180º.
They are found in a C- or U-shape.
CORRESPONDING angles
Find the size of angle x.
a
115º
a
Corresponding angles are the same.
They are found in an F-shape.
EXAM
TIP
b
x
This diagram shows
corresponding angles (spot
the characteristic F-shape).
Corresponding angles are
the same, so x = 115°
Look out for the characteristic Z, C, U and F shapes...
You definitely need to know the ‘proper’ names for these types of angles — you need to use
the proper terms if you’re ever asked to explain how you worked out the size of an angle.
Section Five — Geometry and Measures
118
Geometry Problems
By now, your head’s probably full of angles, rules and parallel lines. Now it’s time to
put all that to use, and use your knowledge to solve some geometry problems.
Try Out All the Rules One by One
1) Don’t concentrate too much on the angle you have been asked to find.
The best method is to find ALL the angles in whatever order they become obvious.
2) Don’t sit there waiting for inspiration to hit you. It’s all too easy to find yourself staring
at a geometry problem and getting nowhere. The method is this:
GO THROUGH ALL THE RULES OF GEOMETRY (including PARALLEL LINES), ONE BY ONE,
and apply each of them in turn in as many ways as possible — one of them is bound to work.
Find the size of angle x.
Angles on a straight line add up to 180º.
So the missing angle in the quadrilateral is
180° – 102º = 78°
x
124°
Angles in a quadrilateral add up to 360º, so
x + 78º + 124º + 84º = 360º
x = 360º – 78º – 124º – 84º = 74º
102°
84°
Find the size of angle x. Give a reason for each step of your working.
B
A
80°
C
x
35°
F
B
A
80°
x
D
E
C
35°
CBD and BDE are alternate angles.
So CBD = 35°
35°
F
1) Notice that AC and FD are parallel.
D
E
Z-shape
2) Now you know two angles
on straight line AC.
Angles on a straight line add up to 180°.
So x = 180° – 80° – 35°
= 65°
Give a reason for each step of your working...
If you’re really stuck, just fill in any angles you can find and see where it gets you.
Watch out for things like hidden parallel lines and isosceles triangles — they can be really helpful.
Section Five — Geometry and Measures
119
Interior and Exterior Angles
You already know about the interior angles of two polygons — triangles and quadrilaterals.
Well, here are some formulas to find interior and exterior angles for all polygons.
Exterior and Interior Angles
You need to know what exterior and interior angles are and how to find them.
For REGULAR POLYGONS only:
For ANY POLYGON (regular or irregular):
Interior angles
Exterior angles
Exterior angle
Interior
angle
SUM OF EXTERIOR ANGLES = 360°
EXTERIOR ANGLE = 360
n
INTERIOR ANGLE = 180° – EXTERIOR ANGLE
(n is the number of sides)
%
Find the exterior and interior angles of a regular hexagon.
360c 360c
Hexagons have 6 sides:
exterior angle = n = 6 = 60°
Use the exterior angle to interior angle = 180° – exterior angle
find the interior angle:
= 180° – 60° = 120°
The Tricky One — Sum of Interior Angles
This formula for the sum of the interior angles works for ALL polygons, even irregular ones.
SUM OF INTERIOR ANGLES = (n – 2) × 180°
(n is the number of sides)
Find the value of x in the diagram on the right.
1) First, find the sum of the interior angles of the 7-sided shape:
Sum of interior angles = (n – 2) × 180º
= (7 – 2) × 180º = 900°
2) Now write an equation and solve it to find x:
x
140°
170°
120°
100°
95°
115°
x + 170° + 95° + 115° + 100° + 120° + 140° = 900°
x = 900° – 170° – 95° – 115° – 100° – 120° – 140° = 160°
ION
RE VIS
TIP
Exterior angles of any polygon always add up to 360°
A common mistake is to think exterior angle = 360° – interior angle. Remember that when
you extend the sides of the polygon, the interior and exterior angles lie on a straight line.
Section Five — Geometry and Measures
120
Warm-Up and Practice Questions
This geometry lark isn’t easy — there are lots of rules to learn, so don’t panic if you’ve forgotten
some. Just go back over this section, then try these warm-up questions to make sure it’s sunk in.
Warm-up Questions
1)
Write down the names of the following angles:
a) 98°b) 234°c) 11°.
2)
Two of the angles in a triangle are 40° and 50°.
What is the size of the third angle?
3)
Find the size of angle x in the diagram on the right.
4)
What type of angles do you find in an F-shape?
5)
Find the size of angle y in the diagram on the right.
6)
Find the sum of the interior angles of a hexagon.
7)
Four interior angles of a pentagon are 90°, 100°,
110° and 140°. Find the size of the fifth angle.
8)
Work out the size of an exterior angle and an interior angle in a regular octagon.
95°
60°
x
y
102º
Practice Questions
If you whizzed through that lot with no trouble, you’re doing well. Best prepare yourself though,
because here come some more questions. But first, a worked example to help you out. (I’m too kind...)
1
Look at the diagram below.
B
E
24°
(not to scale)
80°
Find:
a)
D
C
A
angle ACB
Angles in a triangle add up to 180°,
so ACB = 180° – 90° – 24° = 66°
66°
........................
[1 mark]
b)
angle DCE
Angles on a straight line add up to 180°,
so DCE = 180° – 66° – 80° = 34°
You have to use your
answer from part a) here.
34°
........................
[1 mark]
c)
angle DEC
Isosceles triangles have 2 angles the same, so EDC = DCE = 34°.
So DEC = 180° – 34° – 34° = 112°
112°
........................
[1 mark]
Section Five — Geometry and Measures
121
Practice Questions
2
Find angles c, d and e in the diagram below.
(not to scale)
c
40°
e
d
c = ............................ ,
3
d = ............................ ,
In the diagram, which is not drawn to scale,
the straight lines ABG and DEF are parallel.
The straight lines BEC and GF are also parallel.
e = ............................
[3 marks]
C
52°
D
E
Find the following angles:
110°
A
a)
F
B
G
AGF
........................
[1 mark]
b)
DEB
........................
[1 mark]
c)
ACB
........................
[2 marks]
4
The diagram shows part of a regular polygon.
150
(not to scale)
How many sides does it have?
....................................
[2 marks]
Section Five — Geometry and Measures
122
Transformations
A transformation just means ‘a way of moving a shape about, or changing its size, on a coordinate grid’.
There are 4 transformations you need to know — translation, reflection, rotation and enlargement.
1) Translations
A translation is just a SLIDE around the page. When describing a translation,
you must say how far along and how far up the shape moves using a vector.
Vectors describing translations look like this.
x is the number of spaces right, y is the number of spaces up.
x
e o
y
If the shape moves left x will be negative, and if it moves down y will be negative.
Describe the transformation that maps:
a) triangle A onto triangle B.
To get from triangle A to triangle B you need
to move 5 units left and 6 units down, so it’s:
A translation by the vector e
-5
-6
o
b) triangle B onto triangle C.
It’s a movement of 7 units right and
1 unit down, so it’s:
7
A translation by the vector e o
-1
2) Reflections
A'
D
C'
B'
6 Y
y
5
44
33
22
1
6 yY
A'
55
44
33
22
1
C'
B'
x
0
-6 -5
-5 -4
-1 00 11 2 3 4
-3 -2
-2 -1
4 5 6
-4 -3
-1
-2
-2
-3
-3
-4
-4
-5
-5
C
B
-6
A'
A
B
C
To describe a reflection, you must give the equation of the mirror line.
A'
x=1
See p58 for more on straight line equations.
Triangle D is mapped onto triangle E by a reflection in the line x = 1.
Notice, the matching corners are equal distances from the mirror line.
E
0
x
-5 -4
-3 -2
-6 -5
-2 -1 0 0 1 22 33 44 55 66
-4 -3
-6
-1-1
A
-2
-2
-3
-3
-4
-4
-5
-5
-6
-6
Describe the transformation that maps:
a) Shape F onto shape G.
C A reflection
B
in the x-axis
b) Shape G onto shape H.
A reflection in the line y = x
6 y
5
4
y=x
3
2
1
x
0
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
-1
-2
-3
-4
-5
-6
G
F
H
The only thing a translation changes is the shape’s position...
To describe translations give a vector and to describe reflections give the equation of the mirror line.
Section Five — Geometry and Measures
123
Transformations
3) Rotations
To describe a rotation, you need 3 details:
1) The angle of rotation (usually 90º or 180º).
2) The direction of rotation (clockwise or anticlockwise).
3) The centre of rotation
6 y
A
5
3
centre of
rotation
2
1
0
-6 -5 -4 -3 -2 -1
Shape A is mapped onto Shape B by a
rotation of 90° clockwise about point (–2, 1).
B
4
Shape A is mapped onto Shape C by a
x
0 1 2
3
4
5
-1
For a rotation of 180º, it doesn’t matter
whether you go clockwise or anticlockwise.
-2
-3
C
-4
6 rotation of 180° about point (–2, 1).
-5
-6
Rotate Triangle D 90° clockwise about (0, 0).
The best way to tackle this is with tracing paper:
1) Trace the shape and mark the centre of rotation at (0, 0).
2) Put your pencil point on the centre of rotation
and rotate the tracing paper 90° clockwise.
You’ll know when you’ve gone far enough —
the horizontal side will be vertical, and vice versa.
3) Mark the corners of the shape in their new positions
on the grid, then draw the shape.
6 y
5
4
D
3
2
1
0
-4 -3 -2 -1
0 1 2
3
4
5
x
6
-1
-2
Hold the-3 tracing paper down
with your
-4 pencil point here.
-5
-6
6 y
5
Describe the transformation that maps
Triangle E onto Triangle F.
A rotation of 180º about (–1, 0).
4
3
2
1
0
-3 -2 -1
-6 -5 -4 -3
E
F
x
0 1 2
-1
-2
-3
-4
3
4
5
6
You can use tracing paper to help you find the centre of rotation.
Trace the original shape and then try putting your pencil on
different points until the traced shape rotates onto the image.
When this happens your pencil must be on the centre of rotation.
-5
-6
Use tracing paper to help you see transformations...
ION
RE VIS
TASK
Use some tracing paper to help you rotate shape G on the previous page 90° clockwise about
(0, 0). Draw over shape G, hold the paper in place with your pencil at (0, 0), and turn the
paper 90° clockwise. Then mark where the corners of the new shape are, and join them up.
Section Five — Geometry and Measures
124
Enlargements
You’ve made it to the fourth and final transformation now. Get ready for... enlargements.
4) Enlargements
The scale factor for an enlargement tells you how long the sides of the new shape are compared
to the old shape. E.g. a scale factor of 3 means you multiply each side length by 3.
Enlarge shape X by a scale factor of 2.
2
1) Make each side twice as long as the
matching side on shape X. Start with
the horizontal and vertical sides.
2) Take care with the sloping sides
— they’re much trickier.
X
4
4
6
Describing an Enlargement
1) The scale factor.
2) The centre of enlargement.
For an enlargement, you must specify:
There’s a formula for the scale factor:
scale factor =
new length
old length
Describe the transformation that maps
Triangle A onto Triangle B.
6 y
5
1) Use the formula to find the scale factor.
(Just do this for one pair of sides.)
4
Old length of triangle base = 2 units
New length of triangle base = 4 units
2
1
new length
4
scale factor = old length = 2 = 2
A
-2 -1 0
-1
1 2
2) To find the centre of enlargement, draw
lines that go through matching corners
of both shapes and see where they cross.
So the transformation is an enlargement
of scale factor 2, centre (0, 6).
3
3
4
5
x
6
-2
-3
-4
B
-5
-6
If the shape only changes in size then it’s an enlargement...
Here, you need to give the scale factor and the centre of enlargement to describe the transformation.
Section Five — Geometry and Measures
125
Congruent Shapes
Congruence is another word which sounds really complicated when it’s not.
If two shapes are congruent, they are the same size and have the same shape.
They can however be reflected or rotated — so if a shape is transformed by
a translation, reflection or rotation, the transformed shape will always be
congruent to the original shape.
CONGRUENT
A
B
same size, same shape
Showing Triangles are Congruent
To prove that two triangles are congruent, you have to show that one of the conditions below is true:
1)
2)
3)
4)
SSS
AAS
SAS
RHS
three sides are the same
two angles and a corresponding side match up
two sides and the angle between them match up
a right angle, the hypotenuse and one other side all match up
SSS
AAS
5 cm
16 mm
81°
62°
62°
6 cm
4.5 cm
4.5 cm
6 cm
Make sure the sides match up — here,
the side is opposite the 81º angle.
RHS
2 cm
1.3 cm
81°
16 mm
5 cm
SAS
The hypotenuse is the
longest side of a rightangled triangle — the one
opposite the right angle.
4 cm
5 cm
56°
56°
5 cm
1.3 cm
2 cm
4 cm
Work Out All the Sides and Angles You Can Find
The best approach to proving two triangles are congruent is to write down everything
you can find out, then see which condition they fit.
Show that the triangles ABD and CBD below are congruent.
Write down what you know:
• Both triangles have a right angle (ADB and CDB)
as angles on a straight line add up to 180º.
• AB and CB are the same length (as it’s an isosceles triangle)
— these sides are the hypotenuse of each triangle.
• BD is a side in both triangles, so it’s the same length.
B
A
D
C
The condition RHS holds, so ABD and CBD are congruent triangles.
Congruent shapes are the same size and the same shape...
... but they could be rotated, reflected or translated so that they look different at a glance.
Section Five — Geometry and Measures
126
Similar Shapes
Similar shapes are exactly the same shape but different sizes (they can also be rotated or reflected).
Similar Shapes Have the Same Angles
Two shapes are similar if:
1) All the angles match up.
2) The sides are all enlarged by the same scale factor.
Similar triangles:
2 cm
3 cm
105°
46°
4 cm
29°
4 cm
105°
46°
Each side in the larger triangle is
twice as long as in the smaller triangle.
6 cm
A''
29°
8 cm
Shape B is similar to shape A. Find the
scale factor of enlargement from A to B.
This question is tricky because the shapes
aren’t the same way up. Make sure you’re
comparing matching sides.
Longest side of B = 4 units
Longest side of A = 2 units
A
A'
2
B
4
new length
4
scale factor = old length = 2 = 2
Use the Scale Factor to Find Missing Sides
Exam questions often ask you to find the length of a missing side or
the size of a missing angle in a pair of similar shapes.
Shapes ABCDE and FGHJK are similar. The scale factor of enlargement
from ABCDE to FGHJK is 4.
B
a) Write down the value of x.
The shapes are similar so the angles match up. The angle at C
corresponds with the angle at H, so they both must be 106°.
x = 106°
A
C
85°
x°
D
E
16 cm
G
F
b) Calculate the length of side AB.
The scale factor is 4 so each side in FGHJK must be
4 times as big as the corresponding side in ABCDE.
FG corresponds with AB.
Length of AB = Length of FG ÷ 4
= 16 cm ÷ 4 = 4 cm
106°
K
Not to scale
H
J
Similar shapes are enlargements of each other...
Don’t get ‘similar’ and ‘congruent’ muddled up. Congruent shapes are always exactly the same size,
but similar shapes aren’t. To help you remember it, think ‘similar siblings, congruent clones’.
Section Five — Geometry and Measures
127
Warm-Up and Practice Questions
It’s very easy to get confused with all the different transformations. To get it clear
in your head, here are a few warm-up questions for you to have a go at.
Warm-up Questions
1)
2)
The corners of a triangle have the coordinates (2, 4), (3, 4) and (3, 6).
3
o?
What are the new of the coordinates of the corners after a translation of e
-1
What detail must you give to describe a reflection?
3)
Triangle F has corners (1, 0), (3, 0) and (1, 5) and triangle G has coordinates
(–3, 2), (–3, 4) and (–8, 2). Draw triangles F and G on a graph and describe
the rotation that maps F onto G.
4)
Plot the triangles X (1, 2), (–1, 3), (–1, 5) and Y (7, –4), (1, –1), (1, 5) on a grid and describe
the transformation that maps triangle X onto triangle Y.
5)
What word describes shapes that are the same size and the same shape?
6)
The length of one side of a shape is 4 cm. When the shape is enlarged,
the same side measures 10 cm. What is the scale factor of the enlargement?
Practice Questions
The best way to get to grips with transformations is to practise doing them. Read through this
worked example, make sure you understand it, then have a go at the rest of the practice questions.
1
Quadrilateral B is an enlargement of quadrilateral A.
y
6
B
A
0
a)
Find the scale factor of the enlargement.
new length
Scale factor = old length = 23
6
x
Pick any side from shape A and the
corresponding side from shape B.
3
2
............................
[2 marks]
b)
Find the coordinates of the centre of the enlargement.
Draw lines through matching corners of both
shapes (see diagram) — the lines cross at (4, 2).
(4, 2)
.............................
[2 marks]
Section Five — Geometry and Measures
128
Practice Questions
2
Here is a coordinate grid.
a)
y
C
4
-5
Translate triangle ABC using vector e o .
-1
Label the image A1B1C1.
3
[1 mark]
2
B
A
b)
1
-4
-3
-2
-1
0
2
1
3
4
Translate triangle A1B1C1 onto A2B2C2
using vector e
5 x
-1
6
o.
-2
[1 mark]
-2
c)
Give the translation vector which translates the image of ABC onto A2B2C2.
.............................
[1 mark]
3
Look carefully at the five trapeziums drawn on this coordinate grid.
a)
y
State the mirror lines for the following reflections:
3
A
(i) A to B .............................
2
E1
-3
-2
-1
0
[1 mark]
C
2
1
(ii) A to C .............................
3 x
[1 mark]
-1
(iii) C to E .............................
-2
D
B
[1 mark]
-3
b) Describe completely the transformation which will map C onto B. (It is not a reflection.)
...................................................................................
[2 marks]
4
Look at these four shapes and
complete the sentences:
a)
D
Shapes ….. and ….. are congruent.
[1 mark]
b)
Shapes ….. , ….. and ….. are similar.
[1 mark]
Section Five — Geometry and Measures
A
B
C
129
Triangle Construction
How you construct a triangle depends on what information you’re given about the triangle...
Three Sides — Use a Ruler and Compasses
Construct the triangle ABC where AB = 5 cm, BC = 3 cm, AC = 4 cm.
C
1) First, sketch and label a triangle so you know roughly what's
needed. It doesn’t matter which line you make the base line.
4 cm
A
3 cm
5 cm
B
C
3 cm
4 cm
A
B
5 cm
2) Draw the base line.
Label the ends A and B.
A
A
B
5 cm
3 cm
4 cm
B
5 cm
4) Where the arcs cross is
point C. Now you can
finish your triangle.
3) For AC, set the compasses
to 4 cm, put the point at
A and draw an arc.
For BC, set the compasses
to 3 cm, put the point at
B and draw an arc.
Sides and Angles — use a Ruler and Protractor
Construct triangle DEF. DE = 6 cm, DF = 4 cm, and angle EDF = 40º.
1) Roughly sketch and label the triangle.
F
4 cm
40°
D
6 cm
E
F
4
D
6 cm
E
2) Draw the base line.
D
6 cm
E
3) Draw angle EDF —
place the centre of
the protractor over D,
measure 40º and
put a dot.
D
cm
6 cm
E
4) Measure 4 cm towards the dot
and label it F. Join up D and F.
Now you’ve drawn the two
sides and the angle. Just join up
F and E to complete the triangle.
Triangles with the same side lengths are congruent...
... so it doesn’t matter which line you draw first. You will always end up with the same triangle.
Section Five — Geometry and Measures
130
Constructions
Here are some trickier constructions that you’ll need to learn how to do.
The Perpendicular Bisector of a Line
Step 1
The perpendicular bisector of line segment AB is a line at
right angles to AB, passing through the midpoint of AB.
This is the method to use if you’re asked to draw it.
A
The Bisector of an Angle
Step 2
Step 1
1) Keep the compass setting THE SAME
while you make all four marks.
2) Make sure you leave your
compass marks showing.
3) You get two equal angles.
The
B perpendicular
bisector
Step 2
Keep the compass setting
the same for all of these
construction arcs.
The angle bisector
Second compass marks
First compass marks
The Perpendicular to a Line through a Point
You’ll be given a line and a point, like this...
A
A
B
Step 2
A
4
... or a point that’s on the line, like this:
B
Step 2 — increase your compass
setting a bit for this step
This is the
perpendicular
required
90° angle
created
B
Initial point
90° angle
created
Initial point
Step 1
Step 1
The only way to learn constructions is to try them yourself...
Don’t rub out your compass marks when doing constructions — leave all your working showing.
Section Five — Geometry and Measures
131
Pythagoras’ Theorem
Pythagoras’ theorem sounds scary but it’s actually OK. There’s just one little formula to learn...
Pythagoras’ Theorem — a2 + b2 = c2
1) PYTHAGORAS’ THEOREM only works for RIGHT-ANGLED TRIANGLES.
2) Pythagoras uses two sides to find the third side.
3) The BASIC FORMULA for Pythagoras is a2 + b2 = c2.
4) Make sure you get the numbers in the RIGHT PLACE.
c is the longest side (called the hypotenuse)
and it’s always opposite the right angle.
a
5) Always CHECK that your answer is SENSIBLE.
`
`
c
b
a2 + b2 = c2
Find the length of side PQ in this triangle.
P
Put the numbers into the formula (PQ is the hypotenuse):
6 cm
62 + 82 = PQ2
36 + 64 = PQ2
R
Q
8 cm
100 = PQ2
Check the answer’s sensible — 10 cm is longer than the
PQ = 100 = 10 cm
other two sides, but not too much longer, so it seems OK.
Find the length of SU to 1 decimal place.
Put the numbers into the formula
(this time, ST is the hypotenuse):
SU2 + 52 = 112
SU2 + 25 = 121
SU2 = 121 – 25 = 96
SU = 96 = 9.797... = 9.8 m (to 1 d.p.)
T
11 m
5m
S
U
Check the answer is sensible — yes,
it’s a bit shorter than the longest side.
If you’re getting a bit confused, here’s a quick summary of what you have to do:
1) SQUARE THEM
2) ADD or SUBTRACT
3) SQUARE ROOT
ION
RE VIS
TIP
SQUARE THE TWO NUMBERS that you are given,
(use the x button if you’ve got your calculator).
2
To find the longest side, ADD the two squared numbers.
To find a shorter side, SUBTRACT the smaller one from the larger.
Finally, take the SQUARE ROOT (use the
button on your calculator).
Pythagoras’ theorem only works for right-angled triangles...
... don’t try to use it on other triangles — you will be guaranteed to get the wrong answer.
Section Five — Geometry and Measures
132
Trigonometry
Trigonometry — it’s a scary word and a tough topic. Lots of new words and formulas for you to learn.
The 3 Trigonometry Formulas
There are three trig formulas — each one links two sides and an angle of a right-angled triangle.
Opposite
Sin x = Hypotenuse
Adjacent
Cos x = Hypotenuse
• The Hypotenuse is the LONGEST SIDE.
• The Opposite is the side OPPOSITE the angle being used (x).
• The Adjacent is the other side NEXT TO the angle being used.
Tan x =
Opposite
Adjacent
Hypotenuse
(H)
Opposite
(O)
x
Adjacent (A)
1) Whenever you come across a trig question, work out which two sides of the triangle are involved
in that question — then pick the formula that involves those sides.
2) To find the angle — use the inverse, i.e. press SHIFT or 2ndF , followed by sin, cos or tan (and make
sure your calculator is in DEG mode) — your calculator will display sin–1, cos–1 or tan–1.
Formula Triangles Make Things Simple
A handy way to tackle trig questions is using formula triangles
(there’s more about formula triangles on p95 if you need a reminder).
1) Label the three sides O, A and H
(Opposite, Adjacent and Hypotenuse).
2) Write down from memory ‘SOH CAH TOA’.
3) Decide which two sides are involved: O,H A,H or O,A
and select SOH, CAH or TOA accordingly.
4) Turn the one you choose into a FORMULA TRIANGLE:
SOH
CAH
TOA
O
S×H
A
O
T×A
C×H
O
A
x
H
In the formula triangles,
S represents sin x,
C is cos x, and T is tan x.
5) Cover up the thing you want to find (with your finger), and write down whatever is left.
6) Translate into numbers and work it out.
7) Finally, check that your answer is sensible.
ION
RE VIS
TASK
SOHCAHTOA — three formulas in one made-up word...
You’ll need to know all three of the trig formulas off by heart. Make sure you’ve learnt them
before you move on — cover the page and scribble them down until you get them all right.
Section Five — Geometry and Measures
133
Trigonometry
Now that you’ve learnt the trig formulas it’s time to see how they actually work with some examples.
Examples:
1
Find the length of x in the triangle shown to 1 d.p.
1) Label the sides
x
O
x
H 12 cm
12 cm
40°
40°
A
2) Write down
SOH CAH TOA
3) O and H involved
4) Write down the formula triangle
O
S×H
5) You want O so cover it up to give O = S × H
6) Put in the numbers
2
x = sin 40° × 12
= 7.713...
= 7.7 cm (1 d.p.)
Is it sensible? Yes, it’s shorter than
12 cm, which is the longest side.
Find the angle x in this triangle to 1 d.p.
8 cm
x
8 cm A
x
1) Label the sides
O
2) Write down
18 cm
H
18 cm
SOH CAH TOA
3) A and H involved
4) Write down the formula triangle
A
C×H
5) You want the angle so cover up C to give C = A
H
6) Put in the numbers and find the inverse.
Is it sensible? Yes, the
angle is less than 90°.
8
cos x = 18 = 0.444...
& x = cos –1 (0.444...) = 63.612...°
= 63.6° (1 d.p.)
Make sure you label the sides of the triangle correctly...
...you’d feel a bit silly if you got a question wrong because you’d labelled the sides incorrectly.
Section Five — Geometry and Measures
134
Warm-Up and Practice Questions
This is the last lot of warm-up questions for this section — so make the most of them.
Try them out, then you can move onto the practice questions on the next two pages.
Warm-up Questions
1)
Construct a triangle with sides 3 cm, 4 cm and 4 cm.
Leave your construction marks showing.
2)
Do each of the four constructions from page 130.
Make sure you leave all your construction lines showing.
3)
The two shorter sides of a right-angled triangle measure 5 cm and 12 cm.
Find the length of the other side.
4)
A triangle has sides of length 8 cm, 15 cm and 17 cm.
Use Pythagoras’ theorem to check whether it is a right-angled triangle.
5)
Look at the right-angled triangle shown.
Which of the sides x, y or z is
a) opposite to angle q
b) adjacent to angle q
c) the hypotenuse?
6)
z
x
y
Which trig formula would you use if you knew the adjacent and hypotenuse of a triangle?
Practice Questions
Questions on Pythagoras and trigonometry can be a bit tricky — fortunately, there’s a worked
practice question to show you what to do, followed by some questions to have a go at yourself.
1
A cycle track on the sea wall is 3 m from the ground. A ramp is to be constructed between
the ground and the cycle track on the sea wall. The angle of the ramp is to be 15°.
cycle
track
(not to scale)
H ramp
O 3m
15°
How long is the ramp? Give your answer to 1 decimal place.
Start off by labelling the triangle (see above).
You can draw the formula
triangle here if you need to.
O and H are involved so use SOH
3
S= O
H , so sin 15° = H .
Rearranging gives H = sin315c = 11.591... = 11.6 m (1 d.p.)
11.6 m (1 d.p.)
............................
[3 marks]
Section Five — Geometry and Measures
135
Practice Questions
2
Side BC of the equilateral triangle ABC has been accurately drawn below.
B
3
C
a)
Use a ruler and compasses to complete the accurate drawing of triangle ABC.
b)
Construct the bisector of angle ACB of the triangle.
You must show all your construction lines.
[2 marks]
[2 marks]
A 10 m ladder, standing on horizontal ground, is leaning against a vertical wall.
The base of the ladder is 3 m from the wall.
How far up the wall will the ladder reach? Give your answer to 3 significant figures.
............................ m
[2 marks]
4
ABC is a straight line.
Use a ruler and compasses to construct the perpendicular to the line ABC
that passes through the point B. Show all of your construction lines.
A
B
C
[2 marks]
Section Five — Geometry and Measures
136
Practice Questions
5
A tent is 1.5 m tall. It is anchored to the ground by guy ropes 2.6 m long.
(not to scale)
2.6 m
1.5 m
x
Find the angle, x, between the guy rope and the ground. Give your answer to 1 decimal place.
............................ °
[3 marks]
6
An equilateral triangle has sides of length 4 cm. Find its height to 1 decimal place.
4 cm
............................ cm
[3 marks]
7
The diagram shows a church with a steeple.
(not to scale)
55°
9m
Find the height of the church to 1 decimal place.
............................ m
[3 marks]
Section Five — Geometry and Measures
137
Revision Summary for Section Five
You’ve reached the end of the section — guess what you have to do now...
• Try these questions and tick off each one when you get it right.
• When you’ve done all the questions for a topic and are completely happy with it, tick off the topic.
2D Shapes (p99-101) 
1) Write down the number of lines of symmetry and order of rotational symmetry for a rhombus.
2) Name 2 quadrilaterals that have 2 pairs of equal angles.
3) A regular polygon has 7 sides. What is the name of this polygon? 


Perimeter and Area (p104-106) 

5) Find the area of the shape on the right. 

4 cm
6) Find the area and perimeter of a quarter circle with radius 8 cm.
7 cm
4) Find the perimeter and area of a rectangle that measures 11 cm by 5 cm.
4 cm
3D Shapes (p109-112) 
7) Write down the number of faces, edges and vertices of a triangular prism.
8) Find the surface area and volume of a cuboid measuring 2 cm by 3 cm by 5 cm.
9) Write down the formula for finding the surface area of a cylinder.
10) Find the volume of a cylinder with radius 2 cm and height 8 cm.




Angles (p115-119) 


12) What do the angles in a quadrilateral add up to?
70°
x
13) What type of angles do you find in a Z-shape on parallel lines? 
14) Find the size of angle x in the diagram on the right. 
15) Find the size of an exterior angle and an interior angle of a regular decagon (10-sided shape). 
16) What is the sum of the interior angles in a hexagon? 
11) Give an example of a) an acute angle,
b) an obtuse angle, c) a reflex angle.
Transformations (p122-126) 
17) What 3 details must you give when describing a rotation?

18) A triangle has vertices with coordinates (1, 1), (4, 1) and (3, 4). Write down the coordinates
of the vertices of the triangle after an enlargement with scale factor 3 and centre (–1, 0).

19) Write down the 4 different conditions you can use to show that triangles are congruent. 
20) Write down the two properties that similar shapes have. 
Constructions (p129-130) 
21) Construct triangle ABC, where AB = 3 cm, AC = 3.5 cm and CAB = 50°.
22) Draw an acute angle of any size, then bisect it, leaving your construction marks showing.


Pythagoras and Trigonometry (p131-133) 
10 cm
23) Use Pythagoras’ theorem to find the length of the missing side. 
x

24) Use trigonometry to find the size of angle x in the same triangle.
7 cm
Section Five — Geometry and Measures
Section Six — Probability and Statistics
138
Probability
Believe me, probability’s not as bad as you think it is, but YOU MUST LEARN THE BASIC FACTS.
All Probabilities are Between 0 and 1
Probabilities can only have values from 0 to 1 (including those values).
You can show the probability of any event happening on this scale of 0 to 1.
Impossible
Unlikely
Evens
Likely
Certain
0
0.25
0.5
0.75
1.0
0
¼
½
¾
1
If All the Results are Equally Likely, Use This Formula
Use the formula below to find probabilities for a fair spinner, coin or dice.
A spinner/coin/dice is ‘fair’ when it’s equally likely to land on any of its sides.
Probability =
Number of ways for something to happen
Total number of possible outcomes
Outcomes are just
‘things that could
happen’.
This formula works for other cases where all the possible results are equally likely
(so in the example below, there’s an equal chance of picking any one ball).
A bag contains 6 blue balls, 5 red balls and 9 green balls.
Find the probability of picking out a green ball.
P(green) means ‘the
probability of picking
green’.
Just put the numbers into the formula above:
Number of green balls
9
P(green) =
(or 0.45)
=
Total number of balls 20
Probabilities Add Up to 1
1) If only one possible result can happen at a time,
then the probabilities of all the results add up to 1.
2) So since something must either
happen or not happen (i.e. only one
of these can happen at a time):
Probabilities always ADD UP to 1.
P(event happens) + P(event doesn’t happen) = 1
3) In the example above, the probability of not picking green = 1 – P(green) = 1 – 0.45 = 0.55.
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Probabilities are always between 0 and 1...
The closer a probability is to 1, the more likely it is — the closer it is to 0, the less likely.
Section Six — Probability and Statistics
139
Probability
When an experiment involves more than one event you might need to draw a sample space diagram.
Sample Space Diagrams Show All Possible Outcomes
When there are two things happening (e.g. two spinners being spun),
you can use a table as a sample space diagram.
The fair spinners on the right are spun,
and the outcomes listed.
a) Complete this sample space diagram
showing the possible outcomes.
The possible outcomes for one
spinner go along the top.
The outcomes for the other
spinner go down the side.
1
2
3
Red
1R
2R
3R
Blue Green
1B
1G
2B
2G
3B
3G
Spinning both spinners gives
3 × 3 = 9 different combinations,
so there are 9 outcomes here.
b) Find the probability of spinning a 2 and a green (2G).
ways to spin 2 and green
P(2G) = total number of possible outcomes = 91
c) Find the probability of spinning an odd number and a red.
There are 2 ways of doing
ways to spin odd and red
2
P(odd and red) = total number of possible outcomes = 9
this — 1R and 3R.
Use Probability to Find an “Expected Frequency”
1) Once you know the probability of something, you can predict
how many times it will happen in a certain number of trials.
2) For example, you can predict the number of heads you could expect if you
tossed a fair coin 50 times. This prediction is called the expected frequency.
Expected frequency = probability × number of trials
The probability of someone winning a game at a fair is 0.12.
Estimate the number of times you would expect them to win
if they played the game 50 times.
Expected number of wins = probability of a win × number of trials
= 0.12 × 50
This is an estimate. They might not win
=6
exactly 6 times, but it should be close.
Expected frequencies give the best estimate for an event...
... but it’s only an estimate — the event could still occur more or less frequently.
Section Six — Probability and Statistics
140
Relative Frequency
Relative frequency is a way of working out probabilities after you’ve done an experiment.
Fair or Biased?
‘Fair’ just means that all the possible scores are equally likely.
1
1) You can use the formula on p138 to work out that the probability of rolling a 3 on a dice is 6 .
2) BUT this only works if it’s a fair dice. If the dice is wonky (the technical term is ‘biased’)
then each number won’t have an equal chance of being rolled.
3) This is where relative frequency is useful.
You can use it to estimate probabilities when things are wonky.
Do the Experiment Again and Again and Again...
You need to do an experiment over and over again and count how often a result happens
(its frequency). Then you can find its relative frequency.
Relative frequency =
Frequency
Number of times you tried the experiment
An experiment could
just mean rolling a dice.
You can use the relative frequency of a result as an estimate of its probability.
A dice was rolled 100 times. The results are in the table below.
Estimate the probability of getting each of the scores.
Score
Frequency
1
15
2
13
3
4
4
45
5
16
6
7
The dice was rolled 100 times.
So divide each of the frequencies by 100 to find the relative frequencies.
Score
Relative
Frequency
1
2
3
4
5
6
15
45
7
13
4
16
100 = 0.15 100 = 0.13 100 = 0.04 100 = 0.45 100 = 0.16 100 = 0.07
The MORE TIMES you do the experiment, the MORE ACCURATE your estimate of the probability
will be. If you rolled the above dice 1000 times, you’d get a better estimate of the probabilities.
For a fair dice or spinner, the relative frequencies should all be roughly the same after a large
number of trials. If some of them are very different, the dice or spinner is probably biased.
Do the above results suggest that the dice is biased?
Yes, because on a fair dice, you’d expect all the probabilities to be about
the same (you’d expect each probability to be about 1 ÷ 6 = 0.17(ish)).
These probabilities are very different (the probability of rolling a 4 is
much higher than the probability of rolling a 3), so the dice is biased.
Biased — when some outcomes are more likely than others...
Compare relative frequency with the theoretical probability to see if it’s biased. E.g. when tossing a
fair coin the theoretical probability of a head or tail is 0.5, so compare the relative frequency to this.
Section Six — Probability and Statistics
141
Warm-Up and Practice Questions
Your proficiency on probability problems will probably progress with practice.
So pick up a pen and perfect your performance with this plethora of probability posers.
Warm-up Questions
1)
Today is the 24th November. What is the probability of tomorrow being Christmas day?
2)
There are 16 coloured balls in a bag. The probability of picking a red ball is P(red) = 12 .
How many red balls are there in the bag?
3)
If the probability of a bus arriving on time is 0.7, what is the probability that it is late?
4)
On a fair 10-sided spinner numbered 1-10, what is the probability of spinning:
a) a 7? b) an odd number? c) a number less than 5?
5)
A coin is tossed and a dice is rolled. List all the possible outcomes.
6)
The probability of spinning blue on a spinner is 0.3. If the spinner is spun 40 times,
estimate how many times you would spin blue.
7)
A 3-sided spinner is spun 100 times. It lands on red 43 times, blue 24 times
and green the other times. Calculate the relative frequency of each outcome.
Practice Questions
The probability of you getting a question on probability in a maths test is pretty close to 1,
so have a go and see how you get on. The first one has been done for you.
1
Layla spins the two spinners shown below and adds the numbers to get a score.
1 3
5
9
7
2 4
6 8
a)
Draw a diagram to show all the possible outcomes.
The 4 possible
outcomes for the
first spinner.
b)
2
4
6
8
1
3
5
7
9
3
5
7
9
11
5
7
9
11
13
7
9
11
13
15
9
11
13
15
17
The 5 possible outcomes
for the second spinner.
Find the probability of getting a score less than 10.
There are 10 outcomes that are less than 10,
10 = 1 .
out of 20 possible outcomes, so the probability is 20
2
c)
[3 marks]
1
2
......................
[1 mark]
Find the probability of getting the same number on both spinners.
The numbers on each spinner are different, so getting
the same number on both spinners is impossible.
0
......................
[1 mark]
Section Six — Probability and Statistics
142
Practice Questions
2
The letters that spell the word MATHEMATICS are placed in a bag.
A letter is picked at random. Find:
a)
The probability of picking out an ‘M’.
.........................
[1 mark]
b)
The probability of picking out an ‘A’ or a ‘T’.
.........................
[1 mark]
c)
The probability of picking out a vowel.
.........................
[1 mark]
d)
The probability of picking out an ‘X’.
.........................
[1 mark]
3
When rolling a fair dice, you have an equal chance of getting each number.
a)
What is the probability of rolling a fair dice and getting a 5?
.........................
[1 mark]
b)
A fair dice is rolled 300 times. How many times would you expect it to land on a 5?
.........................
[1 mark]
4
The probability that a train is not on time is 0.22.
What is the probability that the train is on time?
.........................
[1 mark]
5
There are 24 balls in a bag. The bag contains blue, red and green balls.
The probability of picking a blue ball is 1 and the probability of picking a red ball is 1 .
8
4
How many green balls are there in the bag?
.........................
[2 marks]
Section Six — Probability and Statistics
143
Venn Diagrams
Venn diagrams are a way of displaying data in intersecting circles — they’re very useful.
A Set is a Collection of Objects
1) Sets are just collections of things (e.g. numbers) — a pair of curly brackets {}
tells you it’s a set. The things in the set are called elements.
2) The universal set is the group of things that the elements of a set are selected from,
and is shown by this funny symbol: x. So x = {integers from 1 to 10} means
that the universal set is the numbers 1-10.
3) You might have to list the elements in a set — so using the universal set above,
if set A = {prime numbers}, then A = {2, 3, 5, 7}.
4) n(A) just means ‘the number of elements in set A’. So here, n(A) = 4.
Show Sets on Venn Diagrams
1) On a Venn diagram, each set is represented by a circle.
The universal set is a rectangle that goes round the outside of all of the circles.
2) It can show either the actual elements of each set, or the number of elements in each set.
3) If the circles overlap, elements that are in both sets go in the overlap
(this is called the intersection — see next page).
x = {integers from 1 to 10}, P = {factors of 10} and Q = {even numbers}.
Show this information on a Venn diagram.
1) Start by writing out the
P
Q
x
elements of each set:
x = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
3
1
4
2
P = {1, 2, 5, 10}
8
Elements in P
10
Q = {2, 4, 6, 8, 10}
5
6
but not Q
2) Then put the elements into a
7
9
Venn diagram.
Elements of both P and Q
Elements of Q
but not P
Elements of the
universal set that
aren’t in P or Q
The Complement of a Set
1)
2)
3)
4)
The complement of a set is all the members of the universal set that aren’t in the set.
A’ means ‘the complement of set A’.
So for set A above, A’ = {1, 4, 6, 8, 9, 10}.
On a Venn diagram, the complement is everything outside the circle representing the set.
The orange
area is P’...
x
P
Q
x
P
Q
... and the purple
area is Q’.
Section Six — Probability and Statistics
144
Venn Diagrams
You can show more exciting things on Venn diagrams too — like unions and intersections.
Unions and Intersections of Sets
UNIONS
1) The union of two sets contains all the elements that are in either set.
2) Unions are shown by the symbol , , so A , B means ‘the union of sets A and B’.
3) So for sets P and Q on the previous page, P , Q = {1, 2, 4, 5, 6, 8, 10}.
Only list elements once — even if
they appear in both sets.
INTERSECTIONS
1) The intersection of two sets contains all the elements that are in both sets.
2) Intersections are shown by the symbol + , so A + B means ‘the intersection of sets A and B’.
3) So for sets P and Q on the previous page, P + Q = {2, 10}.
Show Unions and Intersections on Venn Diagrams
On a Venn diagram, the union is
all the space covered by the circles.
x
P
The intersection is where
the circles overlap.
x
Q
P
Q
This is P + Q
This is P , Q
If there are no elements
in the intersection, the
circles won’t overlap.
Use Venn Diagrams to Find Probabilities
Venn diagrams can help you find probabilities — the number of elements in the universal set is the
number of possible outcomes, then just find the bit of the diagram you want. It’s easier to use a Venn
diagram showing the number of elements here.
Find the probability of picking a card that is both a factor of 10
and an even number from a set of cards numbered 1-10.
x
P
2
3
Q
2
3
1) Use the Venn diagram from the previous page —
but replace the elements with the number of elements.
2) Then find the bit you want — cards that are
a factor of 10 and even are in the intersection.
2 = 1.
3) There are 2 elements in the intersection, so the probability is 10
5
n(x) = 10, so the denominator is 10.
+ means all elements that are in one set a + d the other set...
The intersection of two sets has fewer elements than their union — unless both sets are the same.
Section Six — Probability and Statistics
145
Warm-Up and Practice Questions
Venn diagrams are a bit different to the rest of the stuff in this section — so check
you’ve got your head round them by having a go at this warm-up question.
Warm-up Question
1)x = {a standard pack of playing cards},
A = {red cards} and B = {kings}.
a) Complete the Venn diagram to show the number
B
A
xof
elements in each set.
b) Indicate on the diagram where the king of clubs
would go.
c) Indicate on the diagram where the queen of hearts
would go.
d) Find the probability of selecting:
(i) a red card
(ii) a red king
(iii) a black king.
Practice Questions
Before you have a go at the trickier questions on the next page, read through this worked question —
it’ll show you how it’s done.
x = {rectangle, square, regular pentagon, kite, scalene triangle,
regular hexagon, irregular octagon, isosceles triangle, rhombus}
A = {quadrilaterals}
B = {regular shapes}
Show this information on the Venn diagram below.
x
B
A
kite
Quadrilaterals
rectangle
rhombus
irregular
octagon
are
a)
squ
1
regular
hexagon
regular
pentagon
isosceles
triangle
Regular shapes
scalene
triangle
Regular quadrilaterals
b)
[3 marks]
Find n(A).
This is just the number of elements in the circle for A
(not forgetting the one in the overlap).
c)
Shapes that aren’t
quadrilaterals or regular
4
.......................
[1 mark]
Find A + B.
This is the element in the intersection of A and B
— a regular quadrilateral.
{Square}
.......................
[1 mark]
Section Six — Probability and Statistics
146
Practice Questions
2
x = {integers from 1-20}
a)
P = {square numbers}
Q = {factors of 20}
Find P and Q.
..............................................................................
[2 marks]
b)
Write down P , Q.
.........................................
[1 mark]
c)
Find n(P + Q).
.......................
[1 mark]
3
On the Venn diagrams below, shade the area that represents:
a) E'b) (E + F)'
x
E
x
F
E
F
[2 marks]
4
There are 300 students in Year 9. 120 students study music, and 30 study music and art.
100 students study neither music nor art.
a)
Show this information on the Venn diagram below,
where M = {music students} and A = {art students}.
x
b)
M
How many students study music or art?
A
[3 marks]
.......................
[1 mark]
Section Six — Probability and Statistics
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Data
Data is what statistics is all about. You’ve got to collect it, process it and then interpret it.
There are Different Types of Data
Data can be primary or secondary...
PRIMARY data is data you’ve collected.
SECONDARY data is collected by someone else.
... and quantitative or qualitative...
Quantitative data tends
to be easier to analyse
than qualitative data.
QUANTITATIVE DATA
measures quantities.
QUALITATIVE DATA
is descriptive.
Quantitative data is anything that
you can measure with a number.
For example, heights of people, the time
taken to complete a task or the mass of things.
Qualitative data is data that uses words to
describe it — it doesn’t use any numbers.
For example, gender, eye colour
or how nice a curry is.
Quantitative Data is Either Discrete or Continuous
DISCRETE DATA is data that
can be recorded exactly.
CONTINUOUS DATA is data that
can take any value in an interval.
If your data is something that’s countable
in whole numbers, or can only take certain
individual values, it’s called discrete data.
Things like the number of points scored in a
game or the number of people going into a
shop on one day are examples of discrete data.
If your data is something that could
always be more accurately measured,
it’s continuous data.
Heights, weights, ages and lengths are
all examples of continuous data.
You can Split Your Data into Classes
1) If you’re collecting lots of data, or your data’s spread out over a large range,
you can make it more manageable by grouping it into different classes.
2) When you do this, it’s important that you define the classes well so none of them overlap
— this means that each bit of data can only be put into one class.
Age in completed years
0 – 19
Number of people
6
20 – 39 40 – 59
13
14
60 – 79
80 – 99
8
9
3) The problem with grouping data is that you lose some accuracy
because you don’t know what the exact data values are any more.
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There are gaps between
these classes, e.g. from 79
to 80. This is because
ages are whole numbers
— you don’t need to fit
in values like 79.5.
Definitions can be dull but you’re going to have to learn them...
...so cover this page and write down as many of the six definitions as you can remember.
Section Six — Probability and Statistics
148
Graphs and Charts
There are lots of different ways of displaying data — let’s start off with bar charts and pie charts.
Bar Charts
1) On a bar chart, the numbers of things are shown by the heights of the different bars.
2) Dual bar charts show two things at once — they’re good for comparing different sets of data.
Bars representing different
categories are separated by gaps.
Draw a dual bar chart to show the
information in the two-way table below.
7
6
5
4
3
2
1
Both axes must 0
Draw a bar for the men
and a bar for the women for
each of the three categories.
Men
Women
Number of people
RightLeftAmbi- Total
handed handed dextrous
Men
7
3
2
12
Women
6
2
0
8
Total
13
5
2
20
be labelled.
Righthanded
Lefthanded
Ambidextrous
Pie Charts
Learn the Golden Rule for Pie Charts:
The TOTAL of Everything = 360º
Some geography students were asked to name their
favourite volcano. The results are displayed in the pie chart.
What fraction of the students chose Etna?
Just remember that ‘everything = 360°’.
angle of Etna
60c = 1
Fraction that chose Etna = angle of everything =
6
360c
The table on the right shows
the number of different types of
sandwiches sold in a cafe. Draw a
pie chart to show this information.
Number
sold
3
6
15
20
16
4) Draw your pie chart
accurately using a protractor.
Egg
18° Cheese
36°
Tuna
3) Now multiply every number by 6
to get the angle for each sector.
Section Six — Probability and Statistics
60°
Etna
egg
2) ‘Everything = 360°’ — so find the multiplier
(or divider) that turns your total into 360°.
Multiplier = 360 ÷ 60 = 6
6 × 6 15 × 6 20 × 6 16 × 6
= 36° = 90° = 120° = 96°
90°
Mount
Fuji
Sandwich cheese ham chicken tuna
1) Find the total by adding.
6 + 15 + 20 + 16 + 3 = 60
Angle
Vesuvius Krakatoa
90°
120°
Ham
96°
90°
120°
3 × 6 Total
= 18° = 360°
Chicken
149
Graphs and Charts
You can display frequencies on graphs (frequency just means how many).
Frequency Tables
Frequency tables show how many
things there are in each category.
Vehicle
Frequency
Car
Bus
Lorry
5
20
31
Grouped frequency tables group
the data into classes (see p147).
Inequalities are often used
if the data isn’t all whole numbers...
Here, 10 goes in the top class,
but 10.1 goes in the next class.
Height (h cm)
Frequency
5 < h £ 10
10 < h £ 15
15 < h £ 20
12
15
11
Frequency Bar Charts
Frequency bar charts are very similar to normal bar charts,
but they show data from a frequency table — usually a grouped data table.
The grouped frequency table shows the weights of 20 micro-sloths.
Draw a frequency bar chart to show this information.
6
Frequency
Weight (w kg) Frequency
1.0 < w ≤ 1.2
3
1.2 < w ≤ 1.4
6
1.4 < w ≤ 1.6
5
1.6 < w ≤ 1.8
4
1.8 < w ≤ 2.0
2
5
There are no gaps
between bars because
there are no gaps
between the classes.
4
3
2
1
0
1.0
1.2
1.6
1.4
Weight (kg)
1.8
2.0
Frequency Polygons
You can also draw frequency polygons from grouped frequency tables — just remember this rule:
Always plot your point at the mid-interval value of a class.
Find the mid-interval value by adding the upper and lower limits and dividing by 2.
6
0 ≤ h < 5 5 ≤ h < 10 10 ≤ h < 15 15 ≤ h < 20
Mid-interval value
2.5
7.5
12.5
17.5
Frequency
3
5
2
2
Plot the mid-interval value of each class on the horizontal axis and its
frequency on the vertical axis and join the points with straight lines.
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5
Frequency
Height (h cm)
4
3
2
1
0
5
10
15
Height (cm)
20
Learn how to draw and read all of these graphs and charts...
...and learn where gaps are needed — if there’s no gap in the values, don’t draw any.
Section Six — Probability and Statistics
150
Warm-Up and Practice Questions
To make sure you’re familiar with all the different types of graphs and charts you might come across,
have a go at these warm-up questions. Make sure you know the different types of data too.
Warm-up Questions
1)
Say whether this data is qualitative, discrete or continuous:
a) the number of spectators at a rugby match
b) the colours of pebbles on a beach c) the weight of eggs in a box
2)
Two bar charts are drawn. One bar chart has shoe size along the horizontal axis and the
other has length of feet along the horizontal axis.
Which bar chart has gaps between the bars? Explain why this is.
3)
A pie chart is drawn, showing the favourite colours of 60 people.
a) 10 people chose blue. Work out the angle for blue on the pie chart.
b) The angle for green is 72°. How many people chose green?
Practice Questions
Practising lots of nice graph questions will help you make sure you can cope with any that you come
across. This first one has been done for you, so look over it before making a start on the next page.
1
A group of students grew sunflowers, and measured the heights of their sunflowers
after 6 weeks. Their results are recorded in the table below.
Draw a frequency bar chart to show this data.
Height in m (h) 0 ≤ h < 0.5 0.5 ≤ h < 1 1 ≤ h < 1.5 1.5 ≤ h < 2 2 ≤ h < 2.5
Frequency
5
12
17
20
11
2.5 ≤ h < 3
10
25
There are no gaps between the bars
because there are no gaps between
the classes (the data is continuous).
Frequency
20
15
10
5
0
0.5
1.0
1.5
2.0
Height (m)
2.5
3.0
[3 marks]
Section Six — Probability and Statistics
151
Practice Questions
2
A class of children were asked to choose their favourite flavour of crisps.
Their results are shown below.
Girls
7
4
5
3
6
Plain
Cheese & Onion
Salt & Vinegar
BBQ Beef
Prawn Cocktail
Boys
3
8
5
9
2
Display this information in a dual bar chart.
[3 marks]
3
45 pupils were asked where they went last year for their summer holiday.
Their responses were organised into this frequency table.
Destination
Frequency
UK & Ireland
Spain
France
USA
Other
12
13
10
7
3
Angle of
sector
a)
Calculate the angle of the sector for each destination and add it to the table.
b)
Draw and label a pie chart showing this information.
[3 marks]
[3 marks]
Section Six — Probability and Statistics
152
Mean, Median, Mode and Range
Mean, median, mode and range pop up all the time — make sure you know what they are.
MODE = MOST common
MEDIAN = MIDDLE value (when values are in order of size)
MEAN = TOTAL of items ÷ NUMBER of items
RANGE = Difference between highest and lowest
REMEMBER:
Mode = most (emphasise the
‘mo’ when you say them)
Median = mid (emphasise the
m*d when you say them)
Mean is the average, but it’s
mean because you have to
work it out.
The Golden Rule
There’s one vital step for finding the median that lots of people forget:
Always REARRANGE the data in ASCENDING ORDER.
(and check you have the
same number of entries!)
You must do this when finding the median, but it’s also really useful for working out the mode.
Find the median, mode, mean and range of these numbers: 6, 4, 7, 1, 2, 6, 3, 5
The MEDIAN is the middle value (when they’re arranged in order of size)
— so first, rearrange the numbers.
Check you have the
1, 2, 3, 4, 5, 6, 6, 7
same number of
When there are two middle
¨ 4 numbers this side 4 numbers this side Æ
values
after you’ve
numbers, the median is
Median = 4.5
rearranged them.
halfway between the two.
MODE (or modal value) is the most common value.
MEAN =
total of items
number of items
Mode = 6
1 + 2 + 3 + 4 + 5 + 6 + 6 + 7 = 34 =
4.25
8
8
RANGE = difference between highest and lowest values.
7–1=6
You might be asked to compare two data sets using the mean, median or mode and the range.
Here is some information about the lengths children
guessed in a ‘guess the length of the snake’ competition.
Compare the lengths guessed by the boys and the girls.
Boys:Mean = 1.2 m,
Range = 0.2 m
Girls:Mean = 2.1 m,
Range = 1.4 m
1) Compare the means: The girls’ mean is higher than the boys’ mean,
so the girls generally guessed longer lengths.
2) Compare the ranges: The girls’ guesses have a bigger range, so the lengths
guessed by the girls are more spread out.
Two final facts to finish: 1) Some data sets have more than one mode, or no mode at all.
2) The range can be distorted by outliers (extremely high or low values).
EXAM
TIP
Put the numbers in ascending order before finding the median
Be sure to read the question carefully — you don’t want to work out the wrong thing.
Section Six — Probability and Statistics
153
Frequency Tables and Averages
The word frequency means how many — so in the table below,
15 people don’t have a cat, 22 have one cat, etc.
Find Averages from Frequency Tables
Categories
You can easily find averages and the range from frequency tables...
1) The MODE is just the CATEGORY with the MOST ENTRIES
(i.e. the highest frequency).
2) The RANGE is found from the extremes of the first column.
3) The MEDIAN is the CATEGORY of the middle value.
How many
Number
Frequency
of cats
0
15
1
22
2
16
3
7
4) To find the MEAN, you use TOTAL OF ITEMS ÷ NUMBER OF ITEMS.
The number of items is the 2nd column total.
You can find the TOTAL of the items by adding a 3rd column.
Mysterious 3rd column...
... here it would contain the values
of ‘Number of cats × frequency’.
Some people were asked how many posters they have
on their bedroom walls. The table shows the results.
Find the mode, the range, the mean and the median of the data.
Number of
posters
0
1
2
3
4
5
1) The MODE is the category with the highest frequency:
The highest frequency is 12 for ‘2 posters’, so MODE = 2
2) The RANGE is the difference between the highest and lowest
numbers of posters — that’s 5 posters and no posters, so:
3) To find the MEAN, add a 3rd column to the
table showing ‘number of posters × frequency’.
Add up these values to find the total number
of posters of all the people asked.
total number of posters
3rd column total
95
MEAN = 2nd column total = 40 = 2.375
total number of people asked
Frequency
1
10
12
9
6
2
RANGE = 5 – 0 = 5
Number of
posters
Frequency
0
1
2
3
4
5
Total
1
10
12
9
6
2
40
No. of
posters
× Frequency
0
10
24
27
24
10
95
4) The MEDIAN is the category of the middle value.
Work out its position, then count through the 2nd column to find it.
There are 40 values, so the middle value is halfway between the 20th and 21st values.
There are a total of (1 + 10) = 11 values in the first two categories, and another 12 in
the third category takes you to 23. So the 20th and 21st values must both be in the
category ‘2 posters’, which means the MEDIAN is 2.
Fill in the third column before finding the mean...
The three averages still mean the same thing, you just find them in different ways.
Section Six — Probability and Statistics
154
Grouped Frequency Tables
You can find averages from grouped frequency tables using similar methods
to the ones used for normal frequency tables (see previous page).
Find Averages from Grouped Frequency Tables
Unlike with frequency tables, you don’t know the actual data values, only the classes they’re in.
So you have to ESTIMATE THE MEAN, rather than calculate it exactly.
Again, you do this by adding columns:
1) Add a 3rd column and enter the mid-interval values for each class.
2) Add a 4th column to show ‘frequency × mid-interval value’ for each class.
3) The estimated mean is then: 4th Column Total ÷ 2nd Column Total
And you’ll be asked to find the MODAL CLASS and the CLASS CONTAINING THE MEDIAN,
not exact values.
This table shows information about the lengths,
in mm, of 30 racing snails.
a) Write down the modal class.
b) Write down the class containing the median.
c) Calculate an estimate for the mean length.
a) The modal class is the one with the highest frequency.
b) Work out the position of the median,
then count through the 2nd column.
=
2280
30
= 76 mm
Modal class is 60 < L ≤ 70
There are 30 values, so the median is halfway
between the 15th and 16th values.
Both these values are in the third class, so the
class containing the median is 70 < L ≤ 80.
c) Add extra columns for ‘mid-interval value‘
and ‘frequency × mid-interval value’.
4th column total
Mean = 2nd column total
Length (L mm) Frequency
50 < L ≤ 60
4
60 < L ≤ 70
8
70 < L ≤ 80
5
80 < L ≤ 90
7
90 < L ≤ 100
6
Length
(L mm)
Frequency
Mid-interval
value
50 < L ≤ 60
60 < L ≤ 70
70 < L ≤ 80
80 < L ≤ 90
90 < L ≤ 100
Total
4
8
5
7
6
30
55
65
75
85
95
—
Frequency ×
mid-interval
value
220
520
375
595
570
2280
The mid-interval values are used to work out the 4th column.
You don’t need to add them up.
You can’t find the range from a grouped frequency table...
... but you can find the three averages. The mean is the tricky one — you’ll need to create two extra
columns: one for the mid-interval values and one for the frequency × mid-interval value.
Section Six — Probability and Statistics
155
Scatter Graphs
Scatter graphs are really useful — they show you if there’s a link between two things.
Scatter Graphs Show Correlation
1) A scatter graph shows how closely two things are related.
The fancy word for this is CORRELATION.
2) If the two things are related, then you should be able to draw a straight line
(called a line of best fit) passing pretty close to most of the points on the scatter diagram.
STRONG correlation is when your
points make a fairly straight line.
WEAK correlation means your points
don’t line up quite so nicely.
WEAK NEGATIVE CORRELATION
+ +
+
+ +
+ +
+
+
+ +
++
+
+
Woolly hat sales
Ice cream sales
STRONG POSITIVE CORRELATION
+
+++
+
++
Temperature
Temperature
If the points form a line sloping downhill
from left to right, then
there is NEGATIVE
++
correlation — as+one quantity
increases,
+
+ +
+ +
the other decreases.
+
+
+
Newspaper sales
3) If the two things are not related, you get a load of
messy points. This scatter graph is a messy scatter
— so there’s no correlation between the two things.
Ice cream sales
If the points form a line sloping uphill from left
to right, then there is POSITIVE correlation —
both things increase or decrease together.
4) You can use a line of best fit to predict other values.
NO CORRELATION
+
+
+
+
Temperature
+
+
+
+
+
+
+
Temperature
a) Describe the strength and type
of correlation shown by the graph.
The age of the car and its value
are strongly negatively correlated.
b) Estimate the value of a 3 year old car.
1. Draw a line of best fit (shown in blue).
2. Draw a line up from 3 years to your line,
and then across to the other axis.
3 years corresponds to roughly £4500.
ION
RE V I S
TIP
Value (£’000s)
This graph shows the value of a car (in £’000s)
plotted against its age in years.
8
7
6
5
4
3
2
1
0
×
×
0
1
×
× ×
×
×
2
3
4
5
Age of car (years)
×
6
Correlations can be strong or weak and negative or positive...
When drawing the line of best fit, aim to have an equal number of points on either side.
Section Six — Probability and Statistics
156
Warm-Up and Practice Questions
You’ve finally made it — this is the last set of questions in the book (well, apart from the revision
questions, but they hardly count). Just get through the next couple of pages, then you can have a rest.
Warm-up Questions
1)
Find the median, mode, mean and range of these numbers: 10, 12, 8, 15, 9, 12, 11.
2)
What is the median of the numbers 10, 20, 12, 21, 19, 9, 5 and 18?
3)
Find the range of the first five prime numbers.
4)
Four numbers have a mean of 8. The first three numbers are 5, 7 and 8.
What is the other number?
5)
A class were asked how many cars their family had. The results are shown in the table
below. Find the mode, median, mean and range for this data.
Number of cars
0
1
2
3
4
Frequency
3
15
12
5
0
6)
In a grouped frequency table, one of the classes is 10 ≤ h < 20.
What is the mid-interval value for this class?
7)
A cafe sells more bowls of soup on cold days than hot days. What type of correlation
is there between temperature and the number of bowls of soup sold?
Practice Questions
Here’s the final set of practice questions — and my last worked example. I’m getting a bit emotional
here... just give me a minute.
1
50 pupils in Year 9 were asked how much TV they usually watched in an evening.
The results are shown in the grouped frequency table below.
Time spent watching
TV in mins (t)
Frequency
0 ≤ t < 60
60 ≤ t < 120
120 ≤ t < 180
180 ≤ t < 240
Total
12
16
15
7
50
Midinterval
value
30
90
150
210
—
Frequency ×
mid-interval
value
360
1440
2250
1470
5520
Add two columns to
the table — one for
the mid-interval values,
one for frequency ×
mid-interval values.
Find an estimate of the mean time spent watching TV in an evening.
th column total
Mean = 24nd
column total
= 5520
50
= 110.4 mins
110.4 mins
.............................
[4 marks]
Section Six — Probability and Statistics
157
Practice Questions
2
The table shows the marks for ten pupils in their Maths and French examinations.
Pupil
Lucy Emily
Jack Ismail Nigel
Steph Rose
Luc
Maths
45
21
39
48
26
38
45
26
18
29
French
48
28
32
44
33
40
40
28
35
34
a)
Adam Samira
Plot the data on the scatter graph below.
50
40
French 30
marks
20
10
0
30 40
10 20
Mathematics marks
50
[3 marks]
b)
Draw the line of best fit on your graph.
c)
Describe the type of correlation this data shows.
[1 mark]
...............................................................................................................................................
[1 mark]
d)
A pupil was absent for the French exam but scored 34 in the Maths exam.
What mark would you expect him to have got in French?
...........................
[1 mark]
3
Matteo is a member of a pub quiz team. His team have played three games.
In the first game they scored 42 points and in the second they scored 49 points.
Their mean score for the three games was 45 points.
How many points did Matteo’s team score in the third game?
.............................
[2 marks]
Section Six — Probability and Statistics
158
Practice Questions
4
15 students were asked how many text messages they had sent the previous day.
These were their responses:
20
18
3
5
15 11
3
9
8
12
22
4
9
12
9
Calculate the following:
a)
median
...........................
[1 mark]
b)
mode
...........................
[1 mark]
c)
mean
...........................
[2 marks]
d)
range
...........................
[1 mark]
5
Lily counts the number of letters in the
names of everyone in her class.
The frequency table on the right
shows the results.
a)
Number of
letters
3
Complete the table.
[2 marks]
Frequency
4
5
6
7
4
3
5
6
4
8
2
9
1
Number of letters
× frequency
Total
b)
Find the mean number of letters per name.
.............................
[2 marks]
c)
Find the median number of letters per name.
.............................
[2 marks]
Section Six — Probability and Statistics
159
Revision Summary for Section Six
You know what’s coming by now — here are some questions to check it’s all sunk in.
• Try these questions and tick off each one when you get it right.
• When you’ve done all the questions for a topic and are completely happy with it, tick off the topic.
Probability (p138-140) 
1) What does a probability of 1 mean? 
2) In a bag of sweets, there are 5 cola bottles, 2 jelly snakes, 3 chocolate buttons
and 2 chocolate mice. Find the probability of picking a cola bottle.
3) In a game, you can either win or lose. If P(win) = 0.1, what is P(lose)?


4) A spinner that is half black and half white is spun twice.
Black White
Fill in the sample space diagram to show all the possible results.

First spin
Second spin
Black
White
BW
5) I roll a fair dice 60 times.
Estimate the number of times it will land on a 5. 
6) Write down the formula for relative frequency. 
Venn Diagrams (p143-144) 
7) Let x = {integers from 1 to 12 inclusive}, X = {prime numbers}, Y = {factors of 8}
a) Find X , Y and X + Y.
b) Draw a Venn diagram and use it to find the probability that a ball selected from a bag
of balls numbered 1-12 is neither a prime number nor a factor of 8.

Types of Data, Graphs and Charts (p147-149) 
8) What is primary data? What is secondary data? 
9) Is ‘favourite flavours of ice cream’ quantitative or qualitative data?
10) I measure the weights of some baby dragons. Is this data discrete or continuous?
11) On a pie chart, the angle representing ‘not if you paid me £1 000 000’ is 30°.
If 120 people took part in the survey, how many gave this answer?
12) What values do you plot when drawing a frequency polygon?
Averages and Frequency Tables (p152-154) 




13) Find the mode, median, mean and range of this data: 2, 8, 7, 5, 11, 5, 4.

14) The frequency table on the right shows the number of TVs
Number
0 1 2 3 4
of TVs
per house in a street. How many houses had 3 TVs?

Frequency 2 10 15 12 1
15) Using the same frequency table, find the mode,
median, mean and range for the number of TVs. 
Number of
16) The grouped frequency table on the right shows the number
Frequency
lengths
of lengths swum by a group of children in a swimming gala.
0-10
3
Find the modal class and the class containing the median
11-20
6

for this table.
21-30
8
Scatter Graphs (p155) 
31-40
3
17) Give an example of data that would show positive correlation on a scatter graph.

Section Six — Probability and Statistics
Section Seven — Exam Practice
160
Mixed Practice Tests
OK, so you’ve done most of the hard work — but are you ready for the big Practice Exam?
To help you decide, here are some brilliant Mixed Practice Tests for you to have a go at.
• Scribble down your answers to the questions in a test. When you’ve answered them all, check your answers
(see p.200-201). Tick the box next to each question you got right. Put a cross in the box if you got it wrong.
• If you’re getting 7 or more out of 10 right on these tests, you should be ready for the Practice Exam on p.170-189.
• If you’re getting less than that, go back and do some more revision. Have another go at the
Revision Summaries — they’re the best way to find out what you know and what you’ve forgotten.
Test 1
ü/û
1. What is a prime number?
2. How many millilitres are in 20 litres?
A
B
C
D
0.02
200
2000
20 000
3. A reflex angle is...
A
B
C
D
...less than 90°.
...exactly 90°.
...between 90° and 180°.
...between 180° and 360°.
4. Give the coordinates of the point where the x- and y-axes cross.
5. What is the next number in this sequence: 1, 7, 13, 19, ... ?
A
B
C
D
13
22
25
29
6. What is discrete data?
7. What is the perimeter of a shape?
8. The statement x < 2 means...
A
B
C
D
...x is less than 2.
...x is less than or equal to 2.
...x is greater than 2.
...x is greater than or equal to 2.
9. What is the median of these numbers: 1 4 5 8 12?
10. Rearrange this equation to make x the subject: y = 2x – 6.
Total (out of 10):
Section Seven — Exam Practice
161
Mixed Practice Tests
Test 2
ü/û
1. The equation of a straight line is y = 4x + 5.
What are the coordinates of the y-intercept of this line?
2. The mean of a set of data is...
A
B
C
D
...the middle value.
...the most common value.
...the sum of all the values, divided by the number of values.
...the difference between the largest and smallest value.
3. What is the formula for Pythogoras’ theorem?
A
B
C
D
y = mx + c
a2 + b2 = c2
sin x = opposite ÷ hypotenuse
a + b + c + d = 360°
4. What is the highest common factor of two numbers?
5. How many significant figures is this number given to: 0.405?
A
B
C
D
one
two
three
four
6. What is primary data?
7. The ratio of red to yellow sweets in a bag is 2:5.
Red sweets make up...
A
B
C
2 of the bag.
5
5 of the bag.
2
2 of the bag.
7
D 5 of the bag.
7
8. A sector of a circle is...
A
B
C
D
...a wedge-shaped area cut from the centre of the circle.
...part of the circumference of a circle.
...a straight line that just touches the outside of the circle.
...the distance from the centre of a circle to its edge.
9. Estimate 102.1 × 3.04.
10. Simplify the following expression: 2a + 7 + a2 – a – 4.
Total (out of 10):
Section Seven — Exam Practice
162
Mixed Practice Tests
Test 3
ü/û
1. What are the coordinates of point A, shown on the graph below?
y
5
4
A
3
2
1
0
1
2
3
4
5
x
2. What is the range of these values: 1, 4, 5, 9?
3. The number 4136 written in standard form is...
A
B
C
D
...0.4136 × 103.
...4.136 × 103.
...0.4136 × 10–3.
...4.136 × 10–3.
4. The complement of set A is...
A
B
C
D
...the group of items that the elements of set A are selected from.
...the number of elements in set A.
...the elements of the universal set that aren’t in set A.
...the sum of the elements in set A.
5. The picture below is a scale drawing of a swimming pool.
What is the length of the swimming pool?
1 cm = 4 m
Tiled area
Pool
length = 6 cm
A
B
C
D
6m
600 m
12 m
24 m
Section Seven — Exam Practice
163
Mixed Practice Tests
6. Round 9.754 to one decimal place.
7. Write 2 as a decimal.
10
8. Congruent shapes are...
A
B
C
D
...the same shape as each other but can be any size.
...shapes whose interior angles always add up to 180°.
...identical to each other.
...shapes in which all the interior angles are equally sized.
9. What is the modal class of the table below?
Class
Frequency
0-5
4
6-10
7
11-15
9
16-20
4
10. Which of these is a correct simplification of (x2)3?
A
B
C
D
x –1
x5
x6
x23
Total (out of 10):
Section Seven — Exam Practice
164
Mixed Practice Tests
Test 4
ü/û
1. What is the formula used for calculating the area of a triangle?
2. Which of the following is an improper fraction?
A
1
2
3
C 3
2
D 11
3
B
3. What must the probabilities of all the possible results of an event add up to?
4. The lowest common multiple of two numbers is...
A
B
C
D
...the lowest number that will divide into both numbers.
...the highest number that will divide into both numbers.
...the lowest number that will divide by both numbers.
...the value of the two numbers multiplied together.
5. What value do all of the interior angles in a quadrilateral add up to?
6. Reduce this ratio to its simplest form: 6:18.
7. Simplify a2 × a19 × a4 as much as possible.
8. In the Venn diagram on the right,
what does the shaded area represent?
A
B
C
D
A
B
A
A’
A,B
,
A B
9. Expand this expression: 5(2x + 5).
10. The price of a sandwich is increased from £1.00 to £1.50.
What is the percentage increase in the price of the sandwich?
A
B
C
D
50%
75%
100%
150%
Total (out of 10):
Section Seven — Exam Practice
165
Mixed Practice Tests
Test 5
ü/û
1. What is next in the sequence: 5, 10, 20, 40, ... ?
2. Which number line shows the inequality 2 ≤ x < 7?
A
B
C
D
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
3. What is the formula used for calculating the volume of a cylinder?
4. What is the order of rotational symmetry of a regular pentagon?
5. The formula for calculating the gradient of a straight line is...
A
B
C
D
...change in y ÷ change in x.
...change in x ÷ change in y.
...change in x × change in y.
...change in y – change in x.
6. What are irrational numbers?
7. A packet of pens contains blue pens and green pens in the ratio 3:2.
There are 12 blue pens in the packet. How many green pens are there?
8. If the possible results of an event have unequal probabilities of occurring, the event is...
A
B
C
D
...accurate.
...fair.
...biased.
...uncertain.
9. What is the formula for working out the sum of the interior angles
for a polygon with n sides?
10. Point A has co-ordinates (3,8). Point B has co-ordinates (7,10).
Find the coordinates of the midpoint of line AB.
Total (out of 10):
Section Seven — Exam Practice
166
Mixed Practice Tests
Test 6
ü/û
1. Write down both square roots of 81.
2. How many lines of symmetry does this shape have?
3. What is 0.42 as a percentage?
4. What is the value of angle a in the diagram below?
a
120°
A
B
C
D
60°
120°
180°
240°
5. In a survey, 100 students were asked how often they go to the cinema.
The pie chart below shows the results of the survey.
What proportion of the students go to the cinema once a week?
Key
45°
A
B
C
D
1
4
1
8
1
10
1
12
Section Seven — Exam Practice
Less than once a month
Once or twice a month
Once a week
More than once a week
167
Mixed Practice Tests
6. Which of the following is a graph of y = x2?
A
-4
B
-4
-2
-2
4
4
2
2
0
2
4
C
-4
0
-2
-2
-2
-4
-4
4
4
2
2
0
2
4
D
-4
0
-2
-2
-2
-4
-4
2
4
2
4
7. When variable x increases, variable y decreases. You can say that x and y are...
A
B
C
D
...simultaneous.
...directly proportional.
...inversely proportional.
...unrelated.
8. Factorise 2p2 + 4pr.
9. A man walks a distance of 30 metres in 20 seconds.
Calculate the man’s walking speed using this formula: speed = distance ÷ time.
10. A completed factor tree for the
number 112 is shown on the right.
Write down the prime factorisation of 112.
112
4
2
28
2
4
2
7
2
Total (out of 10):
Section Seven — Exam Practice
168
Mixed Practice Tests
Test 7
ü/û
1. A scalene triangle...
A
B
C
D
...has three sides of equal length.
...has two sides of equal length.
...has three unequal sides.
...has three equal interior angles.
2. Which of these formulas could you use to calculate A as a percentage of B?
A
A × 100
B
A + B × 100
2
C
B × 100
A
B
D A × B × 100
3. What is the surface area of a shape?
4. Look at the diagram on the right.
Which angle is AED?
A
B
C
D
p
q
r
s
A
D
p
q
E
s
r
B
C
5. If you have the graphs of a pair of simultaneous equations on a single set of axes,
how can you use the graphs to find the solution to both equations?
6. Describe how a graph of x against y will look, if x and y are directly proportional.
7. Write down the formula you would use to calculate the relative frequency of a result.
8. If 3x + 4 = 25, what is the value of x?
9. The triangle to the right is translated by the vector a 74 k .
What are the new coordinates of point A?
A
B
C
D
(4,7)
(5,8)
(–3,–6)
(7,14)
10. Look at the spinner on the right. The probability that it will
land on C is 0.4. Estimate the number of times you would
expect the spinner to land on C if it was spun 20 times.
5
4
3
2
A
1
0 1 2 3 4 5
A
C
B
Total (out of 10):
Section Seven — Exam Practice
169
Mixed Practice Tests
Test 8
ü/û
1. Round 827 682 to the nearest thousand.
2. What is qualitative data?
3. Write down the formula you would use to calculate the area of a parallelogram.
4. A scatter graph has points which make a clear, straight line. As one variable
increases, the other variable increases. The graph is showing...
A
B
C
D
...weak negative correlation.
...strong negative correlation.
...weak positive correlation.
...strong positive correlation.
5. The union of two sets...
A
B
C
D
...contains all the elements that are in both sets.
...contains all the elements that are in either set.
...is the group of items that the elements of the sets are selected from.
...is all the elements from the universal set that are in neither set.
6. Write down the formula you would use to calculate the percentage change in a value.
7. Convert 3 hours 20 minutes into minutes.
8. Which diagram shows the correct construction marks for the bisector of an angle?
A
C
B
D
9. A sequence is given by the rule 2n + 1. Is 18 part of the sequence?
10. Which of these is a correct expansion and simplification of (x + 2)(x – 2)?
A
B
C
D
2x
x2 – 4
x2 + 4
x2 + 4x – 4
Total (out of 10):
Section Seven — Exam Practice
170
Key Stage 3 Mathematics
Practice Exam
CGP
Paper 1
Calculator NOT allowed
Instructions
■
The test is one hour long.
■
Make sure you have these things with you before you start:
pen, pencil, rubber, ruler, angle measurer or protractor,
and a pair of compasses. You may use tracing paper.
■
The easier questions are at the start of the test.
■
Try to answer all of the questions.
■
Don’t use any rough paper — write all your answers and working
in this test paper.
■
Check your work carefully before the end of the test.
You may not
use a calculator
in this test.
Formulas
Trapezium
Prism
b
height (h)
a
(a + b)
Area =
×h
2
length
Volume = area of cross-section × length
Score:
out of 60
171
1.
Write these numbers in order of size, starting with the smallest.
...............
...............
...............
...............
2 marks
ome pupils recorded the number of pets that each girl and boy in their class had.
S
They plotted the information on the bar chart shown below.
8
Girls
Number of pupils
2.
1
0.150.05
4
0.4
Boys
6
4
2
0
1
2
3
Number of pets
4
or more
(a) 2 boys in the class had 3 pets.
Use this information to complete the bar chart.
1 mark
(b) How many girls in the class had fewer than 3 pets?
...................................
1 mark
(c) How many pupils were in the class?
...................................
1
1 mark
Practice Paper 1
172
3.
Fill in the next two numbers in each of the following sequences.
(a)
6
1
16
11
1 mark
(b)
9
16
25
36
1 mark
(c) A sequence is given by the rule 4n + 8.
Find the 9th term in the sequence.
........................
2 marks
4.
Pat is playing a game with two dice.
She works out her score by adding the two numbers rolled together.
(a) Complete the table to show all her possible scores.
Dice two
Dice one
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
.........
7
8
9
10
5
6
7
8
9
10
.........
6
.........
8
9
.........
11
12
1 mark
(b) Use your table to work out the probability that Pat scores a 10.
........................
Practice Paper 1
2
2 marks
173
5.
Shape A has been drawn on this coordinate grid.
3 y
A
2
1
x
-3
-2
-1
0
1
2
3
-1
-2
-3
(a) Rotate shape A 90° clockwise about (0, 0). Label the image shape B.
1 mark
(b) Reflect shape A in the x-axis. Label the image shape C.
1 mark
6.
If n is an integer, write down all the possible values of n
in the following inequalities.
(a) 3 < n < 9
....................................................
(b) –1 ≤ n < 5
....................................................
(c) –4 ≤ n ≤ 0
....................................................
3
1 mark
1 mark
1 mark
Practice Paper 1
174
7.
The prices of some of the souvenirs in a
seaside gift shop are shown on the right.
(a) A customer paid for a
fridge magnet with a £5 note.
How much change did they get?
Postcard
£0.60
Mug
£2.90
Bookmark
£1.45
Fridge magnet
£1.65
£...................................
1 mark
(b) How much would it cost to buy 2 bookmarks, 2 mugs and a postcard?
£...................................
8.
2 marks
Find the missing side, x, in each of the following right-angled triangles.
(a)
(Diagram not
to scale)
x
3 cm
4 cm
x = .................... cm
(b)
2 marks
(Diagram not
to scale)
10 cm
8 cm
x
x = .................... cm
Practice Paper 1
4
2 marks
175
9.
ultiply out the expressions below.
M
Write your answers as simply as possible.
(a) 4(3x + 2) – 6(x – 5)
......................................................
2 marks
(b) (3x + 3)2
......................................................
2 marks
10. A
t a concert the ratio of parents to children was 3 : 5.
120 people were at the concert.
How many parents and how many children went to the concert?
................... parents and .................... children
5
3 marks
Practice Paper 1
176
11. D
ylan has a pencil case containing 24 pens.
14 pens are black, 8 are blue and the rest are red.
Dylan is going to pick a pen at random from the pencil case.
(a) Complete these sentences.
(i) The probability that the pen will be ............................. is
1
.
3
(ii) The probability that the pen will be red is ...................... .
1 mark
1 mark
(b)Before Dylan picks his pen, his cousin Salma borrows two blue pens
and a red pen from the pencil case. If Dylan now chooses a pen at
random from the case, what is the probability that it will be black?
.................
1 mark
12. Here are the numbers of films that eight pupils watched in January and February.
2, 6, 15, 8, 5, 3, 5, 12
(a) Calculate the mean number of films watched.
........................
2 marks
(b) Calculate the range of the number of films watched.
........................
1 mark
(c)Another pupil also watched some films in January and February.
If the number of films she saw was included in the data above, the mean
of the nine values would be 8. How many films did she watch?
........................
Practice Paper 1
6
2 marks
177
13. The population of the town of Greenley Bay is 84 293.
(a) Round this number to 3 significant figures.
...........................
1 mark
(b)The town fair was attended by 17 400 people, to the nearest hundred.
Give the minimum and maximum number of people
who could have gone to the fair.
Minimum: ...........................
Maximum: ...........................
2 marks
14. In the diagram below, the straight lines PS and TW are parallel.
X
U
T
P
60°
(Diagram not
to scale)
V
W
125°
Q
R
S
Find the following angles. Give reasons for your answers.
(a) WVX
....................................................................................................................
....................................................................................................................
1 mark
(b) QXR
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
7
2 marks
Practice Paper 1
178
15. This formula shows the relationship between three variables, p, q and r:
p = 2q + r²
(a) Find the value of p when q = 6 and r = 4.
...................
1 mark
(b) Rearrange the formula to make q the subject.
.................................
2 marks
(c) Rearrange the formula to make r the subject.
.................................
2 marks
16. The diagram shows a trapezium ABCD.
A
3x + 1 cm
B
(Diagram not
to scale)
8 cm
C
4x + 3 cm
D
Side AB is of length 3x + 1 cm. Side CD is of length 4x + 3 cm.
The area of the trapezium is 72 cm².
Find the value of x.
x = ...................
Practice Paper 1
8
3 marks
179
17. These two parallelograms are similar:
(Diagram not to scale)
36 cm
q
12 cm
110°
33 cm
p
(a) Work out the length of side p in the smaller parallelogram.
p = .................... cm
2 marks
(b) What is the size of angle q in the larger parallelogram?
q = .................... °
1 mark
18. Evaluate the following expressions:
(a)
m 4 × ^m 2h5 × m –2
when m = 2.
^m 2h4
............................
(b)
^ y 6h2 × y –4
y0 × y5
2 marks
when y = 4.
............................
2 marks
END OF TEST
9
Practice Paper 1
180
Key Stage 3 Mathematics
Practice Exam
CGP
Paper 2
Calculator allowed
Instructions
■
The test is one hour long.
■
Make sure you have these things with you before you start:
pen, pencil, rubber, ruler, angle measurer or protractor,
pair of compasses and a calculator. You may use tracing paper.
■
The easier questions are at the start of the test.
■
Try to answer all of the questions.
■
Don’t use any rough paper — write all your answers and working
in this test paper.
■
Check your work carefully before the end of the test.
You may use
a calculator
in this test.
Formulas
Trapezium
Prism
b
height (h)
a
(a + b)
Area =
×h
2
length
Volume = area of cross-section × length
Score:
out of 60
181
1.
Complete the following statements.
(a) 750 millimetres = .......................... centimetres = .......................... metres
1 mark
(b) 31 litres = .......................... cm3
1 mark
2.
(a) This diagram has four small squares shaded in.
Shade in four more small squares to make a grey pattern that
has a line of symmetry along the dotted line.
1 mark
(b) Here is another diagram with four small squares shaded in.
Shade in four more small squares to make a grey shape
that has rotational symmetry of order two.
1
1 mark
Practice Paper 2
182
3.
The pie chart shows the proportions of different
sports played by members of a leisure club.
Basketball
Hockey
(Diagram not
to scale)
Rugby
140°
80°
Football
(a) What is the angle of the “Rugby” sector of the pie chart?
...................
°
(b) 84 leisure club members play hockey.
What is the total number of leisure club members?
.......................
4.
1 mark
2 marks
Solve the following equations:
(a) 6x + 9 = 57
x = ....................
(b) 3(3x – 6) = 54
x = ....................
(c) 4x – 7 = 5x – 2
x = ....................
Practice Paper 2
2
1 mark
1 mark
1 mark
183
5.
(a) Express the following fraction in its simplest form.
40
64
...................
(b) Simplify the following expression.
Give your answer as an improper fraction.
1 mark
8
5
÷
9
11
...................
6.
1 mark
Simplify these expressions as much as possible:
(a) 8 – 2s + 9s – 3
..........................
(b) 4t² + 7t + t²
..........................
(c) 5u × 5u
..........................
(d)
27v²
9v
..........................
3
1 mark
1 mark
1 mark
1 mark
Practice Paper 2
184
7.
The diagram below shows a regular 15-sided polygon.
f
(a) Find the value of exterior angle f.
...................°
(b)A regular polygon has an exterior angle of 40°.
Calculate how many sides the polygon has.
...................
8.
1 mark
2 marks
Joe has a circular pond with a diameter of 1.8 m.
(a) Calculate the circumference of the pond to 2 decimal places.
..................... m
(b)Joe wants to put some edging around the pond.
It is sold in 1 m pieces and each piece costs £3.20.
What will it cost Joe to buy enough edging to go around the pond?
£ .....................
9.
2 marks
2 marks
The cost of joining a local gym is £34 plus £11 each time you visit it.
Let C be the total cost in pounds, and n be the number of visits to the gym.
(a) Write a formula to show the cost of joining and using the gym.
.........................................
(b) It cost Amber £210 to join and use the gym.
How many times did she visit it?
............... times
Practice Paper 2
4
1 mark
2 marks
185
10. Some students were investigating how fast people type.
They asked 55 people to type as many words as they could in one minute.
The table below shows the results that they collected.
Words typed (w) in 1 min
Frequency (f)
Midpoint (x)
fx
30 < w £ 40
24
35
840
40 < w £ 50
21
...............
...............
50 < w £ 60
7
...............
...............
60 < w £ 70
3
...............
...............
Total
55
—
2365
(a) Complete the table.
2 marks
(b) Work out an estimate of the mean typing speed
of a person in this investigation.
.................... words per minute
1 mark
11. A school tuck shop sells four different types of sweets. Taj worked out the
probability of a pupil choosing each type of sweet, based on sweet sales
from the last three months. This information is shown in the table below.
Chocolate
Mints
Toffees
Fudge
0.15
0.25
(a) Taj found that a pupil was twice as likely to choose mints as toffees.
What is the probability that a pupil will choose chocolate?
...............................
(b)Taj expects that 320 pupils will buy sweets from the tuck shop next month.
Based on the sales of the last three months, how many more pupils
would you expect to buy fudge than toffees? Show your working.
...................... pupils
5
2 marks
3 marks
Practice Paper 2
186
12. (a) Find the area of the shape below.
4 cm
6 cm
5 cm
8 cm
.................... cm2
2 marks
(b) This prism has the shape from part (a) as its cross-section.
15 cm
Find its volume.
.................... cm3
1 mark
13. The Brown family's farm has an area in m² of 2.45 × 107.
(a) Write 2.45 × 107 as an ordinary number.
..............................................
1 mark
(b) The Black family's farm has an area in m² of 8.33 × 10 .
6
What is the total area of the two farms combined?
Give your answer in standard form.
.............................................. m²
Practice Paper 2
6
2 marks
187
14. P, Q and R are points with the coordinates (–2, 1), (0, 3) and (1, 4).
(a) Plot P, Q and R on the grid below.
Draw a straight line through the points.
4
3
2
1
–3
–2
–1
0
1
2
3
4
–1
–2
2 marks
(b) Find the gradient of the line that you drew in part (a).
Gradient = ...............................
2 marks
(c) Write down the equation of the line that you drew in part (a).
........................................................
7
1 mark
Practice Paper 2
188
15. An ice cream parlour sells ice cream tubs in three different sizes.
A: 490 ml tub at £2.89
B: 1.25 litre tub at £3.55
C: 800 ml tub at £3.10
Which tub of ice cream represents the best value for money?
Show your working.
.................
3 marks
16. Monique travels 34 miles to visit her grandmother.
(a) One day the journey took 50 minutes.
What was her average speed in miles per hour?
.................... mph
2 marks
(b)If she drove at an average speed of 45 miles per hour, how long would
it take her to get to her grandmother's house, to the nearest minute?
.................... minutes
Practice Paper 2
8
2 marks
189
17. In a sale, the price of a washing machine is reduced by 13%.
The price of the washing machine in the sale is £348.
(a) Work out the price of the washing machine before the sale.
£....................
(b)A few weeks later, the manager decreases the sale price of the washing
machine by 10%. Jan says that the product is now 23% cheaper
than the original price. Is Jan correct? Explain your answer.
2 marks
....................................................................................................................
....................................................................................................................
....................................................................................................................
....................................................................................................................
2 marks
18. The diagram below shows the cross-section of the attic room in a doll's house.
roof
wall 7 cm
a°
(Diagram not
to scale)
floor
24 cm
The room is 24 cm wide. The wall is 7 cm high.
(a) Work out the length of the roof.
.................... cm
(b) The roof makes an angle of a ° with the floor.
Work out the size of angle a. Give your answer to 1 decimal place.
.................... °
2 marks
2 marks
END OF TEST
9
Practice Paper 2
190
Answers
Section One — Numbers
3 a)
168
Page 11 (Warm-up Questions)
1 a) 1
b) 21
c) 3
2 a) –9, –7, 45, 54, 304, 442
b) 0.0012, 0.0015, 0.026, 0.056, 0.207, 0.222
3 a) 797
b) 852
c) 18.45
4 a) 490
b) 17290
c) 6600
d) 350 000 e) 0.333
f) 0.08521
g) 13
h) 70
5 a) 990
b) 52570
c) 16
d) 45 remainder 10
6 a) 61.2
b) 5.28 c) 61
7 a) 5
b) –15
c) 18
12
6
3
d) 30
d) –3
Page 12 (Practice Questions)
2 a) 5 – 24 + 5 [1 mark]
= 5 – 19 = –14 [1 mark]
[2 marks available in total — as above]
55 × 2 = 110
b)
[1 mark] = 10 [1 mark]
20 – 9
11
[2 marks available in total — as above]
3 a) 3.09, 3.5, 4.79, 4.8, 5.17, 5.72
[2 marks available — 2 marks for all numbers in the
correct order, lose a mark for missing out a number
or putting one number in the wrong place.]
b) 0.043, 0.303, 0.31, 0.4, 0.43, 0.44
[2 marks available — 2 marks for all numbers in the
correct order, lose a mark for missing out a number
or putting one number in the wrong place.]
4 a) 1200 ÷ 80 is the same as 120 ÷ 8
0 1 5 so one book costs £15.
8 1 240
[2 marks available — 1 mark for a correct method
and 1 mark for the correct answer]
b) 38 – 1 – 1 = 36 mm [1 mark]
0 0 . 1 2 so each page is 0.12 mm thick.
300 3 6 . 0600
[3 marks available — 1 mark as above, 1 mark for the
correct method and 1 mark for the correct answer]
Page 17 (Warm-up Questions)
1 Rational numbers can be given as a fraction and irrational
numbers can’t.
2 Square numbers: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Cube numbers: 1, 8, 27, 64, 125
3 73, 83
4 a) 8, 16, 24, 32, 40
b) 15, 30, 45, 60, 75
c) 21, 42, 63, 84, 105
5 a) 1, 2, 3, 4, 6, 9, 12, 18, 36 b) 1, 2, 4, 7, 8, 14, 28, 56
c) 1, 2, 4, 8, 16, 32, 64
6 22 × 3 × 52
7 36
8 12
Page 18 (Practice Questions)
2 1 × 1 = 1 and 1 × 1 × 1 = 1 [1 mark]
8 × 8 = 64 and 4 × 4 × 4 = 64 [1 mark]
1 and 64 are the first two numbers to be cubes and squares.
[2 marks available in total — as above]
Answers
14
7
2
2
2
[2 marks available in total — 1 mark for correct prime
factors of 12, 1 mark for correct prime factors of 14]
Alternatively, split 12 into 4 and 3, then 4 into 2 and 2.
b) 23 × 3 × 7 or 2 × 2 × 2 × 3 × 7[1 mark]
4 Multiples of 12: 12, 24, 36, 48, 60, 72, 84, ... [1 mark]
Multiples of 15: 15, 30, 45, 60, 75, 90, ... [1 mark]
So Jacob would need to make a minimum of 60 cupcakes
to have nothing left over. [1 mark]
[3 marks available in total — as above]
Page 24 (Warm-up Questions)
1 a)
2 a)
3 a)
4 a)
5 a)
d)
6 a)
7 a)
4
1
b)
5
25
1
b) 0.6
3
5
1
b)
7
4
27
7
b)
4
2
2 = 1
b)
30 15
4
e)
5
196.95
b)
70.41% (2 d.p.)
b)
c)
23
100
d)
999
1000
c) 40%
2
5
71
c)
8
c)
24
10 = 5 = 1
c)
1
35
8
4
4
7
12
436.65
162.16% (2 d.p.)
Page 25 (Practice Questions)
3 Convert all of the numbers to decimals:
1
5
13
0.12, 17% = 0.17,
=
= 0.05,
= 0.13,
20 100
100
9
18
9% = 0.09,
=
= 0.18 [1 mark]
50 100
Now order the decimals from smallest to largest:
0.05, 0.09, 0.12, 0.13, 0.17, 0.18 [1 mark]
Put the numbers back into their original form:
1
13
9
, 9%, 0.12,
, 17%,
[1 mark]
20
100
50
[3 marks available in total — as above]
2
4
of 21 = 21 ÷ 7 × 2 = 3 × 2 = 6 blue marbles.
7
1
of 21 = 21 ÷ 3 = 7 red marbles.
3
21 – 6 – 7 = 8 green marbles.
[2 marks available — 1 mark for finding the number of
blue or red marbles correctly, 1 mark for correct answer]
5 a) 120% as a decimal is 120 ÷ 100 = 1.2 [1 mark]
120% of £4000 = 1.2 × £4000 = £4800 [1 mark]
[2 marks available in total — as above]
Alternatively, 10% = £400, so 20% = £800
120% = £4000 + £800 = £4800.
10 000
b)
× 100 [1 mark]
4 000
10
=
× 100 = 2.5 × 100 = 250% [1 mark]
4
[2 marks available in total — as above]
191
Page 29 (Warm-up Questions)
1
2
3
4
5
6
a) 7.9
b) 7.90
c) 7.895
a) 70
b) 73
c) 72.6
a) 2560
b) 2600
c) 3000
43
2750 ≤ x < 2850
a) E.g. 800
b) E.g. 5
c) E.g. 1000
Page 30 (Practice Questions)
614 600 =
.
4 [1 mark]
154 150
No, Harry is not right, £40 is far too much. [1 mark]
[2 marks available in total — as above]
3 a) The rounding unit is the nearest 10 so the maximum
error can be up to half a rounding unit bigger or
smaller. 10 ÷ 2 = 5 kg. [1 mark]
b) Lower limit of the weight = 510 – 5 = 505 kg
Upper limit of the weight = 510 + 5 = 515 kg
505 kg ≤ weight < 515 kg
[2 marks available — 1 mark for correct values for
upper and lower limits, 1 mark for correct inequality]
4 a) 22.5 metres [1 mark]
b) 9.5 metres [1 mark]
c) The minimum distance for 1 lap of the court is:
22.5 + 9.5 + 22.5 + 9.5 = 64 metres [1 mark]
1000
= 15.625
64
So she would have to run 16 complete laps [1 mark]
[2 marks available in total — as above]
2 E.g.
Page 34 (Warm-up Questions)
1
2
3
4
5
6
7
a) 72
b)
a) 416
b)
12
7
a) 3 and –3 b)
a) 4.36
b)
a) 7.65 × 109
a) 5 600 000
56
72
c) 59
c) y18
Section Two — Algebra
Page 42 (Warm-up Questions)
11 and –11 c) 13 and –13
8.63
c) 4.44
b) 2.4 × 10–5
b) 0.0000111
1
2
3
4
Page 35 (Practice Questions)
3 a)
b)
c)
4 a)
11.92 [1 mark]
4.34 [1 mark]
4.57 [1 mark]
729 = 93, 6561 = 94 and 4 782 969 = 97 [1 mark]
93 × 94 = 97 so 729 × 6561 = 4 782 969 [1 mark]
[2 marks available in total — as above]
98
43 046 721
b)
= 6 = 92 = 81 [1 mark]
531 441
9
1
= 1 × 1 = 1 ÷ 10
5
20 000 2000 10 2000
= 0.0005 ÷ 10 = 0.00005 [1 mark]
1 + 1
= 0.0005 + 0.00005 = 0.00055 [1 mark]
2000 20 000
= 5.5 × 10–4 [1 mark]
[3 marks available in total — as above]
Page 36 (Revision Summary)
1
2
3
4
a) 9
b) 8
c) 10
0.001, 0.02, 0.09, 0.51, 0.9, 11.8, 23.91, 54
a) 611
b) 596
c) 2.673
a) 122.3
b) 15120
c) 0.675
5 a) 2489
b) 48
c) 310.08
d) 5.05
6 a) –2
b) –14 c) 56
d) –9
7 a) Integers are positive and negative whole numbers
(and zero).
b) Rational numbers can be written as fractions.
c) Irrational numbers can’t be written as fractions, e.g. p.
d) All numbers are real numbers.
8 a) 41, 43, 47
b) 83, 89
9 a) 13, 26, 39, 52, 65
b) 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84
10 a) 22 × 32 × 7
b) 2 × 5 × 23
c) 13
11 a) 56
b) 16
3
7
12 a) 5 and 60%
b) 20 and 0.35
2
6
9
13 a) E.g.
and
b) 15
c) 4
10
15
7
8
1
20
7
14 a) 9
b) 12 c) 33
d) 12
15 a) 120
b) 210
16 a) 11.7
b) £132.44
c) £98.01
17 a) 3.74% (2 d.p.)
b) 4.42% (2 d.p.)
18 a) 164.4
b) 76000
c) 765440
19 –45
20 a) 45 ≤ x < 55
b) 1.725 ≤ x < 1.735
21 a) E.g. 50
b) E.g. 100000
22 a) 1000000
b) 1729
c) 49
d) 0.04
23 a) 613
b) 34
c) 9–18
24 a) 16
b) 2.36
c) 4
25 a) 8.56 × 108
b) 0.00005678
a)
a)
a)
a)
c)
5 a)
c)
e)
3a
b) 8d
5
d
b) 16ef
2a + 6
b) 6u – 15
y2 + 3y + 2
6p2 + p – 1
2(2 – 5x)
–4(x + 2) or 4(–x – 2)
3b(3b – c)
c) 6x + 2y
d) –8x + 9y
9
c) g
d) h7
2
c) 3x + 2x d) 4b2 – 6b
b) w2 – 6w + 9
d) q2 – 1
b) 2(3x + y)
d) 2a(5 + 4a)
Page 43 (Practice Questions)
2 a) 7x – y [1 mark]
b) 6x2 + 7x – 5 [1 mark]
3 a) 10x – 2 – 7x [1 mark]
= 3x – 2 [1 mark]
[2 marks available in total — as above]
b) y2 – 6y + 3y – 18 [1 mark]
= y2 – 3y – 18 [1 mark]
[2 marks available in total — as above]
4 a) 36(b + 1) [1 mark]
b) 12(3b + 1) [1 mark]
5 (5x + 2)2 [1 mark]
= 25x2 + 10x + 10x + 4
= 25x2 + 20x + 4 [1 mark]
[2 marks available in total — as above]
d) 0.0062
Answers
192
Page 49 (Warm-up Questions)
4 a)
1 a) x = 3
b) x = 11
c) x = 3
d) x = 21
2 a) x = 1
b) x = 1
c) x = 2
d) x = 2 and x = –2
e) x = 5 and x = –11
3 a) –6
b) 4
c) –4
d) 8
e) 8
f) 6
g) –2.5
h) 2
4 An equation has an equals sign in, an expression doesn’t.
5 3 hours
p–5
6 w=
3
7 v = 3u + 5
[1 mark]
b) In the 5th pattern (above) there are 15 and the number
you add increases by 1 each time.
6th pattern 15 + 6 = 21,
7th pattern 21 + 7 = 28,
8th pattern 28 + 8 = 36
[2 marks available — 1 mark for realising that
the number you add increases by 1 each time,
1 mark for the correct answer]
Page 50 (Practice Questions)
3 a) 7k = 14
k=2
[1 mark]
b) 2t + 5 = 12 [1 mark]
2t = 7
t = 3.5 [1 mark]
[2 marks available in total — as above]
4 a) (2x + 3)2
[2 marks available — 1 mark for 2x + 3,
1 mark for squaring the whole expression]
b) (2x + 3)2 = 81
2x + 3 = 9
2x = 6
x = 3 [1 mark]
If you didn’t know the answer was positive, you’d get
2x + 3 = ±9. Which would give answers of 3 and –6.
5 a) 3V = pr2h [1 mark]
3V
2 = h [1 mark]
πr
[2 marks available in total — as above]
b) 3V = pr2h
3V
= r2 [1 mark]
πh
3V
πh = r [1 mark]
[2 marks available in total — as above]
You could get a negative square root too, but as it’s a
radius, it must be positive, so there’s only one answer.
Page 54 (Warm-up Questions)
1 a) 11, 14
b) 26, 35
c) 108, 324
d) 14, 8
e) 2, 0.5
2 16, 26 — the rule is add the previous two terms together.
3 a) 3n + 2
b) –10n + 7
4 No (because n = 7.2)
5 –3, –2, –1, 0, 1, 2
6
–3
–2
–1
0
1
2
2
1
2
3
4
5
6
7
8
[2 marks for correct answer, lose a mark if
one of the circles is incorrect]
3 a) 2n + 5
[2 marks available — 1 mark for a common
difference of 2n, 1 mark for +5]
b) –4n + 15
[2 marks available — 1 mark for a common
difference of –4n, 1 mark for +15]
Answers
1
2
3
4
5
6
7
a) 5a
b) 9b
a) 6d + e
b) –2f + 9
a) g4
b) 9mn
9
a) r b) s6
a) 3v + 24
b) –14w – 35
4y + 4
Putting in brackets (the opposite of multiplying out
brackets).
8 a) 6(x + 3)
b) 14(x + 2y)
9 a) x = 7
b) x = 22
10 a) x = 4
b) x = 80
11 a) x = 3
b) x = 2
12 a) x = 6
b) x = 4
13 P = 13
14 –4 °F
15 P = 8s + 15t
p+3
16 q = 2
17 a) 26 — Add 6 to the previous term.
b) 243 — Multiply the previous term by 3.
c) 21 — Add the two previous terms together.
18 2n + 3
19 Yes (when n = 6).
20 a) 1, 2, 3, 4, 5, 6
b) 8, 9, 10, 11
21
–4 –3 –2 –1
0
1
2
3 4
Section Three — Graphs
Page 60 (Warm-up Questions)
1
Page 55 (Practice Questions)
0
Page 56 (Revision Summary)
y
10
9
8
7
6
5
4
3
2
1
0
1 2 3 4 5 x
The coordinates of the fourth point are (5, 10).
193
Page 65 (Warm-up Questions)
2 (3, 5)
3 (–1, 0.5)
4 a) Downhill
c) Downhill
5 a) 5
b) –25
6
8
b) Uphill
d) Uphill
c) 155
1 a)
b)
2 a)
3 a)
c)
(i) 1
(ii) –1
(i) y = x – 3
(ii) y = –x + 3
1
b) 1
c) 2
d) –0.5
m = 3, c = 4
b) m = 4, c = –3
m = 5, c = –10
d) m = 0, c = 2
8–2
6
4 First find the gradient, m =
= = 2.
3–0
3
The line passes through (0, 2) so c = 2.
The equation of the line is y = 2x + 2
5
6 y
d) –295
7
6
5
4
3
5
2
4
3
1
2
–2 –1 0
1
2
1
Page 61 (Practice Questions)
2 a)
b)
c)
d)
3 a)
–3 –2 –1 0
–1
y = 2 [1 mark]
x = 4 [1 mark]
y = –2 [1 mark]
x = –4 [1 mark]
x
3
–3
–4
–5
0
3
1
5
2
7
3
9
[2 marks available — 2 marks for all values correct,
or 1 mark for at least four correct values]
b)
2
–2
–3 –2 –1
–3 –1 1
x
y
1
6 a) Yes
b) Yes
Page 66 (Practice Questions)
2 a) 10 y
D
4
9
8
7
6
5
4
C
3
A
2
1
3
0
9 y
8
7
6
5
B
x
2
1
x
-3
-2
-1 0
-1
1
2
3
-2
-3
[2 marks available — 1 mark for plotting at least
3 correct points, 1 mark for the correct straight line]
c)
9 y
8
7
6
5
1 2 3 4 5 6
[2 marks available — 2 marks for all values plotted
correctly, lose 1 mark if one is incorrect]
9–3
6
b) Gradient of AB =
= = 3 [1 mark]
3–1
2
Put point A(1, 3) into the equation y = 3x + c to find c:
3 = (3 × 1) + c = 3 + c
0=c
So the equation of the line is y = 3x [1 mark]
[2 marks available in total — as above]
10 – 3 = 7
c) Gradient of CD =
= 2.333... [1 mark]
3–0
3
No, the lines are not parallel as they have different
gradients. [1 mark]
[2 marks available in total — as above]
3 a) 15 y
4
10
3
2
5
1
x
-3
-2
-1 0
-1
1
2
3
-2
[1
mark]
-3
d) (–0.5, 2) [1 mark]
x
0
1
2
3
4
5
–5
[2 marks available — 1 mark for plotting at least
3 correct points, 1 mark for the correct straight line]
Answers
194
b) 15 y
4 a)
10
b)
x
–3
–2
–1
0
1
2
3
y
12
7
4
3
4
7
12
x
12
10
8
6
x 4
x
2
5
x
x
0
3
2
1
4
5
y
x
x
x
x
–5
[2 marks available — 2 marks for correct parallel
line, 1 mark for any line parallel to y = 3x + 1]
Page 70 (Warm-up Questions)
1
2
3
4
5
a) 600 000
b) 3 days
c) exponential graph
The travelling object is stationary (has stopped).
The speed the object is travelling.
Distance
Draw a straight line from the value you know on one
axis to the graph. Change direction and draw a line
straight to the other axis. Then read off the value.
Cost (£)
6
180
160
140
120
100
80
60
40
20
0
0 2 4 6 8 10 12 14 16
Rental days
–3 –2 –1 0 1 2 3
c) x = 1.7 and x = –1.7
Page 75 (Practice Questions)
2 a)
3 km [1 mark]
30 minutes [1 mark]
14:45 [1 mark]
3 + 5 + 2 = 10 km [1 mark]
Graph B [1 mark]
Graph C [1 mark]
6
5
4
3
[2 marks available in total — 2 marks for the
correct straight line, lose 1 mark if the gradient
or y-intercept are incorrect]
b)
8
–2 –8 –10 –8 –2
b)
10
y=4–x
8
y
5
x
-3
1
2
3
-2
-1
0
-5
1
2
3
-10
y = 2x – 2
–3
–4
b) x = 2 and y = 2
3 It contains an x2 term and no higher power of x.
Answers
x
1 2 3 4 5 6 7
[2 marks available — 2 marks for all values correct,
lose 1 mark for one incorrect value]
x
–2
y = 2x + 1
x+y=7
–2 –1 0
–1
y
1
y
7
6
5
4
3
2
1
[2 marks available in total — 2 marks for the correct
straight line, lose 1 mark if the gradient or y-intercept
are incorrect]
c) x = 2 and y = 5 [1 mark]
3 a)
1 2 3
x –3 –2 –1 0
2
–2 –1 0
–1
1 2 3 4 5 6 7
Page 74 (Warm-up Questions)
1 The solution is found at the point where the two lines
cross.
2 a)
y
y = 2x + 1
x
–2 –1 0
–1
Page 71 (Practice Questions)
2 a)
b)
c)
d)
3 a)
b)
y
7
6
5
4
3
2
1
[2 marks available — 1 mark for plotting all the
points correctly, 1 mark for a smooth curve drawn
through them]
c) x = –2.2 and x = 2.2 [1 mark]
195
Page 76 (Revision Summary)
1
2
3
4
5
6
b)
A(2, 2)
B(3, –2)
C(–2, –3)
D(–2, 1)
(0, –0.5)
(0.5, –0.5)
y = –ax
a) Neither
b) Downhill
c) Uphill
8 y
7
6
5
4
3
2
1
-3 -2 -1 0
-1
-2
-3
-4
1 2 3
11 They are the same.
12 Yes
13 The thing travelling has stopped.
14 a) On his way home.
b) 15 minutes
15 a) £20
b) 4
c) £5
d) £35
y
16
6
5
4
3
2
y=x+2
1
x
2
-3 -2
-1 0
x –2
x
–4
x
1
x
2
x
3
1
2
4
5
6
a) 1 : 9
b) 2 : 5
c) 5 : 3
35 litres
3 £240
Dan = £800, Chris = £2400, Angela = £3200
a) 39
b) t = 3l
4 hours
Page 82 (Practice Questions)
2 1 calculator costs £35 ÷ 14 = £2.50 [1 mark],
so 22 calculators cost £2.50 × 22 = £55 [1 mark].
[2 marks available in total — as above]
2 1 = 4 1
3 a)
[1 mark] = 4 : 1 [1 mark]
:
:
5 10 10 10
[2 marks available in total — as above]
b) Olive oil = 4 parts = 800 ml,
so 1 part = 800 ÷ 4 = 200 ml [1 mark].
Vinegar = 1 part = 200 ml [1 mark]
[2 marks available in total — as above]
Alternatively, scale up the ratio 4 : 1.
Multiply both sides by 200 to get 800 : 200.
4 Total number of parts = 5 + 3 + 2 = 10 [1 mark]
1 part = 50 ÷ 10 = 5 DVDs [1 mark]
Shaun gets 5 parts = 5 × 5 = 25 DVDs [1 mark]
[3 marks available in total — as above]
5 a) Time for 1 person 4 × 5 = 20 hours [1 mark]
Time for 10 people = 20 ÷ 10 = 2 hours [1 mark]
[2 marks available in total — as above]
b) Number of dogs for 1 person = 15 ÷ 5 = 3 [1 mark]
Number of dogs for 11 people = 3 × 11 = 33 [1 mark]
[2 marks available in total — as above]
Page 86 (Warm-up Questions)
1 £3400
4 37.5%
7 £19 000
y = 2x
x
-3 -2 -1 0
-1
-2
1
2
x = 2 and y = 4
17 Symmetrical bucket shape
–3 –2 –1
1
0
2 12.5%
5 £110
3 £60
6 £520
Page 87 (Practice Questions)
3
-3
-4
6
4
Page 81 (Warm-up Questions)
x
1 2 3
x
x
y
x
y
Section Four — Ratio, Proportion
and Rates of Change
7
6
5
4
3
2
1
18 a)
6
7 Gradient of R = 1
Gradient of S = –2
8 m is the gradient and c is the y-intercept
9 y = 10 – 2x or y = –2x + 10
10
8 y
-3 -2 -1 0
-1
-2
-3
-4
-5
x
1
2
3
–2 –3 –2
1
6
2 Mr Patel: £10 extra = £112 per week, and
10% extra = £102 × 1.1 = £112.20 per week [1 mark]
Miss Dalton: £10 extra = £108.50 per week, and
10% extra = £98.50 × 1.1 = £108.35 per week [1 mark]
Mrs Ferrar: £10 extra = £130 per week, and
10% extra = £120 × 1.1 = £132 per week [1 mark]
So Mr Patel and Mrs Ferrar would be better off choosing
10% extra. [1 mark]
[4 marks available in total — as above]
You could have done this one by working out what 10% of
each salary was, and seeing if it was more or less than £10.
Answers
196
3 Single bed = £63.75 ÷ 0.85 = £75
Double bed = £240 ÷ 0.6 = £400
Bunk beds = £168 ÷ 0.8 = £210
[3 marks available — 1 mark for each correct price]
4 a) Increase in height = 1.7 – 1.25 = 0.45 m
Percentage increase = (0.45 ÷ 1.25) × 100 = 36%.
[2 marks available — 1 mark for a correct method,
1 mark for the correct answer]
b) Percentage increase = 36% ÷ 2 = 18% [1 mark]
Height at the end of Year 11 = 1.7 × 1.18 = 2.006 m
= 2.01 m (to the nearest cm) [1 mark]
[2 marks available in total — as above]
Page 92 (Warm-up Questions)
1 a) 35 mm b) 2800 cm3
c) 36 pints
2 a) 20 kg
b) 150 cm
c) 12.5 miles
3 44 mm = 4.4 cm, 0.5 m = 50 cm
Order: 4.2 cm, 44 mm, 0.5 m
4 310 minutes
5 a) 2400 mm3
b) 40 m2
6 96 km/h
7 0.5 cm by 1.5 cm
Page 93 (Practice Questions)
2 a) Conversion factor = 2.2
You’d expect fewer kg than pounds, so divide:
33 ÷ 2.2 = 15 kg [1 mark]
b) 1 stone 8 lb = 14 + 8 = 22 pounds [1 mark]
Conversion factor = 2.2
You’d expect fewer kg than pounds, so divide:
22 ÷ 2.2 = 10 kg [1 mark]
10 kg + 15 kg = 25 kg [1 mark]
[3 marks available in total — as above]
3 Convert all the volumes to the same unit:
2 pints ≈ 2 ÷ 1.75 = 1.14 litres (2 d.p.)
500 cm3 = 0.5 litres
0.5 gallons ≈ 0.5 × 4.5 = 2.25 litres
So the order is 500 cm3, 1 litre, 2 pints, 0.5 gallons
[3 marks for correct order, otherwise 2 marks for all
conversions correct or 1 mark for 2 conversions correct]
4 a) 1 cm = 50 000 cm = 500 m = 0.5 km [1 mark]
6 km ÷ 0.5 = 12 cm on the map [1 mark]
[2 marks available in total — as above]
b) 1 cm = 0.5 km [1 mark]
32 cm = 32 × 0.5= 16 km [1 mark]
[2 marks available in total — as above]
Page 96 (Warm-up Questions)
1 400 ÷ 160 = 2.5 ml per penny
1000 ÷ 300 = 3.33 ml per penny
So 1 litre is the better buy.
2 150 ÷ 100 = 1.5p per gram
275 ÷ 250 = 1.1p per gram
500 ÷ 400 = 1.25p per gram
So 250 g is the best buy.
3 6 cm3
4 0.2 km/min
5
Mass
Volume
Density
10 kg
100 cm3
0.1 kg/cm3
0.02 kg (or 20 g)
1 cm3
0.02 kg/cm3
500 g
4 cm3
125 g/cm3
18 kg
15 cm3
1.2 kg/cm3
Answers
Page 97 (Practice Questions)
2 a) Time = distance ÷ speed = 45 ÷ 60 = 0.75 hours
= 45 minutes [1 mark]
b) Speed = distance ÷ time = 1800 ÷ 2.5 [1 mark]
= 720 km/h [1 mark]
[2 marks available in total — as above]
c) 285 mins = 4.75 hours [1 mark]
Distance = speed × time = 45 × 4.75
= 213.75 km = 214 km to the nearest km [1 mark]
[2 marks available in total — as above]
3 a) Density = mass ÷ volume = 300 ÷ 0.02 = 15 000 kg/m3
[2 marks available — 1 mark for the
correct number, 1 mark for the correct units]
b) Density = mass ÷ volume = 945 ÷ 350 = 2.7 g/cm3
[2 marks available — 1 mark for the
correct number, 1 mark for the correct units]
4 500 ÷ 250 = 2 ml per penny
700 ÷ 420 ≈ 1.67 ml per penny
1200 ÷ 540 ≈ 2.22 ml per penny
So the 1.2 litre bottle is the best value for money.
[2 marks available — 1 mark for finding all three
amounts per penny (or cost per ml), 1 mark for the
correct answer]
5 20 tonnes = 20 000 kg
Volume = mass ÷ density = 20 000 ÷ 800 = 25 m3
[3 marks available — 1 mark for converting the mass
to kg, 1 mark for using the correct formula, 1 mark for
the correct answer including units]
Page 98 (Revision Summary)
1 a) 7 : 8
b) 3 : 2
c) 5 : 2
2 a) 2 : 3
b) 8 : 7
3 1 : 3.6
4 38
5 £105
6 £1.25
7 20 mins
8 £6.72
9 £156
10 12%
11 £90
12 See p88
13 a) 50 miles
b) 75 cm
c) 22 pounds
14 275 minutes
15 a) 60 cm2 b) 0.0078 m3
c) 43.2 km/h
16 4 cm
17 50 m by 25 m
18 300 ÷ 150 = 2 g per penny
450 ÷ 200 = 2.25 g per penny
750 ÷ 300 = 2.5 g per penny
So 750 g is the best buy.
19 0.75 g/cm3
20 75 km
Section Five — Geometry and Measures
Page 102 (Warm-up Questions)
1
3
5
7
8
1
2 2
2
4 5
Kite
6 Parallelogram
Isosceles triangle
Lines of symmetry = 20, order of rotational symmetry = 20
Page 103 (Practice Questions)
2 a) Regular hexagon [1 mark]
b) 6 [1 mark]
c) 6 [1 mark]
197
3 a) A parallelogram has 2 [1 mark] pairs of equal sides
(which are parallel). It has 0 [1 mark] lines of
symmetry and rotational symmetry of order 2 [1 mark].
b) An equilateral triangle has 3 [1 mark] equal sides
and 3 [1 mark] lines of symmetry.
4 a)
(i)
(ii)
(iii)
(iv)
(v)
[3 marks available — 3 marks for all 5 shapes
correct, otherwise 2 marks for 3 or 4 shapes correct,
otherwise 1 mark for 2 shapes correct]
b) (i) 1
(ii) 2 (iii) 2 (iv) 5 (v) 2
[3 marks available — 3 marks for all 5 correct,
otherwise 2 marks for 3 or 4 correct, or 1 mark
for 2 correct]
Page 107 (Warm-up Questions)
1
3
5
7
40 cm2 2 30 cm2
4 cm
4 36 cm2
135 mm2 6 3 cm
Circumference = 62.83 cm, area = 314.16 cm2
(both to 2 d.p.)
8 Tangent
Page 108 (Practice Questions)
2 a) Chord [1 mark]
b) Radius [1 mark]
c) Arc [1 mark]
3 8 × p = 12, so p = 12 ÷ 8 = 1.5 cm [1 mark]
½ × 4 × q = 12, so q = 12 ÷ 4 ÷ ½ = 6 cm [1 mark]
½(6 + r) × 3 = 12, so
½(6 + r) = 4 Þ 6 + r = 8 Þ r = 2 cm [1 mark]
[3 marks available in total — as above]
4 a) Circumference of circle with diameter 6 cm = 6 × p
= 18.849... cm, so curved edge of shape = 18.849... ÷ 2
= 9.424... cm [1 mark]
Perimeter = 14 + 10 + 6 + 9.424... [1 mark]
= 39.424... = 39.42 cm (2 d.p.) [1 mark]
[3 marks available in total — as above]
b) Area of trapezium = ½(14 + 6) × 6 = 60 cm2 [1 mark]
Area of semicircle = (p × 32) ÷ 2 = 14.137... cm2
[1 mark]
Total area = 60 + 14.137... [1 mark] = 74.137...
= 74.14 cm2 (2 d.p.) [1 mark]
[4 marks available in total — as above]
Page 113 (Warm-up Questions)
1 12
2 8
3 a) Regular tetrahedron
b) Faces = 4, edges = 6, vertices = 4
4 96 cm2
5 534.1 cm2 (1 d.p.)
6 168 cm3
Page 114 (Practice Questions)
2 a) 6 [1 mark]
b) 5 [1 mark]
c) 9 [1 mark]
3 Surface area = 2(5 × 3) + 2(8 × 3) + 2(5 × 8)
= 30 + 48 + 80 = 158 cm2
[3 marks available — 1 mark for attempting to find the
area of each rectangle of the net, 1 mark for putting the
numbers in correctly, 1 mark for the correct answer]
4 a) Area = pr 2 = p × (7.4 ÷ 2)2 = 43.008... = 43.01 cm2 (2 d.p.)
[2 marks available — 1 mark for using the correct
formula, 1 mark for the correct answer]
b) Volume = pr2 × h = 43.008... × 11 = 473.092...
= 473.09 cm3 (2 d.p.)
[2 marks available — 1 mark for using the correct
formula, 1 mark for the correct answer]
Page 120 (Warm-up Questions)
1
2
5
8
a) Obtuse
b) Reflex
c) Acute
90°
3 25°
4 Corresponding angles
102°
6 720°
7 100°
Exterior angle = 45° Interior angle = 135°
Page 121 (Practice Questions)
2 Angles on a straight line add up to 180°, so the unlabelled
angle in the triangle = 180° – 90° = 90°. The triangle is
isosceles, so angles c and d are the same, and are equal to
(180° – 90°) ÷ 2 = 45°. So c = d = 45°.
Using angles on a straight line, e = 180° – 40° = 140°
[3 marks available — 1 mark for each correct angle]
3 a) E.g. AGF and GBE are allied angles so
AGF = 180° – 110° = 70° [1 mark]
b) E.g. DEB and GBE are alternate angles so
DEB = GBE = 110° [1 mark]
c) E.g. DEB is 110° (part b) so using angles on a
straight line, DEC = 180° – 110° = 70° [1 mark]
Angles in a triangle add up to 180° so
DCE = 180° – 52° – 70° = 58° = ACB [1 mark]
[2 marks available in total — as above]
Other methods are possible — any sensible method is fine.
4 Exterior angle = 180° – interior angle = 180° – 150° = 30°
Exterior angle = 360 ÷ n, so n = 360° ÷ exterior angle.
n = 360° ÷ 30° = 12, so the polygon has 12 sides.
[2 marks available — 1 mark for a correct method,
1 mark for the correct answer]
Page 127 (Warm-up Questions)
1
2
3
4
5
6
(5, 3), (6, 3), (6, 5)
The equation of the mirror line
Rotation 90° anticlockwise about point (–2, –1)
An enlargement of scale factor 3, centre (–2, 5)
Congruence
Scale factor = 2.5
Page 128 (Practice Questions)
2 a) and b)
y
C
4
C1
3
2
A
C2
1
A1
B1
-4
-1 0
-1
-3 -2
-2
B
x
1
2
A2
3
4
5
B2
[1 mark for
each triangle]
Answers
198
c)
( –31 ( [1 mark]
3 a) (i) y = 0 [1 mark]
(ii) y = x [1 mark]
(iii) x = 1 [1 mark]
b) Rotation 90° clockwise about the origin
[2 marks available — 2 marks for all details correct,
otherwise 1 mark for 2 details correct]
4 a) A and C [1 mark]
b) A, C and D [1 mark]
Page 134 (Warm-up Questions)
1
4 cm
4 cm
6 Split the triangle in
half to give two
right-angled triangles:
x
4 cm
2 cm
Use Pythagoras to find the height (x):
x2 + 22 = 42
x2 = 42 – 22 = 16 – 4 = 12
x = 12 = 3.464... = 3.5 cm (1 d.p.)
[3 marks for correct answer, otherwise 1 mark for
splitting the triangle in half and 1 mark for
substituting the numbers into Pythagoras’ theorem]
7 tan 55° = 9h
h = 9 × tan 55° = 12.853... = 12.9 m (1 d.p.)
[3 marks for correct answer, otherwise 1 mark for
using the correct trig formula and 1 mark for
putting the numbers in correctly]
Page 137 (Revision Summary)
3 cm
2 See p130
3 13 cm
4 If the triangle is right-angled, then 82 + 152 = 172. 82 + 152
= 64 + 225 = 289 = 172, so the triangle is right-angled.
5 a) x
b) y
c) z
6 CAH (cos x = adj ÷ hyp)
Pages 135-136 (Practice Questions)
2
A
B
C
a) [2 marks available — 1 mark for correct arcs, 1 mark
for drawing straight lines joining B and C to the
point where they cross and labelling this point A]
b) [2 marks available — 1 mark for one correct pair of
compass arcs, 1 mark for bisector accurately drawn]
2
3 10 = 32 + x2
x2 = 102 – 32 = 100 – 9 = 91
x = 91 = 9.539... = 9.54 m (3 s.f.)
[2 marks available — 1 mark for using the correct
method, 1 mark for the correct answer]
A
4
1 Lines of symmetry = 2, Order of rotational symmetry = 2
2 Rhombus & parallelogram
3 Heptagon
4 Perimeter = 32 cm, area = 55 cm2
5 30 cm2
6 Area = 50.3 cm2 (1 d.p.), Perimeter = 28.6 cm (1 d.p.)
7 Faces = 5, edges = 9, vertices = 6
8 Surface area = 62 cm2, Volume = 30 cm3
9 Surface area = 2prh + 2pr2
10 100.5 cm3 (1 d.p.)
11 a) E.g. 72° (any value from 0°-89°)
b) E.g. 111° (any value from 91°-179°)
c) E.g. 260° (any value from 181°-359°)
12 360°
13 Alternate angles
14 x = 40°
15 Exterior angle = 36° Interior angle = 144°
16 720°
17 The angle of rotation, the direction of rotation and the
centre of rotation.
18 (5, 3), (14, 3) and (11, 12)
19 SSS, AAS, SAS, RHS
20 All the angles match up.
The sides are all enlarged by the same scale factor.
21 See p129 for method
22 E.g.
×B
C
[2 marks available — 1 mark for one correct pair of
intersecting arcs, 1 mark for perpendicular line]
opp
5 sin x = hyp = 12..56 = 0.576... [1 mark]
x = sin–1(0.576...) [1 mark] = 35.234...°
= 35.2° (1 d.p.) [1 mark]
[3 marks available in total — as above]
Answers
23 7.1 cm (1 d.p.)
24 cos x = 7 ÷ 10 = 0.7
x = cos–1(0.7) = 45.6° (1 d.p.)
Section Six — Probability and Statistics
Page 141 (Warm-up Questions)
1 0 (it is impossible)
1 4 a) 10
2
8
b) 12 3
0.3
c) 52
199
5 H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6
6 12
7 Red = 0.43, blue = 0.24, green = 0.33
4 a)
x
90
Page 142 (Practice Questions)
Page 145 (Warm-up Questions)
1 a)
x
24
2
Page 150 (Warm-up Questions)
1 a) Discrete
b) Qualitative
c) Continuous
2 The bar chart with shoe sizes along the horizontal axis
has gaps between bars because the data is discrete.
3 a) 60°
b) 12
Page 151 (Practice Questions)
2
K
24
b) See K on Venn diagram
c) See Q on Venn diagram
1 d) (i) 12 (ii) 26
9
8
7
6
5
4
3
2
1
0
1
(iii) 26
Page 146 (Practice Questions)
2 a) P = {1, 4, 9, 16} [1 mark]
Q = {1, 2, 4, 5, 10, 20} [1 mark]
[2 marks available in total — as above]
b) P , Q = {1, 2, 4, 5, 9, 10, 16, 20} [1 mark]
c) P + Q = {1, 4}, so n(P + Q) = 2 [1 mark]
x
E
Cheese Salt & BBQ
& Onion Vinegar Beef
Plain
Prawn
Cocktail
Spain
13
(13 ÷ 45) × 360° = 104°
France
USA
Other
Total
10
7
3
45
(10 ÷ 45) × 360° = 80°
(7 ÷ 45) × 360° = 56°
(3 ÷ 45) × 360° = 24°
360°
[3 marks available — 1 mark for using a correct
method to find the angles, 2 marks for all 5 angles
correct, or 1 mark for at least 3 angles correct]
b)
F
USA
r
Othe
3 a)
Girls
Boys
[3 marks available — 1 mark for all bars correct,
1 mark for axes labelled correctly and 1 mark for
gaps between pairs of bars]
3 a)
Destination Frequency
Angle of sector
UK & Ireland
12
(12 ÷ 45) × 360° = 96°
2
Q
80
[3 marks for all 4 numbers correct, otherwise 2 marks
for 3 numbers correct or 1 mark for 2 numbers correct]
b) Number of students studying music or art
= n(M , A) = 90 + 30 + 80 = 200 [1 mark]
B
A
30
100
Number of people
2
2 a) There are 2 Ms out of 11 letters, so P(M) = 11
[1 mark]
b) There are 4 As or Ts out of 11 letters,
4 [1 mark]
so P(A or T) = 11
4
c) There are 4 vowels out of 11 letters, so P(vowel) = 11
[1 mark]
d) There are no Xs, so P(X) = 0 [1 mark]
3 a) P(5) = 16 [1 mark]
b) Expected frequency = P(5) × 300 = 16 × 300 = 50
[1 mark]
4 P(on time) = 1 – P(not on time) = 1 – 0.22 = 0.78 [1 mark]
5 Number of blue balls = 18 × 24 = 3
Number of red balls = 14 × 24 = 6
Number of green balls = 24 – 3 – 6 = 15
[2 marks available — 1 mark for a correct method,
1 mark for the correct answer]
A
M
UK &
Ireland
France
Spain
[1 mark]
b)
x
E
F
[3 marks available — 3 marks for all sectors drawn
correctly, otherwise 2 marks for 3 or 4 sectors drawn
correctly, or 1 mark for 2 sectors drawn correctly]
Page 156 (Warm-up Questions)
[1 mark]
1 Median = 11, mode = 12, mean = 11, range = 7
2 Median is halfway between 4th and 5th numbers:
(12 + 18) ÷ 2 = 15
3 First 5 prime numbers = 2, 3, 5, 7, 11 Range = 11 – 2 = 9
Answers
200
4
5
6
7
Total = 8 × 4 = 32 Missing number = 32 – 5 – 7 – 8 = 12
Mode = 1, median = 1, mean = 1.54 (2 d.p.), range = 3
15
Negative correlation
Pages 157-158 (Practice Questions)
50
French marks
1 A probability of 1 means that a thing is certain to happen.
5
2 12
3 P(lose) = 0.9
Second spin
4
Black White
Black
BB
BW
White
WB
WW
First spin
2 a)
Page 159 (Revision Summary)
40
30
5 10
Frequency
6 Relative frequency = Number of times you tried
20
10
0
7 a) X , Y = {1, 2, 3, 4, 5, 7, 8, 11}
b)
10 20 30 40 50
Mathematics marks
x
[3 marks for all 10 points plotted correctly, otherwise
2 marks for 6-9 points plotted correctly, or 1 mark
for 3-5 points plotted correctly]
b) E.g. See graph [1 mark for an appropriate best fit line]
c) Positive correlation [1 mark]
d) 37 [1 mark for any answer between 36 and 38 that
matches your line of best fit]
3 Total score = 45 × 3 = 135 [1 mark]
Score for first two games = 42 + 49 = 91
Score for third game = 135 – 91 = 44 [1 mark]
[2 marks available in total — as above]
4 a) First line the data up in order:
3, 3, 4, 5, 8, 9, 9, 9, 11, 12, 12, 15, 18, 20, 22
Median is the 8th value = 9 [1 mark]
Hint — find the position of the median by adding on 1,
then dividing by 2, i.e. (15 + 1) ÷ 2 = 8.
b) Mode = 9 [1 mark]
c) Mean = (3 + 3 + 4 + 5 + 8 + 9 + 9 + 9 + 11 + 12 + 12
+ 15 + 18 + 20 + 22) ÷ 15
= 160 ÷ 15 = 10.67 (2 d.p.)
[2 marks available — 1 mark for correct method,
1 mark for correct answer]
d) Range = 22 – 3 = 19 [1 mark]
5 a)
Number of
Number of letters
Frequency
letters
× frequency
6
prime factor
3 5 2 of 8 8
3
7 11
1 4
9
10
12
X + Y = {2}
p(not a prime
number nor a
factor of 8) = 13
8 Primary data is data you’ve collected yourself.
Secondary data is collected by someone else.
9 Qualitative
10 Continuous
30 × 120 = 10
11 360
12 You should plot the mid-interval value on the
horizontal axis and the frequency on the vertical axis.
13 Mode = 5, Median = 5, Mean = 6, Range = 9
14 12
15 Mode = 2, Median = 2, Mean = 2, Range = 4
16 Modal class = 21-30 Median class = 21-30
17 E.g. height and weight of people.
Section Seven — Exam Practice
Page 160 — Mixed Practice Test 1
1
2
4
6
A number that can only be divided by one and itself.
D
3 D
(0, 0)
5 C
Data that can be recorded exactly / can only take
certain values.
7 The total distance around the edge of a shape.
8 A
9 5
y+6
10
2
3
4
12
4
3
12
5
5
25
6
6
36
7
4
28
Page 161 — Mixed Practice Test 2
8
2
16
1
3
4
5
6
7
9
9
1
9
Total
25
138
[2 marks available — 2 marks for all values correct,
otherwise 1 mark for 5-8 values correct]
b) Mean = 3rd column total ÷ 2nd column total
= 138 ÷ 25 = 5.52 letters
[2 marks available — 1 mark for a correct method,
1 mark for the correct answer]
c) (25 + 1) ÷ 2 = 13, so median = 13th value = 6 letters
[2 marks available — 1 mark for a correct method,
1 mark for the correct answer]
Answers
(0, 5)
2 C
B
The biggest number that will divide into both numbers.
C
Data you’ve collected yourself.
C
8 A
300
10 a2 + a + 3
Pages 162-163 — Mixed Practice Test 3
1
3
5
7
9
(4, 3)
B
D
0.2
11-15
2 8
4 C
6 9.8
8 C
10 C
201
Page 164 — Mixed Practice Test 4
1
2
4
6
8
3 a)
b)
c)
A = ½ × b × h (or Area = ½ × base × height)
C
3 1
C
5 360°
1:3
7 a25
D
9 10x + 25
10 A
Page 165 — Mixed Practice Test 5
4 a)
1 80
2 C
2
3 V = pr h (or Volume = p × radius2 × height)
4 5
5 A
6 Numbers that can’t be written as fractions.
7 8
8 C
9 sum of interior angles = (n – 2) × 180°
10 (5,9)
Pages 166-167 — Mixed Practice Test 6
1
3
5
7
9
9 and –9
42%
B
C
1.5 m/s
21, 26 [1 mark]
The sequence is given by the rule ‘add 5 each time’.
49, 64 [1 mark]
The sequence is given by the rule (n + 2)².
(4 × 9) + 8 = 36 + 8 = 44
[2 marks for correct answer, otherwise 1 mark for
correctly substituting 9 into the rule]
2 7
4 A
6 A
8 2p(p + 2r)
10 24 × 7 (or 2 × 2 × 2 × 2 × 7)
Page 168 — Mixed Practice Test 7
C
2 A
The total area of all the faces of the shape added together.
A
The x and y values of the point where the graphs cross is
the solution to both equations.
6 A straight line through the origin.
frequency
7 relative frequency =
number of attempts
8 x = 7
9 B
10 8
1
2
3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
7
8
9
4
5
6
7
8
9
10
5
6
7
8
9
10 11
6 7 8 9 10 11 12
[1 mark]
b) There are 3 ways to score a 10 out of 36 results.
Probability = 3/36 or 1/12
[2 marks for correct answer, otherwise 1 mark for
stating that there are 3 ways to score a 10]
5 a) and b)
3 y
1
3
4
5
828 000
Data that is descriptive and doesn’t use numbers.
A = b × h ( or Area = base × height)
D
5 B
change in value
6 percentage change =
× 100
original value
7 200 minutes
8 A
9 No
10 B
Pages 170-179 — Practice Paper 1
1
1
= 0.25, so the order is: 0.05, 0.15,
, 0.4
4
4
[2 marks for correct answer, lose a mark for missing
a number or putting one number in the wrong place]
2 a)
8
Number of pupils
1
Girls
Boys
6
4
2
2
3
B
1
x
-3
-1
-2
0
1
2
3
-1
-2
-3
1
2
3
4
1
2
C
Page 169 — Mixed Practice Test 8
0
A
4
or more
Number of pets
[1 mark for bar for boys with 3 pets correctly added]
b) 2 + 6 + 2 = 10 [1 mark]
c) 2 + 1 + 6 + 7 + 2 + 2 + 3 + 2 + 1 + 1 = 27 [1 mark]
[1 mark for B correct, 1 mark for C correct]
6 a) 4, 5, 6, 7, 8 [1 mark]
b) –1, 0, 1, 2, 3, 4 [1 mark]
c) –4, –3, –2, –1, 0 [1 mark]
7 a) £5.00 – £1.65 = £3.35 [1 mark]
b) E.g. 2 bookmarks: £1.45 × 2 = £2.90
2 mugs: £2.90 × 2 = £5.80
1 postcard = £0.60
£2.90 + £5.80 + £0.60 = £9.30
[2 marks for correct answer, otherwise 1 mark for
using any correct method]
You could have done this by just adding up all five
prices instead — that would be fine too.
8 a) x2 = 32 + 42
x2 = 9 + 16 = 25
x = 25 = 5 cm
[2 marks for correct answer, otherwise 1 mark for
substituting the numbers into Pythagoras’ theorem]
b) x2 + 82 = 102
x2 = 102 – 82
x2 = 100 – 64 = 36
x = 36 = 6 cm
[2 marks for correct answer, otherwise 1 mark for
substituting the numbers into Pythagoras’ theorem]
Answers
202
9 a) 4(3x + 2) – 6(x – 5)
= 12x + 8 – 6x + 30
= 6x + 38
[2 marks for correct answer, otherwise 1 mark for
correctly multiplying out both brackets]
b) (3x + 3)2
= (3x + 3)(3x + 3)
= 9x2 + 9x + 9x + 9
= 9x2 + 18x + 9
[2 marks for correct answer, otherwise 1 mark for
correctly multiplying (3x +3) by (3x + 3)]
10 3 + 5 = 8, so 1 part = 120 ÷ 8 = 15 [1 mark]
15 × 3 = 45 parents [1 mark]
15 × 5 = 75 children [1 mark]
[3 marks available in total — as above]
1
8
11 a) i) =
3
24
1
The probability that the pen will be blue is .
3
[1 mark]
ii) Number of red pens = 24 – 14 – 8 = 2
The probability that the pen will be red is
2
1
or
. [1 mark]
24
12
b) Salma takes 2 blue pens and 1 red pen, so there are
24 – 2 – 1 = 21 pens left and 14 are black.
14
2
Probability of picking a black pen =
or [1 mark]
21
3
12 a) 2 + 6 + 15 + 8 + 5 + 3 + 5 + 12 = 56
56 ÷ 8 = 7
[2 marks for correct answer, otherwise 1 mark for
adding up all of the values correctly]
b) 15 – 2 = 13 [1 mark]
c) New total films watched
= new mean × number of values = 8 × 9 = 72
New value = difference between old total and new total
= 72 – 56 = 16
[2 marks for correct answer, otherwise 1 mark for
finding the new total number of films watched]
13 a) 84 300 [1 mark]
b) Minimum: 17 350 [1 mark]
Maximum: 17 449 [1 mark]
14 a) WVX = 125° because it is a corresponding angle
with SRX. [1 mark]
b) UVX = 180 – 125 = 55° because the angles along a
straight line add up to 180°.
QXR = UXV = 180° – 55° – 60° = 65° because angles
in a triangle add up to 180°.
[2 marks for correct answer with sensible reasoning,
otherwise 1 mark for correct answer with no reasoning
or for an incorrect answer with some correct working]
15 a) p = (2 × 6) + 42
p = 12 + 16
p = 28 [1 mark]
b) 2q = p – r2 [1 mark]
p – r2
q=
[1 mark]
2
[2 marks available in total — as above]
c) r² = p – 2q [1 mark]
r = p – 2q [1 mark]
[2 marks available in total — as above]
Answers
(a + b)
×h
2
(3x + 1) + (4x + 3)
72 =
× 8 = 4(7x + 4)
2
18 = 7x + 4
14 = 7x
x=2
[3 marks for correct answer, otherwise 1 mark for
correctly substituting the values into the equation
and 1 mark for correctly simplifying the equation]
There are a few ways you could have done the working here.
It’s fine if yours looks different to this, as long as it’s sensible.
17 a) Scale factor: 36 ÷ 12 = 3.
Length of p = 33 ÷ 3 = 11 cm
[2 marks for correct answer, otherwise 1 mark
for correctly working out the scale factor]
b) Angle q = 110° [1 mark]
Similar shapes have the same angles.
m 4 × ^m 2h5 × m –2
m 4 × m10 × m –2 m12
18 a)
=
= 8 = m4 [1 mark]
^m 2h4
m8
m
When m = 2, m4 = 24 = 2 × 2 × 2 × 2 = 16 [1 mark]
[2 marks available in total — as above]
^ y 6h2 × y –4
y12 × y –4
y8
b)
=
= 5 = y3 [1 mark]
5
0
5
1× y
y
y ×y
When y = 4, y 3 = 43 = 4 × 4 × 4 = 64 [1 mark]
[2 marks available in total — as above]
16
A=
Pages 180-189 — Practice Paper 2
1 a) 75 centimetres, 0.75 metres [1 mark]
b) 31000 cm3 [1 mark]
2 a)
b) E.g.
[1 mark]
[1 mark for any correct answer]
3 a) 360 – 140 – 90 – 80 = 50, so the angle is 50° [1 mark]
b) E.g. 140° = 84 members
members represented by 1° = 84 ÷ 140 = 0.6
members represented by 360° = 0.6 × 360 = 216
[2 marks for correct answer, otherwise 1 mark for
correct working]
You could also have done this by finding the angle that
represents 1 member (140 ÷ 84), and then dividing 360
by that to find the total number of members.
4 a) 6x + 9 = 57
6x = 48
x = 8 [1 mark]
b) 3(3x – 6) = 54
9x – 18 = 54
9x = 72
x = 8 [1 mark]
c) 4x – 7 = 5x – 2
–7 = x – 2
x = –5 [1 mark]
40
5
5 a)
=
[1 mark]
64
8
203
8 5 8 11 88
÷
= ×
=
[1 mark]
9 11 9 5 45
6 a) 7s + 5 [1 mark]
b) 5t² + 7t [1 mark]
c) 25u² [1 mark]
d) 3v [1 mark]
7 a) exterior angle = 360° ÷ number of sides
= 360° ÷ 15 = 24° [1 mark]
Remember, this formula only works for regular polygons.
b) number of sides = 360° ÷ exterior angle
= 360° ÷ 40° = 9
[2 marks for correct answer, otherwise 1 mark for
correctly rearranging the formula]
8 a) E.g. circumference = p × diameter = p × 1.8 = 5.65 m
[2 marks for correct answer, otherwise 1 mark for
using a correct formula for circumference]
It’s fine if you said r = d ÷ 2 = 0.9, then used C = 2pr.
b) Joe will need 6 pieces, so 6 × 3.20 = £19.20
[2 marks for correct answer, otherwise 1 mark for
saying that Joe needs six pieces of edging]
9 a) C = 11n + 34 [1 mark]
b) E.g. 210 = 11n + 34
176 = 11n
n = 16, so she goes to the gym 16 times.
[2 marks for the correct answer, otherwise 1 mark for
correctly substituting the numbers into the formula]
10 a) Words typed (w)
b)
in 1 min
b)
11 a)
b)
12 a)
b)
13 a)
b)
Frequency (f) Midpoint (x)
fx
30 < w ≤ 40
24
35
840
40 < w ≤ 50
21
45
945
50 < w ≤ 60
7
55
385
60 < w ≤ 70
3
65
195
Total
55
–
2365
[2 marks available — 1 mark each for Midpoint
column and fx column correctly completed]
2365 ÷ 55 = 43 words per minute [1 mark]
P(mints) = P(toffee) × 2 = 0.15 × 2 = 0.3
P(chocolate) = 1 – 0.3 – 0.15 – 0.25 = 0.3
[2 marks for correct answer, otherwise 1 mark for
correctly finding the probability of choosing mints]
Number of pupils expected to buy:
Fudge = 0.25 × 320 = 80
Toffees = 0.15 × 320 = 48
Difference = 80 – 48 = 32 pupils
[3 marks for correct answer, otherwise 1 mark for
finding pupils expected to buy fudge and 1 mark
for finding pupils expected to buy toffees]
E.g. split the shape into a rectangle and a triangle.
Rectangle: A = l × w = 4 × 5 = 20 cm²
Triangle : A = ½ × b × h = ½ × 6 × 8 = 24 cm²
Total area = 20 + 24 = 44 cm²
[2 marks for correct answers, otherwise 1 mark
for using any correct method]
Volume = cross-section area × length
= 44 × 15 = 660 cm² [1 mark]
24 500 000 [1 mark]
8.33 × 106 = 8 330 000
24 500 000 + 8 330 000 = 32 830 000 = 3.283 × 107
[2 marks for correct answer, otherwise 1 mark
for correct answer not in standard form]
14 a)
4
3
2
1
–3 –2 –1 0
–1
–2
1 2 3 4
[2 marks available —
1 mark for correctly
plotting all 3 points,
and 1 mark for a
correctly drawn line]
b) Gradient of line = change in y ÷ change in x
E.g. gradient = (4 – 1) ÷ (1 – (–2)) = 3 ÷ 3 = 1
[2 marks for correct answer, or 1 mark for using
the correct method]
You could use any 2 points on the line to find the gradient.
c) y = x + 3 [1 mark]
The gradient is 1 and the line crosses the y-axis at y = 3.
15 E.g. A: 490 ÷ 2.89 = 169.55... ml per pound
B: 1.25 litres = 1250 ml
1250 ÷ 3.55 = 352.11... ml per pound
C: 800 ÷ 3.10 = 258.06... ml per pound
B is the best value because you get the most ice
cream per pound. [3 marks available — 1 mark
for converting all the data into the same units,
1 mark for using any correct method to find price
per unit and 1 mark for the correct answer]
There are lots of other ways to give the price per unit here,
e.g. you could say how much each ice cream costs per litre.
As long as you do the same for all three, that’s fine.
16 a) speed = distance ÷ time
34 ÷ 50 = 0.68 miles per minute
0.68 × 60 = 40.8 miles per hour
[2 marks for correct answer, otherwise 1 mark for
correct working]
If you did this by converting 50 mins to hours,
then putting that into the speed formula, that’s fine.
b) time = distance ÷ speed
34 ÷ 45 = 0.755... hours
0.755... × 60 = 45.33... minutes = 45 minutes
[2 marks for correct answer, otherwise 1 mark for
correct working]
If you did this by converting 45 mph to miles per minute,
then putting that into the time formula, that’s fine.
17 a) 100% – 13% = 87%, so £348 is 87% of original price.
1% = 348 ÷ 87 = £4, so 100% = 4 × 100 = £400
[2 marks for correct answer, otherwise 1 mark for
stating that £348 is 87% of the original price]
b) Jan is wrong. E.g. new price is 90% of the sale price.
90% of 87% is 0.90 × 0.87 = 0.783 = 78.3%. The new
price is 100 – 78.3 = 21.7% less than the original price.
[2 marks available — 1 mark for saying Jan is
wrong, 1 mark for any sensible explanation of why]
18 a) (AB)² = 7² + 24²
(AB)² = 49 + 576 = 625
AB = √625 = 25 cm
[2 marks for correct answer, otherwise 1 mark for
substituting the numbers into Pythagoras’ theorem]
b) E.g. tan a = 7 ÷ 24
a = tan–1 (7 ÷ 24) = 16.26... = 16.3°
[2 marks for correct answer given to 1 decimal place,
otherwise 1 mark for correct answer not given to
1 decimal place,]
As you know the length of the hypotenuse (from part a),
you could have used sine or cosine here instead.
Answers
204
Index
acute angles 115
addition 4
algebra 37-41
factorisation 41
multiplying out brackets
39-40
simplifying 37-38
allied angles 117
alternate angles 117
angle bisectors 130
angles 115-119
in a quadrilateral 116
in a triangle 116
on a straight line 116
round a point 116
arcs 106
area 104-106
averages 152-154
B
bar charts 148
best buy 94
bias 140
BODMAS 1-2
C
calculators 2
chords 106
circles 106
circumference 106
complement of a set 143
congruent shapes 125
constructions 129-130
continuous data 147
conversion graphs 68
conversions 19, 88-90
area and volume 90
fractions, decimals and
percentages 19
metric-imperial 88-89
speed 90
coordinates 57
correlation 155
corresponding angles 117
cube numbers 13
cube roots 32
cubes/cuboids 110
cylinders 109, 111
D
data 147
decimal places 26
decimals 19
density 95
diameter 106
direct proportion 79
Index
discrete data 147
division 6, 8-9
decimals 9
without a calculator 8
E
enlargements 124
equilateral triangles 101
estimating 28
expected frequency 139
exterior angles 119
F
factorisation 41
factors 15-16
factor trees 15
formulas 46-48
formula triangles 95, 132
fractions 2, 19-22
frequency bar charts 149
frequency polygons 149
frequency tables
149, 153-154
G
gradients 58, 62-64
graphs 57-76
grouped frequency tables
154
H
highest common factor 16
hypotenuse 131-133
I
imperial units 88
improper fractions 22
inequalities 53
integers 13
interior angles 119
intersections 144
inverse proportion 80
irrational numbers 13
isosceles triangles
101, 116
K
kites 100
L
lowest common multiple
16
line symmetry 99
M
map scales 91
mean 152-154
median 152-154
metric units 88
midpoint of a line 57
mixed numbers 22
mode 152-154
multiples 15-16
multiplication 5-9
decimals 9
without a calculator 7
right-angled triangles
101, 131-133
right-angles 115
roots 32
rotational symmetry 99
rotations 123
rounding errors 28
rounding numbers 26-27
N
S
negative numbers 10, 38
nets 110-111
sample space diagrams
139
scale drawings 91
scale factors 124, 126
scalene triangles 101
scatter graphs 155
sectors 106
segments 106
sequences 51-52
sets 143
significant figures 27
similar shapes 126
simple interest 84
simultaneous equations
72
SOH CAH TOA 132-133
solving equations 44-45
speed 95
square numbers 13
square roots 32
squares 100
standard form 33
straight line graphs 58-64
subtraction 4
surface area 110-111
symmetry 99
O
obtuse angles 115
ordering numbers 3, 20
P
parallel lines 64, 117
parallelograms 100
percentage change 83-85
percentages 19, 23, 83-85
perimeter 104
perpendicular bisectors
130
perpendicular lines 117
pie charts 148
polygons 101, 119
powers 13, 31
prime factors 15
prime numbers 14
prisms 111, 112
probability 138-139, 144
proportional division 78
protractors 115
pyramids 109, 111
Pythagoras’ theorem 131
Q
quadratic graphs 73
quadrilaterals 100, 116
qualitative data 147
quantitative data 147
R
radius 106
range 152-153
rational numbers 13
ratios 77-78
real-life graphs 69
real numbers 13
rearranging formulas 48
rectangles 100
reflections 122
reflex angles 115
regular polygons 101, 119
relative frequency 140
rhombus 100
T
tangents 106
transformations 122-124
translations 122
trapeziums 100
travel graphs 68
triangles 101, 116, 129
triangular prisms 111
trigonometry 132-133
U
unions 144
universal set 143
V
vectors 122
Venn diagrams 143
volume 112
Y
y = mx + c 63, 64
MHS34
A
P
G
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