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Maths for electrotechnology
Copyright B&B Training Associates Ltd Page No.3
No part of this publication may be reproduced, stored in a retrieval system or
transmitted, in any form or by any means, without written prior permission
from B&B Training Associates Ltd.
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For further information contact:
Mrs A Bratley
23 St Pauls Drive,
Tickton, Nr Beverley,
East Yorkshire
HU17 9RN
Tel:-01964 543137
Fax:-01964 544109
Email:- [email protected]
Version 1-2008
mailto:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]7/27/2019 Maths for 2330 2008.pdf
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Disclaimer
The documents and presentations that make up this learning package do not constitute either
fully or partially the full and final information on all aspects of the City and Guilds 2330
Certificate of Electrotechnical Technology. They do not show full representation of Health and
Safety aspects or full knowledge of any job descriptions.
Every effort has been made to ensure accuracy and up-to date information but no responsibility
can be accepted for any errors, misleading information or omissions.
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Table of contents
Table of contents ........................ ......................... ........................ ........................ ......................... ......... ................. 5Table of figures ....................... ....................... ....................... ....................... ...................... ................. ..................... 6Aims and objectives ............................................................................................................................................... 7Multiples and standard form .............................................................................................................................. 8Mathematics .......................................................................................................................................................... 14Basic operations ................................................................................................................................................... 14Fractions .................................................................................................................................................................. 19Adding ..................................................................................................................................................................... 19Subtraction ............................................................................................................................................................. 22Improper fractions ............................................................................................................................................... 24Multiplication ........................................................................................................................................................ 26Division .................................................................................................................................................................... 27Decimals, ratios and percentages .................................................................................................................. 30Decimals .................................................................................................................................................................. 30Ratio and proportion .......................................................................................................................................... 34Percentages ........................................................................................................................................................... 41Algebra .................................................................................................................................................................... 51
Transposing formulae .................... ..................... .................... .................... .................... ..................... .............. 54Indices ...................................................................................................................................................................... 61
Trigonometry ...................... ....................... ...................... ....................... ....................... ................. ...................... . 65Areas and volumes .............................................................................................................................................. 70Areas ......................................................................................................................................................................... 70Rectangle ................................................................................................................................................................ 71Circle ......................................................................................................................................................................... 75
Triangles ..................... ........................ ........................ ....................... ........................ ............. ..................... ............ 77Volume ..................................................................................................................................................................... 84Cubes ........................................................................................................................................................................ 84Cylinders .................................................................................................................................................................. 85
Triangles ...................... ....................... ........................ ....................... ........................ ............. ...................... ........... 87Statistics .................................................................................................................................................................. 91
Tally diagrams .................. ................... ................... ................... .................... ................... ..................................... 92Mode......................................................................................................................................................................... 94Mean ......................................................................................................................................................................... 96Median ..................................................................................................................................................................... 99
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Charts and diagrams ......................................................................................................................................... 101Bar charts .............................................................................................................................................................. 102Pie chart ................................................................................................................................................................. 104Conclusion ............................................................................................................................................................ 106Appendix 1 ........................................................................................................................................................... 107
Table of figures
Figure 1 Engineering standard form .......................... ........................... .......................... .......................... ....... 9Figure 2 Right-angled triangle ........................................................................................................................ 65Figure 3 The square on the hypotenuse is equal to the sum of the squares on the other two sides
........................................................................................................................................................................... 66Figure 4 Relationship of angle to sides ........................................................................................................ 67Figure 2 Rectangle ............................................................................................................................................... 71Figure 3 Area of rectangle ................................................................................................................................ 71Figure 4 Area differences .................................................................................................................................. 72Figure 5 Parallelogram or lozenged rectangle ........................................................................................ 74Figure 6 Square ..................................................................................................................................................... 74Figure 7 Circle ........................................................................................................................................................ 75Figure 8 Triangle .................................................................................................................................................. 77Figure 9 Cube ........................................................................................................................................................ 84Figure 10 Cylinder ................................................................................................................................................ 85Figure 11 Triangular solid-prism .................................................................................................................... 87Figure 12 Bar chart 1 ......................................................................................................................................... 102Figure 13 Bar chart 2 ......................................................................................................................................... 103Figure 14 Pie chart 1 ......................................................................................................................................... 104
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Aims and objectives
By the end of this study book you will have had the opportunity to:
x use the common engineering prefixes milli, micro, kilo, Mega etc.x be able to use the common maths operatorsx add, subtract, multiply and divide fractionsx deal with improper fractionsx use algebrax transpose simple equations and formulaex use indicesx calculate areas and volumes of basic shapes-square, rectangle, circle triangle etc.x
make use of basic statistics, using terms such as mean, median, mode and range.
This booklet is intended to act as support for you as you progress through your course. It would
be well worth your while to attempt each section first, and then try the end assignment before
doing anything else. Dont take any shortcuts, as you need to be able comfortable with each of
the areas that are covered.
I wont say have fun, but do work conscientiously and you will find much of what follows
manageable.
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Chapter
1
Multiples and standard form
In this session you will have the opportunity to:
x understand the basic principles of engineering standard form?
Numbers are a strange concept. They define for us size, quantity, direction and so many other
things that we can forget what it is that we are dealing with. In engineering of type maths is a
prerequisite. You cannot do any form of engineering without a reasonable level of mathematical
knowledge and so this unit is essential.
Numbers use in engineering can often be either very small or very large, and so we can have
currents flowing in the order of 0.000 000 01 amps, or energy transferred in the order of
50 000 000 000 000 joules. You can see that the numbers are either too small or too large to be
easy to use. This is why we start to use prefixes (letters before the numbers) to give us multiples
of a thousand.
In Figure 1 over the page most of the common engineering multiples can be seen.
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Figure 1 Engineering standard form
There are more prefixes than these shown in Figure 1, but these are the most commonly used in
engineering. It is important that you are familiar with these terms. Follow the examples and then
try some for yourself.
Note that the common thing is that they are all based on thepower 10. This is usually
written as:
10
Where any number
any power
nm
m
n
u
Notice how easy it is to change the decimal point and to add the power or to use the prefix
(Remember that the prefix is the letter used to signify the amount).
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A number written with one digit (one and only one digit) to the left of the decimal point,
multiplied by ten which is then raised to some power is said to be written instandard form. So:
i/. 35837 5.837 10 u
ii/. 20.0415 4.15 10 u
iii/. 439400 3.94 10 u
Notice that I have used only one digit. The number of decimal points moved is the number used
for the index. If the decimal point is moved towards the right then we have a negative number,
and if we move it to the left, we have a positive index.
Dont get confused between standard form and the prefixes and numbers that we use in
engineering. With standard form, we have only one number in front of the decimal point. With
engineering form, we often have more than one number in front of the decimal point.
We have looked at some basic standard form conversions. These are less commonly used in an
engineering context. We normally make use of the engineers variant which bases everything on
multiples of 1 000. Refer to Figure 1.
Lets have a look at a few more examples.
With each of the examples shown, convert the number to its whole number.
i/. 22kW; ii/. 340mm; iii/. 220.
i). ;322kW 22 10 W 22000W u
ii). ;3340mm 340 10 m 0.34m u
iii). .6220 22 10 0.00022P : u : :
Remember that each of the prefixes, the letters, have a meaning. Familiarity comes with practice.
It may be a good idea to get into the habit of writing out the table of values each time you come
to a session as a revision aid.
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However, try the following examples, and complete the table.
Name pico Giga milli
Letter k M n
Multiplier6
10
u
If youre not sure of the answers have a look in the appendix.
Well now have a brief review of what you have covered, and what you should have learnt.
Summary
You know the fundamental s.i. units.
You know that s.i. units are used to provide a common basis.You know the common engineering multiples and are able to use them.
You can recognise the difference between standard form and engineering standard form.
Try the exercise below.
Exercise 1.
Express each number in standard form and as a prefix.
1) i) 2 200 000 ii) 10 450 m
iii) 0.020 A
iv) 0.000 034 5 A
v) 567 000 W.
Express the following as whole numbers.
2) i) 0.37 kW
ii) 8 Fiii) 55 TWh
iv) 220 k
v) 0.5 mH.
[Hint:-Dont get confused between the letters used for the unit and the letters that relate to the
number itself]. The answers are in the appendix. Try to work things out first though. Trying
things for yourself helps you with your own self-confidence. Use the answers only as a check.
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If you are using a calculator, use this routine.
1/. Type in the number, remembering to move the decimal point.
5.735
2/. Press EXP or EE. Two zeros should appear in the top right-hand corner of the
calculator.
5.735 00
3/. Type in the number of decimal places moved. In this instance, 3.
5.735 03
4/. If you now press the equals button you should find that the calculator shows 5735.
If you have a calculator with a button labelled ENG follow this routine.
1/. Type in your number just as it is.
5735
2/. Press ENG. The correct numbering should appear.
5.735 03
3/. If you want to return to your original number press INV ENG or 2ndF ENG.
5735 00
Calculatorsare very helpful but they are no substitute for you understanding what ishappening.
Try a few more examples. I have not provided any answers this time.
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3) Convert the following to engineering form to include both powers of ten and the prefix
letter.
i) 12 467 238
ii) 0.000 456 12
iii) 213
iv) 34 673
v) 44 668 822 561 849
vi) 0.000 000 000 005
vii) 0.000 000 003 41
viii) 12 613 790
ix) 45 617 284 935 178 239
x) 0.000 004 5
4) Convert the following to standard form.
i) 234
ii) 45 618
iii) 64 826 438 546 284 928
iv) 0.003 4
v) 0.000 000 000 000 000 54
vi) 0.000 056
vii) 34
viii) 546 283 647
ix) 0.023
x) 0.000 000 000 067 21
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Chapter
2
Mathematics
By the end of this session you will have had the opportunity to:
x understand and use the basic operator rulesx add, subtract, multiply and divide fractionsx use and apply decimals pointsx understand why and where ratios are usedx use and apply percentages and per-unitx understand and apply basic algebraic techniquesx transpose equationsx understand and apply indicesx use Pythagoras theorem.
Basic operationsThis study book is not intended to get you to GCSE level maths. It is an introduction into the
common mathematical skills that you will need on the course. Most of this section therefore, will
be in shorthand and will require you to recognise that Maths is a language like any other, and
that it is merely a means of communicating. Please make ample use of the margins to make any
notes that you feel necessary for your own benefit.
1/. Remember that any sign, such as +, -,q,u, is attached to the number that follows it. So,
in this instance the addition sign is attached to the fourand not to the three.
3 4 7
3 4 1
2/. If no sign is present at the front of a number then we have to assume that it is positive.
Therefore, in this instance the three is positive even though the sign is not seen, as in the
second example shown.
3 4 7
3 4 7
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3/. When adding or subtracting, ifunlike signs are together in a calculation, the overall sign
is negative. [This does not necessarily mean that the answer will be negative].
4/. For addition and subtraction, when like signs are together in a calculation, the overall
sign is positive.
3 4 3 4
5 2 5 2 3
3 4 3 4 7
1
These may appear to be very simple rules but they are very easily forgotten. So DONT!
5/. For multiplication and division, when the numbers have unlike signs, the overall sign is
negative.
4 4
3 3
3 4 12
u
6/. For multiplication and division, when the numbers have like signs, the overall sign is
positive.
4 4
3 33 4 3 4 12
u u
7/. In arithmetic operations, the order in which the operations are performed is critical:
i). to determine the values of operations contained in brackets
ii). Percentages/ratios i.e. "of"
iii). Division
iv). Multiplication
iv). Additionv). Subtraction.
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This can be shortened to an acronym, BODMAS.
Brackets
Of
Division
Multiplication
Addition
Subtraction.
Try to keep an eye on carefully the following rules.
The basic laws governing the use of brackets are as follows:
i). The use of brackets when adding does not affect the result.
2 3 4 2 3 4 2 3 4
ii). The use of brackets when multiplying does not affect the result.
2 3 4 2 3 4 2 3 4u u u u u u
iii). A number placed outside of a bracket indicates that the whole of the contents of the
bracket must be multiplied by that number.
2 3 4 2(3 4) (2 3) (2 4) 6 8 14
or
2 3 4 2 7 14
u u u
u
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iv). Adjacent brackets indicate multiplication.
(2 3)(4 5) (5)(9) 5 9 45
or
2 4 2 5 3 4 3 5
8 10 12 15 45
u
u u u u
v). When an expression contains inner and outer brackets, the inner brackets should be
dealt with first.
2 2 3 3 5
2 2 3 2 2 2 6 2 2 6
2 2 6 2 8 16
Summary
Although you may not be that comfortable with maths, you should have picked up a number of
skills by now.
You should be able to know what sign belongs to which number.
You should be able to know when a sign should change (unlike signs).
You will be able to know that there is an order in the way problems should be solved (BODMAS).
You should be able to use brackets, applying all the previous rules.
Try these problems.
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Exercise 2.
1) 2)7 4 2u 5 2 7 u 3) 7 4 2u
4) 5)4 3 2 4u 9 6 3 y 6) 2 14 3 9u
7) 8) 9 6 3 y 3 7 2 8u u 9) 7 2 3 u
10) 11)2 3 6 4 u 7 2 6 12) 4 2 3 5 6
13) 7 2 4 3 6y 14) 7 2 4 3 6u
15) 4 2 3 5 6 16) 2 3 6 4 u
17) 18)22 2 8 2y u 24 14 5 2 34 4
19) 4 4 6 4 2 y
If you are still not sure about the processes, go back through the material and see what it is that
youre unsure about before you move on. The answers are in the appendix.
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FractionsFractions are often approached with dread by a large number of people. This is a shame, as the
following of some basic steps will allow even those with little confidence to achieve sufficient
levels of proficiency for this and many other courses.
Probably the best way of gaining the necessary skills is to work through a number of examples.
Please take your time with these, as it will be time well spent. There will be plenty of worked
examples over the next few pages. Take your time!
Adding
1/. Simplify:1 3
3 8
The top number in any fraction is always called the numeratorand the bottom number is always
called the denominator. Therefore, in this instance, the numerators are 1 and 3, and the
denominators are 3 and 8.
Step 1: Find the LCM (Lowest Common Multiple) of the two denominators. That is find a number
which both denominators can be divided into. This is often called the Lowest Common
Denominator.
In this case the LCM is: 3 8 24u
Do you see how this works? All you have to do is to multiply the two denominators (bottom two
numbers) together!
What we have now is:24
Do not worry if it looks a little bare now; remember we are taking this one step at a time.
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Step 2: Divide the denominator of the first fraction into the LCM.
The LCM is 24 which is then divided by the 3 from the 13
.
124 3 8
3oo y oo
We then multiply the 8, gained by our division, by the numerator of the fraction, which in this
case is 1.
88 1
24
u oo
Notice that the addition sign can be put in straight away.
Step 3: Repeat this for the second fraction.
324 8 3
8oo y oo
Here, the denominator is 8. However, you still divide the lowest common multiple by the 8. This
gives us an answer of 3.
Then multiply the 3 by the numerator of the fraction.
8 93 3
24
u oo
Step 4: What we have now is the hard work done and the full working out would look like this:
8 1 3 31 3 8 9 173 8 24 24 24
u u
Try working through the example and see how you do. Some more examples will be given at theend of this section.
Well work through another addition before we look at subtraction.
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2/.3 5
4 8
Here the lowest common multiple is even more straightforward.
The 8 can be divided by the 4. This means that rather than multiplying the 4 and the 8 together,
and getting a LCM of 32, we can just use the 8.
8
This is the lowest common multiple. Now for the rest of the process, and remember this is
exactly what it is, a process. It can be followed every time.
This is where we are getting our top line multiplier.
38 4 2
4oo y oo
62 3
8
u oo
58 8 1
8oo y oo
6 51 5
8
u oo
Now to put everything together.
2 3 1 53 5 6 5 11 3184 8 8 8 8
u u
Dealing with improper or top-heavy fractions can seem to be difficult. In our answer we would
have been correct to state that the answer is11
8. However, it is more normal not to express an
answer in an improper form.
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Quite simply:-
3111 3
8 11 8 11 18 8
r
x The numerator (11) is divided by the denominator (8) and the correct quantity of wholenumbers is determined (1).
x The remainder becomes the new numerator (3).x The denominator stays the same (8).x The whole number (1) becomes the standalone number.
The process is always the same. Now try subtracting.
SubtractionThis is very similar to working with addition of fractions. All that you have to do is remember to
subtract rather than add, everything else is the same. Try following the procedure. This time no
explanation will be given.
1/. Simplify:2 1
3 6
Step 1: Find LCM
The lowest common multiple is a number that both the denominators will divide into. Although
3 and 6 do divide into 18 they also divide into 12 and 6. Therefore, the LCM becomes 6.3 6u
Step 2:
6 3 2 2 2 4
4
6
y oo u
Step 3:
6 6 1 1 1 1
4 1
6
y oo u
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Step 4: Again the hard work is done and the full working out should look like this.
Simplifying something is where we reduce the answer to the point where it cannot be made any
simpler.
2 2 2 12 1 4 1 3 13 6 6 6 6 2
u u
Make a careful note that although it is always useful to simplify things it is not always necessary.
In this instance, it is merely preferable.
To simplify the last example we need to find something that is common to both the numerator
and denominator. Well, the numerator is 3 and the denominator is 6, so the common value is 3
because it can divide into both 6 and 3.
1
2
33 3 1 6 3 2
6
3 1
6 2
oo y oo y
Try another example.
2/.3 5
4 8
Well work through this without showing every process. Dont forget to find the lowest common
multiple first however.
2 3 1 53 5 6 5 14 8 8 8 8
u u
Now for improper fractions.
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Improper fractions
Whilst we are still within this area, we will need to consider what is known as improper fractions.
These are fractions that mix whole numbers and fractions of numbers. They are not much more
difficult but do require a little more care. Follow the example below.
1/. Simplify:2 1
3 23 6
There are two possible ways of doing this; the most common way is to convert the whole
numbers to fractions and then to do the sum as shown in the previous two examples.
Step 1: Convert the two fractions into improper fractions.
3Since 1
3
3 9 2 9 2 11then3 3 , and3
3 3 3 3 3 3
u
6Similarly, 1
6
6 12 1 12 1 13and 2 2 = , so 2
6 6 6 6 6 6
u
What we have now is a basic addition of fractions.
Step 2:
2 1 11 13 2
3 6 3 6
3
Step 3: Find the lowest common denominator and sort out the numerators. The LCM is 6 as both
3 and 6 divide into 6.
11 2 13 1
6
u u
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Step 4: Again the hard work is done and all that is necessary is to add up and simplify. The full
working out looks like this:
2 1 11 133 2
3 6 3 62 11 1 13 22 13 35 55
66 6 6
u u
The simplifying may seem complex, but it isnt really. Follow my working out.
5535 5
6 35 6 35 56 6
r
Try another example. This time however, it will be a subtraction.
2/.2 1
3 23 6
Again, you should remember to follow the process. Everything is carried out in the same way as
before.
Convert the fractions into improper fractions.
9 2 9 2Since 3 , then33 3 3 3
11
3
12 1 12 1 13Similarly, 2 , so 2
6 6 6 6
6
We can now carry on as we would for a normal subtraction of a fraction.
12
2 1 11 133 2
3 6 3 6
2 11 1 13 22 13 9 3 11 1
6 6 6 6
u u
2
Now move on and look at multiplication of fractions.
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Multiplication
This is probably the most straightforward way of working with fractions.
1/. Simplify:3 14
7 15u
This type of fraction is probably the most straightforward to solve. You have to multiply the top
line together, and then multiply the bottom line together.
3 14 3 14 42
7 15 7 15 105
uu
u
Try another example.
2/.2 1
23 5u u
This is no more complicated than the first example. Remember to multiply the top line and then
the bottom line. The whole number acts as if it were on the top line.
2 1 2 1 2 42
3 5 3 5 15
u uu u
u
More self-assessment questions will be listed at the end of this section so dont worry about
getting in some practice. Most of you may find that your calculator does this for you. If you do
use your calculator, it will still be worth your while to try to come to terms with the principles
being laid out here.
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Division
Again, this is not a difficult operation to come to terms with. It is very similar to multiplication
but with one little twist. Try and follow the example below.
1/. Simplify:3 12
7 21y
The best way of doing this type of problem is to invert (turn upside down) the second fraction
and then to follow the same procedure as for multiplication.
3 12 3 21
7 21 7 12y u
You can see now how this problem has become less complicated. You now just have to follow
the same technique as if you were multiplying a fraction. The full working out is shown below.
3 12 3 21 63
7 21 7 12y u
3
844
3
4
Try another example.
2/.2 4
5 7y
Remember to follow the process.
2 4 2 7 2 7 14
5 7 5 4 5 4uy u u
7
2010
7
10
Notice that the process is the same.
You should also remember that the application of BODMAS still applies.
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Summary
It is worthwhile just looking at how much you have actually learned by now.
You should be able to add fractions.
You should be able to subtract fractions.
You should be able to manipulate improper (top heavy) fractions.
You should be able to multiply fractions.
You should be able to divide fractions.
You should be able to simplify fractions.
Now try the exercise below.
Exercise 3.
1)1 2
2 5 2)
2 1
9 7 3)
1 2
7 3
4)2 1
9 7 5)
2 1 2
9 7 3 6)
3 210 8
7 3
7)3 3
5 313 4
8)3 5
4 9u 9)
13 7 44 3
17 11 19u u
10)3 5
4 9y 11)
3 41
4 5y 12)
3 45
8 64y
13)1 3 9 1
2 5 15 3 y 14)
2 1 2 1 31 1
3 4 3 4 5
u y
15)
1 12 1
2 3
1 31 2
4 8
16)
1 12 9
2 61 1
3 14 12
y
Check out the appendix for any answers you are unsure about.
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If youve followed this so far, well done! Now move on. There are some more examples below.
The working out is not shown but the answers are shown in the square brackets.
1 2 111/.
3 5 153 2 41
2/.7 9 63
2 1 13/.
5 3 15
4 2 104/.
9 7 63
2 1 135/. 3 2 5
3 5 15
1 1 96/. 4 3 72 7 14
3 1 17/. 3 2 1
4 2 4
2 1 78/. 13 9 4
5 6 30
2 3 69/.
5 7 35
4 1 510/. 1
7 4 7
2 3 1411/.
5 7 15
u
u
y
14
1 2 3912/. 3 4
4 3 56
2 3 1 4 1113/. 3 2 2 3
3 4 5 5 18
2 3 2 3 2 2914/. 4 5
3 4 5 7 5 210
y
y
y
These are just examples. Take your time and try to agree with my answers.
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Decimals, ratios and percentagesIn this section, we will consider three areas that often lead to basic problems with your maths.
DecimalsWe have just finished looking at fractions. When numbers are expressed as fractions then we call
them vulgar fractions, e.g.1 5 12
2 7 13. We can also express these figures and others as decimal
fractions or decimals.
How do we go about this?
We need to understand that decimals are based on the number ten (10), and so we can simplycreate a table and fill in the columns for our decimal.
For example, the number 2459.23 has:
x Two thousandsx Four hundredsx Five tensx Nine onesx Two tenthsx Three hundredths.
1000s 100s 10s 1s 110 1100 11000
2 4 5 9 2 3 0
Here that number is expressed in a table. The decimal point is positioned between the whole
numbers, ones and higher, and the fractions, which are parts of a number. So 210 is 210 of one (1).
Now, the fraction part of that number is the part that exists to the right-hand side of the decimal
point. In the first example of ours, it is the highlighted numbers, 2459.23. Lets look at this in
more detail.
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2 30.23 means and
10 100
2 20multiply numerator and denominator by 10
10 100
20 3 230.23
100 100 100
The number 0.23 is the same as the fraction23
100. This cannot be made any simpler by cancelling
down, and so our initial number becomes23
2459100
Have a look at another example. What will be the vulgar fraction of 0.26?
2 60.26 means and10 100
2 20multiply numerator and denominator by 10
10 100
20 6 260.26 cancel down
100 100 100
26
13
10050
13
50
Therefore, the vulgar fraction of 0.26 is26
100
.
For the following examples, turn the decimal fractions into vulgar fractions.
1/ . 123.4
2 /. 0.156
3 / . 1765.05
4 /. 9876754.006
5/ . 2.5
My working out is shown over the page.
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1/ . 123.4
40.4
2
105
2divide both numerator and denominator by 2
5
2123.4 1235
2 / . 0.156
1560.156
39
1000250
39divide both numerator and denominator by 4
250
390.156
250
3 / . 1765.05
50.05
1
10020
1divide both numerator and denominator by 520
10.05
20
4 /. 9876754.006
60.006
3
1000500
3divide both numerator and denominator by 2
500
39876754.006 9876754
500
5 / . 2.5
50.5
1
102
1divide both numerator and denominator by 5
2
12.5 2
2
Now try the examples for yourself over the page. Remember that the cancelling down is part of
the process. Choose the simple things first to cancel.
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Change the following into vulgar fractions.
My answers are found in the brackets but you do need to be able to convert. I know your
calculators will do it, but there is a need to understand what your answer should be before you
trust in the answer the calculator provides.
1) 0.33
10
2) 0.0011
1000
3) 0.51
2
4) 1.051
120
5) 2.063
250
6) 0.1251
8
7) 0.082
25
8) 1.459
120
9) 0.43434343
10000
10) 7.077
7100
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Ratio and proportionQuite simply ratio means compare.
As an example, the ratio of 100 m to 400 m means 100 m compared to 400 m.
This can then be simplified to express the same ratio in smaller terms and without the units.
100 m compared with 400 m is the same ratio as 1 m compared with 4 m or 1 compared with 4.
This is very wordy and as such we like to simplify. So:
100m compared with 400m is the same as:
1:4
The use of the colon cuts out all the words.
There are many ways in which ratios can be made and simplified. The following examples all
vary slightly, but have the basic principles in mind. We are simply comparing.
Simplify the following ratios.
1/. 32 : 96
2/. 7:21
3/. 16:24
4/. 9 :15:27
5/. 2:8:32
My answers are over the page.
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1/.
32: 96 divide both side by 32
1: 3
2/.
7 : 21 divide both side by 7
1: 3
3/.
16: 24 divide both side by 8
2 : 3
4/.
9 :15: 27 divide all by 3
3 : 5 : 9
5/.
2: 8 : 32 divide all by 2
1:4 :16
It may not be appropriate always to divide to simplify. There will be occasions where it is
necessary to multiply, especially when we are dealing with fractions within ratios.
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Simplify the following ratios.
1/.1
2 :2
2/. 4 : 65
3/.2 4
:3 5
4/.1 3
:4 7
5/.4
3 :3
My working out is shown over the page.
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1/.
12: multiply both side by 2
2
22 2 : 4 :12u
2/.
4: 6 multiply both sides by 5
5
20: 6 5 4
5u
2: 30
15cancel down
4: 6 2 :15
5
3/.
2 4: multiply both side by the common denominator 15
3 5
303
31
60:
12
51
cancel down
31
: 124
cancel down
2 4: 1: 4
3 5
4/.
1 3: multiply both side by the common denominator 28
4 7
287
41
84:
12
71
cancel down
1 3: 7 :12
4 7
5/.
43: multiply both side by 3
3
123 3 :u
4
31
9 : 4
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The following questions relate to specific situations. Read the questions carefully and remember
to convert the description into simple ratios based on what you have already done.
Well work through one together first.
1/. A man earns 18 600 per year. The taxman (bless him) takes more than his fair share at
5 580. What is the ratio of tax to earnings?
The key here is to remove any idea of the context. Simply read the question and look at the
relationship between the numbers.
5580 :18600 remove any extra 0's
5583
: 186010
divide both side by 186
3:10
Now, I know that you are not going to immediately see that 186 divides into 558 and 1 860; but
suck it and see. You will be surprised what drops out when you try. What we see however is that
the taxman takes 3 out of every 10 earned.
2/. What is the ratio of cables of area 1.5 mm2, 2.5 mm2 and 4.0 mm2?
Again, you need simply to deal with the numbers as they are.
1.5: 2.5 : 4 multiply all by 2
3 : 5 : 8
Ratios ideally need to be whole numbers rather than fractions, whether they be vulgar or
decimal fractions.
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Now attempt the following examples. My answers are shown in brackets.
3/. A car park is filled with 420 vehicles. 85 of the vehicles are vans and the rest are cars.
What is the ratio of cars to vans? > @84:17
4/. The ratio of men to women within the engineering industry is approximately 10:1. If
there are 3 500 women engineers, how many men are there? > @35000
5/. The scale on a map is 1:250 000. The distance between Scarborough and Hull is 60 km as
the crow flies. What is the distance on the map? > @0.24m 24cm
The next aspect of ratios is to consider how we can divide an amount between different items.
Lets consider an example.
1/. A lottery win of 125 000 is to be shared in the ratio 2:3:4:5 between four people. How
much will each person get?
The first step is to find out what one portion is. This is reasonably straightforward. We have amaximum amount ( 125 000) and a ratio. In the ratio, there are 14 portions . This
means that we can quite quickly determine how much one portion is worth.
5 4 3 2
Follow my working out.
5 4 3 2 14 Number of portions
1250008928.57 One portion
14
5 8928.57 44642.85 Largest portion
4 8928.57 35714.28 Second largest portion
3 8928.57 26785.71 Third largest portion
2 8928.57 17857.14 Smallest p
u
u
u
u ortion
All we have to do is find how large one portion is, and then multiply it by each of the ratios in
turn.
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Have a look at another example.
2/. Screed is made of a mixture of soft and sharp sand and cement. We will assume that the
proportions are 3:2:1. The volume required for the covering of a conservatory extension
is 3 m3. How much of each material is required?
The process is exactly the same.
3
3
3
3
3 2 1 6 Number of portions
30.5m One portion
6
3 0.5 1.5m Soft sand
2 0.5 1m Sharp sand1 0.5 0.5m Cement
u
u
u
Now work through the following examples. Again, my answers are shown in the brackets.
3/. Copper and tin are mixed in the ratio 3:7. how much copper is needed to mix with 2kg of
tin? > @0.857kg
4/. A machine fills bottle at the rate of 1 200 bottles in 4 minutes. How many bottles are
filled in 1 hour? > @18000 bottles
We will now move on to look briefly at percentages.
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PercentagesPercent is a joining together of two words,perand cent. I know this seems to be a statement of
the blindingly obvious, but it simply means of a hundred.
As an example:-45
45% 0.45100
x To convert a decimal fraction to a percentage we simply need to multiply by 100.x To convert a vulgar fraction to a percentage we need to convert it to a decimal fraction
and then multiply by 100.
Well consider a number of examples.
1/. 0.67
2/. 0.25
3/. 1.45
4/. 0.05
5/. .12.5
The working out is shown below.
1/. 0.67 100 67%u
2/. 0.25 100 25%u
3/. 1.45 100 145%u
4/. 0.05 100 5%u
5/. 12.5 100 1250%u
The following examples look at vulgar fractions being turned into decimals first.
1/. 34
2/.2
3
3/.1
7
4/.4
5
5/.1
16.
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The working out is not really much more complicated, and you can make use of your calculators.
1/.
3 0.754
0.75 100 75%
u
2/.
20.667
3
0.667 100 66.7%
u
3/.
10.1429
7
0.1429 100 14.29%
u
4/.
40.8
5
0.8 100 80%
u
5/.
10.0625
16
0.0625 100 6.25%
u
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6/. Try to complete the table below.
Percentage % Decimal Vulgar fraction
24
0.152
23
6
7
1
1217 %
1.10
11
12
The worked out table is shown over the page.
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Percentage
%
Decimal Vulgar fraction
24 0.24 6
25
15 0.15 3
20
266.7 2.667 22
3
85.7 0.857 6
7
100 1 1
1217 % 0.175 7
40
110 1.10 11
10
108.3 1.083 11
12
My answers are in the shaded boxes. The percentage expressed as a decimal is called theper
unitvalue. This is used almost as often as percentage in engineering.
This is all well and good, but in normal usage we need to be able determine the percentage of a
given quantity. In many areas of engineering it is essential that we use percent values; motor
efficiencies; percentage regulation of transformers all make use of percentage values.
The value of a percentage of a given quantity is given by:
percentquantityu
100
You should be able to see that we have converted the percentage into a decimal before
multiplying by the given quantity. This means that the per unit value is being used.
per unit quantityu
Well look at some examples.
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1/. What is 23 % of 28 000 students?
2/. A 50 trunking has a space factor of 45 %. How much area is that?mm 50mmu
3/. A 200 kW motor is 90 % efficient. How much power is wasted?
4/. You give approximately 30 % of your salary of 14 500 to the taxman. How much do
you keep?
1/.
2323% of 28000 28000
100
0.23 28000 6440
u
u
2/.250 50 2500mm
4545% of 2500 2500
100
0.45 2500 1125
u
u
u
3/.
9090% of 200 200
100
0.9 200 180
200 180 20kW wasted power
u
u
4/.
3030% of 14500 14500
100
0.3 14500 4350
14500 4350 10150 money kept
u
u
Try the examples over the page before looking at the next exercise.
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1/. The volt drop on a length of cable is permitted to be 4 %. The supply voltage is:
a). 230 V
b). 400 V.
What will be the maximum permitted volt drop in each case?
2/. A transformer has a secondary voltage of 18 V. When it is fully loaded 2 % of the voltage
is dropped. How much will the secondary voltage be when it is fully loaded?
3/. What is 27 % of 24 500?
4/. What will be the percentage loss in a motor if the input power is 5 kW and the output
power is 4.6 kW?
5/. A circuit breaker rated at 10 A operates when 50 A flows through it. What is the
percentage increase in current?
My working out is shown over the page. Try things out for yourself first however. Use the box
provided over the page.
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1/.
percenta). quantity
100
4230 9.2
100
4b). 400 16
100
V
V
u
u
u
2/.
percentquantity
100
218 0.36 voltage dropped
100
18 0.36 17.64 actual voltage
V
V V
u
u
3/.
percentquantity
100
2724500 6615
100
u
u
4/.
percentquantity new value
100
new value100 %
quantity
4.6100 92%
5
u
u
u
Be careful here. All I have done is rearrange things a little to get the actual percentage.
5/.
percentquantity new value
100
new value100 %
quantity
50100 500%
10
u
u
u
Dont forget that it is all right to have a percentage increase greater than 100 %. All this means is
that the value is increasing.
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Now attempt the next exercise.
Exercise 4.
1) Divide 96 kg into two parts in the ratio of 5:1.
2) Divide 1hr 20mins into two parts in the ratio of 3:2.
3) Divide 256 into three parts in the ratio of 4:3:1
4) A garage charges an extortionate repair bill of 345 for a car service. The ratio of the
material costs to labour was 3:5. What was the labour cost?
5) Mortar is made from 70 kg of sand to 25 kg of cement.
a) How much sand is needed to mix 35 kg of cement?
b) How much cement is needed to mix with 1 500 kg of sand?
6) Express the following as ratios:
a) 30:85
b) 50:1750
c) 65:26:39
7) Convert the following to vulgar fractions:
a) 1.25
b) 0.14
c) 0.008
d) 12.56
e) 67.89.
8) 240 students took an examination. 65 % of them gained a pass mark, 25 % gained a
credit whilst the remainder failed. How many students fell into each category?
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9) An electrician presents a bill to a customer for 517.79 plus VAT. How much will the
customer have to pay if the rate of VAT is:
a) 17.5 %;
b) 5 %.
10) You are required to prepare a quotation. The following details are available to you:
Description
1st fix 567.23
2nd fix 456.34
Garage extension 122.12
Outside lights 87.76
Less 2.5 % discount
Plus 17.5 % VAT
Total cost
Fill in the gaps.
11) Find the original prices in the following table.
Selling price Profit % Loss % Original price
2345.56 10
12.50 9
24500.00 3
5.67 4
678.78 25
1200.00 23
3400.55 50
500.00 10
256.90 1.5
My answers are found in the appendix. Make sure that you can get the same before you move
on to the next section.
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AlgebraWe now move on to one of the areas that students seem to have most trouble with: that is
algebra and transposing formulae.
It is usually at about this time that you, as the average student, are going to ask why you have to
do all of this, and it is quite an understandable question. After all what does algebra and maths
have to do with electrical installation work etc.?
Sadly for you, algebra has much to do with any engineering work. Without a grasp of algebra,
you will have great difficulty coming to terms with the concepts coming up in the next year or
so.
Without a firm grasp of large areas of background knowledge you will find it almost impossible
to achieve the required level of understanding for the subjects you are to cover.
It is not as if you are going to use all of the information you learn but without it you are not
going to be able to grasp why things happen, for example, when fault finding. So bear with it
and all will come clear.
Algebra is nothing more than a general description of an arrangement of numbers, a type of
coding if you prefer. Instead of numbers we use letters, and so we start using all those things
that you may well have hated at school, and could see no purpose for; letters like a and b andp
and q etc. It can appear to be confusing, but if you remember that we are dealing with real, if
unknown, numbers then things are better understood.
You will already have come across algebraic terms in your school life. You will no doubt
remember that the area of a rectangle is length times breadth, or
l bu . That is an algebraic
expression. We are merely using letters to describe a general truth.
The basic laws introduced earlier, BODMAS etc. are generalized in algebra.
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So:
1/. a b c a b c
In this instance, it doesnt matter whether there are brackets or not. The addition signs mean
that the brackets can all be removed.
2/. ab c a bc abc
The same thing occurs here. All the signs are effectively multiplication signs and so the brackets
can go.
3/. a b b a
It doesnt matter which letter comes first. This is the same as if you were adding 2 to 3 or 3 to 2,
the answer would still be 5.
3/. ab ba
The same thing applies here as it does to the one above. 2 times 3 is the same as 3 times 2, and
both equal 6.
4/. a b c ab ac
Here the letter outside of the bracket is common to either letters or numbers inside the bracket.
5/.a b a b
c c
c
Similarly to the last example, the letter that is the denominator is common to both numerators:
it has no particular desire to attach itself to one or the other of the numbers, and must be
attached to both.
6/. a b c d ac ad bc bd
Here we have a situation where every letter or number in the first bracket must be multiplied byevery other letter or number in the second bracket. The use of the brackets here is the key. Two
brackets placed next to each other tell us that they must be multiplied together.
The laws of precedence, which apply to arithmetic, apply also to algebraic expressions.
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Remember:
BODMAS
Try this example.
1/. Evaluate 3 2ab bc abc when a=1; b=3 and c=5
Replacing a, b, and c with their numerical values gives:
3 1 3 2 3 5 1 3 5u u u u u u
All you need to do is fill in the numbers where the letters are. This is the most common form that
you are likely to see. Most equations and formulae are set up so that you only have to fill in the
numbers.
3 2 9 30 15ab bc abc 6
Try another example.
2/. Evaluate, 5 where, p=1.5; q=2.25.2 2pq pq pqu
As with the first example, just fill in the numbers and carry out the rules.
5 2 2 5 1.5 2.25 2 1.5 2.25 2 1.5 2.25
33.75 3.375 6.75 37.125
pq pq pqu u u u u u u
You should notice that the rules ofBODMAS have been maintained.
You will find more examples at the end of this section, but do not worry about them. The area
that you do need to concentrate on is transposing formulae.
All this is, is the rearranging of the formulae so that other information can be gathered. It is not
complex but it does need care. Again, once the techniques are learned, it is very easy to come toterms with any, and all types of formulae that you will come across.
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Transposing formulaeThe fundamental rules learned with basic operations and with algebra are repeated here. The
main rule however is that:-
WHAT YOU DO TO ONE SIDE OF THE EQUATION MUST BE DONE TO THE
OTHER SIDE.
There are no short cuts and no variations on this!
Have a look at the examples and see how this pans out.
1/. Transpose this formula to make R1 the subject of the formula.
1 2TR R R
This is the formula for series connected resistors and will crop up again later in the course.
The key to this is always to remember that each side of the equation (i.e. each side of the equals
sign) must be balanced. To get R1 on its own it is necessary to somehow remove R2. If we
subtracted R2 from both sides, the equation would still be balanced.
Try the principle of adding two simple numbers.3 2 1
2 3 1
1 3 2
In the above simple example, you can see that 3 2 1 . The other two variations are also true.
You know this without having to be taught it. The same principle is true in transposing formulae.
The only real difference is that we are using letter and number combinations rather than just
numbers.
Now follow the working out of the problem.
1 2
2 1 2
T
T
R R R
R R R R
2R
2
2 1 2 2
1 2
take away from both sides
0T
T
R
R RR R R
R R R
If that seems strange just remember that what is done to one side must be done to another. All
things must remain in balance.
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In this first example, the number (R2), and it is a number remember, (We are using a letter to
express and unknown number) is to be removed from the right-hand side of the equation to the
left hand side. As it has been added to another number on the right-hand side, then it must be
subtracted from that side to remove it, as the opposite of addition is subtraction.
Try this one.
2/. Transpose this so that2
1
Ris the subject of the equation.
1 2
1 1 1
TR R R
This is the equation for resistors connected in parallel, and as such, you will come across this
formula later on in the course. We only need to follow the example above.
1 2
2 1 2
1 1 1
1 1 1 1
T
T
R R R
R R R R
2
1
R
2
2 1 2 2
1 2
1take away from both sides
1 1 1 1 10
1 1 1
T
T
R
R R R R R
R R R
Although it may appear more complex, in reality it is very similar to the example above.
Notice again that we have subtracted the same value from both sides. It matters little if they are
whole numbers or fractions the result is the same.
The following examples are all using multiplication and division.
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3/. The formula that describes power in an electrical circuit is P IU . Transpose this so that I
is the subject.
What you need to recognise is that the opposite of multiplication is division and the opposite of
division is multiplication.
P IU
P I U
U
1
U 1 divide both sides by
1
U
P UI
U U
You may be able to see that the letter Uhas been cancelled out on the right-hand side. You can
only cancel out the same letter or number. You cant just remove letters or numbers just because
they seem to be in the way.
Here is another one.
4/. This formula describes Ohms Law, transpose it for R.U IR
U IR
U I
I
1R
I
1
divide both sides by
1
I
U IR
I I
Notice again how to get rid of the Ion the right-hand side, all we have to do is divide by the I. As
long as we do the same to both sides, we are on the right lines. Again, notice the cancelling out
of the Iin the middle part of the equation.
One more should be enough to make sure that you are able to attempt the examples that
follow, and to be familiar with the required techniques when called on.
5/. This equation describes how resistance is related to length, area and type of material,
lR
A
U . Transpose it for l.
Dont be worried if some of the terms are not familiar. At present we are not concerned with the
actual equation, but with how to manipulate it. Two bites at the cherry are needed to complete
the transforming of this equation.
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lR
A
lR A
A
U
U
u 1
Au 1
multiply both sides by and cancelA
RA l
RA
U
U
U
1l
U 1 divide both sides by and cancel
RAl
U
U
To begin with, both sides we must multiply both sides byA, to get rid of theA on the right-hand
side. The reason for multiplying is that we have initially divided and to cancel out a division we
must multiply. Remember the cancelling out.
We then must divide both sides by (rho). We now divide because we had initially had a
multiplication; again remember to cancel out the on the right-hand side.
Notice that we just had to go through the process twice. As long as you are careful how you
proceed, you should not find things too difficult.
Well work through a few more examples before you attempt the next exercise.
Transpose the following:
6/.2
for4
dA d
S
7/. 2 forC r rS
8/. 2 forP I R I
9/.2
forU
P UR
The answers are found over the page.
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6/.
2
2
4
44
dA
dA
S
S
u 4u
2
multiply both sides by 4 and cancel
4
4
A d
A
S
S
S
2d
S
2
2
divide both sides by and cancel
4
4square root both sides and cancel
4 or 2
Ad
Ad
A Ad d
S
S
S
S S
7/.
2
2
2
C r
C
S
S
S
2
r
S divide both sides by2 and cancel
2
2
Cr
Cr
S
S
S
8/.2
2
P I R
P RI
R
R
2
2
divide both sides by and cancel
square root both sides and cancel
R
PI
R
P IR
PI
R
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9/.2
2
UP
R
UP R R
u Ru
2
2
multiply both sides by and cancel
square root both sides and cancel
R
PR U
PR U
U PR
Summary
After this section you should be able to:
x Recognise that algebra is the use of letters to describe real numbers.x Transpose a variety of types of equation using:
o Additiono Subtractiono Multiplicationo Division.
You see. You know more than you think. Often self-confidence is what you need more than
someone standing over you making sure that you are doing things right. Attempt things and
then see if you were right or not.
Now attempt the next exercise.
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Exercise 5.
Transpose.
1) ; 2)forQ It t forB AA
) ;
3) ; 4)fore Blv v forF BIl B ;
5) 0 forrA
Cd
AH H
; 6) fory mx c m .
7) If and I=2A, whilst R=120, calculate U.U IR
8) If e B , calculate e when B=0.2T, l=5m and v=25ms-1lv
9) If 0 1TR R DT , calculate RT when R0=20; =0.00428/oC and =30oC.
Remember with all of these that BODMAS is always appropriate. Another thing to remember is
that if you are using a calculator, and you should, it will only produce what you put in. The old
adage ofGarbage In=Garbage Out is true. Do not think that the machine naturally knows what
you mean. The calculator will follow BODMAS.
Take your time and re-work those questions that you struggled with.
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IndicesIndices relate back to the first section of this study book. It is a means of simplifying numbers,
quickly and easily. As with so many other areas in maths, there are a number of simple rules to
follow.
1/. When any number is raised to a power, e.g. , the number 2 is called the base and the 8
is the index. Thus
32
32 2 2 2 u u 8 .
Special names may be used when the indices are 2 or 3, these being squared and cubed
respectively. You may well have come across these terms before.
1/. When no index is shown, then the number (base) is assumed to be raised to the power
e.g. 12 2
2/. The reciprocal of a number is when you divide 1 by that number.
1i.e. the reciprocal of 2is
2
The reciprocal of a number, when provided in index form, is when the index is negative.
-1 -4
4
1 1So 2 2
2 2
3/. When the base is raised to a fractional index, then the base is rooted.
21 132 4 3 24e.g. 2 2 81 81 3 4 4
4/. When simplifying calculations involving indices, a number of basic rules or laws can be
applied. These are called the laws of indices. These are:-
a). When multiplying two or more numbers together which have the same base,
then the indices are be added.
2 4 2 4 6Thus 3 3 3 3u
b). When dividing two numbers that have the same base, the indices are
subtracted.
55 2 3
2
3Thus 3 3
3
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c). When a number, which is raised to a power, is raised to a further power, the
indices are multiplied.
25 5 23 3 3u 10
d). When a number has an index of 0, it's value is 1.
03 1
So many rules, but as you can see they are all fairly simple and easy to follow. It is worthwhile
remembering that all these books are resources and can be used again, when you are stuck
somewhere else in the course.
Here are some examples.
Evaluate:
2 3
2 4
2 5
i/. 5 5
ii/. 3 3 3
iii/. 2 2 2
u
u u
u u
All you have to remember is to follow the rules. With each of these, all you have to do is add up
the indices.
i/. 2 3 55 5
ii/. 2 4 1 73 3
iii/. 1 2 5 82 2
Evaluate:
5
3
7
4
7i/.
7
5ii/.
5
i/.5
5 3 2
3
77 7
7
ii/.7
7 4 3
4
55 5
5
Again, see how by following the rules a successful conclusion is reached. Try it with your
calculator and see what you get.
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Try following these.
Simplify:
12
34
i/. 4
ii/. 16
i/.124 4 r2
ii/.34 34 416 16 4096 8
Notice that the terms, simplify and evaluate are very similar, and you should not get confused
about their use.
Summary
You will now be able to:-
Know what squared and cubed means.
Know that indices are powers that a base is raised to.
Add, subtract and multiply indices.
You should be able to surprise yourself as to just how much you do know!
Exercise 6.
1) 2) 3)33 3u 4 4 427 7u 2 34 4 4u u
4)5
2
7
75)
25
24
12
126)
327
7) 4
217 8)2 3
4
2 2
2
u9)
5
2
13
13 13u
Express the following in standard form and engineering form to two decimal places:
10) 2,748 11) 33,170 12) 274,218
13) 0.2401 14) 0.0174 15) 0.00923
16) 1,702.3 17) 0.0109 18) 197.62.
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Have a go at the following examples. Some of the answers are found in the square brackets.
Convert the following to standard form
1/. 234678934
2/. 0.004563/. 43562
4/. 0.0345
5/. 23415
6/. 0.0123
Convert the following to engineering form
7/. 431275
8/. 0.003681
9/. 6453
10/. 0.023411/. 0.0000000000067
12/. 0.000
73 4
4 5 2 11
37
Determine the following fractions
2 4 713/. 1
3 5 15
1 2 514/. 1 2 3
6 3 6
2 3 115/.
7 11 772 4 13
16/. 13 5 15
1 2 12 18417/. 5 1 4 1
5 3 13 195
Simplify using indices
18/. 22 2
19/. 5 5 5 5
2
u
u u 4
1
3
52
27
2
0/. 2 22
1 1321/. 3
3 93
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TrigonometryOriginally trigonometry was the branch of mathematics concerned with solving triangles using
trigonometric ratios which were seen as properties of triangles rather then of angles.
The word Trigonometry comes from a combination of three Greek words:
i) Treis meaning threeii) Gonia meaning angleiii) Metron meaning measure.
The Early Greeks developed the subject and used it in a number of very practical ways. Initially it
was used in Astronomy but was also used in building. In one example a tunnel was dug through
a mountain, with two teams starting at each end and it met perfectly in the middle. Later it was
much used in Architecture, Navigation, Surveying and Engineering, but in the last two centuries
it has been used more for Mathematical Analysis and for repeating Waves and Periodic
Phenomena.
This brief session will act as a simple introduction to the subject.
We will first need to look at what is called a right-angled triangle. Have a look at Figure 2.
Figure 2 Right-angled triangle
Triangles have a number of qualities. For example, all the angles add up to 180q. The extra
qualities of right-angled triangles are of benefit to us in the electrical industry.
A Greek called Pythagoras stated:
The square on the hypotenuse is equal to the sum of the squares on the other two sides.
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So, what does this mean? Have a look at Figure 3.
Figure 3 The square on the hypotenuse is equal to the sum of the squares on the other two sides
Here you can see that I have drawn a square on to each of the sides of the triangle. If you were to
measure the individual areas, you would find that the two areas on the shorter sides of the
triangle added up to the area on the longest side. The longest side is called the hypotenuse.
This relationship can and is described using a formula.
2 2a b c2
This can be transposed to form a slight variant.
2 2 2
2 2
transposea b c
a b c
I have simply square-rooted both sides.
Both of these formulae state the same thing. We dont need to look at these in too much detail
yet as we will make use of them in higher-level courses.
The second set of qualities that we can make use of is the relationships between the sides and
the angles of a right-angled triangle.
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Figure 4 Relationship of angle to sides
Look at Figure 4. In any right-angled triangle, we can describe any angle in terms of a ratio of
two other sides. Assume that we are looking for the angle, (theta). We can find out this angle in
any one of three ways.
It is here that we now have to introduce three terms that you may not have come across before,
these are:
x tangent (tan)x cosine (cos) andx sine (sin).
You dont need to worry about these new terms, although you will come to use them quite
regularly. Have a look at the three formulae below.
opposite side of anglesine
hypotenuse
opp Osin
hyp H
adjacent side of anglecosine
hypotenuse
adj Acos
hyp H
opposite side of angletangent
adjacent side of angle
opp Otan
adj A
T
T
T
T
T
T
Each of the new terms is a relationship between two sides of the triangle. It would be
worthwhile just working through an example to see how it all pans out.
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1/. Based on this triangle determine the values of all the angles and all the sides.
2 2
2 2
1
1
3 4
9 16 25 5
Osin
H
3sin 0.6
5
sin 0.6 36.86
alternatively,
Acos
H
4cos 0.85
cos 0.8 36.86
a b c
a
a
q
q
q
q
q
q
q
q
A few points need to be noticed. Firstly, it doesnt matter which relationship I use. I am allowed
to usesin or cos or tan. The angles will still come out the same.
The second point is that the termssin-1 or cos-1 or even tan-1 all have the same purpose.
What you should do is take your scientific calculator and look at it. You should have a series of
buttons that read sin, cos and tan, and usually above them there are three labels showingsin-1,
cos-1 and tan-1. These are the buttons used to turn the ratios of the sides into real angles. Just
type in your number and press inv sin or 2nd Fsin.
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Exercise 7.
1) The following right-angled triangles have the following dimensions. Fill in the remainder
of the table:
Hypotenuse Side a Side b Angle a Angle b7 4
5 12
16 12
25 16
100 150
230 110
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Chapter
3Areas and volumes
In this session you will have the opportunity to:
x work with the areas of rectangles, circles and trianglesx work with the volumes of cubes, cylinders and pyramids?
AreasThroughout much of the material that we are going to consider, a basic understanding of areasand volumes is required. Although we will take a second look at areas and volumes in the
mechanics study book, a brief introduction here is appropriate. There are three types of shape
that we need to consider when dealing with areas. These are:
x Square or rectanglex Circlex Triangle.
We will consider each of the shapes in turn. Much of the determination of areas etc. is based on
these basic shapes.
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Rectangle
Have a look at Figure 2 below.
Figure 5 Rectangle
You will notice that we have the parallel sides equal in length. If we have labelled each side land
b we can state what the length of the sides, or the perimeter, is.
2 2P l l b b l b
Try an example.
1/. A rectangle has sides of 50mm and 80mm. What is the length of the perimeter?
2 2
2 50 2 80
100 160 260
P l b
P
P mm
u u
You could draw a shape to try and clarify things in your mind.
To consider the area of a rectangle is quite straightforward. Have a look at the shape again.
Figure 6 Area of rectangle
There is a standard formulae used for rectangles.
A l b lb u
The area is dependent on the length and breadth.
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Try another example.
2/. A rectangle has sides of 50mm and 80mm. What is its area?
250 80 4000
A lb
A mm
u
The idea ofmm2 may seem to be confusing. We are not squaring the answer we are merely
expressing the answer in the units ofmm2.
Have a look at Figure 4 below.
Figure 7 Area differences
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The table below shows some of the most common conversions. Become familiar with them.
Conversion Multiply/Divide Standard form
mmm y1000 1q10-3
mmm u1000 1q103
mm2m2 y1000000 1q10-6
m2mm2 u1000000 1q106
mmcm y10 1q10-1
cmmm u10 1q101
cmm y100 1q10-2
mcm u100 1q102
cm2
m2
y10000 1q10-4
mm3m3 y1000000000 1q10-9
m3mm3 u1000000000 1q109
cm3m3 y1000000 1q10-6
m3cm3 u1000000 1q106
You need to become very familiar with all of these. Any slip in your working out and you can be
over 1000 times out.
Try another example.
3/. A rectangle has sides of 120mm by 3m. What is its area?
2
2
120 3000 360000
Alternatively 0.12 3 0.36
A lb
A mm
A m
u
u
You should have noticed that I have either converted everything to metres or to millimetres. You
must have common units. You cannot have a combination of metres and centimetres etc. you
may also notice in this example that the difference between mm2 and m2 is exactly 1000000 (1
million) times.
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You should also note that any shape that looks like a rectangle with two sets of sides parallel
could be considered in exactly the same way.
Look at Figure 5.
Figure 8 Parallelogram or lozenged rectangle
This is a lozenge shape. The rectangle has been distorted; however the area and perimeter are
determined in exactly that same way. Dont get confused. This is true for two sets of sides that
are parallel.
Thesquare is a particular type of rectangle. A square has all four sides the same length.
Have a look at Figure 6.
Figure 9 Square
The perimeter and areas are even more straightforward to determine.
4P l l l l l
In addition, for the area.
2A l l l u
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Circle
Have a look at Figure 7.
Figure 10 Circle
There are a number of labels that you would do well to remember.
x A line drawn from one side of the circle through the centre of the circle connected to theopposite side of the circle is called the diameter. This term is commonly used in
engineering.
x A line drawn from the centre of the circle to any edge of the circle is called the radius.The radius is half the diameter.
x The outer edge of the circle is called the circumference. This is the same as theperimeter.
There are other labels but we will consider them when we need to.
There is a relationship we have to consider when we do any work with circles. This ispi(). Pi is a
relationship formed between the circumference of a circle and its diameter. The value is always
the same irrespective of the different size of circles, and is commonly held to be 3.142 (no units).
Your calculator should have a pi button which will give you a slightly more accurate figure.
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r
The circumference of a circle is found using:-
2C dS S
d is the diameter of the circle and r is the radius of the circle. You can use either. The first is the
most commonly used.
Try an example.
1/. A circle has a diameter of 50 mm. What is its circumference?
3.142 50 157.1
C d
C mm
S
u
2/. The circumference of a circle is 1 m. What is its diameter?
1
1
transpose
10.318
3.142
C d
C d
Cd
d m
S
S
S S
S
The area of the circle can also be found in one of two ways.
2
2
and4
dA
A r
S
S
The first one is the better option, as most measurements of a circle in engineering are diameters.
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A couple more examples should suffice.
3/. For the two last examples, determine the area of the circles.
2
22
2
22
i/.4
3.142 50 78551963.75
4 4
ii/.4
3.142 0.318 0.3180.0795
4 4
dA
A m
dA
A m
S
S
u
u
m
Example (ii) is measured in m2. Dont forget that there is a difference of 1 000 000 (1 million)