M1 Cat-B1 Notes V1.1

Aircraft Maintenance Training Centre
Part-147
Fundamentals
M1 MATHEMATICS
EASA Part-66 Cat-B1
Part-66 Cat-B1 Mathematics M1

____________________________________________________________________________________________________________________________________________
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For training purposes only.
This document is property of MCAST Aircraft Maintenance Training Centre.
Any reproduction or copying of training documents and extracts thereof in any
manner is strictly prohibited.
MCAST Part-147 Aircraft Maintenance Training Centre
Institute of Mechanical Engineering
MCAST Main Campus
Triq Kordin
Paola PLA 9032
Tel:
+356 2398 7450
Fax:
+356 2398 7490
E-mail: ime@mcast.edu.mt
http://www.mcast.edu.mt/institutes_mechanicalengineering.asp
Category B1 Mathematics M1 Notes; May 2009
Version 1.0
Author:
Mr Nicholas Grech, B.Eng. (Hons), M.Sc. (Cran)
Email: nickgrech@gmail.com
Pg 1

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TABLE OF CONTENTS
1.
1.8.2.
Inverse Proportion ..............................................................................13
1.8.3.
Proportional Parts ...............................................................................13
1.8.4.
Constant of Proportionality .................................................................14
Exercises...........................................................................................................14
Arithmetic............................................................................... 1
1.1. Arithmetical Terms and Signs ................................................................... 1
1.1.1.
Addition ................................................................................................ 1
1.1.2.
Subtraction ........................................................................................... 1
1.9. Averages and Percentages ......................................................................15

1.9.1.
Mean ...................................................................................................15
1.9.2.
Median ................................................................................................15
1.9.3.
Mode ...................................................................................................15
1.9.4.
Percentages........................................................................................15
Exercises...........................................................................................................16
1.2. Methods of Multiplication and Division .................................................... 2

1.2.1.
Multiplication......................................................................................... 2
1.2.2.
Division ................................................................................................. 2
Arithmetic Law of Precedence ..................................................................................... 2
Exercises ............................................................................................................ 3
1.3. Fractions ...................................................................................................... 4
1.3.1.
Converting between improper fractions and mixed numbers............... 4
1.3.2.
Multiplication of Fractions..................................................................... 4
1.3.3.
Division of Fractions ............................................................................. 4
1.3.4.
Addition of Fractions............................................................................. 5
1.3.5.
Subtraction of Fractions ....................................................................... 5
1.4. Decimal Numbers........................................................................................ 7
Number of Decimal Places .......................................................................................... 7
Significant Figures ....................................................................................................... 7
1.4.1.
1.4.2.
1.4.3.
Addition and Subtraction of Decimal Numbers .................................... 7

Multiplication of Decimal Numbers....................................................... 8
Division of Decimal Numbers ............................................................... 8
1.10. Areas and Volumes ...................................................................................17

1.10.1. Areas ..................................................................................................17
1.10.2. Volumes..............................................................................................17
Exercises...........................................................................................................18
2.
Algebra..................................................................................19
2.1. Introduction ...............................................................................................19
2.1.1.
Addition and Subtraction ....................................................................19
2.1.2.
Multiplication and Division ..................................................................19
2.1.3.
Brackets..............................................................................................19
Exercises...........................................................................................................20
1.5. Factors and Multiples ................................................................................. 9
2.2. Linear Equations .......................................................................................21

2.2.1.
Transposition of Formulae..................................................................21
1.6. Powers and Roots....................................................................................... 9
2.3. Simultaneous Equations ..........................................................................21
Laws of Indices............................................................................................................ 9
Exercises .......................................................................................................... 10
Standard Form........................................................................................................... 10
1.7. Measures and Conversion Factors ......................................................... 11

Exercises .......................................................................................................... 12
1.8. Ratio and Proportion ................................................................................ 13
1.8.1.
Direct Proportion ................................................................................ 13
Solving simultaneous equations by elimination..........................................................21

Solving simultaneous equations by substitution.........................................................22
2.4. Quadratic Equations .................................................................................23

2.4.1.
Solving by using the formula ..............................................................23
2.4.2.
Solving by factorisation.......................................................................23
2.4.3.
Factorising quadratic equations..........................................................23
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2.5. Division of Algebraic Fractions ............................................................... 24

Exercises .......................................................................................................... 25
2.6. Logarithms................................................................................................. 26
2.7. Number Systems....................................................................................... 27
2.7.1.
The Binary System ............................................................................. 27
2.7.2.
The Hexadecimal System .................................................................. 27
Exercises .......................................................................................................... 28
3.
Geometry.............................................................................. 29
3.1. Graphs........................................................................................................ 29
3.1.1.
Graphical Representation .................................................................. 29
3.1.2.
Graphs of Linear equations ................................................................ 29
Equation of a straight line .......................................................................................... 30
3.1.3.
Graphs of Quadratic equations .......................................................... 31
3.1.4.
Graphical solution of Simultaneous Equations .................................. 32
Exercises .......................................................................................................... 32
3.2. Trigonometry ............................................................................................. 33
3.2.1.
Pythagoras Theorem.......................................................................... 33
3.2.2.
Trigonometrical Ratios ....................................................................... 33
3.2.3.
Use of Tables ..................................................................................... 36
3.2.4.
Rectangular (Cartesian) Co-ordinates ............................................... 36
3.2.5.
Polar Co-ordinates ............................................................................. 36
Converting between Polar and Cartesian .................................................................. 37
Exercises .......................................................................................................... 37
3.3. Geometrical Constructions...................................................................... 38
3.3.1.
The Circle ........................................................................................... 38
Elements and Properties of the circle ........................................................................ 38
Important circle theorems .......................................................................................... 38
3.3.2.
Angles ................................................................................................ 40
Degrees and Radians................................................................................................ 40

Length of Arc ............................................................................................................. 40
3.3.3.
Construction ....................................................................................... 41
Appendix 1 Trigonometric Tables.............................................. 44

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Part-147 Aircraft Maintenance Training Centre
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1. ARITHMETIC
It is important for students to realise that the use of calculators during examinations
is not permitted. Therefore, the mathematics taught is aimed towards using mental
calculations and it is therefore strongly recommended that students practise these
problems without the aid of a calculator.
Example:
hundreds
tens
ones
6
5
5
5
1
______________________
6
1
6
1.1. Arithmetical Terms and Signs

The most common numbers we use are called the natural numbers and these are
the simple 1, 2, 3, 4, etc. These whole numbers are also known as positive
integers. However, we also often use positive fractions, such as 15 . These two
groups together form the positive rational numbers. A rational number is in fact
any number that can be expressed in the form a/b, where a and b represent
integers.
The natural numbers are positive integers, but sometimes we need to quote
negative quantities (numbers less than zero), so in this case we use negative
integers. The number zero is unique and does not fall in any category; in fact, it
has a category of its own.
There are numbers (e.g. 2 ) which are not rational numbers because cannot be
represented by the quotient of two integers. These are called irrational or nonrational numbers. All the above-mentioned categories together form what are
known as real numbers.
These are distinguished from the so-called complex numbers, which will not be
considered in this course.
1.1.1.
Addition
The process of finding the total of two or more numbers is called addition. The
resulting answer is called the sum. When the result is larger than nine, it is
necessary to arrange the numbers in columns so that the last digit of each number
is in the same column.
65 + 511
In the example above, we start first with the ones column. So 5 + 1 gives 6. Same
with the tens column, but in this case the answer is 11. We cannot write down 11 so
we take the ten out of the answer and convert it to a one in the hundreds column.
The remaining 1 from the 11 is written down in the tens column. In this manner, in
the hundreds column we have 5 plus 1 which was added. So the result in the
hundreds column is 6, giving the final answer as 616.
1.1.2.
Subtraction
The process of finding the difference between two numbers is known as

subtraction. The number, which is being subtracted, is the subtrahend and the
number from which the subtrahend is subtracted is the minuend.
In order to simplify the process of subtraction, the numbers are arranged in
columns, similar to the addition process (i.e. hundreds under hundreds, tens under
tens, etc). Then starting from the right, the subtrahend is subtracted from the
minuend.
Example:
783 - 592
hundreds
tens
ones
7
8
3
5
9
2
_______________________
1
9
1
minuend
subtrahend
Starting from the right, (3 2) gives 1. Next column, (8 9) is not possible so 8

becomes 18 by taking 1 from the hundreds column so that 7 becomes 6. (18 9)
gives 9. Now 7 has become 6 so (6 5) gives 1.
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1.2. Methods of Multiplication and Division

1.2.1.
Multiplication
1.2.2.
In multiplication, we are doing addition but repetitively. So (5 x 3 = 5) is similar to (5

+ 5 + 5) or (3 + 3 + 3 + 3 + 3). In this example, 5 and 3 are also called the factors
and 15 is the product. The number to be multiplied (i.e. 5) is the multiplicand and
the number of times the multiplicand is to be added to itself (i.e. 3) is the multiplier.
The order in which numbers are multiplies does not change the product (i.e. 5 x 3 =
3 x 5).
Simple multiplication involves only two numbers and one needs to know the
numerical tables properly in order to work them out easily.
Example:
3 x 5 = 15
9 x 3 = 18
7 x 6 = 42
Division
Division is the reverse of multiplication, finding out how many times a number is
contained in another number. The number divided is called the dividend, the one
dividing with is the divisor and the result is the quotient.
In some problems, the quotient may include a remainder, which represents a
portion of the dividend that cannot be divided by the divisor.
Division can also be represented with a fraction, e.g. 3 4 = 3 4
To divide large quantities, the problem is broken down into a series of operations.
Examples:
12
3 36
Commutative Law of Multiplication: The product of two real numbers is the same no
matter in what order they are multiplied. This means that (a x b) = (b x a).
Associative Law of Multiplication: The product of three or more numbers is the
same no matter in what manner they are grouped, hence a(b x c) = (a x b)c.
573 x
21
573 +
11460
12033
432 x
32
864 +
12960
13824
1471 x
121
1471+
29420
147100
177991
432 x
132
864 +
12960
43200
57024
18
16
10
9
26
24
20
20
When multiplying large numbers, it is important they are aligned vertically, similar to
the addition and subtraction process.
Examples:
33.33
3 100 .0
9
146 .5
4 586 .0
10
9
10
Arithmetic Law of Precedence

It is often confusing, when having different operations to do in the same equation,
which operation goes first. In this case, it is convenient to use the BODMAS rule.
BODMAS stands for Brackets Of Division, Multiplication, Addition, Subtraction.
Hence, the first to be worked out are the brackets of the equation, starting from the
division, then multiplication etc.
Pg 2

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Exercises
Addition:
1.
2.
3.
4.
5.
6.
123 + 4294
2342 + 3939
233.65 + 9483.12
5830.766 + 13339.144
45.34 + 232.5 + 11.89
4
2
23.4 x 10 + 11.8 x 10
Subtraction:
1.
2.
3.
4.
5.
6.
545 232
9833 2334
983.3 843.2
2938 23345
893.45 78.9
7583.5 89921
Multiplication:
1.
2.
3.
4.
5.
6.
54 x 3
345 x 45
234 x 23
2498 x 345
123 x 4592
4592 x 129
Division:
1.
2.
3.
4.
5.
6.
2464 / 4
3486 / 2
8323 / 5
4782 / 9
2922 / 18
4234 / 21
Pg 3

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Lowest Terms
1.3. Fractions
A fraction is a division of one number by another. For example, means one
divided by two. The number above the line is called the numerator and the number
below the line is called the denominator.
Fractions written in this form (e.g. 2/5, 3/8, , etc) are called vulgar fractions.
Those written in decimal form (e.g. 0.5, 3.56, 0.333, etc.) are called decimal
fractions.
When the denominator of a fraction is numerically larger than the numerator (e.g.
4/5), the fraction is said to be a proper fraction. If however the numerator is larger
than the denominator, the fraction is said to be improper. Improper fractions (e.g.
9/4) can be also expressed as what is known as a mixed number. In this case, 9/4
would be expressed as 2 where 2 is whole number and has to be a proper
fraction.
1.3.1.
A fraction is said to be in its lowest terms if its same value cannot be represented by
a fraction with smaller numbers. For example 3 5 cannot be represented with smaller
numbers. But 6 12 can also be written as
denominator by 6.
1.3.2.
Converting between improper fractions and mixed numbers
Converting from an improper fraction to a mixed number is done by simple division.

The answer of the division process becomes the whole number of the mixed
number while the remainder is the new numerator.
Convert
22
if we divide both numerator and
Multiplication of Fractions
2 4 2 4
8
=
=
5 7 5 7 35
11 1 11 1 11
=
=
12 2 12 2 24
into a mixed number.
3
7 22 Therefore the
21
1
1.3.3.
22
can be written as 3 17 .
To convert from a mixed number to an improper fraction, the whole number is

multiplied by the denominator and the result is added to the numerator. This answer
will be the new numerator whilst the old denominator remains unchanged.
Example:
Multiplication of fractions is relatively easy. All one has to do is multiply the

numerator of one fraction with the numerator of the other fraction and the
denominator of the first fraction with the denominator of the second one. These
results will be the numerator and denominator of the answer.
Example:
Example:
31 =
4
(3 4) + 1 12 + 1 13
=
=
4
4
4
52 =
3
(5 3) + 2 15 + 2 17
=
=
3
3
3
Division of Fractions
Division of fractions is similar to multiplication. However the divisor (the fraction we

are dividing with), needs to be inverted. The resulting fractions are then multiplied
normally.
3 1 3 3 33 9
Example:
= =
=
4 3 4 1 4 1 4
2 4 2 7 2 7 14
= =
=
5 7 5 4 5 4 20
Note that this last answer is not in its lowest terms.
Therefore, we can re-write 14 20 as 7 10 .
Pg 4

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1.3.4.
Addition of Fractions
In order to add fractions, we need to determine the least common multiple (LCM).
This means we need to find the smallest possible number, which is a common
multiple of the denominators. For example if we have 3 4 and 1 5 , the LCM would
be 20. Often in order to find the LCM, sometimes it is easier to simply multiply the
denominators, as in the previous example. However, there are cases in which a
smaller number being a multiple of both denominators can be found by intuition,
making the problem easier to solve.
Example:
5 3
+ , the least common multiple would be 28.
7 4
11 5
+ , the least common multiple would be 24, not 96.
12 8
The next step after determining the LCM, is to place the LCM below the
denominators of the fractions we wish to add.
11 5
+
12 8
So we using the second example, we get
24
1.3.5.
Subtraction of Fractions
Subtraction of fractions is carried out similarly to addition of fractions. The LCM still
needs to be determined. However, in the last step, we simply subtract the
numerators instead of adding them.
Examples:
3 1

3 1 4 2 32 1
=
=
=
4 2
4
4
4
1 1

1 1 3 2 23
1
=
=
=
3 2
6
6
6
11 5

11 5 4 3 33 20 13
=
=
=
4 3
12
12
12
Now we divide 24 by 12, which gives 2 and multiply it by the corresponding

numerator i.e. 11, which gives us 22. Again we divide 24 by 8 which gives 3 and
multiply it by its corresponding numerator i.e. 5, which gives 15.
Thus the new fraction will be
Examples:
22 + 15 37
=
24
24
3 1

3 1 4 2 32 1
=
=
=
4 2
4
4
4
3 1

3 1 4 2 32 1
=
=
=
4 2
4
4
4
Pg 5

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Exercises
Mixed: (Remember Arithmetic Law of Precedence)
Convert from improper fractions to mixed numbers:

i) 4
ii) 7
iii) 9
iv) 11
iv) 11 15 3
3
6
4
Convert from mixed number to improper fractions:

i) 3 1
3
ii) 112
i) 1 5 + 7 1
3
6
5
iii) 3 1
2
iv) 7 3
ii) 2 4
5 3
2
v)
15 3
3
9 1
iii) 10
7
16
8
6
5
vi)
3
5
6
4
ii) 2 4
5
3
15
2
v)
3
3
9 1
iii) 7
ii) 3 2 4 + 5 2
5
3
6
3
iii) 10
v) 2 15 3
3
9
1
vi) 6 + 5 3 11
5
6
4
3
16
7 + 95
8
7
Multiplication:
i) 1 5
3 6
11
iv)
15 3
3
6
4
Division:
i) 2 5
3
6
11
iv)
1
3
6
7
16
8
6
5
vi)
3
5
6
4
Addition:
i) 1 + 5
3
6
iv) 11 + 7 + 3
3
6
2
iii) 10
ii) 2 + 4
5
3
v) 2 + 1 + 3
5
9
1
+7
16
8
vi) 1 + 11 + 3
5
6
8
ii) 6 1
5
3
v) 1 5 3
5
6
4
iii) 15
Subtraction:
i) 5 1
3
6
iv) 11 15 3
3
6
4
1
16
8
vi) 17 5 1
5
6
4
Pg 6

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1.4. Decimal Numbers

Decimal numbers are another way to represent fractions. If we write down 0.1, this
can be written in fraction form as 1/10. Similarly, 0.01 is equivalent to 1/100.
Example:
45,200
235,674
1,485
Number of Decimal Places

In many cases, the use of a large numbers of decimal places is unnecessary. For
example a number like 7.3422411343 is too impractical to handle and more often
than not, not required. Therefore, one may choose to what degree of accuracy the
answer is required by specifying the number of decimal places to be used. If for the
previous number, an accuracy of 3 decimal places is required, we only write down
7.342 and forget the rest.
Attention! Had the number been 7.3425, the result to 3 decimal places would have
been 7.343 because the number 5 is too large to ignore. Thus if the number behind
the number of specified decimal places is either 5 or larger, we add 1 to the last
digit of our answer.
Example:
Give the following numbers to 3 decimal places

5.34211
67.8755
34.5625
2.452345
=
=
=
=
5.342
67.876
34.563
2.452
Significant Figures
When asked to write a number down to a certain number of significant figures, we
are (in similar way to the number of decimal points), ignoring the last figures of a
particular number to make it simpler while we lose only little accuracy. For example
if we have 8,432,311, and asked to write it down to 2 significant figures, we take the
first two integers and replace the rest with zeros, thus becoming 8,400,000.
Give the following to 2 significant figures.

=
=
=
45,000
240,000
1,500
Similarly for decimal numbers:

0.004523
0.0000345
0.0783
1.4.1.
=
=
=
0.0045
0.000035
0.078
Addition and Subtraction of Decimal Numbers
Addition and subtraction of decimal numbers is no different from that for whole
numbers. However, in order to avoid confusion because of the decimal, one should
place the numbers under each other with the decimal points beneath each other.
This helps put the powers under each other i.e. tens with tens and hundreds with
hundreds, etc
Example:
1345.65
133.4
1212.25
7823.88 +
1455.31
9279.19
4534
133.4
4400.6
2342.56 +
33.22
2375.78
236.56
184.7
51.86
8356.55 +
452.86
8809.41
Pg 7

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1.4.2.
Multiplication of Decimal Numbers

Example:
254.362 / 12.2
The presence of the decimal point makes multiplication of decimal numbers more
complicated than for whole numbers. Therefore, the decimal point is removed by
shifting it until we have a whole number. The number of times the decimal point is
shifted has to be recorded. This is done for both the multiplicand and the multiplier.
Example:
234.56
8462.768
0.736
0.4664
23456
8462768
736
4664
Then we simply divide 2543.62 by 122. The answer is immediately

correct, without the need to shift decimal points.
(2dp)
(2dp)
(-3dp)
(-3dp)
Thus if we have 17.45 x 11.3, this becomes 1745 x 113 (2 + 1 dp)

Once the decimal point is removed, multiplication is done normally. However, since
we have changed the original multiplicand and multiplier (by moving the decimal
point), the answer needs to have its decimal point moved back. The number of
times the decimal point is shifted is equal to the total number of times it was shifted
before multiplication.
Example:
We cannot divide with 12.2 so this becomes 122 (1dp)

254.362 is also shifted by 1dp, becoming 2543.62
So basically 254.362 / 12.2 = 2543.62 / 122
Exercises
Calculate:
1.
2.
3.
4.
23.76 x 1.5
56.87 x 3.7
46.36 x 45.89
123.78 x 1.83
5.
6.
7.
8.
234.5 / 7.2
786.783 / 8.4
923.67 / 11.3
7362.677 / 20.5
If we have 23.3 x 11.2, this becomes 233 x 112 (1 + 1 = 2dp)

233 x 112 = 26096, but the final answer needs to have its decimal
point shifted by 2dp. Hence, final answer is 260.96.
If we have 132.5 x 87.67, this becomes 1325 x 8767 (1 + 2 = 3dp)
1325 x 8767 = 11616275, but the final answer needs to have its
decimal point shifted by 3dp. Hence, final answer is 11616.275.
1.4.3.
Division of Decimal Numbers
The problem in the case of division is the decimal point within the divisor. Again, the
solution is to move the decimal point, but unlike the multiplication case, we move
only the decimal point of the divisor. Then the number of decimal points shifted in
the divisor, is shifted also for the dividend. In this manner, the operation is not
modified and we do not need to move the decimal point of the final answer.
Pg 8

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1.5. Factors and Multiples

When we multiply 4 x 6, we get 24. In this case, 4 and 6 are factors of 24. But 24
has other factors apart from 4 and 6. For example, 2 and 12 can be multiplied to
give 24 as well. Thus, 2 and 12 are also factors of 24 as are also 3 and 7.
Since any number multiplied by 1 will remain the same, 1 is a factor of all numbers.
In this case, it is called a trivial factor and usually not used since it has little
importance.
Multiples are the resulting numbers when the factors are multiplied. For example if
we have 4 x 6, 4 and 6 are the factors and 24 is the multiple of both 4 and 6.
When a number has an index with no sign ( + or -), it is assumed that it is positive.
Laws of Indices
2.
a m a n = a m+n
a m a n = a m n
3.
(a )
1.
4.
5.
1.6. Powers and Roots
6.
When a number is multiplied by itself, it is raised to a certain power. For example, 5
2
x 5 can be written as 25 or as 5 , where the 2 represents how many times the
3
number is multiplied by itself. So if we have 6 x 6 x 6, this can be written as 6 .
Taking this example, the number 6 is called the base number, and the 3 is called
the index. When written in this form, the number is expressed as an exponent.
2+4
2 x2 =2
4
4-2
=3 =9
2 3
2x3
= 2 = 64
3 /3 = 3
m n
= 2 = 64
= a mn
a0 = 1
(2 ) = 2
am n = n am
a n = 1 n
a
24 2 = 2 24
The root of a number is that value which when multiplied by itself a certain number
of times, produces that number. If 4 x 4 gives 16, then 4 is the root of 16. We often
represent this with the symbol (
), so than
16 =4. In this case, 4 was multiplied

2
We can also say that if for example, we have 2 x 2 x 2 x 2, which can be written as
4
4
2 , 2 is called the fourth power of the base 2.
Example:
and is represented with
64 .
3 is 3 x 3 = 9
6
5 is 5 x 5 x 5 x 5 x 5 x 5 = 15625
Note that for square root, we dont need to write down the 2 next to the root sign. It
is implied that it is the square root of the number. But for cube root and higher
values, the index needs to be clearly written down.
A negative exponent implies a fraction and indicates the inverse (or reciprocal) of
the number.
Example:
by itself only once. We therefore call 4 the square root of 16 since 4 gives 16. If
3
we had 4 x 4 x 4, we get 64, and since 4 gives 64, 4 is called the cube root of 64
1 1 1 1
= 3 = =
2
2 2 8
2
-3
Examples:
27 = 9
since 9 = 27
625 = 5
since 5 = 625
729 = 3
since 3 = 729
4
6
Therefore, 2 is the reciprocal of 2 .

Any number to power zero, is equal to one.
Pg 9

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Exercises
Standard Form
Work out the following:
Often, in both engineering and mathematics, it is convenient to represent numbers

in standard form as shown below, to make calculations easier, and reduce the
possibility of mistakes.
1.
32 35
2.
55 52
3.
(7 )
4.
5.
67 4
4 1
6.
7.
8.
2 3
(3
3 2 3
3
1
2
1
2
1
23
1,000,000
100,000
10,000
1,000
100
10
0
1/10 = 0.1
1/100 = 0.01
1/1,000 = 0.001
1/10,000 = 0.0001
=
=
=
=
=
=
=
=
=
=
=
1 x 10
5
1 x 10
4
1 x 10
3
1 x 10
2
1 x 10
1
1 x 10
0
-1
1 x 10
-2
1 x 10
-3
1 x 10
-4
1 x 10
67 4 62 5
Pg 10

____________________________________________________________________________________________________________________________________________
1.7. Measures and Conversion Factors

Length
1in
=
1m
=
1ft
=
12in
=
3ft
=
1yd
=
1km
=
1mile =
2.54cm
39.37in or 3.281ft
0.3048m
1ft
1yd
0.9144m
0.621miles
1.61km or 5,280ft
Area
2
1m
2
1m
2
1ft
2
1in
10.76ft
2
10,000cm
2
2
0.0929m or 144in
2
6.452cm
=
=
=
=
Volume
3
1m
=
3
1ft
=
1lt
=
3
1ft
=
1gal
=
1gal
=
Mass
1amu
1000kg
1000g
1slug
=
=
=
=
1,000,000cm
3
3
1728in = 0.0283m
3
1000cm = 1.0576qt
7.481gal
8pints
4.546ltrs (3.785ltrs if American gal)
-27
1.66 x 10 kg
1 metric tonne = 0.984tons
1kg
14.59kg
Force and Weight

1N
=
0.2248lb
1lb
=
4.448N
1lb
=
16oz
Velocity
1mph =
1.47ft/sec
1m/s
1knot
1knot
1knot
1knot
1mph
=
=
=
=
=
=
3.281ft/sec
1 nautical mile per hour
1.688ft/sec
1.151mph
1.852km/hr
1.61km/hr
Energy
1J
=
1cal
=
1Btu
=
0.738ft.lb
4.186J
252cal
Time
1year
1day
=
=
365days
24hr = 1,440min
Power
1HP
1HP
1W
1W
1Btu/hr
=
=
=
=
=
550ft.lb/sec
746W
1J/sec
0.738ft.lb/sec
0.293W
Pressure
1atm =
1atm =
1atm =
1atm =
1Pa
=
1bar
=
1bar
=
76.0cmHg
760mmHg
29.92inHg
2
14.7lb/in
2
0.000145lb/in
2
14.5lb/in
100,000Pa
Fundamental Constants
g
=
32lb/slug or 9.81N/kg
Other Useful Data
1 litre water
=
1 pint water
=
1kg
1lb
Pg 11

____________________________________________________________________________________________________________________________________________
Exercises
1. How many centimetres are there in 3.5ft?
2. How many yards are 1.5km?
3. If a table is 3ft wide and 1.6m long, how many square inches is its area?
4. How many pints are there in 5 cubic feet of water?
5. How many litres are there in 4.5 US gallons?
6. If an aircraft travels at 100knots, what is its speed in km/hr?
7. How much will 4.3 cubic feet of water weigh?
8. If a machine produces 670 ft.lb/s, how much is this in horsepower?
9. How many mm of Mercury are 1.7atm?
10. How many slugs are 25kg?
Pg 12

____________________________________________________________________________________________________________________________________________
1.8. Ratio and Proportion

A ratio is a comparison between two similar quantities. For example if we are
comparing a model aircraft to the real thing, we say that their dimensions have a
ratio of 1:25. This means that every centimetre of the model represents 25cm on
the real aircraft. It is important when using ratios that the numbers quoted have all
the same units (e.g. cm to cm, km to km, etc).
Often ratios are used in gearboxes. In this case can represent the ratio of teeth
between gears, the speed ratio of the gearbox, or the torque ratio.
Example: A turboprop gas turbine is attached to the propeller via a gearbox whose
speed ratio is 1:5 so that the propeller rotates slower than the gas turbine. If the
turbine rotates at 20,000rpm, what will be the propeller speed?
The ratios can be written in this form:
( 1 : 5 ) = ( prop rpm : 20,000)
Gas Turbine
20,000rpm
This is a clear case of direct proportion since the more fuel is carried, the farther the
aircraft will travel. Therefore by cross-multiplication:
1 tonne of fuel = 50 nautical miles
4 tonnes of fuel = 200 nautical miles
1.8.2.
In inverse proportion, two quantities are related in a manner that if one increases,
the other has to decrease. Using the car example again, if a car has to travel a fixed
distance, the speed is inversely proportional to the time taken i.e. the faster the
speed, the less time taken to cover the distance.
Example:
The first and fourth numbers (i.e. 1 and

20,000) are called the extremes, while the
second and third numbers are called the
means.
In proportion, the product of means is equal to the product of extremes.

Therefore, (1 x 20,000) = (5 x prop rpm). So the propeller rpm can be calculated by
using {(1 x 20,000) / 5} = 4,000 rpm.
A car travels at 10mph, for 100 miles. How long will it take the car
to cover the same distance if travelling at 20mph?
10 mph for 100 miles = 10 hours
20 mph for 100 miles = 5 hours
Note: There is no mathematical method of defining whether a problem is direct or

inverse proportion. One has to rely on understanding the question properly and
reasoning on it.
1.8.3.
1.8.1.
Inverse Proportion
Proportional Parts
Direct Proportion
Direct proportion means that two quantities are directly related in a manner that if
one increases, the other quantity will increase as well. For example if a car is
travelling at constant speed, the time and distance covered are directly proportional
i.e. if the distance is large, the time taken will also be large, and vice-versa.
Example: A long-range bomber consumes 1 tonne of fuel every 50 nautical miles. If
before the mission it uploads 4 tonnes of fuel, what is the maximum distance it can
travel?
Say that for a particular flight, the aircraft requires 24 tonnes of Jet-A fuel. However
to distribute the weight properly, the captain needs to fill the three available fuel
tanks with a weight ratio of 3 : 4 : 5. How many tonnes will have to be loaded in
each tank?
The first step is to add the ratios i.e. 3 + 4 + 5 = 12 parts. The total tonnage is then
divided by this sum, so 24/12 = 2. This result is then multiplied to each tanks
corresponding ratio number (i.e. 3 x 2 = 6, 4 x 2 = 8 and 5 x 2 = 10). Thus, the
resulting tonnage will be divided into 6 tonnes, 8 tonnes and 10 tonnes.
Pg 13

____________________________________________________________________________________________________________________________________________
1.8.4.
Constant of Proportionality
Exercises
One can write the expression for direct proportion as (y x); meaning y is directly
proportional to x. In cases of inverse proportion, this can be written down as (y
1/x). It is possible to convert this proportional expression to a proper equation by
introducing the constant of proportionality, usually represented by the letter (k).
Hence the direct proportion expression can be written as (y = kx) and inverse
proportion as (y = k/x).
Example:
The rotational speed of a jet engine (w) is directly proportional to

the fuel flow (m). It is also inversely proportional to the square of
the bleed air (b) and the bearing friction (f).
Therefore, this can be written as w m
f b
The proportional sign can be removed by inserting the constant of
proportionality: w = k m
.
f b
a.
A model aircraft has a scale of 1:17. If the wingspan on the model is

25cm, what is the wingspan on the real aircraft in metres?
b.
A gearbox has a speed ratio of 1:5. If the high-speed side is rotating at

25,000rpm, how fast is the slower side rotating?
c.
What is the ratio between the length and the width of a wing if it is 25m
long and 1.25m wide?
d.
During a C-check, an engine check takes 7 hours if performed by 4

maintenance personnel. How long will it take if the number of personnel
is increased to 7?
e.
Five men take 25 hours to produce assemble a certain amount of

engines. How much time would it take if there were eight men on the
job?
f.
Three components on an aircraft require an oil change. Their oil

requirement ratio is 3:5:7 and the first component needs 5 litres. How
much oil is required in total?
g.
If the fuel consumption of an engine is proportional to the thrust and

inversely proportional to the altitude, what is the constant of
proportionality if the fuel consumption is 15kg/s when the thrust is
15,000N and altitude is 20,000 feet?
h.
An electrical resistance of a wire (R) varies directly with the length (L)
and inversely with the square of the radius (r). How can this be
represented?
Pg 14

____________________________________________________________________________________________________________________________________________
1.9. Averages and Percentages

1.9.1.
Mean
1.9.4.
This is the most common average used. To find the mean of a set of values, the
sum of the values has to be divided by the number of values.
Example:
Find the mean of {3, 5, 14, 4, 13, 25}

The sum of these values is [3+5+14+4+13+27] = 66
The number of values we have is 6.
Therefore the mean is 66/6 = 11.
1.9.2.
Sometimes when expressing fractions, it is easier to express them using 100 as

denominator. For example can also be expressed as 25/100. When fractions
have 100 as denominator, they are called percentages.
Example:
Median
Note:
To find the median of a set of values, first the numbers must be placed in
descending (or ascending) order. The median will be the number in the middle.
Example:
Percentages
equivalent to 70 100 . This can be expressed as 70%.

45
0.45 =
= 45%
100
52.5
52.5% =
= 0.525
100
10 is
To convert a vulgar fraction or decimal fraction into a percentage, multiply

by 100. To convert a percentage into a fraction, divide it by 100.
Find the median of {5, 7, 12, 4, 16, 21, 2, 15, 1}

In ascending order, these will become [1, 2, 4, 5, 7, 12, 15, 16, 21]
The median is the middle number i.e. 7.
1.9.3.
Mode
The mode (or modal value) is the value, which appears the most within a set of
values.
Example:
Find the mode of {13, 20, 23, 39, 23, 40, 23, 14, 12, 56, 23, 29}
Putting them in order {12, 13, 14, 20, 23, 23, 23, 23, 29, 39, 40, 56}
It is therefore easy to find that the most common number is 23.
Thus, the mode is 23.
Pg 15

____________________________________________________________________________________________________________________________________________
Exercises
a. Find the mean, mode and median of:
i)
ii)
iii)
{3, 4, 3, 7, 2, 9, 12, 6, 23, 45, 2, 9}

{45, 23, 76, 23, 77, 90, 72, 16}
{34.5, 11.5, 56.2, 8.1, 9.65}
b. Convert to percentage:
i)
ii)
iii)
iv)
v)
4/5
3/7
4/9
13/7.5
7/5
c. Convert to fractions or decimal numbers:

i)
ii)
iii)
iv)
v)
50%
75%
35%
23.5%
54.87%
d. An aircraft travels 150km in 2.7 hours. What is its average speed?

e. A fighter jet consumes 1.2 tonnes of fuel in 2.7 hours. What is the average fuel
consumption per hour?
e. Find 22% of 160.
f. How much oil is required if 20ltrs satisfy only 35% of the required volume.
g. A cargo aircraft is carrying 165 cases. 20% of these weigh 35kg, 30% weigh
45kg, and 25% weigh 50kg. The rest weigh 55kg. How much weight is the
aircraft carrying?
h. How long far would have an aircraft travelled in 3 hours, if for 20% of the time its
average speed was 130km/hr, 25% at 100km/hr and the rest at 150km/hr?
Pg 16

____________________________________________________________________________________________________________________________________________
Triangle:
1.10. Areas and Volumes
Area = b x h
1.10.1. Areas
When we measure the area of something, we are measuring how many square
2
units it contains. This means that if we have an area of 5.5m , then this space
would contain 5 squares, each being 1m in width and 1m in breath.
1 square metre
1 square centimetre
1 square millimetre
1m
2
1cm
2
1mm
1 square inch
1 square foot
1 square yard
1in
2
1ft
2
1yd
or
Area = sqrt s (s-a)(s-b)(s-c)
where s = (a+b+c)
2
Perimeter = a + b + c
Circle:
Rectangle:
Area = r
Area = l x b
Perimeter = 2l x 2b
Perimeter = 2r
1.10.2. Volumes
1 cubic metre
1 cubic centimetre
1 cubic millimetre
Parallelogram:
1m
3
1cm
3
1mm
1 cubic inch
1 cubic foot
1 cubic yard
1in
3
1ft
3
1yd
Cylinder:
Area = b x h
Perimeter = Sum of all 4
sides.
Volume: r h
Surface Area: = 2r (h+r)
r
h
Pg 17

____________________________________________________________________________________________________________________________________________
Exercises
Cone:
2
Volume: 1/3 r h
i)
The passenger section of the aircraft is 4m wide and 30m long. A

centre aisle, 1m wide is installed and the remaining area is to be
carpeted. How many square metres of carpet are required?
ii)
Calculate the surface area of the cockpit having the shape of a cone,
5m side length (i.e. not the vertical height) and base diameter 3m.
iii)
What volume of material is required to create a hollow sphere which

has an outer radius of 2m and internal radius if 1m.
iv)
Calculate the volume of an aircraft having a radius of 3m, 25m long.

The front and rear ends are shaped as a cone 3m long.
v)
A hollow section beam is to be used as part of the wing structure. Its

outer dimensions are 7cm x 7cm and each side is to be 1cm thick.
What volume of aluminium is required if the beam is to be 6m long?
Surface Area: rl
h
r
Sphere:
Volume: 4/3 r
Surface Area: 4r
Solids having uniform cross-section e.g. cube, oblong, etc:

Volume = Cross-sectional area x length of shape
Surface area = sum of all areas enclosing the shape.
Pg 18

____________________________________________________________________________________________________________________________________________
2. ALGEBRA
2.1.2.
2.1. Introduction
In algebra, symbols or letters are used to represent a variable number. So the letter
(a) can represent any number. Coefficients are used in front of the letters and
these show by how much the variable number is to be multiplied. For example (7a)
means any value given to (a) is to be multiplied by 7.
We already mentioned factors in arithmetic. The same principles apply to algebraic
terms. If we have the expression (ab), then a and b are the factors. Again, the
factors 1 and ab are trivial factors and are not used.
Examples:
Find the factors of:

i)
ii)
iii)
8
xy
12
=
=
=
2 and 4
x and y
2, 3, 4, 6
In order to multiply two simple algebraic terms, e.g. (a x b), we only need to write
them down as (ab) and similarly if we have an algebraic term and an integer (e.g. 2
x a = 2a).
Examples:
Addition and Subtraction
Commutative Law of Addition: The sum of numbers is always the same, regardless
the order in which they are written i.e. a + b + c = a + c + b.
Associative Law of Addition: The sum of numbers is always the same, regardless in
which manner they are grouped i.e. (a + b) + c = a + (b + c).
In order to be able to add or subtract algebraic expressions, they need to have the
same terms i.e. have like terms. For example, (7a + 3b) cannot be simplified
further because we have unlike terms. However, (7a + 2a) can be worked out and
gives 9a.
Examples:
5a + 3c -2a = 3a + 3c = 3(a + c)
11b 3c = 11b 3c
8z + 5a z 2a = 7z + 3a
-1
5 x a = 5a
2a x b = 2ab
2
2a x a = 2a
5ab x 2t = 10abt
b xb =b
2
sb x bytu = b styu
Note: The order in which the terms are written down is not important but by
convention, they are usually written down in alphabetical order.
The same rules apply for division of algebraic terms.
Examples:
2.1.3.
2.1.1.
Multiplication and Division
x /2x = x/2
4
2
2
3x /2x = 3x /2
Brackets
When a term is written in the form 5(a + b), the brackets can be opened and each
term inside the brackets has to be multiplied by the outside term.
Example:
7(a + c) = 7a + 7c
2
5a(a + c) = 5a + 5ac
Suppose we are required to multiply (1 + a) and (1 + b). The first step is to multiply
(1 + a) by the first character of the second bracket (i.e. 1). Then, (1 + a) has to be
multiplied by the second character (i.e. b). Hence (1 + a).b gives (b + ab). The two
results are then summed together to obtain (1+a) + (b + ab).
The final answer is thus (1 + a + b + ab)
Example:
(i) (3a + c) (a + c) = (3a + ac) + (3ac + c )

2
2
= 3a + 4ac + c
2
(ii) (2a b) (a + b) = (2a ab) + (2ab b )

2
2
= 2a + ab b
Pg 19

____________________________________________________________________________________________________________________________________________
Exercises
Find:
i)
3x + 5x 7b
ii)
12b 3a + 4x -13b
iii)
-5c 23a + c
iv)
7m n + 5m + 11n
v)
3m(n + z)
vi)
2a(a b )
vii)
6d(2a + c)(c)
viii)
2c (2a c)(5a)
ix)
(3a b)(-a + 2b)
x)
(-2a + 3)(a b)
xi)
(4a 3ab)(a c)(2a)
xii)
(4a + 3c a)(2b + c)
xiii)
(x + 3)(x 2)
xiv)
(x + 5)(x 9)
xv)
(2x + 3)(x 5)
xvi)
(x 3)(3x + 3)
Pg 20

____________________________________________________________________________________________________________________________________________
2.2. Linear Equations
2.3. Simultaneous Equations
Linear equations with one unknown contain just one letter to the power one, e.g.
2x 5 = 7 .
In previous sections, we solved algebraic linear equations. In these cases, we only

had one unknown and solutions were quite straightforward. There cases however
were linear equations have more than one unknown. In such a case, one needs
another equation in order to obtain a solution.
When two different algebraic expressions are equal, we have an equation.

For example:
x x 1
=
is an equation.
3
2
Example:
The solution of an equation is the number, or set of numbers, which the letters
represent in that equation. A problem can be solved if we can use algebra to relate
the given information to form an equation and then solve it.
2.2.1.
Solving simultaneous equations by elimination

Simultaneous can be solved by elimination, meaning eliminating one of the
unknowns. One easy way to do this is to subtract one equation from the other.
Transposition of Formulae
Often it is required to rearrange the formula to change the subject of the formula.
Example:
Whatever we do to the left-hand side of a formula or equation, we must do the

same to the other side, or when we take any term over the equals sign, we change
sign.
Examples:
2 x + y = 8 and x + y = 5
These two linear equations form a set of simultaneous equations.
We have to find a value of x and y which satisfies both equations.
(2 x + y ) (x + y ) = 8 5
x=3
Substituting 3 for x, in any equation gives y = 2.
Make c subject of the formula.
a+c =b
becomes
c=ba
c =b
a
becomes
c = ab
becomes
c=
cx 3
a
=b
bx 3
a
Sometimes one letter can be eliminated by adding the equations.

Example:
or c = bx 3 a 2
(4 x + y ) = 8
2x y = 6
Subtracting one from the other gives 6 x = 14

Hence x = 7 = 2 1
3
3
Substituting this value in any equation, we get:
28 + y = 8
3
y = 4
Pg 21

____________________________________________________________________________________________________________________________________________
Solving simultaneous equations by substitution
Another common method in solving simultaneous equations is by substituting one
of the unknowns.
Example:
Solve these simultaneous equations.
3 x + 4 y = 7 and 2 x + y = 4
nd
Taking the 2 equation, one can re-write it as y = 4 2 x

Substituting this for y in the first equation we get:
3 x + 4(4 2 x ) = 7
3 x + 16 8 x = 7
3 x 8 x = 7 16
5 x = 9
x=9
With this value of x in any of the equations, we get a value for y.

Using the first equation, we get:
( 5 )+ 4 y = 7
39
27 + 4 y = 7
5
4 y = 7 27
4y = 8
y=2
5
Pg 22

____________________________________________________________________________________________________________________________________________
2.4. Quadratic Equations
2.4.2.
Quadratic equations are similar to linear ones, but the highest power found in the
equation is 2, (e.g. 5 x 2 + 3 x + 5 = 0 )
A quadratic equation can usually be expressed as a product of its factors. For

example; x 2 3 x + 2 = 0 can also be expressed as ( x 1)( x 2) = 0 .
There are various methods how to solve such equations. These include by
factorisation, graphically and by completing the square. However, the most
straightforward method is by using the formula.
Therefore it stands to reason than if ( x 1)( x 2) = 0 , then x is either = 1 or 2.

The values of x that satisfy the equation are called the roots of the equation (i.e. 1
and 2). In general, a quadratic equation has two roots, but sometimes these may be
identical.
2.4.1.
Example:
Solving by using the formula
When in this form, the solution can be found by: x =
b b 2 4ac
2a
x 7x 3 = 0
x=
( 7 )
( 7 )2 4(1)(3 )
2 .1
7 49 + 12 7 61
=
2
2
x = 7.405 or -0.405
x 2 + 5 x + 5 = 7 becomes x 2 + 5 x 2 = 0
x=
(5 )
(5 )2 4(1)( 2)
2 .1
4 x 2 + 12 x + 9 = 0 can be expressed as (2 x + 3 )2 = 0
So there is only one root, i.e.
Any quadratic equation can be written in the form ax 2 + bx + c = 0 , where a, b and c

represent numbers.
Examples:
Solving by factorisation
5 25 + 8 5 33
=
2
2
You probably noticed that most times, the square root part of the formula is not
easy to work out, and sometimes may even be negative and so cannot be solved.
Hence it is sometimes convenient to use the factorisation method.
2.4.3.
, called a repeated root.
Factorising quadratic equations
In order to factorise quadratic equations, we need to be aware of the relationship

between what is inside the brackets and the resulting quadratic.
a. The product of the number terms in the two brackets gives the number term
in the expansion.
b. Collecting the number terms gives the coefficient of x.
c. A positive sign throughout the quadratic comes from positive signs in the
both brackets.
d. A positive number term and a negative coefficient of x come from negative
signs in both brackets.
e. A negative number term in the quadratic comes from a positive sign in one
bracket and a negative sign in the other bracket.
Example:
x 2 + 5x + 6
The x term in each bracket is x , as x 2 can only be x x x .
The sign in each bracket is + so (x + )(x + )
The number terms in the brackets can be 6 and 1 or 2 and 3.
The middle term in the quadratic shows that the sum of the number
is 5.
Therefore the factors are:
(x + 2)(x +3 )
Pg 23

____________________________________________________________________________________________________________________________________________
2.5. Division of Algebraic Fractions

3
Given than (a + b) is a factor of (a + b ), how can we find the remaining factors?

We have to be able to divide them first.
This gives a 2 . This result is then multiplied by the

2
+ a 2b b 4
The same process is repeated again with

the remainder. Therefore, a 2 b + b 3 is divided
by a , giving ab . Again, ab is multiplied by
+ a 2b b 4
a 2b + b 3
then subtracted from a + b , giving a b + b .
a 2 ab
a + b a3 + b3
0
3x + 1
3x + 2
2x + 1 6x2 + 7x + 6
x 4 3 x 11x 4
a 3 + a 2b
a + b , giving a 2 b ab 2 . Subtracting this from
a2 + b2
a4 b4
a 4 a 2b 2
a 3 + a 2b
divisor, i.e. a + b , giving a 3 + a 2 b . This is

3
a2 b2
a2
a + b a3 + b3
The first step is to take a 3 + b 3 and divide it by a .
Examples:
a 2 b + b 3 , gives us the remainder, which is
a 2b + b 3
ab 2 + b 3 .
a 2b
ab 2
+ ab 2 + b 3
3 x 2 12 x
x4
6x2 + 3x
4x + 6
x4
4x + 2
The process is repeated again until we get a remainder of zero.
a 2 ab + b 2
a+b
a3 + b3
a 3 + a 2b
a 2b + b 3
a 2b
ab 2
+ ab 2 + b 3
+ ab 2 + b 3
Pg 24

____________________________________________________________________________________________________________________________________________
Exercises
Solve these linear equations:
1.
3.
5.
7.
9.
3x + 5 = 9
x + 5 = 2x 3
2x + x = 4x + 1
12 x + 5 = 4 x
1x + 3 x = 9 x 5
Factorise the following quadratic expressions:

2.
4.
6.
8.
10.
2x 1 = 7
3x + 5 = 9x
6x + 5 = 9x 5
x + 5x = 9 + 3x
9x + 5 = 4
1.
2.
3.
4.
5.
3x + 6
2a + 4b
8q 18 p
x 2 + 3x
x 2 + 2 x 15
6.
7.
8.
9.
10.
x 2 3 x 10
x 2 + 6x 7
x 2 2x 3
2x 2 4x 6
3 x 2 + 16 x + 5
Solve these simultaneous equations:

1.
2.
3.
4.
5.
3x + y = 9
x + 5 y = 2x 3
2x + y = 4 y + 1
x + 5 y = 4x
1y + 3 x = 9
Solve these quadratic equations using the factorisation method:
2 x 1 = y
3x + 5 y = 9
6x + 5 y = 9
x + 5y = 9y + 3
9x + 5 y = 4
1.
2.
3.
4.
5.
6x 2 + 4x 2 = 0
4x 2 4x + 5 = 4
10 x 2 x 3 = 0
2x 2 2x 4 = 8
6 x 2 + 11x = 3
Find the values of x for which:

1.
3.
5.
(x 1)(x 3) = 0
3 x (x 7 ) = 0
(2 x 1)(2 x + 3) = 0
2.
4.
6.
(x 3 )(x + 1) = 0
(3 x 2)(x + 2) = 0
(x 1)(3 x 3 ) = 0
Work out:
1.
2.
3.
(3 x
(8 x
(9 x
2
2
2
)
+ 7 x + 2) (2 x + 7 )
+ 2 x + 1) (3 x + 1)
+ 6 x + 3 (x + 4 )
Solve these quadratic equations using the formula method:

1.
3.
7x 2 + 9x + 2 = 0
2x 2 + 3x 2 = 0
2.
4.
x 2 + 6x + 5 = 0
7x 2 + 8x + 9 = 8
Pg 25

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2.6. Logarithms
Any positive number can be expressed as a power of 10. Thus as 1000 can be
3
1.9138
expressed as 10 , 82 can be written as 10
. These powers of 10 are called
logarithms to the base 10.
This means that the logarithm to base 10 of 1000 is 3. This can be written as
log101000 = 3. Manipulation of numbers, expressions and formulae, which are in
index form, may be simplified by using logarithms. Another use for logarithms is to
be able to reduce sometimes the more difficult arithmetic operations of
multiplication and division to those of addition and subtraction. There are a number
of laws that govern logarithms:
Example:
An equation relates the final velocity v of a machine with the machines variables,
w , p and z .
w
pz
20 .
This is given by v =
Transpose the formula and find the numerical value of
w when v = 15, p = 1.24 and z = 34.65.
One can use log 10 v = log 10
Law 1: If a = b c , then c = log b a

Thus
For example, if we are asked to find x when 750 = 10x, it is convenient to use this
laws so that x = log 10 750 , so that x gives 2.8751 correct to 2 SF.
w

pz
20
w
log 10 v = log10 20 .
pz
Law 2: log a MN = log a M + log a N
w
log 10 v = 1.30103
pz
What this law enables us to do is to convert the multiplication of numbers in index

form into that of addition.
so w =
M
= log a M log a N
N
This law allows conversion from division of a number in index form into that of
subtraction. These laws are helpful when doing transposition of formulae.
Law 3: log a
w=
or
log 10 v
w
=
1.30103 pz
( pz ) log10 v
1.30103
(1.24 )(34.65 )(log10 15 ) (1.24 )(34.65 )(1.17609 )

=
1.30103
1.30103
= 38.84
( )
Law 4: log a M n = n log a M

Law 5: log b M =
log a M
log a b
This law enables us to change the base of a logarithm. This is useful when we have
to deal with logarithms that have a base that is not found on the calculator.
Pg 26

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2.7. Number Systems

2.7.1.
Converting from Binary to Denary

To convert to denary, the numbers are laid out in successive powers.
The Binary System
Example:
Up till now we have been using the decimal system i.e. numbers from 0 to 9 and its
is called the decimal system because we have a total of 10 numbers.
Using this system any number can be represented as follows:
3
Convert 11012 into denary:

3
1101 = (1 x 2 ) + (1 x 2 ) + (0 x 2 ) + (1 x 2 )
= (1 x 8) + (1 x 4) + (0 x 2) + (1 x 1)
=8+4+0+1
= 1310
7892 = (7 x 10 ) + (8 x 10 ) + (9 x 10 ) + (2 x 10 )
This arrangement consists of number from 0 to 9, multiplied by the base (10) which
has a particular power (3, 2, 1 and 0). This is called the denary system because it
is based on the number 10.
2.7.2.
In the binary system, instead of 10, we use 2 as our base number. Therefore, all
numbers are represented in powers of 2.
In a hexadecimal system, the base number is 16. However, since in the decimal
numbering system, we have only 10 numbers (0-9). We make up for this by
introducing capital letters A, B, C, D, E and F which represent 10, 11, 12, 13, 14
and 15 respectively.
Example:
4310 = 2 + 2 + 2 + 2
_____________________________________________________________
7
6
5
4
3
2
1
0
Binary2
2
2
2
2
2
2
2
2
Denary10
128
64
32
16
8
4
2
1
Converting Denary to Binary
To convert from denary to binary, we repeatedly divide by 2 and note the remainder
at each stage.
Example:
Converting 25 to binary:
25/2
12/2
6/2
3/2
1/2
=
=
=
=
=
12 remainder 1
6 remainder 0
3 remainder 0
1 remainder 1
0 remainder 1
Least significant digit
Most significant digit
Thus, the binary equivalent of 2510 is 110012. The order, in which the binary number
is written, is from the MSD (Most significant digit) to LSD (Least significant digit).
The Hexadecimal System
Denary
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Binary
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
Hexadecimal
0
1
2
3
4
5
6
7
8
9
A
B
C
D
E
F
Pg 27

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Converting denary to hexadecimal
Exercises
The process is similar to that of the binary system. We divide the denary number by
16 and use the remainder.
Logarithms
Example:
Evaluate:
Convert 513610 to Hexadecimal:

5136/16
321/16
20/16
1/16
=
=
=
=
321 remainder 0
20 remainder 1
1 remainder 4
0 remainder 1
So the hexadecimal equivalent of 513610 is 141016.
=
=
a. log464
b. log93
c. log12525
d. log12111
Express in terms of log a, log b and log c.

MSD
a. log ab
b. log a/b
c. log a b
d. log a /b
c. 4 2 x +1 = 2
d. 5 x 5 x 1 = 10
Solve the equations:
Similarly to convert 9410 to hexadecimal:

94/16
5/16
LSD
a. 3 x = 6
b. 2 2 x = 5
( )( )
5 remainder 14 (= E16)
0 remainder 5
Number Systems
So the hexadecimal equivalent 9410 is 5E16.
Convert first to binary and then to hexadecimal:
Converting hexadecimal to denary
i) 45
Again in a similar manner to the binary system:
Convert to the denary system:
Example:
i) 10110
iv) EA
Convert to denary system BA45:

2
ii) 7
iii) 56
iv) 89
ii) 0110010
v) 8A
v) 120
iii) 0100101
vi) 2C
BA4 = (B x 16 ) + (A x 16 ) + (4 x 16 )
= (11 x 256) + (10 x 16) + (4 X 1)
= 2980
Thus the denary equivalent to BA416 is 298010.
Pg 28

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3. GEOMETRY
4
2
3.1. Graphs
0
3.1.1.
Graphical Representation
In previous sections, equations were solved analytically. It is also possible to solve

these equations graphically. Sometimes transposition and manipulation of the
equation is required. It is important that the equality sign must always be present.
y-axis
-3
-2
-1
-2
-4
-6
In order to plot a graph, two axes are required which intersect at the point zero
called the origin. It is important that a suitable scale for these axes is chosen and it
need not be the same for both axes. To plot points on the graph, we use coordinates. For example, (2,4) represents 2 units on the x-axis and 4 units on the yaxis. The x-ordinate is always quoted first. The x-ordinate is an independent
variable and is plotted on the horizontal axis. The y-ordinate is a dependent
variable and is plotted on the vertical axis.
3.1.2.
Example:
-1
-2
-4
-6
0
0
-4
-4
x-axis
y = 2 x 4 , m = 2 and c = -4. The m part represents the gradient or slope of the

line. The c part represents the y-intercept, i.e. at which point the line crosses the
y-axis.
Plot y = 2x 4
-2
-4
-4
-8
-10
Every linear equation can be written in the standard form of y = mx + c . So for
Graphs of Linear equations
x
2x
-4
y
-8
1
2
-4
-2
2
4
-4
0
3
6
-4
2
So when x = -2, y = -8, so co-ordinates will be (-2, -8)

These types of equations, where the highest power of x is 1, are known as equation
of the first degree or a linear equation. These produce always straight-line graphs.
Pg 29

____________________________________________________________________________________________________________________________________________
Equation of a straight line
It was already mentioned that a linear equation can be written in the form of
y = mx + c . It was also shown how a graph can be plotted if the equation of the line
is known. However, it is also important to know how to determine the equation of
the line starting from two co-ordinates on the graph. The example below shows
exactly this.
Example:
Find the equation of the line having coordinates (5,4) and (3,2).
m=
y y 2 y1 4 2 2
=
=
= =1
x x 2 x1 5 3 2
Hence the gradient or slope of this line is 1

The y-intercept is found by putting one of the co-ordinates into the
general equation and using the value of the gradient found, i.e. 1
In this case, we can use 4 = 5(1) + c . This gives c = -1
Therefore the final general equation is y = 1x 1
It is also possible to find the values of the slope and the y-intercept by using
simultaneous equations. This is done by putting the co-ordinates into the general
equation. For the previous example, we would get 4 = 5m + c and 2 = 3m + c .
Solving these simultaneously, we get the values of m and c for the general equation
of the line.
Example:
Find the law of the straight line illustrated below.

The y-intercept is easy to find since the graph crosses the y-axis at
y = -4. Thus, c = -4.
The slope m can be found by using the formula:
In this case therefore,
m=
y y 2 y1 B C
=
=
x x 2 x1 D 0
m=
y y2 y1
=
x x2 x1
The gradient can be found by taking any two points and using the
equation shown earlier.
Say we take (2,6) and (5,21) as co-ordinates.
m=
y y 2 y1 21 6 15
=
=
=
=5
x x 2 x1
52
3
Hence the straight-line law is y = 5 x 4
Pg 30

____________________________________________________________________________________________________________________________________________
3.1.3.
Alternatively, using simultaneous equations and using the same
two co-ordinates:
6 = 2m + c
and
21 = 5m + c
Solving these simultaneously, we get m = 5 and c = -4.

Hence the straight-line law would be y = 5 x 4 as before.
45
40
Graphs of Quadratic equations
As already mentioned, quadratic equations are those, whose highest power is 2,

e.g. x 2 + 5 x + 2 = y . There are various methods that can be used to solve these
equations and the formula and factorization method were already mentioned in the
algebra section. Quadratic equations can also be solved graphically.
When a quadratic function of the form ax 2 + bx + c is plotted against x, the resulting
curve is known as a parabola and the sign of the coefficient of a, will determine
which way the curve points.
To plot these curves, a table of values needs to be set up in terms of the values of
the independent and dependent variables.
Example:
35
30
x
2
x
-3x
2
y
25
y - axis
Find x if y = x 2 3 x + 2 .
20
15
0
0
0
2
2
1
1
-3
2
0
2
4
-6
2
0
3
9
-9
2
2
4
16
-12
2
6
10
25
5
0
-5 0
10
20
-10
15
Note: Gradients or slopes are not always positive. It is possible to have negative
gradients so m is a negative number, e.g. y = 5 x + 2
y - axis
x - axis
10
0
0
-5
x - axis
Pg 31

____________________________________________________________________________________________________________________________________________
The points on the curve where the it crosses the x-axis are x = 1 and x = 2. These
are the points on the curve for which y = 0 or x 2 3 x + 2 = 0 . Therefore x = 1 and x
= 2 are the solutions of the quadratic equation.
From this graph, it is also possible to solve any equation of the type x 2 3 x = k ,
where k is a constant. If for example, we wish to solve x 2 3 x + 1 = 0 , then
comparing this equation with the one plotted, one just needs to add 1 to both sides
to acquire the equation y = x 2 3 x + 2 = 1 . So instead of taking the points crossing
the line y = 0 ( x axis), we take points on the line y = 1.
At these points, x = 0.4 or 2.6 which are the solutions to the equation
x 2 3x + 1 = 0 .
3.1.4.
Exercises
Plot the graph of:
a) y = 4x 3
b) 2y = 6x + 8
c) 3x = 6y 9
Determine the equation of the line passing through these co-ordinates:

(try using both methods)
Graphical solution of Simultaneous Equations
It is possible to solve simultaneous equations graphically. This is done by plotting

both equations on the same graph. The solution lies where the plots intersect.
Example:
The graph clearly shows 2 points where the functions intersect. These are the
solutions to the simultaneous equations. Thus x = 2, y = 3 and x = 6 and y = 23 are
solutions.
a) (5,2) and (3,9)
b) (6,3) and (-5,0)
c) (-4,3) and (-9,5)
d) (-5, -2) and (-1,3)
e) (11,5) and (7,-3)
f) (-5,-3) and (-11,2)
y = 5 x 7 and y = x 2 3 x + 5
Solve
Plot the graph and solve for:

25
2
a) y = 3x + 2x 1
b) y = 6x -3x +5
20
Solve these simultaneous equations graphically (plot for 0 < x < 5):
15
A
y - axis
10
a) y = 3x - 7
b) y = 5x 10
and
and
y = 2x 5x - 1
2
y = 3x 10x + 5
5
0
-2
-1
-5
-10
-15
x - axis
Pg 32

____________________________________________________________________________________________________________________________________________
3.2. Trigonometry
3.2.1.
3.2.2.
Pythagoras Theorem
In a right angled triangle, the area of the square on the hypotenuse is equal to the
sum of the area of the squares on the other two sides.
2
Hence for the triangle shown, c = b + a .
Sin =
Trigonometrical Ratios
Opposite
Hypotenuse
Cos =
Opposite
Hypotenuse
Tan =
Opposite
Adjacent
These ratios need to be remembered and it is convenient to remember the

SOHCAHTOA.
B
Hypotenuse
A
Example:
Side
opposite
to angle A
Side adjacent to angle A
Find the length of the remaining side.

Example:
B
opp 3
Sin B =
= = 0 .6
hyp 5
adj 4
Cos B =
= = 0.8
hyp 5
opp 3
Tan B =
= = 0.75
adj 4
2
By Pythagoras: c = b + a
2
2
2
Therefore c = 5 + 12 = 144 + 25 = 169
c=
169 = 13cm
Pg 33

____________________________________________________________________________________________________________________________________________
Consider the equilateral triangle:

B
30
30
3
2
60
2
60
1
Sin 60 o =
opp
3
=
= 0.866
hyp
2
Sin 30 o =
opp 1
= = 0.5
hyp 2
Cos 60 o =
adj 1
= = 0 .5
hyp 2
Cos 30 o =
adj
3
=
= 0.866
hyp
2
Tan 60 o =
opp
3
=
= 1.732
adj
1
Tan 30 o =
opp
1
=
= 0.5774
adj
3
Pg 34

____________________________________________________________________________________________________________________________________________
The Sine Curve
The Tangent Curve
1.5
40
30
20
10
0
0
50
100
150
200
250
300
350
400
Tan X
Sin X
0.5
-0.5
-10
-1
-20
50
100
150
200
250
300
350
400
-30
-1.5
-40
degrees
degrees
The Cosine Curve

1.5
1
Cos X
0.5
0
0
50
100
150
200
250
300
350
400
-0.5
-1
-1.5
degrees
Pg 35

____________________________________________________________________________________________________________________________________________
3.2.3.
Use of Tables
3.2.4.
It is possible to find the Sine, Cosine and Tangent of the most common angles, i.e.
90, 30, 45 etc using the relative curves. However, the Sine, Cosine and Tangent
functions of other angles are not so simply calculated, and without the aid of a
calculator, the use of tables is required.
Example:
Rectangular (Cartesian) Co-ordinates
The most common way of giving the position of a point in a plane is based on using
two fixed perpendicular lines called the axes of coordinates, (x and y axis). The
Cartesian co-ordinates represent the horizontal distance from the y-axis (in direction
Ox) and the vertical distance from the x-axis (in direction Oy).
Use the extract from the table of natural sines to find:

Sin 32, Sin 3124, Sin 3228
For example the point P in the diagram

can be represented by the Cartesian
co-ordinates (3, -2).
3
2
0'
6'
12'
18'
24'
30'
36'
42'
48'
Mean Differences
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
1'
2'
3'
4'
5'
30
5000
5015
5030
5045
5060
5075
5090
5105
5120
10
13
31
5150
5165
5180
5195
5210
5225
5240
5255
5270
10
12
32
5299
5314
5329
5344
5358
5373
5388
5402
5417
10
12
-3
-2
-1
-1
-2
-3
Sin 32 = 0.5299
3.2.5.
Polar Co-ordinates
Sin 3124 = 0.521

If the number of minutes is not exactly a multiple of 6, we use the tables of
differences. Since Sin 3224 = 0.5358 and 28 is 4 more than 24, looking in the
difference table, we find the value of 10. This is added to the Sine of 3224.
So Sin 3228 = 0.5368
Polar co-ordinates are expressed in terms of the direct distance from the fixed point
O, together with the angle between Ox and OP. The position of the point P in the
diagram is given as (6, 60). Anti-clockwise
P
rotation is taken as positive, whilst clockwise is
6
taken as negative.
5
4
3
2
60
1
0
Pg 36

____________________________________________________________________________________________________________________________________________
Converting between Polar and Cartesian
Calculate using tables:

a. Sin 33
e. Cos 4013
b. Cos 25
f. Tan 1520
c. Cos 1524
g. Tan 3012
d. Sin 1232
h. Sin 3520
Exercises
Write down the Cartesian coordinates of the following points:
Write down the polar coordinates of the following points:
Plot the following points:

a. (5,7)
e. (4, 60)
b. (-4,2)
f. (-2, 45)
c. (3,-5)
g. (5, -60)
d. (-2,-4)
h. (-4, -50)
Pg 37

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3.3. Geometrical Constructions

3.3.1.
The Circle
Important circle theorems

Theorem 1:
Theorem 2:
The angle that an arc of a circle subtends at its centre is twice the
angle, which the arc subtends at the circumference.
Angles in the same segment of a circle are equal.
Theorem 3:
Theorem 4:
The triangle in a semi-circle is always a right-angle one.

The opposite angles of any cyclic quadrilateral are equal to 180.
Elements and Properties of the circle

The most important elements of the circle can be seen in the below diagram. A
chord is a straight line, which joins two points on the circumference of a circle. The
diameter is a chord drawn through the centre of the circle.
The constant distance between the centre of the circle and the circumference is
the radius.
A tangent is a line, which just touches the circumference of a circle, at one point
and always has a radius that is perpendicular to it. Where this radius meets the
tangent is called the point of tangency.
A chord line cuts a circle into a minor segment and a major segment. A sector of
a circle is an area enclosed between two radii, and a length of the circumference is
the arc length.
D
C
Pg 38

____________________________________________________________________________________________________________________________________________
Theorem 5:
A tangent to a circle is at right angles to a radius drawn from the

point of tangency.
Theorem 6:
The angle between a tangent and a chord drawn from the point of
tangency equals half of the angle at the centre subtended by the
chord.
Theorem 8:
If two circles touch either internally or externally, then the line that
passes through their centres also passes through the point of
tangency.
Theorem 7:
The angle between the

tangent and a chord that is
drawn from the point of
tangency is equal to the
angle at the circumference
subtended by the chord.
B
Pg 39

____________________________________________________________________________________________________________________________________________
3.3.2.
Angles
Degrees and Radians

There are two ways of measuring angles. The most common way is by using the
degrees scale but one may also use the radians scale.
There are 2 radians in 360.
The significance of the is that the radius of the circle fits around the
circumference 2 times.
n
To convert from degrees to radians use:
2
360
50
Example:
Convert 50 into radians.
2 = 0.8727 radians
360
Vertical Angles: any two lines that

intersect, will produce 4 angles. As in
the diagram below, AOB and COD are
called vertical angles and have the
same degree measurement. Angles
AOC and BOD are also vertical angles.
D
C
Alternate Interior Angles: any pair of parallel

lines that are intersected by a third line produce
angles, such that A and D are equal, as are
angles C and B.
Length of Arc
C
To calculate the length of arc:
n
2r
360
Acute Angles: angles measuring between 0 and 90.
Obtuse Angles: angles measuring between 90 and 180.
Alternate Exterior Angles: similarly to interior

angles, any pair of parallel lines crossed by a
third line will produce angles such that A is
equal to D and angle C is equal to angle B.
Reflex Angles: angles measuring between 180 and 360.

C
Right Angles: angles measuring exactly 90. Two lines that meet at right angles
are said to be perpendicular. Any two right angles are supplementary angles.
Complementary Angles: angles whose sum of their degree measures equal to

90. One of the complementary angles is said to be the complement of the other.
Supplementary Angles: angles whose sum of their degree measures equal to
180. One of the supplementary angles is said to be the supplementary of the other.
Corresponding Angles: any pair of parallel

lines intersected by a third line produce an
angle such that angle D is equal to B and angle
A is equal to C.
Pg 40

____________________________________________________________________________________________________________________________________________
3.3.3.
Construction
To bisect an angle when the arms of the angle meet.

From the centre O, two equal arcs must
be produced to cut the arms at A and B.
With centres A and B, two equal length
arcs then meet at C. The line OC
bisects the angle.
AB becomes 100 units with AC at right angles to AB. BC are joined so than angle
ABC is now 2330.
Similarly for the second figure which shows the sine method for angle = 2836.
Sine of 2836 is 0.4787 which when multiplied by 100 becomes 47.87 units long.
This is our arc length from A and AB as before is 100 units. A line is then drawn
from B that just touches the arc. Angle ABC will be 2836.
To bisect an angle when the arms do not meet.

In order to have a common point, two lines parallel to the given arms need to be
drawn far enough to make them meet at a point and then use the above technique
normally.
Finding the centre of a circle.

The figure shows the circle with three
well-spaced points A, B and C on its
circumference. Two chords can be
produced ( A and B, B and C). When
these are bisected, the lines will
intersect at the centre of the circle.
Set out angles using trigonometric ratios.

For this method, a good knowledge of the trigonometric ratios is important. For the
example, a scale factor of 100 will be used. The figure shows how to set an angle
using the tangent ratio. In this case, the angle is 2330 which gives a value of
0.4348. Using a multiplier of 100 units, the line AC = 43.48 units. The horizontal line
Pg 41

____________________________________________________________________________________________________________________________________________
Drawing a common external tangent to two given circles.
To draw a hexagon given the length of a side.
The first figure shows two circles with radii R and r. With centre O1, draw a circle of
radius R r. Join O1 O2 and bisect to obtain centre C. With C as centre, draw a
semi-circle of radius CO1 to cut the inner circle at T. Draw a line from O1 through T
to locate T1 on the outer circle.
A straight line AF is drawn equal to the given length of the side. With centres A and
F, arcs of radius AF intersect at O. With the centre O, a circle of radius OA cuts the
arcs at B and D. With centres B and E, arcs of radius AF cut the circle at C and D
respectively. The points on the circle are finally joined to obtain the hexagon.
The second figure shows the line O2 parallel to O1T1, drawn to cut the smaller circle
at T2. A line is then drawn through T1 and T2 to obtain the external tangent to the
two circles as shown. This construction is very useful to portray a belt drive around
two pulleys.
To draw the inscribed circle for a given triangle.
The figure shows the given triangle ABC with angles A and B both having been
bisected and the bisectors extended to meet at O. In the second figure, a
perpendicular is constructed from O to cut AB at D. Then with centre O and radius
OD the inscribed circle of triangle ABC is drawn.
To blend an arc in a right angle.

Initially, intersecting lines at right angles need to be drawn. From corner A, set out
AB and AD equal to the required radius. From B and D, arcs of the required radius
are drawn which intersect at O. From O, the arc can be drawn to blend with the
straight lines.
Pg 42

____________________________________________________________________________________________________________________________________________
To draw an arc from a point to a circle of radius r.

Set out a radius R from P and radius R + r from O to meet at C. From C, draw an
arc radius R to touch the circle and point P.
Pg 43

____________________________________________________________________________________________________________________________________________
Appendix 1 Trigonometric Tables

Natural Sines
Natural Cosines
Pg 44

____________________________________________________________________________________________________________________________________________
Natural Tangents
Pg 45

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Aircraft Maintenance Training Centre

Part-66 Cat-B1 Mathematics M1

Part-66 Cat-B1 Mathematics M1

1.9. Averages and Percentages ......................................................................15

1.2. Methods of Multiplication and Division .................................................... 2

Addition and Subtraction of Decimal Numbers .................................... 7

1.10. Areas and Volumes ...................................................................................17

1.5. Factors and Multiples ................................................................................. 9

2.2. Linear Equations .......................................................................................21

1.6. Powers and Roots....................................................................................... 9

2.3. Simultaneous Equations ..........................................................................21

1.7. Measures and Conversion Factors ......................................................... 11

Solving simultaneous equations by elimination..........................................................21

2.4. Quadratic Equations .................................................................................23

Part-66 Cat-B1 Mathematics M1

2.5. Division of Algebraic Fractions ............................................................... 24

Degrees and Radians................................................................................................ 40

Appendix 1 Trigonometric Tables.............................................. 44

Part-66 Cat-B1 Mathematics M1

1.1. Arithmetical Terms and Signs

The process of finding the difference between two numbers is known as

Starting from the right, (3 2) gives 1. Next column, (8 9) is not possible so 8

Part-66 Cat-B1 Mathematics M1

1.2. Methods of Multiplication and Division

In multiplication, we are doing addition but repetitively. So (5 x 3 = 5) is similar to (5

Arithmetic Law of Precedence

Part-66 Cat-B1 Mathematics M1

Part-66 Cat-B1 Mathematics M1

Converting between improper fractions and mixed numbers

Converting from an improper fraction to a mixed number is done by simple division.

if we divide both numerator and

into a mixed number.

To convert from a mixed number to an improper fraction, the whole number is

Multiplication of fractions is relatively easy. All one has to do is multiply the

Division of fractions is similar to multiplication. However the divisor (the fraction we

Part-66 Cat-B1 Mathematics M1

Now we divide 24 by 12, which gives 2 and multiply it by the corresponding

Part-66 Cat-B1 Mathematics M1

Convert from improper fractions to mixed numbers:

Convert from mixed number to improper fractions:

Part-66 Cat-B1 Mathematics M1

1.4. Decimal Numbers

Number of Decimal Places

Give the following numbers to 3 decimal places

Give the following to 2 significant figures.

Similarly for decimal numbers:

Addition and Subtraction of Decimal Numbers

Part-66 Cat-B1 Mathematics M1

Multiplication of Decimal Numbers

Then we simply divide 2543.62 by 122. The answer is immediately

Thus if we have 17.45 x 11.3, this becomes 1745 x 113 (2 + 1 dp)

We cannot divide with 12.2 so this becomes 122 (1dp)

So basically 254.362 / 12.2 = 2543.62 / 122

If we have 23.3 x 11.2, this becomes 233 x 112 (1 + 1 = 2dp)

Division of Decimal Numbers

Part-66 Cat-B1 Mathematics M1

1.5. Factors and Multiples

1.6. Powers and Roots

16 =4. In this case, 4 was multiplied

and is represented with