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Contents
ARITHMETIC TERMS & SIGNS ................................................................... 4
SOME DEFINITIONS ................................................................................... 4
SEQUENCE OF ARITHMETICAL OPERATIONS ................................................ 4
FACTORS & MULTIPLES ............................................................................. 6
LOWEST COMMON MULTIPLE (L.C.M.) ........................................................ 6
HIGHEST COMMON FACTOR (H.C.F.) ......................................................... 6
POWER NUMBERS .................................................................................... 6
SEQUENCES ............................................................................................ 8
FRACTIONS ................................................................................................. 9
VULGAR FRACTIONS ................................................................................. 9
REDUCING A FRACTION TO ITS LOWEST TERMS ........................................... 11
TYPES OF FRACTIONS ............................................................................... 12
LOWEST COMMON DENOMINATOR .............................................................. 14
ADDITION OF FRACTIONS .......................................................................... 15
SUBTRACTION OF FRACTIONS.................................................................... 17
COMBINED ADDITION AND SUBTRACTION .................................................... 18
MULTIPLICATION ....................................................................................... 19
CANCELLING ............................................................................................ 20
DIVISION OF FRACTIONS ............................................................................ 21
OPERATIONS WITH FRACTIONS .................................................................. 22
DECIMALS ................................................................................................... 24
THE DECIMAL SYSTEM .............................................................................. 24
MULTIPLICATION & DIVISION OF DECIMALS .................................................. 25
LONG MULTIPLICATION .............................................................................. 27
LONG DIVISION ......................................................................................... 28
DECIMAL PLACES...................................................................................... 29
SIGNIFICANT FIGURES ............................................................................... 29
ROUGH CHECKS FOR CALCULATIONS ......................................................... 32
FRACTION TO DECIMAL CONVERSION ......................................................... 34
CONVERSION OF DECIMALS TO FRACTIONS................................................. 36
FORMULAE ................................................................................................. 38
EVALUATING FORMULA ............................................................................ 38
TRANSPOSING FORMULAE ........................................................................ 39
WEIGHTS, MEASURES & CONVERSION ................................................... 42
THE INTERNATIONAL SYSTEM OF UNITS ...................................................... 42
S I BASE UNITS ........................................................................................ 42
FACTORS OF MULTIPLES & SUB - MULTIPLES .............................................. 42
SPACE & TIME .......................................................................................... 43
MECHANICS ............................................................................................. 43
HEAT 43
EXPRESSING SI UNITS .............................................................................. 44
CONVERSION FACTORS ............................................................................. 45
RATIO & PROPORTION .............................................................................. 46
PROPORTIONAL PARTS ............................................................................. 47
1
............................................................................... 49
INVERSE PROPORTION ............................................................................. 51
DIRECT PROPORTION
AVERAGES ................................................................................................. 52
AVERAGE SPEED ...................................................................................... 54
PERCENTAGES .......................................................................................... 57
PERCENTAGE OF A QUANTITY ................................................................... 58
AREAS ......................................................................................................... 61
VOLUMES .................................................................................................... 66
UNIT OF VOLUME ..................................................................................... 66
UNIT OF CAPACITY ................................................................................. 67
VOLUMES AND SURFACE AREAS ................................................................ 69
SQUARES & SQUARE ROOTS................................................................... 71
SQUARE NUMBERS ................................................................................... 71
SQUARE ROOTS ....................................................................................... 75
CUBED 79
CUBED ROOTS ......................................................................................... 79
1.2 ALGEBRA ............................................................................................. 80
USE OF SYMBOLS ..................................................................................... 80
SUBSTITUTION ......................................................................................... 81
POWERS ................................................................................................. 83
ADDITION OF ALGEBRAIC TERMS ............................................................... 84
MULTIPLICATION & DIVISION OF ALGEBRAIC QUANTITIES .......................... 84
BRACKETS ............................................................................................... 88
ADDITION & SUBTRACTION OF FRACTIONS .................................................. 91
GRAPHS OF EQUATION............................................................................. 93
THE MEANING OF M & C IN THE EQUATION OF A STRAIGHT LINE ...... 96
THE MEANING OF M & C IN THE EQUATION OF A STRAIGHT LINE .................... 96
INDICES & POWERS................................................................................... 99
LAWS OF INDICES ..................................................................................... 99
Multiplication............................................................................... 99
Powers 99
Negative indices ......................................................................... 99
Fractional indices........................................................................ 100
Zero index .................................................................................. 100
BINARY SYSTEM ........................................................................................ 102
OTHER NUMBER SCALES .......................................................................... 105
OCTAL 106
HEXIDECIMAL .......................................................................................... 106
SIMULTANEOUS EQUATIONS ................................................................... 108
ELIMINATION METHOD IN SOLVING SIMULTANEOUS EQUATIONS ............... 108
INDICES AND LOGARITHMS ..................................................................... 113
LAWS OF INDICES .................................................................................... 113
MULTIPLICATION ...................................................................................... 113
DIVISION ................................................................................................. 113
NEGATIVE INDICES................................................................................... 114
FRACTIONAL INDICES ............................................................................... 114
2
ZERO INDEX ............................................................................................ 114
NUMBERS IN STANDARD FORM .................................................................. 115
LOGARITHMS ........................................................................................... 116
ANTI-LOGARITHMS ................................................................................... 117
RULES FOR THE USE OF LOGARITHMS MUTIPLICATION................................. 118
GEOMETRY ................................................................................................. 122
RADIAN MEASURES .................................................................................. 122
RELATION BETWEEN RADIANS AND DEGREES ............................................. 122
TYPE OF ANGLES ..................................................................................... 124
PROPERTIES OF ANGLES AND STRAIGHT LINES ........................................... 124
GRAPHICAL REPRESENTATION .................................................................. 129
USE OF GRAPHS ...................................................................................... 130
Nomograms ................................................................................ 132
TRIGONOMETRY......................................................................................... 134
THE NOTATION FOR A RIGHT-ANGLED TRIANGLE ......................................... 134
THE TRIGONOMETRICAL RATIOS ............................................................... 134
THE SINE OF AN ANGLE ............................................................................ 135
READING THE TABLE OF SINES OF ANGLES ................................................. 135
THE COSINE OF AN ANGLE ........................................................................ 139
THE TANGENT OF AN ANGLE...................................................................... 142
TRIGONOMETRICAL RATIOS BETWEEN 0 AND 360. .................................. 145
POLAR CO-ORDINATES ............................................................................. 150
3
ARITHMETIC TERMS & SIGNS
SOME DEFINITIONS
The result obtained by adding numbers is called the sum. The sum of 4, 6 and 8
is 4 + 6 + 8 = 18. The order in which numbers are added is not important.
4 + 6 + 8 = 6 + 4 + 8 = 8 + 4 + 6 = 18.
The difference of two numbers is the larger number minus the smaller number.
The difference of 15 and 10 is 15 - 10 = 5. The order in which we subtract is very
important. 7 - 3 is not the same as 3 - 7.
The result obtained by multiplying numbers is called the product. The product of
8 and 7 is 8  7 = 56. The order in which we multiply is not important.
8  7 = 7  8, and
3  4  6 = 4  3  6 = 6  3  4 = 72.
SEQUENCE OF ARITHMETICAL OPERATIONS
Numbers are often combined in a series of arithmetical operations. When this
happens a definite sequence must be observed.
1. Brackets are used if there is any danger of ambiguity. The contents of the
bracket must be evaluated before performing any other operation. Thus:
2  (7 + 4) = 2  11 = 22
15 - (8 - 3) = 15 - 5 = 10
2. Multiplication and division must be done before addition and
subtraction. Thus:
5  8 + 7 = 40 + 7 = 47 (not 5  15)
8 ÷ 4 + 9 =
2 + 9
= 11
(not 8 ÷ 13)
5  4 - 12 ÷ 3 + 7 = 20 - 4 + 7 = 27 - 4 = 23
So far we have used the standard operations of add, subtract, multiply and
divide.
4
Exercise 1 - Questions 1 - 4 Level 1 5 - 10 Level 2
Find values for the following:
1. 3 + 5  2
2. 3  6 - 8
3. 7  5 - 2 + 4  6
4. 8 ÷ 2 + 3
5. 7  5 - 12 ÷ 4 + 3
6. 11 - 9 ÷ 3 + 7
7. 3  (8 + 6)
8. 2 + 8  (3 + 6)
9. 17 - 2  (5 - 3)
10. 11 - 12 ÷ 4 + 3  (6 - 2)
5
FACTORS & MULTIPLES
If one number divides exactly into a second number the first number is said to be
a factor of the second. Thus:
35
= 5  7
240 = 3  8  10
63
= 3  21 = 7  9
… 5 is a factor of 35 and so is 7.
… 3, 8 and 10 are all factors of 240.
… 63 is said to be a multiple of any of the numbers
3, 7, 9 and 21 because each of them divides
exactly into 63.
Every number has itself a 1 as factors. If a number has no other factors apart
from these, it is said to be prime number. Thus 2, 3, 7, 11, 13, 17 and 19 are
all prime numbers.
LOWEST COMMON MULTIPLE (L.C.M.)
The L.C.M. of a set of numbers is the smallest number into which each of the
given numbers will divide. Thus the L.C.M. of 3, 4 and 8 is 24 because 24 is the
smallest number into which the numbers 3, 4 and 8 will divide exactly.
The L.C.M. of a set of numbers can usually be found by inspection.
HIGHEST COMMON FACTOR (H.C.F.)
The H.C.F. of a set of numbers is the greatest number which is a factor of each
of the numbers. Thus 12 is the H.C.F. of 24, 36 and 60. Also 20 is the H.C.F. of
40, 60 and 80.
POWER NUMBERS
The quantity 2  2  2  2 is written 24 and is called the fourth power of 2.
The figure 4, which gives the number of 2's to be multiplied together is called the
index (plural: indices).
55 = 5  5  5  5  5 = 15625
73 = 7  7  7 = 343
6
Exercise 2 - Level 2
1. What numbers are factors of:
a) 24
b) 56
c) 42
2. Which of the following numbers are factors of 12:
2, 3, 4, 5, 6, 12, 18 and 24?
Which of them are multiples of 6?
3. Write down all the multiples of 3 between 10 and 40.
4. Write down the two prime numbers next larger than 19.
5. Find the L.C.M. of the following set of numbers:
a) 8 and 12
b) 3, 4 and 5
c) 2, 6 and 12
d) 3, 6 and 8
e) 2, 8 and 10
f) 20 and 25
g) 20 and 32
h) 10, 15 and 40
i)
j)
12, 42, 60 and 70
18, 30, 42 and 48
6. Find the values of:
a) 25
b) 34
d) 62
e) 83
c) 53
7. Find the H.C.F. of each of the following sets of numbers:
a) 8 and 12
b) 24 and 36
c) 10, 15 and 30
d) 26, 39 and 52
e) 18, 30, 12 and 42
f) 28, 42, 84, 98 and 112
7
SEQUENCES
A set of numbers, which are connected by some definite law, is called a series or
a sequence of numbers. Each of the numbers in the series is called a term of the
series. Here are some examples:
1, 3, 5, 7
… (each term is obtained by adding 2 to the previous term)
2, 6, 18, 54
… (each term is obtained by multiplying the previous term by 3)
Example 1
Write down the next two terms of the following series:
112, 56, 28, …
The second term is found by dividing the first term by 2 and the third term is
found by dividing the second term by 2. Hence:
Fourth term
28
= 2 = 14
Fifth term
14
= 2 = 7
Exercise 3 - Level 1
Write down the next two terms of each of the following series of numbers:
1. 3, 12, 48, …
2. 1, 4, 7, 10, …
3. 5, 11, 17, 23, …
4. 162, 54, 18, …
5. 6, 12, 24
8
FRACTIONS
VULGAR FRACTIONS
The circle in the diagram below has been divided into eight equal parts. Each
1
part is called one-eighth of the circle and written as 8 . The number 8 below the
line shows how many equal parts there are and it is called the denominator.
The number above the line shows how many of the equal parts are taken and it is
called the numerator. If five of the eight equal parts are taken then we have
5
taken 8 of the circle.
From what has been said above we see that a fraction is always a part of
something. The number below the line (the denominator) gives the fraction its
name and tells us the number of equal parts into which the whole has been
divided. The top number (the numerator) tells us the number of these equal
3
parts that are to be taken. For example the fraction 4 means that the whole has
been divided into four equal parts and that three of these parts are to be taken.
The value of a fraction is unchanged if we multiply or divide both its
numerator and denominator by the same amount.
3
12
=
5
20
(by multiplying the numerator (top number) and denominator
(bottom number) by 4).
2
10
7 = 35
(by multiplying the numerator and denominator by 5).
12
3
32 = 8
(by dividing the numerator and denominator by 4).
16
1
64 = 4
(by dividing the numerator and denominator by 16).
9
Example 1
2
Write down the fraction 7 with a denominator (bottom number) of 28.
In order to make the denominator (bottom number) 28, we must multiply the
original denominator of 7 by 4 because 7  4 = 28. Remembering that to
leave the value of the fraction unchanged we must multiply both numerator (top
number) and denominator (bottom number) by the same amount, then
2
2  4
8
7 = 7  4 = 28
Exercise 4 - Level 1
Write down the following fractions with the denominator (bottom number) stated.
1.
3
with denominator 28
4
3
2. 5 with denominator 20
5
3. 6 with denominator 30
1
4. 9 with denominator 63
2
5. 3 with denominator 12
1
6. 6 with denominator 24
3
7. 8 with denominator 64
5
8. 7 with denominator 35
10
REDUCING A FRACTION TO ITS LOWEST TERMS
3 7
3
,
and
are said to be in their lowest terms because it is
8 16
52
impossible to find a number which will divide exactly into both top and bottom
9 8
21
numbers. However, fractions like 18 , 12 and 24 are not in their lowest terms
because they can be reduced further by dividing both the top and bottom
numbers by some number which divides exactly into both of them. Thus,
Fractions like
9
1
=
18
2
(by dividing both top and bottom by 9)
8
2
=
12
3
(by dividing both top and bottom by 4)
21
7
24 = 8
(by dividing both top and bottom by 3)
Sometimes we can divide the top and bottom by the same number several times.
Example 2
Reduce
210
336
Hence;
210
to its lowest terms.
336
=
105
168
(by dividing top and bottom by 2)
=
35
56
(by dividing top and bottom by 3)
=
5
8
(by dividing top and bottom by 7)
210
5
336 reduced to its lowest terms is 8 .
Exercise 5 - Questions 1 - 5 level 1.
Questions 6 - 9 level 2.
Reduce the following fractions to their lowest terms:
1.
8
16
4.
15
25
7.
210
294
2.
9
15
5.
42
48
8.
126
245
3.
8
64
6.
180
240
9.
132
198
10.
210
315
11
TYPES OF FRACTIONS
If the top number of a fraction is less than its bottom number the fraction is called
2 5
3
a proper fraction. Thus, 3 , 8 and 4 are all proper fractions. Note that a
proper fraction has a value which is less than 1.
If the top number of a fraction is greater than its bottom number then the fraction
5 3
9
is called an improper fraction or a top heavy fraction. Thus 4 , 2 and 7 are all
top heavy, or improper fractions. Note that all top heavy fractions have a
value which is greater than 1.
Every top heavy fraction can be expressed as a whole number and a proper
1 1
3
fraction. These are sometimes called mixed numbers. Thus, 12 , 53 and 94
are all mixed numbers. In order to convert a top heavy fraction into a mixed
number it must be remembered that:
top number
top number ÷
bottom number = bottom number
Example 3
15
Express 8 as a mixed number.
15
7
8 = 18 (because 15 ÷ 8 = 1 and remainder 7).
From Example 3 we see that we convert a top heavy fraction into a mixed
number by dividing the bottom number into the top number. Notice that the
remainder becomes the top number in the fractional part of the mixed number.
To change a mixed number into an improper fraction we multiply the whole
number by the bottom number of the fractional part. To this we add the
numerator of the fractional part and this sum then becomes the top number of the
improper fraction. Its bottom number is the same as the bottom number of the
fractional part of the mixed number.
Example 4
Express 3
5
as a top heavy fraction.
8
5
38 =
(8  3) + 5
8
=
24 + 5
29
= 8
8
12
Exercise 6
Express each of the following as a mixed number:
1.
7
2
3.
22
10
2.
8
4
4.
12
11
5.
21
8
Express each of the following as top heavy fractions:
6.
3
28
8.
2
83
7.
1
510
9.
7
620
10.
3
47
Remember (L.C.M.)
The L.C.M. of a set of numbers is the smallest number into which each of the
given numbers will divide. Thus, the L.C.M. of 4, 5 and 10 is 20 because 20 is
the smallest number into which the number 4,5 and 10 will divide exactly.
The L.C.M. of a set of numbers can usually be found by inspection.
Exercise 7 - Questions 1 - 7 level 1.
Questions 8 - 10 level 2.
Find the L.C.M. of the following sets of numbers:
1.
4 and 6
6.
20 and 25
2.
2, 6 and 10
7.
10 and 32
3.
2, 4 and 12
8.
5, 15 and 40
4.
3, 4 and 8
9.
6, 42, 60 and 70
5.
4, 8 and 10
10.
18, 15, 42 and 48
13
LOWEST COMMON DENOMINATOR
When we wish to compare the values of two or more fractions the easiest way is
to express the fractions with the same bottom number. This common
denominator should be the L.C.M. of the denominators of the fractions to be
compared and it is called the lowest common denominator.
Example 5
3 5 7
11
Arrange the fractions 4 , 8 , 10 and 20 in order of size starting with the smallest.
The lowest common denominator of 4, 8, 10 and 20 is 40. Expressing each of
the given fractions with a bottom number of 40 gives:
3
3  10
30
4 = 4  10 = 40
5
5  5
25
8 = 8  5 = 40
7
7  4
28
10 = 10  4 = 40
11
11  2
22
20 = 20  2 = 40
Therefore the order is:
22 25 28 30
40, 40, 40, 40
11 5 7
3
20, 8, 10 and 4
or
Exercise 8 - All level 1
Arrange the following sets of fractions in order of size, beginning with the
smallest:
1.
1
2
5
6
2
3
2.
9
10
3
4
3.
13
16
11
20
7
12
4.
3
4
7
8
5.
11
16
6.
3
8
6
7
7
10
3
5
5
8
3
5
7
10
4
7
13
20
9
14
5
9
3
4
2
5
14
ADDITION OF FRACTIONS
The steps when adding fractions are as follows:
1. Find the lowest common denominator of the fractions to be added.
2. Express each of the fractions with this common denominator.
3. Add the numerators of the new fractions to give the numerator of the answer.
The denominator of the answer is the lowest common denominator found in
(1).
Example 6
Find the sum of
2
3
and .
7
4
First find the lowest common denominator (this is the L.C.M. of 7 and 4).
2
3
It is 28. Now express 7 and 4 with a bottom number of 28.
2
2  4
8
7 = 7  4 = 28
3
3  7 21
4 = 4  7 = 28
Adding the top numbers of the new fractions:
2
3
8
21
29
1
+
=
+
=
= 1
7
4
28
28
28
28
A better way of setting out the work is as follows:
2
3
2  4 + 3  7
8 + 21
29
1
+
=
=
=
=
1
7
4
28
28
28
28
Example 7
3
2
7
Simplify 4 + 3 + 10 .
The L.C.M. of the bottom numbers 4, 3 and 10 is 60.
3
2
7
3  15 + 2  20 + 7  6
+
+
=
4
3
10
60
=
45 + 40 + 42
60
=
127
7
=
2
60
60
15
Example 8
1 2
2
Add together 52, 23 and 35
First add the whole numbers together, 5 + 2 + 3 = 10. Then add the
fractional parts in the usual way. The L.C.M. of 2,3 and 5 is 30.
1
2
2
15  1 + 10  2 + 6  2
52 + 23 + 35 = 10 +
30
= 10 +
15 + 20 + 12
30
= 10 +
47
17
= 10 + 1
30
30
17
= 11
30
Exercise 9 - All level 1
Add together:
1.
1
1
+
2
3
7.
3
9
1 + 3
8
16
2.
2
9
5 + 10
8.
2
3
73 + 65
3.
3
3
+
4
8
9.
3
2
3
3 + 5 + 4
8
7
4
4.
3
1
10 + 4
10.
1
5
1
42 + 36 + 23
5.
1
3
7
2 + 4 + 8
11.
3
3
7
5
78 + 24 + 8 + 16
6.
1
2
3
8 + 3 + 5
12.
2
2
3
1
73 + 5 + 10 + 22
16
SUBTRACTION OF FRACTIONS
The method is similar to that in addition. Find the common denominator of the
fractions and after expressing each fraction with this common denominator,
subtract.
Example 9
5 2
Simplify 8 - 5
The L.C.M. of the bottom numbers is 40.
5 2
5  5 - 8  2
25 - 16
9
=
=
=
8 5
40
40
40
When mixed numbers have to be subtracted the best way is to turn the mixed
numbers into top heavy fractions and then proceed in the way shown in
Example 9.
Example 10
7
3
Simplify 3
- 2
10
4
7
3
37 11
37  2 - 11  5
3
- 2 =
=
10
4
10
4
20
=
74 - 55
19
= 20
20
Example 11
2
7
Simplify 5 - 3
5
8
2
7
55 - 38
=
27 31
27  8 - 31  5
5 - 8 =
40
=
216 - 155
61
21
= 40 = 140
40
17
Exercise 10
1.
1 1
2 - 3
5.
7 5
8 - 6
9.
3
9
58 - 210
2.
1 1
3 - 5
6.
1
3
34 - 28
10.
7
9
432 - 310
3.
2 1
3 - 2
7.
5
3 - 7
11.
5
4
116 - 5
4.
7 3
8 8
8.
4
5 - 3
5
COMBINED ADDITION AND SUBTRACTION
Example 12
3
1
1
7
Simplify 58 - 14 + 22 - 16
3
1
1
7
43 5
5
7
5 - 1 + 2 =
+
8
4
2 16
8
4
2 16
=
43  2 - 5  4 + 5  8 - 7  1
16
=
86 - 20 + 40 - 7
16
=
=
(86 + 40) - (20 + 7)
16
126 - 27
99
3
= 16 = 616
16
Exercise 11 - All level 2
Simplify the following:
1.
1
1
3
22 + 34 - 48
6.
7
1
3
1
1210 - 58 + 320 + 12
2.
1
1
1
510 - 32 - 14
7.
3
3
5
3
216 - 210 + 8 + 14
3.
3
1
48 - 22 + 5
8.
3
7
21
13
124 - 68 + 532 - 216
4.
1
1
1
3
62 - 36 + 212 - 44
9.
9
3
7
3
320 + 18 - 210 + 14
5.
3
2
3
5
116 - 25 + 34 + 58
10.
9
4
7
3
225 + 35 - 210 - 20
18
MULTIPLICATION
When multiplying together two or more fractions we first multiply all the top
numbers together and then we multiply all the bottom numbers together. Mixed
numbers must always be converted into top heavy fractions.
Example 13
5
3
Simplify 8  7
5
3
5  3
15
8  7 = 8  7 = 56
Example 14
Simplify
2
2
 3
5
3
2
2 2 11 2  11 22
7
3  


1
5
3 5 3
5  3 15
15
Example 15
3
1
Simplify 18  14
3
1
11
5
11  5
55
23
18  14 = 8  4 =
= 32 = 132
8  4
Exercise 12 - All level 1
Simplify the following:
1.
2
4
3  5
4.
5
11
9  4
7.
2
2
19  15
2.
3
5
4  7
5.
2
1
15  32
8.
7
4
18  17
3.
2
2
9  13
6.
1
2
22  23
19
CANCELLING
Example 16
2
7
Simplify 3  18
2
7
2
15
2  15
30
5
1
3  18 = 3  8 = 3  8 = 24 = 4 = 14
30
The step to reduce 24 to its lowest terms has been done by dividing 6 into both
the top and bottom numbers.
The work can be made easier by cancelling before multiplication as shown
below.
2/1
15
/ 5
1  5
5
1

3/1
8/4 = 1  4 = 4 = 14
We have divided 2 into 2 ( a top number) and 8 (a bottom number) and also we
have divided 3 into 15 (a top number) and 3 (a bottom number). You will see that
we have divided the top numbers and the bottom numbers by the same amount.
Notice carefully that we can only cancel between a top number and a
bottom number.
Example 17
Simplify
16 7 35
 
20 8 4
2
1 6 1 7 3 5 7 1 7  7 49
9
 


4
2 5 5 8 1 4 2
5  1 2 10
10
Sometimes in calculations with fractions the word 'of' appears. It should always
be taken as meaning multiply. Thus:
4
4
20
/ 4
4  4
of
20
=

5
5/1
1 = 1  1
=
16
= 16
1
20
Exercise 13 - All level 1
1.
3
7

1
4
9
7.
3
3
1
34  15  18
2.
1
10
55  13
8.
15
8
1
32  11  245
3.
5
7
1 
8
26
9.
3
of 16
4
4.
1
2
1
12  5  22
10.
5
7 of 140
5.
5
7
2
8  10  21
11.
2
1
3 of 42
6.
1
1
2  1  1
2
3
12.
4
1
of 2
5
2
DIVISION OF FRACTIONS
To divide by a fraction, all we have to do is to invert it (i.e. turn it upside down)
and multiply. Thus:
3
2
3
7
3  7
21
1
5 ÷ 7 = 5  2 = 5  2 = 10 = 210
Example 18
4
1
Divide 15 by 23
4
1
9
7
9
3
27
15 ÷ 23 = 5 ÷ 3 = 5  7 = 35
Exercise 14 - All level 1
1.
4
1
5 ÷ 13
5.
1
3
22 ÷ 34
2.
1
2 ÷ 4
6.
1
5 ÷ 55
3.
5
15
÷
8
32
7.
1
5
315 ÷ 29
4.
3
1
34 ÷ 22
8.
3
3
210 ÷ 5
21
OPERATIONS WITH FRACTIONS
The sequence of operations when dealing with fractions is the same as those
used with whole numbers. They are, in order:
1st.
nd
Work out brackets.
2
Multiply and divide.
3rd
Add and subtract.
Example 19
Simplify
1
1
1
÷  ÷ 
5
2
3
1
1
1
2
1
1
5 ÷ 3 ÷ 2 = 5 ÷ 3  1
1
2
1
3
3
= 5 ÷ 3 = 5  2 = 10
Example 20
4
1
+ 1
5
4
5
Simplify
- 16 .
3
35
2
With problems of this kind it is best to work in stages as shown below.
4
1
16 + 5
21
1
25 + 14 = 3 20
= 320 = 420
1
420
81
18
81
5
9
3 = 20 ÷ 5 = 20  18 = 8
35
9
5
18 - 5
13
= 16
8 - 16 =
16
22
Exercise 15 - All level 2
1.
3
1
7
314 ÷ 149  10


6.
2
2
4
33 ÷ 3 + 5


2.
1
9
1
4 ÷ 8  10
7.
3
1
2
55 - 32  3
1
2
3
3.
2
3
9
1 ÷  ÷
3
10
5
8.
2
2 1
1
  -  +
5
2
3 4 
4.
17  22 - 32
5
3
 8
9.
9
4
316  9
1
1
2 + 6  1
4
5
5.
2
1
23 + 15
4
55
10.
5
7
9 - 15
5
7
1 - 9  15
23
DECIMALS
THE DECIMAL SYSTEM
The decimal system is an extension of our ordinary number system. When we
write the number 666 we mean 600 + 60 + 6. Reading from left to right each
figure 6 is ten times the value of the next one.
We now have to decide how to deal with fractional quantities, that is, quantities
whose values are less than one. If we regard 666.666 as meaning
6
6
6
600 + 60 + 6 + 10 + 100 + 1000 then the dot, called the decimal point,
separates the whole numbers from the fractional parts. Notice that with the
fractional, or decimal parts, e.g. .666, each figure 6 is ten times the value of the
6
6
following one, reading from left to right. Thus 10 is ten times as great as 100 ,
6
6
and 100 is ten times as great as 1000 , and so on.
Decimals then are fractions, which have denominators of 10, 100, 1000 and so
on, according to the position of the figure after the decimal point.
If we have to write six hundred and five we write 605; the zero keeps the place for
3
5
the missing tens. In the same way if we want to write 10 + 1000 we write .305;
6
7
the zero keeps the place for the missing hundredths. Also 100 + 1000 would
be written .067; the zero in this case keeps the place for the missing tenths.
When there are no whole numbers it is usual to insert a zero in front of the
decimal point so that, for instance, .35 would be written 0.35.
Exercise 16 - All level 1
Read off as decimals:
1.
7
10
5.
3
100
2.
3
7
10 + 100
6.
1
7
100 + 1000
3.
5
8
9
+
+
10
100
1000
7.
8 +
4.
9
1000
8.
2
9
24 + 100 + 10 000
9.
8
50 + 1000
6
100
24
Exercise 17 - All level 1
Write down the values of:
1.
2.375 + 0.625
2.
4.25 + 7.25
3.
3.196 + 2.475 + 18.369
4.
38.267 + 0.049 + 20.3
5.
27.418 + 0.967 + 25 + 1.467
6.
12.48 - 8.36
7.
19.215 - 3.599
8.
2.237 - 1.898
9.
0.876 - 0.064
10.
5.48 - 0.0691
MULTIPLICATION & DIVISION OF DECIMALS
One of the advantages of decimals is the ease with which they may be multiplied
or divided by 10, 100, 100, etc.
Example 3
Find the value of 1.4  10.
1.4  10 = 1  10 + 0.4  10
4
= 10 + 10  10 = 10 + 4 = 14
Example 4
Find the value of 27.532  10.
27.532  10
= 27  10 + 0.5  10 + 0.03  10 + 0.002  10
5
3
2
= 270 + 10  10 + 100  10 + 1000  10
3
2
= 270 + 5 + 10 + 100
= 275.32
In both of the above examples you will notice that the figures have not been
changed by the multiplication; only the positions of the figures have been
changed. Thus in Example 3, 1.4  10 = 14, that is the decimal point has
been moved one place to the right. In example 4, 27.532  10 = 275.32;
again the decimal point has been moved one place to the right.
To multiply by 10, then, is the same as shifting the decimal point one place
to the right. In the same way to multiply by 100 means shifting the decimal
point two places to the right and so on.
25
Example 5
17.369  100 = 1736.9
The decimal point has been moved two places to the right.
Example 6
0.07895  1000 = 78.95
The decimal point has been moved three places to the right.
Exercise 18 - All level 1
Multiply each of the numbers in questions 1 to 6 by 10, 100 and 1000.
1.
4.1
6.
0.001753
2.
2.42
7.
0.4853  100
3.
0.046
8.
0.009  1000
4.
0.35
9.
170.06  10
5.
0.1486
10.
0.56396  10000
When dividing by 10 the decimal point is moved one place to the left, by 100, two
places to the left and so on. Thus:
154.26 ÷ 10 = 15.426
The decimal point has been moved one place to the left.
9.432 ÷ 100 = 0.09432
The decimal point has been moved two places to the left.
35 ÷ 1000 = 0.035
The decimal point has been moved three places to the left.
In the above examples note carefully that use has been made of zeros following
the decimal point to keep the places for the missing tenths.
26
Exercise 19
Divide each of the numbers in questions 1 to 5 by 10, 100 and 1000.
1.
3.6
6.
5.4 ÷ 100
2.
64.198
7.
2.05 ÷ 1000
3.
0.07
8.
0.04 ÷ 10
4.
510.4
9.
0.0086 ÷ 1000
5.
0.352
10.
627.428 ÷ 10000
LONG MULTIPLICATION
Example 7
Find the value of 36.5  3.504.
First disregard the decimal points and multiply 365 by 3504.
365
3504
1095000
182500
1460
1278960
Now count up the total number of figures following the decimal points in both
numbers (i.e. 1 + 3 = 4). In the answer to the multiplication (the product), count
this total number of figures from the right and insert the decimal point. The
product is then 127.8960 or 127.896 since the zero does not mean anything.
Exercise 20 - All level 1
Find the values of the following:
1.
25.42  29.23
4.
3.025  2.45
2.
0.3618  2.63
5.
0.043  0.032
3.
0.76  0.38
27
LONG DIVISION
Example 8
Find the value of 19.24 ÷ 2.6.
First convert the divisor (2.6) into a whole number by multiplying it by 10. To
compensate multiply the dividend (19.24) by 10 also so that we now have
192.4 ÷ 26. Now proceed as in ordinary division.
26)192.4(7.4
- this line 26  7
182
10 4
- 4 brought down from above. Since 4 lies to the
10 4
right of the decimal point in the dividend insert a
.. .
decimal point in the answer.
Notice carefully how the decimal point was obtained. The 4 brought down from
the dividend lies to the right of the decimal point. Before bringing this down put
a decimal point immediately following the 7.
The division in this case is exact (i.e. there is no remainder) and the answer is
7.4. Now let us see what happens when there is a remainder.
Example 9
Find the value of 15.187 ÷ 3.57.
As before make the divisor into a whole number by multiplying it by 100 so that it
becomes 357. To compensate multiply the dividend also by 100 so that it
becomes 1518.7. Now divide.
357)1518.7(4.25406
- this line 357  4
1428
907
- 7 brought down from the dividend. Since it
714
1930
lies to the right of the decimal point insert a
decimal point.
1930
- bring down a zero as all the figures in the
1785
dividend been used up.
1450
- bring down a zero.
1428
2200
2142
58
- bring down a zero.
the divisor will not go into 220 so bring down
another zero.
The answer to 5 decimal places is 4.25406. This is not the correct answer
because there is a remainder. The division can be continued in the way shown to
give as many decimal places as desired, or until there is no remainder.
28
DECIMAL PLACES
It is important to realise what is meant by an answer given to so many decimal
places. It is the number of figures which follow the decimal point which give the
number of decimal places. If the first figure to be discarded is 5 or more then the
previous figure is increased by 1.
Thus:
85.7684
= 85.8 correct to 1 decimal place
= 85.77 correct to 2 decimal places
= 85.768 correct to 3 decimal places
Notice carefully that zero must be kept:
0.007362 = 0.007 correct to 3 decimal places
= 0.01 correct to 2 decimal places
7.601
= 7.60 correct to 2 decimal places
= 7.6 correct to 1 decimal place.
If an answer is required correct to 3 decimal places the division should be
continued to 4 decimal places and the answer correct to 3 decimal places.
SIGNIFICANT FIGURES
Instead of using the number of decimal places to express the accuracy of an
answer, significant figures can be used. The number 39.38 is correct to 2
decimal places but it is also correct to 4 significant figures since the number
contains four figures. The rules regarding significant figures are as follows:
1. If the first figure to be discarded is 5 or more the previous figure is
increased by 1.
8.1925
= 8.193 correct to 4 significant figures.
= 8.19 correct to 3 significant figures.
= 8.2 correct to 2 significant figures.
29
Exercise 21- All level 2
1.
18.89  14.2
correct to 2 decimal places
2.
0.036  2.51
correct to 3 decimal places
3.
7.21  0.038
correct to 2 decimal places
4.
13.059  3.18
correct to 4 decimal places
5.
0.1383  0.0032
correct to 1 decimal places
30
Zeros must be kept to show the position of the decimal point, or to indicate that
the zero is a significant figure.
24392
= 24390 correct to 4 significant figures.
= 24400 correct to 3 significant figures.
0.0858
= 0.086 correct to 2 significant figures.
425.804
= 425.80 correct to 5 significant figures.
= 426 correct to 3 significant figures.
Exercise 22 - All level 2
Write down the following numbers correct to the number of significant figures
stated:
1.
24.865 82
(i) to 6
(ii) to 4
(iii) to 2
2.
0.008 357 1
(i) to 4
(ii) to 3
(iii) to 2
3.
4.978 48
(i) to 5
(ii) to 3
(iii) to 1
4.
21.987 to 2
5.
35.603 to 4
6.
28 387 617
(i) to 5
(ii) to 2
7.
4.149 76
(i) to 5
(ii) to 4
8.
9.204 8 to 3
(iii) to 3
31
ROUGH CHECKS FOR CALCULATIONS
The worst mistake that can be made in a calculation is that of misplacing the
decimal point. To place it wrongly, even by one place, makes the answer ten
times too large or ten times too small. To prevent this occurring it is always worth
while doing a rough check by using approximate numbers. When doing these
rough checks always try to select numbers which are easy to multiply or which
will cancel.
Example 10
1.
0.23  0.56
For a rough check we will take 0.2  0.6.
Product roughly = 0.2  0.6 = 0.12.
Correct product
= 0.1288.
(The rough check shows that the answer is 0.1288 not 1.288 or 0.01288.)
2.
173.3 ÷ 27.8.
For a rough check we will take 180 ÷ 30.
Answer roughly = 6
Correct answer = 6.23.
3.
8.198  19.56  30.82  0.198
.
6.52  3.58  0.823
Answer roughly =
Correct answer
8  20  30  0.2
= 40.
6  4  1
= 50.94
(Although there is a big difference between the rough answer and the
correct answer, the rough check shows that the answer 50.94 and not
509.4.)
32
Exercise 23 - All level 1
1.
223.6  0.004 8
2.
32.7  0.259
3.
0.682  0.097  2.38
4.
78.41 ÷ 23.78
5.
0.059 ÷ 0.002 68
6.
33.2  29.6  0.031
7.
0.728  0.006 25
0.028 1
8.
27.5  30.52
11.3  2.73
33
FRACTION TO DECIMAL CONVERSION
We found, when doing fractions, that the line separating the numerator and the
denominator of a fraction takes the place of a division sign. Thus:
17
80 is the same as 17 ÷ 80.
Therefore to convert a fraction into a decimal we divide the denominator into the
numerator.
Example 11
27
Convert 32 to decimals.
27
32 = 27 ÷ 32
32)27.0(0.843 75
25 6
1 40
1 28
120
96
240
224
160
160
...
27
Therefore 32 = 0.843 75.
Example 12
9
Convert 216 into decimals.
When we have a mixed number to convert into decimals we need only deal with
9
the fractional part. Thus to convert 2
into decimals we only have t deal with
16
9
16 .
9
= 9 ÷ 16
16
16)9.0(0.562 5
80
1 00
The division shows that
96
9
hence 216 = 2.562 5.
40
32
9
= 0.562 5 and
16
Sometimes a fraction will not divide out exactly
as shown in Example 13.
80
80
34
Example 13
1
Convert 3 to decimals.
1
3 = 1 ÷ 3
3)1.0(0.333
9
10
9
10
9
1
It is clear that all we shall get from the division is a succession of threes.
This is an example of a recurring decimal and in order to prevent endless
1
repetition the result is written 0.3 Therefore
= 0.3.
3
Further examples of recurring decimals are:
2

3 = 0.6
(meaning 0.666 6 … etc.)
1

6 = 0 .1 6
(meaning 0.166 6 … etc.)
5

11 = 0.45
(meaning 0.454 545 … etc.)
3


7 = 0.428 57 1 (meaning 0.454 545 … etc.)
For all practical purposes we never need recurring decimals; what we need is an
answer given to so many significant figures or decimal places. Thus:
2
= 0.67
3
(correct to 2 decimal places).
5
11 = 0.455
(correct to 3 significant figures).
35
Exercise 24 - Questions 1 - 6 level 1. Questions 7 - 10 level 2.
Convert the following to decimals correcting the answers, where necessary, to 4
decimal places:
1.
1
4
5.
1
2
9.
5
16
2.
3
4
6.
2
3
10.
7
216
3.
3
9
7.
21
32
4.
11
16
8.
29
64
Questions 11 - 16 level 1. Questions 17 - 20 level 2.
Convert the following to three decimal places:
11. 0.3
15. 0.3 5
12.
0.7
16.
13.
0.1 3
0.1 8
17.
0.2 3
0.5 2
18.
0.3 6
14.
19.
20.
0.3 28
0.5 671
CONVERSION OF DECIMALS TO FRACTIONS
We know that decimals are fractions with denominators 10, 100, 1000, etc.
Using this fact we can always convert a decimal to a fraction.
Example 14
Convert 0.32 to a fraction.
32
8
0.32 = 100 = 25
When comparing decimals and fractions it is best to convert the fraction into a
decimal.
Example 15
3
Find the difference between 116 and 1.163 2.
3
116 = 1.187 5
3
116 - 1.163 2
= 1.187 5 - 1.163 2
= 0.024 3
36
Exercise 25 - Questions 1 - 2 level 1. Questions 3 - 8 level 2.
Convert the following to fractions in their lowest terms:
1.
0.2
3.
0.312 5
5.
0.007 5
2.
0.45
4.
2.55
6.
2.125
7.
9
What is the difference between 0.281 35 and 32 ?
8.
19
What is the difference between 64 and 0.295?
37
FORMULAE
EVALUATING FORMULA
A formula is an equation, which describes the relationship between two or more
quantities. The statement that I  PRT is a formula for I in terms of P, R and T .
The value of I may be found by substituting the values of P, R and T . The value
of I may be found by substituting the values of P, R and T .
Example 1
(a) If I = PRT find the value of I when P =20, R =2 and T =5.
Substituting the given values of P, R and T and remembering that multiplication
signs are omitted in formulae, we have
I
= 20 x 2 x 5
=200
(b) The formula v  u  at is used in physics. Find the value of v when u = 8,
a =3 and t =2.
v
=8+3x2
=8+6
= 14
Exercise 1-3 Level 1
1.
If V  Ah , find the value of V when A =6 and h =3.
2.
The formula P 
3.
If a  b  cx, , find the value of a when b =32, c =3 and x =7
4.
The formula V 
RT
is used in connection with the expansion of gases.
V
Find the value of P when R =25, T =230 and V =5
2 gh is used in physics. Calculate the value of V when
g  9.8 and h  7 .
5. Calculate d from the formula d 
2( S  an)
when S  12, a  2, n  5 and
n( n  p )
p  3.
38
TRANSPOSING FORMULAE
The formula y  ax  b has y as is subject. By rearranging this formula we
could make x the subject.
The rules for transforming a formula are:
(1) Remove square roots or other roots
(2) Get rid of brackets
(3) Clear brackets
(4) Collect together the terms containing the required subject
(5) Factorise if necessary
(6) Isolate the required subject
These steps should be performed in the order given.
Example 2
(a) Transpose the formula V 
2R
to make R the subject.
Rr
Step 1 Since there are no roots get rid of the fraction by multiplying both sides of
the equation by ( R  r )
V (R  r )  2R
Step 2 Clear the bracket
VR  Vr  2R
Step 3 Collect the terms containing R on the LHS.
VR  2R  Vr
Step 4 Factorise the LHS.
R(V  2)  Vr
Step 5 Isolate R by dividing both sides of the equation by (V  2).
R
Vr
V 2
Although we used five steps to obtain the required subject, in very many cases
far fewer steps are needed. Nevertheless, you should work through the steps in
order given.
(b) Transpose d 
2hr to make h the subject.
Step 1 Remove the square root by squaring both sides.
2
d  2hr
Step 2 Since there are no fractions or brackets and factorisation is not needed we
can now isolate h by dividing both sides of the equation by 2r
2
d2
 h or h  d since it is usual to position the subject on the LHS.
2r
2hr
39
Exercise 1-10 Level 1 11-30 Level 2
Transpose each of the following formulae:
1. C  ad for d. 2. PV=c for P.
3. x =
b
for y.
y
4. I 
5. S =
ta
for a.
p
6. a =b + 8 for b.
7. y =
7
for x.
4x
E
for E.
R
8. 3k = kx + 5 for k.
9. E =½ mv² for m.
10. y =ms + c for x.
11. v = u + at for t.
12. V =
abh
for h.
3
13. M = 5 (x + y) for y.
14. C =
N n
for n.
2p
15. S = ar(r + h) for h.
16. t = a + (n - 1)d for n.
17. A = 3(x – y) for y.
2
18. d= v + k for k.
200
19. 6x + 2y = 8 for y.
20. y =
2 - 5x
for x.
2  3x
21. k =
3n  2
f or n.
n3
R H
for R.
g
22. T + 2
23. a =
b
for b.
bc
2
24. K = mv for v.
2g
25. r =
26. q =
A
for k.
4
m
p
for p.
27. x = (x - a) (x + b) for a.
40
28. y +
ax
for x.
5  bx
29. x =
5  4y
for y.
3y  2
2

30. T = 2 k h
gh
2
41
WEIGHTS, MEASURES & CONVERSION
THE INTERNATIONAL SYSTEM OF UNITS
Together with major metric countries Britain has adapted the International
System of Units known worldwide as the S I Units.
(Système International d' Unit). The effect of this system is to introduce standard
units for many of the quantities for which a multitude of units exist as present.
S I BASE UNITS
Quantity
Unit
Symbol
Length
metre
m
Mass
kilogram
kg
Time
second
s
Electric Current
Ampere
A
Thermodynamic
Temperature
Kelvin
K
Plain Angles
Radians
Rad
Luminous Intensity (i.e.
brilliance)
candela
cd.
FACTORS OF MULTIPLES & SUB - MULTIPLES
Multiple
Prefix
Symbol
106
Mega
M
103
kilo
k
10
milli
m
10-6
micro
µ
10-9
nano
n
pico
p
-3
10-
12
There are others extending beyond this range both greater and smaller.
42
SPACE & TIME
Quantity
Unit
Symbol
Area
square metre
m2
Volume
cubic metre
m3
Velocity
metre per second
m/s
Acceleration
metre per second squared
m/s2
Angular Velocity
radian per second
rad/s
Angular Acceleration
radian per second squared
rad/s2
Frequency
Herts
Hz = 1/s
MECHANICS
Quantity
Unit
Symbol
Density
Kilogram per cubic metre
kg/m3
Momentum
Kilogram metre per second
kg m/s
Force
Newton
N = kgm/s2
Torque or Moment of Force
Newton metre
N m
Energy, work
Joule
J = Nm
Power
watt
W = J/s
Pressure & Stress
Newton per square metre or
Pascal
N/m2 = Pa
HEAT
Quantity
Unit
Symbol
Celsius temperature
Degrees Celsius
°C
43
EXPRESSING SI UNITS
The symbol for SI units and the conventions which govern their use should be
strictly followed.
1.
Use the correct symbols used in the foregoing lists.
2.
Never use a prefix without a unit either in writing or speech, e.g.
Kilogram or kilometre not kilo.
Millimetre or millilitre not mil.
3.
4.
Always put a zero before a decimal quantity less than Unit, e.g. 0.705 m.
When two units are multiplied together use a small space between the
symbols as the multiplier, e.g.
Kilogram metre squared kg m2.
Newton metre N m.
5.
When dividing, use an oblique stroke to separate the numerator and
denominator.
e.g. metre per second
Joule per second
6.
7.
m/s
J/s
Use a space as a thousands marker not the comma. The comma is used
as a decimal marker in most countries using the metric system and its use as a
thousand marker will cause confusion. Up to four figures may be blocked
together but five or more figures should be grouped in threes, e.g.
1000 mm
=
1m
1 000 000 J
=
1 MJ (MegaJoule)
0.000 000 001 s
=
1 ns (nanosecond).
Leave a small space between figures and symbols.
44
CONVERSION FACTORS
The units which it is thought most likely you will be required to know are set out
below with appropriate conversion factors.
To go from the first quantity into the second multiply by the number given.
Inches
Millimetres
25.4
m
Inches
39.37
Pounds
Kilograms
0.4536
Kilograms
Pounds
2.205
Imp. Galls
Litres
4.546
bar
p.s.i.
14.5
p.s.i.
Pa (Pascal)
6895
bar
Pa
105
N/m2
Pa
1
1bf
N (Newton)
4.45
horsepower
W (Watt)
746
B.Th.U.
KJ
1.055
ft 1bf
J (Joule)
1.356
Knot
=
0.5148 m/s
Knot
=
1.85 Kilometres / hour
Knot
=
1.15 MPH.
Relative Density =
Density of a Substance
Density of Water (at the same temperature)
Density
Mass
Volume
=
Density of a/c fuel is typically
Density of water is
(Units: kg/m3)
800 kg/m3
1000 kg/m3
 R.D. or S. G. (specific gravity) of a/c fuel
= 0.8
R.D. = S.G.
45
RATIO & PROPORTION
A ratio is a comparison between two similar quantities. If the length of a certain
aircraft is 20 metres and a model of it is 1 metre long then the length of the model
1
is 20 th of the length of the aircraft. In making the model the dimensions of the
aircraft are all reduced in the ratio of 1 to 20. The ratio 1 to 20 is usually written
1 : 20.
As indicated above a ratio may be expressed as a fraction and all ratios may be
2
looked upon as fractions. Thus the ratio 2 : 5 = 5 . The two terms of a ratio
may be multiplied or divided without altering the value of the ratio.
1
Hence 6:36 = 1:6 = . Again, 1:5 = 0.20.
6
Before a ratio can be stated the units must be the same. We can state the ratio
between 7 pence and £2 provided both sums of money are brought to the same
units. Thus if we convert £2 to 200p the ratio between the two amounts of money
is 7 : 200.
Example 1
Express the ratio 20p to £4 in its simplest form.
£4
= 4  100p = 400p
20 : 400 =
20
1
=
400
20
Example 2
Express the ratio 4 :
1
in its lowest terms.
4
1
1
4:4 = 4 ÷ 4
4
= 4  1
=
16
1
1
4 : 4 = 16:1
Example 3
Two lengths are in the ratio 8:5. If the first length is 120 metres, what is the
second length?
5
5
The second length = 8 of the first length = 8  120 = 75 metres.
Example 4
Two amounts of money are in the ratio of 12 : 7. If the second amount is £21
what is the first amount?
12
First amount = 7  £21 = £36.
46
Exercise 29 - Question 1 - 7m level 1. Question 8 - 10, level 2.
Express the following ratios as fractions in their lowest terms:
1.
8:3
4.
9 : 15
2.
4:6
5.
8 : 12
3.
12 : 4
6.
Express the ratio of 30p to £2 as a fraction in its lowest terms.
7.
Express the ratio £5 : 80p as a fraction in its lowest terms.
8.
Two lengths are in the ratio 7 : 5. If the first length is 210 metres, what is
the second length?
9.
Two lengths of money are in the ratio 8 : 5. If the second is £120, what is
the first amount?
10.
1
Express 3 : 2 in its lowest terms.
PROPORTIONAL PARTS
The following diagram shows a line AB whose length is 16 centimetres divided
into two parts in the ratio 3 : 5. As can be seen in the diagram the line has been
divided into a total of 8 parts.
The length AC contains 3 parts and the length BC contains 5 parts.
Each part is
16
= 2 centimetres long; hence AC is
8
3  2 = 6 centimetres long, and BC is
5  2 = 10 centimetres long.
47
We could tackle the problem in this way:
Total number of parts = 3 + 5 = 8 parts.
Length of each parts
16
= 8 = 2 centimetres.
Length of AC
= 3  2 = 6 centimetres.
Length of BC
= 5  2 = 10 centimetres.
Example 5
Divide £1100 into two parts in the ratio 7:3.
Total number of parts
= 7 + 3 = 10
Amount of each part
=
Amount of first part
= 7  110 = £770
Amount of second part
= 3  110 = £330
1100
10 = £110
Example 6
An aircraft carries 2880 litres of fuel distributed in three tanks in the ratio 3 : 5 : 4.
Find the quantity in each tank.
Total number of parts = 3 + 5 + 4 = 12.
2880
12
Amount of each part
=
= 240 litres.
Amount of 3 parts
= 3  240 = 720 litres.
Amount of 4 parts
= 4  240 = 960 litres.
Amount of 5 parts
= 5  240 = 1200 litres.
The three tanks contain 720, 1200 and 960 litres.
48
Exercise 30 - Questions 1 - 3, level 1. Questions 4 - 8, level 2.
1.
Divide £800 in the ratio 5 : 3.
2.
Divide £80 in the ratio 4 : 1.
3.
Divide £120 in the ratio 5 : 4 : 3.
4.
A sum of money is divided into two parts in the ratio 5 : 7. If the smaller
amount is £200, find the larger amount.
5.
A alloy consists of copper, zinc and tin in the ratios 2 : 3 : 5. Find the
amount of each metal in 75 kilograms of the alloy.
6.
A line is to be divided into three parts in the ratios 2 : 7 : 11. If the line is
840 millimetres long, calculate the length of each part.
7.
Two plane maintenance hangers have a work force of 336 and 240
respectively. The two hangers are to share a bonus of £10 728 in
proportion to their work force. Calculate how much each hanger will
receive.
8.
Four friends contribute sums of money to a charitable organisation in the
ratio of 2 : 4 : 5 : 7. If the largest amount contributed is £1.40, calculate the
total amount contributed by the four people.
DIRECT PROPORTION
Two quantities are said to vary directly, or be in direct proportion, if they
increase or decrease at the same rate. Thus the quantity of fuel used and the
distance travelled by an aircraft are in direct proportion. Again if a company buys
sorbsil at 20 pence for 2 kilograms then we expect to pay 40 p for 4 kilograms
and 10 p for 1 kilogram. That is if we double the amount bought then we double
the cost; if we halve the amount bought we halve the cost.
In solving problems on direct proportion we can use either the unitary method or
the fractional method.
Example 7
If 25 kilograms of dry powder fire extinguishant cost £17, how much does 8
kilograms cost?
1.
Using the unitary method:
25 kilograms cost £17 or 1700 pence.
1 kilograms cost
1700
25 = 68 pence.
8 kilograms cost 8  68
= 544 pence or £5.44.
49
2.
Using the fractional method:
Cost of 8 kilograms.
8
8  1700
= 25  1700 =
25
= 544 pence or £5.44
Example 8
A recipe for Beef Stroganoff quotes the following amounts to serve four people:
450 grams of rump steak, 3 tablespoons flour, 4 tablespoons butter, 50 grams of
onion, 75 grams of mushrooms, 140 grams of sour cream. What amounts should
be used for six people?
The quantities required and the number of people are in direct proportion. Hence
the amounts must be increased in the ratio of 6 : 4 or 3 : 2.
3
 450
2
Amount of rump steak.
=
Amount of flour.
3
= 2  3
Amount of butter.
=
Amount of onion.
3
= 2  50
= 75 grams.
Amount of mushrooms.
3
= 2  75
1
= 1122 grams.
Amount of sour cream.
3
= 2  140
= 210 grams.
3
 4
2
= 675 grams.
1
= 42 tablespoons.
= 6 tablespoons.
Exercise 31 - All level 1.
1.
If 7 kilograms of silica gel cost £2.80, how much do 12 kilograms cost?
2.
If 74 tech logs cost £5.92, how much do 53 cost?
3.
If 40 cost rivets cost £35, how much does 1 cost? What is the cost of 55
rivets.
4.
Split pins cost 70 p per 10. How much will 25 split pins cost?
5.
A towing tractor travels 20 kilometres on 20 litres of petrol. How much
petrol is need for a journey of 35 kilometres.
50
6.
If 9 metres of asbestos tape cost £21, how much will 96 metres cost?
7.
An aircraft flies 2000 kilometres in 4 hours. How long will it take to
complete a journey of 3500 kilometres?
INVERSE PROPORTION
Suppose that 8 fitters working on an aircraft 'C' check takes 10 days to complete
it. If we double the number of men then we should halve the time taken. If we
halve the number of men then the job will probably take twice as long. This is an
example of inverse proportion.
Example 9
20 men working at BA, Filton produce 3000 components in 12 working days.
How long will it take 15 men to produce the 3000 components.
15
3
The number of men is reduced in the ratio 20 = 4 .
Since this is an example of inverse proportion the number of days required must
4
be increased in the ratio 3 .
Number of days required
=
4
 12.
3
= 16 days.
Exercise 32 - All level 1
1.
Bristol Flying Centre employs 12 builders to extend the hanger. They take
9 days to do the job. If they had employed 8 men how long would it have
taken them?
2.
10 men produce 500 composite panels in 5 working days. How long would
it take 15 men to produce the same amount?
3.
Two gear wheels mesh together. One has 40 teeth and the other has 25
teeth. If the larger wheel makes 100 revolutions per minute how many
revolutions per minute does the smaller wheel make?
4.
4 men can do a piece of work in 30 hours. How many men would be
required to do the work in 6 hours?
51
AVERAGES
To find the average of a set of quantities, ass the quantities together and divide
by the number of quantities in the set. Thus,
sum of the quantities
Average = number of quantities
Example 1
1.
A student falls asleep in every lesson, the following number of times: 8, 20,
3, 0, 5, 9, 15 and 12. What is his average per lesson?
Average score
2.
=
8 + 20 + 3 + 0 + 5 + 9 + 15 + 12
8
=
72
8 = 9
1
A 2 “ ‘Taper Lock Fastners’ in a box have a mass of 4680 gm. If the
average mass of one fastners is 97.5 gm find the number of fastners in the
box.
Total mass = average mass of an fastner
 number of fastners in the box.
 Number of fastners in the box =
3.
4680
= 48
97.5
Find the average age of a team of boys given that four of them are each 15
years 4 months old and the other three boys are each 14 years 9 months
old.
Total age of 4 boys at 15 years 4 months
= 61 years 4 months.
Total age of 3 boys at 14 years 9 months
= 44 years 3 months.
Total age of all 7 boys
= 105 years 7 months.
Average age
=
105 years 7 months
7
= 15 years 1 month.
52
4.
the average age of the lecturers in the faculty is 39 years and their total age
is 1170 years, whereas the pupils whose average age is 14 years have a
total age 6580 years. Find the average age of all the people in the faculty.
The first step is to find the number of teachers:
Number of teachers:
total age of the teachers
= average age of the teachers
=
1170
39
= 30
We now find the number of pupils:
=
6580
14
= 470
We can now find the average age of people in the faculty.
Total age of all the people in the faculty:
= 1170 + 6580 = 7750 years
Total number of people in the faculty:
= 30 + 470 = 500
Average age of all the people in the faculty:
=
7750
500
= 15.5 years
53
Exercise38 - Question 1 - 4, level 1. 5 - 8 level 2.
1.
Find the average of the following readings: 22.3 mm, 22.5 mm, 22.6 mm,
21.8 mm and 22.0 mm.
2.
Find the average mass of 22 boxes if 9 boxes each have a mass of 12 kg,
1
3
8 boxes each have a mass of 122 kg and 5 have a mass of 114 kg.
3.
4 kg of engine blanks costing 20 p per kg are mixed with 8 kg costing 14 p
per kg. What is the average price per kg?
4.
30 litres of Mogas costing 8 p per litre is mixed with 40 litres costing 9 p per
litre. Find the average price of the mixture.
5.
The average of nine numbers is 72 and the average of four of them is 40.
What is the average of the other five?
6.
Find the average age of a team of men if 5 of them are each 25 years old
and 6 of them are 24 years 1 month old.
7.
The average mark of 24 candidates taking an examination is 42. Find what
the average mark would have been if one candidate, who scored 88, had
been absent.
8.
The average of three numbers is 58. The average of two them is 49. Find
the third number.
AVERAGE SPEED
The average speed is defined as total distance travelled divided by the total
time taken. The unit of speed depends on the unit of distance and the unit of
time. For instance, if the distance travelled is in kilometres (km) and the time
taken is in hours (h) then the speed will be stated in kilometres per hour (km/h).
If the distance is given in metre (m) and the time in seconds (s) then the speed is
in metres per second (m/s).
Example 2
1.
A car travels a total distance of 200 km in 4 hours. What is its average
speed?
Average speed =
distance travelled
time taken
=
200
4
= 50 km/h
2.
A car travels 30 km at 30 km/h and 30 km at 40 km/h. Find its average
speed.
Time taken to travel 30 km at 30 km/h.
30
= 30
= 1 hour
Time taken to travel 30 km at 40 km/h.
=
30
40
= 0.75 hour
54
Total distance travelled
= 30 + 30 = 60 km.
Total time taken = 1 + 0.75 = 1.75 hour.

3.
Average speed
60
= 1.75
= 34.3 km/h
A train travels for 4 hours at an average speed of 64 km/h. For the first 2
hours its average speed is 50 km/h. What is its average speed for the last
2 hours
Total distance travelled in 4 hours
= average speed  time taken = 64  4
= 256 km
Distance travelled in first two hours
= 50  2 = 100 km
 Distance travelled in last two hours
= 256 - 100 = 156 km
Average speed for the last two hours
=
distance travelled
156
= 2
time taken
= 78 km/h
Exercise - Questions 1 - 3, Level 1 Level 2
1.
A train travels 300 km in 4 hours. What is its average speed?
2.
A car travels 200 km at an average speed of 50 km/h. How long does it
take?
3.
If a car travels for 5 hours at an average speed of 70 km/h how far has it
gone?
4.
For the first 1½ hours of a 91 km journey the average speed was 30 km/h.
If the average speed for the remainder of the journey was 23 km/h,
calculate the average speed for the entire journey.
5.
A motorist travelling at a steady speed of 90 km/h covers a section of
motorway in 25 minutes. After a speed limit is imposed he finds that, when
travelling at the maximum speed allowed he takes 5 minutes longer than
before to cover the same section. Calculate the speed limit.
55
6.
In winter a train travels between two towns 264 km apart at an
average speed of 72 km/h. In summer the journey takes 22 minutes less than in
the winter. Find the average speed in summer.
7.
A train travels between two towns 135 km apart in 4½ hours. If on the
return journey the average speed is reduced by 3 km/h, calculate the time taken
for the return journey.
8.
A car travels 272 km at an average speed of 32 km/h. On the return
journey the average speed is increased to 48 km/h. Calculate the average speed
over the whole journey.
56
PERCENTAGES
When comparing fractions it is often convenient to express them with a
denominator of a hundred. Thus:
1
50
2 = 100
2
40
=
5
100
Fractions with a denominator of 100 are called percentages. Thus:
1
4
25
= 100
= 25 per cent
3
30
10 = 100
= 30 per cent
The sign % is usually used instead of the words per cent.
To convert a fraction into a percentage we multiply it by 100.
Example 1
3
4
3
= 4
 100 = 75
17
17
=
20
20  100 = 85
Exercise 34 - All type A
Convert the following fractions to percentages:
1.
7
10
4.
4
5
7.
7
10
2.
11
20
5.
31
50
8.
19
20
3.
9
25
6.
1
4
Decimal numbers may be converted into percentages by using the same rule.
Thus:
0.3 =
3
3
=
 100 = 30%
10
10
The same rule result is produced if we omit the intermediate step of turning 0.3
into vulgar fraction and just multiply 0.3 by 100. Thus:
0.3 = 0.3  100 = 30
57
Exercise 35 - level 1
Convert the following decimal numbers into percentages:
1.
0.7
4.
0.813
2.
0.73
5.
0.927
3.
0.68
6.
0.333
7.
0.819
To convert a percentage into a fraction we divide by 100.
Example 3
45%
45
= 100
3.9
3.9% = 100
= 0.45
= 0.039
Note that all we have done is to move the decimal point 2 places to the left.
Exercise 36 - Level 1
Convert the following percentages into decimal fractions:
1.
32%
5.
31.5%
9.
3.95%
2.
78%
6.
48.2%
10.
20.1%
3.
6%
7.
2.5%
4.
24%
8.
1.25%
PERCENTAGE OF A QUANTITY
It is easy to find the percentage of a quantity if we first express the percentage as
a fraction.
Example 4
1.
What is 10% of 40?
Expressing 10% as a fraction it is
10
and the problem then becomes:
100
10
what is 100 of 40?
10% of 40
10
= 100  40
= 4
58
2.
What is 25% of £50?
25
= 100  £50
25% of £50
3.
= £12.50
22% of a certain length is 55 cm. What is the complete
length?
55
1% of the length = 22 cm = 2.5 cm
now the complete length will be 100%, hence:
Complete length = 100  2.5 cm = 250 cm
Alternatively,
22% of the length
= 55 cm
Complete length
=
=
4.
100  55
22
100
22  55
= 250 cm
What percentage is 37 of 264? Give the answer correct to
5 significant figures.
Percentage
=
37
 100
264
=
37  100
264
= 14.015%
59
Exercise 37 - Question 1 - 6, level 1. Question 7 - 10, level 2.
1.
2.
What is:
a.
20% of 50
d.
12% of 20
b.
30% of 80
e.
20.3% of 105
c.
5% of 120
f.
3.7% of 68
What percentage is:
a.
25 of 200
d.
29 of 178
b.
30 of 150
e.
15 of 33
c.
25 of 150
Where necessary give the answer correct to 3 significant figures.
3.
A student scores 36 marks out of 100 in an examination. What is her
percentage mark? If the percentage needed to pass the examination is
75% how many marks are needed to pass?
4.
If 20% of a length is 23 cm, what is the complete length.
5.
Given that 13.3 cm is 15% of a certain length, what is the complete length?
6.
What is:
7.
a.
9% of £80
b.
12% of £110
c.
75% of £250
Express the following statements in the form of a percentage:
a.
3 light bulbs are broken in a box containing 144 light bulbs.
b.
In a school of 650 students, 20 are absent.
c.
In a school of 980 students, 860 each school lunches.
8.
in a certain county the average number of children eating lunches at school
was 29 336 which represents 74% of the total number of children attending
school. Calculate the total number of children attending school in that
country.
9.
23% of a consignment of bananas is bad. There are 34.5 kg of bad
bananas. How many kilograms were there in the consignment?
10.
A retailer accepts a consignment of 5000 ball point pens. He finds that
12% are faulty. How many faulty pens are there?
60
AREAS
We are already familiar with the concept of length, e.g. the distance between 2
points, we express length in some chosen unit, e.g. in meters, and if I want to fit a
picture - rail along a wall, all I need to known is the length of the wall, so that I
can order sufficient rail. But if I wish to fit a carpet to the room floor, the length of
the room is insufficient. I obviously need to know the width. This 2-dimensional
concept of size is termed Area.
Consider a room 4m by 3m as shown above. Clearly it can be divided up into 12
equal squares, each measuring 1m by 1m. Each square has an area of 1 square
meter. Hence, the total area is 12 square meters (usually written as 12m2 for
convenience). So, to calculate the area of a rectangle, multiply length of 1 side
by the length of the other side.
Note. 4m x 3m = 12m2 (Don't forget the m2).
Example: An office 8.5m by 6.3m is to be fitted with a carpet, so as to leave
surround 600mm wide around the carpet. What is the area of the
surround?
With a problem like this, it is often helpful to sketch a diagram.
The area of the surround = office area - carpet area.
=
(8.5 x 6.3) - (8.5 - 2 x 0.6) (6.3 - 2 x 0.6)
=
53.55
- (7.3) (5.1)
=
53.55
- 37.23
=
16.32m2
Note that 600mm had to be converted to 0.6m. Don't forget to include units in the
answer e.g. m2.
61
The following table shows the formulaes for the more common shapes. Students
will require a knowledge of these formulae and attain a JAR 66 Level 2 in this
topic.
or d
or  d x
0
360
62
1).The cross section of a block of metal is shown. Find its area.
Area of trapezium
= ½ x 40 x (30 + 50)
= ½ x 40 x 80
= 1600 mm²
2). A hollow shaft has an outside diameter of 2.5cm. Calculate the cross-sectional
area of the shaft.
Area of cross-section
= area of outside circle – area of inside circle
=  x 1.626² -  x 1.25²
=  (1.625² - 1.25²)
= 3.142 x (2.640 – 1.563)
= 3.142 x 1.077
=3.388cm²
(3) Calculate:
(a) the length of arc of a circle whose radius is 8m and which subtends an
angle of 56° at the centre, and
(b) the area of the sector so formed.

Length of arc = 2r x 0
360
63
=2xx8x
56
360
= 31.28 m²
Exercise 1-3 Level 1 Level 2 remainder
1) The area of a rectangle is 220mm². If its width is 25mm find its length.
2) A sheet metal plate has a length of 147.5mm and a width of 86.5mm find its
length to the nearest four decimal places.
3) Find the areas of the sections shown in
4)Find the area of a triangle whose base is 7.5cm and whose altitude is 5.9cm.
5) Find the area of a trapezium whose parallel sides are 75mm and 82mm long
respectively and whose vertical height is 39mm.
6) The parallel sides of a trapezium are 12cm and 16cm long. If its area is
220cm², what is its altitude?
7) Find the areas of the shaded portions in each of the diagrams.
64
8) Find the circumference of a circle whose radii are:
(a) 3.5mm
(b) 13.8mm
(c) 4.2cm
9) Find the diameter of a circle whose circumference is 34.4mm.
10) How many revolutions will a wheel make in travelling 2km if its diameter is
700mm?
11) If r is the radius and 0 is the angle subtended at the centre by an arc find the
length of arc when:
(a) r = 2cm, 0 =30°
65
VOLUMES
The concept and calculation of volume in the logical extension of length and
area.
Instead of squares, we now consider cubes. This is a 3-dimensional concept and
the typical units of volume are cubic metres (m3).
If we have a box, length 4m, width 3m and height 2m, we see that the total
volume=24 cubic metres (24m3).
Each layer contains
4 x 3 = 12 cubes.
There are 2 layers.
Hence the volume is
12 x 2 = 24m3.
Basically, therefore, when calculating volume, it is necessary to look for 3
dimensions, at 90º to each other, and then multiply them together. For a box type shape, multiplying length x width x height = volume.
UNIT OF VOLUME
The volume of a solid figure is measured by seeing how many cubic units it
contains. A cubic metre is the volume inside a cube which has a side of 1 metre.
Similarly a cubic centimetre is the volume inside a cube which has a side of 1
centimetre. The standard abbreviations for units of volume are:
cubic metre
m³
cubic centimetre cm³
cubic millimetre mm³
Example
(1) How many cubic centimetres are contained in 1 cubic metre?
1m = 10² cm
6
1m³ = (10² cm)³ = 10 cm³
= 1 000 000 cm³
66
UNIT OF CAPACITY
The capacity of a container is usually measured in litres (  ), such that
1 litre = 1000cm³
Example
A tank contains 30 000 litres of liquid. How many cubic metres does it contain?
30 000 litres = 30 000 x 1 000 cm³
= 3 x 107 cm³
1cm = 102 m
6
1cm³ = (10 2 m) ³ = 10 m3
 3  107 cm3 = 3  107  10 6 m2
= 3 x 10 = 30m²
Exercise- All Level 1
Convert the following volumes into the units stated:
1) 5 m² into cm³
2) 0.08 m³ into mm³
3) 18 m³ into mm³
4) 830 000 cm³ into m³
5) 850 000 mm³ into m³
6) 78 500 cm³ into m³
7) A tank contains 5000 litres of petrol. How many cubic metres of petrol does
it contain?
8) A small vessel contains 2500mm³ of oil. How many litres does it contain?
9) A tank holds, when full, 827m³ of water. How many litres does it hold?
10) A container holds 8275cm² when full. How many litres does it hold?
Example
A steel section has the cross-section shown. If it is 9m long calculate its volume
and total surface area.
To find the volume we use the formulae given on page 138.
Area of cross- section
67
= ½  x 75² + 100 x 150
= 23 836 mm²
2
=
23836 mm
2
= 0.023 836 m²
(1000)
Volume of solid
= 0.023 836 x 9
= 0.214 5m³
To find the surface area:
Perimeter of cross-section
= x 75 + 2 x 100 +150
=585.5mm
=
585.5
= 0.585 5m
1000
Lateral surface area
=0.585 5 x 9 = 5.270 m²
Surface area of ends
= 2 x 0.024 = 0.048 m ²
Total surface area
= 5.270 + 0.048
= 5.318 m²
68
VOLUMES AND SURFACE AREAS
The following table gives volumes and surface areas of some simple solids
69
Exercise- Questions 1- Level 2
1) A steel ingot whose volume is 2 m³ is rolled into a plate 15mm thick and 1.75m
wide. Calculate the length of the plate in m.
2) A block of lead 2.0 m x 1m x 0.72m is hammered out to make a square sheet
10mm thick. What are the dimensions of the square?
3) The volume of a small cylinder is 180 cm³. If the radius of the cross-section is
25mm find its height.
4)A cone has a diameter of 28mm and a height of 66mm. What is its volume?
5) Calculate the diameter of a cylinder whose height is the same as its diameter
and whose volume is 220 cm³.
6) An ingot whose volume is 12320 mm² is to be made into ball bearings whose
diameters are 12mm. How many ball bearings will be produced from the ingot?
7) A water tank with vertical sides has a horizontal base in the shape of a
rectangle with semi-circular ends as illustrated in Fig. 26.14. The total inside
length of the tank is 7m, its width 4m and its height 2m.
Calculate:
(a) the surface area of the vertical walls of the tank in m²
(b) the area of the base in m²
(c) the number of litres of water in the tank when the depth of water is 1.56m.
70
SQUARES & SQUARE ROOTS
SQUARE NUMBERS
When a number is multiplied by itself the result is called the square of the
number. The square of 9 is 9  9 = 81. Instead of writing 9  9, it is usual to
write 92 which is read as the square of 9. Thus;
122
= 12  12 = 144
(1.3)2 = 1.3  1.3 = 1.69
The square of any number can be found by multiplication but a great deal of time
and effort is saved by using printed tables. Either three or four figure table may
be used. In the three figure tables the squares of numbers are given correct to
three significant figures, but in the four figure tables the square are given correct
to four significant figures. Hence the four figure table are more accurate.
Although the tables only give the squares of numbers from 1 to 10 they can be
used to find the squares of numbers outside this range. The method is shown in
the examples which follows.
Example 1
Find (168.8)2.
(168.8)2
= 168.8  168.8
or
= 1.688  100  1.688  100
or
= (1.688)2  1002
From the tables of squares,
(1.688)2
= 2.848
Hence
(168.8)2
= 2.848  1002 = 28 480
71
Example 2
Find (0.2388)2.
(0.2388)2
1
1
= 2.388  10  2.388  10
1
= (2.388)2  100
= (2.388)2 ÷ 100
From the tables,
(2.388)2
= 5.702
Hence
(0.2388)2
= 5.702 ÷ 100
= 0.057 02
Example 3
Find the value of
.
 0.9 


 0.15 
 0.9 


 0.15 
2
2
 6 2  36
Exercise - Question 1 - 12, level 1. Question 13 - 20, level 2.
Find the square of the following numbers.
1.
1.5
11.
23
2.
2.1
12.
40.6
3.
8.6
13.
3093
4.
3.15
14.
112.3
5.
7.68
15.
98.12
6.
5.23
16.
0.019
7.
4.263
17.
0.729 2
8.
7.916
18.
0.004 219
9.
8.017
19.
0.283 4
10.
8.704
20.
0.000 578 4
72
21.
Find the value of (3.142)2 correct to 2 places of decimal.
22.
Find the value of:
a.
 0.75 


 0.15 
b.
 0.8 


 0.2 
2
2
c.
 0.25 


 2 
 0.36 


 6 
2
d.
2
73
74
SQUARE ROOTS
The square roots of a number is the number whose square equals the given
number. Thus since 52 = 25, the square root of 25 = 5.
is used to denote a square root and hence we write 25  5 .
The sign
Similarly, since 92 = 81,
81  9 .
The square root of a number can usually be found with sufficient accuracy by
using the printed tables of square roots. There are two of these tables. One
gives the square roots of numbers 1 to 10 and the other gives the square roots of
numbers from 10 to 100. The reason for having two tables is as follows:
2.5  1.581
25
 5
Thus there are two square roots for the same figures, depending upon the
position of the decimal point. The square root tables are used in the same way
as the tables of squares.
Example 4
1.
2.748  1.657
(directly from the tables from 1 to 10).
2.
92.65  9.626
(directly from the tables from 1 to 100 ).
3.
To find
836.3 .
Mark off the figures in pairs to the left of the decimal point. Each pair of
figures is called a period. Thus 836.3 becomes 8'36.3. The first period is 8
so we use the table of square roots from 1 to 10 and look up
8.363  2.892 . To position the decimal point in the answer remember
that for each period to the left of the decimal point in the original number
there will be one figure less to the left of the decimal point in the answer.
Thus:
836.3  28.92
4.
To find
836.3 is three figures to the left of the decimal
point. One less is two figures hence 2.892 is 28.92
173 900 .
Marking off in periods 173 900 becomes 17'39'00. The first period is 17 so
we use the table of square roots from 10 to 100 and look up.
17.39  4.17
173900 is four figures to the left of the decimal
point. One less is three so the answer will be
417.0
173 900  417.0
75
5.
To find
0.000 094 31 .
In the case of numbers less than 1 mark off the periods to the right of the
decimal point. 0.000 094 31 becomes 0.00'00'94'31. Apart from the zero
pairs the first period is 94 so we use the tables from 10 to 100 to look up
94.31  9.712. For each zero pair in the original number there will be
one zero following the decimal point in the answer. Thus:
0. 0 0 9 712
0.00'00'94'31
0.000 094 31  0.009 712
6.
To find
0.07365 .
Marking off in periods to the right of the decimal point 0.073 65 becomes
07'36'50. Since the first period is 07 we use the tables between 1 and 10
and look up 0.07365  2.714 .
Exercise - Questions 1 - 12, level 1. Questions 13 - 23, level 2.
Find the square root of the following numbers.
1.
3.4
13.
900
2.
8.19
14.
725.3
3.
5.264
15.
7142
4.
9.239
16.
89 000
5.
7.015
17.
3945
6.
3.009
18.
893 400 000
7.
35
19.
0.153 7
8.
89.2
20.
0.001 698
9.
53.17
21.
0.039 47
10.
82.99
22.
0.000 783 1
11.
79.23
23.
0.001 978
12.
50.01
76
77
78
CUBED
When a number is multiplied by itself , i.e. 3  3 = 9, it is usual to write it as 32
or 3 squared. We can take this a stage further and multiply by another 3, i.e.
3  3  3 = 27, it is usual to write it as 33 or 3 cubed.
CUBED ROOTS
The cubed root of a number is the number which cubed equals the number. E.g.
the cubed root of 64 = 4 (4  4  4).
The sign
3
is used to denote a cubed root and hence we write
3
64  4 .
79
1.2 ALGEBRA
The methods of Algebra are an extension of those used in arithmetic. In algebra
we use letters and symbols as well as numbers to represent quantities. When we
write that a sum of money is £50 we are making a particular statement but if we
write that a sum of money is £P we are making a general statement. This
general statement will cover any number we care to substitute for P.
USE OF SYMBOLS
The following examples will show how verbal statements can be translated into
algebraic symbols. Notice that we can chose any symbol we like to represent the
quantities concerned.
1.
The sum of two numbers.
Let the two numbers be x and y.
Sum of the two numbers = x + y.
2.
Three times a number.
Let the number be N.
Three times the number = 3  N.
3.
One number divided by another number.
Let one number be a and the other number be b.
One number divided by another number =
4.
a
b
Five times the product of two numbers.
Let the two numbers be m and n.
5 times the product of the two numbers = 5  m  n.
80
Exercise - All level 1.
Translate the following into algebraic symbols:
1.
Seven times a number, x.
2.
Four times a number x minus three.
3.
Five times a number x plus a second number, y.
4.
The sum of two numbers x and y divided by a third number, z.
5.
Half of a number, x.
6.
Eight times the product of three numbers, x, y and z.
7.
The product of two numbers x and y divided by a third number, z.
8.
Three times a number, x, minus four times a second number, y.
SUBSTITUTION
The process of finding the numerical value of an algebraic expression for given
values of the symbols that appear in it is called substitution.
Example 1
If x = 3, y = 4 and z = 5, find the values of:
a.
2y + 4
d.
y
x
b.
3y + 5z
e.
3y + 2z
x + z
c.
8 - x
Note that multiplication signs are often missed out when writing algebraic
expressions so that, for instance, 2y means 2  y. These missed multiplication
signs must reappear when the numbers are substituted for the symbols.
a.
2y + 4 = 2  4 + 4 = 8 + 4 = 12
b.
3y + 5z = 3  4 + 5  5
c.
8 - x = 8 - 3 = 5
81
d.
y
x
e.
3y + 2z
x + z
=
4
3
1
= 1
3
=
3  4 + 2  5
3 + 5
=
12 + 10
8
=
22
8
3
= 24
Exercise - All level 1
If a = 2, b = 3 and c = 5, find the values of the following:
1.
9 + 7
10.
4c + 6b
2.
c - 2
11.
8c - 7
3.
6 - b
12.
a + 2b + 5c
4.
6b
13.
8c - 4b
5.
9c
14.
2 ÷ a
6.
ab
15.
ab
8
7.
3bc
16.
abc
6
8.
abc
17.
2c
a
9.
5c - 2
18.
5a + 9b + 8c
a+b+c
82
POWERS
The quantity a  a  a or aaa is usually written as a3. a3 is called the third power
of a. The number 3 which indicates the number of a's to be multiplied together is
called the index (plural: indices).
24
= 2  2  2  2 = 16
y5
= y  y  y  y  y
Example 2
Find the value of b3 when b = 5.
b3
= 53 = 5  5  5 = 125
When dealing with expressions like 8mn4 note that it is only the symbol n which
is raised to the fourth power. Thus:
8mn4 = 8  m  n  n  n  n
Example 3
Find the value of 7p2q3 when p = 5 and q = 4.
7p2q3
= 7  52  43 = 7  25  64
= 11 200
Exercise - All level 1
If a = 2, b = 3 and c = 4 find the values of the followings:
1.
a2
5.
ab2c3
9.
3a4
c2
2.
b4
6.
5a2 + 6b2
10.
c5
ab3
3.
ab3
7.
a2 + c2
4.
2a2c
8.
7b3c2
83
ADDITION OF ALGEBRAIC TERMS
Like terms are numerical multiplies of the same algebraic quantity. Thus:
7x, 5x and -3x
are three like terms.
An expression consisting of like terms can be reduced to a single term by adding the
numerical coefficients together. Thus:
7x - 5x + 3x
= (7 - 5 + 3)x = 5x
3b2 + 7b2
= (3 + 7)b2 = 10b2
-3y - 5y
= (-3 -5)y = -8y
q - 3q
= (1 - 3)q = -2q
Only like terms can be added or subtracted. Thus 7a + 3b - 2c is an expression
containing three unlike terms and it cannot be simplified any further.
Similarly with 8a2b + 7ab3 + 6a2b2 which are all unlike terms.
It is possible to have several sets of like terms in an expression and each set can then be
simplified.
8x + 3y - 4z - 5x + 7z - 2y + 2z
= (8 - 5)x + (3 - 2)y + (-4 + 7 + 2)z
= 3x + y + 5z
MULTIPLICATION & DIVISION OF ALGEBRAIC QUANTITIES
The rules are exactly the same as those used with directed numbers:
(+ x)(+ y)
= + (xy) = + xy = xy
5x  3y
= 5  3  x  y = 15xy
(x)(-y)
= - (xy) = - xy
(2x)(- 3y)
= - (2x)(3y) = -6xy
(- 4x)(2y)
= - (4x)(2y) = 8xy
(- 3x)(- 2y)
= + (3x)(2y) = 6xy
+x
+y
x
= +y
x
= y
84
- 3x
2y
3x
= - 2y
- 5x
- 6y
5x
= + 6y
4x
- 3y
4x
= - 3y
5x
= 6y
When multiplying expressions containing the same symbols, indices are used:
m  m = m2
3m  5m = 3  m  5  m = 15 m2
(- m)  m2 = (- m)  m  m = - m
5m2n  3mn3
= 5  m  m  n  3  m  n  n  n
= 15m3n4
3mn  (-2n2)
= 3  m  n  (- 2)  n  n = - 6mn3
When dividing algebraic expressions, cancellation between numerator and
denominator is often possible. Cancelling is equivalent to dividing both
numerator and denominator by the same quantity:
pq
p  q

 q
p
p
3p 2 q  3  p  p  q
3p
p


6pq 2
6  p  q  q
6q
2q
2
18 x 2 y z
18  x  x  y  y  z

 3xy
6 xyz
6  x  y  z
85
Exercise - All level 1
Simplify the following:
1.
7x + 11x
2.
7x - 5x
3.
3x -6x
4.
- 2x - 4x
5.
- 8x + 3x
6.
- 2x + 7x
7.
8a - 6a - 7a
8.
5m + 13m - 6m
9.
6b2 - 4b2 + 3b2
10.
6ab - 3ab - 2ab
11.
14xy + 5xy - 7xy + 2xy
12.
- 5x + 7x - 3x - 2x
13.
- 4x2 - 3x2 + x2
14.
3x - 2y + 4z - 2x2 - 3y + 5z + 6x + 2y - 3z
15.
3a2b + 2ab3 + 4a2b3 - 5ab3 + 11b4 + 6a2b
16.
1.2x3 - 3.4x2 + 4a2b2 - 3.7x2 + 3.6x - 2.8
17.
pq + 2.1pr - 2.2rq + 8pq
18.
2.6a2b2 - 3.4b3 - 2.7a3 - 3a2b2 - 2.6b3 + 1.5a3
19.
2x  5y
20.
3a  4b
21.
3  4m
22.
1
4 q  16p
86
23.
x  (- y)
24.
(- 3a)  (- 2b)
25.
8m  (- 3n)
26.
(- 4a)  3b
27.
8p  (- q)  (- 3r)
28.
3a  (- 4b)  (- c)  5d
29.
12x ÷ 6
30.
4a ÷ (- 7b)
31.
(- 5a) ÷ 8b
32.
(- 3a) ÷ (- 3b)
33.
4a ÷ 2b
34.
4ab ÷2a
35.
12x2yz2 ÷ 4xz2
36.
(- 12a2b) ÷ 6a
37.
8a2bc2 ÷ 4ac2
38.
7a2b2 ÷ 3ab
39.
a  a
40.
b  (- b)
41.
(- m)  m
42.
(- p)  (- p)
43.
3a  2a
44.
5X  X
45.
5q  (- 3q)
87
46.
3m  (- 3m)
47.
(- 3pq)  (- 3q)
48.
8mn  (- 3m2n3)
49.
7ab  (- 3a2)
50.
2q3r4  5qr2
51.
(- 3m)  2n  (- 5p)
52.
5a2  (- 3b)  5ab
53.
m2n  (- mn)  5m2n2
BRACKETS
Brackets are used for convenience in grouping terms together. When removing
brackets each term within the bracket is multiplied by the quantity outside the
bracket:
3(x + y) = 3x + 3y
5(2x + 3y) = 5  2x + 5  3y = 10x + 15y
4(a - 2b) = 4  a - 4 2b = 4a - 8b
m(a + b) = ma + mb
3x(2p + 3q) = 3x  2p + 3x  3q = 6px + 9qx
4a(2a + b) = 4a  2a + 4a b = 8a2 + 4ab
When a bracket has a minus sign in front of it, the signs of all the terms inside the
bracket are changed when the bracket is removed. The reason for this rule may
be seen from the following example:
- 3(2x - 5y)
= (- 3)  2x + (- 3)  (- 5y)
= - 6x + 15y
- (m + n) = - m - n
- (p - q) = -p + q
- 2(p + 3q) = - 2p - 6q
88
When simplifying expressions containing brackets first remove the brackets and
then add the like terms together:
(3x + 7y) - (4x + 3y)
= 3x + 7y - 4x - 3y
= - x + 4y
3(2x + 3y) - (x + 5y)
= 6x + 9y - x - 5y
= 5x + 4y
x(a + b) - x(a + 3b)
= ax + bx - ax - 3bx
= - 2bx
2(5a + 3b) + 3(a - 2b)
= 10a + 6b + 3a - 6b
= 13a
Exercise - Questions 1 - 20, Level 1. Remainder Level 2.
1.
3(x + 4)
9.
- (3p - 3q)
2.
2(a + b)
10.
- (7m - 6)
3.
3(3x + 2y)
11.
- 4(x + 5)
4.
1
(x - 1)
2
12.
- 2(2x - 5)
5.
5(2p - 3q)
13.
- 5(4 - 3x)
6.
7(a - 3m)
14.
2k(k - 5)
7.
- (a + b)
15.
- 3y(3x + 4)
8.
- (a - 2b)
16.
a(p - q - r)
17.
4xy(ab - ac + d)
18.
3x2(x2 - 2xy + y2)
19.
- 7P(2P2 - P + 1)
20.
- 2m(- 1 + 3m - 2n)
89
Remove the brackets and simplify:
21.
3(x + 1) + 2(x + 4)
22.
5(2a + 4) - 3(4a + 2)
23.
3(x + 4) - (2x + 5)
24.
4(1 - 2x) - 3(3x - 4)
25.
5(2x - y) - 3(x + 2y)
26.
1
1
2 (y - 1) + 3 (2y - 3)
27.
- (4a + 5b - 3c) - 2(2a + 3b - 4c)
28.
2x(x - 5) - x(x - 2) - 3x(x - 5)
29.
3(a - b) - 2(2a - 3b) + 4(a - 3b)
30.
3x(x2 + 7x - 1) - 2x(2x2 + 3) - 3(x2 + 5)
90
ADDITION & SUBTRACTION OF FRACTIONS
The method for algebraic fractions is the same as for arithmetical fraction, that is:

Find the L.C.M. of the denominators.

Express each fraction with the common denominators.

Add or subtract the fractions.
Example 3
1.
a
b c
Simplify 2 + 3 - 4 .
The L.C.M. of 2,3 and 4 is 12.
a
b c
2 + 3 - 4
6a
4b 3c
= 12 + 12 - 12
=
2.
Simplify
6a + 4b - 3c
12
2
3
4
+
+
.
x
2x
3x
The L.C.M. of x, 2x and 3x is 6x.
2
3
4
x + 2x + 3x
=
12 + 9 + 8
6x
29
= 6x
The sign in front of a fraction applies to the fraction as a whole. The line which
separates the numerator and denominator acts as a bracket.
Example 4
m
2m + n m - 2n
Simplify 12 +
.
4
3
The L.C.M. of 12, 4 and 3 is 12.
m
2m + n m - 2n
+
12
4
3

=
m + 3(2m + n) - 4(m - 2n)
12
=
m + 6m + 3n - 4m + 8n
12
=
3m + 11n
12
91
Exercise - All level 2
Simplify the following:
1.
x
x
x
+
+
3
4
5
8.
2x
x
1 - 5 + 8
2.
5a 7a
12 - 18
9.
3m -
3.
2
3
q - 2q
10.
3a + 5b a - 3b
4
2
4.
3
5
4
y - 3y + 5y
11.
3m - 5n 3m - 7n
6
2
5.
3
2
5p 3q
12.
x - 2
2
+
4
5
6.
3x 5y
2y - 6x
13.
x - 5 x - 2
3
4
2m + n
7
92
GRAPHS OF EQUATION
One of the most important applications of the straight-line equation is the
determination of an equation connecting two quantities when values have been
obtained from an experiment.
Example
In an experiment carried out with a lifting machine the effort E and the load W
were found to have the values given in the table below:
W (kg)
15
25
40
50
60
E (kg)
2.75
3.80
5.75
7.00
8.20
Plot these results and obtain the equation connecting E and W which is thought
to be of the type E = aW + b.
If E and W are connected by an equation of the type E = aW + b then the
graph must be a straight line. Note that when plotting the graph, W is the
independent variable and must be plotted on the horizontal axis. E is the
dependent variable and must be plotted on the vertical axis.
On plotting the points (see diagram below) it will be noticed that they deviate only
slightly from a straight line. Since the data are experimental we must expect
errors in measurement and observation and hence slight deviations from a
straight line must be expected. Although the straight line will not pass through
some of the points an attempt must be made to ensure an even spread of the
points above and below the line.
To determine the equation we choose two points which lie on the straight line.
Do not use any of the experimental results from the table unless they happen to
lie exactly on the line. Choose the points as far apart as is convenient because
this will help the accuracy of your result.
The point W = 55, E = 7.5 lies on the line. Hence,
7.5
= 55a + b
[1]
The point W = 20, E = 3.3 also lies on the line, Hence,
3.3
= 20a + b
[2]
93
Subtracting equation [2] from equation [1],
4.2
= 35a
a
= 0.12
Substituting for a = 0.12 in equation [2],
3.3
b
= 20  0.12 + b
= 0.9
The required equation connecting E and W is therefore;
E
= 0.12W + 0.9
Exercise - Questions 1 - 8, type A. Remainder type B.
Draw graphs of the following simple equations:
1.
y = x + 2 taking values of x between -3 and 2.
2.
Y = 2x + 5 taking values of x between -4 and 4.
3.
Y = 3x - 4 taking values of x between -4 and 3.
4.
Y = 5 - 4x taking values of x between -2 and 4.
The following equations represent straight lines. State in each case the gradient
of the line and the intercept on the y-axis.
5.
Y = x + 3
7.
Y = -5x - 2
6.
Y = -3x + 4
8.
Y = 4x - 3
9.
Find the values of m and c if the straight line y = mx + c passes
through the point (-2,5) and has a gradient of 4.
10.
Find the values of m and c if the straight line y = mx + c passes
through the point (3,4) and the intercept on the y-axis is -2.
In the following find the values of m and c if the straight line y = mx + c
passes though the given points:
11.
(-2, -3) and (3,7)
12.
(1,1) and (2,4)
13.
(-2,1) and (3,-9)
14.
(-3,13) and (1,1)
94
15.
(2,17) and (4,27)
16.
The following table gives values of x and y which are connected by an
equation of the type y = ax + b. Plot the graph and from it find the
values of a and b.
17.
18.
19.
X
2
4
6
8
10
12
y
10
16
22
28
34
40
The following observed values of P and Q are supposed to be related by
the linear equation P = aQ + b, but there are experimental errors. Find
by plotting the graph the most probably values of a and b.
Q
2.5
3.5
4.4
5.8
P
13.6
17.6
22.2
28.0
Q
7.5
9.6
12.0
15.1
P
35.5
47.4
56.1
74.6
In an experiment carried out with a machine the effort E and the load W
were found to have the values given in the table below. The equation
connecting E = aw + b. By plotting the graph check if this is so and
hence find a and b.
W (kg)
10
30
50
60
80
100
E (kg)
8.9
19.1
29
33
45
54
A test on a metal filament lamp gave the following values of resistance
(R ohms) at various voltages (V volts).
V
62
75
89
100
120
R
100
117
135
149
175
These results are expected to agree with an equation of the type
R = mV + c where m and c are constants. Test this by drawing the
graph and find suitable values for m and c.
20.
During an experiment to verify Ohm's Law the following results were
obtained.
E (volts)
0
1.0
2.0
2.5
3.7
I (amperes)
0
0.24
0.5
0.63
0.92
E (volts)
4.1
5.9
6.8
8.0
I (amperes)
1.05
1.48
1.70
2.05
Plot these values with I horizontal and find the equation connecting
E and I.
95
THE MEANING OF M & C IN THE EQUATION OF A
STRAIGHT LINE
Every linear equation may be written in the STANDARD FORM:
Y = mx + c
Hence y = 2x - 5 is in the standard form with m = 2 and c = -5.
The equation y = 4 - 3x is in standard form, if we rearrange it to give
y = -3x + 4. We then see that m = -3 and c = 4.
THE MEANING OF M & C IN THE EQUATION OF A STRAIGHT LINE
The point B is any point on the straight line shown in the diagram below, and it
has the co-ordinates x and y. Point A is where the line cuts the y-axis and it
has co-ordinates x = 0 and y = c.
BC
AC is called the gradient of the line,
now
BC
BC
= AC  AC = AC  gradient of the line
y
= BC + CD = BC + AO
= AC  gradient of the line + AO
= x  gradient of the line + c
96
But
y = mx + c
Hence it can be seen that:
m
= gradient of the line
c
= intercept on the y-axis
The diagram below shows the difference between positive and negative
gradients.
Example 1
Find the law of the straight line shown in the following diagram.
Since the origin is at the intersection of the axes, c is the intercept on the y axis.
From the diagram it will be seen that c = - 4. We now have to find m. Since this
is the gradient of the line we draw
QPN making the sides reasonably long
since a small triangle will give very inaccurate results. Using the scales of x and
y we see that QP = 2 units and PN = 10 units.

m
NP
= QP
=
10
2
= 5
 The standard equation of a straight line y = mx + c becomes y = 5x -4.
97
Example 2
Find the values of m and c if a the straight line y = mx + c passes through the
point (-1,3) and has a gradient of 6.
Since the gradient is 6 we have m = 6.

y = 6x + c
Since the line passes through the point (-1,3) we have y = 3 when x = -1. By
substitution,
3 = 6  (-1) + c
3 = -6 + c

c = 9
Hence
y = 6x + 9
98
INDICES & POWERS
LAWS OF INDICES
The laws of indices are as shown below.
MULTIPLICATION
When multiplying powers of the same quantity together add the indices.
x5
x2
a3  a4  a8
a5  a7
= x5-2 = x3
=
a3 + 4 + 8
a5 + 7
a15
= a12
3y2  2y5  5y4
6y3  4y4
= a15 - 12 = a3
=
30y2 + 5 + 4
24y3 + 4
=
30y11
5y11 - 7
5y4
=
=
7
24y
4
4
POWERS
When raising the power of a quantity to a power multiply the indices together.
(3x)3
= 31  3  x1  3 = 33x3 = 27x3
(a2b3c4)2
= a2  2b3  2c4  2 = a4b6c8
 3m 3

 5n 2





2
=
32m3  2
52n2  2
=
9m6
25n4
NEGATIVE INDICES
A negative index indicates the reciprocal of the quantity.
a-1
5x-3
a2b-2c-3
=
1
a
5
= x3
a2
= b2c3
99
FRACTIONAL INDICES
The numerator of a fractional index indicates the power to which the quantity
must be raised; the denominator indicates the root which is to be taken.
2
x3

3
ab 4
 a 
1
a2

3
x2
4
b3
a
(Note that for square roots the number indicating the root is usually omitted.)

 64a 6
64a 6
 8
2
1
2


1
2
a
 8 2 a6
6
1
2

1
2
 8a 3
ZERO INDEX
Any quantity raised to the power of zero is equal to 1.
a0  1
x
 
y
0
 1
Example 1
 1
1.  
3
4

14
3 4

2 
5
2.
42
3.
9x 2

 81
2
5
2 2
 25
44
14

 21

5
2
 32
3 x 
2
1
2 2
2
 31

1
2
2
x1

1
2
 31 x 1  3 x
100
Example 2
If 3p + 4 = 9p - 2 find the value of p.
3p + 4 = (32)p - 2
3p + 4 = 32p - 4
Since (p + 4) and (2p - 4) are both powers of 3, they must be equal.

p + 4
p
= 2p - 4
= 8
Exercise - Questions 1 - 7
Simplify the following:
1.
35  32  37
2.
b2  b4  b5  b8
3.
57
52
4.
23  24  27
22  25
5.
(72)3
6.
(3x2y3)4
7.
(a2b3c)5
101
BINARY SYSTEM
In the ordinary decimal system the digits 0 to 9 are used.
Consider the number 23. It means:
2  10 + 3  1 = 23
Now remembering that 100 = 1, 101 = 10.
We may write 23 as follows, using decimal to base 10.
2  101 + 3  100
Now lets consider 5623 to the base 10.
5623
5  103 + 6  102 + 2  101 + 3  100
Thus:
80, 321 = 8  104 + 0  103 + 3  102 + 2  101 + 1  100
It is perfectly possible to have a number system which works on the powers of
any number. The most popular of these systems is the Binary (Bi means two),
which operates with the powers of 2 instead of 10 as in the decimal system.
It will be noticed in the decimal system that the greatest digit used is 9 which is
one less than 10. Thus, in the binary system the greatest digit that can be used
is 1 which is one less than 2.
A number written in binary consists only at the digits 0 and 1.
The number
10111
means
1  24 + 0  23 + 1  22 + 1  21 + 1  20
16 + 0 + 4 + 2 + 1 = 23 in decimal 10
102
To convert from a Base 10 system to Binary the following method may be used.
To convert 23 to Binary:
26
25
24
23
22
21
20
64
32
16
8
4
2
1
7
3
1
1
1
1
23
1
0
- Binary
Step 1.
Look for the largest number that is equal to or just under the
decimal number you want to convert. In this case 16.
Step 2.
Put a 1 in the box below.
Step 3.
Take 16 from 23.
Step 4.
Look for the largest number that is equal to or just under 7. In this
case 4.
Step 5.
Put a 1 in the box below.
Step 6.
Take 7 from 4.
Step 7.
Look for the largest number that is equal to or just under 3. In this
case 2.
Step 8.
Put a 1 in the box below.
Step 9.
Take 3 from 2.
Step 10.
Look for the largest number that is equal to or just number 3. In this
case 1.
Step 11.
Put 1 in the box below. This final step finishes the conversion. All
gaps between the digit 1 and the extreme left are filled in with 0.
23 - 16 = 7
7 - 4 = 3
3 - 2 = 1
Thus 23 in Binary is 10111.
Note: The above conversion is not limited to 64, there is no limit. The next
number would be 128, the next 256, the next 512 etc.
103
Example
The following decimal numbers have been converted to binary:
Decimal
Binary
1)
18
10010
2)
32
100000
3)
40
101000
4)
43
101011
Exercise
Convert the following decimal numbers into binary.
1.
11
2.
29
3.
30
4.
111
5.
90
6.
3
7.
48
8.
61
9.
119
10.
127
As well as whole numbers being expressed in Binary we can also express
decimal fractions in Binary but this will be covered in Module 5.
The binary system is used on computers and other calculating machines. Since
only the digits 0 and 1 are used in thee system this is equivalent to a two-state
system. For instance if a device is off it represents an 0 and if it is on a 1 is
represented. The figure shows how the number 10110 can be represented by 5
electric light bulbs.
104
OTHER NUMBER SCALES
In the scale of 5, powers of 5 are used. Only the digits 0,1,2,3, and 4 are
available because the greatest digits used must be one less than 5. If you are
told that the number 3412 is in the scale of 5 it means that the number is based
upon the powers of 5. To show that this is so we write 3412 5 . The suffix 5
indicates that the number of 5 is being used. The number scale is called the
BASE. We say 3412 5 is to the BASE 5.
Example 1
3412 5 = (3 x 53 )  (4  52)  (1 52)  (2  50 )
= (3  125)  (4  25)  (1 5)  (2  1)
= 375 + 100 + 5 + 2
= 48210.
Example 2:
463 8 is a number of BASE 8.
2
1
0
463 8 = (4  8 )  (6  8 )  (3  8 )
= (4  64)  (6  8)  (3  1)
= 256 + 48 + 3
= 30710
Converting from a number in BASE 10 to a number in any other BASE use the
table shown below
ExampleConvert 41310 into BASE 8
BASE 10
 BY BASE 
REMAINDER
POWERS OF 8
413
= 51
5
8º
51
=6
3
8¹
105
6
=0
6
8²
Read the remainder from
Bottom up
=
635
Hence 41310
=
635 8
Check:
6358  (6  82)  (3  81)  (5  80 )
= 384 + 24 + 5
= 41310
Exercise:
Convert the following numbers to BASE 8
a) 390
b) 495
c) 1102
d) 80
e) 772
OCTAL
As well as Binary (Base 2) Base 8 OCTAL and Hexidecimal Base 16 is also used
in computer technology, though OCTAL and Hexidecimal would ultimately be
converted to Binary as all internal computer operations are binary. JAR 66
Module 5 will cover more on this topic.
HEXIDECIMAL
As previously mentioned the hexidecimal is to Base 16. It differs from other
systems in using a combination of both numbers and letters. The rules for
manipulation of the arithmetic are similar to those for decimal. The chart that
follows is only an introduction. Conversions and arithmetic calculations will be
practised in Module 5.
106
107
SIMULTANEOUS EQUATIONS
Consider the two equations:
2x + 3y = 13
[1]
3x + 2y = 12
[2]
Each equation contains the unknown quantities x and y. The solutions of the
equations are the value if x and y which satisfy both equations. Equations
such as these are called simultaneous equations.
ELIMINATION METHOD IN SOLVING SIMULTANEOUS EQUATIONS
The method will be shown by considering the following examples.
Example 1
1.
Solve the equations:
3x + 4y
= 11
[1]
x + 7y
= 15
[2]
If we multiply equation [2] by 3 we shall have the same coefficient of x in both
equations:
3x + 21y = 45
[3]
We can now eliminate x by subtracting equation [1] from equation [3].
3x + 21y = 45
[3]
3x + 4y
= 11
[1]
17y
= 34
y = 2
To find x we substitute for y = 2 in either of the original equations. Thus,
substituting for y = 2 in equation [1],
3x + 4  2
= 11
3x + 8
= 11
3x
= 11 - 8
3x
= 3
x
= 1
Hence the solutions are:
x = 1 and y = 2
Hence the solutions are correct since the L.H.S. and R.H.S. are equal.
108
2.
Solve the equations:
5x + 3y
= 29
[1]
4x + 7y
= 37
[2]
The same coefficient of x can be obtained in both equations if
equation [1] is multiplied by 4 (the coefficient of x in equation
[2]) and equation [2] is multiplied by 5 (the coefficient of x in
equation [1]).
Multiply equation [1] by 4,
20x + 12y
= 116
[3]
Multiply equation [2] by 5,
20x + 35y
= 185
[4]
Subtracting equation [3] from equation [4],
23y = 69
y = 3
Substituting for y = 3 in equation [1],
5x + 3  3 = 29
5x + 9 = 29
5x = 20
x = 4
Hence the solutions are:
y = 3 and x = 4
Check in equation [2],
L.H.S. = 4  4 + 7  3
= 16 + 21 = 37 = R.H.S.
109
3.
Solve the equations:
7x + 4y
= 41
[1]
4x - 2y
= 2
[2]
In these equations it is easier to eliminate y because the same coefficient of y
can be obtained in both equations by multiplying equation [2] by 2.
Multiplying equation [2] by 2,
8x - 4y
= 4
[3]
Adding equation [1] and [3],
15x
x
= 45
= 3
Substituting for x = 3 in equation [1],
7  3 + 4y = 41
21 + 4y = 41
4y = 20
y = 5
Hence the solutions are:
x = 3 and y = 5
Check in equation [2],
L.H.S. = 4  3 - 2  5
= 12 - 10 = 2 = R.H.S.
110
4.
Solve the equations:
2x y
7
=
3
4
12
[1]
3x 2y
3
4 - 5 = 10
[2]
It is best to clear each equation of fractions before attempting to
solve.
In equation [1] the L.C.M. of the denominators is 12. Hence by
multiplying equation [1]by 12,
8x - 3y
= 7
[3]
In equation [2] the L.C.M. of the denominators is 20. Hence by
multiplying equation [2] by 20,
15x - 8y
= 6
[4]
We now proceed in the usual way. Multiplying equation [3] by
8,
64x - 24y = 56
[5]
Multiplying equation [4] by 3,
45x - 24y = 18
[6]
Subtracting equation [6] from equation [5],
19x = 38
x = 2
Substituting for x = 2 in equation [3],
8  2 - 3y = 7
16 - 3y = 7
- 3y = -9
y = 3
Hence the solutions are:
x = 2 and y = 3
111
Exercise - Questions 1 - 5, Level 1.
Solve the following equations for x and y and check the solutions:
1.
2.
3.
3x + 2y
= 7
x + y
= 3
x - 3y
= 1
x + 3y
= 19
x + 3y
= 7
2x - 2y
= 6
4.
5.
7x - 4y
= 37
6x + 3y
= 51
4x - 6y
= -2.5
7x - 5y
= -0.25
112
INDICES AND LOGARITHMS
LAWS OF INDICES
The laws of indices are shown below.
MULTIPLICATION
When multiplying power of the same quality together add the indices.
x 6  x7  x6 7  x13
y2  y3  y 4  y5  y23 45  y14
DIVISION
When dividing powers of the same quantity subtract the index of the denominator
(bottom part) from the index of the numerator (top part).
x 5  x 5 2  x 3
a3  a4  a8  a3 48
a5  a7
a5 7
15
 a12  a1512  a3
a
3y2  2y5  5y 4
3
6y  4y
4

30 y 25 4
24 y
3 4
117
11

30 y
7
24 y
5y

4
4
5y

4
POWERS
When raising the power of a quantity to a power multiply the indices together.
3
(3x)  313  x 33  27x 3
(a2 b2 c 4 )2  a22 b32 c 42  a4 b6 c8
 3m 
 5n 
 
3
2
2
2 32
 3 2m22  9m 4
5 n
25n
6
113
NEGATIVE INDICES
A negative index indicates the reciprocal of the quantity
1
1
a a
5
3
5x  x 3
a2
2  2 3
a b c  b2 c 3
FRACTIONAL INDICES
The numerator of a fractional index indicates the power to which the quantity
must be raised; the denominator indicates the root which is to be taken.
2
3
3
x  x
2
3
4
3
4
ab  a  b
1
2
a  a
(Note that for square roots the number indicating the root is usually omitted).
1
6 2
6
1
6 2
64a  ( 64a )  (8 a )
8
2
1
2
6
a
2
1
2
 8a
3
ZERO INDEX
Any quantity raised to the value of zero is equal to 1.
0
a 1
0
x
  1
y
 
Example 1
4
4
4
 
(1)  1   1 4  34  81
 3
3
1
(2)
 
5
4
5
2 5

2
2 2
1
4  2 2
 25  32
(3)
 
9x  3 x
2
2
1
2 2
2 1

2
 31
x
2 1

1 2
 31 x1  3x
114
Example 2
If 3p  4  9p  2 find the value of p.
3 
p2
p 4

p 4
 32p 4
3
3
2
Since (p + 4) and (2p-4) are both powers or 3, they must be equal.
 p + 4=2p – 4
p=8
Exercise- Questions 1-7 Level 1
Simplify the following:
1) 35  32  37
2) b2  b4  b5  b8
7
3) 52
5
3 4 7
4) 2 2 2
2
5
2 2
5)
7 
2
3

6) 3x 2 y3
7)

4
a b c 
2
5
3
NUMBERS IN STANDARD FORM
n,
A number expressed in the form A  10 where A is a number between 1 and 10
and n is an integer is said to be in standard form.
Example 3
50 000  5  10 000  5  10
0.003 
4
3
3
 3
1000 10
Exercise- Level 1
Express each of the following in standard form:
1) 8000
5) 0.0035
2) 92 500
6) 0.7
3) 893
7) 0.000 365
4) 5 600 000
8) 0.007 12
115
LOGARITHMS
Any positive number can be expressed as a power of 10. For instance:
1000  10
3
1.869 2
74  10
These powers of 10 are called logarithms to the base 10.
log arithm
That is:
number= 10
The log tables at the end of this book give the logarithms of numbers between
1 and 10.
Thus,
log 5.176=0.714 0
To find out the logarithms of numbers outside this range we make use of
numbers in standard form and the multiplication law of indices. For example:
324 .3  3.243  10
2
log 3.243  0.510 9
324 .3  10
 10
0.510 9
 10
2
2.510 9
log 324 .3  2.510 9
A logarithm therefore consists of two parts:
(1) A whole number part called the characteristic.
(2) A decimal part called the mantissa which is found directly from the log
tables
For a number, 10 or greater, the characteristic is found by subtracting 1 from
the number of figures to the left of the decimal point in the given number.
In the number 825.7 the characteristic is 2.

log 825.7- 2.916 8
In the number 18 630 the characteristic is 4.

log 18 630 = 4.270 2
NEGATIVE CHARACTERISTICS
0.632 0  6.321  10
1
log 6.321  0.800 0
0.632 1  100.800 8  101
 1010.800 0
116
The characteristic is therefore –1 and the mantissa is 0.800 8. However writing –
1+0.800 0 for the logarithm of 0.632 1 would be awkward and we therefore write:
log 0.632 1  1.8008
Note that the minus sign has been written above the characteristic but it must be
clearly understood that
and
2.735 6  2  0.735 6
4.067 3  4  0.067 3
All numbers between 0 and 1 have negative characteristics which are found by
adding 1 to the number of zeros following the decimal point.
In the number 0.073 58 the characteristic is 2 .

log 0.073 58  2.866 8
In the number 0.000 612 3 the characteristic is 4.

log 0.000 612 3  4.787 0
ANTI-LOGARITHMS
The table of antilogs at the end of this book contains the numbers which
correspond to the given logarithms. In using these tables remember that only the
decimal part of the log is used.
Example 4
(1)To find the number whose log is 2.531 2. Using the mantissa .531 2, we find
3398 as the number corresponding. Since the characteristic is 2 the number
must be 339.8.
(Note the log 339.8=2.531 2.)
(2)To find the number whose log is 3.617 8. Using the mantissa .617 8 we find
4148 as the number corresponding. Since the characteristic is 3 the number
must be 0.004 148.
(Note that log 0.004 148  3.617 8. )
Exercise 106- Level 1
Write down the following numbers:
1) 7.263
7) 70.01
2) 8.197
8) 176 300
3) 63.25
9) 0.178 6
4) 716.4
10) 0.006 341
5)1823
11) 0.068 91
6) 78 640
12) 0.000 718 2
Write down the antilogs of the following
13) 2.618 3
17) 1.234 5
14) 1.735 8
18) 2.600 8
15) 0.628 8
19) 4.631 8
117
16) 3.105 8
20) 3.555 7
RULES FOR THE USE OF LOGARITHMS MUTIPLICATION
Find the logs of the numbers and add them together. The antilog of the sum
gives the required answer.
Example 5
19.63 x 0.067 34 x 0.918 7
number
logarithm
19.63
1.292 9
0.067 34
2.828 3
0.918 7
1.963 2
Answer  1.215
DIVISION
0.084
Find the log of each number. Then subtract the log of the denominator (bottom
number) from the log of the numerator (top number).
Example 6
17.63
0.038 62
number
logarithm
17.63
1.246 3
0.038 62
2.586 8
Answer  456.6
2.659 5
Example 7
0.617 8  20.31
136.5  0.092 73
In problems where there is multiplication and division a table layout like the one
below is helpful.
Numerator
number
Denominator
logarithm
number
1.790 8
136.5
1.307 7
0.097 73
numerator
1.098 5
denominator 1.102 3
denominator
1.102 3
0.617 8
20.31
Answer  0.991 3
logarithm
2.135 1
2.967 2
1.996 2
118
119
120
121
GEOMETRY
RADIAN MEASURES
We have seen that an angle is measured in degrees. There is however a second
way of measuring an angle. In this second system the unit is known as the
radian. Refering to
RELATION BETWEEN RADIANS AND DEGREES
If we make the arc AB equal to a semi-circle then,
= r
Length of arc
Angle in radians =
r
r

But the angle at the centre subtended by a semi-circle is 180 and hence
 radians  180ο
ο
1 radian  180  57.3 ο

It is worth remembering that
ο
0
0  180 radians
ο

ο

ο

ο

60  3 radians
45  4 radians
90  2 radians
30  6 radians
Example 4
(1) Find the angle in radians subtended by an arc 12.9 cm long whose radius is
4.6 cm.
122
Angle in radians 
length of arc
radius of circle
12.9
(2) Express an angle of 1.26 radians in
4.6
 2.804 radians

degrees and minutes.
Angle in deg rees


180  angle in radians

180  1.26

 72.18
ο
Now
ο
0.18  0.18  60 minutes  11 minutes Angle=7211'
(3) Express an angle of 104 in radian
Angle in radians


  angle in deg rees
180
  104
180
 1.815 radians
Exercise- Level 1
(1) Find the angle in radians subtended by the following arcs:
(a) arc = 10.9cm, radius = 3.4cm
(b) arc = 7.2m, radius = 2.3m
(2) Express the following angles in degrees and minutes:
(a) 5 radians
(b) 1.73 radians
(c) 0.159 radians
(4)Express the following angles in radians:
(a) 83
(b) 189
(c) 295
(d) 5.21
123
TYPE OF ANGLES
An acute angle is less than 90.
A right angle is equal to 90.
A reflex angle is greater than 180.
An obtuse angle lies between 90 and 180.
Complementary angles are angles whose sum is 90.
Supplementary angles are angles whose sum is 180
PROPERTIES OF ANGLES AND STRAIGHT LINES
(1)The total angle on a straight line is 180. The angles A and B are called
adjacent angles. They are also supplementary.
(2) When two straight lines intersect the
opposite angles are equal. The angles A and C are called the vertically opposite
angles. Similarly the angles B and D are also vertically opposite angles.
(4) When two parallel lines are cut by a transversal
(a) The corresponding angles are equal a=1; b=m; c=p;d=q.
(b) The alternate angles are equal d=m; c=l.
(d) The interior angles are supplementary d + l = 180; c + m= 180.
124
Conversely if the two straight lines are cut by a transversal the lines are parallel if
any one of the following is true:
(a) Two corresponding angels are equal.
(b)Two alternate angles are equal.
(c) Two interior angles are supplementary .
Example
(1)Find the angle A shown in Fig.
ο
ο
ο
B  180  138  42
B  A corresponding angles 
A  42
ο
(2) In Fig. the line BF bisects ABC. Find the value of the angle  .
The lines AX, BZ and EY are all parallel
because they lie at right-angles to the line XY.
125
c  b alternate angles : BZ || EY 
ο

b  38 since
c  38
ο

a  d alternate angles :
ο

d  80 sin ce a  80
ο
XD || BZ 

ο
ο
ABC  b  d  80  38  118
ο
FBC  118ο  2  59ο sin ce BF bi sec ts ABC 
b    59
ο
ο
ο
38    59
ο
ο
ο
  59  38  21
Exercise-All Level 1
1) Find x in Fig.
2) Find A in Fig.
3) Find x in Fig.
4) In Fig. find a, b ,c & d.
126
5)Find the angle x in Fig.
6) Find x in Fig.
7) A reflex angle is:
a less than 90
b greater than 90
c greater than 180
d equal to 180
8) Angles whose sum is 180 are called:
a complementary angles
b alternate angles
c supplementary angles
d corresponding angles
9) In Fig. find A
127
10) In Fig., AB is parallel to ED. Find the angle x.
11) Find A in Fig.
12) In Fig. the lines AB, CD and EF are parallel. Find the values of x and y.
13) In Fig.
a q=p + r
b p + q + r = 360
cq=r–p
d q = 360 – p – r
128
GRAPHICAL REPRESENTATION
Charts and Graphs are pictorial representations of date. They enable you to
quickly visualise certain relationships, completer complex calculations and predict
trends. Furthermore, charts allow you to see the rate and magnitude of changes.
Information is presented graphically in many different forms. Graphs are often
found in the form of bar charts, pictographs, broken line graphs (or continuous
curve graphs) and the circular or pie chart. Another type of graph that you will
meet in aircraft maintenance if the nomogram.
Many of the graphs that you will meet will conform to a standard layout of two
variables displayed on adjacent axes, normally vertical and horizontal. This
layout is described as Cartesian and usually has the two axes, labelled x and y
which intersect at the zero point.
129
USE OF GRAPHS
You will find many graphs also produce a straight line, which may, or may not
pass through the origin. A graph of this type is formed when load is plotted
against extension for an ‘elastic’ material subjected to a tensile test.
For such a graph, it is evident that the load value is directly proportional to the
extension that the load produces.
If you plot a graph, which represents the compression of a gas in a close d
cylinder, it takes the form as shown. If the temperature of the gas remains
constant during the compression, then P x volume = constant, produces a curve
known as a Hyperbola.
130
131
Graphs of sine and Cosine Waves
Alternating voltages and currents are often represented by sine and cosine
waves. These are the result of plotting the path of a rotating output along a
straight axis.
The only difference between them is that the sine wave always has its zero value
at the start and completion of each rotation. The cosine wave however,
begins and finishes its rotation with the output at its maximum value.
Nomograms
The need to show how two or more variables affect a value is common in the
maintenance of aircraft. Nomograms also known as an alignment chart, are a
special type of graph that enables you to solve complex problems involving more
than one variable.
Most nomogram charts contain a great deal of information and require the use of
scales on three sides of the chart, as well as diagonal lines. In fact, some
charts contain so much information, that it can be very important for you to
carefully read the instructions before using the chart and to show care
when reading information from the chart itself.
Illustrated below is a graph of three variables, distance, speed and time, the
resulting distance can be extracted from the graph at the point where these
two dashed lines meet. A speed of 375 knots for 2.5 hours would result in
a distance of approximately 950 nautical miles.
132
133
TRIGONOMETRY
THE NOTATION FOR A RIGHT-ANGLED TRIANGLE
The sides of a right-angled triangle are given special names. The side AB lies
opposite the right-angle and it is called the hypotenuse. The side BC lies
opposite to the angle A and it is called the side oppostite to A. The side AC is
called the side adjacent to A.
When we consider the angle B the side AB is still the hypotenuse but AC is now
the side opposite to B and BC is te side adjacent to B.
THE TRIGONOMETRICAL RATIOS
Consider any angle 0 which is bounded by the lines OA and OB as shown. Take
any point P on the boundary line OB. From P draw line PM perpendicular to OA
to meet it at the point M. Then
the ratio
MP
is called the sine of AOB
OP
the ratio
OM
is called the cosine of AOB
OP
and
the ratio
MP
is called the tangent of AOB
OM
134
THE SINE OF AN ANGLE
The abbreviation ‘sin’ is usually used for sine. In any right-angled triangle the
sine of an angle

side opposite the angle
hypotenuse
BC
AC
AB
sin C 
AC
sin A 
Example 1
Find by drawing a suitable triangle the value of sine 30.
ο
Draw the lines AX and AY which intersect at A so that the angle YAX  30 as
shown. Along AY measure off AC equal to 1 unit (say 10cm) and from C draw
CB perpendicular to AX. Measure CB which will be found to be 0.5 units (5cm in
this case).
ο
Therefore sin 30 
5
 0.5.
10
Although it is possible to find the sines of the angles by drawing, this is
inconvienient and not very accurate. Tables of sines have been calculated which
allow us to find the sine of any angle. Part of this table is reproduced below and
in full, with the other trigonometrical tables, at the end of the book.
READING THE TABLE OF SINES OF ANGLES
(1) To find sin 12. The sine of an angle with an exact number of degrees is
shown in the column headed 0. Thus sin 12=0.2079.
(2) To find sin 1236’. The value will be found under the column headed 36’.
Thus sin 12= 0.2181.
(3) To find sin 1240’. If the number of minutes is not an exact multiple of 6
we use the table of mean differences. Now 1236’=0.2181 and 40’ is 4’
more than 36’. Looking in the mean difference headed 4 we find the
135
value 11. This is added on to the sine of 1236’ and we have sin
1240’=0.2181 + 0.0011= 0.219 2.
(4) To find the angle whose sine is 0.1711. Look in the table of sines to find
the nearest to find the nearest number lower than 0.1711. This is found to
be 0.1702 which corresponds to an angle of 948’. Now 0.1702 is 0.000 9
less than 0.1711 so we look in the mean difference table in the row
marked 9 and find 9 in the column headed 3’. The angle whose sine is
0.1711 is then 948’ + 3’ =951’ or sin 951’ = 0.1711.
Example 2
(1) Find the length of AB.
AB is the side opposite ACB. AB is the hypotenuse since it is opposite to the
right angle.
Therefore
136
AB
ο
 sin 22
BC
ο
AB  BC  sin 22  80  0.3746
 29.97 mm
(2) Find the length of AB
BC is the side opposite to BAC and AB is the hypotenuse.
ο
BC
 sin 23 35'
AB
BC
60
AB 

ο
sin 23 35' 0.400 0
 150 mm
(3) Find the angles CAB and ABC in  ABC which is shown below..
AC 20

 0.333 3
AB 60
From the sine tables
ο
'
 B  19 28
sin B 
ο
ο
ο
A  90  19 28'  70 32'
Exercise- All Level 1
1)Find, by drawing, the sines of the following angles:
(a) 30
(b)45
(c)68
2) Find, by drawing, the angles whose sines are:
(a)
1
3
(b)
3
4
(c) 0.72
3)Use the tables to write down the values of:
(a) 0.156 4 (b) sin 1812
(c) sin 7442
(d) sin 723
137
(e) sin 8735
(f) sin 011
4)Use the tables to write down the angles whose sines are:
(a) 0.156 4 (b) 0.913 5 (c) 0.988 0
(d) 0.080 2 (e) 0.981 4 (f) 0.739 5
(g) 0.050 0 (h) 0.270 0
5)Find the lengths of the sides marked x
6)Find the angles marked 0
7) In  ABC,  C=90,  B=2317 and AC=11.2cm. Find AB.
8) In  ABC,  B=90, A=6728 and AC=0.86 m. Find BC.
138
THE COSINE OF AN ANGLE
In any right-angled triangle the cosine of an angle

side adjacent to the angle
hypotenuse
AB
AC
BC
cos C 
AC
cos A 
The abbreviation ‘cos’ is usually used for cosine.
The cosine of an angle may be found by drawing, the construction being similar
to that used for the sine of an angle. However, tables of cosines are available
and these are used in a similar way to the table of sines except that the mean
differences are now subtracted.
Example
Find the length of the side BC
BC is the side adjacent to BCA and AC is the hypotenuse.
ο
BC
 cos 38
AC
ο
BC  AC  cos 38  120  0.788 0
 94.56mm
(2) Find the length of the side AC
AB is the side adjacent to BAC and AC is the hypotenuse.
139
Therefore
ο
AB
 cos 60
AC
AB
28
AC 

 56 cm
ο
0.5000
cos 60
(3) Find the angle 0 shown
Since  ABC is isosceles the perpendicular AD bisects the base BC and hence
BD=15mm.
cos 0 
BD 15

 0.3
AB 50
ο
0  72 32'
Exercise-All type Level 1
1) Use the tables to write down the values of:
(a) cos 15
(b) cos 2418
(c) cos 7824
(d) cos 011
(e) cos 7322
(f) cos 3959
2) Use the tables to write down the angles whose cosine are:
(a) 0.913 5 (b) 0.342 0 (c) 0.967 3
(d) 0.428 9 (e) 0.958 6 (f) 0.008 4
(g) 0.261 1 (h) 0.470 0
140
3 )Find the lengths of the sides marked x
4) Find the angles marked 0 , the triangles being right angled.
5) An isosceles triangle has a base of 3.4cm and the equal sides are each 4.2cm
long. Find the angles of the triangle and also its altitude.
6)In  ABC,  C=90,  B=33 and BC-2.4cm. Find AB.
7) In  ABC,  =90,  A=6245 and AC=4.3cm. Find AB.
8) Calculate  BAC and the length BC.
9)Calculate BD, AD, AC and BC.
141
THE TANGENT OF AN ANGLE
In any right-angled triangle
the tangent of an angle

side opposite to the angle
side adjacent to the angle
BC
AB
AB
tan C 
BC
tan A 
The abbreviation ‘tan’ is usually used for tangent. From the table of tangents the
tangents of angles from 0 to 90 can be read directly.
For example:
ο
tan 37  0.753 6
and
ο
tan 62 29'  1.919 6
Example 4
(1) Find the length of the side AB
AB is the side opposite C and AC is the side adjacent to  C. Hence,
AB
 tan C
AC
AB
ο
 tan 42
AC
ο
AB  AC  tan 42  40  0.9004
 36.02 mm
(2)Find the length of the side BC
142
There are two ways of doing this problem
(a)
ο
AB
AB
 tan 38 or BC 
ο
BC
tan 38
Therefore BC 
32
 40.96 mm
0.781 3
(b) Since C  38
ο
ο
ο
A  90  38  52
ο
now
BC
 tan A or BC  AB  tan A
AB
BC  32  1.280  40.96 mm
Both methods produce the same answer but method (b) is better because it is
quicker and more convenient to multiply than divide. Whenever possible the ratio
should be arranged so that the quantity to be found is the numerator of the ratio.
Exercise- All Level 1
1)
Use tables to write down the values of:
(a) tan 18
(b) tan 3224
(c) tan 5342
(d) tan 3927
(e) tan 1120
(f) tan 6923
2)
Use tables to write down the angles whose tangents are:
(a) 0.445 2 (b) 3.270 9 (c) 0.076 9
(d) 0.397 7 (e) 0.356 8 (f) 0.826 3
(g) 1.925 1 (h) 0.016 3
3)
Find the lengths of the sides marked y in the triangles being right-angled.
143
4)
Find the angles marked  , the triangles being right-angled.
5)
An isosceles triangle has a base 10cm long and the two equal angles are
each 57. Calculate the altitude of the triangle.
6) In  ABC,  B=90,C=49 and AB=3.2cm. Find BC.
7) In  ABC, A=1223, B=90 and BC=7.31cm. Find AB.
8) Calculate the distance x
9) Calculate the distance d
144
TRIGONOMETRICAL RATIOS BETWEEN 0 AND 360.
Previously the definitions for the sine, cosine and tangent of an angle between 0
and 90 were given. In this chapter we show how to deal with angles between 0
and 360.
In Fig., the axes XOX and YOY, have the four quadrants In each of these four
quadrants we make use of the sign convention used when drawing graphs.
Now an angle, if positive, is always measured in an anti-clockwise direction for
OX and an angle is formed by rotating a line (such as OP) in an anti-clockwise
direction. It is convenient to make the length of OP equal to 1 unit. Referring to
Fig. we see
In the first quadrant,
sin 0  P1M1  P1 M1
OP1
 y co  ordinate of P1
cos 01  OM1  OM1
OP1
 x co  ordinate of P1
tan 01 
P1M 1
OM1

y co  ordinate of P
x co  ordinate of P
145
Hence in the first quadrant all the trigonometrical ratios are positive.
In the second quadrant
sin 02 
P2M 2
 P2 M2
OP2
 y co  ordinate of P2
The y co-ordinate of P 2 is positive and hence in the second quadrant the sine of
an angle is positive.
cos 02 
OM2

Op2 OM2
 x co  ordinate of P2
The x co-ordinate of P 2 is negative, and hence in the second quadrant the cosine
of an angle is negative. tan 02  P2M 2  y co  ordinate of P2 But the y co-ordinate
OM2
x co  ordinate of P2
of P2 is positive and the x co  ordinate of P2 is negative, hence the tangent of an
angle in the second quadrant is negative.
146
The trigonometrical tables usually give values of the trigonometrical ratios for
angles between 0 and 90. In order to use these tables for angles greater than
90 we make use of the triangle OP2M 2
,
where we see that
P2M
2

y co  ordinate of P2
x co  ordinate of P2
But
P2 M2  sin 02

sin 02  sin (180  02 )
Also
OM2   OP2 cos (180 02 )
 - cos (180  02 )

cos 02  cos (180  02 )
Similarily
tan 02   tan(180  02 )
In the third quadrant by similar considerations
sin 03  sin (03  180ο )
cos 03  cos (03  180ο )
tan 03  tan (03  180ο )
In the fourth quadrant,
sin 04   sin (360 ο  04 )
cos 04  cos (360 0  04 )
tan 04   tan (360  04 )
The results are summarised in Fig.
147
Example
Find the values of sin 158, cos 158 and tan 158. Referring to Fig.
sin 158 ο 
MP
 sin  POM
OP
ο
ο
 sin (180  158 )
ο
 sin 22  0.3746
ο
OM
cos 158  OP  cosPOM
ο
ο
 cos (180  158 )
ο
  cos 22  0.9272
ο
MP
tan 158  OM   tan POM
ο
ο
  tan (180  158 )
ο
  tan 22  0.4040
The table below may be used for angles in any quadrant.
148
Example
(1) Find the sine and cosine of the following angles:
(a) 171
(b) 216
(c) 289º
(a) sin 171º = sin (180º-171º)= sin 9º
= 0.1564
cos 171º = -cos (180º-171º)
= -cos 9º = - 0.9877
(b) sin 216º = -sin (216º - 180º)
= - sin 36º = - 0.5878
cos 216º = - cos (216º-180º)
= - cos 36º= -0.8090
(c) sin 289º = - sin (360º-289º)
= - sin 71º= -0.9455
cos 289º = cos(360º-289º)
= cos 71º = 0.3256
(2) Find all the angles between 0º and 180º:
(a) whose sine is 0.4676;
(b)whose cosine is –0.357 2.
(3) Is sin A 
3
find the values of cos A .
5
In the second quadrant:
2
OM  4
cos A 
4
OM2

5
M 2P2
5)Copy and complete the following table.

sin 
cos 
tan 
108º
163º
207º
320º
134º
168º
225º
286º
300º
95º
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POLAR CO-ORDINATES
It was shown that a point on a graph may be positioned by using rectangular coordinates (sometimes called Cartesian co-ordinates). Hence if P is the point (3,4)
its position is as shown in Fig.
However, the position of P may also be indicated by stating the length OP and
the angle . Thus in Fig.
Op  3 2  4 2  25  5
(by u sin g Pythagoras ' theorem )
tan 0 
and
4
 1.333
3
ο
0  53 7 '
P is then said to have the polar co-ordinates (5,537). The angle  may be
 
expressed in degrees or in radians. If Q is the point  7,  the angle  is
 3

3
radians or 60.
Example
(1) A point P has Cartesian co-ordianates (5, -7). State the polar co-ordinates of
P.
150
From Fig.
OP  52  72  74  8.602
7
tan φ  1.4
5
φ  54ο28'
θ  360ο  54ο 28'  305ο 32'
Hence the polar co  ordinates of P are
(8.602,305ο 32')
(2) A point A has the polar co  ordinates
 5π 
 8,
. Determine the Cartesian co  ordinates

6 

of A.
In Fig.
5π
5  180
radians 
 150 ο
6
6
φ  180 ο  150 ο  30ο
ο
OB  OA cos 30  8  0.8660  6.928
θ
AB  OA sin 30ο  8  0.5000  4
Hence the Cartesian co  ordiantes of A
are ( 6.928,4).
Exercise-All Level 2
1)Calculate the polar co-ordinates for the following points:
(a) (3,2)
(b) (5,8)
(c) (-4,8)
(d) (-3,-5)
(e) (6,-4)
(f) (-4,-6)
(g) (8,-7)
(h) (-1,3)
2) Calculate the Cartesian co-ordinates of the following points (polar):
(a) (5,30)
(b) (7,65)
(c) (2,112)
(d) (4,148)
(e) (7,198)
(f) (3,265)
(g) (5,297)
(h) (3,330)
3)Calculate the Cartesian co-ordinates for the following points:
151
 
(a)  5, 
 3
 
(b)  4, 
 2
 3 
(c)  6, 
 4 
 5 
(d) 10, 
3 

4)Calculate the polar co-ordinates for the following points, stating the angle in
radian measure:
(a) (2,1)
(b) (-3,5)
(c) (-2,-4)
(d) (4,-2)
5)In Fig., with origin O, the polar co-ordinates of the point X are (5,40). YXP is
a straight line parallel to the x-axis.
Find:
(a)the polar co-ordinates of Y
(b)the polar co-ordinates of P
152