M1 Cat-B1 Notes V1.1

Aircraft Maintenance Training Centre Part-147

Fundamentals M1 MATHEMATICS EASA Part-66 Cat-B1

Part-66 Cat-B1 Mathematics M1 ____________________________________________________________________________________________________________________________________________

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


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TABLE OF CONTENTS

1. Arithmetic............................................................................... 1 1.1. Arithmetical Terms and Signs ................................................................... 1

1.1.1. Addition ................................................................................................ 1 1.1.2. Subtraction ........................................................................................... 1

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

1.2.1. Multiplication......................................................................................... 2 1.2.2. Division................................................................................................. 2

Arithmetic Law of Precedence .....................................................................................2 Exercises ............................................................................................................ 3

1.3. Fractions ...................................................................................................... 4

1.3.1. Converting between improper fractions and mixed numbers............... 4 1.3.2. Multiplication of Fractions..................................................................... 4 1.3.3. Division of Fractions ............................................................................. 4 1.3.4. Addition of Fractions............................................................................. 5 1.3.5. Subtraction of Fractions ....................................................................... 5

1.4. Decimal Numbers........................................................................................ 7

Number of Decimal Places ..........................................................................................7 Significant Figures .......................................................................................................7

1.4.1. Addition and Subtraction of Decimal Numbers .................................... 7 1.4.2. Multiplication of Decimal Numbers....................................................... 8 1.4.3. Division of Decimal Numbers ............................................................... 8

1.5. Factors and Multiples ................................................................................. 9 1.6. Powers and Roots....................................................................................... 9

Laws of Indices............................................................................................................9 Exercises .......................................................................................................... 10

Standard Form...........................................................................................................10 1.7. Measures and Conversion Factors ......................................................... 11

Exercises .......................................................................................................... 12 1.8. Ratio and Proportion ................................................................................ 13

1.8.1. Direct Proportion ................................................................................ 13

1.8.2. Inverse Proportion ..............................................................................13 1.8.3. Proportional Parts ...............................................................................13 1.8.4. Constant of Proportionality .................................................................14 Exercises...........................................................................................................14

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

1.9.1. Mean...................................................................................................15 1.9.2. Median ................................................................................................15 1.9.3. Mode...................................................................................................15 1.9.4. Percentages........................................................................................15 Exercises...........................................................................................................16

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

1.10.1. Areas ..................................................................................................17 1.10.2. Volumes..............................................................................................17 Exercises...........................................................................................................18

2. Algebra..................................................................................19

2.1. Introduction ...............................................................................................19

2.1.1. Addition and Subtraction ....................................................................19 2.1.2. Multiplication and Division ..................................................................19 2.1.3. Brackets..............................................................................................19 Exercises...........................................................................................................20

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

2.2.1. Transposition of Formulae..................................................................21 2.3. Simultaneous Equations ..........................................................................21

Solving simultaneous equations by elimination..........................................................21 Solving simultaneous equations by substitution.........................................................22

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

2.4.1. Solving by using the formula ..............................................................23 2.4.2. Solving by factorisation.......................................................................23 2.4.3. Factorising quadratic equations..........................................................23


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

Exercises .......................................................................................................... 25 2.6. Logarithms................................................................................................. 26 2.7. Number Systems....................................................................................... 27

2.7.1. The Binary System............................................................................. 27 2.7.2. The Hexadecimal System .................................................................. 27 Exercises .......................................................................................................... 28

3. Geometry.............................................................................. 29

3.1. Graphs........................................................................................................ 29

3.1.1. Graphical Representation .................................................................. 29 3.1.2. Graphs of Linear equations................................................................ 29

Equation of a straight line ..........................................................................................30 3.1.3. Graphs of Quadratic equations .......................................................... 31 3.1.4. Graphical solution of Simultaneous Equations .................................. 32 Exercises .......................................................................................................... 32

3.2. Trigonometry............................................................................................. 33

3.2.1. Pythagoras Theorem.......................................................................... 33 3.2.2. Trigonometrical Ratios ....................................................................... 33 3.2.3. Use of Tables ..................................................................................... 36 3.2.4. Rectangular (Cartesian) Co-ordinates ............................................... 36 3.2.5. Polar Co-ordinates ............................................................................. 36

Converting between Polar and Cartesian ..................................................................37 Exercises .......................................................................................................... 37

3.3. Geometrical Constructions...................................................................... 38

3.3.1. The Circle ........................................................................................... 38 Elements and Properties of the circle ........................................................................38 Important circle theorems ..........................................................................................38

3.3.2. Angles ................................................................................................ 40 Degrees and Radians................................................................................................40 Length of Arc .............................................................................................................40

3.3.3. Construction ....................................................................................... 41 Appendix 1 – Trigonometric Tables.............................................. 44

Part-66 Cat-B1 Mathematics M1 Part-147 Aircraft Maintenance Training Centre ____________________________________________________________________________________________________________________________________________

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1. ARITHMETIC It is important for students to realise that the use of calculators during examinations is not permitted. Therefore, the mathematics taught is aimed towards using mental calculations and it is therefore strongly recommended that students practise these problems without the aid of a calculator.

1.1. Arithmetical Terms and Signs The most common numbers we use are called the natural numbers and these are the simple 1, 2, 3, 4, etc. These whole numbers are also known as positive

integers. However, we also often use positive fractions, such as 51 . These two

groups together form the positive rational numbers. A rational number is in fact any number that can be expressed in the form a/b, where a and b represent integers. The natural numbers are positive integers, but sometimes we need to quote negative quantities (numbers less than zero), so in this case we use negative integers. The number zero is unique and does not fall in any category; in fact, it has a category of its own.

There are numbers (e.g. 2 ) which are not rational numbers because cannot be

represented by the quotient of two integers. These are called irrational or non-rational numbers. All the above-mentioned categories together form what are known as real numbers. These are distinguished from the so-called complex numbers, which will not be considered in this course.

1.1.1. Addition

The process of finding the total of two or more numbers is called addition. The resulting answer is called the sum. When the result is larger than nine, it is necessary to arrange the numbers in columns so that the last digit of each number is in the same column.

Example: 65 + 511

hundreds tens ones

6 5 + 5 5 1 ______________________

6 1 6 In the example above, we start first with the ones column. So 5 + 1 gives 6. Same with the tens column, but in this case the answer is 11. We cannot write down 11 so we take the ten out of the answer and convert it to a one in the hundreds column. The remaining 1 from the 11 is written down in the tens column. In this manner, in the hundreds column we have 5 plus 1 which was added. So the result in the hundreds column is 6, giving the final answer as 616.

1.1.2. Subtraction

The process of finding the difference between two numbers is known as subtraction. The number, which is being subtracted, is the subtrahend and the number from which the subtrahend is subtracted is the minuend. In order to simplify the process of subtraction, the numbers are arranged in columns, similar to the addition process (i.e. hundreds under hundreds, tens under tens, etc). Then starting from the right, the subtrahend is subtracted from the minuend. Example: 783 - 592 hundreds tens ones 7 8 3 - minuend 5 9 2 subtrahend _______________________ 1 9 1 Starting from the right, (3 – 2) gives 1. Next column, (8 – 9) is not possible so 8 becomes 18 by taking 1 from the hundreds column so that 7 becomes 6. (18 – 9) gives 9. Now 7 has become 6 so (6 – 5) gives 1.


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1.2. Methods of Multiplication and Division

1.2.1. Multiplication

In multiplication, we are doing addition but repetitively. So (5 x 3 = 5) is similar to (5 + 5 + 5) or (3 + 3 + 3 + 3 + 3). In this example, 5 and 3 are also called the factors and 15 is the product. The number to be multiplied (i.e. 5) is the multiplicand and the number of times the multiplicand is to be added to itself (i.e. 3) is the multiplier. The order in which numbers are multiplies does not change the product (i.e. 5 x 3 = 3 x 5). Simple multiplication involves only two numbers and one needs to know the numerical tables properly in order to work them out easily. Example: 3 x 5 = 15 9 x 3 = 18 7 x 6 = 42 Commutative Law of Multiplication: The product of two real numbers is the same no matter in what order they are multiplied. This means that (a x b) = (b x a). Associative Law of Multiplication: The product of three or more numbers is the same no matter in what manner they are grouped, hence a(b x c) = (a x b)c. When multiplying large numbers, it is important they are aligned vertically, similar to the addition and subtraction process. Examples: 573 x 432 x 21 32 573 + 864 + 11460 12960 12033 13824

1471 x 432 x 121 132 1471+ 864 + 29420 12960 147100 43200 177991 57024

1.2.2. Division

Division is the reverse of multiplication, finding out how many times a number is contained in another number. The number divided is called the dividend, the one dividing with is the divisor and the result is the quotient. In some problems, the quotient may include a remainder, which represents a portion of the dividend that cannot be divided by the divisor.

Division can also be represented with a fraction, e.g. 4343 =÷

To divide large quantities, the problem is broken down into a series of operations. Examples: Arithmetic Law of Precedence It is often confusing, when having different operations to do in the same equation, which operation goes first. In this case, it is convenient to use the BODMAS rule. BODMAS stands for Brackets Of Division, Multiplication, Addition, Subtraction. Hence, the first to be worked out are the brackets of the equation, starting from the division, then multiplication etc.

5146

20

20

24

26

16

18

4

05864

.

.

3333

10

9

10

9

10

9

01003

.

.

12

363


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Exercises Addition:

1. 123 + 4294 2. 2342 + 3939 3. 233.65 + 9483.12 4. 5830.766 + 13339.144 5. 45.34 + 232.5 + 11.89 6. 23.4 x 10

4 + 11.8 x 10

2

Subtraction:

1. 545 – 232 2. 9833 – 2334 3. 983.3 – 843.2 4. 2938 – 23345 5. 893.45 – 78.9 6. 7583.5 – 89921

Multiplication:

1. 54 x 3 2. 345 x 45 3. 234 x 23 4. 2498 x 345 5. 123 x 4592 6. 4592 x 129

Division:

1. 2464 / 4 2. 3486 / 2 3. 8323 / 5 4. 4782 / 9 5. 2922 / 18 6. 4234 / 21


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1.3. Fractions A fraction is a division of one number by another. For example, ½ means one divided by two. The number above the line is called the numerator and the number below the line is called the denominator. Fractions written in this form (e.g. 2/5, 3/8, ½, etc) are called vulgar fractions. Those written in decimal form (e.g. 0.5, 3.56, 0.333, etc.) are called decimal fractions. When the denominator of a fraction is numerically larger than the numerator (e.g. 4/5), the fraction is said to be a proper fraction. If however the numerator is larger than the denominator, the fraction is said to be improper. Improper fractions (e.g. 9/4) can be also expressed as what is known as a mixed number. In this case, 9/4 would be expressed as 2¼ where 2 is whole number and ¼ has to be a proper fraction.

1.3.1. Converting between improper fractions and mixed numbers

Converting from an improper fraction to a mixed number is done by simple division. The answer of the division process becomes the whole number of the mixed number while the remainder is the new numerator.

Example: Convert 722 into a mixed number.

3

1

21

227 Therefore the 722 can be written as 7

13 .

To convert from a mixed number to an improper fraction, the whole number is multiplied by the denominator and the result is added to the numerator. This answer will be the new numerator whilst the old denominator remains unchanged.

Example: 4

13

4

112

4

1434

13 =+

=+×

=)(

3

17

3

215

3

2353

25 =+

=+×

=)(

Lowest Terms A fraction is said to be in its lowest terms if its same value cannot be represented by

a fraction with smaller numbers. For example 53 cannot be represented with smaller

numbers. But 126 can also be written as 2

1 if we divide both numerator and

denominator by 6.

1.3.2. Multiplication of Fractions

Multiplication of fractions is relatively easy. All one has to do is multiply the numerator of one fraction with the numerator of the other fraction and the denominator of the first fraction with the denominator of the second one. These results will be the numerator and denominator of the answer.

Example: 35

8

75

42

7

4

5

2=

×

×=×

24

11

212

111

2

1

12

11=

×

×=×

1.3.3. Division of Fractions

Division of fractions is similar to multiplication. However the divisor (the fraction we are dividing with), needs to be inverted. The resulting fractions are then multiplied normally.

Example: 4

9

14

33

1

3

4

3

3

1

4

3=

×

×=×=÷

20

14

45

72

4

7

5

2

7

4

5

2=

×

×=×=÷

Note that this last answer is not in its lowest terms.

Therefore, we can re-write 2014 as 10

7 .


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1.3.4. Addition of Fractions

In order to add fractions, we need to determine the least common multiple (LCM). This means we need to find the smallest possible number, which is a common

multiple of the denominators. For example if we have 43 and 5

1 , the LCM would

be 20. Often in order to find the LCM, sometimes it is easier to simply multiply the denominators, as in the previous example. However, there are cases in which a smaller number being a multiple of both denominators can be found by intuition, making the problem easier to solve.

Example: 4

3

7

5+ , the least common multiple would be 28.

8

5

12

11+ , the least common multiple would be 24, not 96.

The next step after determining the LCM, is to place the LCM below the denominators of the fractions we wish to add.

So we using the second example, we get 24

8

5

12

11

+

Now we divide 24 by 12, which gives 2 and multiply it by the corresponding numerator i.e. 11, which gives us 22. Again we divide 24 by 8 which gives 3 and multiply it by its corresponding numerator i.e. 5, which gives 15.

Thus the new fraction will be 24

37

24

1522=

+

Examples: 4

1

4

23

4

2

1

4

3

2

1

4

3=

−=

−

=−

4

1

4

23

4

2

1

4

3

2

1

4

3=

−=

−

=−

1.3.5. Subtraction of Fractions

Subtraction of fractions is carried out similarly to addition of fractions. The LCM still needs to be determined. However, in the last step, we simply subtract the numerators instead of adding them.

Examples: 4

1

4

23

4

2

1

4

3

2

1

4

3=

−=

−

=−

6

1

6

32

6

2

1

3

1

2

1

3

1−=

−=

−

=−

12

13

12

2033

12

3

5

4

11

3

5

4

11=

−=

−

=−


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Exercises Convert from improper fractions to mixed numbers:

i) 3

4 ii) 3

7 iii) 2

9 iv) 4

11

Convert from mixed number to improper fractions:

i) 3

13 ii) 3

211 iii) 2

13 iv) 4

37

Multiplication:

i) 6

53

1 × ii) 3

45

2 × iii) 8

716

10 ×

iv) 4

36

153

11 ×× v) 1

39

153

2 ×× vi) 4

36

55

6 ××

Division:

i) 6

53

2 ÷ ii) 3

45

2 ÷ iii) 8

716

7 ÷

iv) 6

13

11 ÷ v) ( )1

39

153

2 ÷÷ vi) ( )4

36

55

6 ÷÷

Addition:

i) 6

53

1 + ii) 3

45

2 + iii) 8

716

10 +

iv) 2

36

73

11 ++ v) 1

39

15

2 ++ vi) 8

36

115

1 ++

Subtraction:

i) 6

13

5 − ii) 3

15

6 − iii) 8

116

15 −

iv) ( )4

36

153

11 −− v) ( )4

36

55

1 −− vi) ( )4

16

55

17 −−

Mixed: (Remember Arithmetic Law of Precedence)

i) ( )5

176

53

1 +× ii) 3

26

53

45

23 −+× iii) 7

598

716

10 +×

iv) ( )4

36

153

11 −× v) 1

39

153

2 ÷× vi) 3

114

36

55

6 ×−+


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1.4. Decimal Numbers Decimal numbers are another way to represent fractions. If we write down 0.1, this can be written in fraction form as 1/10. Similarly, 0.01 is equivalent to 1/100. Number of Decimal Places In many cases, the use of a large numbers of decimal places is unnecessary. For example a number like 7.3422411343 is too impractical to handle and more often than not, not required. Therefore, one may choose to what degree of accuracy the answer is required by specifying the number of decimal places to be used. If for the previous number, an accuracy of 3 decimal places is required, we only write down 7.342 and forget the rest. Attention! Had the number been 7.3425, the result to 3 decimal places would have been 7.343 because the number 5 is too large to ignore. Thus if the number behind the number of specified decimal places is either 5 or larger, we add 1 to the last digit of our answer. Example: Give the following numbers to 3 decimal places 5.34211 = 5.342 67.8755 = 67.876 34.5625 = 34.563 2.452345 = 2.452 Significant Figures When asked to write a number down to a certain number of significant figures, we are (in similar way to the number of decimal points), ignoring the last figures of a particular number to make it simpler while we lose only little accuracy. For example if we have 8,432,311, and asked to write it down to 2 significant figures, we take the first two integers and replace the rest with zeros, thus becoming 8,400,000.

Example: Give the following to 2 significant figures. 45,200 = 45,000 235,674 = 240,000 1,485 = 1,500 Similarly for decimal numbers: 0.004523 = 0.0045 0.0000345 = 0.000035 0.0783 = 0.078

1.4.1. Addition and Subtraction of Decimal Numbers

Addition and subtraction of decimal numbers is no different from that for whole numbers. However, in order to avoid confusion because of the decimal, one should place the numbers under each other with the decimal points beneath each other. This helps put the powers under each other i.e. tens with tens and hundreds with hundreds, etc… Example: 1345.65 – 7823.88 + 133.4 1455.31 1212.25 9279.19

4534 – 2342.56 + 133.4 33.22 4400.6 2375.78

236.56 – 8356.55 + 184.7 452.86 51.86 8809.41


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1.4.2. Multiplication of Decimal Numbers

The presence of the decimal point makes multiplication of decimal numbers more complicated than for whole numbers. Therefore, the decimal point is removed by shifting it until we have a whole number. The number of times the decimal point is shifted has to be recorded. This is done for both the multiplicand and the multiplier. Example: 234.56 23456 (2dp) 8462.768 8462768 (2dp) 0.736 736 (-3dp) 0.4664 4664 (-3dp) Thus if we have 17.45 x 11.3, this becomes 1745 x 113 (2 + 1 dp) Once the decimal point is removed, multiplication is done normally. However, since we have changed the original multiplicand and multiplier (by moving the decimal point), the answer needs to have its decimal point moved back. The number of times the decimal point is shifted is equal to the total number of times it was shifted before multiplication. Example: If we have 23.3 x 11.2, this becomes 233 x 112 (1 + 1 = 2dp)

233 x 112 = 26096, but the final answer needs to have its decimal point shifted by 2dp. Hence, final answer is 260.96.

If we have 132.5 x 87.67, this becomes 1325 x 8767 (1 + 2 = 3dp) 1325 x 8767 = 11616275, but the final answer needs to have its decimal point shifted by 3dp. Hence, final answer is 11616.275.

1.4.3. Division of Decimal Numbers

The problem in the case of division is the decimal point within the divisor. Again, the solution is to move the decimal point, but unlike the multiplication case, we move only the decimal point of the divisor. Then the number of decimal points shifted in the divisor, is shifted also for the dividend. In this manner, the operation is not modified and we do not need to move the decimal point of the final answer.

Example: 254.362 / 12.2 We cannot divide with 12.2 so this becomes 122 (1dp) 254.362 is also shifted by 1dp, becoming 2543.62

Then we simply divide 2543.62 by 122. The answer is immediately correct, without the need to shift decimal points.

So basically 254.362 / 12.2 = 2543.62 / 122

Exercises Calculate:

1. 23.76 x 1.5 5. 234.5 / 7.2 2. 56.87 x 3.7 6. 786.783 / 8.4 3. 46.36 x 45.89 7. 923.67 / 11.3 4. 123.78 x 1.83 8. 7362.677 / 20.5


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1.5. Factors and Multiples When we multiply 4 x 6, we get 24. In this case, 4 and 6 are factors of 24. But 24 has other factors apart from 4 and 6. For example, 2 and 12 can be multiplied to give 24 as well. Thus, 2 and 12 are also factors of 24 as are also 3 and 7. Since any number multiplied by 1 will remain the same, 1 is a factor of all numbers. In this case, it is called a trivial factor and usually not used since it has little importance. Multiples are the resulting numbers when the factors are multiplied. For example if we have 4 x 6, 4 and 6 are the factors and 24 is the multiple of both 4 and 6.

1.6. Powers and Roots When a number is multiplied by itself, it is raised to a certain power. For example, 5 x 5 can be written as 25 or as 5

2, where the 2 represents how many times the

number is multiplied by itself. So if we have 6 x 6 x 6, this can be written as 63.

Taking this example, the number 6 is called the base number, and the 3 is called the index. When written in this form, the number is expressed as an exponent. We can also say that if for example, we have 2 x 2 x 2 x 2, which can be written as 2

4, 2

4 is called the fourth power of the base 2.

Example: 3

2 is 3 x 3 = 9

56 is 5 x 5 x 5 x 5 x 5 x 5 = 15625

A negative exponent implies a fraction and indicates the inverse (or reciprocal) of the number.

Example: 8

1

2

1

2

1

2

1

2

12

3

3=××==

−

Therefore, 2

-3 is the reciprocal of 2

3.

Any number to power zero, is equal to one.

When a number has an index with no sign ( + or -), it is assumed that it is positive. Laws of Indices

1. nmnm

aaa+

=× 22 x 2

4 = 2

2+4 = 2

6 = 64

2. nmnm

aaa−

=÷ 34/3

2 = 3

4-2 = 3

2 = 9

3. ( ) mnnm aa = (22)3 = 2

2x3 = 2

6 = 64

4. 10=a

5. n mnm

aa = 2 424 22 =

6. nn

aa 1=

−

The root of a number is that value which when multiplied by itself a certain number of times, produces that number. If 4 x 4 gives 16, then 4 is the root of 16. We often

represent this with the symbol ( ), so than 16 =4. In this case, 4 was multiplied

by itself only once. We therefore call 4 the square root of 16 since 42 gives 16. If

we had 4 x 4 x 4, we get 64, and since 43 gives 64, 4 is called the cube root of 64

and is represented with 3 64 .

Note that for square root, we don’t need to write down the 2 next to the root sign. It is implied that it is the square root of the number. But for cube root and higher values, the index needs to be clearly written down.

Examples: 9273 = since 93 = 27

56254 = since 54 = 625

37296 = since 36 = 729


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Exercises Work out the following:

1. 52 33 ×

2. 25 55 ÷

3. ( )327

4. 476

5. 14−

6. ( )

2

23

3

333 ××−

7. 313 2

1

2

1

2

1−+

−−

8. 5247 66 ×

Standard Form Often, in both engineering and mathematics, it is convenient to represent numbers in standard form as shown below, to make calculations easier, and reduce the possibility of mistakes. 1,000,000 = 1 x 10

6

100,000 = 1 x 105

10,000 = 1 x 104

1,000 = 1 x 103

100 = 1 x 102

10 = 1 x 101

0 = 0 1/10 = 0.1 = 1 x 10

-1

1/100 = 0.01 = 1 x 10-2

1/1,000 = 0.001 = 1 x 10

-3

1/10,000 = 0.0001 = 1 x 10-4


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1.7. Measures and Conversion Factors Length 1in = 2.54cm 1m = 39.37in or 3.281ft 1ft = 0.3048m 12in = 1ft 3ft = 1yd 1yd = 0.9144m 1km = 0.621miles 1mile = 1.61km or 5,280ft Area 1m

2 = 10.76ft

2

1m2 = 10,000cm

2

1ft2 = 0.0929m

2 or 144in

2

1in2 = 6.452cm

2

Volume 1m

3 = 1,000,000cm

3

1ft3 = 1728in

3 = 0.0283m

3

1lt = 1000cm3 = 1.0576qt

1ft3 = 7.481gal

1gal = 8pints 1gal = 4.546ltrs (3.785ltrs if American gal) Mass 1amu = 1.66 x 10

-27kg

1000kg = 1 metric tonne = 0.984tons 1000g = 1kg 1slug = 14.59kg Force and Weight 1N = 0.2248lb 1lb = 4.448N 1lb = 16oz Velocity 1mph = 1.47ft/sec

1m/s = 3.281ft/sec 1knot = 1 nautical mile per hour 1knot = 1.688ft/sec 1knot = 1.151mph 1knot = 1.852km/hr 1mph = 1.61km/hr Energy 1J = 0.738ft.lb 1cal = 4.186J 1Btu = 252cal Time 1year = 365days 1day = 24hr = 1,440min Power 1HP = 550ft.lb/sec 1HP = 746W 1W = 1J/sec 1W = 0.738ft.lb/sec 1Btu/hr = 0.293W Pressure 1atm = 76.0cmHg 1atm = 760mmHg 1atm = 29.92inHg 1atm = 14.7lb/in

2

1Pa = 0.000145lb/in2

1bar = 14.5lb/in2

1bar = 100,000Pa Fundamental Constants g = 32lb/slug or 9.81N/kg Other Useful Data 1 litre water = 1kg 1 pint water = 1lb


Pg 12

Exercises

1. How many centimetres are there in 3.5ft? 2. How many yards are 1.5km?

3. If a table is 3ft wide and 1.6m long, how many square inches is its area?

4. How many pints are there in 5 cubic feet of water?

5. How many litres are there in 4.5 US gallons?

6. If an aircraft travels at 100knots, what is its speed in km/hr?

7. How much will 4.3 cubic feet of water weigh?

8. If a machine produces 670 ft.lb/s, how much is this in horsepower?

9. How many mm of Mercury are 1.7atm?

10. How many slugs are 25kg?


Pg 13

1.8. Ratio and Proportion A ratio is a comparison between two similar quantities. For example if we are comparing a model aircraft to the real thing, we say that their dimensions have a ratio of 1:25. This means that every centimetre of the model represents 25cm on the real aircraft. It is important when using ratios that the numbers quoted have all the same units (e.g. cm to cm, km to km, etc). Often ratios are used in gearboxes. In this case can represent the ratio of teeth between gears, the speed ratio of the gearbox, or the torque ratio. Example: A turboprop gas turbine is attached to the propeller via a gearbox whose speed ratio is 1:5 so that the propeller rotates slower than the gas turbine. If the turbine rotates at 20,000rpm, what will be the propeller speed?

The ratios can be written in this form: ( 1 : 5 ) = ( prop rpm : 20,000) The first and fourth numbers (i.e. 1 and 20,000) are called the extremes, while the second and third numbers are called the means.

In proportion, the product of means is equal to the product of extremes.

Therefore, (1 x 20,000) = (5 x prop rpm). So the propeller rpm can be calculated by using {(1 x 20,000) / 5} = 4,000 rpm.

1.8.1. Direct Proportion

Direct proportion means that two quantities are directly related in a manner that if one increases, the other quantity will increase as well. For example if a car is travelling at constant speed, the time and distance covered are directly proportional i.e. if the distance is large, the time taken will also be large, and vice-versa. Example: A long-range bomber consumes 1 tonne of fuel every 50 nautical miles. If before the mission it uploads 4 tonnes of fuel, what is the maximum distance it can travel?

This is a clear case of direct proportion since the more fuel is carried, the farther the aircraft will travel. Therefore by cross-multiplication: 1 tonne of fuel = 50 nautical miles 4 tonnes of fuel = 200 nautical miles

1.8.2. Inverse Proportion

In inverse proportion, two quantities are related in a manner that if one increases, the other has to decrease. Using the car example again, if a car has to travel a fixed distance, the speed is inversely proportional to the time taken i.e. the faster the speed, the less time taken to cover the distance. Example: A car travels at 10mph, for 100 miles. How long will it take the car

to cover the same distance if travelling at 20mph? 10 mph for 100 miles = 10 hours 20 mph for 100 miles = 5 hours Note: There is no mathematical method of defining whether a problem is direct or inverse proportion. One has to rely on understanding the question properly and reasoning on it.

1.8.3. Proportional Parts

Say that for a particular flight, the aircraft requires 24 tonnes of Jet-A fuel. However to distribute the weight properly, the captain needs to fill the three available fuel tanks with a weight ratio of 3 : 4 : 5. How many tonnes will have to be loaded in each tank? The first step is to add the ratios i.e. 3 + 4 + 5 = 12 parts. The total tonnage is then divided by this sum, so 24/12 = 2. This result is then multiplied to each tank’s corresponding ratio number (i.e. 3 x 2 = 6, 4 x 2 = 8 and 5 x 2 = 10). Thus, the resulting tonnage will be divided into 6 tonnes, 8 tonnes and 10 tonnes.

Gas Turbine 20,000rpm


Pg 14

1.8.4. Constant of Proportionality

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

the fuel flow (m). It is also inversely proportional to the square of the bleed air (b) and the bearing friction (f).

Therefore, this can be written as bf

mw α

The proportional sign can be removed by inserting the constant of

proportionality:

=

bfmkw .

Exercises

a. A model aircraft has a scale of 1:17. If the wingspan on the model is 25cm, what is the wingspan on the real aircraft in metres?

b. A gearbox has a speed ratio of 1:5. If the high-speed side is rotating at

25,000rpm, how fast is the slower side rotating?

c. What is the ratio between the length and the width of a wing if it is 25m long and 1.25m wide?

d. During a C-check, an engine check takes 7 hours if performed by 4

maintenance personnel. How long will it take if the number of personnel is increased to 7?

e. Five men take 25 hours to produce assemble a certain amount of

engines. How much time would it take if there were eight men on the job?

f. Three components on an aircraft require an oil change. Their oil

requirement ratio is 3:5:7 and the first component needs 5 litres. How much oil is required in total?

g. If the fuel consumption of an engine is proportional to the thrust and

inversely proportional to the altitude, what is the constant of proportionality if the fuel consumption is 15kg/s when the thrust is 15,000N and altitude is 20,000 feet?

h. An electrical resistance of a wire (R) varies directly with the length (L)

and inversely with the square of the radius (r). How can this be represented?


Pg 15

1.9. Averages and Percentages

1.9.1. Mean

This is the most common average used. To find the mean of a set of values, the sum of the values has to be divided by the number of values. Example: Find the mean of {3, 5, 14, 4, 13, 25} The sum of these values is [3+5+14+4+13+27] = 66 The number of values we have is 6. Therefore the mean is 66/6 = 11.

1.9.2. Median

To find the median of a set of values, first the numbers must be placed in descending (or ascending) order. The median will be the number in the middle. Example: Find the median of {5, 7, 12, 4, 16, 21, 2, 15, 1} In ascending order, these will become [1, 2, 4, 5, 7, 12, 15, 16, 21] The median is the middle number i.e. 7.

1.9.3. Mode

The mode (or modal value) is the value, which appears the most within a set of values. Example: Find the mode of {13, 20, 23, 39, 23, 40, 23, 14, 12, 56, 23, 29} Putting them in order {12, 13, 14, 20, 23, 23, 23, 23, 29, 39, 40, 56} It is therefore easy to find that the most common number is 23. Thus, the mode is 23.

1.9.4. Percentages

Sometimes when expressing fractions, it is easier to express them using 100 as denominator. For example ¼ can also be expressed as 25/100. When fractions have 100 as denominator, they are called percentages.

Example: 107 is equivalent to 100

70 . This can be expressed as 70%.

%. 45100

45450 ==

5250100

552552 .

.%. ==

Note: To convert a vulgar fraction or decimal fraction into a percentage, multiply

by 100. To convert a percentage into a fraction, divide it by 100.


Pg 16

Exercises a. Find the mean, mode and median of:

i) {3, 4, 3, 7, 2, 9, 12, 6, 23, 45, 2, 9} ii) {45, 23, 76, 23, 77, 90, 72, 16} iii) {34.5, 11.5, 56.2, 8.1, 9.65}

b. Convert to percentage: c. Convert to fractions or decimal numbers:

i) 4/5 i) 50% ii) 3/7 ii) 75% iii) 4/9 iii) 35% iv) 13/7.5 iv) 23.5% v) 7/5 v) 54.87%

d. An aircraft travels 150km in 2.7 hours. What is its average speed? e. A fighter jet consumes 1.2 tonnes of fuel in 2.7 hours. What is the average fuel

consumption per hour? e. Find 22% of 160. f. How much oil is required if 20ltrs satisfy only 35% of the required volume. g. A cargo aircraft is carrying 165 cases. 20% of these weigh 35kg, 30% weigh

45kg, and 25% weigh 50kg. The rest weigh 55kg. How much weight is the aircraft carrying?

h. How long far would have an aircraft travelled in 3 hours, if for 20% of the time its

average speed was 130km/hr, 25% at 100km/hr and the rest at 150km/hr?


Pg 17

r

1.10. Areas and Volumes

1.10.1. Areas

When we measure the area of something, we are measuring how many square units it contains. This means that if we have an area of 5.5m

2, then this space

would contain 5½ squares, each being 1m in width and 1m in breath. 1 square metre 1m

2 1 square inch 1in

2

1 square centimetre 1cm2 1 square foot 1ft

2

1 square millimetre 1mm2 1 square yard 1yd

2

Rectangle: Area = l x b Perimeter = 2l x 2b Parallelogram: Area = b x h Perimeter = Sum of all 4 sides.

Triangle: Area = ½ b x h or Area = sqrt s (s-a)(s-b)(s-c) where s = (a+b+c) 2 Perimeter = a + b + c Circle: Area = πr

2

Perimeter = 2πr

1.10.2. Volumes

1 cubic metre 1m

3 1 cubic inch 1in

3

1 cubic centimetre 1cm3 1 cubic foot 1ft

3

1 cubic millimetre 1mm3 1 cubic yard 1yd

3

Cylinder: Volume: πr

2h

Surface Area: = 2πr (h+r)

b

l

h

b

h

b

a b

c

h

r


Pg 18

Cone: Volume: 1/3 πr

2h

Surface Area: πrl Sphere: Volume: 4/3 πr

3

Surface Area: 4πr

2

Solids having uniform cross-section e.g. cube, oblong, etc: Volume = Cross-sectional area x length of shape Surface area = sum of all areas enclosing the shape.

Exercises

i) The passenger section of the aircraft is 4m wide and 30m long. A centre aisle, 1m wide is installed and the remaining area is to be carpeted. How many square metres of carpet are required?

ii) Calculate the surface area of the cockpit having the shape of a cone,

5m side length (i.e. not the vertical height) and base diameter 3m.

iii) What volume of material is required to create a hollow sphere which has an outer radius of 2m and internal radius if 1m.

iv) Calculate the volume of an aircraft having a radius of 3m, 25m long.

The front and rear ends are shaped as a cone 3m long.

v) A hollow section beam is to be used as part of the wing structure. Its outer dimensions are 7cm x 7cm and each side is to be 1cm thick. What volume of aluminium is required if the beam is to be 6m long?

l h

r

r


Pg 19

2. ALGEBRA

2.1. Introduction In algebra, symbols or letters are used to represent a variable number. So the letter (a) can represent any number. Coefficients are used in front of the letters and these show by how much the variable number is to be multiplied. For example (7a) means any value given to (a) is to be multiplied by 7. We already mentioned factors in arithmetic. The same principles apply to algebraic terms. If we have the expression (ab), then a and b are the factors. Again, the factors 1 and ab are trivial factors and are not used. Examples: Find the factors of: i) 8 = 2 and 4 ii) xy = x and y iii) 12 = 2, 3, 4, 6

2.1.1. Addition and Subtraction

Commutative Law of Addition: The sum of numbers is always the same, regardless the order in which they are written i.e. a + b + c = a + c + b. Associative Law of Addition: The sum of numbers is always the same, regardless in which manner they are grouped i.e. (a + b) + c = a + (b + c). In order to be able to add or subtract algebraic expressions, they need to have the same terms i.e. have like terms. For example, (7a + 3b) cannot be simplified further because we have unlike terms. However, (7a + 2a) can be worked out and gives 9a. Examples: 5a + 3c -2a = 3a + 3c = 3(a + c) 11b – 3c = 11b – 3c 8z + 5a – z – 2a = 7z + 3a

2.1.2. Multiplication and Division

In order to multiply two simple algebraic terms, e.g. (a x b), we only need to write them down as (ab) and similarly if we have an algebraic term and an integer (e.g. 2 x a = 2a). Examples: 5 x a = 5a b

-1 x b

2 = b

2a x b = 2ab sb x bytu = b2styu

2a x a = 2a2

5ab x 2t = 10abt Note: The order in which the terms are written down is not important but by convention, they are usually written down in alphabetical order. The same rules apply for division of algebraic terms. Examples: x

2/2x = x/2

3x4/2x

2 = 3x

2/2

2.1.3. Brackets

When a term is written in the form 5(a + b), the brackets can be opened and each term inside the brackets has to be multiplied by the outside term. Example: 7(a + c) = 7a + 7c 5a(a + c) = 5a

2 + 5ac

Suppose we are required to multiply (1 + a) and (1 + b). The first step is to multiply (1 + a) by the first character of the second bracket (i.e. 1). Then, (1 + a) has to be multiplied by the second character (i.e. b). Hence (1 + a).b gives (b + ab). The two results are then summed together to obtain (1+a) + (b + ab). The final answer is thus (1 + a + b + ab) Example: (i) (3a + c) (a + c) = (3a

2 + ac) + (3ac + c

2)

= 3a2 + 4ac + c

2

(ii) (2a – b) (a + b) = (2a

2 – ab) + (2ab – b

2)

= 2a2 + ab – b

2


Pg 20

Exercises Find: i) 3x + 5x – 7b ii) 12b – 3a + 4x -13b iii) -5c – 23a + c iv) 7m – n + 5m + 11n v) 3m(n + z) vi) 2a(a – b

2)

vii) 6d(2a + c)(c) viii) 2c

2(2a – c)(5a)

ix) (3a – b)(-a + 2b) x) (-2a

2 + 3)(a – b)

xi) (4a – 3ab)(a – c)(2a) xii) (4a

2 + 3c – a)(2b + c)

xiii) (x + 3)(x – 2) xiv) (x + 5)(x – 9) xv) (2x + 3)(x – 5) xvi) (x – 3)(3x + 3)


Pg 21

2.2. Linear Equations Linear equations with one unknown contain just one letter to the power one, e.g.

752 =−x .

When two different algebraic expressions are equal, we have an equation.

For example: 2

1

3

−=

xx is an equation.

The solution of an equation is the number, or set of numbers, which the letters represent in that equation. A problem can be solved if we can use algebra to relate the given information to form an equation and then solve it.

2.2.1. Transposition of Formulae

Often it is required to rearrange the formula to change the subject of the formula. Whatever we do to the left-hand side of a formula or equation, we must do the same to the other side, or when we take any term over the equals sign, we change sign. Examples: Make ‘c’ subject of the formula.

bca =+ becomes abc −=

ba

c = becomes abc =

ba

cx=

−

2

3

becomes 2

3

−=

a

bxc or 23abxc =

2.3. Simultaneous Equations In previous sections, we solved algebraic linear equations. In these cases, we only had one unknown and solutions were quite straightforward. There cases however were linear equations have more than one unknown. In such a case, one needs another equation in order to obtain a solution.

Example: 82 =+ yx and 5=+ yx

These two linear equations form a set of simultaneous equations. We have to find a value of x and y which satisfies both equations. Solving simultaneous equations by elimination Simultaneous can be solved by elimination, meaning eliminating one of the unknowns. One easy way to do this is to subtract one equation from the other.

Example: ( ) ( ) 582 −=+−+ yxyx

3=x

Substituting 3 for x, in any equation gives y = 2.

Sometimes one letter can be eliminated by adding the equations.

Example: ( ) 6284 =−=+ yxyx

Subtracting one from the other gives 146 =x

Hence 3

123

7 ==x

Substituting this value in any equation, we get:

348

328 −==+ yy


Pg 22

Solving simultaneous equations by substitution Another common method in solving simultaneous equations is by substituting one of the unknowns. Example: Solve these simultaneous equations.

743 =+ yx and 42 =+ yx

Taking the 2nd

equation, one can re-write it as xy 24 −=

Substituting this for y in the first equation we get:

( ) 72443 =−+ xx

78163 =−+ xx

16783 −=− xx

95 −=− x

5

9=x

With this value of x in any of the equations, we get a value for y. Using the first equation, we get:

( ) 745

93 =+ y

745

27 =+ y

52774 −=y

5

84 =y

52=y


Pg 23

2.4. Quadratic Equations Quadratic equations are similar to linear ones, but the highest power found in the

equation is 2, (e.g. 0535 2=++ xx )

There are various methods how to solve such equations. These include by factorisation, graphically and by completing the square. However, the most straightforward method is by using the formula.

2.4.1. Solving by using the formula

Any quadratic equation can be written in the form 02=++ cbxax , where a, b and c

represent numbers.

When in this form, the solution can be found by: a

acbbx

2

42−±−

=

Examples: 0372=−− xx

x = 7.405 or -0.405

7552=++ xx becomes 0252

=−+ xx

You probably noticed that most times, the square root part of the formula is not easy to work out, and sometimes may even be negative and so cannot be solved. Hence it is sometimes convenient to use the factorisation method.

2.4.2. Solving by factorisation

A quadratic equation can usually be expressed as a product of its factors. For

example; 0232=+− xx can also be expressed as 021 =−− ))(( xx .

Therefore it stands to reason than if 021 =−− ))(( xx , then x is either = 1 or 2.

The values of x that satisfy the equation are called the roots of the equation (i.e. 1 and 2). In general, a quadratic equation has two roots, but sometimes these may be identical.

Example: 09124 2=++ xx can be expressed as ( ) 032

2=+x

So there is only one root, i.e. 23− , called a repeated root.

2.4.3. Factorising quadratic equations

In order to factorise quadratic equations, we need to be aware of the relationship between what is inside the brackets and the resulting quadratic.

a. The product of the number terms in the two brackets gives the number term in the expansion.

b. Collecting the number terms gives the coefficient of x. c. A positive sign throughout the quadratic comes from positive signs in the

both brackets. d. A positive number term and a negative coefficient of x come from negative

signs in both brackets. e. A negative number term in the quadratic comes from a positive sign in one

bracket and a negative sign in the other bracket.

Example: 652++ xx

The x term in each bracket is x , as 2x can only be x x x .

The sign in each bracket is + so ( )( )++ xx

The number terms in the brackets can be 6 and 1 or 2 and 3. The middle term in the quadratic shows that the sum of the number is 5.

Therefore the factors are: ( )( )32 ++ xx

( ) ( ) ( )( )

2

617

2

12497

12

314772

±=

+±=

−−±−−=

.x

( ) ( ) ( )( )

2

335

2

8255

12

214552

±−=

+±−=

−−±−=

.x


Pg 24

2.5. Division of Algebraic Fractions Given than (a + b) is a factor of (a

3 + b

3), how can we find the remaining factors?

We have to be able to divide them first.

The first step is to take 33 ba + and divide it by a .

This gives 2a . This result is then multiplied by the

divisor, i.e. ba + , giving baa 23+ . This is

then subtracted from 33 ba + , giving 32 bba +− .

The same process is repeated again with

the remainder. Therefore, 32 bba +− is divided

by a , giving ab− . Again, ab− is multiplied by

ba + , giving 22 abba −− . Subtracting this from 32 bba +− , gives us the remainder, which is

32 bab + .

The process is repeated again until we get a remainder of zero.

Examples:

32

23

2

33

bba

baa

a

baba

+−

+

++

32

22

32

23

2

33

bab

abba

bba

baa

aba

baba

++

−−

+−

+

−

++

0

32

32

22

32

23

22

33

bab

bab

abba

bba

baa

baba

baba

++

++

−−

+−

+

+−

++

0

42

42

224

22

4422

bba

bba

baa

ba

baba

−+

−+

−

+

−−

0

4

4

123

13

41134

2

2

−

−

−

+

−−−

x

x

xx

x

xxx

4

24

64

36

23

67612

2

2

+

+

+

+

+++

x

x

xx

x

xxx


Pg 25

Exercises Solve these linear equations:

1. 953 =+x 2. 712 =−x

3. 325 −=+ xx 4. xx 953 =+

5. 142 +=+ xxx 6. 5956 −=+ xx

7. xx 4512 −=+ 8. xxx 395 +=+

9. 5931 −=+ xxx 10. 459 =+x

Solve these simultaneous equations:

1. 93 =+ yx yx =−12

2. 325 −=+ xyx 953 =+ yx

3. 142 +=+ yyx 956 =+ yx

4. xyx 45 =+ 395 +=+ yyx

5. 931 =+ xy 459 =+ yx

Find the values of x for which:

1. ( )( ) 031 =−− xx 2. ( )( ) 013 =+− xx

3. ( ) 073 =−xx 4. ( )( ) 0223 =+− xx

5. ( )( ) 03212 =+− xx 6. ( )( ) 0331 =−− xx

Solve these quadratic equations using the formula method:

1. 0297 2=++ xx 2. 0562

=++ xx

3. 0232 2=−+− xx 4. 8987 2

=++ xx

Factorise the following quadratic expressions:

1. 63 +x 6. 1032−− xx

2. ba 42 + 7. 762−+ xx

3. pq 188 − 8. 322−− xx

4. xx 32+ 9. 642 2

−− xx

5. 1522−+ xx 10. 5163 2

++ xx

Solve these quadratic equations using the factorisation method:

1. 0246 2=−+ xx

2. 4544 2=+− xx

3. 0310 2=−− xx

4. 8422 2=−− xx

5. 3116 2=+ xx

Work out:

1. ( ) ( )4363 2+÷++ xxx

2. ( ) ( )72278 2+÷++ xxx

3. ( ) ( )13129 2+÷++ xxx


Pg 26

2.6. Logarithms Any positive number can be expressed as a power of 10. Thus as 1000 can be expressed as 10

3, 82 can be written as 10

1.9138. These powers of 10 are called

logarithms to the base 10. This means that the logarithm to base 10 of 1000 is 3. This can be written as log101000 = 3. Manipulation of numbers, expressions and formulae, which are in index form, may be simplified by using logarithms. Another use for logarithms is to be able to reduce sometimes the more difficult arithmetic operations of multiplication and division to those of addition and subtraction. There are a number of laws that govern logarithms:

Law 1: If cba = , then ac blog=

For example, if we are asked to find x when 750 = 10x, it is convenient to use this

laws so that 75010log=x , so that x gives 2.8751 correct to 2 SF.

Law 2: NMMN aaa logloglog +=

What this law enables us to do is to convert the multiplication of numbers in index form into that of addition.

Law 3: NMN

Maaa logloglog −=

This law allows conversion from division of a number in index form into that of subtraction. These laws are helpful when doing transposition of formulae.

Law 4: ( ) MnM an

a loglog =

Law 5: b

MM

a

ab

log

loglog =

This law enables us to change the base of a logarithm. This is useful when we have to deal with logarithms that have a base that is not found on the calculator.

Example: An equation relates the final velocity v of a machine with the machines variables,

w , p and z .

This is given by

=pz

w

v 20 . Transpose the formula and find the numerical value of

w when v = 15, p = 1.24 and z = 34.65.

One can use

=pz

w

v 201010 loglog .

Thus 201010 loglog

=

pz

wv .

pz

wvor

pz

wv =

=

301031301031 10

10.

log.log

so ( )

301031

10

.

log vpzw =

( )( )( ) ( )( )( )

301031

1760916534241

301031

156534241 10

.

...

.

log..==w

= 38.84


Pg 27

2.7. Number Systems

2.7.1. The Binary System

Up till now we have been using the decimal system i.e. numbers from 0 to 9 and its is called the decimal system because we have a total of 10 numbers. Using this system any number can be represented as follows: 7892 = (7 x 10

3) + (8 x 10

2) + (9 x 10

1) + (2 x 10

0)

This arrangement consists of number from 0 to 9, multiplied by the base (10) which has a particular power (3, 2, 1 and 0). This is called the denary system because it is based on the number 10. In the binary system, instead of 10, we use 2 as our base number. Therefore, all numbers are represented in powers of 2. Example: 4310 = 2

5 + 2

3 + 2

1 + 2

0

_____________________________________________________________ Binary2 2

7 2

6 2

5 2

4 2

3 2

2 2

1 2

0

Denary10 128 64 32 16 8 4 2 1 Converting Denary to Binary To convert from denary to binary, we repeatedly divide by 2 and note the remainder at each stage. Example: Converting 25 to binary: 25/2 = 12 remainder 1 Least significant digit 12/2 = 6 remainder 0 6/2 = 3 remainder 0 3/2 = 1 remainder 1 1/2 = 0 remainder 1 Most significant digit Thus, the binary equivalent of 2510 is 110012. The order, in which the binary number is written, is from the MSD (Most significant digit) to LSD (Least significant digit).

Converting from Binary to Denary To convert to denary, the numbers are laid out in successive powers. Example: Convert 11012 into denary: 1101 = (1 x 2

3) + (1 x 2

2) + (0 x 2

1) + (1 x 2

0)

= (1 x 8) + (1 x 4) + (0 x 2) + (1 x 1) = 8 + 4 + 0 + 1 = 1310

2.7.2. The Hexadecimal System

In a hexadecimal system, the base number is 16. However, since in the decimal numbering system, we have only 10 numbers (0-9). We make up for this by introducing capital letters A, B, C, D, E and F which represent 10, 11, 12, 13, 14 and 15 respectively.

Denary Binary Hexadecimal 0 0000 0 1 0001 1 2 0010 2 3 0011 3 4 0100 4 5 0101 5 6 0110 6 7 0111 7 8 1000 8 9 1001 9 10 1010 A 11 1011 B 12 1100 C 13 1101 D 14 15

1110 1111

E F


Pg 28

Converting denary to hexadecimal The process is similar to that of the binary system. We divide the denary number by 16 and use the remainder. Example: Convert 513610 to Hexadecimal: 5136/16 = 321 remainder 0 LSD 321/16 = 20 remainder 1 20/16 = 1 remainder 4 1/16 = 0 remainder 1 MSD So the hexadecimal equivalent of 513610 is 141016. Similarly to convert 9410 to hexadecimal: 94/16 = 5 remainder 14 (= E16) 5/16 = 0 remainder 5 So the hexadecimal equivalent 9410 is 5E16. Converting hexadecimal to denary Again in a similar manner to the binary system: Example: Convert to denary system BA45: BA4 = (B x 16

2) + (A x 16

1) + (4 x 16

0)

= (11 x 256) + (10 x 16) + (4 X 1) = 2980 Thus the denary equivalent to BA416 is 298010.

Exercises Logarithms Evaluate: a. log464 b. log93 c. log12525 d. log12111 Express in terms of log a, log b and log c. a. log ab b. log a/b c. log a

2b d. log a

2/b

Solve the equations:

a. 63 =x b. 522

=x c. 24 12

=+x d. ( )( ) 1055 1

=−xx

Number Systems Convert first to binary and then to hexadecimal: i) 45 ii) 7 iii) 56 iv) 89 v) 120 Convert to the denary system: i) 10110 ii) 0110010 iii) 0100101 iv) EA v) 8A vi) 2C


Pg 29

3. GEOMETRY

3.1. Graphs

3.1.1. Graphical Representation

In previous sections, equations were solved analytically. It is also possible to solve these equations graphically. Sometimes transposition and manipulation of the equation is required. It is important that the equality sign must always be present. In order to plot a graph, two axes are required which intersect at the point zero called the origin. It is important that a suitable scale for these axes is chosen and it need not be the same for both axes. To plot points on the graph, we use co-ordinates. For example, (2,4) represents 2 units on the x-axis and 4 units on the y-axis. The x-ordinate is always quoted first. The x-ordinate is an independent variable and is plotted on the horizontal axis. The y-ordinate is a dependent variable and is plotted on the vertical axis.

3.1.2. Graphs of Linear equations

Example: Plot y = 2x – 4 x -2 -1 0 1 2 3 2x -4 -2 0 2 4 6 -4 -4 -4 -4 -4 -4 -4 y -8 -6 -4 -2 0 2 So when x = -2, y = -8, so co-ordinates will be (-2, -8) These types of equations, where the highest power of x is 1, are known as equation of the first degree or a linear equation. These produce always straight-line graphs.

-10

-8

-6

-4

-2

0

2

4

-3 -2 -1 0 1 2 3 4

x-axis

y-a

xis

Every linear equation can be written in the standard form of cmxy += . So for

42 −= xy , m = 2 and c = -4. The ‘m’ part represents the gradient or slope of the

line. The ‘c’ part represents the y-intercept, i.e. at which point the line crosses the y-axis.


Pg 30

Equation of a straight line It was already mentioned that a linear equation can be written in the form of

cmxy += . It was also shown how a graph can be plotted if the equation of the line

is known. However, it is also important to know how to determine the equation of the line starting from two co-ordinates on the graph. The example below shows exactly this.

The slope m can be found by using the formula: In this case therefore,

Example: Find the equation of the line having coordinates (5,4) and (3,2). Hence the gradient or slope of this line is 1

The y-intercept is found by putting one of the co-ordinates into the general equation and using the value of the gradient found, i.e. 1

In this case, we can use ( ) c+= 154 . This gives c = -1

Therefore the final general equation is 11 −= xy

It is also possible to find the values of the slope and the y-intercept by using simultaneous equations. This is done by putting the co-ordinates into the general

equation. For the previous example, we would get cm += 54 and cm += 32 .

Solving these simultaneously, we get the values of m and c for the general equation of the line. Example: Find the law of the straight line illustrated below.

The y-intercept is easy to find since the graph crosses the y-axis at y = -4. Thus, c = -4. The gradient can be found by taking any two points and using the equation shown earlier. Say we take (2,6) and (5,21) as co-ordinates.

Hence the straight-line law is 45 −= xy

12

12

xx

yy

x

ym

−

−=

∆

∆=

012

12

−

−=

−

−=

∆

∆=

D

CB

xx

yy

x

ym

12

2

35

24

12

12 ==−

−=

−

−=

∆

∆=

xx

yy

x

ym

53

15

25

621

12

12 ==−

−=

−

−=

∆

∆=

xx

yy

x

ym


Pg 31

Alternatively, using simultaneous equations and using the same two co-ordinates:

cm += 26 and cm += 521

Solving these simultaneously, we get m = 5 and c = -4.

Hence the straight-line law would be 45 −= xy as before.

-10

-5

0

5

10

15

20

25

30

35

40

45

0 2 4 6 8 10

x - axis

y -

ax

is

Note: Gradients or slopes are not always positive. It is possible to have negative

gradients so m is a negative number, e.g. 25 +−= xy

3.1.3. Graphs of Quadratic equations

As already mentioned, quadratic equations are those, whose highest power is 2,

e.g. yxx =++ 252 . There are various methods that can be used to solve these

equations and the formula and factorization method were already mentioned in the algebra section. Quadratic equations can also be solved graphically.

When a quadratic function of the form cbxax ++2 is plotted against x, the resulting

curve is known as a parabola and the sign of the coefficient of a, will determine which way the curve points. To plot these curves, a table of values needs to be set up in terms of the values of the independent and dependent variables.

Example: Find x if 232+−= xxy .

x 0 1 2 3 4 x

2 0 1 4 9 16

-3x 0 -3 -6 -9 -12 2 2 2 2 2 2 y 2 0 0 2 6

-5

0

5

10

15

20

25

0 1 2 3 4 5 6 7

x - axis

y -

ax

is


Pg 32

The points on the curve where the it crosses the x-axis are x = 1 and x = 2. These

are the points on the curve for which y = 0 or 0232=+− xx . Therefore x = 1 and x

= 2 are the solutions of the quadratic equation.

From this graph, it is also possible to solve any equation of the type kxx =− 32 ,

where k is a constant. If for example, we wish to solve 0132=+− xx , then

comparing this equation with the one plotted, one just needs to add 1 to both sides

to acquire the equation 1232=+−= xxy . So instead of taking the points crossing

the line y = 0 ( x – axis), we take points on the line y = 1. At these points, x = 0.4 or 2.6 which are the solutions to the equation

0132=+− xx .

3.1.4. Graphical solution of Simultaneous Equations

It is possible to solve simultaneous equations graphically. This is done by plotting both equations on the same graph. The solution lies where the plots intersect.

Example: Solve 75 −= xy and 532+−= xxy

-15

-10

-5

0

5

10

15

20

25

-2 -1 0 1 2 3 4 5 6 7

x - axis

y -

axis

A

B

The graph clearly shows 2 points where the functions intersect. These are the solutions to the simultaneous equations. Thus x = 2, y = 3 and x = 6 and y = 23 are solutions.

Exercises Plot the graph of: a) y = 4x – 3 b) 2y = 6x + 8 c) 3x = 6y – 9 Determine the equation of the line passing through these co-ordinates: (try using both methods) a) (5,2) and (3,9) b) (6,3) and (-5,0) c) (-4,3) and (-9,5) d) (-5, -2) and (-1,3) e) (11,5) and (7,-3) f) (-5,-3) and (-11,2) Plot the graph and solve for: a) y = 3x

2 + 2x – 1 b) y = 6x

2 -3x +5

Solve these simultaneous equations graphically (plot for 0 < x < 5): a) y = 3x - 7 and y = 2x

2 – 5x - 1

b) y = 5x – 10 and y = 3x2 – 10x + 5


Pg 33

3.2. Trigonometry

3.2.1. Pythagoras Theorem

In a right angled triangle, the area of the square on the hypotenuse is equal to the sum of the area of the squares on the other two sides. Hence for the triangle shown, c

2 = b

2 + a

2.

Example: Find the length of the remaining side.

By Pythagoras: c

2 = b

2 + a

2

Therefore c2 = 5

2 + 12

2 = 144 + 25 = 169

c = 169 = 13cm

3.2.2. Trigonometrical Ratios

Hypotenuse

OppositeSin =

Hypotenuse

OppositeCos =

Adjacent

OppositeTan =

These ratios need to be remembered and it is convenient to remember the SOHCAHTOA. Example:

7504

3

805

4

605

3

.

.

.

===

===

===

adj

oppBTan

hyp

adjBCos

hyp

oppBSin

Hypotenuse

Side adjacent to angle A A

B

C

Side opposite to angle A

A

B

C

5

4

3


Pg 34

Consider the equilateral triangle:

86602

360 .===

hyp

oppSin o 50

2

130 .===

hyp

oppSin

o

73211

360 .===

adj

oppTan o 57740

3

130 .===

adj

oppTan

o

2

2 2

30°

60°

B

C

3

2

D

B

C

30°

1

60°

502

160 .===

hyp

adjCos

o

86602

330 .===

hyp

adjCos

o


Pg 35

The Sine Curve

-1.5

-1

-0.5

0

0.5

1

1.5

0 50 100 150 200 250 300 350 400

degrees

Sin

X

The Cosine Curve

-1.5

-1

-0.5

0

0.5

1

1.5

0 50 100 150 200 250 300 350 400

degrees

Co

s X

The Tangent Curve

-40

-30

-20

-10

0

10

20

30

40

0 50 100 150 200 250 300 350 400

degrees

Tan

X


Pg 36

3.2.3. Use of Tables

It is possible to find the Sine, Cosine and Tangent of the most common angles, i.e. 90°, 30°, 45° etc using the relative curves. However, the Sine, Cosine and Tangent functions of other angles are not so simply calculated, and without the aid of a calculator, the use of tables is required. Example: Use the extract from the table of natural sines to find: Sin 32°, Sin 31°24’, Sin 32°28’

0' 6' 12' 18' 24' 30' 36' 42' 48' Mean Differences

0.0° 0.1° 0.2° 0.3° 0.4° 0.5° 0.6° 0.7° 0.8° 1' 2' 3' 4' 5'

30 5000 5015 5030 5045 5060 5075 5090 5105 5120 3 5 8 10 13

31 5150 5165 5180 5195 5210 5225 5240 5255 5270 2 5 7 10 12

32 5299 5314 5329 5344 5358 5373 5388 5402 5417 2 5 7 10 12

Sin 32° = 0.5299 Sin 31°24’ = 0.521 If the number of minutes is not exactly a multiple of 6, we use the tables of differences. Since Sin 32°24’ = 0.5358 and 28’ is 4’ more than 24’, looking in the difference table, we find the value of 10. This is added to the Sine of 32°24’. So Sin 32°28’ = 0.5368

3.2.4. Rectangular (Cartesian) Co-ordinates

The most common way of giving the position of a point in a plane is based on using two fixed perpendicular lines called the axes of coordinates, (x and y axis). The Cartesian co-ordinates represent the horizontal distance from the y-axis (in direction Ox) and the vertical distance from the x-axis (in direction Oy).

For example the point P in the diagram can be represented by the Cartesian co-ordinates (3, -2).

3.2.5. Polar Co-ordinates

Polar co-ordinates are expressed in terms of the direct distance from the fixed point O, together with the angle between Ox and OP. The position of the point P in the

diagram is given as (6, 60°). Anti-clockwise rotation is taken as positive, whilst clockwise is taken as negative.

1

2

3

4

1 2 43 5 6-1

-2

-3

-1-2-3

P

6

1

2

4

3

5

P

0

60°


Pg 37

Converting between Polar and Cartesian

Exercises Write down the Cartesian coordinates of the following points:

Write down the polar coordinates of the following points:

Plot the following points: a. (5,7) b. (-4,2) c. (3,-5) d. (-2,-4) e. (4, 60°) f. (-2, 45°) g. (5, -60°) h. (-4, -50°)

Calculate using tables: a. Sin 33° b. Cos 25° c. Cos 15°24’ d. Sin 12°32’ e. Cos 40°13’ f. Tan 15°20’ g. Tan 30°12’ h. Sin 35°20’


Pg 38

3.3. Geometrical Constructions

3.3.1. The Circle

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

Important circle theorems Theorem 1: The angle that an arc of a circle subtends at its centre is twice the

angle, which the arc subtends at the circumference. Theorem 2: Angles in the same segment of a circle are equal.

Theorem 3: The triangle in a semi-circle is always a right-angle one. Theorem 4: The opposite angles of any cyclic quadrilateral are equal to 180°.

OBA

C

D


Pg 39

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

point of tangency. Theorem 6: The angle between a tangent and a chord drawn from the point of

tangency equals half of the angle at the centre subtended by the chord.

OA

B

Theorem 7: The angle between the

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

Theorem 8: If two circles touch either internally or externally, then the line that

passes through their centres also passes through the point of tangency.

A

B


Pg 40

3.3.2. Angles

Degrees and Radians There are two ways of measuring angles. The most common way is by using the degrees scale but one may also use the radians scale. There are 2π radians in 360°. The significance of the π is that the radius of the circle fits around the circumference 2π times.

To convert from degrees to radians use: π2360

×°n

Example: Convert 50° into radians. radians872702360

50.=×

°π

Length of Arc

To calculate the length of arc: rn

π2360

×°

Acute Angles: angles measuring between 0° and 90°. Obtuse Angles: angles measuring between 90° and 180°. Reflex Angles: angles measuring between 180° and 360°. Right Angles: angles measuring exactly 90°. Two lines that meet at right angles are said to be perpendicular. Any two right angles are supplementary angles. Complementary Angles: angles whose sum of their degree measures equal to 90°. One of the complementary angles is said to be the complement of the other. Supplementary Angles: angles whose sum of their degree measures equal to 180°. One of the supplementary angles is said to be the supplementary of the other.

Vertical Angles: any two lines that intersect, will produce 4 angles. As in the diagram below, AOB and COD are called vertical angles and have the same degree measurement. Angles AOC and BOD are also vertical angles.

Alternate Interior Angles: any pair of parallel lines that are intersected by a third line produce angles, such that A and D are equal, as are angles C and B. Alternate Exterior Angles: similarly to interior angles, any pair of parallel lines crossed by a third line will produce angles such that A is equal to D and angle C is equal to angle B. Corresponding Angles: any pair of parallel lines intersected by a third line produce an angle such that angle D is equal to B and angle A is equal to C.

A

O

B

C

D

A B

C D

A B

C D

A B

C D


Pg 41

3.3.3. Construction

To bisect an angle when the arms of the angle meet. From the centre O, two equal arcs must be produced to cut the arms at A and B. With centres A and B, two equal length arcs then meet at C. The line OC bisects the angle. To bisect an angle when the arms do not meet. In order to have a common point, two lines parallel to the given arms need to be drawn far enough to make them meet at a point and then use the above technique normally.

Set out angles using trigonometric ratios. For this method, a good knowledge of the trigonometric ratios is important. For the example, a scale factor of 100 will be used. The figure shows how to set an angle using the tangent ratio. In this case, the angle is 23°30’ which gives a value of 0.4348. Using a multiplier of 100 units, the line AC = 43.48 units. The horizontal line

AB becomes 100 units with AC at right angles to AB. BC are joined so than angle ABC is now 23°30’. Similarly for the second figure which shows the sine method for angle θ = 28°36’. Sine of 28°36’ is 0.4787 which when multiplied by 100 becomes 47.87 units long. This is our arc length from A and AB as before is 100 units. A line is then drawn from B that just touches the arc. Angle ABC will be 28°36’.

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


Pg 42

Drawing a common external tangent to two given circles. The first figure shows two circles with radii R and r. With centre O1, draw a circle of radius R – r. Join O1 O2 and bisect to obtain centre C. With C as centre, draw a semi-circle of radius CO1 to cut the inner circle at T. Draw a line from O1 through T to locate T1 on the outer circle.

The second figure shows the line O2 parallel to O1T1, drawn to cut the smaller circle at T2. A line is then drawn through T1 and T2 to obtain the external tangent to the two circles as shown. This construction is very useful to portray a belt drive around two pulleys. To draw the inscribed circle for a given triangle. The figure shows the given triangle ABC with angles A and B both having been bisected and the bisectors extended to meet at O. In the second figure, a perpendicular is constructed from O to cut AB at D. Then with centre O and radius OD the inscribed circle of triangle ABC is drawn.

To draw a hexagon given the length of a side. A straight line AF is drawn equal to the given length of the side. With centres A and F, arcs of radius AF intersect at O. With the centre O, a circle of radius OA cuts the arcs at B and D. With centres B and E, arcs of radius AF cut the circle at C and D respectively. The points on the circle are finally joined to obtain the hexagon.

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


Pg 43

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


Pg 44

Appendix 1 – Trigonometric Tables

Natural Sines

Natural Cosines


Pg 45

Natural Tangents

M1 Cat-B1 Notes V1.1

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