1.Easa Part 66 Mod.1 1

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TTS Integrated Training System Module 1 Licence Category B 1 and 82 Mathematics 1.1 Arithmetic

~.

Module 1.1 Arithmetic use and/or disclosure is ,----.,

governed by the statement on page 2 of this chapter

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Knowledge Levels - Category A, 81, 82 and C Aircraft Maintenance Licence Basic knowledge for categories A, 81 and 82 are indicated by the allocation of knowledge levels indicators (1, 2 or 3} against each applicable subject. Category C applicants must meet either the category 81 or the category 82 basic knowledge levels. The knowledge level indicators are defined as follows:

LEVEL 1 A familiarisation with the principal elements of the subject. Objectives: The applicant should be familiar with the basic elements of the subject. The applicant should be able to give a simple description of the whole subject, using common words and examples. The applicant should be able to use typical terms.

LEVEL 2 A general knowledge of the theoretical and practical aspects of the subject. An ability to apply that knowledge. Objectives: The applicant should be able to understand the theoretical fundamentals of the subject. The applicant should be able to give a general description of the subject using, as appropriate, typical examples. The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. The applicant should be able to read and understand sketches, drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using detailed procedu res.

LEVEL 3 A detailed knowledge of the theoretical and practical aspects of the subject. A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: The applicant should know the theory of the subject and interrelationships with other subjects. The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. The applicant should understand and be able to use mathematical formulae related to the subject. The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.


Module 1.1 Arithmetic

Use and/or disclosure is



Table of Contents Module 1.1 Arithmetic_ _ _ __ _ _ _ __ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ ?

Fractions Types of Fractions Working with Fractions

7 7 7

Decimals Working with Decimals

17 17

Conversion Between Fractions and Decimals Convert a Decimal to a Fraction Convert a Fraction to a Decimal

23 23 23

Percentages Definition Changing a Fraction to a Percentage Changing a Percentage to a Fraction Changing a Percentage to a Decimal Changing a Decimal to a Percentage Values of a Percentage of a Quantity Expressing one Quantity as a Percentage of Another

29 29 29 29 29 29 30 30

Rounding, Significant Figures, and Decimal Places Rounding Significant Figures Decimal Places

35 35 36 37

Mean, Median, Mode and Range Definitions Calculating Mean Calculating Median Calculating Mode Calculating Range

39 39 39 40 40 41

Angles Definitions and Conversions Degrees and Radians: Measuring Angles Acute Angles Obtuse Angles

47 47 48 49 49

Reflex angles

49

Right Angles Complementary Angles Supplementary Angles Perpendicular Lines

50 50 51 51

Triangles Properties of shapes. Definitions

57 57 57

Module 1.1 Arithmetic Use and/or disclosure is governed by the statement

on page 2 of this Chapter

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Areas and Volume of Common Shapes_ _ _ _ _ _ _ _ _ __ _ __ _ _ _ _ 61 Rectangle 61 Square 61 Triangle 62 Parallelogram 62 Rhombus 63 Trapezium 64 Kite 65 Circle 65 Other Regular Polygons 66 Summary of Quadrilaterals 67 Surface Area and Volume of Common Solids Introduction Common Solids

69 69 69

Common Conversions Length

79 79

A~

~

Volume Mass

79 79


Modu le 1.1 Arithmetic

Use and/or disclosure is governed by the statement on page 2 or this chapter


Module 1.1 Enabling Objectives and Certification Statement Certification Statement These Study Notes comply with the syllabus of EASA Regulation 2042/2003 Annex Ill (Part-66) . d Knowe I d1ge Leve Is as spec1T1ed beow: I A\ppen d"IX I, an d th e associate EASA66 Level Objective Reference 81 82 Arithmetic 1.1 Arithmetical terms and signs, methods of 2 2 multiplication and division, fractions and decimals, factors and multiples, weights, measures and conversion factors, ratio and proportion, averages and percentages, areas and volumes, squares, cubes, square and cube roots .

Module 1 .1 Arithmetic Use and/or disclosure is governed by the statement

on page 2 of this chapter



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Module 1 .1 Arithmetic

Use and/or disclosure is governed by the statement on page 2 of this chapter


Module 1.1 Arithmetic ,--_

Fractions Types of Fractions 1.

Proper Fractions. Proper fractions may be defined as fractions less than 1. For example:

1

2' 2.

1 2 11 3 ' 5 ' 17

Improper Fractions. These are fractions which are greater than 1. For example:

7

8

17

5

3'3'11'5 3.

etc

etc

Mixed Numbers. These include whole numbers and vulgar fractions. For example:

2~

1 _!_

2 '

5 '

6_i_

11 '

27 ~ etc 7

4.

For all fractions, the number above the bar is called the numerator and the number below the bar is called the denominator.

5.

Simplest Form. The simplest form of

30

60

is _!_. Fractions can be expressed in simplest

2

form by dividing numerator and denominator by equal numbers until they will not divide further. For example:

8

2

in simplest form (after dividing numerator and 12 3 denominator by 4).

=

Working with Fractions 6.

Cancelling. The process of dividing numerator and denominator by equal values is called cancelling. For example: 27

81

=

9 27

3

1

= 9 = 3

Module 1.1 Arithmetic Use and/or disclosure is gov e rned by the str\lernent


1-7 TIS Integrated Training System ©Copyright 201 1


7.

Converting. To convert mixed numbers to improper fractions, multiply the whole number by the denominator and add to the numerator. For example:

2~ = 13 5 5 To convert improper fractions to mixed numbers, divide the numerator by the denominator to give a whole number - the remainder gives a new numerator. For example:

25 4 8.

4

Cancelling. Cancelling improper fractions involves exactly the same process as cancelling vulgar fractions. For example:

28 4 9.

= 6_!_

=

7 1

=7

45 6

and

=

15 2

= 7~ 2

Multiplication (a) Express all mixed numbers as improper fractions (b)

Cancel vertically if possible

(c)

Cancel across the multiplication sign if possible

(d)

Multiply numerators together, multiply denominators together

(e)

If the result is an improper fraction, convert to a mixed number

(f)

Check that your answer is in the simplest form

Examples: (1)

2 4 2 8 - x4 = - X-= 9 1 9 9 3

(2)

10.

14

X

2..!_

5

3

X

i_ 14

1

=%X7Xjl ?;' /1 ~

1

=

3

2

= 1..!_ 2

Division (a)

Convert all mixed numbers to improper fractions

(b)

Invert the fraction you are dividing by

(c)

Proceed as for multiplication.





Examples:

----(1)

,........,.

1~ = 3

3

4

12

4

7

~ 7 -X-

=

7

4

Jl'4

=

7 16

-

,~

,.--..,

(2)

3

4

3

1

4

=

7

=

X -

7

3 28

~.

~

(3)

r--..

5 16

3

=

-

8

3

~

;62

=

X-

5

6 5

-

=

1_!_

5

1 11.

Mixed Multiplication and Division (a) Invert all the tractions preceded by a division sign (b)

Treat the calculations as multiplication only.

Example:

1~ 4

4_.!. 2

= =

12.

X

7 4

/1

1~ 7

=

7 4

2

12

9

7

9 2

X

12 7

-

You Q!!!Y..Jurn upside down the fraction you are dividing by, i.e. the fraction after the division sign

X- X-

2

;d_z-1

-x-x-

~ ~ 71

=

2 3

Addition (a) Express all fractions as mixed numbers in lowest terms (b)

Add the whole numbers together

(c)

To add the vulgar fractions, you must convert each fraction so that their denominators are all the same. This is done by finding the lowest common

multiple (LCM) of the denominators.

Module 1.1 Arithmetic Use and/or disclosure is gove rned by the state men t on page 2 of this chapter



Examples:

(1)

(2)

1

1

3

5

6

10

9 5 + - + 13 4 12 8 1 4

= 3

20 30

=

2

3

3 2_:!_ + 5- + 14 12 8

=

6 + 10 + 9 24

3 +

=

5 3 + 12 8

= 3 + -+ =

6+5+9 30

=

+- +-

=

25 24

3 +

= 4-1 24

+ 1-124

Note: If your addition of fractions results in an improper fraction, you must convert this to a mixed number as shown in example (2).

13.

Subtraction The same basic procedure should be used for subtraction as for addition. Examples:

(1) (2)

2

8

- -

9

8 6 -

=

3

8

14

3

7

=

=

9 4 22 - 13 7

2 9

=

1 +

14 12 = 1 + 21 (3)

4_:!_ 3

1~ 4

=

3 +

1 3

4

2 +

= 2 +


4

_....,_

7

-,

= 1~ 21

3 4

= 3 + 12 =

2 3

-

9

12+ 4 9 12 16 9 12

As numerator (4- 9) give a negative value, one whole unit has to be

12 before the 12

converted to -

subtraction of fractions is carried out.


Use and/or disclosure is governed by the statement on page 2 or this chapter

~


= 2!_ 12

14.

Mixed addition and subtraction can be carried out exactly as above. Examples:

,........,

r"""'

(1)

,...--..

5!_ + 3~ = 2 + 6 ft. 9

4.2_ 2

12

4

12

=

2 +

8

12

= 2.3.3 ,.--.

(2)

~

4

8

= 4

r--"""\

15.

1~

2.2_

+

1 + 4-

= 3

5 +

24+3 181- 8 24

1S- 8

3

24 4.!.Z. 24

=

Remember that your final step in any calculation must be to simplify (cancel fractions). Example:

3~ 5

=

+

2~ 20

1~ 10

2~ = 4

=

2 +

12 + 18 15 20

2~ 4

Module 1 .1 Arithmetic Use and/or disclosure is governed by the statement on page 2 of this Chapter

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U se and/or disclosure is

governed by the statement on page 2 of this ctlapter

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Worksheet 1.

Convert the following mixed numbers to improper fractions:

2~

(a)

2.

11

7.

s3.2.

2_!_ 7

(e)

25

53

(c)

5

(d)

7

210

99

(e)

4

8

6

14

7

15

(b)

-X-

2 3

X

1 27

1_g_ 5

X

3

1_g_ 5

(c)

10

X-

X -

21

8

Divide and simplify the following: 3

-

10

9

3_g_ 3

(b)

25

3_!_ 7

9_!_ 3

(c)

3_!_ 9

Evaluate the following: (a)

6.

(d)

5

Multiply and simplify the following:

(a)

5.

21

(b)

3

(a)

4.

21~

(c)

Convert the following improper fractions to mixed numbers: (a)

3.

34 9

(b)

7

7

5

3

10

6

14

-X- X -

(b)

7_!_ 5

X

3

15 24

17

(c)

-

8

-

29

1 1 2- x74 8

Add the following fractions and mixed numbers: 2

(a)

-

(d)

-

+

3 1

3 7 1

3

-

(b)

1

1

+

4

1

2

+ -

2

3

2

1

+- +- +- + 4 2 3 5 6

3

-

(c)

+ -

3

(e)

4

4

+5

3_!_ + 716 8

Subtract the following: (a)

,....-

(d)

1_!_

3

2

5

4

2

11

7

(b)

1~ 4

2~

(c)

5

(e)

74 9

Module 1.1 Arithmetic Use and/or disclosure is governed by the statement


1~ 5

2.2. 7

82 3



8.

Evaluate the following :

(a)

2~+3~ 1~ 3

8

1

1

9

(b)

1..! 2

2~ 5 3

(c)

42

1 2 5 1-+3 7 6 3 3 -x4 7

4

1-x2- (d)

5

3

1

5

5

2-+2 6

3 -+ 16 4 2 1 -+3 4

5

(g)

( 2~x_?_) + ( 3_! 7 24 2

(3 (i)

~)

(h)

~ + ~)(1 ~ 1 ~)

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4

1

3 1 6

3 1 -+ 7 3



,.----,--.

~ -

,--.

,--,

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Answers

146 25

e)

15 7

d)

52_!_ 2

e)

12~

c) 2~

d)

1~

e)

3~

13 20

26 c) - -

d)

6 77

e) -1-

b)

5 7

c) 2~

d)

g)

1~

31

108 5

1.

a) 20

2.

a) 32

3.

a)

4.

a) 5

b) 1_!_

c) 3

5.

a) -

1 8

b) 12

c) 2

6.

a) 1~

b) 1_!_!

7.

a)

8.

a) 4~ 5 6

7

3

4 5

-

6

21

9

10

72

f)

b)

9

c)

-

b) 4_!_

c) 74

b) 2

c)

5

6

12

b) - -

12

7

1 4

35

9

5 36

i)

20

3

16 2

9

e)

5

1~ 55

1..!.:!_ 24

Module 1 .1 Arithmetic Use and/or disclosure is oovemed by the statement on page 2 of this chapter

8

-

60

h)

d) -



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Use and/or disclOsure is governed by the statement on page 2 of this chapter


Decimals Working with Decimals 1.

Decimals are a very important and particular set of fractions . They are fractions whose denominators are powers of 10, i.e. 10, 100, 1000, 10000 etc (do not be concerned about the meaning of 'powers of 10', you will deal with this later in the course). Decimals are not written in the usual fraction form, but in shorthand using a decimal point.

Examples: a)

1 = 0.1 10

b)

-

d) 52_= 5.7 10 r--.

2.

e)

1 = 0.01 100

c)

1 = 0.001 1000

63_2_ = 63.07 100

If you have difficulty in relating decimals to fractions, the following table may help. THOUSANDS

HUNDREDS

TENS

UNITS

1000

100

10

1

5

3

4

6

TENTHS

HUNDREDTHS

THOUSANDTHS

1 10

1 100

-1-

7

9

2

-

1000

The number in the table is 5346.792; it consists of 5 thousands, 3 hundreds, 4 tens, 6 units, 7 tenths, 9 hundredths and 2 thousandths .

.-

3.

The number of digits after the decimal point is called decimal places.

Examples:

4.

a)

27.6 has one decimal place

b)

27.16 has two decimal places

c)

27.026 has three decimal places

d)

101.2032 has four decimal places

In addition of decimals, particular care must be taken to ensure that decimal points are in line.

Example:

Evaluate 27.3 + 0.021 + 68.3

+

Module 1.1 Arithmetic use and/or disclOsure is oovemM by tha statenl9nt

on page 2 of this Chapter

27.3 0.021 68.3 95.621

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5.

Similarly, in subtraction, ensure that decimal points are in line. Example:

6.

Evaluate 27.3 - 4.36

27.3 4.36 22.94

27.3 can also be written as 27.30

When multiplying decimals, ignore the decimal point until the final answer is obtained, then count the number of decimal places in both the numbers being multiplied. This is the number of decimal places in the answer. Example:

Evaluate 27.3 x 9.31 273 931 245 700 8 190 273 254163

Note: Total of 3 decimal places.

Answer = 254.163 (3 dpl.)

Note: It does not matter which you multiply first (i.e. the 9, the 3 or the 1) providing '00' is placed before the answer when multiplying out the 1OO's (in this case the 9) and '0' is placed before the answer when multiplying out the 1O's (in this case the 3) So the above calculation could have looked like this: 273 931 273 8 190 245 700 254 163 7.

The answer is the same

In division, it is easier to divide by a whole number than by a decimal. To make the divisor (the number you are dividing by) into a whole number, move the decimal point a specific number of places to the right. You must then also move the decimal point in the dividend (the number you are dividing into) to the right by the same number of decimal places. Example:

Evaluate 24.024

2402.4

=

4.62

462

We have moved the decimal point 2 places in both the divisor and the dividend, but the answer is unaltered

5.2 462)2402.4 2310 924 924

An approximate answer could be calculated as follows:

1-18 TTS Integrated Training System

24.024 4.62

24 5

5

Use and/or disclosure is govemed by the statement on page 2 of this chapter


Worksheet

r--

1.

Calculate the sum of the following: a)

2.

b)

18.098 + 210.099

c)

0.025 + 10.995

b)

32.76- 20.086

c)

10.75- 19.999

b)

1.27 X 0.871

c)

-1.01

b)

3.375

Evaluate: a)

3.

0.251 + 10.298

21.76- 18.51

Find the product of: a)

5.05

X

13.8

X

0.89

,----._

4.

Calculate: a)

42.39 0.09

1.5

c)

Module 1.1 Arithm etic

.-


0.002

0.8

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Answers 1.

a) 10.549

b) 228.197 c) 11.02

2.

a) 3.25

b) 12.674

3.

a) 69.69

b) 1.10617 c) -0.8989

4.

a) 471

b) 2.25

c) -9.249

c) 0.0025

Module 1.1 Arithmetic Use and/or disclosure is governed by the statement on page 2 of tllis chapter



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Use and/or disdosure is govemed by the statement on page 2 of this chapter


Conversion Between Fractions and Decimals Convert a Decimal to a Fraction Write the number over the appropriate power of 10 and, if possible, cancel to lowest terms.

Examples:

~

=

a)

0 .8

b)

6.25

c)

0.037

8

=

10

=6 =

4 5

25

+

100

= 6..!.4

37 1000

Convert a Fraction to a Decimal Divide the numerator by the denominator.

Examples: ,...--..

a)

-

=

0.80 5)4.00

b)

3 8

=

0.375 8)3.000

c)

5 6

=

r--

4

5

,...--

,...--

r-

~

. re-occur f or ever ) 0.8333 . .. .... the 3 wtll

6 5.0000

Here, we cut off the result to the number of decimal places required.

5 Thus -

6

or

5

6

=

0.83 correct to 2 decimal places

= 0.8333 correct to 4 decimal places

Module 1.1 Arithmetic Use and/or disclosure is governed by the statement on page 2 of this chapter



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Use and/or disdosure is

governed by the statement on pzago 2 of this chaptor


Worksheet

1.

Convert the following decimals to proper fractions in their simplest form : a)

2.

---..

3.

b)

0.02

c)

0.004

Convert the following proper fractions to decimals to 2 decimal places:

a) ,...---.,

0.73

5

-

8

b)

13

-

c)

15

3

-200

Place in ascending order of magnitude:

r-.,

,.---..

a)

1

6'

3 0 .167 and 20

b)

2

5'

0.44 and -

7

16

,...--., ~-

Module 1.1 Arithmetic Use and/or disclosure is governed b y the statement on page 2 of this chapter



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Use and/or disdosure is governed by the statement

on page 2 of !his chapter


Answers 73

1.

a)

2.

a) 0.63

3.

a)

100

b)

1 50

0 .87

1 3 and 0.167 20' 6

c)

1 250

0.02

b)

2

.!_ and 0.44

5' 16

Module 1.1 Arithmetic Use and/or disclosure is governed by the statement on p age 2 o f this chapter

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-.

1-28 TTS Integrated T raining System © Copyright 2011


Use and/or disclosure is g ov erned by th e sta te ment



Percentages Definition. A percentage is a fraction whose denominator is 100.

Example:

3%means

~ 100

Changing a Fraction to a Percentage To change a fraction to a percentage, multiply by 100%.

Example:

3 - as a percentage =

5

3

5

X

100%

=

60%

Changing a Percentage to a Fraction To change a percentage to a fraction, divide by 100%.

Examples:

=

8% 8 2 = -- - 100% 100 25

a)

8% as a fraction

b)

12 Y2 % as a fraction =

12~% 25 1 25 1 -=---- =-X- - = - - = 100% 2 100 200 8

Changing a Percentage to a Decimal To convert a percentage to a decimal, firstly, convert the percentage to a fraction, then convert the fraction to a decimal.

Examples:

=

65 . I --,as a d ec1ma 100

=

a)

65% as a fraction

b)

32..!.2 32 Y2 % as a fraction = - , as a decimal = 0.325 100

0.65

Changing a Decimal to a Percentage To convert a decimal to a percentage, firstly, convert the decimal to a fraction, then convert the fraction to a percentage.

,. . . . . __

Examples: a)

21

2.1

0.021 as a fraction = - - ,= --,as a percentage= 2. 1% 1000 100


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b)

0.037 as a fraction

=

37 3.7 )/ , = - - , as a percentage = 3.7~o 1000 10 0

Values of a Percentage of a Quantity To find the value of a percentage of a quantity, express the percentage as a fraction and multiply by the quantity.

Examples: 2

a)

4% of 60

.K xoo = -tOO'

= 2~5

12

=

5

5

b)

3 Y2% of 1500

=

3..!.2_

_

100

X

=

1500

7

-X

~

1!)e1;Y =

105 2

= 52 Y2

Expressing one Quantity as a Percentage of Another To express one quantity as a percentage of another, make a fraction of the 2 quantities and multiply by 100.

Examples:

a)

12 as a percentage of 50

b)



=

=

12 X

50

4 60

X

100

100

=

24%

= 6.67%




Worksheet r----

1. ~-

Calculate: a)

4% of 30

b)

0.8% of 360

d)

120% of 75

e)

80% of 90

c)

1.5% of 60

r-.. r -. ~

,...-,

2.

r-·-,-.....

,..-...,

Express: a)


b)


c)

0.5 as a percentage of 12.5

d)


e)


,..--. ~

,-.....

3. .---.,

,........._

Express as a proper fraction: a)

0.6

b)

0.35

c)

0.48

d)

0.05

e)

0.325

f)

25%

g)

13%

h)

4.5%

i)

16%%

0.025

c)

1.25

d)

,........._ ,...-.

4.

Express as a percentage: a)

,--. r----

e)

0.43 3 7

b) f)

1 12

g)

g o ve rned by the st::dema nt


3

3

-

8

Module 1.1 Arithm etic Use and/or disclosure is

2



Intentionally Blank

~.



Use and/or disclosure is govemed by the statement



Answers r--~'·'

1.

1 a) 1- or 1.2 5

2.

a) 60%

3.

a)

b)

2.88

c) 0 .9

d) 90

b) 150%

c) 4%

d)

7 b) 20

c)

4

13 g) - 100

h)

a) 43%

b) 2.5%

c) 125%

e) 72

,...-.

,.........

f) r-..

---,-....,

4.

f) r-..

3 5 1

8~% 3

g)

12

-

25 9 200

133~% 3

d) -

e)

11~% 9

13 e) 40

I

20 49

i)

--

d)

66~%

300

3

e)

42~% 7

37~% 2

,-....,

Module 1.1 Arithm etic Use and/or disclosure is governed by the statement on page 2 of this chapter



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Use and/or disclosure is governed by the statement on page 2 of tllis chapter

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Rounding, Significant Figures, and Decimal Places Rounding Rounding is the process of reducing the number of significant digits in a number. The result of rounding is a "shorter" number having fewer non-zero digits yet similar in magnitude. The result is less precise but easier to use. For example: 73 rounded to the nearest ten is 70, because 73 is closer to 70 than to 80. Methods of Rounding Common Method: This method is commonly used in mathematical applications, for example in accounting. It is the one generally taught in elementary mathematics classes. This method is also known as Symmetric Arithmetic Rounding or Round-Half-Up (Symmetric Implementation) Decide which is the last digit to keep. Increase it by 1 if the next digit is 5 or more (this is called rounding up) Leave it the same if the next digit is 4 or less (this is called rounding down)

Examples: 3.044 rounded to hundredths is 3.04 (because the next digit, 4, is less than 5). 3.045 rounded to hundredths is 3.05 (because the next digit, 5, is 5 or more). 3.0447 rounded to hundredths is 3.04 (because the next digit, 4, is less than 5). For negative numbers the absolute value is rounded.

Examples: 2.1349 rounded to hundredths is 2.13 2.1350 rounded to hundredths is 2.14 Round to Even Method: This method is also known as unbiased rounding, convergent rounding, statistician's rounding or bankers' rounding. It is identical to the common method of rounding except when the digit(s) following the rounding digit start with a five and have no nonzero digits after it. The new algorithm is:

Decide which is the last digit to keep. Increase it by 1 if the next digit is 6 or more, or a 5 followed by one or more non-zero digits. Leave it the same if the next digit is 4 or less Otherwise, all that follows the last digit is a 5 and possibly trailing zeroes; then change "the last digit to the nearest even digit. That is, increase the rounded digit if it is currently odd; leave it if it is already even. With all rounding schemes there are two possible outcomes: increasing the rounding digit by one or leaving it alone. With traditional rounding, if the number has a value less than the halfway mark between the possible outcomes, it is rounded down; if the number has a value exactly half-way or greater than half-way between the possible outcomes, it is rounded up. The round-

Module 1.1 Arithmetic Use and/or disclosure is g ovgmed b y t h9 state me nt

on page 2 of this ch apter



to-even method is the same except that numbers exactly half-way between the possible outcomes are sometimes rounded up - sometimes down. Although it is customary to round the number 4 .5 up to 5, in fact 4.5 is no nearer to 5 than it is to 4 (it is 0 .5 away from both). When dealing with large sets of scientific or statistical data, where trends are important, traditional rounding on average biases the data upwards slightly. Over a large set of data, or when many subsequent rounding operations are performed as in digital signal processing, the round-to-even rule tends to reduce the total rounding error, with (on average) an equal portion of numbers rounding up as rounding down. Th is generally reduces the upwards skewing of the result. Round-to-even is used rather than round-to-odd as the latter rule would prevent rounding to a result of zero.

Examples: 3.016 rounded to 3.013 rounded to 3.015 rounded to digit (1) is odd) 3.045 rounded to digit (4) is even) 3 .04501 rounded non-zero digits)

hundredths is 3.02 (because the next digit (6) is 6 or more) hundredths is 3.01 (because the next digit (3) is 4 or less) hundredths is 3.02 (because the next digit is 5, and the hundredths hundredths is 3.04 (because the next digit is 5, and the hundredths to hundredths is 3.05 (because the next digit is 5, but it is followed by

Significant Figures Rounding to n significant figures is a form of rounding. Significant figures (also called significant digits) can also refer to a crude form of error representation based around significant figure rounding. Rounding to n significant figures is a more general-purpose technique than rounding to n decimal places, since it handles numbers of different scales in a uniform way.

Rules of Significant Figures All non-zero digits are significant. Example: '123.45' has five significant figures: 1 ,2,3,4 and 5. Zeros appearing in between two non-zero digits are significant. Example: '1 01.12' has five significant figures : 1 ,0, 1,1 ,2. All zeros appearing to the right of an understood decimal point or non-zeros appearing to the right of a decimal after the decimal point are significant. Example: '12.2300' has six significant figures: 1 ,2,2,3,0 and 0. The number '0.001-22300' still only has six significant figures (the zeros before the '1' are not significant). In addition, '12.00' has four significant figures. All zeros appearing in a number without a decimal point and to the right of the last nonzero digit are not significant unless indicated by a bar. Example: '1300' has two significant figures: 1 and 3. The zeros are not considered significant because they don't have a bar. However, 1300.0 has five significant figures.



Use and/or disclosure is governed by the statement on page 2 ollhis chapoer


However, this last convention is not universally used; it is often necessary to determine from context whether trailing zeros in a number without a decimal point are intended to be significant. Digits may be important without being 'significant' in this usage. For instance, the zeros in '1300' or '0.005' are not considered significant digits, but are still important as placeholders that establish the number's magnitude. A number with all zero digits (e.g. '0.000') has no significant digits, because the uncertainty is larger than the actual measurement. r----

Examples: Rounding to 2 significant figures: 12,300 becomes 12,000 13 stays as 13 0.00123 becomes 0.0012 0.1 becomes 0.1 0 (the trailing zero indicates that we are rounding to 2 significant figures). 0.02084 becomes 0.021 0.0125 becomes 0.012 in unbiased rounding, while it is 0.013 in biased. One issue with rounding to n significant figures is that the value of n is not always clear. This occurs when the last significant figure is a zero to the left of the decimal point. For example, in the final example above, when 19 800 is rounded to 20 000, it is not clear from the rounded value what n was used - n could be anything from 1 to 5. The level of rounding can be specified explicitly. The abbreviation s.f. is sometimes used, for example "20,000 to 2 s.f." Scientific notation could be used to reduce the ambiguity, as in (2.0 x 104 ). As always, the best approach is to state the uncertainty separately and explicitly, as in 20,000 ± 1%, so that significant-figures rules do not apply. A less common method of presenting ambiguous significant figures is underlining the last significant figure of a number, for example "2QOOO"

Decimal Places The precision of a value describes the number of digits that are used to express that value. In a scientific setting this would be the total number of digits (sometimes called the significant digits) or, less commonly, the number of fractional digits or places (the number of digits following the point). This second definition is useful in financial and engineering applications

where the number of digits in the fractional part has particular importance. In both cases, the term precision can be used to describe the position at which an inexact result will be rounded. For example, in floating-point arithmetic, a result is rounded to a given or fixed precision, which is the length of the resulting significand. In financial calculations, a number is often rounded to a given number of places (for example, to two places after the point for many world currencies). As an illustration, the decimal quantity 12.345 can be expressed with various numbers of significant digits or decimal places. If insufficient precision is available then the number is rounded in some manner to fit the available precision. The following table shows the results for Module 1.1 Arithmetic Use and/or disclosure is g ovArnAd by tho s t nhHnent




various total precisions and decimal places, with the results rounded to nearest where ties round up or to an even digit (the most common rounding modes). Note that it is often not appropriate to display a figure with more digits than that which can be measured. For instance, if a device measures to the nearest gram and gives a reading of 12.345 kg, it would create false precision if you were to express this measurement as 12.34500 kg.

Rounded to 'n' significant digits

Rounded to 'n' decimal places

Five

12.345

12.34500

Four

12.35

12.3450

Three

12.3

12.345

12

12.35

n

Two One Zero

1

X

10

1

12.4

n/a

12

(2 3'-L \

~Q,

\'L.o.:>

I~ o 0



Use and/or disclosure is governed by the stat ement on page 2 of this chapter

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Mean, Median, Mode and Range Definitions The Mean, Median and Mode of a set of numbers are three types of "average" of the set. However, the "Mean" is the term most commonly taken as the average.

Mean:

The sum of a set of data divided by the number of data

Median:

The middle value or the mean of the middle two values, when the data is arranged in numerical order.

Mode:

The value (number) that appears the most. It is possible to have more than one mode, and it is possible to have no mode.

Calculating Mean To find the mean, you need to add up all the data, and then divide this total by the number of values in the data.

Example 1: Find the Mean of

2, 2, 3, 5, 5, 7, 8

Adding the numbers up gives: 2 + 2 + 3 + 5 + 5 + 7 + 8 There are 7 values, so you divide the total by 7:

32 + 7

= 32 = 4.57 ...

So the mean is 4.57 (2 d.p.)

Example 2: Find the Mean of

2, 3, 3, 4, 6, 7

Adding the numbers up gives:

2+ 3+3+4+6+ 7

There are 6 values, so you divide the total by 6:

= 25

25 + 6 = 4.33 ...

So the mean is 4.33 (2 d.p.)




Calculating Median To find the median , you need to put the values in order, then find the middle value. If there are two values in the middle then you find the mean of these two values.

Example 1: Find the median of

2, 2, 3, 5, 5, 7, 8

The numbers in order: 2 , 2 , 3 , (5) , 5 , 7 , 8 The middle value is marked in brackets, and it is 5. So the median is 5

Example 2: Find the median of The numbers in order:

2, 3, 3, 4, 6, 7

2 , 3, (3 , 4), 6, 7

This time there are two values in the middle. They have been put in brackets. The median is found by calculating the mean of these two values: (3 + 4) -;. 2 =3.5 So the median is 3.5

Calculating Mode The mode is the value which appears the most often in the data. It is possible to have more than one mode if there is more than one value which appears the most.

Example 1: Find the mode of The data values:

2, 2, 3, 5, 5, 7, 8

2 ,2 ,3 ,5 ,5 ,7 ,8

The values which appear most often are 2 and 5. They both appear more time than any of the other data values. So the modes are 2 and 5

Example 2: Find the mode of

2, 3, 3, 4, 6, 7

The data values: 2 , 3 , 3 , 4 , 6 , 7 This time there is only one value which appears most often- the number 3. It appears more times than any of the other data values. So the mode is 3

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U se a nd/or disclosure is



Calculating Range To find the range, you first need to find the lowest and highest values in the data. The range is found by subtracting the lowest value from the highest value. Example 1: Find the range of

2, 2, 3, 5, 5, 7, 8

The data values: 2 , 2 , 3 , 5 , 5 , 7 , 8 The lowest value is 2 and the highest value is 8. Subtracting the lowest from the highest gives: 8-2

=6

So the range is 6 Example 2: Find the range of

2, 3, 3, 4, 6, 7

The data values: 2 , 3 , 3 , 4 , 6 , 7 The lowest value is 2 and the highest value is 7. Subtracting the lowest from the highest gives: ... ?- 2 = 5 ~

So the range is 5

Module 1.1 Arithmetic Use and/or disclosure is governed by the stat ement




Intentionally Blank





Worksheet 1.

A data set contains these 12 values: 3, 5, 9, 4, 5, 11, 10, 5, 7, 7, 8, 10 (a) (b) (c) (d)

2.

What is the What is the What is the What is the

mean? median? mode? range?

Calculate the mean, median, mode and range for each set of data below:

(a)3, 6,3, 7, 4, 3, 9 (b) 11, 10, 12, 12, 9, 10, 14, 12, 9 (c) 2, 9, 7, 3, 5, 5, 6,5, 4, 9




Intentionally Blank

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Answers ,..-.,

1

(a)

7

2

(a)

Mean= 5 Median= 4 Mode= 3 Range= 6

(b)

Mean= 11 Median= 11 Mode= 12 Range= 5

(b)

7

(c)

5

(d)

8

,..-.,

,--,

,---..

r---.

.,....--. ,--,

(c)

Mean= 5.5 Median= 5 Mode= 5 Range= 7

,----.

Module 1 .1 Arithmetic Use and/or disclosure is govQm e d by tha state men t on page 2 of this chapter

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Use and/or disclosure is governed by the statement on page 2 of thi s chapter


Angles Definitions and Conversions We can specify an angle by using a point on each ray and the vertex. The angle below may be specified as angle ABC or as angle CBA; you may also see th is written as L ABC or as L CBA. Note how the vertex point is always given in the middle.

A

c B Example: Many different names exist for the same angle. For the angle below, L PBC, L PBW, L CBP, and L WBA are all names for the same angle.

p

A

B

w

c

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Degrees and Radians: Measuring Angles We measure the size of an angle using degrees. We can also use radians to measure angles.

There are 2

radians in 360°

The radius of a circle fits around the circumference 6.28 (or 2 ) times. 1 radian= 57.3 degrees.

no

To convert from degrees to radians, use - - x 2 360

Note:

where n° is the number of degrees

Degrees can be further subdivided into minutes and seconds.

60 seconds= 1 minute 60 minutes= 1 degree

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Acute Angles - .

An acute angle is an angle measuring between 0 and 90 degrees. Example: The following angles are all acute angles.

88 °

Obtuse Angles An obtuse angle is an angle measuring between 90 and 180 degrees. Example: The following angles are all obtuse.

117 .8 °

Reflex angles . . . . . .,

A reflex angle is an angle measuring between 180 and 360 degrees

Module 1.1 Arithm etic Use and/or disclosure is governed by the statement on page 2 of this chapter


© Copyright 2011


Right Angles A right angle is an angle measuring 90 degrees. Two lines or line segments that meet at a right angle are said to be perpendicular. Note that any two right angles are supplementary angles (a right angle is its own angle supplement).

Complementary Angles Two angles are called complementary angles if the sum of their degree measurements equals 90 degrees. One of the complementary angles is said to be the complement of the other.

Example: These two angles are complementary.

I Note that these two angles can be "pasted" together to form a right angle!





Supplementary Angles Two angles are called supplementary angles if the sum of their degree measurements equals 180 degrees. One of the supplementary angles is said to be the supplement of the other.

Example: These two angles are supplementary.

Note that these two angles can be "pasted" together to form a straight line!

139°

41°

Perpendicular Lines Two lines that meet at a right angle are perpendicular. They are also said to be "normal" to each other.




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governed by th e statement on page 2 of this chapter


Worksheet

1.

r-

~

Convert the following angles to radians

£

90° e)

d)

,_....

~

c)

b)

f)

72° 120° )

rr-

Module 1.1 Arithmetic Use and/or disclosure is gov9 mod by thQ statQITIOnt on page 2 of this chapter



Intentionally Blank



Use and/or disclosure is governed by the statement on page 2 of thrs chapter

,-


Answers ,-------.

1.

1A

a) d)

23

180

radians b)

1A

radians e) 2/5

radians

c)

Yz

radians

radians

f) 2/3

radians

Module 1.1 Arithmetic Use and/or disclosure is governed by the state ment



©Copyright 2011


Intentionally Blank






Triangles Properties of shapes. The 3 properties of shapes that we are going to look at are:

1. The number of sides 2. The interior angles (the angles inside). 3. The length of the sides. These properties help use to remember which shapes are which and why they are so called (in some cases). Let's start with a shape that has 3 sides: TRIANGLES {tri- means 3). Triangles ALWAYS have 3 sides. The interior angles of a triangle add up to 180 degrees.

Definitions r ~

Here are the triangles you are expected to know about: 1. 2. 3. 4.

2.

Equilateral Triangle Isosceles Triangle Right- Angled Triangle Scalene Triangle

Congruent and Similar Triangles

Congruent and Similar are two words usually applied to triangles but can equally be applied to other shapes. Congruent triangles are two triangles which have equal angles and are the same size (i.e. identical in every way) but may be oriented differently. Similar triangles are two triangles which have the same angles but are of different size.

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3.

Equilateral Triangle

An equilateral triangle has got 3 sides of equal length and 3 angles that are equal. Since ALL the angles in a triangle add up to 180Q then 180 divided by 3 must be 60Q.

The clue is in the name EQUILateral.

4.

Isosceles Triangle

An Isosceles triangle has got two sides of equal length and 2 angles equal.

What is the value of the angle at the top of this Isosceles triangle?

The answer is 80 ~ All angles in a triangle add up to 1ao·so 180- (50+ 50)= so· So an isosceles triangle has only got two sides of equal length and two angles the same.

----

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.

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5.

Right- Angled Triangle

The right angled triangle contains a right angle (an angle of 90)

The

Shows a right ang)e

In a right angled triangle what must the other two angles add up to 90obecause all the angles in a triangle add up to 180oand a Right Angled Triang le has got one angle of 90 ~

6.

Scalene Triangle

A scalene triangle is the easiest of them all. The scalene triangle has got NO sides of equal length and NO angles the same.


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Areas and Volume of Common Shapes Rectangle The area A of any rectangle is equal to the product of the length I and the width w. Formula: A = lw

A

w

~

Square The area A of any square is equal to the square of the length

,. . . . .,

Formula: A

s of a side.

=!I

L

A

s

0




Triangle The area A of any triangle is equal to one-half the product of any base band corresponding height h. Formula: A = "'hbh

b

Parallelogram A parallelogram is a 4 sided shape with the 2 opposing sides parallel to each other. The area A of any parallelogram is equal to the product of any base b and the corresponding height h. Formula: A = bh

A b



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Rhombus A rhombus is a parallelogram with all4 sides equal length. The diagonals bisect the interior angles equally and the diagonals intersect each other at right angles. The area A of any rhombus is equal to one-half the product of the lengths d 1 and d2 of its diagonals. Formula: A = Y2d 1d2 or Formula: A

=bh

as in the parallelogram

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Trapezium A trapezium has only 2 sides parallel. (UK definition) The area A of any trapezium is equal to one-half the product of the height hand the sum of the bases, bt and b2. Formula: A = %h (b 1 + b2)

A

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Kite A Kite shape has no sides parallel. The area can be found by Formula: A= V2 d1d2

Circle The area A of any circle is equal to the product of Formula: A

=

and the square of the radius r.

r2

Sector Area Theorem The area A of any sector with an arc that has degree measure n and with radius r is equal to the product of the arc's measure divided by 360 multiplied by

times the square of the radius.

Formula: A = (n/360)( ,2)

Module 1.1 Arithmetic Use and/or disclosure is governed by the slalemenl on page 2 of !his chapter



Other Regular Polygons Regular polygons are any polygons that are equilateral and equiangular. The area A of any regular polygon with perimeter P and apothem of measure a is equal to onehalf the product of the perimeter and the apothem. This formula can be derived if you make 5 triangles inside the shape. The area of each triangle is Y2 Sa ( 1/2 base x height) . The total area is therefore 5 x Y2 Sa (in the case of the pentagon shown - the 5 only represent the number of sides). But 5 x Sis the total perimeter of the shape, so: Formula: A = !haP

Regular Pentagon (shape could be any regular shape)

s P ::;:: 5s

5 could be anything, depending on no. of sides

The angle shown is one equal portion of 360 degrees. In the case of the pentagon , it is 360/5 = 72°. Hence the internal angle of any polygon can be found by calculating the supplement of the external angle.





Summary of Quadrilaterals SQUARE 4 Sides equal 4 right angles Diagonals bisect each other at right angles Diagonals are equal

RECTANGLE 2 pairs of opposite sides equal and parallel 4 right angles Diagonals are equal and bisect each other

'.

RHOMBUS 4 sides equal , opposite sides parallel Diagonals bisect each other but are not of equal length

PARALLELOGRAM 2 pairs opposite sides equal and parallel . '

Diagonals bisect each other but are not of equal length

TRAPEZIUM I TRAPEZOID 1 pair opposite sides parallel or

KITE 2 pairs of adjacent sides equal Longer diagonal bisects shorter at right angles




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Surface Area and Volume of Common Solids ,-...

Introduction There are special formulas that deal with solids, but they only deal with right prisms. Right prisms are prisms that have two special characteristics - all lateral edges are perpendicular to the bases, and lateral faces are rectangular. The figure below depicts a right prism.

Common Solids 1.

,-...

Right Prism Area

The lateral area L (area of the vertical sides only) of any right prism is equal to the perimeter of the base times the height of the prism (L =Ph). The total area T of any right prism is equal to two times the area of the base plus the lateral area.

=

Formula: T 28 + Ph

B : ; ;: lw P = 21+ 2w

h

(The base·s formula could chan.ge depending on the base's shape.} (The perimeter's formula oould change depending on the base·.s shape.)


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2.

Right Prism Volume

The volume V of any right prism is the product of 8, the area of the base, and the height h of the prism. Formula: V =Bh

B ·.;.,.,lw

h '\//~ " (

l-

't\

t

~

.

($s )

w (~ · (Ttle base·s formula coold ct)ange dependng on the base's srape.)



Use and/or disclosure is governed by the statement



3.

Pyramid Volume

A pyramid is a polyhedron with a single base and lateral faces that Q.re all triangular. All lateral edges of a pyramid meet at a single point, or vertex. The volume V of any pyramid with height h and a base with area 8 is equal to one-third the product of the height and the area of the base. This applies even if the prism is not a 'rightprism' i.e. the axis is not perpendicular to the base. The height however, is still measured perpendicular to the base as shown below. Formula: V =

X Bh B = wl (Base's formula couLd change depending on tis shap$ .)

A regular pyramid is a pyramid that has a base that is a regular polygon and with lateral faces that are all congruent isosceles triangles.

Module 1.1 Arithmetic Use and/or disclosure is governed Oy tne statement on page 2 of this chapter

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4.

Cylinder Volume

The volume V of any cylinder with radius rand height h is equal to the product of the area of a base and the height. Formula: V =

lh

II

5.

Cylinder Surface Area

For any right circular cylinder with radius rand height h, the total surface area Tis two times the area of the base (2 Formula: T = 2 IC r h

I) plus the curved surface area (2

rh).

+ 2 ?t l

b



Use and/or disclosure is govemed by the statement on page 2 of this chapter


6.

r--

Cone Volume

The volume V of any cone with radius rand height h is equal to one-third the product of the height and the area of the base. Formula: V =

7.

X lh

Cone Surface Area

The total surface area T of a cone with radius rand slant height I is equal to the area of the base ( lT I) plus TC times the product of the radius and the slant height. Formula: T = 7\T I +

Tr r2




8.

Sphere Volume and Surface Area

The volume Vfor any sphere with radius ris equal to four-thirds times the product of cube of the radius. The area A of any sphere with radius r is equal to 4 the radius. Volume Formula: V

=){

TTS Integrated Training System © Copyright 2011

times the square of

r

Surface Area Formula: A= 4

1-74

and the

I




Worksheet ~

1.

A grave is dug 2m x 1m x 1m deep. The earth removed is piled into a pyramid of circular base 2 m diameter. Find the height of the pyramid (in terms of Ti). Give the answer in m, em and mm.

2.

A right prism has ends 10 em x 10 em and is 50 em long. It is drilled lengthwise with an 8 em drill through its full length. Find

a) b)

~.

3.

Find the surface area of a right cone with base radius 3 inches and perpendicular height of 4 inches. Leave the answer in terms of 1T and include the base area.

4.

Find the ratio of the 'surface area to volume' of spheres of the following diameters: a) b) c)

.----.. ,--...

the remaining volume of the prism material. Give the answer in terms of 7\ and in 3 mm the surface area of the inside of the hole. Give the answer in terms of 7[ and in 2 mm

2m 4m 8m

Module 1.1 Arithmetic Use and/or disclosure is governed by t he stat eme nt




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governed by the statement


Integrated Training System Designed in association with !he club66pro.com question practice aid

Answers 1.

6/rrm, 600/1fcm, 6000/IT""mm

2.

a)

(5

b)

40011 cm ,

a) b) c)

3:1

x 10 6 ) -71: (8 x 105 ) mm 3 2

40,00011 mm

2

3. 4.

.,....-.._ ,..---...,

1112:1 %:1

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Common Conversions .------

Length metric >

imperial >

imperial

1 millimetre [mm]

metric

1 inch [in)

0.0394 in

1 centimetre [em]

10mm

0.3937 in

1 metre [m]

100 em

1.0936 yd

1 kilometre [km]

1000 m

0.6214 mile

2.54 em

1 foot [ft]

12in

0.3048 m

1 yard [yd]

3ft

0.9144 m

1 mile

1760 yd

1.6093 km

1 nautical mile

1.1 5 mile

1.852 km

Area imperial> 2

1 sq inch [in

6.4516 cm

]

1 sq foot [tel 1 sq yd (yd

2

metric

2

1 sq em (cm

2

)

100 mm

0.1550 in2 1.1960 yd 2

144 in2

0.0929 m

2

1 sq m [m2 )

10,000 cm 2

2

0.8361 m

2

1 hectare [ha]

10,000 m

2

9 ft

]

imperial

metric>

2

2

1 acre

4840 yd

4046.9 m

1 sq mile [mile2]

640 acres

2.59 km 2

1 sq km [km

2

2

2.4711 acres 0.3861 mile2

100 ha

)

Volume metric > 1 cu em [cm1

metric

imperial >

imperial 0.0610 in3

1 cu inch [in1

16.387 cm 3 0.0283 m

1,000 cm 3

0.0353 ft3

1 cu foot [ft"]

1 cu metre [m"J

1,000 dm3

1.3080 yd 3

1 fluid ounce [fl oz]

1 litre [I]

1 dm 3

1.76 pt

1 pint [pt]

20fl oz

0.5683 1

1 hectolitre [hi)

100 I

21.997 gal

1 gallon [gal]

8 pt

4.5461 I

USA measure >

28.413 ml

1 fluid ounce

1.0408 UK fl oz

29.574 ml

1 pint (16 fl oz)

0.8327 UK pt

0.4731 I

1 gallon

0.8327 UK gal

3.78541

metric

imperial > imperial

metric>

0.0154 grain

1 milligram [mg] 1 gram [g]

1,000 mg

0.0353 oz

1 kilogram [kg]

1,000 g

2.2046 1b

1 kilogram [kg]

1,000 g

0.068 slug

1.000 kg

(UK)


1 ounce [oz]

437.5 grain 28.35 g

1 pound [lb)

16 oz

0.4536 kg

1 stone

141b

6.3503 kg

1 hundredweight [cwt]

11 21b

50.802 kg 14.6 kg

1 slug

0.9842 long ton

3

metric

Mass

1 tonne [t)

1,728 in

3

1 cu decimetre [dm1

1 long ton (UK)

20 cwt

1.016 t

1 short ton (US)

2,000 lb

0.907 t



Intentionally Blank




1.Easa Part 66 Mod.1 1

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