HSC Year 12 Mathematics Standard 2 Complete Course Notes (2022 - 2024) 1922394882, 9781922394880

Table of contents :
I Topic 1: Algebra
Formulae and Equations
Evaluating expressions by substitution of known pronumerals
Changing the subject of a formula
Solving linear equations
Solving an equation from a worded description
Applications of algebraic expressions
Conversion
Calculating speed, distance, and time
Calculating stopping distance
Calculating blood alcohol content (BAC)
Calculating required medical dosages
Linear Relationships
Straight line graphs
Finding the gradient
Finding the y-intercept
Reviewing the linear function bold0mu mumu y=mx+cy=mx+cy=mx+cy=mx+cy=mx+cy=mx+c
Direct variation relationships
Constant of variation
Types of Relationships
Simultaneous linear equations
Solving simultaneous equations by substitution
Solving simultaneous equations by elimination
Applications of simultaneous equations to practical problems
Non-linear relationships
Modelling non-linear relationships
II Topic 2: Measurement
Applications of Measurement
Practicalities of measurement
Conversions between common units of measurement
Absolute error
Significant figures
Scientific notation
Perimeter, area, and volume
Perimeter and area of common shapes
Pythagoras' theorem
Similar figures
Surface area
Volume
Units of energy and mass
Metric units of mass
Metric units of energy
Working with Time
Latitude and longitude
Times and time differences around the world
Interpreting timetables
Non-right-angled Trigonometry
Trigonometric ratios
Finding unknown sides (lengths)
Finding unknown angles
Finding the area of non-right-angled triangles
Sine rule
Cosine rule
Bearings
Compass bearings
True bearings
Angles of elevation and depression
Radial surveys
Navigational methods used by different cultures
Rates and Ratios
Rates
Ratios
III Topic 3: Financial Mathematics
Money Matters
Interest and depreciation
Simple interest
Applying percentage increases and decreases (GST)
Straight line method of depreciation
Earning and managing money
Calculating rates of pay
Budgeting and household expenses
Household expenses
Budgeting
Investments and Loans
Investments
Compound interest
Comparing growth of simple and compound interest investments
Shares and dividends
Depreciation and loans
Depreciation
Reduced-balance loans
Credit cards
Annuities
Future value annuities
Present value of annuities
IV Topic 4: Statistical Analysis
Data Analysis
Classifying and representing data (grouped and ungrouped)
Data collection methods: sampling
Classifying data
Organising and displaying data
Summary statistics
Features of a population and sample
Measures of central tendency
Measures of spread
Identifying modality
Applications of central tendency and spread
Relative Frequency and Probability
Theoretical probability
Conducting experiments
Tree diagrams and multistage events
Relative frequency
Expected frequencies based on existing probabilities
Bivariate Data Analysis
Scatterplots
Correlation coefficient
Line of best fit
Least-squares line of best fit
Making predictions by interpolation or extrapolation
The Normal Distribution
z-scores
V Topic 5: Networks
Network Concepts
Networks
Drawing network diagrams
Shortest paths
Minimum spanning trees
Critical Path Analysis
Duration and interdependencies of activity
Critical path analysis
Earliest start time
Earliest finishing time
Performing a forward scan
Latest finish and start times
Performing a backward scan
The critical path
Float times
Maximum-flow, minimum-cut theory
Maximum flow
Minimum cut
VI Exam and Revision Tips
Advice for studying
Advice for exams

Citation preview

Year 12 Mathematics Standard 2 Complete Course Notes 2022–2024

Krystelle Vella and Stephanie Azzopardi

Published by InStudent Publishing Pty Ltd 91a Orrong Cres Caulfield North, Victoria, 3161 Phone (03) 9916 7760 www.atarnotes.com As and when required, content updates and amendments will be published at: atarnotes.com/product-updates Copyright © InStudent Publishing Pty Ltd 2022 ABN: 75 624 188 101 All rights reserved. These notes are protected by copyright owned by InStudent Publishing Pty Ltd and you may not reproduce, disseminate, or communicate to the public the whole or a substantial part thereof except as permitted at law or with the prior written consent of InStudent Publishing Pty Ltd. We acknowledge the Wurundjeri People of the Kulin nation as the traditional owners of the land on which this text was created. We pay our respects to Elders past, present, and future and acknowledge that this land we work on is, and always will be, Wurundjeri land. Title: Year 12 Mathematics Standard 2 Complete Course Notes ISBN: 978-1-922394-88-0 Disclaimer No reliance on warranty. These ATAR Notes materials are intended to supplement but are not intended to replace or to be any substitute for your regular school attendance, for referring to prescribed texts or for your own note taking. You are responsible for following the appropriate syllabus, attending school classes and maintaining good study practices. It is your responsibility to evaluate the accuracy of any information, opinions and advice in these materials. Under no circumstance will InStudent Publishing Pty Ltd (“InStudent Publishing”), its officers, agents and employees be liable for any loss or damage caused by your reliance on these materials, including any adverse impact upon your performance or result in any academic subject as a result of your use or reliance on the materials. You accept that all information provided or made available by InStudent Publishing is in the nature of general information and does not constitute advice. It is not guaranteed to be error-free and you should always independently verify any information, including through use of a professional teacher and other reliable resources. To the extent permissible at law InStudent Publishing expressly disclaims all warranties or guarantees of any kind, whether express or implied, including without limitation any warranties concerning the accuracy or content of information provided in these materials or other fitness for p urpose. InStudent Publishing shall not be liable for any direct, indirect, special, incidental, consequential or punitive damages of any kind. You agree to indemnify InStudent Publishing, its officers, agents and employees against any loss whatsoever by using these materials. Trademarks "ATAR" is a registered trademark of the Victorian Tertiary Admissions Centre ("VTAC"); "HSC" is a registered trademark of the Board of Studies Teaching and Educational Standards (“BOSTES”). VTAC and BOSTES have no involvement in or responsibility for any material appearing in these guides. Nor does BOSTES endorse or make any warranties regarding the material in these books or sold by InStudent Publishing Pty Ltd. HSC syllabuses and related content can be accessed from the BOSTES website.

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Preface Welcome to Standard Mathematics 2! My name is Krystelle, and I studied this course (under the old syllabus) over 2016 and 2017 as part of my HSC. Although the new syllabus has been slightly altered to include new concepts (particularly Networks), it remains an interesting, rewarding, and enjoyable subject. I hope this guide will help share my passion and love for Maths with you all, and inspire you to enjoy this course as much as I did. When studying Maths, you must have patience, practice, and persistence. These would have to be my three key tips for completing this course in the best way you can. As a Maths student, it is important to realise that key concepts will not just jump out at you, and often there are times when extra work must be completed to ensure you clearly understand the syllabus requirements. There is nothing wrong with asking for extra help from teachers or peers, or looking at online videos and tutorials about how to successfully change the subject of a formula, or find the area using the trapezoidal rule! Patience is so vital in this subject, and will see you through to the end. Moreover, practice and persistence are key! Unlike some other subjects, typically the only way to successfully study for this course is to complete as many practice questions as you can. But you need to study efficiently and maximise your time. Tackle the concepts you struggle with, and go over worked solutions if you still do not understand anything. By breaking down each question into smaller steps, you’re more likely to feel more confident when tackling the exam. Most importantly, if you don’t understand something, don’t just brush over it. There is no way you can predict what will be in the final exam, and you definitely do not want to be left regretting having skipped over parts of the syllabus. Remaining persistent, not only in Maths but in all your subjects, allows you to thrive as a motivated HSC student! I have also included a list of exam and study tips at the end of this book – please refer to these for further guidance. It is important to also realise that this study guide has been created for both Years 11 & 12. In reality, your final HSC exam is comprised of all the content studied over the duration of your senior years. Therefore, if there is any concept in the course you are unsure of, you can just look over these explanations and sample questions! I wish each of you the best for your HSC studies and life beyond school. Remember to keep a balance between your study, social life, and personal life – you need to maintain your health and wellbeing above all else. Good luck and best wishes! — Krystelle Vella

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Contents I

Topic 1: Algebra

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Formulae and Equations 1.1 Evaluating expressions by substitution of known pronumerals 1.2 Changing the subject of a formula . . . . . . . . . . . . . . . 1.3 Solving linear equations . . . . . . . . . . . . . . . . . . . . 1.4 Solving an equation from a worded description . . . . . . . . 1.5 Applications of algebraic expressions . . . . . . . . . . . . . 1.5.1 Conversion . . . . . . . . . . . . . . . . . . . . . . 1.5.2 Calculating speed, distance, and time . . . . . . . . 1.5.3 Calculating stopping distance . . . . . . . . . . . . . 1.5.4 Calculating blood alcohol content (BAC) . . . . . . . 1.5.5 Calculating required medical dosages . . . . . . . .

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Linear Relationships 2.1 Straight line graphs . . . . . . . . . . . 2.1.1 Finding the gradient . . . . . . . 2.1.2 Finding the y -intercept . . . . . 2.1.3 Reviewing the linear function y = 2.2 Direct variation relationships . . . . . . 2.2.1 Constant of variation . . . . . .

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Types of Relationships 3.1 Simultaneous linear equations . . . . . . . . . . . . . . . . . . . . 3.1.1 Solving simultaneous equations by substitution . . . . . . . . 3.1.2 Solving simultaneous equations by elimination . . . . . . . . 3.1.3 Applications of simultaneous equations to practical problems 3.2 Non-linear relationships . . . . . . . . . . . . . . . . . . . . . . . . 3.2.1 Modelling non-linear relationships . . . . . . . . . . . . . .

II

Topic 2: Measurement

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Applications of Measurement 1.1 Practicalities of measurement . . . . . . . . . . . . . . . . . 1.1.1 Conversions between common units of measurement 1.1.2 Absolute error . . . . . . . . . . . . . . . . . . . . . 1.1.3 Significant figures . . . . . . . . . . . . . . . . . . . 1.1.4 Scientific notation . . . . . . . . . . . . . . . . . . . 1.2 Perimeter, area, and volume . . . . . . . . . . . . . . . . . . 1.2.1 Perimeter and area of common shapes . . . . . . . . 1.2.2 Pythagoras’ theorem . . . . . . . . . . . . . . . . . 1.2.3 Similar figures . . . . . . . . . . . . . . . . . . . . . 1.2.4 Surface area . . . . . . . . . . . . . . . . . . . . . . 1.2.5 Volume . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Units of energy and mass . . . . . . . . . . . . . . . . . . . 1.3.1 Metric units of mass . . . . . . . . . . . . . . . . . . 1.3.2 Metric units of energy . . . . . . . . . . . . . . . . .

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Working with Time 2.1 Latitude and longitude . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Times and time differences around the world . . . . . . . . . . . . . . . . . . . . . 2.1.2 Interpreting timetables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Non-right-angled Trigonometry 3.1 Trigonometric ratios . . . . . . . . . . . . . . . . . . . 3.1.1 Finding unknown sides (lengths) . . . . . . . . 3.1.2 Finding unknown angles . . . . . . . . . . . . 3.2 Finding the area of non-right-angled triangles . . . . . . 3.2.1 Sine rule . . . . . . . . . . . . . . . . . . . . . 3.2.2 Cosine rule . . . . . . . . . . . . . . . . . . . 3.3 Bearings . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Compass bearings . . . . . . . . . . . . . . . 3.3.2 True bearings . . . . . . . . . . . . . . . . . . 3.3.3 Angles of elevation and depression . . . . . . . 3.3.4 Radial surveys . . . . . . . . . . . . . . . . . . 3.3.5 Navigational methods used by different cultures

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Rates and Ratios 4.1 Rates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Topic 3: Financial Mathematics

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Money Matters 1.1 Interest and depreciation . . . . . . . . . . . . . . . . . . . 1.1.1 Simple interest . . . . . . . . . . . . . . . . . . . . 1.1.2 Applying percentage increases and decreases (GST) 1.1.3 Straight line method of depreciation . . . . . . . . . . 1.2 Earning and managing money . . . . . . . . . . . . . . . . . 1.2.1 Calculating rates of pay . . . . . . . . . . . . . . . . 1.3 Budgeting and household expenses . . . . . . . . . . . . . . 1.3.1 Household expenses . . . . . . . . . . . . . . . . . 1.3.2 Budgeting . . . . . . . . . . . . . . . . . . . . . . .

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Annuities 3.1 Future value annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Present value of annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Investments and Loans 2.1 Investments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1.1 Compound interest . . . . . . . . . . . . . . . . . . . . . . . . 2.1.2 Comparing growth of simple and compound interest investments 2.1.3 Shares and dividends . . . . . . . . . . . . . . . . . . . . . . . 2.2 Depreciation and loans . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Depreciation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2.2 Reduced-balance loans . . . . . . . . . . . . . . . . . . . . . . 2.2.3 Credit cards . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Topic 4: Statistical Analysis Data Analysis 1.1 Classifying and representing data (grouped and ungrouped) 1.1.1 Data collection methods: sampling . . . . . . . . . 1.1.2 Classifying data . . . . . . . . . . . . . . . . . . . 1.1.3 Organising and displaying data . . . . . . . . . . . 1.2 Summary statistics . . . . . . . . . . . . . . . . . . . . . 1.2.1 Features of a population and sample . . . . . . . . 1.2.2 Measures of central tendency . . . . . . . . . . . . 1.2.3 Measures of spread . . . . . . . . . . . . . . . . . 1.2.4 Identifying modality . . . . . . . . . . . . . . . . . 1.2.5 Applications of central tendency and spread . . . .

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Relative Frequency and Probability 2.1 Theoretical probability . . . . . . . . . . . . . . . . . . . . . 2.1.1 Conducting experiments . . . . . . . . . . . . . . . 2.1.2 Tree diagrams and multistage events . . . . . . . . . 2.2 Relative frequency . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Expected frequencies based on existing probabilities

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The Normal Distribution 4.1 z-scores . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Bivariate Data Analysis 3.1 Scatterplots . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Correlation coefficient . . . . . . . . . . . . . . . . . . . . 3.3 Line of best fit . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Least-squares line of best fit . . . . . . . . . . . . 3.3.2 Making predictions by interpolation or extrapolation

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Topic 5: Networks

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Network Concepts 1.1 Networks . . . . . . . . . . . . . . 1.1.1 Drawing network diagrams 1.2 Shortest paths . . . . . . . . . . . 1.2.1 Minimum spanning trees .

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Critical Path Analysis 2.1 Duration and interdependencies of activity 2.2 Critical path analysis . . . . . . . . . . . . 2.2.1 Earliest start time . . . . . . . . . 2.2.2 Earliest finishing time . . . . . . . 2.2.3 Performing a forward scan . . . . 2.2.4 Latest finish and start times . . . . 2.2.5 Performing a backward scan . . . 2.2.6 The critical path . . . . . . . . . . 2.2.7 Float times . . . . . . . . . . . . 2.3 Maximum-flow, minimum-cut theory . . . . 2.3.1 Maximum flow . . . . . . . . . . . 2.3.2 Minimum cut . . . . . . . . . . . .

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Exam and Revision Tips

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Part I

Topic 1: Algebra

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1

Formulae and Equations

Section 1

Formulae and Equations Understanding how to interpret and solve algebraic expressions comprises a significant part of this course. The following section covers the content learned in Year 11. This should briefly refresh your memory on the ‘assumed knowledge’ required for your HSC learning.

1.1

Evaluating expressions by substitution of known pronumerals

K EY P OINT :

When completing these questions, particularly when they include various signs or operations (e.g. x 5 ), ensure you carefully complete each step and substitute the correct values into the right spots!

√

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Example 1.1 For a = 6, b = –3, and c = 1, find the value of 3c – a. To solve this, we need to substitute the known pronumerals: 3(1) – 6 = 3 – 6 = –3

Section 1 – Formulae and Equations

Example 1.2 Given that x = a + bt, find x when a = 25, b = 2, and t = 4. Again, substituting the known pronumerals:

x = a + bt x = 25 + (2 × 4) x = 25 + 8 x = 33

1.2

Changing the subject of a formula

This is just a simple matter of rearranging information, as we’ll see in the following examples. Example 1.3

x

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Make x the subject of the formula p =

K EY P OINT :

Again, when changing the subject of a linear formula, it is important to complete each operation one step at a time. These questions often become lengthy and confusing in their working, so it is a great idea to set them out step-by-step, helping both you and the marker to understand your solution! Example 1.4 Make y the subject of the formula k = t – ry . First, we subtract t from both sides (using subtraction, since the t is initially positive, so we need to complete the opposite operation): k – t = –ry Now we have to remove the –r to get y on its own. As the –r and y are multiplied by each other, we use division (the opposite operation) to rearrange the equation: k –t

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–ry

–r –r Cancelling out –r on the right side, we are left with an equation in terms of y : y =

2

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1.3 Solving linear equations

1.3

Solving linear equations

When evaluating equations, both sides must balance at all times! Therefore, if you complete an operation on one side, you must do the same operation on the other side. Example 1.5 Solve the equation 7x + 3 = 24 for x. First, we subtract three from both sides, then divide by 7: 7x = 24 – 3 21 x = 7 x =3 As we did in this question, the most efficient way to solve equations is to add or subtract the number not attached to a pronumeral, before multiplying and dividing.

1.4

Solving an equation from a worded description

Example 1.6 Mary bought 4 pairs of equally priced shoes for a total of $325, which included a delivery fee of $12. What is the cost of each pair of shoes?

$325 = 4x + $12 ...where $325 is the total price, 4x represents 4 pairs of shoes at $x each, and $12 is the delivery fee. Now using our knowledge of equations, we can continue solving this problem: 325 – 12 = 4x 313 =x 4 x = $78.25 Thus, the cost of each pair of shoes is $78.25.

1.5 1.5.1

Applications of algebraic expressions Conversion

Example 1.7 Convert 9 metres/second to kilometres/hour. Answer correct to one decimal place. First, convert the metres/second to metres/hour: 9 × 60 [seconds] × 60 [minutes] = 32, 400 m/hour Net, convert metres to kilometres: 32, 400 m 1, 000 m

= 32.4 km

Therefore, our answer is 9 m/s = 32.4 km/h.

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3

Section 1 – Formulae and Equations

In this question, we are trying to find the price of one pair of shoes. Let the price of one pair of shoes be x. Using the information provided, we can write an equation to help solve this problem:

1.5 Applications of algebraic expressions

1.5.2

Calculating speed, distance, and time

Below are the formulas you need to know: average speed = s=

distance travelled time taken d t

distance = speed × time time =

distance speed

Example 1.8 How far has Jackson travelled if he is travelling at a speed of 75 km/h for 5.5 hours? Answer to nearest kilometre. To find the distance travelled, we use the formula distance = speed × time. Substituting the known values: d = 75 × 5.5 d = 412.5 km Section 1 – Formulae and Equations

Example 1.9 A motorcyclist travels at 107 km/h from Sydney to Melbourne. How long will it take her to complete the trip if it is 890 km. Answer to the nearest minute. distance Here, we need to formula for time, which is: time = . speed t=

890

107 = 8.31776...

Now we need to convert this into hours and minutes. K EY P OINT :

Calculator tip: Press SHIFT

to change your answer into hours/minutes/seconds.

This gives us 8 hours and 19 minutes. Because this was a worded question, our answer should be in words too, so our answer is: it will take the motorcyclist approximately 8 hours and 19 minutes to complete the trip from Sydney to Melbourne.

1.5.3

Calculating stopping distance stopping distance = reaction time distance + braking distance

...where reaction time distance is the distance travelled in the time it takes a driver to react to a situation (i.e. put their foot on the brake), and braking distance is the distance the car travels after the brakes have been applied. The average reaction time is 2.5 seconds, and for a car with good brakes and tyres, travelling in dry conditions on a good road: d = 0.01 v2 ... where v is the speed of the car in km/h, and d is the braking distance in metres. For the same car travelling on a slippery or wet road: d = 0.014 v2

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1.5 Applications of algebraic expressions

Example 1.10 Calculating the total stopping distance of a car travelling at 45 km/h in good conditions, assuming a reaction time of 2.5 seconds. Answer correct to the nearest metre. To solve this question, we must find the reaction time distance and the braking distance. For the reaction time distance, we have to find how far the car travels in 2.5 seconds when its speed is 45 km/h. In one hour, the car travels 45 kilometres, so in one minute, the car travels: 45 km 60 minutes

= 0.75 km/minute = 750 m/minute

In one second, the car travels: 750 m

= 12.5 m/second (this is our reaction time distance!)

60 seconds To calculate the braking distance:

d = 0.01 v2 = 0.01 × 452 Section 1 – Formulae and Equations

= 20.25 m Thus: stopping distance = reaction time distance + braking distance = 31.25 + 20.25 = 51.5 m

≈ 52 m ∴ The stopping distance of this car is 52 m.

1.5.4

Calculating blood alcohol content (BAC)

To calculate the BAC of a male or female: BACmale = BACfemale =

10 N – 7.5 H 6.8 M 10 N – 7.5 H 5.5 M

...where N is the number of standard drinks consumed, H is the number of hours drinking, and M is the person’s weight in kilograms. To calculate the number of hours it would take to reach a BAC of zero after the person has stopped drinking alcohol: BAC time = 0.015 To calculate the number of standard drinks in a container: N = 0.789 × V × A ...where N is the number of standard drinks, V is the volume of the container in litres, and A is the percentage of alcohol (% alc/vol) in the drink. One standard drink contains 10 g of alcohol.

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1.5 Applications of algebraic expressions

Example 1.11 17-year-old Jemma attended a party, where she consumed 5 standard drinks in 2 hours. She weighs 64 kilograms. Calculate her blood alcohol content to 3 decimal places.

Example 1.12 Using the BAC calculated previously, determine the number of hours Jemma must spend not consuming alcohol to reach a BAC of zero. Answer correct to the nearest hour.

Using the formula for the BAC of a female, and substituting our known values: BACfemale = =

time =

10 N – 7.5 H =

5.5 M 10(5) – 7.5(2)

BAC 0.015 0.099

0.015 = 6.6 hours

5.5(64)

≈ 7 hours

= 0.099 K EY P OINT :

When answering BAC questions, it is a good idea to highlight the information in the question prior to substituting the values into the equation. This ensures you only focus on the necessary correct values, and do not input the wrong information into your equations.

Section 1 – Formulae and Equations

However, there are some limitations of methods of estimating BAC. Firstly, each method is an approximation based on average values, and therefore do not apply equally to everyone. Secondly, BAC is dependent on other factors apart from weight, length of time drinking, and the number of standard drinks, including liver function, a person’s mood, and whether the person is a regular drinker.

1.5.5

Calculating required medical dosages

There are three formulas used to calculate the correct dosage of medication for children. These all depend on the age of the child. K EY P OINT :

You do not need to memorise these formulas in an exam situation – they will be provided on the formula sheet. Also, the question will always tell you which formula to apply! Fried’s formula: dosage for children aged 1 - 2 years =

age (in months) × adult dosage 150

Young’s formula: dosage for children aged 1 - 12 years = Clark’s formula: dosage for children =

age (in years) × adult dosage age (in years) + 12

weight (in kg) × adult dosage 70

Example 1.13 A doctor places a patient on a drip for four hours, releasing 2 litres over this time. If 1 ml contains 4 drops, how many drops is the patient receiving per minute? Answer correct to the nearest whole number. There are numerous ways to complete this style of question. If you find this explanation does not work for you, apply your own technique to achieve the same answer. Begin by converting the 2 L over four hours to mL per minute: 2 L × 1000mL = 2000 mL 4 hours × 60 minutes = 240 minutes 2000 mL = 8.33 mL per minute 240 minutes Now we can conver the mL per minute result into drops per minute (since we are told 1 mL contains 4 drops): 8.33 mL × 4 = 33.33 ≈ 33, ∴ the patient is receiving approximately 33 drops per minute. 6

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Linear Relationships

Section 2

Linear Relationships 2.1

Straight line graphs

In this course, we often model practical situations using straight-line graphs. Using two variables, points are plotted on a number plane, and the linear relationship between them produces a straight line in the form y = mx + c. K EY P OINT :

It is important to recognise that the values of the dependent variable depend on the values of the independent variable. The dependent variable is found on the y -axis and the independent variable is found on the x-axis. Example 2.1 George purchases Granny Smith apples at $4.50 per kilogram. The table of values below shows the weight versus cost for the apples. 1

2

5

10

20

Cost ($)

4.50

9

22.50

45

90

Section 2 – Linear Relationships

Weight (kg)

Using the information above, draw the graph of weight versus cost.

For all graphs, make sure you have the correct heading, label the independent variable on the xaxis, and the dependent variable on the y -axis. Also, make sure you include the correct units on each axis (and use accurate intervals, e.g. 5 kg on the x-axis). When drawing straight line graphs, we must recognise the gradient and the vertical intercept. These two figures enable us to draw straight line graphs without using a table of values.

2.1.1

Finding the gradient

The gradient is the slope of the graph. It can be calculated using the formula: gradient = or

rise

run change in y (y2 – y1 ) change in x(x2 – x1 )

For instance, let’s say we wanted to find the gradient of the previous example. First, we select two convenient points on the line – in this case, we will use (20, 90) and (10, 45).

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2.1 Straight line graphs

Using these points, we need to determine the ‘rise’ and ‘run’ of the graph. To rise from 45 to 90 (units on y -axis), we move 45 units vertically. To run from 10 to 20 (units on x-axis), we move 10 units horizontally. Now we can insert this data into our gradient formula: gradient = Section 2 – Linear Relationships

=

rise run 45

10 = 4.5 This graph has a positive gradient of 4.5. We can tell this as the line slopes ‘uphill.’ If the line slopes ‘downhill,’ the graph has a negative gradient.

2.1.2

Finding the y -intercept

The vertical intercept is the point that the straight line graph passes through the y -axis. In our previous example, the line passes through the y -axis at 0 (also known as the origin). To find the y -intercept, we make x = 0.

2.1.3

Reviewing the linear function y = mx + c

Now that we have an understanding of how to find the gradient and vertical intercept of straight-line graphs, we can apply this information to an equation. This equation can therefore be used to draw straight line graphs, connecting every point to form a relationship: y = mx + c ...where x is the independent variable, y is the dependent variable, m is the gradient of the line, and c is the vertical intercept. Example 2.2 Draw the graph of y = –5x + 9. Some questions may ask you to complete a table of values prior to drawing a straight line graph. These are often helpful when completing these questions for the first time, but we’re just going to use the formula! Since the information is given to us in the form y = mx + c, we already know that the gradient is –5, and that the y-intercept occurs at –9, or (0, –9) in co-ordinate form. K EY P OINT :

If the gradient is a whole number (as is –5) we can imagine it as –5 1 . This may be an easier way of remembering how to plot the graph!

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2.2 Direct variation relationships

So, from our y -intercept of –9 we can use the gradient to plot further points. As our gradient is negative, we need to ensure our line slopes downhill. Note that the line passes through the vertical axis at the point (0, –9). Since the gradient is –5, we can start at y = –9 and move 5 spaces upwards, and once to the left. This ensures the line slopes downhill and reflects the negative gradient. Once we have our second point (–1, –4) we simply join the dots and draw the rest of the line!

2.2

Direct variation relationships

In the first example, we determined that the cost of apples varied with the weight. This is known as a direct variation relationship.

K EY P OINT :

Any straight line that passes through the origin indicates a direct variation relationship. In these situations, the y -intercept is equal to 0 (the origin), and therefore the equation is of the form y = mx.

2.2.1

Constant of variation

Direct variation problems always include a constant of variation (k ). This constant is the rate at which the quantities vary (e.g. the hourly rate of pay is the constant of variation when determining a person’s weekly wage). y = kx where y is directly proportional to x y k = to find the value of the constant x In a direct variation straight line graph, the gradient is the constant of variation. Example 2.3 It is known that a is directly proportional to b. When a = 64, b = 32. a) Write an equation in the form a = kb to describe this example. Our equation should look like:

c) Find the value of a if b is 50. All we need to do is make b = 50 and solve for a: a = kb a = 2 × 50

a = kb

a = 100

64 = k × 32 b) Find the value of the constant. To find the constant, we need to divide both sides by 32 to get k on its own: 32k

d) Find the value of b if a is 84. Here, we solve for b instead: a = kb 84 = 2 × b 84 =b 2 b = 42

64

= 32 32 k =2

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Section 2 – Linear Relationships

This relationship is also applicable to real-world situations (e.g. a person’s weekly wage and the number of hours worked; the distance travelled at a constant speed and the time taken, etc.) It expresses that one quantity is a constant (e.g. hourly rate) and is multiplied by the other (e.g. number of hours) to produce a final answer (e.g. weekly wage). Therefore, an increase/decrease in one variable causes a proportional increase/decrease in the other.

Types of Relationships

Section 3

Types of Relationships 3.1

Simultaneous linear equations

Simultaneous equations are used to solve problems with more than one equation and multiple unknowns. There are two methods to solve simultaneous linear equations: substitution and elimination. When solving simultaneous equations, we are looking to find the value of two unknown pronumerals. On a graph, these unknowns will form the point of intersection, in which both lines intersect each other. Example 3.1 Solve the simultaneous equations 3x + y = 1 and 5x + y = 7 by drawing a graph. To do this, we must use our knowledge of graphing straight-line graphs using the formula y = mx + c. We must rearrange each equation to match this formula and therefore help us complete our graph. 3x + y =1 y = – 3x + 1...(1) Section 3 – Types of Relationships

5x + y = 7 y = –5x + 7...(2) Now we can plot these points on the number plane, and find the point of intersection as the solution of these simultaneous equations:

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3.1 Simultaneous linear equations

It is always important to test your answers by substituting the values of the point of intersection! y = – 3x + 1 –8 = – 3(3) + 1 –8 = – 8

(correct answer!)

y = – 5x + 7 –8 = – 5(3) + 7 –8 = – 8

3.1.1

(correct answer!)

Solving simultaneous equations by substitution

Example 3.2 Solve the following simultaneous equations by substitution: x + y = 15...(1) y = x + 8...(2)

Next, we substitute the expression for this subject into the other equation (in this case, substituting the second equation into the first): x + y = 15 x + (x + 8) = 15 Solving this new equation will let us find the value of one pronumeral (x): 2x + 8 = 15 2x = 7 7 x = 2 x = 3.5 Now we can substitute this value into one of the equations to find the value of the second pronumeral (y ): y =x + 8 y =3.5 + 8 y =11.5 Now let’s do a quick check by substituting these values into one of our equations: x + y =15 3.5 + 11.5 =15 (correct answer!)

∴ x = 3.5 and y = 11.5 K EY P OINT :

The only way to be sure you have solved the simultaneous problem correctly is to test the pronumerals in both equations. In an exam situation, doing this quick test of equations could save you marks!

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Section 3 – Types of Relationships

Usually, the first thing we need to do is make one pronumeral the subject in one of the equations. In this case, the second equation has y as the subject of the formula – therefore, this step is already completed.

3.1 Simultaneous linear equations

3.1.2

Solving simultaneous equations by elimination

Example 3.3 Solve the following simultaneous equations by elimination: 2x – y =1 ...(1) x + y =14 ...(2) When solving simultaneous equations via elimination, firstly, we have to make sure that the two coefficients of one pronumeral are the same. This may require multiplying or dividing one or both equations by a number. In this example, we will make the y have the same coefficient (that is, the same sign, + or –, before it). To do so, we will multiply each part of the first equation by –1. This will make the negative y positive, and therefore match the coefficient in equation 2. (2x × –1) – (y × –1) = (1 × –1) –2x + y = –1

Section 3 – Types of Relationships

Now we have to eliminate one pronumeral by adding or subtracting two equations. In this case, by subtracting the first equation from the second, we can eliminate the pronumeral y . Splitting the equations up into sections can make this easier: (a)

(b)

(c)

–2x

+y

= –1

x

+y

= 14

(a) – 2x – x = –3x (b) y – y = 0 (eliminates the pronumeral) (c) – 1 – 14 = –15 Therefore, our new equation now looks like this: –3x = –15 We can use this to find the vale of the other pronumeral (x): x =

–15

–3 x =5 Substituting this value into one of the equations will let us find the value of the remaining pronumeral: 5 + y = 14 y =9

∴ x = 5, y = 9

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3.1 Simultaneous linear equations

3.1.3

Applications of simultaneous equations to practical problems

Determining the break-even point We often use simultaneous equations to determine the break-even point for financial problems. In such cases, we determine the point when costs equal income. Therefore, after this point, profit will be earned. Profit = Income – Costs To determine the break-even point, we require two linear equations: • Income: I = mx where m is the selling price per item and x is the number of items sold. • Cost: C = mx + b where m is the cost price per item manufactured, x is the number of items sold, and b is the fixed costs of production. The point at which these two lines intersect is called the break-even point. K EY P OINT :

If a business does not reach the break-even point, they will make a loss. If a business exceeds the break-even point, they will make a profit. Example 3.4 Archie organises a fundraiser for Beyond Blue. The fixed cost is $2,000, as well as an additional $20 per person who attends. Each person attending pays $60 per ticket. Draw a graph to represent these equations. a) What is the break-even point? What does this mean?

The equation for the cost of the fundraiser is therefore: C = 20x + 200 The income received from the fundraiser is $60 per person. This is the gradient. If no-one attends the fundraiser, the income will be zero. Therefore, the y -intercept is 0. If we refer to our learning of direct variation, we will remember that “one quantity is a constant (in this case, the payment of $60/pp attending), and is multiplied by the other (again in this case, the number of people attending). Direct variation equations are in the form y = mx.” The equation for the income (payment per person attending) is therefore I = 60x. So we can plot these equations on our graph to answer the question:

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13

Section 3 – Types of Relationships

Before attempting to draw our graph, we need to find the straight-line equation of both sets of information. The cost of the fundraiser for Archie will form the first equation. If we observe carefully, we will note that it will cost him $2,000 regardless of how many people are attending. This is the y-intercept. We should then recognise that it costs an additional $20 per person who attends. This is the gradient.

3.2 Non-linear relationships

K EY P OINT :

Ensure you use an accurate graph to represent the data. This requires you to determine an appropriate scale, and to label your graph appropriately. The break-even point occurs when there are 50 people attending. When 50 people attend, the cost is $3,000 and the income is $3,000. Therefore, if more than 50 people attend, Archie will make a profit. If less than 50 people attend, Archie will make a loss. K EY P OINT :

It is always important to test the accuracy of your break-even point. You can do this using the equations of the graphs (i.e. like simultaneous equations): C = (20 × 50) + 2000 C = 3000 I = 60 × 50 I = 3000 Example 3.5 b) How much profit is made when 80 people attend the fundraiser? Section 3 – Types of Relationships

To determine the amount of profit made when 80 people attend, we can either use the graph or the equations. As this graph does not provide information on the income when 80 people attend, we will have to use the equations: First, find the cost of 80 people attending: C = (20 × 80) + 2000 C = 3600 Now find the income when 80 people attend: I = 60 × 80 I = 4800 Calculate the profit: Profit = Income – Costs = 4800 – 3600 = 1200

∴ Archie raises $1200 when 80 people attend the fundraiser.

3.2 3.2.1

Non-linear relationships Modelling non-linear relationships

Exponential functions An exponential function is a non-linear curve that has an equation with x as the power (e.g. 2x ). The general rules for this function are y = ax and y = a–x (a > 0).

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3.2 Non-linear relationships

K EY P OINT :

In this topic, there are various types of functions to remember. If you are more of a visual learner, it may be helpful to memorise the shape of the graph. You can do this by drawing each on labelled flashcards and testing your understanding of each! Exponential functions are often used to solve growth or decay problems. Example 3.6 The growth of bacteria is estimated according to the formula b = 10(2.2)t , where b is the number of bacteria after t hours. a) Draw the graph of this equation. Construct a table of values to assist you, using ordered pairs of 0, 2, 4... up to 14 for values of t. Calculate the number of bacteria to the nearest whole number. To construct a table of values, we must recognise the x (independent) and y (dependent) values in the equation. In this question, we are trying to find the growth of bacteria, and this is reliant on the number of hours. Therefore, our b is the dependent variable, and our t is the independent variable. 0

2

4

6

8

10

12

14

b

10

48

234

1,134

5,488

26,660

128,550

622,182

Now we can graph this data. The intervals between the values for bacteria must be chosen appropriately. However, markers will recognise that you may have to estimate where the point shall go. Try to plot each point as accurately as possible.

The initial number of bacteria is the value of b when t = 0. Therefore, the initial number of bacteria is 10. Example 3.7 b) What is the number of bacteria after 7 hours? It is often difficult to interpret this information from the graph. To calculate this, use the exponential formula provided and input 7 as the number of hours: b = 10(2.2)7 = 2494

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Section 3 – Types of Relationships

t

3.2 Non-linear relationships

Example 3.8 c) Estimate the time taken for the number of bacteria to reach 350,000. To answer this question, we can use the graph. Observe the point where 350,000 bacteria have been grown, and line this up with the number of hours on the x-axis. This is approximately 13 hours. Quadratic functions A quadratic function is a non-linear curve with an equation that has an x 2 . The general rule for this function is y = ax 2 + bx + c where a, b, and c are numbers. To distinguish a quadratic function from the other non-linear relationships, we recognise the shape of the graph as a parabola. A parabola (as displayed on the right) always has a turning point and an axis of symmetry. These key aspects allow you to recognise a quadratic equation, thus helping you remember it more clearly! Example 3.9 Complete a table of values from –1 to 5 and graph the quadratic function y = x 2 – 4x. Section 3 – Types of Relationships

x

–1

0

1

2

3

4

5

y

5

0

–3

–4

–3

0

5

The features of parabolas include the following: • Turning points: the highest or lowest point on a graph where the graph changes direction. This point represents the minimum or maximum value of a parabola. In the example above, the turning point occurs at (2, –4). • Intercepts: – x-intercept: parabolas can have zero, one, or two x-intercepts. Not all parabolas have an x-intercept (as the graph can turn above or below the x-axis). – y -intercept: all parabolas have one y -intercept.

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3.2 Non-linear relationships

K EY P OINT :

,

A parabola that looks like a happy face ( ) has a positive gradient. A parabola that looks like a sad face ( ) has a negative gradient.

/

It’s important to recognise the range of values for x and y for which the quadratic model makes sense. Example 3.10 A rectangular deck is constructed so that the length is 6 metres less than the width. Give an example of a length and width that would be possible for this deck. Here, we must determine values that fit the required measurements, but actually make sense! For example, it is not possible for the width to be 6 metres, as this would make the length 0 metres. In a similar way, the width cannot be less than 6 metres, as the length would be a negative number, which is also impossible. However, if the width was 8 metres, the length could be 2 metres. This is possible, and makes sense in a practical context. Therefore, when solving these questions, we must consider the range of values for which the equation actually makes sense. While this may seem simple, students will often work through these questions too quickly, and make easily avoidable mistakes. Cubic functions

Section 3 – Types of Relationships

A cubic function is a non-linear curve with an equation that has an x 3 . The general rule for these functions is y = ax 3 + c where a and c are numbers. Example 3.11 A bee population can be predicted using the formula N = 500t 3 , where N is the number of bees and t is the time in days. a) Draw the graph of N = 500t 3 . Construct a table of values to assist you. To construct a table of values, we must recognise the x (independent) and y (dependent) values in the equation. In this question, we are trying to find the number of bees in the population, and this is reliant upon the number of days. Therefore, N is the dependent variable and t is the independent variable. t

2

4

6

8

10

N

4,000

32,000

108,000

256,000

500,000

Now we can use this data to construct a graph.

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3.2 Non-linear relationships

Example 3.12 b) How many bees were present after one week? Use the graph to estimate. To estimate the number of bees after one week (i.e. 7 days), we need to determine where the non-linear curve passes 7 days, and match this to the number of bees it corresponds with. Drawing a line up from the x-axis at 7 days, until the line touches the curve, and then across to the y -axis will allow you to get an accurate approximation of the number of bees after 7 days (as shown above). Thus, your answer should be approximately 175,000 bees. K EY P OINT :

When a question asks you to use a graph to estimate, you must show your working! The marker wants to see that you understand how to interpret the graph correctly, and therefore will reward you with marks for appropriate working where need be. It is also important that you complete such working in lead pencil, just in case you have to erase it and start again. You can also test these answers using the formula above to ensure you did not interpret the graph incorrectly. Inverse variation We have previously explored direct variation. This concept involves the use of a constant (k ), and occurs when one variable depends directly on another. Inverse variation also uses a constant, but uses a reciprocal function that takes the form y = kx . Section 3 – Types of Relationships

In inverse variation questions, one variable increases while the other decreases. This is used in many practical problems (i.e. the amount of pizza received when sharing a pizza between increasing numbers of people – the smaller your slice of pizza gets, the larger the amount of people!). We can recognise inverse variation problems in the form of hyperbolic functions. This is where x is the denominator of the fraction (bottom), with the constant k as the numerator (top). Example 3.13 The cost per person (c) to hire a holiday house varies inversely with the number of people staying (n). If the number of people attending is 10, the cost per person is $85. a) What is the cost per person if there are 5 people attending? To solve this question, we need to substitute our known values into the formula to find the constant: k c= n k 85 = 10 k = 850 Therefore, if 5 people are attending, c = 850 5 = 170, the cost per person is $170. Example 3.14 b) How many people are required to reduce the cost to $60 per person? 850 60 = x x = 14.166...

∴ 14 people would be required. It is important to recognise what the question is asking you to find. This is why we state that inverse variations use reciprocals of the form y = kx , as we are constantly rearranging the formula to suit what we are trying to solve. 18

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Part II

Topic 2: Measurement

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19

Applications of Measurement

Section 1

Applications of Measurement 1.1

Practicalities of measurement

1.1.1

Conversions between common units of measurement

While this part of the syllabus is seemingly simple, it is crucial that you understand and apply these conversions properly. A great way to ensure this information is properly understood is to practise various questions until you can do them without looking at a chart. Be aware that questions may not simply ask you to convert between units close together (e.g. centimetres to metres) but further apart (e.g. centimetres to kilometres). In this case, you have to convert the unit several times, so ensure you take your time to do this carefully. Metric units of measurement If you are getting smaller, you multiply. If you are getting bigger, you divide. Length: Section 1 – Applications of Measurement

Capacity:

Area:

Volume:

To properly understand this syllabus requirement, we must be able to calculate conversions between all common units of measurement. 20

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1.1 Practicalities of measurement

Example 1.1 Steven measured his backyard swimming pool with a 30cm ruler. He gained measurements of 247cm for the width and 650cm for the length. What are the measurements of the pool in metres? Now, to answer this question, we are asked to convert cm into m – when looking at the conversion charts, this is only asking us to complete one conversion. Which conversion will we use?

width = length =

247 cm 100 650 cm 100

= 2.47 m = 6.50 m

∴ The measurements of Steven’s pool are 2.47 m width and 6.50 m length. K EY P OINT :

If you’re confused by the conversions, try to memorise how the arrows on the chart go, and keep this handy in exam situations. Then, to make sure you’re always getting the questions right, draw the diagram before you complete the question, and then make the conversion from there. Remember, these take plenty of practice before being able to do them without referencing any notes! Also, when answering these worded questions, ensure that you clearly write the answer at the end by circling or highlighting it. Do not leave the marker to infer the answer through your working out! Let’s try two conversions of the same metric unit within the same question. 0.06 kL = _____ mL Observing the conversion chart for capacity, we realise that converting millilitres to kilolitres will require two steps. Let’s do these one at a time. 1. Convert kL to L (multiply by 1000): 0.06 kL × 1000 = 60 L 2. Convert L to mL (multiply by 1000): 60 L × 1000 = 60000 mL

1.1.2

Absolute error

To calculate absolute error, we use the following formula: absolute error = ±

1 2

× precision

This formula indicates that the absolute error (the difference between the measured and actual value) can be determined by either adding or subtracting ½ the limit of reading – that is, the precision of the measuring instrument used. Copyright © 2021 InStudent Publishing Pty. Ltd.

21

Section 1 – Applications of Measurement

Following the arrows, we are converting from centimetres to metres, so we will divide by 100.

1.1 Practicalities of measurement

Limits of accuracy The limits of accuracy in measurements account for the error that can occur when conducting measurements, namely due to the instrument/tool used. We state that any particular measurement can have an upper and lower bound, found using the formulas: upper bound = measurement + absolute error lower bound = measurement – absolute error As humans, we are limited in our ability to accurately measure lengths due to device imprecision (e.g. inaccurate rulers) as well as parallax error (eyesight). For these reasons, we apply absolute error and percentage error to recognise the possibility that our measurements are slightly inaccurate. Think of some examples where measurements could be inaccurate. What is the smallest unit on the ruler you use? Compare a measurement you determine with the same measurement your friend observes. It is the same or different? Percentage error To calculate percentage error, we use the following formula: percentage error =

absolute error

Section 1 – Applications of Measurement

measurement

× 100%

Example 1.2 The length of a football field is measured to be 58.6 m in length and 23.4 m in width. a) Calculate the absolute error in this question, and state the corresponding limits of accuracy for both the length and width of the football field. b) Calculate the percentage error in the length and width of the football field, correct to two decimal places. To calculate the absolute error in the measurements of the football field for part a), we first must determine the accuracy of the measuring instrument used. As represented by the question, the measurements are rounded to the nearest 0.1 m, so therefore the precision is 0.1. Therefore, to determine the absolute error of the length measurement, we calculate: 1 2

× 0.1 = 0.05

Now we can determine the limits of accuracy for each measurement: • Length: – Upper bound: – Lower bound: • Width: – Upper bound: – Lower bound:

58.6 + 0.05 = 58.65 m 58.6 – 0.05 = 58.55 m 23.4 + 0.05 = 23.45 m 23.4 – 0.05 = 23.35 m

To calculate the percentage error of the length and width for part b): 0.05

× 100% 58.6 = 0.09% 0.05 width percentage error = × 100% 23.4 = 0.21%

length percentage error =

22

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1.2 Perimeter, area, and volume

1.1.3

Significant figures

Significant figures are the digits that contribute to the meaning and accuracy of a number. K EY P OINT :

When rounding to a specified number of significant figures, remember that zeros at the start of a number or at the end of a number are placeholders – they indicate the size of the number. For example: 37.30 has three significant figures (3, 7, and 3), whereas 0.00032 has two significant figures (3 and 2) but the zeros must remain there as they are placeholders that indicate the size of the number. Example 1.3 Round 234,789 to two significant figures. To round this number, we must observe the significant figures already present, and recognise where we must round. In this number, we only need the first two digits to be significant. We can therefore round the following digits to zeros, therefore acting as placeholders to indicate the size of the number. The 4 (indicating 4,000) was the deciding number as to whether we round up to 24,000 or down to 23,000. If the deciding number is 5 or larger, we round the required figure up. If the deciding number is 4 or smaller, then we round down. Thus, our answer is 230,000

Again, we must remember the value of the zeros in this number. To round to one significant figure, we can only look at the ‘6’ and beyond, as these contribute to the accuracy of the number. In this example, the 6 is the first significant figure, with the 9 being the deciding figure. As the 9 is larger than 5, we must round the 6 up, therefore correctly rounding to 1 significant figure gives us 0.007. We do not have to include any zeros after the 7 as these values do not contribute to the size of the number, and therefore are insignificant to this question.

1.1.4

Scientific notation

We use scientific notation to write very large or very small numbers in a more convenient and understandable way. It involves us determining how many times a number has been multiplied by 10, and expressing this as its power. To express a number in scientific notation (using 7,400,000,000 as an example): 1. Put a decimal point in between the first two numbers – i.e. 7.4. These numbers contribute to the meaning and accuracy of the large number. 2. Next, count the number of times the value has been multiplied by 10 – this will become the power of ten. We can do this by counting the placeholders from the decimal point. Here, we have 9 places after the decimal point. Therefore, this number in scientific notation would be 7.4 × 109 .

K EY P OINT :

Large numbers (i.e. anything above zero) have a positive power. Small numbers (i.e. anything below zero) have a negative power.

1.2

Perimeter, area, and volume

Let’s start with some quick definitions: • Perimeter: the distance around the edges of the shape. • Area: the amount of space inside the shape. • Volume: the amount of space inside a three-dimensional object (i.e. how much liquid can fit inside it)

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Section 1 – Applications of Measurement

Example 1.4 Round 0.006928 to 1 significant figure.

1.2 Perimeter, area, and volume

1.2.1

Perimeter and area of common shapes

When learning this content, it is often helpful to practise a range of questions to ‘memorise’ the formulae for each. In this way, you will not have to keep referring to a formula list to solve the questions – this wastes precious time in exams! Rectangle • Area: A = l × h where l is the length and h is the height of the rectangle. • Perimeter: add all four sides together (or add 2 × length and 2 × height) together Triangle • Area: A =

b×h

where b is the base and h 2 is the perpendicular height of the triangle. • Perimeter: add all three sides together (this may require a further understanding of trigonometric ratios with right-angled and nonright-angled triangles) Section 1 – Applications of Measurement

Parallelogram (any four-sided figure with opposite sides parallel e.g. rectangle, rhombus, square, etc.) • Area: A = b × h where b is the base and h is the height of the parallelogram. • Perimeter: add all four sides together. Trapezium (a quadrilateral with one pair of parallel sides) 1 • Area: A = h(a + b) where h is the perpen2 dicular height of the trapezium and a and b are the two parallel sides. • Perimeter: add all four sides together – be careful to add the external sides and not the height used to calculate area. K EY P OINT :

Often, exam-style questions will have the trapezium on a different side to confuse you! Make sure you recognise a and b as the two parallel sides, as this will ensure that the calculation of area is always correct. Circle • Area: A = π r 2 where r is the radius. • Perimeter (or circumference): 2π r or π d

24

Sectors of circles A sector is a part of a circle (think of a piece of pizza!). The area of a sector is calculated using: A=

θ 360

× πr 2.

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1.2 Perimeter, area, and volume

K EY P OINT :

It is important to recognise that there are various types of sectors which look relatively different. However, by applying the same principle, you will always gain the correct answer! To determine the perimeter, we calculate the arc length of a sector. The arc is the curved edge connecting the two radii. When calculating the perimeter of a sector, we must find the arc length as well as the two radii to obtain the correct answer. To do so, we use the formula r is the radius of the circle.

θ 360

× 2π r where θ is the centre angle and

Example 1.5 Calculate the perimeter of this sector.

Section 1 – Applications of Measurement

Always make sure you are calculating the perimeter of the sector and not the entire circle! To find the perimeter, we first need to find the arc length. 24

× 2 × π × 12 360 = 5.027... cm

arc length =

Now, we need to add the two radii to complete the perimeter. perimeter = arc length + radius + radius = 5.027... + 12 + 12 = 29.03 cm (correct to two decimal places) Compound shapes These are shapes that are made up of two or more regular shapes. • Area: to calculate the area of composite shapes, we will either have to add or subtract various areas. • Perimeter: again, this requires us to use measurements of length given to find the total perimeter. You may have to find some of these yourself, however the steps of the question will help you to do this. For example, to find the area of this composite shape, we need to add the various shapes together. This can be done by splitting the composite shape into recognisable areas, and then adding the areas of each shape together.

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1.2 Perimeter, area, and volume

Example 1.6 Determine the area of the composite shape. The shape is made up of the shaded area only.

We must subtract the 4 circles from the square. Using our previous knowledge of how to derive the area of circles and squares, an appropriate solution to this question would look like this: total square = 16 × 16 = 256 cm2 Section 1 – Applications of Measurement

each circle = π × 42 = 50.2655 cm2

∴ area = 256 – (4 × 50.2655) = 54.94 cm2 Annulas This is where we calculate the area between two circles with the same centre, but different radii. The area is calculated by A = π (R 2 – r 2 ).

1.2.2

Pythagoras’ theorem

Pythagoras’ theorem is a method used to determine the side lengths of right-angled triangles. It can be used to solve various problems involving right-angled triangles, such as finding the perimeter and area of such shapes. Pythagoras’ theorem: c 2 = a2 + b2

The side labelled c is always the longest side (hypotenuse) and is found opposite the right angle – the decision where to place a and b is entirely up to you. 26

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1.2 Perimeter, area, and volume

Example 1.7 Determine the perimeter of the triangle below.

Using Pythagoras’ theorem, we can determine the missing side denoted by x. c 2 = a2 + b 2 c 2 = 25

√

c=

25

c = 5 cm

perimeter = a + b + c =3+4+5 = 12 cm K EY P OINT :

It is important that you read all parts of the question to ensure you get full marks (i.e. not just finding the length of x, but calculating the entire perimeter of the triangle). Examiners often use these questions to trick students!

1.2.3

Similar figures

Similar figures are figures with exactly the same shape but are different sizes. By using a scale factor, we can use the measurements of one figure to determine unknown lengths in similar figures. Example 1.8 Find the unknown length (x) using a scale factor.

These questions often attempt to trick students by providing shapes that appear to be different when they are actually similar (as above). If such questions confuse you, it is a great exam strategy to label the sides as a, b, and c, matching the sides of both shapes correctly.

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27

Section 1 – Applications of Measurement

After finding the missing side x to equal 5 cm, we must add all the sides together to determine the perimeter.

1.2 Perimeter, area, and volume

To find the unknown length (x) using a scale factor: 13 cm

=

x cm

12 cm 24 cm 12 cm × x cm = 13 cm × 24 cm 312 x = 12 x = 26 cm Alternatively, if we know the scale factor (which in this case is 1.08333...): 13 cm 12 cm

=

1.08333... =

x cm 24 cm x cm

24 cm x = 26 cm

K EY P OINT :

Section 1 – Applications of Measurement

Always place the unknown length you are trying to find as the numerator of the equation. Just remember: what we don’t know goes on top because it’s our top priority!

1.2.4

Surface area

Surface area is the total area of the external faces of a solid. Right prisms Right prisms include rectangular prisms, cubes and triangular prisms. As there is no set formula to determine the surface area of right prisms, you must calculate the area of each face and then add them together. You may find it helpful to draw the net of the prism to ensure you do not miss any faces. Example 1.9 Determine the surface area of the rectangular prism.

To determine the surface area of this rectangular prism, we must recognise the 6 faces of this solid, and calculate the area of each. It is clear that each face has an exact replica on the opposite side, so we only have to find 3 faces, and multiply each by 2 to save us time! surface area = 2(2 × 4) + 2(2 × 7) + 2(7 × 4) = 16 + 28 + 56 = 100 m2

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1.2 Perimeter, area, and volume

Cylinders and spheres Solid

Surface area formula

Closed cylinder

SA = 2π r 2 + 2π rh area of top: π r 2 area of bottom: π r 2 area of curved surface: 2π rh

Open cylinder (curved surface)

SA = 2π rh

Sphere

SA = 4π r 2

Diagram

Similar to prisms, we must ensure we include all faces when calculating the surface area of pyramids. Pyramids include square pyramids, rectangular pyramids and triangular pyramids. The name of the pyramid refers to the shape of the base.

Square pyramid (square base)

Triangular pyramid (triangle base)

Example 1.10 Find the surface area of the square pyramid that has a side length of 8 cm with triangles of height 10 cm. Note that perpendicular height refers to the length from the apex (point) of the triangle straight down to the base. Don’t get this confused with the side height (i.e. height of the triangles). As we haven’t been given a diagram, we should draw one to accurately include all faces and determine the surface area correctly. area of square base = 8 cm × 8 cm = 64 cm2 1 area of each triangle = × 8 cm × 10 cm = 40 cm2 2 total area of four triangles = 4 × 40 = 160 cm2 total surface area of pyramid = 64 cm2 + 160 cm2 = 224 cm2

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29

Section 1 – Applications of Measurement

Pyramids

1.2 Perimeter, area, and volume

Composite solids Similar to determining the area of compound shapes, calculating the surface area of composite solids requires us to determine the area of all faces and add these together. Calculating perimeters and areas of irregular shapes For this section, it is very important to label all parts of the diagram prior to beginning calculations. The question may require you to use previous knowledge, such as Pythagoras’ theorem or the area of trapeziums, in order to correctly answer the question. Example 1.11 Determine the area and perimeter of the irregularly shaped block of land below. Answer correct to two decimal places where necessary.

Section 1 – Applications of Measurement

Notice that various side lengths of the shape are missing. To find the area of this irregular block, we only need the sides given, however this may differ across other questions. The best way to calculate the total area of this shape is to split it into recognisable shapes, and add these all together at the end.

triangle AGB = = = triangle CBG = = = triangle CHD = = =

1 2 1

× AG × 28 × 15

2 210 m2 1 × CG 2 1 × 19 × 28 2 266 m2 1 × CH × HD 2 1 × 7 × 23 2 80.5 m2

trapezium EFHD = = = triangle AFE = = =

1 2 1

× FH(DH + EF )

× 17(12 + 23) 2 297.5 m2 1 × AF × FE 2 1 × 10 × 12 2 60 m2

∴ total area = 914 m2

To calculate the perimeter, we will need to use Pythagoras’ theorem. Before we start, it is best to check that each shape will allow you to find any missing side lengths. Here, the trapezium (DEFH) has been split into a triangle and rectangle, so we need to use Pythagoras’ theorem to find the missing length DE. 30

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1.2 Perimeter, area, and volume

length AB = AG2 + BG2 = AB 2 = 152 + 282 = 1009 = 31.765... m length BC = CG2 + BG2 = BC 2 = 192 + 282 = 1145 = 33.848... m length CD = CH 2 + DH 2 = CD 2 = 72 + 232 = 578 = 24.042... m length DE = a2 + b2 = DE 2 = 112 + 172 = 410 = 20.248... m length AE = AF 2 + EF 2 = AE 2 = 102 + 122 = 244 Section 1 – Applications of Measurement

= 15.620... m

∴ total perimeter = 125.52 cm

Trapezoidal rule The trapezoidal rule is used to estimate an area bounded by a curved edge. The area is approximated by replacing the curved edge with a straight line, thus creating a trapezium. To use the trapezoidal rule for a single application, we use the formula: h A ≈ (df + dl ) 2 ...where df (first length) and dl (last length) are lengths of the parallel sides of the trapezium and h is the perpendicular distance between them.

1.2.5

Volume

K EY P OINT :

Volume is measured in units3 . Ensure you always include the units when answering these question to achieve full marks! Solid

Volume formula

Cube

side length3 , or V = s3

Rectangular prism

length × breadth × height, or V = l × b × h

Closed cylinder

π × radius2 × height, or V = π r 2 h 1

Pyramid

Cone

3

1 3

× area of base × perpendicular height, or V =

1 3

× Abase × h

× area of circular base × perpendicular height or V =

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1 3

× πr 2 × h

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1.2 Perimeter, area, and volume

K EY P OINT :

As there are different types of pyramids, the calculation of the base will differ according to the base shape. Be careful to calculate this area properly to ensure your volume calculations are correct! Volume of composite solids Again, much like calculating the area of composite shapes, calculating the volume of composite solids just involves breaking things up into two or more regular solids and adding them together. K EY P OINT :

To convert between units of volume and capacity: 1 cm3 =1 mL 1 m3 =1000 L = 1 kL Using the trapezoidal rule to find volume

Section 1 – Applications of Measurement

For these questions (which typically involve finding the volume of dams, reservoirs, and pools), we apply a similar trapezoidal rule, however remembering to include the appropriate measurements to calculate the volume. Example 1.12 a) Use the trapezoidal rule to find the area of the lake shown below.

First, we need to calculate the whole area (of both the lake and the land): A = 60 × 50 = 3000 m2 Now we will use our understanding of the trapezoidal rule to find the area of the land (green section). To do so, we need to calculate the area of each section as labelled below:

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1.3 Units of energy and mass

h

area 1 ≈

area 3 ≈

(df + dl )

(df + dl ) 2 30 ≈ (42 + 3) 2 ≈ 675 m2

2 30

≈

(8 + 12) 2 ≈ 300 m2

area 2 ≈

≈

h

h

area 4 ≈

(df + dl )

h

(df + dl ) 2 30 ≈ (3 + 3) 2 ≈ 90 m2

2 30

(12 + 47) 2 ≈ 855 m2

total area of land ≈ 300 + 885 + 675 + 90 = 1950 m2 area of lake ≈ area of land and lake – area of land

≈ 3000 – 1950 ≈ 1050 m2 Section 1 – Applications of Measurement

Example 1.13 b) If the average depth of the lake is 27 m, calculate an estimate for the volume of water in the lake. Now, using our area finding, we can calculate the volume and capacity of the lake. V = Ah = 1050 m2 × 27 m = 28350 m3 or 20,350 kL

1.3 1.3.1

Units of energy and mass Metric units of mass

Example 1.14 Convert the following masses: • 2.4 kg = ______ grams • 120,000 mg = ______ kilograms • 2 t = ______ milligrams

Solutions: • 2.4 × 1000 = 2400 grams • 120000 ÷ 1000 ÷ 1000 = 0.12 kilograms • 2 × 1000 × 1000 × 1000 = 2, 000, 000, 000 milligrams Copyright © 2021 InStudent Publishing Pty. Ltd.

33

1.3 Units of energy and mass

Example 1.15 Susie is asked to add the following masses together and give her final answer in kilograms: 5.7 kg, 234 g, 1.2 kg, 0.5 t and 740 g. She struggles to correctly convert the measurements. Can you help her calculate the correct answer? Examiners will often use these questions to confuse students – make sure you read these carefully and circle or highlight important details before you start your calculations! First, we convert each of the measurements to kilograms, as this is what the question is asking us to do: 234

= 0.234 kg 1000 0.5 t = 0.5 × 1000 = 500 kg 740 740 g = = 0.740 kg 1000 234 g =

Now, add all converted measurements together and calculate the final answer: 5.7 + 0.234 + 1.2 + 500 + 0.740 = 507.87 kg (to two decimal places) Section 1 – Applications of Measurement

1.3.2

Metric units of energy

Energy is measured in kilojoules (kJ) or calories (cal). The joule (J) is the international standard (SI) unit of energy. It is a very small unit, so the kilojoule is often used when calculating energy conversions. In nearly every country today, the kilojoule is used as the official unit for the energy value of food or drink. Calories are also used, often in a bracket after the kilojoule amount. When describing the energy content of food, the calorie is too small – hence another energy unit called the Calorie (capital C) represents the energy contained in the food humans eat. 1 kJ = 1000 J 1 cal = 4.184 J 1 Cal = 1000 cal To convert Calories to kilojoules: 1 Cal = 1 kcal = 1000 cal = 4184 J = 4.184 kJ Remember: when completing any conversions between metric units of energy, make sure you use the correct conversion factor to ensure your answer is accurate. Calculating energy intake from foods When we consume food, our body is provided with the energy it requires to function. According to Nutrition Australia, most of this energy can be obtained from carbohydrates, protein, and fat dietary fibre. The average amount of energy contained in each of these nutrients is:

34

Carbohydrates

17 kJ/gram (1 gram = 17 kJ)

Protein

17 kJ/gram (1 gram = 17 kJ)

Fat

38 kJ/gram (1 gram = 138 kJ)

Dietary fibre

8 kJ/gram (1 gram = 8 kJ)

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1.3 Units of energy and mass

Example 1.16 In one 85 g serve of chicken-flavoured noodles, there is 8.7 g of protein, 16.6 g of fat, 52.7 g of carbohydrate and 2.5 g of dietary fibre. Calculate the total energy intake from consuming one serve of these noodles (answering to the nearest kilojoule). For each nutrient in the noodles, we need to convert the grams into kilojoules, to obtain the total energy intake. We will do this using the conversion table above. protein = 8.7 g × 17 kJ = 147.9 kJ fat = 16.6 g × 38 kJ = 630.8 kJ carbohydrate = 52.7 g × 17 kJ = 895.9 kJ dietary fibre = 2.5 g × 8 kJ = 20 kJ Now total these kilojoule amounts to find the total energy intake: 147.9 + 630.8 + 895.9 + 20 = 1695 kJ To apply our knowledge of conversion to this question, we are now going to convert the amount of kilojoules in the noodles into calories. According to our conversion diagrams, 1 calorie contains 4.184 kJ. Let’s apply this conversion factor to calculate the total energy intake in calories. 4.184 kJ

= 405 cal (to the nearest calorie)

K EY P OINT :

Be aware of what the rounding questions ask you to do. By making note of whether it is asking for decimal places, significant figures, or the nearest whole number, you will maximise the amount of marks you receive for your answer. Remember, every mark counts! Calculating energy expenditure Our energy expenditure refers to the amount of energy our body uses to perform basic physiological functions (e.g. breathing, maintaining body temperature) and daily physical activities. These energy needs are influenced by individual factors, including gender, age, weight and level of fitness. These amounts are estimated to be 10,300–109,000 kJ for males aged 16–18, and 8,400–8,500 kJ for females aged 16–18. The table below shows the estimated number of kilojoules burned per kilogram of body weight for every 30 minutes of activity. Activity

kJ/kg/30 minutes

Aerobics

10.8

Cycling

20.3

Dancing

9.1

Jogging

17.8

Running

27.9

Walking

8.1

Example 1.17 Use the table above to calculate the energy expenditure of a 75 kg male cycling for 30 minutes. To solve this question, we need to note the weight, activity and length of time of the individual at hand: weight = 75 kg activity = cycling length of time = 30 minutes

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35

Section 1 – Applications of Measurement

1695 kJ

1.3 Units of energy and mass

Now, to solve this question, we need to determine the amount of energy expenditure per 30 minutes, and apply this to the details given. So, when cycling, this individual expends 20.3 kJ per kilogram of body weight per 30 minutes: 75 kg × 20.3 kJ = 1, 522.5 kJ

K EY P OINT :

Be aware that each time period may differ from 30 minutes, so this number needs to be altered to gain accurate answers. (e.g. 1 hour = 2 × 30 minute periods, 45 minutes = 1.5 × 30 minute periods). Calculating the consumption of electricity As a form of energy, electricity is consumed by appliances in homes and businesses, and can be used for heating, cooling, lighting and running machinery. All electrical appliances have a power rating – the rate at which the appliance consumes electricity. Power is measured in units called watts (W). one kilowatt (kW) = 1, 000 W To measure the power used by an appliance, we use a unit called the kilowatt-hour (kWh). This is the energy consumed by a one kilowatt appliance in one hour. To calculate energy consumption: Section 1 – Applications of Measurement

energy (kWh) = power (kW) × time (h) Example 1.18 Determine the cost of running a 20W LED light bulb for 30 days if the average rate for electricity is $0.14 per kWh. The easiest way to correctly complete these questions is to initially convert the watts used by the light-bulb in 30 days into kilowatt-hours. As electricity is charged in kWh, this is the unit we need to convert our light-bulb consumption into. The light bulb uses 20 watts per hour. To calculate the watt-hour usage, we multiply 20 watts per hour by 24 hours per day, and then multiply this by 30 days: 20 W × 24 hours = 480 Wh per day 480 W × 30 days = 14, 400 Wh Now our light bulb usage is in watt-hours. However, we must convert to kilowatt-hours in order to fulfil the requirements of the question. If we observe the diagram above, we recognise that there are 1,000 Wh per 1 kWh, thus, we divide our current answer by 1,000 to convert our answer into kWh: 14, 400 W 1, 000 W

= 14.4 kWh

Finally, using the average cost of $0.14 per kWh, we calculate the cost of using the light bulb for 30 days: 14.4 kWh × 0.14 = $2.02 (to two decimal places) You could also solve this question by first converting 20 W into kilowatts and then finding the kilowatt-hours per 30 days. While these questions require various conversions and steps before gaining a final answer, it is important that each step is completed carefully, helping you to understand your solution and calculate a correct answer.

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Working with Time

Section 2

Working with Time 2.1

Latitude and longitude

When determining positions on Earth’s surface, we use the terms latitude and longitude. Latitude measures the angular distance between Longitude measures the angular distance between great circles that lie north and south of the Equator great circles that lie east and west of the Greenwich at 0˚ (the great circle at the centre of Earth). (prime) meridian at 0˚. Remember: latitude = flat!

Times and time differences around the world

12-hour and 24-hour time K EY P OINT :

To convert between 12 hour and 24 hour time, remember that 24 hour time always has 4 digits. For example, 1300 is 1 p.m., 0035 is 12:35 a.m., etc. If a time is before 10:00 a.m., a zero acts as a placeholder to ensure 4 digits (e.g. 0800 is 8 a.m.). Example 2.1 Convert 1845 into 12-hour time. We need to convert the 18 into a number below 12. Since 18 is larger than 12, we subtract 12 from the hours and add ‘p.m.’ to the end. This leaves us with 6:45 p.m. If the 24-hour time is not larger than 12, we simply remove the zero from the front and make it ‘am.’ Time intervals

Calculating time difference When calculating the time difference between countries, cities and regions, we must remember: 1◦ of longitude = 4 minutes Longitude is measured in degrees east and west. If we are finding the local time in a location east of the given time, we add the time difference. If we are finding the local time in a location west of the given time, we subtract the time difference. Copyright © 2021 InStudent Publishing Pty. Ltd.

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Section 2 – Working with Time

2.1.1

2.1 Latitude and longitude

Example 2.2 Calculate the time difference between Rome (42◦ N, 12◦ E) and Sydney (34◦ N, 151◦ E). Answer in hours and minutes. To find the time difference between cities, we must simply find the difference in longitude (◦ E) and multiply this by 4 minutes. 151◦ – 12◦ = 139◦ 139◦ × 4 minutes = 556 minutes 556 = 9 hours 16 minutes 60 Remember to convert a decimal into hours by pressing the degrees/minutes button on your calculator! Example 2.3 Kyle lives in New York (41◦ N, 76◦ W) and calls his mother in Brazil (13◦ S, 50◦ W) at 8 p.m. NY local time. What time is it in Brazil when he makes the phone call? Before calculating the time difference between these areas, we must determine whether we will be adding or subtracting the time.

Section 2 – Working with Time

We are moving to the east, so we need to add the time difference. time difference = 76◦ W – 50◦ W = 26◦ = 26◦ × 4 minutes = 104 minutes or 1 hour 44 minutes local time in Brazil = time in NY + time difference = 8 p.m. + 1 hour 44 minutes = 9 : 44 p.m.

∴ When Kyle calls his mother at 8 pm in New York, it is 9:44 p.m. in Brazil. Time zones A time zone is a certain area of the Earth with one standard local time. There are 12 time zones to the east and 12 to the west of the prime meridian, and each zone covers approximately 15◦ of longitude. Time zones often conveniently follow country or state borders (i.e. Australia has three different time zones). The numbers of each zone (e.g. UTC+6) indicate the number of hours to be added or subtracted from Coordinated Universal Time (UTC) – previously known as Greenwich Mean Time (GMT).

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2.1 Latitude and longitude

Example 2.4 Cheryl is planning a trip from Abu Dhabi (UTC +4) to Alaska (UTC –9).

To answer part a), we must recognise that Abu Dhabi is 4 hours ahead of Coordinated Universal Time, and Alaska is 9 hours behind Coordinated Universal Time. Therefore, there is a 13 hour time difference. Next, part b) asks us to find the time and day in Alaska using a specific time in Abu Dhabi. We need to subtract the time difference: 8 p.m. in Abu Dhabi – 13 hours = 7 a.m. Tuesday 9 June in Alaska K EY P OINT :

When solving these questions, it is important to note whether the question asks you to work forwards or backwards. In this question, we were given the time in Abu Dhabi to solve the time in Alaska, and therefore we had to work backwards to find the time in Alaska. As the 13 hour difference did not affect the day, it remained the same. However, examiners will often include the time-change to also alter the day, so be conscious of this as you are solving these questions! Finally, let’s solve part c). Here is one way of finding the answer, though there are multiple methods that would work for these questions. We will use the boarding time in Abu Dhabi to determine the local time and date in Alaska at that moment (remembering the time difference we calculated earlier!) 4:30 p.m. 24 September in Abu Dhabi – 13 hours = 3:30 a.m. 24 September in Alaska Since the flight lasts 15 hours, we must add 15 hours to the boarding time equivalent in Alaska to account for the duration of the trip. 3:30 a.m. 24 September in Alaska + 15 hour flight time = 6:30 p.m. 24 September in Alaska

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Section 2 – Working with Time

• a) What is the time difference between these cities? • b) What is the time and day in Alaska if it is 8.00 p.m. on Tuesday 9 June in Abu Dhabi? • c) Cheryl boards a plane at 4.30 pm on 24 September Abu Dhabi time, and the flight lasts 13 hours. What is the local time and date in Alaska when she arrives?

2.1 Latitude and longitude

2.1.2

Interpreting timetables

We use timetables as part of everyday life: catching the bus to school, the train to outings, and so on. As simple as it may seem to interpret them, you’ve probably had an experience where you were totally confused! This is why it’s important to understand how they work and how to use them.

Section 2 – Working with Time

The important things to note when reading timetables are: • Understand which timetable you have to read – you may be given more than one (e.g. trains into and out of the city), so ensure you have the correct day, travel service, and direction (e.g. Sydney trains to the city on Tuesday morning). • Find the correct station and follow it across horizontally. This will help you see the different service times. • Note any abbreviations or symbols that may indicate disruptions to normal services. • Highlight the departure and arrival times/locations – this will help you read the timetable more clearly. Example 2.5 Using the timetable above, answer the following questions: • a) Katie lives in St Marys and wants to arrive at Town Hall before 12 p.m. What time should she board the train? • b) How long will it take Katie to arrive in Redfern if she leaves Richmond at 10 a.m?

To solve part a), we need to determine a time suitable for Katie to leave St Marys and arrive at Town Hall before 12 p.m. To do so, begin by finding the closest time the train arrives at Town Hall before 12 p.m. On this timetable, the closest time is 11.54 a.m. However, we now need to check if this train departs from St Marys. If we follow the same line up (use a ruler to do this!), we can see that it does not, therefore, Katie cannot catch this train. We continue doing this until we find an appropriate train that departs from St Marys. You should find that the train arriving at Town Hall at 11.53 a.m departs St Marys at 10.57 a.m. Therefore, Katie should board the 10:57 a.m. train to arrive at Town Hall before 12 p.m. For part b), we apply a similar method to first determine the first train to depart Richmond after 10 a.m. On the timetable, this is 10:11 a.m. Using this line, we can now follow the route to Redfern. The train will arrive there at 11:29 a.m. Therefore, Katie’s journey on the train is 1 hour and 18 minutes.

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Non-right-angled Trigonometry

Section 3

Non-right-angled Trigonometry 3.1

Trigonometric ratios

Trigonometry is useful in finding the length of an unknown side or the size of an unknown angle in a right-angled triangle. Note that these ratios only work for right-angled triangles. Trigonometric ratios are defined using the three sides of a right-angled triangle. Observe the diagram below:

Trigonometry questions will ask you to find either the unknown side or angle. Regardless of which it asks, you need to know how the ratios work. These ratios must be used carefully, as the simple mistake of using the wrong sides will cost you marks. sin θ =

opposite

hypotenuse o sin θ = (SOH) h

cos θ =

adjacent

hypotenuse a cos θ = (CAH) h

tan θ =

opposite

adjacent o tan θ = (TOA) a

If you struggle to remember the ratios, a helpful mnemonic is to remember SOH CAH TOA (’so-ka-toa’). As evident above, this mnemonic match the ratios. Even in an exam situation, it can be a great idea to write this at the top of your page so you know you are using the right ratio at the right time.

3.1.1

Finding unknown sides (lengths)

When finding unknown sides in right-angled triangles, the question will give you one angle (other than the right angle), and one known side to help you answer the question. Example 3.1 Find the length of the unknown side x in the triangle below correct to two decimal places.

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Section 3 – Non-right-angled Trigonometry

The hypotenuse is opposite the right angle, the opposite side is opposite the angle θ , and the adjacent is the remaining side.

3.1 Trigonometric ratios

First, it is a great technique to always label the sides of the diagram (O = opposite, A = adjacent, H = hypotenuse). By doing so, you are able to more clearly understand which ratio to use, thus ensuring you minimise the chance of a silly mistake!

In this case, the length of x is the adjacent side, so we will use the cosine ratio as it involves the adjacent side and the hypotenuse. We can’t use the opposite side as we don’t know its length. cos θ = cos 42◦ =

adjacent hypotenuse x 18

Section 3 – Non-right-angled Trigonometry

Having substituted the known values into the equation, we need to rearrange the formula so that x is the subject and solve on the calculator. x = 18 × cos 42◦ = 13.38 cm (correct to two decimal places)

3.1.2

Finding unknown angles

Like finding unknown lengths of right-angled triangles, we need to use the information provided in the question to calculate the answer. Each question will provide you with an unknown angle and two sides relevant to solving the question. Whilst we will use the same trigonometric ratios, their application differs slightly. For this reason, it is important to differentiate the steps to finding unknown angles from unknown sides, ensuring that each time you complete a question you are completely aware of the differences. Example 3.2 Find the size of the unknown angle below to the nearest minute.

Note that angles of a triangle are measured in degrees and minutes – you must read the question to make sure you are answering to the required unit.

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3.2 Finding the area of non-right-angled triangles

To find θ , we must first label the sides of the triangle:

Now, recognising our two labelled sides as the opposite and adjacent, we need to determine the relevant ratio to solve the question, that being the tangent ratio. tan θ = tan θ =

opposite adjacent 12 14

From here, we use our calculator using the tan–1 function. Our answer will be 40.601...

K EY P OINT :

When inputting such numbers into your calculator, remember to press the degrees and minutes button after each necessary number (i.e. For 85° 27’, you would press it after both 85 and 27) to ensure your calculations are correct. When rounding to the nearest degree or minute, we must remember that one degree is equal to 60 minutes, and so forth. Therefore, when rounding, if the deciding number is below 30 we round down, and if the number is above 30, we round up.

θ = 40◦ 36’4.66” ≈ 40◦ 36’ (to the nearest minute)

3.2

Finding the area of non-right-angled triangles

We can also solve problems related to the area of non-right-angled triangles using trigonometry. To solve these problems, we are required to know two sides and their included angle. Before we learn how to determine the area using this information, we need to understand how to label a non-right-angled triangle. The sides of the triangle are named according to the opposite angle. To find the area of a non-right-angled triangle, we use the formula: A=

1 2

ab sin C

This means that the area of a non-right-angled triangle is half the product of two sides multiplied by the sine of the angle between the two sides. Copyright © 2021 InStudent Publishing Pty. Ltd.

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Section 3 – Non-right-angled Trigonometry

But we are asked for our final answer in minutes, so we need to convert this on our calculator using the degrees and minutes button:

3.2 Finding the area of non-right-angled triangles

Example 3.3 Find the area of the triangle to the nearest square centimetre.

The best way to begin these questions is to label the diagram according to the formula:

Now, we can substitute our known values into the formula: A= =

1 2 1

ab sin C

× 24 × 17 × sin 23◦

Section 3 – Non-right-angled Trigonometry

2 = 79.709 cm2

≈ 80 cm2 Some questions will not give you a diagram, but rather an explanation of the triangle for you to infer. In these questions, it is vital that you draw a diagram using the information, giving yourself a clearer understanding of what the question is asking and showing the marker you know your stuff! Example 3.4 In triangle CDE, side c is 33 cm, side d is 21 cm, and angle E is 40◦ . Find the area of the triangle correct to one decimal place. In exam conditions, it is okay to quickly sketch a triangle and label sides without drawing it perfectly – the diagram will not be marked unless clearly stated in the question.

Now we can continue completing the question as per the above description. Substitute our known values into the formula: A=

1 2 1

ab sin C

× 33 × 21 × sin 48◦ 2 = 257.499 cm2 =

≈ 257.5 cm2

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3.2 Finding the area of non-right-angled triangles

3.2.1

Sine rule

The sine rule is used in non-right-angled trigonometry, and relates the sides and angles in a triangle. The sine rule can be used to find both sides and angles. To use the sine rule, you must be given either two angles and one side, or two sides and one angle opposite one of the given sides. To find a side, we use the formula:

a

=

b

c

=

. sin A sin B sin C sin A sin B sin C To find an angle, we use the formula: = = a b c Finding an unknown side Example 3.5 Find the value of x correct to the nearest whole number.

When starting this question, we must check if we have the correct information necessary to apply the sine rule. By drawing arrows between the sides and angles, we will know if they are opposite, and thus useful in answering the question.

a sin A x sin 60

= =

b sin B 15 sin 42

The side you are trying to find should always be side a, as this makes the workings easier. Now, we use our algebraic knowledge to solve the equation. To make x the subject of the formula, we need to multiply each side by sine 60: x = x =

15

× sin 60

sin 42 15 sin 60 sin 42

To complete the question, we must input these calculations into the calculator which will produce our final answer. Remember, we have been asked to round to the nearest whole number. x = 19.414... cm x ≈ 19 cm

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Section 3 – Non-right-angled Trigonometry

Clearly, the angles and sides are opposite each other, therefore will substitute perfectly into the formula:

3.2 Finding the area of non-right-angled triangles

K EY P OINT :

Some questions will purposely leave out information to confuse you! For example, if you got the following triangle:

K EY P OINT :

Section 3 – Non-right-angled Trigonometry

...you would have to recognise that the required angle is missing, and you must use your knowledge of triangles (i.e. the sum of all angles is 180◦ ) to correctly find the angle: 180◦ =60◦ + 78◦ + θ

θ =180◦ – 138◦ θ =42◦ ... and then complete the question as normal! Be aware that some questions may not include a diagram for your use. It is best practice to always draw a quick diagram using the information you are given. This exam technique is guaranteed to aid your understanding of the question and therefore provide a correct answer. Finding an unknown angle Before we begin our working, we need to check that we have the relevant information to do so. To solve an unknown angle using the sine rule, we require two sides and one angle opposite one of those sides. Let’s begin by checking we have this information:

Example 3.6 Find the value of the missing angle, correct to the nearest degree.

Now we need to insert the information above into the formula. Remember, we always place the value we are trying to find (i.e. the value of θ ) as the first numerator (in this case, θ will be sin A). sin A a sin θ 43 46

= =

sin B b sin 100 45

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3.2 Finding the area of non-right-angled triangles

We now need to make θ the subject of the formula. To do so, we need to multiply both sides by 43, (the opposite action cancels the division sign): sin θ = 43 ×

sin 100

45 = 0.9410385195

To get this as an angle, use ‘SHIFT → SIN =’ on your calculator, and convert to degrees and minutes:

θ = 70.22669998 = 70◦ 13’36.12”

≈ 70◦

3.2.2

Cosine rule

The cosine rule is a formula used to find both sides and angles in non-right-angled triangles, using three sides and one angle. K EY P OINT :

Section 3 – Non-right-angled Trigonometry

In applications of the cosine rule, the angle is being ‘hugged’ by two sides.

This is a trick you can use to work out whether you need to use the sine or cosine rule in an exam! To find an unknown side given two angles and the included angle, we use: c 2 = a2 + b2 – 2ab cos C. To find an unknown angle given three sides: cos C =

a2 + b 2 – c 2 2ab

Finding an unknown side Example 3.7 Find the value of x correct to two decimal places.

To solve this question, we need to match the labels to the sides and angles of the triangle according to the formula: c 2 = a2 + b2 – 2ab cos C.

x 2 = 92 + 82 – 2 × 9 × 8 × cos 69◦ x 2 = 93.39501527

√

x =

93.39501527

x = 9.664... x ≈ 9.67 cm

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3.3 Bearings

Finding an unknown angle The cosine rule is only used to find unknown angles when you are provided with all three sides. Like the previous equations, we are going to begin by labelling the sides of the triangle according to the relevant formula.

Example 3.8 Find the value of θ to the nearest degree.

Now we can insert the information into the formula and use our calculator: cos θ = Section 3 – Non-right-angled Trigonometry

cos θ =

52 + 82 – 42 2×5×8 73 80 –1

θ = cos



73



80

θ = 24.146848 θ = 24◦ 3’48.65” θ ≈ 24◦

3.3

Bearings

A bearing is the direction of one object from another object. There are two types of bearings: compass bearings and true bearings.

3.3.1

Compass bearings

Compass bearings use the four directions of the compass – north, south, east, and west.

• The north-south line is vertical, and the east-west line is horizontal. • In between the four cardinal directions are another four directions (denoted by dotted lines): northeast, south-east, south-west, and north-west. • Each of these directions makes an angle of 45◦ with the vertical and horizontal lines. 48

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3.3 Bearings

To use a compass bearing, we state the angle as either north or south, followed by the degrees of the angle to the east or west. For example, a compass bearing of S30◦ W is found by measuring an angle of 30◦ from the south direction towards the west side. Example 3.9 Jackson leaves his campsite in the Blue Mountains and travels 25 km on a bearing of N45◦ E. How far did he travel north, to the nearest kilometre? To answer this question, we first need to draw a diagram with all parts labelled as detailed by the question. It is very important that these diagrams are drawn accurately, as the incorrect placement of an angle or side will cause miscalculations. However, you do not have to measure the angle – you can simply estimate where it will fall and write the label above.

We need to find the angle inside the triangle that falls between the compass bearing and the east line. Knowing that the bearing from north to east is 90◦ (a right angle), we can subtract 54 from 90 and conclude that the inside angle is 36◦ . We have to apply one of the three trigonometric ratios for right-angled triangles to find the value of x. Looking at the placement of the angle, we must decide which side the x is on:

Now we basically have a triangle:

The x is on the opposite side and the 25 km is on the hypotenuse (the longest side). Therefore, we will use the sine ratio to find the value of x: sin θ = sin 36 =

opposite hypotenuse x

25 x = 25 × sin 36

x = 14.695... x ≈ 15 km

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Section 3 – Non-right-angled Trigonometry

Now that we have a visual understanding of the question, we can begin our calculations. As evident in the diagram above, we can recognise the distance we are trying to find as denoted by the dotted line. If we closely look at this triangle, we can see it is a right-angle. Therefore, we can apply our knowledge of right-angled trigonometry to solve this question.

3.3 Bearings

3.3.2

True bearings

A true bearing is an angle measured clockwise from north around to the required direction. It is always written with 3 digits (e.g. 040◦ T or 120◦ T).

Section 3 – Non-right-angled Trigonometry

140◦ T is the direction measured 140◦ clockwise from north; it is the same bearing as S40◦ E. Example 3.10 A boat travels north for 52 km and then west for 24 km. What is its true bearing from its starting point, to the nearest degree? To begin, we must draw a diagram. These questions are quite wordy, and to properly understand and comprehend it, we must sketch an accurate diagram. We always start at the centre.

The diagram displays the information presented by the question. To find the value of the bearing, we must first calculate the value of θ , and subtract it from 360◦ . First, we have to recognise the right-angled triangle present, and the ratio we will use to find the value of θ . The side labelled 52 km is adjacent to the angle we need to find, and the side labelled 24 km is opposite the angle. Therefore, we will use the tangent ratio: tan θ = tan θ =

opposite adjacent 24

52 θ = 24.77514057

θ = 24◦ 47’ θ ≈ 25◦ 50

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3.3 Bearings

Now that we have the value of θ , we can determine the bearing in the question: bearing = 360◦ – 25◦ = 335◦ T Thus the true bearing of the boat from the starting point is 335◦ T (to the nearest degree). K EY P OINT :

True bearings questions can seem very tricky, however, it is important that you complete each as carefully as possible. To ensure you always solve them correctly, keep in mind that: • You always start from the centre. • The bearings are measured clockwise from north • True bearings have three-figures (even if the bearing is 20◦ from north, you would write it as 020◦ T)

3.3.3

Angles of elevation and depression

Angles of elevation and depression are measured from the horizontal.

3.3.4

Radial surveys

A radial survey involves measuring angles and sides from a central point. In this course, we use compass radial surveys to apply our knowledge of triangles to practical situations (i.e. calculating angles, unknown sides and the area of non-right-angled triangles). Therefore, it is crucial that you have proficient knowledge of the concepts previously covered, ensuring that these questions do not confuse or trick you. Compass radial surveys involve measuring the true bearing of each corner with a compass. Example 3.11 Determine the area of 4ABC displayed in the compass radial survey below, correct to the nearest m2 .

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Section 3 – Non-right-angled Trigonometry

As evident in the diagram on the right above, the angle of elevation is equal to the angle of depression. This is because they form alternate angles between two parallel lines. This information can be useful to solve problems involving this concept.

3.3 Bearings

Triangle 2: These questions require a significant amount of working, but don’t worry – they are quite simple once you get the hang of them! The explanations above tell us that compass radial surveys have the angles measured from the north, therefore, we must calculate the area of the triangles using these figures. First, we will split the radial survey into two triangles. This will help us determine the angles we need to find. Triangle 1: In this triangle, we are attempting to solve the included angle. However, we are only provided with enough details to find the largest angle, therefore, we will begin our calculations from there. Angle COA is comprised of the 13◦ from north to point A, and the difference between 360◦ and 243◦ between point C and north. Section 3 – Non-right-angled Trigonometry

θ = 117◦ + 13◦ = 130◦ To find the area of this triangle, we have to substitute our known values into the formula A = 1 2 ab sin C. However, we must find the value of the included angle first. We know that between north and point B is 88◦ . However, our triangle only starts from point A. Therefore, by subtracting the 13◦ gap between north and point A, we will be able to determine that θ = 75◦ . Knowing this angle, we can now complete our calculations for this triangle: A=

1 2 1

sin A a sin θ 6.3

= =

sin B b sin 130 7.2

sin θ = 6.3 ×

sin 130

sin θ = 42◦

ab sin C

× 6.3 × 6.8 × sin 75 2 = 20.6901... =

But we still don’t have enough information! By finding the size of the final angle (∠ACO) in the triangle, we will be able to subtract our known angles from 180° (the total of all angles in a triangle) and conclude with the size of θ . To find ∠ACO, we use the sine rule:

Now we can determine the size of θ .

θ = 180◦ – 42◦ – 130◦ = 8◦

We now have enough information to determine the area of the triangle: 1 A = × 6.3 × 7.2 × sin 8 2 = 3.1564... total area of radial survey = 20.6901... + 3.1564... = 23.8465

≈ 24 m2 (to the nearest square metre)

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7.2

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3.3 Bearings

3.3.5

Navigational methods used by different cultures

It is important to understand the navigational methods used by different cultures as part of this topic. We will primarily focus on Aboriginal and Torres Strait Islander people, but it is definitely rewarding to investigate other cultures, and how they compare and contrast to the Indigenous population of Australia. Aboriginal and Torres Strait Islander people used star maps as a means of teaching navigation outside of one’s own local area. As explained by one Indigenous man [Michael], “he pointed out a pattern of stars to the south-east,” which were used to teach travellers how to navigate outside their own country during the summer travel season. The pattern of stars showed the “waypoints” on a particular route. These waypoints were usually particular landmarks, such as waterholes or turning places on the landscape. These waypoints were used in a very similar way to navigating with a GPS. Interestingly, the star maps used by the Indigenous Australians are reflected in the highway networks today. By memorising the maps through song, Aboriginal and Torres Strait Islander people effectively used this method to navigate the Australian land.

Section 3 – Non-right-angled Trigonometry

The following resources are recommended as further reading:

• https://www.sbs.com.au/topics/life/culture/article/2016/04/11/how-ancient-aboriginal-star-maps-have-shaped-a • http://www.australiangeographic.com.au/topics/history-culture/2016/09/how-oral-cultures-memorise-so-much-i

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Rates and Ratios

Section 4

Rates and Ratios 4.1

Rates

A rate is used to make comparisons of amounts with different units. We use rates to solve a range of practical problems, such as comparing the price of strawberries per kilogram across a range of supermarkets. It is important to note the order of a rate – that is, which unit comes first and second. For example, 45 km/h represents 45 kilometres travelled per one hour. To convert rates, we need to make sure they are in the same units. Example 4.1 Convert $2.50/kg to c/g. We first have to convert $2.50 into cents: $2.50 × 100 c = 250 c We know that 1 kg is equal to 1000 grams, therefore: Section 4 – Rates and Ratios

250 c 1000 g

4.2

= 0.25 c/g

Ratios

A ratio is used to make comparisons of the same units in a definite order. A ratio is expressed as a fraction, and can be used to solve many practical problems, such as in the purchasing of assets, map scales or determining the fuel consumption of a vehicle. Example 4.2 Express the ratio 14:60 in simplest form. To solve this question, we must divide the ratio by a common denominator to obtain the simplest answer. In this case, 14 and 60 only have a common denominator of 2, therefore the simplest form of this ratio is 7 : 30. K EY P OINT : 14 To save time completing these questions, you can simply input the ratio into your calculator as 60 , and it will give you the simplest form!

Example 4.3 a) Business owners Andy, Byron, and Cassidy invest in the ratio 6:7:2. The total amount invested in the business is $450,000. How much was invested by each individual? These questions require a bit more thought, but are still fairly straightforward. First, we have to find the total amount of parts that the investment is divided into. To do this, we must add the numbers of the ratio together: total parts = 6 + 7 + 2 = 15 Now we can determine the value of one part of the investment: $450000 15

54

= $30000/part

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4.2 Ratios

Now, using the ratio, we can determine how much each individual invested in the business by simply multiplying the value of one part by the number of parts invested. Andy invested 6 parts: 6 × 30000 = $180, 000 Byron invested 7 parts: 7 × 30000 = $210, 000 Cassidy invested 2 parts: 2 × 30000 = $60, 000 To ensure our calculations are correct, always total the single investments to check it matches the total investment of the business: 180, 000 + 120, 000 + 60, 000 = 450, 000 Furthermore, we can use these calculations to complete subsequent questions (often involving fuel consumption ratios, or income distribution). Example 4.4 b) Determine the value of income earned by Byron if the business received profits of $60,000 in the first six months. Since Byron invested 7 parts, he should receive 7 parts of the profit. As we did before, we must divide the total profit by 15 parts, and then multiply the value of one part by Byron’s 7 parts: 60000

Energy efficient housing (BASIX) In NSW, BASIX sets energy targets for housing in order to reduce greenhouse gas emissions by the NSW community. These energy targets vary depending on the type and location of the home you are building. BASIX suggests tips to meet these energy targets, including: • Choosing a gas instantaneous hot water system • Improving water efficiency of hot water fixtures (showerheads and taps) to reduce hot water consumption • Install a washing line to reduce dependence on electric clothes dryers • Using direct natural lighting where appropriate and installing energy-efficient lighting (e.g. LED) • Use natural ventilation where possible (opening windows and doors) Scale drawings We also use ratios to describe map scales and obtain measurements from scale drawings. By using ratios, we can accurately measure real-world distances but scale these down to a ratio which can be interpreted easily by those reading it. The ratios may be presented with or without units. For example, a map scale may be expressed as 1cm:1m, or 1:100. For example, a map scale may be in the ratio 1:100,000. This means that for every one unit on the map (drawing length), there are 100,000 of these units in real life (actual length).

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Section 4 – Rates and Ratios

= $4000/part 15 Byron’s profit: 7 parts × $4000 = $28000

4.2 Ratios

Example 4.5 Using the map and scale provided, answer the following questions.

a) Determine a scale for the map using the measurements provided. Calculate your answer as 1cm:___m. b) How far is it in metres from Win Stadium to the train station via Burelli St? To solve these questions, we need to determine the scale of the map, and then apply this to the relevant parts of the map. Section 4 – Rates and Ratios

As you can see, the map has provided us with a measurement of 1500 m. This line indicates that in ‘real life,’ this section of the map is equal to 1500 m. However, the actual line only measures approximately 6 cm. Therefore, by determining a scale, we will be able to apply this measurement to the other roads of the map, helping us to answer part b). So if 6 cm is equal to 1500 m on the map, what is the scale? 6 cm = 1500 m (or 1.5 km)

∴ 1 cm = 250 m To solve part b), we need to use this scale to determine the actual distance between Win Stadium and the train station. If we draw a rough path from Wollongong Station to Win Stadium along Burelli Street, we’ll find that it measures approximately 7.5 cm in total. By applying the scale of 1 cm:250 m to this line, we are able to calculate the actual distance of this measurement as: 1 cm = 250 m 7.5 × 250 m = 1875 m (or 1.875 km) Therefore, the distance from the train station to Win Stadium is 1875 m. Example 4.6 A scale on a map is given as 1:100. On the map, the playground at a park is 3.7 cm from the park gates. What is the actual length of this distance? On the map, 1 cm of drawing length is equal to 100 cm (or 1m) of actual length. Therefore, by multiplying 3.7 cm by 100, we will be able to obtain the actual length of the distance: 3.7 cm × 100 cm = 370 cm or 3.7 m

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Part III

Topic 3: Financial Mathematics

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57

Money Matters

Section 1

Money Matters 1.1 1.1.1

Interest and depreciation Simple interest

To calculate this, we use the formula I = Prn, where I is the interest earned, P is the principal, r is the rate, and n is the number of periods. Example 1.1 Ryan borrowed $150,000 at 5.4% simple interest over 10 years. Calculate the interest paid on this investment. To solve this problem, we need to identify the principal, rate, and number of periods, and substitute these values into the formula: I =Prn I =150, 000 ×

5.4 100

× 10

I =81, 000 Section 1 – Money Matters

K EY P OINT :

You may have to rearrange the simple interest formula to suit what the question is asking you to solve. Alternative arrangements of this formula are used when finding the interest rate, principal or number of periods.

1.1.2

Applying percentage increases and decreases (GST)

Example 1.2 Susan bought groceries at a value of $35.87 including GST. Calculate the amount of GST paid if it is calculated at 15% of the purchase price. groceries + GST = 115% 115% = 35.87 115 35.87 = 115 115 1% = $0.31 GST = 15% = $0.13 × 15 GST = $4.65

K EY P OINT :

In these questions, you must recognise that the total value of the groceries is equal to the purchase price without GST (100%) + the value of GST (in this case, 15%), to equal a total of 115%. This is often a common mistake that students make, so take note of the working of this answer and apply to your own practice!

1.1.3

Straight line method of depreciation

The straight-line method of depreciation is an application of the simple-interest formula used to determine the rate at which certain goods lose their value. To calculate this, we use the formula S = V0 – Dn where S is the salvage value of an asset after n periods, V0 is the initial value of the asset, D is the amount of depreciation per period, and n is the number of periods. 58

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1.2 Earning and managing money

Example 1.3 A MacBook Air depreciates at a value of $180 per year. Calculate its salvage value after 5 years if it was originally bought for $1,300. S = V0 – Dn S = 1300 – (180 × 5)

∴ The salvage value is $400.

1.2

Earning and managing money

1.2.1

Calculating rates of pay

K EY P OINT :

Keep in mind for this section that in a year there are: • • • •

365 days 52 weeks 26 fortnights 12 months

Example 1.4 Tom earns a salary of $42,000 per annum. He is paid weekly. How much does he receive each week? 42000

52 = 808

∴ Tom is paid $808 per week. A wage is a payment for work calculated on an hourly basis. Example 1.5 Jessica works as a factory assistant and is paid $1,238 for a 38-hour week. How much does she earn per hour? wage per hour =

1238

38 = 32.58

∴ Jessica is paid $32.58 per hour. Common overtime rates are as follows: time and a half rate = normal pay rate × 1.5 double time rate = normal pay rate × 2

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Section 1 – Money Matters

weekly pay =

1.2 Earning and managing money

Example 1.6 Alan works at a fast-food restaurant. Find Alan’s wage during one week where he works 38 hours at the normal rate of $14.50 per hour, 5 hours at time-and-a-half rates, and 2 hours at double time rates. When solving these questions, it is important to account for all figures included. This can be done by writing out the solution as the question states the certain figures (e.g. starting with the normal pay, followed by time-and-a-half and then double-time). normal pay = 38 × 14.50 = 551 time and a half pay = 5 × 14.50 × 1.5 = 108.75 double time pay = 2 × 14.50 × 2 = 58

∴ total wages = 551 + 108.75 + 58 = 717.75

∴ Alan’s wage is $717.75. Special allowances are paid to employees if they have expenses related to their work (e.g. uniform, meals, travel) or if they work under dangerous or difficult conditions. This allowance provides compensation for the expenses related to these conditions.

Section 1 – Money Matters

• Commission is paid as a percentage of the value of goods sold. This is typically earned by employees working as real-estate agents or salespeople. Commission can be paid on a sliding scale (e.g. 5% on the first $100,000, 2% from thereon, etc.) – Calculating commission based on a sliding scale often includes an individual being paid a minimum amount (retainer) regardless of sales, and then commission based on the value of sales made. • Piecework: a fixed payment for units of work completed. • Royalties: calculated as a percentage of the revenue or profit received from the goods sold (e.g. books or songs). Example 1.7 Paul works at a car yard as a salesman. He is paid $700 per week plus commission on his sales. His commission is calculated using the scale shown in the table below. Sales value

Commission rate

Up to $30,000

No commission

$30,001 to $45,000

3% of sales over $30,001

$45,001 to $60,000

$400 plus 2.5% of sales over $45,001

$60,001 and over

$790 plus 1.8% of sales over $60,001

Calculate Paul’s earnings if his sales for this week total $47,500. As you can see, the table indicates specific amounts Paul must reach in order to earn commission (i.e. the sliding scale). In this question, we are asked to find Paul’s earnings if his sales total $47,500. To do so, we must find where this sales value sits within the table, and calculate the commission accordingly. Paul earns $400 for making sales over $45,000, and then 2.5% on sales above that amount: sales over $45000 = 47400 – 45000 = 2500 commission = 2.5 × 2500 = 62.50 total amount earned = 700 + 400 + 62.50 = 1162.50

∴ Paul’s earnings are $1,162.50. 60

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1.2 Earning and managing money

K EY P OINT :

When solving sliding scale questions, ensure you read the requirements of the question! Here, we are asked to find the total amount of Paul’s earnings – this includes his initial retainer and the commission earned from his sales. Annual leave loading annual leave loading = 17.5% of 4 weeks of normal pay Example 1.8 Liam works a 38-hour week at a rate of $22.70 per hour. He receives 17.5% of 4 weeks’ normal pay as annual leave loading. What is Liam’s total pay for the holiday? As always, make sure you include all parts of the question in your working (i.e. annual leave loading and the 4 weeks of pay). Highlight or circle the important parts of the question to make sure you don’t forget! 4 weeks of pay = 22.70 × 38 × 4 = 3450.40

total holiday pay = 3450.40 + 603.82 = 4054.22

∴ Liam receives $4052.22 in holiday pay. K EY P OINT :

Pay attention to the wording of the question – it may only ask for the annual leave loading, as opposed to the total holiday pay, meaning you will only need to work out how much extra the person will get paid for going on holiday! Government allowances and pensions Example 1.9 A youth allowance aids people who are studying, undertaking training, or in an apprenticeship to afford costs of living. Use the table below to answer the following question. Status

Allowance per fortnight

Under 18, at home

$135.50

Under 18, away from home

$345.80

18 or over, at home

$265.90

18 or over, away from home

$365.80

Jessica is 19 and lives away from home while studying. How much youth allowance does she receive in a year? youth allowance rate = $365.80 per fortnight $365.80 × 26 = $9510.80

∴ Jessica receives $9,510.80 in youth allowance per year. Copyright © 2021 InStudent Publishing Pty. Ltd.

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annual leave loading = 17.5% of $3450.40 17.5 = × 3450.40 100 = 603.82

1.2 Earning and managing money

Income tax Income tax is tax paid based on the total income earned by an individual. This may include a salary, interest, or money earned on investment properties. Allowable deductions are deductions allowed by the Australian Taxation Office (ATO). • Specific deductions that can be claimed: – Vehicle and travel expenses – Clothing, laundry, and dry-cleaning expenses – Gifts and donations – Self-education expenses – Tools and equipment • To claim a work related deduction: – You must have spent the money yourself and were not reimbursed. – It must be directly related to earning your income. – You must have a record to prove it. Calculating taxable income taxable income = gross income – allowable deductions

Section 1 – Money Matters

Example 1.10 Jeremy is a tradesman who earns $90,000 gross pay per annum. He has allowable deductions of $57/week in vehicle and travel expenses, $150/quarter in self-education expenses and $10/fortnight in clothing and laundry expenses. Calculate his taxable income.

allowable tax deductions = ($57 × 52) + ($150 × 4) + ($10 × 26) = $3824 taxable income = $90000 – $3824 = $86176

∴ Jeremy’s taxable income is $86,176 per annum. Medical levy Medicare gives Australian residents access to health care. It is partly funded by taxpayers who pay a Medicare levy of 2% of their taxable income. Calculating net pay Net pay is the final amount received by an individual after deductions have been taken from their gross pay. People have different deductions subtracted from their gross salary including: • • • •

Income tax Superannuation Health insurance Union fees net pay = gross pay – deductions

K EY P OINT :

The way I like to remember this is: ‘gross pay is what you find; net pay is what you catch!’

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1.2 Earning and managing money

Example 1.11 Lara is a sales assistant who receives a gross weekly wage of $985. She has the following deductions subtracted from her pay: • Income tax: $233 • Health insurance: $32.80 • Superannuation: $58.55 What is Lara’s net pay? net pay = gross pay – deductions = 985 – (233 + 32.80 + 58.55) = $660.65

∴ Lara’s net pay is $660.65 per week. Calculating PAYG tax using tax tables The amount of tax paid by an individual depends on the amount of money they earn. Personal income tax rates: are regularly changed to account for current government policies and inflation. The table below lists the resident tax rates for 2017–2018. Tax on this income

0 – $18,200

0

$18,201 - $37,000

19c for each $1 over $18,200

$37,001 - $87,000

$3,572 plus 32.5c for each $1 over $37,000

$87,001 - $180,000

$19,822 plus 37c for each $1 over $87,000

$180,001 and over

$54,232 plus 45c for each $1 over $180,000

As evident in the table, the tax payable depends on the taxable income. If your taxable income is $18,200 or less, you don’t have to pay any tax. The amount of tax you have to pay increases progressively with the taxable income. Most individuals have Pay As You Go (PAYG) tax deducted from their wage or salary regularly throughout the year. This PAYG tax may be greater or less than the actual amount of tax you should be paying – this is where tax refunds or tax owing is calculated. • Tax refund = tax paid – tax payable • Tax owing = tax payable - tax paid Example 1.12 a) Using the table above, calculate Kristy’s tax payable on her taxable income of $63,700. rate of tax to be paid =$3572 + 32.5c for each $1 over $37,000 amount over $37,000 =$63700 – $37000 = 26700 tax payable =3572 + ($26700 × $0.325) =$12249.50 Example 1.13 b) Kristy paid $240 per week in PAYG tax. Is she entitled to a tax refund, or does she have tax owing? amount of PAYG tax paid =$240 × 52 = 12480

∴ Kristy paid $12,480 in tax and was only supposed to pay $12,249.50. Therefore, she is entitled to a tax refund of $230.50 If you are asked to calculate income tax in an exam, a taxable income table will always be provided. Copyright © 2021 InStudent Publishing Pty. Ltd.

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Section 1 – Money Matters

Taxable income

1.3 Budgeting and household expenses

1.3 1.3.1

Budgeting and household expenses Household expenses

In order to understand electricity, water, or gas expenses of a household, we must learn how to interpret household bills. These statements provide information on the current charges of the utility, the total amount due, any discounts that may be applied, as well as the due date. K EY P OINT :

If you are given a household bill to examine during an exam, it is vital that you take the time to read all components of both the question and the utility statement provided. This will ensure you have a thorough understanding of what the question is asking, and therefore avoid mistakes. Example 1.14 Observe the electricity bill below and answer the following questions:

Section 1 – Money Matters a) What is the opening balance? b) What is the due date of the account? c) What is the value of the discount applied if the amount is paid by the due date? To answer these questions, we must be aware of the different parts of a statement. This can be done by highlighting or circling the sections relevant to answering the question, as shown above. • a) opening balance = $169 • b) due date = 16th of July 2016 • c) value of discount = $6.29 K EY P OINT :

While these questions may seem simple, exam questions may ask you to determine values not explicitly stated on the bill. This is why it is so important to read the entire account statement prior to attempting the questions, ensuring you take note of any fine print or important details!

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1.3 Budgeting and household expenses

Purchasing a car The cost of purchasing a motor vehicle differs according to various factors, including whether it is new or used, manual or automatic, the number of kilometres travelled, and the model and make. The additional installed equipment used to increase the power or running of the vehicle also has an effect on the purchase price. A motor vehicle decreases in value, thus it is not an investment. Example 1.15 A new vehicle is bought for $38,000 and sold one year later for $31,000. Calculate the percentage decrease in the value of the new vehicle, correct to one decimal place. amount decrease = $38000 – $31000 = $7000 percentage decrease =

$7000 $38000

× 100% = 18.4%

Loan repayments Many car dealers offer buyers finance plans for the purchase of a motor vehicle. The buyer pays a deposit and then makes regular repayments according to the dealer’s finance arrangements.

total cost = deposit + total repayments total repayments = repayment × number of repayments interest paid = total cost – sale price Registration The registration of a vehicle in New South Wales is comprised of two fees: the registration fee at $65 per annum, and the motor vehicle tax, based on the tare (empty) weight of the vehicle – the more it weighs, the higher the vehicle tax. Additionally, vehicles used for business purposes attract a higher vehicle tax than those for private use. You can visit the following site for a closer look at the current rates for vehicles at various tare weights: http://www.rms.nsw.gov.au/roads/registration/fees/registration-costs.html

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The total cost of purchasing a motor vehicle using finance is always greater than the sale price paid upfront (i.e. you always end up paying more in the end).

1.3 Budgeting and household expenses

Insurance Insurance is a major cost in allowing a motor vehicle to stay on the road. There are three types of insurance. You must know the differences between them, and be able to define each type. • Green Slip or Compulsory Third Party Insurance (CTP): – Required of every registered driver in Australia. – Protects vehicle drivers/owners who are legally liable for personal injury to another party (e.g. other drivers, passengers, cyclists, and pedestrians) in the event of a car accident or other claim • Third Party Property Insurance (non-compulsory): – Protects drivers against claims for damage that their car has caused to another person’s vehicle or property • Comprehensive Insurance: – Protects drivers for damage to their own vehicle, as well as damage their car has caused to another person’s vehicle or property Insurance companies offer online calculators to compare the costs of insurance according to the vehicle model/make, number of claims, age of drivers etc. K EY P OINT :

It is best to memorise the definition of each type of insurance in case a question does not specify (e.g. remembering that CTP only covers injury to people, whereas Third Party Property Insurance covers damage to vehicles and property). Section 1 – Money Matters

Stamp duty Stamp duty is a tax imposed by state and territory governments on certain transactions, including when you transfer a motor vehicle into your name. Stamp duty is calculated on the market value of the vehicle or purchase price, whichever is higher. You can use the calculator on the following site to observe the rates of stamp duty on different motor vehicles: https://www.service.nsw.gov.au/transaction/check-motor-vehicle-stamp-duty Sustainability of motor vehicles fuel consumption =

amount of fuel (L) × 100 distance travelled (km)

Example 1.16 Lisa travelled 500 km using 35 L of petrol. What was the fuel consumption?

fuel consumption =

35 × 100

500 =70 L/100 km

∴ Lisa’s car consumes 7 L of petrol per 100 km. Other running costs • Servicing a car: depends on the make and model of the car, timing of service (i.e. how many kilometres the car has done), and other repairs that are required. • Tyres: mileage of tyres depends on various factors, including design, the driver’s habits, road conditions, and the care of the tyre. Tyres should be changed approximately once every 5 years, but this varies according to a combination of the factors above. • General repairs

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1.3 Budgeting and household expenses

1.3.2

Budgeting

When preparing a budget, we must account for both the income and expenses of the individual being evaluated. Income can include the wage, interest earned or income from investment properties. Expenses vary greatly, and can include electricity, food and motor vehicle costs, as well as clothing, gifts, and workrelated expenses. Example 1.17 Below is Jasmine’s yearly budget. Use this to answer the following questions. Income

Expenses $54,948.80

Food

$10,456.30

Interest

$755.90

Electricity

$4,327.70

Income from rent

$15,600

Water

$2,543.80

Insurance

$3,788.90

Telephone and internet

$1,240.40

Council rates

$1,567.00

Motor vehicle costs

$3,715.30

Loan repayments

$20,540.50

Work-related costs

$645

Gifts and Christmas

$4,780.65

Clothing

$1,072.88

Entertainment

$5,800

Total

Section 1 – Money Matters

Wage

Total a) Calculate the total income. To calculate the total income, we need to add each value underneath the income heading. income = 54948.80 + 755.90 + 15600 = 71304.70 b) Calculate the total expenses. To calculate the total expenses, we need to add each value underneath the expenses heading. expenses = 60478.43 c) Assess the budget and determine whether any changes should be made to create savings. Provide reasons for your changes. When evaluating a budget, we need to target certain areas where spending is excessive or unnecessary. This may include ‘wants,’ such as gifts, clothing, and entertainment. It is important that you consider ‘needs’ that cannot be significantly changed, for example, the cost of loan repayments or council rates.

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Investments and Loans

Section 2

Investments and Loans 2.1

Investments

An investment is used to grow money. A loan is where an individual borrows money and must pay interest on top. Methods of calculating interest are detailed below. • Simple interest is paid according to a flat-rate for funds that have been invested, or is charged for funds that have been borrowed. • Compound interest (which uses the future value formula) is calculated on the initial amount invested plus any interest that has been charged or earned.

2.1.1

Compound interest

Finding the future value

Section 2 – Investments and Loans

To calculate the future value of an investment, we use the formula FV = PV (1 + r )n where FV is the future value of the investment (final balance), PV is the present value of the principal (initial quantity of money), r is the rate of interest per compounding time period (expressed as a decimal number), and n is the number of compounding time periods. Example 2.1 Lukas invests $12,000 over 3 years compounding quarterly. Calculate the future value of the investment if the compound interest rate is 3.5% p.a. To calculate these questions, we simply have to identify the components required by the formula, and input these figures to gain an answer. I have created a table below to make this question easier. You do not need to do this each time you complete these questions, but it can be an effective way to ensure you have each figure required before you begin. Figures required Present value

Rate

Calculation The present value of the investment is the initial quantity of money invested. In this example, the present value of Lukas’ investment is $12000. ∴ PV = 12000 The interest rate is 3.5% - however, we have to calculate the rate in a quarter (the time period used per annum). Therefore, we must convert the percentage to a decimal, and then divide by four: r =

Number of periods

3.5 100

÷ 4 = 0.00875

The question states that Lukas invests for 3 years compounding quarterly. Therefore, he invests his money four times a year for three years (four quarters make up one whole). n = 3 × 4 = 12

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2.1 Investments

K EY P OINT :

It is very important to note the number of time periods presented by the question. Other common phrases used include: • • • • •

Annually: once per year Bi-annually: twice per year Monthly: 12 times per year (every month) Fortnightly: 26 times per year Weekly: 52 times per year

Now that we have the required figures, we can insert this information into the formula to calculate the future value: FV = PV (1 + r )n = 12000(1 + 0.00875)12 = $13, 322.44 Finding the future value

Figures required

Calculation

Future value

The future value of the investment as stated by the question is $30,000.

∴ FV = 30000

Rate

The interest rate per annum is 7.8%. As this investment has been compounded bi-annually, we must find the interest used per 6 months (twice a year). r =

Number of periods

7.8 100

÷ 2 = 0.039

The question states the investment as compounded bi-annually for 10 years. Therefore: n = 10 × 2 = 20

Now we can calculate our answer: PV = =

FV (1 + r )n 30000 (1 + 0.039)20

= $13, 957.59

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Section 2 – Investments and Loans

We can also use the future-value formula to find the present value of an investment, given the required FV . numbers. To find present value: PV = (1 + r )n Example 2.2 Calculate the present value of an investment that has a future value of $30,000 after 10 years and earns 7.8% p.a. compound interest, compounded bi-annually.

2.1 Investments

Finding the interest earned Alternatively, we may be asked to only find the interest earned on an investment, and not the total value. To do so, we must find the future value as a whole, and then subtract the initial investment. This will therefore find the interest earned on the investment: I = FV – PV . This is not to be confused with the interest rate. We will learn how to calculate this in the next section! Finding the interest rate In some questions, we may be asked to find the interest rate used to calculate the present or future value of an investment. When completing these questions, we are simply rearranging the future value formula. Observe the example and solution below. Example 2.3 The present value of Brendan’s investment is $55,000. After three years, this has grown to $60,000. Interest is compounded monthly. Calculate the compound interest rate at which Brendan’s investment is growing, to the nearest two decimal places.

Section 2 – Investments and Loans

Figures required

Calculation

Future value

The future value of the investment as stated by the question is $60,000. Remember, the question may not always explicitly state “the future value is...” It is important to learn what the question is inferring through the wording used (i.e. “has grown to...”)

∴ FV = $60, 000 Present value

The present value of the investment is $55,000.

Number of periods

The question states the investment as compounded monthly for 3 years. Therefore: n = 12 × 3 = 36

However, our solution is not as simple as inserting data into a formula: FV = PV (1 + r )n 60000 = 55000(1 + r )36 To find the rate, we will first remove the 55,000 from the right-hand side: 60000

= (1 + r )36 55000 1.090909091 = (1 + r )36 Now we will remove the power of 36. Remember, as this is a power, we must remove it by cancelling the power of 36:

√

36

1.090909091 = 1 + r

1.0024199906 = 1 + r Now we are left with a simple equation. To solve the value of r , we can now minus 1 from both sides: r = 0.00241990595 70

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2.1 Investments

As we can see, the rate is an extremely long decimal number. To convert this decimal into a percentage, we must multiply our answer by 100%: r = 0.00241990595 × 100% = 0.24% per month

K EY P OINT :

Remember to round to two decimal places, but do not round off until the question is completed! Rounding off too early could cause your answer to be incorrect. As these questions require a significant amount of working out, it is always a great idea to test your final answer to ensure it is actually correct. There are often situations where students get lost amongst their working, and therefore result with an incorrect answer. In this case we will be testing our answer using a rounded number. We must check that our answer is close enough to the actual answer to be correct. Alternatively, you could test the answer using the long decimal number, just to be sure. FV = PV (1 + r )n = 55000(1 + 0.0024)36

2.1.2

Comparing growth of simple and compound interest investments

Simple and compound interest investments grow at different rates due to their different calculations. In this topic, we will be able to compare which investment grows more effectively, displaying this both numerically and graphically. Example 2.4 Steven and Jackie invest the same amount of money for the same period of time, however, each use different methods to earn interest. Steven invests $6,000 for two years, with a simple interest rate of 9% per annum. Jackie invests $6,000 for two years compounding annually, with a compound interest rate of 9% per annum. Using worked calculations, identify which person will earn the most interest, and which method is more effective for investors. Let’s work through this question slowly. Often, these types of questions require knowledge from a range of areas. Therefore, we will complete each step one at a time, avoiding silly mistakes. First, we will calculate the interest earned on Steven’s investment: I = Prn I = 6000 ×

9 100

×2

I = $1080 Next, we will calculate the interest earned on Jackie’s investment: FV = PV (1 + r )n FV = 6000(1 + 0.09)2 FV = $7128.60 I = FV – PV I = 7128.60 – 6000 I = $1128.60 Copyright © 2021 InStudent Publishing Pty. Ltd.

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Section 2 – Investments and Loans

= $59, 957.12 (close enough to the expected $60,000!)

2.1 Investments

Now that we have calculated the interest of both investments, we can determine which is more effective. Thus, our answer is that Jackie’s compound interest investment earns a higher amount of interest with Jackie earning $48.60 more than Steven. We can now represent this information graphically.

Section 2 – Investments and Loans

By representing this information graphically, we are able to easily interpret and compare the investments. While both investments initially make a similar amount of interest, the difference is noticeable by the end of three years, with the compound interest investment having a higher total value. It is also important to note the types of relationship represented by each graph. Steven’s investment (simple interest) is a linear model. His investment follows a steadily increasing value, therefore in the shape of a straight line. Jackie’s investment (compound interest) is an exponential model. Her investment is to the power of x, therefore corresponding with exponential growth. This is evident as the non-linear graph curves slightly. Appreciation and inflation Appreciation is the increase in value of goods including antiques, land, or gold. We also use the compound interest formula (future value) to calculate how much an item will appreciate to after a period of time. Example 2.5 Joanna bought a block of land in 2009 for $100,000. If the land appreciates at 15% p.a., calculate its value after 8 years. Answer to the nearest dollar. FV = PV (1 + r )n FV = 100000(1 + 0.15)8 FV = $305, 902 The value of the land is $305,902 to the nearest dollar.It has therefore appreciated by $205,902 over the past 8 years. Example 2.6 In 2008 the average cost of a loaf of bread was $0.80. Accounting for an inflation rate of 12% p.a., determine the cost of bread in 2018. FV = PV (1 + r )n FV = 0.80(1 + 0.12)10 FV = $2.48

∴ The average cost of bread in 2018 is $2.48.

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2.1 Investments

K EY P OINT :

A word of advice – the most common mistake when answering compound interest questions is determining the rate of interest and/or number of periods to use. It is vital that you calculate these on the basis of how many periods there are in total, accounting for whether the interest is calculated monthly, bi-annually etc. Both figures rely on this information, so please ensure you calculate these numbers carefully and correctly! For example, if an investor is paid 4% compound interest p.a. which is compounded bi-annually, you would have to divide 4% by 2 to get the percentage per 6 months.

2.1.3

Shares and dividends

A share is when an individual/company owns part of another company. These financial tools are bought and sold on the stock market, including the ASX (Australian Securities Exchange). To buy and sell shares, we use a broker. A broker receives a brokerage fee for their work – a percentage of the transaction. A goods and services tax (GST) of 10% is also charged for buying and selling shares; this is often a part of the brokerage fees. When calculating the cost of shares, we must look at their original market value, and how many are bought. Additionally, we must consider the brokerage fees. Example 2.7

To solve this question, we need to identify the key components needed. First, we calculate the total cost of the shares: cost of shares =$15.95 (market value) × 200 (quantity) cost of shares =$3, 190 Next, we need to calculate the brokerage fees. As stated by the question, this is to be calculated as 7% of the cost of the shares. 7 × 3, 190 brokerage fees = 100 =$223.30

∴ total cost of shares =$3, 190 + $223.30 =$3, 413.30 A shareholder is entitled to a share in the company’s profits, known as a dividend. Dividends are a payment given as a percentage of the original market value of the share. To calculate the value of a dividend, we use the dividend yield. This financial ratio indicates how much a company will distribute to its shareholders in relation to the market price of the share. annual dividend dividend yield = × 100% market price dividend = dividend yield × market price Example 2.8 The share price of Woolworths is currently $27.10. a) The predicted dividend yield is 4.2%. What would be the dividend? b) Woolworths decides to pay a dividend of $3.05. What is the dividend yield? For part a), we need to multiply the dividend yield by the market price. dividend = 4.2% × $27.10 = $1.14 For part b), we need to calculate the dividend yield using the annual dividend price and market price. dividend yield =

$305 $27.10

× 100% = 11.25%

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Section 2 – Investments and Loans

Polly bought 200 shares at a market value of $15.95 each. She incurred brokerage fees, which were charged at 7% of the price of the shares. Calculate the total cost of purchasing the shares.

2.2 Depreciation and loans

2.2 2.2.1

Depreciation and loans Depreciation

In a similar way to calculating the future value of investments, we can solve problems regarding the depreciation of an asset. To do so, we use the declining-balance method formula S = V0 (1 – r )n where S is the salvage value of the asset after n periods, V0 is the initial value of the asset, r is the depreciation rate per period, and n is the number of periods. This method is an application of the compound interest formula! Example 2.9 Isla bought a Toyota Corolla in 2012 for $15,000. The car depreciates at 15% per annum. Calculate the value of the asset after 5 years to the nearest dollar. S =V0 (1 – r )n S =15000(1 – 0.15)5 S =$6656

2.2.2

Reduced-balance loans

Section 2 – Investments and Loans

When calculating reduced-balance loans, we calculate the balance owing as each payment is made – unlike a flat-rate loan. Therefore, as the balance owing is reduced, so is the interest charged. As this process can be confusing, we use a table of loan repayments to easily identify the balance owing, interest charged, and number of periods remaining. Example 2.10 A small-business took out a loan on a reduced-balance schedule. They borrowed $170,000 with interest charged at 6% per annum. The business is repaying $1,000 per month. a) Complete a following table of loan repayments for the reduced-balance loan for the first three months. Months (n)

Principal (P)

Interest (I)

P+I

P+I–R

1

$170,000

$850

$170,850

$169,850

2

$169,850

$849.25

$170,699.25

$169,699.25

3

$169,699.25

$848.50

$170,547.75

$169,547.75

Note: • The interest must be recalculated each time! As the loan reduces, so too does the interest charged. • Interest is charged one month at a time, so make sure you divide your annual interest rate by 12. For instance, in the above example, the monthly interest rate would be 0.06 ÷ 12 = 0.05. So to solve for interest in the first row, you would simply enter 170000 × 0.05 × 1 into your calculator. • The figure in the final row (principal + interest – repayment) is the balance owing, or the principal of the next month. • Examiners may ask questions based on the loan repayment table completed – therefore, ensure you complete it carefully to maximise your marks! b) What is the balance owing at the end of the second month? The balance owing at the end of the second month is $169,699.25 (as indicated in the last box of the second months’ row). c) How much has the principal reduced in the first three months? To calculate how much the principal has been reduced, we need to minus the balance owing at the end of the third month from the initial loan. $170, 000 – $169, 547.75 = $452.25

∴ The balance has been reduced by $452.25 over three months. 74

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2.2 Depreciation and loans

2.2.3

Credit cards

Credit cards act as a loan – they are used to buy goods and services now and pay for them later. We say credit cards are a reducing balance loan, as interest is paid only on the outstanding balance. This fluctuates as the individual repays their purchases. Like a normal loan, interest is charged on credit card purchases. The only time when interest is not charged is the interest-free period. This occurs if the payment is received before the due date. If the payment is not received by its due date, interest is charged from the date of purchase. K EY P OINT :

Understanding what is meant by the interest-free period is crucial when completing credit card questions. This minor section often causes a lot of confusion for HSC students, and therefore needs to be understood thoroughly! Interest on credit cards is paid on the outstanding balance using compound interest. This is usually calculated daily. annual interest rate daily interest rate = 365 Let’s calculate the cost of using a credit card.

a) What is the total amount she paid for the clothing, including the interest charged? b) How much interest did Grace pay on her credit-card purchase? To solve these question, we will use our knowledge of compound interest and apply our new understanding relating to credit cards. Using the compound interest formula, insert the figures relevant: A = P(1 + r )n

 A = 175

1+

0.17

12

365

Note that the interest rate (r) is the initial interest rate of 17% (0.17) over 365 days. Now we can calculate the total: A = $175.98

∴ Grace paid $175.98 in total for her new shoes. To solve part b), we need to find only the interest paid, and not the total amount. Therefore, by subtracting the original transaction price from the final total, we will obtain an answer of: I =A–P I = 175.98 – 175 I = 0.98

∴ Grace paid $0.98 in interest.

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Section 2 – Investments and Loans

Example 2.11 Grace has a credit card with a compound interest rate of 17% p.a. and no interest free-period. She uses her credit card to pay for new shoes, costing her $175, and paid the account 12 days later.

2.2 Depreciation and loans

Understanding credit card statements Credit card statements are issued monthly and contain several pieces of important information, shown below:

Section 2 – Investments and Loans

K EY P OINT :

You may be asked questions about credit card statements provided for you. In these questions, analyse the statement carefully, noting the important details. Highlight any key points (particularly those asked by the question). Read each question slowly and ensure you place the right values in the right places. You don’t want to lose marks on relatively simple questions such as these. Comparing interest rates of credit cards with other loans When comparing the cost of credit cards with other loans, it is important to recognise that credit card interest rates are generally higher due to the risk of the credit provided. A regular personal or home loan uses the asset as security, whereas a credit card loan is ‘unsecured.’ It may be difficult however to feasibly compare interest rates of credit cards with other loans due to the specific terms and conditions regulating each loan. For example, the MyBank credit card may have an interest rate of 16% p.a., with a $40 annual fee, whereas another may have an interest rate of 18% p.a. with no annual fee. This is similar among other loans too, and makes it increasingly difficult to conclude which credit card is more cost-effective. Financial websites may provide a comparison rate to easily compare particular loans, assuming all fees and related charges into a single percentage. This is a useful tool to accurately compare interest rates of credit cards, personal loans, and home loans. See https://www.moneysmart.gov.au/borrowing-and-credit/home-loans/interest-rates for more information about comparing interest rates of loans.

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Annuities

Section 3

Annuities Annuities are basically financial instrument that involve an investment account with regular, equal contributions. Interest compounds at the end of each period.

3.1

Future value annuities

The future value of annuity is the total value of the investment at the end of the specified term. To calculate the future value of an annuity, we use a table of future value interest factors. These tables, when used correctly, allow students to easily answer annuity questions. You will never be asked to calculate the future value of an annuity without a table. The calculations are quite difficult, and therefore (thankfully!) not in the syllabus of this course. Example 3.1 The table below shows the future value of annuity with a contribution of $1. Future value of $1 1.0%

1.5%

2.0%

2.5%

3.0%

3.5%

4.0%

1

1.000

1.000

1.000

1.000

1.000

1.000

1.000

2

2.010

2.015

2.020

2.025

2.030

2.035

2.040

3

3.030

3.045

3.060

3.076

3.091

3.106

3.122

4

4.060

4.091

4.122

4.153

4.184

4.215

4.246

5

5.101

5.152

5.204

5.256

5.309

5.362

5.416

6

6.152

6.230

6.308

6.388

6.468

6.550

6.633

7

7.214

7.323

7.434

7.547

7.662

7.779

7.898

8

8.286

8.433

8.583

8.736

8.892

9.052

9.214

Section 3 – Annuities

Period

Use the table to calculate the future value of the following annuities: a) $570 per month for 6 months at 18% p.a. compounded monthly b) $13 000 per quarter for 9 months at 4% p.a. compounded quarterly To solve these questions, we need to ensure that we are using the right figures. In question a), we are asked to calculate the future value of an annuity that: • Is paid at $570 per month • Is paid for a period of 6 months • Has a compound interest rate of 18% p.a. compounded monthly To find the appropriate interest factor, we need to determine the rate of interest earned on this annuity. 10% ÷ 12 months = 1.5% per compounding period K EY P OINT :

It is often helpful to use a ruler when matching the two factors. Often, students misread the table and end up using the wrong number. As in the previous section, finding the interest rate is often where mistakes are made. Ensure you complete this calculation prior to selecting the interest factor. Now we can complete the question. With 6 periods at 1.5%, our future value interest factor is 6.230. Copyright © 2021 InStudent Publishing Pty. Ltd.

77

3.2 Present value of annuities

Future value of $1 Period

1.0%

1.5%

2.0%

2.5%

3.0%

3.5%

4.0%

1

1.000

1.000

1.000

1.000

1.000

1.000

1.000

2

2.010

2.015

2.020

2.025

2.030

2.035

2.040

3

3.030

3.045

3.060

3.076

3.091

3.106

3.122

4

4.060

4.091

4.122

4.153

4.184

4.215

4.246

5

5.101

5.152

5.204

5.256

5.309

5.362

5.416

6

6.152

6.230

6.308

6.388

6.468

6.550

6.633

7

7.214

7.323

7.434

7.547

7.662

7.779

7.898

8

8.286

8.433

8.583

8.736

8.892

9.052

9.214

Now that we have our interest factor of 6.230, we need to find the future value of our annuity. As this graph is only for the future value of $1, we need to use this number to find the future value of an annuity of $570. Therefore, by multiplying our interest factor by the annuity, we will find the future value: 570 × 6.230 = 3551.10

Section 3 – Annuities

This future value includes both the annuity paid per month and interest. If a question asks you to find the interest earned on an annuity, you find the future value first, and then subtract the number of periods by the amount of the annuity. 3551.10 – (6 × 570) = 131.10

3.2

Present value of annuities

The present value of an annuity is a single amount that could be invested now to equal the future value of the annuity. We also use a table of present value factors as a simpler way to calculate these questions. Example 3.2 Use the table provided to determine the present value of an annuity if $9000 is contributed bi-annually for 2 years at 16% p.a. compounded six-monthly. Present value of $1 End of year

6%

8%

10%

12%

1

0.94

0.93

0.91

0.89

2

1.83

1.78

1.74

1.69

3

2.67

2.58

2.49

2.40

4

3.47

3.31

3.17

3.04

Like completing future value questions, we must first determine the correct period and interest rate: period = 2 (times per year) × 2 years =4 16% interest rate = = 8% 2 Now we can find the intersection of these from the table (i.e. 3.31). Finally, we multiply the value of the intersection by the money invested: $9000 × 3.31 = 29790

∴ The present value of the annuity is $29,970. 78

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Part IV

Topic 4: Statistical Analysis

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Data Analysis

Section 1

Data Analysis 1.1 1.1.1

Classifying and representing data (grouped and ungrouped) Data collection methods: sampling

It is often difficult to distinguish between the various sampling methods, especially when dealing with exam questions. The table below will define and review the different kinds of sampling methods. Use this to help you remember these definitions. It will likely save you a few marks in an exam! Method

Description

Advantages

Disadvantages

Sampling is systematic: the population is divided structurally and selected according to intervals e.g. every 4th person

3 Simple

7 Chance that the

method evenly sampled

process of selection can follow the tested ‘trait’ of the populations, so the sample is no longer random

Selfselected sampling

The sample is created by volunteers: the participants choose to be part of the sample by accepting/responding to a request for volunteers

3 Saves

7 Often a degree of

time/money in coordinating sample groups

bias in who decides to take part, so opinions sway toward one side

Simple random sampling

Each member of the relevant population has an equal chance of being selected

3 Eliminates

7 May not accurately

sampling bias

reflect the entire population (e.g. in a school population, you may randomly select all students from the same year group instead of a spread across a range)

Stratified sampling

The population is divided into strata (groups according to certain characteristics i.e. year group) and then a random sample is taken from each; the size of each random sample is proportional to the size of the subgroup within the population

3 Allows for a

7 Members can only

proportional mix of the population

be classified into one group – this cannot be applied to all sampling groups

Systematic sampling

3Population is

Section 1 – Data Analysis

3 Ensures entire population is accurately reflected

K EY P OINT :

Stratified = STRATA = random selection from groups Systematic = divided structurally in INTERVALS = works as a system

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1.1 Classifying and representing data (grouped and ungrouped)

Note that there are some potential faults in data collection processes. Responses to questions may vary in their usefulness – they rely on the relevance, clarity, and structure of the question being asked. Also, questions must be free from bias – this is a common issue with survey questions.

1.1.2

Classifying data

A categorical variable describes a characteristic that can be divided into categories. For example, eye colour, gender, type of car, or country. Categorical variables can be ordinal or nominal: • An ordinal variable has an order. For example, the satisfaction of a customer on their shopping experience may be poor, satisfactory, good, or excellent. • A nominal variable categorises the data into particular groups – however, there is no order. For example, children in a kindergarten classroom may be sorted according to their eye colour – blue, green, or brown. K EY P OINT :

ORDINAL = ORDER NOMINAL = NO ORDER A numerical variable has a numerical value – it can be counted or measured. Numerical variables can be discrete or continuous:

1.1.3

Organising and displaying data

When representing data collected from the various sampling or census methods above, we need to choose an appropriate tabular or graphical representation in order to effectively communicate this information to those it concerns. • Tables: are a simple method of displaying certain information, and can be read easily (provided the headings and data are in the right place). • Pareto charts: are tools used in quality control to analyse data about the frequency of problems in a process or their causes. They combine a vertical column graph and a cumulative relative frequency graph.

• The key elements are: – Statistical variable (problem) on the x-axis – Frequency on the y -axis – The problems/causes are arranged in descending order (highest frequency on the left and the lowest frequency on the right) allows the problem creating the most impact to be clearly shown – The purpose of the cumulative frequency graph is to help judge the added contribution of each problem Copyright © 2021 InStudent Publishing Pty. Ltd.

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Section 1 – Data Analysis

• A discrete variable only takes exact numbers – i.e. when counting. For example, when counting the number of children in a family, there can only be whole numbers (e.g. 6 children). • A continuous variable can take any number within a certain range – i.e. when measuring. For example, when measuring the height of a student, the numerical value could be 161 cm, or 161.1 cm.

1.1 Classifying and representing data (grouped and ungrouped)

Frequency distribution tables and histograms Example 1.1 The temperatures for 30 days in Sydney are shown below. Construct a frequency distribution table, and use this to create a histogram and polygon.

32

40

45

31

31

43

30

34

32

30

30

41

33

33

32

36

35

36

40

33

38

32

37

34

31

40

42

38

42

39

A frequency table would look like the following:

lowest score −→

Section 1 – Data Analysis

Score

Tally

Frequency

30

III

3

31

III

3

32

IIII

4

33

III

3

34

II

2

35

I

1

36

II

2

37

I

1

38

II

2

39

I

1

40

II

2

41

II

2

42

II

2

43

I

1

44 highest score −→

45

0 I

←− highest frequency

←− lowest frequency

1

K EY P OINT :

In some cases, you may need to use a grouped frequency table. These tables are used when you have data with a large range of values, so grouping conveniently allocates the numbers into smaller intervals (class intervals) to assist in completing the question. It is useful to choose an interval that is easy to understand, e.g. 5 units, 10 units.

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1.1 Classifying and representing data (grouped and ungrouped)

A histogram (a column or bar graph of a frequency table) would look like the following:

A polygon is a line graph joining the midpoints of the tops of the columns of the frequency histogram.

Cumulative frequency distribution tables Cumulative frequency is the frequency of the score plus the frequency of all the scores less than that score. To follow on from the previous example, inserting the cumulative frequency in a new column on the right: Score Tally Frequency Cumulative frequency III

3

3

31

III

3

6

32

IIII

4

10

33

III

3

13

34

II

2

15

35

I

1

16

36

II

2

18

37

I

1

19

38

II

2

21

39

I

1

22

40

II

2

24

41

II

2

26

42

II

2

28

43

I

1

29

0

29

1

30

44 45

I

Cumulative frequency graphs These histograms use the scores or classes on the horizontal axis and the cumulative frequencies associated with these scores on the vertical axis.

Section 1 – Data Analysis

30

A polygon/ogive is a line graph joining the top right-hand corner of the columns in a cumulative frequency histogram

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1.1 Classifying and representing data (grouped and ungrouped)

Dot plots A dot plot is a number line with the frequency of each data point marked by a dot. These graphs are useful in conveniently and efficiently representing small data sets. They allow readers to easily recognise clusters, gaps and outliers in the data. For example, the following box plot shows the number of hours per day a student spent doing their homework.

Stem-and-leaf plots These plots group and rank data to show range and distribution. The leaf is the final digit of a number – it is always one number. The stem is any number/s that precede the leaf – this can be more than one number. Section 1 – Data Analysis

Example 1.2 The ages of residents in a retirement village in Sydney are listed below. Construct a stem-and-leaf plot to represent this data. 73

84

88

90

72

74

78

81

85

87

92

98

76

83

84

89

88

90

85

82

When constructing stem and leaf plots, try to arrange the numbers in order – this will help in solving questions further on. Stem

Leaf

7

23468

8

12344557889

9

0028

We can also use back-to-back stem-and-leaf plots to compare two sets of comparable data. For example, if we were given the ages of residents in another retirement home, we could compare both homes and determine which has the larger range, higher median and other important information. Melbourne retirement home

Stem

Sydney retirement home

976320

7

23468

99877742

8

12344557889

996432

9

0028

We can see that these data sets share the same stem. It is also important to note that on each side, the numbers ascend from the middle outwards. 84

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1.2 Summary statistics

1.2 1.2.1

Summary statistics Features of a population and sample

Population

Sample

Population: the entire group at hand

Sample: a selection of individuals from a group at hand

Population values (parameters): a measurable characteristic of a population

Sample-based estimates (statistics): a measurable characteristic of a sample

Population mean (µ): the mean of the entire population

Sample mean (x¯): the mean of a sample from a population

Population standard deviation (σ ): a measure of the spread of data around the mean; only used when we have data for the entire population

Sample standard deviation (s): a measure of spread of data around the mean; only used when a sample is taken from a large population

1.2.2

Measures of central tendency

The mean (x¯) is the sum of all scores divided by the number of scores. x¯ =

sum of scores Section 1 – Data Analysis

number of scores

Let’s go through an example of calculating the mean from grouped data. Class

Class centre (x)

Frequency (f )

fx

1–5

3

2

6

6–10

8

9

72

11–15

13

11

143

16–20

18

13

234

21–25

23

10

230

26–30

28

5

140

50

825

Total: Using the fx and f columns, we can calculate the mean: x¯ = x¯ =

fx (frequency × score) f (frequency) 825

50 x¯ = 16.5 The median is the middle score or value of a data set. To calculate the median: 1. Arrange all the scores in ascending order. 2. Count the total number of scores. This is represented by the letter n. n+1 3. Odd number of scores: median is score. 2 n n+1 4. Even number of scores: median is the average of the and the score. 2 2 Copyright © 2021 InStudent Publishing Pty. Ltd.

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1.2 Summary statistics

Class

Class centre (x)

Frequency (f )

Cumulative frequency

1–5

3

4

4

6–10

8

1

5

11–15

13

7

12

16–20

18

9

21

21–25

23

5

26

26–30

28

6

32

32

32

Total:

In this example, both the 16th and 17th scores are between the 12th and 21st scores in the cumulative frequency, therefore both scores are 18. median = = =

32

score +

32 + 1

score 2 2 16th score + 17th score 2 18 + 18 2

Section 1 – Data Analysis

= 18 The mode is the score that has the highest frequency (i.e. occurs most often) in a data set. Choosing the appropriate measure of central tendency • Mean: – Easy to understand and calculate – Dependent on every score in the data set (so can be easily distorted by outliers) – Not suitable for categorical data • Median: – Easy to understand – its central position means there are an equal number of scores above and below it – Not affected by outliers – Not suitable for categorical data • Mode: – Easy to determine; not affected by outliers – May not be a central value, but rather which score has the highest frequency – Only measure suitable for categorical data

1.2.3

Measures of spread

The range is the difference between the highest and lowest scores in a data set. range = highest score – lowest score The interquartile range is the difference between the upper quartile and lower quartiles in a data set. interquartile range = upper quartile – lower quartile IQR = Q3 – Q1

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1.2 Summary statistics

To calculate the interquartile range: 1. 2. 3. 4. 5.

Arrange the data in ascending order. Divide the data into two equal-sized groups using the median. If n is odd, omit the median. Find Q1 , the median of the first group. Find Q3 , the median of the second group. Calculate the interquartile range.

Example 1.3 Calculate the interquartile range of the data set below.

Q1 =

3+6

Q3 =

2 =4.5

12 + 15

2 =13.5

IQR =13.5 – 4.5 =9

• Deciles: separate the data set into 10 equal parts. – Decile 1 is the score that has 10% of the scores below it and 90% above it – Decile 3 has 30% of the scores below it and 70% above it, etc. – Decile 5 is the median. • Percentiles: separate the data set into 100 equal parts. – The 45th percentile (P45 ) is the score that has 45% of the scores below it (and 55% above it) – The 80th percentile (P80 ) has 80% of scores below it and 20% above it, and is also equal to decile 8 The standard deviation (σ ) measures the spread of data around the mean – calculated by finding the distances of each score from the mean and calculating an ‘average’ of these distances. We cal determine this using our calculator: K EY P OINT :

Calculator steps: 1. 2. 3. 4. 5. 6. 7.

Enter STATS mode > MODE > 2:STAT > 1:1-VAR Get frequency column (if it does not appear automatically) > SHIFT > MODE > 3:STAT > 1:ON Enter data Click > AC Click > 3:σ x (population standard deviation) Click > 4:sx (sample standard deviation) Click > =

When completing questions involving statistics, you are often changing your calculator into statistics mode. While this mode is useful for these statistics questions, ensure you reset your calculator to the normal mode when moving onto new questions, as it will not calculate your answers correctly! To reset calculator: > SHIFT > 3: All > = YES > AC

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Section 1 – Data Analysis

For larger data sets, we are able to divide the scores into either deciles or percentiles. This concept is an extension of the concept of quartiles.

1.2 Summary statistics

If a set of data has a low standard deviation, the results are closer to the mean. If the data set has a high standard deviation, the results are further from the mean. Calculating the outlier An outlier is a score separated from the majority of the data. We can calculate this using either the median and quartiles, or the mean and standard deviation. Using the median and quartiles: Using the mean and standard deviation: A score is classified as an outlier if it is 1.5 × A score is classified as an outlier if it lies outside the IQR above the third or upper quartile or 1.5 × range, which is calculated as the mean – 3 x standard IQR below the lower or third quartile. deviation to the mean + 3 x standard deviation. This method is preferred as the median and This method is less favoured because the outlier afquartiles are not affected by outliers. This fects the mean and standard deviation. This method method can therefore be used for skewed is therefore not suitable for skewed data. data. The outlier predominantly affects the mean, as this calculation requires all scores to be included. The median and mode are less affected. K EY P OINT :

If questions ask you to determine which measure is most impacted by the outlier, ensure that you show all your working. Markers are looking for your understanding of these concepts.

1.2.4

Identifying modality

Section 1 – Data Analysis

Unimodal: one mode

Bimodal: two modes

Multimodal: multiple modes

If there is no skew, the graph is symmetrical. A positive skew where there is a long tail on the right side means there is more data on the left side. A negative skew where there a long tail on the left side means there is more data on the right side.

Box and whisker plots In box-and-whisker plots, each section contains 25% of the data.

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1.2 Summary statistics

Example 1.4 Using the double box-and-whisker plot below, determine which class has performed better in their recent Maths exam. Justify your answer using relevant information.

When comparing two sets of data in a box-and-whisker plot, it is important to first establish the five-number summary for each set. By doing so, we are able to determine further measures of central tendency and spread, and thus compare the data at hand.

Thus, our answer would be something along the lines of: class B performed better as they have a lower IWR of 4 in comparison to class A at 9, as well as a higher median of 9 in comparison to 7.

1.2.5

Applications of central tendency and spread

When media articles are published, they often use statistics and data displays to accompany the text. While this information may be seen as contributing to the authority of the article, it is important to question the use of the data. Kew questions to ask yourself when observing media-published data: • Which organisation collected the data? • Is the data displayed reasonably? Has it been shaped to convey a biased opinion? • Are the conclusions made in the article feasible based on the data? By asking these questions, we can quickly determine if measures of central tendency and spread have been used or misused. By doing so, we can ultimately ‘re-write’ sections of the article to make them more accurate, helping us to recognise the common misrepresentation of information by media outlets in order to persuade a particular view to the audience.

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Section 1 – Data Analysis

A: B: Five number summary: Five number summary: • Lower extreme: 1 • Lower extreme: 1 • Q1 : 3 • Q1 : 7 • Median: 7 • Median: 9 • Q3 : 12 • Q3 : 12 • Upper extreme 16 • Upper extreme 16 Measures of spread: Measures of spread: • Range 15 • Range 15 • IQR: 9 • IQR: 5 Now to compare these data sets, we could say that both sets have the same range, but class B has a lower interquartile range, indicating that the scores are closer together. Class B has a higher median, indicating their performance is better than class A.

Relative Frequency and Probability

Section 2

Relative Frequency and Probability 2.1

Theoretical probability

Probability is the measure of the likely chance of an occurrence. The probability scale is: for each event A: 0 ≤ P(A) ≤ 1 if A is an impossibility: P(A) = 0 if A is a certainty: P(A) = 1 To calculate the probability of an event where outcomes are equally likely: Pr (event) = Section 2 – Relative Frequency and Probability

2.1.1

number of favourable outcomes total number of outcomes

Conducting experiments

When calculating probability of simple games and experiments, the sample space lists all the possible outcomes of the experiment. For example, when rolling a die with six sides, the possible outcomes are 1, 2, 3, 4, 5, or 6. Example 2.1 A card is chosen from 3 black cards and 9 red cards. List the sample space and use this to determine the number of outcomes. • Sample space: – Selecting a black card – Selecting a red card • Number of possible outcomes = 2 When calculating the probability of an event occurring, we can simultaneously calculate the probability of the complement, P(eventc ), which is the probability of an event not occurring. P(an event does not occur) = 1 – P(the event does occur) Following on from the previous example, we need to find the probability of selecting a red card, and then find the complement of this event (selecting a black card). P(an event does not occur) = 1 – P(the event does occur) 9 P(selecting black card) = 1 – 12 3 P(selecting black card) = 12

2.1.2

Tree diagrams and multistage events

An effective way to determine the probability of multi-stage events is through the completion of tree diagrams. Tree diagrams make it easier to understand the likelihood of events occurring, and provide a list of possible outcomes. Some students approach multi-stage events without drawing tree diagrams to save time. However, it is always best to draw such diagrams, as you will be guaranteed to understand the question’s requirements and thus answer correctly. 90

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2.2 Relative frequency

Example 2.2 Jemimia rolled a die and tossed a coin. Determine the probability of gaining an even number and a tail.

To draw a tree diagram we need to draw one event at a time and list the probability of each event on each branch. To solve this question, we must calculate the probability of getting an even number and a tail.

 P=

1 6

 ×3 ×

1 2

= 0.25 Thus, the probability of getting an even number and a tail is 0.25 or 25%.

If the questions asks you to find the probability of one event occurring or another event occurring (e.g. the probability of getting an even number or a tail) we add the events.

2.2

Relative frequency

Relative frequency is another estimate of probability. To determine relative frequency, we perform an experiment a number of times, and find the event’s relative frequency. For this reason, it is often termed experimental probability. relative frequency =

frequency of the event total frequency

f = P × 100% f

P

... where f is the frequency and f is the total number of possibilities. Example 2.3 A beach club has 600 members. Of these, 240 live within 1 km of the beach. Based on this information, estimate the probability that a person selected at random lives more than 1 km away from the beach.

frequency = 600 – 240 = 360 total number of possibilities = 600 360 relative frequency = 600 = 0.6 or 60% Increasing the number of trials in an experiment produces relative frequencies that gradually become closer to the value of the theoretical probability. Copyright © 2021 InStudent Publishing Pty. Ltd.

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Section 2 – Relative Frequency and Probability

K EY P OINT :

It is often confusing as to which operation to use when calculating probability. If the question asks you to find the probability of one event occurring and another event (like the example above), we multiply the events by eachother.

2.2 Relative frequency

2.2.1

Expected frequencies based on existing probabilities

We can also determine the expected frequency of certain events based on existing known probabilities. For example, using the percentage breakdown of blood-types published by the Australian Red Cross Blood Service, we can estimate the frequency of these blood types in certain populations. Example 2.4 Penrith has a total population of 180,000 people. Using the percentages below, estimate the probability the number of individuals with the blood type: a) AB b) O – c) B Percentage breakdown of blood types

Section 2 – Relative Frequency and Probability

AB

3%

A

38%

B

10%

O–

9%

O+

40%

To solve these questions, we must apply the percentage breakdown of blood types in Australia to the population at hand. • a) AB = 3% of the population number of individuals with blood type AB =

3 100

× 180, 000 = 5, 400 individuals

• b) O – = 9% of the population number of individuals with blood type O – =

9 100

× 180, 000 = 16, 200 individuals

• a) B = 10% of the population number of individuals with blood type B =

92

10 100

× 180, 000 = 18, 000 individuals

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Bivariate Data Analysis

Section 3

Bivariate Data Analysis 3.1

Scatterplots

Scatterplots are used to determine the relationship between two numerical values (bivariate meaning two). When completing bivariate data analysis, data is collected for each variable and displayed in a table of ordered pairs. For example, we could test students’ height and weight to determine if a relationship is present. From this information displayed in a scatterplot, we can determine if there are dependent or independent variables, the strength of the relationship and other key features. As seen below, each ordered pair is a dot on the graph. To describe the patterns we see, we use the following terminology: • • • •

Positive linear Negative linear Non-linear No correlation

Section 3 – Bivariate Data Analysis

K EY P OINT :

For a linear relationship, the dots should approximate a straight line. For a non-linear relationship, the dots should approximate a curve. We can also describe the correlation as strong, moderate, or weak.

The strength of the correlation is represented by the correlation coefficient. Copyright © 2021 InStudent Publishing Pty. Ltd.

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3.2 Correlation coefficient

3.2

Correlation coefficient

We use Pearson’s correlation coefficient, denoted by the pronumeral r , to determine a number which represents the strength of the relationship between two variables. The correlation coefficient refers to how the points on the scatterplot lie. For example, if the points lie exactly in a straight line, we can describe the relationship as a perfect linear relationship. If the strength of the relationship is not perfect (i.e. not 1 or –1) or no correlation (0), the correlation coefficient r has a value between –1 and +1. To correctly quantify the strength, use the following table as a guide: Coefficient (r ) Positive

Negative

Strong

0.8 to 1

–0.8 to –1

Moderate

0.5 to 0.8

–0.5 to –0.8

Weak

0.3 to 0.5

–0.3 to –0.5

No correlation

0 to 0.3

0 to –0.3

Section 3 – Bivariate Data Analysis

K EY P OINT :

You are only expected to determine the correlation coefficient using your calculator! This is a very advanced formula that you are not expected to ever calculate by hand. We must also note the meaning of the relationships we determine: • Positive correlation (0 to +1): both variables increase or decrease at the same time • No correlation (0): no relationship between the variables. • Negative correlation (–1 to 0): one variable increases while the other variable decreases. Some students confuse high correlation with causation (e.g. the height of a student directly impacts their weight. However, this is not always true. It is important to remember that while these two variables seem to correlate, one variable is not responsible for the other entirely. K EY P OINT :

To calculate the correlation coefficient given a data set: 1. MODE > 2: STAT > 2: A + BX 2. Enter data in x and y columns 3. AC > SHIFT > 1 > 5: reg > 3: r > = Example 3.1 A teacher measured her students’ marks in an examination and their height in cm, and presented the information in a table. Heart rate (r)

61

62

63

67

69

71

72

75

78

Height (h)

145

172

151

163

180

165

168

170

158

a) Draw a scatterplot using the data above. b) State whether the relationship is positive or negative. c) Describe the strength of the relationship (i.e. strong, moderate, weak). d) Calculate the value of the correlation coefficient, correct to four decimal places.

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3.3 Line of best fit

For part a), our scatterplot should look like:

For part b), although the data is fairly scattered, we can say the relationship is positive. This is indicated by the tendency of the points to begin from the bottom left of the graph and move their way to the top right of the graph. For part c), the strength of this relationship is fairly weak. This will be further supported in the calculation of the correlation coefficient.

correlation coefficient = 0.3318 (weak correlation)

3.3

Line of best fit

After drawing a scatterplot using the value of ordered pairs, we may be able to draw a line of best fit. If the points have a linear correlation, we can approximate a straight-line graph that relates the two variables. This is called linear regression. Linear regression is a measure of the relationship between the mean value of one variable and the corresponding values of other variables. K EY P OINT :

As with a normal straight line graph, the aim of linear regression is to model the gradient-intercept formula y = mx + c. Example 3.2 Use the points below to draw a scatterplot and line of best fit: (0,6) (2,24) (3,39) (4,44) (5,59) (6,64) (7,79) (8,84) First let’s plot these points on a graph:

As we can see, the points do not exactly line up. Therefore, we will draw a line of best fit. Copyright © 2021 InStudent Publishing Pty. Ltd.

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Section 3 – Bivariate Data Analysis

For part d) using the calculator steps provided above, the correlation coefficient is:

3.3 Line of best fit

K EY P OINT :

It is often tricky to determine exactly where the line should be placed. It is important to remember that the line should pass through as many points as possible. If the line does not pass through a point, it must have an even amount of points on each side. This ensures it evenly crosses the data, and assumes its role as line of best fit. The line of best fit in this example does not pass through many points. However, there are approximately 4 points on either side of the line. Therefore, this is the line of best fit.

Section 3 – Bivariate Data Analysis

3.3.1

Least-squares line of best fit

To find the least-squares line of best fit, we use the formula y = mx + b where: sy • m is the gradient: r = sx – r is the correlation coefficient – sy and sx are the standard deviations of x and y • b is the y -intercept: y¯ = mx¯ – x¯ and y¯ are the means of x and y K EY P OINT :

If we are given the data set, we can use this to calculate the gradient and y-intercept: 1. 2. 3. 4. 5.

MODE > 2: STAT : 2: A + BX Enter data in x and y columns AC > SHIFT > 1 > 5:reg 1:A > = 2:B > =

Example 3.3 Find the equation of the least-squares line of best fit for the data set provided, correct to two decimal places. x¯ = 12.76

sx = 1.92

y¯ = 224.9

sy = 95.43

r = 0.87

Using this information, we must find the value of the formula y = mx + b. First, we need the gradient m: m=r

sy sx

= 0.87 ×

95.43

1.92 = 43.24171875

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3.3 Line of best fit

We do not round the value of the gradient yet, as we must still use the full figure to determine the value of b. It is only once these final calculations are done that the rounding occurs. Now we can determine the value of the y -intercept: b = y¯ – mx¯ = 224.9 – (43.24171875 × 12.76) = –326.8643313 Now that we have found the value of m and b, we can determine the equation of the least squares line of best fit: y = mx + b y = 43.24x – 326.86 Example 3.4 The heights and foot length of high school students are measured and recorded in a table. 150

152

155

160

165

167

169

Foot length (l) cm

25

23

28

26

30

29

27

a) Determine the equation of the least-squares line of best fit. To complete this question, we must use our calculators to determine the value of m and b, using the data provided. When inputting this into our calculator, note that the height is the x data and foot length is the y data. Your answers should come to be: A : y-intercept = –8.354409318 B : gradient = 0.2204658902 Now we can insert these figures into our equation. We must also change the x and y pronumerals into their actual measurements (i.e. in this case, y refers to foot length and x refers to height. l =(0.2204658902 × h) – 8.354409318 We can now use the equation found to answer subsequent questions. Example 3.5 b) What is the expected foot length of a student given their height is 180cm? Answer correct to two decimal places. Before you begin calculations for this question, change your calculator from STAT to normal mode! Here, we must change the value of h to match the information provided. l = 0.2204658902 × 180 – 8.354409318 = 31.33 cm

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Section 3 – Bivariate Data Analysis

Height (h) cm

3.3 Line of best fit

3.3.2

Making predictions by interpolation or extrapolation

When determining the equation of the line of best fit, we are using the values of a specific data set. We can use this line to then make predictions about other values, which may be in the data set or outside the data range, as completed in the question previous. When predicting values within the data set, it is called interpolation. When predicting values outside the data range, it is called extrapolation. This process must be used very carefully, as the line of best fit determined may not apply and therefore skew the information gathered. Example 3.6 The equation relating a basketball player’s height to number of hoops scored per game is given below (for heights between 180 and 196 cm): hoops scored = 0.34 × height – 61.62 a) What is the number of hoops (to the nearest whole number) scored for a person 185 cm tall? Is this interpolation or extrapolation? b) What is the number of hoops (to the nearest whole number) scored for a person 204 cm tall? Is this interpolation or extrapolation? For part a), to determine the number of hoops scored, input the relevant information into the formula: Section 3 – Bivariate Data Analysis

s = 0.34 × 185 – 61.61 = 1.28

≈ 1 This is interpolation as the x value (185 cm) is inside the data range provided. The equation therefore appropriately applies to this question, providing us with an accurate answer. For part b), again we input the relevant information into the formula: s = 0.34 × 204 – 61.62 = 7.74

≈ 8 This is extrapolation as the x value (204 cm) is outside the data range provided. The equation therefore does not account for this value, and may skew the answer calculated.

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The Normal Distribution

Section 4

The Normal Distribution In a normal distribution, the distribution of data is symmetrical about the mean. The mean and median are approximately equal. A bell-shaped curve represents a normal distribution.

Section 4 – The Normal Distribution

When observing this curve, we must note various properties: • Approximately 68% of data will have z-scores between –1 and 1. • Approximately 95% of data will have z-scores between –2 and 2 • Approximately 99.7% of data will have z-scores between –3 and 3 This empirical rule of distribution (the 68 – 95 – 99.7 rule) helps to remember the percentages. If you do not remember the empirical rule, you can also remember the percentages of the bell-curve on one side. As the curve is symmetrical, remembering the digits as 34%, 13.5%, 2.35%, and 0.15% can also help you apply this concept to z-score questions. Either way, it is vital that you have memorised at least one way of doing these questions, as examiners will not provide the percentages for you.

4.1 z-scores The z-score (standardised score) is used as a tool of comparison in a normal distribution. It is the number of standard deviations the score is from the mean. To calculate the z-score: z= • • • •

x – x¯ s

z is the z-score or standardised score x is the score x¯ is the mean of a set of scores s is the standard deviation

Therefore, we are subtracting the mean from the score, and dividing the outcome by the standard deviation.

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4.1 z-scores

K EY P OINT :

Many students find the concept of z-scores very confusing! Just remember: a z-score is the number of standard deviations from the mean. As we are about to explore, these scores allow us to easily compare data in a normal distribution. • A z-score of 0 indicates that the score is equal to the mean. • A z-score of 1 is one standard deviation above the mean. • A z-score of –1 is one standard deviation below the mean. The larger the z-score is, the further it is from the mean. This applies to both positive and negative numbers (i.e. –4 is further away from the centre than –3). Example 4.1 Miss Coleman checked the results of her students’ Maths test. She determined the mean of the class as 67% with a standard deviation of 5. What is Russell’s z-score if he obtained a score of 85? Answer correct to one decimal place. For this question, x = 85, x¯ = 67, and s = 5. z=

85 – 67

Section 4 – The Normal Distribution

5

= 3.6 To explain this result, we would say that Russell obtained a z-score of 3.6. This means that his test score is 3.6 standard deviations from the mean of 67. He therefore performed at a significantly higher rate than most students in his Maths class. Example 4.2 Markus is is also in the same Maths class as Russel. He obtained a score of 58 in the Maths test. Find Markus’ z-score and use both students’ z-scores to compare their results to the data and against each other. Markus’ z-score is: z=

58 – 67

5 = –1.8

In this Maths test, Markus obtained a z-score of –1.8. This indicates that his score is 1.8 standard deviations below the mean, and therefore, his performance is lower than the mean of the class. In comparison to Russell’s z-score of 3.6, it is clear that Markus performed worse than Russell.

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Part V

Topic 5: Networks

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Network Concepts

Section 1

Network Concepts This topic is new to the current HSC Mathematics Standard 2 syllabus. The knowledge gained from networks provides you with a clear understanding of the logical sequence of tasks and connections between people or items. Some aspects of this topic may be difficult to pick up straight away, so make the most of practice questions and extra resources!

1.1

Networks Network terminology

Definition

Network

A set of objects/tasks connected together to create a sequence

Vertices

The objects of a network (normally drawn as points)

Edges

Connect the objects of a network together

Section 1 – Network Concepts

Walk

A sequence of vertices and the edges between them

Path

A walk (sequence of events through a network) which does not visit any vertex more than once

Cycle

A walk with the same start and end vertex, which does not visit any vertex more than once (apart from the start/end)

Directed network Undirected network Degree of a vertex Weighted edge

When the edges in a network point in only one direction, typically indicated by an arrow When the edges in a network are bidirectional (or undirected) The number of edges which protrude from each vertex When the connection/edge have weights (numbers) assigned to them

Below is an example of a network diagram. It has several objects connected together to create a sequence. This network is directed as each edge only points in one direction (i.e. the edge connecting A to B only travels from A to B).

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1.1 Networks

Network diagrams are very useful when planning a sequence of events, tasks, or activities to achieve a final objective. For example, a university may use a network diagram to determine the cost of connecting various locations on a university campus with computer cables, or a local council may use a network diagram to determine the most cost-efficient way to connect water to each house.

1.1.1

Drawing network diagrams

We are going to begin this topic by learning how to draw simple network diagrams from information given in a table. This activity will enable you to understand the need for specific sequences of tasks. In the table below, we have objects A, B, C and D and an explanation of the connections between them. A

B

C

D

A

-

2

4

-

B

2

-

3

-

C

4

3

-

1

D

-

-

1

-

Before drawing a network diagram from this table, we need to understand a few key points:

Now we can attempt to represent this information in a network diagram. There are, of course, many different ways to interpret and visualise this information. If this explanation doesn’t work for you, take an approach which best suits your learning! 1. Begin by creating a vertex for object A. Note which vertices also connect to A, and use edges to connect these vertices together. In this case, objects B and C also connect to A. Ensure you label the weighted edges of these connections!

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Section 1 – Network Concepts

• The same object (vertex) does not connect to itself (e.g. there is a ‘-’ in the A row and A column, meaning that A does not link to A). • The values in the boxes represent the weighted edge between the two vertices (i.e. the edge between A and B is assigned a value of 2. • Not all vertices are connected to each other (e.g. vertex D does not connect to vertex B). • The network is undirected. We can determine this as the edge connecting A to B is presented two times, (A, B) and (B, A).

1.2 Shortest paths

2. Continue to connect the vertices together according to the table. For example, C connects to B with a weighted edge of 3, and C connects to D with a weighted edge of 1.

Now we have a simple network diagram reflecting the information in the table. Remember, there is no one way to draw this diagram. Have a look at the network diagram below – it is representing the same information!

Section 1 – Network Concepts

1.2

Shortest paths

We have previously learned that a path is a sequence of events through a network which does not visit any vertex more than once. In this section, we will be looking at paths, and will learn to determine the shortest path in a network. K EY P OINT :

The shortest path is a path between two vertices in a network for which the sum of the weights of its edges is minimised. For example, let’s look at the network diagram below to determine the shortest path from A to B:

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1.2 Shortest paths

First, let’s look at our potential options when travelling from A to B: 1. Start at A, follow the edges to C and D before arriving at B (ACDB = 4 + 6 + 1 = 11). 2. Start at A, follow the edges straight through the network to vertices E and F before arriving at B (AEFB = 6 + 5 + 1 = 12). 3. Start at A and work down, connecting vertices G and H before arriving at B: (AGHB = 4 + 6 + 2 = 12). After considering the length of each path connecting A to B, we need to identify which path has the weights of its edges minimised (i.e. is the smallest). From our calculations, we can see that the path ACDB has the smallest value of 11, and is therefore the shortest path. K EY P OINT :

When calculating the shortest path, be very careful to consider all potential paths connecting the two vertices identified. This may require you to double check the network to make sure you have not missed any smaller edges.

1.2.1

Minimum spanning trees

A tree is a network, or part of a network, that contains no circuits or cycles. Hence, a minimum spanning tree is a set of edges connecting every vertex in a network. Examples of using minimum spanning trees may involve things like connecting electricity or phone connections throughout a housing estate in the most efficient and cost-effective way possible. Example 1.1 There was once a city with no roads. During the wet season, getting around the city was difficult – the ground became very muddy and cars were getting stuck in the mud. The mayor decided that paved roadways were necessary in the city, but did not want to spend excessively as there were other aspects of the city that needed work. The mayor therefore specified two conditions: 1. Enough pathways must be built so that each person could travel from their house to anyone else’s house using the paths (this may be via other houses). 2. The paving should be completed at a minimum total cost.

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Section 1 – Network Concepts

To understand this concept, we are going to use the Muddy City Problem.

1.2 Shortest paths

As you can see, this problem involves the use of minimum spanning trees. To find the shortest path, we will draw a diagram of the muddy city. This will be an undirected network including the vertices and weighted edges shown on the previous page.

Section 1 – Network Concepts

When solving minimum spanning trees, there are two algorithms that we can use: 1. Prim’s algorithm: to find the minimum spanning tree, select the shortest edge and continue by selecting the shortest attached edge from there. This continues until all vertices are included, and helps avoid any possible loops. 2. Kruskal’s algorithm: to find the minimum spanning tree, first select the shortest edge, and then select the next shortest edge, regardless of whether it is attached to the previous edge or not. This continues until all vertices are connected within the spanning tree. Here are our two solutions, using Prim’s and Kruskal’s algorithms to find the minimum spanning tree. Please note that whichever algorithm is used, you should still get the same value from the spanning tree, even if the trees look slightly different!

These diagrams represent the minimum spanning trees of the muddy city. Using both Prim’s and Kruskal’s algorithms, the shortest path enabling the connection of each vertex is 23. 2 + 3 + 2 + 3 + 3 + 3 + 2 + 3 + 2 = 23 As we can see, solving the minimal spanning tree of a network can be a simple process, but requires you to be very careful! The two algorithms provide you with effective methods to find the minimum spanning tree of any given network diagrams. Use these to your advantage! 106

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Critical Path Analysis

Section 2

Critical Path Analysis 2.1

Duration and interdependencies of activity

Network diagrams represent the duration and interdependencies of activities that must be completed during a particular project (e.g. preparing a meal or completing an assessment task). In the table below, we are provided with activities, as well as which activities they must be preceded by. Using this information, we will construct a network diagram to represent the importance of sequencing tasks in the correct order, so that the interdependency of each activity allows the overall objective to be completed efficiently. Redecorating a bedroom Activity

Duration (hrs)

Must be preceded by

Paint woodwork

8

-

B

Assemble drawers and tallboys

4

-

C

Lay carpet

5

A, D

D

Hang wallpaper

12

A

E

Hang curtains

2

A, D

1. We start by drawing the activities that must occur before other activities can commence. For example, painting the woodwork (A) must be finished before activities C, D, and E can occur. 2. Now we can input the activities that must occur after A. We must note, however, that activity D must be completed before C and E can commence. Draw this on the diagram. Activity B does not have any predecessors and therefore can be completed at any stage during the sequence. Notice the direct line connecting activity B from the start vertex to the finish vertex.

This network diagram now represents the sequence, duration, and interdependencies of these activities in order to re-decorate the bedroom in the correct order. By constructing network diagrams correctly, we can apply more complex concepts to the diagram such as finding the critical path.

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Section 2 – Critical Path Analysis

A

2.2 Critical path analysis

2.2

Critical path analysis

An application of network diagramming is the use of a critical path analysis. A critical path represents the shortest length of time to complete all tasks in a particular project. It also helps us identify the latest start time for certain steps without delaying the overall project completion. • The weight of the critical path is the minimum length of time required to complete the project. • Increasing the time for any critical activity will also increase the time necessary to complete the project.

2.2.1

Earliest start time

The earliest starting time (EST) of an activity refers to the earliest time the activity can commence after completing all prerequisite activities. • Activities that share the same start vertex also share the same EST. • The EST of the start vertex is always zero. • The EST of the finish vertex is the time it takes to complete all activities in the project. This is called the critical time of the project. • The EST of each activity is recorded on the forward scan (which will be discussed later). Example 2.1 Calculate the EST for each activity in the following activity chart. Section 2 – Critical Path Analysis

• Activity A: EST = 0 Remember that any vertex connected to the start vertex will have an EST of 0. Recall that these activities have no prerequisites and can start immediately. • Activity B: EST = 2 Edges that share the same vertex also share the same EST. This activity can start after A has been completed. • Activity C: EST = 2 As with activity B, this can start after activity A has been completed. • Activity D: EST = 6 The EST of activity D must be 6 as it has to account for the completion of A, B, and C, which has a maximum path of 6.

2.2.2

Earliest finishing time

The earliest finishing time (EFT) of an activity refers to the soonest an activity can be completed. This is the time it takes to complete the activity as well as all prerequisite activities. Earliest finishing time = weight + EST Unlike the EST, the EFT is not shared by all activities with the same vertex. Instead, the EFT depends on the weight of the individual activity.

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2.2 Critical path analysis

2.2.3

Performing a forward scan

A forward scan helps us to find the EST for all activities. This algorithm begins at the start vertex and moves along the edges of the activity chart. It calculates the EST at each vertex as the maximum EFT of all activities that end at that vertex. Example 2.2 Perform a forward scan to determine the EST for each activity in the following activity chart.

As you can see on the diagram above, each vertex is labelled with two boxes: one for the EST (left) and one for the LST or LFT (right). In this forward scan, we are just finding the earliest start time.

(This example is continued when we cover completing backward scans later – see page 110 for a completed forward scan for this network!)

2.2.4

Latest finish and start times

The latest finish time (LFT) of an activity is the latest time an activity can be completed without causing a delay in the critical time of the project. • Activities the share the same end vertex are all part of the same collection of prerequisites, and so will have the same LFT. • The LFT of the start vertex is always zero. • The LFT of the finish vertex is the critical time of the project. • The final activities in a project all have an LFT equal to the critical time of the project. Consider the activity chart below. Weights are in minutes, and the critical time is 5 minutes.

Let’s work backwards! Since activity C is the final activity, it must be completed by the critical time of 5 minutes. Therefore, the LFT is 5. Activities A and B share the same end vertex, and so must share the same LFT. They must be finished by 3 minutes to allow activity C to commence.

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Section 2 – Critical Path Analysis

1. The start vertex always has an EST of 0. This means that activities A, B, and C can begin straight away. 2. To find the EST of the middle vertex, we must find the maximum EFT of all edges which end at this vertex. This number represents the earliest time that all activities A, B, and C can be completed, which is also the soonest that D, E, and F can begin. In this example, the EST of this vertex is 6. 3. We use the EST of the middle vertex to find the EST of the finish vertex. Again, find the maximum time for all activities to be completed. The EFT for the finish vertex is therefore equal to 6 + 8 = 14. Remember that the EST of the finish vertex is the critical time for the project.

2.2 Critical path analysis

The latest start time (LST) is the latest time an activity can begin. This is calculated as the latest time it can be finished (LFT) minus the time taken to complete the activity. Latest starting time = LFT – weight Unlike the LFT, the LST is not shared by all activities with the same end vertex. Instead, it depends upon the weight of the individual activity.

2.2.5

Performing a backward scan

A backward scan helps us to find both the latest start and finish times for all activities. The backward scan algorithm starts at the finish vertex and works backwards along the edges of the activity chart. It calculates the LFT at each vertex as the smallest LST of all activities that start from that vertex. So, ultimately – our backward scan helps us find both the LFT and LST, but of different vertices. Example 2.3 Perform a backward scan to determine the LFT for each activity in the following activity chart.

Section 2 – Critical Path Analysis

1. Begin at the finish vertex and copy the EST over to the box on the right. This number is now the LFT for every activity that points towards the finish vertex (i.e. D, E, and F). 2. Move towards the left and consider the LST of every edge that starts from the middle vertex. Activity

D

E

F

LFT

14

14

14

Activity

A

B

C

Weight

3

8

6

LFT

6

6

6

LST

11

6

8

Weight

4

5

6

LST

2

1

0

As we can see, the LFT at this middle vertex is the smallest LST – therefore the LFT is 6. This means that the latest finish time for activities A, B, and C is 6, without delaying the rest of the project.

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3. Now consider the start vertex. Calculate the LST of every edge that begins at this vertex.

The LFT at the start vertex is again the smallest LST, which is 0.

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2.2 Critical path analysis

K EY P OINT :

The LFT of the start vertex should always be equal to 0. This is an easy way to check that your working throughout the backward scan has been correct!

2.2.6

The critical path

The critical path is the longest path from the start vertex to the finish vertex. It represents the sequence of activities that are time-critical, meaning any delay in these activities will cause a delay to the entire project. Any activity on the critical path is called a critical step. There will always be at least one critical path, and there may sometimes be more than one. Let’s look at an example to see these concepts in practice.

Different paths throughout this network could include: • Start – A – B – E – Finish = 35 • Start – A – B – D – Finish = 26 • Start – C – D – Finish = 6

2.2.7

Float times

We can use the float time to identify the critical activities, and thus the critical path. K EY P OINT :

A critical activity is any task that, if delayed, will hold up the earliest project. If LST = EST , the activity is said to be critical. The gloat time measures the variability in the start time of an activity. It is calculated as the difference between the latest possible time (LST) and the earliest possible start time (EST). Float time = LST – EST Every activity on the critical path must have a float time of zero! Let’s go back to our previous example to check the float time of each activity. Observe the annotated diagram below to understand where we get each value for activity C.

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Section 2 – Critical Path Analysis

However, it is important to prove that each activity is critical. This can be determined by calculating the float time of each activity.

2.3 Maximum-flow, minimum-cut theory

When calculating the float time, it’s a great idea to set up a table which clearly displays each of the key findings after completing a forward and backward scan. This process takes practice, so try different examples to test your knowledge! Calculating the float time Activity

A

B

C

D

E

Weight

10

15

5

1

10

EST

0

10

0

25

25

LFT

10

25

34

35

35

LST

10 – 10 = 0

25 – 15 = 10

34 – 5 = 29

35 – 1 = 34

35 – 10 = 25

Float time (LST – EST)

0

0

29

9

0

critical

critical

critical

As calculated using our float times, we know that our three critical activities are A, B, and E, as they each have a float time of 0. This means that these activities cannot be delayed without affecting the completion time of the entire network. Section 2 – Critical Path Analysis

Our other two activities, C and D are not critical. Activity C can be delayed by up to 29 units (e.g. minutes, hours), without delaying the completion of the entire project. Activity D can be delayed by up to 9 units without delaying the project. Therefore, our critical path is Start – A – B – E – Finish, with a value of 35. K EY P OINT :

Although this is a long process, it is very effective in correctly calculating the float time for each activity! With practice, you’ll be able to do this quickly and hopefully enjoy the process!

2.3

Maximum-flow, minimum-cut theory

When solving small-scale network flow problems, we can use the maximum-flow, minimum-cut theory to determine the maximum flow of a variable (e.g. water, electricity) throughout a network. For example, a telephone company may want to investigate the maximum number of simultaneous phone calls between two points via the operating land-lines and towers in place.

2.3.1

Maximum flow

There are many ways to calculate the maximum flow through a network diagram. The maximum flow is the maximum amount of a variable that can be ‘pushed’ through the network. To explain this theory, observe the diagram and explanations below. First, we will walk through a simple example to explain the concept. Then, we will complete a more complicated, practical example that could be asked in an exam situation.

In this example, we have a network diagram of the flow of water throughout a network. It is important to recognise the source and the sink. This is where the flow of water must follow – leaving the source (S) to reach the sink (T). As we are finding the maximum flow of this network, we must find the maximum volume of water that can flow through the network. 112

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2.3 Maximum-flow, minimum-cut theory

K EY P OINT :

In network flow diagrams, the objective is to discover how much can actually flow from the source to the sink by assigning a flow to each edge. The maximum flow from the source to sink is called the flow capacity. In order to find the maximum flow, we must note the capacity of each edge. For example, the edge connecting the source to A in the above network can carry a capacity of 3. We begin by trying to push as much water through one set of edges as possible. Once the capacity has been pushed through, we will write this next to the edge. It is very important that we are aware of how much capacity can be pushed through the edges connected to the sink – these are the final edges and must be considered as the points the capacity has to be able to flow through.

K EY P OINT :

The excess flow capacity of an edge is the capacity of the edge minus the flow throughout the edge. Edges that have had all capacity used (i.e. there is zero excess flow capacity) are called ‘saturated.’ Example 2.4 Determine the maximum flow of the network diagram below.

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Section 2 – Critical Path Analysis

• 1st flow: S–A–B–T. This flow will start at the source (S) and travel through A and B to T. As we can see, we have pushed the maximum capacity through the edges (S to A) and (A to B), however the edge (B to T) has only used 3 of 5 capacity, therefore still allowing 2 to flow through. • 2nd flow: S–C–B–T. This flow will start at the source (S) and travel through C and B to T. While the edge S to C has a maximum capacity of 4, we can only push 2 through as this is what the other edges in the flow can cater for. It is in this flow that we will use the remaining 2 in the edge B to T, as revealed in the 1st flow. I have crossed the original capacities out as they have been used. The circled numbers at the end will contribute to the total maximum flow of the network. • 3rd flow: S–C–T. This flow will start at the source (S) and travel through C to T. If we observe the flow once completed, we can determine that the maximum flow of this network is the sum of the circled numbers: 3, 2, and 2. maximum flow = 3 + 2 + 2 = 7

2.3 Maximum-flow, minimum-cut theory

Following the same process as we went through earlier:

• 1st flow: A – B – C – D – G = 370 • 2nd flow: A – E – F – G = 400 After completing these first two flows, we can recognise that the final edges (D–G) and (F–G) have not been used to their full capacity. Therefore, we will attempt to maximise their flow by finding other flow networks. • 3rd flow: A – B – F – G = 200 • 4th flow: A – B – C – F – D – G = 30 After completing the third and fourth flows, we must recognise that the edge F–G still has 50 of its capacity that has not been used. Let’s try to maximise the flow by using this edge. • 5th flow: A – B – C – F – G = 50 Section 2 – Critical Path Analysis

Yay! We have used the full capacity of the last two edges connecting the sink, therefore have achieved the maximum flow of the network! We need to add our flow values to determine the maximum flow: maximum flow = 370 + 400 + 200 + 30 + 50 = 1050

2.3.2

Minimum cut

The minimum cut theory is a useful way to easily check the maximum flow of a network diagram. It is often a quicker method to do so. Cuts are used to prevent all flow from the source to the sink – therefore, a valid cut must completely isolate these two points. To calculate the minimum cut, we add the weights of the cut edges. K EY P OINT :

When completing minimum cuts, there are a few key points to remember: • The cut must isolate the source from the sink. • Always work from above to below. • Make enough cuts to find the minimum cut. Let’s use the same network diagram to determine the minimum cut:

To find the minimum cut, we must cut through the edges and find the minimum value possible to isolate the source from the sink. 114

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2.3 Maximum-flow, minimum-cut theory

We will begin by completing as many cuts as possible through the diagram. The value of each cut is calculated by adding the weighted edges the cut goes through.

As we can see, each of the cuts in this diagram is equal to 7. Therefore, the minimum cut, and maximum flow in this network is 7. K EY P OINT :

There are often very complex cuts that can be made in a network diagram. However, as we are trying to find the minimum cut, these confusing cuts are not necessary. We are trying to find the lowest value!

Below is another example of finding a minimum cut:

Thus, our minimum cut is 5.

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Section 2 – Critical Path Analysis

We can check this answer by completing the maximum flow calculations (as we did previously!) to find that our minimum cut is 7.

Part VI

Exam and Revision Tips

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Advice for studying

Section 1

Advice for studying Complete as many practice papers as you can! I know, this can be draining. Sitting down for two-and-a-half hours to complete a whole exam is definitely not fun. But, with hard work, you will achieve the best result possible. Prior to my HSC exams, I began completing each previous past HSC paper, answering each question, and then going over them using the worked solutions. In this way, I was able to find which concepts I struggled with, and refined my knowledge in these areas. This is a little bit tougher for the new syllabus, but you can still skim through past HSC examinations (which can be found on the NESA website, or many school websites) and find questions that are relevant to the current syllabus. For the new topic (Networks) there are a heap of online resources for you to use. Expose yourself to a wide variety of questions, solutions, and techniques used by high-achieving students, and you will reap the benefits in refining your own skills! Make a study routine

If you know you will be studying Maths first thing in the morning, prepare what you will be practising the night before. Print off any worksheets, exam papers, or resources that will be useful, and absolutely maximise your time. Remove distractions, annoying people, or unnecessary noise, and create a space that suits you! This will not only enable you to feel more comfortable during your study period, but will help you to recreate this perception in an exam situation – therefore allowing you to remain focused and thorough in your work. Yes, they give you the formula sheet in an exam! But really, you should know these off-by-heart already! Throughout the year, you will learn a whole range of formulas to use across different topics. Whilst it may be hard to remember these all, try to commit them to memory! Use flashcards, create a Quizlet, ask your friend to test you, whatever it takes. The formula sheet is always a saviour, but it doesn’t give you any context as to how to use the formulas. This is based purely on your own learning.

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Section 1 – Advice for studying

You will have to coordinate your study routine for all of your subjects, and sticking to some kind of schedule is a highly valuable technique to maximise your study time. Allocate yourself a specific slot on each day, or create a timetable to ensure you have devoted enough time to Maths – don’t ignore it just because you don’t enjoy it.

Advice for exams

Section 2

Advice for exams Let the marker know you know exactly what you are doing. In your exams, each question is allocated a number of marks. You should look at these marks as ‘steps’ you need to take. For example, if a question is worth 4 marks, assume that there are around 4 steps you need to complete in order to receive full marks. Set your work out clearly, showing each step you took to determine your final solution. By guiding the marker through your work, you are proving yourself as a confident, intelligent student, and that’s exactly what they want to see! Even if you don’t think part of your working is relevant, show it anyway. You can always maximise marks if you include diagrams, small workings out on the side, or anything related to solving the question (as long as it is calculated correctly). This is one simple way to gain as many marks as possible! Break down each question carefully

Section 2 – Advice for exams

In an exam situation, your mind will be racing, and you may think that if you’re not going fast, you won’t finish everything. Most often, this is not the case. Going in with a rough idea of how much time you will dedicate to each section, or even just following the suggested time printed clearly on the front page will allow you to complete each question thoroughly and within the time frame. Look at each question carefully. Use a highlighter, red pen, or ruler to find the key points. What is the question asking you to find? What are the initial measurements? How many marks are allocated to the question? Then, once you are confident, begin your working. There is no point rushing through the questions, and then receiving a poor mark. You have the time – use it wisely! Avoid losing marks through unnecessary mistakes! Ask yourself: did I put the units on the end of a question like it asked me to? Did I actually read the calculator properly and write 3.79 instead of 3.97? Did I round off too early and distort my final answer? These are all common mistakes HSC students make – and yes, I did too! But by addressing each of these mistakes and being fully aware of what you are doing, you can easily maximise your marks!

My final note to you... To every student reading these notes: please try to enjoy the HSC as much as you can! These years are so rewarding – they can solidify your work ethic, ready you for what life may throw at you after Year 12, and connect with others in your year level to make lifelong friends with whom you can reminisce about these tiring experiences! And most importantly, the HSC gives you an opportunity to thrive as a student. You know your capabilities, so it’s up to you to do everything you can to reach (or perhaps even exceed) them. If you dedicate yourself to practice, patience, and persistence, I promise you will reach your goals! Good luck! 118

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