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Reg. No.: 2011/011959/07
2509-E-MAM-SG01
CAPS-aligned
Prof. C Vermeulen, Lead author
H Otto M Sherman
Exercise
Exercise 65: Relationship between exterior angle of a triangle and opposite interior angles.
SAMPLE
5. Minimum conditions for two triangles to be congruent
Exercise 66: Investigating the conditions for two triangles to be congruent ...................................................................................................123 6. Sides, angles and diagonals of quadrilaterals........................................125
Exercise
Investigating the sides, angles and diagonals of
Exercise
PREFACE
SAMPLE
Grade 9 is the last year of the General Education and Training (GET) phase, in which Mathematics is a compulsory subject. From Grade 10 you must choose between Mathematics and Mathematical Literacy. Sometimes you choose Mathematical Literacy for the wrong reasons. These include finding Mathematics up to Grade 9 difficult and that Mathematical Literacy in Grades 10 to 12 is less challenging. However, before you choose at the end of Grade 9, you must carefully consider your plans for the future – especially if you wish to obtain tertiary (university or college) qualifications. For many tertiary courses Mathematics, and not Mathematical Literacy, at Grade 12 level is a prerequisite for admission. For this reason, it is important that you should work hard in Mathematics in Grade 9, enabling you to continue with Mathematics in Grades 10 to 12 to keep later options open.
Optimi’s Grade 9 Mathematics offering consists of two study guides and two facilitator’s guides, which are based on the concepts of Optimi’s GuidEd Learning™ model to help you and your facilitator achieve success in the study of mathematics. These books cover all work required for Grade 9 mathematics and have been compiled in accordance with the CAPS guidelines as required by the Department of Basic Education.
The study guides are supported by supplementary lesson structures on the online Optimi Learning Platform (OLP), available at optimi.learning.co.za/login. These lesson structures offer continuous guidance to support and enrich the learning process. This guidance is based on the latest insights in education, cognitive psychology and neuroscience. Note that the study guides can also be used independently of the OLP.
In the following section, we explain how the study guides and facilitator’s guides have been compiled and how you and facilitators can use these to achieve success in mathematics.
The study guides and facilitator’s guides are divided into 21 themes. Study guide 1/2 and facilitator’s guide 1/2 cover themes 1 to 12 (term 1 and 2) and study guide 2/2 and facilitator’s guide 2/2 cover themes 13 to 21 (term 3 and 4). The themes
correspond with the CAPS guidelines with regard to content and time allocation and represent the year plan.
Time allocation
According to the CAPS requirements, at least 4,5 hours should be spent on teaching mathematics per week. For example, 13,5 hours (three weeks × 4,5 hours per week) will be spent on teaching Theme 1 (whole numbers and integers). Themes have not been subdivided into lessons; you and your facilitator are at liberty to complete as much content per session and per week as your progress allows.
If you work at a slower pace, the necessary adaptations should be done so that you will still be able to master all the work in time.
Note that the teaching time referred to above does not include the time during which you should apply and practise the knowledge and concepts you have learned. For this purpose, various exercises are provided throughout each theme.
These exercises involve different ways of applying and practising new knowledge and cover various degrees of difficulty. You should try to do all these exercises. Complete solutions are provided in the facilitator’s guide.
Structure of themes
Learning is a complex process. Millions of cells and neural pathways in our brains work together to store new information in the long-term memory so that we will be able to remember it later on.
Long-term memory is not our only type of memory and when we learn, our working memory is just as important. Working memory is different from long-term memory and has a limited capacity. This means that one’s working memory can only handle a small amount of new information at a time.
Each theme has the same structure to make it easier for you to navigate through the study guide. The structure is as follows:
Introduction
SAMPLE
When one learns mathematics, there is a lot of new information your brain needs to process, which can easily exhaust your working memory. The study guides have been written and compiled in such a way that it does not overtax the working memory and therefore simplifies the process of learning mathematics.
This briefly tells you what the theme is about without providing details or using ‘difficult’ or unknown concepts. A comprehensive list of the learning outcomes you need to master in a specific theme is given as a summary at the end of the theme.
Prior knowledge
This section tells you what existing knowledge you need to master the theme involved.
Revision
This may involve one of the following:
1. revision of the concepts, definitions and procedures required as previous knowledge
2. an exercise or activity with solutions so that you can test your prior knowledge by yourself, or
3. a combination of the above.
Do not neglect this revision. It is important to work through this section thoroughly. Mathematical concepts often follow on one another and if basic knowledge is lacking or has not been mastered sufficiently, this will handicap the formation of new knowledge.
Following the introductory part of the theme, new knowledge is dealt with in subthemes.
Each subtheme has the following structure:
SUBTHEME
Introduction
New concepts and procedures are explained. Relevant previous knowledge is also dealt with here if necessary.
Worked examples
Worked examples show you how the new concepts and procedures are applied and help you understand and apply the newly taught concepts and procedures.
Exercises
The exercises give you the opportunity to practise the concepts and procedures taught. It is important for you to try and complete all exercises. Complete solutions are provided in the facilitator’s guides.
Mixed exercises are also provided, where you get the opportunity to practise different concepts and procedures and integrate these with previous themes.
Summary of theme
Here you will find a summary of what you should have mastered in the theme. This is expressed in more formal mathematical language to be in keeping with the CAPS learning objectives.
End of theme exercise
SAMPLE
This is a mixed exercise involving all concepts and procedures dealt with in the theme and can also be integrated with previous work. The degree of difficulty of this exercise varies. It is important that you try and complete all the exercises. Complete solutions can be found in the facilitator’s guides.
Mixed exercises such as these in this textbook form a very important component of mastering mathematics. There is a big difference between the ability to recognise one’s work and the ability to recall it. When you are able to recognise your work, you will often say ‘Oh, of course!’ but you may struggle to remember this when writing an
examination.
When you can recall your work, this means that you have captured that knowledge in your long-term memory and are able to remember and use it. Mixed exercises enable you to not only recognise the work, but also recall it from your long-term memory.
When you practise the same type of sum or problem over and over, you often get lazy and do not reflect upon the exercise anymore. You are convinced that you know exactly what type of sum or problem you need to solve. But in a test or exam, all these problems are mixed up and then it might be difficult to know what to do.
When mixed exercises form part of the learning process, you learn to identify and complete a sum or problem correctly. This means that you are truly prepared for tests or exams, because you can recall your work instead of merely recognising it.
Self-evaluation
In each theme, and usually following each sub-theme, there is an activity where you need to reflect critically about the extent to which you have mastered certain concepts and procedures. This activity has the following format:
Use the following scale to determine how comfortable you are with each topic in the table below:
1. Alarm! I don’t feel comfortable, but I just need more time to work through the topic again.
2. Help! I don’t feel comfortable with the topic at all. I need help.
3. OK! I feel moderately comfortable with the topic, but I still struggle sometimes.
4. Sharp! I feel comfortable with the topic.
5. Party time! I feel totally comfortable with the topic and can even answer more complicated questions about it.
Complete the table:
Tip: Complete each self-evaluation as honestly as possible. If there are aspects which you have not mastered, revisit these and make sure that you do master them. Ask the facilitator for help. It is important not to move on to a next theme or sub-theme before you have mastered the topic involved, even if this means that you spend more time on a specific theme than recommended by the CAPS.
Assessment criteria
Visit Impaq’s online platform (OLP) for the assessment plan and comprehensive information about the compilation and mark allocation of tests, assignments and examinations. The number of assignments, mark allocation and relative weighting are subject to change.
SAMPLE
*To be completed before the end of the year examination
The two papers in the middle and at the end of the year are compiled as follow:
Mid-year examination
Tip: Mark allocation per topic ≈ percentage weighting per topic × total mark for the paper.
SAMPLE
Tip: Mark allocation per topic ≈ percentage weighting per topic × total mark for the paper.
End-of-year examination
Numeric and geometric patterns
SAMPLE
Supplementary books
Any other books can be used along with this textbook for extra exercises and explanations, including:
• Maths 4 Africa, available at www.maths4africa.co.za
• The Si��avula textbook, available online for free at www.siyavula.com
• P��thagoras, available at www.fisichem.co.za.
Calculator
We recommend the CASIO fx-82ES (Plus) or CASIO fx-82ZA. However, any scientific, non-programmable and non-graphing calculator is suitable.
Tip: Ensure that you have a suitable calculator.
Theorem of Pythagoras
Theorem
Numbers, operations and relationships
THEME 1
WHOLE NUMBERS AND INTEGERS
Introduction
In this theme we will revise and extend work covered in Grade 8 regarding:
• numbers that form the real number system and their properties
• calculations with whole numbers using all four operations, different calculation techniques and strategies for calculations with whole numbers
• prime factorising of numbers to find the lowest common multiple (LCM) and highest common factor (HCF)
• problems which involve ratio and rate, as well as direct and indirect (inverse) proportion
• problems which involve money
• calculations with integers using all four operations.
1. C LASSIFICATION OF NUMBERS
Our number system consists of different types of numbers. Numbers can be classified into different groups.
• Natural numbers
Natural numbers are 1 ; 2 ; 3 ; ...
The symbol for natural numbers is ℕ.
Natural numbers include
• Even numbers (numbers divisible by 2): 2; 4; 6; …
SAMPLE
• Odd numbers (numbers not divisible by 2): 1; 3; 5; …
• Prime numbers (numbers with only two factors namely the number itself and 1): 2; 3; 5; …
Rational numbers are numbers that can be written as a b , with b ≠ 0.
The symbol used for rational numbers is ℚ.
Examples of rational numbers:
◦ Natural numbers such as 1; 2; 3, because we can write them as ratios or fractions a b . For example, 1 = 6 6 ; 2 = 12 6 ; 3 = 15 5 .
◦ 0, because we can write 0 as 0 5 , for example.
◦ Integers such as –3; –2; –1, because we can write them as ratios or fractions a b . For example, 1 = 6 6 ; 2 = 12 − 6 ; 3 = 15 5 .
◦ Ordinary fractions, for example 3 4 ; 2 3 ; 5 6 .
◦ Improper fractions, for example 3 2 ; 5 4 ; 25 3 .
◦ Mixed numbers like 1 1 2 ; 3 5 12 ; − 5 3 4 , because we can write them as improper fractions a b where a > b. For example, 1 1 2 = 3 2 .
◦ Terminating decimal fractions like 0,5; 0,25 or 0,875, because they can be written as ordinary fractions or ratios a b : 0,5 = 1 2 ; 0,25 = 1 4 or 0,875 = 7 8 .
◦ Recurring decimal fractions like 0, 3 ˙ or 0, 3 ˙ 7 ˙ , because they can be written as ordinary fractions a b : 0, 3 = 1 3 or 0, 37 ˙ = 37 99 .
• Irrational numbers
Irrational numbers are numbers that cannot be written as a b , with b ≠ 0. The symbol used for irrational numbers is ℚ′. Examples of irrational numbers:
◦ π
◦ Any non-recurring, non-terminating decimal fraction, for example 0,3876519...
◦ Any root that cannot be determined exactly, for example √ 2 ; 3 √ 20 ; 4 √ 36
Summary: Types of decimal fractions
The fractions named in the shaded rectangles above are rational numbers.
• Real numbers
• Non-real numbers
All numbers that are not real numbers, for example √ 4 ; 4 √ 16 or 6 √ 10 . The symbol used for non-real numbers is ℝ′ .
SAMPLE
All the numbers listed above are called real numbers.
The symbol used for real numbers is ℝ.
Representation of the number system
3 √ 27 and 5 √ 32 are real numbers, because 3 √ − 27 = − 3 and 5 √ 32 = − 2
We can represent the number system and the types of numbers it consists of in two ways, namely from the inside to outside or from top to bottom.
• From the inside to the outside 1; 2; 3; ... Real number (ℝ) Non-real numbers (ℝ')
Complete this table by making a cross in the appropriate block or blocks if the given number belongs to that specific type.
Rational numbers (ℚ) Fractions
Integers (ℤ) Ordinary
Decimal fractions
Positive integers, i.e. natural numbers (ℕ)
Whole numbers (ℕ 0 )
Terminating
Non-terminating recurring
Explanations
• 22 7 is a ratio and an improper fraction, and therefore a rational number that is only an approximation of π. Rational numbers form part of real numbers.
• π is always an irrational number and irrational numbers form part of real numbers.
NOTE
• π is the circumference of a circle divided by its diameter and the answer is a non-recurring, non-terminating decimal. So, it is an irrational number.
• 22 7 is an approximation of π.
• 3,56 is a terminating decimal fraction and therefore a rational number. Rational numbers form part of real numbers.
• 3 + √ 49 2 = 3 + 7 2 = 10 2 = 5. This number is a natural number, a whole number, an integer, a rational number and a real number.
• 1 2 3 is a mixed number and therefore a rational number and a real number.
• √ 7 is an irrational number because it cannot be written as a ratio or fraction. It is also a real number because irrational numbers form part of real numbers.
• − 100 is a negative integer and a rational number (it can be written as a ratio or fraction), and rational numbers form part of real numbers.
• √ 10 × √ 10 = 10, so this is a natural number, a whole number, an integer, a rational number and a real number.
• 1,010010001... is a non-terminating, non-recurring decimal fraction and can therefore not be written as a ratio or fraction. That means it is an irrational, real number.
• √ 8 is a non-real number.
• 0, 3 ˙ 1 ˙ 2 ˙ is a recurring decimal fraction and can therefore be written as an ordinary fraction. It is therefore a rational, real number.
Exercise 1: Classification of numbers
1. Complete this table by only making a cross in the appropriate block if the number is an element of the specific number type.
3. You are given the following list of numbers:
SAMPLE
3.1 Natural numbers
3.2 Integers which are not natural numbers
3.3 Irrational numbers
3.4 Rational numbers which are not integers
2. PR OPERTIES OF WHOLE NUMBERS
In this subtheme we briefly review the properties of whole numbers.
• Commutative property for addition and multiplication
If we add or multiply two whole numbers, we can change the order of the numbers. Examples include:
3 + 2 = 2 + 3 = 5 and 3 × 2 = 2 × 3 = 6.
• Associative property for addition and multiplication
If we add or multiply three or more whole numbers, we can group the numbers differently and still get the same answer. Examples are 5 + (2 + 3) = 5 + 5 = 10 and (5 + 2) + 3 = 7 + 3 = 10. So, 5 + (2 + 3) = (5 + 2) + 3.
If we add 0 to any number, the number stays the same. For example, 5 + 0 = 0 + 5 = 5. We say that 0 is the identity element of addition.
If we multiply any number by 1, the number stays the same. For example, 1 × 32 = 32 × 1 = 32. We say that 1 is the identity element of multiplication.
But this is impossible, since any number multiplied by 0 will be 0. Therefore, division by 0 is not defined.
REMEMBER
Division by 0 is not defined. Example: 32 8 = 4 because 8 × 4 = 32.
Now if we divide 32 by 0 then 32 0 = a certain number. That number multiplied by 0 must again give 32.
Exercise 2: Properties of whole numbers and using them for calculations
1. Match each of the following calculations with one of the number properties shown. Only write down the letter of the correct property next to the number.
SAMPLE
1.1 10 + 0 = 10 A. Commutative property of addition
1.2 3 × (8 + 2) = 3 × 8 + 3 × 2 B. Commutative property of multiplication 1.3 16 + 20 = 20 + 16 C. Associative property of addition 1.4 (18 + 8) + 2 = 18 + (8 + 2) D. Associative property of multiplication
1.5 306 × 1 = 306 E. Identity element of addition 1.6 (9 × 10) × 2 = 9 × (10 × 2) F. Identity element of multiplication 1.7 100 × 301 = 301 × 100 G. Distributive property
2. Use the properties of whole numbers to make it easy to determine the answers of the following. State which property you used and show all your work to prove you used the property.
C ALCULATIONS WITH WHOLE NUMBERS AND DIFFERENT CALCULATION TECHNIQUES
In this subtheme we revisit doing calculations with whole numbers using all four operations.
3.1 U se estimation to determine if answers are realistic
Worked example 2: Use of estimation when adding numbers
Use estimation to calculate 433 + 237. Complete a table like the one given here:
Notes on calculator use
• Calculators should be used for calculations with large numbers only.
• For smaller numbers, use mental or pen-and-paper methods.
• Do not become dependent on your calculator for all calculations.
• Calculators remain a useful tool for checking solutions.
• When you use a calculator to find an answer, estimate the value of your answer. You can sometimes get the most absurd answers if you type in the wrong numbers on your calculator. Estimation prevents this.
SAMPLE
Note when we add
If the quantity that we round one number down is the same as the quantity that we round the other number up, the estimated answer is equal to the calculated answer.
Worked example 3: Use of estimation when subtracting numbers
Use estimation to calculate 852−425. Complete a table like the one given here:
Theme 1: Whole numbers and integers
Worked example 4: The use of estimation when multiplying numbers
Use estimation to calculate 78 × 102. Complete a table like the one given here: Operation on whole numbers
to the nearest
78 × 102 Solution
on whole numbers
(round 2 down)
(round 2 down)
Exercise 3: Using estimation
1. Use tables like the one given below to estimate and then calculate, using a calculator, the following:
on whole numbers
3.2 A ddition, subtraction and multiplication in columns and long divi sion with whole numbers
In this subtheme you get the chance to revise addition, subtraction and multiplication in columns and long division. In each case, use a calculator to check your answer.
Exercise 4: Addition, subtraction and multiplication in columns and long division with whole numbers
1. Write the following in columns and calculate without using a calculator:
1.1 87 769 + 29 944
1.2 Increase 528 296 by 296 992
1.3 3 999 878 + 4 956 987 + 882 798
1.4 1 234 898 + 23 795 + 2 895 + 823 597
2. Write the following in columns and calculate without using a calculator:
2.1 1 000 000−892 901
2.2 Decrease 7 899 589 by 2 523 589
2.3 What is the difference between 5 000 000 and 2 859 985?
2.4 How much is 5 345 893 less than 9 000 000?
3. Calculate the following without using a calculator, then check your answer using your calculator:
3.1 54 × 73 3.2 543 × 85
3.3 The product of 5 234 and 76
4. Use long division to calculate the following:
REMEMBER
When rounding a number:
Digits 0 to 4 = round down
Digits 5 to 9 = round up
Examples:
84,143… to 2 decimal places = 84,14
84,145… to 2 decimal places = 84,15
84,147… to 2 decimal places = 84,15
5. Use a calculator to calculate the following. Where necessary, give your answer correct to two decimal places.
REMEMBER
Order of calculations using BODMAS B Brackets
SAMPLE
O Other operations like squares, square roots, cubes, etc.
D and M Division and multiplication from left to right
A and S Addition and subtraction from left to right
5.1 1 000 523 ÷ 22
5.2 734 + 562 × 45
5.3 783 + 6 583 ÷ 83
5.4 45 854 + 34 123 73 123−54 789
5.5 (243 123−124 789) × (4 186 + 789)
4. MULTIPLES AND FACTORS
A factor of a number is a number which can divide into that number without leaving a remainder. For example, 6 is a factor of 18 because 18 ÷ 6 = 3. All the factors of 18 are 1, 2, 3, 6, 9 and 18.
Any number can be written as the product of its factors. For example, 18 = 3 × 6.
If a number has two factors only, itself and 1, it is called a prime number.
Examples are 2 with factors 2 and 1, and 3 with factors 3 and 1.
REMEMBER
1 is not a prime number because it has 1 factor only, namely 1.
Numbers with more than two factors, for example 18, are called composite numbers. Factors of 18 are 1, 2, 3, 6, 9 and 18. Prime factors of 18 are 2 and 3, because 18 is divisible by 2 and 3 and 2 and 3 are prime numbers.
In this subtheme we are going to write composite numbers as the product of their prime factors. We will then use this to find the lowest common multiple (LCM) and highest common factor (HCF) of groups of numbers.
4.1 P rime factorising of numbers to determine the LCM and HC F of numbers
Worked example 5: Using prime factors to determine the LCM and HCF
a) Find the prime factors of 450; 725 and 1 500.
b) Find the LCM of these numbers.
c) Find the HCF of these numbers.
NOTE
Here are three rules of divisibility. Knowing these rules can make your work much easier.
• A number is divisible by 2 if it ends in an even number or 0.
• A number is divisible by 3 if its digits add to a number which is divisible by 3.
• A number is divisible by 5 if it ends in a 0 or in 5.
Solutions
a) Start with the lowest prime numbers: 2, then 3, then 5, etc.
SAMPLE
by 3 – digits add to 9
by 3 – digits add to 12
25 Divisible by 5 – number ends in 5
5 Divisible by 5 – number ends in 5
5 725 Divisible by 5 – number ends in 5
5 145 Divisible by 5 – number ends in 5
29 29 29 is a prime number 1
2 1 500 Divisible by 2 – number ends in 0
2 750 Divisible by 2 – number ends in 0
3 375 Divisible by 3 – digits add to 15
5 125 Divisible by 5 – number ends in 5
5 25 Divisible by 5 – number ends in 5
5 5 Divisible by 5 – number ends in 5
2 × 3 × 5 3
b) The LCM is the lowest number into which all the given numbers can divide without leaving a remainder. For example, the LCM of 6; 18 and 36 is 36, because 36 can divide into 36, 18 can divide into 36 and 6 can divide into 36.
NOTE
Finding the LCM without inspection
Step 1: Write all the numbers of which you want to find the LCM as the product of their prime factors.
Step 2: Write down all the bases of the prime factors.
Step 3: Write down the highest power of each of these bases.
Step 4: Calculate the answer of step 3. 450 = 2 × 3 2 × 5
Highest power of each base of prime factors
c) The HCF (highest common factor) is the highest number that can divide into all the given numbers.
For example, the HCF of 12; 18 and 24 is 6, because 6 is the highest number that can divide into 12, 18 and 24.
NOTE
Finding the HCF without inspection
SAMPLE
Step 1: Write all the numbers of which you want to find the HCF as the product of their prime factors.
Step 2: Write down the base of the factors that are common to all these numbers.
Step 3: Write down the lowest power of each of these bases.
Step 4: Calculate the answer of step 3.
Worked example 6: Word problem where the LCM must be determined
The traffic lights at three different road crossings change to green after 60 seconds, 72 seconds and 108 seconds. If they change simultaneously to green at 15 : 00, when will they again change simultaneously to green?
Solution
Let us try to understand the problem, using a simpler case. Say we have two traffic lights.
The one light changes to green every 3 seconds. So it will change after 3 ; 6 ; 9 ; 12 ; … seconds.
The other light changes to green every 4 seconds. So it will change after 4 ; 8 ; 12 ; … seconds.
So, if the two lights change simultaneously to green it will be after 12 seconds, as 12 is the lowest common multiple (LCM) of these two lights.
To solve the given problem, we have to find the LCM of 60, 72 and 108.
2 60 Divisible by 2 – number ends in 0
2 30 Divisible by 2 – number ends in 0
3 15 Divisible by 3 – digits add to 6
5
2 108 Divisible by 2 – number ends in even number 2 54 Divisible by 2 – number ends in even number
3 27 Divisible by 3 – digits add to 9
SAMPLE
John has 120 litres of apple juice and 180 litres of orange juice. He wants to fill containers of equal volume with apple juice and orange juice respectively so that no juice remains. What is the biggest volume of such a container, and how many containers of apple juice and orange juice respectively will he have?
Solution
The given quantities 120 and 180 can be divided by 10 and 20 exactly. He wants to use containers with the biggest volume for both juices. So, we have to find the biggest number that will divide into 120 and 180 exactly. We therefore have to find the HCF of 120 and 180.
120 = 2 3 × 3 × 5 180 = 2 2 × 3 2 × 5
HCF = 2 2 × 3 1 × 5 1 = 4 × 3 × 5 = 60
The biggest volume of a container is 60 litres.
John will have two containers of apple juice and three containers of orange juice.
= 1 080 60 minutes = 18 minutes. Therefore, the lights will change simultaneously to green at 15 : 18. Worked example 7: Word problem where the HCF must be determined
Exercise 5: Use prime factorising to determine the LCM and HCF
1. Write each number as the product of its prime factors and then determine the LCM and HCF of the numbers.
1.1 100 and 132
1.2 378 and 252
1.3 175 and 63
1.4 625 and 195
1.5 120, 150 and 135
1.6 102; 170; 136
1.7 1 323, 5 292 and 12 348
1.8 430, 516 and 817
1.9 632, 790 and 869
1.10 291, 582 and 776
2. A winemaker has three kinds of wine. He has 400 litres of chenin blanc, 440 litres chardonnay and 460 litres of sauvignon blanc. Calculate the smallest possible number of barrels with equal volumes which can be filled with the different wines without mixing them so that no wine is left.
5. RATIO AND RATE
The following about ratios is important:
• A ratio is formed when two or more quantities with the same units are compared. For example, if John is 14 years old and his mother is 42 years old, their ages are in the ratio 14 : 42.
• A ratio does not have units.
• A ratio can also be written as a fraction and can therefore be simplified, 14 : 42 = 14 42 = 1 3 = 1 : 3.
• The units of the quantities which are compared must be the same before a ratio can be written in its simplest form.
Rate is a special kind of ratio in which two quantities with different units are compared. The use of the word “per” indicates rate and is always “per one” of the second quantity. Examples include:
• population growth is 10 000 per year
• a horse eats 11 kg grass per day
• money is invested at an interest rate of 10% per year
• a car travels at 120 kilometers per hour (or in short 120 km/h).
Speed is the rate at which distance increases over time. The diagram below shows that, when working with speeds, there are three quantities involved, namely speed (S), distance (D) and time (T). If we know two of the quantities, we can work out the missing one.
Speed = Distance Time
Distance = Speed × Time
Time = Distance Speed
Worked example 8: Writing ratios in their simplest form
Write the following ratios in their simplest form:
a) 200 g : 1 kg
b) 3 8 : 2 5
Solutions
a) 200 g : 1 000 g
= 1 : 5
Convert ratio to same units
Divide both sides by 200 b) 3 8 : 2 5
= 40( 3 8 ) : 40 ( 2 5 ) Multiply both sides by 40, LCM of 8 and 5
= 5(3) : 8(2) = 15 : 16
Worked example 9: Division of amount in given ratios
Municipal taxes for Bob’s house are R5 400 per year. Bob sells his house to Peter at the end of May. The taxes must be divided pro rata (in the same ratio) between the two. Calculate how much each of them must pay at the end of that year.
Solution
Bob paid until the end of May. So, he paid taxes for five months. Peter is responsible to pay taxes for the remaining seven months. So, the R5 400 must be shared in the ratio 5 : 7, which means there are twelve parts.
∴ Bob pays 5 12 of R5 400
= 5 12 × R5 400
= R2 250
Peter pays 7 12 × R5 400
= R3 150 or R5 400−R2 250 = R3 150.
SAMPLE
Worked example 10: Calculate distance if speed and time are known
Mr. Smith travelled by car at an average speed of 100 km/h for three hours. Then he travelled at an average speed of 115 km/h for the next two hours. Calculate the total distance Mr. Smith covered.
Solution
First distance = Speed × time = 100 km / h × 3 h = 300 km
Second distance = Speed × time = 115 km / h × 2 h = 230 km
Total distance = 300 + 230 = 530 km
Worked example 11: Increase and decrease in a certain ratio
The cost price, R360, is increased in the ratio 5 : 12 to find the marked price. On a sale this marked price is decreased in the ratio 4 : 3. Calculate the:
a) marked price
b) sale price.
Solutions
a) Let the marked price be equal to x. Write as ratios
360 : x = 5 : 12
∴ 360 x = 5 12
∴ 5x = 360 × 12
∴ x = 360 × 12 5 = R864
b) Let the sale price be equal to y. Write as ratios
864 : y = 4 : 3
∴ 864 y = 4 3
∴ 4y = 864 × 3
∴ y = 864 × 3 4 = R648
Exercise 6: Ratio and rate
From these solutions we see that
• if we increase a quantity in the ratio 5 : 12, we multiply by the improper fraction 12 5 .
• if we decrease a quantity in the ratio 4 : 3, we multiply by the proper fraction 3 4 .
3. A fertilizer for roses, 2 : 3 : 2 (22), consists of nitrogen, phosphorus and potassium in the ratio 2 : 3 : 2. The 22 means that 22% of the total mass in the bag of 5 kg consists of fertilizer. The remainder is just fillers that help to spread the fertilizer evenly over the ground. Calculate the mass of nitrogen in the bag of fertilizer. Give the answer in gram correct to one decimal place.
1. Express each of the following ratios in simplest form:
1.1 20 cm : 5 m
1.2 218 g : 3 kg
1.3 36 s : 24 h
1.4 3 cm : 15 km
2. The term carat defines how much gold there is in gold jewellery. If you buy an 18 carat ring, it means that 18 of 24 parts are gold. The other six parts consist of other metals.
SAMPLE
2.1 Calculate the mass of gold in several 18 carat pieces of jewellery pieces with a total mass of 48 g.
2.2 What is the total mass of other metals that the jewellery contains?
4. Concrete is made by mixing gravel, sand and cement in the ratio 3 : 2 : 1 by volume. How much gravel will be needed to make 12 m 3 concrete?
5. The shop manager in a certain shop increases the cost price, R240, of an article in the ratio 5 : 7. If a customer pays cash, he decreases the marked price in the ratio 6 : 5. Calculate the following:
5.1 marked price
5.2 price if the customer pays cash
6. Mr. Johnson drives for 2 1 2 hours and covers a distance of 240 km. How long in hours and minutes will it take him to cover a distance of 320 km at the same average speed?
7. A bus travels at a constant average speed of 80 km/h for 3 hours and 20 minutes. At what constant speed must it travel to cover the same distance in 2 hours 40 minutes?
8. Anne, Sophy and Jessica want to start a business selling muffins in the tuck shop of their school. The amounts that Anne, Sophy and Jessica invest are in the ratio 8 : 5 : 6. The largest amount is R60 more than the smallest amount.
8.1 Calculate the total amount which is invested. (Hint: Let the smallest part be equal to R5x)
8.2 The total income after sales is R1 330. Calculate the amount of money that each of them receives if the profit is divided according to the ratio of the amounts that each of them initially invested.
6. DIRE CT AND INDIRECT PROPORTION
In this subtheme we deal with two important applications of ratio, namely direct and indirect (or inverse) proportion.
6.1 Direct proportion
Suppose one bag of lemons costs R12. The table below shows the price for various bags of lemons.
• If one quantity (x) in the ratio is divided by a number, the other quantity (y) is divided by the same number. For example:
The table shows if x represents the number of bags and y represents the price, then
. Therefore, the value of x y remains constant at
. We can also say that x : y remains constant at 1 : 12.
REMEMBER
Two quantities (x and y) form a direct proportion if the ratio x y (or x : y) remains constant.
Properties of a direct proportion
• The ratio x y (or x : y) remains constant.
• If one quantity (x) in the ratio is multiplied by a number, the other quantity (y) is multiplied by the same number. For example:
• From the previous two bullets it follows that if the one quantity:
◦ increases, the other quantity increases in the same ratio (for example 1 : 3).
◦ decreases, the other quantity decreases in the same ratio (for example 3 : 1).
• If one quantity is 0, the other quantity is also 0 (for example, 0 bags of lemons will cost R0).
• The graph is always a straight-line graph that increases (rises) and passes through 0 on both axes.
NOTE
Note that the graph consists of separate points because the lemons are only sold in whole bags.
6.2 Indirect proportion
Suppose several workers are working on a task. The table below shows how long it will take different numbers of workers to complete the task:
Number of workers (x) 3 4 6 8 12 16
Number of days (y) 16 12 8 6 4 3
The table shows that if x represents the number of workers, and y represents the number of days, then
x × y = 3 × 16 = 4 × 12 = 6 × 8 = 8 × 6 = 12 × 4 = 16 × 3 = 48. Therefore, the value of x × y remains constant at 48.
REMEMBER
Two quantities (x and y) form an indirect (or inverse) proportion if x × y remains constant.
Properties of an indirect proportion
• x × y remains constant.
• If one quantity (x) in the ratio is multiplied by a number, then the other quantity (y) is divided by the same number. For example:
• None of the quantities can be equal to 0 (for example, if there are 0 workers, the task will never be completed. Also, it is impossible to complete the task in 0 days regardless of how many workers there are).
SAMPLE
So, 4 × 12 = 48, and (4 × 3) × (12 ÷ 3) = 12 × 4 = 48
• From the previous two bullets it follows that if the one quantity:
◦ increases, the other decreases in the same ratio (for example 1 : 3 and 3 : 1).
◦ decreases, the other increases in the same ratio (for example 3 : 1 and 1 : 3).
• The graph is always a decreasing (falling) graph that never touches either of the axes.
NOTE
Note that the graph consists of separate points, because we cannot get fractions of workers.
Worked example 12: Direct proportion problem
Johnny earns R594 if he works for 22 hours. How much will he earn if he works 37 hours?
Solution
Method 1
Recognise that this is an example of direct proportion:
• If the number of hours increase, the earnings will also increase.
• If he works twice as long, he will earn twice as much (both quantities increase in the same ratio).
• If he works for 0 hours, he will earn R0.
That means that the ratio x y (= Earnings Time worked ) remains constant.
Let the amount he earns for 37 hours of work be x rand.
∴ x 37 = R594 22 Always put unknown above line to simplify calculation
∴ x = ( R594 22 ) × 37 = R999
So, he will earn R999.
Method 2
First work out the rate per hour. This is called the unitary method.
Rate per hour = R594 22 = R27
So, for 37 hours he will earn R27 × 37 = R999
Worked example 13: Inverse proportion problem
It takes 35 workers 16 days to build a house. How many days will it take 28 workers working at the same rate to build the same house?
Solution
First recognise that this is a case of indirect or proportion because:
• If the number of workers increase, the time required to build the house decrease.
• If there are twice as many workers, the house will be built in half the time.
• None of the two quantities can be 0.
That means that the product (x × y) remains constant.
Method 1
Let a be the number of days.
Method 2
35 workers take 16 days.
∴ 1 worker will take 16 × 35 days
SAMPLE
∴ 28 workers will work 28 times shorter
Work out how long 1 worker takes (unitary method)
Worked example 14: Scale as direct proportion (length on plan)
On a plan of the school grounds the scale is 1 : 2 000. Calculate the length on the plan which will represent 40 m.
NOTE
If the scale on a map is 1: 500 000, it can also be written as 1 500 000 .
For such a scale the numerator (number above the line) represents the distance on the map or plan. The denominator (number below the line) represents the distance on the ground or the real distance.
Map distance
1 500 000 OR
Solution
This is an example of direct proportion. If the distance on the ground increases, the length on the plan increases in the same ratio. We need to find the distance on the plan, so we make the plan distance the numerator.
Plan distance
Real distance = 1 2 000
Plan distance 40 m = 1 m 2 000 m Real distance is in m; make all units m
Plan distance = 40 × 1 2 000 m
= 40 2 000 = 0,02 m = 100 × 0,02 cm = 2 cm
Multiply both sides by 40
Worked example 15: Scale as direct proportion (distance on the ground)
On a map the scale is 1 : 500 000. If you measure the distance between two towns on the map, the distance is 2,5 cm. What is the actual distance between the two towns in kilometre?
Solution
Real distance
Map distance = 500 000 cm 1 cm Map distance is in cm; make all distances cm
Real distance 2,5 cm = 500 000 cm 1 cm
∴ Real distance = 2,5 cm × 500 000 cm = 1 250 000 cm = 1 250 000 ÷ 100 m
= 12 500 ÷ 1 000 km
= 12,5 km
Worked example 16: Time if speed and distance are given
SAMPLE
Annandale is 60 km from Bedford. John cycles from Annandale to Bedford at an average speed of 20 km/h. Andrew starts at the same time and cycles from Bedford to Annandale at an average speed of 25 km/h. How long will it be before they meet each other? Give the answer in hours and minutes.
Solution
Let the time each of them cycled be equal to t hours.
Speed × time = distance Distance remains constant
∴ Distance that John cycles = 20 × t = 20t Distance that Andrew cycles = 25 × t = 25t
Because the distance remains constant, we can say that speed and time form an inverse proportion, or that they are inversely proportional.
Worked example 17: Distance if speed and time are given
A family drives to a picnic spot at an average speed of 80 km/h. They return using the same road at an average speed of 100 km/h. Calculate the distance to the picnic spot if the total travel time is four hours. Give the answer to one decimal place.
Solution
Let the distance be equal to d.
Distance
Speed = Time
∴ d 80 + d 100 = 4
400( d 80 + d 100 ) = 400(4)
∴ 5d + 4d = 1 600
∴ 9d = 1 600
∴ d = 177,7777…km
∴ d = 177,8 km
80 = 2 4 × 5
100 = 2 × 5 2
LCM of 80 and 100
= 2 4 × 5 2
= 16 × 25
= 400
Total travel time is 4 hours
Multiply by 400, the LCM
6. The scale of a map is 1 : 200 000. On the map the distance between two towns is 8,5 cm. What is the actual distance between the two towns?
7. A soccer field has a length of 105 m and has a width of 68 m. A scale drawing of the field is drawn by using the scale: 1 : 1 000. Calculate the lengths on the scale drawing, in cm, of the following:
7.1 length
7.2 width
Correct to one decimal place
Because the time remains constant, we can say that distance and speed form a direct proportion, or that they are directly proportional.
Exercise 7: Direct and indirect proportion
1. Five boxes of mandarins cost R400. Calculate the cost of 12 boxes of mandarins.
2. The length of a spring is 12 cm if an object with a mass of 1 1 2 kg is hung from it. What is the length of the spring if the mass of an object hung from it is 4 kg?
3. The cost to fill a 60 l petrol tank is R1 260. What will be the cost of 25 l of petrol?
SAMPLE
4. A truck driver travels for five hours at an average speed of 80 km/h. If he must complete the journey in four hours, at what average speed should he travel?
5. There are 70 students in a university residence. There is enough food to last them 30 days. Calculate the number of students for which the same quantity of the food will last for 28 days.
8. A vehicle uses 5,5 l of fuel to cover a distance of 82,5 km. Calculate the distance it can cover using 13,2 l of fuel if the speed stays constant.
9. If 270 kg of alfalfa is enough to feed 42 horses for 36 days, for how many days would 360 kg feed 42 horses?
10. An army camp had provisions for 300 men for 90 days. After 20 days, 50 men leave the camp. How long will the food last if it was used at the same rate?
11. A gas has a volume of 2 litres and exerts a pressure of 400 kPa on the walls of its container. What would the pressure be if all the gas is completely transferred into a new container with a volume of 3 litres? Assume that the temperature of the gas remains the same.
REMEMBER
Pascal is a unit of pressure. One kilopascal (kPa) is 1 000 Pascal (“kilo” means “thousand”).
12. Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours they have travelled a total of 72 km. Find the speed of each cyclist.
13. Danny cycles to school at an average speed of 20 km/h. When returning home he uses the same road. He then travels at an average speed of 25 km/h. Calculate the distance to school if the total time to school and back is 2 hours. Give the answer correct to one decimal place.
• Revision exercises to refresh prior knowledge.
• Detailed explanations of concepts and techniques.
• Worked examples help learners to better understand new concepts.
• Varied exercises to entrench theory and practise mathematical skills.
• Test papers and memorandums for exam preparation
• Formula sheets and accepted geometrical reasons for quick reference.
• Index of mathematical terms.
• The facilitator’s guide contains step-by-step calculations and answers
