HIGH SCHOOL EQUIVALENCY PROGRAMME
(HISEP)
MATHEMATICS
S T AG E 2
Published in Jamaica
© NCTVET, 2007, All rights reserved.
HISEP is a registered trademark of the National Council on Technical and Vocational
Education and Training (NCTVET)
First published as a pilot edition in 2005
Revised edition June 2007
No part of this document may be reproduced, stored in a retrieval system or transmitted
without prior written permission of the National Council on Technical and Vocational
Education and Training
ACKNOWLEDGEMENTS
We would like to acknowledge and thank the following persons for their contribution to
the development and production of the HISEP Mathematics Course.
Writers
· Norma Lee
· Lurline Bannister
· Margaret Gordon-Rowe
· Verlaine Henry
· Joyce McKenzie
· Lola McKinley
· Cynthia Williams-Cooke
· Janice Steele
· Edgar Cato
· Michelle Anderson
Reviewers
· Kevin Williams
· Dawn Thompson-Clarke
· Lurline Bannister
Photographers
· Michelle Anderson
· Judith Williams
NCTVET
·
·
·
·
·
Paulette Dunn-Smith, Senior Director
Michelle Anderson, Subject Officer
Pauline M. Bain, Materials Development Coordinator
Clover Barnett, Director, Learning Management Services Department
Jennifer Walker, Director, Quality Assurance
VISION STATEMENT
A certified, empowered and confident
individual, committed to lifelong learning,
contributing to nation building and the global
marketplace
MISSION STATEMENT
To provide an opportunity for adult learners to
access secondary level certification and so
promote lifelong learning and improve
employability
Welcome to the High School Equivalency Programme!
Heartiest congratulations to you for taking on the challenge to do this High School
Equivalency Course. The National Council on Technical and Vocational Education and
Training (NCTVET) is pleased to be involved in the development of the HISEP courses.
We hope you will enjoy every moment of this programme and become a lifelong learner.
The five (5) subjects in the High School Equivalency Programme (HISEP) are: ·
·
·
·
·
Language and Communication
Literature, Culture and the Arts
Mathematics
Science and Technology, and
Society and Citizenship
Each subject is divided to represent five (5) Stages. Each Stage is sub-divided into
Modules and Units to make it easier for you to learn. The Language and
Communication and the Literature, Culture and the Arts have been packaged in two
parts. Part 1 is equivalent to Stages 1, 2, and 3 and Part 2 to Stages 4 and 5. For each
of the other three subjects there are five books: one book for each stage.
As a learner, you may take one or any number of these courses. You will receive credit
for each subject you have successfully completed. However, you will only qualify to
receive a High School Equivalency Diploma when you successfully complete the
course material and or the final assessments in all five (5) subjects.
Happy learning!!!
TABLE OF CONTENTS
MODULE
PAGE
INTRODUCTION TO STAGE 2 ............................................................................. i
MODULE 1: NUMBER SENSE AND MONEY MANAGEMENT 3 ........................ 1
UNIT 1: INTEGERS.............................................................................................. 1
UNIT 2: MONEY, INCOME AND FOREIGN EXCHANGE ................................... 10
UNIT 3: BEST BUYS ............................................................................................ 17
UNIT 4: RATED EARNINGS AND BUDGETING .................................................. 25
ANSWERS ............................................................................................................ 31
MODULE 2: ALGEBRA PART 2 ......................................................................... 1
UNIT 1: FURTHER ALONG THE ALGEBRA TRAIL ............................................. 1
UNIT 2: ALGEBRAIC EXPRESSIONS ................................................................. 11
UNIT 3: EQUATIONS AND INEQUALITIES ......................................................... 16
UNIT 4: SIMULTANEOUS EQUATIONS (1) ......................................................... 24
UNIT 5: SIMULTANEOUS EQUATIONS (2) ......................................................... 30
ANSWERS ............................................................................................................ 38
MODULE 3: GEOMETRY PART 3
INTRODUCTION TO MODULE 3 .......................................................................... i
UNIT 1: LINES AND THEIR RELATIONSHIPS .................................................... 1
UNIT 2: ANGLES AND THEIR RELATIONSHIPS ................................................ 5
UNIT 3: TRIANGLES AND THEIR PROPERTIES ................................................ 19
UNIT 4: QUADRILATERALS AND THEIR PROPERTIES .................................... 30
ANSWERS ............................................................................................................ 43
MODULE 4: MEASUREMENT PART 2
UNIT 1: UNITS OF MEASURE ............................................................................. 1
UNIT 2: PERIMETER OF PLANE SHAPES ......................................................... 8
UNIT 3: AREA OF PLANE SHAPES .................................................................... 13
UNIT 4: CIRCUMFERENCE AND AREA OF THE CIRCLE .................................. 30
ANSWERS ............................................................................................................ 35
MODULE 5: INTRODUCTION TO RELATIONS, FUNCTIONS AND GRAPHS
UNIT 1: THE CARTESIAN PLANE ....................................................................... 1
UNIT 2: INTRODUCTION TO RELATIONS .......................................................... 8
UNIT 3: INTRODUCTION TO FUNCTIONS ......................................................... 15
ANSWERS ............................................................................................................ 20
MODULE 6: HANDLING DATA PART 2 ............................................................. 1
UNIT 1: REPRESENTING AND INTERPRETING DATA (2) ................................. 1
UNIT 2: COLLECTING AND ORGANISING DATA (1) ......................................... 9
UNIT 3: ANALYSING AND INTERPRETING DATA (1) ........................................ 26
UNIT 4: PROBABILITY (1) ................................................................................... 39
END OF MODULE TEST ...................................................................................... 51
ANSWERS ............................................................................................................ 54
ANSWERS TO END OF MODULE TEST.............................................................. 66
BIBLIOGRAPHY ................................................................................................... 69
INTRODUCTION TO STAGE 2
Congratulations! You have made it through Stage 1. Good for you! I hope you are now
ready for a new challenge. This stage will take you further along the road to
understanding money and knowing how to manage it. We will start off with a new set of
numbers called Integers and then move right into some practical ways of dealing with
money. These ways of dealing with money will be our introduction to a topic in
Mathematics called Consumer Arithmetic. We will continue with Algebra, Geometry
and Measurement. Module 5 is an introduction to Relations, Functions and Graphs and
Handling Data 2 is the final module in this stage.
Having completed Stage 1, you have gained knowledge and understanding of certain
concepts. Be encouraged by this. You have come a far way and you are now one step
closer to your goal. You have what it takes to make it through all the stages and
modules. Believe it and keep up the good work!
i
Mathematics – Stage 2
STAGE 2
MODULE 1
MODULE 1:
NUMBER SENSE AND MONEY MANAGEMENT 3
UNIT 1:
INTEGERS
Introduction
This module introduces you to the concept of negative numbers and continues to teach
about money management. It is made up of four units, each of which will help you to
build your knowledge and therefore your confidence concerning mathematical concepts
and ideas. As was said before and probably will be said again, you are expected to work
through the exercises and then check your answers. Do not move on until you have
understood the current topic. Mathematics is a subject where the topics are inter-related
and failure to master one section could result in failure to understand another topic in a
later module or stage. This is the reason for prerequisites at the beginning of each unit.
This is a guide to tell you what previous knowledge you need to have in order to grasp
the lesson. As always, important words and concepts will be in bold type. Please pay
special attention to them.
By the end of this unit you will be able to:
·
carry out all operations with positive and negative integers.
·
compare and order two or more integers.
Prerequisites
Before you begin you should know:
·
operations with whole numbers
·
operations with decimals
·
operations in algebra.
1
Mathematics - Stage 2 Module 1
Integers
Integers are all the negative numbers, positive numbers and zero together in one
number group. We can carry out the four mathematical operations (addition, subtraction,
multiplication and division) using integers.
Addition and Subtraction of Integers
The sign of a number, is the sign directly in front of that number. In the case of a number
that starts a series of numbers, if there is no sign, the number is assumed to be positive.
6 + 7 (both are positive)
- 3 + 4 (the 3 is negative)
When there is one sign between two integers it is an operational sign (plus, minus)
16 − 12
sixteen minus twelve
8+9
eight plus nine
When there are two signs between two integers, the first one is the operation sign
and the second is the integer sign.
Example 1.1
3 + ( − 1)
three plus negative one
5 + ( + 9)
five plus positive nine
14 −(+ 2)
fourteen minus positive two
8 − (− 4)
eight minus negative four
Now you must be saying that this is getting a bit complicated but stay with me and see
where we can go.
2
Mathematics - Stage 2 Module 1
When two signs are right beside each other between two integers, the
mathematical sentence or statement can be simplified by converting the two signs to
one sign according to specific rules.
The rules are:
(a)
If the signs are the same, change to one positive sign
5− (− 9) = 5 + 9
(b)
8 + (+ 7) = 8 + 7
If the signs are different, change to one negative sign.
9 − (+ 3) = 9 − 3
5 + (− 1) = 5 − 1
To carry out addition or subtraction between two integers we have to consider whether
their signs are the same or different.
If the signs are the same, sum them (same, sum) and use the given sign.
− 2 − 4 (both negative) sum = − 6
3 + 5 (both positive), sum = 8
− 2 + (− 4) (both negative) sum = − 6
3 + (+ 5) (both positive, sum = 8
The examples above can be shown on the number line. Remember you were introduced
to the number line in stage 1 module 2. We can make use of it here.
-7
-6
-5
-4
-3
-2
-1
-0
1
2
3
4
5
6
7
Take two of the examples above, 3 + 5 and −2 −4. To do the first one on the number
line, we would go to the number three, then move five spaces to the right because we
are doing addition. This would give us an answer of eight. For the second one, we would
go to negative two then move four spaces to the left because we are doing subtraction.
3
Mathematics - Stage 2 Module 1
This would give us an answer of negative six. The direction of the movement depends
on the operation being carried out. Generally, if we are doing addition then we move to
the right. If we are doing subtraction, then we move to the left.
Take the next two examples, 3 + (+ 5) and − 2 +( − 4). If we wanted to do these
questions on the number line without first changing the double signs to one sign, we
would proceed as follows:
-7
6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
For the first one, we would go to three then add positive five which means move five
places to the right. The answer is positive eight. For the second one, we would go to
negative two then add negative four which means add four in the negative direction
which is left. This would then give us negative 6. The procedure and results for the two
sets of examples would be the same.
If the signs are different, find the difference (different, difference) and use the sign of
the “bigger” one. At this point, I believe it would be helpful to stop and check your
dictionary for the meanings of the words ‘different’ and ‘difference’.
− 6 + 5 (bigger negative) difference = − 1
− 6 + (+ 5)
7 − 2 (bigger positive) difference = 5
7 − (+ 2)
Doing the first two (-6 + 5 and 7 – 2) on the number line would be as follows:
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
4
Mathematics - Stage 2 Module 1
For the first one, go to negative six then move five places to the right. This would give us
an answer of negative one. For the second one, go to positive seven then move two
places to the left. This would give us an answer of positive five.
Doing the second set -6 + (+5) and 7 – (+ 2) ) would be as follows:
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
For the first one, -6 +(+5) go to negative six then add positive five by going five places to
the right. This would give an answer of negative one. For the second one, 7 - (+ 2) go to
positive seven then subtract a positive two which means going two places to the left.
This would give an answer of positive five. The procedure and results for the two sets of
examples would be the same.
Example 1.2
(1) 24 − (+ 16)
(2) 15 −(− 7)
= 24 − 16
= 15 + 7
=8
= 22
(3) 43 + (+ 6)
(4) 17 + (− 5)
= 43 + 6
= 17 − 5
= 49
= 12
5
Mathematics - Stage 2 Module 1
Exercise 1.1
Work out the following.
(1)
13 + 24
(6) 12 + (+16)
(2)
12 + (− 2)
(7) 18 − (+ 6)
(3)
− 9 + 14
(8)
33 − 11
(4)
−3−4
(9)
- 17 + 8
(5)
5 − (− 6)
(10) − 76 + (− 34)
Example 1.3
When there are more than two integers to add and subtract, do them two at a time.
3 +( − 8) − (− 4) + 2
= 3 − 8 + 4 + 2 (changing double signs between integers)
= −5 + 4 + 2
= −1 + 2
= 1
Exercise 1.2
Work out the following
(1)
3 − (− 4) + 16 − 2
(2) 56 − (+ 67) + 23
(3)
− 5 + (− 3) + 8 − 11
(4) − 72 + (+ 54) + (− 19)
(5)
− 1 − (− 14 )− (+ 7)
(6) 29 − (− 15) + 30 − 42
(7)
21 + (+ 6) − (− 2) + 4
(8) 65 + 39 + 40 − (− 81)
(9)
16 + 13 − 18 − 12 + 27
(10) 37 − (+ 61) + (− 22)
6
Mathematics - Stage 2 Module 1
Multiplication and Division with Integers
The sign of the answer to multiplication of integers and the sign of the answer to division
of integers depend on the signs of the numbers being multiplied and the signs of the
numbers being divided.
If the signs are the same, use the sum sign (+) for the answer
Example 1.4
(a) 2 x 25 = 50
(b) − 7 x (− 13) = 91
(c) − 40 ÷ (− 5) = 8
(d) 42 ÷ 3 = 14
If the signs are different, use the difference sign (−) for the answer
Example 1.5
(a) 4 x (− 16) = − 64
(b) − 9 x 12 = −108
(c) 45 ÷ (− 15) = − 3
(d) − 56 ÷ 7 = −8
Example 1.6
(Multiplication and division cannot be clearly shown on the number line.)
(1) − 3 x 14 = − 42
(6) − 24 ÷ 3 = − 8
(2) 13 x 5 = 65
(7) 225 ÷ 5 = 45
(3) 25 x (− 7) = − 175
(8) 120 ÷ (− 4) = − 30
(4) − 15 x (− 8) = 120
(5) − 55 ÷ (− 11) = 5
Exercise 1.3
Work out the following
(1) 23 x (− 2)
(6) 69 ÷ (− 3)
(2) 12 x (− 6)
(7)
(3) 128 ÷ 8
(8) −39 ÷ 13
(4) 64 ÷ 16
(9) 27 x (− 32)
(5) − 19 x (− 6)
(10) − 204 ÷ (− 17)
− 91 ÷ 7
7
Mathematics - Stage 2 Module 1
The Number Line
Now it is time for you to learn about comparing and ordering integers. This will be
achieved, again with the aid of the number line. Looking at the number line below we
see how integers relate to each other. A number is larger than the one to its left and
smaller than the one to its right.
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
So zero (0) is smaller than four (4) but larger than negative three (− 3), and negative one
(− 1) is bigger than negative four (− 4) but smaller than two (2). Remember that the
number line goes on indefinitely in both directions.
Inequalities
An inequality is a sign used to compare quantities to each other. There are four such
signs. ( <, >, ≤, ≥ ). From left to right they mean “less than”, “more than”, “less than or
equal to”, “more than or equal to”.
Example 1.7
2<6
−3 > − 7
means - 2 is less than 6
means - − 3 is more than − 7
x ≤ 14
means - x is less than or equal to 14 (x = 14 or 13 or 12)
y≥ 4
means - y is more than or equal to 4 ( y = 4 or 5 or 6…)
Example 1.8
For our purposes right now, we will only be using the less than and more than signs to
compare integers.
− 16 < 4
9 > −1
8
Mathematics - Stage 2 Module 1
Exercise 1.4
Place the sign < or > between the following pairs of numbers to make true statements.
9
(6)
11
− 23
(2) − 4
2
(7)
−6
−1
(3) − 3
−20
(8)
8
14
(4) 7
−7
(9)
13
9
(5) 15
10
(10)
5
−2
(1)
5
Congratulations! You made it to the end of the unit. That was good. Wasn’t it? Press on!
9
Mathematics - Stage 2 Module 1
MODULE 1:
NUMBER SENSE AND MONEY MANAGEMENT 3
UNIT 2:
MONEY, INCOME AND FOREIGN EXCHANGE
Introduction
As a person living in Jamaica, I am sure that you have had many opportunities to use
money and so you already know some of its main uses. We use money to buy goods,
pay for services and use as collateral. Though there are other uses of money, these are
the uses on which we will focus.
By the end of this unit you will be able to:
·
list three functions of money
·
explain the need for different currencies
·
convert money from one currency to another
·
identify sources of income
·
calculate total income over a given period
·
calculate percent and ratios of parts of that income that is spent on items
such as transportation, food, savings and recreation.
Prerequisites
Before you begin you should know:
·
operations with decimals
·
finding percentages
·
ratios
Functions of Money
Buying Goods
Goods are the things we buy that are tangible. That means we can see and touch them.
For example, we can see and touch food, clothes, appliances, cars, and houses.
10
Mathematics - Stage 2 Module 1
Paying for services
Services are the things we buy from persons and organisations. Unlike food and clothes,
services are mostly intangible. We cannot see and touch many of them. For example,
electricity, water, telephone, carpentry, masonry, plumbing, learning, hairdressing.
11
Mathematics - Stage 2 Module 1
You might be saying that we can touch the water so it can be considered goods, but
listen to this: when we pay our water bills we are not paying for the water, we are paying
for the use of the water. At the hairdresser, we are not paying for the hair, (that is
already ours unless we are buying hair to do extensions) we are paying for the services
of the person giving us our new hairstyle.
I hope you now understand the difference between goods and services. Try to think of
some other examples of goods and services and discuss them with a family member or
someone from your group.
Using money as collateral
When we want to buy a house and need a loan from the bank, they usually ask us if we
have something of value that they could keep if we do not meet the loan repayments.
This is called collateral. Sometimes a sum of money can be used as collateral.
†Activity 2.1
Make a list of at least five goods and five services which can be bought or paid for using
money. Try not to use any of the examples given before in the text. Put these lists
together for your facilitator to examine.
†Activity 2.2
Although we are dealing with money, it would be useful for your learning experience to
find out what other forms of collateral there are. Go to a bank or lending institution and
find out what other forms of collateral they use. Make a list of these and show them to
your facilitator.
12
Mathematics - Stage 2 Module 1
Currency
Currency is the type of money used in a particular country. Jamaica uses a different kind
of currency from the United States of America (USA) and also a different kind of
currency from the United Kingdom (UK). Although we use dollars and cents, they are not
the same as those used in the USA. The UK uses pounds and pence.
When tourists from the USA or the UK come to Jamaica, they usually change their
foreign currency. They exchange US dollars or UK pounds for Jamaican dollars because
Jamaican dollars are our primary legal tender. We therefore buy and sell mainly in
Jamaican currency. So the tourists receive a certain amount of Jamaican money for
each US dollar or UK pound they exchange at the bank or cambio. This is known as the
exchange rate which varies depending on the strength of the Jamaican dollar. At
present the rates are approximately J$60.00 to US$1.00 and J$100.00 to UK₤1.00.
Example 2.1
A tourist with US$500.00 dollars who wanted to change that into J$, would get
500 x 60 = J$30000.00
If when that person is leaving the island they have Jamaican money they need to
change then to convert Jamaican dollars into US dollars we would do the opposite
operation. If the person had J$300.00, they would get
300.00÷ 60 = US$5.00
13
Mathematics - Stage 2 Module 1
So to convert from US$ to JA$ we multiply and to convert from J$ to US$ we divide. We
would use the same procedure of multiplication and division for changing to and from the
UK₤ to J$ and from J$ to UK₤.
Exercise 2.1
Convert the following Jamaican currencies to American dollars then convert these same
Jamaican dollars to English pounds. (Use the suggested rates of 60:1 for US and 100:1
for UK.)
(1) J$3,600.00
(6) J$690.00
(2) J$780.00
(7) J$9,450,120.00
(3) J$450.00
(8) J$1,500,000.00
(4) J$16,000.0
(9) J$13,500.00
(5) J$7,200.60
(10) J$828,000.00
Exercise 2.2
Convert the following US and UK currencies to Jamaican dollars. (Use the suggested
rates of 1:60 for US and 1:100 for UK)
(1) US$139.00
(6) US$42.75
(2) UK₤72.00
(7) US$592.30
(3) US$1,500.00
(8) UK₤19.80
(4) US$16.25
(9) UK₤69.85
(5) UK₤27.52
(10) US$5,763.23
You are encouraged to explore various texts in order to supplement your knowledge and
understanding of this topic.
14
Mathematics - Stage 2 Module 1
Income
Income is the money that someone earns for doing work. Examination of this topic will
expose you to the various commodities on which people spend their money.
Whenever we get paid, whether it is at the end of the month, at the end of two weeks or
at the end of a week, there are some things on which we usually spend our money. This
is called our expenditure. The more common items on which we spend our money are
food, transportation, school fees, rent, utilities, and clothes.
Each of these things uses up a part of our money and we say that we spend a
percentage of our income on each of them. (Review the unit on percentages in stage 1
module 6).
Example 2.2
Mr. Jacobs received $45,000.00 for his monthly salary. He spent $20,000.00 on rent,
$10,000.00 on food, $5,000.00 on utilities, $8,000.00 on school fees and the rest he
saved.
(a) How much money did Mr. Jacobs save?
(b) What percentage of his salary did Mr. Jacobs spend on rent?
(c) What is the ratio of his expenditure on rent to his expenditure on food?
Solutions
(a) Mr. Jacobs saved
45000 − (20000 + 10000 + 5000 + 8000)
= 45000 − 43000
= $2,000.00
15
Mathematics - Stage 2 Module 1
(b) Percentage of salary spent on rent
20000 x 100
45000
=
20000
1
=
x 1
450
2000
= 44
20
45
45
=
(c)
44
4
%
9
Ratio of rent expenditure to food expenditure
20000 : 10000
= 2:1
Exercise 2.3
(1)
Mr. Harris earns $22,000.00 per month. How much money will he earn in six
months?
(2)
If Mr. Harris gets a 15% raise in pay, how much will he be earning each
month?
(3)
Mr. Guinness spends 30% of his $34,000.00 per month salary on rent.
(a) How much does he pay for rent?
(b) If he spends 40% of what is left on food? How much does he spend on
food?
(c) After rent and food, how much money does he have left?
(4)
If a man and his wife together earn $42,000.00 per month and his portion is
2
of the amount, how much does each of them earn monthly?
3
(5)
If a man’s earnings for three months is $90,000.00 and he spends $22,500.00
on rent during this period, what is the ratio of his rent to his salary?
(6)
Miss Kidd told her friend that she spent a third of her monthly salary on
utilities. If her utilities cost $17,480.00, how much is her monthly salary?
(7)
Devon and Deborah had a combined monthly income of $128,402.00. Out of
this they had to pay school fees for three children $16,500.00, $10,850.00 and
$15,550.00. How much money is left after paying school fees?
16
Mathematics - Stage 2 Module 1
MODULE 1:
NUMBER SENSE AND MONEY MANAGEMENT 3
UNIT 3:
BEST BUYS
Introduction
We all go to the grocery to buy food and to clothing stores or arcades to buy clothes. We
all want the best value for our money or the best buy. Sometimes when we are buying
more than one item we have to check to see if we have enough money. We also search
for discounts and deals. This unit will teach us how to calculate these discounts and
deals.
By the end of this unit you will be able to:
·
calculate total price given the unit price.
·
calculate unit price given the total price.
·
calculate “best buys”.
·
determine if certain amounts of money are adequate to purchase certain
goods.
·
calculate profit and loss in monetary terms.
·
calculate percentage profit and loss and discount.
·
calculate cost price given selling price and percentage discount, profit or loss.
Prerequisites
Before you begin you should know:
·
operations with whole numbers.
·
percentages
17
Mathematics - Stage 2 Module 1
Unit Price / Total Cost
The unit price is the price we pay for one item. The total cost is the price paid for more
than one of the same item when the price quoted is for one. If a box of 10 pens costs
$240.00 then the unit price would be the price for 1 pen. This would be found by dividing
$240.00 by 10. Using the same amount for one pen we would then be able to work out
the cost for 30 or 40 pens and so on.
Designer Brands: $642 $624
$246 $264 $462 $426
" I planned to spend $500. Which brands can I afford to buy?"
Example 3.1
(a) If the cost of a dozen sandwich biscuits is $84.00 then what is the cost of one?
Cost for one
$84.00 ÷ 12 = $7.00
(b) One doughnut costs $25.00 and Janine wanted to buy twenty of them. How much
money would she need?
Cost for twenty
$25.00 x 20 = $500.00
18
Mathematics - Stage 2 Module 1
Best Buys
In our efforts to spend our money wisely, we sometimes check around to find
“Best Buys”. A “Best Buy” occurs when we encounter similar items for sale at different
prices in different stores and purchase the one which gives us the best value for money.
Sometimes if we examine the items closely we will find that one of them is of inferior
quality. Purchasing this item might not be a good idea or a “Best Buy” even though it
might cost less. The item of better quality will most likely be useful for a longer time.
Buying the item of better quality for more money is therefore an example of a
“Best Buy”.
If the two items being compared are of the same quality, then buying the cheaper one is
therefore the “Best buy”.
Example 3.2
Two shoe stores, Leather Sole and Never Tanned had shoes of similar style. The shoes
from Leather Sole cost $2745.00 while the shoes from Never Tanned cost $2550.00.
Both stores sell only Genuine leather shoes. Which is the better buy and by how much?
Solution
The better buy in this case will be the cheaper one. That is, the shoe being sold by
Never Tanned because there is no consideration of which has better quality. The quality
at both stores is the same as they both sell only genuine leather.
Difference in cost =
=
2745.00 - 2550.00
$195.00.
The shoe from Never Tanned is the better buy by $195.00.
19
Mathematics - Stage 2 Module 1
Exercise 3.1
1. Joyann wanted to buy three items each costing $15.97. She had $50.00 and knew
that was enough but she was not sure she would have bus fare left. If her bus fare
was $2.50, would she have enough? If not, how much more would she need?
2.
Find the unit cost of a packet of soup if a box of 144 packets costs $4604.80.
3.
What is the difference in cost of fifteen brooms for $89.65 each at 16.5% tax and
fifteen brooms for $95.00 each at 15% tax?
4.
One bulla costs $30.00 and six cost $150.00. Which is the best buy?
5. An item of clothing is being sold in one store for $2,250.00 with a discount of 10%
and for $2390.00 with a discount of 15%. Which is the best buy?
6. Two companies which manufacture vacuum cleaners are having promotional sales.
One company (Quality Manufacturers) is selling their vacuums for 3 payments of
$5,431.00 while the other company (Goal Manufacturers) is selling their vacuums for
4 payments of $4,201.00. Which manufacturer is giving the best deal?
7. Mrs. Bernard sent her son to buy 2 litres of ice cream for a party. She sent him to the
corner shop where it is sold for $267.50 per litre. Instead he went to the supermarket
where he bought a two litre container for $550.00. Did he spend more than his
mother bargained for?
8.
An appliance store was having a sale on video tapes. There were two offers.
(a)
tapes that originally cost $375.00 each were now being sold for 3 for
$1000.00.
(b)
tapes that originally cost $420.00 each were now being sold for 3 for
$1200.00. Which is the best buy?
Profit and Loss
When stores or individuals sell goods to us, they make a profit or a loss on the goods.
They make a profit if they sell the goods for more than it was bought for. They make a
loss if they sell for a price less than what was spent to buy the goods.
Most stores usually try to make a profit.
20
Mathematics - Stage 2 Module 1
Example 3.3
A wholesaler bought 15 electric fans for $1,370.00 each. He then sold them all for
$23,925.00. How much profit or loss did he make?
Solution
Total cost of fans = $1,370.00 x 15
= $20,550.00
Difference between cost price and selling price
= $23,925.00 - $20,550.00
= $3,375.00
The selling price was higher than the cost price so the wholesaler made a profit of
$3,375.00.
Example 3.4
A retailer bought 12 bath towels for $465.00 each then sold them all for $4,980.00. How
much profit or loss did he make?
Solution
Total cost of towels = $465.00 x 12
= $5,580.00.
Difference between cost price and selling price
= $5,580.00 - $4,980.00
= $600.00
The selling price was lower than the cost price so the retailer made a loss of $600.00.
21
Mathematics - Stage 2 Module 1
Exercise 3.2
1.
A baker wanted to buy a new oven so he sold the old one for $5,367.42. He had
bought it for $7,500.00 How much money did he lose on the sale?
2.
A video store owner wanted to sell out all her 6 year-old videos. There were 360 of
them. She sold them for $120 each. If she had bought half of them for $22,500.00
and the other half for $15,600.00, how much profit or loss would she be making?
3.
A hotel waitress bought the wrong size uniform and the store would not take it back.
She decided to sell it to her co-worker. She had paid $1,235.00 for the skirt and
$1,642.00 for the blouse. How much profit did she make in selling them both for
$3,295.00?
4.
A garment factory was selling out their sergers because they were closing down.
There were 32 sergers and they had cost $10,335.00 each. Now they were being
sold for $359,680.00. What was the profit made?
5.
The management of a hotel was refurbishing and sold 6 pieces of bedroom furniture
for $94,550.00. There were 2 beds costing $20,000.00 each, 2 chairs costing
$6,000.00 each, a table costing $12,000.00and a chest of drawers costing
$40,000.00. How much of a loss did they take?
6.
An electronics store was selling a DVD player for $7,592.00 after buying it for
$6,525.00. How much profit did they make?
7.
A lighting store had a certain kind of lamp for sale. They bought the lamp for
$1,172.00 and sold it for $1,500.00. How much profit did they make on the sale?
8.
Denzel wanted to sell a used exercise machine to his cousin. He paid $15,805.56
for it. He asked his cousin to pay him $2,840.00 each month for 6 months. How
much profit or loss would he make?
9.
A book store had a sale in which their customers would get a free book if they
bought 2 books costing at least $1,500.00 together. A man chose 2 books he
wanted plus an extra one and went to the cashier. She told him his two books came
up to $1,693.00. Normally the 3 books would cost $491.00, $1,202.00 and $677.00
each. How much money did the store lose?
10. A mother was asked to help raise funds for her son’s school. She decided to bake
cakes and sell them. She spent $1,100.00 on ingredients and made 8 cakes. How
much would she have to sell each for in order to make a profit of $5,000.00?
22
Mathematics - Stage 2 Module 1
Percentage Profit or Loss
The profit or loss made by someone selling goods may sometimes be written as a
percentage of the cost price. This is called percentage profit or loss.
Example 3.5
A box containing a gross (144) of pens was bought for $3,600.00. The pens were then
sold for $30.00 each. What is the percentage profit or loss for the transaction?
Solution
Total selling price of pens = 144 x $30
= $4,320.00
Difference in prices
= $4,320.00 - $3,600.00
= $720.00
The selling price was greater than the cost price so a profit was made.
Percentage profit
=
100
720
x
1
3600
= 20%.
Example 3.6
A box containing 25 bottles of correcting fluid was bought for $1,687.50. The bottles
were then sold for $57.38 each. How much percentage profit or loss was made? Give
your answer to the nearest whole number.
Solution
Total selling price of bottles = $57.38 x 25
= $1,434.50
Difference between prices = $1,687.50 - $1,434.50
= $253.00
The selling price was less than the cost price so a loss was made.
23
Mathematics - Stage 2 Module 1
Percentage loss
=
100
253
x
1
1687.5
=
14.99%
=
15% to the nearest whole number.
Exercise 3.3
Give answers to the nearest whole number where necessary.
1.
What is the percentage profit on an item bought for $362.50 and sold for $384.25?
2.
What is the percentage loss on a watch bought for $5,025.00 and sold for
$4,673.25?
3.
A food scale was sold for $3,093.41 after being bought for $3,362.40. What is the
percentage loss?
4.
Miss Ida bought 40 pounds of yams to resell at the market. She paid $1,680.00 and
sold it for $1,848.00. What was the percentage profit?
5.
If Mr. Isaac bought a car for $800,000.00 and sold it for $700,000.00, what is the
percentage loss?
6.
A hairdresser bought 15 packets of hair to sell to her customers. She wanted to
make a profit but something went wrong with her calculations. What was her
percentage loss if she bought them for $8,400.00 and sold them for $532.00 each?
7.
A calculator was being sold for $975.00. The seller had bought 12 of them for
$10,263.12. What is the percentage profit or loss?
8.
A supermarket bought a box containing a gross of oranges for $4,320.00. They sold
them for $432.00 per dozen. What is the percentage profit or loss?
9.
A hardware store was overstocked with bathtubs. They decided to have a sale.
Each bathtub was bought for $3,724.00. How much would they sell for if there is to
be a 25% loss?
10. If a cloth store bought 100 yards of a certain material for $27,300.00, what would be
the percentage profit if they sold it for $313.95 per yard?
24
Mathematics - Stage 2 Module 1
MODULE 1:
NUMBER SENSE AND MONEY MANAGEMENT 3
UNIT 4:
RATED EARNINGS AND BUDGETING
Introduction
There are people who get paid according to the number of hours they work. They are
called hourly paid workers. Most hourly paid workers are allowed to work for forty hours
per week. This is called a forty hour week. Any time they work beyond this is counted as
overtime. For overtime they are usually paid a higher hourly rate. Examples of this new
rate are (a) Time and a half: one and a half times the rate per hour. (b) Double time: two
times the rate, per hour.
By the end of this unit you will be able to:
·
calculate commissions and salary rates.
·
make simple budgets.
·
identify ways to stick to a budget.
Prerequisites
Before you begin you should know:
·
operations with whole numbers
·
operations with decimals
·
operations with fractions
Example 4.1
A waitress at a restaurant was being paid $250.00 per hour for a forty hour work week.
She was paid time and a half for overtime during the week and double time on the
weekends. How much money did she earn if she worked for forty-eight hours, three of
which were on Saturday?
25
Mathematics - Stage 2 Module 1
5 hours overtime @ $250 x 1.5
Basic 40 hours =$ 250 x 40
= 5 x 250x 1.5
= $10,000.00
= $1,875.00
3 hours overtime @ $250 x 2
= 3 x 250 x 2
Total earnings for that week
= $1,500.00
$10,000.00 + $1,875.00 + $1,500.00
= $13,375.00
Commission
A commission is a percentage of the money earned from the sale of goods or services.
Some people are paid only by commission while others get a fixed salary and the
commission is extra. This method of payment is used mainly for Insurance salespersons
or salespersons in stores.
Example 4.2
Mrs. King, earned $34,000.00 per month plus 3% commission on all sales she made.
How much was her total salary in January of last year when she sold goods amounting
to $114,458.70 that month.
Amount of commission
114458.70 x 3%
= 114458.7 x 0.03
= $3,433.76
Total salary = 34000 + 3433.76 = $37,433.76
Exercise 4.1
1. George needs to pay back a loan of $15,000.00. He wants to pay by salary
deductions over a period of twelve months. How much money would he have to pay
each month to achieve his goal?
2. Frank normally works eight hours per day, five days per week. For this he is paid
$415.00 per hour. If he works overtime, he is paid time and a half. How much money
does Frank make in a week when he works ten hours per day for five days?
3. For forty hours of work, Marty receives $5400.00. What is his hourly rate?
26
Mathematics - Stage 2 Module 1
4. One week, Kyle received a pay cheque for $6,300.00. Normally he gets $6,000.00
for a forty hour work week. He was told he got a raise in pay. How much per hour
was this raise?
5
Amy Sousa, a sales clerk, is paid $230.00 per hour for a forty hour week from
Monday to Friday. For overtime during the week she gets time and a half. For
overtime on the weekend she gets double time. How much money would Amy make
if she worked five hours overtime each week and five hours overtime each weekend
for a month?
6. Sales clerks at “Your Taste Boutique”, earn a monthly salary of $23,456.37 plus
3% commission on all sales. Janice had sales of $27,992.00 in the month of
February. How much money did she make in February?
7. An insurance salesman makes 8% commission on all his sales. If his sales
amounted to $42,300.00 how much was his commission?
8.
A 5% commission is paid to workers who sell at least $500,000.00 worth of
insurance premiums for the month. If Miss Defoe sells $757,000.00 worth of
premiums, how much is her commission?
9.
A real estate agent gets 3% of the value of the house she sells for commission. If
she sold a house for $12,345,000.00, how much was her commission?
10. A car salesman gets 2.5% of the value of his sales per month. In June he sold a car
for $985,000.00 and in July he sold a car for $1,478,765.00. How much commission
did he make for the two months?
BUDGET
A budget is the money needed or available for expenditure. When we make a budget,
we decide on which things we will spend our available money. The most common type of
budget among individuals is the monthly budget. In this budget, we decide what to do
with our monthly salaries.
27
Mathematics - Stage 2 Module 1
Example 4.3
Mr. Kincaid gets a monthly salary of $60942.58 to take home. He needs to make sure
that all his monthly bills are paid and that he puts some money towards savings. This is
how he allots his money.
Amount available for spending:$60942.58
Tithe and offering
$ 6294.30
Rent
$ 25000.00
Food
$ 12000.00
Savings
$ 5000.00
Utilities
$ 6000.00
Total
$54294.30
Miscellaneous
$ 6648.28
The miscellaneous amount in his budget is very important. It covers anything he has
forgotten to put in the budget and anything unexpected which comes up.
The savings amount is also very important. It is the part of his budget which helps to
secure his future. Even if the amount he saves is not as much as he would like to save, it
is important that he does save something every month.
†Activity 4.1
Make a list of possible items or events that would require Mr. Kincaid to use his
miscellaneous fund. Discuss this list with your facilitator or another student from your
group to make sure that your list reflects the realities of Jamaican life.
28
Mathematics - Stage 2 Module 1
†Activity 4.2
1.
Make up your own budget for a month with five or six items. These items must I
include savings and miscellaneous.
2.
Make a list of three items which could be miscellaneous for you.
3.
Make up another budget with a smaller savings amount and a larger
miscellaneous amount.
4.
Work out how long it would take you to save $50,000.00 using each of the
budgets you have made up.
5.
Make a list of three things this $50,000.00 could buy for you and your family.
†Activity 4.3
Find five organizations and or institutions which make up a yearly budget. From these
institutions, find out the following:
(a) what span of months make up their financial year?
(b) do these companies allocate funds for donations? (Do they give money away?)
(c) the names of three organizations, clubs, or groups that they have donated to every
year for the past three years.
Write this information down on separate pieces of paper, show them to your facilitator
and add them to your portfolio.
How to stick to your budget
(1) Be realistic in making up the budget in the first place. (Do not allocate $4,000.00
for food if you usually spend closer to $6,000.00).
(2) Make your savings amount the maximum you can realistically afford but be
willing to adjust this amount either up or down from month to month.
(3) Decide on a minimum amount for savings and never go below it.
(4) Take control of your utilities! Do not waste electricity and water and do not spend
unnecessary lengths of time on the telephone.
29
Mathematics - Stage 2 Module 1
(5) Compare prices in supermarkets and find a supermarket that best suits your
pocket.
(6) Take into consideration as well, the distance of the various supermarkets in your
area. (Is the money you saved at the supermarket later spent on transportation?)
(7) Remember this is a process of trial and error. It might not work the first time but
keep at it!
Guess what! You have made it to the end of another module! I knew you could do it!
Press on!
30
Mathematics - Stage 2 Module 1
ANSWERS
Exercise 1.1
7) − 13
1) 37
8) − 3
2) 14
9) − 864
3) 5
10) 12
UK₤94,501.20
8) US$25,000.0
UK₤15,000.00
9) US$225.00
4) − 7
UK₤135.00
5) 11
10)US$13,800.00
UK8,280.00
6) 28
Exercise 1.4
7) 12
1) 5 < 9
8) 22
2) −4 < 2
Exercise 2.2
9) − 9
3) − 3 > −20
1) $8,340.00
10) − 110
4) 7 > −7
2) $7,200.00
5) 15 > 10
3) $90,000.00
6) 11 > −23
4) $1,625.00
Exercise 1.2
7) −6 < −1
5) $2,752.00
1) 21
8) 8 < 14
6) $2,565.00
2) 12
9) −13 < 9
7) $59,230.00
3) − 11
10) 5 > −2
8) $1,980.00
4) −37
9) $6,985.00
5) 6
10)$345,793.80
6) 32
Exercise 2.1
7) 33
1) US$60.00
8) 225
9) 26
10) − 46
UK₤36.00
2) US$13.00
UK₤7.80
3) US$7.50
UK₤4.50
Exercise 1.3
4) US$266.67
Exercise 2.3
1) $132,000.00
2) $25,300.00
3) (a) $10,200.00
(b) $9,520.00
(c) $14,280.00
4) Husband $28,000.00
1) − 46
UK₤160.00
Wife
2) − 72
5) US$120.01
5) 1: 4
3) 16
4) 4
5) 114
6) − 23
$14,000.00
UK₤72.01
6) $52,440.00
6) US$11.50
7) $85,500.00
UK₤6.90
7) US$157,502.00
31
Mathematics - Stage 2 Module 1
Exercise 3.1
8)
20% profit
1) No. 41¢
9)
$2793.00
2) $31.98
10) 15% profit
3) $3,562.99
4) $101.55
5) 15%
Exercise 4.1
6) $1,169.40
1) $1,250.00
7) $101.30
2) $22,825.00
8) $1,968.75
3) $135.00 per hr.
9) $20,668.73
4) $7.50 per hr.
10)$3,310.19
5) $25,300
6) $24,296.13
7) $3,384.00
Exercise 3.2
8) $37,850.00
1)
$2,132.58
9) $370,350.00
2)
$5,100.00 loss
10)$61,594.13
3)
$418.00
4)
$28,960.00
5)
$9,450.00
6)
$1,067.00
7)
$328.00
8)
$1,234.44 profit
9)
$677.00
10) $762.50
Exercise 3.3
1)
6%
2)
7%
3)
8%
4)
10%
5)
12.5%
6)
5%
7)
14%
32
Mathematics - Stage 2 Module 1
STAGE 2
MODULE 2
MODULE 2:
ALGEBRA PART 2
UNIT 1:
FURTHER ALONG THE ALGEBRA TRAIL
Introduction
Now that you have successfully completed stage 1 algebra, this stage will take you
further along the algebraic journey. As you travel you will find that what you learnt in
stage 1 fits neatly into what you will learn in stage 2. Each stage acts as a building block
for future stages, and as you explore the concepts further you may find it necessary to
revisit the previous stage to remind yourself of what went before. The journey gets more
interesting as you travel so be on the alert for more discoveries. Bear in mind that to
gain mastery in algebra the magic word is ‘Practice”.
In this unit you will investigate some properties of numbers specifically as they apply to
algebra.
You will discover the interesting ways in which numbers are used in
Mathematics and particularly in algebra. The focus of this unit will be on the use of the
distributive property. You will appreciate the importance of using brackets to simplify
what seem to be complex problems, you will do more work with signed numbers and you
will discover the strong relationship between algebra and arithmetic.
By the end of this unit, you will be able to:
·
recognize and apply number properties to algebraic expressions.
·
demonstrate by examples the Commutative, Associative and Distributive
properties to simplify arithmetic and algebraic expressions.
·
investigate the properties of addition and apply them to algebraic
expressions.
·
investigate the properties of multiplication and apply them to algebraic
expressions.
·
use variables to explain general ideas e.g. a + b = b + a for any number a
and b.
1
Mathematics – Stage 2 Module 2
Prerequisites
Before you begin you should know:
·
how to add and subtract positive and negative numbers.
·
how to multiply and divide positive and negative numbers.
·
how to combine like terms and simplify algebraic expressions.
Properties of Addition
¨ Commutative property
Arithmetic
6 + 5 = 5 + 6 = 11
Algebra
a+b=b+a
The order in which two numbers are added does not affect the sum.
¨ Additive Inverse
Arithmetic 7 + (- 7) = 0
Algebra
a + (-a) = 0
There exists a unique number (-a) which when it is added to (+a) the result is 0.
¨ Additive Identity
Arithmetic 6 + 0 = 6
Algebra
a+0=a
When zero (0) is added to any number the number does not lose its identity. In
other words the number does not change. Zero is the additive identity.
¨ Associative property
Arithmetic 6 + (4 + 9) = (6 + 4) + 9 = 19
Algebra
a + (b + c) = (a + b) + c
In addition, the order in which numbers are grouped does not affect the sum.
NB. We can group or associate only two numbers at a time.
Let us see how well we can identify these properties now.
2
Mathematics – Stage 2 Module 2
Example 1.1
Identify the property used in each statement below:
(a) 3x + y = y + 3x
(b) 5a + 0 = 5a
(c) 2y + (3x + 4y) = (2y + 4y) + 3x
(d) 2x + (-2x) = 0
Solution
(a) Commutative property
(b) Additive identity
(c) Associative property
(d) Additive inverse
Properties of Multiplication
¨ Commutative property
Arithmetic 6 x 3 = 3 x 6 = 18
Algebra
axb=bxa
The order in which two numbers are multiplied does not affect the product.
¨ Associative property
Arithmetic 7 x (5 x 2) = (7 x 5) x 2 = 70
Algebra
a x b x c = (a x b) x c
In multiplication the order in which numbers are grouped does not affect the
product.
¨ Multiplicative Inverse
Arithmetic 6 ¸ 6 = 6 x
Algebra a ¸ a = a x
1
6
1
a
=1
=1
3
Mathematics – Stage 2 Module 2
The multiplicative inverse of any number a is 1. The product of any number a
a
and its multiplicative inverse is 1. The multiplicative inverse is also called the
reciprocal of the number.
¨ Multiplicative identity
Arithmetic 4 x 1 = 4
Algebra
ax1=a
The number 1 is the multiplicative identity. No number multiplied by 1 loses its
identity. In other words, if a number is multiplied by 1 it remains the same.
Example 1.2
State the property used in each statement below
((a) 2 ( 3b) = (2 x 3 ) x b
(b) ab = ba
(c) 2b x 1 = 1
2b
(d) 7v x 1 = 7v
Solution
(a) Associative property
(b) Commutative property
(c) Multiplicative inverse
(d) Multiplicative identity
We got them all right. Did you know that we perform these operations in our daily lives
without even making the connection with Mathematics? Most interesting, however, is
the use of variables to explain these general truths. Note also that we have added new
words to our algebra vocabulary. Please write them down and get familiar with them.
4
Mathematics – Stage 2 Module 2
The Distributive Property
In addition to the properties of addition and multiplication we will now investigate another
property that involves both operations.
Consider these examples:
(a) Arithmetic
5( 3 + 2) = (5 x 3) + (5 x 2)
OR
5 (5) = 25
= 15
+
10
= 25
Algebra 5(3x + 2y) = (5 x 3x) + (5 x 2y)
= 15x + 10y (multiplication is distributed over addition)
(b) Arithmetic
2(6 + 3 - 1) = (2 x 6) + 2 (3) - (2 x 1)
= 12 + 6 -2
OR 2 (6 + 3 -1)
= 2(8)
= 16
Algebra 2(6x + 3y - c) = (2 x 6x) + (2 x 3y) - (2 x c)
=
12x + 6y - 2c
(multiplication of a positive and negative number)
(c) Arithmetic
OR
-3(7 - 9) = -3(-2) = 6
-3(7 - 9) = -3 (7) -3 (-9) (multiplication of 2 negative numbers)
=
-21 + 27
= 6
Algebra -3(7x -9y)
= -3(7x) -3(-9y) (multiplication of 2 negative numbers)
= -21x + 27y
5
Mathematics – Stage 2 Module 2
Let us list all we have learnt about using the distributive property
(a) Multiplication is distributed over addition and subtraction
(b) All the terms in the bracket are multiplied by the factor outside the bracket
(c) When a positive and a negative number are multiplied the product is
negative
(d) When a negative number is multiplied by a negative number the product is
a positive number
Subtraction and Division
So far we have been investigating addition and multiplication. We will now examine
subtraction and division.
Consider this
Is (a) 6 - 1 = 1 - 6?
(b) 3 - 1 = 1 - 3?
(c) 7 - 5 = 5 - 7?
(a) 6 - 1 = 5 but 1 - 6 = -5 so the answer is no. Complete the others. Try other
examples.
The Commutative property does not hold for subtraction
Consider this
Is (a) 5 ¸ 2 = 2 ¸ 5?
(b) 6 ¸ 3 = 3 ¸ 6?
(c) 8 ¸ 4 = 4 ¸ 8?
The answer is no. 6 ¸ 3 = 2 but 3 ¸ 6 = 1. Complete the others. Try other examples
2
The Commutative property does not hold for division
In general a - b ¹ b - a and a ¸ b ¹ b ¸ a
6
Mathematics – Stage 2 Module 2
Significance of Zero (0)
(a)
6 x 0=0
(b)
5 x 0=0
(c)
4 x 0=0
In general any number multiplied by zero (0) is 0. If a x b = 0 either a = 0 or b = 0
or a and b = 0.
Division by zero is impossible
(a)
0
6
= 0 but
is impossible.
0
6
Exercise 1.1
In each question apply the distributive property to rewrite the expression. Combine like
terms and simplify where possible. The first one is done for you
1.
3(2a + b) - (a + b - c)
= 6a + 3b -a - b + c
(remove brackets)
= (6a - a) + (3b - b) + c (combine like terms)
= 5a + 2b + c (add and subtract terms)
= 5a + 2b + c
2.
2x + (b + c + x)
3.
3(x - y + z) -2(x + 4y +z)
4.
(3s + 2v - b) + ( s + v - b)
5.
5x - (x + y - z)
State the property used to justify each of the following statements.
6.
3( x + 6) = 3x + 18
7.
7 + (- 7) = 0
8.
6x + 0 = 6x
9.
5m + 2m = m(5 + 2)
10. 4 ¸ 4 = 1
7
Mathematics – Stage 2 Module 2
11. (x + 3) ( x + 2) = x( x + 2) + 3( x + 2)
12. (x - y) + z = x + (z - y)
13. 1 x 3 = 1
3
1
14. 5w + ( 2 + 7w) = ( 5w + 7w) + 2
15. ( 2) + ( - 2) = 0
5
5
16. b x 0 = 0
Simplify each expression below by applying the appropriate technique.
17. 4a + 3b - 2a - 2b
18. x + ( 1 - a) x - a
19. 2 ( 2a - a ) + b ( 1 - 4)
20. 12x - ( 3x - 7x + g ) + 3
21. 3 ( 4 - 3x) - 7x + 1
22. -2 ( 3x - 5 ) - 3 ( x - 1 )
23. 4 ( 8c + 4d ) - 2 ( 3c - d )
24. 5 ( x + y ) - 6 ( x - y )
25. x ( 2 - 3 ) +y ( 6 - 5 ) - z ( 4 - 3 )
26. x
-9 ( 5a - 6b ) + 7( 2a - 4b )
Substitution
As letters are used to represent numbers in algebra, these letters can therefore be
replaced by numbers. When this is done, a value can be found for a particular algebraic
expression. We use the word evaluate, which means to find the value of something and
in this case to find the value of a particular algebraic expression.
8
Mathematics – Stage 2 Module 2
Example 1.3
Evaluate the following when a= 2, b = 3, c = 1
Solution
6abc = 6 x a x b x c
= 6x2x3x1
= 36
NB I have found the number value for the algebraic expression.
Exercise 1.2
1.
Evaluate the following when a = 3, b = 4 and c = 0
(a) 3ab
(f)
(b) 2a - b
(g) a + 3( 2b -c )
(c) b - 2c
(h) ( a - 1 ) - ( 2 + b )
(d) 2 ( a + c)
(i)
- 4a + 3b - 2cb
(e) 2bc ( abc)
(j)
5a (2b + 4c)
2.
3ac + ab
In the spaces provided give examples of the properties stated. Use algebraic
expressions for your examples
(a)
Commutative property of addition_________________________
(b)
Multiplicative Inverse___________________________________
(c)
Additive Identity_______________________________________
(d)
Multiplication by zero___________________________________
(e)
Associative property of addition___________________________
(f)
Multiplicative Identity___________________________________
(g)
Distributive property____________________________________
(h)
Commutative property of multiplication_____________________
(i)
Associative property of multiplication_______________________
9
Mathematics – Stage 2 Module 2
3.
If a, b and c represent whole numbers state whether each of the statements
below is True or False
(a)
a x b = b x a
(b)
a ¸ b = b ¸ a
(c)
(a x b) x c = (c x b) x a
(d)
a - c = c - a
(e)
a + (b x c ) = ( a + b) x c
(f)
( a x b ) ¸ c = a x ( b ¸ c)
10
Mathematics – Stage 2 Module 2
MODULE 2:
ALGEBRA PART 2
UNIT 2:
ALGEBRAIC EXPRESSIONS
Introduction
Your work in this unit will expand your experience with algebraic expressions. You will
observe many similarities to your work with numbers. Your ability to apply the four basic
arithmetic operations of addition, subtraction, multiplication and division will be very
useful. It is important that you know how to manipulate algebraic expressions because
in your later work in algebra you will need to know how. Further, many of the technical
trades and professions depend on your ability to solve real world problems dealing with
algebraic expressions.
By the end of this unit you will be able to:
·
add and subtract algebraic expressions.
·
multiply and divide algebraic expressions.
·
apply the distributive property to simplify algebraic expressions.
·
insert and remove brackets appropriately to simplify algebraic expressions.
·
identify a common factor in an algebraic expression.
Prerequisites:
Before you begin you should know:
·
how to apply the four arithmetic operations with signed numbers.
·
how to identify a common factor in a given set of numbers.
·
how to differentiate between a factor and a term.
Adding and Subtracting Algebraic Expressions
You have been studying the properties of numbers. Much of that work can be extended
to algebraic expressions where you will apply the four arithmetic operations (+, x, -, ¸)
11
Mathematics – Stage 2 Module 2
Example 2.1
Find the sum of 2a + 3b + 4 and 3a - 2b + 4c
Solution
Note that there are two expressions. Follow the steps
2a + 3b + 4c and 3a - 2b + 4c
=
( 2a + 3b + 4c ) + ( 2a - 2b + 4c )
® insertion of brackets
=
2a + 3b + 4c + 2a - 2b + 4c
® removal of brackets
=
2a + 2a + 3b - 2b + 4c + 4c
® combining like terms
=
a ( 2 + 2 ) + b (3 - 2 ) + c ( 4 + 4 )
® distributive property
=
4a + b + 8c
Example 2.2
Subtract x + 2xy - y from 2 (x + xy + y)
Solution
=
2 (x + xy + y) - ( x + 2xy - y)
=
2x + 2xy + 2y - x - 2xy + y) ® removing brackets
=
2x - x + 2xy -2xy + 2y + y ® combining like terms
=
x (2 - 1) + xy (2 - 2) + y (2 + 1) ® distributive property
=
x
=
x + 3y
0 + 3y ® property of zero
+
When you are removing brackets remember the following:
¨ All the terms of the expression within the bracket must be multiplied by
the factor outside the bracket
¨ If there is a minus sign outside the bracket all the signs within the
bracket change
¨ Like terms differ only by their numerical coefficients
¨ When zero is multiplied by any number the product is zero
12
Mathematics – Stage 2 Module 2
Review the examples given and make sure that you understand them then do the
exercise following.
Exercise 2.1
Find the sum of:
1.
3x + 4y + 4z and x + 3y + 8z
2.
a + b - c and a - b + c
3.
x ( a - b) and x ( a + b)
4.
4ax - 3by + 5cx and 7ax + 8by - 2cx
5.
4 ( a + b + c), 3 (2a - b - c), and 8 ( b - a + 2c)
6.
x2 + 2xy + y2 and x2 - 2xy - y2
Subtract:
7.
x + 3y + 3z from 5x + 7y + 2z
8.
3x - 4a + 11b from 5x -8a -2b
9.
5a - 3c + 4d from 6a - 2c - 2d
10. 3x2 -2xy -3y2 from x2 + 2xy + 5y2
11. By how much does 3a - 4x exceed a + 7x
12. Find the excess of 3(a + b) over 2( a - b)
13
Mathematics – Stage 2 Module 2
Multiplying Algebraic Expressions
Multiplication of algebraic terms mostly involves extending the distributive law.
Example 2.3
Remove the brackets and simplify the expression.
5( 9 + 2x ) + 2 ( x + 4 )
Solution
5 (9) + 5 (2x) + 2 (x) + 2 (4)
multiply each term in brackets by the
number outside the bracket
45 + 10x + 2x + 8
45 + 8 + 10x + 2x
group like terms
53 + 12x
add like terms and simplify
Example 2.4
Remove the brackets and simplify.
4 ( 3x + 6) - 6 (x – 2)
Solution
4(3x) + 4(6) - 6 (x) + 6 (2)
apply the distributive law to both sets of
brackets
12x + 24 - 6x + 12
12x – 6x + 24 + 12
group like terms
6x + 36
add like terms and simplify
14
Mathematics – Stage 2 Module 2
Exercise 2.2
Remove the brackets and simplify each expression.
1.
2 ( x - 1) + 3( 1 - x) -2( 2 - 3x)
2.
3 (2 - a) - 7( a + b) + 6 ( 2a + 7)
3.
3 ( 2a - c) - 5 (c - 3a) - 4 ( 5a - 2c)
4.
6 ( a -b + c ) - 4 ( b -a + c ) -2 ( c - a - b)
5.
2 ( 3x + 12) + 3 ( x - 4 ) - 4 ( 2x + 3)
Example 2.5
Identify the common factor and reduce each expression to its lowest term.
Solution
24ab
=
4a ( 6b )
4a
4a
= 6b
( 4a is common to both the term in the numerator
and the term in the denominator)
Exercise 2.3
1.
12a ¸ 4
2.
49pq ¸ 7p
3.
72ab ¸ 8b
4.
2yz ¸ 2z
5.
8st ¸ t
6.
36xy ¸ 6xy
7.
9ab ¸ ab
15
Mathematics – Stage 2 Module 2
MODULE 2:
ALGEBRA PART 2
UNIT 3:
EQUATIONS AND INEQUALITIES
Introduction
In this unit you will expand your knowledge in working with equations and inequalities.
You will recall your work with arithmetic operations and your previous knowledge about
equations and inequalities. This should not pose a challenge for you but if it does,
appreciate the fact that you are being called upon to think your way through a challenge
in a structured manner. Every challenge presents an opportunity for you to do better
By the end of this unit you will be able to:
·
use the additive inverse to solve equations and inequalities.
·
use the multiplicative inverse to solve equations and inequalities.
·
solve equations and inequalities by removing brackets.
·
solve inequalities involving the symbols ³, £, <, >.
·
solve inequalities algebraically and graphically using the number line.
·
translate verbal sentences into equations.
Prerequisites
Before you begin you should know:
·
how to add, subtract, multiply and divide using signed numbers.
·
how to apply the additive and multiplicative inverses.
·
how to graph inequalities involving x > 0 and x< 0 on the number line.
·
how to apply the distributive property in removing brackets.
·
how to find the perimeter of a rectangle.
16
Mathematics – Stage 2 Module 2
Removing Brackets in an Equation
When brackets are included in an equation, follow the steps below in order to solve the
equation.
(a) Apply the distributive property to remove the brackets.
(b) Combine like terms and apply the arithmetic operation.
(c) Isolate the variable to one side of the equation.
(d) Solve the equation.
Example 3.1
Solve the following equation
x + 3 ( 3x - 4) = 4 ( x + 2 ) + 4
Solution
x + 9x - 12
= 4x + 8 + 4
® remove brackets
10x - 12
= 4x + 12
® combine like terms and add
10x - 12 + 12
= 4x + 12 + 12 ® Add 12 to both sides
10x
= 4x + 24
10x - 4x
= 4x - 4x + 24 ® isolate the variable to the LHS of the equation
6x
= 24
x
= 4
Check: Substitute x = 4 in the equation
4 + 3 (12 - 4) = 4 (4 + 2) + 4
4 + 3 (8)
= 4 (6) + 4
4 + 24
= 24 + 4
28
= 28
17
Mathematics – Stage 2 Module 2
Review the steps and make sure you understand them, before doing the problems
that follow.
Exercise 3.1
Solve the following equations and check your answers.
1.
8x + 4 = 3x + 24
6.
9 - 4( 3x + 1) = 9 - 9x
2.
3x – 3 = 2x + 3
7.
2(2x – 5) = 6x – 5 + x
3.
6x – 6 = 3x + 3
8.
8x – 10 – 3(2x – 8) = 16
4.
3x – 2x + 6 = 2(x + 1) + 2
9.
7 + 4(x +2) = 8x – 5
5.
5x – (x + 4) = 8x – 8 +4
10. 9 -4(x – 3) = 3x – 28
Application and Problem Solving
Recall the steps involved in solving a word problem. Review stage 1 module 2 unit 2.
Example 3.2
Translate the following story into an algebraic equation and solve.
The length of a rectangular plot is 4cm more than 3 times its width. The perimeter of the
rectangle is 56cm, what are the dimensions of the rectangle?
Solution
The perimeter of a rectangle is equal to twice the length plus twice the width
2 (L + B) = 56
Let the Width be x cm
\ Length is (3x + 4) cm
2 (3x + 4 + x) = 56
® Substitute the values for L and B, insert brackets
6x + 8 + 2x
= 56
® Remove brackets, apply the distributive property
8x + 8
= 56
® Combine like terms and add
8x + 8 - 8
= 56 - 8
® subtract 8 from both sides of the equation
8x
= 48
® isolate the variable on the LHS
8x
8
=
48
8
x
=6
18
Mathematics – Stage 2 Module 2
Width = x = 6 cm
Length
= ( 3x + 4 )
= 3(6) + 4
= 18 + 4 = 22cm
Check: substitute x
= 6 in equation
2 (18 + 4 + 6 )
= 56
56
= 56
Exercise 3.2
Translate into algebraic equations and solve. Check your results.
1.
You have $80 to share for Tamara, Leslie-Ann and yourself. You are to get $2
more than 3 times what Tamara gets and Leslie-Ann gets twice Tamara’s share.
How much must each person get?
2.
The perimeter of one of Mr. Brown’s rectangular gardens is 35m more than its
width. The length is 15m. Find the width of the garden and hence find the
perimeter.
3.
The sum of three consecutive numbers is 96. What are the numbers?
4.
Miss Pat sent her son to buy 3 curling irons for her beauty salon. She knows the
cost of each was $x and so she gave him $400.00 extra to cover taxi fare and
lunch. She gave him $1450.00 in all. What is the cost of 1 curling iron?
5.
Three consecutive multiples of 3 sum to give 81. Find the three numbers.
6.
Three consecutive multiples of 5 sum to give 105. Find the three numbers.
7.
A barber wanted to buy 6 towels. He gave the shopkeeper $200.00 and got back
$14.00 change. How much did he pay for each towel?
8.
The three interior angles in a triangle are x, (2x + 20) and (3x + 10). Find the size
of each angle if the sum of the angles is 180°
9.
Joanie paid $161.00 inclusive of $14.00 tax, for 3 bars of chocolate. How much
did each chocolate cost?
10. The angles of a triangle are x, 2x and ( 2x - 5 ). Find the size of each angle if the
sum of the angles is 180°
19
Mathematics – Stage 2 Module 2
Inequalities Represented on the Number line
The symbol ³ (read from left to right) means greater than or equal to, so the statement
x ³ 4 reads x is greater than or equal to 4. Recall that the statement x > 2 (read from
left to right) means x is greater than 2 and cannot have the value 2. Both these
statements can be represented on the number line.
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
12
X ³4
-4
-3
-2
-1
X
0
1
2
3
4
5
6
7
8
9
>2
Note the filled in circle over the 4. Note also the open circle over 2. This shows that x =
4 while x ¹ 2
Exercise 3.3
Represent each of the following on a different number line
1.
x ³ 5
5.
y £ - 2
2.
x £ 4
6.
y < 0
3.
x > -2
7.
x £ 6
4.
x ³ - 1
8.
x ³ -3
You have seen that when both sides of an inequality are multiplied by the same positive
number the inequality sign remains unchanged.
If both sides of an inequality are multiplied or divided by the same negative
number the inequality sign is reversed.
20
Mathematics – Stage 2 Module 2
Consider this:
-5 < - 2 multiply both sides by - 1 or - 2 the results are (a) 5 > 2 and (b) 10 > 4.
Notice that the inequality sign changes direction. Remember when 2 negative signs
are multiplied or divided the result is a positive sign.
3x + 7 > 8x - 18 then graph the solution on the number line.
Solve:
Solution
-6
-5
3x + 7 - 7
> 8x - 18 - 7
3x
> 8x - 25
3x - 8x
> 8x - 8x - 25
- 5x
> - 25(divide both sides by - 5)
x
< 5 ( inequality sign reverses)
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
x < 5
Exercise 3.4
Solve each of the following inequalities, then graph the solution set on the number line
1.
5x - 3 £ 17
6.
8x - 2 ³ 2x + 10
2.
3x - 4 > 2x + 5
7.
9x + 2 ³ 2x - 19
3.
5x - 1 > 2 x + 8
8.
4x - 3 £ 6x + 5
4.
4x + 3 £ 3x + 11
9.
7 + 5x > 3x + 9
5.
7 - 5x > 3x - 9
10. 7x - 1 £ 10x - 7
21
Mathematics – Stage 2 Module 2
Removing Brackets in an Inequality
Solve each inequality and graph the solution set on the number line
Example 3.3
(a) 5m - 7
> 3 (m - 5)
5m - 7
> 3m - 15
5m - 7 + 7
> 3m - 15 + 7
5m
> 3m - 8
5m - 3m
> 3m - 3m - 8
2m
>-8
m
> -4
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
m > -4
³ 3(y - 1) + 4
(b) 2y + 7
2y + 7
³ 3y - 3 + 4
2y + 7
³ 3y + 1
2y + 7 - 7
³ 3y + 1 - 7
2y
³ 3y - 6
2y - 3y
³ 3y - 3y - 6
-y
³ - 6 (multiply both sides of the inequality by -1)
y
£ 6
-2
-1
0
1
( the inequality symbol is reversed)
2
3
4
5
6
7
8
9
10
11
12
y £ 6
22
Mathematics – Stage 2 Module 2
Exercise 3.5
Remove the brackets and solve each inequality. Graph the solution set on the number
line.
1.
3x + 2 < 2 ( x + 3)
2
3 ( x + 3 ) ³ 2 ( 4x - 8 )
3.
4k - 2 < 2 ( 3k + 2 )
4.
5 ( 2p + 3 ) > 3( 4p - 7 )
5.
4g - 9 £ 4 ( 2 - g ) - 1
6.
2 ( 3x - 5) £ 8
7.
6( 3x - 4 ) < 12x + 6
8.
7x + 3 > 3( x + 9 )
9.
-5( x + 4 ) > 6x + 2
10. 8x + 16 < 5x + 1
23
Mathematics – Stage 2 Module 2
MODULE 2:
ALGEBRA PART 2
UNIT 4:
SIMULTANEOUS EQUATIONS (1)
Introduction
So far you have been solving linear equations in one variable. The focus of this unit
is to introduce you to a new technique - the technique of solving linear equations in
two variables, and finding the solution set that will make both equations true at the
same time.
simple.
Are you asking yourself why you should learn this? The answer is
Learning to solve simultaneous equations will help you to answer some
personal questions especially when you have to make choices that can save you
money, or make the right decision whenever you have to. This first technique is
called the substitution method.
By the end of this unit you will be able to:
·
explain the meaning of simultaneous equations.
·
solve a set of simultaneous equations by the substitution method.
Prerequisites
Before you begin you should know:
·
how to solve linear equations in one variable
·
how to add, subtract, multiply and divide signed numbers
·
how to substitute given values for variables in an algebraic expression
Remember
¨ The product of two numbers with the same sign is always positive e.g.
2x2=4
-2 x ( -2 ) = 4 hence -a x ( -b ) = a b, for any two numbers a and b
¨ The product of two numbers with opposite signs is always negative e.g.
- 2 x 3 = - 6 hence -a x b = -ab
24
Mathematics – Stage 2 Module 2
¨ The quotient of two numbers with the same sign is always positive e.g.
-6 ¸ ( -2 ) = 3
Hence, -4a ¸ ( -2 ) = 2a
¨ The quotient of two numbers with opposite signs is always negative e.g.
6 ¸ ( -3 ) = 2
Hence, - 4a ¸ 2 = -2a
Exercise 4.1
Using the above information, simplify the following expressions
1. 8x - 3 ( 2x + 5 ) - 7 ( 3x + 4 )
6. 5 (2x + 3) -4 ( 4x + 3) - 3( 5x + 1)
2. 3x + 4 ( 3x - 2 ) + 3 ( 2x + 3 )
7. 4( 5x + 1) - 3(8x - 3) + 4x
3. 6x - 2 ( 3x + 7 ) - 8 ( 1 - 4x )
8. 3(3b + 7) + 2(2b - 1) - 6( 4b - 3)
4. - 2 ( 1 - 5x ) - 3 ( 2x + 7 ) + 9x
9. 6c + 7(2c + 5) - 3(c - 3) +4
5. 2x - 3 ( 4x + 3) + 4 (5x + 1)
10. 4y - 6 (y - 2) + 8 ( y + 1)
Simultaneous Equations
Whenever two linear equations are considered together, they form a system of linear
equations and may be referred to as simultaneous equations. These equations involve
the use of two variables and in order to solve them it is necessary that you find two
numbers that will satisfy both equations at the same time hence the word simultaneous.
Consider this equation:
x+y=6
I am sure you could think of several pairs of numbers that could satisfy that equation e.g.
x = 4 and y = 2 or x = 5 and y = 1 or x = 12 and y = - 6 and so on.
However if you had to solve a pair like this together: x + y = 6 and x = y + 4, only one
pair of numbers would satisfy both equations at the same time. Here is one of the ways
we solve two equations at the same time. This method is called the substitution method
because the first step in solving for the two variables is substitution.
25
Mathematics – Stage 2 Module 2
Example 4.1 (a)
x + y = 6…………… (1)
x = y + 4…………… (2)
Step 1
From equation (2) x = y + 4
Substitute x = y + 4 in equation (1)
(y + 4) + y = 6
y+4+y
=6
(apply the distributive property to remove brackets)
2y + 4 = 6
(addition of like terms)
2y + 4 - 4 = 6 - 4
(subtract 4 from both sides)
2y
=2
(isolate the variable on one side)
y
=1
(solve for y)
Step 2 Substitute y = 1 in equation (2) to find the value of x
x=1+4
x=5
Solution : x = 5,
Check
y=1
Substitute the values for x and y in both equations
x + y = 6 ………….. (1)
5+1=6
(equation is satisfied)
x
= y + 4 ………….. (2)
5
= 1 + 4
(equation is satisfied)
Example 4.1 (b)
x - 2y = 7………………. (1)
y = x - 5………..……… (2)
Step 1
Substitute y = x - 5 in equation (1)
x - 2 ( x - 5 ) = 7 ( substitute x - 5 for y)
x - 2x + 10
- x + 10
= 7 ( distributive property)
=7
26
Mathematics – Stage 2 Module 2
-x
+ 10 – 10 = 7 – 10 (subtract 10 from both sides)
-x
= -3
x = 3
Step 2
(divide both sides by -1)
Substitute x = 3 in equation (2)
y=x -5
y =3 -5
y= -2
Solution : x = 3,
y =-2
Check Substitute x = 3 and y = - 2 in both equations
x - 2y = 7
3 -2 ( -2) = 7
3+ 4 = 7
7 = 7 (equation satisfied)
y = x -5
-2 = 3 -5
-2 = - 2 (equation satisfied)
Example 4.1(c)
Solve
3x + 2y = - 18…………… (1)
X = 3y + 5………….. (2)
Step 1 Substitute x = 3y + 5 in equation (1)
3 (3y + 5) + 2y = - 18
9y + 15 + 2y
= - 18
11y + 15
= -18
11y = -18 - 15
11y = - 33
y = -3
27
Mathematics – Stage 2 Module 2
Step 2 Substitute y = -3 in equation (1)
3x + 2( -3) = -18
3x - 6 = -18
3x
= -18 + 6
3x = -12
x
Solution : x = - 4,
= -4
y=-3
Check to see if this solution satisfies both equations.
Review the above examples and try some for yourself from the recommended
texts.
Exercise 4.2
Solve the following simultaneous equations for x and y by the substitution method.
Verify your results.
1.
2x + y = 7
6.
x=y-2
x = y + 2
2.
3x + 2y = 12
7.
y=3
3.
5x - 6y = 21
5.
4x + 5y = 6
-4x + 2y = 12
y = -4x
8.
x = 2y + 5
4.
3x - y = 4
3y + 6 = 30
y = x +5
9.
8x + 5 = 61
y = 2x - 10
y=x-1
2x - 5y = 15
10. 9x + 20 = -25
x=y-7
x = 7y + 3
28
Mathematics – Stage 2 Module 2
Substitution Method for Solving Simultaneous Equations in Two Unknowns
¨ Identify the equation that gives a value for one of the variables in terms of the
other
¨ Substitute this value for the variable wherever it appears in the other equation
¨ The result will be a single linear equation with one unknown
¨ Solve for this variable
¨ Substitute this value in one of the original equations to solve for the other
variable
¨ Verify your results.
Think about this
Farmer Brown has a total of 33 cows and chickens. Altogether the animals have
90 feet. How many cows and chickens does farmer Brown have?
29
Mathematics – Stage 2 Module 2
MODULE 2:
ALGEBRA PART 2
UNIT 5:
SIMULTANEOUS EQUATIONS (2)
Introduction
You have learnt to solve simultaneous equations by substitution. This unit will introduce
you to another method that you can use as an alternative. The method is the elimination
method. Here you will eliminate one of the variables, before you can find the value of the
other. Then, by substituting the value of the variable into the equation, you find the value
of the second variable. It is always very good to be able to solve problems in different
ways. Mathematics is no exception.
By the end of this unit you will be able to:
·
solve simultaneous equations in 2 variables by the method of elimination.
·
write an equivalent equation, given a linear equation in 2 variables.
·
translate word problems into simultaneous equations and solve them.
Prerequisites:
Before you begin you should know:
·
how to solve linear equations.
·
how to add and subtract signed numbers.
·
how to substitute numbers for variables in an expression.
·
how to evaluate algebraic expressions.
The Elimination Method
The elimination method is sometimes referred to as the addition method because both
equations have to be added in order to eliminate one variable. Do you remember the
concept of equivalence? You met this concept in your work with sets, fractions and
ratios. Generally if 2 expressions are equivalent they have the same value.
30
Mathematics – Stage 2 Module 2
The set F{ red, blue, green } is equivalent to the set G { child, adult, teenager }
because they have equal number of members.
2 Û 6 because they have the same value. By what is the numerator and
3
9
denominator of the 1st fraction multiplied to get the 2nd ?
Similarly x + y = 5 Û 2x + 2y = 10. What is the multiplier in this example?
In solving simultaneous equations by the addition method, there are some steps to
commit to memory:
Step 1
The coefficients of one of the variables in the equations must be equal.
Step 2
Identify which variable will be the first to be eliminated.
Step 3
If necessary, multiply one or both equations by a constant (a number) to
get equivalent system(s) so that one variable can be eliminated.
Step 4
The signs of the variable to be eliminated in the equivalent systems must
be the inverse of each other.
Step 5
Add the equations of the equivalent system to eliminate one variable.
Step 6
Solve the equation and find the value of the other variable.
Step 7
Substitute this value into the original equation to solve for the first
variable.
Step 8
Check the solution by substituting the values in the original equations.
As we go through the example, you should use the steps that are necessary for the
solution.
Example 5.1
Solve the following simultaneous equations.
x + 3y = 12
2x - 3y = 6
31
Mathematics – Stage 2 Module 2
Solution
x + 3y = 12…………. (1)
2x
- 3y = 6…………. (2)
Notice that the coefficients of y in both equations is 3 and the signs are inverses of each
other. Therefore the first thing to do is to add both equations and eliminate ‘y’ (Step 5).
Add equations (1) and (2)
3x
= 18
=6
x
Substitute x = 6 in equation 1
x + 3y = 12
6 + 3y
6 – 6 + 3y
= 12
:
= 12 - 6
3y
=6
y
= 2
Solution : x = 6
y = 2
Example 5.2
x + y = 3
3x - 2y = 4
Solution
x + y = 3……………. (1)
3x
- 2y = 4……………. (2)
The coefficients of the variables in both equations are different. The signs of the
variable ‘y’ are inverses of each other, so eliminate ‘y’. Multiply (1) by 2 ® (step3)
2x + 2y = 6…………. (3)
3x - 2y = 4………….. (4)
Add (3) and (4) ® (step 5)
5x
x =
= 10
10
5
x=2
32
Mathematics – Stage 2 Module 2
Substitute x = 2 in equation (1)
x + y = 3
2 +y = 3
y = 3 - 2
y = 1
Solution: x = 2,
y = 1
Follow this example very carefully and ensure that you understand each step
3x - 2y = 14, 2x - 5y = 2
Solve:
Solution:
3x - 2y = 14………….. (1)
2x - 5y = 2……….. … .(2)
Multiply (1) by 2 and (2) by - 3 ® (step 3)
6x - 4y = 28……………… (3)
-6x
+ 15y = -6…………….. (4)
Add (3) and (4)
11y = 22
y = 2
Substitute y = 2 in (1)
3x - 2y = 14
3x - 4 = 14
3x = 18
x = 6
Solution: x = 6, y = 2
Check: Substitute x = 6 and y = 2 in (2)
2x - 5y = 2
12 - 10 = 2
2 = 2
33
Mathematics – Stage 2 Module 2
Exercise 5.1
Apply the elimination method to solve the following simultaneous equations. Check your
solutions.
1.
x + y = 6
6.
5x - y = 18
2.
x + y = 4
3y + x = 9
7.
y - x = 5
8.
3x - 4y = 0
4.
2x + 3y = 7
3x - y = 5
7x - 2y = 29
x=y+7
9.
x - y = 1
5.
6x + y = -15
x = y - 13
2x + y = 15
3.
4x + 3y = 27
3x - y = 31
x + 2y = 22
10. 5x + 4y = -11
x=y-4
x + 3y = 35
Solving Word Problems
Solving simultaneous equations gives you another tool for solving problems. It enables
you to solve problems dealing with two variables. The method used in solving word
problems remain basically the same no matter which approach is used. Remind yourself
of the method.
Step 1
Read the problem carefully to determine the unknown quantities.
Step 2
Choose variables to represent the unknowns.
Step 3
Translate the problem to the language of algebra to form an equation or a
system of equations.
Step 4
Solve the equation or system of equations.
Step 5
Verify your solution by substituting the values in the original problem.
In the previous unit you were asked to think about a problem. Let us examine the
problem and using the method above, find a solution.
34
Mathematics – Stage 2 Module 2
Example 5.3
Farmer Brown has a total of 33 cows and chickens. Altogether the animals have 90 feet.
How many cows and chickens does farmer Brown have?
Solution:
You are required to find the number of chickens and cows the farmer has. This is a
simultaneous equation problem. Two different variables are necessary. Let ‘b’ represent
the number of chickens and ‘c’ represent the number of cows. You are told that there is
a total of 33 cows and chickens. Translated into an equation, this information is:
b + c = 33
Altogether the animals have 90 feet. You know that each chicken has 2 feet while each
cow has 4 feet. Translated into an equation, the information is:
2b + 4c = 90
Solving for ‘b’ and ‘c’ will give the solution to the problem.
Solution
b+c
= 33……………… (1)
2b + 4c
= 90…………….. (2)
Multiply (1) by -2
-2b - 2c
= - 66…………. (3)
2b + 4c
=
90………….. (4)
Add (3) and (4)
2c
= 24
c
= 12
Substitute c = 12 in equation (1)
b + c
= 33
35
Mathematics – Stage 2 Module 2
b + 12 = 33
b
= 33 - 12
b
= 21
Check: Substitute b = 21 and c = 12 in (1)
b + c
= 33
21 + 12 = 33
33
= 33
So there are 12 cows and 21 chickens
Example 5.4
Find 2 numbers such that twice the first added to 3 times the second is equal to 45, and
such that 5 times the first added to 4 times the second equals 74.
Solution:
2 numbers are required. Let ‘x’ be the first number and ‘y’ the second.
twice the first added to 3 times the second equal 45
2x + 3y = 45
and 5 times the first added to 4 times the second equal 74
5x + 4y = 74
Solve:
2x + 3y = 45 and 5x + 4y = 74
2x + 3y = 45…………….. (1)
5x + 4y = 74…………….. (2)
Multiply line (1) by - 4 and line (2) by 3
-8x - 12y = - 180…………. (3)
15x + 12y =
222…………. (4)
Add lines (3) and (4)
7x = 42
x=6
36
Mathematics – Stage 2 Module 2
Substitute x = 6 in line (1)
2x + 3y = 45
Check: Substitute x = 6 and y = 11 in (1)
12 + 3y = 45
2x + 3y = 45
3y = 33
12 + 33 = 45
y = 11
First number = 6, second number = 11
Review the examples and ensure you understand the method.
Exercise 5.2
For each of the following problems write the equation by choosing a variable to
represent each unknown quantity, and then solve the equation.
1. An apple and a mango together cost $40.00. 3 apples and 4 mangoes cost
$130. 00. What is the cost of each mango and each apple?
2. You need a rectangular box to send a package. The length of this box must be 5cm
more than 3 times its width. The perimeter of the box must be 74cm. Find the
dimensions of the box. (Hint: use the formula for the perimeter of a rectangle)
3. One number is 5 more than twice another. If the sum of the two numbers is 29, find
the 2 numbers.
4. The average of two numbers is 25, and their difference is 8. What are the numbers?
(Hint: use the formula to find average)
5. Separate 20 plums into 2 groups such that twice the larger group is equal to 3 times
the smaller group. How many plums are in each group?
6. One number is 3 more than 4 times the other. The sum of the numbers is 38, what
are the numbers.
7. Kai is 4 times as old as his baby sister. In 3 years time the sum of their ages will be
26 years. Write an equation for this story and find out how old Kai is now.
8. One number is 5 less than twice the other. The sum of the two numbers is 19. What
are the numbers?
You now have two methods of solving simultaneous equations. Use the one with which
you are more comfortable. These equations can also be solved graphically. You will
meet this method later in the course.
37
Mathematics – Stage 2 Module 2
ANSWERS
Exercise 1.1
(d) 6
2.
3x + b + c
(e) 0
3.
x – 11y + z
(f)
4.
4s + 3v – 2b
(g) 27
5.
4x – y + z
(h) -4
6.
distributive
(i)
0
7.
additive inverse
(j)
120
8.
additive identity
9.
distributive
12
2.
10. multiplicative inverse
(a)
Answers will vary
11. distributive
(b)
Answers will vary
12. commutative
(c)
Answers will vary
13. multiplicative inverse
(d)
Answers will vary
14. commutative
(e)
Answers will vary
15. additive inverse
(f)
Answers will vary
16. zero property
(g)
Answers will vary
17. 2a + b
(h)
Answers will vary
18. 2x – ax – a
(i)
Answers will vary
19. 2a – 3b
20. 16x – g + 3
3.
21. 13 – 16x
(a)
True
22. -9x + 13
(b)
False
23. 26c + 18d
(c)
True
24. -x + 11y
(d)
False
25. -x + y – z
(e)
False
26. x - 31a + 26b
(f)
True
Exercise 2.1
Exercise 1.2
1.
4x + 7y + 12z
1(a)
36
2.
2a
(b)
2
3.
2ax
(c)
4
4.
11ax + 5by + 3cx
5.
2a + 9b + 17c
38
Mathematics – Stage 2 Module 2
6.
2x 2
7.
4x + 4y – z
8.
2x – 4a – 13b
9.
a + c -6d
10.
-2x 2 + 4xy + 8y 2
11.
2a – 11x
12.
a + 5b
5x – 3
2.
2a – 7b + 48
3.
a
4.
12a – 8b
5.
x
x=
-4
3
7.
x=
-5
3
8.
x=1
9.
x=5
10.
x=7
Exercise 3.2
Exercise 2.2
1.
6.
1.
x + (3x +2) + 2x = 80
Tamara gets $13.00
Leslie-Ann gets $26.00
You get $41.00
2.
30 + 2w = w + 35
width = 5m, perimeter = 40m
3.
Numbers : 31, 32, 33.
Exercise 2.3
1.
3a
2.
7q
3.
9a
4.
y
5.
8s
6.
6
7.
9
x + (x+1) + (x + 2) = 96
4.
3x + 400 = 1450
1 curling iron cost $350.00
5.
x + (x + 3) +(x + 6) = 81
Numbers : 24, 27, 30
6.
x + (x + 5) +(x + 10) = 105
Numbers : 30, 35, 40
7.
6x + 14 = 200
Each towel cost $31.00
8.
Angles : 25°, 70°, 85°.
Exercise 3.1
1.
x=4
2.
x=6
3.
x=3
4.
x=2
5.
x=0
x + (2x + 20) + (3x + 10) = 180
9.
3x + 14 = 161
1 chocolate bar cost $49.00.
10.
x + 2x + (2x – 5) = 180
Angles : 37°, 74°, 69
39
Mathematics – Stage 2 Module 2
Exercise 3.3
1.
x³5
0
2.
4
5
6
7
8
9
10
11
12
13 14
-3
-2
-1
0
1
2
3
-1
0
1
2
3
4
5
-2
-1
0
1
2
3
4
5
6
-7
-6
-5
-3
-2
-1
0
1
4
5
6
7
8
9
10
-3
-2
6
7
8
9
10
7
8
9
10
2
3
4
5
x ³ -1
-4
5.
3
x > -2
-4
4.
2
x£4
-4
3.
1
-3
y £ -2
-9
-8
-4
40
Mathematics – Stage 2 Module 2
6.
x<0
-9
7.
-8
-6
-5
-4
-3
-2
-1
0
-4
-3
-2
-1
0
1
2
-3
-2
-1
0
1
2
3
4
-4
-3
-2
-1
0
1
2
3
8
9
10
11
1
2
3
4
5
4
5
6
7
8
7
8
9
7
8
x£6
-6
8.
-7
-5
3
x ³ -3
-5
-4
5
6
Exercise 3.4
1.
x£4
-6
2.
-5
4
5
6
16
17
18
x>9
6
7
12
13
14
15
19 20
41
Mathematics – Stage 2 Module 2
3.
x>3
0
1
0
1
6
7
8
9
10
11
12
13
14
2
3
4
5
6
7
8
9
10
11
12
13
14
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
0
1
2
3
4
5
6
7
8
9
10
11
12
13
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
x ³ -3
-6
8.
5
x³2
-1
7.
4
x<2
-9
6.
3
x£8
4.
5.
2
-5
x ³ -4
-7
-6
42
Mathematics – Stage 2 Module 2
9.
x>1
-3
-2
-1
0
1
2
3
4
5
6
7
8
9
10
11
1
2
3
4
5
6
7
8
9
10
11
12
13
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
-4
-3
-2
-1
0
1
2
3
4
5
6
7
8
x³2
10.
-1
0
Exercise 3.5
1.
x<4
-6
-5
x£5
2.
-6
3.
-5
k > -3
-6
-5
43
Mathematics – Stage 2 Module 2
4.
p < 18
9
-7
12
13
14
15
16
17
18
19
20
21
22
23
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
7
x£3
6.
-7
-6
x<5
-7
8.
11
g£2
5.
7.
10
-6
-5
-4
-3
-2
-1
5
6
7
8
9
0
1
2
3
4
5
6
7
10
11
12
13
14
15
16
17
x>6
3
4
44
Mathematics – Stage 2 Module 2
9.
x<2
-7
10.
-6
-5
-4
-3
-2
-1
0
1
2
3
4
-10 -9
-8
-7
-6
-5
-4
-3
-2
-1
0
9.
x = 7, y = 6
10.
x = -5, y = 2
5
6
7
x < -5
-11
Exercise 4.1
1
2
3
1.
-19 - 43
2.
21x + 1
3.
32x - 22
4.
13x - 23
Exercise 5.1
5.
10x - 5
1.
x = 4, y = 2
6.
-21x
2.
x = 11, y = -7
7.
13
3.
x = -20, y = -15
8.
-11b + 43
4.
x = 2, y = 1
9.
17c + 48
5.
x = 5, y = 10
10.
6y + 20
6.
x = 6, y = 1
7.
x = -4, y = 9
8.
x = 3, y = -4
Exercise 4.2
9.
x = 12, y = 5
1.
x = 3, y = 1
10.
x = -3, y = 1
2.
x = 2, y = 3
3.
x = 3, y = -1
4.
x = 4, y = -2
Exercise 5.2
5.
x = 10, y = 1
1.
6.
x = 3, y = 5
3a + 4m = 130
7.
x = -1, y = 4
Cost of each apple = $30.00
8.
x = 3, y = 8
Cost of each mango = $10.00
a + m = 40
45
Mathematics – Stage 2 Module 2
2.
l = 3w + 5
2 ( l + w) = 74
w = 8 cm
l = 29 cm
3.
x + y =29
2x + 5 = y
x=8
y = 21
4.
(x + y ) ÷ 2 = 25
x-y=8
x = 29
y = 21
5.
2y = 3x
x + y = 20
x=8
y = 12
6.
x = 4y + 3
x + y = 38
x = 31
y=7
7.
y = 4x
y + 3 + x + 3 = 26
x = 4, y = 16
Kai is 16 years old.
8.
2x – 5 = y
x + y = 19
x=8
y = 11
46
Mathematics – Stage 2 Module 2
STAGE 2
MODULE 3
INTRODUCTION TO MODULE 3
The quilt pattern shown below demonstrates how your knowledge of lines, angles and
shapes can be used effectively to create designs. In this module you will continue your
study of lines, angles, and plane shapes. You will explore the relationships that exist with
lines and angles, and the types of angles that result.
You are reminded that a good knowledge and understanding of the characteristics and
properties of plane shapes will enable you to develop an appreciation for their use in
day-to-day activities. For example, in marketing a product, the shape of the package is
considered important.
The work done in Stage 1 will be useful in this module. Do not hesitate to turn back to
what was previously done to remind you of the work. Be patient with yourself and learn
all you can. You will reap the benefits and will be happy for your achievements.
i
Mathematics – Stage 2 Module 3
MODULE 3:
GEOMETRY PART 3
UNIT 1:
LINES AND THEIR RELATIONSHIPS
Introduction
As discussed earlier in the course, lines, line segments and rays are fundamental
concepts in geometry. Lines intersect or meet at a point to form angles, plane shapes
and three-dimensional shapes.
By the end of this unit you will be able to:
·
identify lines that are parallel, perpendicular or oblique.
·
recognize these lines where they exist in the environment.
·
identify and describe the different types of lines according to their position in
the plane.
Prerequisites
Before you begin you should know:
·
the different types of lines.
†Activity 1.1
Here is a simple activity for you. You could even use it as a game with your friends.
Collect about 24 matchsticks, toothpicks or straws. Hold them in your hand about 40 cm
above a flat surface, and let them fall freely from your hand to the flat surface. Spend a
few minutes to observe how the matchsticks are positioned on the surface. You should
not try to rearrange them. Observe them as they appear. Make a few sketches to show
the formation of the matchsticks on the flat surface. Do this activity a second time, or if
you have a friend with you allow the friend to do the activity.
1
Mathematics – Stage 2 Module 3
It is likely that some of your matchsticks could look like this on the surface.
The position of these lines on the surface and how they relate to each other
communicate important geometric concepts. You should be able to identify horizontal
(running in the same direction as the horizon, on our paper this would be straight left to
right.), vertical (running in the same direction as a house column, on our paper this
would be straight top to bottom) and oblique (running from left to right or from right to
left slanted up or down). We can also look at the positions of the lines as they relate to
each other.
We can identify lines that are parallel to each other,
and also lines that are
perpendicular to each other. Notice that unlike the
parallel lines, the perpendicular lines meet. There are other lines that meet and/or
intersect but they are not perpendicular. Here is an example.
Have you observed that the parallel lines are different from the others? What is the
difference? There are many more things you can learn about lines and how they relate
to each other.
2
Mathematics – Stage 2 Module 3
†Activity 1.2
You have learned that lines are comprised of a series of points along a straight path.
You also know that a line is formed where two planes intersect. Use this information to
identify some of the places where these different lines exist in your home, at your
workplace etc. Make a list of at least 10 places (doors, windows, bookshelves etc.) and
say which lines you found (vertical, oblique, horizontal etc.)
Lines/ Line Segments in the Environment
straight
oblique
curved
horizontal
vertical
Line segments like these are seen in two- and three- dimensional figures. They also
form the sides of angles (except for the curved line).
Exercise 1.1
1. What type of lines can be paired to have parallel and perpendicular
relationships?
2. What line relationship will the intersection of a horizontal and a vertical line yield?
3. Do 5 different combinations of lines. Illustrate and name the line relationships.
4. What types of lines could meet to form acute angles?
5. What types of lines would not meet to form acute angles?
6. Different types of angles could be formed from a pair of lines that meet, if one is
oblique and the other is vertical, what type of angle is formed?
7. Which types of angles could not be formed from a vertical line and an oblique line
meeting?
3
Mathematics – Stage 2 Module 3
Here are some possibilities for pairs of lines for the acute angle mentioned before.
a pair of oblique lines
a horizontal and an
a vertical and an
oblique line
oblique line
Which angle would be formed when a vertical line and a horizontal line meet?
You would be right if you answered right angle.
Here it is.
Could these lines be used to form an acute angle?
It should not take long for you to realize the answer is “No”.
It is therefore correct to conclude that an acute angle cannot be formed from the
intersection of a vertical and a horizontal line.
However, it is very important to note that a right angle can be formed from a pair of
oblique lines.
Here are some.
Observe the different positions of the right angle. This is so because oblique lines go in a
variety of directions. You should learn to recognize angles pointing in different directions.
These are all right angles.
4
Mathematics – Stage 2 Module 3
MODULE 3:
GEOMETRY PART 3
UNIT 2:
ANGLES AND THEIR RELATIONSHIPS
Introduction
As you know, when two lines or two rays meet, they have a common P
endpoint. An angle is formed. In this figure, the angle is PQR (written
R
ˆ R ). The angle could also be named angle Q. The use
< PQR or PQ
of one letter to name an angle is done only when one angle is formed
at the vertex.
Q
By the end of this unit you will be able to:
·
identify and describe angles that are acute, right, obtuse, straight, and reflex.
·
estimate the size of angles with reasonable accuracy.
·
measure angles accurately using the protractor.
·
identify angles in the environment.
·
recognize ways angles are used in everyday life.
·
identify various pairs of angles – complementary, supplementary, adjacent,
vertically opposite.
·
identify angles associated with parallel lines – corresponding, alternate, cointerior or allied.
·
recognize angles at a point
Prerequisites
Before you begin you should know:
·
the different types of lines, and how they relate to each other in terms of their
position in the plane.
·
the relationship between points and lines.
·
the concept of angles.
·
parts of angles.
5
Mathematics – Stage 2 Module 3
P
S
Q
R
In this case three letters should be used to name each angle to avoid confusion. For
example If you say angle Q, which angle would it be? You could be describing < PQS or
<SQR .
< PQS + < SQR = < PQR.
This means that the size of the
two angles < PQS and < SQR together is the same as < PQR.
Remember! The measure of an angle can be thought of as the amount of turning about
a point. The movement of the hands of the clock demonstrates very well the turning as it
is taking place. The angle is formed between the two hands of the clock.
Look at the diagram below. The size of this angle is determined by the amount of turning
from the position of QR to PQ about the point Q. A number is assigned to the spread
between the two sides (arms).
You may say < PQR measures 50° or < PQR = 50°.
P
R
Q
As mentioned earlier, the movement of the hands of a clock is a good demonstration of
the formation of angles.
6
Mathematics – Stage 2 Module 3
†Activity 2.1
1. Use cardboard to make a clock face with the numbers clearly marked.
2. Use split pins to attach the hands to the face.
3. Unlike the real clock, allow the hour hand to remain stationary at 12 while you move
the minute hand. You can use this to form acute, obtuse and reflex angles.
Angles on Clock Faces – Types of Angles
Here are pictures of some clock faces. The position of the hands of each clock
represents a different type of angle.
Straight angle
Right angle
Obtuse angle
Acute angle
Activity 2.2
For each type of angle, write down two other times. Look at a real clock in order to help
you get accurate answers.
7
Mathematics – Stage 2 Module 3
From Activity 2.2
·
a half turn measures 180°
·
a quarter turn measures 90°.
Looking at the clock faces on the previous page, we can make the following
statements.
1. A straight angle measures 180°.
2. A right angle measures 90°.
3. An acute angle measures less than 90°.
4. An obtuse angle measures more than 90°, but less than 180°.
We did not represent the reflex angle on the clock face. However a reflex angle is
represented as the exterior angle formed, for all the times except 6:00 o’clock. A
reflex angle is more than 180° but less than 360°.
reflex
reflex
†Activity 2.3
Now draw your own examples of these types of angles. Draw at least two of each
type.
8
Mathematics – Stage 2 Module 3
†Matchstick Activity
This time we will look at the angles formed as a result of the formation of the lines
(sticks). You could also reorganize the sticks when necessary, to model certain angles.
Types of Angles
After dropping my matchsticks and arranging some of them, these are some of the
angles that were formed.
Observe the angles that are marked. They have special
relationships. Some angles are named according to the position of the angles in relation
to parallel lines. These are corresponding angles, alternate angles and co-interior
angles (seen below).
Corresponding angles
Alternate angles
Co-interior angles
x
y
p
q
Vertically opposite angles
c
d
Supplementary angles
complementary angles
a
b
Adjacent angles
Angles at a point
9
Mathematics – Stage 2 Module 3
There are other pairs of angles which have special relationships to each other. The
angles are vertically opposite angles, the complementary angles, the adjacent
angles, and the supplementary angles.
Angles at a Point
The last set of angles to be mentioned are all drawn from the same point, hence they are
classified as angles at a point. When the arms of many different angles all radiate from
one point, the arms sometimes look like the spokes of a bicycle. The angles around the
point form a circle and the sum of the angles is 360°. (on previous page).
All the angles formed from one common point and forming a circle, together add
up to 360°.
Remember now!
Angles at a point add up to 360°.
All the angles mentioned above can be seen in some form in the environment.
We will refer to these angle relationships again when we discuss the properties of plane
shapes.
10
Mathematics – Stage 2 Module 3
Angles involving parallel lines
1. Corresponding angles are two or more angles which are found in the same
kind of positions in relation to the parallel lines and the transversal (the line that
cuts through both parallel lines).
a
b
Angles a and b are both above one of the parallel lines and to the right of the
transversal. They are therefore in the same kind of position and are said to be
corresponding. If there were two angles that were both below one of the parallel lines
and to the left of the transversal, these too would be said to be corresponding angles.
Corresponding angles are equal to each other.
2. Alternate angles are found where the parallel lines and transversal form an N or
a Z. These angles are found in the corners. The N or the Z does not have to be
right side up or facing front. Alternate angles are equal.
t
b
x
a
c
d
11
Mathematics – Stage 2 Module 3
e
f
3. Co-interior angles are formed where the parallel lines and transversal form an
F. The F could be backward or up side down. The angles are both on what could
be called the inside or interior of the F, in the corner. Co-interior angles sum to
give 180°, that is, they are supplementary.
h
k
Angle pairs
1.
Complementary angles are two angles that sum to give 90°. So an angle of
27° is complementary to an angle of 63° and 35° is the complement of 55°.
2.
Supplementary angles are two angles which sum to give 180°. So 97° and 83°
are supplementary to each other and 46° is the supplement of 134°.
3.
Adjacent angles are just simply angles that are beside each other. They share
a common arm and are also both part of a common vertex. When adjacent
angles are on a straight line they are said to be supplementary (ie. they sum to
give 180°).
12
Mathematics – Stage 2 Module 3
C
A
B
G
D
F
K
H
In the diagrams above, the common arms are AB and HG respectively. The shared
vertices are A and H. In the second diagram the adjacent angles (FHG and GHK) are
supplementary.
Measuring Angles
Angles are most commonly measured in degrees, denoted °. Radians are also used to
record the size of an angle. The use of radians will be discussed at another stage. A
protractor is the tool used to measure angles in degrees. There are two types of
protractors - the 180° protractor and the 360° protractor. Your geometry set should have
an example of the 180° protractor.
Using the protractor to measure angles
Take a good look at the protractor from your geometry set (the semi-circle). You will
notice that it has two rows of numbers following the curve. These are called the scales.
Parallel to the straight part of the protractor, is a line which we will call the zero line. Find
90° on the scales and trace the line that runs straight down from the 90° angle to the
zero line. Where this line crosses the zero line, we will call the centre.
13
Mathematics – Stage 2 Module 3
Steps (Ask your facilitator to help you with this).
1.
Place one arm of the angle directly under the zero line with the vertex at the
centre.
2.
Along the zero line where it covers the arm, find the zero.
3.
Starting with that zero, follow the scale until you get to the place where the
second arm exits or would exit from under the semi-circular edge of the
protractor.
4.
Note the numbers on both sides of the line and take a reading.
NB. There are ten spaces between each pair of numbers. If the line is halfway between
40° and 50°, then the size of the angle is 45° and so on.
†Activity 2.4
1. a) Draw a pair of lines that are not parallel, and a transversal to cut them. Measure
the angles. Are there any angles in this figure that could be called corresponding
angles or alternate angles?
b) Write a statement to explain what you notice.
How aware are you of some of the mathematical ideas that are displayed in the
environment, portrayed in the objects you see and interact with from day to day?
Perhaps after this unit you will agree that a lot is often ignored and taken for granted.
The use of lines and angles in our everyday activities is evident in our homes,
workplaces, community, in short, our environment. Take a look at some of the furniture
in your home.
14
Mathematics – Stage 2 Module 3
·
Look at the pictures below. They display different types of lines and angles.
There is an angle between the chair back and its seat. There is also an angle between
the chair seat and its feet. This angle varies with the type of chair and the purpose for
which the chair is used. The angle determines how erect the chair back is.
·
There are some stools below.
15
Mathematics – Stage 2 Module 3
The legs of the stool are set at a particular angle to the seat and to the crossbar. The
type of angle formed between the seats and the legs of the stools seen here are obtuse.
The angles formed by the legs and the crossbars are acute angles and the angles
formed within the crossbars themselves are right angles. A good carpenter would be
able to explain to you why some angles are used instead of others. Engineers and
builders also use angles when they design and build structures.
·
Take a look at the trees around. You should notice that the angles at which the
branches are set to the main trunk, vary with the kind of tree. Find out more
about this from your study of Science.
Lines and angles work hand-in-hand. To obtain certain angles the lines must be drawn in
a particular way. Equally, if the objective is to have the lines intersect in a certain way,
then a certain kind of angle will be formed in the process.
Exercise 2.1
1.
Think about the distance you would see between the arms of a 60° angle, then
sketch the angle. Use your protractor to measure the angle you sketched to see
how close you were!
2.
What kind of angle could be formed if a horizontal line meets a vertical line?
3.
Could an angle be formed if the two lines are parallel?
4.
Could an angle be formed if the two lines are skewed?
How many of us give thought to questions like these. Investigations of this nature
provide useful information about outcomes before tasks are done. In a home
improvement project, or any similar task where lines and angles are used, this
information is quite useful. When one is aware of what the outcome should be, or the
options that are available, it will be easier to identify errors and make corrections early. A
window frame incorrectly made, or a piece of grillwork poorly designed and made, are
costly errors which need to be avoided. The use of accurate measurement could
sometimes prevent these costly errors.
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Mathematics – Stage 2 Module 3
What is your response to Activity 2.3? Match your answers with these diagrams.
Obtuse angle
Acute angle
Reflex angle
You should realize by now that neither a right angle nor a straight angle can be formed
from the combination of a vertical line and an oblique line. The position of lines in the
plane will determine the resulting angle where these lines meet. This must always be
remembered by those who use lines and angles in their job.
Example 2.1
A vertical line and an oblique line used to
form a 70° angle.
70°
17
Mathematics – Stage 2 Module 3
Exercise 2.2
1.
Identify the different pairs of lines that can be used to form the following angles.
Include a sketch of each angle.
(a) An angle of 50°
(c) An angle of 90°
(b) An angle of 110°
(d) An angle of 320°
(e) An angle of 180°
2.
Copy and complete the following.
(a) A complete turn (one revolution) = ----------- degrees
(b) A half turn = ------------ degrees
(c) A third of a turn = ------------- degrees
(d) A quarter of a turn = ------------ degrees
(e) A fifth of a turn = ----------- degrees
(f) One sixth of a turn = ------------ degrees
(g) One eighth of a turn = ----------- degrees
(h) One ninth of a turn = ----------- degrees
(i) One tenth of a turn = ------------ degrees
18
Mathematics – Stage 2 Module 3
MODULE 3:
GEOMETRY PART 3
UNIT 3:
TRIANGLES AND THEIR PROPERTIES
Introduction
It’s all about lines and angles ‘shaping’ shapes. In the previous unit, you were told that
the angle/line relationships are also evident in plane shapes. The position of intersecting
lines in the plane determines the angle that is formed at the vertex, and therefore the
type of plane shape that emerges.
If you look around your community or any other community, you are likely to find some
homes that are being worked on. There are many things you could notice about those
homes. If I were to do the same in my community here are some of the things I am likely
to observe:
·
windows
·
doors
·
landscaping ideas – the garden, the lawn, the hedge, trees and shrubs, the
walkway
·
the fence or wall
·
the grillwork
If possible, I would try to get some information about:
·
the fixtures (ceiling lights, door handles etc.)
·
furniture
·
furnishings and decorations on the inside.
By the end of this unit you will be able to:
·
state the properties of triangles according to their sides.
·
state the properties of triangles according to their angles.
·
differentiate between different types of triangles.
·
determine the sum of the angles in a triangle.
19
Mathematics – Stage 2 Module 3
Prerequisites:
Before you begin you should know:
·
the different types of angles.
In any home improvement project you are likely to see a wide variety of shapes as you
observe the objects in that environment. For this part of the course the focus will be on
the polygons that could be used in some of those designs.
For the next few units, the focus will be on the properties of different two-dimensional
shapes, as we look at the lines that represent their sides and the angles that are formed.
As you make a study of the angles and sides of the polygons, you will observe that the
lines and angles relationships are as different as the shapes themselves.
The more you know about plane shapes -- their characteristics and their properties, the
more you discover about how you can use them.
Here are some pictures that portray these geometry ideas.
Polygons
A polygon is a closed figure that has three or more sides. A triangle is a three-sided
polygon and a quadrilateral is a four-sided polygon.
20
Mathematics – Stage 2 Module 3
The triangle and the quadrilateral are singled out here as special polygons.
Triangles and their Names
Equilateral
Right-angled
Isosceles
Acute-angled
Scalene
Obtuse-angled
Much of the information on polygons that you will learn, in the remainder of the module
will focus on triangles and quadrilaterals. Much of your work at the higher level will
feature more in-depth study of the triangle.
Unlike the other polygons, triangles and quadrilaterals are groups of polygons with many
variations in shape. These variations result from the relationship between the sides and
the angles of these polygons. In the units that will follow, you will have the opportunity to
study the variations in features. These polygons are classified based on their properties/
characteristics relating to their angles and sides. These will be explored in detail in
another unit. In this unit you will be introduced to the basic features of these polygons.
You will learn their names and get to know what each type looks like.
SEE A PART OF THE POLYGON FAMILY ON THE NEXT PAGE
21
Mathematics – Stage 2 Module 3
Polygons
Triangles
Obtuseangled
Scalene
Isosceles
Right-angled
Quadrilateral
Equilateral
Kite
Acuteangled
Trapezium
Parallelogram
Rhombus
Rectangle
Square
22
Mathematics – Stage 2 Module 3
Triangles
Triangles are classified and named according to their sides or angles. There is a
correlation between the lengths of the sides and the sizes of the angles. The longest
side is opposite the largest angle and the smallest side is opposite the smallest angle.
†Activity 3.1
Using a ruler and pencil, draw a large triangle on plain paper. The type of triangle does
not matter.
(a) Cut off the three angles of the triangles.
(b) Align the cut angles along a straight edge (like one side of a ruler) so that all
three angles are beside each other along the straight edge.
(c) What kind of angle is formed from the three angles of the triangle?
(d) Since the type of triangle drawn for part (a) did not matter, it means that what
we learn can be used to make a statement about all triangles.
(e) What is the general statement that can be made about sum of the angles of all
triangles?
You were already introduced to the different types of triangles in a previous unit.
However not much was mentioned about the properties that distinguish them from each
other and from other shapes.
Scalene triangle
A scalene triangle is a triangle with all its sides having different lengths.
23
Mathematics – Stage 2 Module 3
Isosceles triangle
The isosceles triangle has two or more sides measuring the same length.
Equilateral triangle
An equilateral triangle is one which has all its sides the same length.
Acute-angled triangle
The acute angled-triangle has only acute angles as its interior angles. (An acute angle is
one that measures between 0° and 90°).
24
Mathematics – Stage 2 Module 3
Obtuse-angled triangle
The obtuse-angled triangle has one of its interior angles being an obtuse angle. (An
obtuse angle is one which measures between 90° and 180°).
Obtuse angle
Right-angled triangle
The right–angled triangle has one angle measuring 90°. (A 90° angle is called a right
angle).
There will be some triangles which can fit two or more of these classifications at the
same time. For instance, an equilateral triangle is an isosceles triangle and it is also an
acute-angled triangle. In your future lessons with triangles, look out for these triangles
with multiple classifications.
25
Mathematics – Stage 2 Module 3
Example 3.1
a)
Two angles of a triangle were measured and found to be 34° and 69°. What is
the size of the third angle?
Solution
180 − ( 34 + 69)
= 180 − (103)
= 77
The third angle measures 77°
(b)
In a right-angled triangle, one angle measures 63°. What is the size of the third
angle?
Solution
Since it is a right angled triangle, we already have two angles (The right angle and the
63° angle). The third angle would therefore be equal to:
180 − (90 + 63 )
= 180 − ( 153)
= 27 °
Exercise 3.1
1. Name two triangles that fit into more than one classification.
2. What can be said about at least two angles of an isosceles triangle?
3. What is the size of each angle in an equilateral triangle?
4. In a right-angled triangle, one angle measures 67°, what is the size of the third
angle?
5. In an obtuse-angled isosceles triangle, one angle measures 140°. What is the
size of the other two angles?
26
Mathematics – Stage 2 Module 3
6. An obtuse-angled triangle has a 16° angle and a 59° angle. What is the size of
the third angle?
7. A right-angled triangle has one angle as 19°. What is the size of the third angle?
8. An isosceles triangle has a top angle measuring 67°. What is the size of each
base angle?
9. A scalene triangle has angles measuring 96° and 27°. State the size of the third
angle.
10. An isosceles triangle has base angles of 73°. What is the magnitude of its top
angle?
†Activity 3.2
(a) Select an equilateral triangle from the set of triangles provided at the back of the
module. Cut it out and place it on a sheet of thin paper on a flat surface and carefully
trace around it. Cut out this shape. You now have two equilateral triangles.
(b) Take the triangle cut from thin paper and fold it so that two of the sides lie on top of
each other as shown below.
Equilateral triangle folded
(c) Unfold the triangle. Fold a second time, but this time choose a different pair of sides
to lie on top of each other, edge to edge.
(d) Unfold and repeat a third time with a different pair of sides. What do you notice?
27
Mathematics – Stage 2 Module 3
In this activity, all the sides and angles fitted exactly over each other each time the
triangle was folded. What does this tell you about the size of the angles and the sides?
(e) Take the other equilateral triangle and use a ruler to measure the length of each
side. Record the measurements.
(f) Now use your protractor to measure each angle. Record the measurements you get.
Do you notice anything interesting having done the two activities?
If you measured the angles and the sides correctly you would discover that all three
sides are the same length, and all three angles are the same size. This is the reason all
the sides and angles fitted exactly over each other when you folded the other triangle.
Select other equilateral triangles of different sizes and repeat the measuring activity.
Again, if accurately measured, you should observe that all the angles and all the sides
are the same size.
All the sides of the equilateral triangle are equal.
All the angles are equal and such a triangle is also called equiangular.
Exercise 3.3
Select each of the other triangles from the set you have.
Measure the sides and the angles.
Copy and complete the table on the following page and record your measurements.
28
Mathematics – Stage 2 Module 3
Length of Sides
Name of Triangle
a
b
Size of Angles
c
A
B
C
†Research Activity
Use the information provided in the recommended texts to enhance your information on
the properties of each type of triangle. With the aid of diagrams, show how three of these
shapes could be used in a home improvement project. Identify the shapes you use.
29
Mathematics – Stage 2 Module 3
MODULE 3:
GEOMETRY PART 3
UNIT 4:
QUADRILATERALS AND THEIR PROPERTIES
Introduction
In the previous unit you measured the sides and angles of triangles to establish their
properties. You also did some paper-folding activities. In this unit you will carry out a
similar exercise on quadrilaterals to establish their properties. You may need to look
back at the previous unit to remind yourself about what was done. Explore the shapes
and see what you can find out about them.
We will be studying the six quadrilaterals mentioned in unit 3. They all have one property
in common. The sum of their angles is 360°. Four of them can be classified as
parallelograms. In fact, one of the quadrilaterals we will study is called a parallelogram.
By the end of this unit you will be able to:
·
identify the properties of figures by measuring and by folding.
·
state the properties of the square.
·
state the properties of a parallelogram.
·
state the properties of the rectangle.
·
state the properties of the rhombus.
·
identify the properties of the trapezium.
·
identify the properties of the kite.
·
determine the sum of the angles of a quadrilateral.
Pre-requisites
Before you begin you should know:
·
the properties of parallel lines.
·
different types of quadrilateral.
30
Mathematics – Stage 2 Module 3
Types of Quadrilaterals
Parallelogram
A parallelogram is a quadrilateral with its opposite sides parallel and equal. Parallel
means that they are running in the same direction without meeting. There will be more
discussion on parallelism later.
Rhombus
A rhombus is a parallelogram with all four sides equal in length.
†Activity 4.1
Draw the diagonals (a diagonal is a line that cuts across a figure, from one corner or
vertex to another) of the rhombus and measure the angles formed at the centre. You will
notice these are right angles.
31
Mathematics – Stage 2 Module 3
Square
A square is a parallelogram with four equal sides and has diagonals that bisect each
other at right angles. (A diagonal is a line that cuts across a figure, from one corner or
vertex to another). Its four interior angles are right angles.
Rectangle
A rectangle is a parallelogram with four 90° angles and its diagonals are equal in length.
Its opposite sides are also equal in length.
Trapezium
A trapezium is a quadrilateral with one pair of parallel sides.
32
Mathematics – Stage 2 Module 3
Kite
A kite has two pairs of adjacent sides equal, its diagonals cut each other at right angles
and it has one pair of opposite equal angles.
†Activity 4.2
Select the quadrilaterals from the set of shapes that you have at the back of the module.
Your set of quadrilaterals should include the following shapes: kite, trapezium, rhombus,
parallelogram, rectangle, square. Do the folding activity with each of the quadrilaterals
and make notes. Measure the angles and sides of each shape and record the
measurements.
Identify the quadrilaterals with parallel lines. Identify the quadrilaterals with right angles.
Measure the diagonals and record the lengths.
Record the information about these shapes based on the measurements that you have.
Now read about the properties of quadrilaterals in your recommended texts, and
compare with the information you wrote down and what has been given above.
†Activity 4.3 -
Getting to Know the Family of Quadrilaterals
Use the references provided in the recommended texts to obtain information on the
properties of each type of quadrilateral. Support your work with diagrams.
1. Write a brief description about each member of the quadrilateral family highlighting
the properties that make them different from each other, to ensure that each member
can be readily identified.
33
Mathematics – Stage 2 Module 3
2. Compare features (properties) of the different quadrilaterals and record the
similarities.
3. With the aid of diagrams, show how three of these shapes could be used in a home
improvement project. Identify the shapes you use.
†Activity 4.4
For this Activity you will be using the shapes on the next two pages.
Tangram
A tangram is a Chinese puzzle square cut into 7 pieces to be combined to form various
figures. Look at your tangram pieces (on the following pages) and identify the types of
lines and angles. Discuss these with your facilitator. Also identify the various types of
plane figures.
Now that you have identified the shapes, let us now make a tangram. As you will notice,
the pages have only 5 pieces. To do the tangram exercise, you must have seven pieces:
2 x triangle 1
2x triangle 2
1 x triangle 3
1 x parallelogram 4
1x square 5
You will need to copy the pages because you need two each of triangles 1 and 2. Cut
out the 7 pieces and try to use all of them to make a square. After you have succeeded
in making the square, try to make other figures. They do not have to be plane figures.
34
Mathematics – Stage 2 Module 3
35
Mathematics – Stage 2 Module 3
36
Mathematics – Stage 2 Module 3
TRIANGLES 1
Equilateral triangle
Right-angled triangle
Obtuse-angled triangle
Isosceles triangle
37
Mathematics – Stage 2 Module 3
EQUILATERAL TRIANGLES
38
Mathematics – Stage 2 Module 3
QUADRILATERALS
Square
Rectangle
Kite and Rhombus
Trapezium
39
Mathematics – Stage 2 Module 3
Rhombus
Parallelogram
40
Mathematics – Stage 2 Module 3
Worksheet 2.1
Measure and compare the sizes of the marked angles shown in each diagram. Write
down the sum of the angles in each of the diagram, Nos.3, 5, 6, 7, 8.
1
2
Alternate angles
Corresponding angles
a
b
c
d
3
e
4
x
f
5
Co-interior angles
x
y
p
q
6
complementary angles
Vertically opposite angles
b
a c
f
d
e
7
8
Angles at a point
r
s
h
Adjacent angles
i
Supplementary angles
41
Mathematics – Stage 2 Module 3
This is what the tangram pieces would look like when put together to form a square.
This was a long one but you have made it! Congratulations! Give yourself a pat on the
back, take a break and then PRESS ON!
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Mathematics – Stage 2 Module 3
Exercise 2.2
ANSWERS
1 (a) 1 vertical and 1 oblique
2 oblique lines
1 horizontal and 1 oblique
Exercise 1.1
(1) Parallel - 2 vertical lines
2 horizontal lines
2 oblique lines
Perpendicular 1 vertical and 1 horizontal
2 oblique lines
2.
(2) Perpendicular lines
(3) Answers will vary
(4) Acute angles 2 oblique lines
1 vertical and 1 oblique line
1 horizontal and 1 oblique
(5) Vertical and horizontal
(6) Oblique and vertical gives:
(a) acute angle
(b) obtuse angle
(c) reflex angle
(b)
(c)
(d)
(e)
Same as (a)
Vertical and horizontal
Same as (a)
2 vertical lines
2 horizontal lines
2 oblique lines
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
360°
180°
120°
90°
72°
60°
45°
40°
36°
Exercise 3.1
(1) Right-angled triangle
- Scalene, isosceles
Isosceles triangle
Isosceles triangle
- Equilateral triangle
- Right-angled triangle
Scalene triangle
- Right- angled triangle
- Obtuse- angled triangle
- Acute-angled triangle
(7) Oblique and vertical does
not give:
(a) right angle
(b) straight angle
Exercise 2.1
(2) A right angle
(3) No. The lines would not meet
(4) Yes.
(2) They are equal
(3) 60°
(4) 23°
(5) 20°
(6) 105°
(7) 71°
(8) 56.5°
(9) 57°
(10)34°
43
Mathematics – Stage 2 Module 3
STAGE 2
MODULE 4
MODULE 4
MEASUREMENT PART 2
UNIT 1:
UNITS OF MEASURE
Introduction
The Metric System of Length
The metric system is essentially a decimal system. The standard unit for measuring
length is the metre. However, there are times when the object being measured is less
than or more than a metre. The following are the units of measuring length as they relate
to the metre.
Kilometre (Km)
10
D
I
Hectometre (Hm)
U
10
Decameter (Dm)
V
10
Metre
I
L
T
(m)
I
10
Decimetre (dm)
D
M
10
P
L
Centimetre (cm)
E
Y
10
Millimetre
(mm)
By the end of this unit you will be able to:
·
convert from one metric unit of length to another.
·
convert from one metric unit of mass to another.
·
carry out the four mathematical operations with metric units.
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Mathematics - Stage 2 Module 4
Prerequisites
Before you begin you should know:
·
operations with whole numbers.
·
operations with decimals.
Since the metric system is essentially a decimal system we can easily convert from one
unit to another by multiplying or dividing by 10, 100,1000 or higher multiples of 10.
The Greek prefixes deca, hecto, and kilo signify the multiplication by 10, 100 or 1000,
respectively, of the unit to which they are prefixed; the Latin prefixes deci, centi, and
milli, signify the division by 10,100 or 1000, respectively, of the unit to which they are
prefixed.
Thus a decametre is 10 metres, but a decimetre is one tenth of a metre. Using the
diagram on the previous page, we see that to convert from a smaller unit to a higher unit
(e.g millimetre to decimetre) we would divide (follow the arrow).
Conversely, to convert from a larger unit to a smaller unit (e.g kilometre to metre) we
would multiply. The number of 10s between the two units indicate by what multiple of 10
we either multiply or divide.
Example 1.1
(a) Convert 15.36 metres to millimetres.
15.36m = 15.36 x 1000 = 15,360 mm
There are three 10s between metre and millimetre. The multiple of 10 needed
therefore is 10 x 10 x 10 = 1000. The operation would be multiplication. (The
arrow that shows movement from metre to millimetre is the down arrow and the
down arrow indicates multiplication).
(b) Convert 936 cm to m
936cm = 936 ÷ 100 = 9.36m
Centimetre to metre is two units up. There are two 10s between them. This
means we divide by 100.
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Mathematics - Stage 2 Module 4
(c) Convert 92,352 m to Km
92,352 m = 92,352 ¸1000 = 92.352 km
Three units up. Divide by 1000.
The Metric System of Mass /weight
The standard unit of measurement for measuring mass is the gram. Light objects are
weighed using grams or milligrams but heavy objects are measured in kilograms or
tonnes.
tonne (t)
D
1000
M
I
Kilogram (Kg)
U
1000
L
gram (g)
T
I
1000
I
D
milligrams (mg)
P
V
L
E
Y
Using the diagram above, we see that to convert from a smaller unit to a larger unit,
(grams to Kilograms) we divide. To convert from a larger unit to a smaller one, (grams to
milligrams) we multiply.
Example 1.2
Follow the examples below by using the diagram above.
(a) How many grams are equivalent to 8945 mg?
Here we need to convert milligrams to grams
8,945 mg = 8,945 ÷ 1000 = 8.945g
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Mathematics - Stage 2 Module 4
(b) How many grams are there in 19 kg?
Here we need to convert kilograms to grams
19 kg = 19 x 1000 = 19,000 g
(c) Convert 15 tonnes to kilograms
15 tonnes = 15 x 1000 = 15,000 kg
Addition and Subtraction of Metric Quantities
Metric quantities are added and subtracted in the same way as decimal numbers.
However, it is important that all the quantities be in the same units.
Example 1.3
(a) Find the sum of 3.8 cm, 15.47 cm and 96.08 cm
Solution
Since the units are the same, we simply add the numbers as decimals.
3.8
15.47
96.08
115.35 cm
(b)
Add together 15.2 m, 39.2 cm and 150.2 mm and state the answer in metres.
Solution:
15.2 m
39.2 cm =
150.2 mm =
0.392 m
(is already in m)
(change cm to m by dividing by 100)
0.1502 m (change mm to m by dividing by 1000)
15.7422 m
Therefore 15.2 m + 39.2 cm + 150.2 mm = 15.7422 m
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Mathematics - Stage 2 Module 4
Multiplication and Division of Metric Quantities
Metric quantities are multiplied and divided in exactly the same way as decimal
numbers.
Example 1.4
(a)
57 lengths of wood, each 95 cm long, are required by a builder. What is the total
length of wood needed? Give the answer in metres.
Solution
95 cm = 0.95 m (change cm to m by dividing by 100)
\ the total length required = 57 x 0.95 m
= 54.15 m
(b)
Frozen peas are packed in bags containing 450 g. How many packets can be
filled from 2000 kg of peas?
Solution
450 g = 450 ÷ 100 = 0.45 kg (change g to kg by dividing by 1000)
\ the number of bags = 2000 ÷
0.45
= 4444 packets
(c) A certain cloth cost $36.80 per metre. How much will 90m of this cloth cost?
Solution
the total cost = 90 x $36.80
= $3312.00
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Mathematics - Stage 2 Module 4
Exercise 1.1
1.
2.
3.
4.
5.
Convert to centimetres
(a)
17.15 m
(c) 0.395 m
(b)
1.78 km
(d) 864 mm
Convert to kilometres:
(a) 9375 m
(c) 73,568 cm
(b) 2.75 m
(d) 3941000 mm
Convert to tones. Give your answer to 2 places of decimal where possible.
(a) 27,000 kg
(c) 7824139 g
(b) 600049700 mg
(d) 4008 kg
Convert to grams.
(a)
7800 mg
(c) 19.1 kg
(b)
0.59 kg
(d) 45 mg
A greengrocer starts the day with 127 kg of apples. He sells 3.5 kg, 450 g and 25
kg. What mass of apples remains?
6.
A certain type of tablet has a mass of 8.2 mg. What is the mass, in kilograms, of
5,000,000 of these tablets?
7.
Add together the following sets of lengths, stating your answers in metres.
(a)
47 cm, 5.83 m, and 15 mm
(b)
93 km, 462 m and, 500 cm
(c)
0.185 m, 7.36 cm, and 8.2 mm
(d)
8.
6.2km, 9.7 m, 143.4 mm
Find the differences in the following pairs of measurements.
(a)
37 km and 5024 m
(b)
698.09 mm and 147.2 mm
(c)
0.16 km and 9746.99 cm
(d)
5342.33 cm and 740.07 mm
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Mathematics - Stage 2 Module 4
9.
A concrete mixer mixed 1056 kg, 2641 kg and 1498 kg of concrete in
3 days. How many tonnes of concrete were mixed?
10. Maurice works in a supermarket warehouse. On a certain day, he lifted 20 bags
each containing 2250 g of sugar. What is the total weight that Maurice lifted that
day?
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Mathematics - Stage 2 Module 4
MODULE 4:
MEASUREMENT PART 2
UNIT 2:
PERIMETER OF PLANE SHAPES
Introduction
The distance around a plane figure is called the perimeter. In this unit you will learn to
find the perimeter of plane shapes and how to change simple formulae to find length,
and width. You will also learn to do other applications when the perimeter is given.
By the end of this unit you will be able to:
·
calculate the perimeter of triangles.
·
calculate the perimeter of quadrilaterals.
·
calculate the perimeter of other polygons.
Prerequisites:
Before you begin you should know:
·
how to do all operations with metric units.
·
the difference between vertical and horizontal lines.
·
how to convert from one metric unit to another.
Example 2.1
(a) An equilateral triangle has side measuring 5 cm. What is its perimeter?
Solution
Remember an equilateral triangle has all sides equal.
Perimeter = (5 + 5 + 5) cm = 15cm
(b) A rectangle has length and width 5cm.
Calculate its perimeter.
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Mathematics - Stage 2 Module 4
Solution
Remember, a rectangle has two sets of parallel and equal sides.
Perimeter = (8 +5 + 8+ 5) cm = 26 cm
(c) A regular hexagon (six sided figure) has sides measuring 8cm. Find its perimeter.
Solution
The word regular here means that all the sides are equal in length. A hexagon has six
sides.
Perimeter = 8 + 8+ 8+ 8+ 8+ 8
= (8 x 6) cm = 48 cm
(d) A trapezium has sides measuring 5cm, 8cm, 7cm and 9cm. Find its perimeter.
Solution
Perimeter = 5+ 8 + 7+ 9
= 29 cm
Example 2.2
Find the perimeter of the shape shown below
3cm
A
B
7cm
C
D
2cm
E
F
6cm
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Mathematics - Stage 2 Module 4
Solution
To find the perimeter, we first have to count the number of sides the figure has. This will
tell us the number of lengths we need to add in order to find the perimeter. The figure
has 6 sides. We have been given the lengths of only 4 sides. We therefore need to find
the lengths of the other 2 sides.
The vertical short side without a measurement is equal to the difference between the
vertical long side and the vertical short side which do have measurements.
7 cm – 2 cm = 5 cm.
The horizontal short side that does not have a measurement is equal to the difference
between the horizontal long side and the horizontal short side which do have
measurements.
6 cm – 3 cm = 3 cm
Perimeter therefore = 3 + 5 + 3+ 2 + 6 +7
= 26 cm
NB It is wise to start adding from the top and move in a clockwise direction, so that
none of the measurements are missed.
Example 2.3
For the figure below, find the length of the fifth side if the perimeter is 36cm.
P
6.5cm
T
Q
6.9cm
6.5cm
S
7.2cm
R
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Mathematics - Stage 2 Module 4
Solution
The sum of the four sides given is 6.5 + 6.5 + 7.2 + 6.9 = 27.1 cm
The length of the fifth side would be the difference between the perimeter and the sum of
the 4 sides.
Length of fifth side = 36 - 27.1 = 8.9 cm.
Exercise 2.1
1.
The figure below shows a scalene triangle. Calculate its perimeter.
L
8.2 mm
6.3 mm
M
N
2.
7.1 mm
A rectangle has long sides 8.2 cm in length and short sides 7.5 cm in length.
Find the perimeter of the rectangle.
3.
The figure below shows an isosceles triangle. Calculate its perimeter.
F
8.2 cm
H
G
7.8 cm
4.
A kite has long sides 7.3 cm in length. If its short sides are 3.8 cm long, what is the
perimeter of the kite?
5.
A rhombus has sides 5 cm long. What is its perimeter?
6.
A trapezium has sides of lengths 13 cm, 17 cm, 8 cm, 10 cm. Find its perimeter.
7.
An equilateral triangle has sides measuring 13 cm. Find its perimeter.
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Mathematics - Stage 2 Module 4
8.
Find the length of the fourth side of the figure below if its perimeter is 28 cm.
H
9 cm
I
4 cm
J
K
9.
8 cm
Find the perimeter of the compound figure below.
M
3 cm
N
2 cm
2 cm
R
O
4 cm
Q
P
6 cm
10. Find the length of the fifth side of the pentagon (five sided figure) below if the
perimeter is 15.4 cm.
C
3 cm
3 cm
D
G
2.5 cm
2.5 cm
F
E
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Mathematics - Stage 2 Module 4
MODULE 4:
MEASUREMENT PART 2
UNIT 3:
AREA OF PLANE SHAPES
Introduction
The area of a plane shape is measured by seeing how many square units it contains.
The area can be measured by counting the number of equal squares in the shape. If the
length of each side of the equal squares is 1 cm then the area of each square is 1cm2
and the area of the figure is given as a number of centimetre squares. If the equal
squares measure 1 metre by 1 metre, then the area is given as a number of metre
squares etc. As seen in the first lesson, units can also be converted from one to another.
There may be instances where the measurements are given in one unit and the answer
is to be given in another unit.
By the end of this unit you will be able to:
·
calculate the areas of quadrilaterals.
·
calculate the areas of triangles.
·
calculate the areas of compound figures.
Prerequisites:
Before you begin you should know:
·
how to do operations with decimal numbers.
·
how to convert from one metric unit to another.
Example 3.1
Look at the shapes shown on the following page. By counting the number of equal
squares in the shaded regions in each figure, we can find which has the greatest area.
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Mathematics - Stage 2 Module 4
A
B
C
D
Figure A contains 12 equal squares
Figure B contains 14 equal squares
Figure C contains 14 equal squares
Figure D contains 16 equal squares
Therefore, figure D has the greatest area.
Area of Rectangle
The rectangle shown in the figure below contains 4 x 2 = 8 equal squares, each of which
has an area of 1 cm2 . Hence the area of the rectangle is 8 cm2. We can shorten this by
multiplying the length by the width in order to find the area of the rectangle. This applies
to any rectangle.
Hence area of rectangle = length x width
If we let A=area of the rectangle = length of the rectangle x width of the rectangle, then
A=LxW
2 cm
4 cm
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Mathematics - Stage 2 Module 4
Note that when using this formula, the units of the length and the width must be the
same. That is, they both must be in metres, centimetres or millimetres or any other
measurement given.
Example 3.2
A carpet measures 6 m by 5 m. What is the area of the carpet?
Solution
Remember, unless told otherwise we will assume that the shape of the carpet is
rectangular.
Length of the carpet = 6 m
Width of carpet
=5m
Area of the carpet
= (6 x 5) m2
= 30 m2
Example 3.3
A room 9 m long by 7 m wide is to be carpeted so as to leave a surround of 50 cm wide,
as show in diagram below. Find the area of the carpet.
(a)
50 cm
7m
9m
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Mathematics - Stage 2 Module 4
The carpet is represented by the inside rectangle. The best way to solve this problem is
to subtract the width of the surround from both ends of the length and from both ends of
the width then find the product of the lengths left.
Solution
50 cm = 50 ÷ 100 = 0.5 m
Length of carpet
= 9 m - (2 x 0.5)
=9-1=8m
width of carpet = 7 m – (2 x 0.5)
= 7 - 1 = 6m
Area of carpet
= (8 x 6) m2
= 48 m2
(b) Find the area of the shape shown below
3 cm
A
7 cm
B
2 cm
6 cm
The shape can be divided into two rectangles, as shown. Call one rectangle A and the
other one B.
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Mathematics - Stage 2 Module 4
Solution
Area of the rectangle A =
(7 x 3 ) cm2
=
21 cm2
Area of rectangle B
=
( 3 x 2) cm2
=
6 cm2
Total area of shape
=
( 21 + 6) cm2 = 27 cm 2
Since the bottom of the figure measures 6cm and the top piece measures 3cm (because
it does not go all the way across), the top piece of rectangle B = 6 − 3 = 3cm.
Example 3.4
A piece of wood is 85 cm long and 30 cm wide. What is its area in square metres?
In a question of this kind it is best to express each dimension in metres before
attempting to find the area. Therefore 85 cm = 85 ¸ 100 = 0.85 m and 30 cm = 30 ¸ 100
= 0.3 m.
Solution
Area of wood = (0.85 x 0.3)m2 = 0.255 m2
Sometimes we are given the area of a rectangular object or a rhombus or a
parallelogram and its measurement (such as width) and we have to find the other
dimension (such as length). This is done by dividing the area by the given dimension.
The formula A = LW can be changed or transposed to give either:
L = A
W
or
W =
A
L
Note carefully that the units for each term must be of the same kind. Thus if the area is
given in square metres, the given dimension (length or width) must be in metres and the
remaining dimension will also be in metres.
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Mathematics - Stage 2 Module 4
Example 3.5
The area of a rectangle is 44 cm2 and its width is 5.5 cm. Find its length.
Solution
Area of rectangle = length x width
A=LxW
L= A ¸W
L = 44 ¸ 5.5
L = 8 cm
The length of the rectangle is 8 cm.
Area of a Square
Since the square is a rectangle with all its sides equal in length, the formula for finding
the area of a square is written as:
Area of a square = side x side = side2.
A = L x L = L2
Example 3.6
(a) A square has side 9 cm long, Calculate its area
Solution:
Area = (9 x 9) cm2 = 81 cm2
(b)
A square plate has an area of 25 m2. What is the length of its side?
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Mathematics - Stage 2 Module 4
Solution:
Area = L2 (because the square has equal sides)
Length of side = Ö Area
Length of side = Ö25 =5m
Area of Parallelogram
A parallelogram is, in effect, a rectangle pushed out of shape, as shown below. The
vertical sides of the equivalent rectangle are shown as dotted lines. Hence the formula
Area of a Parallelogram = Length of base x vertical height (altitude) or
A = bh
Where A = area, b= length of base and h = vertical height.
altitude
base
Example 3.7
Find the area of a parallelogram whose base is 15 cm long and altitude is 8 cm.
Solution
Area of Parallelogram = Base x Altitude
= (15 x 8) cm2
= 120 cm2
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Mathematics - Stage 2 Module 4
Exercise 3.1
Find the area of the following parallelograms:
1. Base 7 cm, altitude 5 cm.
2. Base 15 mm, altitude 12 mm.
3. Base 250 mm, height 4 cm.
4. Base 5000 mm, height 2m.
5. room 8.5 m long and 6.3 m wide is to be carpeted to leave an area around the
carpet 60 cm wide.
Find (a) The area of the room
(b) The area of the carpet
(c) The area of the section around the carpet
6.
(a)
(b)
2 cm
1.2 cm
1.2 cm
12 cm
8 cm
8 cm
3 cm
Figure 1
Figure 2
Find the areas of the figures above.
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Mathematics - Stage 2 Module 4
2 cm
2 cm
2 cm
2 cm
8.5 cm
8 cm
10.5 cm
2 cm
2 cm
8 cm
Figure 3
12 cm
Figure 4
7. Find the areas of the shapes shown above.
8. The area of a room is 84 m2 and its width is 7 m, calculate its length.
9. A room is 5.4 m long and 4.2 m wide. It takes 1,575 square tiles to cover the floor.
(a) Calculate the area of the room
(b) Work out the area of each tile
10. The area of a parallelogram is 54 m2. If its base is 16 m long, calculate its height.
11.
7cm
2cm
12cm
5cm
2cm
Figure 5
The diagram above shows a steel section. Calculate its area in square centimetres.
12. A parallelogram has an area of 90 cm². If its vertical height is 6 cm, find its base
length.
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Mathematics - Stage 2 Module 4
Area of a triangle
The diagonals of the parallelogram shown below, split the parallelogram into two equal
triangles, hence the area of the triangle is
Area of triangle =
1
the area of a parallelogram.
2
1
x base x altitude
2
As a formula, the statement becomes
A=
1
bh
2
Where A = area, b = base and h = height (altitude)
Height
Height
Base
Base
Height
Base
Example 3.8
A triangle has a base 8 cm long and a height of 5 cm. Calculate its area.
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Mathematics - Stage 2 Module 4
Solution
Area of triangle = ½ x base x height
= (1/2 x 8x 5) cm²
=20 cm²
When we are given the length of the three sides of a triangle, we can find its area by
using the following formula:
A = S ( S - a)( S - b)( S - c)
Where S stands for half the perimeter of the triangle and a, b, and c are the lengths of
the sides of the triangle.
Example 3.9
The sides of a triangle are 7 cm, 8 cm, and 13 cm long. Calculate the area of the
triangle.
Solution
S= a +b+c
2
= 7 + 8 + 13
2
= 14
A=
=
14(14 - 7)(14 - 8)(14 - 13)
14(7)(6)(1)
=
588
A = 24.2 cm2
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Mathematics - Stage 2 Module 4
As with the quadrilaterals done before, we can work backwards and find height when
given the base and the area, or find base when given the height and area.
Example 3.10
The area of a triangle is 72cm2. If its base is 8cm, find its height.
Solution
Area
=
1
base x height
2
1
bh
2
1
72 =
x8xh
2
A
=
72 = 4 h
72 = h
4
18cm = h
18cm = height.
Exercise 3.2
Work out the areas of the following triangles
1. (a)
Base 6 cm, height 3 cm
(b)
Base 4 ½ cm, altitude 1 1/3 cm
(c)
Base 3.2 cm, vertical height 4 cm
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Mathematics - Stage 2 Module 4
2. Four right-angled triangles are shown below. Work out the area of each one.
(a)
9cm
(b)
12cm
4cm
5cm
15cm
3cm
(c)
(d)
5cm
26cm
12cm
10cm
24cm
13cm
3. A triangle has an area of 60 cm² and a base 12 cm long. What is the height of the
triangle?
4. A triangle has sides 4 cm, 5 cm and 7 cm long. Calculate its area.
5. A triangle has an area of 80 cm2 and base of 5 cm. Find its height.
6. A triangle has an area of 198 cm2 and a height of 22 cm. Calculate its base.
Area of a trapezium
A trapezium is a quadrilateral with one pair of parallel sides.
The area of the trapezium is found by using the formula:
Area of trapezium = (½ x sum of the parallel sides x the perpendicular distance between
the parallel sides)
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Mathematics - Stage 2 Module 4
A = ½ (a + b) h
Perpendicular distance between
parallel lines (height)
Example 3.11
The parallel sides of a trapezium are 12 cm and 16 cm long. The perpendicular distance
between them is 9 cm. Calculate the area of the trapezium
Solution
Area = [
=(
1
x (12 + 16) x 9] cm²
2
1
x 28 x 9) cm²
2
= (14 x 9) cm²
= 126 cm²
Example 3.12
The area of a trapezium is 220 cm². The parallel sides are 26 cm and 14 cm long, Find
the perpendicular distance between the parallel sides.
As with all the other figures examined so far, we can work backwards with the trapezium
formula.
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Mathematics - Stage 2 Module 4
Solution
1
sum parallel sides x height = 220
2
[
1
(26+ 14) x h] = 220
2
(
1
x 40) h = 220
2
20 h = 220
h = (220 ¸ 20)
distance/height = 11 cm
Area of a kite
The area of a kite is given by the formula
1
d1 d2 , where d1 and d2 are the diagonals
2
of the kite.
Example 3.13
The diagonals of a kite are 16 cm and 8cm. Find the area of the kite.
Solution
1
d1d2
2
1
= x 16 x 8
2
Area of kite =
=8x8
= 64 cm2.
Exercise 3.3
1. Find the area of a trapezium, which has parallel sides of 12 cm and 8 cm if the
distance between them is 6 cm.
2. A square has a perimeter of 40 cm. What is its area?
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Mathematics - Stage 2 Module 4
3. A rectangular piece of cardboard 10 cm by 6 cm has equal squares of side 2 cm cut
from two of its corners. The final shape is shown below.
Calculate: (a) the perimeter of the shape.
(b) the area of the shape.
2cm
2cm
4cm
10cm
Figure 6
A
B
²
Area of ABCD
=72cm2
E
D
8cm
C
16cm
Figure 7
4. From the figure above calculate:
(a) the length AB
(b) the area of the triangle ADE.
(c) hence, the area of the trapezium ABCDE.
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Mathematics - Stage 2 Module 4
5. Find the cost of a rectangular sheet of glass 0.75 m long and 1.08 m wide when the
glass costs $500 per square metre.
6. Find the area of the parallelogram shown below.
2cm
1.6cm
3.6cm
Figure 8
7. Find the cost of covering a kite-shaper frame with silver paper if the diagonals of the
kite are 86cm and 34cm and the silver paper costs $0.09 per square centimetre.
8. The diagonals of a kite are 27cm and 13cm. Find the area of the kite.
9. The area of a rhombus is 35cm2. If the base is 7cm, find the height of the rhombus.
10. The area of a kite is 114cm2. Find the length of the shorter diagonal if the longer one
is 19cm.
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Mathematics - Stage 2 Module 4
MODULE 4:
MEASUREMENT PART 2
UNIT 4:
CIRCUMFERENCE AND AREA OF THE CIRCLE
Introduction
In units 2 and 3 we were looking at the perimeter and area of plane shapes. The circle is
treated as a special case because it is the only plane shape that has a constant
associated with finding its perimeter and area. The same constant pi (p) is used in
finding the perimeter and area of a circle. The value of pi is
22
7
or 3.142 or 3.14.
By the end of this unit you will be able to:
·
find the circumference of a circle
·
find the radius of a circle given the circumference
·
find the length of the arc of a circle
·
find the area of a circle
·
find the area of the sector of a circle
Prerequisites:
Before you begin you should know:
·
the parts of a circle
·
how to carry out mathematical operations with decimals
·
how to carry out mathematical operations with fractions
Circumference
The formula for the perimeter or circumference of a circle is C = 2 p r or p d
The formula for the area is A = p r2
As you learned before from a previous module, r represents the radius, which is half the
diameter.
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Mathematics - Stage 2 Module 4
Example 4.1
(a)
A circle has a diameter of 9.8 cm, what is its area?
Solution
The diameter is twice the radius, so radius = 9.8 ÷ 2
r = 4.9
A = p r2
A = 3.142 x 4.9 x 4.9
Area = 75.4 cm2.
(b)
A circle has a radius of 14 cm. Find its perimeter and its area.
Solution
Circumference = 2 p r
Area = p r2
= 2 x 22 x 14
7
= 22 x 14 x 14
7
= 88 cm
= 616 cm2
Knowing how to find the circumference and area of the circle can help us to find the
length of an arc or the area of a sector of the circle. The angle at the centre of the circle
formed by the radii at either end of the arc, helps us to find what fraction of the
circumference the arc is.
Example 4.2
Find the length of the minor arc of the figure below. The radius is 7 cm and the angle at
the centre is 30°.
7cm
30°
O
31
Mathematics - Stage 2 Module 4
Solution
Length of arc = a fraction of the circumference.
Length of arc = 30 x 2 p r
360
= 30 x 2 x 22 x 7
360
=
11
3
7
= 3
2
cm.
3
The angle at the centre also helps us to find the area of the sector. The area of the
sector is a fraction of the area of the circle. The angle at the centre divided by 360° gives
us the fraction.
Example 4.3
The sector angle of a circle is 40°. Its radius is 35cm. Find the area of the sector.
Solution
Area of sector = fraction of area of circle
Area of sector = 40 x p r2
360
= 40 x 22 x 35 x 35
35cm
40°
360 x 7
= 3850 ÷ 9
= 427.8 cm2 .
32
Mathematics - Stage 2 Module 4
The circumference formula can be manipulated to find the radius.
Example 4.4
The circumference of a circle is 132 cm. Find its radius. Use p =
22
7
.
Solution
C = 2pr
132 = 2pr
132
= pr
2
132 22
=
xr
2
7
132 7
x
=r
2
22
21 cm = r
The radius of the circle is 21 cm.
The area formula can be manipulated to find the radius
Example 4.5
The area of a circle is 314.2 cm2. Find its radius by using p = 3.142.
Solution
A = pr2
314.2
= 3.142 x r2
314.2 ¸ 3.142 = r2
100 = r2
100
= r
10 cm = r.
The radius of the circle is 10 cm.
33
Mathematics - Stage 2 Module 4
Exercise 4.1
1.
A circle has a radius of 3.5 cm. What is its circumference?
2.
What is the circumference of a circle of diameter 21 cm?
3.
Find the circumference and area of a circle whose radius is 11cm.
4.
Calculate the length of an arc which has an angle of 72° at the centre of a circle with
radius 12 cm.
5.
Calculate the area of the minor sector of a circle of radius 5 cm and angle at
the centre of 55 °.
6.
If a circle has a radius of 600 cm, what is its area in m2 ?
7.
The area of a circle is 314 cm2. Find its radius. Use p = 3.14.
8.
Find the diameter of a circle of circumference 50.29 cm2.
9.
Use p =
22
7
to find the area of a circle with diameter 5 m. Give the answer in square
centimetres.
10. Find the lengths of the arcs and the areas of the minor sectors in the following
circles.
(a)
(b)
14cm
O
O
35°
6cm
120°
(c)
15cm
95°
O
34
Mathematics - Stage 2 Module 4
ANSWERS
Exercise 2.1
1.
21.6 mm
2.
31.4 cm
3.
24.2 cm
4.
22.2 cm
5.
20 cm
6.
48 cm
7.
39 cm
8.
7 cm
9.
21 cm
10.
4.4 cm
Exercise 1.1
1(a)
(b)
(c)
(d)
1715 cm
178000 cm
39.5 cm
86.4 cm
2(a)
(b)
(c)
(d)
9.375 km
0.00275 km
0.73568 km
3.941 km
3(a)
(b)
(c)
(d)
27 tonnes
0.60 tonnes
7.82 tonnes
4.01 tonnes
4(a)
(b)
(c)
(d)
7.8 g
590 g
19100 g
0.045 g
5.
98.05 kg
6.
41 kg
7(a)
6.315 m
(b)
(c)
(d)
93467 m
0.2668
6209.84
8(a)
(b)
(c)
(d)
31976 m or 31.976 km
550.89 mm
6253.01 cm
5268.323 cm
9.
5.2 tonnes
10.
45 kg
Exercise 3.1
1.
2.
3.
4.
5.
35 cm2
180 mm2
10,000 mm2 / 100 cm2
10 m2
(a) 53.55 m2
(b) 37.23 m2
(c) 16.32 m2
6.
(a) 34.8 cm2
(b) 96 cm2
7.
(a) 128 cm2
(b) 58 cm2
8.
9.
12 m
(a) 22.68 m2
(b) 144 cm2
10.
11.
12.
3.375 m
52 cm2
15 cm
35
Mathematics - Stage 2 Module 4
Exercise 4.1
Exercise 3.2
1.
(a) 9 cm2
(b) 3 cm2
(c) 6.4 cm2
2.
(a) 6 cm2
(b) 54 cm2
(c) 30 cm2
(d) 120 cm2
3.
4
5.
6.
10 cm
9.8 cm2
32 cm
18 cm
1.
2.
3.
22 cm
66 cm
(a) 69.14 cm
(b) 380.29 cm2
4.
5.
6.
7.
8.
9.
10.
15.1 cm
12 cm2
113.1 m2
10 cm
16 cm
196428.6 cm2
(a) (i)
8.56 cm
(ii) 59.9 cm2
(b) (i)
12.6 cm
(ii) 37.7 cm2
(c) (i) 24.9 cm
(ii) 186.6 cm2
Exercise 3.3
1.
2.
3.
60 cm2
100 cm2
(a) 32 cm
(b) 52 cm2
4.
(a) 9 cm
(b) 28 cm2
(c) 100 cm2
5.
6.
7.
8.
9.
10.
$405.00
5.76 cm2
$131.58
175.5 cm2
5 cm
12 cm
36
Mathematics - Stage 2 Module 4
STAGE 2
MODULE 5
MODULE 5:
INTRODUCTION TO RELATIONS, FUNCTIONS AND GRAPHS
UNIT 1:
THE CARTESIAN PLANE
Introduction
Relations and functions is a topic that contains both geometry and algebra. We will be
investigating how variables relate to each other and we will be displaying these
relationships in various ways including using graphs. In this module you will need graph
paper, ruler and pencil so make sure that you are properly equipped before you begin.
By the end of this unit you will be able to:
·
define a Cartesian plane.
·
identify and label the four quadrants of the Cartesian plane.
·
plot points on the Cartesian plane.
·
read coordinates for a point from a graph.
Prerequisites
Before you begin you should know:
·
operations with whole numbers.
·
operations with integers.
The Cartesian Plane
The Cartesian plane is made up of a horizontal line and a vertical line which
intersect each other at right angles. The vertical line is called the y- axis and the
horizontal line is called the x-axis. Where the two axes meet is called the origin. The
axes are numbered from the origin. The x-axis is numbered to the left and right of the
origin and the y-axis is numbered above and below the origin. To the right of the origin,
the numbers are positive and to the left they are negative. Above the origin, the numbers
are positive and below it they are negative. (See the page following).
1
Mathematics - Stage 2 Module 5
2
Mathematics - Stage 2 Module 5
Notice that the two lines divide the paper into four regions. These are called
quadrants.
For the Cartesian plane to be very useful to us there must be grid lines on it to provide
guidance and greater accuracy. Grid lines are vertical and horizontal lines drawn across
a page to intersect each other at regular intervals. They are the lines seen on a graph
paper. This is where the graph paper comes in. The grid lines divide the graph paper
into squares. The smallest size square that can be found on the graph paper is the 2
millimetre square. Each side measures 2 mm. Five rows of five of these give a 1cm
square. Each side measures 1cm. (See graph paper).
The 1cm square is the standard used in drawing graphs. It is used to represent specific
quantities according to the needs of the graph.
Take out a sheet of your own graph paper and familiarize yourself with the 2 millimetre
square and the centimetre square.
The Scale
You will notice on the example, that the numbers are equally spaced. This is how it is
done on the graph paper. One centimetre will represent 1, 2, 3, 5, 10, or some other
number of centimetres. This is called a scale. The scale for the y-axis can be the same
as that for the x-axis or it can be different. However, the scale cannot differ along the
same axis. This means that if 1cm is being used for the x-axis, then it must be used for
all of the x-axis, both positive and negative sides. Equally, the scale being used for the yaxis must be the same throughout the y-axis.
Plotting points
You may have seen graphs in the newspaper or in magazines. You may have noticed
that sometimes these graphs have lines to represent the quantity being studied. On
these lines there are points indicating various values. These points are called
coordinates. A coordinate is given by using two values. An x value and a y value. The x
value is always given first. These values are called x and y values because they
3
Mathematics - Stage 2 Module 5
indicate how the point relates to the x-axis and the y-axis respectively. A coordinate is
written as (2,3) for example, showing an x value of 2 and a y value of 3. Written like this,
the x and y values are called an ordered pair.
To plot points on a graph, we always start at the origin. The origin is the point (0,0) and
as said before, it is the point where the two axes meet. The point (2,3) for example is
plotted by starting from the origin then moving two places to the right and three places
up. At the end of this movement the coordinate is usually marked by using a dot or a
very small x. Below are the basic rules for plotting points.
(a)
A positive x value means move right (from the origin)
(b)
A negative x value means move left (from the origin)
(c)
A positive y value means move up
(d)
A negative y value means move down (from the x value).
( from the x value)
Therefore to plot the points (2,7), (-1,5), (6, -4), (-3, -8):
Starting at the origin for each one:
(2,7) means 2 places right
(-1,5) means 1 place left
then 7 places up
then 5 places up
(6,-4) means 6 places right
(-3,-8) means 3 places left
then 4 places down
then 8 places down.
Example 1.1
From the graph paper on the following page, find and write down the coordinates of the
points A, B, C and D.
4
Mathematics - Stage 2 Module 5
5
Mathematics - Stage 2 Module 5
Solution
To get to A, start at the origin and move left to −3 then up 2 places. Therefore, the
coordinates of A = (−3, 2)
To get to B, start at the origin and move right to 1 then up 9 places. Therefore, the
coordinates of B = (1,9).
To get to C, start at the origin and move right to 7 then 4 places up. Therefore, the
coordinates of C = (7, 4).
To get to D, start at the origin and move left to -6 then 5 places down. Therefore, the
coordinates of D = (-6,- 5).
Exercise 1.1
1. Following the example above, write out the meanings of the following coordinates:
(5,6), (-3,7), (9,-5), (-8, -1)
2. (a)
Using graph paper, draw up x and y axes in the middle of the graph paper.
Make sure that the lines go to the ends of the graph paper. Label the axes as
shown in the diagram under “The Cartesian Plane” using 1cm as the distance
between the numbers.
(b)
The quadrant where x and y are both positive is quadrant 1. Starting with
quadrant 1 and moving in an anticlockwise direction, label the quadrants 1,2,3
and 4.
(c)
Using this graph paper, plot the points from question 1. Use a small x to
indicate each position and label them A,B,C, and D respectively.
3. Ask someone to mark four or more points on your graph paper and label them E,F, G
H. There must be at least one in each quadrant. Find the coordinates for these
points. (Remember, start at the origin).
6
Mathematics - Stage 2 Module 5
4. Plot the following points in order and join them closing the figure at the end. V(3,2)
B(2,3), N(3,5), M(6,5), P(8,3), T(9,2).
5.
Write coordinates for the following.
(a) Six places left and eight places up.
(b) Three places right and nine places down.
(c) Ten places left and five places down.
(d) Two places right and nine places up.
(e) Eleven places left and seven places up.
6. The coordinates of the corners of a parallelogram are Q(-3,1), R(4,1), S(x,y) and
T(-6,-5) . Find the values of x and y.
Summary
The point of intersection of the x-axis and the y-axis is called the origin. The x-axis runs
horizontally and the y-axis runs vertically. A point on the Cartesian plane is defined by its
coordinates. Coordinates are written as an ordered pair of numbers in brackets, the x
coordinate being stated first, that is, (x,y).
7
Mathematics - Stage 2 Module 5
MODULE 5:
INTRODUCTION TO RELATIONS, FUNCTIONS AND GRAPHS
UNIT 2:
INTRODUCTION TO RELATIONS
Introduction
In this unit we will be exploring different types of relations. We will also be learning to
represent relations in different ways. We will use ordered pairs, arrow diagrams,
algebraic equations and graphs.
By the end of this unit you will be able to:
·
define a relation
·
name and identify the four types of relations
·
recognise a relation
·
describe a relation as a set of ordered pairs
·
represent relations as arrow diagrams
·
represent relations as algebraic equations
·
represent relations on the Cartesian plane
·
deduce the relation from a given set of domain and range.
Prerequisites
Before you begin you should know:
·
the concept of the number line
·
how to plot points on the Cartesian plane
8
Mathematics - Stage 2 Module 5
Definition of a Relation
A relation is defined as a set of ordered pairs. The same connection exists between
each pair of the set of ordered pairs. The connection describes how the first element in
each pair relates to the second element in each pair.
(Mrs. Beecher, Mary), (Mrs. King, Anthony), (Mrs. Thomas, Jane),
( Mrs. Peters, Claire)
The relation for the ordered pairs above could be “is the mother of”. This means that
Mrs. Beecher is the mother of Mary, Mrs. King is the mother of Anthony and so on.
(1,3), (2,4), (3,5), (4,6), (5,7)
The relation for the ordered pairs above could be y = x +2. This means that the first
number in each pair, plus 2 is equal to the second number in each pair. That means
1 + 2 = 3, 2 + 2 =4 and so on.
The first elements in a set of ordered pairs are called the domain. The second elements
are called the range. So for the first example above, the domain is (Mrs. Beecher, Mrs.
King, Mrs. Thomas, Mrs. Peters) and the range is (Mary, Anthony, Jane, Claire).
Can you name the domain and range for the second example?
You are right if you said:
Domain: (1,2,3,4,5)
Range:
(3,4,5,6,7)
The domain for y = x + 2 could also have been written as 1≤ x ≤ 5. This reads: x is
greater than or equal to 1 and less than or equal to 5 which means x is between 1 and 5
inclusive.
9
Mathematics - Stage 2 Module 5
Representing Relations
Given a set of ordered pairs, we can represent relations in a number of ways. They can
be represented as arrow diagrams or graphs. An arrow diagram is a diagram with the
domain and range in two different figures with arrows going from each member of the
domain to its corresponding member or members in the range. (See below)
Domain
Range
Mrs. Beecher
Mary
Mrs. King
Anthony
Mrs. Thomas
Jane
Mrs. Peters
Claire
7
6
5
4
3
2
1
0
1
2
3
4
5
Series 1
Above is the graphical representation of the ordered pairs (1,3), (2,4), (3,5), (4,6), (5,7).
This set of ordered pairs could also have been represented like the set before, as an
arrow diagram.
10
Mathematics - Stage 2 Module 5
Types of Relations
There are four types of relations. These are: (i) one to one, (ii) one to many,
(iii) many to one, and (iv) many to many. They are also called mappings.
One to one means that each member in the domain only relates to one member in the
range. The two examples above are both one to one relations / mappings.
One to many means that one or some members in the domain relates to or maps to
more than one member in the range.
For the relation “is the mother of” we could have as our domain (Mrs. Nash, Mrs. Robb,
Mrs. Jones and Mrs. Bell). Our range could then be (Janet, Arlene, Kenneth, Joseph.
Carol, David, Glenroy). If Mrs. Nash is the mother of both Janet and Kenneth, Mrs. Robb
is the mother of Arlene and Glenroy, Mrs. Jones is the mother of Joseph and David and
Mrs. Bell is the mother of Carol. The arrow diagram to illustrate this is seen below.
“is the mother of ”
Domain
Range
Janet
Mrs. Nash
Kenneth
Mrs. Robb
Arlene
Glenroy
Mrs. Jones
Joseph
Mrs. Bell
David
Carol
11
Mathematics - Stage 2 Module 5
A many to one arrow diagram can be formed from the relation y = x2 . Two or more
members of the domain relate to one member of the range. Using the domain
(-3,-2,--1,0,1,2,3) or ( -3 ≤ x ≤ 3) and the range (0,1,4,9), the arrow diagram is seen
below.
Domain
Range
-3
9
-2
-1
4
0
1
1
2
0
3
y = x2
The arrow diagram turns out like this because 32 and -32 are both equal to 9, 22 and -22
are both equal to 4, 12 and -12 are both equal to 1.
Many to many means that two or more members of the domain relate to two or more
members of the range. An example of this type of relation is seen in the arrow diagram
below. The relation is “y is a multiple of x”.
Domain
Range
2
4
5
3
6
7
4
8
9
5
10
"Y is a multiple of x"
12
Mathematics - Stage 2 Module 5
The number 2 in the domain maps to 4, 6, 8 and 10 in the range because they are all
multiples of 2. The number 3 in the domain maps to 6 and 9 in the range because they
are multiples of 3. The number 4 in the domain maps to 4 and 8 in the range because
they are multiples of 4. The number 5 in the domain maps to 5 and 10 in the range
because they are multiples of 5.
The ordered pairs for this relation would be (2,4), (2,6), (2,8), (2, 10), (3,6), (3,9), (4, 4),
(4,8), (5,5) and (5,10)
Exercise 2.1
1. Using the following set of ordered pairs, draw an arrow diagram and say what type of
relation it shows. (1,2), (2,3), (3,4), (4,5), (5,6), (6,7).
2. Draw an arrow diagram to illustrate the following ordered pairs and identify the type
of relation.
(a,1), (b, 1), (c,1), (d,2), (e,2), (f,2).
3. A relation is given by the equation y = 4x + 3. Using the domain (1,2,3,4,5,6,7) or
1 ≤ x ≤ 7, write out the ordered pairs for this relation.
4. Draw the graph of the relation y = 5x −2, on graph paper. Use 1cm to represent
1 unit on each axis. Use the domain (-2, -1, 0, 1, 2 ) or −2 ≤ x ≤ 2. Show your
answer to your facilitator.
5. For the relation “is a member of” draw an arrow diagram using the following domain
and range. Domain : (apple, orange, peach, lettuce, tomato, carrot),
Range: (fruits, vegetables, dairy products.). Which element in the range does not
have corresponding elements in the domain?
6. Find the relation for the following domain and range. Domain: (2, 3, 4, 5, 6) or
2 ≤ x ≤ 6 and Range: (5, 10, 17, 26, 37).
7. Mrs. Lawson has two children, Matthew and Lisa. Mrs. Latham has three children,
Carly, Machelle and Sophia. Mrs. Lindsay has two children, Mona and David.
(a)
Draw an arrow diagram to show the relation using the mothers as the domain.
(b)
Draw an arrow diagram to show the relation using the children as the domain.
(Indicate the relation above or below the diagram).
13
Mathematics - Stage 2 Module 5
8. Copy and complete the table below which represents the relation y = 3x -1.
x
-4 -3
-2
-1 0
1
2
3
4
-1 -1
-1- 1- 1- 1- -1 -1 -1
3x
-1
y
-10
-1
11
Plot this relation on graph paper. Show it to your facilitator.
9. Make up your own example of a one to one relation. Using a domain of about five
numbers between 0 and 10, write down the ordered pairs for this relation.
(a) Draw an arrow diagram to represent the relation.
(b) Represent the relation on graph paper.
10. For the relation y = 2x2 − 4, copy and complete the table below.
NB. To do 2x2, first square the x, then multiply the answer by 2.
x
-4
2
2x
32
−4
-4
y
-3 -2 -1 0
1
2
2
3
4
18
-4 -4 -4 -4 -4 -4 -4
-2
-4
14
14
Mathematics - Stage 2 Module 5
MODULE 5:
INTRODUCTION TO RELATIONS, FUNCTIONS AND GRAPHS
UNIT 3:
INTRODUCTION TO FUNCTIONS
Introduction
Now that we have an understanding of how relations work, we will be building on this
knowledge. We will be exploring the topic of functions which is a more in depth part of
relations. As always, knowledge gained in modules before will be very useful.
By the end of this unit you will be able to:
·
define functions
·
differentiate between a relation and a function
·
draw graphs of functions.
Prerequisites
Before you begin you should know:
·
what a relation is
·
how to draw arrow diagrams
·
how to draw the graph of a relation
Definition of a function
A function is a relation in which each member in the domain is mapped to one and
only one member in the range. Therefore, the only relations that can be functions are
those that are either one to one or many to one.
We can differentiate between relations that are functions and those that are not by
looking at the ordered pairs, the arrow diagram or the graph.
A set of ordered pairs that represent a function would not have a repeat of x values.
15
Mathematics - Stage 2 Module 5
Example 3.1
For the ordered pairs following, state whether they represent a normal relation or a
function.
(0, -7), (1, -3), (2, 1), (3, 5), (4, 9), (5, 13)
Solution
Since there is no repeat of x values, the relation is a function.
To decide whether a relation is a function from looking at an arrow diagram we need to
see if there is more than one arrow going from any member of the domain. If there is
then it is not a function.
Example 3.2
From the arrow diagram below determine whether the relation is a function.
x
y = 2x − 5
y
-4
-13
-3
-11
-2
-9
-1
-7
0
-5
1
-3
2
-1
3
1
4
3
5
5
16
Mathematics - Stage 2 Module 5
Solution
The arrow diagram on the previous page represents a function as there is only one
arrow going from each member of the domain. This means that each member of the
domain maps to only one member of the range.
Another way of determining whether a relation is a function, is from its graph.
Example 3.3
Draw the graph of y = x2 − 7 for the domain (-4, -3, -2, -1, 0, 1, 2, 3, 4) or −4 ≤ x ≤ 4.
Solution
X -4 -3 -2 -1 0
y
9
2
1
2
3 4
-3 -6 -7 -6 -3 2 9
10
8
6
4
2
0
-4
-3
-2
-1
0
1
2
3
4
-2
-4
-6
-8
Series 1
17
Mathematics - Stage 2 Module 5
If there is any place on the graph that a vertical line will cut it more than once then it is
not a function. The grid lines running parallel to the y- axis provide an easy answer for
us. None of them cut the graph more than once. The graph therefore represents a
function.
Exercise 3.1
1.
Using the domain (-3, -2, -1, 0, 1, 2, 3), draw the graph of the function y = x2.
(Do a table for the graph first).
2.
Draw arrow diagrams for the following relations and say whether they are
functions and why.
3.
4.
(a)
y = 3 x2 − 9
domain = ( -3 ≤ x ≤ 3)
(b)
y=5x+4
domain = ( 0 ≤ x ≤ 5)
(c)
y=
1
x
2
domain = (2, 4, 6, 8, 10, 12)
Find the relation that has domain (3, 4, 5, 6, 7) and range (7,14, 23, 34, 47).
(a)
Is the relation a function? Explain.
(b)
Draw an arrow diagram to show the relation.
(c)
Draw a graph to represent the relation. Show the graph to your facilitator.
Say whether the following are graphs of functions. Explain your answer.
SEE NEXT PAGE FOR GRAPHS
18
Mathematics - Stage 2 Module 5
a)
8
6
4
2
0
0
1
2
3
4
5
-2
-4
-6
-8
19
Mathematics - Stage 2 Module 5
ANSWERS
Please do the graphs and make sure your facilitator sees them.
Exercise 1.1
1. (5,6) = 5 places right then 6 places up
(-3, 7) = 3 places left then 7 places up
(9, -5) = 9 places right then 5 places down
(-8, -1) = 8 places left then 1 place down.
4.
5. (a)
Six places left and eight places up = (-6,8)
(b)
Three places right and nine places down = (3, -9)
(c)
Ten places left and five places down = (-10, -5)
(d)
Two places right and nine places up = (2, 9)
(e)
Eleven places left and seven places up = (-11, 7)
20
Mathematics - Stage 2 Module 5
6. x = 1, y = - 5
S = (1, -5)
Exercise 2.1
1.
Domain
Range
1
2
2
3
3
4
4
5
5
6
6
7
The relation is a one to one relation.
2.
Domain
Range
a
1
Many to one relation.
b
c
d
2
e
f
3.
y = 4x + 3
(1,7) (2, 11) (3 ,15) (4, 19) (5, 23) (6, 27) (7, 31).
21
Mathematics - Stage 2 Module 5
5.
“ is a member of”
Domain
Range
Apple
Fruit
Orange
Peach
Vegetable
Lettuce
Tomato
Carrot
Dairy
Dairy does not have a connection in the domain.
6.
y = x2 + 1
7. (a)
“ Is the Mother of “
Mrs. Lawson
Matthew
Lisa
Mrs. Latham
Carly
Machelle
Sophia
Mrs. Lindsay
Mona
David
22
Mathematics - Stage 2 Module 5
(b)
“Is a child of”
Matthew
Mrs. Lawson
Lisa
Carly
Mrs. Latham
Machelle
Sophia
Mona
Mrs. Lindsay
David
8.
x
9.
-4
-3
-2
-1 0
1
2
3
4
3x -12 -9
-6
-1 0
3
6
9
12
-1
-1
-1- 1- 1- 1- -1 -1 -1
y
-13 -10 -7
-1
-2 -1 2
5
8
11
Answers will vary.
10.
x
-4
2
-3
-2 -1 0
2x
32 18 8
−4
-4
-4
2
0
1
2
3
2
8
18 32
-4 -4 -4 -4 -4 -4
4
-4
y
28 14 4
x
-3
-2
-1
0
1
2
3
y
9
4
1
0
1
4
9
-2 -4 -2 4
14 28
Exercise 3.1
1.
23
Mathematics - Stage 2 Module 5
y = 3x2 − 9
2 (a)
-3
18
It is a function.
-2
It is a many to one relation.
-1
3
There are no repeats of
domain values.
0
1
-6
2
3
(b)
-9
y = 5x + 4
0
4
It is a function
1
9
It is a one to one relation
2
14
There are no repeats of domain
values.
3
19
4
24
5
29
(c)
y=
1
x
2
2
1
4
2
6
3
8
4
10
5
12
6
24
Mathematics - Stage 2 Module 5
The Relation is y = x2 − 2
3.
(a) It is a function because it is a one to one mapping.
y = x2 − 2
(b)
3
7
4
14
5
23
6
34
7
47
y = x2 - 2
(c)
50
45
40
35
30
25
Series1
20
15
10
5
0
0
4. (a)
(b)
2
4
6
8
It is a function because a vertical line at any point only cuts it once.
It is not a function because a vertical line will cut it twice
25
Mathematics - Stage 2 Module 5
STAGE 2
MODULE 6
MODULE 6:
HANDLING DATA PART 2
UNIT 1:
REPRESENTING AND INTERPRETING DATA (2)
Introduction
At Stage 1 you had many experiences with collecting, organising, displaying and
interpreting data. As a result, you are probably much more aware of all the data around
you and of the many ways in which you use, and will continue to use data. That's good.
Keep looking out for tables, bar graphs, pictograms and pie charts, and keep analysing
the information that they represent. Now you must build on the knowledge and skills that
you have already developed. The work at this Stage, then, will help you explore and
acquire additional skills at collecting and organising data. It will also introduce you to
other ways of displaying or representing data and help you improve your ability to
analyse and interpret data.
There are four units in this module. Each unit has its own set of Specific Learning
Objectives and selected activities to help you achieve those objectives. Please work
through each unit carefully. If you take short cuts or skip some of the activities, you will
miss some of the important steps that you need to help you grasp the new concepts and
prepare you for further work at the next Stage. Note, too, that some activities require
you to do individual work while others must be done in collaboration with other learners.
Both types of experiences will contribute to your learning.
By the end of this unit you will be able to:
·
select from a given set of problems or questions, the ones which should have
numerical information.
·
list possible sources of existing data.
·
prepare and use simple interview schedules for gathering data.
·
identify and describe situations where it is better to use extended interviews
rather than observations to collect data.
1
Mathematics - Stage 2 Module 6
Prerequisites
Before you begin you should know:
·
how to determine when a sample rather than the entire population may be
used for the data you need.
·
how to design and use an Observation Check List to collect data.
·
how to use the tally system to record numerical information.
·
how to read and make frequency tables to organise raw data.
·
how to represent data given in tables by pictograms, bar graphs and pie
charts.
·
how to calculate fractions, ratios and percentages.
·
how to estimate numerical values to the nearest whole number.
Let's make sure we understand why, from time to time, we need to collect useful,
meaningful data, that is, information in numerical form, and how we may collect that
data.
Exercise 1.1
Consider the following situations and in each case state, giving reasons, whether or not
you would need numerical information to help you answer the question that follows the
statement.
(Although this is meant for you to do by yourself in the first instance, you should
also talk about these situations with members of your study group. Hear what
they think. Share your thoughts with them.)
1. It is commonly believed that more male than female infants die within two weeks of
their birth. Is this true?
2. Each year less than 40% of the boys who sit the Grade Six Achievement Test
(GSAT) gain places in the secondary schools of their choice. Would you say this is
really so?
2
Mathematics - Stage 2 Module 6
3. When James Brown, a famous footballer, was interviewed he said that all things
being considered he would be better off playing for The Caresum Football Club than
for The Giant Scorers. What might be the ‘things’ which he considered?
4. Some boys who visited the Safe Way parking lot last Monday afternoon reported that
of the 50 cars that were there, red cars were by far the most prevalent. Were they
correct?
5. Jamaican cricket fans say that curried goat and rice is the favourite dish of most
members of the present West Indies cricket team. Do you agree with them?
6. As the three girls were leaving the bus shelter, one of them, Tanya, offered to share
her raincoat with the others. Was this because Tanya was kind and thoughtful?
7. Consider again #s 1, 2, 4 and 5 above.
a. If you think or know that the data already exists, say where and how you might
find it.
b. For each of the situations for which you must do your own survey:
·
describe the population for your survey;
·
explain how you would gather and record the data.
c. Who do you think might be interested in the results of the survey at number 4?
Why?
3
Mathematics - Stage 2 Module 6
While doing these Exercises, you no doubt recalled some of what you had already
discovered in Stage 1 about the importance of data and of how we might collect the data
we need. Especially, you should have recalled that:
a) numerical information is usually more specific than information given in words
only;
b) there are many sources of stored data including reports of surveys and research
studies, as well as documents in libraries and in planning and financial
institutions;
c) if the data we need does not already exist, it is necessary and possible to collect that
data, often by doing our own survey(s);
d) in collecting data, we must be careful to identify the population or an appropriate
sample of it;
e) raw data should be organised in tables so they may be more easily read and
interpreted and/or displayed graphically.
Keep these in mind as you continue to collect and represent data.
The Interview Schedule
So far, the means we have used for gathering data in a survey are:
·
on - site observations
·
simple interviews
We have found that in some cases where an observation cannot give the information we
need, the interview is able to provide it. For example, trying to find out, by observation,
what is a person’s favourite fruit is a more challenging task than just asking that person
directly, “What is your favourite fruit?” Trying to get this same information about a group
of persons, by observation, would be almost impossible. We know, though, that we may
ask the same question to each member of a group, record the results, analyse them and
be able to make a statement about the whole group. Moreover, we may use the same
interview to get information on many aspects of the same topic or on related topics. In
this case, we develop a series of questions or an Interview Schedule.
4
Mathematics - Stage 2 Module 6
We put the questions in writing to ensure that each participant in the survey is asked the
same questions, in the same order. Consider, for example, the amount of information
that can be gathered from answers to the following set of questions or interview
schedule:
1. What is your favourite fruit?
2. Is the size of the fruit important to you? Must it be large, medium or small?
3. When and at what time of day do you most enjoy eating this fruit?
4. How many of these fruits do you usually eat at a time?
5. Would you be willing to pay more for this fruit when it is not ‘in season’?
Exercise 1.2
Imagine that the 5 questions in the interview schedule above were posed to the 30
workers at Best Fit Factory by Lady Whiz who was planning to operate a fruit stall near
the gate of the factory. She wanted to get information which would help her decide what
fruits to sell and the best time of day to have them available for her expected customers.
The responses to each question were tabulated and the information from
the five tables combined as shown below.
FAVOURITE
FRUIT (FF)
PREFERRED
SIZE OF FF
Orange
12
Naseberry
5
Watermelon
2
Pawpaw
3
Mango
8
Medium
9
Large
5
Small
3
Any size
8
TIME WHEN FF
IS MOST
ENJOYED
NUMBER OF WILLINGNESS
FF EATEN
TO PAY
AT A TIME
INCREASED
PRICE FOR FF
6 a.m. - 10 a.m.
Two
Yes
4
10
9
8 a.m. - 12 noon
Two
Yes
5
4
4
2 p.m. - 4 pm.
Part of one Yes
0
2
2
6 a.m. - 12 noon
One
Yes
0
3
3
2 p.m. - 8 p.m.
Three
Yes
8
8
6
5
Mathematics - Stage 2 Module 6
With the help of the information given in the tables answer the questions that follow.
1. Lady Whiz decided that she would begin by operating her stall each day between
the hours of 6:00 a.m. and 12:00 noon.
Why do you think she made this
decision?
2. What percentage of the workers said that the orange was their favourite fruit?
3. At the start of a day, which of these fruits would you expect to see in greatest
number on Lady Whiz’s stall? Explain your answer.
·
oranges and mangoes.
·
oranges and naseberries
·
naseberries and pawpaws.
·
naseberries and mangoes
4. What is the probability that naseberry customers will buy them when they are ‘out
of season’?
5. Is there additional information that you think Lady Whiz would find useful? If so,
what question might she have asked to get that information?
6. Was there, in your opinion, any unnecessary question in this schedule? Give a
reason for your answer.
Exercise 1.3
1. Ten ‘returning residents’ have just moved to your community. It is rumoured that
they are all retired persons who left Jamaica at the same time, lived in the same part
of England, did the same type of work while they were there, and returned to
Jamaica within the same year. It is your responsibility to find out whether all of this is
true and you have decided to interview them. Prepare the interview schedule that
you will use.
Before you start, there are some guidelines on the following page. Read them carefully.
6
Mathematics - Stage 2 Module 6
Constructing and Using the Interview Schedule
In preparing the schedule:
(i)
ensure that the questions are directly related to the purpose of the survey
(ii) use as few questions as will give the information you need
To get maximum information from the answers:
(i)
Avoid as far as possible, questions which may be answered by simply saying, “Yes”
or “No”
In conducting the interview here are five points to remember:
(i)
Select a mutually convenient time and place. There should be no need to
rush. The whole atmosphere should be calm and relaxed.
(ii) Greet the interviewee cheerfully and introduce yourself by name.
(iii) Tell the interviewee the objective of the interview and emphasise the extent to
which his/her answers will help achieve this objective.
(iv) Pose the questions in the planned order. Be prepared to get information which
you might deem irrelevant; continue to listen graciously; you should have space
on your recording sheet for this information too.
(v) Close with expressions of thanks and, where this is applicable, indicate how the
interviewee will know about the outcome of the survey.
2. Pretend that you have conducted the interviews and that all the rumours have turned
out to be facts.
a. What information do you think you will now have in addition to what was
already rumoured?
b. To whom would you give all this information? Why?
c. How could you, yourself use the information?
7
Mathematics - Stage 2 Module 6
3. If you had to collect information from a large population,
a.
What do you think would be the main advantages of using an interview
schedule?
b.
What do think would be the main disadvantages of using an interview
schedule?
4. Go over all the work you have done so far then go on to unit 2.
8
Mathematics - Stage 2 Module 6
MODULE 6:
HANDLING DATA PART 2
UNIT 2:
COLLECTING AND ORGANISING DATA (1)
Introduction
You have already seen that data are more easily understood and interpreted when they
are represented in pictorial or graphical form. You have already met three types of
pictures or graphs that are commonly used to display or represent data: the pictogram,
the bar chart/graph and the pie chart.
It is time to jog your memory:
-
what does each of them look like?
-
when do you use one or the other?
-
how do you make these pictures/graphs?
By the end of this unit you will be able to:
·
draw line graphs to represent data;
·
interpret line graphs.
·
choose graphs which are most suitable to represent a given set of data for a
particular purpose.
Prerequisites
Before you begin you should know:
·
how to use pictograms, bar charts and pie charts to represent data.
·
how to represent points on a Cartesian plane.
·
how to use basic units of measurement.
·
how to convert whole numbers to percentages;
·
how to calculate percentage increase and/or decrease.
·
how to round/approximate numbers as required
9
Mathematics - Stage 2 Module 6
Exercise 2.1
1.
In his Budget presentation, the Minister of Finance announced that of the $500m
to be spent on Education, Early Childhood Education would receive $200m,
Primary Education would receive $150m, Secondary Education, $75m and
Special Education, $ 25m. Other levels and types of education would share the
remainder.
Mr. Show-Tell, the Minister of Information, decided to put this information in
pictorial/graphic form for use at the press briefing that would follow the
announcement. First he made a pictogram. Then he made a bar chart. Finally,
he made a pie chart. A reduced copy of each is shown below.
1. Distribution of Education Budget
TYPE/LEVEL OF
AMOUNT IN $m
EDUCATION
Early Childhood
$$$$$$$$
Primary
$$$$$$
Secondary
$$$
Special
$
Other
$$
10
Mathematics - Stage 2 Module 6
2. Distribution of Education Budget.
Amount in $m
250of
Amount in
Type/level
$m
Education
Early Childhood
200
200
Primary
150
Secondary
75
Special 150
25
Other
50
100
50
O
th
er
Pr
im
ar
y
Se
co
nd
ar
y
Sp
ec
ia
l
Ea
rly
C
hi
ld
ho
od
0
Types/level of Education
3. Distribution of Education Budget
Education
$m
Other
Early Childhood
10%
Primary
Special
Secondary
5%
Special
Other
Secondary
200
150
75
25 Early
Childhood
50
40%
15%
Primary
30%
11
Mathematics - Stage 2 Module 6
The Minister of Information felt that all three graphs represented the information equally
well, but wanted to use just one of them.
Make the choice for him, and give the
reason(s) for your choice.
2. Select from the messages below the one which you think the graphs are meant
to carry.
(a)
Primary education still gets massive support.
(b)
No type or level of education has been neglected.
(c)
Early Childhood Education, a number 1 priority.
(d)
Education for all is our motto this year.
Explain your answer.
Discuss your responses with members of your Study Group and compare them with the
comments at the end of the module
†ACTIVITY 2.1
Do you remember how well the Minister of Information prepared for the press
briefing? (Look again at the previous page).
His presentation was well received and a lively discussion followed in which many
persons expressed their satisfaction at seeing that great attention would be paid to Early
Childhood Education. Then Mr. Speak-up, a concerned father, made his contribution.
“Minister”, he said, “all that you have said sounds good, but why is there
such a paltry sum for Special Education?”
“Do you call $25 million paltry?” asked the Minister. “This amount”, he continued,
“represents a massive 400% increase over what it used to be 5 years ago. Since
1999, there has been a steady increase each year in the amount allocated to
Special Education and you may expect a similar rate of increase over the next 5
years.”
12
Mathematics - Stage 2 Module 6
The father was not convinced and the Minister wished he had a graph to support his
argument.
Neither the pictogram, the bar graph nor the pie chart could show changes made over
time. He wished he had a line graph which he knew could do that job well.
Back in his office, he examined the Table which showed the amount allocated to Special
Education in each of the 5 years, 1999 to 2003.
AMOUNT ALLOCATED TO SPECIAL EDUCATION
YEAR OF
ALLOCATION
AMOUNT OF
ALLOCATION
BY $ m
(1999)
Year 1
(2000)
Year 2
(2001)
Year 3
(2002)
Year 4
(2003)
Year 5
Year 10
5
10
15
20
25
?
He prepared a line graph to display this data.
Making and using the line graph
In a previous Module which dealt with Relations and Functions, you learned how to draw
line graphs on a Cartesian plane to represent the relation between a set of ordered
pairs.
You have already seen, then, that the set of ordered pairs can be taken from a table of
values which shows how one thing is related to another.
In the case of the allocation for Special Education in each of 5 consecutive years,
the table shows the relationship between time and money allocated.
13
Mathematics - Stage 2 Module 6
Below, the corresponding values are shown as points plotted on a Cartesian plane. The
points, when joined together, form a straight, continuous line.
AMOUNT BUDGETED FOR SP. ED.
SCALE: 2 cm = 1year on the x-axis
2 cm = $5million on the y-axis
50
45
AMOUNT SPENT IN $M
40
35
30
25
20
15
10
5
0
1
2
3
4
5
6
7
8
9
YEARS
TAKE NOTE
1.
The graph has a title.
2.
The scale used on each axis is given.
3.
It is easy to see how the expenditure is changing or increasing over time. The
straight line indicates that the rate of increase is constant.
4.
When the rate of increase or decrease is constant, it is possible to predict what is
likely to happen in the future.
14
Mathematics - Stage 2 Module 6
Exercise 2.2
1. On graph paper, using a scale of 2cm to represent 1 year on the x-axis and 2cm to
represent $5 million on the y-axis, plot the ordered pairs shown in the table for the
preceding line graph and join the points.
2. Use the graph to help you answer the questions below.
a. In which year might the amount budgeted for Special Education be $45
million?
b. At the present rate of increase, what is likely to be the total amount spent on
Special Education in Years 6 and 7?
c. Was there ever a time when there was no money allocated to Special
Education?
3. Do the necessary calculations to show whether a change from $5 million to $25
million represents a 400% increase.
Ask your facilitator to check your work before you go on.
Be on the look-out for line graphs in newspapers and magazines. Notice how much they
are used for reports and advertisements. Start a collection and keep comparing them.
You will no doubt have observed that not all line graphs are single straight lines,
but they all show movement of a variable over time, and are
therefore useful for showing upward or downward trends.
In the Exercise that follows you will see a graph that is made up of a series of straight
lines joined together to display data about the monthly profit made by a Company over a
period of 6 months.
15
Mathematics - Stage 2 Module 6
Exercise 2.3
1.
The GURU SOFT DRINK COMPANY is proud of its performance over the last 6
months of the past year. The Company has used a line graph to tell its
success story. Look at the graph below.
COMPANY PROFITS JULY – DECEMBER
SCALE: On x – axis, 2cm represent 1 month
On y-axis, 2cm represent $50,000
250
Profit in $'000s
200
150
Look out for:
100
·
·
·
50
0
J
A
S
O
N
D
the title of the graph
the scale on each axis
times when profits
increase or decrease
MONTHS
2.
Read the graph then copy and complete the table below.
MONTHS JULY AUGUST SEPTEMBER OCTOBER NOVEMBER DECEMBER
PROFIT
($000s)
3.
50
115
200
Using a scale of 2cm to one month on the x–axis and a scale of 2cm to $50,000 on
the y–axis, draw your own line graph to represent the data.
16
Mathematics - Stage 2 Module 6
4.
(a) From your graph, or otherwise, find out how much more profit the Company
made in December than it did in October.
(b) Which was more the percentage increase in profit from August to September or
from November to December, and by how much?
(c) What do you think will be the profit in January of the next year? Explain your
answer.
5.
The marks gained by two students, Jack and Jill, in six consecutive end-of-week
tests were displayed on a graph by their class teacher. A copy of the graph is
shown below.
†ACTIVITY 2
Now that you have been introduced to the line graph, it is important that you
become familiar with the many purposes for which it is used and be aware of the
ways in which it is sometimes misused.
Double Line Graphs
Often two line graphs, representing two sets of data are presented at the same time so
that the data may be easily compared.
WEEKLY TEST MARKS FOR JACK & JILL
SCALE: On x – axis, 2cm represent 1 month
On y-axis, 2cm represent $50,000
25
MARKS
20
15
Jack's
Jill's
10
5
0
1
2
3
4
5
6
WEEKS
17
Mathematics - Stage 2 Module 6
Examine the graph. Think and talk about:
·
the teacher’s reason for wanting to display these two sets of marks;
·
what the teacher might say to Jack about his performance;
·
how Jill might assess her own performance;
·
the ease or difficulty of predicting each student’s mark for the test at the end of
the seventh week.
Exercise 2.4
1. ‘It is commonly believed that more male than female infants die within two weeks of
birth’. In unit 1, you realised that you needed numerical information to determine
whether or not this common belief is true.
Imagine that you examined the records, collected the relevant data for the past five
years, and entered them on a frequency table like the one shown here.
RECORD OF INFANT DEATHS FOR THE PAST FIVE YEARS
YEAR
Year 1
Year 2
Year 3
Year 4
Year 5
Male Infants
90
110
100
90
80
Female Infants
80
70
60
50
40
Use the records above to answer the questions below.
a. How many male infants, aged 2 weeks or under, died during the 5-year period?
b. What percentage of all the deaths does this number represent?
c. Are you now able to state whether or not the commonly held belief is true?
Explain your answer.
d. On a sheet of graph paper, using the same axes and a convenient scale, make two
line graphs to show the information that is recorded in the table above.
e. With the help of your graphs predict how many infants are likely to die in Year 7.
18
Mathematics - Stage 2 Module 6
Starting the Vertical Scale at Zero
Sometimes a zero reading is meaningless. The size of the variable often influences the
numbering on the vertical axis. You need to decide when it is appropriate to start the
numbering at zero. At the same time, you should be careful not to come to wrong
conclusions when reading a line graph where the zero has been omitted.
Katie was in the hospital for ten days, and each morning her temperature was read and
recorded by two different nurses. As the line graphs below show, the readings were
similar, but there was a slight difference in the way they were presented.
A.
KATIE’S TEMP. OVER TEN DAYS
41
TEMPERATURE IN DEGREE C
40
39
38
37
36
35
34
1
2
3
4
5
6
7
8
9
10
DAYS
19
Mathematics - Stage 2 Module 6
B.
—
KATIE’S TEMP. OVER TEN DAYS
Think and talk about these graphs. Compare your ideas with those at the
end of the module.
-
In graph A, the reading on the vertical scale begins at 34°C.
Can you
suggest why it did not begin with 0°C?
-
In graph B, there is a zero at the start of the vertical scale and then a zig-zag
line between 0° and 35°. What work do you think this zig-zag line is doing?
-
If you had to choose one of these methods for presenting similar types of
data, which would you choose and why?
20
Mathematics - Stage 2 Module 6
Adjusting the Scale to Change the Message
When the scale is changed on one or both axes, the ‘look’ of the line graph changes and
so does the impression it gives. The reader has to look carefully at the scales in order to
interpret the graph accurately.
The graphs below were used by a manufacturing company to give the same information
to different audiences.
X.
YEARLY PROFITS OF COMPANY ABC FOR PAST SIX YEARS
40
PROFIT IN YEARS
35
30
25
20
Series1
15
10
5
0
95
96
97
98
99
100
YEARS
21
Mathematics - Stage 2 Module 6
Y.
YEARLY PROFIT OF COMPANY ABC FOR SIX YEARS
Exercise 2.5
1.
Make sure that both graphs are displaying the same data. Copy and complete the
table below using the information from ABC manufacturing company.
YEAR
1995
PROFIT IN $M SHOWN IN X
15
1996
1998
1999
2000
27.5
20
PROFIT IN $M SHOWN IN Y
2.
1997
30
If the data in both cases is the same, what is responsible for the difference in the
appearance of the graphs?
[Hint:
Look carefully at the scale used on each of the vertical axes].
22
Mathematics - Stage 2 Module 6
3.
a. What do you think is the message that Graph Y wants to carry, and to whom?
b. Does the omission of zero at the start of the vertical axis help to convey this
message? Explain.
4.
Go over all that has been discussed in this unit so far, make sure you understand
the new ideas and get ready to apply them in new or different situations.
Exercise 2.6
1.
On each birthday, from his first to his eighth, Kevin’s weight was read and recorded.
The record is shown in the table below.
RECORD OF KEVIN’S WEIGHT
AGE IN YEARS
1
2
3
4
5
6
7
8
WEIGHT IN KG
10
14
17
20
23
19
24
30
a. Using a vertical scale of 2 cm to 10 kg and a horizontal scale of 2 cm to 1
year, draw a line graph to represent the information given in the table.
b. During which period did Kevin’s weight increase the most?
c. Suggest a reason for Kevin’s weight loss between his 5th and 6th birthdays.
d. Can you use the graph to tell his weight at birth? Explain.
2.
The External Sector Statistics (2002) show that for the period 1995 – 2000 private
inflows of US$ through Remittance Companies far exceeded inflows through
Commercial Banks. The figures are presented in the table below. Each year, the
number of dollars is rounded to the nearest $10 million
23
Mathematics - Stage 2 Module 6
PRIVATE INFLOWS OF US$ (IN MILLIONS)
YEAR
1995
1996
1997
1998
1999
2000
COMMERCIAL
310
280
260
250
210
190
150
200
260
340
380
470
BANKS
REMITTANCE
COMPANIES
Before you continue, make sure you understand the information that is being
presented here.
Talk with members of your Study Group or community
members about ‘private inflows of US$’. Name the Commercial Banks and the
Remittance Companies that are near to you.
Do you know someone who
receives US$ through a Commercial Bank or a remittance company? Have you,
yourself ever been sent money this way?
How does one get the money once he/she knows that it has been sent?
[Read the table again and then proceed.]
a. On the same page, using the same axes, draw two line graphs to
represent the given data. [Remember, you should decide whether or not
you will start the vertical axis at zero].
b. What is the total amount of money that came through Remittance
Companies?
c. By how much does it exceed the total amount that came through
Commercial Banks?
d. What percentage of the total amount of inflows came through Commercial
Banks?
e. Suggest a reason for the increased use of Remittance Companies over
the period.
24
Mathematics - Stage 2 Module 6
3.
“We must increase our prices “, claimed the chicken producers. “Widespread
drought in Grainland since April has increased the price of corn by some 60%
between April and August. See for yourself”
PRICE OF CORN FOR JANUARY 2000 TO AUGUST 2000
US CENTS PER BUSHEL
300
250
200
150
100
50
0
J
F
A
M
M
J
J
A
JANUARY -AUGUST 2000
a) Describe two ways in which this graph is deceptive.
b) What is the actual percentage increase in the price of corn over the period
April to August?
c) Draw a line graph illustrating the same data in what you consider a more
accurate manner.
4. Continue to collect statistical graphs from newspapers and magazines and discuss
them.
a) Say what kinds of graphs they are.
b) Decide whether each is the best kind to do the job it is meant to do.
c) Determine how clearly and honestly they represent the data.
25
Mathematics - Stage 2 Module 6
MODULE 6:
HANDLING DATA PART 2
UNIT 3:
ANALYSING AND INTERPRETING DATA (1)
Introduction
Let us look again at the role of ‘averages’ when we want to analyse and interpret data.
In analysing data, one approach is to examine the measures of central tendency, that
is, the extent to which there is a pull towards the average or middle value of the set of
data. The question to be answered is: ‘Is there a single value that characterises the
group?’
You will recall that there are three ‘averages’ or measures of central tendency that are
commonly used:
·
the arithmetic mean which is often referred to simply as the mean
·
the median
·
the mode
By the end of this unit you will be able to:
·
use the mean, median and mode to describe the value that is typical of a set
of data.
·
use the most appropriate ‘average’ measure for a given purpose.
·
calculate and use the range, the interquartile range and the semiinterquartile range as measures of variability among members of a set of
data.
·
determine when the average measure and the measure of dispersion of a set
of data are misleading.
Prerequisites
Before you begin you should know:
·
how to compute the mean, median and mode for raw data or data given in
frequency distribution tables.
·
how to do simple addition and subtraction operations accurately, with or
without the help of a calculator.
26
Mathematics - Stage 2 Module 6
The Mean
Given a set of values, the mean is the sum of all the values divided by the number of
values.
Example 1a
The heights, in cm, of 10 girls are:
125 152 139
154 177
149
140 177 124 143
Calculate the mean height of the girls.
Solution
The mean height =
=
sum of heights
1480 cm
number of girls
Note :
=
148 cm
10
All the values are used to compute the mean, therefore the mean is affected by
all the measures. The mean, then could be misleading if there are values that are much
greater or much less than the rest.
Pause now and learn some special symbols and a formula
that are associated with the mean.
X = ΣΧ
[å is the Greek letter, sigma]
N
where X = the sample mean
Σ = the sum of
Χ = the single measures
N = the number of measures
27
Mathematics - Stage 2 Module 6
The Median
When the values are arranged in ascending or descending order, the median is the
middle value or the sum of the two middle values divided by 2.
Example 1b.
The heights, in cm, of 10 girls are the same as those given in Example 1a. Find the
median height of the girls.
Solution
The arrangement of the heights in ascending order gives:
124
125
139
140
There are 2 ‘middle values’,
The median height
149
152
154
177
177
143 and 149
= (143 + 149)
2
Note :
143
= 292=
146 cm
2
The median is unaffected by abnormally high or low values.
It sometimes
represents an actual member of the set of values. Where there is an odd pattern of
distribution of values, however, the median may not be characteristic of the group.
28
Mathematics - Stage 2 Module 6
The Mode
This is the value that occurs most often in a set of values. This may be observed directly
from an orderly arrangement such as that shown in Example 1b.
Example 1c
If the heights of 10 girls are those given in Example 1b, find the modal height of the
girls.
Solution
From observation, the height, 177 cm, occurs twice, while each of the others occurs only
once.
The mode of the measures or the modal height = 177 cm
Note:
This is a particularly helpful average to manufacturers and service providers
who need to know which goods and/or services are most in demand.
It could be
misleading if extremely high or low scores are the ones that occur most frequently.
Sometimes there is no modal value; at other times there are two or more.
Exercise 3.1
1. Although all three, the mean, the median and the mode, are used as representative
values of a set of data, for a given set of data one of the three averages may be
more appropriate than the other two to describe the distribution, or to give
information for a special purpose. In each of the following situations, give the type of
average that you think would be most appropriate and say why.
(a)
During the summer holidays Brad worked as a handyman at a Biscuit
Company for 5 days and was paid the following amounts at the end of each
day: Monday : $450, Tuesday : $550, Wednesday : $1 800, Thursday
$500 and Friday : $600. What was his average daily pay?
29
Mathematics - Stage 2 Module 6
(b)
The owners of Buy Right Supermarket, as they were getting ready to buy
new stock, carried out a survey of the number of persons who bought
different types of laundry detergent during the week. The findings are
recorded below.
BRAND OF DETERGENT NUMBER OF PURCHASES
Breeze
1,150
Dismiss
1,280
Sudsil
1,300
Tide
850
Washo
1,280
What is the average number of purchasers per brand?
(c) Brenda wants to know what is the average temperature in Bridgetown, Barbados so
she may pack suitable clothes for her trip. The temperatures, in ºC, for the past 6
days were: 29.5, 31.5, 29, 30, 32, and 31. What was the average temperature in
Bridgetown last week?
(d) A car dealer put up a poster in the parking lot to show the amount of savings on a
new car purchased from her. Read the poster.
30
Mathematics - Stage 2 Module 6
NUMBER of
AMOUNT of SAVINGS
SALES
(i)
2
$ 5 000
1
$ 3 500
8
$ 1 000
What is the average saving that you might expect if you bought a new car from this
dealer?
(ii) Compare your responses with those given at the end of the module.
Go over all your work carefully before you go on.
As we have seen, averages give us a good idea of what is typical of a set of data.
Taken by themselves, however, they do not tell us enough about the entire distribution.
In addition, we need to know how the rest of the data is grouped around the average.
Are they clustered around the centre (average), or are they scattered widely?
For more careful analysis of the data, we need a measure of their spread or dispersion.
Besides, we need to be able to summarise or describe how the individual values differ
from one another. We need a measure of their variability. Small variability indicates that
the values do not differ greatly, so they must be rather similar. On the other hand, large
variability indicates that the values are inconsistent, that is, they differ. Additionally, the
measure of variability enables you to look more critically at the average and to judge the
extent to which the average value truly represents all the data.
31
Mathematics - Stage 2 Module 6
One simple way of measuring the variability is to consider the range of values.
The range is calculated by finding the difference between
the largest and smallest values in the distribution.
Example 2a
Look back quickly at unit 2. There you will be reminded of how a teacher used double
line graphs to display the marks gained by two students, Jack and Jill, in six consecutive
end-of-week tests. The graphs showed clearly that Jill’s marks were about the same for
each test while Jack’s fluctuated.
He got relatively low marks sometimes and high
marks at other times.
An examination of the range of the scores in each case would give us more specific
information about the students’ performances.
The table following, gives the actual scores of the two students.
WEEKLY SCORES, EACH OUT OF 20
STUDENTS
WK. 1
WK. 2
WK. 3
WK. 4
WK. 5
WK.6
Jack
10
15
18
12
15
20
Jill
14
14
15
15
16
16
Let us compare their average scores.
Jack’s Case:
(a)
X =
∑X
N
[This is a common practice, isn’t it?]
= 10 + 15 + 18 + 12 + 15 + 20
6
= 90 = 15
6
The mean score = 15
32
Mathematics - Stage 2 Module 6
(b) Arrangement of the scores in numerical order gives:
10
12
15
15
The median score =
18
20
= 15 [There are 2 middle scores]
15 + 15
2
(c) From observation, 15 is the most frequently occurring score.
The modal score = 15
Jill’s Case: (a) X = ∑X = 14 + 14 + 15 + 15 + 16 + 16
N
6
The mean score
=
90
6
= 15
= 15
(b) Arrangement of the scores in numerical order gives:
14
14
15
15
16
16
The median score = 15 + 15 = 15
2
(c) From observation, there are 3 modes: each of the scores, 14, 15, 16 appears
twice.
One of the modal scores = 15
Obviously, the average score does not help us to differentiate
between the test by test performances of the 2 students.
Let us see whether the range will.
In Jack’s case, the range of the scores = 20 - 10 = 10 marks
In Jill’s case, the range of the scores
= 16 - 14 = 2 marks
33
Mathematics - Stage 2 Module 6
What is now highlighted that you had not observed before?
REMEMBER! The range measures how the scores vary their spread or dispersion.
For Jack, there is a spread of 10, so some scores are as many as
10
or 5 marks from
2
the average score, on either side of that average. There are relatively large differences
between the scores.
For Jill, there is a spread of 2, indicating a clustering of marks around the average.
Actual marks are no more than
2
or 1 mark from the average, on either side of it.
2
The scores are somewhat similar to one another.
With this additional information:
(i)
you are now able to describe, in some detail, the pattern of each student’s
performance, not just that both had the same average score;
(ii)
you are now able to predict, with some accuracy, the performance on similar
tasks in the future.
In interpreting the results,
― which student’s work would you describe as ‘inconsistent’?
― what would be your prediction about their results on another series of similar
tests?
Comment: Jack’s work can certainly be described as ‘inconsistent’; the range or spread
of the scores supports this judgement. It would be difficult to predict what Jack’s work in
the future will be like since his test results vary so much.
You could with some
confidence, however, predict that Jill will gain about 15 marks on each test.
34
Mathematics - Stage 2 Module 6
Although the range tells about the variability or spread in a set of data, since it involves
only the two most extreme values, the highest and the lowest, its use is somewhat
limited. Example 3.1. illustrates this.
Example 3.1.
Here are two sets of scores:
(a) 12, 6, 7, 3, 15, 10, 18, 5;
(b) 9, 4, 8, 10, 9, 8, 9, 19.
For each set find (i) the mean score
(ii) the range of the scores
Compare the results.
Solution
Set (a):
X =
(i)
åX = 12 + 6 + 7 + 3 + 15 + 10 + 18 + 5
N
=
8
76 = 9 . 5
8
(ii) Arrangement of the values in ascending order gives:
3, 5, 6, 7, 10, 12, 15, 18
The range of values = 18 – 3 = 15.
Set (b):
(i)
X = åX = 9 + 4 + 8 + 10 +9 +8 + 9 + 19
N
8
=
76 = 9.5
8
(ii) Arrangement of the values in ascending order gives: 4, 8, 8, 9, 9, 9, 10, 19
The range of values = 19 - 4 = 15
35
Mathematics - Stage 2 Module 6
For both sets the mean score is the same, but it is easily observed that the way the
scores are spread in relation to the mean is entirely different. In (a) the scores are
scattered widely, while in (b) they are closely gathered around the mean. It is in a
situation such as this that the range of the values should confirm the observation and
show the difference in spread.
But what do we find? For both sets, the range is the same. The extreme values have
influenced the range and rendered it useless.
Since the range indicates no difference between the sets, it is not a good measure
of variability or spread in this case.
In general, when extreme values are present
the range is a poor measure of variability or spread.
What then do we do when there are extreme values as there sometimes will be?
Although they tend to distort the whole picture, we cannot discard them. Instead, we can
use two other measures to help us analyse the given data. They are the Interquartile
Range and the Semi-interquartile range.
36
Mathematics - Stage 2 Module 6
The Interquartile and Semi-interquartile Range.
To begin, an ordered set of data is divided into four equal parts.
A quartile is the name given to one of three values that indicate where the
divisions occur.
The median, second or middle quartile, Q2, is the middle value of the whole set of
data and divides the set into two equal parts. One part naturally has values which are
lower than the median and is described as forming the lower half of the data, while the
other part, with values which are higher than the median, form the upper half of the
data.
The lower quartile, Q1, is the middle value of the lower half of the data.
The upper quartile, Q3 is the middle value of the upper half of the data.
Example 3.2.
Given the set of scores: 43, 38, 34, 41, 36, 39, and 35, find
(a) the middle quartile or median
(b) the lower quartile
(c) the upper quartile.
Solution
The scores, in ascending order of magnitude, are written: 34, 35, 36, 38, 39, 41 , 43
(a) the middle quartile, Q2 = 38 [It is the median or middle value of all the data]
(b) the lower quartile, Q1 = 35 [It is the middle value of the lower half of the
data]
(c) the upper quartile, Q3 = 41 [It is the middle value of the upper half of the
data]
37
Mathematics - Stage 2 Module 6
The interquartile range of a distribution is the difference between
its upper and lower quartiles or Q3 - Q1.
The semi – interquartile range is half the inter – quartile range or Q3 - Q1
2
Using the data in Example 3.2:
the inter quartile range
=
41 - 35 = 6
the semi – inter quartile range
=
41 - 35 =
3
2
The range of these scores is
43 – 34 or 9. We are now able
to include the extreme scores
and yet have a more accurate
picture of the spread of all the
data.
Practice finding the range, interquartile range and semi – interquartile range of sets
of raw data and of data given in frequency distribution tables. Use the recommended
texts in the bibliography. It is important that you be very familiar with these measures of
dispersion and that you be able to calculate them speedily and accurately.
Exercise 3.2
1.
For the set of measures, 5, 7, 2, 3, 8, 10, 14, calculate:
(a) the median
(b) the range
(c) the interquartile range
(d) the semi-interquartile range
38
Mathematics - Stage 2 Module 6
MODULE 6:
HANDLING DATA PART 2
UNIT 4:
PROBABILITY (1)
Introduction
At Stage 1 you performed a number of experiments and discovered that it is possible to
find experimentally, the probability or likelihood of a particular event occurring when you
performed an activity.
By observing and comparing the results of your experiments, you discovered that there
was a constant relationship between the number of favoured or desired or successful
outcomes of your activity and all the possible number of outcomes of that activity.
That relationship is given by the formula:
Pr (event) = number of successful outcomes
number of possible outcomes
The result or relationship can be left as a common fraction or be converted to a decimal
fraction or to a percentage, whatever the appropriate form for the given situation.
You discovered too, that it is not always possible or necessary to perform actual
experiments. By applying the same formula to the results of surveys or other reports you
can find the relationship between the successful and possible outcomes of an event.
By the end of this unit you will be able to:
·
find the probability of single events by using either an experimental or a
theoretical approach
·
distinguish between and/or identify combined events which are independent,
dependent and mutually exclusive
39
Mathematics - Stage 2 Module 6
Prerequisites
Before you begin you should know:
·
how to apply the probability ideas that you have already met.
·
how to carry out all operations with whole numbers.
·
how to carry out all operations with decimals.
·
how to carry out all operations with fractions.
·
set theory
Example 4.1
Ten cards are numbered from 1 – 10. They are shuffled and one is drawn from the set.
Calculate the probability that:
(a) it is the 7
(b) it is an even number
Solution
(a) The number of possible outcomes
= 10
[Each of the 10 cards has an equal
chance of being drawn].
The (desired) successful outcome
= ‘7’
The probability of drawing the 7
=
[Only one card has this number]
number of successful outcomes
number of possible outcomes
\ Pr (7)
=
1
or 0 .1 or 10%
10
40
Mathematics - Stage 2 Module 6
(b) The number of possible outcomes
= 10
The number of successful outcomes = 5
[The desired draws are: 2, 4, 6, 8, 10]
The probability of drawing an even number =
number of successful outcomes
number of possible outcomes
\ Pr (even number)
=
5
or 0 .5 or 50%
10
The set of possible outcomes is called the sample space and is denoted by S.
Using set language, the number of all possible outcomes is the universal set, (U), for a
particular event, while the number of successful outcomes is a subset of U.
Example 4.2
GET YOUR FERTILISED EGGS TODAY!
90% chance of hatching.
Members of the “Try R Die Club” bought 3000 eggs for their chicken-rearing
project. How many chickens can they expect to have?
Solution.
Pr (hatched eggs) = number of successful outcomes = 90% = 90
number of possible outcomes
\ Number of chickens they may expect
100
=
2700
3000
= 2700
41
Mathematics - Stage 2 Module 6
Exercise 4.1
Discuss each of the following in your study group and, where calculations are required,
do them orally.
1.
At the Producer’s Maternity Hospital, the midwives estimate that the probability of a
male birth is 0 .52. What is the probability of a baby girl being born there?
2.
In 150 rolls of a die, a 6 is thrown 120 times. Do you think the die is fair? Explain
your answer.
3.
On a local ‘Wheel of Fortune’ television show, the stations are marked as follows:
$10, $20, $50, $100, $10, $500, $50, $10. When the wheel is spun, all stations
are equally likely to have the pointer. Tanya is on the show.
Calculate the
probability that, when the wheel is spun she will win:
(i) $10
4.
(ii) $500
(iii) $1000
Explain why the probability of an event that is certain to happen is 1, while the
probability of an event which is certain not to occur is 0.
5.
Some people say, “Probability is no better than guessing”. What do you think and
why?
NB. Write suitable statements and set out your work in an orderly manner.
Combining Probabilities
So far you have calculated probabilities for single events. But events do not always
happen singly or in isolation. Often two or more events are happening simultaneously or
one may succeed the other(s). According to the relationship between them, they are
classified as independent events, dependent events, or mutually exclusive events.
42
Mathematics - Stage 2 Module 6
Independent Events and their Probabilities
Events are said to be independent when the occurrence of one of them in no way affects
or excludes the occurrence of the other(s).
Example 4.3.
If a coin and a die are tossed at the same time, find the probability of getting a head with
the coin and a six with the die.
Solution
Note: 1. The probability of getting a head with the coin in no way affects the
probability of obtaining a six on the die. These are therefore considered
independent events.
2. The new challenge is to find all the possible outcomes for both activities.
One way of doing this is to write down, in order, all the pairs that could
possibly show when the coin and the die are tossed at the same time.
The pairs are written below. The pair highlighted in the table, shows the desired
combination or outcome.
H, 1
H, 2
H, 3
H, 4
H, 5
H, 6
T, 1
T, 2
T, 3
T, 4
T, 5
T, 6
The number of possible outcomes
= 12
The number of desired outcomes
=
1
The probability of getting a head and a six
=
number of successful outcomes
number of possible outcomes
\ Pr (H and 6)
=
1
12
43
Mathematics - Stage 2 Module 6
Alternatively, treating the events singly:
Pr (Head)
=
number of successful outcomes
number of possible outcomes
\ Pr (Head) =
1
2
Numbers on die
Faces
on coin
1
2
3
4
5
6
Head
x
x
x
x
x
x
Tail
x
x
x
x
x
x
Pr (Six)
=
number of successful outcomes
number of possible outcomes
=
1
6
Combining the probabilities by multiplication
Pr ( Head and Six)
= Pr (Head) x Pr (Six)
=
1
1
x
2
6
=
1
12
In general, if A and B are independent events, Pr(A and B) = Pr(A) x Pr(B)
N.B. Another way of displaying all the possible outcomes is by making a table
with one activity shown along the horizontal axis and the other along the
vertical axis.
In the table, each X represents a possible outcome (12 altogether). The desired
outcome is marked. The probability of the event happening is
1
12
44
Mathematics - Stage 2 Module 6
Example 4.4
On a particular day the probability of my alarm clock not going off is
probability of a bus arriving on time is
1
, and the
12
3
5
a. Do you agree that these are independent events? Explain.
b. Find the probability that my alarm does go off and the bus is late.
Solution
a. Yes, these are independent events. The incident of my alarm going off or not, in
no way affects the time of arrival of the bus.
b. The probability that my alarm goes off, Pr (alarm) = 1 The probability that the bus is late,
Pr (lateness)
11
1
=
12 12
= 1 -
3
5
=
2
5
The probability of both, (independent events)
Pr (alarm and lateness)
=
=
11 2
x
12 5
22
11
=
60
30
Dependent Events and their Probabilities
In the case of dependent events, all events are possible, but the outcome of the first
affects the outcome of the second. This often occurs when the first item selected from a
set is not replaced before the second item is selected.
45
Mathematics - Stage 2 Module 6
Example 4.5
A class consists of 25 boys and 15 girls. Two students are selected at random. What is
the probability that both will be boys? Do you remember the meaning of random?
In this example, random means that every student has an equal chance of being
selected.
Solution
Note:
Both outcomes are possible that is, it is quite possible to select one boy and
then to select a second boy. However, the probability of selecting a second boy
is affected by the first selection. These, then are dependent events.
Pr (1st choice is boy)
=
25
40
[There are 25 desired outcomes (25 boys in the class)
and 40 possible outcomes (40 students altogether)]
Pr (2nd choice is boy)
=
24
39
[There are now 24 desired outcomes (1 boy has
already been chosen, and now only 39 possible outcomes]
Pr ( both boys)
=
25 24
x
40 39
=
5
13
Mutually Exclusive Events and their Probabilities
Events are mutually exclusive when one of them happening excludes or prevents the
other(s) from happening at the same time.
Example 4.6
Calculate the probability of picking either a King or a Queen from a full pack of cards.
46
Mathematics - Stage 2 Module 6
Solution
Note: The successful outcomes are: picking a King or picking a Queen. Either may
happen but not both at the same time. The events are therefore exclusive.
=
4
1
=
52 13
Pr (Queen) =
4
1
=
52 13
Pr (K or Q) =
1
1
+
13
13
Pr (King)
=
2
13
In general, if A and B are mutually exclusive events, Pr(A or B) = Pr(A) + Pr(B)
Note:
A Venn diagram helps illustrate the relationship between mutually exclusive
events.
The diagram below shows the information given in Example 4.6.
U = full pack of 52 cards
U
L
K
M
K
K
K
Q
Q
Q Q
47
Mathematics - Stage 2 Module 6
The set of Kings and the set of Queens are two separate subsets of U. [The other cards
are not shown in the diagram]. The events are represented by disjoint sets. The
probability in this case is given as,
Pr (K or Q) = Pr (K) + Pr (Q)
Exercise 4 .2
The problems in this exercise bring together all the concepts and ideas that have been
dealt with in this unit. They provide, therefore, an opportunity for revision and application
of what you have learnt so far. Read each one carefully, think logically and work neatly.
1. In each case say whether two events A and B are mutually exclusive or not.
a. Two coins are tossed. A is getting 2 heads, B is getting 2 tails.
b. Two dice are rolled. A is getting a total score that is an odd number. B
is getting a total score that is divisible by 2.
c. Two dice are rolled. A is getting a total score of eight. B is getting a total
score that is a multiple of four.
2. A box contains 10 black pens and 12 blue pens. Two pens are selected at
random from the box. What is the probability that both pens are black?
3. If the probability of school children having defective hearing is 0.03, how many
children in a school of 1200 would you expect to be hearing impaired?
4. In an experiment there are 3 possible outcomes, R, S or T.
If Pr (R) =
1
2
and Pr (S) = ,
3
5
a. Calculate Pr (T)
b. What is the probability that either R or T occurs?
48
Mathematics - Stage 2 Module 6
5. The table below shows the sample space for two dice, one black, and the other
white, rolled at the same time.
Each X represents a possible outcome.
Numbers on white die
1
Numbers
2
3
4
5
6
on
1
x x x x x x
black
2
x x x x x x
die
3
x x x x x x
4
x x x x x x
5
x x x x x x
6
x x x x x x
a.
What is the probability of scoring a total of seven?
b.
Find the probability of getting an even number on both dice.
49
Mathematics - Stage 2 Module 6
LET’S SUMMARISE!
What does data – handling at this Stage involve?
Identifying the problem to be
solved, the question to be
answered, the decision to be
taken, or the prediction to be
justified…for which you need
numerical information
Searching for and accessing existing
information; collecting relevant
information from the appropriate
population or a representative sample
of it; using direct observations or
interview schedules, where the
choice of instrument is influenced
by the type of information
needed as well as by the size and
location of the population; sorting,
organising and tabulating the data
collected to make them more easily
managed and/or visually represented
Making guesses even before
formal investigation begins;
recognising the degree of uncertainty
that exists even when the collection,
sorting, and analysis of data are all
done with efficiency and care; using
probability ideas to help predict the
outcomes of single and compound
events; comparing research findings
with guesses and predictions.
and analysed.
Analysing the data: what do they tell?
What do they reveal when they are
compared to other data?; finding what is
typical of the collection (the average
measure); finding the measure of
variability or spread (the range,
interquartile and semi – interquartile
range; increasing the possibility of drawing
logical conclusions, making suitable
inferences and reasonably accurate
predictions.
Displaying the data by using tables,
pictograms, bar charts, pie charts or line
graphs; ensuring that the type of graph
chosen is the one best suited for the
audience to be reached and the message to
be communicated; ensuring also that enough
information is presented, that it is not
misleading, and that the graphical
presentation will enable the decision –
making/ problem – solving process.
50
Mathematics - Stage 2 Module 6
END OF MODULE TEST
Attempt all items.
Time allowed: 1 hour
Show all working. Where needed, write meaningful statements and draw clearly
labelled graphs.
1.
The following marks were scored by 25 students in a Mathematics test which was
marked out of 10:
9
5
2
7
8
3
7
6
10 5
3
1
6
6
7
4
7
2
7
2
6
6
5
2
4
a. Copy and complete the frequency table given below.
Marks Frequency
1
1
2
4
3
4
5
6
7
8
9
10
Table 1
b. Find: (i) the mean
(ii) the median
(iii) the range of the distribution.
c. What is the probability that a student chosen at random received a mark of 6?
51
Mathematics - Stage 2 Module 6
2.
MR. JONES’ SALARY
Savings
5%
Food
40%
Rent
30%
Utilities
10%
Other
15%
Figure 1
a.
For which item does Mr. Jones budget the most money?
b.
What fraction of his salary does he budget for “other”?
c.
If the amount budgeted for savings is $800.00, how much is his monthly salary?
d.
What is the size of the angle of the sector representing “Rent”?
3.
The body temperatures of Joseph at different times of the day are given in the
table below.
Time of day
6:00 8:00 10:00 12:00 2:00 4:00 6:00 8:00 10:00
(am) (am) (am) noon (pm) (pm) (pm) (pm) (pm)
Temperature 36
36.5 37
38.5
37.5 37
36.5 36
35.5
(°C)
Table 2
52
Mathematics - Stage 2 Module 6
a. Draw a line graph to represent this information.
b. Use the graph to answer the following questions:
(i) What is the difference in body temperature between 11:00 a.m. and mid-day?
(ii) What period shows the greatest rise or fall in temperature?
c. What do you think will be the temperature at midnight? Give a reason for your
answer.
4. The ages of a class of 30 boys in grade V at Mount Carey High School are given
below.
Age in years
11 14 15 16
Number of boys 2
6
20 2
Table 3
(a) What is the mean age of the class?
(b) What is the median age of the class?
(c) Find the upper and lower quartiles, and then write down the interquartile
range of the boy’s ages.
5.
A bag contains 40 marbles: 25 black ones and 15 red ones. A marble is chosen at
random from the bag.
a. What is the probability of drawing a red marble?
b. What is the probability of drawing a blue marble?
c. Calculate the probability of drawing a black marble or a red marble.
Go over your work carefully then check your answers against those given at the end of
the module. Use the mark scheme and give yourself a score out of 25. Use this score
alongside the results of all the previous exercises to help make a judgement about the
extent to which you have mastered the work done so far. Treat yourself fairly and
honestly.
Remember! You should be striving for 100% mastery.
Compare your self-assessment with that of your facilitator.
53
Mathematics - Stage 2 Module 6
ANSWERS TO EXERCISES
Exercise 1.1
About the situation
Numerical information is needed in question numbers 1, 2, 4 and 5. In each of these
cases you are being asked to make decisions based on facts. In particular, you are
required to compare numbers: the number of male versus the number of female infant
deaths; the number of boys (over time) who, having sat the GSAT examination, are
placed in secondary schools of their choice versus the number who are not; the number
of team members who claim curried chicken to be their favourite dish versus those who
do not.
No doubt data already exists at the Ministry of Health and the Registrar General’s office
for the situation in question 1 and at the Ministry of Education and Culture for question
number 2.
For the situations in questions 4 and 5 you would have to do your own survey.
In question number 4 the population is all the cars in the parking lot while for question 5
the population is all the members of the West Indies Cricket Team.
For question number 4 the data could be gathered by making on-the-spot observations,
while for question number 5, a simple interview could provide the answer to the
question: “What is your favourite food?”
In both cases, the responses could be recorded on prepared tally sheets, and the tally
marks subsequently counted and arranged in an orderly manner for use at a later date.
Answers will vary but possibilities include:
(a) car dealers who might be doing a series of surveys to identify the most popular
colour in cars and use this information to influence the colours of cars which they
would supply;
54
Mathematics - Stage 2 Module 6
(b) garages where cars are repaired, so they would have paints in stock to meet
the needs of car owners.
Exercise 1.2
1.
Answers may vary but basically, Lady Whiz realised that 20 out of 30 or
2
of her
3
customers most enjoyed their fruit in the morning.
12
x 100 or 40%.
30
2.
Percentage that had orange as their favourite fruit =
3.
Oranges and naseberries, most of the persons who preferred their favourite fruit in
the mornings also said their favourite fruit was orange or naseberry.
5
= 100%
5
4.
Pr (buying naseberries out of season) =
5.
Answers will vary but might include preferences for colour, texture, and degree of
ripeness of fruit.
6.
Answers may vary.
Exercise 1.3
1. Questions asked will vary, but they should have a direct relationship to the
information needed. [Be guided by the hints given in the text].
2. Ask your facilitator to comment on your answers here.
3. (i) Main advantages:
ability to reach a much larger population than by direct
observation and to reach persons in many different locations simultaneously.
(ii) Main disadvantages: there is less control over the quality of data collected and if
the population or sample is large, the hiring of interviewers could prove costly.
Exercise 2.1
1.
In most cases, data that are represented by pictograms may be represented equally
by bar graphs and by pie charts. An examination of the three in this case shows
each of them presenting the data effectively.
In the pictogram, although the actual sums of money are not mentioned, the dollar
signs are directly related to the proposed expenditure for each type or level of
55
Mathematics - Stage 2 Module 6
education. The key provides a ready help to calculate what the actual amounts of
money are.
In the bar chart, too, the tallest bar indicates where there will be
greatest expenditure and the shortest, where there will be least expenditure. The
scale on the vertical axis gives the reader the actual amount without much effort.
The pie chart, by giving percentages on each ‘slice,’ shows where the greatest
expenditure will be. It is the pie chart, however, that seems to ‘have the edge’ in this
case.
In his representation, the Minister of Finance described how the entire
education budget would be divided. The pie chart does the best job by showing
how the whole is shared.
2.
There is a case to be made for all four messages. It seems, however, that the one
that is predominant is the Early Childhood Education will receive priority attention
and that there will be funding in place to ensure this.
Exercise 2.2
1.
ALLOCATION FOR SPECIAL EDUCATION
56
Mathematics - Stage 2 Module 6
2. (a) In 2007, Year 9
(b) The total amount of money spent on Special education in years 6 and 7 was
$65m ($ 30m in Yr 6 + $ 35m in Yr 7).
(c) The graph suggests that there was a time when no money was allocated Special
Education (1998).
3. Increase = ($ 25m - $ 5m) = $ 20m;
percentage increase =
20
x 100 = 400%
5
Exercise 2.3
1.
Completed table.
Months
Profit
($’000s)
2.
July
50
August
75
September October
125
115
November
135
December
200
See graph in Exercise 3.2.
3.(a)
The difference in profit between December and October was :
$200,000.00 - $115,000.00 = $85,000.00
(b)
The percentage increase in profit August to September was:
50
x 100 = 66 23 %
75
The percentage increase in profit from November to December was:
65
x 100 = 48%
135
The percentage increase from August to September was more by:
66 23 - 48 = 18 23 %.
57
Mathematics - Stage 2 Module 6
(c)
It is difficult to predict what will be the profit in January of next year. The rate of
increase in profits has not been regular over the period July to December and
although profits have moved up significantly since November, there could well
be a drop in consumption or sales in soft drinks and a resultant drop in profits
after December.
Exercise 2.4
1.
(a) 470 male infants
(b) Percentage =
470
x 100 = 61%
770
(c) The records for 5 consecutive years certainly show a trend in keeping with the
belief, but it is usual for statisticians to observe a phenomenon of this kind and
keep records for at least 10 years before confirming a ‘belief’
58
Mathematics - Stage 2 Module 6
d)
GRAPH OF INFANT DEATHS over a FIVE - YEAR PERIOD
NUMBER OF INFANT DEATHS
120
100
80
60
40
20
0
1
2
3
4
5
YEARS
Female
Male
e) Since Year 2 the rate of decline in male infant deaths has been constant. A
cautious prediction is that there will be 60 male infant deaths in Year 7.
There has been a steady rate of decline in female infant deaths for the entire
5-year period. A safe prediction is that there will be 20 deaths in Year 7.
Total number of infant deaths predicted = 80.
Exercise 2.5
Completed table
YEAR
1995
1996
1997
1998
1999
2000
15
20
22 . 5
27 . 5
30
35
15
20
22 . 5
27 . 5
30
35
PROFITS in $M
SHOWN in X
PROFITS in $M
SHOWN in Y
59
Mathematics - Stage 2 Module 6
2. In graph X, 1 unit represents $ 10 m on the vertical axis, while in graph Y,
the same unit represents $ 5m. Values in X will therefore be more ‘squashed’
than those in Y.
3. (a) Graph Y wants to communicate that profits are growing by ‘leaps and
bounds’. This would be good news to persons who want to invest in this
Company.
(b) The omission of zero certainly helps the financial perception of the
company, for although the symbol is missing, the visual impression is that
profits have risen from $ 0 to $200m in the time.
Exercise 2.6
1.
(a)
RECORD of KEVIN’s WEIGHT
35
WEIGHT IN KG
30
25
20
15
10
5
0
1
2
3
4
5
6
7
8
9
AGE IN YEARS
60
Mathematics - Stage 2 Module 6
(b) Between his 7th and 8th birthdays
(c) He might have had a severe illness or a series of illnesses that
prevented him from eating normally.
(d) The rate of change of weight over the years is not sufficiently constant
to enable a reasonable prediction. Moreover, it is known that the rate
of change of weight during the first year of life is more erratic than the
rate during the succeeding 3 years. The graph is not helpful in this
situation.
(a)
PRIVATE INFLOWS OF US$ (IN MILLIONS)
$US (in millions)
2.
500
450
400
350
300
250
200
150
100
50
0
1994 1996 1998 2000 2002
Comercial Banks
Remittancee
Companies
Years
(b) Amount through Remittance Companies = US$ 1 800M
(c) Excess = US $ 1 800 M - US$ 1 500M = US $ 300M
(d) Required percentage = 1 500/3 300 x 100 = 45 . 5%
(e) Answers will vary and could include:
(i) the increased number of such Companies in recent times;
(ii) the location of the branches so they are more readily accessible than most
Commercial Banks.
61
Mathematics - Stage 2 Module 6
3.
(a) On the horizontal axis 2 different scales are used to represent the months.
The smaller scale representing the months April through August makes the
cost values bunch together to give the impression of a very steep rise
in price. On the vertical axis also there is a misrepresentation. No zero
value is indicated, but at the beginning of the axis 2cm represent 180 cents,
while further up the axis, 2 cm represent 20 cents.
(b) Actual percentage increase April to August =
80
x 100 = 40%
200
(c) Do graph and take to facilitator for discussion.
RESPONSES TO EXERCISE 3.1 #1
In each case a comparison of the actual averages helps the decision-making process.
a. The mean pay = $ ( 450 + 550 + 1800 + 500 + 600) = $ 3900 = $ 780
5
5
When arranged in numerical order, the payments are:
$450, $500, $550, $600, $1800
The median = $550
There is no modal value.
One large figure, $1 800, has pushed up the value of the mean pay, so it is not
typical of the set of daily earnings. The median is the most appropriate
measure.
b. Neither the mean nor the median gives the information that the owners need. In
preparing to buy new stock, the fact that there were 1172 purchasers per brand
62
Mathematics - Stage 2 Module 6
of detergent or that a certain number of purchasers fell in the middle of the group,
is not what will help them make a decision about what to buy.
The mode
indicates that there were two brands of detergent that were equally liked by
the purchasers.
c. The mean temperature = 30 .5°C, and so, also is the median temperature. There
is no mode. Either the mean or the median is a suitable ‘average’. Since
there are no extreme temperatures, however, one could use the mean
without making a comparison with the median.
d.
(i)
The average savings expected if purchasing a car from the particular
dealer is:
2 x 5000 + 1 x 3500 + 8 x 1000 = 21500
11
11
= $1,954.55
Exercise. 3.2
1.
The set of measures arranged in numerical order : 2, 3, 5, 7,
8, 10, 14.
(a) The median = 7
(b) The range = 14 - 2 = 12
(c) The interquartile range = Q3 - Q1, where Q3 = 10 and Q1 = 3
= 10 -
3 = 7
(d) The semi-interquartile range
=
7
= 3 .5
2
63
Mathematics - Stage 2 Module 6
Exercise 4.1
1.
Your discussions should help clarify misunderstandings and deepen
understanding of the new concepts.
(a) Pr (Girl) = 1 - 0 .52 = 0 . 48
(b) From experiments done, for any one roll of a die, Pr (6) =
From 150 rolls, then, the expectation is that Pr (6) =
1
6
1
x 150 = 25.
6
When Pr (6) gives 120, it is clear that the die is biased…weighted so there is
a greater chance that it will show a ‘6’ when it is rolled.
(c) (i)
(ii)
Pr (10) =
3
8
Pr (500) =
1
8
(iii) Pr (1 000) = 0
(d)
When an event is certain the number of expected or desired or successful
outcomes is the same as the total number of possible outcomes of the activity.
For example, if there is a box of n blue pens, then success at selecting a blue
pen can happen n times. Pr (blue pen) =
n
= 1. At the same time, an event is
n
impossible if there can be no successful outcome of the activity. There is no
chance whatsoever of getting a red pen from a box of n blue pens.
Pr (red pen) =
0
= 0.
n
(e) Support your argument with the results of your own experiments.
64
Mathematics - Stage 2 Module 6
Exercise 4.2
1.
(a)
2.
Pr (1st. black pen) =
10
5
or
11
22
Pr ( both choices) =
5
3 15
x
=
11
7
77
3.
Yes
(b)
Yes
(c)
Pr (children hearing impaired)
No
Pr (2nd. black pen) =
= 0 .03 or
9
3
or
7
21
3
36
=
100
1200
Number expected to be hearing impaired in school of 1,200 = 36
4.
5.
1 2
+ )
3 5
(a)
Pr (T) = 1 - (
(b)
Pr (R or T) =
(a)
From the table, Pr (7) = 6
1
9
4
+
=
or
3
15
15
36
(b)
11
4
=
15
15
= 1 -
=
3
5
1
6
From the table, Pr (even number on both dice) =
9
1
or
4
36
65
Mathematics - Stage 2 Module 6
ANSWERS TO END OF MODULE TEST
1.
(a) Completed frequency table:
MARKS
FREQUENCY
1
1
2
4
3
2
4
2
5
3
6
5
7
5
8
1
9
1
10
1
[2 marks]
(b) (i)
The mean = 5 .2
(ii)
The median = 6
(iii)
The range = 9
(c) Pr (getting a mark of 6) =
2.
[3 marks]
1
5
[1 mark]
(a) Food
(b)
3
20
(c) $16,000.00
(d) 108°
[4 marks]
66
Mathematics - Stage 2 Module 6
3.
(a)
BODY TEMPERATURES AT DIFFERENT TIMES OF THE DAY
[3 marks]
(b) (i)
(ii)
0 .7°
Between 10: a.m. and noon
[2 marks]
(c) The graph shows a steady decline in temperature of 0 .5° every 2 hours from
2:00 p. m to 10 p. m. Following this trend, by midnight the temperature should
be down another 0 .5° to 35°
4.
(a)
Mean age = 14 years
(b)
Median age (Q2) = 15 years
(c)
Upper quartile, (Q3) = 15
[2 marks]
[1 mark]
Lower quartile, (Q1) = 14
Interquartile range
=
1
[3 marks]
67
Mathematics - Stage 2 Module 6
5.
(a) Pr (drawing a red marble) =
3
8
(b) Pr (drawing a blue marble) = 0
(c) Pr (drawing a black or a red marble) = 1
[3 marks]
Total possible score: 25 marks
68
Mathematics - Stage 2 Module 6
BIBLIOGRAPHY FOR STAGE 2 MATHEMATICS
Cox, C. and Bell, D. (1989). Understanding Mathematics Book 1. London:
John Murray Publishers Ltd.
Foster, A. and Tomlinson, T. (1994). Mathematics for Caribbean Schools Book 1.
Boston: Addison Wesley Longman.
Foster, Althea A. and Rose, Gerald et. al. (2001) Success in Maths. For the
Caribbean Book 1 (student’s book). London: Longman.
Foster, Althea A. and Rose, Gerald et. al. Success in Maths. For the Caribbean
Book 2 (student’s book). London: Longman.
Greer, A. and Layne, C.E. (2001). Certificate Mathematics: A Revision Course for
the Caribbean (4th edition). Cheltenham: Nelson Thornes Ltd.
James, T., Johnson, P. and Rowe, A. (1995). Count me in Book 1. Educational
Publishing Enterprises.
Layne, C. E. and Ali, F. W. et. al. (1997). STP Mathematics Book 1. Cheltenham:
Stanley Thornes (Publishers) Ltd.
Layne, C.E. and Ali, F. W. et. al. (1997). STP Mathematics Book 2. Cheltenham:
Stanley Thornes (Publishers) Ltd.
Layne, C. E. and Ali F. W. et. al. (1997). STP Mathematics Book 3.
Stanley Thornes (Publishers) Ltd.
Cheltenham;
Subnaik, H. (1996). Active Mathematics Enrichment Exercises Book 5. Port of
Spain: Charran Educational Publishers Ltd.
Toolsie, Raymond. (1996). Mathematics: A Complete Course Volume 1. San
Fernando: Caribbean Educational Publishers Ltd.
Toolsie, Raymond. (2004). Mathematics: A Complete Course Volume 1 (2nd edition).
San Fernando: Caribbean Educational Publishers Ltd.
Toolsie, Raymond. (2004). Mathematics: A Complete Course Volume 2. (2nd
edition). San Fernando: Caribbean Educational Publishers Ltd.
Zawojewski, J. et. al. (1991) Dealing with Data and Chance. Reston: National
Council Of Teachers of Mathematics.
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Mathematics - Stage 2 Module 6