Department of DECISION SCIENCES
Elementary
Quantitative Methods
Only study guide for
QMI1500
University of South Africa
Pretoria
ELEMENTARY
QUANTITATIVE
METHODS
The purpose of this module is
to enable a student to become conversant
with a variety of basic quantitative techniques.
©2019 Department of Decision Sciences,
University of South Africa.
All rights reserved.
Printed and published by the
University of South Africa,
Muckleneuk, Pretoria.
QMI1500/1
Contents
1 Revision of prior knowledge
1.1
1
Priorities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.1.1
Order of operations
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.1.2
Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.2
Variables
1.3
Laws of operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.4
1.5
1.6
1.7
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.3.1
Commutative laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.3.2
Associate laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.3.3
Distributive law
1.3.4
Summary of laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
Fractions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.4.1
Types of fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
1.4.2
Reducing fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
1.4.3
Multiplication of fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
1.4.4
Division of fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
1.4.5
Adding fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
1.4.6
Subtracting fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
1.4.7
Decimals
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
Powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
1.5.1
Positive powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
1.5.2
Negative powers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
1.5.3
Rules and properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
Roots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
1.6.1
Square root . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
1.6.2
Cube and higher order roots
1.6.3
Fractional exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
. . . . . . . . . . . . . . . . . . . . . . . . . . 54
Ratios, proportions and percentages . . . . . . . . . . . . . . . . . . . . . . . . . . 59
1.7.1
Ratios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
iii
CONTENTS
1.8
1.9
1.7.2
Proportion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
1.7.3
Percentages . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
Signs, notations and counting rules . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
1.8.1
The number line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
1.8.2
Positive and negative numbers . . . . . . . . . . . . . . . . . . . . . . . . . 82
1.8.3
Summation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
1.8.4
Multiplication rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
Units and measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
1.9.1
SI system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
1.9.2
Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99
1.9.3
Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102
1.9.4
Volume . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
2 Functions and representations of functions
2.1
2.2
2.3
iv
111
What is a function? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
2.1.1
Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112
2.1.2
Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
2.1.3
Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
2.1.4
Function notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120
2.1.5
Graphing functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
2.1.6
The Cartesian plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
2.1.7
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
Linear functions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
2.2.1
Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126
2.2.2
Draw the graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
2.2.3
The slope of the straight line . . . . . . . . . . . . . . . . . . . . . . . . . . 134
2.2.4
Using two points to determine the equation of a straight line . . . . . . . . 136
2.2.5
Special case: b = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141
2.2.6
Special case: a = 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
2.2.7
Special case: Parallel to y-axis . . . . . . . . . . . . . . . . . . . . . . . . . 148
2.2.8
Special case: Parallel lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
2.2.9
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
Quadratic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156
2.3.1
Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157
2.3.2
Turning point – vertex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159
2.3.3
Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
2.3.4
Discriminant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163
2.4
2.3.5
Draw the graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
2.3.6
Slope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
2.3.7
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 175
Exponential and logarithmic functions . . . . . . . . . . . . . . . . . . . . . . . . . 177
2.4.1
Exponential function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177
2.4.2
Logarithmic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
2.4.3
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
3 Linear systems
3.1
3.2
3.3
3.4
Linear equations in one variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190
3.1.1
What is an equation in one variable? . . . . . . . . . . . . . . . . . . . . . . 190
3.1.2
Solving linear equations in one variable algebraically . . . . . . . . . . . . . 190
3.1.3
Solve by x-intercept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
3.1.4
Word problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
Systems of linear equations in two variables . . . . . . . . . . . . . . . . . . . . . . 199
3.2.1
Systems of equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
3.2.2
Solving systems of equations . . . . . . . . . . . . . . . . . . . . . . . . . . 200
3.2.3
Word problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205
3.2.4
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207
Linear inequalities in one variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
3.3.1
Introduction
3.3.2
Solving an inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
3.3.3
Miscellaneous . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212
3.3.4
Word problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214
3.3.5
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208
Systems of linear inequalities in two variables . . . . . . . . . . . . . . . . . . . . . 217
3.4.1
Linear inequalities in two variables . . . . . . . . . . . . . . . . . . . . . . . 217
3.4.2
Systems of two linear inequalities . . . . . . . . . . . . . . . . . . . . . . . . 219
3.4.3
Systems of more linear inequalities . . . . . . . . . . . . . . . . . . . . . . . 222
3.4.4
Word problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
3.4.5
No solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224
3.4.6
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225
4 Mathematics of finance
4.1
189
227
Simple interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228
4.1.1
How it works and formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . 228
4.1.2
Calculations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232
v
CONTENTS
4.1.3
4.2
Simple discount . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238
4.2.1
4.3
4.4
4.5
4.6
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244
Compound interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
4.3.1
Compound interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245
4.3.2
Compounding periods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248
4.3.3
Calculate P . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
4.3.4
Calculate n and i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252
4.3.5
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253
The time value of money . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255
4.4.1
Time value of money . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255
4.4.2
Replacing financial obligations . . . . . . . . . . . . . . . . . . . . . . . . . 261
4.4.3
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265
Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266
4.5.1
Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266
4.5.2
Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267
4.5.3
Future value S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269
4.5.4
Present value P . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273
4.5.5
The relationship between P and S . . . . . . . . . . . . . . . . . . . . . . . 276
4.5.6
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278
Amortisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279
4.6.1
Calculate payments
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279
4.6.2
The amortisation schedule . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281
4.6.3
Interest rate (i) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284
4.6.4
Period and payments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286
4.6.5
Present value (P) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290
4.6.6
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291
5 Collection, presentation and description of data
5.1
Statistics: An introduction
5.2
Data collection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295
5.3
vi
293
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
5.2.1
Simple random sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298
5.2.2
Stratified random sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . 300
5.2.3
Systematic sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
Presentations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305
5.3.1
Bar representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307
5.3.2
Frequency table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310
5.4
5.5
5.3.3
Pie chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315
5.3.4
Cumulative frequency . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318
5.3.5
Stem-and-leaf diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320
Measures of locality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
5.4.1
Introduction
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
5.4.2
The mean (raw data) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
5.4.3
The mean (interval data) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327
5.4.4
The median . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332
5.4.5
The mode . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335
5.4.6
Distribution of data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337
Measures of dispersion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339
5.5.1
Introduction
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339
5.5.2
Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340
5.5.3
Standard deviation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340
5.5.4
Quartiles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 348
5.5.5
Coefficient of variation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349
5.5.6
Box-and-whiskers diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352
6 An application of differentiation
357
6.1
Total, fixed and variable costs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358
6.2
Marginal cost . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360
6.3
Derivatives and slope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 368
6.4
Marginal profit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375
A Answers to exercises
383
Topic 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383
Topic 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400
Topic 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410
Topic 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417
Topic 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 431
Topic 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 437
vii
Topic
1
Revision of prior knowledge
The focus of this topic is on numbers and working with numbers, that is, basic numeracy skills
learnt at school level.
CONTENTS
Theme
1.1
Priorities
1.2
Variables
1.3
Laws of operations
1.4
Fractions
1.5
Powers
1.6
Roots
1.7
Ratios, proportions and percentages
1.8
Signs, notations and counting rules
1.9
Units and measures
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Theme 1.1
Priorities
Learning objective
On completion of this theme you should be able to apply
the priority rules in solving problems.
If you see the four symbols #, *, @ and , your first thought might be “social media”! These
symbols are used daily in the digital communication space, but they are also used daily in the
mathematical sciences.
Have a careful look at the four expressions below:
*
+
*
+
*
+
*
=
40
#
+
#
+
#
−
*
=
2
#
+
@
+
@
+
*
=
20
*
+
#
+
@
×
#
=
Your very first challenge in module QMI1500 is to figure out what numerical values the four
symbols #, *, @ and represent now.
If you find that = 26, browse through Topic 1 to refresh your memory, do Assignment 01,
submit it and proceed to Topic 2.
If you find that = “any number other than 26”, study Topic 1 carefully and in detail, work
through many examples and exercises, and only then do Assignment 01, submit it and proceed
to Topic 2.
1.1.1
Order of operations
What are the values of the expressions
2 × 3 + 4 × 5 and 8 ÷ 2 − 6 ÷ 3?
I hope you are able to answer 26 and 2 respectively, without using your calculator. How do you
derive these results? In the first case, 2 × 3 is equal to 6; 4 × 5 is 20; and 6 + 20 is 26. In the
second case 8 ÷ 2 is equal to 4; 6 ÷ 3 is 2; and 4 − 2 is 2.
This is elementary, but why?
It is essential that when we see an expression, we all understand the same thing and obtain the
same result, otherwise all our numbers and operations will lead to confusion. Therefore, we all
agree to perform the operations in a specific order.
Looking at the above two expressions and their accepted results, it is evident that multiplication
takes precedence over addition, and division takes precedence over subtraction. But what about
more complex expressions like
2 × 3 − 4 × 5 + 6 ÷ 3?
Here, too, multiplication and division take precedence over addition and subtraction, but whether
you do the multiplication before the division and the subtraction before the addition, or vice versa,
does not matter. Check for yourself.
The commonly accepted convention is thus: multiplication and division have a higher priority
than addition and subtraction.
2
1.1. PRIORITIES
You may wonder whether there are operations that have a higher priority than multiplication
and division. Yes, there are. Specifically, the operation “change sign”, which changes a positive number to a negative number and vice versa, has the highest priority of all. Next comes
exponentiation in its various forms.
Evaluating mathematical expressions can be a simple process, but mathematicians need to follow
an order of operations to get the correct answer. The following sequence describes the order you
follow to add, subtract, multiply and divide. The order is:
B. E. M. D. A. S
Brackets (B)
Exponents (E)
Multiplication (M)
& division (D)
Perform the operations inside the brackets first.
Then exponents.
Then multiplication and division, from left to right.
If more than one operation have the same priority, operate
from left to right.
× and ÷ have the same priority.
Addition (A)
& subtraction (S)
Then addition and subtraction, from left to right.
If more than one operation have the same priority, operate
from left to right.
+ and − have the same priority.
You can also create a little phrase to memorise the sequence:
Bongi Exercises Most Days After School.
At school you probably came across:
B. O. M. D. A. S or B. O. D. M. A. S,
which stands for
Brackets
Order (which represents exponents)
Multiplication
Division
Addition
Subtraction
This represents the same order of operations.
3
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
When you have more than one operation with the same priority, you just operate from left to
right. For instance,
15 ÷ 3 × 4
is not 15 ÷ 12, but is rather 5 × 4. This gives
5 × 4 = 20
because, going from left to right, you get to the division symbol first.
Video:
Watch the video “Priorities” to make sure that you understand the concept of the
order of operations.
Activity
Calculate the value of each of the following expressions. State the order in which the
operations are performed, assuming that you operate from left to right when priorities are
equal:
1. 4 × 3 ÷ 2 + 5 × 6 − 20 ÷ 4 × 2
√
2. 32 × 7 − 9 × 8 + 22 ÷ 4 × 3
Answer
Simplifying the expressions gives:
4 × 3 ÷ 2 + 5 × 6 − 20 ÷ 4 × 2 = 12 ÷ 2 + 30 − 5 × 2
1.
= 6 + 30 − 10
= 26
32 × 7 −
2.
√
9 × 8 + 22 ÷ 4 × 3 = 9 × 7 − 3 × 8 + 1 × 3
= 63 − 24 + 3
= 42
1.1.2
Brackets
Say, for instance, that in the example
2×3+4×5
you actually want to add 3 and 4 first and then multiply by 2 and 5. How can we achieve this?
The answer is to use brackets. We therefore write
2 × (3 + 4) × 5
and understand that the expression in brackets, namely (3 + 4), must be calculated first, followed
by the other operations. In this case we obtain, 3 + 4 is 7; 2 × 7 is 14; and 14 × 5 is 70.
4
1.1. PRIORITIES
Another example is the use of brackets to indicate fractional exponents, for example
2
8 3 is the same as 8(2÷3) .
Brackets are used to override the preconceived priorities.
It is evident that brackets are a very powerful means of casting expressions in a form that conveys
the specific meaning that we want. The point is that any expression between brackets is regarded
as a number that must be determined first before proceeding with the calculation.
Note that an expression itself may contain brackets. Does this mean that we can have brackets
within brackets? Yes, it does. That is, in fact, the beauty and the power of brackets. Let’s look
at a more complex example.
Consider
3×
(42 + 32 ) + (4 + 3)2 .
This means that you take the square root of the expression between the outermost brackets. However, this cannot be done until we know the value of the expression contained by the innermost
brackets. Thus, the innermost brackets are calculated first. Note that the usual priority rules
apply. For example, exponentiation is done before addition in the case of the first inner bracket.
Step by step, the calculation will take place as follows:
3×
Step 1
(42 + 32 ) + (4 + 3)2
Calculate the two sets of innermost
brackets first. It does not matter
which one is done first.
Step 2
3×
=
2
(16 + 9) + 7
Within a set of brackets, the usual
priority rules apply: in the first set,
addition happens after exponentiation.
Step 3
=
3×
√
25 + 72
Next do the exponentiation
√
1
25 = 25 2
and 72 .
Step 4
=
Step 5
=
Step 6
=
Step 7
=
(3 × 5 + 49)
Multiplication (3 × 5) is done before addition.
√
8
(15 + 49)
Addition.
64
Now we can take the square root.
Result of simplification.
The simplification of the mathematical expression therefore results in 8.
5
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Activity
Set out the steps that will be followed in the calculation of the expression below. Check
your result with your calculator:
42 − (6 × 3 − 80 ÷ 5)2 ÷ 92 − 77
Answer
The answer is
42 − (6 × 3 − 80 ÷ 5)2 ÷ 92 − 77
=
=
42 − (18 − 16)2 ÷ 92 − 77
42 − 22 ÷ 92 − 77
= (16 − 4) ÷ (81 − 77)
= 12 ÷ 4
= 3.
This exercise illustrates that any expression between brackets is simply a number that has to be
determined before you can proceed.
For every left bracket in an expression there must be a corresponding right bracket and vice
versa, otherwise the expression is not uniquely specified. Always work from the innermost bracket
outwards.
Lastly, a useful hint about brackets and priorities: If you are unsure about the relative priorities
of different operations, use brackets to enforce the priority order you want. A set or two of extra
brackets, even if they are redundant, will enhance the readability of an expression.
6
1.2. VARIABLES
Theme 1.2
Variables
Learning objective
On completion of this theme you should be able to express
relations between numbers in general terms, by making use of
symbols or letters.
If you want to explain to someone how to find the area of a rectangle 4 cm by 3 cm, you can tell
them to multiply 4 by 3. However, if you do not have specific values for the rectangle, you can
call the length of the rectangle L and its width W - then you can say that the area is L × W .
This will be true for any value of L and W. Consider the rule for expressing one quantity as a
percentage of another: “Put the first quantity on top of the second and multiply by 100.” What
a mouthful! However, if the first number is denoted by x and the second by y, then the rule boils
down to
x
× 100.
y
This is be true for any values of x and y. This is the great strength of using letters or symbols
to represent numbers – we are then able to write down rules, expressions and so on that are
completely general, so that to find the answer in a particular case all we need to do is substitute
our particular values of x and y, or whatever, into the appropriate expression.
The main purpose of using letters or symbols to represent numbers or quantities is that it enables
us to express practical truths about the real world neatly, succinctly and in more general terms
than we could if we insisted on sticking to definite numbers all the time.
Definition of a variable
A variable can be considered to be a box in which values can be placed.
Sometimes there is only one value that can be placed in the box and you have to determine what
that value is. At other times you get to pick a value that can be put into the box. These two
different situations are explained below.
Do you remember when you were in primary school and you were learning about addition? The
teacher would hand you worksheets with instructions like the following:
Fill in the box:
+4=6
Variables can be used to represent the box. Suppose we use an x to replace the box. Now we can
say:
Find the value of x for which
x + 4 = 6.
Why did we switch from boxes to letters? We did this because letters are more practical. For
instance, the formula for finding the area of a square is:
A = ×
= 2 .
The letter A represents the area and the letter the length of a side of the square.This formula
makes more sense than, for example,
= 2 .
The two formulas say exactly the same thing, but using A for area and for length is more useful
than using a circle and a triangle. Boxes are fine, but letters are better.
7
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Symbols
nnn
A variable is something that can assume any one of several numeric values.
The symbols most often used to represent variables are letters of the alphabet, both upper case
letters such as A, B and C, . . . and lower case letters such as x, y and z.
The symbol or letter that is used to represent a variable is simply a label. Its specific value has
to be assigned directly or be determined by calculation in each relevant instance.
Sometimes we choose a letter that reminds us of the quantity it represents, such as t for time, v
for voltage or b for bacteria. Always choose a symbol that is short and that makes sense.
How do we convert phrases to mathematical expressions?
Example 1
Convert the following statement to a mathematical expression:
“The sum of a number and 6.”
Solution
Step 1
“A number” means a variable. Pick a letter, for example x.
Step 2
“The sum” means you add.
Step 3
“And” means what you add comes next, namely 6.
Step 4
We can write this expression as x + 6.
So, the mathematical expression is
x + 6.
Example 2
Convert the following statement to a mathematical expression:
“The sum of a man’s age and half his son’s age.”
Solution
Step 1
There are two variables here, because we know neither the man’s age nor his
son’s. Let’s use different variables than x and y. Let’s call the man’s age
m and the son’s age s.
Step 2
The word “sum” means we are adding the two terms.
Step 3
The son’s age s needs to be divided by two before it gets into the expression.
Step 4
1
We can write this expression as m + s.
2
So, the mathematical expression is:
1
m + s.
2
8
1.2. VARIABLES
Exercise 1.1
1. Determine the numerical value of the following:
(a) 12x + 17
if x = 2
(b) x2 − 3
if x = 4
(c) 2x2
if x = 3
(d) 4x − 1
x+7
+3
(e)
4
(f) (x + 3) (x − 2)
if x = 5
(g) 10 − 3x
x x
+
(h)
2
3
if x = 1
if x = 5
if x = 6
if x = 12
2. Substitute x with 3 and calculate:
(a) 5x + 7
(b) x + 3x − 1
(c) 5x2 − 9
x+4
(d)
7
(e) 2 (x + 4)
(f) 7 − x + 2
3. If a = 2, b = 1 and c = 7, then determine the value of the following:
(a) 2(a + b − c) + c(b − a)
(b) a2 + b2 + c2
(c) (a + b)(b − c)
c+3
+ b2
(d) 2b −
a
4. Write the following as a mathematical expression:
(a) the sum of x and y
(b) subtract the sum of a and b from 8
(c) three times x added to two times y
(d) Robert’s age in seven years’ time if he is now y years old
9
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Theme 1.3
Laws of operations
Learning objective
On completion of this theme you should be able to apply
the laws of operations in solving problems.
There are specific laws that apply to combinations of the basic operations.
1.3.1
Commutative laws
Commutative law of addition
The sum of two numbers is unique. It does not matter which number
is placed first and which is placed second, the result is the same.
Suppose we use variables A and B. Then this law can be written as:
A+B =B+A
5+2 =2+5
Commutative law of multiplication
The product of two numbers is unique.
For variables A and B this law can be written as:
A×B =B×A
2×3 = 3×2
10
1.3. LAWS OF OPERATIONS
Examples
The two most basic laws are illustrated in the following results:
For addition:
6
+
2
=
2
+
6
(and both are equal to 8)
3,1
+
9,3
=
9,3
+
3,1
(= 12,4)
−7,6
+
2,3
=
2,3
+
−7,6
(= −5,3)
−11,1
+
−4,4
=
−4,4
+
−11,1
(= −15,5)
5
×
2
=
2
×
5
(= 10)
2,2
×
3,5
=
3,5
×
2,2
(= 7,7)
−8,1
×
9,4
=
9,4
×
−8,1
(= −76,14)
−11
×
−13
=
−13
×
−11
(= 143)
or
or
or
For multiplication:
or
or
or
What about subtraction and division? Does the commutative law apply to them too? A counter
example shows us that it does not, for example
7 − 3 = 3 − 7 and
10 ÷ 2 = 2 ÷ 10.
The sign = is read as “is not equal to”. In the first case, subtraction, commutation (i.e. changing
the order) leads to the negative of the initial result (i.e. −4 instead of 4); whereas in the second
case, division, commutation leads to the inverse of the initial result (i.e. 1 ÷ 5 instead of 5).
The conclusion is that, whereas for addition and multiplication we can change the order of the
factors, for subtraction and division we cannot - or rather, if we do, we must do so with care.
1.3.2
Associate laws
Suppose for two days you have worked five hours per day at R16 per hour. How much money did
you receive?
You have to calculate
2 × 5 × 16.
Using the associative law of multiplication, we obtain
2 × 5 × 16 = (2 × 5) × 16
= 10 × 16
= 160.
You therefore received R160.
Next we consider the associative laws that apply to addition and multiplication.
11
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Associate law of addition
The sum of three numbers does not depend on
which two are added first and the result is unique.
For variables A, B and C this law can be written as:
(A + B) + C = A + (B + C)
(5 + 2) + 3 = 5 + (2 + 3)
Associate law of multiplication
The product of three numbers does not depend on
which two are multiplied first and the result is unique.
For variables A, B and C this law can be written as:
(A × B) × C = A × (B × C)
(3 × 5) × 2 = 3 × (5 × 2)
Examples
Consider the following, where we use brackets to indicate the sequence in which the operations
are to be performed.
For addition:
12
7 + (3 + 2)
=
(7 + 3) + 2
(= 12)
6,1 + (−5,1 + 3,7)
=
(6,1 + −5,1) + 3,7
(= 4,7)
−9,3 + (2,2 + 4,5)
=
(−9,3 + 2,2) + 4,5
(= −2,6)
1.3. LAWS OF OPERATIONS
For multiplication:
5 × (4 × 2)
=
(5 × 4) × 2
(= 40)
6,1 × (7,3 × 5,5)
=
(6,1 × 7,3) × 5,5
(= 244,915)
−9 × (5 × −7)
=
(−9 × 5) × −7
(= 315)
This implies, for example, that when adding long columns of numbers, it does not matter which
numbers we add first and which last. A similar remark applies to multiplication, but when it
comes to subtraction and division, we must be very careful, as the following examples show:
10 − (5 − 2) = (10 − 5) − 2
and
(8 ÷ 4) ÷ 2 = 8 ÷ (4 ÷ 2)
This means that an expression like 8 ÷ 4 ÷ 2 is highly ambiguous because the result depends on
the order in which the operations are performed. It is therefore preferable that you indicate and
enforce a specific order using brackets.
1.3.3
Distributive law
So far, we have stated laws for expressions containing one type of operation only – either addition
or multiplication. Let’s take a look at combinations of these two operations.
Activity
Confirm if the expressions to the left and the right of the equal sign yield the same results:
1. 6 × (2 + 3) = 6 × 2 + 6 × 3
2. 7,2 × (2,2 + 3,3) = 7,2 × 2,2 + 7,2 × 3,3
3. −5 × (−4 + 6) = −5 × −4 + −5 × 6
4. 12,3 × (−3,4 + −6,6) = 12,3 × −3,4 + 12,3 × −6,6
What we observe in this activity is an illustration of the distributive law of multiplication over
addition.
13
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Distributive law of multiplication over addition
The product of a number with the sum of two other
numbers is equal to the sum of the products of the
first number with each of the other two numbers.
For variables A, B and C this law can be written as:
A × (B + C) = A × B + A × C
3 × (2 + 4) = 3 × 2 + 3 × 4
So, the 3 × can be distributed across the 2 + 4 into 3 × 2 and 3 × 4.
Examples
1. Sometimes it is easier to break up a difficult multiplication:
Solving 5 × 206 gives
5 × 206 = 5 × (200 + 6)
= 5 × 200 + 5 × 6
= 1 000 + 30
= 1 030.
2. Or it is easier to combine:
Solving 13 × 7 + 13 × 3 gives
13 × 7 + 13 × 3 = 13 × (7 + 3)
= 13 × 10
= 130.
3. You can use it with subtraction too:
Solving 36 × 8 − 34 × 8 gives
36 × 8 − 34 × 8 = (36 − 34) × 8
= 2×8
= 16.
14
1.3. LAWS OF OPERATIONS
1.3.4
Summary of laws
A+B = B+A
Commutative laws:
A×B = B×A
(A + B) + C = A + (B + C)
Associative laws:
Distributive law:
(A × B) × C = A × (B × C)
A × (B + C) = A × B + A × C
But be careful!
The commutative law does not work for subtraction and division. For example
7 − 3 = 3 − 7
10 ÷ 2 = 2 ÷ 10.
The equals sign with a line through is read as “does not equal”.
Note that we can transform expressions containing subtraction and division to ones containing
only addition and multiplication.
In the subtraction calculation, commutation (changing the order) leads to the negative of the
initial result, that is −4 instead of 4:
7 − 3 = 4, while 3 − 7 = −4
To be specific, we change the sign of the number being subtracted and then add, for example:
7 − 3 = 7 + −3 = −3 + 7 = 4
1
In the division calculation, commutation leads to the inverse of the initial result, that is instead
5
of 5:
1
10 ÷ 2 = 5, while 2 ÷ 10 =
5
We can replace the number being divided by with its inverse and multiply, for example:
10 ÷ 2 = 10 × 2−1 = 5
15
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
The associative law does not work for subtraction and division, for example:
10 − (5 − 2) = (10 − 5) − 2
(8 ÷ 4) ÷ 2 = 8 ÷ (4 ÷ 2)
We know that when you have more than one operation with the same priority, you just operate
from left to right. This means that an expression like 8 ÷ 4 ÷ 2 is highly ambiguous, because the
result depends on the order in which the operations are performed. If you want a specific order,
you must enforce that order using brackets.
The conclusion is that, whereas for addition and multiplication we can change the order of the
factors, for subtraction and division we cannot. Or rather, if we do, we must do so with care.
Identity properties
You must also take notice of the following four fundamental rules:
The product of any number with zero is zero:
A×0=0
The product of any number with one is the number:
A×1=A
The sum of any number and its negative is zero:
A + (−A) = 0
The product of any number and its inverse is one:
A × A−1 = 1
Exercise 1.2
1. Apply the distributive law of multiplication over addition to expand the following expression
to one containing the sum of four terms, each term being the product of three numbers. Do
not actually calculate the result.
The expression is 7 × (6 × (5 + 4) + 3 × (2 + 1)) .
2. Which of the commutative or associative laws are associated with the following expressions?
(a) 2 + (5 + 4) = (2 + 5) + 4
(b) (3 + 7) + 4 = (7 + 3) + 4
(c) (7 × 5) × 2 = 2 × (7 × 5)
(d) 2 × (7 × 4) = (4 × 2) × 7
16
1.4. FRACTIONS
Theme 1.4
Fractions
Learning objectives On completion of this theme you should be able to
• identify a proper fraction, improper fraction, mixed number,
equivalent fraction and a reciprocal of a fraction
• reduce fractions to their lowest terms
• divide and multiply fractions
• determine the place value in decimal numbers
• write a decimal number as a fraction
• write a fraction as a decimal number
• round numbers
1.4.1
Types of fractions
Terminology
In a fraction, the denominator tells us how many parts the whole is divided into and the numerator
tells us how many of those parts we are dealing with, for example:
3
4
numerator
denominator
Take note of some other important fraction terms to remember:
Proper fraction
Improper fraction
Mixed number/fraction
Equivalent fractions
Reciprocal
The numerator is less than the denominator,
2
1
for example and .
2
3
The numerator is greater than or equal to the denominator,
9
5
for example and .
3
9
A whole number and a fraction,
4
1
for example 2 and 5 .
3
9
Fractions that represent the same number,
2
4
for example and .
6
3
The multiplicative inverse of a number. For a fraction it
means “turn the fraction around”,
1
and 10.
for example
10
17
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Definition
One way to think of a fraction is as a division that has not been done yet. Why do we even use
fractions? Why do we not just divide the two numbers and use the decimal instead? In this day
of cheap calculators and cell phones, this is a very good question.
Fractions were invented long before decimal numbers as a way of showing portions less than one
- and they are still around. They are used in cooking, building, sewing, the stock market - they
are everywhere and we need to understand them.
Remember, the number above the bar is called the numerator and the number below the bar is
called the denominator:
numerator
3
4
denominator
We can read this fraction as “three fourths”, “three over four” or “three divided by four”.
Every fraction can be converted to a decimal by dividing. If you use a calculator to divide 3 by
4, you will find that it is equal to 0,75:
3
= 0,75
4
Remember, you can find the decimal equivalent of any fraction by dividing. Here are some other
fractions and their decimal equivalents:
1
= 0,5
2
1
= 0,25
4
1
= 0,125
8
Proper fractions
When the numerator is less than the denominator,
we call the expression a proper fraction.
3
4
2
3
1
9
97
100
Activity
At Bergville Secondary School, there are 14 female teachers and 11 male teachers. What
fraction of the teachers is male?
Answer
The numerator of the fraction is the number of male teachers. The denominator of the
fraction is the total number of teachers at the school.
The total number of teachers is
14 + 11 = 25.
The fraction of teachers that is male is
18
11
.
25
1.4. FRACTIONS
Improper fractions
An improper fraction occurs when the numerator
is greater than or equal to the denominator.
19
7
4
3
9
1
6
6
Mixed number
When a number consists of a whole number and a proper
fraction, we call it a mixed number or mixed fraction.
3
1
3
9
3
7
1
7
8
Conversions: Mixed number to improper fraction
We can convert a mixed number to an improper fraction, using the following method:
Step 1
Multiply the whole number by the denominator of the fraction.
Step 2
Add the numerator of the fraction to the product.
Step 3
Write the sum over the original denominator.
Example
Convert 1
1
to an improper fraction.
3
1
3
Start with the the mixed number.
1
Step 1
Multiply the number (1) by the denominator (3).
(1 × 3)
Step 2
Now add the numerator (1) to the product.
(1 × 3) + 1
Step 3
Write the sum over the original denominator (3).
4
3
1
2
3
4
4
1
Therefore, 1 = .
3
3
In this example, since three thirds are a whole, the whole number one is equal to three thirds.
Then add one more third to obtain four thirds.
19
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Activity
Change the mixed number 5
3
to an improper fraction.
4
Answer
Multiply the integer by the denominator and add the numerator:
5 × 4 = 20
Then add 3. Therefore,
5
3
4
=
=
(5 × 4) + 3
4
23
.
4
Conversions: Improper fraction to mixed number
We can convert an improper fraction to a mixed number using the following method:
Step 1
Divide the numerator by the denominator to get an integer and a remainder.
Step 2
Write down the integer and next to it the remainder over the original denominator.
Example
Convert
13
to a mixed number.
3
Step 1
Divide 13 by 3.
13
= 4, remainder 1
3
Step 2
Write the integer with the remainder over the denominator.
4
Therefore,
20
1
13
=4 .
3
3
1
3
1.4. FRACTIONS
Activity
Change the improper fraction
7
to a mixed number.
2
Answer
Divide the numerator by the denominator (to get the integer). The remainder is then the
new numerator:
7 ÷ 2 = 3,
with a remainder of 1. The integer part is the 3 and the remainder (the new numerator) is
1. Therefore,
1
7
=3 .
2
2
More examples of improper fractions and mixed numbers:
Improper fraction to mixed number
5
3
7
4
13
5
=
=
=
2
3
3
1
4
3
2
5
1
5 ÷ 3 = 1 with remainder of 2
7 ÷ 4 = 1 with remainder of 3
13 ÷ 5 = 2 with remainder of 3
Mixed number to improper fraction
2
3
3
1
4
3
2
5
1
=
=
=
5
3
7
4
13
5
[1 × 3 + 2 = 5]
[1 × 4 + 3 = 7]
[2 × 5 + 3 = 13]
Equivalent fractions
There are many ways to write a fraction. Fractions that
represent the same number are called equivalent fractions.
Example
4
2
and
4
8
are equivalent fractions. Equivalent fractions are basically the same as equivalent ratios. To find
out if two fractions are equivalent, use a calculator and divide. If the answer is the same, then
they are equivalent.
Activity
Determine the value of a in
16
a
= .
3
24
21
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Answer
The fraction
16
can be rewritten as
24
16 ÷ 8
2
16
=
= .
24
24 ÷ 8
3
16
, we divide the numerator (16) and denominator
To construct an equivalent fraction for
24
(24) by the same number, for example 8.
2
The fraction can be rewritten as
3
2
2×8
16
=
= .
3
3×8
24
2
To construct an equivalent fraction for , we multiply the numerator (2) and denominator
3
(3) by 8.
8
The number is equal to 1. It is clear that an equivalent fraction has the same value as
8
the original fraction, because it has been multiplied or divided by a value of 1.
Thus, if
16
a
= ,
3
24
then the value of a is equal to 2.
Reciprocal
When the product of two fractions equals one, the fractions are
reciprocals.
Every non-zero fraction has a reciprocal. It is easy to determine the reciprocal of a fraction,
because all you have to do is switch the numerator and denominator. That means, just turn the
fraction around.
Examples
3
To find the reciprocal of , just switch the numerator and denominator around. So the reciprocal
4
4
3
4
3
of is . We know and are reciprocals because
4
3
4
3
12
3 4
× =
= 1.
4 3
12
To find the reciprocal of a whole number, just put one over the whole number. For example, the
1
1
reciprocal of 2 is . We know 2 and are reciprocals because
2
2
2
2 1
× = = 1.
1 2
2
22
1.4. FRACTIONS
Activity
Find the reciprocal of
110
.
7
Answer
The fractions
7
110
and
are reciprocals because
7
110
7
110
×
= 1.
7
110
1.4.2
Reducing fractions
Greatest common factor (GCF)
We reduce a fraction to its lowest terms by finding an equivalent fraction
in which the numerator and denominator are as small as possible.
This means that when a fraction has been reduced, there is no number except one that can be
divided exactly into both the numerator and the denominator.
To reduce a fraction to its lowest terms, divide the numerator and denominator by their greatest
common factor (GCF).
The GCF is the largest number that divides
two or more numbers evenly.
This is also called simplifying the fraction.
Example
Simplifying the fraction
20
by reducing it to its lowest terms gives
60
20 ÷ 20
1
20
=
= .
60
60 ÷ 20
3
The GCF of 20 and 60 is 20.
A fraction is reduced to its lowest terms, or simplified, when its numerator and denominator
have no common factors. It is easier to multiply, divide, add and subtract fractions when they
are simplified. To simplify a fraction we find an equivalent fraction where the numerator and
denominator have no common factors. To do this we need to use prime factors.
23
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Prime factors
nnn
The prime factors of a number are the smallest divisors of that number.
For example, the prime factors of 12 are
12 = 2 × 2 × 3.
By smallest divisors we mean that a prime factor has no divisors other than itself and one:
2 can only be divided by itself, namely two, and one.
3 can only be divided by itself, namely three, and one.
Method to simplify (reduce) fractions
Follow these steps to simplify a fraction:
Step 1
Rewrite both the numerator and denominator as the product of its prime
factors respectively.
Step 2
Find the factors common to both the numerator and denominator (i.e. the
GCF), then divide by all the common factors. This is called cancelling.
Step 3
Find the products of the remaining factors, if there are more than one, in both the
numerator and the denominator.
Example
Reduce
6
to its lowest terms.
8
6
8
Simplify the fraction:
Step 1
Rewrite both the numerator and the
denominator as the product of their prime
=
2×3
2×2×2
=
2×3
2×2×2
=
3
4
factors.
Step 2
Divide by, or cancel, the factor of two that is
common to both the numerator and
denominator, that is the GCF. Since two
divided by two equals one, the twos cancel
each other out.
Step 3
You are now left with only a 3 in the
numerator. Multiply the remaining factors
in the denominator to get 4.
24
1.4. FRACTIONS
nActivity
Reduce
30
to its lowest terms.
45
Answer
Step 1
30
45
=
2×3×5
3×3×5
=
Step 2
Step 3
2× 3× 5
3× 3× 5
=
2
3
The steps are:
1.
Rewrite each of the numerator and denominator as the product of its prime factors.
2.
Cancel the common factors, 3 and 5.
3.
The only factors remaining are 2 in the numerator and 3 in the denominator,
2
therefore, the final answer is .
3
If you can see this beforehand, you can divide the numerator and denominator directly by
15:
30 ÷ 15
2
30
=
=
45
45 ÷ 15
3
Video:
Watch the video “Factors” on the prime factors of a number.
Revision
30
to its lowest terms.
Reduce
45
30
45
Simplify the fraction:
Step 1
Rewrite both the numerator and denominator as the
product of its prime factors. The prime factors of 30 are
=
2×3×5
3×3×5
=
2× 3× 5
3× 3× 5
=
2
3
2, 3 and 5. The prime factors of 45 are 3, 3 and 5.
Step 2
Find the factors common to both the numerator and denominator.
Then divide by all these common factors. This is called
cancelling. The greatest common factor of 30 and 45 is 15.
Step 3
Find the products of the remaining factors, if more than one, in
both the numerator and the denominator, respectively.
So, to simplify a fraction, we rewrite both the numerator and the denominator as the product of
their prime factors. The prime factors of 30 are 2, 3 and 5. The prime factors of 45 are 3, 3 and
5.
Then we divide the numerator and denominator by the greatest common factor (GCF). The GCF,
that is the largest integer that divides exactly into the numbers 30 and 45, is 15. We can also
say that we cancel out the 3 and 5 on top of and below the line.
25
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
1.4.3
Multiplication of fractions
Video:
Watch the video “Fraction_Part” on fractions as parts of a whole.
Video:
Watch the video “Fraction_Multiply” on multiplying fractions.
There are a few steps to follow when multiplying fractions. Let’s use the following example to
illustrate these steps.
Example
Calculate the following:
9
2 1
2 × ×
3 4 10
First method
Simplify the fractions by dividing a numerator and a denominator by the same number. Then
multiply the numerators together and multiply the denominators together.
9
2 1
2 × ×
3 4 10
Calculate the following:
Step 1
=
9
8 1
× ×
3 4 10
=
8
1 9 3
× ×
3 1 4 10
Note that 3 ÷ 3 = 1 and 9 ÷ 3 = 3.
=
3
8 1
× ×
1 4 10
Simplify again by dividing a numerator and a
=
1
3
8 2
×
×
1
4 1 10
4 ÷ 4 = 1 and 8 ÷ 4 = 2.
=
3
2 1
× ×
1 1 10
Simplify again by dividing a numerator and a
=
3
2 1 1
× ×
5
1
1 10
2 ÷ 2 = 1 and 10 ÷ 2 = 5.
=
1 1 3
× ×
1 1 5
Multiply the numerators and place the answer
=
1×1×3
1×1×5
=
3
5
Convert any mixed numbers to improper
fractions.
Step 2
Simplify by dividing a numerator and a
denominator by the same number.
Step 3
denominator by the same number. Note that
Step 4
denominator by the same number. Note that
Step 5
above the line. Then multiply the denominators
and place the answer below the line.
26
1.4. FRACTIONS
Second method
Alternatively you could simply multiply the numerators together and place the answer above the
line; then multiply the denominators together and place the answer below the line. Then simplify
the fraction.
9
2 1
2 × ×
3 4 10
Calculate the following:
Step 1
=
9
8 1
× ×
3 4 10
=
8×1×9
3 × 4 × 10
and place the answer below the line.
=
72
120
Simplify the fraction. Note that 72 and 120
=
72 ÷ 4
120 ÷ 4
=
18
30
=
18 ÷ 6
30 ÷ 6
=
3
5
Convert any mixed numbers to improper
fractions.
Step 2
Multiply the numerators and place the answer
above the line. Then multiply the denominators
Step 3
are both divisible by 4.
Step 4
Simplify the fraction again. Note that 18 and 30
are both divisible by 6.
Activity
Calculate the following:
15
5
×
12 45
21
2. 7 ×
54
13
4
×2
3. 2 ×
34
16
1.
27
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Answer
Simplifying the expressions gives:
1.
5
15
5 1
15
5
×
=
×
4
9
12 45 12
45
2.
3.
7×
2×
=
1 5
×
4 9
=
5
36
7 21
21
= ×
54
1 54
147
=
54
147 ÷ 3
=
54 ÷ 3
49
=
18
13
=2
18
13
2
13
4
×2
=2×
×2
34
16
17
16
36 45
×
=
17 16
=
9
36
17
405
68
65
=5
68
=
28
Simplify: 5 ÷ 5 = 1,
45 ÷ 5 = 9,
12 ÷ 3 = 4,
15 ÷ 3 = 5.
×
45
4
16
Simplify: 36 ÷ 4 = 9,
16 ÷ 4 = 4.
1.4. FRACTIONS
1.4.4
Division of fractions
Video:
Watch the video “Fraction_Divide1” on dividing fractions.
Video:
Watch the video “Fraction_Divide2” on dividing fractions.
Activity
Calculate the following:
1.
2 4
÷
9 5
2.
15
5
÷2
15
20
3.
4
16
÷
24 20
4.
10
37
÷1
42
14
5.
22
÷3
25
Answer
Simplifying the expressions gives:
1.
2 4
÷
9 5
=
=
=
=
2 5
×
9 4
5
2 1
×
9
4 2
1 5
×
9 2
5
18
2.
15
5
÷2
15
20
=
=
=
=
1
3
÷2
3
4
1 11
÷
3
4
4
1
×
3 11
4
33
29
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
3.
4
16
÷
24 20
2 1
÷
3 5
2 5
×
=
3 1
10
=
3
1
= 3
3
=
4.
10
37
÷1
42
14
=
=
=
=
=
37
5
÷1
42
7
37 12
÷
42
7
7
37
×
42 12
37
7 1
×
6
42
12
37
72
5.
22
÷3 =
25
=
=
30
22 3
÷
25 1
22 1
×
25 3
22
75
1.4. FRACTIONS
Multiplying and dividing fractions
Activity
Calculate
4
5
4
2 ×1 ÷
5
21 6
Answer
4
5
4
2 ×1 ÷
5
21 6
Simplify the expression.
Rewrite all the mixed numbers as
=
14 25 5
×
÷
5
21 6
=
14 25 6
×
×
5
21 5
improper fractions.
Change ÷
6
5
to × .
6
5
Divide 14 and 21 by 7:
14 25
× .
5
21
=
2 25 6
×
×
5
3
5
Divide 5 and 25 by 5:
2 25
× .
5
3
=
2 5 6
× ×
1 3 5
Divide 5 and 5 by 5
and 6 and 3 by 3:
5 6
× .
3 5
Multiply the numerators and place the
answer above the line. Then multiply
the denominators and place the answer
below the line.
=
=
=
2 5 6
× ×
1 3 5
2 1 2
× ×
1 1 1
4
31
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
1.4.5
Adding fractions
Video:
Watch the video “Fraction_AddYT” on adding fractions.
The following example illustrates the method of addition and subtraction of fractions.
Example
Calculate the following:
5
1
3
5 +2 −7
4
6
3
Method
5
1
3
5 +2 −7
4
6
3
Calculate the following:
Step 1
Convert any mixed numbers to improper
=
23 17 22
+
−
4
6
3
=
23 × 3 17 × 2 22 × 4
+
−
4×3
6×2
3×4
=
69 34 88
+
−
12 12 12
fractions.
Step 2
Change each fraction into an equivalent fraction
so that all the fractions have the same
denominator. The number 12 is the smallest
number divisible by the denominators 4, 6 and 3.
Step 3
Add or subtract the numerators.
=
=
Step 4
=
15 ÷ 3
12 ÷ 3
in this case 3.
=
5
4
Write the improper fraction as a mixed number.
=
1
Simplify the fraction by dividing both the
numerator and denominator by the GCF,
Step 5
32
69 + 34 − 88
12
15
12
1
4
1.4. FRACTIONS
nnn
Activity
Calculate the following:
1. Add
2
5
and
.
6
30
2. Calculate the sum of 11
3. Calculate
12
7
and
.
12
30
34
6
+7 .
88
64
Answer
Simplifying the expressions gives:
1.
5
2
+
6 30
=
=
=
2.
11
12
7
+
12 30
7
12
+
12 30
12 × 2
7×5
+
11 +
12 × 5 30 × 2
35 24
+
11 +
60 60
59
11 +
60
59
11
60
= 11 +
=
=
=
=
3.
25
2
+
30 30
27
30
9
10
34
3
17
6
+7 =
+7+
88
64
44
32
17 × 44
3 × 32
+
=7+
44 × 32 32 × 44
748
96
+
=7+
1 408 1 408
=7+
844
1 408
Simplify: ÷4,
÷4.
211
352
211
=7
352
=7+
33
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
1.4.6
Subtracting fractions
Video:
Watch the video “Fraction_Sub” on subtracting fractions.
Activity
Calculate the following:
1. Subtract
3
15
from
.
40
16
2. Determine the difference between 14
3. Calculate 12
58
74
and 8
.
90
120
1
10
−8 .
30
12
Answer
Simplifying the expressions gives:
1.
15
3
−
16 40
2.
14
74
58
−8
90
120
3
3
−
16 8
6
3
−
=
16 16
3
= −
16
=
= 14
=
=
=
=
=
34
29
37
−8
45
60
29 37
−
(14 − 8) +
45 60
29 × 4 37 × 3
−
6+
45 × 4 60 × 3
116 − 111
6+
180
5
6+
180
1
6
36
1.4. FRACTIONS
3.
12
1
10
−8
30
12
1
1
= 12 − 8
3
12
=
=
=
=
=
1
1
−
(12 − 8) +
3 12
1×4
1
−
4+
3 × 4 12
4−1
4+
12
3
4
12
1
4
4
Adding and subtracting fractions
Activity
Calculate
1
3
3
3 +2 −1 .
5
10
10
Answer
1
3
3
3 +2 −1
5
10
10
Simplify the expression.
Rewrite all the mixed numbers as
=
18 21 13
+
−
5
10 10
=
36 21 13
+
−
10 10 10
=
44
10
=
22
5
=
4
improper fractions.
Rewrite the first fraction as a
fraction with a denominator of 10.
Add/subtract the numerators of
the three fractions.
Simplify the fraction by dividing the
numerator and denominator by 2.
Rewrite the improper fraction as
a mixed number.
2
5
35
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
1.4.7
n
Decimals
A decimal number, based on the number 10, contains a decimal comma.
The word “decimal” really means “based on 10”
(from the Latin word decima meaning “a tenth part”).
Place value
To understand decimal numbers, you must first know about place value.
When we write numbers, the position or placement of each digit is important.
In the number 634:
• the “4” is in the units position, meaning just 4 (or four “1”s),
• the “3” is in the tens position, meaning 3 tens (or thirty),
• and the “6” is in the hundreds position, meaning 6 hundreds (or six hundred).
As we move left, each position is 10 times bigger!
n
From units, to tens, to hundreds.
n
. . . and . . .
n
as we move right, each position is 10 times smaller!
n
From hundreds, to tens, to units.
But what if we continue past units as we move right?
36
1.4. FRACTIONS
What is 10 times smaller than units?
We must write a decimal comma so that we know exactly where the units are positioned.
n
This can be illustrated as follows with the number 634,5:
Now we can say “six hundred and thirty four and five tenths”
but we usually just say
“six hundred and thirty four comma five”.
Decimal comma
The decimal comma is the most important part of a decimal number. It is exactly to the right
of the units position. Without it we would be lost because we would not know what each position
meant.
Now we can continue with smaller and smaller values, from tenths to hundredths and so on.
This can be illustrated as follows with the number 18,692:
37
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Representation of decimal numbers
A decimal number can be written as a whole number plus tenths, hundredths and so forth.
For example, what is 3,24?
• On the left side of the decimal comma is 3, that is the whole number part.
• There are two digits on the right side, the 2 is in the “tenths” position,
2
, and the 4 is in the “hundredths” position,
meaning “2 tenths” or
10
4
.
meaning “4 hundredths” or
100
• So, 3,24 is “3 and 2 tenths and 4 hundredths”.
Written as a decimal fraction
A decimal number can also be written as a decimal fraction.
A decimal fraction is a fraction where the denominator
(the bottom number) is a number such as 10, 100, 1 000
and so forth, in other words a power of ten.
324
.
100
24 752
.
• The decimal number 24,752 can be written as
1 000
• The decimal number 3,24 can be written as
Written as a whole number and a decimal fraction
We can also think of a decimal number as a whole number plus a decimal fraction.
24
24
or 3
.
100
100
752
752
or 24
.
• The decimal number 24,752 can be written as 24 and
1 000
1 000
• The decimal number 3,24 can be written as 3 and
Examples
Consider the following decimal fractions:
1
= 0,1
10 has one zero, therefore one digit after the decimal comma.
10
1
= 0,01
100 has two zeros, therefore two digits after the decimal comma.
100
1
= 0,001
1 000 has three zeros, therefore three digits after the decimal comma.
1 000
1
= 0,0001
10 000 has four zeros, therefore four digits after the decimal comma.
10 000
1
= 0,00001
100 000 has five zeros, therefore five digits after the decimal comma.
100 000
38
1.4. FRACTIONS
Convert decimals to fractions
Follow the following steps to convert a decimal to a fraction:
decimal
.
1
Step 1
Write down the decimal divided by one, like this:
Step 2
Multiply both the numerator and denominator (top and bottom) by 10 for
every number after the decimal comma.
For example, if there are two numbers after the decimal comma, use 100;
if there are three, then use 1 000, and so on.
Step 3
Simplify (or reduce) the fraction.
Example
Express 0,625 as a fraction.
Step 1
Write down the decimal divided by one.
=
0,625
1
Step 2
Multiply both the numerator and the
=
0,625 1 000
×
1
1 000
=
625
1 000
=
625 ÷ 25
1 000 ÷ 25
factor 25.
=
25
40
Simplify the fraction again by dividing both the
=
25 ÷ 5
40 ÷ 5
=
5
8
denominator by 1 000. There were three
digits after the decimal comma, so that is
10 × 10 × 10 = 1 000.
Step 3
Simplify the fraction by dividing both the
numerator and denominator by the common
numerator and denominator by the common
factor 5.
Activity
Convert the decimal number 7,65 to a fraction.
Answer
Write 7,65 as a whole number plus a decimal number, which is less than one:
7,65 = 7 + 0,65.
39
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Write the decimal number as a fraction. Multiply both the numerator and denominator of
the fraction by 100 (there were 2 digits after the decimal comma so that is 10 × 10 = 100):
65
0,65 100
×
=7+
.
7+
100
100
100
Simplify the fraction
13
65 ÷ 5
=7 .
7+
100 ÷ 5
20
Convert fractions to decimals
Follow these steps to convert a fraction to a decimal:
Step 1
Find a number you can multiply the denominator (bottom) of the fraction by
to make it 10, 100, 1 000, or any 1 followed by zeros.
Step 2
Multiply both the numerator and the denominator (top and bottom) by that
number.
Step 3
Then write down just the numerator (top number), putting the decimal
comma in the correct place, that is one space from the right-hand side for
every zero in the denominator (bottom number).
Example
Express
3
as a decimal.
16
Step 1
We have to multiply the denominator 16 by 625 for it
to become 10 000.
Step 2
Multiply both the numerator and the denominator
16 × 625 = 10 000
3 × 625
16 × 625
(top and bottom) by 625.
Step 3
Write down the numerator (top number) 1 875 with
the decimal comma four spaces from the
right, because 10 000 has four zeros.
The fraction
=
1 875
10 000
=
0,1875
3
expressed as a decimal is therefore 0,1875.
16
Calculator method
The example above has been given to explain the mathematics behind the scenes to you. The
good news is that there is an easier way to express a fraction as a decimal number: just use your
calculator. Using your calculator, key in
3 ÷ 16 =
and 0,1875 will be displayed.
40
1.4. FRACTIONS
nnn
Activity
Convert the fraction
12
to a decimal number.
50
Answer
Converting
12
to a decimal number gives
50
12
= 12 ÷ 50 = 0,24.
50
You could also follow these steps:
Step 1
Multiply the denominator 50 by 2 for it
to become 100.
Step 2
Multiply both the numerator and the denominator
50 × 2 = 100
12 × 2
50 × 2
(top and bottom) by 2.
Step 3
Write down the numerator (top number) 24 with
the decimal comma two spaces from the
right, because 100 has two zeros.
=
24
100
=
0,24
Rounding
Rounding means reducing the digits in a number while trying to keep
its value similar. The result is less accurate, but easier to use.
There are various methods of rounding. Follow these steps to round a number according to the
“round half up” method commonly used:
Step 1
Decide which is the last digit to be kept.
Step 2
Leave it the same if the next digit is less than 5. This is called rounding down.
Increase it by one if the next digit is 5 or more. This is called rounding up.
Step 3
If necessary, replace removed digits with zeros, for example when rounding
whole numbers.
41
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Example of rounding whole numbers
n
Round 74 to the nearest 10.
Step 1
We want to keep the 7 as it is in the tens’ position.
Step 2
The next digit is 4, which is less than 5, so no change
is needed to 7.
Step 3
74
≈
Replace the removed digits with zeros, so the 4 is
replaced by a 0.
74
70
70
So, 74 was rounded down and the answer is 70.
Example of rounding to units
Round 8,6 to the nearest unit.
Step 1
We want to keep the 8 as it is in the units position.
8,6
Step 2
The next digit is 6, which is 5 or more, so increase the
8 by one to 9.
8,6
9
Step 3
Since the removed digits came after the decimal
comma, it is not necessary to replace them with zeros.
So, 8,6 was rounded up and the answer is 9.
Examples of rounding to different digits
Rounding
Rounding
238,7248
238,7248
to the nearest hundredth gives
to the nearest tenth gives
238,72.
238,7.
Rounding
Rounding
238,7248
238,7248
to the nearest unit gives
to the nearest ten give
239.
240.
Rounding
Rounding
238,7248
238,7248
to three decimal places gives
to hundred gives
238,725.
200.
Activity
Round 14,327 off to the nearest tenth (one decimal place).
Answer
Rounding 14,327 off to the nearest tenth gives
14,327 ≈ 14,3.
42
≈
9
1.4. FRACTIONS
Non-terminating decimal
22
to a decimal number and round off your answer to three decimal places.
7
Convert the fraction
The fraction
22
converted to a decimal number is 3,1428571 . . .
7
and 3,1428571 · · · ≈ 3,143.
The symbol ≈ is read as “approximately equal to”.
The decimal 3,1428571 . . . is an example of a non-terminating decimal
because the answer goes on and on and is never completed.
Terminating decimal
3
to a decimal number and round off your answer to one decimal place.
8
Convert the fraction
The fraction
3
converted to a decimal number is 0,375
8
and 0,375 ≈ 0,4.
The decimal 0,375 is an example of a terminating decimal because it
is exact and complete.
Recurring decimal
Convert the fraction
7
to a decimal number and round off your answer to two decimal places.
11
The fraction
7
converted to a decimal number is 0,636363 . . .
11
and 0,636363 . . . ≈ 0,64.
The decimal 0,636363 . . . is an example of a recurring decimal
because there are repeating digits.
This decimal 0,636363 . . . can also be written as 0,6̇3̇ and is pronounced as “nought comma six
three recurring”. The dots on the 6 and the 3 tell us that the 6 and the 3 are recurring.
The above recurring decimal can also be written as 0,63.
43
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Exercise 1.3
1.
2.
Calculate the answers to the following expressions:
8
1
4
12
1
5
3.
8
13
12
25
4.
3 12
7
5.
8 14
1
2
6.
17 1 3
+ −
20 4 5
7.
6
2
+5−
3
7
2
7
1
8. 5 + 3 − 6
2
5
12
9.
44
5 1
3
÷ 1 −
4
6 2
+
3
5
1.5. POWERS
Theme 1.5
Powers
Learning objective
On completion of this theme you should be able to solve problems
containing powers.
Although powers are not as much part of our everyday lives as percentages, they are an extremely
important concept of mathematics.
1.5.1
Positive powers
Start with multiplying 10 by itself:
10 × 10 = 100
10 × 10 × 10 = 1 000
10 × 10 × 10 × 10 = 10 000
It seems a very long way of doing it. Fortunately a shorter notation has been developed:
10 × 10 = 102 = 100
(10 is multiplied 2 times by itself)
10 × 10 × 10 = 103 = 1 000
(10 is multiplied 3 times by itself)
10 × 10 × 10 × 10 = 104 = 10 000
(10 is multiplied 4 times by itself)
The number of times that 10 is multiplied by itself is the power of 10 involved. For example, 103
is pronounced as “ten to the power three”.
Note that 10 = 101 or 10 to the power one. And what about 100 or 10 to the power zero? By
definition, any number to the power zero is one. So
100 = 1, 0x = 0 for x = 0 and 00 = 1.
Consider 103 again, where 10 is called the base and 3 is called the exponent.
Video:
Watch the video “Powers” on powers and exponents.
Activity
Which one is the base and which one is the exponent in the following?
64
;
32
;
108
Answer
For 64 : 6 is the base and 4 is the exponent. It is read as “six to the power four”.
For 32 : 3 is the base and 2 is the exponent. It is read as “three to the power two”.
For 108 : 10 is the base and 8 is the exponent. It is read as “ten to the power eight”.
45
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Product property
The calculation
100 × 100 = 10 000
can also be written as
102 × 102 = 104 .
If the bases are the same, we can just add the exponents to get the answers. But beware! You
cannot do this if the bases are different.
Activity
Using 23 × 32 as an example, decide whether the above statement is correct or not.
Answer
Is
23 × 32 = 23+2 = 25 = 32 or 33+2 = 35 = 243?
We have that
23 = 2 × 2 × 2 = 8
and
32 = 3 × 3 = 9.
Therefore
23 × 32 = 8 × 9 = 72.
It is clear that you can only
add exponents if the bases are the same.
This can also be illustrated if variables are used instead of actual numbers. For example
x5 × x4 = x5+4
= x9
and
x3 y 5 × x2 y 3 = x3+2 y 5+3
= x5 y 8 .
We can also use actual numbers and variables in the same calculation:
4abc × 3a2 bc × 2a3 b2 c = 4 × 3 × 2 × a1+2+3 × b1+1+2 × c1+1+1
= 24a6 b4 c3 .
Simplifying
2 x2
4 w3
3 x7
2
w
gives
2 x2
x2
x2
x2
w3
= 2x2+2+2+2+7+7 w3+3+3+1
= 2x22 w10 .
46
w3
w3
x7
x7 w
1.5. POWERS
Another way to do this calculation, is
2 x2
4 w3
3 x7
2
w
= 2x2×4 w3×3 x7×2 w
= 2x8 w9 x14 w
= 2x8+14 w9+1
= 2x22 w10 .
Activity
Calculate the value of the following:
3
1
2. 5p2 q 4 p3
6
1. 2a 4 × 3a 4
q2
4
Answer
Simplifying gives the following:
3
3+1
4
1
2a 4 × 3a 4
1.
= 2 × 3 × a4
4
= 6a 4
2.
2 4
5p q
p
3
6 = 6a
q
2
4
= 5p2 q 4 p3×6 q 2×4
= 5p2 q 4 p18 q 8
= 5p2+18 q 4+8
= 5p20 q 12
1.5.2
Negative powers
What is the meaning of a negative power?
If 102 = 100, what is 10−2 ?
To deal with this, we introduce the concept of the inverse or reciprocal of a number. The
inverse or reciprocal of a number is the result obtained when 1 is divided by that number.
1
.
The inverse of 10 is
10
If
10 = 101 ,
then
1
1
10−1
10−1
10−1
1
= 1 = 1 × −1 =
= 10−1 .
=
10
10
10
10
100
1
The superscript −1 is used to indicate the inverse of a number.
Therefore
1
1
10−2
10−2
10−2
10−2
1
= 2 = 2 × −2 = 2
= 10−2 .
=
=
100
10
10
10
10 × 10−2
100
1
47
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Activity
Derive a rule to find the inverse of any number written as a base number with an exponent.
Answer
If
10 = 10+1 ,
then
1
= 10−1 .
10
If
100 = 10+2 ,
then
1
= 10−2 .
100
Then for any number written as a base with an exponent, the inverse is found by using
the same base, but changing the sign of the exponent.
Let’s look at a few examples:
7−2
10−3
=
=
10ab4
5a2 b2
103
72
1 000
49
= 2ab4 a−2 b−2
= 2a1−2 b4−2
= 2a−1 b2
2b2
=
a
or
10ab4
5a2 b2
=
=
Activity
Calculate the value of the following:
1. 2 2−3
2.
48
43x × 2x+2
82x
10abbbb
5aabb
2b2
a
1.5. POWERS
Answer
Simplifying gives the following:
2 2−3
1.
2 2−3
or
1
23
1
= 2
8
1
=
4
= 2
= 21 × 2−3
= 21−3
= 2−2
1
=
4
2.
43x × 2x+2
82x
× 2x × 22
(23 )2x
26x × 2x × 22
=
26x
6x
= 2 × 2x × 22 × 2−6x
=
22
3x
= 26x+x−6x × 22
= 2x × 22
= 4 × 2x
1.5.3
Rules and properties
Rules for exponents
1. Any number raised to the power of one simply equals that number:
21 = 2
2. Any number raised to the power zero equals one:
20 = 1
3. Numbers with bases and exponents can only be added or subtracted if they have the same
base and the same exponent:
6y 3 + 2y 3 = 8y 3 ,
6y 3 − 2y 3 = 4y 3 ,
but
6y 3 + 2y 4 = 6y 3 + 2y 4 .
49
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
4. If the bases are the same, then numbers are multiplied by adding the exponents:
23 × 24 = 27
5. If the bases are the same, then numbers are divided by subtracting the exponents:
25
23
= 25−3
= 22
6. A negative exponent is equivalent to the reciprocal of that number:
2−5 =
1
25
7. When a number with a base and an exponent is raised to a power, the exponents are
multiplied:
(y a )b = y ab
x3 y 4
2
= x6 y 8
Properties summary
Below is a list of the properties of exponents:
am × an = a(m+n) = am+n
(am )n = a(m×n) = amn
(ab)n = an bn
n
a
b
=
xa
xb
an
bn
= xa−b
a−n =
1
, a = 0
an
x−a
y −b
yb
,
xa
=
a0 = 1,
a1 = a
50
x = 0, y = 0
a = 0
1.5. POWERS
Exercise 1.4
1. Determine the value of the following:
(a) 23
(b) 42
(c) 3−1
(d) 5−3
(e) 32 × 30
(f) 23 × 22
(g) 32 × 42
(h) 5−1 × 52
51
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Theme 1.6
Roots
Learning objective
1.6.1
On completion of this theme you should be able to solve problems
containing roots.
Square root
We know that 102 = 10 × 10 = 100 and we say “10 to the power 2 is 100”. What about going the
other way, that is, what number raised to the power 2 is 100? This problem is called determining
the root of a number.
We reverse the power story by using a root sign
102 = 100
√
2
100 = 10
and 10 is called the “square root” of 100.
Most calculators have a square root key. Convention has it that the little 2 is not written - only
√
sign is required.
the
Activity
What is the square root of 9?
How would you describe the square root of 9 in words?
Answer
The answer is
√
9 = 3.
In words we say, “three is the number when raised to the power two that will give an answer
of nine”, that is
32 = 9.
√
NOTE! The
sign is taken away by squaring the 3. How does this work?
1
√
We see that 9 is actually 9 2 (or 9 to the power half). Then
√
9 = 3
1
92
9
1×2
2 1
= 3
2
= 31
9 = 32 .
What about something like
√
25 − 16 and
√
16 + 9?
Firstly
√
25 − 16 =
√
= 3.
52
9
1.6. ROOTS
Be careful:
√
√
25 = 5
16 = 4
5 − 4 = 1,
which is not the correct answer as obtained above (3 = 1).
Secondly
√
16 + 9 =
√
25
= 5.
We also see that
√
16 = 4
√
9 = 3
4 + 3 = 7,
which is also not the correct answer as obtained above (5 = 7).
The moral of the story is:
First finish all the calculations inside the square root before taking the
square root away by squaring.
At this time, it goes without saying that the square root of a number, x, is that number, r,
which, when squared, becomes x:
r2 = x
Consider
62 = 36.
Then
√
6×6 =
√
36
1
= 36 2
= 6.
In this case, x is the number, namely 36, and r is the square root, namely 6.
Video:
Watch the video “Square roots(1)” on square roots and perfect squares.
Video:
Watch the video “Square roots(2)” on square roots and perfect squares.
If you still find these challenging, try to visualise what happens if we apply the square root.
53
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Visualisation
When a number is a perfect square, the square root of that number can easily be determined:
Make two similar groups
Take one of the similar
of factors.
groups out of the root
to obtain the answer.
If a calculator is not used and you are calculating square roots of numbers that are not perfect
squares, part of the answer can be given as a root:
Make two similar groups
Take one of the similar groups
and a remainder.
out of the root. The remainder
is left under the root.
Properties
The properties of square roots are:
√
√
√
ab =
a× b
√
a
√
b
√
a×
√
=
a
b
for a ≥ 0 and b ≥ 0
for a ≥ 0 and b > 0
a = a
√
√
ab = a × b only applies to numbers in roots that are multiplied.
√
√
√
√
√
√
a − b = a − b.
Remember that a + b = a + b and
The property
1.6.2
√
Cube and higher order roots
Cube root
The cube root of a number, x, is a number, r, whose cube is x:
r3 = x
Consider
23 = 8.
54
1.6. ROOTS
Then
√
3
2×2×2 =
√
3
8
1
= 83
= 2.
In this case, x is the number, namely 8, and r is the cube root, namely 2.
Video:
Watch the video “Cube roots(1)” on cube roots.
Video:
Watch the video “Cube roots(2)” on cube roots.
The nth root
The nth root of a number, x, is a number, r that, when raised to the power n is x:
rn = x
Consider the case where n = 5:
25 = 32.
Then
√
5
2×2×2×2×2 =
√
5
32
1
= 32 5
= 2.
In this case, x is the number, namely 32, and r is the fifth root, namely 2.
Activity
Determine the following and describe each answer in words:
1.
2.
3.
4.
√
√
16
25
√
4
√
3
16
125
55
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Answer
Simplifying gives the following:
1.
√
16 = 4,
that is, 4 raised to the power two is 16.
2.
√
25 = 5,
that is, 5 raised to the power two is 25.
3.
√
4
16 = 2,
that is, 2 raised to the power four is 16 (24 = 2 × 2 × 2 × 2 = 16).
4.
√
3
125 = 5,
that is, 5 raised to the power three is 125 (53 = 5 × 5 × 5 = 125).
1.6.3
Fractional exponents
Another way to write roots is to write them as fractional exponents.
A fractional exponent is a way to write a root without the root sign. Then the expression is
evaluated using the exponent rules.
1
For example, the fourth root is denoted by an exponent of
4
1
and the fifth root is denoted by an exponent of .
5
Every root is denoted by an exponent with 1 in the numerator and the order of the root in the
denominator.
A fractional exponent like
1
means to take the nth root:
n
A number to the power of a half is the square root:
√
n
√
1
x = xn
1
x = x2
A number to the power of a third is the cube root:
√
3
x = x3
A number to the power of a seventh is the seventh root:
√
7
x = x7
56
1
1
1.6. ROOTS
Examples
nnn
Write the following two roots as fractional exponents:
1.
3
x9 y 6
2.
4
x4 y 8
Writing these roots as fractional exponents gives:
3
1. n
x9 y 6
=
9 6
x y
9×1
3
= x1
= x3 y 2
4
2. n
x4 y 8
=
4 8
x y
4×1
4
= x1
13
6×1
3
y1
14
8×1
4
y1
= xy 2
A fractional exponent with one in the numerator is a root of some sort, for example a number to
the power of a third is the cube root.
2
But what would an exponent of mean?
3
2
The number y 3 can also be written as 3 y 2 .
What would an exponent of
5
mean?
2
5
The number x 2 can also be written as
√
x5 .
In a fractional exponent, the numerator is the power to which the number should be taken and
the denominator is the root that should be taken.
Example
2
Calculate 64 3 .
2
The number 64 3 means you need to square 64 and take the cube root of the result:
2
64 3
=
=
=
√
3
√
3
√
3
642
4 096
16 × 16 × 16
= 16
57
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Exercise 1.5
1. Calculate:
√
√
(a) 196 + 144
√ 2
5
(b)
(c)
(d)
√
64
2
√ 2
1
2. Simplify:
√
(a) 100 ÷ 4
36
4
√
(c) 2 × 144
√
(d) 25 − 4
(b)
3. Calculate:
√
(a) 18
√
45
(b) −2
3
(c)
58
9 − x2
x2
1.7. RATIOS, PROPORTIONS AND PERCENTAGES
Theme 1.7
Ratios, proportions and percentages
Learning objective
On completion of this theme you should be able to relate numbers
to each other using ratios, proportions and percentages.
In our day-to-day lives we are constantly confronted with situations in which we have to compare
numbers: “One out of every three South Africans prefers sugar-free soft drinks”, “Twenty percent
of our national budget is spent on education and welfare”, and so on.
1.7.1
Ratios
Writing ratios
A ratio says how much of one thing there is compared to another thing.
4 : 1
4 squares : 1 square
In this case, 4 : 1 means there are four gray squares to one white square.
A ratio can be written in different ways. Note the following three ways:
Use the colon (:) to separate the values:
4 : 1
Use the word to instead of the colon:
4 to 1
Write the ratio as a fraction:
4
1
Activity
Write the ratios of each of the following pairs of numbers in each of the three ways:
7 and 16;
22 and 11;
9 and 12.
Answer
The ratios, written in each of the three ways, are the following:
7 to 16
or
7 : 16
or
22 to 11
or
22 : 11
or
9 to 12
or
9 : 12
or
7
16
22
11
9
21
59
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Scaling ratios
A ratio can be scaled up. See below how 4 : 1 is scaled up.
4 : 1
16 squares : 4 squares
4 big squares : 1 big square
Here the ratio is still four gray squares to one white square, even though there are more squares.
There are 16 small gray squares to four small white squares, which is the same ratio as four big
gray squares to one big white square.
Parts, wholes and one
Part to part
A part to part ratio means one part of a ratio is compared to another part.
The trick with ratios is always to multiply or divide the numbers by the same value.
Example
The numbers of the ratio
3
multiplied by the same value 2
×2
scales the ratio up to
6
:
4
×2
:
8
So, 3 : 4 is the same as 3 × 2 : 4 × 2 = 6 : 8.
Part to whole
A part to whole ratio means a ratio can show a part compared to the whole lot.
Example
There are five children - three are girls and two are boys. Determine the part to part ratios and
the part to whole ratios.
The part to part ratios compare the number of boys and number of girls with each other:
60
The ratio of boys to girls is
2 : 3
or
The ratio of girls to boys is
3 : 2
or
2
.
3
3
.
2
1.7. RATIOS, PROPORTIONS AND PERCENTAGES
The part to whole ratios compare the number of boys or girls with the number of children:
The ratio of boys to children is
2 : 5
or
The ratio of girls to children is
3 : 5
or
2
.
5
3
.
5
Part to one
Consider the ratio
9
.
12
Note the last form of the ratio is nothing more than a fraction that can be reduced:
9 to 12
or
9 : 12 or
9÷3
3
9
=
= .
12
12 ÷ 3
4
Thus, the ratio 9 : 12 is the same as 3 : 4.
Another way in which a ratio is reduced is by expressing the fraction in decimal form, in which
case it is compared to one:
3÷4
0,75
3
=
=
or 0,75 : 1
4
4÷4
1
Be very careful that you place the quantity that you are comparing to, second.
Example
A room is 4 m wide and 6 m long. What is the ratio of the width to the length, and the ratio of
the length to the width? Also express both ratios in the “comparison-to-one” form, working to
three significant figures.
The ratio of the width to the length is 4 to 6. Reducing the fraction gives
4÷2
2
4
=
= .
6
6÷2
3
Expressing the fraction as a comparison-to-one ratio, gives
0,667
2÷3
=
, that is 0,667 : 1.
3÷3
1
The ratio of the length to the width is 6 to 4. Reducing the fraction gives
6÷2
3
6
=
= .
4
4÷2
2
Expressing the fraction as a comparison-to-one ratio, gives
1,50
3÷2
=
, that is 1,50 : 1.
2÷2
1
This example illustrates that we must be very careful to place the quantity that we are comparing
to, second. In other words, the ratio of width to length is not the same as that of length to width:
width : length = length : width
61
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Activity
1. Last year the total sales for Water Walker Sailboards amounted to R240 000, while the gross
profit was R30 000. Determine the ratio of gross profit to total sales, reduce it, and express
it as a comparison-to-one ratio.
2. Percy’s Programming Persons, a placement bureau for data processing personnel, has 3
female and 21 male job seekers on its books. What is the ratio of women to men? Reduce
this ratio and express it as a comparison-to-one ratio.
Answer
1. The ratio of gross profit to total sales is 30 000 to 240 000 or
1
30 000
= ,
240 000
8
that is, 1 : 8 or 0,125 : 1.
2. The ratio of women to men is
3 to 21 or
1
3
= ,
21
7
that is, 1 : 7 or 0,143 : 1 (to three decimals).
Example
Benjamin and Ernest start a transport service, BETS. Benjamin invests R25 000 and Ernest
R20 000, and they agree to divide any profits in the same ratio as their capital investments. At
the end of the first month, the business shows a net profit of R9 000. What is each partner’s
share of the profit?
The ratio of their investments is 25 to 20. Now we must divide the profit into two parts in order
to satisfy this ratio. This is done as follows:
Add 25 and 20 together to obtain 45 parts in total.
Benjamin’s share is R5 000:
25
× 9 000 = 5 000
45
Earnest’s share is R4 000:
20
× 9 000 = 4 000
45
Why specifically “45 parts” in total? Would we get the same results if we worked with a reduced
ratio? Yes, we would, but let’s check:
5
1,25
25
= =
20
4
1
Consider
62
5
first. Add 5 and 4 together to obtain 9 parts in total.
4
1.7. RATIOS, PROPORTIONS AND PERCENTAGES
For Benjamin
5
× 9 000 = 5 000
9
and for Ernest
4
× 9 000 = 4 000
9
as before.
Or, working with 1,25 to 1, we add 1,25 and 1 together to obtain 2,25 parts in total.
Benjamin’s and Ernest’s shares are respectively
1,25
× 9 000 = 5 000
2,25
and
1
× 9 000 = 4 000.
2,25
In other words, it does not matter whether we work with the given ratio, or a reduced ratio,
or a comparison-to-one ratio. We may choose whichever is most convenient.
Activity
Cleansweep, an office cleaning company, lands a big new contract and has to expand its
staff from a total of 240 cleaners and 10 supervisors to a total of 400. If the same ratio of
supervisors to cleaners is to be maintained, how many supervisors are required?
Answer
The ratio of supervisors to cleaners is
10 to 240, or 1 to 24
in reduced form (or 0,04167 to 1).
Thus we add 1 and 24 together to obtain 25 parts in total.
The number of supervisors required, is
1
× 400 = 16.
25
The number of cleaners is
24
× 400 = 384.
25
Note that we could have worked with the original ratio:
10 + 240 = 250
parts in total or the comparison-to-one ratio and obtained the same results.
To check we calculate the ratio of supervisor to cleaners. This is
16 to 384,
which is
0,04167 to 1.
63
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Ratios are not just used to compare two numbers. Three or more numbers can also be compared
using ratios, as the next example shows.
Example
Benjamin and Ernest’s Transport Service is doing very well, and when they recognise a need for
a courier service for sensitive documents, they decide to establish a new firm, Discreet Deliveries.
They go into partnership with Salina. Their initial investments are as follows: Benjamin R7 500,
Ernest R10 000 and Salina R2 500. If they agree to divide profits in the same ratio as their capital
investments, how will they share out the first month’s profit of R8 000?
The ratio of their investments is
7 500 to 10 000 to 2 500
or, in reduced form,
3 to 4 to 1,
which we can also write as
3 : 4 : 1.
We add 3, 4 and 1 together to obtain 8 parts in total.
Thus, Benjamin’s share is R3 000,
3
× 8 000 = 3 000,
8
Ernest’s share is R4 000,
4
× 8 000 = 4 000
8
and Salina’s share is R1 000,
1
× 8 000 = 1 000.
8
The next activity requires the same approach.
Activity
Much to the chagrin of all his relatives, Uncle Wilfred’s will stipulates that R240 000 of his
estate should be divided between his three pets, Percy the parrot, Bozo the bulldog and
Sullivan the Siamese cat, in the ratio 7 : 5 : 4. How much did each pet receive?
Answer
Add 7, 5 and 4 together to obtain 16 parts in total.
Thus, Percy receives R105 000,
7
× 240 000 = 105 000,
16
Bozo receives R75 000,
5
× 240 000 = 75 000
16
and Sullivan receives R60 000,
4
× 240 000 = 60 000.
16
64
1.7. RATIOS, PROPORTIONS AND PERCENTAGES
To check, note that
nn
reduces to
105 : 75 : 60
1,75 : 1,25 : 1
(after division by 60), which is the comparison-to-one form of
7 : 5 : 4.
1.7.2
Proportion
In ordinary language the words ratio and proportion are often used as synonyms and, indeed,
for most purposes this slight confusion is acceptable and of little consequence. Mathematically
speaking, however, there is a subtle difference.
A proportion states that two ratios are equal.
compared with
means
or
1 out of 3
is the same as
2 out of 6
1
3
=
2
6
The ratios are the same, so they are in proportion.
When things are in proportion, then their relative sizes are the same.
In the two drawings below you can see that the ratios of the horse’s head length to head width
are the same in both drawings. So they are proportional.
Both have the width to length ratio of 1 : 2,
meaning the length is double the width of the head.
Making the head too long or too short would look distorted.
10
20
=
15
30
10 : 20
=
15 : 30
65
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Statements of proportionality are used all the time to scale up, or down, the values for a known
or accepted situation to new values for new situations.
Example
If you are paid R1 580 for 5 working days, how much would you earn for 17?
You will probably do this calculation without thinking, but I would like you to take a few minutes,
as the basic step is the same for all proportion-type problems.
The ratio of rand to days is
1 580 to 5,
which reduces to
316 : 1
(rand per day or rand to days).
This is, of course, nothing more than the rate of pay. To obtain the wage for 17 days simply
multiply this rate by 17 to obtain
316 × 17 = 5 372.
The wage for 17 days is thus R5 372. Finally, check your answer by making sure that the two
ratios are equal, that is,
1 580
5 372
=
.
17
5
Let’s consider again the steps that were followed to do the above calculation.
Let x be the wage for 17 days.
wage
days
1 580
5
wage : days
or
1 580 : 5
or
x : 17
or
x
17
1 580
5
=
x
17
In a proportion the cross products are equal,
which gives
5×x
=
1 580 × 17
Now solve for x
5x
5
=
x
=
The ratio of wage to days is
If x is wage for 17 days, then in this case the
ratio of wage to days is
The two ratios in a proportion are
equal, which means 1 580 is to 5
as x is to 17
=
Therefore, the wage for 17 days is R5 372,00.
66
1 580 × 17
5
1 580 × 17
5
5 372
1.7. RATIOS, PROPORTIONS AND PERCENTAGES
Activity
Property tax is often assessed on a proportional basis. Suppose that a certain municipality
charges R350,00 per year for every R10 000,00 assessed valuation. What would the annual
tax be on a property valued at R950 000,00?
Answer
Let x be the yearly tax on this property valued at R950 000,00.
tax
value
350
10 000
tax : value
or
350 : 10 000
or
x : 950 000
or
x
950 000
350
10 000
=
x
950 000
In a proportion the cross products
are equal, which gives
10 000 × x
=
350 × 950 000
Now solve for x
10 000x
10 000
=
x
=
The ratio of tax to value is
If x is the yearly tax, then in this case
the ratio of tax to value is
The two ratios in a proportion are
equal, which means 350 is to 10 000
as x is to 950 000
=
350 × 950 000
10 000
350 × 950 000
10 000
33 250
Therefore, the yearly tax on this property valued at R950 000,00 is R33 250,00.
So, in this method the basic trick is to determine the comparison-to-one ratio. This gives us the
rate.
But how do you know to which number you must compare? In other words, which number comes
second in the ratio? The answer is simple: write down the information that is given. In this case:
tax
:
value
350
:
10 000
?
:
950 000
The second number in the ratio is always the one with two given values. In this case it is the
valuation (with R10 000,00 and R950 000,00).
Remember always check to your answer as shown in the example above.
67
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Activity
1. You undertake a trip in your new car and find that for the first 300 km you use 24 litres of
petrol. How many litres do you anticipate you will need for the next 375 km?
2. You obtain the licence for Gobbling Goblin, the latest computer game, and make a profit
of R52 000 on the first 650 games sold. Assuming all prices remain the same, how much do
you expect to make on the anticipated sales of 3 000 games in the next quarter?
Answer
1. The information that is given, is
litres
:
kilometres
24
:
300
?
:
375.
The ratio of litres to kilometres is
24
.
300
24 to 300 or
Suppose you will need x litres of petrol for the next 375 km.
The ratio of litres to kilometres is
x to 375 or
x
.
375
Set the ratios of the proportion equal to each other and solve for x:
24
300
=
x
375
300 × x = 24 × 375
x =
24 × 375
300
= 30
You will need 30 litres for the next 360 km.
Check:
68
30
24
=
300
375
1.7. RATIOS, PROPORTIONS AND PERCENTAGES
2. The information that is given, is
profit
:
sales
52 000
:
650
?
:
3 000.
The ratio of profit to number of games sold, is
52 000 to 650 or
52 000
.
650
Suppose you will make x rand if you sell 3 000 games.
The ratio of profit to number of games sold is
x to 3 000 or
x
.
3 000
Set the ratios of the proportion equal to each other and solve for x:
52 000
650
=
x
3 000
650 × x = 52 000 × 3 000
x =
52 000 × 3 000
650
= 240 000
The anticipated profit is R240 000.
Check:
240 000
52 000
=
650
3 000
69
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
1.7.3
Percentages
What is a percentage?
The words per cent come from the Latin per centum. The Latin
word centum means 100, for example a century is 100 years.
When you say “per cent” you are really saying “per 100”.
For example, a percentage of 25% means 25 per 100. Therefore, we can say 25% (or a quarter)
of the block below is shaded gray:
Here are three ways to write the same thing:
nnn
70
A quarter can be written as a percentage:
25%
A quarter can be written as a decimal:
0,25
A quarter can be written as a fraction:
1
4
1.7. RATIOS, PROPORTIONS AND PERCENTAGES
Conversion
To convert a decimal number or a fraction to a percentage,
multiply by 100. The answer is expressed as a percentage (%).
Examples
Converting 0,093 to a percentage gives
0,093 × 100 = 9,3%.
Converting 3
1
to a percentage gives
5
1
3 × 100 =
5
16
× 100
5
= 320%.
To convert a percentage to a fraction or a decimal number,
divide by 100 and simplify.
Examples
2
Converting 16 % to a fraction gives
3
2
16 % =
3
=
=
=
=
50
%
3
50 100
÷
3
1
1
50
×
3
100
50
300
1
.
6
Converting 83% to a decimal number gives
83% =
83
100
= 0,83.
71
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Representation
A percentage of 100% means all. For example, calculate 100% of 60:
nnn
100% of 60
is
100
× 60
100
=
60.
50% means half. For example, calculate 50% of 60:
nnn
50% of 60
5% means
is
50
× 60
100
=
30.
5
ths (five hundredths). For example, calculate 5% of 60:
100
nnn
5% of 60
72
is
5
× 60
100
=
3.
1.7. RATIOS, PROPORTIONS AND PERCENTAGES
Three types of calculations
The following sentence describes the calculation of a percentage in general terms:
A certain percentage of a base number is equal to a final number.
Let’s say
40% of a base number = final number
or
40
× base number = final number
100
or
0,40 × base number = final number.
Set the base number for example equal to 160. The final number is then calculated as
0,40 × 160 = 64.
Therefore,
40% of 160 is 64.
As this example illustrates, in order to use per cent in computations we need to express it as a
decimal (or a fraction). Fortunately, in the case of per cent this is much simpler than for ratios in
general. You will recall that division by 100 simply implies shifting the decimal point two places
to the left. For example:
6% =
6
= 0,06
100
10,5% =
10,5
= 0,105
100
122,7% =
122,7
= 1,227
100
Of course, if you are not sure in any specific instance you can always use your calculator, or you
can always enter a per cent as a number followed by ÷ 100, if you want to be on the safe side.
Next, three types of calculations that involve percentages are considered.
73
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
First type
In the first type of calculation you are given a base number and a percentage, and you are asked
to calculate the final number.
Example
1
A salesman receives a commission that is 2 % of his sales. If his sales for the month total
2
R320 000,00, what is his commission? In this case we have
2,5% of a base number = final number
2,5
× 320 000 = commission
100
commission = 0,025 × 320 000
= 8 000.
His commission is R8 000,00.
Activity
During a sale at Wendy’s Wardrobe, goods are marked down by 30%. How much will a
dress that normally costs R500 now cost?
Answer
Converting 30% to a decimal gives
30
= 0,30.
100
The reduction in price is calculated as
0,30 × 500 = 150.
The reduction in price is R150,00. The reduced price is calculated as
500 − 150 = 350.
The dress now costs R350.
74
1.7. RATIOS, PROPORTIONS AND PERCENTAGES
Second type
In the second type of calculation you are asked to determine what percentage one amount is of
another. In other words, you are given a base number and a final number, and you must then
calculate the percentage rate.
Example
Frederick had R2 024,00 income tax deducted from his gross monthly pay. If his gross pay was
R8 800,00, what percentage of this is his income tax? Express your answer to two decimal places.
Let x represent the percentage amount. In this case we have
x% of a base number
x
× 8 800,00
100
x
100
x
100
= final number
= 2 024,00
=
2 024,00
8 800,00
= 0,2300
x = 0,2300 × 100
= 23,00%.
Thus, 23,00% was paid as income tax.
Alternatively, you can consider the ratio
2 024 to 8 800
or
2 024 : 8 800
or
2 024
.
8 800
We reduce the last form of the ratio by expressing the fraction in decimal form, in which case it
is compared to one:
2 024 ÷ 8 800
0,2300
2 024
=
=
8 800
8 800 ÷ 8 800
1
or
0,2300 : 1
However, per cent means parts per hundred, so multiply by 100 to obtain 23,00%. Thus, 23,00%
was paid as income tax.
Activity
1. A survey reveals that in a small town, with 11 275 families, 3 608 families own television
sets. What percentage of families owns sets?
2. Sky High, a construction company, last year realised a net profit of R552 500 on a contract
for R8 500 000. What percentage of the contract value was the net profit?
75
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Answer
1. Let x represent the percentage amount. In this case we have
x
× 11 275 = 3 608
100
3 608
x
=
100
11 275
x
= 0,3200
100
x = 0,3200 × 100
= 32,00%.
Thus, 32,00% of families own sets.
2. Consider the ratio
552 500
.
8 500 000
Reduce the ratio by expressing the fraction in decimal form, in which case it is
compared to one:
552 500 ÷ 8 500 000
0,0650
552 500
=
=
8 500 000
8 500 000 ÷ 8 500 000
1
or
0,0650 : 1
However, per cent means parts per hundred, so multiply by 100 to obtain 6,50%.
Thus, 6,50% of the contract value was the net profit.
Third type
In the third type of calculation you know both the percentage and the final number, but not the
initial or base number, and you want to work back to it.
Example
A television set is offered as a special for R3 600 and it is stated that this is 75% of the usual
price. What was the price before discount?
In this case we have
75% of a base number = final number
75
× price = 3 600
100
0,75 × price = 3 600
price =
3 600
0,75
= 4 800.
The normal price is R4 800.
76
1.7. RATIOS, PROPORTIONS AND PERCENTAGES
To check, calculate the ratio of
3 600 : 4 800,
which reduces to
0,75 : 1.
Multiplying by 100 gives 75%.
Activity
1. Patrick’s Paintshop advertises “all goods at 80% of their normal price”. For a quantity of
paint and brushes you are charged R422,40. What would the normal price have been?
2. Benjamin and Ernest calculate that 55% of last year’s total expenses was for salaries. If
the salary account was R825 000, what were the total expenses?
Answer
1. The percentage 80% is 0,8. Dividing 422,40 by 0,8 gives 528, which would have been the
normal price:
80% of the price
= 422,40
0,80 × price = 422,40
422,40
price =
0,80
= 528,00
The normal price is R528,00.
To check, calculate the ratio of
422,40 : 528,
which reduces to
0,80 : 1.
Multiplying by 100 gives 80%.
2. The percentage 55% is 0,55. Dividing 825 000 by 0,55 gives 1 500 000 for the total expenses:
55% of expenses
= 825 000
0,55 × expenses = 825 000
825 000
expenses =
0,55
= 1 500 000
The total expenses are R1 500 000.
To check, calculate the ratio of
825 000 : 1 500 000,
which reduces to
0,55 : 1.
Multiplying by 100 gives 55%.
77
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
% Change
% change =
change
× 100
original
The change can be an increase or a decrease.
Example
Mandile buys a handbag on a sale. It is marked down from R780,00 to R650,20.
Calculate the percentage decrease in price on the handbag.
The original price was R780,00 and the new price is R650,20.
The decrease in price is calculated as
decrease = original price − new price
= 780,00 − 650,20
= 124,80.
The decrease in price is R124,80.
The percentage decrease in price is calculated as
% change =
change
× 100
original
=
124,80
× 100
780,00
= 16%.
The handbag was marked down 16%.
Activity
If the price of an item increases from R215,26 to R247,55, what is the percentage
increase in price?
Answer
The original price was R215,26.
The new price is R247,55.
The increase in price is R32,29.
(247,55 − 215,26)
The percentage increase in price is calculated as
% increase =
increase
× 100
original
=
32,29
× 100
215,26
= 15%.
There is a 15% increase in price.
78
1.8. SIGNS, NOTATIONS AND COUNTING RULES
Theme 1.8
Signs, notations and counting rules
Learning objectives On completion of this theme you should know and be able to use
• all the basic signs and notations
• the counting rules
1.8.1
The number line
You can visualise negative and positive numbers using the number line:
negative numbers
−6
−5
−4
−3
−2
positive numbers
−1
0
1
2
3
4
5
6
It is important to understand the number line because it shows you that every number has an
opposite.
Why do we need negative numbers? As it turns out, there are many everyday problems where
negative numbers are useful, for example:
• An investment could increase or decrease in value.
• Temperature can rise or fall.
• Locations on the earth can be above or below sea level.
A number that is greater than zero is called positive. If it is less than zero it is called negative.
Zero is neither positive nor negative. Two integers that are the same distance from the origin in
opposite directions are called opposites, for example, negative 5 is the opposite of positive 5 .
The arrows on each end of the number line show us that the line stretches to infinity in both the
negative and positive direction. We do not have to include a positive sign (+) when we write
positive numbers. However, we do have to include the negative sign (−) when we write negative
numbers. Zero is called the origin and is neither negative nor positive.
Examples
2<5
5>2
2 is less than 5 because 2 lies to the left of 5.
5 is greater than 2 because 5 lies to the right of 2.
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
79
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
−3 < −1
−1 > −3
−3 is less than −1 because −3 lies to the left of −1.
−1 is greater than −3 because −1 lies to the right of −3.
−6
−5
−4
−3
−2
−1
0
−4 < 1
1
2
3
4
5
6
1 > −4
−4 is less than 1 because −4 lies to the left of 1.
1 is greater than −4 because 1 lies to the right of −4.
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
Signs
There are signs in mathematics without which we cannot live.
Sign
80
In a statement
Example
=
We use the equal sign if the two values are equal.
3×2 = 6
=
We use the not equal sign to indicate that two values are
definitely not equal.
3 = 5
<
We use a less than sign if the value on the left is smaller
than the one on the right.
3<5
>
We use a greater than sign if the value on the left is bigger
than the one on the right.
3>1
≤
The less than or equal to sign is used if the value on the left
is less than or equal the one on the right. In the example, a
can be any value less than 4. It can also be equal to 4.
a≤4
≥
The greater than or equal to sign is used if the value on the
left is bigger than or equal the one on the right. In the
example, b can be any value bigger than 6. It can also be
equal to 6.
b≥6
1.8. SIGNS, NOTATIONS AND COUNTING RULES
The less than sign and the greater than sign look like a V on its side, does it not?
To remember which way around the < and > signs go, just remember:
BIG > small
small < BIG
The small end always points to the smaller number, like this:
Phrases
Many times mathematical signs are not given, but only phrases like “at least”, “no more than”
and “at the most” are used. How can we write the following using only signs?
At least 2 means two or more, two included. We write it as x ≥ 2.
−2
−1
x
0
1
2
3
4
5
6
7
8
9
10
No more than 3 and 3 at the most both mean everything less than 3 and 3 included. We write
these statements as x ≤ 3.
−6
−5
−4
−3
−2
−1
x
0
1
2
3
4
5
6
Less than 3 means everything less than 3 and 3 excluded. We write it as x < 3. Note the open
circle above the 3 on the number line, because 3 is excluded in this case.
−6
−5
−4
−3
−2
−1
x
0
1
2
3
4
5
6
The words “between 2 and 5” are often confusing. It must be stated clearly whether the endpoints
2 and 5 are included or not.
The number x is less than 5 and greater than or equal to 0. It is written as 0 ≤ x < 5.
−3
−4
−1
−0
x
1
2
3
4
5
6
7
8
9
81
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
1.8.2
Positive and negative numbers
nnn
Up to now we worked with positive integers only to illustrate the symbols and signs, for example, 1 + 3. Next we need to consider what happens if we have something like 1 − 3. Think
first about 1 + 3 and indicate the two situations on number lines as below. I trust that we agree
on answers of 4 and −2 respectively as shown by the positions of the pointers on the number lines.
1+3
A:
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
1−3
B:
or
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
1 − (+3)
In the examples above, we started in both cases with +1. What happens when se start with −1?
Consider −1 + 3 and −1 − 3:
−1 + 3
C:
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
−1 − 3
D:
or
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
−1 − (+3)
In the examples above, we added and subtracted a positive number (+3), starting with either a
positive or a negative number (+1 or −1) on the number line. Finally we need to look at adding
and subtracting negative numbers, like −3. Consider 1+(−3), 1−(−3), −1+(−3) and −1−(−3):
1 + (−3)
E:
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
1 − (−3)
F:
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
−1 + (−3)
G:
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
−1 − (−3)
H:
82
−6
−5
−4
−3
−2
−1
0
1
2
3
4
5
6
1.8. SIGNS, NOTATIONS AND COUNTING RULES
The table below gives a summary of the results illustrated on the previous page
Diagram
Operation
Result
A
B
1+3
1 − 3 or 1 − (+3)
+4
−2
C
D
−1 + 3
−1 − 3 or −1 − (+3)
+2
−4
E
1 + (−3)
−2
F
G
1 − (−3)
−1 + (−3)
+4
−4
H
−1 − (−3)
+2
Which operations are giving the same results? We trust you agree with A and F, B and E, C
and H and D and G. Most important here is for you to note the results of F and H!
Add and subtract
If a number has no sign, it usually means that it is a positive number: 5 is actually +5.
Addition
• Adding a positive number is just simple addition:
+ and + gives +
(+5) + (+3) = (+8)
or
5+3 = 8
• Subtracting a negative number is the same as addition:
− and − gives +
(+5) − (−3) = (+8)
or
5+3 = 8
Subtraction
• Subtracting a positive number is just simple subtraction:
− and + gives −
(+5) − (+3) = (+2)
or
5−3 = 2
• Adding a negative number is the same as subtraction:
+ and − gives −
(+5) + (−3) = (+2)
or
5−3 = 2
83
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Multiply and divide
What happens to plus signs and minus signs in multiplication and division?
+
×
+
=
+
+
÷
+
=
+
−
×
−
=
+
−
÷
−
=
+
−
×
+
=
−
−
÷
+
=
−
+
×
−
=
−
+
÷
−
=
−
Two positives make a positive.
Two negatives make a positive.
A negative and a positive make a negative.
A positive and a negative make a negative.
Examples
3
×
2
=
6
−3
×
−2
=
6
−3
3
×
×
2
−2
=
=
−6
−6
What about multiplying three or more numbers? That is easy: you multiply two at a time and
follow the rule. (Keep in mind that dividing is like multiplying with the inverse.)
Examples
3
×
2
×
5
=
6
×
5
=
30
3
×
−2
×
5
=
−6
×
5
=
−30
3
×
2
×
−5
=
6
×
−5
=
−30
−3
×
2
×
5
=
−6
×
5
=
−30
−3
×
−2
×
5
=
6
×
5
=
30
−3
×
2
×
−5
=
−6
×
−5
=
30
3
×
−2
×
−5
=
−6
×
−5
=
30
−3
×
−2
×
−5
=
6
×
−5
=
−30
When there is an even number of minuses in multiplication or division (that is 2, 4, 6 and so
forth), the sign becomes a plus (+) and the answer is a positive number.
When there is an uneven number of minuses in multiplication or division (that is 1, 3, 5, 7 and
so forth), the sign becomes a minus (−) and the answer is a negative number.
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1.8. SIGNS, NOTATIONS AND COUNTING RULES
1.8.3
Summation
The summation symbol is an important symbol when working with numbers. It is a very useful
sign, because it is the mathematical shorthand sign for adding.
and it is pronounced “sigma”.
The Greek capital letter for S is
Sign for sum
The summation symbol is the mathematical
shorthand sign for adding.
What must
we add?
Whatever appears after the sigma is added.
Here we will add n.
What is the
The values are shown below and above the
n
4
value of n?
n
n=1
sigma. In this case, n starts at 1 and ends at 4,
so that is 1, 2, 3 and 4.
4
Have a look
So now we add up 1, 2, 3 and 4.
n
n=1
=
1+2+3+4
=
10
Instead of adding numbers, we can also add variables.
Examples
Suppose we have the following variables:
x1 = 4, x2 = 9, x3 = 2,
x4 = 5
and x5 = 8
5
xi gives
Calculating the value of
i=1
5
xi = x1 + x2 + x3 + x4 + x5
i=1
= 4+9+2+5+8
= 28.
4
xi you will calculate x2 + x3 + x4 . Substitute the value of i in xi for the specified range.
For
i=2
The variable i starts at 2 and ends at 4:
4
xi = x2 + x3 + x4
i=2
= 9+2+5
= 16
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TOPIC 1: REVISION OF PRIOR KNOWLEDGE
2
For
xi you will calculate x1 + x2 . Substitute the value of i in xi for the specified range.
i=1
The variable i starts at 1 and ends at 2:
2
xi = x1 + x2
i=1
= 4+9
= 13
3
For
i=1
xi 2 you will calculate x12 + x22 + x32 . Substitute the value of i in xi 2 for the specified
range. The variable i starts at 1 and ends at 3:
3
i=1
xi 2 = x12 + x22 + x32
= 42 + 92 + 22
= 16 + 81 + 4
= 101
86
1.8. SIGNS, NOTATIONS AND COUNTING RULES
1.8.4
Multiplication rules
The registration numbers of cars in Gauteng consist of two letters, two numeric figures (numbers),
two letters and GP (for Gauteng Province). An example is
BC 38 YM GP
(with Gauteng’s coat of arms in front of the GP).
How many registrations are possible if no vowels (a, e, i, o and u) may be used? Remember that
a letter and a number may be used more than once.
There are 21 letters that may be used if no vowel is allowed.
For the first two-letter part:
The first position can be filled in 21 different ways.
The second position can be filled in 21 different ways.
For the numeric part we may use any of the 10 figures, 0, 1, 2, . . . , 9.
The first position can be filled in 10 ways and the second position can be filled in 10 ways.
For the last two-letter part:
The first position can be filled in 21 different ways.
The second position can be filled in 21 different ways.
The total number of registrations possible is
21 × 21 × 10 × 10 × 21 × 21 = 19 448 100,
which is quite a lot!
If an operation can be performed in n1 ways, and thereafter it is performed in any one of these
ways, a second operation can be performed in n2 ways, and after this second operation has been
performed in any one of these ways a third operation can be performed in n3 ways, and so on for
k operations, then the k operations can be performed in
n1 × n2 × n3 × . . . × nk
ways.
Activity
If a parking garage has five entrances and three exits, in how many ways can a motorist
enter and leave the garage?
Answer
The motorist can enter in 5 different ways.
The motorist can leave in 3 different ways.
The total number of ways in which the motorist can enter and leave is
5 × 3 = 15.
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TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Factorial
nnn
The factorial (denoted by !) of a natural number is the product of
all natural numbers that are less than or equal to that number.
In other words, the factorial of a non-negative integer n, denoted by n!, is the product of all
positive integers less than or equal to n:
n! = n × (n − 1) × (n − 2) × · · · × 3 × 2 × 1
Examples
3! is shorthand for 3 × 2 × 1. 3! is usually pronounced “three factorial”.
nnn
nnnnnnnnnnn
!
3! = 3 × 2 × 1 = 6
2! = 2 × 1 = 2
5! = 5 × 4 × 3 × 2 × 1 = 120
You can easily calculate the factorial of a natural number from the factorial of the preceding
natural number:
n
n!
1
1
2
2×1
=
2 × 1!
=
2
3
3×2×1
=
3 × 2!
=
6
4
4×3×2×1
=
4 × 3!
=
24
5
5×4×3×2×1
=
5 × 4!
=
120
1
1
Examples
What is 9! if you know that 8! = 40 320? The answer is calculated as follows:
9! = 9 × 8!
9! = 9 × 40 320
= 362 880
So the rule is:
n! = n × (n − 1)!
This just means that the factorial of any natural number is that number times the factorial of
one smaller than that number. Therefore,
9! = 9 × 8!
and
130! = 130 × 129!
88
1.8. SIGNS, NOTATIONS AND COUNTING RULES
What about zero factorial?
Zero factorial is interesting. It is generally agreed that:
0! = 1.
It may seem funny that multiplying no numbers together gets you one, but it helps to simplify a
lot of calculations.
Where are factorials used?
Factorials are used in many areas of mathematics, but particularly in combinations and permutations.
8!
What is ?
3!
Let’s write the factorials out in full:
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
3! = 3 × 2 × 1
Therefore,
8!
3!
=
8×7×6×5×4×3×2×1
3×2×1
=
8 × 7 × 6 × 5 × 4 × 3 × 2 × 1
3 × 2 × 1
= 8×7×6×5×4
= 6 720.
The 3 × 2 × 1 cancelled out, leaving only 8 × 7 × 6 × 5 × 4.
Activity
What is
7!
?
5!
Answer
The answer is calculated as
7!
5!
=
7×6×5×4×3×2×1
5×4×3×2×1
= 7×6
= 42,
or
7!
5!
=
7 × 6 × 5!
5!
= 7×6
= 42.
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TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Permutations
Video:
Watch the video “Permutations” on permutations and ways of ordering objects.
The two formulas used in the video to calculate permutations are the following:
The number of permutations of n objects out of n objects is:
n Pn = n!
The number of permutations of r objects out of n objects is:
n Pr =
n!
(n − r)!
The notations n Pr or P (n, r) are also used.
Activity
Seven horses run in a race. What is the total number of ways in which they can complete
the race?
Answer
In this case the value of n is 7. The number of permutations of 7 objects out or 7 objects
has to be calculated
The first horse can take any of seven places.
The next horse can take any of six places, and so forth.
The total number of ways is
7P 7
= 7!
= 7×6×5×4×3×2×1
= 5 040.
Example
How do we handle the following situation?
A list of ten investment possibilities are presented to the directors of a company. Each director
must order the five projects he considers to be the best in order of importance. How many
different arrangements are possible?
It is clear that order of placement is important and that only five must be chosen out of the ten
possibilities. When order is of importance, we use permutations.
We want to determine the number of permutations of five out of ten objects.
90
1.8. SIGNS, NOTATIONS AND COUNTING RULES
When r = 5 objects are chosen out of n = 10 objects, the formula to use is
10 P 5
=
10!
(10 − 5)!
=
10!
5!
=
10 × 9 × 8 × 7 × 6 × 5!
5!
= 30 240.
Activity
How many arrangements are possible for the first three places in a race with eight horses?
Answer
When r = 3 objects are chosen out of n = 8 objects, the formula to use, is
8 P3
=
8!
(8 − 3)!
=
8!
5!
=
8 × 7 × 6 × 5!
5!
= 8×7×6
= 336.
There are 336 arrangements possible for the first three places in a race with eight horses.
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TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Combinations
Video:
Watch the video “PermuvsComb(1)”, the first part of this video on permutations
versus combinations.
Video:
Watch the video “PermuvsComb(2)”, the second part of this video on permutations
versus combinations.
The two formulas used in the videos to calculate combinations are the following:
The number of combinations of n objects out of n objects is:
n Cn = 1
The number of combinations of r objects out of n objects is:
n Cr
n Pr
=
r!
n!
(n − r)! × r!
=
The notations n Cr or C(n, r) or
n
r
are also used.
Example
When order of placement is not important, we use combinations instead of permutations.
Suppose we have four workers of equal competence. In how many different ways can we select
two workers?
Suppose the workers are A, B, C and D. The possible choices of two out of four is
A B
A C
A D
B C
B D
C D
The formula used here is
4 C2
=
=
=
=
=
92
4!
2! 2!
4 × 3 × 2!
2! × 2!
4×3
2!
12
2
6.
1.8. SIGNS, NOTATIONS AND COUNTING RULES
If order of placement were important, then we would have:
A B
A C
A D
B A
C A
D A
B C
B D
C B
D B
C D
D C
Activity
In how many ways can a police captain choose any three of his seven detectives for a special
assignment?
Answer
Order of placement is not important. The possible number of combinations is
7 C3
=
7!
(7 − 3)! 3!
=
7!
4! 3!
=
5 040
24 × 6
= 35.
The difference between permutations and combinations
We sometimes get confused between “permutation” and “combination” – which one is which?
Here is an easy way to remember: permutation sounds complicated, doesn’t it? And it is.
With permutations, every little detail matters. Alice, Bob and Charlie are different from Charlie,
Bob and Alice.
Combinations, on the other hand, are pretty easy. The details do not matter. Alice, Bob and
Charlie is the same as Charlie, Bob and Alice.
Permutations are for lists (when order matters) and combinations are for groups (when
order does not matter).
Recap on permutations
Let’s start with permutations, or all the possible ways of doing something. We are going to
care about every last detail, including the order of items. Let’s say we have eight people:
1: Alice (A = 1)
2: Bob
3: Charlie
4: David
5: Eve
6: Frank
7: George
8: Horatio
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TOPIC 1: REVISION OF PRIOR KNOWLEDGE
How many ways can we pick a gold, silver and bronze medal for “Best friend in the world”?
Let’s look at the following representation:
A
B
C
D
E
F
G
H
8 choices
B
C
D
E
F
G
H
7 choices
C
D
E
F
G
H
6 choices
We’re going to use permutations since the order in which we hand out these medals matters.
Here is how it breaks down:
•
Gold medal:
8 choices:
A B C D E F G H.
Let’s say A wins the gold.
•
Silver medal:
7 choices:
B C D E F G H.
Let’s say B wins the silver.
•
Bronze medal:
6 choices:
C D E F G H.
Let’s say C wins the bronze.
We had eight choices at first, then seven, then six. The total number of options was
8 × 7 × 6 = 336.
Let’s look at the details. We had to order three people out of eight. To do this, we started with
all options (eight) and then took them away one at a time (seven, then six) until we ran out of
medals.
We know the factorial is
8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.
Unfortunately, that does too much. We only want 8 × 7 × 6. How can we “stop” the factorial at
5?
This is where permutations get cool: notice how we want to get rid of 5 × 4 × 3 × 2 × 1. What’s
another name for this? Five factorial!
8!
we get
So, if we do
5!
8!
5!
=
8×7×6×5×4×3×2×1
5×4×3×2×1
= 8 × 7 × 6.
And why did we use the number 5? Because it was left over after we picked three medals from
eight. So, a better way to write this would be
8!
(8 − 3)!
8!
is just a fancy way of saying: “Use the first three numbers of eight.” If we have n
(8 − 3)!
items in total and want to pick r of them in a certain order, we get
where
n!
.
(n − r)!
94
1.8. SIGNS, NOTATIONS AND COUNTING RULES
This just means: “Use the first r numbers of n!”, which is the permutation formula. Suppose you
have n items and want to find the number of ways r items can be ordered, then
n Pr =
n!
.
(n − r)!
Recap on combinations
Combinations are easy. Order does not matter. Let’s say that I cannot afford separate gold,
silver and bronze medals. In fact, I can only afford empty tin cans.
In how many ways can I give three tin cans to eight people?
Well, in this case, the order in which we pick people does not matter. If I give a can to Alice,
Bob and then Charlie, it is the same as giving one to Charlie, Alice and then Bob. Either way,
they are going to be equally disappointed.
This raises an interesting point – we have some redundancies here. Alice Bob Charlie is equal to
Charlie Bob Alice. For a moment, let’s just figure out in how many ways we can rearrange three
people.
Well, we have three choices for the first person, two for the second, and only one for the last. So
we have 3 × 2 × 1 ways to rearrange three people.
Wait a minute . . . this looks a bit like a permutation.
It is indeed! If you have m people and you want to know how many arrangements there are for
all of them, you simply work out m factorial or m!
So, if we have three tin cans to give away, there are 3! or 6 variations for every choice of three
picked out of eight. If we want to figure out how many combinations we have, we just create
all the permutations and divide by the number of variations for each permutation.
In our case, we get 336 permutations (from above) and we divide by the 6 variations for each
permutation and get
336 ÷ 6 = 56.
The general formula is
n Pr
,
r!
which means: “Find all the ways to pick r people from n, and divide by the r! variations.” Writing
this out, we get our combination formula, or the number of ways to combine r items from a
set of n:
n Cr =
n Cr
=
n Pr
r!
=
n!
÷ r!
(n − r)!
=
1
n!
×
(n − r)! r!
=
n!
(n − r)! r!
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TOPIC 1: REVISION OF PRIOR KNOWLEDGE
A few examples
Here are a few examples of combinations (order does not matter) and permutations (order matters).
1. Combination: Picking a team of three people from a group of ten gives
10 C3
=
10!
(10 − 3)! 3!
=
10!
7! 3!
= 120.
Permutation: Picking a president, a vice-president and a secretary from a group of ten gives
10 P3
=
10!
(10 − 3)!
=
10!
7!
= 720.
2. Combination: Choosing three desserts from a menu of ten.
Permutation: Listing your three favourite desserts, in order of preference, from a menu of
ten.
96
1.8. SIGNS, NOTATIONS AND COUNTING RULES
Exercise 1.6
1. Use the symbols <, > or = to make the following true:
−2
(a) −5
−2
(b) 9
(c) −100
7
−12
(d) −6
0
(e) 2
3
(f) +3
2. Write down the following and complete the missing parts:
x
5.
(a) x < 5 and x ≥ 0 can also be written as 0
x
3.
(b) x ≥ −3 and x < 3 can also be written as −3
and
.
(c) −6 < x ≤ 5 can also be written as
and
.
(d) 0 ≤ x < 6 can also be written as
3. Graph the following inequalities on a number line (x is an integer):
(a) x ≥ −3
(b) x < 5
(c) −3 < x ≤ 7
(d) x ≥ −4 and x < 5
4. Solve the following:
7!
5!
(b) (14 − 11)! + 2! × 4!
(a)
5. If x1 = 3, x2 = 5, x3 = 4 and x4 = 2, solve the following:
4
xi
(a)
i=1
3
(b)
xi
i=2
4
(c)
i=1
x2i
6. (a) How many four-letter words (including those not making sense) are possible if a character may appear more than once in the same word?
(b) How many meals are possible if there is a choice of four starters, ten main courses and
six desserts?
(c) Any three people out of twelve can be chosen for a committee. How many possible
arrangements are there?
(d) How many four-letter words are possible if a letter may not occur more than once in
the same word (including words not making sense)?
97
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Theme 1.9
Units and measures
Learning objectives On completion of this theme you should be able to
• convert units of length/distance (e.g. mm to m)
• convert units of area (e.g. mm2 to m2 )
• convert units of volume (e.g. mm3 to m3 )
1.9.1
SI system
It is difficult to imagine a world without a system for measuring things. You would not know the
distance to another town, the capacity in litres of a fuel tank, whether your body mass is within
limits and so on.
The earliest measuring systems originated in the barter system and units corresponded with things
like the length or size of a hand or a foot. Many countries had their own measuring systems.
In the modern world, which is characterised by international trade and extended industrial and
technological development, it became necessary to have a common system. Consequently, a modern international system, the Système Internationale d’Unités, also known as SI, was developed.
South Africa was one of the first countries to accept this system.
The following table contains information on the SI sytem:
Number
10
100
1 000
1 000 000
1 000 000 000
1 000 000 000 000
Power
101
102
103
106
109
1012
Common name
ten
hundred
thousand
million
milliard∗
billion∗
SI name
deca
hecta
kilo
mega
giga
tera
SI abbreviation
D
h
k
M
G
T
0,1
0,01
0,001
0,000001
0,000000001
0,000000000001
10−1
10−2
10−3
10−6
10−9
10−12
tenth
hundredth
thousandth
millionth
milliardth
billionth
deci
centi
milli
micro
nano
pico
d
c
m
μ
n
p
* The American word for milliard is billion, and for billion is trillion.
98
1.9. UNITS AND MEASURES
1.9.2
Length
Length refers to measurement in one dimension.
The standard unit for length is the metre.
The SI abbreviation for metre is m.
Therefore we write ten metres as 10 m.
Lengths in the SI system:
10 millimetres
(mm)
= 1 centimetre
(cm)
10 centimetres
(cm)
= 1 decimetre
(dm)
10 decimetres
(dm)
= 1 metre
(m)
10 metres
(m)
= 1 decametre
(dam)
10 decametres
(dam) = 1 hectometre
(hm)
10 hectometres
(hm)
(km)
= 1 kilometre
For everyday purposes, we normally use km, m, cm and mm. Then:
1 km = 1 000 m
1 m = 100 cm = 1 000 mm
1 cm = 10 mm
In the textile industry, where clothes are manufactured, centimetres are used for technical reasons.
As a result, all body lengths are measured in centimetres.
Distance is another word for length and it is convention that distances between places are measured in kilometres.
Activity
Convert the following:
1. 24 cm to mm
2. 416 m to km
3. 20 km to m
4. 8 214 mm to m
5. 12,4 m to cm
6. 1 932,3 m to km
7. A km to cm
8. 50 m + 0,5 km to m
99
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Answer
1. We know that 1 cm = 10 mm.
Thus,
24 cm = 24 × 1 cm
= 24 × 10 mm
= 240 mm.
2. We know that 1 000 m = 1 km or 1 m =
1
km.
1 000
Thus,
416 m = 416 × 1 m
= 416 ×
1
km
1 000
= 0,416 km.
3. We know that 1 km = 1 000 m.
Thus,
20 km = 20 × 1 km
= 20 × 1 000 m
= 20 000 m.
4. We know that 1 000 mm = 100 cm = 1 m.
Thus,
8 214 mm = 8 214 × 1 mm
= 8 214 ×
1
m
1 000
= 8,214 m.
5. We know that 1 m = 100 cm.
Thus,
12,4 m = 12,4 × 1 m
= 12,4 × 100 cm
= 1 240 cm.
100
1.9. UNITS AND MEASURES
6. We know that 1 000 m = 1 km.
Thus,
1 932,3 m = 1 932,3 × 1 m
= 1 932,3 ×
1
km
1 000
= 1,9323 km.
7. We know that 1 km = 1 000 m and 1 m = 100 cm.
Thus,
A km = A × 1 km
= A × 1 000 m
= A × 1 000 × 1 m
= A × 1 000 m × 100 cm
= 100 000A cm.
8. We know that 1 km = 1 000 m.
Thus,
50 m + 0,5 km = 50 m + 0,5 × 1 km
= 50 m + 0,5 × 1 000 m
= 50 m + 500 m
= 550 m.
Perimeter
The perimeter is the length of the outline of a shape.
To find the perimeter of a rectangle or square you have to add the lengths of all four the sides.
In the case illustrated below, l is the length of the rectangle while w is the width.
The perimeter, P , is
P
= l+l+w+w
= 2l + 2w
w
= 2(l + w).
l
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TOPIC 1: REVISION OF PRIOR KNOWLEDGE
1.9.3
Area
A two-dimensional shape can be defined as a flat plane figure or a shape that has two dimensions
− length and width. Two-dimensional shapes do not have any thickness and can be measured in
only two faces.
The area is a measurement of the surface of a shape. The surface area is the area that describes
the material that will be used to cover a geometric solid.
The SI unit most often used for area is square metre or m2 .
To find the area of a rectangle or a square you need to multiply the length and the width of the
rectangle or the square. In the case illustrated below, l is the length while w is the width of the
rectangle.
The area, A, is
A = l×w
= lw.
w
l
Square metre
One square metre (1 m2 ) is the area size of a square where each side of the square is one metre
in length:
nn
A = l×w
= 1m × 1m
= 1 m2
Note that (m × m = m2 ).
102
1.9. UNITS AND MEASURES
Example
The sketch below shows a rectangular metal plate. The plate has a length of 5 m and a width of
3 m.
The area of the plate is the number of squares, each 1 m2 , that can fit onto the plate.
The area, A, is
A = l×w
= 5m × 3m
= 15 m2 .
That means the number of 1 m2 squares that can fit onto the plate is 15. You can count them to
check.
Square metre versus square centimetre
If 1 m = 100 cm, then the area can also be calculated as
A = l×w
= 1m × 1m
= 100 cm × 100 cm
= 10 000 cm2 .
Thus, 1 m2 is equal to 10 000 cm 2 .
Hectare versus square metre
When we are talking about land and its size, then something like km2 and m2 can be difficult
to visualise. Instead we use a hectare. A hectare is the area of a square where each side has a
length of 100 m. The area, A, of a hectare, in square metres, is thus
A = l×w
= 100 m × 100 m
= 10 000 m2 .
Thus, one hectare equals 10 000 m2 .
103
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Activity
How many hectares are in one square kilometre?
Answer
The area, A, of a square piece of land where each side is one kilometre long is calculated as
A = 1 km2
= 1 km × 1 km
= 1 000 m × 1 000 m
= 10 × 100 m × 10 × 100 m
= 10 × 10 × (100 m × 100 m)
= 100 × 1 ha
= 100 ha.
There are 100 hectares in one square kilometre (1 km2 ).
A last note
Instead of saying that the width is x and the length is y, we talk about a rectangle
of size x by y and write x × y or xy.
104
1.9. UNITS AND MEASURES
1.9.4
Volume
We can describe length as being one dimensional and area as two dimensional. But for an object
to take up space in the real world, it needs a third dimension. We use volume to measure three
dimensional spaces.
The standard way to measure liquid is in volume. Volume can be given in litres.
Questions such as “How much water is in that cup, that aquarium or that pool?” are examples
of volume problems. Volume is the space that an item or substance occupies.
We generally measure liquids in litres () or millilitres (m). The volume of solids is generally
measured in cubic metres m3 or cubic centimetres cm3 .
To determine the volume (space inside) of a rectangular object, we measure the length, l, the
width, w, and the height, h, and multiply them:
The volume, V , is
V
= l×w×h
= lwh.
The SI unit for volume is the cubic metre or m3 .
Cubic metre
The volume of a cube where each side is one metre in length, is one cubic metre:
V
= l×w×h
1m
= 1m × 1m × 1m
= 1 m3
1 m3
1m
1m
105
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Cubic metre versus cubic centimetre
We know from before that 1 m is equal to 100 cm. Then the volume of the cube can also be
written as
V
= l×w×h
= 1m × 1m × 1m
= 100 cm × 100 cm × 100 cm
= 1 000 000 cm 3
= 106 cm3 .
Thus, 1 m3 is equal to 1 000 000 cm 3 .
To change from m3 to cm3 , we must multiply by 106 ; and
to change from cm3 to m3 , we must divide by 106 .
Cubic centimetre versus litre
The litre is frequently used as unit of volume. Say you have a plastic cubic container with sides
(length, width and height) of 10 cm each. Then you can pour one litre of fluid into that container.
The volume of 1 is calculated as
V
= l×w×h
= 10 cm × 10 cm × 10 cm
= 1 000 cm3 .
10 cm
1l
10 cm
= 1 000 cm3
= 1 000 ml
10 cm
Thus, 1 equals 103 cm3 or 1 000 cm3 . One litre can also be written as 1 000 m (1 cm3 = 1 m).
To change from litres to cm3 (or m), we must multiply by 103 ;
and to change from cm3 (or m) to litres, we must divide by 103 .
106
1.9. UNITS AND MEASURES
Cubic metre versus litre
Consider the previous figure where the lengths of the sides of a cube were 10 cm each. We know
that 10 cm is equal to 0,1 m. Thus, the cube can also be represented as
V
= l×w×h
0,1 m
= 0,1 m × 0,1 m × 0,1 m
= 0,001 m3 .
1l
0,1 m
0,1 m
Thus, 1 equals 0,001 m3 .
To change from litres to m3 , we must divide by 1 000 or 103 ; and
to change from m3 to litres, we must multiply by 1 000 or 103 .
Activity
Convert the following:
1. 2 386 cm 3 to 2. 283 to m3
3. 2 m3 to cm3
4. 146 cm3 to m3
5. 2,4 to cm3
6. 1 800 cm 3 to Answer
1. We know that 1 = 1 000 cm3 and 1 cm3 =
1
.
1 000
Thus,
2 386 cm3 = 2 386 × 1 cm3
= 2 386 ×
1
1 000
= 2,386 .
107
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
2. We know that 1 m3 = 1 000 and 1 =
1
m3 .
1 000
Thus,
283 = 283 × 1 = 283 ×
1
m3
1 000
= 0,283 m3 .
3. We know that 1 m = 100 cm.
Thus,
2 m3 = 2 × 1 m 3
= 2 × 1m × 1m × 1m
= 2 × 100 cm × 100 cm × 100 cm
= 2 × 10 6 cm3 .
4. We know that 1 cm =
1
m.
100
Thus,
146 cm3 = 146 × 1 cm3
= 146 × 1 cm × 1 cm × 1 cm
= 146 ×
1
1
1
m×
m×
m
100
100
100
= 146 ×
1
m3
10 6
= 1,46 × 10−4 m3 .
5. We know that 1 = 1 000 cm3 .
Thus,
2,4 = 2,4 × 1 = 2,4 × 1 000 cm3
= 2 400 cm3 .
108
1.9. UNITS AND MEASURES
6. We know that 1 cm3 =
1
.
1 000
Thus,
1 800 cm3 = 1 800 × 1 cm3
= 1 800 ×
1
1 000
= 1,8 .
Distance, area and volume – a summary:
Distance
Area
Volume
[103 m]
1 km
=
1 000 m
1m
=
1 000 mm
1 m2
=
104 cm2
=
106 mm2
1 cm2
=
100 mm2
=
102 mm2
1 m3
=
106 cm3
=
109 mm3
1 m3
=
103 l
=
1 000 l
1l
=
103 cm3
1 ml
=
1 cm3
109
TOPIC 1: REVISION OF PRIOR KNOWLEDGE
Exercise 1.7
1. Determine the area of each of the following rectangles:
(a) 25 mm × 24 mm
(b) 1,2 km × 375 m
(c) 4,4 m × 450 mm
(d) 225 mm × 122 mm
2. Complete the following:
(a) 1 km2 equals
m2 .
(b) 1 m2 equals
mm2 .
(c) 1 cm2 equals
mm2 .
(d) 1 ha equals
m2 .
(e) 1 m2 equals
cm2 .
(f) 24,6 cm2 equals
(g) 24 869,3 mm2 equals
m2 .
m2 .
3. The length of a rectangle is 9 m and the area is 45 m2 . Determine the width of the rectangle.
4. A rectangular container has a length of 1 m, a width of 1 m and a depth of 1 m. Calculate
in cubic metres the volume of the container. Calculate the number of litres that can be
poured into the container.
5. The length, width and depth of a fuel tank are 60 cm, 50 cm and 20 cm respectively. The
price of fuel is R9,16 per litre. How much will it cost to fill the tank?
110
Topic
2
Functions and representations of
functions
On completion of this topic you should be able to
• explain the concept of a function
• differentiate among linear, quadratic, exponential and logarithmic functions
• represent the functions graphically
CONTENTS
Theme
2.1
What is a function?
2.2
Linear functions
2.3
Quadratic functions
2.4
Exponential and logarithmic functions
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Theme 2.1
What is a function?
Learning objectives On completion of this theme you should know and be able to
• explain the concepts of a formula and a function
• plot coordinates on a Cartesian plane
2.1.1
Variables
In mathematics, variables are used in mathematical expressions and are most commonly denoted
by a letter.
In elementary mathematics, a variable is a symbol
representing a number that is arbitrary or unknown.
Variables are usually letters or other symbols that represent unknown numbers or values.
Any letter you can think of can be used as a variable. It can be uppercase or lowercase. When
you get into more complex mathematical applications, you might use several variables at a time.
Normally we use a letter that reminds us of the item we are discussing in the problem. So if it is
the number of cars, we may use either n for number or c for cars.
Example
Let the number of people employed by a firm be n.
So, the number of people employed by the firm = n.
Mathematical expression
A mathematical expression is a combination of variables, numbers and arithmetic
symbols that work together to solve mathematical problems.
A mathematical expression may be made up of one mathematical term or the algebraic addition
of more than one mathematical term. It also includes the addition and subtraction of integers.
A mathematical expression can contain only numbers or only variables, or both numbers and
variables.
Examples
7
x
y
5+2
3 + 9a − y
(6p − 5) + q − 2
10 − x(z + 6)
112
2.1. WHAT IS A FUNCTION?
Variables in expressions
When working with variables, remember the following:
No matter how many times a variable is used in
an expression, the value is always the same.
Example with one variable used more than once
If you have one variable in an expression that is used more than once, then the variable represents
the same value in each instance.
In the expression 3k − 2 + 8k, the variable k is used twice. You cannot substitute two different
numbers for k. If k = 6, then 6 must be substituted for k in both places in the expression, as
illustrated below:
3k − 2 + 8k
Mathematical expression
Substitute 6 for k
=
3(6) − 2 + 8(6)
Calculate
=
(3 × 6) − 2 + (8 × 6)
Simplify
=
=
18 − 2 + 48
16 + 48
=
64
Example with two different variables
If you have two different variables in an expression, then most likely they will represent different
values.
Evaluate the mathematical expression 2r 2 + 1 + s with r = 4 and s = 7. Substitute 4 for r and 7
for s in the expression, as illustrated below:
2r 2 + 1 + s
Mathematical expression
Substitute 4 for r and 7 for s
=
2(4)2 + 1 + (7)
Calculate
=
(2 × 16) + 1 + 7
Simplify
=
32 + 1 + 7
=
=
33 + 7
40
Variables in formulas
Why use a letter when you can just write the number in a mathematical expression? That is a
good question.
The term “variable” means to change. This is why we use variables. If a number can change
based on the situation, then we can use a variable in its place.
113
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
This is better represented when we talk about formulas. Think of any familiar mathematical
formula, for example the formula for the area of a rectangle:
A = w×h
= wh
The letter A represents the area, the letter w the width and the letter h the height of the
rectangle. The variable A (area) changes every time you substitute the length of the side of a
different rectangle.
2.1.2
Formulas
What exactly is a formula?
A formula is a special type of equation that shows the relationship between
different variables. It is in fact a recipe for calculating the value
of some desired variable, called the dependent variable, from
the values of the relevant independent variables.
Example using chart
Suppose Mary has a house cleaning business and gets paid R285,00 per house in her neighbourhood. She wants to calculate the amount of money she can make cleaning three houses as opposed
to five houses. She might start by creating a chart:
Number of
houses
Amount of money
earned (R)
1
285
2
570
3
855
4
1 140
5
1 425
So, she can earn a total amount of R855,00 for cleaning 3 houses and R1 425,00 for cleaning 5
houses in her neighbourhood.
Example using formula
Since Mary multiplies the number of houses by R285,00 each time, she can write a formula in
words to make it easier to calculated her income:
income = 285 times the number of houses cleaned
She can rewrite this formula, using variables instead of words.
Let the variable h represent the number of houses cleaned.
Let the variable i represent income earned.
114
2.1. WHAT IS A FUNCTION?
Her income is therefore calculated using the following formula:
i = 285 × h
Using this formula, she can substitute any number for h (the number of houses cleaned) and
determine the amount of her earnings. In this way, she can always change h and substitute the
number of houses that she cleaned that week, and easily determine her earnings.
To calculate the amount of money she can make cleaning 3 houses as opposed to 5 houses, use
the formula in the following way:
Number of houses cleaned
h
=
3
h
=
5
Income formula
i
=
285 × h
i
=
285 × h
Substitute h
=
285 × 3
=
285 × 5
Income earned
=
855
=
1 425
So, using the formula we get the same result as using the chart, namely she can earn a total amount
of R855,00 for cleaning 3 houses and R1 425,00 for cleaning 5 houses in her neighbourhood.
Notice how the variables i and h change based on the number of houses cleaned. Using a formula
is an easy way to calculate her earnings without the added work of creating a chart.
Advantage
Note that there is no binding reason to use specific symbols in formulas. Instead of using the
formula
i = 285 × h,
where i represents Mary’s income in rand and h the number of houses cleaned, the formula can
also be written as
R = 285 × N.
Here R represents her income in rand and N the number of houses cleaned.
So, what are some advantages of using formulas?
A major advantage of formulas is their brevity. In addition, they can be easily manipulated.
Dependence
The independent variable is the variable you change, and the
dependent variable changes because of it (cause and effect).
Consider again the income formula used by Mary. A change in the value of h will change the
value of i. Therefore, i is dependent on h:
i = 285 × h
i is the dependent variable
h is the independent variable
115
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
2.1.3
Functions
When the value of one variable is dependent on the value of another variable (or several other
variables), say that the dependent variable is a function of the independent variable.
Value
The independent variable determines the value of the dependent variable.
• The way in which the value of the dependent variable is determined must be clearly stated
and unambiguous.
• Only one single value for the dependent variable must result.
Relationship
A function is a relationship between two variables.
• The independent variable determines the value of the dependent variable.
• A function relates an input (independent variable) to an output (dependent variable).
• This may result in only one value for the dependent variable.
Parts of a function
A function has three main parts:
• Input
• Relationship
• Output
Example
“Add four” is an example of a very simple function. Look at the three parts of this function:
Input
Relationship
(independent variable)
116
Output
(dependent variable)
0
+4
4
1
+4
5
2
+4
6
5
+4
9
9
+4
13
..
.
..
.
..
.
2.1. WHAT IS A FUNCTION?
The name of a function
nnn
It is useful to give a function a name. The most common name is f for function, but you can
have other names, like g or h. For now we will use f .
A function is written in a special way. The classic way of writing a function is f (x) = . . ..
“Add four” is an example of a very simple function. This function is written as: f (x) = x + 4
where
and
f( )
x
x+4
represents the function name,
represents the input
represents the output, or what to do.
What goes into the function is put inside brackets () after the name of the function:
f (x) shows you the function is called f and x goes into the function.
Then what the function does with the input, namely the output, appears:
f (x) = x + 4 shows that the function f takes x and adds 4.
Examples
Consider the function f (x) = 3x. An input of 5 gives an output of 15. Can you see why?
Function
f (x)
=
3x
Input 5
x
=
5
Substitute x in function
f (5)
=
3(5)
Calculate and simplify
=
3×5
Obtain output
=
15
Do not get too concerned about naming the input variable x. It is just a placeholder that is used
to show you where the input goes and what happens to it. The input variable could be named
anything. The name of the function as well as the input variable can be changed. Consider the
following function:
Function f and variable x
f (x)
=
x + 2x − 3
are the same as function
f (r)
=
r + 2r − 3
which is the same as function
g(d)
=
d + 2d − 3
which is the same as function
k(p)
=
p + 2p − 3
Calculate the value of this function with an input of 2, so calculate the value of k(2):
Function
Input 2 to calculate k(2)
Calculate and simplify
Obtain output
k(p)
=
p + 2p − 3
k(2)
=
(2) + 2(2) − 3
=
2+2×2−3
=
=
2+4−3
6−3
=
3
117
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Does a function always have a name, such as f (x) or k(p)? Sometimes a function has no name,
and you might just see something like
y = 3x + 2.
However, we still have:
an input
x
a relationship
multiplying by 3 and adding 2
and an output
y
A single value result
There is another important concept about a function we need to consider: a function may only
result in a single value for the dependent variable.
So, does the formula y = 3x + 2 represent a function? Yes it does, because every time that you
substitute a number for x, you will only get a single value for y.
To illustrate, we can substitute different values for x and get different single values for y:
Formula
y
=
3x + 2
Let x = 3, then
y
=
=
3(3) + 2
3 × 3 + 2 = 9 + 2 = 11
When x = 3, then
y
=
11
Let x = 7, then
y
=
3(7) + 2
=
3 × 7 + 2 = 21 + 2 = 23
=
23
When x = 7, then
y
So, we see that when x equals 3, y equals 11 every time. No other number than 11 will correspond
with 3 when we use this formula.
If we input another number, like 7, we get a different output. When x equals 7, y equals 23 every
time. No other number than 23 will correspond with 7 when we use this formula.
Therefore this formula can be labelled a function.
This seems straightforward. It would seem that all formulas could be considered functions.
However, this is not always the case.
When is a formula not a function?
A function may only result in one single
value for the dependent variable.
118
2.1. WHAT IS A FUNCTION?
It might seem that all formulas could be considered functions. However, this is not always the
case, as can be seen in the following illustration:
Number of houses cleaned
y2
=
or
y
=
which is
y
=
√
± x
√
x
If x = 4, we get
y2
=
4
or
y
=
which is
y
=
So when x = 4, we get
y
x
and
y
=
and
y
=
=
=
√
± 4
√
4
√
2×2
2
=
2
and
y
√
− x
=
=
√
− 4
√
− 2×2
−2
=
−2
√
Look at the three parts of the formula y 2 = x (or y = ± x) for different values of x:
Input
Relationship
(independent variable)
(dependent variable)
√
± 4
√
± 6
√
± 16
4
6
16
Output
2 and −2
2,45 and −2,45
4 and −4
Each time we input a value of x > 0, we get two different answers for y. As we can see, when x
equals 4, y equals 2 and −2 and so forth.
Therefore the formula y 2 = x does not represent a function, because for each positive x there are
two values for y.
Functions with multiple independent variables
Functions can have more than one independent variable. The following function is a function of
one independent variable:
y = f (x) = x + 3
where
x
is the independent variable
and
y
is the dependent variable
The following functions are all functions of more than one independent variable:
y = f (x, z)
=
5x − 4z
w = g(x, y, z)
h(x1 , x2 , x3 , x4 )
=
=
3xy 2 z 3
4x1 − 2x2 − 5x3 + x4
119
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Note that the notation used for functions of several variables is similar to the notation used for
single variable functions.
The following function is a function of two variables:
y = f (x, z) = 5x − 4z
where
x and z
are the independent variables
and
y
is the dependent variable
Likewise, the following function is a function of three variables:
w = g(x, y, z) = 3xy 2 z 3
where
and
x, y and z
w
are the independent variables
is the dependent variable
The function h is similar, except that there are four independent variables.
2.1.4
Function notation
How do you write a function?
Is the following function written as
z = 6x − 3y
or
f (x, y) = 6x − 3y
or
z = f (x, y) = 6x − 3y?
In function notation all independent variables are listed in the brackets, separated by commas or
semicolons. In this notation the function is written as
z = f (x, y),
which is read as “z is a function of x and y”.
However, note that with this notation no information about the specific form of the function is
conveyed. It is merely stated that z is the dependent variable that is a function of x and y. In
this sense the notation does not discriminate between the above function and any other function
of x and y.
The function is only completely specified once the functional form or relationship, that is
the rule for determining the value of the dependent variable from the values of the independent
variables, is given.
Thus a complete specification of this function is written as follows: If the function is
z = 6x − 3y,
then the complete specification is
z = f (x, y) = 6x − 3y.
120
2.1. WHAT IS A FUNCTION?
What is the purpose of function notation?
Why then do we use this notation z = f (x, y) if it does not convey complete information about
the function?
The answer is that the notation provides us with a concise and clear way of indicating that in
a particular function, which has been previously defined, particular numerical values are to be
substituted for the independent variables.
Function notation shows us how to substitute particular numerical values for each independent
variable in a specific function.
Examples
If we consider the function z = f (x, y) = 6x − 3y, then f (2, 3) means that 2 is substituted for x
and 3 is substituted for y. The value of f (2, 3) is then calculated as follows:
If the function is
z
=
Then f (2, 3) is
f (x, y)
=
6x − 3y
f (2, 3)
=
6(2) − 3(2)
=
6×2−3×3
=
=
12 − 9
3
Calculate and simplify
Similarly, the value of f (5,5; 4,5) is calculated as follows:
If the function is
Then f (5,5; 4,5) is
Calculate and simplify
z
=
f (x, y)
=
6x − 3y
f (5,5; 4,5)
=
6(5,5) − 3(4,5)
=
6 × 5,5 − 3 × 4,5
=
=
33 − 13,5
19,5
Note that confusion may arise when using the decimal, so we use a semicolon (; ) instead of a
comma (,) to separate variables.
121
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
2.1.5
Graphing functions
We have all heard bold statements and predictions such as the following on the news and in
advertisements:
• “Fluoride toothpaste prevents cavities.”
• “45 000 university graduates will buy new cars in December.”
Have you ever stopped to wonder how these claims were reached? How was it concluded that
smoking and cancer are related? How can someone predict how many new cars will be sold?
We can use functions to help us make such predictions. We can also graph these functions.
Graphing functions is a visual way of representing relationships
between variables. A graph basically summarises how one
quantity changes if other quantities related to it also change.
Graphs and functions are very useful in the financial world. A function is the mathematical expression that describes the relationship between two or more variables. By graphing the function
you show the relationship in a picture.
In real life, whenever a relationship between two variables exists and is known, one variable can
be used to guess the other. In the media we sometimes see a graph with a straight line that is
going up or down. What does this mean? A straight line shows you a picture of the relationship
between two variables and helps you to make predictions.
An example is the relationship between the price of a product and the quantity that consumers
are willing to buy. At lower prices, consumers may be willing to buy more, but they buy less as
prices increase.
Quantity
D
Price (R)
122
2.1. WHAT IS A FUNCTION?
2.1.6
The Cartesian plane
René Descartes was a Frenchman who lived in the 1600s. He invented the rectangular coordinate
system or plane. In fact, this coordinate system is sometimes called the Cartesian plane in his
honour. The Cartesian plane has two perpendicular number lines. The x-axis is the horizontal
line and the y-axis is the vertical line. The origin is the point where the axes intersect. The
coordinates of the origin are zero: (0; 0). Using this plane, we can describe any point in the plane
using an ordered pair of numbers.
The position of any point of the Cartesian plane is described by using two numbers (x; y). The
first number, x, is the horizontal position of the point from the origin. It is called the x-coordinate.
The second number, y, is the vertical position of the point from the origin. It is called the ycoordinate. Since a specific order is used to represent the coordinates, they are called ordered
pairs.
y
(x; y)
x
(0; 0)
For example, the ordered pair (−4; 3) represents a point 4 units to the left of the origin in the
direction of the x-axis and 3 units above the origin in the direction of the y-axis, as shown in the
following diagram:
y
4
(−4; 3)
3
2
1
−4
−3
−2
−1
−1
x
1
2
3
4
−2
123
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Quadrants
We want to represent the linear function y = f (x) graphically. The function notation y = f (x)
means that y is a function of x. The dependent variable is y and the independent variable is x.
A linear function is graphically represented in the Cartesian plane.
The x-axis or horizontal axis is generally used for the independent variable and the y-axis or
vertical axis for the dependent variable. A scaled set of axes introduced in this way is referred to
as a rectangular coordinate system or the Cartesian plane.
The x-axis and the y-axis divide the Cartesian plane into four sections. These areas are called
quadrants.
• Quadrant 1 (the upper right-hand box) is the section where both x and y are positive. Then
we rotate counterclockwise to number the other quadrants.
• In quadrant 2, x is negative and y is positive.
• In quadrant 3, both x and y are negative.
• Finally, in quadrant 4, x is positive and y is negative.
y or B or x2
dependant
variable
Quadrant 2
Quadrant 1
independant
variable
x or A or x1
Quadrant 3
Quadrant 4
• Always label the axes clearly. Remember the variables you draw on the axes need not be x
and y, but could be any variables, for example A and B or x1 and x2 .
• Since most business problems deal with positive quantities, we are mainly concerned with
points in the first quadrant. However, if we regard losses as negative profits, deductions
as negative additions, deficits as negative income, and so on, we will have the occasion to
work with points in the other three quadrants.
• Whenever you read a graph, carefully establish the variables represented on each axis and
the relevant scales.
124
2.1. WHAT IS A FUNCTION?
• According to the definition of a function, a function assigns one value of y to each value of
x for all the values that x may assume. Thus an ordered pair of data that we can write as
(x; y) is established. Each of these pairs corresponds to a point in the plane. If we plotted
all these points, we would obtain what is called a graph of a given function.
In other words, the graph of the function y = f (x) consists of all ordered
pairs (x; y) that satisfy y = f (x). We speak of the coordinates (x; y)
of each point P , and call x the abscissa and y the ordinate of P .
• It is not only points on the graph of y = f (x) that may be referred to in this way. Any
point in the plane is located by the specification of an ordered pair of numbers (x; y). That
is in fact why we refer to a rectangular coordinate system.
2.1.7
Summary
• An equation is a statement in which two algebraic expressions are equal.
• A linear equation is a simple equation of degree one; that is, the highest exponent of the
unknown variable is one.
• Solving a linear equation in one variable means finding the value of the variable. This
involves performing the same operations to both sides of an equation to maintain equality
while working to isolate the variable on one side of the equation.
• Given any equation, if you multiply, divide, add or subtract the same number on both sides,
the equality of the new equation will hold. You can also raise both sides to the same power
and keep equality.
• To get the variable you are solving alone on one side and everything else on the other side,
you use inverse operations.
125
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Theme 2.2
Linear functions
Learning objectives On completion of this theme you should be able to
• make a graphical representation of a linear function
• determine the equation of a linear function
• describe the special cases of linear functions
2.2.1
Characteristics
Linear functions are used in many real-world situations. Graphing the functions can give an
unbelievable amount of information and help you to solve problems more easily
The equation of a linear function
An equation is defined as linear when its graph presents a linear function, in other words, a
straight line. The equation of a linear function is
y = f (x) = ax + b
where a and b are constants.
The intercepts of a straight line
Let’s look at the general properties of a straight line. We are going to talk about x- and yintercepts.
The x-intercept of a straight line is the point where it crosses the x-axis.
The y-intercept of a straight line is the point where it crosses the y-axis.
Example
y
5
The y-intercept is the
point where the line
crosses the y-axis.
→4 (0; 4)
3
The x-intercept is the
point where the line
2
crosses the x-axis.
(2; 0)
↓
1
2
1
−5
−4
−3
−2
−1
−1
−2
−3
126
x
3
4
5
2.2. LINEAR FUNCTIONS
The x-intercept is the point (2; 0). This is the point where y = 0. Sometimes the x-intercept is
defined as the x-coordinate of the point where the line crosses the x-axis.
The y-intercept is the point (0; 4). This is the point where x = 0. Sometimes the y-intercept is
defined as the y-coordinate of the point where the line crosses the y-axis.
The equation for this straight line is
y = f (x) = −2x + 4.
From this, we see that if the general equation for a straight line is
y = f (x) = ax + b,
then the value of a is equal to −2 and the value of b is equal to 4.
The y-intercept of a straight line
The point where the line cuts the y-axis is called the y-intercept. At this point, where x = 0, the
value of y is calculated as follows:
In general
Example (a = −2, b = 4)
Equation
y = ax + b
y = −2x + 4
Substitute
x=0
y =a×0+b
=b
y = −2 × 0 + 4
=4
Conclusion
The intercept on the y-axis is equal
to the constant term b in the
equation for the straight line.
That is the point (0; b).
The intercept on the y-axis is equal
to 4.
That is the point (0; 4).
The straight line y = ax + b cuts the y-axis at the point (0; b).
Therefore, the line y = −2x + 4 cuts the y-axis at the point y = b = 4.
y
5
y-intercept:
This is where
x = 0.
→4 (0; 4)
3
2
1
(2; 0)
−5
−4
−3
−2
−1
−1
1
x
2
3
4
5
−2
−3
127
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
The x-intercept of a straight line
The point where the line cuts the x-axis is called the x-intercept or root. At this point, where
y = 0, the value of x can be determined by following the steps below:
In general
Example (a = −2, b = 4)
Equation
y = ax + b
y = −2x + 4
Substitute
y=0
0 = ax + b
or
ax + b = 0
0 = −2x + 4
or
−2x + 4 = 0
Solve for
x
Subtract b both sides:
ax + b − b = 0 − b
ax = −b
Subtract 4 both sides:
−2x + 4 − 4 = 0 − 4
−2x = −4
Divide by a (a = 0):
b
ax
=−
a
a
b
x=−
a
Divide by −2:
−4
−2x
=
−2
−2
Conclusion
x=2
The x-intercept is the point
b
− ;0
a
The x-intercept is the point
b
−4
; 0 = (2; 0).
− ;0 =
a
−2
b
The straight line y = ax + b cuts the x-axis at the point − ; 0 .
a
−4
b
= 2.
Therefore, the line y = −2x + 4 cuts the x-axis at x = − =
a
−2
y
5
4
(0; 4)
3
x-intercept:
2
This is where
y = 0.
1
(2; 0)
−5
−4
−3
−2
−1
−1
−2
−3
128
1
x
2
3
4
5
2.2. LINEAR FUNCTIONS
Therefore, we can conclude that the straight line
nnn
y = ax + b
b
cuts the x-axis at the point − ; 0
a
and the y-axis at the point (0; b).
y
(0; b) coordinates of y-intercept
b
b
− ; 0 coordinates of x-intercept
a
x
−
2.2.2
b
a
Draw the graph
Only two points are needed to determine an equation of a straight line. If you are not convinced
of this, just mark two points on a piece of paper and try to put more than one straight line
through them. To draw the graph of a straight line, we use the general method:
General method to draw a straight line
Step 1
Draw the axes and label them.
Step 2
Choose the scale of the axes.
Step 3
Step 4
Plot the two given points.
Draw a line through the two plotted points.
Example
The graph of the straight line that passes through the points (1; 4) and (4; 2) is shown below:
y
5
4
(1; 4)
3
(4; 2)
2
1
−4
−3
−2
−1
−1
1
2
3
4
5
6
7
x
−2
How do we draw the graph of a straight line if we do not have two points but the
equation of the line?
Video:
Watch the video “Linedraw” on graphing a straight line using the equation of the line.
129
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Activity
The equations that describe four straight lines are given below. Determine the intercepts
on the x- and y-axes for each of the following straight lines. Then draw the graph of each
straight line:
1. y = 1 + x
2. y = 2 − 4x
3. y = −6 + 9x
4. y = −25 − 5x
Answer
1. The equation is
y = 1 + x.
Compared to the general equation for a straight line, y = ax + b, the value of a is 1 (which
is > 0) and the value of b is 1 (which is > 0).
Determine the y-intercept.
The value of x is equal to 0:
y = 1+0
= 1
The coordinates are (0; 1).
Determine the x-intercept.
The value of y is equal to 0:
0 = 1+x
1+x = 0
x = −1
The coordinates are (−1; 0).
The graph is shown below:
y
2
1
−2
−1
−1
−2
130
x
1
2
2.2. LINEAR FUNCTIONS
2. The equation is
y = 2 − 4x.
Compared to the general equation for a straight line, y = ax + b, the value of a is −4 (which
is < 0) and the value of b is 2 (which is > 0).
Determine the y-intercept.
The value of x is equal to 0:
y = 2−4×0
= 2
The coordinates are (0; 2).
Determine the x-intercept.
The value of y is equal to 0:
0 = 2 − 4x
−2 = −4x
−4x = −2
−2
−4x
=
−4
−4
1
x =
2
1
;0 .
2
The graph is shown below:
The coordinates are
y
2
1
−2
−1
−1
x
1
2
−2
131
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
3. The equation is
y = −6 + 9x.
Compared to the general equation for a straight line, y = ax + b, the value of a is 9 (which
is > 0) and the value of b is −6 (which is < 0).
Determine the y-intercept.
The value of x is equal to 0:
y = −6 + 9 × 0
= −6
The coordinates are (0; −6).
Determine the x-intercept.
The value of y is equal to 0:
0 = −6 + 9x
6 = 9x
9x = 6
=
6
9
x =
2
3
9x
9
2
;0 .
3
The graph is shown below:
The coordinates are
y
6
4
2
−3
−2
−1
−2
−4
−6
132
x
1
2
3
2.2. LINEAR FUNCTIONS
4. The equation is
y = −25 + −5x.
Compared to the general equation for a straight line, y = ax + b, the value of a is −25
(which is < 0) and the value of b is −25 (which is < 0).
Determine the y-intercept.
The value of x is equal to 0:
y = −25 − 5 × 0
= −25
The coordinates are (0; −25).
Determine the x-intercept.
The value of y is equal to 0:
0 = −25 − 5x
25 = −5x
−5x = 25
−5x
−5
25
−5
=
x = −5
The coordinates are (−5; 0).
The graph is shown below:
y
15
10
5
−5
−4
−3
−2
−1
−5
x
1
2
3
4
5
−10
−15
−20
−25
133
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
2.2.3
The slope of the straight line
Video:
Watch the video “Slope1” on the slope of a straight line.
Video:
Watch the video “Slope2” on the ratio of change or slope of a line.
Video:
Watch the video “Slope3” on the negative slope of a straight line.
The steepness with which straight lines ascend or descend is called
the slope. The slope, a, is the ratio of the change in
y-values to a given change in x-values.
In terms of two arbitrary points, P1 with coordinates (x1 ; y1 ) and P2 with coordinates (x2 ; y2 )
on the straight line, we can write:
slope = a =
y2 − y1
x2 − x1
This is depicted in the following graph:
y
P2
y2
y2 − y1
y1
P1
x2 − x1
b
x1
x2
x
Now it is clear that
nn
nnnna: is a measure of the steepness of a straight line.
nnnnThe greater the change in y for a given change in x, the steeper the line.
We also know the following:
nnnnIf a > 0, then the line is ascending from left to right.
nnnnIf a < 0, then the line is descending from left to right.
134
2.2. LINEAR FUNCTIONS
Four specific cases
Considering the slope, a, and the y-intercept, b, in the equation of the straight line
y = ax + b,
there are four specific cases that can occur. These cases are illustrated in the following graphs:
y
y
Line ascends from left to right,
Line ascends from left to right,
slope is positive.
slope is positive.
a>0
a>0
←y-intercept is positive.
b>0
x
x
←y-intercept is negative.
b<0
y
y
Line descends from left to right,
Line descends from left to right,
slope is negative.
slope is negative.
a<0
a<0
y-intercept is positive.→
b>0
x
x
←y-intercept is negative.
b<0
135
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
2.2.4
Using two points to determine the equation of a straight line
The general expression for a straight line is y = ax + b
y2 − y1
with the slope as a =
x2 − x1
for any two points (x1 ; y1 ) and (x2 ; y2 ).
With this knowledge on straight lines, we are able to determine the specific equation for the
straight line passing through two given points.
Method to determine equation
Step 1
Use the general equation for a straight line y = ax + b and the slope a =
for any two points (x1 ; y1 ) and (x2 ; y2 ).
y2 − y1
x2 − x1
Step 2
Find the value of a, the slope of the straight line.
Step 3
Find the value of b.
Step 4
Substitute a and b into the general equation y = ax + b to get the specific
equation for the straight line.
Example
Determine the expression for the straight line passing through the points (1; 3) and (3; 7).
Find the value of slope a
Use the two points on the line
(x1 ; y1 )
=
(1; 3)
and
(x2 ; y2 )
=
Formula for slope of line
a
=
Value of slope a is
a
=
(3; 7)
y2 − y1
x2 − x1
7−3
3−1
4
=2
2
2
General equation for straight line
y
=
ax + b
Substitute a = 2
y
=
2×x+b
General equation reduces to
y
=
2x + b
136
(x1 ; y1 ) and (x2 ; y2 )
=
Calculate and simplify
=
2.2. LINEAR FUNCTIONS
Find the value b
Use the either point on the line
(x1 ; y1 )
=
(1; 3)
or
(x2 ; y2 )
=
(3; 7)
Use for example first point, i.e.
(x; y)
=
(1; 3)
y
=
2x + b
Substitute point in equation
3
=
2×1+b
Calculate and simplify
3
=
2+b
2+b
=
3
Subtract 2 both sides
2−2+b
=
3−2
Value of b is
b
=
1
Substitute b = 1 into the equation
y
=
2x + 1
Equation with slope a = 2
The equation for the line passing through the two points (1; 3) and (3; 7) is therefore
y = 2x + 1.
Note:
1. It does not matter which point we call (x1 ; y1 ) and which (x2 ; y2 ). Had we numbered them
the other way around above we would have found that (x1 ; y1 ) = (3; 7) and (x2 ; y2 ) = (1; 3).
The slope would be still calculated as before:
3−7
1−3
−4
=
−2
= 2
a =
2. We could also just as well have used the point (3; 7) instead of (1; 3) to find the value of b.
Using point (3; 7), we would calculate the value of b as follows:
7 = 2×3+b
7 = 6+b
7−6 = 6+b−6
1 = b
b = 1
137
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
nnn
Activity
Determine the expression for the straight line passing through the points (−2; 8) and (4; 1).
Answer
The general expression is
y = ax + b
with
y2 − y1
x2 − x1
for any two points. Taking (x1 ; y1 ) = (−2; 8) and (x2 ; y2 ) = (4; 1), we find that
a=
a =
=
1−8
4 − (−2)
−7
.
6
The general expression thus reduces to
−7
x + b.
6
How do we find the value of b? Since the line must pass through both given points, either
−7
x + b. This
point can be used. Substitute the x- and y-values of the first point in y =
6
gives
−7
× −2 + b
8 =
6
7
+ b.
8 =
3
7
Subtract from both sides to find that
3
17
b= .
3
The expression for the line passing through the two points (−2; 8) and (4; 1) is therefore
y=
y=
17
−7
x+ .
6
3
Word problems
Straight line graphs can be used to describe how a lot of things behave in real life. In many cases
the two points that are needed to draw the line are not given to you, but you must unravel them
from the information given.
Example
Betty’s Phone Service has been calling you to join their service to enjoy low rates. Clients on her
plan pay R90,00 per week for 60 minutes of airtime. If you use 152 minutes, your weekly charge
is only R159,00. The weekly charge is a linear function of the number of minutes you use per
week.
Determine the equation for the straight line that describes this linear relationship.
138
2.2. LINEAR FUNCTIONS
The weekly charge is a linear function of the number of minutes used per week. Therefore, the
charge per week depends on the number of minutes used per week:
y = f (x) = ax + b,
where y is the dependent variable and x is the independent variable.
Therefore, let
x
represent the number of minutes you use per week
and let
y
represent the weekly charge in rands.
The following data for x and y are given:
Number of minutes
x
Weekly charge (R)
y
60
90
152
159
Therefore, the two data points that satisfy this linear relationship are
(60; 90) and (152; 159).
Taking (x1 ; y1 ) = (60; 90) and (x2 ; y2 ) = (152; 159), we can calculate the slope, a, of the straight
line as
y2 − y1
a =
x2 − x1
159 − 90
=
152 − 60
69
=
92
3
.
=
4
The general equation thus reduces to
3
y = x + b.
4
3
Substitute the x- and y-values of the first point in (60; 90) in y = x + b. This gives
4
3
x+b
y =
4
3
× 60 + b
90 =
4
90 = 45 + b
90 − 45 = 45 − 45 + b
b = 90 − 45
b = 45.
The equation for the line passing through the two points (60; 90) and (152; 159) is therefore
3
y = x + 45.
4
139
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Activity
Mr BR Wash sells BB (Brighter and Better) washing powder. If he charges R49 per box,
he has a weekly demand of 26 000 boxes. If he charges R51 per box, the weekly demand is
16 000. If p is the price per box and d is the weekly demand, derive an expression for the
linear weekly demand.
Answer
Let d represent the weekly demand and p the price per box.
Thus, the weekly demand, d, can be written as a linear function in terms of the price, p.
The general expression is
d = ap + b.
But
a=
d2 − d1
p2 − p1
for any two points. Taking (p1 ; d1 ) = (49; 26 000) and (p2 ; d2 ) = (51; 16 000), we find
16 000 − 26 000
51 − 49
−10 000
=
2
= −5 000.
a =
The general expression thus reduces to
d = −5 000p + b.
How do we find the value of b? Since the line must pass through both given points, either
point can be used. Substitute the p- and d-values of the first point, namely, (49; 26 000) in
d = −5 000p + b:
26 000 = −5 000 × 49 + b
26 000 = −245 000 + b
26 000 + 245 000 = b
b = 271 000.
The expression for the line passing through the two points (49; 26 000) and (51; 16 000) is
therefore
d = −5 000p + 271 000.
140
2.2. LINEAR FUNCTIONS
2.2.5
Special case: b = 0
Consider the special case for which the constant term is zero, which is b = 0.
A straight line y = ax + b has slope a and y-intercept b. If a straight line passes through the
origin, then its y-intercept is 0, which means b = 0.
The equation of a straight line passing through the origin is
y = ax
where a is the slope of the line.
Because the line goes through the origin, the intercept on the x-axis is also zero, as shown in the
following graph:
y
y = ax, a > 0
x
y = ax, a < 0
Example
Determine the equation of the following straight line:
y
7
6
5
4
3
2
1
−5
−4
−3
−2
−1
−1
x
1
2
−2
141
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Step 1
Identify any two points on the line.
For this example we select two sets of points, namely Set A and Set B. Note that it is
not necessary to use two sets of points. It is done to show that there will be only one
value for the slope, no matter which two points on the line are chosen.
Step 2
Set A:
Take (x1 ; y1 ) = (0; 0) and (x2 ; y2 ) = (−3; 6).
Set B:
Take (x1 ; y1 ) = (−2; 4) and (x2 ; y2 ) = (−1; 2).
Determine the slope, a, by substituting the two identified points into
y2 − y1
.
a=
x2 − x1
The slope for Set A is
y2 − y1
x2 − x1
6−0
=
−3 − 0
6
=
−3
= −2.
a=
The slope for Set B is
y2 − y1
x2 − x1
2−4
=
−1 − −2
−2
=
−1 + 2
−2
=
2
= −2.
a=
Note that for both sets of points the calculated slope is the same. The general
equation of a straight line y = ax + b is thus reduced to y = −2x + b.
Step 3
Determine the constant b by substituting any of the two identified points into the
equation y = −2x + b.
Substituting the point (−3; 6) from Set A gives
y = −2x + b
6 = −2 × (−3) + b
6=6+b
b= 6−6
b = 0.
142
2.2. LINEAR FUNCTIONS
Substituting the point (−1; 2) from Set B gives
y = −2x + b
2 = −2 × (−1) + b
2=2+b
b=2−2
b = 0.
Note that for both sets of points the calculated constant b is the same. The general
equation thus reduces to y = −2x.
Note also that this step is actually not required because, for a straight line
passing through the origin, the value of b will always be equal to zero.
Step 4
The equation for the line passing through the two points (0; 0) and (−3; 6) and
through the two points (−2; 4) and (−1; 2) is therefore
y = −2x + 0 or y = −2x.
Until now, we have determined the equation for a straight line by using two points on the line.
There is however another method that can be used. This method determines the ratio between
the change in y-values and the change in x-values.
Method using changes in x- and y-values
Step 1
Start with the general equation for a straight line y = ax + b.
Determine the value of b, that is the y-intercept of the line, where x = 0.
Step 2
Determine whether the slope a has a positive or negative value. The slope
is positive if the line ascends and negative if the line descends when you
move from left to right on the x-axis.
Step 3
Choose any two points on the line. Calculate the slope as the change in
y-value divided by the change in x-value.
So, a =
Step 4
change in y-value
.
change in x-value
Substitute the values of a and b into the general equation for a straight line
y = ax + b to obtain the specific equation for the line.
143
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Example
Determine the equation for the straight line in the following graph:
y
7
Change in y-value = 4
6
5
4
3
2
Change in
x-value
=2
−5
−4
−3
−2
1
−1
1
2
x
−1
−2
Step 1
The general equation for the line is y = ax + b. Since the line goes through the
origin, the y-intercept is 0. Therefore, b = 0.
So, in this case the equation becomes y = ax.
Step 2
The slope will be negative because the line descends when you move from left to
right on the x-axis.
So, we know that a will have a negative value.
Step 3
Take any two points on the line, such as (−1; 2) and (−3; 6) as indicated on the
graph. Calculate the slope as
a=
Step 4
4 units
change in y-value
=
= −2. (Negative because line descends.)
change in x-value
2 units
Substitute a = −2 into the equation y = ax. The equation for the straight line
is therefore
y = −2x.
144
2.2. LINEAR FUNCTIONS
2.2.6
Special case: a = 0
If the slope of a line is zero, then y does not increase, no matter how much x increases. How does
the zero slope of a line affect the line’s equation and graph?
A line with zero slope is perfectly flat in the horizontal direction.
No matter what value of x you have, you get the same y-value.
It does not increase or decrease.
Most of the time we think of slope as the amount that y changes when x changes some amount.
Say you are riding your bicycle along a straight path. Suppose that you are not a very experienced
cyclist and do not want to break a sweat. Without actually riding downhill, the easiest ride would
obviously be on a perfectly flat road. So what you want is to ascend zero metres, no matter how
far you ride.
Change in
} y-value
=0m
Change in x-value = 1 000 m
Say you ride a distance of 1 kilometre or 1 000 metres on this flat road. The slope is calculated
as
a =
change in y-value
change in x-value
=
y2 − y1
x2 − x1
=
0
1 000 − 0
=
0
1 000
= 0.
The slope is zero, meaning there is no slope at all. You rise not one centimetre no matter how
far you ride your bike.
The important thing to notice is that zero slope means horizontal.
145
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Examples
When a line with zero slope is graphed, the result is a horizontal line. One thing to be aware
of is that horizontal lines can be at different heights. For example, in the sketch below you see
three horizontal lines. In each case the slope is zero:
y
5
y=4
4
3
y=2
2
1
−5
−4
−3
−2
x
−1
−1
1
2
3
4
5
−2
y = −3
−3
−4
−5
Note the equations of the lines on the left-hand side of the sketch above. They are all similar in
that they all look like y = some value. Just like a flat road, a horizontal line does not progress,
or change in the vertical, or y, direction. Every point on the line has the same y-value.
y
5
4
3
y=2
2
(−2; 2)
(1; 2)
(3; 2)
1
−5
−4
−3
−2
−1
−1
x
1
2
3
4
5
−2
−3
−4
−5
Note the horizontal line in the sketch above. The y-coordinate of every point on the horizontal
line is 2. The x-coordinate can have any value.
146
2.2. LINEAR FUNCTIONS
Parallel to x-axis - Zero slope
The general equation for a straight line is y = ax + b. When the slope of the line is zero, the
value of a is zero. This causes the x to drop out of the equation altogether. The equation of the
line then becomes y = b.
Equation of a straight line with slope 0
The expression for a becomes
a =
y2 − y1
x2 − x1
=
0
x2 − x1
= 0.
The numerator (top part) of the fraction can only be equal to zero if
y2 = y1 ,
that is, if the function values (values of y) are the same. In this case the value of y is not
dependent on x. It is a constant. The equation of the line becomes
y = ax + b
= 0×x+b
= b.
This is represented by a straight line parallel to the x-axis, which is a horizontal line.
y
y=b
b
x
The equation for this line is y = b, where b is the intercept
on the y-axis. There is no intercept on the x-axis.
147
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
2.2.7
Special case: Parallel to y-axis
Video:
Watch the video “Paralleltoy” on the special case of a straight line parallel to
the y-axis.
Examples
When a line with undefined slope is drawn, it will look vertical, just like a steep cliff. The sketch
below shows three vertical lines, all with undefined (infinite) slope:
y
5
4
3
2
x = −4
−5
−4
−3
x=2
1
−2
x=3
x
−1
−1
1
2
3
4
5
−2
−3
−4
−5
Note that all the lines have similar equations in that they all look like x = some value. Just like
a vertical cliff face, a vertical line does not progress, or change, in the horizontal, or x, direction.
Every point on the line has the same x-value.
y
5
(2; 5)
4
(2; 4)
3
(2; 2)
2
1
−5
−4
−3
−2
−1
−1
x=2
x
1
2
3
4
5
(2; −1)
−2
−3
−4
−5
Note the vertical line in the sketch above. The x-coordinate of every point on the vertical line is
2. The y-coordinate can have any value.
148
2.2. LINEAR FUNCTIONS
Parallel to y-axis - Infinite slope
y
x=c
c
x
Consider the vertical line above. The general equation for a straight line is y = ax + b. Suppose
(x1 ; y1 ) and (x2 ; y2 ) are two points on the vertical line above.
In this case we would have x2 = x1 and a would become
a =
y2 − y1
x2 − x1
=
y2 − y1
.
0
Division by zero is not defined. We say that the slope becomes infinite and the line is vertical.
The equation for this line is x = c, where c is the intercept
on the x-axis. There is no intercept on the y-axis.
149
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
2.2.8
Special case: Parallel lines
Note the following special case: two straight lines have the same slope but different y-intercepts.
Parallel lines are lines that will go on and on forever without ever intersecting. This is because
they have the same slope. If you have two lines that have the same slope but different y-intercepts,
then these lines are parallel to each other.
When two parallel lines are drawn, they must always have the same slope or steepness.
Example
The two lines shown in the graph below are equally steep. They decrease, or slope downward, at
the same rate. The straight lines
1
y =− x+2
2
1
y = − x + 4 and
2
1
have the same slope, namely − , but different intercepts on the y-axis, namely 4 and 2 respec2
tively.
y
1
y =− x+4
2
1
y =− x+2
2
4
1
(x; − x + 4)
2
3
2
1
(x; − x + 2)
2
1
x
−4
−3
−2
−1
−1
1
2
3
4
5
6
7
8
9
−2
−3
Consider two points, one on each of the lines, with the same x-coordinate, namely x. To obtain
the vertical distance between the two lines, subtract the y-coordinate of the point on the lower
line from the y-coordinate of the point on the upper line:
y-coordinate of
point on upper line
Vertical distance
=
=
2
=
150
1
− x+4
2
1
1
− x+4+ x−2
2
2
4−2
=
−
y-coordinate of
point on lower line
−
1
− x+2
2
2.2. LINEAR FUNCTIONS
This indicates that the vertical distance between the two lines does not depend on the value of
x, it is a constant value of 2.
y
1
y =− x+4
2
4
3
1
y =− x+2
2
2
2
2
1
x
−4
−3
−2
−1
−1
1
2
3
4
5
6
7
8
9
2
−2
−3
As can be seen in the graph above, no matter what the value of x is, the vertical distance between
these specific two lines will always be 2.
Activity
Draw the following lines on the same set of axes:
1. y = 4
2. x = 6
3. y = 2x
4. y = 2x + 4
5. y = 2x − 1
Do you notice anything about 3, 4 and 5?
Answer
1. The line
y=4
is a horizontal line at y = 4.
2. The line
x=6
is a vertical line at x = 6.
151
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
3. Consider the line
y = 2x.
If x = 0, then
y = 2×0
= 0.
The point is (0; 0).
If y = 0, then
0 = 2x
2x = 0
2x
2
=
0
2
x = 0.
The point is (0; 0).
Thus, we still have only one point. Take any other y-value and determine a x-value.
If y = 1, then
1 = 2x
1
.
x =
2
Thus, two points on the line are (0; 0) and
1
;1 .
2
4. Consider the line
y = 2x + 4.
If x = 0, then
y = 2×0+4
= 4.
If y = 0, then
0 = 2x + 4
2x + 4 = 0
2x = −4
2x
2
=
−4
2
x = −2.
Thus, two points on the line are (0; 4) and (−2; 0).
152
2.2. LINEAR FUNCTIONS
5. Consider the line
y = 2x − 1.
If x = 0, then
y = 2×0−1
= −1.
If y = 0, then
0 = 2x − 1
1
.
x =
2
Thus, two points on the line are (0; −1) and
1
;0 .
2
Plot the data points for the different graphs and draw the necessary lines through them.
The following graph is obtained:
y
7
6
(4)
(3)
(5)
(2)
5
(1)
4
3
2
1
−6
−5
−4
−3
−2
−1
−1
x
1
2
3
4
5
6
7
−2
−3
−4
Lines (3), (4) and (5) are parallel. The three lines have the same slope.
153
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
2.2.9
Summary
• The general expression for a straight line or linear function is
y = ax + b,
where b is the intercept on the y-axis and a is the slope of the line.
• The x-intercept is
b
− ;0
a
and the y-intercept is (0; b).
• The formula for the slope a in terms of two points (x1 ; y1 ) and (x2 ; y2 ) on the line is
a=
y2 − y1
.
x2 − x1
• Special cases
y = ax
The constant term b is zero.
The line passes through the origin.
y=b
The case of a zero-valued slope.
Straight line parallel to the x-axis through any point (x; b).
x=c
The case of an undefined slope.
Straight line parallel to the y-axis through any point (c; y).
Parallel lines
Two or more straight lines.
The slopes of the lines are the same.
The y-intercepts of the lines are different.
154
2.2. LINEAR FUNCTIONS
Exercise 2.1
1. (a) Determine the equation of the straight line through the points (1; 2) and (3; 3).
(b) Find the intercepts on the x- and the y-axes of the line in (a).
(c) Is the line in (a) parallel to the line
y = 2 + x?
Why or why not?
(d) Draw the lines of (a) and (c) on one graph.
2. Consider the lines
y = 5 + 2x
and
y = 2 + x.
What are their intercepts on the axes? Are they parallel or not? What is the vertical
distance between the lines at
x = 3,5?
3. Draw the following lines on one graph:
(a) x = 2
(b) y = 4x
(c) y = −2x − 3
4. A bus agency has room for 60 people on a bus tour. If they charge R6 000 per person, they
will be able to fill the bus. They know from experience that if they increase the price of
the tour by R500, they will lose three customers. Determine the price function if the price
p (in rand) is a linear function of the demand (number of customers).
155
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Theme 2.3
Quadratic functions
Learning objectives On completion of this theme you should be able to
• represent a quadratic function graphically
• explain the different characteristics of the graph of a quadratic
function
• determine the intercepts and the vertex of the graph of a quadratic
function
A parabola is the curve that is produced by graphing a quadratic or second-degree equation.
Parabolas have many practical applications.
Parabolas occur frequently in the real world, for example, in suspension bridges, projectile motions, satellite dishes and reflectors.
The appearances of parabolic shapes in the physical world are abundant.
Normally a thrown object, like a ball, will follow a parabolic path of motion as it leaves the hand
of the thrower, flies and returns to earth. From basketballs to high jumpers to shot puts, nearly
all objects travel through the air in the same kind of path. This path is called a parabolic curve
or a parabola.
We can describe the position and speed of the object at any point of the parabola, or at any point
in time, using some simple equations.
All parabolas have a common shape. An example is a stream of water from a hose, or fountain,
which starts upward, curves as it nears the peak, and straightens out somewhat as it heads back
down. It is the same path followed by any thrown object, but it is easiest to see with water.
Quadratic equations are actually used in everyday life, as when calculating areas, determining a
product’s profit or formulating the speed of an object. Because the quantity of a product sold
often depends on the price, we also sometimes use a quadratic equation to represent revenue as
a product of the price and the quantity sold.
156
2.3. QUADRATIC FUNCTIONS
2.3.1
Characteristics
Equation
Parabolas are described by quadratic equations.
A quadratic equation is one in which the highest power of the independent variable x is a square.
The general equation of a quadratic function is where the relationship between the dependent
variable, y, and the independent variable, x, has the form:
y = f (x) = ax2 + bx + c,
where a, b and c are constants, and a = 0.
The three values a, b and c are also called coefficients. A coefficient is a number used to multiply
with a variable.
For example, 3x means “3 times x”, so 3 is a coefficient.
Similarly, ax2 means “a times x2 ”, so a is a coefficient.
The coefficient c in the quadratic equation is also called the constant.
The term a = 0 means that a may not be zero and is read as “a is not equal to 0”. If a = 0, the
expression becomes:
Quadratic equation
y
=
ax2 + bx + c
If a = 0
y
=
=
0 × x2 + bx + c
bx + c
Linear equation
y
=
bx + c
If a = 0, the quadratic equation becomes a linear equation.
157
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Shape of the graph
nnn
The graph of a quadratic function is called a parabola.
A parabola contains a point called a vertex. This is the point at which the curve turns, and is
thus also known as the turning point. A parabola can
open upward (smiling face)
y = 2x2 − 2
or
open downward (sad face).
y = −2x2 + 2
y
−4
−3
−2
y
4
4
3
3
2
2
1
1
−1
−1
−2
−3
x
1
2
3
4
−4
−3
−2
x
1
2
3
4
−2
Vertex: y-coordinate is the
−4
−1
−1
Vertex: y-coordinate is the
maximum value.
a<0
−3
minimum value.
a>0
−4
The vertex is the lowest point.
It is the point at which the curve turns.
The vertex is the highest point.
It is the point at which the curve turns.
The function has a minimum value at
this point.
The function has a maximum value at
this point.
Identifying the form of a parabola without graphing
How do we know which one of the two forms we are dealing with without actually drawing the
curve?
The answer lies in the value of a, the coefficient of x2 (the number in front of the x2 ), as we can
see illustrated below:
For y = 2x2 + 0x − 2,
For y = −2x2 + 0x + 2,
which is actually y = 2x2 − 2,
a = 2, b = 0 and c = −2.
which is actually y = −2x2 + 2,
a = −2, b = 0 and c = 2.
So, the term a is positive and
the graph opens upward.
So, the term a is negative and
the graph opens downward.
Symmetry of a parabola
A parabola is symmetric with respect to a vertical line called the axis of symmetry. The vertex
of the parabola will lie on the axis of symmetry. In the two graphs above, the axis of symmetry
is the y-axis.
158
2.3. QUADRATIC FUNCTIONS
2.3.2
Turning point – vertex
nn
The vertex of the graph of a quadratic function is the turning point of the parabola. The vertex
is defined as the point where the graph changes from
decreasing to increasing if the
parabola opens upward or
or
increasing to decreasing if the
parabola opens downward.
The vertex
is called
the maximum.
The vertex
is called
the minimum.
Axis of symmetry
The axis of symmetry is the line that splits the parabola into two
separate branches that mirror each other.
The axis of symmetry is a vertical line that goes through the vertex and is defined by “x = some
value”. The number that replaces “some value” will be the x-coordinate of the vertex.
The x-coordinate of the vertex
The x-coordinate of the vertex defines the axis of symmetry. The x-coordinate of the vertex is
calculated by the formula
b
xm = − .
2a
Example
y
x = −1
2
1
−4
−3
−2
−1
−1
x
1
2
−2
−3
−4
For the quadratic function
with constants
the graph will open upward because
y = x2 + 2x − 3
a = 1, b = 2 and c = −3,
a>1
The x-coordinate of the vertex is
xm = −
Equation for axis of symmetry
x = −1
Minimum value of −4 at
x = −1
2
b
=−
= −1
2a
2×1
159
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
The y-coordinate of the vertex
nn
The y-coordinate of the vertex tells us
how high or low the parabola sits.
To find the y-coordinate of the vertex, we substitute the x-coordinate of the vertex into the original general equation for the quadratic function, as illustrated below:
Quadratic function equation
y
=
f (x) = ax2 + bx + c
y-coordinate of the vertex
xm
=
−
y-coordinate of the vertex
y
=
f −
b
2a
b
2a
=
b
a× −
2a
=
−b2 + 4ac
4a
2
b
+b× −
2a
+c
y
Example
2
x = −1
1
−4
−3
−2
−1
−1
x
1
2
−2
−3
(−1, − 4)
−4
y = −4
For the quadratic function or parabola
y
=
f (x) = x2 + 2x − 3
the x-coordinate of the vertex is
xm
=
−
Substitute the x-coordinate of the vertex
x
=
−1
into the quadratic function
y
=
f (x)
to calculate the y-coordinate of the vertex:
y
=
f (−1)
=
=
=
(−1)2 + 2(−1) − 3
1−2−3
−4
(−1; −4)
So, the vertex is the point
The vertex is the point −
160
2
b
=−
= −1
2a
2×1
b
b
;f −
2a
2a
.
2.3. QUADRATIC FUNCTIONS
2.3.3
Intercepts
The intercepts on the y-axis
The y-intercept is the point where the parabola crosses the y-axis. The y-coordinate of this point
is simply the value of the quadratic function for x = 0, namely f (0).
So, the intercept on the y-axis is calculated as follows:
Quadratic function equation
y
=
f (x) = ax2 + bx + c
The y-intercept is where
x
=
0
The intercept on the y-axis is
y
=
f (0)
=
=
a(0)2 + b(0) + c
c
The y-intercept of the parabola is (0; c).
Example
y
2
1
−4
−3
−2
−1
−1
x
1
2
−2
−3
−4
(0, − 3)
x=0
For the quadratic function or parabola
y
=
f (x) = x2 + 2x − 3
the y-intercept is where
x
=
0
Substitute the value of x = 0 into y = f (x)
y
=
f (0)
=
=
=
(0)2 + 2(0) − 3
0+0−3
−3
to calculate the y-intercept:
The y-intercept is
(0; −3)
161
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
The intercepts on the x-axis
nn
The x-intercepts are the points where the parabola crosses the x-axis. These are the values of x
for which the quadratic function is zero:
y = ax2 + bx + c = 0
The values of the x-intercepts are difficult to calculate, so the formulas are given. The x-intercepts
are
−b + b2 − 4ac
−b − b2 − 4ac
and x =
.
x=
2a
2a
Example
y
2
y=0
x-intercept
(−3, 0)
−4
−3
−2
1
x-intercept
(1, 0)
−1
−1
1
x
2
−2
−3
−4
In our example,
y = x2 + 2x − 3
there are two intercepts on the x-axis.
Substitute a = 1, b = 2 and c = −3 into the given formulas:
x
=
=
=
=
=
=
=
=
−b −
b2 − 4ac
2a
−(2) −
−2 −
−2 −
and
=
(2)2 − 4(1)(−3)
2(1)
4 − (−12)
√
2
4 + 12
2
√
−2 − 16
2
−2 − 4
2
−6
2
−3
So, the x-intercepts are (−3; 0) and (1; 0).
162
x
=
=
=
=
=
=
=
−b +
b2 − 4ac
2a
−(2) +
−2 +
−2 +
(2)2 − 4(1)(−3)
2(1)
4 − (−12)
√
2
4 + 12
2
√
−2 + 16
2
−2 + 4
2
2
2
1
2.3. QUADRATIC FUNCTIONS
2.3.4
Discriminant
The x-coordinates of the x-intercepts of a parabola are
x=
−b −
−b + b2 − 4ac
b2 − 4ac
and x =
.
2a
2a
The quantity b2 − 4ac in the equations above is known as the discriminant and it determines the
number of x-intercepts that a parabola has.
Discriminant = b2 − 4ac
A parabola can have two or one or no x-intercepts.
Parabola with two x-intercepts
When the discriminant is positive,
b2 − 4ac > 0,
the parabola has two x-intercepts. We can evaluate the square root of the discriminant and
determine the two separate x-intercepts using the given equations. Note that the x-intercepts are
equally far from the x-coordinate of the vertex. In fact, they are often written in the form
x=
−b ±
−b
b2 − 4ac
=
±
2a
2a
b2 − 4ac
2a
to emphasise this fact. This means that when b2 − 4ac > 0, there are always two x-values that
give the same y-value.
Example
y
−
b
= −1
2a
2
1
(−3, 0)
−4
−3
2
−2
2
−1
−1
(1, 0)
1
x
2
−2
Each x-intercept is
2 units from the
x-coordinate of the vertex.
−3
−4
163
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
y
=
x2 + 2x − 3
a
=
1, b = 2 and c = −3
In the equations
x
=
−b
±
2a
the discriminant is
b2 − 4ac
=
(2)2 − 4(1)(−3)
=
16
>
0
=
4
=2
2
For the parabola
nn
The discriminant is positive.
√
Note the value of
16
√
16
2(1)
b2 − 4ac
2a
So, the distance from the x-intercepts
b2 − 4ac
2a
2 units
to the x-coordinate of the vertex is
Note the value of −
b
2a
−
2
2(1)
=
−
2
= −1
2
−1
x-coordinate of vertex is
x
=
x
=
−b
b2 − 4ac
±
2a
2a
−1 ± 2
The two x-coordinates are
x1
=
−1 + 2 = 1
and
x2
=
−1 − 2 = −3
Determine the two x-intercepts
The x-intercepts are (−3; 0) and (1; 0).
Parabola with one x-intercept
When the discriminant is equal to zero, the parabola has only one x-intercept, as is illustrated
below.
If b2 − 4ac = 0, the values of the x-intercepts are calculated as
b2 − 4ac
2a
√
0
b
= − ±
2a
2a
b
= − ±0
2a
b
= − .
2a
x =
−b ±
Both expressions reduce to the same, namely
x=−
b
.
2a
This is also the x-coordinate of the vertex. In other words, in this case the vertex of the parabola
just touches the x-axis at this point.
164
2.3. QUADRATIC FUNCTIONS
Example
y
5
−
4
b
=2
2a
3
2
1
−2
−1
−1
x
1
2
3
4
5
The vertex just
touches the
x-axis.
−2
y
=
x2 − 4x + 4
a
=
1, b = −4 and c = 4
In the equations
x
=
−b
±
2a
the discriminant is
b2 − 4ac
=
=
(−4)2 − 4(1)(4)
16 − 16
=
0
=
4
=2
2
For the parabola
b2 − 4ac
2a
The discriminant is zero.
Note the value of −
b
2a
−
−4
2(1)
x-coordinate of vertex is
There is only one x-intercept
So, the x-coordinate is
2
x
x
=
−b
±
2a
√
=
2±
=
0
2a
2±0
=
2
=
2
b2 − 4ac
2a
The x-intercept is (2; 0). Note that the vertex just touches the x-axis.
165
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Parabola with no x-intercepts
In this case,
b2 − 4ac < 0.
We usually determine the x-intercepts using the given equations, which could be written as
x=
−b ±
b2 − 4ac
.
2a
In this case, b2 − 4ac < 0, which means that we cannot evaluate the square root since the square
root of a negative number is not a real number. Therefore, the parabola has no intercepts on the
x-axis, nor does it touch it.
Example
y
2
1
−2
−1
−1
There are no intercepts
on the x-axis.
x
1
2
3
4
5
−2
−3
−4
−5
b
=1
2a
y
=
−x2 + 2x − 2
a
=
−1, b = 2 and c = −2
In the equations
x
=
−b
±
2a
the discriminant is
b2 − 4ac
=
(2)2 − 4(−1)(−2)
=
=
4−8
−4
<
0
For the parabola
The discriminant is negative.
There are no x-intercepts.
166
−
−4
b2 − 4ac
2a
2.3. QUADRATIC FUNCTIONS
Specific cases
Considering the discriminant, d, and the shape of the parabola, (whether it has a minimum or a
maximum value), there are specific cases that can occur.
In the graphs d = b2 − 4ac (discriminant = d). For each one of the following graphs, indicate the
correct values for a and d.
d>0
√
−b − d
2a
d>0
×
−b
2a
×
×
×
√
−b + d
2a
x
√
−b − d
2a
×
×
×
−b
2a
a>0
×
√
−b + d
2a
x
a<0
d=0
d=0
×
−b
2a
×
x
−b
2a
x
a<0
a>0
d<0
d<0
−b
2a
×
x
×
×
×
x
−b
2a
a<0
a>0
167
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
2.3.5
Draw the graph
nn
Note that when you are asked to draw the graph of a quadratic function, it is not necessary
to plot it in great detail. Just indicate the intercepts, the vertex and the approximate shape.
Parabolas can be drawn in eight easy steps:
Step 1
Step 2
Step 3
Write the function in the form
y = ax2 + bx + c
and determine the values of
a, b and c.
Determine whether function has a
a>0
minimum or maximum value:
a<0
Determine the x-coordinate of the vertex:
xm = −
b
2a
Step 4
Determine the y-coordinate of the vertex:
b
f −
2a
Step 5
Determine the value of the discriminant:
b2 − 4ac
Step 6
Determine the x-coordinates of the
x=
−b −
b2 − 4ac
2a
−b +
b2 − 4ac
2a
x-intercepts, if any:
and
Step 7
Determine the y-coordinate of the
x=
y = f (0) = c
y-intercept:
Step 8
Draw the graph.
Example
Draw the graph of the quadratic function
y = −2x2 + 8x − 6.
Step 1
Write the function in the form
and determine the values of
y
a
=
=
ax2 + bx + c
−2, b = 8 and c = −6.
Step 2
The function has a maximum value
because
a
<
0.
Step 3
The x-coordinate of the vertex is
x
=
−
b
2a
=
−
8
2 × −2
=
=
168
8
4
2.
2.3. QUADRATIC FUNCTIONS
Step 4
The y-coordinate of the vertex is
f (2)
=
=
−2 × 22 + 8 × 2 − 6
−8 + 16 − 6
=
2.
The vertex is the point
Step 5
Step 6
(2; 2).
b2 − 4ac
The value of the discriminant
b2 − 4ac
The two x-intercepts exist because
=
82 − 4 × −2 × −6
=
64 − 48
=
16.
>
0.
The two x-intercepts are
b2 − 4ac
2a
√
−8 − 16
=
2 × (−2)
−8 − 4
=
−4
−12
=
−4
=3
x=
−b −
x=
The x-intercepts are
Step 7
Step 8
−b +
b2 − 4ac
2a
√
−8 + 16
=
2 × (−2)
−8 + 4
=
−4
−4
=
−4
= 1.
and
(3; 0) and (1; 0).
The y-intercept is
y
=
=
f (0) = c
−6
=
(0; −6).
Draw the graph.
y
3
2
1
−2
−1
−1
x
1
2
3
4
5
6
7
−2
−3
−4
−5
−6
169
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Activity
Determine the intercepts on the axes and the vertices of the following quadratic functions,
and draw their graphs:
1. y = 2x2 − x − 3
2. y = 4x2 − 16x + 16
3. y = −3x2 + 3x − 2
Answer
1. From
y = 2x2 − x − 3
we have that
a = 2, b = −1 and c = −3.
Since a > 0, the function has a minimum.
The value of x at the vertex is
−b
2a
xm =
= −
−1
2×2
1
.
4
=
The value of the function at the vertex, which is the minimum, is
1
4
2 1
1
= 2
−
−3
4
4
1
= −3 .
8
y = f
1
1
; −3 .
4
8
Determine the intercept on the y-axis as
The vertex has coordinates
f (0) = c
= −3.
The y-intercept has coordinates (0; 3).
Before calculating the intercepts on the x-axis, we first determine the value of the discriminant. The discriminant is
b2 − 4ac = (−1)2 − 4 × 2 × (−3)
= 1 + 24
= 25.
Since this is greater than 0, two intercepts exist.
170
2.3. QUADRATIC FUNCTIONS
Thus the intercepts are
x
=
=
=
=
=
b2 − 4ac
2a
√
−(−1) − 25
2×2
1−5
4
−4
4
−b −
and
x
−b +
b2 − 4ac
2a
√
−(−1) + 25
2×2
1+5
4
6
4
1
1 .
2
=
=
=
=
−1
=
The graph of the quadratic function is shown below:
y
y = 2x2 − x − 3
3
2
1
0,25
−4
−3
−2
−1
−1
x
1
2
3
4
−2
The vertex has coordinates
−3
−4
y = −3
1
1
; −3 .
4
8
1
8
171
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
2. From
y = 4x2 − 16x + 16
we have that
a = 4, b = −16 and c = 16.
Since a > 0, the function has a minimum.
The value of x at the vertex is
−b
2a
−16
= −
2×4
= 2.
xm =
The minimum value of the function is thus
y = f (2)
= 4 × 22 − 16 × 2 + 16
= 0.
The vertex has coordinates (2; 0).
Determine the intercept on the y-axis as
f (0) = c
= 16.
The y-intercept has coordinates (0; 16).
The discriminant is
b2 − 4ac = (−16)2 − 4 × 4 × 16
= 256 − 256
= 0.
Because the discriminant is equal to zero, the parabola just touches the x-axis at x = 2.
The graph of the quadratic function is shown below:
y
y = 4x2 − 16x + 16
16
12
8
4
−1
−4
172
x
1
2
3
4
2.3. QUADRATIC FUNCTIONS
3. nnn
From
y = −3x2 + 3x − 2
we have that
a = −3, b = 3 and c = −2.
The value of x at the vertex is
−b
2a
−3
=
2 × −3
1
.
=
2
Since a < 0, the function has a maximum. The maximum value is
1
y = f
2
2
1
1
= −3
+3
−2
2
2
1
= −1 .
4
1
1
; −1 .
The vertex has coordinates
2
4
Determine the intercept on the y-axis as
xm =
f (0) = c
= −2.
The y-intercept has coordinates (0; −2).
The discriminant is
b2 − 4ac = 32 − 4 × (−3) × (−2)
= 9 − 24
= −15.
Since the discriminant is less than zero, there are no intercepts on the x-axis.
The graph of the quadratic function is shown below:
y
0,5
−1
−1
−1,25 →
x
1
2
3
4
−2
−3
−4
y = −3x2 + 3x − 2
173
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
2.3.6
Slope
The slope of a parabola cannot be determined in only one calculation. Looking at graphs of
parabolas, we notice that a parabola has a slope at a point, but it is constantly changing as we
move from point to point. This is unlike a straight line, which has a constant slope.
This is illustrated in the graph below with the use of tangents. A tangent is a straight line that
just touches the graph in one point, as illustrated.
The slope of a parabola at a specific point is actually the slope of the tangent to the parabola at
that point. The slopes of the tangents vary at different points and hence the slope of a parabola
is never the same at different points.
y
x
We will not calculate the slope of a parabola at different points. The main point to note is that
the slope is ever-changing from point to point on the parabola, in contrast to the straight line,
where it is a constant.
Video:
174
Watch the video “Tangent” on the slope of a parabola.
2.3. QUADRATIC FUNCTIONS
2.3.7
Summary
We can recognise, even if only intuitively, that the potential uses for the quadratic function are
considerable.
It is evident that they can be used to model situations in which the dependent variable is expected
to peak, or pass through a dip, for some value of the independent variable, for example in demand,
supply and profit functions.
• The general equation for a parabola or quadratic function is
y = ax2 + bx + c,
where a, b and c are constants and a = 0.
• If a > 0, the parabola opens upwards (smiling face) and
if a < 0, it opens downwards (sad face).
• The coordinates of the vertex of the parabola are
b
b
− ;f −
2a
2a
.
• The x-intercepts (if any) are found by solving f (x) = 0 with
x=
−b −
−b + b2 − 4ac
b2 − 4ac
and x =
.
2a
2a
• The y-coordinate of the y-intercept is
y = f (0) = c,
so the y-intercept has coordinates (0; c).
• The discriminant, d, is calculated as
b2 − 4ac.
Specific cases of the discriminant:
d>0
Two separate x-intercepts exist.
The intercepts are equally far from the x-coordinate of the vertex.
d=0
There is only one x-intercept and it is also the vertex.
The parabola just touches the x-axis at the vertex.
d<0
There are no intercepts on the x-axis.
The parabola also does not touch the x-axis.
• The slope of a parabola is ever-changing in contrast to a straight line, which has a constant
slope.
175
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Exercise 2.2
1. Determine the intercepts on the axes and the vertices of each of the following quadratic
functions, and sketch the curves:
(a) y = −0,4x2 + 0,2x + 1,2
(b) y = −x2 − 2x − 1
(c) y = x2 + 4x + 5
(d) y = −x2 + 9
2. If
d = p2 − 45p + 520
describes a weekly demand for a certain ice cream in litres, with p the price per litre and
d the demand, what is the price per litre that minimises the weekly demand? What is the
minimum weekly demand?
176
2.4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Theme 2.4
Exponential and logarithmic functions
Learning objectives On completion of this theme you should know and be able to explain
• what an exponential function is
• what a logarithm to base 10 is
• what a logarithm to base e is
2.4.1
Exponential function
Exponential functions look somewhat similar to quadratic functions in that they involve exponents, but there is a big difference: the variable rather than the base is now the exponent.
Quadratic functions
Quadratic functions are functions such as f (x) = x2 , where the variable x is the base and a fixed
number, such as 2, is the exponent.
y
8
7
6
5
4
3
2
1
−4
−3
−2
−1
1
2
3
x
4
Exponential functions
Exponential functions, however, are functions such as f (x) = 2x , where the base is a fixed number,
such as 2, and the exponent is the variable x.
y
8
7
6
5
4
3
2
1
−6
−5
−4
−3
−2
−1
−1
1
2
3
x
177
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Equation
An exponential function derives its name from the
independent variable x that appears as an exponent.
The exponential function is one of the most useful functions and is found in virtually every field
where mathematics is applied. The exponential function with base a is the function defined by
y = f (x) = ax where
a > 0,
a = 1 and
a is a real constant.
We require a = 1
because if
a
=
1
then
y
=
f (x) = 1x = 1
We just get the horizontal line
y
=
1
because if
a
<
0
for example
a
=
−4
then
y
=
f (x) = (−4)x
Calculating f (x) when
x
=
1
2
would mean calculating
y
=
f
which gives
y
=
(−4) 2 =
We require a > 0
1
2
1
which is not a real number
178
√
−4
2.4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
The best thing about exponential functions is that they are so useful in real-world situations. For
example, exponential functions are used to model compound interest, radioactive decay and various types of growth (populations, bacterial or animal). This explains the expression “exponential
growth”.
Example of compound interest
An initial investment of R1 000 with a 5% annual interest rate compounded annually will grow
according to the formula
S(n) = P (1 + i)n
= 1 000(1 + 0,05)n ,
where n is the number of years the money has been invested. Such an investment will double
roughly every 14 years.
Example of population growth
In 2013, the population of a town was estimated to be 35 000 people with an annual growth rate
of about 2,4%. The population will grow according to the formula
y = 35 000(1,024)x ,
where x is the number of years since 2013 and y is the size of the population at that time.
Using this equation to estimate the population in 2017 to the nearest hundred people will give
y = 35 000(1,024)4
≈ 38 500.
The estimated population in 2017 is 38 500 people.
Positive and negative values of x
Consider the exponential function
y = f (x) = ax ,
where a > 1. For example, consider the exponential function
f (x) = 2x .
To evaluate this function, choose a numerical value for x and calculate the function’s value at
this x.
If x has positive values, for example
if x = 3, then
f (3)
If x has negative values, for example
if x = −3, then
=
23
=
2×2×2
=
=
8.
=
f (−3)
=
2−3 =
1
23
1
2×2×2
1
.
8
179
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
The results for positive, zero and negative values of x for the function f (x) = 2x are given in the
following table:
x
y = f (x) = 2x
x
y = f (x) = 2x
0
20 = 1
1
21 = 2
−1
2−1 =
2
22 = 4
−2
3
23 = 8
−3
4
24 = 16
−4
5
25 = 32
−5
1
1
= = 0,5
1
2
2
1
1
2−2 = 2 = = 0,25
2
4
1
1
−3
2 = 3 = = 0,125
2
8
1
1
≈ 0,0625
2−4 = 4 =
2
16
1
1
≈ 0,03125
2−5 = 5 =
2
32
From the table, the following is noted:
• If x > 0, then y > 1.
• If x < 0, then 0 < y < 1.
• If x = −6, then y will be less than 0,03.
y
The graph of y = 2x is shown below:
1.8
1.6
1.4
1.2
1
1.0
0.8
0.6
0.4
0.2
−6
−5
−4
−3
x<0
−2
−1
−0.2
−0.4
1
2
3
x
x>0
We can see that when x < 0, the function values are between zero and one, but never equal to
zero. Also, when x > 0, the function values are greater than one.
Video:
180
Watch the video “Exponential” on exponential functions where x > 0 and x < 0.
2.4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Different values of a
Consider the exponential function
y = f (x) = ax .
The graphs for exponential functions with different values of a will look different.
Examples where a > 1
Examples where 0 < a < 1
Consider the following exponential function
Consider the following exponential function
with the form y = f (x) = ax . The variable,
with the form y = f (x) = ax . The variable,
1
1 1
a, takes on the values , and :
2 3
5
a, takes on the values 2, 3 and 5:
y=5
y
x
y=3
y=
x
1.8
1.6
−2
nnnnn
y=2
1 x
y=
2
x
1 x
3
y
1.8
1.6
1.4
1.4
1.2
1.2
1.0
1.0
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
x
−1
1
−0.2
nnnnn
1 x
5
y=
x
−1
1
2
−0.2
−0.4
−0.4
All the graphs go through point (0; 1).
All the graphs go through point (0; 1).
If you move from left to right on the x-axis,
If you move from left to right on the x-axis,
the y-values of the three graphs, where
the y-values of the three graphs, where
where a > 1, are increasing.
where 0 < a < 1, are decreasing.
181
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Revision
Consider the exponential function
y = f (x) = ax ,
where a > 0, a = 1 and a is a real constant.
There are two cases for the base a, namely if a > 1 and if 0 < a < 1:
a>1
0<a<1
The exponential function is an increasing
function of x. As x increases from large
The exponential function is a decreasing
function of x. As x increases from large
negative values to large positive values,
the function values constantly increase.
negative values to large positive values,
the function values constantly decrease.
In both cases the functions are positive for all values of x
and they all pass through the point (0; 1).
Thus, graphically we have the following:
y
y = ax
0<a<1
y = ax
a>1
1
x
Exponential functions with base a > 1 are known as growth curves.
On the other hand, exponential functions with base 0 < a < 1 are known as decay functions
because of their decreasing property. They are useful to model certain “negative growth” phenomena such as radioactive decay processes, price-demand curves and non-linear depreciation.
182
2.4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
2.4.2
Logarithmic functions
nn
We need to understand the term “inverse function” before we look at logarithmic functions.
Consider the function f (x):
y = f (x) = x + 2
x
x+2
3
x goes in
→
3+2
y
→
5
y comes out
The inverse function just goes the opposite way. The inverse function is written as f −1 (y):
x = f −1 (y) = y − 2
y
5
→
y−2
5−2
x
3
→
y goes in
x comes out
The inverse function gives back the
original value.
Definition
The logarithmic function is very closely related to the exponential function. Logarithms are
simply another way to write exponents.
Exponential function
Logarithmic function
f (x) = ax = y
with a > 0, a = 1
if and
only if ∗
a to the power of x is
equal to y.
loga y = x
with y > 0.
The logarithm of y to the
base a is x.
The logarithmic function (or log as it is often referred to) is completely
determined by the corresponding exponential function.
Likewise, the exponential function is completely determined by the corresponding
logarithmic function.
They are inverse functions.
Read from x to y.
Read back from y to x.
∗ “If and only if” means that
if y = ax then loga y = x
and
if loga y = x then y = ax .
183
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
Example
Evaluate the expression log10 100 without using a calculator. Note that the base is equal to 10.
Step 1
Set the log equal to x.
Step 2
Use the definition of log
to write the equation in
log10 100
=
x
10x
=
100
10x
=
100
10x
=
10 × 10
10x
=
102
x
=
2.
exponential form.
Step 3
Solve for x.
From this
Therefore, log10 100 = 2.
So, a logarithm gives you the exponent as its answer.
Activity
In each case state the meaning of
x = loga y
and determine x:
1. log10 10 000
2. log10 0,01
√
3. log5 5
184
To what power must 10
be raised to get 100?
If loga y = x with a > 0
and a = 1, then y = ax .
2.4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Answer
1. nn
Step 1
Set the log equal to x.
log 10 10 000 = x
To what power
must 10 be raised
to get 10 000?
Step 2
Use the definition
of log to write
10x
=
10 000
and a = 1,
then y = ax .
the equation in
exponential form.
Step 3
Solve for x.
From this
If loga y = x
with a > 0
10x
=
10 000
10x
=
10 × 10 × 10 × 10
10x
=
104
x
=
4.
Therefore, log 10 10 000 = 4.
2. nn
Step 1
Set the log equal to x.
log 10 0,01 = x
To what power
must 10 be raised
to get 0,01?
Step 2
Use the definition
of log to write
10x
=
0,01
and a = 1,
then y = ax .
the equation in
exponential form.
Step 3
Solve for x.
From this
If loga y = x
with a > 0
10x
=
0,01
10x
=
10−1 × 10−1
10x
=
10−2
x
=
−2.
Therefore, log 10 0,01 = −2.
185
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
3. nn
Step 1
Set the log equal to x.
Step 2
Use the definition
of log to write
the equation in
√
log5 5 = x
√
5x
=
5x
=
5x
=
52
x
=
1
.
2
5
To what power must 5
√
be raised to get 5?
If loga y = x with a > 0
and a = 1, then y = ax .
exponential form.
Step 3
Solve for x.
From this
√
5
1
√
1
Therefore, log5 5 = .
2
In general
Regarding exponents, any number raised to the power of one is equal to the number, and any
number raised to the power of zero is equal to one.
Therefore, for logarithms there are two general results in particular:
Because a1 = a, it follows that loga a = 1,
and
because a0 = 1, it follows that loga 1 = 0.
• Note that the base in the two cases above is equal to a.
• When the base is 10, you get the common logarithm log10 y, which is sometimes just
written as log y.
• When the base is e, you get the natural logarithm. The calculator has a key for loge , which
is usually denoted by the symbol ln. (The symbol e is a famous number in mathematics.
It is called Euler’s number and approximates to 2,718.)
• In computer science we have lg, which is the logarithm base 2.
186
2.4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Graph
What does the graph of the logarithmic function look like?
Consider the exponential function
y = f (x) = 2x .
In this case the base is equal to 2. The logarithmic function is the inverse of the exponential
function. Remember that the inverse function should give you back the original value that was
put into the function. Therefore, if (4; 16) is a point on the graph of an exponential function,
then (16; 4) would be the corresponding point on the graph of the logarithmic function.
Look at the following explanation, using this exponential function:
Exponential function
Logarithmic function
x
→
f
→
y
→
f −1
→
x
−1
→
2−1
→
0,5
→
log2 0,5
→
−1
−1 goes in and 0,5 comes out.
0,5 goes in and −1 comes out.
The point is (−1; 0,5).
The point is (0,5; −1).
This is illustrated in the following graph:
y
Exponential function
3
Logarithmic function
2
2−1 = 0,5
−3
−2
1
0
−1
0
−1
x
1
2
3
log2 0,5 = −1
−2
−3
187
TOPIC 2: FUNCTIONS AND REPRESENTATION OF FUNCTIONS
2.4.3
Summary
• The general equation for an exponential function is
y = f (x) = ax ,
where a > 0, a = 1 and a is a real constant.
• If a > 1, the exponential function is an increasing function of x and
if 0 < a < 1, the exponential function is a decreasing function of x.
• Exponential functions are used to model compound interest, radioactive decay and various
types of growth (populations, bacterial or animal).
• An inverse function is a function that reverses the action of another function: If the function applied to an input x gives a result of y, then applying the inverse function to y
gives the result of x.
• The logarithmic function and corresponding exponential function are inverse functions, that is
f (x) = ax = y if and only if loga y = x.
Specific cases of logarithmic functions:
loga a = 1
This is because a1 = a.
loga 1 = 0
This is because a0 = 1.
log10 y
The base is 10 and the logarithm is called
the common logarithm.
ln
The base is e and the logarithm is called
the natural logarithm.
lg
The base is 2 and the logarithm is called
the logarithm base 2 in computer science.
188
Topic
3
Linear systems
The focus of this topic is systems of linear equations and inequalities.
CONTENTS
Theme
3.1
Linear equations in one variable
3.2
Systems of linear equations in two variables
3.3
Linear inequalities in one variable
3.4
Systems of linear inequalities in two variables
TOPIC 3: LINEAR SYSTEMS
Theme 3.1
Linear equations in one variable
Learning objective
3.1.1
On completion of this theme you should be able to solve an
equation in one variable algebraically.
What is an equation in one variable?
An equation in one variable is a statement containing an = sign, with algebraic expressions
to the left and right of the sign, using only one variable. The values of the variable that make
the statement true are called solutions or roots of the equation.
A few examples should make this definition clear. In each case the values on the right (the
solutions) make the statements on the left (the equations) true:
Equation
Solution(s) or root(s)
2x + 3 = x + 4
x=1
A+1=3
A=2
The process of determining which values of the variables make the statement true is known as
solving the equation.
3.1.2
Solving linear equations in one variable algebraically
Video:
Watch the video “LinEqV1” on the first part of solving linear equations in one
variable.
Video:
Watch the video “LinEqV2” on the second part of solving linear equations in one
variable.
Linear equations are equations in which the unknown variable only appears to the power one:
3x + 1
=
4x + 3
5A − 40
−5s + 2
=
=
0
s+8
Solving linear equations is simply a matter of juggling and manipulating the equation
until the variable is alone on the left-hand side.
The golden rule is that we can perform the same operation on the expressions on both sides of
the equal sign without altering the solution.
More specifically, we may
1. add (or subtract) the same number or expression to (or from) both sides of the equation;
2. multiply (or divide) both sides of the equation by the same non-zero number or expression.
The new equation obtained by any one of these operations is equivalent to the original equation.
190
3.1. LINEAR EQUATIONS IN ONE VARIABLE
Steps
The following eight steps can be used to solve a linear equation in one variable:
Step 1
Remove the brackets, if any.
Step 2
Add like terms.
Step 3
Eliminate any constants on the left-hand side of the equation
by adding its additive inverse to both sides of the equation.
Step 4
Add like terms.
Step 5
Eliminate any variable on the right-hand side of the equation
by adding its additive inverse to both sides of the equation.
Step 6
Add like terms.
Step 7
If the coefficient of the variable is not one, multiply each side
of the equation by the reciprocal of the coefficient.
Step 8
Solve the equation.
Examples
1. Solve for x in
3x + 5 = 2x − 3.
Step 1
Remove the brackets, if any.
Step 2
Add like terms.
Step 3
Eliminate any constants on the left-hand
side of the equation by adding its additive
inverse to both sides of the equation.
Step 4
Add like terms.
Step 5
Eliminate any variable on the right-hand
side of the equation by adding its additive
inverse to both sides of the equation.
Step 6
Add like terms.
3x + 5 − 5
=
2x − 3 − 5
3x
=
2x − 8
3x − 2x
=
2x − 8 − 2x
x
=
−8
Consider step 5. Because 2x is a positive number, the phrase “adding the additive inverse
of 2x to both sides” of the equation has the same meaning as “subtracting 2x from both
sides” of the equation.
Note that the associative and commutative laws of addition allow us to enter the term that
we are adding at any position on each side of the equation. For example, whether we write
2x − 8 − 2x or − 2x + 2x − 8 or 2x − 2x − 8,
the result is the same.
191
TOPIC 3: LINEAR SYSTEMS
2. Solve for A in
4A − 25 = 0.
Step 1
Remove the brackets, if any.
Step 2
Add like terms.
Step 3
Eliminate any constants on the left-hand
side of the equation by adding its additive
inverse to both sides of the equation.
Step 4
Add like terms.
Step 5
Eliminate any variable on the right-hand
side of the equation by adding its additive
inverse to both sides of the equation.
Step 6
Add like terms.
Step 7
If the coefficient of the variable is not one,
multiply each side of the equation by the
reciprocal of the coefficient.
Step 8
4A − 25 + 25
=
0 + 25
4A
=
25
1 4
× A
4 1
=
1
× 25
4
A
=
Solve the equation.
=
25
4
1
6
4
Consider step 3. Because −25 is a negative number, the phrase “adding the additive inverse
of −25 to both sides” of the equation has the same meaning as “adding 25 to both sides”
of the equation.
From the previous examples it is clear that not all eight steps are always followed when solving
an equation in one variable. It depends on the nature of the problem. It is also not necessary to
write down the steps. The steps were only given for explanation purposes.
Example
Solve
5s − 6 = 10 −
Add 6 to both sides:
s
.
6
s
+6
6
5s − 6 + 6 = 10 −
5s = 16 −
Add
s
6
s
to both sides:
6
s
6
5s +
= 16 −
s s
+
6 6
1
s = 16
6
where we have used the distributive law to add the two terms containing the s.
5+
192
3.1. LINEAR EQUATIONS IN ONE VARIABLE
30
. Therefore, the equation becomes
Writing 5 as a fraction with 6 as the denominator, gives
nnn
6
30 1
+
s = 16
6
6
31
s = 16.
6
6
:
Multiply both sides by
31
31
6
16
6
× s =
×
31
6
31
1
6 × 16
s =
31
96
=
31
Activity
Solve the following:
1. p in 6 − 4(p + 3) = 2(p − 1)
2. h in
13
4 − 5h 1 − 2h
−
=
6
3
42
3. q in
25
5
=
3q
27
Answer
1. The equation is
6 − 4(p + 3) = 2(p − 1).
Remove the brackets:
6 − 4 × p − 4 × 3 = 2 × p + 2 × −1
6 − 4p − 12 = 2p − 2.
−6 − 4p = 2p − 2
Add 6 to both sides:
−6 − 4p + 6 = 2p − 2 + 6
−4p = 2p + 4
Subtract 2p from both sides:
−4p − 2p = 2p + 4 − 2p
−6p = 4
Divide both sides by −6:
−6p
−6
4
−6
2
p = −
3
=
193
TOPIC 3: LINEAR SYSTEMS
Although we have been working step by step here, there is no reason why you cannot use
short cuts and add several terms at once:
6 − 4(p + 3) = 2(p − 1)
6 − 4p − 12 = 2p − 2
−6 − 4p = 2p − 2
−6 − 4p + 6 − 2p = 2p − 2 + 6 − 2p
−6p = 4
4
−6p
=
−6
−6
2
p = −
3
However, if you are at all unsure, rather be on the safe side and work step by step.
2. The equation is
13
4 − 5h 1 − 2h
−
= .
6
3
42
Multiply all the terms by 42 because 42 is the smallest number that is divisible by 42, 6
and 3 (42 = 7 × 6 = 7 × 3 × 2):
42 13
42 4 − 5h 42 1 − 2h
×
−
×
=
×
1
6
1
3
1
42
7(4 − 5h) − 14(1 − 2h) = 13
28 − 35h − 14 + 28h = 13
14 − 7h = 13
14 − 7h − 14 = 13 − 14
−7h = −1
−1
−7h
=
−7
−7
1
h =
7
3. The equation is
25
5
= .
3q
27
Multiply both sides by 27q because 27q is the smallest number that is divisible by both 3q
and 27:
5
27q 25
27q
×
=
×
1
3q
1
27
9 × 5 = 25q
45 = 25q
25q = 45
45
25q
=
25
25
9
q =
5
4
= 1
5
194
3.1. LINEAR EQUATIONS IN ONE VARIABLE
3.1.3
Solve by x-intercept
Linear equations can be solved with a step by step method, but there is no reason why you cannot
use short cuts and add several terms at once. However, if you are unsure about using alternative
methods, rather be on the safe side and work step by step.
There is another method of solving linear equations in one variable. Any linear equation,
ax + b = 0,
can be solved by determining the x-intercept of the line y = ax + b.
Step 1
Manipulate the linear equation, using operations, until it is in the form
ax + b
Step 2
0.
This equation is solved by adding −b to both sides to get
ax + b − b = −b
ax
Step 3
=
−b.
=
Divide by a, assuming a = 0, to obtain the root or solution
b
x = − .
a
b
The equation x = − is the equation for the intercept on the x-axis of the linear function
a
y = ax + b. This intercept occurs at y = 0.
In other words, we can always interpret the root of a linear equation in one variable as the
x-intercept of the corresponding line as shown below:
y
The equation of this line is
y = ax + b.
At this point, y = 0, therefore
ax + b = 0 and the value of x is
−b
x=
.
a
x
Activity
Solve the following by writing them in the form ax + b = 0:
1. 5x + 6 = −3x − 10
2. 1,1x + 3,4 = 2,5x + 1,3
195
TOPIC 3: LINEAR SYSTEMS
Answer
1. The equation is
5x + 6 = −3x − 10.
Add 3x + 10 to both sides:
y
5x + 3x + 6 + 10 = 0
8x + 16 = 0
20
y = 8x + 16
16
We see that a = 8 and b = 16. Therefore,
nnn
x = −
12
nnn
b
a
8
8x + 16 = 0
4
16
= −
8
−3
= −2.
−2
x
−1
−4
1
2
3
−8
−12
2. The equation is
1,1x + 3,4 = 2,5x + 1,3.
Add −(2,5x + 1,3) to both sides:
1,1x − 2,5x + 3,4 − 1,3 = 2,5x − 2,5x + 1,3 − 1,3
(1,1 − 2,5)x + 3,4 − 1,3 = 0
−1,4x + 2,1 = 0
y
We see that a = −1,4 and b = 2,1. Therefore,
5
nnn
nnn
x = −
x =
4
b
a
3
2
−2,1
−1,4
= 1,5.
1
−3
−2
−1
−1
−2
−3
196
−1.4x + 2.1 = 0
x
1
2
3
4
5
y = −1.4x + 2.1
3.1. LINEAR EQUATIONS IN ONE VARIABLE
3.1.4
Word problems
nnn
The following example demonstrates setting up your own linear equation. A linear equation can
be formed by converting a word statement to a linear equation:
“The first number is three less than two times the second number. If their sum is increased by
seven, the result is 37. Find the numbers.”
Determine the second number
Let x be the second number:
x
“The first number is three less than two
times the second number.”
So, the first number is three less than two times x,
giving the first number as
2x − 3
“If their sum...”
The sum of the first and second number is
(2x − 3) + x
=
=
2x − 3 + x
3x − 3
“If their sum is increased by seven, the result is 37.”
This gives the equation
(3x − 3) + 7
=
37
3x − 3 + 7
3x + 4
=
=
37
37
3x + 4 − 4
3x
3x
3
x
=
=
=
37 − 4
33
33
3
11
=
=
2 × 11 − 3
22 − 3
=
19
=
So, the second number is 11.
Determine the first number
(2x − 3)
The first number is
So, the first number is 19.
Check the numbers
These numbers can be checked as follows:
It is true that 19 = 2(11) − 3,
→
The first requirement, “the first
number is three less than two times
the second”, is satisfied.
it is true that 19 + 11 + 7 = 37.
→
The second requirement, “if
their sum is increased by seven,
the resultis 37”, is satisfied.
and
197
TOPIC 3: LINEAR SYSTEMS
Activity
Rewrite the following word statements in the form of a linear equation and solve for the
unknown variable.
1. If one half and one third of a number are added to the number, the result is 44. Find the
number.
Suppose the number is x. Add one half and one third of x to x and solve for x.
1
5
2. A person travelled of the distance by train, by bus and the remaining 15 kilometres by
8
4
boat. Find the total distance that he travelled.
Start by setting the total distance travelled by the person equal to d and solve for d.
Answer
1. Setting and solving the equation gives
nnn
1
1
× x + × x + x = 44
2
3
6 1
6 1
× × x + × × x + 6 × x = 6 × 44
1 2
1 3
3x + 2x + 6x = 264
The number is 24.
To check:
One half of 24 is 12,
one third of 24 is 8 and
12 + 8 + 24 = 44.
11x = 264
264
11x
=
11
11
x = 24.
2. Setting and solving the equation gives
nnn
1
5
× d + × d + 15 = d
8
4
8 1
8 5
× × d + × × d + 8 × 15 = 8 × d
1 8
1 4
5d + 2d + 120 = 8d
7d + 120 = 8d
7d + 120 − 120 − 8d = 8d − 120 − 8d
−d = −120
−120
−d
=
−1
−1
d = 120.
198
The total distance travelled is 120 km.
To check:
5 120
600
×
=
= 75,
8
1
8
120
1 120
×
=
= 30 and
4
1
4
75 + 30 + 15 = 120.
3.2. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
Theme 3.2
Systems of linear equations in two variables
Learning objective
On completion of this theme you should be able to solve
simultaneous linear equations in two variables algebraically.
In this theme we consider linear equations in two variables. Such an equation has the form
ax + by = c.
The graph of a linear equation in two variables is a straight line. A solution of a linear equation
will be any (x; y) point that will make the equation true.
To illustrate, consider the graph of the line:
3
y =− x+2
4
y
6
3
y =− x+2
4
5
Consider point (0; 2).
Substitute x = 0 into equation
3
y = − (0) + 2
4
= 2.
If x = 0, then y = 2, which is true.
4
Consider point (2; 2).
Substitute x = 2 into equation
3
y = − (2) + 2
4
= 0,5.
If x = 2, then y = 0,5, which is not
equal to 2 and therefore, not true.
3
2
1
−2
−1
−1
x
1
2
3
4
5
6
7
−2
Consider the point (0; 2) on the line:
Equation of the line
y
=
Substitute x = 0 into the equation
y
=
=
3
x+2
4
3
×0+2
4
2
So, if x = 0, then y = 2, which is true.
Consider the point (2; 2) not on the line:
Equation of the line
y
=
Substitute x = 2 into the equation
y
=
3
x+2
4
=
3
×2+2
4
−1,5 + 2
=
0,5
So, if x = 2, then y = 0,5, which is not equal to 2 and therefore not true.
199
TOPIC 3: LINEAR SYSTEMS
3
In fact, all the points on the line y = x + 2 are solutions of the linear equation. There is actually
4
no limit to the number of solutions, since there is no limit to the number of points on the line.
Every point on the line graph is a solution of the linear equation and
any solution of the linear equation is a point on the line graph.
3.2.1
Systems of equations
A system of linear equations in two variables is a set or collection of linear equations whose
common solutions are looked for. The simplest linear system is one with two equations and two
variables. We can also refer to it as simultaneous linear equations or a linear system.
Consider the following system with two linear equations and two variables:
5
x−5
2
y = −x + 1
y =
Since the two equations are in a system, we work with them at the same time. In particular, we
can graph them together on the same set of axes.
3.2.2
Solving systems of equations
A solution for a single equation is any point that lies on the line for that equation. A solution
for a system of equations is any point that lies on each line in the system. For example, the
point (2; −3) is not a solution of the system, because it is not on either line.
y
5
y = x−5
2
2
1
−2
−1
−1
x
1
2
3
4
5
−2
−3
−4
−5
200
(2; −3)
y = −x + 1
3.2. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
The point (0,5; 0,5) is also not a solution of the system, because it lies on only one of the lines,
not on both of them.
y
5
y = x−5
2
2
(0,5; 0,5)
1
−2
x
−1
−1
1
2
3
4
5
−2
y = −x + 1
−3
−4
−5
The point (1,71; −0,71) is a solution of the system, because it lies on both of the lines.
y
5
y = x−5
2
2
1
−2
−1
−1
x
1
2
3
4
5
(1,71; −0,71)
−2
−3
y = −x + 1
−4
−5
In particular, this point is the intersection of the two lines. Since this point is on both lines, it
thus satisfies (or is a solution of) both equations, so it satisfies the entire system of equations.
For systems of equations, the solutions are the points of
intersections of the corresponding lines.
You can confirm the solution by plugging it into the system of equations, and confirming that
the solution satisfies each equation.
201
TOPIC 3: LINEAR SYSTEMS
Three methods can be used to solve linear equations in two variables. To learn about them, watch
the following videos:
Method of solution
Video to watch
1. Graphing
SystemsSolveGraphs
2. Substitution
SystemsSub
3. Elimination
SystemsElimV1_HD
SystemsElimV2_HD
The method for solving a systems of linear equations using substitution, is as follows:
Step 1
Isolate a variable in an equation.
Step 2
Substitute the result of step 1 into the other equation and solve
for one variable.
Step 3
Substitute the result of step 2 into the result of step 1 and solve
for the other variable.
Step 4
State the solution.
Example
Use the method of substitution to solve the following system of linear equations:
5x − y = 1
4x − y = −1
Step 1
Step 2
5x − y
=
1
(Subtract 5x both sides.)
−y
=
−5x + 1
(Divide both sides by −1.)
y
=
5x − 1
4x − y
=
−1
(Substitute y = 5x − 1 in.)
4x − (5x − 1)
=
−1
(Subtract 1 both sides.)
4x − 5x + 1
−x + 1 − 1
=
=
−1
−1 − 1
−x
=
−2
x
=
2
into the result of step 1, (y = 5x − 1),
and find the value of y:
y
=
5x − 1
(Substitute x = 2 in.)
(Calculate and simplify.)
y
=
=
5(2) − 1
10 − 1
=
9
Isolate y in the first equation:
Substitute result of step 1, (y = 5x − 1),
into the second equation and find value of x:
(Divide both sides by −1.)
Step 3
Step 4
202
Substitute the result of step 2, (x = 2),
Since x = 2 and y = 9, the solution is
(2; 9)
3.2. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
When the lines are drawn you can see that they intersect at (x; y) = (2; 9):
y
11
10
(2; 9)
9
8
7
6
5
4
3
2
5x − y = 1
1
−2
4x − y = −1
−1
−1
x
1
2
3
4
5
6
7
−2
−3
Now, the method of elimination will be demonstrated on the same system of linear equations:
5x − y
=
1
(1)
4x − y
=
−1
(2)
Multiply each term in equation (1) by −1:
−5x + y = −1
to obtain equation (3). Consider equations (3) and (2):
−5x + y
4x − y
=
=
−1
−1
(3)
(2)
Adding equations (3) and (2) gives
−5x + 4x + y − y = −1 − 1
−x = −2
x = 2.
To obtain the value of y, the value of x = 2 can be substituted back into equation (1) or equation
(2).
203
TOPIC 3: LINEAR SYSTEMS
To demonstrate that is does not matter which one of the two equations is used, both will be used:
nnn
Substituting x = 2 into equation (1) gives
Substituting x = 2 into equation (2) gives
5x − y = 1
4x − y = −1
5×2−y = 1
4 × 2 − y = −1
10 − y = 1
8 − y = −1
−y = 1 − 10
−y = −1 − 8
−y = −9
−y = −9
y = 9.
y = 9.
We have the same solution as obtained previously, namely (x; y) = (2; 9).
Activity
Use the method of elimination to solve the system of linear equations:
2x + 5y
=
13
(1)
3x + 4y
=
9
(2)
Answer
It does not matter which terms are eliminated from the equations first. In this case the xterms will be eliminated first. The coefficients of the x-terms need to be the same (but with
different signs), in order for the x-terms to be eliminated from the equations. Therefore,
multiply each term in equation (1) by 3 and multiply each term in equation (2) by −2:
2x × 3
3x × −2
+
+
5y × 3
4y × −2
=
=
13 × 3
9 × −2
(1)
(2)
to obtain equations (3) and (4). The x-coefficient of equation (3) is 6 and of equation (4)
is −6:
6x + 15y =
39
(3)
−6x − 8y
= −18
(4)
Add equations (3) and (4) and solve for y:
6x − 6x + 15y − 8y = 39 − 18
7y = 21
21
y =
7
= 3
Substitute y = 3 back into any one of the original equations, (1) or (2), to obtain the value
of x. Choose equation (2):
3x + 4y = 9
3x + 4 × 3 = 9
3x + 12 = 9
3x = −3
x = −1
The solution is (x; y) = (−1; 3).
204
3.2. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
3.2.3
Word problems
There are many situations or problems in the business environment where the values of some
unknown quantities have to be found. We can represent the unknown quantities by the names of
two variables and form two linear equations involving the variables.
Example
A company publishes two magazines, namely Fresh Food and Big Bite. It costs R4 to print a
copy of Fresh Food and R6 to bind it. It costs R5 to print a copy of Big Bite and R3 to bind it.
The budget of the company only allows R21 000 for binding and R20 000 for printing.
Set up two linear equations that will determine how many copies of each magazine can be published.
Let x represent the number of Fresh Food magazines that are published.
Let y represent the number of Big Bite magazines that are published.
Printing
The budget of the company only allows R20 000 for printing. It costs R4 to print a copy of Fresh
Food and R5 to print a copy of Big Bite.
The equation that represents the cost and budget for printing is
4 × Fresh Food printed
+
5 × Big Bite printed
=
20 000
4x
+
5y
=
20 000.
So, the equation that represents the cost and budget for printing is
4x + 5y = 20 000.
Name this equation one.
Binding
The budget of the company only allows R21 000 for binding. It costs R6 to bind a copy of Fresh
Food and R3 to bind a copy of Big Bite.
The equation that represents the cost and budget for binding is
6 × Fresh Food bound
+
3 × Big Bite bound
=
21 000
6x
+
3y
=
21 000.
So, the equation that represents the cost and budget for printing is
6x + 3y = 21 000.
Name this equation two.
205
TOPIC 3: LINEAR SYSTEMS
Activity
The price of a pen is R11 more than the price of a pencil.
The price of a pen is represented by P and the price of a pencil is represented by R.
If you add R3 to the price of the pen and R3 to the price of a pencil, the price of the pen
will be twice the price of the pencil.
Without solving, write down two linear equations that will determine the values of P and
R.
Answer
If the price of a pen is R11 more than the price of a pencil, then
price of pencil
R
+
+
R11
11
=
=
price of pen
P
R
P
−
−
P
R
=
=
−11
11.
Add R3 to the price of a pen and R3 to the price of a pencil. Therefore, the price of a pen
is equal to P + 3 and the price of a pencil is equal to R + 3. But the price of a pen will
be double the price of a pencil, or the price of two pencils will be equal to the price of one
pen. Thus, the mathematical expression is
new price of pen = 2 × the new price of a pencil
P + 3 = 2 (R + 3)
P + 3 = 2R + 6
P + 3 − 2R = 6
P − 2R = 6 − 3
P − 2R = 3.
The two equations are
P − R = 11
P − 2R =
206
3.
3.2. SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES
3.2.4
Summary
• A system of linear equations in two variables is just a set of two or more equations.
The simplest case is two linear equations in two variables. The graph of such a system is a
pair of lines in the Cartesian plane.
• The solution of a system of two linear equations can be classified as follows:
No solution
The slopes of the lines are the same, but their
y-intercepts are different.
The lines are parallel.
The lines do not intersect at any point.
One solution
Infinitely
many solutions
The slopes of the lines are different.
The lines intersect at exactly one point.
The slopes of the lines are the same, but their
y-intercepts are also the same.
The two equations represent the same line.
The lines intersect at infinitely many points.
• Word problems. Sometimes the equations are not given directly, but instead a description
of a real life problem is given. The given sentences need to be translated into more than
one equation. If there are two unknown variables then at least two equations are needed to
solve the variables.
Exercise 3.1
Solve the following systems of equations:
1. 7x + 5y = −4 and 3x + 4y = 2
1
2. 2x + 2y = 3 and 5x + y = −6
2
3. x + 4y = 49 and − 2x + y = 1
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TOPIC 3: LINEAR SYSTEMS
Theme 3.3
Linear inequalities in one variable
Learning objective
3.3.1
On completion of this theme you should be able to solve a linear
inequality in one variable.
Introduction
Greater than/less than
The concept
The concept
“greater than”
“less than”
is denoted by the symbol
is denoted by the symbol
>.
<.
We say b is greater than a if and only if the difference b − a is positive.
Thus,
b>a
For example
4 > 2,
means
b − a > 0.
4 − 2 = 2 and 2 > 0.
We say b is less than a if and only if the difference b − a is negative.
Thus,
b<a
For example
2 < 4,
means
b − a < 0.
2 − 4 = −2 and −2 < 0.
If b = a or a = b, then a and b are the same number.
Greater than or equal to/less than or equal to
Sometimes the symbol = is combined with > or < as follows:
b≥a
means
b − a ≥ 0,
b
is greater than or equal to
a,
b≤a
means
b − a ≤ 0,
b
is less than or equal to
a.
If we replace the = sign in any equation, or system of equations, with a
>, <, ≥ or ≤,
then we obtain an inequality or system of inequalities.
The inequality, or system of inequalities, is linear or non-linear depending on whether the functions
involved are linear or non-linear.
3.3.2
Solving an inequality
As is the case with equations, a solution to an inequality is a value that makes the inequality
true.
For example, if the inequality
5x − 15 < 0
had to be solved, it means that you have to determine the values of x for which this statement is
true.
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3.3. LINEAR INEQUALITIES IN ONE VARIABLE
How is this done? You can solve inequalities in the same way you can solve equations, by following
these rules:
Rules to solve inequalities
Rule 1
If b is greater than a, then a is less than b.
Rule 2
If b is greater than a and a is greater than c, then b is greater than c.
Rule 3
You may add the same number to both sides of an inequality.
Rule 4
You may multiply or divide both sides of an inequality by any positive
number.
Rule 5
When you multiply or divide both sides of an inequality by a negative
number, reverse the direction of the inequality sign.
Rule 1
If b is greater than a, then a is less than b.
If
b>a
then
a < b.
Example:
4>1
and
1<4
Although you might say that this is obvious, it is nevertheless very useful at times when we want
to interchange the left- and right-hand sides.
Rule 2
If b is greater than a and a is greater than c, then b is greater than c.
If
b>a
and
a>c
then
b > c.
Example:
23 > 21
and
21 > 20
therefore
23 > 20
If your sister is older than you and you are older than 20 years, then your sister must also be
older than 20.
Rule 3
You may add the same number to both sides of an
inequality. This number can be positive or negative.
For any c, if
b>a
then
b + c > a + c.
Example:
4>1
then
4+3 >1+3
or
7>4
Also:
4>1
then
4−5 >1−5
or
−1 > −4
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TOPIC 3: LINEAR SYSTEMS
Rule 4
You may multiply or divide both sides of
an inequality by any positive number.
For any c > 0, if
b>a
then
b × c > a × c.
Example:
Also:
4>1
4>1
then
then
4×8>1×8
4 ÷ 20 > 1 ÷ 20
or
or
32 > 8
0,2 > 0,05
Rule 5
Regarding multiplying or dividing both sides of an inequality by a negative number, consider the
following:
On the number line, the rule of order states that any number on the number line that is to the
right of another number, is greater than that number that is to the left of it.
Example
Consider the inequalities 3 > 2 and −1 < 4:
−4
−3
3 is to the right of 2.
3 is greater than 2.
−1 is to the left of 4.
−1 is less than 4.
3>2
−1 < 4
−2
−1
1
2
3
4
−4
−3
−2
−1
1
2
3
4
Multiplying or dividing by a negative number switches the numbers to the opposite side of the
number line relative to zero:
3 × −1 = −3
4 × −1 = −4
−1 × −1 = 1
2 × −1 = −2
−4
−3
−2
−1
1
2
3
4
−4
−3
−2
−1
1
2
3
4
3
4
The numbers end up on opposite sides of each other:
−4
−3
−3 is to the left of −2.
1 is to the right of −4.
−3 is less than −2.
−3 < −2
1 is greater than −4.
1 > −4
−2
−1
1
2
3
4
−4
−3
−2
−1
1
2
When you multiply or divide both sides of an inequality by a negative number,
reverse the direction of the inequality sign.
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3.3. LINEAR INEQUALITIES IN ONE VARIABLE
Example
Solve the inequality 5x − 15 < 0.
Inequality in one variable
5x − 15
<
0
Add +15 both sides
5x − 15 + 15
5x
<
<
0 + 15
15
5x
5
x
Divide both sides by 5
<
<
15
5
3
This is indicated on the number line by highlighting all the points for which the inequality is true.
The arrow shows that all the values on the number line less than 3 are in the solution. Since the
point x = 3 is not included, it is shown by means of an open circle:
x<3
−6
−5
−4
−3
−2
−1
1
2
3
4
5
6
x
If the solution had been x is less than or equal to 3, the circle would be a dark or a shaded
circle:
x≤3
−6
−5
−4
−3
−2
−1
1
2
3
4
5
6
x
How can we check our answer?
We cannot use 3 to substitute in the inequality, because it lies outside our solution. To check, we
can choose any value that lies in the solution and substitute it into the inequality. Let’s use −2:
5x − 15 < 0
5 × −2 − 15 < 0
−10 − 15 < 0
−25 < 0,
which is correct. Our substitution gave a correct result, so −2 is a solution of the inequality.
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TOPIC 3: LINEAR SYSTEMS
3.3.3
Miscellaneous
Greater than
With a greater than inequality, all numbers larger than x, but not equal to x, satisfy the
inequality. Consider
x > 2.
Any number larger than 2 satisfies the inequality.
For example 2,1; 2,2; 3; 4; 17; . . .
−6
−5
−4
−3
−2
−1
1
2
3
4
5
6
x
Less than
With a less than inequality, all numbers smaller than x, but not equal to x, satisfy the
inequality. Consider
x < 2.
Any number smaller than 2 satisfies the inequality.
For example 1,9; 1,8; 1; 0; −1; −100; . . .
−6
−5
−4
−3
−2
−1
1
2
3
4
5
6
x
Greater than or equal to
With a greater than or equal to inequality, all numbers larger than x, and equal to x, satisfy
the inequality. Consider
x ≥ 2.
Any number that is 2 or is larger than 2 satisfies the inequality.
For example 2; 2,1; 2,2; 3; 4; 17; . . .
−6
−5
−4
−3
−2
−1
1
2
3
4
5
6
x
Less than or equal to
With a less than or equal to inequality, all numbers smaller than x, and equal to x, satisfy
the inequality. Consider
x ≤ 2.
Any number that is 2 or is smaller than 2 satisfies the inequality.
For example 2; 1,9; 1,8; 1; 0; −1; −100; . . .
−6
212
−5
−4
−3
−2
−1
1
2
3
4
5
6
x
3.3. LINEAR INEQUALITIES IN ONE VARIABLE
Variable on right side
nnn
If you need to solve an inequality, but the variable is on the wrong side of the inequality sign,
what do you do?
Solve 10 > m + 3.
Solve the inequality
10
>
m+3
Add −3 both sides
10 − 3
7
>
>
m+3−3
m
Rewrite this inequality
m
<
7
Reason
If b is greater than a, then a is less than b.
If the variable is on the wrong side of the inequality,
swop the sides and reverse the inequality sign.
Multiply or divide by a variable
We can multiply and divide both sides of an inequality by any positive or negative number.
However, when we multiply or divide both sides of an inequality by a variable (an unknown
value) we need to be more careful.
Solve hx < 4h.
Solve the inequality for x
hx
<
4h
Divide both sides by h, but h > 0 or h < 0
If h is positive
Say h = 2
If h is negative, reverse the inequality sign
Say h = −2
hx
h
x
2x
2
x
hx
h
x
−2x
−2
x
<
<
<
<
>
>
>
>
4h
h
4
4×2
2
4
4h
h
4
4 × −2
−2
4
You cannot divide an inequality by a variable unless you are sure of its sign
since you do not know whether you must reverse the sign of the inequality.
When dealing with an inequality, you must always ask yourself whether you know what the sign
of the variable is before dividing or multiplying.
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TOPIC 3: LINEAR SYSTEMS
3.3.4
Word problems
Example
Jacob has R2 000 in his savings account at the beginning of the holiday. He wants to have at
least R600 in his account by the end of the holiday. He withdraws R280 each week for snacks
and entertainment. For how many weeks can he withdraw money from his account?
Write an inequality that represents Jacob’s situation.
Let n be the number of weeks.
Unravel the inequality from the information given:
Information
Inequality
He started with R2 000.
2 000
He withdraws money from R2 000.
2 000 −
He withdraws R280 every week.
2 000 − 280 × n
He wants to have R600 left in his account by the end
2 000 − 280 × n
600
of the holiday.
2 000 − 280 × n ≥ 600
It must be at least R600. There must be R600 or more
than R600 left in his account.
How many weeks can Jacob withdraw money from his account?
Solve the inequality to find the value of n:
Solve the inequality for n
2 000 − 280n
≥
600
Add −2 000 both sides
−280n
≥
600 − 2 000
−280n
≥
−1 400
Divide by −280, so reverse sign to get ≤
−280n
−280
n
≤
≤
−1 400
−280
5
Therefore, Jacob can withdraw money from his account for five or less weeks and will still have
R600 or more left in his account.
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3.3. LINEAR INEQUALITIES IN ONE VARIABLE
Key words
When you are solving word problems, pay close attention to the key words given below. These
keywords will help you to write the inequality.
Once the inequality is written, you can solve it using the following steps:
Step 1
Read through the entire problem.
Step 2
Highlight the important information and key words that you need to
solve the problem.
Step 3
Identify your variable.
Step 4
Write the inequality.
Step 5
Solve.
Here are a few key words that we associate with inequalities:
Key words
at least
is not less than
Meaning
greater than or equal to
≥
less than or equal to
≤
greater than
>
less than
<
is not smaller than
minimum
at most
is not more than
is not greater than
maximum
is more than
is greater than
is larger than
above
is less than
is smaller than
below
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TOPIC 3: LINEAR SYSTEMS
3.3.5
Summary
• Many simple inequalities can be solved using the techniques of adding, subtracting, multiplying or dividing both sides of the inequality until you are left with the variable on its
own.
• We discussed the following rules:
If b > a
then a < b.
If b > a
and if a < c,
then b > c.
If b > a
then b + c > a + c
for any c.
If b > a
then b × c > a × c
and b ÷ c > a ÷ c
for any c > 0.
If b > a
then b × c < a × c
and b ÷ c < a ÷ c
for any c < 0.
If you multiply or divide both sides of an inequality by a negative
number, reverse the direction of the inequality sign.
• Word problems. Sometimes the inequality is not given directly, but instead a description
of a real-life problem is given. The given sentences need to be translated into an inequality.
• If you want to swop the expressions on either sides of an inequality, remember to reverse
the inequality sign.
• You cannot divide an inequality by a variable unless you are sure of its sign since you do
not know whether you must reverse the sign of the inequality.
Exercise 3.2
Solve the following:
1. 11 ≥ 6 − 4x
2. 4x + 4 < 1,5x − 6
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3.4. SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES
Theme 3.4
Systems of linear inequalities in two variables
On completion of this theme you should be able to solve a system
of linear inequalities in two variables.
Learning objective
3.4.1
Linear inequalities in two variables
Although the rules in the previous theme may be applied to linear inequalities in two or more
variables, they are not of much use when it comes to solving a system of linear inequalities in
two or more variables. The trouble is that the solution is not generally a single point, but usually
an infinite sequence of points. In fact, in the two variable cases it is usually even more than that
– it is a whole area or region in the x–y plane. This means that the most successful approach to
solving linear inequalities in two variables is by means of graphs.
Suppose we have to draw a graph with a set of points that obey the inequality
−x + 2y − 2 ≥ 0.
Now, if we ignore the > sign for a moment and just consider the = sign, we can easily draw a
graph of
−x + 2y − 2 = 0.
1
This is the straight line y = x + 1 depicted in the figure below:
2
y
−x + 2y − 2 = 0
or
1
y = x+1
2
4
(3; 3)
3
(−1; 2)
(3; 2)
2
(1; 1)
1
(−5; 0)
−5
(4; 1)
(−2; 0)
−4
−3
(−4; −1)
−2
−1
−1
x
1
2
3
4
5
−2
Now any point on the line satisfies the original inequality (or rather the = part of it). Thus, all
points on the line are in the solution space. What about points not on the line? Well, the simple
approach is simply to test a few and see whether they satisfy the inequality or not.
Consider (1; 1). Substitute it into the left-hand side of the inequality, −x + 2y − 2 ≥ 0. It does
not satisfy the inequality since
−1 + 2 × 1 − 2 = −1
and
−1 0
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TOPIC 3: LINEAR SYSTEMS
(read this as “is not greater than or equal to 0”). Also note that it lies below the straight line
on the graph. So, too, do the points (4; 1) and (3; 2). Nor do they satisfy the inequality. On
the other hand, the points (3; 3), (−1; 2) and (−5; 0) all satisfy the inequality and all lie above
the straight line. For example, when (3; 3) is substituted into the left-hand side of the inequality,
−x + 2y − 2 ≥ 0, we find
−3 + 2 × 3 − 2 = 1
and
1 ≥ 0.
If we carry on in this fashion we will find that all points above the line satisfy the inequality,
whereas all points below the line do not. Thus, the line divides the x–y plane into two regions
– those that satisfy the inequality and those that do not. In this case the points on the line,
for example (−2; 0) and (−4; −1), also satisfy the inequality (since the = sign is included in the
statement). We can use this result as the basis of a prescription for graphing inequalities.
Video:
Watch the video “LinIneq_myU” on linear inequalities in two variables.
Rules for graphing linear inequalities
Rule 1
Graph the line that results when the inequality is changed to an equality.
Rule 2
Select any point not on the line.
Rule 3
If the coordinates of the point satisfy the inequality, then all points on the same
side of the line satisfy the inequality.
Rule 4
If the coordinates of the point do not satisfy the inequality, then all points on the
opposite side of the line satisfy the inequality.
Rule 5
If the = sign is part of the inequality, then the points on the line also satisfy
the inequality, otherwise they do not.
This means that all you have to do in order to solve a single inequality in two variables is to draw
the corresponding straight line and to examine a single point in the plane.
Boundary line
Using the Cartesian plane is especially helpful for understanding the range of possible solutions
for inequalities with two variables. When a linear inequality with two variables is graphed in the
Cartesian plane, the full range of possible solutions is represented as a shaded area in the plane.
The boundary line for the inequality is drawn as a solid line if the points on the line itself do
satisfy the inequality, as in the cases of ≤ and ≥.
The boundary line for the inequality is drawn as a dashed line if the points on the line do not
satisfy the inequality, as in the cases of < and >.
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3.4. SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES
3.4.2
Systems of two linear inequalities
Just as systems of linear equations can be formulated, so can systems of linear inequalities. When
solving a system, we must determine all points that simultaneously satisfy all linear inequalities
in the system.
A system of inequalities means that there is more than one line involved. The lines must
be drawn on one set of axes and therefore there is only one solution space.
Once again, the solution is generally a region in the x–y plane. This is demonstrated by an
example.
Video:
Watch the video “Systems” on solving systems of linear inequalities in two variables.
Consider the following system of inequalities:
−x + y − 1 ≤ 0 and 2x + y − 4 < 0
If we examine each inequality separately, we can graph its solution along the lines discussed in
the previous section. The solution of the first inequality is the region including all points on the
line −x + y − 1 = 0, and all those below and to the right of this line. The solution of the second
inequality is all points below and to the left of the line 2x + y − 4 = 0, but not the points on the
line.
The two solutions are depicted graphically below, the first with vertical lines and the second with
horizontal lines.
y
5
−x + y − 1 = 0
4
3
2
(1; 2)
↓
1
−5
−4
−3
−2
−1
−1
x
1
2
3
4
−2
−3
−4
2x + y − 4 = 0
Now the region in the figure where the horizontal and vertical lines cross, the so-called crosshatched or grey space, is the region in which both inequalities are satisfied. It is, in other words,
the solution of the system of inequalities. Note that the first line is included in the solution
while the second is not, as is indicated by dashing the second line. An important point is the
219
TOPIC 3: LINEAR SYSTEMS
corner of the region where the two lines intersect at (1; 2). This is known as an extreme point
of the region of solution.
To determine the solution of a system of inequalities, solve each inequality
separately and determine the region that is common to all solutions. This
region is known as the solution space of the system of inequalities.
Activity
Solve the system of inequalities graphically:
x + y + 2 ≥ 0 and −x + 2y + 2 < 0
Answer
The inequalities are
x + y + 2 ≥ 0 and −x + 2y + 2 < 0.
The solution of the first inequality is the region including all the points on, above and to
the right of the line
x + y + 2 = 0.
This is the region indicated by the horizontal lines in the sketch below. The solution of the
second inequality is the region including all points below and to the right of the line
−x + 2y + 2 = 0,
but not the points on the line. This region is indicated by the vertical lines in the sketch
below. The solution of the system of two inequalities is the cross-hatched region.
y
3
2
1
−4
−3
−2
4
2
− ;−
3
3
−x + 2y + 2 = 0
−1
−1
x
1
2
3
4
5
→
−2
−3
−4
−5
−6
x+y+2=0
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3.4. SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES
Determine whether a point is a solution of the system of inequalities
nnn
A system of linear inequalities is a set of linear inequalities whose common solutions need to be
determined.
When solving a system of linear inequalities, you must determine all points that simultaneously
satisfy all linear inequalities in the system.
This is done by graphing each individual inequality and then finding the overlap of the various
solution spaces.
Any point that satisfies all the inequalities simultaneously will be in the area of this overlap.
Example
Determine if the point (3; −1) is a solution of the system of inequalities:
x+y ≤ 6
2x − y > 4
Substituting (3; −1) into
Substituting (3; −1) into
the first inequality gives
x+y ≤ 6
3 + (−1) ≤ 6
3−1 ≤ 6
2 ≤ 6.
the second inequality gives
2x − y > 4
2 × 3 − (−1) > 4
6+1 > 4
7 > 4.
This is true.
This is true.
Therefore, the point (3; −1) is a solution of the system of inequalities.
The graph shows that point (3; −1) lies in the area of the overlap of the solutions of the inequalities
(the area that is shaded gray). Therefore, it is a solution of the system of inequalities.
y
7
6
5
x+y ≤6
2x − y > 4
4
3
2
1
−1
−1
−2
x
1
2
3
4
5
6
7
(3; −1)
−3
−4
−5
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TOPIC 3: LINEAR SYSTEMS
3.4.3
Systems of more linear inequalities
So far we have only considered systems of two inequalities in two unknowns. However, it is
quite feasible, and in practice very often necessary, to consider systems with a greater number of
inequalities than the number of unknowns. This is illustrated below, where we consider a system
of five inequalities in two unknowns.
The procedure for solving the system is exactly as before, namely solve each inequality separately
and then determine which region is common to all solutions.
Example
Solve the system of five inequalities graphically:
−x + y − 3 ≤ 0
x+y−5 ≤ 0
x−3 ≤ 0
x ≥ 0
y ≥ 0
We look at the last two of these inequalities first. The inequality x ≥ 0 simply means the region
to the right of and including the y-axis, whereas y ≥ 0 means the region above and including the
x-axis. Taken together, these two inequalities imply the first quadrant of the x–y plane, so we
can restrict our considerations to this region.
The first inequality is satisfied by all points on, below and to the right of the line
or
−x + y − 3
=
0
y
=
x + 3.
The second is satisfied by all points on, below and to the left of the line
or
x+y−5
y
=
=
0
−x + 5.
x−3
≤
0
x
≤
3
Finally, the third inequality
or
is satisfied by all points on and to the left of the vertical line x = 3.
The solution is the overlap region of the solutions of all these inequalities, that is the crosshatched area shown in the graph on the following page.
Note that the solution is called a closed or bounded solution, because there are lines on all sides
of the solution (i.e. the overlap region as indicated on the graph).
222
3.4. SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES
y
x=3
6
5
(1; 4)
4
3
(3; 2)
2
1
−4
−3
−x + y − 3 = 0
−2
−1
−1
x
1
2
3
4
5
x+y−5=0
−2
3.4.4
Word problems
Example
The Fit For Life running club is trying to raise money for new equipment. The men are selling
water bottles for R20 a bottle and the women are selling sweat bands for R40 each. The women
expect to sell at least 100 sweat bands. Jointly they must raise more than R8 000.
Let x represent the number of water bottles sold.
Let y represent the number of sweat bands sold.
Write a system of inequalities that represents this situation.
Write the inequality representing the income:
or
20x + 40y
2x + 4y
>
>
8 000
800
(More than R8 000 implies >.)
Write the inequality representing the sales of the sweat bands:
y
≥
100
(At least 100 sweat bands implies ≥ 100.)
Write the inequalities representing positive or zero sales:
x
≥
0
(You cannot sell a negative amount of water bottles.)
y
≥
0
(You cannot sell a negative amount of sweat bands.)
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TOPIC 3: LINEAR SYSTEMS
Activity
Katie works part-time at the Fallbrook Riding Stable. She makes R50 an hour for exercising
horses and R30 an hour for cleaning stalls. Because Katie is a full-time student, she cannot
work more than 12 hours per week. She plans to earn not less than R450 per week.
Let x represent the number of hours Katie exercises horses.
Let y represent the number of hours Katie cleans stalls.
Write a system of inequalities that illustrates how many hours Katie needs to work at each
job.
Answer
The correct system of inequalities is represented by
≥
≤
≥
≥
5x + 3y
x+y
x
y
3.4.5
45
12
0
0.
No solution
Find the solution of the following system of inequalities:
≤
≥
x+y
x+y
−3
3
Drawing the lines and their solution spaces gives the following:
y
5
x+y ≥3
4
3
2
1
−5
−4
−3
−2
−1
−1
x
1
2
3
4
5
−2
−3
x + y ≤ −3
−4
−5
There is no place where the individual solution spaces overlap. The lines x+y = −3 and x+y = 3
never intersect because they are parallel lines with different y-intercepts.
Since there is no intersection, there is no solution.
224
3.4. SYSTEMS OF LINEAR INEQUALITIES IN TWO VARIABLES
3.4.6
Summary
• A system of linear inequalities is a set of linear inequalities whose common solutions are
looked for.
• The steps for graphing systems of inequalities are as follows:
Step 1
Graph the boundary line for the first inequality.
Step 2
Use a test point to determine which half plane to shade. Shade the
half plane that contains the solutions to the first inequality.
Step 3
Graph the boundary line for the second inequality.
Step 4
Use a test point to determine which half plane to shade. Shade the
half plane that contains the solutions to the second inequality.
Step 5
Do this for all the inequalities.
Step 6
Analyse your system of inequalities and determine which area
is shaded by the solution spaces of all the inequalities.
This area is the solution for the system of inequalities.
Step 7
Any point that satisfied all the inequalities at once will be in
the area of this overlap.
• If the inequality symbol is greater than (>) or less than (<), then you will use a dashed
boundary line. This means that the solutions are not included on the boundary line.
• If the inequality symbol is greater than or equal to (≥) or less than or equal to (≤), then
you will use a solid line to indicate that the solutions are included on the boundary line.
• If the solution space for the system continues forever in at least one direction, it is called
an unbounded solution.
• If the solution space for the system has lines on all sides, it is called a closed or bounded
solution.
• If the solution spaces for the individual lines do not overlap, the system has no solution.
225
TOPIC 3: LINEAR SYSTEMS
Exercise 3.3
1. Draw the linear inequality
3x + y − 3 > 0
on a graph.
2. Draw the linear inequality
2x + 4y + 1 ≤ x + y − 2
on a graph.
3. Solve the following system of inequalities graphically:
2x + y − 5 ≤ 0
x−2 ≤ 0
y−4 ≤ 0
x ≥ 0
y ≥ 0
4. Solve
3x − 7 ≤ 5x + 2
and indicate your solution on the number line.
5. Solve
5x + y + 1 < −x − y − 1
graphically.
226
Topic
4
Mathematics of finance
On completion of this topic you should be able to
• calculate simple interest and compound interest
• determine the present value and future value of an annuity
• set up an amortisation schedule for a loan
• reschedule payments on a loan for changes in interest rate or term
CONTENTS
Theme
4.1
Simple interest
4.2
Simple discount
4.3
Compound interest
4.4
The time value of money
4.5
Annuities
4.6
Amortisation
TOPIC 4: MATHEMATICS OF FINANCE
Theme 4.1
Simple interest
Learning objectives On completion of this theme you should be able to
• calculate simple interest
• manipulate the formula in order to obtain expressions for the
respective variables
Money is not free to borrow.
People can always find a use for money, so it costs money to borrow money.
How much does it cost to borrow money?
Different places charge different amounts at different times, but they usually charge it as a
percentage (%) per year of the amount borrowed.
Interest is the price paid for the use of borrowed money.
Think of interest as the rent you have to pay for using someone else’s money.
Did you know that there is a cost of money even if you are using only your own money? How
does it work? Since you are using your own money instead of putting it into a savings account,
you are foregoing the 8% interest you could make on it. In this case the cost of money is actually
8%!
4.1.1
How it works and formulas
Interest is paid by the party who uses or borrows the money to the party who lends the money.
Interest is calculated as a fraction of the amount borrowed or saved (the principal amount) in a
time period. The fraction, also known as the interest rate, is usually expressed as a percentage
per year, but it must be reduced to a decimal fraction for calculational purposes. For example, if
we have borrowed an amount from the bank for a period of one year at an interest rate of 12%
per year, we can express the interest as
12% of the amount borrowed × number of periods borrowed,
that is,
12
of the amount borrowed × one year,
100
that is,
0,12 × the amount borrowed × one year.
When and how interest is calculated give way to different types of interest. For example, simple
interest is interest that is calculated at the end of the entire term for which the money was used,
on the original amount (principal amount) used (borrowed or saved).
Now let’s look at an example.
228
4.1. SIMPLE INTEREST
How much simple interest will be paid on a loan of R10 000 borrowed for a year at an interest
rate of 15% per year?
Now 15% of the R10 000 must be paid as interest per year for the use of the money. Thus, the
interest per year is calculated as
0,15 × 10 000 = 1 500.
The interest per year is R1 500.
Suppose we are using the money for two years. Thus, the interest for the full loan period or term
is
1 500 × 2 = 3 000.
The interest for the full load period is thus R3 000. In this case we have multiplied the amount
used by the interest rate per year, multiplied by the number of years used.
Formula
Simple interest is interest which is computed, on the principal, for the entire term for which it
is borrowed, and which is therefore due at the end of the term. Simple interest is given by
I = P in
where
I
is the simple interest (in rand) paid for the use of the money at the end of the term;
P
is the principal, or total amount borrowed (in rand) that is subject to interest; this
i
is also known as the present value of the loan;
is the rate of interest, that is, the fraction of the principal that must be paid at the
n
end of each period (say, year) for the use of the principal; and
is the term or time, that is, the number of periods for which the principal is borrowed.
There are four variables needed when determining simple interest.
If you know any three of these, the fourth can be determined.
Annum or year
• Note the use of the word annum or year:
How long is an annum?
It is one year.
per annum
per annum
per year
each year
annually
yearly
• The units used for the rate of interest and the term must be consistent:
If the interest rate is per year (sometimes called “per annum”), then the term must be in
years or a fraction thereof.
If the interest rate is expressed for a shorter period (e.g. per month) then the term must
be expressed accordingly (e.g. in months).
229
TOPIC 4: MATHEMATICS OF FINANCE
• A distinction is sometimes made between the so-called ordinary interest year and the exnnn interest year:
act
The ordinary interest year is based on a 360-day year.
The exact interest year is based on a 365-day year.
In an ordinary interest year, each month has 30 days, each quarter 90 days, and so on.
In an exact interest year, the exact number of days or months of the loan is used.
Unless otherwise stated, we always work with exact periods.
• Other textbooks may use different symbols for the different variables. This should not be
a source of confusion for you.
Activity
Calculate the simple interest to be paid on a loan of R10 000 at an interest rate of 12,5%
per annum over 90 days.
Answer
The interest rate is expressed “per annum” and we have to express the term, n, in years.
The following is given:
P = 10 000
12,5
= 0,125 per year
100
90
years
n = 90 days =
365
i = 12,5% =
The interest is calculated as
I = P in
= 10 000 × 0,125 ×
90
365
= 308,22.
The simple interest to be paid is R308,22.
As has been pointed out above, reference is occasionally made to a 360-day year. This has its
origin in pre-calculator days when sums of the above type were tedious. In the above example,
the effect of this would be that
90
n =
360
1
,
=
4
which obviously makes manual calculation a lot easier. However, unless the contrary is clearly
stated, you should assume that the “exact” year (365 days) is used.
Note that the units used for the rate of interest and the term must be consistent; that is, if the
interest rate is per year (sometimes called annum), then the term must be in years or a fraction
thereof.
In the activity above the interest rate is given per year (or annum), therefore the term is also
converted from months to a year.
230
4.1. SIMPLE INTEREST
Time line
nnn
A very useful way of representing interest rate calculations is with the aid of a so-called time line.
For a simple interest rate calculation, the timeline is as follows:
P
n = term
@ i = interest rate
The symbol @ is used for “at”.
S
Accumulated sum formula
At the beginning of the term, the principal, P (or present value), is put down (or borrowed). At
the end of the term, the amount or accumulated sum, S (or future value), is received (or paid
back). The accumulated sum includes the interest received.
The accumulated sum at the end of the term n is calculated as
accumulated sum = principal value + interest received.
Let
S
be the accumulated sum,
P
I
be the principal value and
be the interest received.
We arrive at a formula for the accumulated sum as follows:
The accumulated sum is calculated as
S
=
P +I
The interest is calculated as
I
=
P in
Substitute I into the formula
S
=
P + P in
Rewrite the two terms
S
=
(P × 1) + (P × in)
So, the first term is
P ×1
and the second term is
P × in
The common factor of the terms is
Rewrite the two terms
P
S
=
P × (1 + in)
=
P (1 + in)
So, the formula for the accumulated sum is
S = P (1 + in).
Four variables are needed when determining simple interest.
If you know any three of these, the fourth can be determined.
231
TOPIC 4: MATHEMATICS OF FINANCE
The accumulated sum is also known as the maturity value, the future value or the accrued
principal. The date, at the end of the term, on which the debt is to be paid is known as the due
date or maturity date.
4.1.2
Calculations
Calculate S
If R5 000 is invested for five years at a simple interest rate of 7,5% per annum, what is the amount
that will be received at the end of the period?
The following is given:
P = 5 000
i = 7,5% =
7,5
= 0,075 per year
100
n = 5 years
Two methods can be used to obtain the answer:
First method
Second method
Calculate I, then S = P + I.
Calculate S = P (1 + in).
The accumulated sum at the end of
The interest is calculated as
I
=
P in
=
=
=
P (1 + in)
5 000 × 0,075 × 5
=
5 000 × (1 + 0,075 × 5)
1 875.
=
6 875.
The interest is R1 875.
The accumulated sum at the end of the period is
S
=
=
P +I
5 000 + 1 875
=
6 875.
The accumulated sum is R6 875.
232
the period is
S
The accumulated sum is is R6 875.
4.1. SIMPLE INTEREST
Calculate P
Sometimes, we do not only consider the basic formulas, but also rearrange them in order to obtain
formulas for each variable respectively in terms of the others. We call this “changing the subject
of the formula”.
It is possible to obtain the present value P from the basic formula for
the interest amount, I, in the
following way:
Basic formula
Divide by (in)
I
I
in
I
in
P
=
=
P in
P in
in
=
P
=
I
in
the accumulated sum, or future value
S, in the following way:
Basic formula
Divide by (1 + in):
S
S
(1 + in)
S
(1 + in)
P
=
=
P (1 + in)
P (1 + in)
(1 + in)
=
P
=
S
(1 + in)
How do we interpret this result?
P is the amount that must be borrowed now to accrue
the amount of interest I after a term n
at interest rate i per time unit. As such
it is known as the present value of the
the interest amount I.
to the sum S after a term n at interest rate
i per time unit. As such it is known as the
present value of the sum S.
Stated formally, the present value of a debt
that accrued an interest amount of I
on a date prior to the due date is the
value P of the debt on the date in
question, and is given by the formula
P =
I
in
S on a date prior to the due date is the value
P of the debt on the date in question, and
is given by the formula
P = S ÷ (1 + in) =
S
(1 + in)
where i is the rate of interest and n the “time to run to maturity”.
Activity
1. Calculate the principal that must be invested over four years at 9,5% per annum to earn
R779 simple interest.
2. What principal will accumulate to R5 100 in seven months if the simple interest rate is 9%
per year?
233
TOPIC 4: MATHEMATICS OF FINANCE
Answer
1. The following is given:
I = 779
9,5
= 0,095 per year
100
n = 4 years
i = 9,5% =
The principal to be invested is calculated as
P
=
I
in
779
0,095 × 4
= 2 050.
=
The principal is R2 050.
2. The interest rate is expressed “per year” and we have to express the term, n, in years.
The following is given:
S = 5 100
9
= 0,09 per year
100
7
years
n = 7 months =
12
i = 9% =
The principal to be invested is calculated as
P
=
S
(1 + in)
=
5 100
7
1 + 0,09 ×
12
= 4 845,61.
An amount of R4 845,61 will accumulate to R5 100,00 in seven months at a simple interest
rate of 9% per year.
234
4.1. SIMPLE INTEREST
Calculate n
It is possible to obtain the term or time, n, from the basic formula for
the interest amount, I, in the
following way:
the accumulated sum, or future value
S, in the following way:
Basic formula
Basic formula
Divide by (P i)
I
I
Pi
I
Pi
=
=
=
n
=
S
=
P × (1 + in)
S
=
P + P in
n
S−P
=
P in
I
Pi
S−P
Pi
S−P
Pi
=
P in
Pi
=
n
=
S−P
Pi
P in
P in
Pi
n
Activity
How long will Lucy have to wait before her R2 500 invested at 6% simple interest per year
accumulates R600 interest?
Answer
The following is given:
P = 2 500
I = 600
i = 0,06 per year
The term is calculated as
n =
I
Pi
600
2 500 × 0,06
= 4,00.
=
Lucy will have to wait four years.
235
TOPIC 4: MATHEMATICS OF FINANCE
Calculate i
It is possible to obtain the yearly interest rate, i, from the basic formula for
the interest amount, I, in the
following way:
Basic formula
Divide by (P n)
I
I
Pn
I
Pn
=
=
=
i
=
the accumulated sum, or future value
S, in the following way:
S
=
P × (1 + in)
S
=
P + P in
i
S−P
=
P in
I
Pn
S−P
Pn
S−P
Pn
=
P in
Pn
=
i
=
S−P
Pn
P in
P in
Pn
Basic formula
i
Activity
A client borrows money from a microlender who offers short-term loans. The client borrows
R2 175 and has to repay R2 349 after six months. What yearly simple interest rate does the
microlender use to calculate the interest?
Answer
The following is given:
P = 2 175
S = 2 349
n = 6 months = 0,5 year
The yearly interest rate is calculated as
S−P
Pn
2 349 − 2 175
=
2 175 × 0,5
= 0,1600.
i =
Multiply 0,1600 by 100 to get 16%. The microlender uses an interest rate of 16% to calculate
interest.
236
4.1. SIMPLE INTEREST
4.1.3
Summary
• Interest is the price paid for the use of borrowed money.
• Simple interest implies the following:
You make one investment and leave it to gain the same amount of interest per year, or
you borrow one amount of money that accumulates the same amount of interest per year.
• The formulas used for simple interest are
where
I
=
P in
S
S
=
=
P +I
P × (1 + in)
I
is the simple interest (in rand) paid for the use of the money at the end
of the term;
P
is the principal, or total amount borrowed (in rand) that is subject to
interest. This is also known as the present value of the loan;
S
is the accumulated sum (in rand). This is also known as the future value
of the loan;
i
is the rate of interest, that is, the fraction of the principal that must be
paid at the end of each period (for example, each year) for the use of the
principal; and
n
is the term or time, that is, the number of periods for which the principal
is borrowed.
• Sometimes the above formulas are rearranged in order to obtain formulas for each variable
respectively. It is called “changing the subject of the formula”.
Exercise 4.1
1. Calculate the simple interest and sum accumulated when R5 000 is borrowed for 90 days at
12% simple interest per annum.
1
2. Determine the principal required to yield R300 in 18 months at 12 % simple interest per
2
annum. (Remember: yield = return on money = interest earned.)
3. How much must be paid back on a loan of R3 000 for six months if simple interest of 12%
per annum is charged?
4. You invest R1 000 at a simple interest rate of 10% per annum for four years. What is the
total interest that you receive? If the interest is paid monthly, how much do you receive
each month?
5. You borrow R800 and three months later pay back R823 in full settlement of your debt.
What is the annual simple interest charged?
237
TOPIC 4: MATHEMATICS OF FINANCE
Theme 4.2
Simple discount
Learning objective
Video:
On completion of this theme you should be able to calculate simple
discount.
Watch the video “Simple discount” on discount loans.
Interest calculated on the principal value but payable at the beginning of the term is called
discount.
Previously, emphasis was placed on the interest that has to be paid at the end of the term for
which the loan (or investment) is made. On the due date, the principal borrowed plus the interest
earned is paid back (or received).
In practice, there is no reason why the interest cannot be paid at the beginning rather than at
the end of the term.
A discount loan is a loan on which the interest and financing charges are deducted from the
face/future value when the loan is first taken out. The borrower only receives the discounted
value after the financing charges and interest are subtracted, but must repay the full amount of
the loan.
A discount loan is a short-term lending arrangement in which the interest amount
for the entire loan period is calculated on and deducted from the face/future
value at the time when a loan is disbursed. The borrower pays off the full
face value as arranged.
The discount D, on the face value S, is then simply the difference between the face value S and
the discounted value P :
P
238
D
S
4.2. SIMPLE DISCOUNT
Formula for discount, D
The discount D, on the face value S, is the difference between the face value S and the discounted
value P . Thus the discount, D, is given by
D = S − P.
Finding the discounted or present value is not simply the reverse of finding the future value by the
interest formula. Financial institutions introduced a simple discount rate, d, which, by analogy
with the interest rate, is the fraction of the face value, S, per time period which must be paid.
This rate, d, is expressed as a percentage.
The discount D is given by
D = Sdn
where
D
is the discount;
S
is the face value (the ending amount);
d
is the discount rate (%), divided by 100; and
n
is the number of time periods.
Here d is the simple discount rate.
Formula for discounted value, P
The discounted (or present) value of S is the difference between the face value and the simple
discount:
The formula for the discounted value, P, is
P
=
S−D
The formula for the discount, D, is
D
=
Sdn
P
=
S − Sdn
=
S × 1 − S × dn
S × (1 − dn)
S (1 − dn)
So, the formula for the discounted value is
P = S × (1 − dn).
239
TOPIC 4: MATHEMATICS OF FINANCE
Time line
The discounted (or present) value, P , of the face value, S, may be expressed in the form of the
following timeline:
P = Discounted or
present value
n = Term of discount
@ d = Discount rate
S = Face value or
future value
Note two important aspects:
• This timeline is structurally very similar to the timeline for simple interest.
• Since we are working with simple discount and the discount rate is now expressed as a
percentage of the future value and not as a percentage of the present value, a negative sign
appears in the formula.
This means that the discount, D, calculated as S × d × n, is subtracted from the face value, S,
to obtain the discounted/present value, P :
P
= S−D
= S − Sdn
= S (1 − dn)
Activity
Determine the simple discount on a loan of R3 000 due in eight months at a discount rate of
11% per annum. What is the discounted value of the loan? What is the equivalent simple
interest rate R?
Answer
It is given that
S = 3 000
d = 0,11
n = 8 months =
240
2
8
year = year.
12
3
4.2. SIMPLE DISCOUNT
This may be expressed in the form of the following time line:
P =?
2
year
3
@ 11% per annum
n=
S = 3 000
The discount is calculated as
D = Sdn
= 3 000 × 0,11 ×
2
3
= 220.
That is, the simple discount is R220.
The discounted value is calculated as
P
= S−D
= 3 000 − 220
= 2 780.
The discounted value is R2 780.
In order to determine the equivalent interest rate i, we note that R2 780 is effectively the
price now and that R3 000 is paid back eight months later. The interest is thus
I = S−P
= 3 000 − 2 780
= 220.
The question can thus be restated: What simple interest rate, when applied to a principal
of R2 780, will yield R220 interest in eight months? In this case
I = P in,
that is
220 = 2 780 × i ×
2
3
220
3
×
2 2 780
= 0,1187.
i =
Thus the equivalent simple interest rate is 11,87% per annum.
Note the considerable difference between the interest rate of 11,87% and the discount rate of
11,00%. This emphasises the very important fact that the interest rate and the discount rate are
not the same thing. The point is that they act on different amounts and at different times – the
former acts on the present value; whereas the latter acts on the future value. This is illustrated
on the next page in two case studies.
241
TOPIC 4: MATHEMATICS OF FINANCE
Simple interest
Simple discount
Interest is calculated on the present value or
principal amount.
Interest is calculated on the future/face value.
The loan is an add-on loan in which the
interest is added to the principal:
The interest, called the discount or discounted
amount, is subtracted from the future value
and only the proceeds are received by the
borrower:
P =S−D
S =P +I
Example: Happy Homes Furniture Store wants to borrow R38 000 for two years.
Community Bank offers them a 10% simple
interest loan.
National Bank offers them an 8% simple
discount loan.
The store wants to compare the implications of these two loans.
The following is given:
The following is given:
P
i
n
P
d
n
=
=
=
38 000
0,10 per year
2 years
=
=
=
38 000
0,08 per year
2 years
The interest that Happy Homes will have
to pay on the loan is
The amount that Happy Homes must borrow
in order to receive proceeds of R38 000 now, is
I
=
P in
P
=
=
38 000 × 0,10 × 2
S
=
=
7 600.
=
=
The amount that Happy Homes will have
to repay after two years is
S = P +I
=
=
38 000 + 7 600
45 600.
S (1 − dn)
P
1 − dn
38 000
1 − 0.08 × 2
45 238,10.
The discount subtracted from the future value
of the loan is
D = S−P
=
=
45 238,10 − 38 000,00
7 238,10.
To determine the equivalent interest rate:
P = 38 000,00
I = 7 238,10
n = 2 years
The equivalent simple interest rate of this
loan can now be determined as
I
i =
P ×n
7 238,10
=
38 000,00 × 2
= 0,0952
=
242
9,5%.
4.2. SIMPLE DISCOUNT
Consider the timelines of the two cases on the previous page:
Simple interest timeline
Simple discount timeline
P = discounted or present value
P = 38 000
P = 38 000
n=2
n=2
10% interest rate
8% discount rate
S =P +I
S = 45 600
S = face or future value
S = 45 238,10
Differences between simple interest and simple discount:
I
S
=
=
P in
P +I
D
P
=
=
Sdn
S−D
S
=
P × (1 + in)
P
=
S × (1 − dn)
Note that a simple interest rate of 10% (calculated on the discounted/present value) is not the
same as a simple discount rate of 10% (calculated on the face/future value).
Similarities between simple interest and simple discount:
The borrower receives money at the beginning of the term.
It is repaid with a single payment at the end of the term.
243
TOPIC 4: MATHEMATICS OF FINANCE
4.2.1
Summary
• For a discount loan, the interest and any other related charges are calculated and paid at
the time when the loan is granted.
• The borrower receives the reduced amount, called the “present value”, and must repay the
face value or future value of the discount loan.
• The formulas used for simple discount are
where
D
D
=
Sdn
P
P
=
=
S−D
S × (1 − dn)
is the simple discount (in rand) paid for the use of the money at the
beginning of the term;
P
is the present value of the loan;
S
is the face or future value of the loan;
d
is the discount rate, that is, the fraction of the future value to be paid at
the beginning of the period; and
n
is the term of the loan.
Exercise 4.2
1. A bank’s simple discount rate is 10% per annum. If you take a loan and have to repay
R4 000 in six months’ time, how much would you receive from the bank now? What is the
equivalent simple interest rate?
2. Determine the simple interest rate that is equivalent to a discount rate of
(a) 12% for three months;
(b) 12% for nine months.
Hint: Let S = 100 and use the appropriate formulas to set up an equation for i.
3. Mary needs R750 to buy a calculator. June is prepared to lend her the money on condition
that Mary repays her in ten months’ time. Calculate the future value of the loan if June
charges a 16% discount rate.
244
4.3. COMPOUND INTEREST
Theme 4.3
Compound interest
Learning objective
On completion of this theme you should be able to calculate compound interest.
Simple and compound interest
What is the difference between simple interest and compound interest?
Simple interest is interest paid on the original principal only.
Suppose Paul invests R600,00 for three years at a simple interest rate of 10% per annum. Determine the simple interest that he earns.
Interest formula
I
=
P in
Substitute the variables
I
=
600 × 0,10 × 3
Calculate
I
=
180
At the end he will have earned R180,00 in interest.
The accrued principal (accumulated sum or future value) will be calculated as follows:
Accrued principal
S
=
P +I
Substitute the variables
S
=
600 + 180
Calculate
S
=
780
The accrued principal (accumulated sum or future value) is R780,00. So actually, every year the
bank started with R600,00 to calculate the interest. This is, however, not the case with compound
interest.
4.3.1
Compound interest
Compound interest is the interest earned on the original
principal, as well as on all interests earned previously.
In other words, with compound interest, at the end of each year the interest earned is added to
the original amount and the money is reinvested.
Consider again Paul’s investment of R600,00 for three years at a simple interest rate of 10% per
annum. This time, determine both the simple interest and compound interest that he earns.
The results are shown on the following page:
245
TOPIC 4: MATHEMATICS OF FINANCE
Simple interest (R)
Compound interest (R)
Interest earned on
Interest earned on original principal
original principal
+ all interests earned previously
Original balance
600
600,00
Interest end Year 1
60
Balance end Year 1
660
660,00
Balance beginning Year 2
600
660,00
Interest end Year 2
60
(600 × 0,1)
66,00
Balance end Year 2
660 + 60 = 720
726,00
Balance beginning Year 3
600
726,00
Interest end Year 3
60
(600 × 0,1)
72,60
Balance end Year 3
720 + 60 = 780
798,60
600 + (3 × 60) = 780
798,60
780 − 600 = 180
798,60 − 600 = 198,60
Value of S
Interest
(600 × 0,1)
60,00
(600 × 0,1) leave in account
(660 × 0,1) leave in account
(726 × 0,1) leave in account
For simple interest, the interest amount never changes; it remains the same for every year.
For compound interest, the interest amount increases every year; it is different every year.
For compound interest, Paul’s investment can be illustrated as shown below. Remember, the
interest earned at the end of each year is left in the account:
600 + (600 × 0,10 × 1)
= 660
600
660
Year 1
660 + (660 × 0,10 × 1)
= 726
726 + (726 × 0,10 × 1)
= 798,6
726
Year 2
Year 3
It sounds a better proposition to leave the interest in the bank and earn “interest on interest”
or, in other words, “compound the interest”. This is the basis for compound interest. Compound
interest is just the repeated application of simple interest to an amount that is increased at each
stage by the simple interest earned in the previous period. However, as the investment term
stretches over a long period, we will have to do a lot of calculations.
246
4.3. COMPOUND INTEREST
Formula
Compound interest yields better results than simple interest, so you make more money.
Because a lot of calculations are required if the investment term stretches over a long period, a
formula is needed for the general situation.
Suppose you deposit a principal amount, P , at a certain annual interest rate, i,
but compounded after every time period of some specified length. How much
is your accumulated sum S after n time periods?
We let i denote the percentage interest your money earns each time period. After each time period
your money is multiplied by (1 + i), where i is expressed as a decimal. After n compoundings,
your money has been multiplied by (1 + i) a total of n times.
Your accumulated sum S is given by the formula
S = P (1 + i)n
where
S
is the accrued principal (total money at end of period);
P
is the principal (original amount invested or borrowed);
i
n
is the rate of interest per period, given as a decimal fraction; and
is the number of periods of investment.
Four variables are needed when determining compound interest.
If you know any three of these, the fourth can be determined.
The formula is not difficult to remember if you understand how it is derived. At each compounding, the amount is multiplied by (1 + i), so that after n compoundings the original principal P
has been multiplied by (1 + i)n .
Example
1
Calculate the accumulated sum on R500,00 invested for 10 years at 7 % per annum and com2
pounded annually.
The following is given:
P = 500,00
7,5
1
= 0,075 per year
i=7 %=
2
100
n = 10 years
Use the general formula to calculate the accumulated sum, S, as
S = P (1 + i)n
= 500,00(1 + 0,075)10
= 500,00 × 1,07510
= 1 030,52.
The accumulated sum/future value is R1 030,52.
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TOPIC 4: MATHEMATICS OF FINANCE
4.3.2
Compounding periods
Interest is not always compounded annually (once a year). Sometimes it is compounded semiannually, quarterly, monthly, daily or even continuously. In these cases interest is added more
frequently than once a year. As a result, there is more opportunity to earn interest on interest.
The mathematics remains the same, but you must be careful about what you substitute into the
equation, S = P (1 + i)n , for i and n.
If interest is compounded:
annually
then interest is added∗
once a year
semi-annually
then interest is added∗
every six months
quarterly
then interest is added∗
every three months
daily
then interest is added∗
every day
∗ Note that the interest is added to the total of the original principal plus all interests earned
previously.
The focus in this module is on cases where the interest is compounded annually, semi-annually,
quarterly and monthly.
Yearly
Consider again the case where Paul invested R600,00 for three years at 10% interest per year,
compounded yearly.
Interest is compounded yearly.
Interest rate %
Compounding periods
Interest rate is 10%.
There are 3 compounding periods in 3 years.
=
P (1 + i)n
Substitute
=
600,00(1 + 0,10)3
Calculate
=
798,60
Formula
S
Semi-annually
Calculate Paul’s accumulated sum of his investment when the interest is compounded semiannually. This means his interest is added to the principal every six months, not every year (the
interest is added twice a year).
For a half-year period, his interest percentage is cut in half, from 10% to 5%. This interest rate of
5% is called the periodic interest rate for Paul’s loan. It represents the percentage of the principal
his money earns over the time period between compoundings.
248
4.3. COMPOUND INTEREST
For interest compounded semi-annually, the periodic interest
rate is half the annual interest rate.
Interest is compounded semi-annually.
Interest rate %
For a half-year period, the yearly interest %
is cut in half (÷2), from 10% to 5%.
Compounding periods
There are 6 compounding periods in 3 years.
The compounding periods are doubled (×2).
=
P (1 + i)n
Substitute
=
600,00 1 +
Calculate
=
=
600,00 (1 + 0,05)6
804,06
Formula
S
0,10
2
3×2
Quarterly
Calculate Paul’s accumulated sum of his investment when the interest is compounded quarterly.
This means his interest is added to the principal every three months, not every year (the interest
is added four times a year).
For a period of three months, his interest percentage is cut in four, from 10% to 2,5%. This
interest rate of 2,5% is called the periodic interest rate for Paul’s loan.
For interest compounded quarterly, the periodic interest rate is
a quarter of the annual interest rate.
Interest is compounded quarterly.
Interest rate %
For a three month period (quarter), the yearly interest %
is cut in four (÷4), from 10% to 2,5%.
Compounding periods
Formula
There are 12 compounding periods in 3 years.
The compounding periods are multiplied by four (×4).
S
=
P (1 + i)n
=
600,00 1 +
Substitute
Calculate
=
=
0,10
4
0,10
600,00 1 +
4
806,93
3×4
12
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TOPIC 4: MATHEMATICS OF FINANCE
Monthly
nnn
Calculate Paul’s accumulated sum of his investment when the interest is compounded monthly.
This means his interest is added to the principal every month, not every year (the interest is
added twelve times a year).
For a period of a month, his interest percentage is cut in twelve:
10%
= 0,83% = 0,0083
12
Consequently, every month (twelve times a year) his principal is multiplied by the factor
(1 + 0,0083) = 1,0083.
For interest compounded monthly, the periodic interest rate is a
twelfth of the annual interest rate.
Interest is compounded monthly.
Interest rate %
Compounding periods
For a period of a month, the yearly interest %
is cut in twelve (÷12), from 10% to 0,83%.
There are 36 compounding periods in 3 years.
The compounding periods are multiplied by 12 (×12).
=
P (1 + i)n
Substitute
=
600,00 1 +
Calculate
=
600,00 1 +
=
808,91
Formula
S
0,10
12
0,10
12
3×12
36
All together
To compute the periodic interest rate, you divide the annual interest rate by
the number of compounding periods each year.
To compute the number of time periods, you multiply the number of years
by the number of compounding periods each year.
Compounding
period (interval)
Compounding
period
Fraction of
one year
Interest rate per
period/interval
Number of
time periods
1 month
monthly
1
12
interest % ÷ 100
12
no. of years ×12
3 months
quarterly
1
4
interest % ÷ 100
4
no. of years ×4
6 months
semi-annually /
1
2
interest % ÷ 100
2
no. of years ×2
1
interest % ÷ 100
1
no. of years ×1
half-yearly
1 year
250
annually / yearly
4.3. COMPOUND INTEREST
4.3.3
Calculate P
Sometimes, we not only consider the basic formula for compound interest, but also turn it inside
out and upside down, as it were, in order to obtain formulas for each variable respectively in
terms of the others.
Of particular importance is the concept of present value, P , which is obtained from the basic
formula for the accumulated sum or future value S.
Dividing the accumulated sum, S, by the factor (1 + i)n gives the formula for the present value,
P , of the accumulated sum, S:
Accumulated sum formula
S
=
P × (1 + i)n
Divide by factor (1 + i)n
S
(1 + i)n
=
P × (1 + i)n
(1 + i)n
S
(1 + i)n
=
P
P
=
S
(1 + i)n
Present value formula
Activity
Alex wishes to have R30 000,00 available to buy a car in four years’ time. The bank pays
9,75% interest per year, compounded semi-annually. How much should he invest in a savings
account now so that he will be able to do this?
Answer
Interest is compounded semi-annually, therefore, we have to express the term, n, in half
years.
The following is given:
S = 30 000,00
0,0975
per half year
2
n = 4 × 2 = 8 half years
i=
The present value/principal is calculated as
P
=
S
(1 + i)n
30 000,00
0,0975 8
1+
2
= 20 499,60.
=
The principal/present value is R20 499,60.
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TOPIC 4: MATHEMATICS OF FINANCE
4.3.4
Calculate n and i
nnn
The formulas for calculating i or n when the other variables are given are rather difficult to derive
and will therefore just be provided.
Formula to calculate the term
To calculate the term when the present value, future value and interest rate are given, use the
formula
S
ln
P
n=
ln(1 + i)
where ln (pronounced as “Lynn”) is the logarithm to the base e. Use the ln key on your calculator.
Formula to calculate the interest rate
To calculate the interest rate when the present value, future value and term are given, use the
formula
1
S n
− 1.
i=
P
Activity
1. Anna receives R6 000,00 as a gift from her rich aunt. How long will she have to wait for
this amount to grow to R10 906,00 if she can earn 10% interest per annum, compounded
monthly, on an investment with the Big Bank of Southern Africa?
2. You put R2 400,25 into an account and three years later had R3 522,83. If the account
earned interest compounded quarterly, what was the annual interest rate?
Answer
1. Interest is compounded monthly, therefore, we have to express the term, n, in months.
The following is given:
S = 10 906,00
P = 6 000,00
i=
0,10
per month
12
The term is calculated as
S
P
ln(1 + i)
ln
n =
10 906,00
ln
6 000,00 =
0,10
ln 1 +
12
= 72.
Anna will have to wait 72 months or 6 years for R6 000,00 to accumulate to R10 906,00 at
an interest rate of 10% per annum, compounded monthly.
252
4.3. COMPOUND INTEREST
2. Interest is compounded quarterly, therefore, we have to express the term, n, in quarters.
The following is given:
S = 3 522,83
P = 2 400,25
n = 3 × 4 = 12 quarters
The interest rate is calculated as
i =
=
S
P
1
n −1
1
3 522,83 12
−1
2 400,25
= 0,0325.
The interest rate as calculated is the interest rate per quarter. To calculate the annual
interest rate, multiply the answer by 4:
0,0325 × 4 = 0,13
The annual interest rate is 13%, compounded quarterly.
4.3.5
Summary
• Compound interest is the interest earned not only on the original principal, but also on all
interests earned previously.
• Compound interest implies that you make one investment and leave it to gain an increased
amount of interest per year, or you borrow one amount of money that accumulates an
increased amount of interest per year.
• The formula used for compound interest is
S = P (1 + i)n
where
P
is the principal or present value (original amount invested or borrowed);
S
is the accumulated sum/accrued principal/future value (total money at
end of period);
i
is the rate of interest per period, given as a decimal fraction; and
n
is the number of periods of investment (term).
• Sometimes the above formula is rearranged in order to obtain formulas for each variable
respectively. This is known as “changing the subject of the formula”.
253
TOPIC 4: MATHEMATICS OF FINANCE
Exercise 4.3
1. Determine the interest earned if R1 000 is invested for one year at 8% per annum and the
interest is compounded semi-annually (i.e. every six months).
1
2. What is the total amount available after 2 years if R2 000 is invested at an interest rate
2
of 12% per annum compounded quarterly?
3. You wish to invest R1 000 for two years. Which of the following investment opportunities
will give you the best return on money invested?
(a) 10% simple interest per annum
1
(b) 9 % interest per annum compounded semi-annually
2
(c) 9% interest per annum compounded quarterly
254
4.4. THE TIME VALUE OF MONEY
Theme 4.4
The time value of money
Learning objectives On completion of this theme you should be able to
• give and apply the equations and the corresponding time lines
relating to present and future values of money when compounding
is applicable
• state and apply the two rules for moving money backwards and
forwards in time
• use the rules to replace one set of financial obligations with another, that is, to reschedule debts
4.4.1
Time value of money
Would you prefer R10 000 today
or
R10 000 in five years?
Obviously, R10 000 today.
You already recognise that there is
TIME VALUE TO MONEY.
A rand received today is worth more than a rand received tomorrow:
• This is because a rand received today can be invested to earn interest.
• The amount of interest earned depends on the rate of return that can be earned on the
investment.
The time value of money quantifies the value of a rand through time.
A rand received today is worth more than a rand promised at
some time in the future.
In a financial transaction involving money due on different dates, every sum of money should
have an attached date, the date on which it falls due.
It means that the mathematics of finance deals with dated values. This is one of the most
important facts in the mathematics of finance.
Present and future values
To understand the time value of money, we need to consider the present and future value concepts.
If P is invested at an interest rate of i per period for a term of length n periods, it accumulates
to
S = P (1 + i)n
255
TOPIC 4: MATHEMATICS OF FINANCE
and we call P the present value and S the future value of the investment.
Sometimes we are given the future value S and wish to know the present value P . To do this we
can rearrange the above formula as follows:
S = P (1 + i)n
P (1 + i)n = S
P
= S ÷ (1 + i)n
P
= S × (1 + i)−n
In this latter case of finding the present value of some future value, the present value P is often
referred to as the discounted value of the future sum and the factor (1 + i)−n is termed the
discount factor.
This must not be confused with simple discounting for which a discount rate is defined. The rate
used in compound discounting is simply the relevant interest rate. It would indeed be possible
to introduce a compound discount rate, but this is not used in practice since it does not have a
readily understood meaning.
The important concepts are those of present value and future value, and the process of determining
present value is often referred to as discounting.
Example
You want to have an amount of R10 000,00 in your account one year from now. You can earn 8%
interest per year, compounded monthly. Calculate the amount that you need to invest now.
We have that the future value S = R10 000,00 and the interest rate written as a decimal i = 0,08
compounded monthly, for a one-year term of n = 12 time periods. We need to calculate the
present value P .
The money needed now is
P
=
=
=
S
(1 + i)n
or
P
=
S × (1 + i)−n
10 000,00
0,08 12
1+
12
9 233,61
=
0,08
10 000,00 × 1 +
12
=
9 233,61.
−12
You need to invest R9 233,61 now.
Now suppose you were offered R10 000,00 after one year or R10 830,00 after two years. The
interest you can earn is 8% per year, compounded monthly. Which one would you choose? Most
people would probably take the R10 000,00 after one year because it is money in their hands.
Take the R10 000,00 after one year and invest it for another year at 8%:
S = P (1 + i)n
0,08
= 10 000,00 1 +
12
= 10 830,00
At the end of two years you will have R10 830,00.
256
12
4.4. THE TIME VALUE OF MONEY
We see that the two values are the same, since you have taken the R10 000,00 and invested it for
another year at 8% after which it had also accumulated to R10 830,00.
Note that the dated values are not equivalent at some other rate of interest.
Time line
The following time diagram illustrates the dated values equivalent to a given dated value of
R10 000:
10 000 1 +
0,08 −12
12
= 9 233,61
10 000
12 months at
0
now
0,08
12
12 months at
0,08
12
2 years
1
10 000
10 000 1 +
0,08 12
12
= 10 830,00
The following time diagram illustrates dated values equivalent to a given dated value P :
P (1 + i)−n1
P
n1 at i
earlier date
n2 at i
current date
later date
P
P (1 + i)n2
257
TOPIC 4: MATHEMATICS OF FINANCE
Moving repayments forwards
nnn
Consider an amount of R100,00 that is due eight months from now. Suppose interest is compounded monthly.
P
S
100
0
1
2
3
4
5
6
7
8
9
10
11
12
month
Suppose you do not have money to repay the R100,00 after eight months and rather want to
repay it after a year.
If you move R100,00 forwards, for example from month eight to month twelve, you will pay more
than R100,00. You will pay interest on the R100,00 because you pay back the debt later than
originally planned.
The amount of R100,00 at month eight is the present value, P . You want to determine the
amount at month twelve, which is the future value, S. The value of n is four because the R100,00
is moved forwards by four months.
You can calculate the future value using the formula S = P (1 + i)n .
Moving repayments backwards
P
S
100
0
1
2
3
4
5
6
7
8
9
10
11
12
month
Suppose you obtain money from winning a competition and rather want to repay the R100,00 at
the end of month six.
If you move R100,00 backwards, for example from month eight to month six, you will pay less
than R100,00. You will save on interest because you pay back the debt earlier than originally
planned.
The amount of R100,00 at month eight is the future value, S. You want to determine the amount
at month six, which is the present value, P . The value of n is two because the R100,00 is moved
backwards by two months.
You can calculate the present value using the formula P = S(1 + i)−n .
Rules
We can formulate two simple rules for moving money forwards and backwards:
Rule 1
To move money forwards (determine a future value), inflate the relevant sum
by multiplying by the accumulation factor (1 + i)n .
Rule 2
To move money backwards (determine a present value), deflate the relevant
sum by dividing by the accumulation factor (1 + i)n . That is equivalent to
multiplying by the discount factor (1 + i)−n .
The point is that the mathematics of finance deals with dated values of money. This fact is
fundamental to any financial transaction involving money due on different dates. In principle,
every sum of money specified should have an attached date. Fortunately in practice the implied
date is often clear from the context.
258
4.4. THE TIME VALUE OF MONEY
Activity
An obligation of R50 000,00 falls due in three years. The interest is credited quarterly at an
interest rate of 12,5% per annum. What amount will cover the debt if it is paid as follows?
1. At the end of six months
2. At the end of four years
Answer
1. Draw the relevant time line.
50 000
0
2
4
6
8
10
12
14
16 quarters
10
quarters
X
1
@ 12 % per annum, compounded quarterly
2
To determine the debt if it is paid at the end of six months (i.e. two quarters from now),
we must discount the debt back two and a half years (i.e. 2,5 × 4 = 10 quarters) from the
due date to obtain the amount due.
The following is given:
S = 50 000,00
0,125
per quarter
4
n = 10 quarters
i=
The present value is calculated as
P
= S (1 + i)−n
= 50 000,00 × 1 +
0,125
4
−10
= 36 756,18.
The amount due after six months is R36 756,18.
2. Draw the relevant time line.
50 000
0
2
4
6
8
10
12
14
4
quarters
16 quarters
X
1
@ 12 % per annum, compounded quarterly
2
259
TOPIC 4: MATHEMATICS OF FINANCE
To determine the debt if it is allowed to accumulate until one year after the due date, we
must move the money forwards one year (four quarters) to obtain the future value, S. The
present value, P is now R50 000,00.
The following is given:
P = 50 000,00
0,125
per quarter
4
n = 4 quarters
i=
The future value is calculated as
S = P (1 + i)n
= 50 000,00 × 1 +
0,125
4
4
= 56 549,12.
The amount due after four years is R56 549,12.
Consider again the previous activities:
36 756,18
0
2
50 000,00
4
6
8
10
56 549,12
12
10
quarters
14
16
quarters
4
quarters
1
@ 12 % per annum, compounded quarterly
2
When the R50 000,00 is moved backwards from quarter twelve to quarter two, the amount to be
repaid is less than R50 000,00. There is a saving on interest because the debt is paid back earlier
than originally planned. The amount of R50 000,00 at quarter twelve is the future value, S. The
amount at quarter two was determined; this is the present value, P .
The present value is calculated as P = S (1 + i)−n .
When the R50 000,00 is moved forwards from quarter 12 to quarter 16, the amount to be repaid
is more than R50 000,00. Interest is paid on the R50 000,00 because the debt is paid back later
than originally planned. The amount of R50 000,00 at quarter 12 is the present value, P . The
amount at quarter 16 was determined; this is the future value, S.
The future value is calculated as S = P (1 + i)n .
260
4.4. THE TIME VALUE OF MONEY
4.4.2
Replacing financial obligations
From time to time a debtor (the person who owes money) may wish to replace his/her set of
financial obligations with another set. On such occasions he/she must negotiate with his/her
creditor (the person who is owed money) and agree upon a new due date and a new interest rate.
This is generally achieved by evaluating each obligation in terms of the new due date, and equating
the sum of the old and the new obligations on the new date. The resultant equation of value is
then solved to obtain the new future value that must be repaid on the new due date.
Example
Mr Murray owes R5 000,00 due in one year’s time and R2 000,00 due in two years’ time. Money
is worth 12% per annum, compounded monthly.
He decides to reschedule his payments by paying R3 000,00 six months from now and the rest
fifteen months from now. What will his last payment be?
This example illustrates the replacement of one set of financial obligations with another equivalent
set. Note that in time value of money, payments equal obligations.
First, we represent the money on the timeline:
5 000
2 000
3
months
0
3
6
9
3 000
12
9
months
15
18
21
24
months
9
months
X
@ 12% per annum, compounded monthly
Secondly, we determine the size of each payment and the obligation at the new due date 15
months hence.
Obligations: There are two obligations, namely R5 000,00 and R2 000,00.
For the R5 000,00 obligation:
We must move this R5 000,00 from month 12 (when it was originally due) to the new due date
of 15 months.
Thus n = 15 − 12 = 3 months.
The future value is
S = P (1 + i)n
0,12
= 5 000,00 1 +
12
= 5 151,51.
3
The R5 000,00 obligation is worth R5 151,51 at month 15.
For the R2 000,00 obligation:
We are discounting the R2 000,00 back (from month 24 to month 15) nine times.
Thus n = 24 − 15 = 9 months.
261
TOPIC 4: MATHEMATICS OF FINANCE
The present value is calculated as
S = P (1 + i)n
0,12 9
2 000,00 = P 1 +
12
2 000,00
P = 0,12 9
1+
12
= 2 000,00 1 +
−9
0,12
12
= 1 828,68.
The R2 000,00 obligation is worth R1 828,68 at month 15.
Payments: There are two payments, namely R3 000,00 and Rx.
For the R3 000,00 payment:
We move the R3 000,00 to the new date.
Then n is equal to nine months (n = 15 − 6 = 9).
The future value is
S = P (1 + i)n
0,12
= 3 000,00 1 +
12
= 3 281,06.
9
The R3 000,00 payment is worth R3 281,06 at month 15.
For the Rx payment at month 15:
As this is the last payment, no interest is involved and the payment remains Rx.
Thirdly, we determine the final payment.
Remember the total amount to be paid is equal to the total obligations.
Therefore, at month 15,
payments = obligations
3 000,00 1 +
0,12
12
9
+ x = 5 000,00 1 +
0,12
12
3
+ 2 000,00 1 +
0,12
12
3 281,06 + x = 5 151,51 + 1 828,68
3 281,06 + x = 6 980,19
x = 6 980,19 − 3 281,06
= 3 699,13.
Therefore he will pay R3 699,13 to settle his account at the end of fifteen months.
262
−9
4.4. THE TIME VALUE OF MONEY
The replacement of one set of financial obligations with another equivalent set is really just a case
of
• applying the rules for moving money backwards and forwards,
• keeping a clear head, and
• remembering that, at all times, the only amounts that may be added together (or subtracted) are those with a common date.
Activity
Maxwell owes Bernard Broker money. An amount of R5 000,00 is due three months from
now, and R2 000,00 is due six months from now. Maxwell offers to pay R3 000,00 immediately if he can pay the balance in one year’s time. Bernard agrees, on condition that they
use an interest rate of 16% per annum, compounded monthly. They also agree that, for
settlement purposes, the R3 000,00 paid now will also be subject to the same rate. How
much will Maxwell have to pay at the end of the year?
Answer
All amounts are shown on the following time line, with debts above the line and payments
below. The new due date is month 12. Determine the values of the debts and payments
at month 12.
5 000
9 months
2 000
0
3
6 months
6
9
12 months
3 000
12
months
X
@ 16% per annum, compounded monthly
Debts: The numerical values of the R5 000,00 and R2 000,00 debts must be moved to 12
months.
For the R5 000,00 debt:
The future value is calculated as
0,16
S = 5 000,00 × 1 +
12
= 5 633,02.
9
The amount due at the end of the year is R5 633,02.
For the R2 000,00 debt:
The future value is calculated as
0,16
S = 2 000,00 × 1 +
12
= 2 165,43.
6
The amount due at the end of the year is R2 165,43.
263
TOPIC 4: MATHEMATICS OF FINANCE
Payments: There are two payments, namely R3 000,00 at month zero and Rx at month
twelve.
For the R3 000,00 payment:
The future value is
0,16
S = 3 000,00 × 1 +
12
= 3 516,81.
12
The value of the R3 000,00 at the end of the year is R3 516,81.
For the Rx payment:
As this is the last payment, no interest is involved and the payment remains Rx.
All the values of the debts and payments at month 12 are now determined:
Remember, at month 12, the total amount to be paid is equal to the total debts. Therefore
3 000,00 × 1 +
payments = debts
0,16 12
0,16
+ X = 5 000,00 × 1 +
12
12
3 516,81 + X = 5 633,02 + 2 165,43
9
+ 2 000,00 × 1 +
0,16
12
6
3 516,81 + X = 7 798,45
X = 7 798,45 − 3 516,81
= 4 281,64.
That is, under the agreement, Maxwell will have to pay Bernard R4 281,64 in one year’s
time.
264
4.4. THE TIME VALUE OF MONEY
4.4.3
Summary
• A rand received today is worth more than a rand received at some time in the future because
it can be invested to earn interest.
• Mathematics of finance deals with dated values of money:
To move money forwards, inflate the relevant sum by multiplying by the accumulation
factor
(1 + i)n .
To move money backwards, deflate the relevant sum by multiplying by the discount factor
(1 + i)−n .
• When one set of financial obligations is replaced with another equivalent set, the
rules for moving money backwards and forwards are applied. Each obligation and payment
is evaluated in terms of the new due date. At the new due date all the obligations are equal
to all the payments.
Exercise 4.4
1. Melanie owes R500 due in eight months. The interest rate is 15% per annum, compounded
monthly. What single payment will repay her debt in the following cases?
(a) Now
(b) Six months from now
(c) In one year’s time
2. Mary-Jane must pay the bank R2 000, which is due in one year (interest is included). She
is anxious to lessen her burden in advance and therefore pays R600 after three months, and
another R800 three months later. If the bank agrees that both payments are subject to the
same interest rate as the loan, namely 12% per annum compounded quarterly, how much
must she pay at the end of the year to settle her outstanding debt?
265
TOPIC 4: MATHEMATICS OF FINANCE
Theme 4.5
Annuities
Learning objectives On completion of this theme you should be able to
• explain the basic structure and elements of an annuity
• determine the future value of an annuity
• determine the present value of an annuity
4.5.1
Definition
An annuity is a sequence of equal payments at equal intervals of time.
Equal cash flows each
one period apart
0
R100
R100
R100
1
2
3
years
The payment interval (or period)
of an annuity is the time
between successive payments.
The term is the time from the beginning
of the first payment interval to the end
of the last payment interval.
Payments
Equal cash flows one period apart.
Payment interval/Period
Time between successive payments.
Term
Examples of annuities
• Student loan payments
• Car loan payments
• Insurance premiums
• Mortgage payments
• Retirement savings
266
Time from beginning of first payment interval to
end of last payment interval.
4.5. ANNUITIES
4.5.2
Types
Ordinary annuity
An annuity is termed an ordinary annuity when payments are made
at the same time that interest is credited, that is at the end of the
payment intervals.
R100
R100
R100
1
2
3
0
End of
period 1
End of
period 2
years
End of
period 3
Annuity due
By contrast, an annuity whose periodic payment is made at the
beginning of each payment interval is known as an annuity due.
R100
R100
R100
0
1
2
Beginning of
period 1
Beginning of
period 2
Beginning of
period 3
3
years
Annuity certain
If the payments begin and end on fixed dates, the annuity is known
as an annuity certain.
0
R100
R100
R100
1
2
3
Begin on
fixed date
years
End on
fixed date
267
TOPIC 4: MATHEMATICS OF FINANCE
Perpetuity
On the other hand, if the payments continue forever, the annuity is
known as a perpetuity.
0
R100
R100
R100
1
2
3
R100
4
5
6
7
years
The basic concepts are illustrated by the following examples (notice how time lines, with the
interest period as the unit, are used).
Examples
The premium on an endowment policy is R2 400 per year payable for 20 years, and payable at
the end of each period.
R2 400
0
1
R2 400
2
3
4
20
years
Here, the payment interval is one year and the term is 20 years. Since the payments are made at
the end of each period, this is an ordinary annuity certain.
The monthly rent for a shop is R1 000, payable in advance.
R1 000
0
R1 000
1
2
3
n
months
Here, the payment interval is one month, while the term will be determined by the contract.
Since the payment is made at the beginning of each period, this is an annuity due.
A company is expected to pay R180 indefinitely every six months on a share of its preferred stock.
This is an example of a perpetuity with a payment interval of six months. In this case, no term
is defined, since payments continue indefinitely.
We will concentrate mainly on the ordinary annuity certain. From now on, we will simply speak
of “an annuity”, which, unless otherwise stated, means an ordinary annuity certain.
268
4.5. ANNUITIES
4.5.3
Future value S
Video:
Watch the video “AnnuitiesFV” on calculating the future value of an annuity.
We have seen in the video that if R is the payment in rand made at each payment interval in
respect of an annuity at interest rate i per payment interval, then the amount or future value of
the annuity after n intervals is given by the formula
(1 + i)n − 1
S=R
i
where
S
R
i
n
is the future value of the annuity;
is the regular payment;
is the interest rate per payment interval; and
is the number of payment intervals.
Four variables are needed when determining the future
value of an annuity, namely S, R, i and n.
If you know any three of these, the fourth can be determined.
Activity
Determine the accumulated amount of an annuity after four payments of R1 000,00 each,
paid annually, and at an interest rate of 15,8% per annum, compounded annually.
Answer
At the end of the term, the first payment of R1 000,00 will have accumulated interest for
three years, compounded at 15,8% per annum as indicated by the following time line:
R1 000
3 years @ 15,8% per annum
0
1
2
3
4
years
S1
Its accumulated value, S1 , at the end of the term is
S1 = 1 000,00 (1 + 0,158)3
= 1 552,84.
The accumulated value is R1 552,84.
269
TOPIC 4: MATHEMATICS OF FINANCE
This is obviously a straightforward application of the compound interest formula.
Similarly, the accumulated values of the second, third and fourth payments at the end of
the term are respectively
S2 = 1 000,00 (1 + 0,158)2
= 1 340,96;
S3 = 1 000,00 (1 + 0,158)1
= 1 158,00; and
S4 = 1 000,00.
The accumulated amount of the annuity after four payments is
S = S1 + S2 + S3 + S4
= 1 000,00 (1 + 0,158)3 + 1 000,00 (1 + 0,158)2 + 1 000,00 (1 + 0,158) + 1 000,00
= 5 051,80.
The accumulated amount is thus R5 051,80.
The result is represented by the following time line:
0
R1 000
R1 000
R1 000
R1 000
1
2
3
4
years
R5 051,80
@ 15,8% per annum
Using the formula for the future value of an annuity, the calculation can be done as follows:
(1 + i)n − 1
S = R
i
(1 + 0,158)4 − 1
= 1 000,00
0,158
= 5 051,80
The following two types of calculations are relevant:
1. Calculate S when R, i and n are known.
Use the formula for the future value of an annuity:
S=R
270
(1 + i)n − 1
i
4.5. ANNUITIES
2. Calculate R when S, i and n are known.
Manipulate the formula for the future value of an annuity to solve for R:
S
Future value of
annuity
i
S×
(1 + i)n − 1
Multiply both sides
S×
R alone on right
i
(1 + i)n − 1
=
(1 + i)n − 1
R
i
=
=
R
=
S×
i
(1 + i)n − 1
R
Simplify
R on left
(1 + i)n − 1
i
R
×
i
(1 + i)n − 1
R
=
i
S
(1 + i)n − 1
Activity
1. Ken turned 35 on 31 December 2004. He then decided to deposit R400,00 into his savings
account at the end of every month. The first deposit was made on 31 January 2005. The
savings account paid interest at a rate of 18% per annum, compounded monthly.
What is the amount of money that Ken will have in his account just after his 50th birthday?
2. The Collins family are planning a vacation and want to have R45 500,00 in their account in
three years’ time. The account will pay 14% interest per annum, compounded half-yearly.
What amount must they deposit into their account at the end of every six months?
Answer
1. Ken turns 50 on 31 December 2019. He will then have made monthly payments for 15 years,
that is 180 payments. It can be illustrated with the following time line:
0
400 400 400 400 400
400 400 400
1
178
2
3
4
5
@ 18% per annum, compounded monthly
179
180
months
S=?
The following is given:
R = 400,00
0,18
= 0,015 per month
12
n = 15 × 12 = 180 months
i=
(50 − 35 = 15) years
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TOPIC 4: MATHEMATICS OF FINANCE
The future value is calculated as
S = R
(1 + i)n − 1
i
(1 + 0,015)180 − 1
= 400,00
0,015
= 362 249,81.
The amount in his account will be R362 249,81.
2. This can be illustrated with the following time line:
0
?
?
?
?
?
?
1
2
3
4
5
6
@ 14% per annum, compounded half-yearly
half years
S = 45 500
The following is given:
S = 45 500,00
0,14
= 0,07 per half year
2
n = 3 × 2 = 6 half years
i=
The size of the half-yearly deposits is calculated as
(1 + i)n − 1
S = R
i
i
R = S
(1 + i)n − 1
0,07
= 45 500,00
(1 + 0,07)6 − 1
= 6 360,71.
The half-yearly deposits are R6 360,71 each.
272
4.5. ANNUITIES
4.5.4
Present value P
Video:
Watch the video “AnnuitiesPV” on calculating the present value of an annuity.
We have seen that if R is the payment in rand made at each payment interval in respect of an
annuity at interest rate i per payment interval, then the present value of the annuity after n
intervals is given by the formula
(1 + i)n − 1
P =R
i (1 + i)n
where
P
R
i
n
is the present value of the annuity;
is the regular payment;
is the interest rate per payment interval; and
is the number of payment intervals.
Four variables are needed when determining the present
value of an annuity, namely P , R, i and n.
If you know any three of these, the fourth can be determined.
Activity
Determine the present value of an annuity that provides R1 000,00 per year for five years if
the interest rate is 12,5% per annum, compounded annually.
Answer
The present value of the first payment is P1 :
P1 = 1 000,00 ÷ (1 + 0,125)
= 888,89
1
Remember, this means that R888,89 invested now at 12 % will yield R1 000,00 in one year’s
2
time.
Similarly, the present values of the other four payments are
P2 = 1 000,00 ÷ (1 + 0,125)2
= 790,12;
P3 = 1 000,00 ÷ (1 + 0,125)3
= 702,33;
P4 = 1 000,00 ÷ (1 + 0,125)4
= 624,30; and
P5 = 1 000,00 ÷ (1 + 0,125)5
= 554,93.
273
TOPIC 4: MATHEMATICS OF FINANCE
Thus, the present value of the annuity is
P
= P1 + P2 + P3 + P4 + P5
= 1 000 (1,125)−1 + 1 000 (1,125)−2 + 1 000 (1,125)−3 + 1 000 (1,125)−4 + 1 000 (1,125)−5
= 3 560,57.
1
Thus, R3 560,57 must be invested now at 12 % to provide for five payments of R1 000,00
2
each at yearly intervals, commencing one year from today.
The result is represented by the following time line:
0
R3 560,57
R1 000
R1 000
R1 000
R1 000
R1 000
1
2
3
4
5
years
@ 12,5% per annum
Using the formula for the present value of an annuity, the calculation can be done as follows:
P
= R
(1 + i)n − 1
i (1 + i)n
(1 + 0,125)5 − 1
= 1 000,00
0,125 (1 + 0,125)5
= 3 560,57
The following two types of calculations are relevant:
1. Calculate P when R, i and n are known.
Use the formula for the present value of an annuity:
P =R
(1 + i)n − 1
i (1 + i)n
2. Calculate R when P, i and n are known. Manipulate the formula for the present
value of an annuity to solve for R:
P
Present value of
annuity
Multiply both sides
i (1 + i)n
P×
(1 + i)n − 1
R alone on right
P×
i (1 + i)n
(1 + i)n − 1
=
R
=
(1 + i)n − 1
i (1 + i)n
R
×
i (1 + i)n
(1 + i)n − 1
=
R
R on left
R
=
i (1 + i)n
P×
(1 + i)n − 1
Simplify
R
=
P
274
(1 + i)n − 1
i (1 + i)n
i (1 + i)n
(1 + i)n − 1
4.5. ANNUITIES
Activity
1. Sarah borrows money at an interest rate of 15% per annum, compounded monthly, to buy
a second-hand car. Every month for a term of five years she has to repay R950,00.
What is the price of the car?
2. A loan of R12 000,00 with an interest rate of 16% per annum, compounded quarterly, is to
be repaid by equal quarterly payments over six years.
What is the size of the quarterly payments?
Answer
1. This can be illustrated with the following time line:
0
950 950 950 950 950
950 950 950
1
58
2
3
4
5
59
60
months
@ 15% per annum, compounded monthly
P =?
For the calculation of the present value of an annuity, we have the following information:
R = 950,00
0,15
= 0,0125 per month
12
n = 5 × 12 = 60 months
i=
The present value is calculated as
P
(1 + i)n − 1
= R
i (1 + i)n
(1 + 0,0125)60 − 1
= 950,00
0,0125 (1 + 0,0125)60
= 39 932,86.
The price of the car (present value) is R39 932,86.
2. This can be illustrated with the following time line:
0
?
?
?
?
?
?
?
?
1
2
3
4
5
22
23
24
P = 12 000
quarters
@ 16% per annum, compounded quarterly
275
TOPIC 4: MATHEMATICS OF FINANCE
For the calculation of the present value of an annuity, we have the following information:
P = 12 000,00
0,16
= 0,04 per quarter
4
n = 6 × 4 = 24 quarters
i=
The size of the quarterly payments is calculated as
P
= R
R = P
(1 + i)n − 1
i (1 + i)n
i (1 + i)n
(1 + i)n − 1
0,04 (1 + 0,04)24
= 12 000,00
(1 + 0,04)24 − 1
= 787,04.
The payment every quarter will be R787,04.
4.5.5
The relationship between P and S
How do you distinguish between the present and future value of an annuity?
Future value
Calculating the future value of an annuity allows you to see what
the value of the sum of individual amounts will be in the future.
Example
Suppose you want a lump sum of R1 000 000,00 in 50 years’ time, the future value calculation
will tell you how much you need to save each month at a given interest rate. In other words,
R1 000 000,00 is the future value of your regular monthly investment.
Present value
As an annuity is a series of equal payments or receipts, the present value
of an annuity is the sum of the present values of the individual amounts.
If you can afford a mortgage repayment of a certain amount per month, calculating the present
value of the repayments at a fixed rate of interest over a given number of years, will tell you the
principal sum you can afford to borrow.
Example
Suppose you can afford a monthly payment of R3 000,00 on a loan to buy a car, taken out on
30 May 2019 for 48 months, beginning on 30 June 2019, at an interest rate of 14% per annum,
276
4.5. ANNUITIES
compounded monthly. If you now calculate the present value, this would be the principal sum
you can afford to borrow.
Present value
The present value is also the amount of money you need to invest now
in order to withdraw a certain amount of money for a specified period
of time in the future.
Example
Suppose you invested a certain amount on 1 January 2010 in such a way that it generated a
periodic payment of R1 500,00 at the end of each month of the calendar year 2010 at an interest
rate of 13,2% per year, compounded monthly. If you now calculate the present value, this would
be the amount you needed to invest.
Present value of annuity
P
=
(1 + i)n − 1
R
i (1 + i)n
Future value of annuity
S
=
(1 + i)n − 1
R
i
Relationship between present and future values of annuity
The present value P , is determined by dividing
the future S, value by the accumulation factor
(1 + i)n , which is equivalent to multiplying
1
:
by
(1 + i)n
P
=
S ÷ (1 + i)n
=
R
S
=
P × (1 + i)n
=
R
(1 + i)n − 1
÷ (1 + i)n
i
=
1
(1 + i)n − 1
R
×
i
(1 + i)n
=
R
The future value S, is determined by
multiplying the present value P , by the
accumulation factor (1 + i)n :
(1 + i)n − 1
i (1 + i)n
(1 + i)n − 1
× (1 + i)n
i (1 + i)n
=
(1 + i)n
(1 + i)n − 1
R
×
n
i (1 + i)
1
=
R
(1 + i)n − 1
i
277
TOPIC 4: MATHEMATICS OF FINANCE
4.5.6
Summary
• An annuity is a sequence of equal payments at equal intervals of time. We looked at
annuities where the payments are made at the end of each period for a fixed period of time,
known as an ordinary annuity certain.
• The amount or future value of an annuity is the sum of all payments made and of all
accumulated interest at the end of the term.
If the payment in rand made at each payment interval in respect of an annuity, at interest
rate i per payment interval, is R, then the amount or future value, S, of the annuity after
n intervals is given by the formula
S=R
(1 + i)n − 1
.
i
• The present value of an annuity is the sum of all payments, each discounted to the
beginning of the term; that is, the sum of the present values of all payments.
If the payment in rand made at each payment interval in respect of an annuity, at interest
rate i per payment interval, is R, then the present value, P , of the annuity after n intervals
is given by the formula:
(1 + i)n − 1
.
P =R
i (1 + i)n
• The relationship between the present value P , of an annuity and the future value S, of
an annuity is
or S = P × (1 + i)n .
P = S ÷ (1 + i)n
Exercise 4.5
1. Determine the accumulated amount of an annuity after five payments of R600, each paid
annually, at an interest rate of 10% per annum.
2. Mrs Dooley decides to save for her daughter’s higher education, and every year, from the
child’s first birthday onwards, puts away R1 200. If she receives 11% interest, what will the
accumulated amount be after her daughter’s eighteenth birthday?
3. What is the accumulated amount of an annuity with a payment of R600 four times per
year, and an interest rate of 13% per annum compounded quarterly, at the end of a term
of five years?
4. Max pays R1 000 down on a second-hand motorbike and contracts to pay the balance in 24
monthly instalments of R200. If interest is charged at a rate of 12% per annum, payable
monthly, how much did the motorbike originally cost when Max purchased it? How much
interest does he pay?
5. Determine the present value of an annuity with half-yearly payments of R800, compounded
half-yearly at an interest rate of 12,5% per annum, and with a term of ten years.
278
4.6. AMORTISATION
Theme 4.6
Amortisation
Learning objectives On completion of this theme you should be able to
• calculate the payments on a mortgage loan
• set up an amortisation schedule for a loan
Not many people can afford to buy a house or brand new car for cash. Most of us have to borrow
money from a bank to make such a big purchase. We usually repay such a loan to the bank over a
long period of time in fixed payments that include interest on the amount still owed to the bank.
The loan is said to be amortised when all liabilities, that is both the principal (the amount of the
loan) and interest, are paid by a sequence of equal payments made at equal intervals of time.
We may ask the following typical questions about this process:
• What should the amount of the equal payments be?
• What is the remaining value of the loan still to be repaid at any stage during its repayment?
• What is the total interest payable on the loan?
These may sound like difficult questions to answer, but if you consider that this way of paying
back a loan is just an ordinary annuity, the good news is that you already have the tools to answer
these questions.
4.6.1
Calculate payments
Video:
Watch the video “Amortisation1” on calculating payments on amortisation.
Example
In the video, a loan of R10 000,00 with interest of 16% compounded quarterly is to be amortised
by equal quarterly payments over five years. The first payment is due at the end of the first
quarter. The payment has to be calculated.
The representation time line is as follows:
0
R10 000
?
?
?
?
?
?
1
2
3
18
19
20
quarters
@ 16% per annum, compounded quarterly
It is evident that the 20 payments form an ordinary annuity with a present value of R10 000,00
and an interest rate of 16% ÷ 4 = 4%.
279
TOPIC 4: MATHEMATICS OF FINANCE
The following is given:
P = 10 000,00
0,16
= 0,04 per quarter
4
n = 5 × 4 = 20 quarters
i=
Thus
R = P×
i (1 + i)n
(1 + i)n − 1
0,04 × 1,0420
= 10 000,00 ×
1,0420 − 1
= 735,82.
Thus, the quarterly payment is R735,82.
The example illustrates the typical amortisation problem.
A loan of present value P rand must be amortised over n payments at interest
rate i per payment interval. What is the amount of the payment R?
As pointed out in the video example, the n payments form an ordinary annuity, so we can use
the formula for the present value of such an annuity to calculate R:
Present value formula
P
=
(1 + i)n − 1
R
i (1 + i)n
therefore
R
=
(1 + i)n − 1
P÷
i (1 + i)n
or simplified
R
=
P×
Video:
i (1 + i)n
(1 + i)n − 1
Watch the video “Amortisation2” on the typical amortisation problem.
Consider the mechanics of amortisation. Initially, the total amount loaned (i.e. the present
value at instant 0) is owed. However, as payments are made, the outstanding principal, or
outstanding liability as it is also known, decreases until it is eventually zero at the end of
the term. At the end of each payment interval, the interest on the outstanding principal is first
calculated. The payment R is then first used to pay the interest due. The balance of the payment
is thereafter used to reduce the outstanding principal. Since the outstanding principal decreases
with time, the interest owed at the end of each period also decreases with time. This means that
the fraction of the payment that is available for reducing the principal increases with time.
At any stage of the term, the amount outstanding just after a payment has been made is the
present value of all payments that still have to be made.
The amortisation schedule is a table indicating the distribution of each payment with regard to
interest and principal reduction.
280
4.6. AMORTISATION
We have also seen that
amortisation is the process of gradually reducing a debt
through instalment payments of principal and interest.
Activity
An amortisation schedule has to be drawn up for a loan of R5 000,00 that is repaid in annual
payments over five years at an interest rate of 15% per annum, compounded annually.
First, calculate the size of the yearly payments.
Answer
The following is given:
P = 5 000,00
i = 0,15 per year
n = 5 years
Thus
i (1 + i)n
R = P×
(1 + i)n − 1
0,15 × 1,155
= 5 000,00 ×
1,155 − 1
= 1 491,58.
Thus, the size of the yearly payment is R1 491,58.
4.6.2
The amortisation schedule
Activity
Draw up an amortisation schedule for a loan of R5 000,00 that is repaid in annual payments
over five years at an interest rate of 15% per annum, compounded annually.
Answer
Draw a table with five columns. The number of rows is equal to the number of years it will
take to repay the loan.
281
TOPIC 4: MATHEMATICS OF FINANCE
Year
Outstanding
principal at year
beginning
Interest due at
year end
(simple)
Payment
1
5 000,00
750,00 (b)
1 491,58
2
3
4 258,42 (d)
Principal
repaid
(a)
741,58 (c)
1 491,58
1 491,58
4
5
1 491,58
1 491,58
(a) In the previous activity, the yearly payments were calculated as R1 491,58 each.
(b) The interest due at the end of the first year is
I = P in
= 5 000,00 × 0,15 × 1
= 750,00.
The interest due at the end of the first year is R750,00. Note that you have to keep n
equal to 1 when you calculate interest at the end of the second to the fifth year. P is
the outstanding amount at the start of the year.
(c) The principal repaid at the end of the first year is
principal repaid = payment − interest due
= 1 491,58 − 750,00
= 741,58.
The principal repaid at the end of the first year is R741,58.
(d) The outstanding principal at the beginning of the second year is
= outstanding principal from previous year − principal repaid from previous year
= 5 000,00 − 741,58
= 4 258,42.
The outstanding principal at the beginning of the second year is R4 258,42.
Repeating steps (b) to (d) for each year, gives the complete amortisation schedule:
(d)
Outstanding
principal at year
beginning
(b)
Interest due at
year end
(simple)
(a)
Payment
(c)
Principal
repaid
1
2
5 000,00
4 258,42
750,00
638,76
1 491,58
1 491,58
741,58
852,82
3
4
3 405,60
2 424,86
510,84
363,73
1 491,58
1 491,58
980,74
1 127,85
5
1 297,01
194,55
1 491,58
1 297,03
2 457,88
7 457,90
5 000,02
Year
Total
282
4.6. AMORTISATION
Note:
(a) The interest due at the end of each year is simply 15% of the outstanding principal.
(b) The principal repaid is the difference between the payment and the interest due.
(c) The outstanding principal at the beginning of the year is equal to the outstanding principal at the beginning of the previous year minus the principal repaid of the previous
year.
(d) Also note that, due to rounding errors, the total principal repaid is in error by two
cents.
We have seen that the amortisation schedule is a complete schedule of periodic (in this case
yearly) loan payments, showing the amount of principal and the amount of interest that comprise
each payment (R1 491,58) in order that the loan will be paid off at the end of its term. Early
in the schedule, the majority of each payment is interest. Later in the schedule, the majority of
each payment is put towards the principal.
Rand
1 400
1 200
1 000
741,58
852,82
980,74
1 127,85
1 297,03
Principal repaid
750,00
638,76
510,84
363,73
194,55
Simple interest due
at end of year
1
2
3
4
5
700
600
400
200
Year
Year
Outstanding principal
at year beginning
Interest due at
year end (simple)
Payment
Principal
repaid
1
5 000,00
750,00
1 491,58
741,58
2
3
4 258,42
3 405,60
638,76
510,84
1 491,58
1 491,58
852,82
980,74
4
2 424,86
363,73
1 491,58
1 127,85
5
1 297,01
194,55
1 491,58
1 297,03
283
TOPIC 4: MATHEMATICS OF FINANCE
4.6.3
Interest rate (i)
A loan can have a fixed interest rate or a variable interest rate.
Variable interest rate
A variable interest rate loan is a loan in which the interest rate charged on the outstanding
balance varies as market interest rates change. As a result, your payments vary as well if they
consist of a blend of principal and interest payments.
Fixed interest rate
Fixed interest rate loans are loans in which the interest rate charged on the loan will remain fixed
for that loan’s entire term, no matter what market interest rates do. This will result in your
payments being the same over the entire term. If interest rates fall, you do not stand to benefit
as your rates remain fixed.
Whether a fixed-rate loan is better for you will depend on the interest rate environment when
the loan is taken out and on the duration of the loan.
Which one?
When a loan is fixed for its entire term, it will be fixed at the prevailing market interest rate at
the time the loan is taken out. Generally speaking, if interest rates are relatively low, but are
about to increase, then it will be better to lock in your loan at that fixed rate. Depending on
the terms of your agreement, your interest rate on the new loan will remain fixed, even if interest
rates climb to higher levels.
On the other hand, if interest rates are on the decline, then it would be better to have a variable
rate loan. As interest rates fall, so will the interest rate on your loan.
This discussion is simplistic, but the explanation will not change in a more complicated situation.
It is important to note that studies have found that over time, the borrower is likely to pay less
interest overall with a variable rate loan than a fixed rate loan. However, the borrower must
consider the amortisation period of a loan. The longer the amortisation period of a loan, the
greater the impact a change in interest rates will have on your payments.
Therefore, variable interest rate loans are beneficial for a borrower in a decreasing interest rate
environment, but when interest rates rise, loan payments will rise sharply.
Video:
Watch the video “Amortisationpayment1” on calculating the outstanding principal
when interest rates change.
Video:
Watch the video “Amortisationpayment2” on calculating the new payments when
interest rates change.
284
4.6. AMORTISATION
Variable interest rate – Example
Jonathan purchases an apartment by making a down payment of R40 000,00 and obtaining a
1
20-year loan for the balance of R280 000,00 at 15% per annum, compounded monthly. After 4
2
years, the bank adjusts the interest rate to 16%. What is the new amount that he must pay if
the term of the loan remains the same?
Initially, the interest rate was
15% ÷ 12 = 1,25% = 0,0125
per month, and the number of payments to be made was
20 × 12 = 240.
Thus, initially the size of his payments was
0,0125 × 1,0125240
R = 280 000,00 ×
1,0125240 − 1
= 3 687,01.
His initial payments were R3 687,01.
1
1
After 4 years, 54 payments 4 × 12 have been made. The value of n is now calculated as
2
2
240 − 54 = 186.
There are 186 months of payments still remaining. The present value of the loan at this stage is
P
1,0125186 − 1
= 3 687,01 ×
0,0125 × 1,0125186
= 265 699,85.
1
years, the present value of the loan is R265 699,85.
2
The outstanding principal of R265 699,85 must be amortised over the remaining 186 months at
4
16
% = % per month. Thus, the new payments are calculated as
an interest rate of
12
3
After 4
⎡
⎤
0,16 186
0,16
1
+
⎢
⎥
12
⎢ 12
⎥
R = 265 699,85 × ⎢ ⎥
186
⎣
⎦
0,16
1+
−1
12
= 3 872,30.
The new payments are now R3 872,30.
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TOPIC 4: MATHEMATICS OF FINANCE
4.6.4
Period and payments
Example
In order to buy a house, Sam takes out a loan of R600 000,00 at 17% interest per year, compounded
monthly. This loan has to be repaid in equal monthly payments over a period of 15 years.
What is the size of the monthly payments?
The following is given:
P = 600 000,00
0,17
per month
12
n = 15 × 12 = 180 months
i=
Thus
i (1 + i)n
R = P×
(1 + i)n − 1
⎡
⎤
0,17 180
0,17
1+
⎢
⎥
12
⎢ 12
⎥
= 600 000,00 × ⎢ ⎥
180
⎣
⎦
0,17
1+
−1
12
= 9 234,03.
Therefore the payments are R9 234,03 per month for 15 years.
What is the outstanding amount on the loan at the end of the first year (i.e. after
12 payments)?
It was stated earlier that at any stage of the term, the amount outstanding just after a payment
has been made is the present value of all payments that still have to be made.
There are 180 payments in total. After 12 payments have been made, there are still 168 payments
left to be made. Calculate the present value of the 168 payments that are still to be made as
P
(1 + i)n − 1
= R×
i (1 + i)n
⎡
⎤
0,17 168
1
+
−1 ⎥
⎢
12
⎢
⎥
= 9 234,03 × ⎢
168 ⎥
⎣ 0,17
⎦
0,17
1+
12
12
= 590 472,18.
The outstanding amount of the loan at the end of the first year is R590 472,18.
This amount can also be calculated using the financial keys of the recommended calculator, or
by calculating the values in the amortisation schedule by hand. The resulting value is slightly
different from the value calculated using the present value formula. This is due to different
rounding techniques. The values in the amortisation schedule are shown below:
286
4.6. AMORTISATION
Month
Outstanding principal
at beginning of month
Simple interest due
at end of month
Payment
Principal
repaid
1
600 000,00
8 500,00
9 234,03
734,03
2
..
.
599 265,97
..
.
8 489,60
..
.
9 234,03
..
.
744,43
..
.
12
13
..
.
591 328,74
590 471,87
..
.
8 377,16
8 365,02
..
.
9 234,03
9 234,03
..
.
856,87
869,01
..
.
The outstanding amount of the loan at the end of the first year is R590 471,87. It is quite shocking
to see that after a year’s payments of 12 × R9 234,03 have been made, less than R10 000,00 of the
outstanding principal has been repaid!
What is the outstanding amount on the loan at the end of ten years (i.e. after 120
payments)?
After 120 payments have been made, there are still 60 payments left to be made. Calculate the
present value of the 60 payments that are still to be made as
P
(1 + i)n − 1
= R×
i (1 + i)n
⎡
⎤
0,17 60
1
+
−1 ⎥
⎢
12
⎢
⎥
= 9 234,03 × ⎢
60 ⎥
⎣ 0,17
⎦
0,17
1+
12
12
= 371 552,23.
The outstanding amount of the loan at the end of ten years is R371 552,23.
The values in the amortisation schedule are shown below:
Month
Outstanding principal
at beginning of month
Simple interest due
at end of month
Payment
Principal
repaid
1
2
..
.
12
600 000,00
599 265,97
..
.
591 328,74
8 500,00
8 489,60
..
.
8 377,16
9 234,03
9 234,03
..
.
9 234,03
734,03
744,43
..
.
856,87
13
..
.
590 471,87
..
.
8 365,02
..
.
9 234,03
..
.
869,01
..
.
120
121
..
.
375 465,83
371 550,90
..
.
5 319,10
5 263,64
..
.
9 234,03
9 234,03
..
.
3 914,93
3 970,39
..
.
287
TOPIC 4: MATHEMATICS OF FINANCE
After 120 months or ten years, the outstanding principal on the loan is R371 550,90. It is quite
shocking to see that after 10 out of the 15 year’s payments have been made, more than half of
the loan still has to be repaid!
The values of the last month’s payment are shown in the amortisation schedule below:
Month
Outstanding principal
at beginning of month
Simple interest due
at end of month
Payment
Principal
repaid
1
600 000,00
8 500,00
9 234,03
734,03
2
..
.
599 265,97
..
.
8 489,60
..
.
9 234,03
..
.
744,43
..
.
12
13
..
.
591 328,74
590 471,87
..
.
8 377,16
8 365,02
..
.
9 234,03
9 234,03
..
.
856,87
869,01
..
.
120
121
..
.
180
375 465,83
371 550,90
..
.
9 101,93
5 319,10
5 263,64
..
.
128,94
9 234,03
9 234,03
..
.
9 234,03
3 914,93
3 970,39
..
.
9 105,09
After 15 years, 180 payments of R9 234,03 each have been made. The total amount paid on the
loan over the 15 years is
9 234,03 × 180 = 1 662 125,40.
The total amount of interest paid is
1 662 125,40 − 600 000,00 = 1 062 125,40.
The total amount of interest paid is R1 062 125,40. This is almost double the amount of the loan!
The example was probably a big eye-opener for you.
Most people do not realise what they really are paying in interest on a big long-term loan. It is
also quite interesting to see what a big difference a little extra on your payment can make.
If Sam decides to pay R10 073,00 a month instead of R9 234,03, how long will it take
him to repay his loan?
The financial keys of the recommended calculator are used to calculate the number of months
it will take Sam to repay his loan. The calculation is not required in this module and only the
result will be given.
It will take him 132 months or 11 years to repay his loan.
288
4.6. AMORTISATION
What is the total amount of interest paid on the loan after 11 years?
After 11 years, 132 payments of R10 073,00 each have been made. The total amount paid on the
loan over the 11 years is
10 073,00 × 132 = 1 329 636,00.
The total amount of interest paid is
1 329 636,00 − 600 000,00 = 729 636,00.
Compare the two situations
Sam borrowed R600 000,00 at an interest rate of 17% per year, compounded monthly.
If he repays the loan at R9 234,03 per month, it will take him 15 years to repay the loan.
The total interest paid on the loan is R1 062 125,40.
If he repays the loan at R10 073,00 per month, it will take him 11 years to repay the loan.
The total interest paid on the loan is R729 636,00.
The difference in the two payments is calculated as
10 073,00 − 9 234,03 = 838,97.
It is amazing to see what a difference an extra payment of 838,97 can make in the term of a loan
and in the total interest paid on the loan.
Comparing values at different periods
A mortgage of R200 000,00 has to be amortised in monthly payments at an
interest rate of 16% per year, compounded monthly. Four periods for
repaying the loan were considered: 10, 15, 20 and 25 years.
The following table shows how much interest you would have to pay depending on the monthly
payment and amortisation period chosen:
Amortisation
period (years)
Monthly payment
(R)
Total amount paid
(R)
Total interest paid
(R)
10
15
2 220,41
1 687,71
266 449,20
303 788,46
66 449,20
103 788,46
20
25
1 432,86
1 288,60
343 886,91
386 580,84
143 886,91
186 580,84
Another way to look at it is to compare how much of the amount borrowed (the principal) would
be paid off in the first few years, depending on the amortisation period.
Using the above mortgage as an example, the table below shows how much of the principal would
be paid off in the first five years. The last column represents the percentage of the original amount
289
TOPIC 4: MATHEMATICS OF FINANCE
borrowed, paid back to the lender. The percentage is calculated by dividing the rand amount of
principal paid back to the lender after five years by the original amount borrowed (R200 000,00)
and multiplying the result by 100.
4.6.5
Amortisation
period (years)
Principal paid back to
lender after five years (R)
Principal paid back to
lender after five years (%)
10
15
85 148,05
47 981,80
43
24
20
25
30 200,80
20 135,83
15
10
Present value (P)
Sometimes you have to determine the present value of the loan before you can calculate the
payments.
The present value of the loan is the amount that is
borrowed and has to be repaid in equal payments.
Examples
Sithuli wants to buy a flat screen television for R8 000,00. He agrees to pay a deposit of R600,00
and a monthly instalment for two years at an interest rate of 12% per annum, compounded
monthly. What is the amount he borrows? In other words, what is the present value of the loan?
Price of television is
8 000
Deposit is
600
Present value of loan is
price of television − deposit
=
=
8 000 − 600
7 400
The present value of the loan is R7 400,00. This is the amount that he borrows after the deposit
is subtracted.
Jonathan purchases an apartment by making a down payment of R40 000,00 and obtaining a
20-year loan for the balance of R280 000,00 at 15% per annum, compounded monthly. What is
the price of the apartment and the present value of the loan?
40 000
Deposit is
Price of apartment is
present value of loan + deposit
=
=
280 000 + 40 000
320 000
The price of the apartment is R320 000,00. He has R40 000,00, which he uses as a deposit, and
he borrows the rest, namely R280 000,00. The present value of the loan is R280 000,00. This is
the amount that he borrows after the deposit is subtracted.
290
4.6. AMORTISATION
Sam wants to buy a new fridge for R6 500,00. He agrees to pay a deposit of 15% of the price and
a monthly instalment for one year at an interest rate of 11% per annum, compounded monthly.
What is the amount he borrows? In other words, what is the present value of the loan?
6 500
Price of fridge is
Deposit is
Present value of loan is
15% of 6 500
price of fridge − deposit
=
=
15
× 6 500
100
0,15 × 6 500
975
=
=
6 500 − 975
5 525
=
The present value of the loan is R5 525,00. This is the amount that he borrows after the deposit
is subtracted.
4.6.6
Summary
• Amortisation is the process of gradually reducing a debt through instalment payments
of principal and interest.
• The loan is said to be amortised when all liabilities, that is both the principal (the amount
of the loan) and the interest, are paid by a sequence of equal payments made at equal
intervals of time.
• The size of the payments is calculated using the formula for the present value of an
annuity. P is known and R must be calculated:
R=P×
i (1 + i)n
.
(1 + i)n − 1
• The amortisation schedule is a table indicating the distribution of each payment with
regard to interest and principal reduction. The table has five columns:
Column 1
Payment interval
Column 2
Outstanding principal at the beginning of the payment interval
Column 3
Simple interest due at the end of the payment interval
Column 4
Payment
Column 5
Principal repaid at the end of the payment interval
• Interest rates change from time to time. If the interest rate changes, the payments
change accordingly. After the change in interest rate, the present value is recalculated for
the remaining period of time. The size of the payments is also recalculated according to
the new interest rate.
291
TOPIC 4: MATHEMATICS OF FINANCE
Exercise 4.6
1. You purchase a house for R270 000 with a down payment (often referred to as a deposit) of
R45 000. You secure a mortgage bond with a building society for the balance at 11,5% per
annum, compounded monthly, with a term of 20 years. What are the monthly payments?
2. Your Great-aunt Agatha dies and leaves you an inheritance of R60 000, which is to be paid
to you in ten annual payments at the end of each year. If the money is invested at 12% per
annum, how much do you receive each year?
3. Draw up an amortisation schedule for a loan of R4 000 for three years at 15% per annum
compounded half-yearly and repayable in six half-year payments.
292
Topic
5
Collection, presentation and
description of data
After completion of this topic you should be able to explain how to collect data, how to present
data visually and how to calculate simple measurements of data.
CONTENTS
Theme
5.1
Statistics: An introduction
5.2
Data collection
5.3
Presentations
5.4
Measurements of locality
5.5
Measurements of dispersion
TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Theme 5.1
Statistics: An introduction
Statistics are all around us. In fact, it would be difficult to go through a full week without using
statistics. Imagine watching a football game where no one kept score. The action itself might
provide enough excitement to hold your attention for a while, but think of all the drama that
would be lost if winning and losing were not an issue.
Imagine going to the grocery store and trying to find the best buy on a box of treats for your
dog, Fluffy. Without statistics this task would come down to simple guess work. You could never
know for sure if you get the best (or cheapest) treats for your rand.
Without statistics we could not plan our budgets, pay our taxes, enjoy games to their fullest, or
evaluate classroom performance. Are you beginning to get the picture? We need statistics.
The study of statistics can be categorised into two main branches. These branches are descriptive statistics and inferential statistics.
Descriptive statistics
The most basic form of statistics is known as descriptive statistics. This branch of statistics
lays the foundation for all statistical knowledge. This makes it quite important, but it is not
something that you should learn simply so you can use it in the distant future. Descriptive
statistics can be used now, in English class, in Mathematics class, at the football stadium and in
the grocery store. You probably already know more about statistics than you think.
Descriptive statistics refer to the collecting, summarising and processing of data to transform
the data into information. Data can be collected through surveys or by taking samples. It can
be presented visually in graphs. It can also be summarised in tables and processed by calculating
certain measures like the mean.
Inferential statistics provide the basis for predictions, forecasts and estimates that are used to
transform data into knowledge.
This theme only focuses on descriptive statistics.
Video:
294
Watch the video “Data1_V1” on the importance of statistics.
5.2. DATA COLLECTION
Theme 5.2
Data collection
Learning objectives On completion of this theme you should know and be able to explain
• what a sample and a population are
• what simple random sampling, stratified random sampling and
systematic sampling are
Key definitions
Radial is a company that advertises that its XXX tyres, generally known as T riple X, will
complete at least 65 000 kilometres before one of the four tyres will no longer meet the minimum safety requirements. However, lately several complaints have been received that the tyres
completed only 50 000 kilometres and thus the minimum requirements were not met.
Radial sells directly to the public and it is company policy to keep a record of customers. They
have been manufacturing tyres for the past two years. In the last six months, they sold 2 600
sets of XXX tyres. Radial feels that they do not have the time, personnel or money to locate
and question all 2 600 of their customers. They feel that if they could question 80 customers, it
should give them a good idea of what the actual situation is.
In other words, they take a sample of 80 out of the population of 2 600. Sample and population
are words that are always used when we wish to obtain data.
Consider the following definitions:
A population is the set of all the elements or items being studied.
In Radial’s case the set of all 2 600 sets of XXX tyres sold is the population.
A sample is a representative group or a subset of the population.
The sets of 80 tyres that Radial will investigate form the sample.
A variable is any property or characteristic
that can be measured or observed.
A variable can take on a range of different values. For example, the distance completed with
a set of tyres is different for each customer and therefore the observations vary continually. The
distance that has been completed is a variable.
The sample unit is the item that is measured or
counted with respect to the variable being studied.
We see that the members of a sample are the sample units. Radial’s sample unit is a set of
tyres to be measured for the minimum safety requirements.
295
TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Sampling
A population is a collection of all items of interest or under investigation.
Typically there are too many items in a population to consider every one. Examples of populations
include the following:
• Names of all registered voters in South Africa
• Incomes of all families living in Cape Town
• Annual returns of all stocks traded on the Johannesburg Stock Exchange
• Semester marks of all the students at Unisa
A sample is the observed subset of a population.
It is very important that this sample must be representative of the population. How does one
manage this? There are several methods that can be used to choose a representative group.
296
5.2. DATA COLLECTION
Sampling is the process of choosing a number of individuals for a study in such a way
that the individuals represent the population from which they have been chosen.
Generalise
conclusions from
the sample to the
population.
Choose a sample
from the
population.
The theory of sampling is as follows:
1. Researchers want to gather information about a whole group of people – the population.
2. Researchers can only observe a part of the population – the sample.
3. The findings from the sample are generalised, or extended, back to the population.
Types of sampling
The types of sampling methods that will be discussed are given in the diagram below:
There are also several other types of sampling techniques, but they fall outside the scope of this
module.
297
TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
5.2.1
Simple random sampling
Video:
Watch the video “Data1_V2(1)”, the first part of the video on simple random
sampling.
Video:
Watch the video “Data1_V2(2)”, the second part of the video on simple random
sampling.
A good sample requires that each item in the population has an equal and independent chance
to be included in the sample.
A simple random sample is a sample that has been chosen in such a way that each possible sample
containing the same number of observations has the same chance of being drawn.
One method for drawing a simple random sample is to allocate a number to each item in the
population. Then use a computer to generate a sequence of random numbers and use these
numbers to identify items in the population to be included in the sample.
Activity
A printing company, Printapage, has 30 clients with the following outstanding balances (in
rand):
Account No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Balance
25
0
605
1 010
527
34
245
59
667
403
918
801
227
0
47
Account No.
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Balance
0
102
215
429
197
159
279
115
27
27
291
16
0
402
570
The following random numbers are available:
84322; 27217; 11683; 34957; 46227; 76654; 62919; 93151; 37339; 42659; 17784; 14620
Use these numbers to draw a random sample of size 5 out of the 30 customer accounts.
298
5.2. DATA COLLECTION
Answer
The numbers in the list of customer accounts only have two digits, that is 01 to 30. Therefore, choose the last two digits of the given random numbers:
84322; 27217; 11683; 34957; 46227; 76654; 62919; 93151; 37339; 42659; 17784; 14620
Since the total number of elements in the population is 30, a number larger than 30 is of
no use:
22; 17; 83; 57; 27; 54; 19; 51; 39; 59; 84; 20
The sample units are the numbers of the accounts to be drawn. These are
22; 17; 27 19; and 20.
The first random number that we can use is 22.
The outstanding debt for account 22 is R279.
The second random number that we can use is 17.
The outstanding debt for account 17 is R102.
The third random number that we can use is 27.
The outstanding debt for account 27 is R16.
The fourth random number that we can use is 19.
The outstanding debt for account 19 is R429.
The fifth random number that we can use is 20.
The outstanding debt for account 20 is R197.
Activity
A political candidate wishes to determine the opinions of the voters in his ward. He decides
on a sample of size 20. Using random numbers, he chooses 20 telephone numbers from the
telephone directory for a telephonic survey. Is this procedure correct? Give a reason for
your answer.
Answer
All residents may not have telephones, and all the numbers of those who do have telephones may not be included in the telephone directory. Therefore, such a sample cannot
be considered to be random.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
5.2.2
Stratified random sampling
Video:
Watch the video “Data1_V2(3)”, on stratified random sampling.
Simple random sampling requires no prior (a priori) knowledge of the population and can therefore
be done with relatively little effort. It can, however, happen that all the elements drawn for the
sample are nearly homogeneous or alike. This may cause biased conclusions about the population.
If, however, you have prior information about the population, you can rule out this problem
to some degree and consider more correct information about the population by making use of
stratified random sampling.
The population is divided into mutually exclusive sets or strata. This means that a specific
element may belong to one group or stratum only.
The strata must be chosen in such a way that there will be large differences between the strata, but
small differences between the elements within the same stratum. Now simple random samples are
taken from each stratum. Often the number of elements taken from each stratum is proportional
to the size of that stratum.
Example
There are 44 employees at a consulting firm. There are 15 male and 29 female employees. A
sample of seven employees is to be taken to obtain their views on the quality of service that they
deliver.
Calculate the number of male and female employees that should be included in the sample.
Solution
The population is the group of 44 employees at the firm. The sample size is given as 7. Stratum
1 is the group of female employees and stratum 2 is the group of male employees of the firm.
To draw a proportional sample of size 7, the size of the sample that should be taken from each
stratum and the proportions are calculated as follows:
Stratum 1
sample size
=
=
Stratum 2
sample size
=
=
Stratum 1
proportions
300
number of observations in stratum 1
× size of sample
number of observations in population
29
×7
44
4,6
number of observations in stratum 2
× size of sample
number of observations in population
15
×7
44
2,4
Stratum 1 size
29
Size of sample in stratum 1
4,6
:
:
:
:
Population size
44
Size of total sample
7
5.2. DATA COLLECTION
Stratum 2
proportions
Stratum 2 size
15
Size of sample in stratum 2
2,4
:
:
:
:
Population size
44
Size of total sample
7
One cannot choose 4,6 females and 2,4 males. The number 4,6 rounded to the nearest integer is
5 and 2,4 rounded to the nearest integer is 2.
So, randomly choose five female employees from stratum 1 and two male employees from stratum
2 to be included in the sample.
Population
Stratum 1: females
Sample
Stratum 2: males
Activity
Divide Printapage’s 30 customers into three strata as follows:
Stratum
1
2
3
Balance (rand)
< 200
200 – 600
> 600
A proportional sample of size 12 must be drawn from the population. How would you do
it?
Answer
The data divided into three strata look as follows:
301
TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
nnn
Stratum 1
Stratum 2
Stratum 3
Account number
Balance
1
2
6
8
14
15
16
17
20
21
23
24
25
27
28
25
0
34
59
0
47
0
102
197
159
115
27
27
16
0
5
7
10
13
18
19
22
26
29
30
527
245
403
227
215
429
279
291
402
570
3
4
9
11
12
605
1 010
667
918
801
The number of elements (frequency) in each stratum is as follows:
Stratum
1
2
3
Frequency
15
10
5
30
To draw a proportional sample of size 12, the number of items that must be drawn from
stratum 1 is
15
× 12 = 6,
30
the number of items that must be drawn from stratum 2 is
10
× 12 = 4, and
30
the number of items that must be drawn from stratum 3 is
5
× 12 = 2.
30
302
5.2. DATA COLLECTION
5.2.3
Systematic sampling
Systematic sampling starts at a randomly chosen starting point in the population. Then each
subsequent kth element is chosen.
Step 1
Define the population.
Step 2
Determine the desired sample size.
Step 3
Obtain a list, preferably a randomised list of the population.
Step 4
Determine what k is equal to by dividing the size of the population
by the desired sample size.
Step 5
Start at some random place at the top of the population list.
Step 6
Starting at that point, take every kth name on the list until the desired
sample size is reached.
Step 7
If the end of the list is reached before the desired sample is reached,
go back to the top of the list.
In this case, every third item in the population is systematically selected.
Example
Suppose you want a sample of 9 houses from a street with 64 houses.
Systematic sampling starts at a randomly chosen starting point in the population. Then each
subsequent kth element is chosen. Use the following method for systematic sampling to determine
the sample:
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Step 1
Define the population.
The population is the group
of 64 houses.
Step 2
Determine the desired sample size.
The sample size is given as 9.
Step 3
Obtain a list, preferably a randomised
list of the population.
The list will be the group of
64 houses on the street.
Step 4
To determine the value of k, divide the
k=
size of the population by the sample
64
9
= 7,11
size.
So, each 7th house will be chosen.
Step 5
Start at some random place at the top
of the population list.
Here we started at house number
two.
Step 6
Starting at that point, take
Starting at the second house, take
every kth element on the list until the
every 7th house on the list until the
desired sample size of 9 is reached.
desired sample size is reached.
Suppose a political candidate wishes to determine the opinions of the voters in his ward. He has
a list of voters available. He could, for example, start with voter number six and thereafter select
every tenth voter to complete a questionnaire.
Disadvantages of systematic sampling
What is the disadvantage of such a method? If the variable being considered is periodic in nature,
systematic sampling may lead to misleading results.
The disadvantage is that the system may interact with some hidden pattern in the population,
for example every third house along a street might always be the middle one of a terrace of three.
Or if, for example, we are to estimate a shop’s sales and use a 1-in-7 systematic sampling design,
it could happen that only sales figures for Saturdays are chosen. Sales will then be overestimated.
Advantages of systematic sampling
The advantage of systematic sampling is that the sample is spread evenly over the population. It
is convenient, especially when the size of the population is not known. It is also easier to conduct
than a random sample.
304
5.3. PRESENTATIONS
Theme 5.3
Presentations
Learning objectives On completion of this theme you should be able to
• distinguish between qualitative and quantitative data
• draw up and interpret a frequency table
• draw and interpret a histogram
• draw and interpret a pie chart
• draw and interpret a cumulative frequency polygon
• draw and interpret a stem-and-leaf diagram
Consider the tyre company Radial again. It takes a sample of 80 out of the population of 2 600
sets of tyres that have been sold during the past six months. Radial is happy that its sample
is representative. The variable in this sample is the number of thousands of kilometres that a
customer had driven until one of the four tyres no longer met the minimum safety requirements.
In short, the variable is the number of kilometres driven. The sample is given below:
61
42
59
98
61
82
38
35
72
64
77
16
19
77
46
64
29
78
58
70
50
72
26
34
66
56
37
59
62
70
64
69
78
88
80
50
72
45
66
75
22
69
66
46
75
67
83
54
64
59
66
55
53
78
75
14
67
61
51
77
51
32
86
50
62
70
80
66
45
62
45
58
86
62
50
56
58
40
67
90
What should we do with this sample? It is just a list of numbers of which one can make neither
head nor tail! Before we try doing something with these numbers, let’s first consider the different
types of data one may get.
305
TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Data are facts, such as values or measurements. Data can be numbers, words,
measurements, observations or even just descriptions of things.
To summarise the diagram on the previous page:
There are two main groupings of data, namely qualitative data and quantitative data.
Qualitative data are data that are not given numerically,
for example example your favourite colour, place of birth, favourite food and type
of car.
Quantitative data are numerical. There are two types of quantitative data:
(1)
Discrete data can only be specific numeric values. They include everything that can be
considered as separate units because of their nature,
for example shoe size, number of brothers and number of cars in a car park.
(2)
Continuous data can be any numerical values. Continuous data are usually the result
of a measurement and are not fixed isolated points. There can be a whole range of
values between any two values,
for example height, mass and length.
Activity
Classify the data collected in each of the following questions:
1. Do you own a TV set?
Yes Yes
No Yes
2. How many TV sets do you own?
3. How many kilometres did you drive with your set of Radial tyres?
4. What was your electricity bill last month?
Answer
1. The data is qualitative.
2. This is discrete quantitative data.
3. This is continuous quantitative data.
4. This is continuous quantitative data.
What is the nature of Radial’s data? We could identify Radial’s data as quantitative and continuous.
If we could envisage the data, we would be able to form a better idea of what it represents.
306
5.3. PRESENTATIONS
5.3.1
Bar representations
Bar graph
Suppose Radial also asked their customers in the sample which brand of tyre they had before
they switched to the XXX tyres. The results are listed in the following table.
Note: The variable under investigation is the brand of tyre.
Brand of tyre
Number of customers
BFGoodrich
7
Dunlop
17
Continental
22
Goodyear
8
Bridgestone
15
Firestone
11
This information is presented in a bar graph:
Brand of tyres driven by customers before XXX
24
22
20
Number of customers
18
16
14
12
10
8
6
4
2
0
BFGoodrich
Dunlop
Continental Goodyear Bridgestone
Firestone
Brand of tyre
307
TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Histogram
Consider the first variable that Radial examined, namely the number of kilometres driven with
a set of XXX tyres. The data are grouped into a set of intervals, with each interval of equal
width. The results are listed in the following table.
Note: The variable under investigation is the number of kilometres driven.
Number of thousands of
kilometers driven
Number of customers
13,5 - 24,5
4
24,5 - 35,5
5
35,5 - 46,5
9
46,5 - 57,5
11
57,5 - 68,5
25
68,5 - 79,5
17
79,5 - 90,5
8
90,5 - 101,5
1
This information is presented in a histogram:
Number of customers
35
30
25
20
15
10
5
13,5
24,5
35,5
46,5
57,5
68,5
Number of thousands of
kilometres driven
308
79,5
90,5
101,5
5.3. PRESENTATIONS
Comparison
What is the difference between a bar graph and a histogram?
There are two differences: one difference is the type of data that are presented and the other is
the way in which bar graphs and histograms are drawn.
Bar graph
Histogram
Displays categorical data, that is data that
Presents continuous data, that is data that
fits into categories.
represent a measured quantity.
The numbers can take on any value in a
certain range.
The bars are usually drawn with spaces
between them.
The bars are never drawn with spaces
between them because there should not be
any gaps between the intervals.
309
TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
5.3.2
Frequency table
Did we miss something about Radial’s case?
How did we get from the raw data...
61
42
59
98
61
82
45
51
38
35
72
64
77
16
62
32
19
77
46
64
29
78
45
86
58
70
50
72
26
34
58
50
66
56
37
59
62
70
90
62
64
69
78
88
80
50
86
70
72
45
66
75
22
69
62
40
66
46
75
67
83
54
50
67
64
59
66
55
53
78
56
80
75
14
67
61
51
77
58
66
...to data organised into intervals?
Number of thousands of
kilometers driven
Number of customers
13,5 - 24,5
4
24,5 - 35,5
5
35,5 - 46,5
9
46,5 - 57,5
11
57,5 - 68,5
25
68,5 - 79,5
17
79,5 - 90,5
8
90,5 - 101,5
1
We identified Radial’s data as quantitative and continuous data.
If we could now only envisage the data, we would be able to form a better idea of what is going
on.
The histogram is one of the most common ways to represent data visually. A histogram is a
graphical presentation of a frequency table.
A frequency table is an organised count of items or values. The raw data are organised into
intervals and counted to show how many of the raw data lie in each interval.
These tables are often useful for discovering basic statistical information about a data set.
Video:
310
Watch the video “Data2_V1” on how to construct a frequency table.
5.3. PRESENTATIONS
Number of intervals
According to the video, the number of intervals to use when constructing a frequency table is
determined by using the guideline
maximum value − minimum value
.
10
no. of intervals =
We can also use Sturge’s Rule as a guideline to select the number of intervals for a histogram.
If we have a number n of data points, then the number of intervals, k, is given by
k = 1 + 3,3 log n.
If the value of k is not an integer, the value is increased to the next integer just larger than the
calculated value. This will ensure that all the data points are included on the histogram. Sturge’s
Rule works best for values of n larger than 30.
Example
If the number of data points is 52, then the number of intervals is calculated as
k = 1 + 3,3 log n
= 1 + 3,3 log(52)
= 1 + 3,3 × 1,716
= 1 + 5,663
= 6,663
≈ 7.
The number of intervals for 52 data points is 7.
The table below gives an indication of the values of k for values of n from 10 to 2 154:
Value of
n
No. of intervals
k
10 – 16
5
17 – 32
6
33 – 65
7
66 – 132
8
133 – 265
9
266 – 533
10
534 – 1 072
11
1 073 – 2 154
12
311
TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
The complete frequency table for Radial is the following:
Interval
13,5 – 24,5
Frequency
4
24,5 – 35,5
5
35,5 – 46,5
9
46,5 – 57,5
11
57,5 – 68,5
25
68,5 – 79,5
17
79,5 – 90,5
8
90,5 – 101,5
1
80
It is clear that the highest frequency is in the interval 57,5 to 68,5. This shows that most of the
customers got between 57 500 and 68 500 kilometres per set of tyres.
Now we can graphically represent the frequency table by drawing the interval lengths on a horizontal axis and the frequencies on a vertical axis:
Number of customers
35
30
25
20
15
10
5
13,5
24,5
35,5
46,5
57,5
68,5
79,5
90,5
101,5
Number of thousands of
kilometres driven
(Notice that the horizontal axis starts at 0 and that the zigzag line is there to break the line in
order to prevent a huge space from appearing to the left of the actual data shown on the graph.)
312
5.3. PRESENTATIONS
Relative frequency
We might not just be interested in the absolute frequencies, but also in how these numbers relate
to each other. In other words, we might want to know the proportion of distances in each interval
in order to understand how many distances from each interval make up the whole. This is called
the “relative frequency”.
For example, 4 out of 80 distances are in the first interval:
4
= 0,0500
80
The proportion of distances for the first interval is 0,05.
Activity
Calculate the proportion of distances for all the intervals of Radial’s frequency table.
Answer
Interval
Frequency
13,5 - 24,5
4
24,5 - 35,5
5
35,5 - 46,5
9
46,5 - 57,5
11
57,5 - 68,5
25
68,5 - 79,5
17
79,5 - 90,5
8
90,5 - 101,5
1
Relative
frequency
4
= 0,0500
80
5
= 0,0625
80
9
= 0,1125
80
11
= 0,1375
80
25
= 0,3125
80
17
= 0,2125
80
8
= 0,1000
80
1
= 0,0125
80
All proportions are always between 0 and 1.
For any frequency table, the relative frequencies should add up to 1.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Percentages
Another way to present the frequencies is with percentages. Because relative frequencies have
decimals, it is easier to interpret percentages. If we use percentages instead of relative frequencies
we may be able to work with whole numbers.
A percentage is basically a proportion, except we multiply by 100 and we call that per cent.
The proportion of the first interval converted to a percentage is
0,05 × 100 = 5%.
Activity
Calculate the percentage of distances for all the intervals of Radial’s frequency table.
Answer
Interval
Frequency
Relative
frequency
13,5 - 24,5
4
0,0500
24,5 - 35,5
5
0,0625
35,5 - 46,5
9
0,1125
46,5 - 57,5
11
0,1375
57,5 - 68,5
25
0,3125
68,5 - 79,5
17
0,2125
79,5 - 90,5
8
0,1000
90,5 - 101,5
1
0,0125
Total
80
1,0000
Percentage
%
4
× 100 = 0,0500 × 100 = 5,00%
80
5
× 100 = 0,0625 × 100 = 6,25%
80
9
× 100 = 0,1125 × 100 = 11,25%
80
11
× 100 = 0,1375 × 100 = 13,75%
80
25
× 100 = 0,3125 × 100 = 31,25%
80
17
× 100 = 0,2125 × 100 = 21,25%
80
8
× 100 = 0,1000 × 100 = 10,00%
80
1
× 100 = 0,0125 × 100 = 1,25%
80
100,00%
To turn a proportion (relative frequency) into a percentage, multiply the proportion by 100.
The percentages range from 0% to 100%.
The percentages of Radial add up to 100%.
314
5.3. PRESENTATIONS
nnn
Activity
What percentage of the customers got
1. 80 000 kilometres or more per set of tyres?
2. 46 000 kilometres or less per set of tyres?
Answer
1. The distance of 80 000 kilometres or more is represented by the intervals
79,5 – 90,5
and
90,5 – 101,5.
The sum of the percentages is
10% + 1,25% = 11,25%.
Therefore, 11,25% of the customers got 80 000 kilometres or more per set of tyres.
2. The distance of 46 000 kilometres or less is accounted for by the first three intervals. The
sum of the percentages is
5,00% + 6,25% + 11,25% = 22,50%.
Therefore, 22,50% of the customers got 46 000 kilometres or less per set of tyres.
Frequency tables and interval size
Video:
5.3.3
Watch the video “Data2_V2” on frequency tables, histogram and interval size.
Pie chart
Another way of representing data is the pie chart.
Example
Look at this traffic record of vehicles travelling down a particular road:
Traffic survey 31 January 2019
Type of vehicle
Number of vehicles
Cars
140
Motorbikes
70
Vans
55
Buses
5
Total number of vehicles
270
315
TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
To draw a pie chart, we need to represent each part of the data as a proportion of 360, because
there are 360 degrees in a circle.
Calculate the degrees that each type of vehicle will take up. For example, if 55 out of 270 vehicles
are vans, we represent this on the circle as a segment with an angle of
55
× 360 = 73.
270
Therefore, 73 degrees out of 360 are represented by the vans. We have the following results:
Traffic survey 31 January 2019
Type of
vehicle
Number of
vehicles
Calculation
Degrees of
a circle
Cars
140
140
× 360
270
187
Motorbikes
70
70
× 360
270
93
Vans
55
55
× 360
270
73
Buses
5
5
× 360
270
7
Before you draw the pie chart, remember to check that the angles you have calculated add up to
360 degrees.
This data set is represented in the pie chart below:
Cars
140
Buses
5
Motorbikes
70
316
Vans
55
5.3. PRESENTATIONS
nn
Activity
Calculate the number of degrees for Radial’s data.
Answer
Interval
Frequency
Calculation
13,5 - 24,5
4
0,0500
24,5 - 35,5
5
0,0625
35,5 - 46,5
9
0,1125
46,5 - 57,5
11
0,1375
57,5 - 68,5
25
0,3125
68,5 - 79,5
17
0,2125
79,5 - 90,5
8
0,1000
90,5 - 101,5
1
0,0125
Degrees of a
circle
4
× 360 = 18,0
80
5
× 360 = 22,5
80
9
× 360 = 40,5
80
11
× 360 = 49,5
80
25
× 360 = 112,5
80
17
× 360 = 76,5
80
8
× 360 = 36,0
80
1
× 360 = 4,5
80
The pie chart for Radial is as follows:
35,5 – 46,5
46,5 – 57,5
13,75%
24,5 – 35,5
11,25%
6,25%
5,00%
57,5 – 68,5
31,25%
10,00%
13,5 – 24,5
→ 1,25%
90,5 – 101,5
79,5 – 90,5
21,25%
68,5 – 79,5
We have calculated that 22,5% of the customers drove 46 000 kilometres or less with a set of tyres.
Next we present this information graphically. First we obtain the “cumulative less than” table.
Such a table is set up from the frequency table, setting the upper limits to “less than . . . ”.
317
TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
nn
5.3.4
Cumulative frequency
In a data set, the cumulative frequency for a value is the total number
of scores that are less than or equal to that value.
The charts below illustrate the difference between frequency and cumulative frequency. Both
charts show scores for a test administered to 300 students.
Graphical presentation of
frequency table (histogram)
Cumulative frequency chart
Column height shows frequency –
the number of students in each test
Column height shows cumulative frequency –
the number of students up to and including
score grouping.
each test score.
24 students received a test score
between 51 and 60.
25 students received a test score of at most 50.
49 students received a test score of at most 60.
109 students received a test score of at most 70.
350
Number of students (frequency)
Number of students (frequency)
100
90
80
70
60
50
40
30
20
10
40,5
50,5
60,5
70,5
80,5
90,5
300
250
200
150
100
50
100,5
40,5
Test score
50,5
60,5
70,5
80,5
90,5
100,5
Test score
The total frequencies of all intervals less than the upper class boundary of a specified interval is
called the cumulative frequency of that interval.
Interval
Frequency
Upper
limit
Cumulative
frequency
40,5 – 50,5
50,5 – 60,5
60,5 – 70,5
70,5 – 80,5
80,5 – 90,5
90,5 – 100,5
25
24
60
92
64
35
< 50,5
< 60,5
< 70,5
< 80,5
< 90,5
< 100,5
25
49
109
201
265
300
318
Cumulative means
“added up”
(25 + 24 = 49)
(25 + 24 + 60 = 109)
(25 + 24 + 60 + 92 = 201)
(25 + 24 + 60 + 92 + 64 = 265)
(25 + 24 + 60 + 92 + 64 + 35 = 300)
5.3. PRESENTATIONS
The “cumulative less than” table for Radial is obtained by using the frequency table and setting
the upper limits to “less than . . . ”.
The cumulative frequency table for Radial is as follows:
Upper limit
Cumulative frequency
< 24,5
< 35,5
4
9
(4 + 5 = 9)
< 46,5
< 57,5
18
29
(4 + 5 + 9 = 18)
(4 + 5 + 9 + 11 = 29)
< 68,5
< 79,5
54
71
(4 + 5 + 9 + 11 + 25 = 54)
< 90,5
< 101,5
79
80
Cumulative means “added up”.
Ogive
The graph of the cumulative frequency distribution is better known as the cumulative frequency
curve or Ogive. The term Ogive is used in architecture to describe curves or curved shapes.
An Ogive is a graph that represents the cumulative frequencies of the intervals in a
frequency distribution. It shows the data below or above a particular value.
Ogive is pronounced as “O-jive”. Ogive is a graph of a cumulative distribution that shows data
values on the horizontal axis and the cumulative frequencies on the vertical axis. The Ogive is
constructed by plotting a point corresponding to the cumulative frequency of each interval.
The Ogive for Radial is given below. Previously we calculated that 22,5% of Radial’s customers
drove 46 000 kilometres or less with a set of tyres. Such information can be presented graphically
once the cumulative frequencies are known:
Cumulative Frequency (Cum f )
90
80
70
60
50
40
30
20
18 ÷ 80 × 100 = 22,5%
18
10
24,5
35,5
46,5
57,5
68,5
79,5
90,5 101,5
Distance in thousands of kilometres
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
5.3.5
Stem-and-leaf diagram
Stem-and-leaf diagrams are very useful and easy to set up.
Data can be shown in a variety of ways, including graphs, charts and tables. A stem-and-leaf
diagram is a type of graph that is similar to a histogram but shows more information. A stemand-leaf diagram summarises the shape of a set of data (the distribution) and provides extra
information about individual values.
The data are arranged by place value. Each data value is split into a leaf (usually the last digit)
and a stem (the other digits). The stem values are listed down and the leaf values are listed next
to them.
In this way the stem groups the values and each leaf indicates a particular value within the group.
The leaves are displayed to the right of the stem. Stem-and-leaf diagrams are great organisers
for large amounts of information.
Example
The first step in setting up this diagram is to decide how to separate each observation into two
parts – the stem and the leaf.
Let’s separate Radial’s data in such a way that the first digit of each number is the stem and the
second digit is the leaf. We already know that the smallest number is 14 and the largest is 98.
Thus 14 has stem 1 and leaf 4.
Thus 98 has stem 9 and leaf 8.
All the other numbers lie between these two and we can therefore set up the stem from 1 to 9.
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5.3. PRESENTATIONS
Now the second digit of each number is written next to its stem.
Stem
Leaf
Frequency
1
9 6 4
3
2
9 6 2
3
3
8 5 7 4 2
5
4
2 5 6 6 5 5 0
7
5
8 6 9 9 0 9 5 3 1 0 4 8 0 6 8 1 0
17
6
1 6 4 6 4 9 6 6 7 4 4 7 1 1 2 9 2 2 2 7 6
21
7
2 5 7 0 2 8 5 2 5 7 8 0 8 7 0
15
8
8 0 3 2 6 6 0
7
9
8 0
2
To make it more readable, we can sort the data for each stem.
Radial’s sorted stem-and-leaf diagram is as follows:
Stem
Leaf
Frequency
1
4 6 9
3
2
2 6 9
3
3
2 4 5 7 8
5
4
0 2 5 5 5 6 6
7
5
0 0 0 0 1 1 3 4 5 6 6 8 8 8 9 9 9
17
6
1 1 1 2 2 2 2 4 4 4 4 6 6 6 6 6 7 7 7 9 9
21
7
0 0 0 2 2 2 5 5 5 7 7 7 8 8 8
15
8
0 0 2 3 6 6 8
7
9
0 8
2
Now turn the page on its side, and it is easy to see that most of the customers drove sixty
thousand kilometres with a set of tyres.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Alternative presentation
Looking at the stem-and-leaf diagram from its side, it is easy to see that most of the customers
drove between sixty and seventy thousand kilometres with a set of tyres.
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5.3. PRESENTATIONS
Exercise 5.1
As the manager of an insurance claims division, you have to set up performance levels.
You have asked 30 of your experienced claims processing personnel to record the number
of claims that they processed during a specific week.
The following data set was collected:
Claims processed by 30 claims processors in a week
31
37
30
36
28
38
33
38
35
39
37
36
38
34
39
31
30
34
40
46
41
44
40
39
41
34
48
40
45
42
1. Display the data in the form of a histogram.
2. Use the frequency table to set up a cumulative frequency table.
3. Set up a stem-and-leaf diagram.
4. Now answer the following questions:
(a) During the past month, an experienced worker has processed only 26 claims per week.
Do you sense a problem? Give reasons for your answer.
(b) Information obtained from a competitor indicates that 50% of his workers can process
36 to 39 claims per week. What is happening here?
(c) You decide to transfer some of the workers to other divisions if you find that less than
36 claims are processed per week by half of the workers. What is you decision?
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Theme 5.4
Measures of locality
Learning objective
5.4.1
On completion of this theme you should be able to calculate the
mean, the mode and the median of a data set.
Introduction
It is often not possible to list all the data or to draw a histogram. It would be nice to have one
number that best represents a data set. Such a number, or a measure of location, is useful to
indicate where the data lie.
A measure of central tendency is a single value that attempts to describe a set
of data by identifying the central position within that set of data.
As such, measures of central tendency are sometimes called measures of central location.
The mean (often called the average) is most likely the measure of central tendency that you are
most familiar with, but there are others, such as the median and the mode.
The following are valid measures of central tendency:
• mean
• median
• mode
Under different conditions, some measures of central tendency become more appropriate to use
than others.
It is important to consider the mean, mode and median, how to calculate them and under what
conditions they are most appropriate to be used.
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5.4. MEASURES OF LOCALITY
5.4.2
The mean (raw data)
If the mayor of a city is asked to provide a single value that best describes the income level of
the city, he or she would answer with a measure of central tendency.
The mean (or average) is the most popular and well-known measure of central tendency. It can
be used with both discrete and continuous data.
Definition
The mean is equal to the sum of all the values in the data
set divided by the number of values in the data set.
This is called the arithmetic mean. To calculate the arithmetic mean, add up the values and
divide their sum by their quantity.
The first three examples below show higher values, dragging the mean upward. The final example
piles on some low values to drag the mean downward. What you see is characteristic of the
arithmetic mean. Any value added above or below the mean will move the mean in the direction
of the new value. The move may be slight, or it may be large, but it will be non-zero.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Radial advertises that its XXX tyres will travel at least 65 000 kilometres before one of the
four tyres will no longer meet the minimum safety requirements. What is the mean number of
kilometres that can be driven with a set of XXX tyres? Radial has only the following sample of
80 observations available to estimate the mean:
61
42
59
98
61
82
45
51
38
35
72
64
77
16
62
32
19
77
46
64
29
78
45
86
58
70
50
72
26
34
58
50
66
56
37
59
62
70
90
62
64
69
78
88
80
50
86
70
72
45
66
75
22
69
62
40
66
46
75
67
83
54
50
67
64
59
66
55
53
78
56
80
75
14
67
61
51
77
58
66
If we consider the sample as being representative of the population, we use the sample mean as
an estimator of the population mean.
To obtain the sample mean, we add up all the observations and divide the result by the number
of observations.
Activity
Add up all the observations in Radial’s sample and divide the sum by the number of
observations.
Answer
The answer is
4 783
= 59,79.
80
We can therefore expect a set of tyres to travel
59,79 × 1 000 = 59 790
kilometres on average.
Formula
If we have n values in a data set and they have values x1 , x2 , . . ., xn , the sample mean, usually
denoted by x (pronounced “x bar”), is
(x1 + x2 + . . . + xn )
.
n
A formula for the mean is written in a different way:
x=
x=
where
326
1 n
xi ,
n i=1
x
(read as “x bar”) is the generally accepted symbol for the arithmetic mean,
n
is the number of observations,
(pronounced “sigma”) is the Greek letter for S and means “sum of . . .”, and
xi
represents the ith observation.
5.4. MEASURES OF LOCALITY
The given formula refers to the sample mean. So, why have we called it a sample mean? This
is because, in statistics, samples and populations have very different meanings, which are very
important, even if they are calculated in the same way, as in the case of the mean. We will only
consider the sample mean.
You will notice that the mean is not often one of the actual values that you observe in your data
set. However, one of its important properties is that it minimises error in the prediction of any
one value in your data set.
Example
The mean is the measure of locality used most often. However, sometimes it can be misleading.
Consider the monthly salaries (in R1 000,00) of ten staff members at a factory:
Staff
1
2
3
4
5
6
7
8
9
10
Salary
15
18
16
14
15
15
12
17
90
95
The mean is
x =
15 + 18 + 16 + 14 + 15 + 15 + 12 + 17 + 90 + 95
10
= 30,70.
The mean salary for these ten staff members, in rand, is therefore,
1 000,00 × 30,70 = 30 700,00.
However, inspecting the data suggests that this mean value of R30 700,00 might not be the best
way to reflect the typical salary of a staff member accurately, since most staff members have
salaries in the R12 000,00 to R18 000,00 range.
The mean is rather sensitive to outliers and may often be misleading. Outliers are values that
are unusual compared to the rest of the data set by being especially small or large in numerical
value. On its own, without any additional information, the arithmetic mean may often lead to
incorrect conclusions. In this situation, the mean is being skewed by the two large salaries. We
would therefore like to have a better measure of central tendency. Taking the median would be
a better measure of central tendency in this situation.
A big advantage of the arithmetic mean is that it uses all the available data. This is not the
case for the other measures of locality.
Since the mean can be calculated exactly, it forms the basis for many advanced analyses and is
not only descriptive in nature.
5.4.3
The mean (interval data)
Sometimes we are given a chart showing frequencies of certain groups instead of the actual values.
We can still come up with a good estimate of a typical value for the set of data, provided that
we make some assumptions.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Intervals
We assume that the values in each interval are
spread evenly throughout the interval.
If this is the case, then the mean for each interval should be approximately equal to the
midpoint for each interval.
The midpoint of an interval divides the interval into two equal parts. It is
obtained by adding the upper and lower limits of each interval,
and dividing the result by two.
This middle value represents the class interval in calculations.
The tyre company Radial advertises that their XXX tyres will travel at least 65 000 kilometres
before one of the four tyres will no longer meet the minimum safety requirements. A frequency
table providing data about intervals has previously been drawn up.
The first interval in Radial’s frequency table is 13,5 – 24,5.
The midpoint is found by adding the lower interval boundary to the higher interval boundary
and then dividing that sum by 2:
13,5 + 24,5
2
=
38,0
2
= 19,0
The midpoint for the interval 13,5 – 24,5 is 19,0.
Formula
The tyre company Radial advertises that their XXX tyres will travel at least 65 000 kilometres
before one of the four tyres will no longer meet the minimum safety requirements.
The midpoints for the intervals of Radial’s frequency table have been calculated. So, for each
interval we have the number of values (frequency) and the midpoints:
328
Interval
Frequency
fi
Midpoint
xi
Percentage
%
13,5 – 24,5
4
19
5,00
24,5 – 35,5
5
30
6,25
35,5 – 46,5
9
41
11,25
46,5 – 57,5
11
52
13,75
57,5 – 68,5
25
63
31,25
68,5 – 79,5
17
74
21,25
79,5 – 90,5
8
85
10,00
90,5 – 101,5
1
96
1,25
5.4. MEASURES OF LOCALITY
If we multiply each midpoint by its frequency, add them all up and then divide their sum by the
nn
total number of values in the frequency table, we have an estimate of the mean.
Method to determine estimated mean
So, the method to determine an estimate of the mean is as follows:
Step 1
Determine the total number of values by finding the sum of the frequency values.
Step 2
Multiply each midpoint by its frequency.
Step 3
Find the sum of all these products.
Step 4
Divide the sum of these products by the total number of values.
Step 1
Determine the total number of values as
fi
n =
= 4 + 5 + 9 + 11 + 25 + 17 + 8 + 1
= 80.
Step 2
Multiply each midpoint by its frequency:
Interval
Frequency
fi
Midpoint
xi
13,5 – 24,5
4
19
4 × 19
=
76
24,5 – 35,5
5
30
5 × 30
=
150
35,5 – 46,5
9
41
9 × 41
=
369
46,5 – 57,5
11
52
11 × 52
=
572
57,5 – 68,5
25
63
25 × 63
=
1 575
68,5 – 79,5
17
74
17 × 74
=
1 258
79,5 – 90,5
8
85
8 × 85
=
680
90,5 – 101,5
1
96
1 × 96
=
96
fi × xi
Step 3
Find the sum of the products of the midpoints and their frequencies. Add these products together:
fi xi = 76 + 150 + 369 + 572 + 1 575 + 1 258 + 680 + 96
= 4 776
The sum of the products is 4 776.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Step 4
Divide this sum of the fi × xi products by the total number of values:
fi xi
n
4 776
=
80
= 59,70
x =
Therefore, the estimate of the mean is 59,70.
The mean calculated for the raw data (the original 80 values) of Radial is 59,79. We did not lose
too much information by classifying the data into a frequency table.
Activity
The time between equipment breakdowns in a factory was recorded over a period of several
months. During this period, breakdowns were observed. The times are shown in the
following frequency table:
Time between
breakdowns (days)
Number of breakdowns
frequency (f )
−0,5 – 4,5
6
4,5 – 9,5
10
9,5 – 14,5
14
14,5 – 19,5
6
19,5 – 24,5
4
Total
40
Determine the mean number of days between breakdowns.
Answer
First determine the midpoint of each interval. Remember, the midpoint of an interval
divides the interval into two equal parts. It is obtained by adding the upper and lower
limits of each interval, and dividing the result by two.
The midpoint of the first interval is
−0,5 + 4,5
= 2.
2
The midpoint of the second interval is
4,5 + 9,5
= 7.
2
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5.4. MEASURES OF LOCALITY
After the midpoints for all the intervals have been determined, it is necessary to add a
column to the table and to calculate the values of fi × xi .
Interval
Frequency
Midpoint
fi
xi
−0,5 – 4,5
6
2
6×2
=
12
4,5 – 9,5
10
7
10 × 7
=
70
9,5 – 14,5
14
12
14 × 12
=
168
14,5 – 19,5
6
17
6 × 17
=
102
19,5 – 24,5
4
22
4 × 22
=
88
fi × xi
5
5
n=
fi
fi xi
i=1
i=1
The sum of the frequencies is calculated as
5
fi
n =
i=1
= 6 + 10 + 14 + 6 + 4
= 40.
Note that there are five intervals, therefore n = 5.
Find the sum of the products of the midpoints and their frequencies. Add these products
together:
5
fi xi = 12 + 70 + 168 + 102 + 88
i=1
= 440
The mean is calculated as
5
fi xi
x =
i=1
n
440
=
40
= 11.
The mean number of days between breakdowns is 11 days.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
5.4.4
The median
The mean is sensitive to extreme values and may often give rise to misleading conclusions. Therefore, the median is often preferred as a measure of locality.
Definition
The median (M e) is the score value that cuts the distribution in half, such
that half the scores fall above the median and half fall below it.
Computation of the median is relatively straightforward.
1.
The first step is to order the data values from the smallest to largest.
one way if there are an odd number
of data values in the sample
2.
The procedure branches at this step:
another if there are an even number
of data values in the sample
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5.4. MEASURES OF LOCALITY
Odd number of data values
When computing the median of an odd number of data values in a sample, the first step is to
order the data values from the smallest to the largest.
Refer to the example below to discover the rest of the procedure to compute the median of an
odd number of data values.
A survey of 15 employees’ performance scores out of 50 gave the following results, ordered from
lowest to highest:
32
32
35
36
36
37
38
38
39
39
39
40
40
45
46
The median is simply the middle number. In the case above the median would be number 38,
because there are 15 scores all together, with 7 scores to the left and 7 to the right of the median.
How do we know where the middle number is?
Given a data set of size n sorted in ascending sequence,
the position of the median (M e) is the
n+1
2
th
value.
Our data set has 15 values, so the position of the median is
n+1
2
15 + 1
2
16
=
2
= 8.
=
The median is the 8th value in the ordered data set.
1st
2nd
3rd
4th
5th
6th
7th
32
32
35
36
36
37
38
8th
9th
10th
11th
12th
13th
14th
15th
39
39
39
40
40
45
46
38
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Even number of data values
When computing the median of an even number of data values in a sample, the first step is to
order the data values from the smallest to the largest.
Refer to the example below to discover the rest of the procedure to compute the median of an
even number of data values.
A survey of 14 employees’ performance scores out of 50 gave the following results, ordered from
lowest to highest:
32
35
36
36
37
38
38
39
39
39
40
40
42
45
The median is the midpoint between the two middle scores. In this case the median is the value
38,5. It is found by adding the two middle scores together and dividing their sum by two:
38 + 39
= 38,5
2
If the two middle scores are the same value, then the median is that value.
How do we know where the middle number is?
Given a data set of size n sorted in ascending sequence,
the position of the median (M e) is the
n+1
2
th
value.
Our data set has 14 values, so the position of the median is the following value:
n+1
2
14 + 1
2
15
=
2
= 7,5
=
The median is the 7,5th value in the ordered data set. So, the median is 38,5.
7,5th
1st
2nd
3rd
4th
5th
6th
7th
8th
9th
10th
11th
12th
13th
14th
32
35
36
36
37
38
38
39
39
39
40
40
42
45
38,5
Note that it is not taken into account whether there is a duplication of scores around the median.
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5.4. MEASURES OF LOCALITY
Sensitivity
The median is not sensitive to extreme scores.
Compare the medians of the following two similar ordered data sets. The two sets are the same,
except for the highest score in each case. In the first set the highest score is not extreme, but in
the second it is extreme.
In the first instance, a survey of 15 employees’ performance scores out of 50 gave the following
results, ordered from lowest to highest, with 42 as the highest score:
32
32
35
36
36
37
38
38
39
39
39
40
40
41
42
The median is the middle number. In the case above, the median is the number 38.
In the second instance, a survey of 15 employees’ performance scores out of 50 gave the following
results, ordered from lowest to highest, with 50 as an extreme score:
32
32
35
36
36
37
38
38
39
39
39
40
40
41
50
In this case with the extreme score, the median is still the value of 38.
5.4.5
The mode
Definition
The mode (M o) of a data set is that value that occurs most often.
A survey of 14 employees’ performance scores out of 50 gave the following results, ordered from
lowest to highest:
33
33
35
36
37
38
38
39
39
40
39
39
40
42
43
44
40
42
45
Let’s look at the data in a different way:
33
34
35
36
37
38
41
45
The mode of the data is 39 because 39 occurs three times, which is more than any other score.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
More than one mode
A data set may have more than one mode if the most frequently occurring scores occur the same
number of times.
A survey of 14 employees’ performance scores out of 50 has these results, ordered from lowest to
highest:
33
33
33
36
37
38
38
39
39
39
40
42
43
44
40
42
45
Let’s look at the data in a different way:
33
34
35
36
37
38
39
40
41
45
There are two modes, namely 33 and 39. Such distributions are called bimodal distributions.
No mode
A data set may have no mode.
A survey of 14 employees’ performance scores out of 50 has these results, ordered from lowest to
highest:
32
33
34
35
36
37
38
39
40
41
42
42
43
43
44
Let’s look at the data in a different way:
32
33
34
35
36
37
No value occurs more than any other value.
336
38
39
40
41
44
45
45
5.4. MEASURES OF LOCALITY
Sensitivity of mode
The mode is not sensitive to extreme scores.
Compare the modes of the following two similar ordered data sets. The two sets are the same,
except for the highest score in each case. In the first set the highest score is not extreme, but in
the second set it is extreme.
In the first instance, a survey of 14 employees’ performance scores out of 50 gives the following
results, ordered from lowest to highest, with 45 as the highest score:
33
33
35
33
36
37
35
36
34
38
37
38
39
39
40
38
39
39
40
42
43
44
41
40
42
45
45
The mode of the data is 39 because it occurs three times, which is more than any other score.
In the second instance, a survey of 14 employees’ performance scores out of 50 gives the following
results, ordered from lowest to highest, with the highest score of 50 as an extreme score:
33
33
34
33
35
35
36
36
37
37
38
39
40
38
38
41
42
39
39
43
44
39
45
40
46
47
40
42
48
49
50
50
The mode of the data is still 39.
In any case, the mode is a quick and dirty measure of central tendency: quick, because it is easily
and quickly computed; and dirty, because it is not very useful, meaning it does not give much
information about the data.
It may happen that there is no value that occurs more than any other value, or that there is more
than one value with the same maximum number of occurrences. The only thing counting in the
mode’s favour is that it is easy to understand.
5.4.6
Distribution of data
Video:
Watch the video “Data3_V2” on the distribution of data.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Exercise 5.2
Calculate the
1. mean,
2. median, and
3. mode
for the following data:
190;
338
104;
135;
314;
179;
175;
170;
146;
127;
131
5.5. MEASURES OF DISPERSION
Theme 5.5
Measures of dispersion
Learning objectives On completion of this theme you should be able to
• calculate the variance of a data set
• explain the notion of a standard deviation
• calculate and interpret the quartile deviation of a data set
• calculate and interpret a coefficient of variation for a data set
• present and interpret a box-and-whiskers diagram
5.5.1
Introduction
Problems can occur when we work with a measure of central tendency only. Without knowing
something about how data are spread out, measures of central tendency may be misleading.
Measures of dispersion provide a more complete picture. For example, a residential street, say
Street A, with 20 homes on it having a mean home value of R950 000,00, with little variation
from the mean, would be very different from a street, say Street B, with the same mean home
value, but with three homes having a value of R3 500 000,00 and the other 17 clustered around
R500 000,00.
Street A
Street B
20 homes at around R950 000,00
20 × 950 000,00 = 19 000 000,00
17 homes at R500 000,00
17 × 500 000,00 = 8 500 000,00
3 homes at around R3 500 000,00
3 × 3 500 000,00 = 10 500 000,00
Total value of 20 homes is R19 000 000,00.
8 500 000 + 10 500 000 = 19 000 000
Total value of 20 homes is R19 000 000,00.
19 000 000,00 ÷ 20 = 950 000,00
Mean value of homes is R950 000,00.
19 000 000,00 ÷ 20 = 950 000,00
Mean value of homes is R950 000,00.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
5.5.2
Range
A measure of location, such as the mean or the median, only describes the centre of the data. It
is valuable from that standpoint, but it does not tell us anything about the spread of the data.
The arithmetic mean, even though it uses all the values in the data set, does not give much
information about what the data set really looks like. What we also need is information on the
spread of the data around the mean.
The range is the simplest measure of spread. It is the difference between the largest and the
smallest values in the data set:
range = largest value − smallest value
This measure of spread does not take into account anything about the distribution of the data
other than the extremes.
10
11
12
13
14
15
The range is 15 − 10 = 5.
10
11
12
13
14
15
The range is 15 − 10 = 5.
The range ignores how the data are distributed.
5.5.3
Standard deviation
The standard deviation is the measure of spread most commonly used in statistical practice when
the mean is used to calculate central tendency.
Spread around the mean
The standard deviation can be difficult to interpret as a single number on its own. Basically, a
small standard deviation means that the values in a statistical data set are on average close to
the mean of the data set. A large standard deviation means that the values in the data set are
on average farther away from the mean.
The standard deviation measures how concentrated the data are around the mean – the more
concentrated, the smaller the standard deviation.
The standard deviation measures the spread of data around the mean.
340
5.5. MEASURES OF DISPERSION
Compare data sets
The normal (bell) curve
Frequency
A
B
Data
Another reason for studying the dispersion in a set of data is to compare the spread in two or
more sets of data.
When comparing the spread of the two separate data sets above, namely A and B, which have
approximately the same mean, we see that data set A has a smaller standard deviation. It has
a narrower spread of measurements around the mean and therefore usually has comparatively
fewer high or low data values. Data set B has a larger standard deviation. It has a wider spread
of measurements around the mean and therefore usually has more high and low data values.
An item selected at random from data set A, whose standard deviation is low, has a better chance
of being close to the mean than an item from data set B, whose standard deviation is higher.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Shape
If two sets of data are plotted in bar graphs, the data set with a low standard deviation would be
displayed as a tall narrow shape, while a large standard deviation would be indicated by a lower,
wider shape.
n
Below is a bar graph of data with a small
standard deviation
Below is a bar graph of data with a large
standard deviation
Standard deviation generally does not indicate right or wrong, or better or worse. A lower
standard deviation is not necessarily more desirable. It is used purely as a descriptive statistic.
It describes the distribution in relation to the mean.
Video:
Watch the video “Data4_V1” on the standard deviation.
How to calculate standard deviation
Video:
342
Watch the video “Data4_V2” on how to calculate the standard deviation.
5.5. MEASURES OF DISPERSION
Activity
After watching the video “Data4_V2”, calculate the standard deviation for the scores of
the second class:
n
Class 1
10
11
12
13
14
15
Class 2
16
17
18
19
10
11
12
13
14
15
16
17
18
19
n
Answer
xi
xi − x
(xi − x)2
xi
xi − x
(xi − x)2
13
13 − 15 = −2
(−2)2 = 4
10
10 − 15 = −5
(−5)2 = 25
14
14 − 15 = −1
(−1)2 = 1
12
12 − 15 = −3
(−3)2 = 9
14
14 − 15 = −1
(−1)2 = 1
14
14 − 15 = −1
(−1)2 = 1
15
15 − 15 = 0
02 = 0
14
14 − 15 = −1
(−1)2 = 1
15
15 − 15 = 0
02 = 0
15
15 − 15 = 0
02 = 0
15
15 − 15 = 0
02 = 0
15
15 − 15 = 0
02 = 0
15
15 − 15 = 0
02 = 0
16
16 − 15 = 1
12 = 1
16
16 − 15 = 1
12 = 1
17
17 − 15 = 2
22 = 4
16
16 − 15 = 1
12 = 1
18
18 − 15 = 3
32 = 9
17
17 − 15 = 2
22 = 4
19
19 − 15 = 4
42 = 16
10
10
xi
(xi − x)2
10
10
xi
i=1
i=1
i=1
i=1
= 150
= 12
= 150
= 66
10
xi
x
=
i=1
n
S
=
10
(xi − x)2
i=1
n−1
10
xi
x
=
i=1
n
S
=
(xi − x)2
10
(xi − x)2
i=1
n−1
=
150
10
=
12
9
=
150
10
=
66
9
=
15
=
1,15
=
15
=
2,71
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Variance
nnn
Compare the standard deviations for the scores of two classes, Class 1 and Class 2:
Class 1
Mean:
Standard deviation:
Class 2
x = 15,00
S = 1,15
Mean:
Standard deviation:
x = 15,00
S = 2,71
Although the mean scores of the two classes are the same, the scores of Class 2 are clearly more
spread out. The standard deviations give the teacher some idea about the spread of the marks
of these two classes. Although the two classes have the same average scores, the teacher cannot
conclude that the levels of understanding are the same for the two classes.
This is even more visible when we consider the variance of each data set. The variance is simply
the square of the standard deviation; and the standard deviation is the positive square root of
the variance.
Variance is S 2 .
√
Standard deviation is S 2 .
Variance for scores of Class 1:
Variance for scores of Class 2:
S 2 = 1,152 = 1,32
S 2 = 2,712 = 7,34
Activity
Calculate the standard deviation of the following sample:
Selling price of a specific share over 14 working days (in cents)
1 630
1 550
1 430
1 440
1 390
1 400
1 480
1 490
1 410
1 905
1 540
1 890
1 900
1 900
Answer
First calculate the mean, x. Then calculate the deviation from x for each observation and
square it. Divide the sum of the squares by n − 1 = 13. The standard deviation is the
square root of the variance.
14
The mean is
xi
x =
i=1
n
22 355
=
14
= 1 596,79
≈ 1 597.
344
5.5. MEASURES OF DISPERSION
The following table will help you with the calculations for the standard deviation:
xi
(xi − 1 596,79)
(xi − 1 596,79)2
1
2
1 630
1 550
33,21
−46,79
1 102,90
2 189,30
3
4
1 430
1 440
−166,79
−156,79
27 818,90
24 583,10
5
6
1 390
1 400
−206,79
−196,79
42 762,10
38 726,30
7
1 480
−116,79
13 639,90
8
9
1 490
1 410
−106,79
−186,79
11 404,10
34 890,50
10
11
1 905
1 540
308,21
−56,79
94 993,40
3 225,10
12
13
1 890
1 900
293,21
303,21
85 972,10
91 936,30
14
1 900
303,21
91 936,30
Total
565 180,30
The variance is
14
S2 =
(xi − x)2
i=1
13
565 180,30
=
13
= 43 475,41.
The standard deviation is
S =
43 475,41
= 208,51.
Standard deviation for data in frequency table
It is possible to calculate the mean, variance and standard deviation if we have not been provided
with the raw data, but with data grouped into intervals and with frequencies associated with
each interval.
The assumption we make is that each interval is represented by its midpoint x.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Example
The time between breakdowns for equipment in a factory was recorded over a period of several
months. During this period, 40 breakdowns were observed. The mean number of days between
breakdowns is 11 days. Complete the fi × x2i column in the following table:
Interval
Frequency
Midpoint
fi
xi
−0,5 – 4,5
6
2
6 × 22
=
24
4,5 – 9,5
10
7
10 × 72
=
490
9,5 – 14,5
14
12
14 × 122
=
2 016
14,5 – 19,5
6
17
6 × 172
=
1 734
19,5 – 24,5
4
22
4 × 222
=
1 936
=
6 200
fi × x2i
5
5
n=
fi = 40
i=1
i=1
The variance of the data is calculated by the formula
5
S 2 = i=1
fi x2i − n × x2
n−1
.
Substitute the values in the formula and calculate the variance:
5
S2 =
i=1
fi x2i − n × x2
n−1
6 200 − 40 × 112
40 − 1
= 34,87
=
The standard deviation is
S =
=
√
S2
34,87
= 5,91.
Formulas
Standard deviation
The standard deviation is the positive square root of the variance:
√
S = S2
346
fi x2i
5.5. MEASURES OF DISPERSION
Variance
The formulas for finding the variance for both types of data are as follows:
Raw data
n
S 2 = i=1
Data in frequency distribution
n
(xi − x)2
n−1
S 2 = i=1
fi x2i − nx2
n−1
k
where n is the
fi and k is the
where n =
i=1
number of observations
number of intervals.
in the data set.
Normal curve
The standard deviation is defined as the square root of the variance. But what does it tell us?
It tells us how far away the observations are from the mean. The larger the standard deviation,
the further away are the data points from the mean.
The following schematic presentation shows how many of the data points are between the standard
deviation to the left and to the right of the mean:
For data sets that have a normal distribution, the standard deviation (SD) can be used to determine the proportion of values that lie within a particular range of the mean value.
For such normal distributions the following is always true:
• About 68% of values are less than one standard deviation (1 SD) away from the mean value.
• About 95% of values are less than two standard deviations (2 SD) away from the mean.
• About 99,7% of values are less than three standard deviations (3 SD) away from the mean.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
5.5.4
n
Quartiles
The median (M e) is that value that separates a sorted data set into two equal parts:
If we divide a sorted data set into four equal sized parts we get quartiles.
A quartile divides a sorted data set into four equal parts so that
each part represents a quarter of the data set.
Q1
The first quartile represents the value indicating the end of the first 25% of
the data values.
25% of data ≤ Q1
348
Q2
The second quartile represents the value indicating the end of the second 25%
of the data values (or the value that divides the data set into two equal
parts). The second quartile has the same value as the median.
50% of data ≤ Q2
Q3
The third quartile represents the value indicating the end of the third 25% of
the data values.
75% of data ≤ Q3
5.5. MEASURES OF DISPERSION
Quartile deviation
Another measure of dispersion is the range of the middle half of the data.
The middle 50% of the data lies between Q1 and Q3 .
The measure, QD , is called the quartile deviation and is the measurement of the dispersion of the
data around the median.
Q3 − Q1
QD =
2
As with the median, the quartile deviation does not use all the observations. It ignores outliers
since the top 25% and the bottom 25% of the data values are not taken into account.
An outlier is a data point that is significantly different numerically from the other data points in
a sample. The term is used in statistical studies and can point to abnormalities in the data set
studied or to errors in the measurements taken.
• If a measurement falls to the right of Q3 of a set of data, then we know that it is in the top
25% of the data.
• If a measurement falls to the left of Q1 of a set of data, then we know that it is in the
bottom 25% of the data.
5.5.5
Coefficient of variation
The coefficient of variation is a statistical measure of the dispersion of data points in a data set
around the mean. It is calculated as follows:
Coefficient of variation
=
CV
=
standard deviation
mean
S
x
The coefficient of variation represents the ratio of the standard deviation to the mean. It is a
useful statistic for comparing the degree of variation from one data set to another, even if the
means are drastically different from each other.
The coefficient of variation provides a way to compare the dispersion of values of samples or populations around their means. The following graphics show two normally distributed populations,
349
TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
both with a mean of 1 000. The one on the left has a standard deviation of 20 and the one on the
right has a standard deviation of 100, giving a coefficient of variation of 0,02 and a coefficient of
variation of 0,10 respectively.
mean = 1 000
standard deviation = 20
coefficient of variation = 0,02
mean = 1 000
standard deviation = 100
coefficient of variation = 0,10
In cases where the means are the same, as in the cases illustrated above, the coefficients of
variation provide no more information than the standard deviations themselves.
The coefficient of variation is of more use when populations
have different means and standard deviations.
The following graph is loosely based on the yearly precipitation at two places in areas with
significantly different climates:
Semi-arid area
mean = 543 mm
standard deviation = 214 mm
coefficient of variation = 0,39
Humid, subtropical area
mean = 1 490 mm
standard deviation = 458 mm
coefficient of variation = 0,31
14
Frequency
12
10
9
6
4
2
200
400
600
800
1 000
1 200
1 400
1 600
1 800
2 000
2 200
2 400
Interval (mm)
While there are marked differences in the means and standard deviations of the precipitation at
the two places, the coefficients of variation are of similar magnitude.
350
5.5. MEASURES OF DISPERSION
Variability of data sets
n
It is often difficult to use the standard deviation formula to compare measurements from different
populations, therefore statisticians produced the coefficient of variation (CV ). The coefficient
of variation expresses the standard deviation as a percentage of what is being measured relative
to the sample or population mean.
The coefficient of variation is useful because the standard deviation of data must always be
understood in the context of the mean of the data. In contrast, the actual value of the CV
is independent of the unit in which the measurement has been taken, so it is a dimensionless
number.
The numerator and denominator of the CV have the same units, so the CV itself has no units
of measurement. If a data set consists of the monthly salaries of employees, the mean salary will
be in rand and the standard deviation will also be in rand, but the CV will not be.
What we mean when we say that the CV has no units of measurement, is that when the CV
is calculated, the units cancel out. This can be illustrated as follows, using rand (R) as unit of
measurement:
S [R]
S [in R]
=
CV =
x [in R]
x [R]
The CV is sometimes expressed as a percentage, in which case the CV is multiplied by 100:
S
× 100 = %
x
The advantage is that we can compare the variability of two different populations directly using
the coefficient of variation.
CV =
For comparison between data sets with different units or widely different means, we must use
the coefficient of variation instead of the standard deviation. We can also use the coefficient of
variation if we wish to compare the variability of two or more data sets.
The data set for which the coefficient of variation is large indicates
that the group is more variable, and it is less stable or less uniform.
Example
A stockbroker wants to compare two unit trusts. He has the annual return rates of the two unit
trusts for the past ten years available, and calculates the mean and variance of each:
Fund A
Fund B
Mean x
Variance Sx2
16
12
280,34
99,37
We see that the variance for Fund A is higher than the variance of Fund B, and we draw the
conclusion that the risk associated with Fund A is higher than the risk associated with Fund B.
However, Fund A has displayed a higher mean return over the past ten years than Fund B. This
feels intuitively right – an investment with a higher associated risk should have a higher mean
rate of return.
But what if Fund A had a mean rate of return of 21% with the same variance? Would we then
still be able to say that Fund A is subject to higher fluctuation than Fund B?
It is only when the two means are close together that we may compare the variances or standard
deviations.
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
5.5.6
Box-and-whiskers diagram
Some presentations incorporate all the original data observed, while others only use certain statistics.
The box-and-whiskers diagram is a very useful presentation that does not use all the data. It
shows only certain statistics rather than all the data.
The box-and-whiskers diagram uses only five observations from a data set, namely:
smallest value
first quartile
median
third quartile
largest value
M in
Q1
Me
Q3
M ax
In the diagram below the median, first quartile and third quartile form the box portion. They
show the range of the middle 50% of the data values:
This plot allows people to explore the data and to draw informal conclusions when two or more
variables are present. Compared to a normal distribution, the box aligns with the middle portion
of the data.
352
5.5. MEASURES OF DISPERSION
Example
Draw a box-and-whiskers diagram for the following data:
6, 3, 9, 8, 4, 10, 8, 4, 15, 8, 10
Order the data in ascending order. There are 11 values in the data set, so n = 11:
1st
2nd
3rd
4th
5th
6th
7th
8th
9th
10th
11th
3
4
4
6
8
8
8
9
10
10
15
Then, determine the Q1 , M e and Q3 :
Q1 is the value of the
For Q1
1
(11 + 1)
4
1
(n + 1)th observation:
4
12
=
4
= 3
Q1 is the third observation, so Q1 = 4.
For M e/Q2
M e/Q2 is the value of the
1
(11 + 1)
2
=
=
1
(n + 1)th observation:
2
12
2
6
M e is the sixth observation, so M e = 8.
For Q3
Q3 is the value of the
3
(11 + 1)
4
3
(n + 1)th observation:
4
36
=
4
= 9
Q3 is the ninth observation, so Q3 = 10.
The positions of these values can be illustrated in the ordered data set as follows:
1st
2nd
3rd
4th
5th
6th
7th
8th
9th
10th
11th
3
4
4
6
8
8
8
9
10
10
15
Q1
Me
Q3
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TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
The box-and-whiskers diagram is given below:
Interpretation
Various types of box-and-whisker diagrams
When graphed with real data, box-and-whiskers diagrams can take on many different forms. How
the diagram looks depends on the distribution of the underlying data. Exceptionally high and
low maximums and minimums can stretch out the whiskers. Diagram C in the figure above is an
example of a very low minimum and diagram D is an example of a very high maximum.
354
5.5. MEASURES OF DISPERSION
Depending on the data distribution, the box part can be stretched or compressed. This is called
the spread and indicates whether the middle 50% of the data are spread out over a large range of
values or compressed over a small range of values. The corresponding bell curve for a compressed
box diagram would have a sharper and higher peak, while a stretched out box diagram would
have a rounder and lower peak.
Compressed box diagram
gives a sharper, higher peak:
Stretched out box diagram
gives a rounder, lower peak:
The box can also lie at various positions between the whiskers. This is called the skewness. A
diagram such as diagram D in the figure on the previous page is a good example of a skewed data
set where the box is close to the minimum value.
Diagram of a skewed data set:
355
TOPIC 5: COLLECTION, PRESENTATION AND DESCRIPTION OF DATA
Exercise 5.3
1. Consider the selling price of a specific share over 14 working days (in cents):
1 630;
1 550;
1 430;
1 440;
1 390;
1 400;
1 480;
1 490;
1 410;
1 905;
1 540;
1 890;
1 900;
1 900
(a) Calculate the mean of the data.
(b) Calculate the variance of the data.
(c) Calculate the standard deviation of the data.
2. The following frequency table shows the time (in minutes) taken to travel to work for a
sample of 25 people living in Emalahleni. The mean time to travel to work is 31,7 minutes.
Interval
Frequency
fi
15,5 – 21,5
2
21,5 – 27,5
6
27,5 – 33,5
8
33,5 – 39,5
4
39,5 – 45,5
4
45,5 – 51,5
1
6
(a) Calculate n =
fi .
i=1
6
(b) Calculate
i=1
fi x2i .
(c) Calculate the variance.
(d) Calculate the standard deviation.
3. Rainfall measurements in regions A, B and C yielded the following information:
Region
A
B
C
Smallest value
Q1
0
50
200
300
0
200
Me
Q3
200
300
350
550
800
900
Largest value
500
1 000
950
Graphically represent the data and interpret the graphs.
356
Topic
6
An application of differentiation
After completion of this topic you should be able to apply some rules of differentiation to calculate
marginal costs and marginal profits, and to interpret these.
CONTENTS
Theme
6.1
Total, fixed and variable costs
6.2
Marginal cost
6.3
Derivatives and slopes
6.4
Marginal profit
TOPIC 6: AN APPLICATION OF DIFFERENTIATION
Theme 6.1
Total, fixed and variable costs
Learning objective
On completion of this theme you should be able to calculate fixed,
variable and total costs
Cost, revenue and profit are the three most important factors in determining the success of a
company. A company can have high revenue, but if its costs are higher, it will show no profit and
is destined to go out of business when available capital runs out. Managing costs and revenue to
maximise profits is the key for any company.
The total cost faced by companies can be divided into two main categories: fixed costs and
variable costs.
total cost = fixed cost + variable cost
Fixed costs
Fixed costs are costs that are independent of output. These remain constant throughout production. It is the upfront cost a business must pay to create an infrastructure.
Examples of fixed costs are rent, buildings and equipment.
The best way to identify fixed costs is to produce zero output.
Fixed cost is incurred whether or not any output is produced.
Variable costs
Variable costs may fluctuate during production. During production these costs may change.
Examples of variable costs are labour directly involved in production, raw materials, fuel expenses
and packaging.
Variable costs can be calculated using the following formula:
variable cost = number of units produced ×
variable cost to produce a single unit
Say, for example, a factory manufactures office desks. The manufacturing costs for a month are
given below:
The cost of wood and other materials amount to R105 per desk,
direct labour costs R175 per desk,
variable factory overheads amount to R70 per desk, and
fixed factory overheads amount to R100 000 for the month.
Calculate the total monthly cost of the factory.
358
6.1. TOTAL, FIXED AND VARIABLE COSTS
The variable cost per desk per month is calculated from three costs:
Wood and other materials
Labour
105
175
Variable factory overheads
70
350
Let q represent the number of desks (quantity) the factory manufactures per month.
The variable cost is therefore R350 per desk or
350 × q = 350q.
The fixed cost per month is R100 000.
The total cost (T C) is calculated as
T C = fixed cost + variable cost
= 100 000 + 350q.
The total monthly cost of the factory is a function of the number of desks manufactured (q) and
can also be written as
C(q) = 100 000 + 350q.
A function can be written in many different ways. Using x instead of q to represent the number
of desks, the cost function can be written as
C(x) = 100 000 + 350x.
Instead of using C, we could also use f and write the function as
f (x) = 100 000 + 350x
or use any other letter we want to.
359
TOPIC 6: AN APPLICATION OF DIFFERENTIATION
Theme 6.2
Marginal cost
Learning objectives On completion of this theme you should be able to
• calculate the marginal cost at a specific production level
• interpret the marginal cost at a specific production level
The concept of marginal cost is very important in the theory of economics. Marginal cost is the
change in total cost that comes from making or producing one additional item.
Marginal cost is defined as the cost that results
from a one-unit increase in the production rate.
If a company has already produced 13 cars, the marginal cost of the 14th car is only the additional
costs that the company incurs in making that car.
Expanding scale of production
A company can realise cost advantages by expanding their scale of production.
The cost of producing something can fall as the volume of output increases. Hence it might cost
R3 000 to produce 100 copies of a magazine, but only R4 000 to produce 1 000 copies. In this
case the cost has fallen from R30 to R4 a copy, because the main costs involved in producing a
magazine, that is editorial and design, are unrelated to the number of magazines produced.
The aim is to reduce the production cost per unit as production increases. In other words:
The cost of producing an additional unit, the marginal
cost, decreases as the volume of production increases.
Optimum production level
The purpose of analysing marginal cost is to determine at what point a company can operate at
an optimum production level.
Companies constantly examine the cost implication of adding one more unit to their production
schedules. This is because at some point, the benefit of producing one additional unit and
generating revenue from that item will bring the overall cost of producing the product line down.
The key to optimising manufacturing costs is to
find this point or level as quickly as possible.
360
6.2. MARGINAL COST
Three methods used to calculate marginal cost
nnn
When asked to compute marginal cost (M C), the data you need is likely to come in one of three
forms:
1.
In a table that gives you total cost and quantity produced.
2.
A linear equation relating total cost (T C) and quantity produced (q). This would
be an equation like
T C = 400q + 20
T C = 50 + 6q.
or
3.
A non-linear equation relating total cost (T C) to quantity produced (q). This
would be an equation like
T C = 34q 2 − 24q + 9.
Each of these situations has to be dealt with differently.
Using a table to calculate marginal cost
The total cost of producing one pen is R5 and the total cost of producing two pens is R9. The
marginal cost of expanding output by one unit is R4 only, and is calculated as
9 − 5 = 4.
The marginal cost of the second unit is the difference between the total cost of producing two
units and the total cost of producing one unit.
Marginal cost (M C), which is really an incremental cost, can be written as
MC =
change in total cost (T C)
.
change in number of units produced
The total cost of producing four pens is R16 and the total cost of producing five pens is R21.
The marginal cost of the fifth unit is R5. It is the difference between the total cost of producing
five units and the total cost of producing four units, and is calculated as
T C of producing 5 units − T C of producing 4 units
change in number of units produced
21 − 16
=
1
= 5.
MC =
The following table explains how the marginal cost is calculated:
Units of output
Total cost (R)
Marginal cost (R)
1
5,00
5,00
2
9,00
4,00
3
12,00
3,00
4
16,00
4,00
5
21,00
5,00
6
29,00
8,00
361
TOPIC 6: AN APPLICATION OF DIFFERENTIATION
Using a linear function to calculate marginal cost
You can calculate marginal cost when you have a linear relationship between total cost and
quantity.
Consider again the example of the factory that manufactures office desks. The total monthly cost
(T C) of the factory is calculated as
T C = fixed cost + variable cost
= 100 000 + 350q.
The total cost is a linear function of the number of desks (q) manufactured.
Marginal cost is the additional cost we incur when we manufacture one more desk. With the
equation T C = 100 000 + 350q, our total cost goes up by R350 whenever we manufacture an
additional desk, as shown by the coefficient in front of the q.
So we have a constant marginal cost of R350 per desk produced.
Let’s return to linear functions in general. The equation for a linear function, or straight line, is
y = ax + b
where a is the slope of the line and b is the intercept of the line on the y-axis.
In terms of two arbitrary points on the straight line, P1 with coordinates (x1 ; y1 ) and P2 with
coordinates (x2 ; y2 ), the slope is defined as
a =
=
change in y
change in x
y2 − y1
.
x2 − x1
y
y2
y2 − y1
y1
x2 − x1
x1
362
x2
x
6.2. MARGINAL COST
The office desk factory has a constant marginal cost of R350 per desk manufactured. Consider
the graph of this linear function:
TC
Constant marginal
T C = 100 000 + 350q
101 750
350
101 500
cost of R350 per
desk manufactured
101 400
1
350
101 050
101 000
1
350
100 700
1
100 500
350
100 350
1
350
100 000
1
.
.
.
q
1
2
3
4
5
The slope is the change in total cost (T C) that corresponds to a change of one unit in the value
of the quantity (q):
change in T C
change in q
350
=
1
= 350
slope =
Change consists of two components, namely size and direction.
The value of the slope gives the size of the change. The sign of the slope gives the direction
of the change.
In this case the slope is positive, indicating an increase. If the slope has a negative sign it indicates
a decrease.
Video:
Watch the video “Derivative1” on marginal cost and linear relationships.
363
TOPIC 6: AN APPLICATION OF DIFFERENTIATION
Activity
Sweet Tooth, which specialises in homemade chocolates, packs the chocolates in 100 g gift
packets. The fixed production cost is R400,00 per day, while the production cost per 100 g
packet amounts to a further R9,00 per packet. Sweet Tooth cannot produce more than 200
packets per day.
Let x represent the number of packets of chocolates produced per day.
Calculate the marginal cost if Sweet Tooth produces 150 packets per day.
Answer
Determine the total cost function, C(x), of Sweet Tooth.
The variable cost is calculated as
= variable cost to produce one packet × number of packets produced per day
= 9×x
= 9x.
The fixed cost per day is R400,00.
The total cost, as a function of x, is given by
= fixed cost + variable cost
= 400 + 9x.
(If the cost is written as C(x), then the marginal cost is written as C (x)).
Calculate C (x) to determine the marginal cost:
If f (x) = n then f (x) = 0
If f (x) = xn then f (x) = n × xn−1
If C(x) = 400 + 9x then C (x) = 0 + 9 × 1 × x1−1
= 9 × x0
= 9×1
= 9
C (x) = 9, therefore C (150) = 9.
In the activity above we determined the total cost and marginal cost function of Sweet Tooth,
which specialises in home-made chocolates, as
C(x) = 400 + 9x
C (x) = 9.
The cost function is a straight line. The derivative of the function is 9, which is a constant. For
a straight line, the slope remains constant. Therefore the marginal cost remains constant.
364
6.2. MARGINAL COST
Remember, Sweet Tooth cannot produce more than 200 packets of chocolates per day. If it
produces 150 packets per day, the marginal cost is R9,00. The marginal cost is always R9,00 a
day no matter how many packets of chocolates are produced (as long as no more than 200 packets
are produced).
Marginal cost implies that the change in the total cost, resulting from
producing one additional packet of chocolates, is R9,00.
Note that the fixed cost reduced to zero when the marginal cost was calculated.
The marginal cost is only influenced by the variable cost;
it is unaffected by the fixed cost.
This is because fixed costs, such as rent or other overheads, generally remain the same, while
variable costs correlate with the number of products manufactured. The number of products
manufactured in turn affects the marginal cost.
Using a non-linear function to calculate marginal cost
Video:
Watch the video “Derivative2” on marginal cost and non-linear relationships.
Activity
1. If C(x) is the total cost of producing x items, then C (x) represents the
a.
b.
average cost of producing x items.
approximate cost of producing one additional item after the xth item is produced.
c.
number of items to produce to minimise cost.
2. Total cost is the sum of
a. average cost and marginal cost.
b.
c.
variable cost and marginal cost.
fixed cost and variable cost.
3. Marginal cost is
a.
the change in the total cost resulting from a one-unit change in output.
b.
c.
all the costs that vary with output.
all the costs of producing of goods.
4. A company produces 149 items for R3 160 and 150 of the same items for R3 200. The
marginal cost of the 150th item is
a.
R21,21.
b.
c.
R21,33.
R40,00.
365
TOPIC 6: AN APPLICATION OF DIFFERENTIATION
5. Suppose the cost function of a company can be written as
C(x) = 250 + 10x.
The marginal cost is equal to
a.
b.
10x.
10.
c.
260.
6. The amount by which total cost rises when the firm produces one additional item, is called
a.
b.
variable cost.
average cost.
c.
marginal cost.
7. The marginal cost of the 8th item that is produced, is
a.
the difference between the total cost of producing 7 items and 8 items.
b.
c.
the difference between the fixed cost of producing 7 items and 8 items.
the difference between the total cost of producing 8 items and 9 items.
Answer
1. If C(x) is the total cost of producing x items, then C (x) represents the
approximate cost of producing one additional item after the xth item is produced.
2. Total cost is the sum of
fixed cost and variable cost.
3. Marginal cost is
the change in the total cost resulting from a one-unit change in output.
4. A company produces 149 items for R3 160 and 150 of the same items for R3 200. The
marginal cost of the 150th item is
R40,00.
5. Suppose the cost function of a company can be written as
C(x) = 250 + 10x.
The marginal cost is equal to
10.
6. The amount by which total cost rises when the firm produces one additional item, is called
the marginal cost.
7. The marginal cost of the 8th item that is produced, is
the difference between the total cost of producing 7 items and 8 items.
366
6.2. MARGINAL COST
Exercise 6.1
The total cost C(x), in thousands of rand, to manufacture x small sailing boats is given by
the function
x2
C(x) = 575 + 25x − .
4
1. Calculate the marginal cost if x boats are manufactured.
2. Calculate the marginal cost if 40 boats are manufactured and interpret the result.
3. Calculate C (30) and interpret it.
367
TOPIC 6: AN APPLICATION OF DIFFERENTIATION
Theme 6.3
Derivatives and slope
On completion of this theme you should be able to calculate the
derivative of simple functions.
Learning objective
Differentiation is the method used to determine the derivative of a function. The derivative of a
function is used to determine the slope of the graph of a function. There are quite a number of
differentiation rules, but in this module we consider only four of them.
Differentiation rule 1
The derivative of any constant term, that is a term that consists of a number only, is zero:
da
= 0,
dx
where a is a constant.
For example: If f (x) = 4 then f (x) = 0.
Differentiation rule 2
If f (x) = xn , then f (x) = nxn−1 .
If f (x) equals x to the power of n, then f (x) (pronounced f accent x) is equal to n times x to
the power of n minus one.
For example: If f (x) = x7 , then f (x) = 7x6 .
Differentiation rule 3
d
[af (x)] = af (x)
dx
The derivative of a constant times a function is equal to the constant times the derivative of the
function.
If f (x) = axn , then f (x) = anxn−1 .
For example: If f (x) = 7x5 , then f (x) = 7 × 5x4 = 35x4 .
Differentiation rule 4
If f (x) = g(x) + h(x), then f (x) = g (x) + h (x).
When a function consists of the sum of two other functions, then the derivative of this function
is the sum of the derivatives of the other two functions.
If f (x) = a + g(x), then f (x) = g (x).
When a function consists of the sum of a constant and another function, then the derivative of
this function is zero plus the derivative of the other function.
For example: If f (x) = x2 − 2x − 3, then f (x) = 2x − 2.
368
6.3. DERIVATIVES AND SLOPE
Activity
Use rule 2 to give the derivative of the function
f (x) = xn
for n = 0; 1; 2; 3 and 4.
Answer
x
f (x) = xn
f (x)
0
x0 = 1
0
1
x1 = x
1
2
x2
2x
3
x3
3x2
4
x4
4x3
Examples
Consider the following functions. For each function, the derivative is also given:
Function
Derivative
f (x)
f (x)
f (x) = x5
f (x) = x−4
2
f (x) = x 3
f (x) =
√
1
x = x2
dy
= f (x) = 5x4
dx
dy
= f (x) = −4x−5
dx
2 2 3
2 1
dy
= f (x) = x 3 − 3 = x− 3
dx
3
3
1 1 2
1 1
dy
= f (x) = x 2 − 2 = x− 2
dx
2
2
f (x) = 2x4
dy
= f (x) = 2 × 4x4−1 = 8x3
dx
f (x) = −6x2
dy
= f (x) = −6 × 2x2−1 = −12x
dx
2
3
3
f (x) = x
2
1
3 2 2 3
dy
= f (x) = × x 3 − 3 = x− 3
dx
2 3
369
TOPIC 6: AN APPLICATION OF DIFFERENTIATION
Function
Derivative
f (x)
f (x)
f (x) =
−x
1
=− x
2
2
f (x) = 12x2 − 24x + 8
1
1
dy
= f (x) = − × 1x1−1 = − × x0
dx
2
2
1
1
=− ×1=−
2
2
Determine the derivative of each term:
d 12x2 = 12 × 2x2−1 = 24x1 = 24x
dx
d
(−24x) = −24 × 1x1−1 = −24x0 = −24
dx
d
(8) = 0
dx
dy
= f (x) = 24x − 24.
Therefore
dx
f (x) =
−x4
+ 3x3 − 2x
2
Determine the derivative of each term:
1
d −x4
= − × 4x4−1 = −2x3
dx
2
2
d 3
3x = 3 × 3x3−1 = 9x2
dx
d
(−2x) = −2 × 1x1−1 = −2x0 = −2
dx
dy
= f (x) = −2x3 + 9x2 − 2.
Therefore
dx
√
4
f (x) = 12x + 6 x −
x
Determine the derivative of each term:
d
(12x) = 12 × 1x1−1 = 12x0 = 12
dx
⎛
⎞
1
−1
1 − 12
3
d ⎜
⎟
= 3x 2 = √
⎝6x 2 ⎠ = 6 × x
dx
2
x
4
d −4x−1 = −4 × −1x−1−1 = 4x−2 = 2
dx
x
4
dy
3
= f (x) = 12 + √ + 2 .
Therefore
dx
x x
370
6.3. DERIVATIVES AND SLOPE
Non-linear function
Consider, for example, a farmer who produces peanuts. His non-linear cost function in rand is
given by
x2
+ 100
C(x) =
100
where x represents the number of tons of peanuts produced. (1 ton = 1 000 kg.)
The graph of the total cost function is given as follows:
Cost
55 000
50 000
45 000
40 000
35 000
30 000
25 000
20 000
15 000
10 000
5 000
400
800
1 200
1 600
2 000
2 400
Number of kilograms of peanuts produced
We can clearly see that the cost function is continuously increasing until it reaches a point where
it will be extremely expensive for the farmer to produce another kilogram of peanuts. The slope
of the function is not constant and there is no constant rate of change. Thus the slope changes
continuously.
The farmer can see how the cost is increasing, but he cannot see how a change in production level
will influence the cost. In other words, if he produces at a certain production level, what will it
cost to produce one extra kilogram?
For instance, if the farmer produces 1 000 kilograms of peanuts and he wants to calculate how
the cost will change if one extra kilogram is produced, then he wants to determine the marginal
cost at a production level of 1 000 kilograms.
371
TOPIC 6: AN APPLICATION OF DIFFERENTIATION
He uses the following procedure:
The derivative of the 1st term of the cost function is
x2
100
d
dx
=
=
2x2−1
100
x
50
The derivative of the 2nd term of the cost function is
d
(100)
dx
=
0
x2
+ 100, is
100
C (x)
=
x
50
Substitute x = 1 000 into C (x) to calculate the
C (1 000)
=
The derivative of the cost function, C(x) =
=
marginal cost at a production level of 1 000 kilograms:
1 000
50
20
This means that for an extra kilogram produced at a production level of 1 000 kilograms, the cost
will increase by R20,00.
Activity
For the peanut farmer, calculate the cost and the marginal cost for the number of kilograms
of peanuts in the table:
Kilograms
C(x) =
(x)
x2
+ 100
100
x
50
C (x) =
400
1 200
1 800
Interpret the results.
Answer
372
x2
+ 100
100
C (x) =
4002
+ 100
100
= 1 700
C (400) =
Kilograms
(x)
C(x) =
400
C(400) =
x
50
400
50
=8
1 200
14 500
24
1 800
32 500
36
6.3. DERIVATIVES AND SLOPE
• If the farmer produces 400 kilograms, the marginal cost function approximates that it
will cost him an extra R8 to produce the next kilogram, because C (400) = 8. The
exact change in cost can be determined by calculating C(401) − C(400):
C(401) − C(400) =
=
=
4012
+ 100 −
100
4002
+ 100
100
4002
4012
+ 100 −
− 100
100
100
4012 4002
−
100
100
= 1 708,01 − 1 700,00
= 8,01
The approximated cost of R8,00 is very close to the exact cost of R8,01.
• At a production level of 1 200 kilograms, the marginal cost function approximates that
it will cost him an extra R24 to produce the next kilogram, because C (1 200) = 24.
The exact change in cost can be determined by calculating C(1 201) − C(1 200), which
is equal to R24,01.
• At a production level of 1 800 kilograms, the marginal cost function approximates that
it will cost him an extra R36 to produce the next kilogram, because C (1 800) = 36.
The exact change in cost can be determined by calculating C(1 801) − C(1 800), which
is equal to R36,01.
In the activity you have calculated that at a production level of 400 kilograms, the marginal cost
function approximates that it will cost the farmer an extra R8,00 to produce the next kilogram of
peanuts. The graphs of the cost function as well as the marginal cost function are drawn below:
Cost
x2
+ 100
C(x) =
100
Marginal cost
x
C (x) =
50
Cost
55 000
50 000
45 000
40 000
Rand
35 000
30 000
60
25 000
50
20 000
40
15 000
10 000
1 708,01
1 700,00
30
The difference
between C(401) and
C(400) is R8,01.
20
5 000
8
400
800
1 200
1 600
2 000
2 400
Number of kilograms of peanuts produced
10
400
800
1 200
1 600
2 000
2 400
Number of kilograms of peanuts produced
373
TOPIC 6: AN APPLICATION OF DIFFERENTIATION
Exercise 6.2
Calculate the derivatives of the following functions:
1. f (x) = 6x
2. f (x) = 3 + 5x
3. f (x) = 3 + 2x2
374
6.4. MARGINAL PROFIT
Theme 6.4
Marginal profit
Learning objectives On completion of this theme you should be able to
• calculate the marginal profit at a specific production level
• interpret the marginal profit at a specific production level
Profit is a very important concept for any business. Profit earned by a business can be used
to measure the success of the business. Profit is also an important signal to other providers of
finance to a business. Banks, suppliers and other lenders are more likely to provide finance to a
business that can demonstrate that it makes a profit and that it can pay debts as they fall due.
Profits earned that are kept in the business are an important source of finance for the business.
The moment a product is sold for more than it cost to produce, a profit is earned that can be
reinvested.
Profit can be measured and calculated using this formula:
profit = total revenue − total costs
Profit maximisation is the process by which a firm determines the price and output level that
returns the greatest profit.
Consider again the peanut farmer’s cost function, which is given by
C(x) =
x2
+ 100
100
where x represents the number of kilograms of peanuts produced. The cost is in rand.
The revenue function is given by
R(x) = 31x,
that is, each kilogram of peanuts produced is sold at R31,00.
The farmer’s profit is the difference between income earned by selling (revenue) and the cost
incurred for the production of the peanuts.
The profit is now given by
P (x) = R(x) − C(x)
= 31x −
= 31x −
= −
x2
+ 100
100
x2
− 100
100
x2
+ 31x − 100.
100
375
TOPIC 6: AN APPLICATION OF DIFFERENTIATION
A graph of the profit function is given below:
Profit
25 000
20 000
15 000
10 000
5 000
400
800
1 200
1 600
2 000
2 400
Number of kilograms of peanuts produced
We can clearly see that the profit increases to a point where the profit is at a maximum, after
which the profit decreases. The slope of the function is not constant and there is no constant
rate of change.
The farmer can see where the profit will increase and where it will decrease, but he cannot see
how the change in production level will influence the profit. He wants to determine the marginal
profit at a specific production level.
The marginal profit is the change in profit due
to a change in the number of units produced.
This means that if the farmer produces 1 000 kilograms and he is interested in knowing how the
profit will change if one extra kilogram is produced, he has to determine the marginal profit at a
production level of 1 000 kilograms.
In order to know the change in profit at different production levels, the slope of the function at
the different production levels must be determined. The method of differentiation will once
again be used.
The point of calculating marginal profit is that the costs and revenue of each unit produced, are
not always consistent. This can have a significant effect on marginal costs and revenue. The
marginal calculations are thus based on the specific level of production that a company is on at
the time of calculation.
The profit function is
P (x) = −
376
x2
+ 31x − 100.
100
6.4. MARGINAL PROFIT
The farmer uses the following procedure:
The derivative of the 1st term of the profit function is
d
dx
x2
−
100
=
=
2x2−1
100
x
−
50
−
The derivative of the 2nd term of the profit function is
d
(31x)
dx
=
31
The derivative of the 3rd term of the profit function is
d
(−100)
dx
=
0
P (x)
=
−
P (1 000)
=
−
The derivative of the profit function,
P (x) = −
x2
100
+ 31x − 100, is
Substitute x = 1 000 into P (x) to calculate the
=
marginal profit at a production level of 1 000 kg
x
+ 31
50
1 000
+ 31
50
11
This means that the profit will increase by R11,00 for an extra kilogram produced at a production
level of 1 000 kilograms. It is illustrated in the following graphs of the profit and marginal profit
functions:
Profit
P (x) = −
Marginal profit
x2
+ 31x − 100
100
P (x) = −
x
+ 31
50
Rand
Profit
60
50
25 000
20 910,99
20 900,00 20 000
40
15 000
30
The difference
between P (1 001) and
P (1 000) is R10,99.
10 000
20
11 10
5 000
400
800
1 200
1 600
2 000
Number of kilograms of peanuts produced
Video:
400
2 400
800
1 200
1 600
2 000
Number of kilograms of peanuts produced
Watch the video “Derivative3” on marginal functions.
377
TOPIC 6: AN APPLICATION OF DIFFERENTIATION
Marginal profit is the profit that would be made by producing and selling one additional unit of
a product or service. Therefore, the
marginal profit = marginal revenue − marginal cost
of the extra unit. All the functions discussed so far in the example of the peanut farmer are listed
below:
Cost
Marginal cost
C(x) =
x2
100
C (x) =
+ 100
x
50
Revenue
Marginal revenue
R(x) = 31x
R (x) = 31
Profit
Marginal profit
P (x) = −
x2
+ 31x − 100
100
P (x) = −
x
+ 31
50
Observe the linear graphs of the marginal revenue function, the marginal cost function and the
marginal profit function below:
Rand
50
40
Marginal Revenue
30
Marginal Profit
20
10
Marginal Cost
400
800
1 200
1 600
2 000
2 400
Number of kilograms of peanuts produced
If marginal revenue > marginal cost, marginal profit is positive;
if marginal revenue < marginal cost, marginal profit is negative; and
if marginal revenue = marginal cost, marginal profit is zero.
378
6.4. MARGINAL PROFIT
Since total profit increases when marginal profit is positive and total profit decreases when
marginal profit is negative, total profit must reach a maximum where marginal profit is zero
or where marginal cost equals marginal revenue.
This is because the producer has collected positive profit up until the intersection of marginal
revenue and marginal cost (where zero profit is collected and any further production will result
in negative marginal profit as marginal cost will be larger than marginal revenue).
A company should continue increasing production
and sales until the marginal profit falls to zero.
To find the maximum profit, let P (x) = 0 and solve for x;
at the x-value where P (x) = 0, profit is maximised.
Activity
1. When the marginal revenue equals marginal cost, the firm is
a.
establishing its shutdown point.
b.
c.
maximising its profit.
maximising its revenue.
d.
setting price.
2. A firm should expand output as long as its
a.
b.
average total revenue exceeds its average variable cost.
average total revenue exceeds its average total cost.
c.
d.
marginal cost exceeds its marginal revenue.
marginal revenue exceeds its marginal cost.
3. A company manufactures and markets a new LED light. The number of lights manufactured and sold is represented by x. The financial department of the company provides the
following cost function:
C(x) = 4 000 + 4x
where R4 000 is the estimate of fixed costs and R4 is the estimate of variable costs per light.
The revenue is the amount of money received by the company for manufacturing and selling
x lights:
R(x) = 11x − 0,001x2
Determine the number of lights to manufacture and sell to maximise profit by setting
P (x) = 0 and solving for x.
Answer
1. When the marginal revenue equals marginal cost, the firm is
maximising its profit.
379
TOPIC 6: AN APPLICATION OF DIFFERENTIATION
2. A firm should expand output as long as its
marginal revenue exceeds its marginal cost.
3. The profit function is determined as
P (x) = R(x) − C(x)
= 11x − 0,001x2 − (4 000 + 4x)
= 11x − 0,001x2 − 4 000 − 4x
= −0,001x2 + 7x − 4 000.
The marginal profit function is
P (x) = −0,001 × 2x + 7
= −0,002x + 7.
Set the marginal profit function equal to zero and solve for x:
P (x) = 0
−0,002x + 7 = 0
−0,002x = −7
−7
−0,002x
=
−0,002
−0,002
x = 3 500
The company should manufacture and sell 3 500 lights.
8
Thousands of rands
7
6
5
4
3
2
1
0
1
2
3
4
5
6
Thousands of lights
380
7
6.4. MARGINAL PROFIT
Exercise 6.3
1. The profit (in rand) yielded by the sales of x knives is given by the function
P (x) = 5x −
x2
− 450.
200
Calculate the marginal profit at x = 450 and x = 750, and interpret the results.
2. A manufacturer of scissors estimates his revenue function (in rand) to be
R(x) = 10x −
x2
,
1 000
where x is the number of scissors produced. The cost function (in rand) is
C(x) = 7 000 + 2x.
Calculate and interpret the marginal profit when 2 000, 4 000 and 5 000 scissors are produced.
381
Topic
Appendix
A
Answers to exercises
Topic 1. Revision of prior knowledge
Exercise 1.1:1(a)
Substituting x with 2 gives
12 × 2 + 17 = 24 + 17
= 41.
Exercise 1.1:1(b)
Substituting x with 4 gives
42 − 3 = 16 − 3
= 13.
Exercise 1.1:1(c)
Substituting x with 3 gives
2 × 32 = 2 × 9
= 18.
Exercise 1.1:1(d)
Substituting x with 5 gives
4 × 5 − 1 = 20 − 1
= 19.
383
Exercise 1.1:1(e)
Substituting x with 5 gives
12
5+7
+3 =
+3
4
4
= 3+3
= 6.
Exercise 1.1:1(f)
Substituting x with 6 gives
(6 + 3) (6 − 2) = 9 × 4
= 36.
Exercise 1.1:1(g)
Substituting x with 1 gives
10 − 3 × 1 = 10 − 3
= 7.
Exercise 1.1:1(h)
Substituting x with 12 gives
12 12
+
2
3
= 6+4
= 10.
Exercise 1.1:2(a)
Substituting x with 3 gives
5 × 3 + 7 = 15 + 7
= 22.
Exercise 1.1:2(b)
Substituting x with 3 gives
3+3×3−1 = 3+9−1
= 11.
Exercise 1.1:2(c)
Substituting x with 3 gives
5 × 32 − 9 = 5 × 9 − 9
= 45 − 9
= 36.
Exercise 1.1:2(d)
Substituting x with 3 gives
3+4
7
384
7
7
= 1.
=
TOPIC 1
Exercise 1.1:2(e)
Substituting x with 3 gives
2 (3 + 4) = 2 × 7
= 14.
Exercise 1.1:2(f)
Substituting x with 3 gives
7 − 3 + 2 = 6.
Exercise 1.1:3(a)
Substituting a with 2, b with 1 and c with 7 gives
2(2 + 1 − 7) + 7(1 − 2) = 2(−4) + 7(−1)
= 2 × −4 + 7 × −1
= −8 − 7
= −15.
Exercise 1.1:3(b)
Substituting a with 2, b with 1 and c with 7 gives
22 + 12 + 72 = 4 + 1 + 49
= 54.
Exercise 1.1:3(c)
Substituting a with 2, b with 1 and c with 7 gives
(2 + 1)(1 − 7) = 3 × −6
= −18.
Exercise 1.1:3(d)
Substituting a with 2, b with 1 and c with 7 gives
10
7+3
+ 12 = 2 −
+1
2×1−
2
2
= 2−5+1
= −2.
Exercise 1.1:4(a)
The mathematical expression is
x + y.
Exercise 1.1:4(b)
The mathematical expression is
8 − (a + b).
385
Exercise 1.1:4(c)
The mathematical expression is
3 × x + 2 × y = 3x + 2y.
Exercise 1.1:4(d)
The mathematical expression is
y + 7.
Exercise 1.2:1
Applying the distributive law of multiplication over addition gives
7 × (6 × (5 + 4) + 3 × (2 + 1))
= 7 × (6 × 5 + 6 × 4 + 3 × 2 + 3 × 1)
= 7 × 6 × 5 + 7 × 6 × 4 + 7 × 3 × 2 + 7 × 3 × 1.
Exercise 1.2:2(a)
The associative law of addition is associated with
2 + (5 + 4) = (2 + 5) + 4.
Exercise 1.2:2(b)
The commutative law of addition is associated with
(3 + 7) + 4 = (7 + 3) + 4.
Exercise 1.2:2(c)
The commutative law of multiplication is associated with
(7 × 5) × 2 = 2 × (7 × 5).
Exercise 1.2:2(d)
The associative law of multiplication is associated with
2 × (7 × 4) = (4 × 2) × 7.
Exercise 1.3:1
Simplifying gives
Exercise 1.3:2
Simplifying gives
386
8×
4
= 32.
1
12 ×
5
= 60.
1
TOPIC 1
Exercise 1.3:3
Simplifying gives
25
8
×
13 12
200
156
50
=
39
11
= 1 .
39
=
Exercise 1.3:4
Simplifying gives
7
2
7
=
=
7 1
×
2 7
1
.
2
Exercise 1.3:5
Simplifying gives
33
4
1
2
33 2
×
4
1
66
=
4
33
=
2
1
= 16 .
2
=
Exercise 1.3:6
Simplifying gives
17 1 3
+ −
20 4 5
=
=
=
=
Exercise 1.3:7
Simplifying gives
6
2
+5−
3
7
17 + 1 × 5 − 3 × 4
20
17 + 5 − 12
20
10
20
1
.
2
2 × 7 + 5 × 21 − 6 × 3
21
14 + 105 − 18
=
21
101
=
21
17
= 4 .
21
=
387
Exercise 1.3:8
Simplifying gives
2
7
1
5 +3 −6
2
5
12
11 17 79
=
+
−
2
5
12
=
11 × 30 + 17 × 12 − 79 × 5
60
First multiply the two larger denominators
by each other and check to see if the value
is divisible by the other denominators.
330 + 204 − 395
60
139
=
60
19
= 2 .
60
=
Exercise 1.3:9
Simplifying gives
=
=
=
=
=
=
=
=
=
=
Exercise 1.4:1(a)
Simplifying gives
23 = 8.
Exercise 1.4:1(b)
Simplifying gives
42 = 16.
388
5 1
3
3
÷ 1 −
+
4
6 2
5
3
11 1
3
÷
−
+
4
6
2
5
3
11 − 3
3
÷
+
4
6
5
3 8 3
÷ +
4 6 5
3 6 3
× +
4 8 5
18 3
+
32 5
3
9
+
16 5
9 × 5 + 3 × 16
80
45 + 48
80
93
80
13
1 .
80
TOPIC 1
Exercise 1.4:1(c)
Simplifying gives
1
3−1 = .
3
Exercise 1.4:1(d)
Simplifying gives
1
53
1
.
125
5−3 =
=
Exercise 1.4:1(e)
Simplifying gives
32 × 30 = 32 × 1
= 9.
Exercise 1.4:1(f)
Simplifying gives
23 × 22 = 23+2
= 25
= 32.
Exercise 1.4:1(g)
Simplifying gives
32 × 42 = 9 × 16
= 144.
Exercise 1.4:1(h)
Simplifying gives
5−1 × 52 = 5−1+2
= 5.
Exercise 1.5:1(a)
Calculating gives
√
196 +
√
144 =
√
14 × 14 +
√
12 × 12
= 14 + 12
= 26.
Exercise 1.5:1(b)
Calculating gives
√ 2
5
=
=
√
√
5×
√
5
5×5
= 5.
389
Exercise 1.5:1(c)
Calculating gives
√
64
2
=
√
2
8×8
= (8)2
= 8×8
= 64.
Exercise 1.5:1(d)
Calculating gives
√ 2
1
= 12
= 1.
Exercise 1.5:2(a)
Simplifying gives
√
√
100 ÷ 4 =
√
=
25
5×5
= 5.
Exercise 1.5:2(b)
Simplifying gives
36
4
=
6×6
2×2
6
2
= 3
=
or
36
4
=
=
√
√
9
3×3
= 3.
Exercise 1.5:2(c)
Simplifying gives
2×
√
144 = 2 ×
√
12 × 12
= 2 × 12
= 24.
390
TOPIC 1
Exercise 1.5:2(d)
Simplifying gives
25 −
√
√
4 = 25 − 2 × 2
= 25 − 2
= 23.
Exercise 1.5:3(a)
Calculating gives
√
√
18 =
2×9
√
2×3×3
=
√
= 3 2.
Exercise 1.5:3(b)
Calculating gives
√
45
−2
3
√
3 × 3 × 5 × 32
√
= 3 5 × 32
=
1
= 3 × 5 2 × 32
1
= 31+2 × 5 2
1
= 33 × 5 2
√
= 27 5.
Exercise 1.5:3(c)
Calculating gives
Exercise 1.6:1(a)
The correct symbol is
Exercise 1.6:1(b)
The correct symbol is
Exercise 1.6:1(c)
The correct symbol is
Exercise 1.6:1(d)
The correct symbol is
9 − x2
9 − x2
=
x2
x
1
2
.
−5 < −2.
9 > −2.
−100 < 7.
−6 > −12.
391
Exercise 1.6:1(e)
The correct symbol is
2 > 0.
Exercise 1.6:1(f)
The correct symbol is
+3 = 3.
Exercise 1.6:2(a)
The correct answer is
0 ≤ x < 5.
Exercise 1.6:2(b)
The correct answer is
−3 ≤ x < 3.
Exercise 1.6:2(c)
The correct answer is
x ≤ 5 and x > −6.
Exercise 1.6:2(d)
The correct answer is
x < 6 and x ≥ 0.
Exercise 1.6:3(a)
The correct graphical representation is as follows:
−6
−5
−4
−3
−2
−1
x
0
1
2
3
4
5
6
1
2
3
4
5
6
2
3
4
5
6
7
Exercise 1.6:3(b)
The correct graphical representation is as follows:
−6
−5
−4
−3
−2
−1
x
0
Exercise 1.6:3(c)
The correct graphical representation is as follows:
−5
392
−4
−3
−2
−1
x
0
1
TOPIC 1
Exercise 1.6:3(d)
The correct graphical representation is as follows:
−5
−4
−3
−2
x
−1
0
1
2
3
4
5
6
7
Exercise 1.6:4(a)
The solution is
7!
5!
=
7×6×5×4×3×2×1
5×4×3×2×1
= 7×6
= 42
or
7!
5!
=
7 × 6 × 5!
5!
= 7×6
= 42.
Exercise 1.6:4(b)
The solution is
(14 − 11)! + 2! × 4! = 3! + 2! × 4!
= 3×2×1+2×1×4×3×2×1
= 6 + 2 × 24
= 6 + 48
= 54.
Exercise 1.6:5(a)
If x1 = 3, x2 = 5, x3 = 4 and x4 = 2, then the solution is
4
xi = x1 + x2 + x3 + x4
i=1
= 3+5+4+2
= 14.
Exercise 1.6:5(b)
If x1 = 3, x2 = 5, x3 = 4 and x4 = 2, then the solution is
3
xi = x2 + x3
i=2
= 5+4
= 9.
393
Exercise 1.6:5(c)
If x1 = 3, x2 = 5, x3 = 4 and x4 = 2, then the solution is
4
i=1
x2i = x21 + x22 + x23 + x24
= 32 + 52 + 42 + 22
= 9 + 25 + 16 + 4
= 54.
Exercise 1.6:6(a)
There can be
26 × 26 × 26 × 26 = 456 976
words.
Remember: A letter may appear more than once.
Exercise 1.6:6(b)
The number of different possible meals is
4 × 10 × 6 = 240.
Exercise 1.6:6(c)
Because the order of placement is not important, the answer is a combination:
12 C3
=
12!
9! 3!
=
12 × 11 × 10 × 9!
9! × 3!
=
12 × 11 × 10
3!
= 220
Exercise 1.6:6(d)
In this case the order of placement is important. For example, ABCD, BCAD and DBCA all
form different words. If the question had stated groups instead of words, it would have been a
combination. Then ABCD, BCAD and DBCA would have been the same, because the order of
placement would not have been important. The answer is
26 P4
=
26!
(26 − 4)!
=
26!
22!
=
26 × 25 × 24 × 23 × 22!
22!
= 26 × 25 × 24 × 23
= 358 800.
394
TOPIC 1
Exercise 1.7:1(a)
The area of the rectangle is calculated as
area = length × width
= 25 × 24
mm × mm
= 600.
mm2
The area is 600 mm2 .
Exercise 1.7:1(b)
The area of the rectangle is calculated as
area = length × width
= 1 200 × 375
m×m
= 450 000.
m2
area = 1,2 × 0,375
km × km
The area is 450 000 m2 .
Or
= 0,45.
km2
The area is 0,45 km2 .
Exercise 1.7:1(c)
The area of the rectangle is calculated as
area = length × width
= 4,4 × 0,45
m×m
= 1,98.
m2
area = 4 400 × 450
mm × mm
The area is 1,98 m2 .
Or
= 1 980 000.
mm2
The area is 1 980 000 mm2 .
Exercise 1.7:1(d)
The area of the rectangle is calculated as
area = length × width
= 225 × 122
mm × mm
= 27 450.
mm2
The area is 27 450 mm2 .
395
Exercise 1.7:2(a)
The area of 1 km2 converted to m2 is
area = length × width
=1×1
km × km
= 1 000 × 1 000
Convert km to m to get m × m.
= 1 000 000.
m2
Therefore, 1 km2 equals 1 000 000 m2 .
Exercise 1.7:2(b)
The area of 1 m2 converted to mm2 is
area = length × width
=1×1
m×m
= 1 000 × 1 000
Convert m to mm to get mm × mm.
= 1 000 000.
mm2
Therefore, 1 m2 equals 1 000 000 mm2 .
Exercise 1.7:2(c)
The area of 1 cm2 converted to mm2 is
area = length × width
=1×1
cm × cm
= 10 × 10
Convert cm to mm to get mm × mm.
= 100.
mm2
Therefore, 1 cm2 equals 100 mm2 .
Exercise 1.7:2(d)
The area of 1 ha converted to m2 is
area = length × width
= 100 × 100
m×m
= 10 000.
m2
Therefore, 1 ha equals 10 000 m2 .
396
TOPIC 1
Exercise 1.7:2(e)
The area of 1 m2 converted to cm2 is
area = length × width
=1×1
m×m
= 100 × 100
Convert m to cm to get cm × cm.
= 10 000.
cm2
Therefore, 1 m2 equals 10 000 cm2 .
Exercise 1.7:2(f)
The area of 1 cm2 converted to m2 is
area = length × width
=1×1
cm × cm
= 0,01 × 0,01
Convert cm to m to get m × m.
= 0,0001.
m2
Therefore, 1 cm2 equals 0,0001 m2 . Consider the ratio
cm2
:
m2
1
:
0,0001
1 × 24,6 = 0,0001 × 24,6
24,6 = 0,00246.
Therefore, 24,6 cm2 equals 0,00246 m2 .
Exercise 1.7:2(g)
The area of 1 mm2 converted to m2 is
area = length × width
=1×1
mm × mm
= 0,001 × 0,001
Convert mm to m to get m × m.
= 0,000001.
m2
Therefore, 1 mm2 equals 0,000001 m2 . Consider the ratio
mm2
:
m2
1
:
0,000001
1 × 24 869,3 = 0,000001 × 24 869,3
24 869,3 = 0,0248693.
Therefore, 24 869,3 mm2 equals 0,0248693 m2 or 2,48693 × 10−2 m2 .
397
Exercise 1.7:3
A sketch of the rectangle (where the width is x m) is given below:
9m
xm
45 m2
Use the area formula to determine the value of x:
length × width = area
m × m = m2
9 × x = 45
9x = 45
45
9x
=
9
9
m2
m
x=5
m
The width is 5 m.
Exercise 1.7:4
The volume of the container is calculated as
volume = length × width × height
=1×1×1
m×m×m
= 1.
m3
The volume is 1 m3 .
To convert from m3 to litres we must multiply by 1 000 or 103 :
m3 × 1 000 = l
1 × 1 000 = 1 000
Thus 1 000 l can be poured into the container.
This can also be illustrated as follows:
1m
1 000 l
1m
1m
398
TOPIC 1
The volume of the cube (converted from m3 to cm3 ) is calculated as
volume = length × width × height
=1×1×1
m×m×m
= 100 × 100 × 100
Convert m to cm to get cm × cm × cm.
= 1 000 000.
cm3
The volume of the cube is 1 000 000 cm 3 or 1 000 000 ml. (Remember 1 cm3 = 1 ml.)
There are 1 000 ml in one litre.
Thus, the volume of the cube (in litres) is calculated as
1 000 000
= 1 000.
1 000
ml
=l
1 000
Therefore, the volume of the cube is 1 000 l.
Exercise 1.7:5
Determine the volume of the fuel tank:
volume = length × width × height
= 60 × 50 × 20
cm × cm × cm
= 60 000
cm3
The volume of the fuel tank is 60 000 cm3 .
Remember: 1 l = 103 cm3 or 1 000 cm3 . Therefore, the volume of the fuel tank is 60 l.
The fuel costs R9,16 per litre. The cost to fill the tank is calculated as
9,16 × 60 = 549,60.
rand/litre × litres = rand
It will cost R549,60 to fill the tank.
399
Topic 2. Functions and representations of functions
Exercise 2.1:1(a)
The equation of the straight line is y = ax + b.
Determine a:
y2 − y1
a=
x2 − x1
Two points (1; 2) and (3; 3) are given. Select any one of the two points to be (x1 ; y1 ) and the
other one to be (x2 ; y2 ). Let (1; 2) = (x1 ; y1 ) and (3; 3) = (x2 ; y2 ).
Then
3−2
3−1
1
=
2
= 0,5.
a =
Thus
y = 0,5x + b.
Determine b.
Take any one of the two points and substitute the values for x and y into the equation
y = 0,5x + b. Say we choose the point (1; 2):
y = 0,5x + b
2 = 0,5 × 1 + b
2 = 0,5 + b
2 − 0,5 = b
b = 1,5
Thus the equation of the line that cuts through the points (1; 2) and (3; 3) is
y = 0,5x + 1,5.
Exercise 2.1:1(b)
Determine the intercepts on the x- and y-axes for y = 0,5x + 1,5.
The intercept on the x-axis is where the line cuts through the x-axis, therefore the x-value where
the y-value is 0, is
0 = 0,5x + 1,5
0 − 1,5 = 0,5x
0,5x = −1,5
−1,5
x =
0,5
= −3.
The point is (−3; 0).
400
TOPIC 2
The intercept on the y-axis is where the line cuts through the y-axis; where x = 0 is
y = 0,5 × 0 + 1,5
y = 1,5.
The point is (0; 1,5).
Exercise 2.1:1(c)
Two lines are parallel if the slopes of the two lines are the same.
The line in (a) is
y = 0,5x + 1,5
with slope 0,5.
The line in (c) is
y =2+x
with slope 1.
The two lines are not parallel.
Exercise 2.1:1(d)
For the line in (a): Plot the two points (1; 2) and (3; 3) and draw a line through them.
For the line in (c): Calculate two points on the line y = 2 + x.
If x = 0, then
y = 2+0
= 2.
The point is (0; 2).
If y = 0, then
0 = 2+x
x = −2.
The point is (−2; 0).
Plot the two points and draw a line through the points. The graph is given below:
y
(0; 2)
(c)
(a)
3
(3; 3)
2
(1; 2)
1
−3
−2 −1
x
1
2
3
(−2; 0)
401
Exercise 2.1:2
For the line y = 5 + 2x the intercepts are as follows:
On the y-axis (where x = 0):
y=5
On the x-axis (where y = 0):
x=
The points are (0; 5) and
−5
2
1
−2 ; 0 .
2
For the line y = 2 + x the intercepts are (0; 2) and (−2; 0), as seen in 2.1:1(d).
The lines are not parallel since the slopes are different, namely 2 and 1.
At x = 3,5 the y-values for the two lines are respectively
y = 5 + 2 × 3,5
= 12
and
y = 2 + 3,5
= 5,5.
The vertical distance between the two lines at x = 3,5 is
12 − 5,5 = 6,5.
This is illustrated on the graph below:
y
12
y = 5 + 2x
10
6,5
8
6
4
2
−4
−3
−2
−1
−2
−4
402
y =2+x
x
1
2
3
4
TOPIC 2
Exercise 2.1:3
The line x = 2 is a straight line through 2 on the x-axis parallel to the y-axis.
For the line y = 4x,
if x = 0, then
y = 4×0
= 0
and if x = 1, then
y = 4×1
= 4.
Two points on the line are (0; 0) and (1; 4).
For the line y = −2x − 3,
if x = 0, then
y = −2 × 0 − 3
= −3
and if y = 0, then
0 = −2x − 3
Two points on the line are
2x = −3
3
x = − .
2
3
− ;0
2
and (0; −3).
The graph is given below:
y
(c)
4
(b)
(a)
3
2
1
−4
−3
−2
−1
−1
x
1
2
3
4
−2
−3
−4
403
Exercise 2.1:4
Let p represent the price and x the number of customers.
The point (x; p), that is (60; 6 000), is given.
The general expression is
y = ax + b
or, written in terms of our variables,
p = ax + b.
The slope a gives you the rate of change in the p-value for a unit change in x-value. It is given
that if the price increases by R500, the number of customers decreases by 3. Thus
a=
The general expression reduces to
p=
500
.
−3
500
x + b.
−3
How do we find the value of b? Since the line must pass through the given point (60; 6 000),
substitute the x- and p-value, into the last expression. This gives
500
× 60 + b
−3
6 000 = −10 000 + b
6 000 =
b = 16 000.
The expression for the line is therefore
p=
500
x + 16 000
−3
or
p = −166,67x + 16 000.
404
TOPIC 2
Exercise 2.2:1(a)
From
y = −0,4x2 + 0,2x + 1,2
we have that
a = −0,4, b = 0,2 and c = 1,2.
Since a < 0, the function has a minimum.
The value of x at the vertex is
xm =
=
−b
2a
−0,2
2(−0,4)
−0,2
−0,8
= 0,25.
=
The maximum value of the function is
y = f (0,25)
= −0,4 (0,25)2 + 0,2 (0,25) + 1,2
= 1,225.
The intercept on the y-axis is at
c = 1,2.
The discriminant is
b2 − 4ac = 0,22 − 4 × (−0,4) × (1,2)
= 1,96.
The intercepts on the x-axis are
x
=
−0,2 − 1,96
2(−0,4)
=
2
and
x
=
−0,2 + 1,96
2(−0,4)
=
−1,5.
The quadratic is sketched as follows:
y
1,2 2
1
−2
−1
−1
(0,25; 1,225)
×
x
1
2
y = −0,4x2 + 0,2x + 1,2
405
Exercise 2.2:1(b)
From
y = −x2 − 2x − 1
we have that
a = −1, b = −2 and c = −1.
Since a < 0, the function has a maximum.
The value of x at the vertex is
−b
2a
−(−2)
=
2(−1)
= −1.
xm =
The maximum value of the function is
y = f (−1)
= −(−1)2 − 2(−1) − 1
= −1 + 2 − 1
= 0.
The intercept on the y-axis is at
c = −1.
The discriminant is
b2 − 4ac = 0.
The intercepts on the x-axis are the same as the value of x at the vertex:
√
−(−2) ± 0
x =
2(−1)
2
=
−2
= −1
Thus the parabola just touches the x-axis at xm = −1. The sketch appears below:
y
2
1
−4
−3
−2
−1
−1
x
1
2
−2
−3
−4
406
y = −x2 − 2x − 1
TOPIC 2
Exercise 2.2:1(c)
From
y = x2 + 4x + 5
we have that
a = 1, b = 4 and c = 5.
Since a > 0, the function has a minimum.
The value of x at the vertex is
−b
2a
−4
=
2(1)
= −2.
xm =
The minimum value is
y = f (−2)
= (−2)2 + 4(−2) + 5
= 1.
The intercept on the y-axis is
c = 5.
The discriminant is
b2 − 4ac = 42 − 4(1)(5)
= 16 − 20
= −4,
which is < 0. There are no intercepts on the x-axis. The sketch is given below:
y
5
y = x2 + 4x + 5
4
3
2
1
−4
−3
−2
−1
−1
x
1
2
−2
407
Exercise 2.2:1(d)
From
y = −x2 + 9
we have that
a = −1, b = 0 and c = 9.
Since a < 0, the function has a maximum value.
The value of x at the vertex is
−b
xm =
2a
−0
=
2(−1)
= 0.
The maximum value of the function is
y = f (0)
= − (0)2 + 9
= 9.
The intercept on the y-axis is at
c = 9.
The discriminant is
b2 − 4ac = 02 − 4(−1)(9)
= 36.
The intercepts on the x-axis are
x
√
0 − 36
2(−1)
−6
−2
=
=
=
and
x
√
0 + 36
2(−1)
6
−2
=
=
3
−3.
=
The quadratic is sketched below:
y
10
8
y = −x2 + 9
6
4
2
−4
408
−3
−2
−1
−2
x
1
2
3
4
TOPIC 2
Exercise 2.2:2
From
d = p2 − 45p + 520
we have that
a = 1, b = −45 and c = 520.
The price at the minimum is
p =
−(−45)
2(1)
= 22,5.
Thus, the price that minimises the weekly demand is R22,50 per litre.
The demand at the minimum is
d = f (p)
= f (22,5)
= 22,52 − 45(22,5) + 520
= 13,75.
The minimum demand will be 13,75 litres at R22,50 per litre.
409
Topic 3. Linear systems
Exercise 3.1:1
The equations are
7x + 5y = −4
(1)
3x + 4y = 2.
(2)
and
From (1):
7x = −4 − 5y
4 5
x = − − y
7 7
Substitute (3) into (2):
(3)
4 5
3 − − y + 4y = 2
7 7
−
12 15
− y + 4y = 2
7
7
−
28
12
15
y+ y = 2+
7
7
7
13
y =
7
26
7
13y = 26
y = 2
Substitute y = 2 into (3):
4 5
x = − − ×2
7 7
=
−14
7
= −2
The solution is the point
(−2; 2).
Exercise 3.1:2
The equations are
2x + 2y = 3
and
5x +
y
= −6.
2
(1)
(2)
From (1):
2x = 3 − 2y
x =
=
410
3 2y
−
2
2
3
−y
2
(3)
TOPIC 3
Substitute (3) into (2):
3
y
−y +
5
2
2
y
15
− 5y +
2
2
10y y
−
+
2
2
9y
−
2
−9y
= −6
= −6
= −6 −
15
2
27
2
= −27
= −
y = 3
Substitute y = 3 into (3):
3
−3
2
3 6
−
=
2 2
3
= −
2
x =
The solution is
3
− ;3 .
2
Exercise 3.1:3
The equations are
x + 4y = 49
(1)
−2x + y = 1.
(2)
x = 49 − 4y
(3)
and
From (1):
Substitute (3) into (2):
−2 (49 − 4y) + y = 1
−98 + 8y + y = 1
9y = 98 + 1
9y = 99
y = 11
Substitute y = 11 into (3):
x = 49 − 4 × 11
= 49 − 44
= 5
The solution is
(5; 11) .
411
Exercise 3.2:1
The inequality is
11 ≥ 6 − 4x.
Subtract 6:
5 ≥ −4x
Divide by −4:
−
5
4
≤ x
x ≥ −
5
4
5
(that is −1,25), as shown:
4
Thus, the solution is all values of x greater or equal to −
−1,5
−2
−0,5
−1
x
0,5
0
1
Exercise 3.2:2
The inequality is
4x + 4 < 1,5x − 6.
Subtract 4:
4x < 1,5x − 10
Subtract 1,5x:
2,5x < −10
Divide by 2,5:
x < −4
Thus, the solution consists of all values of x less than −4, as shown:
−10
−9
−8
−7
−6
−5
−4
−3
Exercise 3.3:1
The inequality is
3x + y − 3 > 0.
The corresponding straight line is
3x + y − 3 = 0
or
y = −3x + 3.
412
−2
−1
x
0
1
2
TOPIC 3
This is depicted below:
y
4
3
(2; 2)
2
1
−1
−1
x
1
2
3
4
5
−2
−3
−4
3x + y − 3 = 0
Now consider the point (2; 2), which is to the right of and above the line. Substituted in the
left-hand side of the inequality 3x + y − 3 > 0, it gives
3×2+2−3 = 5
and
5 > 0.
Thus all the points to the right and above the line satisfy the inequality. Since the inequality
contains no = sign, the points on the line do not satisfy it. We indicate this by drawing a dashed
line.
Exercise 3.3:2
The inequality is
2x + 4y + 1 ≤ x + y − 2.
Add −x − y + 2 to both sides:
2x − x + 4y − y + 1 + 2 ≤ 0
x + 3y + 3 ≤ 0
The corresponding straight line is
x + 3y + 3 = 0
or
1
y = − x − 1.
3
413
This is shown in the graph:
y
2
1
(0; 0)
−5
−4
−3
−2
−1
−1
1
x
2
3
4
5
x + 3y + 3 = 0
−2
−3
Consider the point (0; 0). Substitute it in the inequality x + 3y + 3 ≤ 0. This gives
0+3×0+3 = 3
and
3 0.
That is, it does not satisfy the inequality. Thus all the points below and to the left of the line,
as well as those on the line (why?), satisfy the inequality.
Exercise 3.3:3
The system of inequalities is
2x + y − 5 ≤ 0
x−2 ≤ 0
y−4 ≤ 0
x ≥ 0
y ≥ 0.
The solution of the first inequality is the region on, below and to the left of the line
2x + y − 5 = 0.
The second implies all x-values on or to the left of the line
x = 2,
while the third implies all y-values on or below the line
y = 4.
414
TOPIC 3
The last two inequalities imply the first quadrant, including the axes. The solution is graphed
below and indicated by the grey area:
y
x=2
6
5
4
1
;4
2
y=4
3
2
(2; 1)
1
−2
−1
−1
x
1
2
3
4
5
2x + y − 5 = 0
−2
Exercise 3.3:4
The inequality is
3x − 7 ≤ 5x + 2.
Add +7:
3x ≤ 5x + 9
Subtract 5x:
−2x ≤ 9
Divide by −2:
x≥−
9
2
Thus the solution is all values of x greater than or equal to −
−6
−5
−4
−3
−2
9
(i.e. −4,5), as shown:
2
−1
x
0
Exercise 3.3:5
The inequality is
5x + y + 1 < −x − y − 1.
Add x + y + 1:
6x + 2y + 2 < 0
415
The corresponding straight line is
6x + 2y + 2 = 0
or
y = −3x − 1.
Graphically it is represented in the following graph:
y
4
3
2
1
− ;0
3
−4
−3
−2
1
−1
−1
x
1
2
3
−2
−3
y = −3x − 1
−4
−5
The point (0; 0) does not satisfy the inequality. Substitute (0; 0) into the left-hand side of the
inequality 6x + 2y + 2 < 0. That is,
6×0+2×0+2=2
and
2 ≮ 0.
Thus all the points to the left and below the line satisfy the inequality, but the points on the line
are excluded. (Why?)
416
TOPIC 4
Topic 4. Mathematics of finance
Exercise 4.1:1
The interest rate is expressed “per annum” and we have to express the term, n, in years. The
following information is given:
P = 5 000,00
n = 90 days =
90
18
=
year
365
73
i = 12% = 0,12
The interest is calculated as
I = P in
= 5 000,00 × 0,12 ×
18
73
= 147,95
and the sum accumulated as
S = P +I
= 5 000,00 + 147,95
= 5 147,95.
The simple interest is R147,95 and the accumulated sum is R5 147,95.
As pointed out above, reference is occasionally made to a 360-day year. This
has its origin in pre-calculator days when sums of the above type were tedious. In the above ex1
90
= , which obviously makes manual calculation
360
4
a lot easier. However, unless informed otherwise, you should assume that the “exact” year (365
ample, the effect of this would be that T =
days) is used.
(See Tutorial Letter 101: Using recommended calculator.)
Exercise 4.1:2
The following is given:
I = 300,00
1
n = 1 year
2
1
i = 12 % = 0,125
2
From
I = P in
417
we have that
P
=
I
in
300,00
0,125 × 1,5
300,00
=
0,1875
= 1 600,00.
=
The principal required is R1 600,00.
Exercise 4.1:3
The following is given:
P = 3 000,00
n = 6 months =
6
= 0,5 year
12
i = 12% = 0,12
The amount is
S = P +I
= P + P in
= P × (1 + in)
= 3 000,00 × (1 + 0,12 × 0,5)
= 3 000,00 × 1,06
= 3 180,00.
The total amount to be paid back is R3 180,00.
Exercise 4.1:4
The following is given:
P = 1 000,00
n = 4 years
i = 10% = 0,10
The interest is
I = P in
= 1 000,00 × 0,10 × 4
= 400,00.
The total interest received is R400,00.
The amount received per month is calculated as
400,00
12 × 4
400,00
48
= 8,33.
=
If the interest is paid monthly, you will receive R8,33 per month.
418
TOPIC 4
Exercise 4.1:5
The following is given:
P = 800,00
S = 823,00
n = 3 months =
3
= 0,25 year
12
From
S = P (1 + in)
we have that
S
P
= 1 + in
S
− 1 = in
P
1
S
−
= i
Pn n
1
823,00
−
800,00 × 0,25 0,25
1
823,00
−
=
200,00 0,25
= 4,115 − 4
i =
= 0,115.
The interest rate is 11,5% per annum.
Alternatively, the interest earned is
I = 823,00 − 800,00
= 23,00.
Then
I = P in
I
i =
Pn
23,00
800,00 × 0,25
= 0,115.
=
The percentage interest is then 0,115 × 100 = 11,5%.
Exercise 4.2:1
The following is given:
S = 4 000,00
d = 0,10
n = 6 months =
1
6
= year
12
2
419
The discount is calculated as
D = Sdn
= 4 000,00 × 0,10 ×
1
2
= 200,00.
That is, the simple discount is R200,00.
The discounted value is calculated as
= S−D
P
= 4 000,00 − 200,00
= 3 800,00.
One will therefore receive R3 800,00.
Since the interest, I, paid is R200,00, we use
I = P in
to get
200,00 = 3 800,00 × i ×
i = 2×
1
2
200,00
3 800,00
= 0,1053.
Thus, the equivalent simple interest rate is 10,53% per annum.
Exercise 4.2:2(a)
The following is given:
S = 100,00
d = 0,12
n = 3 months =
1
3
= year
12
4
The discount is calculated as
D = Sdn
= 100,00 × 0,12 ×
= 3,00.
That is, the simple discount is R3,00.
The discounted value is calculated as
P
= S−D
= 100,00 − 3,00
= 97,00.
One will therefore receive R97,00.
420
1
4
TOPIC 4
Since the interest, I, paid is R3,00, we use
I = P in
to get
3,00 = 97,00 × i ×
i = 4×
1
4
3,00
97,00
= 0,1237.
Thus the equivalent simple interest rate is 12,37% per annum.
Exercise 4.2:2(b)
The following is given:
S = 100,00
d = 0,12
n = 9 months =
3
9
= year
12
4
The discount is calculated as
D = Sdn
= 100,00 × 0,12 ×
3
4
= 9,00.
That is, the simple discount is R9,00.
The discounted value is calculated as
P
= S−D
= 100,00 − 9,00
= 91,00.
One will therefore receive R91,00.
Since the interest, I, paid is R9,00, we use
I = P in
to get
9,00 = 91,00 × i ×
i =
3
4
9,00
4
×
3 91,00
= 0,1319.
Thus, the equivalent simple interest rate is 13,19% per annum.
421
Exercise 4.2:3
The following is given:
P = 750,00
d = 0,16
n = 10 months =
5
10
= year
12
6
From
P = S × (1 − dn)
we calculate the future value of the loan as
S = P ÷ (1 − dn)
750,00
= 5
1 − 0,16 ×
6
= 865,38.
Mary will have to repay June R865,38 in ten months’ time.
Exercise 4.3:1
The following is given:
P = 1 000,00
i=
0,08
= 0,04 because it is compounded semi-annually
2
n = 2 × 1 = 2 half years
Then
S = P (1 + i)n
0,08
= 1 000,00 1 +
2
2
= 1 000,00 × 1,04
2
= 1 081,60.
The compounded amount is R1 081,60.
The interest earned is
1 081,60 − 1 000,00 = 81,60.
The interest earned is R81,60.
Exercise 4.3:2
The following is given:
P = 2 000,00
0,12
= 0,03 because it is compounded quarterly
4
1
n = 2 × 4 = 10 quarters
2
i=
422
TOPIC 4
Then
S = P (1 + i)n
= 2 000,00(1 + 0,03)10
= 2 000,00 × 1,0310
= 2 687,83.
The total amount available is R2 687,83.
Exercise 4.3:3(a)
In all three cases we have P = 1 000,00.
We have that
n = 2 and
i = 0,10 simple interest.
Then
I
= P ×i×n
= 1 000,00 × 0,10 × 2
= 200,00.
The interest earned is R200,00.
Exercise 4.3:3(b)
We have that
n = 2 × 2 = 4 half years and
0,095
.
i=
2
Then
S = P (1 + i)n
= 1 000,00 × 1 +
0,095
2
4
= 1 203,97.
The interest earned is R203,97.
Exercise 4.3:3(c)
We have that
n = 2 × 4 = 8 quarters and
0,09
.
i=
4
Then
S = 1 000,00 × 1 +
0,09
4
8
= 1 194,83.
423
The interest earned is R194,83.
1
The best investment is 9 % interest per year, compounded semi-annually.
2
Exercise 4.4:1
The three problems can be represented as follows on a time line (all at 15% interest per annum,
compounded monthly):
500
8 months
2
months
0
X0
2
4
6
4 months
8
10
X6
@ 15% per annum, compounded monthly
12
months
X12
Exercise 4.4:1(a)
We have to determine the size of payment X0 . From month zero to month eight there are eight
months. We have to discount the debt eight months backwards to determine the present value.
From
S = P (1 + i)n
we have that
P
= S ÷ (1 + i)n
= 500,00 ÷ 1 +
0,15
12
8
= 452,70.
The single payment that will repay her debt now (X0 ) is R452,70.
Exercise 4.4:1(b)
We have to determine the size of payment X6 . From month six to month eight there are two
months. We have to discount the debt two months backwards to determine the present value:
P
= 500,00 ÷ 1 +
0,15
12
2
= 487,73
The single payment that will repay her debt six months from now (X6 ) is R487,73.
Exercise 4.4:1(c)
We have to determine the size of payment X12 . From month eight to one year there are four
months. We must move the debt forwards four months:
0,15 4
S = 500,00 × 1 +
12
= 525,47
The single payment that will repay her debt one year from now (X12 ) is R525,47.
424
TOPIC 4
Exercise 4.4:2
In the following diagram, debts are shown above the line and payments below the line:
2 000
0
3
6
9
12
months
2 quarters
R800
3 quarters
X
R600
@ 12% per annum, compounded quarterly
Payments: The numerical values of the R600 and R800 payments must be moved to the end of
the year. There is also an Rx payment at month twelve.
For the R600 payment:
The money is moved forwards nine months (i.e. three quarters). The future value is
0,12
S = 600,00 × 1 +
4
= 655,64.
3
The value of the R600,00 payment at the end of the year is R655,64.
For the R800 payment:
The money is moved forwards six months (i.e. two quarters). The future value is
0,12
S = 800,00 × 1 +
4
= 848,72.
2
The value of the R800,00 payment at the end of the year is R848,72.
For the Rx payment:
As this is the last payment, no interest is involved and the payment remains Rx.
Debt: There is only one debt, namely R2 000 at the end of the year.
For the R2 000 debt:
No interest is involved for the debt of R2 000 at the end of the year.
Remember, the total amount to be paid is equal to the total debt. Therefore
payments = debt
655,64 + 848,72 + x = 2 000,00
1 504,36 + x = 2 000,00
x = 2 000,00 − 1 504,36
= 495,64.
Thus she must pay R495,64 at the end of the year to settle her debt.
425
Exercise 4.5
(See Tutorial Letter 101: Using the recommended calculator.)
Exercise 4.5:1
The accumulated amount is
S = S1 + S2 + S3 + S4 + S5
= 600,00 × (1,1)4 + 600,00 × (1,1)3 + 600,00 × (1,1)2 + 600,00 × (1,1)1 + 600,00
= 3 663,06.
The accumulated amount is thus R3 663,06.
Alternatively you could have used the formula for the future value of the annuity, giving
(1 + i)n − 1
S = R
i
(1 + 0,10)5 − 1
= 600,00 ×
0,10
600,00 × (1,15 − 1)
0,10
= 3 663,06.
=
Exercise 4.5:2
The time line is given below:
R1 200
R1 200
R1 200
R1 200
R1 200
R1 200
1
2
3
16
17
18
0
@ 11% per annum
The following is given:
R = 1 200,00
i = 0,11 per year
n = 18 years
Thus
(1 + i)n − 1
S = R
i
1,1118 − 1
= 1 200,00 ×
0,11
= 60 475,12.
When her daughter is 18 years old, the amount will be R60 475,12.
426
S=?
years
TOPIC 4
Exercise 4.5:3
The time line is given below:
0
R600
R600
R600
R600
R600
R600
1
2
3
18
19
20
@ 13% per annum, compounded quarterly
quarters
S=?
The following is given:
R = 600,00
0,13
= 0,0325 per quarter
4
n = 5 × 4 = 20 quarters
i=
Thus
1,032520 − 1
S = 600,00 ×
0,0325
= 16 538,55.
Thus, the accumulated amount at the end of the term is R16 538,55.
Exercise 4.5:4
The time line is given below:
0
R200
R200
R200
R200
R200
R200
1
2
3
22
23
24
P =?
months
@ 12% per annum, compounded monthly
The following is given:
R = 200,00
i=
0,12
= 0,01 per month
12
n = 24 months
The present value of the 24 payments is
P
= R
(1 + i)n − 1
i (1 + i)n
1,0124 − 1
= 200,00 ×
0,01 × 1,0124
= 4 248,68.
427
Thus the present value of the payments is R4 248,68. The cost of the motorbike is calculated as
1 000,00 + 4 248,68 = 5 248,68.
The cost of the motorbike is R5 248,68. The interest paid is the difference between the sum of all
payments and the cost, namely
I = (1 000,00 + 24 × 200,00) − 5 248,68
= 551,32.
The interest that he paid is R551,32.
Exercise 4.5:5
The relevant time line is given below:
0
R800
R800
R800
R800
R800
R800
1
2
3
18
19
20
P =?
half years
@ 12,5% per annum, compounded half-yearly
The following is given:
R = 800,00
i=
0,125
= 0,0625 per half year
2
n = 10 × 2 = 20 half years
The present value is thus
P
1,062520 − 1
= 800,00 ×
0,0625 × 1,062520
= 8 992,58.
Thus the amount of R8 992,58 must be invested now at 12,5% per annum, compounded halfyearly, to provide half-yearly payments of R800 for ten years.
Exercise 4.6:1
The present value of the loan is R225 000 (i.e. 270 000 − 45 000).
The following is given:
P = 225 000,00
i=
0,115
per month
12
n = 12 × 20 = 240 months
428
TOPIC 4
The payments are
i (1 + i)n
R = P×
(1 + i)n − 1
⎡
⎤
0,115 240
0,115
1
+
⎢
⎥
12
⎢ 12
⎥
= 225 000,00 × ⎢ ⎥
240
⎣
⎦
0,115
1+
−1
12
= 2 399,47.
The monthly payments are R2 399,47.
Exercise 4.6:2
The following is given:
P = 60 000,00
i = 0,12 per year
n = 10 years
The payments are
R = P
i (1 + i)n
(1 + i)n − 1
0,12 (1 + 0,12)10
= 60 000,00 ×
(1 + 0,12)10 − 1
= 10 619,05.
Thus, you will receive R10 619,05 at the end of each year for ten years.
Exercise 4.6:3
The following is given:
P = 4 000,00
i=
0,15
= 0,075 per half year
2
n = 3 × 2 = 6 half years
The payments are
0,075 (1 + 0,075)6
R = 4 000,00 ×
(1 + 0,075)6 − 1
= 852,18.
The payments are R852,18.
429
The amortisation schedule is as follows:
Period (i.e.
half years)
Outstanding
principal at half
year beginning
Interest due at
end of
half year
Payment
Principal
repaid
1
4 000,00
300,00
852,18
552,18
2
3
3 447,82
2 854,23
258,59
214,07
852,18
852,18
593,59
638,11
4
5
2 216,12
1 530,15
166,21
114,76
852,18
852,18
685,97
737,42
6
792,73
59,45
852,18
792,73
1 113,08
5 113,08
4 000,00
Total
430
TOPIC 5
Topic 5. Collection, presentation and description of data
Exercise 5.1:1
The smallest value is 28 and the largest is 48. The range is
R = 48 − 28 = 20.
The measure for the number of intervals is
R
20
=
= 2,
10
10
but, in my opinion, a frequency table with only two intervals is not very useful. If I use five
intervals, the width of each interval is
20
= 4.
5
The lower bound of the first interval is half a unit less than the smallest value of 28. Then the
first interval is 27,5 – 31,5. A quick calculation
(5(intervals) × 4(width) = 20 and 20 + 27,5 = 47,5)
gives an upper limit for the last interval of 47,5, which poses a problem. So I will use six intervals:
1
20
=3 ,
6
3
hence I choose the width as four. Now the first interval is 27,5 – 31,5. My quick calculation now
gives
6 × 4 + 27,5 = 51,5
for the last interval’s upper bound.
The frequency table of the data is given below:
Interval
Frequency
27,5 – 31,5
5
31,5 – 35,5
35,5 – 39,5
5
10
39,5 – 43,5
43,5 – 47,5
6
3
47,5 – 51,5
1
30
431
The histogram of the insurance claims data is given below:
Frequency (f )
10
8
6
4
2
27,5
31,5
35,5
39,5
43,5
47,5
51,5
Claims
Exercise 5.1:2
The cumulative frequency table is as follows:
Upper limit Cumulative frequency
< 31,5
< 35,5
5
10
< 39,5
< 43,5
20
26
< 47,5
29
< 51,5
30
(With the cumulative frequency table we may use either “smaller than” or “smaller than or equal
to” (< or ≤).)
Exercise 5.1:3
The stem-and-leaf diagram looks like this:
Stem
Leaf
Frequency
2
8
1
3
4
0 0 1 1 3 4 4 4 5 6 6 7 7 8 8 8 9 9 9
0 0 0 1 1 2 4 5 6 8
19
10
The values (or leaves) within a stem may be separated so that the values from 0 to 4 are separated
from the values from 5 to 9.
This gives more information and looks like this:
Stem
432
Leaf
Frequency
2
8
1
3
3
0 0 1 1 3 4 4 4
5 6 6 7 7 8 8 8 9 9 9
8
11
4
4
0 0 0 1 1 2 4
5 6 8
7
3
TOPIC 5
Exercise 5.1:4(a)
The minimum number of claims processed per week is 28. If a worker processes only 26 claims
per week for a whole month, the chances are good that there is a problem.
Exercise 5.1:4(b)
In the interval 35,5 to 39,5 the frequency is 10. Therefore,
1
10
× 100 = 33 %
30
3
of the workers can process 36 to 39 claims per week. Is the competitor more productive? This
depends on whether the claims being processed are similar. More information should be obtained
before I take action.
Exercise 5.1:4(c)
Considering the frequency table, the number of claims of 36 or less is accounted for by the first
two intervals. The sum of the frequencies is
5 + 10 = 15.
Expressed as a percentage it gives
15
× 100 = 50%.
30
Therefore, 50% of the workers processes less than 36 claims per week. I would transfer some of
the personnel to other divisions.
Exercise 5.2:1
The mean is
10
xi
i=1
x =
n
1 671
=
10
= 167,1.
Exercise 5.2:2
The data must be in ascending order before the median can be determined:
104;
127;
131;
135;
146;
170;
175;
179;
190;
314
The median is the observation in the
n+1
2
=
10 + 1
2
= 5,5th
position, that is the 5,5th observation. Because there is no such value, we calculate the mean of
the fifth and sixth values. The median is
146 + 170
2
= 158.
Me =
433
Exercise 5.2:3
The mode is the value that occurs the most often. There is no value that occurs more than once
and there is no mode.
Exercise 5.3:1(a)
The mean is
14
xi
x =
=
i=1
n
22 355
14
= 1 596,79.
Exercise 5.3:1(b)
Use a table to help you with the calculations for the standard deviation. Calculate the deviation
from x for each observation and square it. Divide the sum of the squares by n − 1 = 13. The
standard deviation is the square root of the variance.
434
xi
xi − 1 596,79
(xi − x)2
1
1 630
33,21
1 102,90
2
1 550
−46,79
2 189,30
3
1 430
−166,79
27 818,90
4
1 440
−156,79
24 583,10
5
1 390
−206,79
42 762,10
6
1 400
−196,79
38 726,30
7
1 480
−116,79
13 639,90
8
1 490
−106,79
11 404,10
9
1 410
−186,79
34 890,50
10
1 905
308,21
94 993,40
11
1 540
−56,79
3 225,10
12
1 890
293,21
85 972,10
13
1 900
303,21
91 936,30
14
1 900
303,21
91 936,30
Total
565 180,30
TOPIC 5
The variance of the data is
14
i=1
S2 =
(xi − x)2
n−1
565 180,30
13
=
= 43 475,41.
Exercise 5.3:1(c)
The standard deviation of the data is
43 475,41
S =
= 208,51.
Exercise 5.3:2(a)
Calculate the values in the table:
Interval
Frequency
fi
Midpoint
xi
15,5 – 21,5
2
18,5
2 × 18,52
=
684,50
21,5 – 27,5
6
24,5
6 × 24,52
=
3 601,50
27,5 – 33,5
8
30,5
8 × 30,52
=
7 442,00
33,5 – 39,5
4
36,5
4 × 36,52
=
5 329,00
39,5 – 45,5
4
42,5
4 × 42,52
=
7 225,00
45,5 – 51,5
1
48,5
1 × 48,52
=
2 352,25
=
26 634,25
6
n=
fi × x2i
6
fi = 25
i=1
i=1
There are 6 intervals:
fi x2i
6
fi = 25
n=
i=1
Exercise 5.3:2(b)
6
i=1
fi x2i = 26 634,25
435
Exercise 5.3:2(c)
The variance of the data is calculated as
6
S2 =
i=1
fi x2i − n × x2
n−1
26 634,25 − 25 × 31,72
25 − 1
= 63,00.
=
Exercise 5.3:2(d)
The standard deviation is
S =
=
√
S2
63,00
= 7,94.
Exercise 5.3:3
The box-and-whiskers diagram is as follows:
1 000
900
800
700
600
500
400
300
200
100
A
B
C
It is clear that region C has the highest median, but also the largest interquartile deviation. The
inner 50% of the rainfall figures is spread rather widely.
The spreads for A and B are quite similar. However, whereas the data for A are more or less
symmetrical to the median, the distribution for B is rather skew. In B, 25% of the figures fall
between 300 and 350, and 25% of the figures fall between 350 and 550. The figures are more
“concentrated” between 300 and 350. The distribution for C is also very skew, with 25% of the
figures falling between 800 and 900, and 25% of the figures falling between 200 and 800.
436
TOPIC 6
Topic 6. An application of differentiation
Exercise 6.1:1
The cost function is
C(x) = 575 + 25x −
x2
.
4
Then
d
(575) = 0
dx
d
(25x) = 25 × 1 × x1−1
dx
= 25 × x0
= 25 × 1
d
dx
−
x2
4
= 25
2 × x2−1
4
x
= − .
2
= −
The marginal cost to manufacture x boats is
C (x) = 25 −
x
.
2
Exercise 6.1:2
If 40 boats are manufactured, the marginal cost is
C (40) = 25 −
40
2
= 5.
At a production level of 40 boats, the cost to manufacture one additional boat is R5 000,00.
Exercise 6.1:3
If 30 boats are manufactured, the marginal cost is
C (30) = 25 −
30
2
= 10.
At a production level of 30 boats it costs R10 000,00 to manufacture one additional boat.
Exercise 6.2:1
The function is
f (x) = 6x.
The derivative of 6x is the derivative of 6 × x1 , which is
6 × 1x1−1 = 6 × x0
= 6×1
= 6.
437
Remember that any number to the power of zero equals one. Thus
f (x) = 6.
Exercise 6.2:2
The function is
f (x) = 3 + 5x.
The derivative of 3, which is a constant, is zero.
The derivative of 5x is the derivative of 5 × x1 , which is equal to
5 × x1−1 = 5 × x0
= 5×1
= 5.
Thus
f (x) = 0 + 5
= 5.
Notice that f (x) = 3 + 5x is nothing more than the familiar straight line or linear function. Its
intercept on the y-axis is 3 and its slope is 5. Its derivative is also 5, which therefore confirms
that the derivative represents the slope.
Exercise 6.2:3
The function is
f (x) = 3 + 2x2 .
The derivative of 3, which is a constant, is zero.
The derivative of x2 is
2x2−1 = 2x1
= 2x.
The derivative of 2x2 is
2 × 2x2−1 = 2 × 2x1
= 4x.
Thus
f (x) = 4x.
Exercise 6.3:1
The profit function is
P (x) = 5x −
438
x2
− 450.
200
TOPIC 6
The derivatives of the terms of the profit function are
d
5x = 5 × x1−1
dx
= 5x0
d
dx
−
x2
200
= 5
2x
200
x
= −
100
= −
d
(−450) = 0.
dx
The derivative of the profit function is
P (x) = 5 −
x
.
100
The marginal profit when 450 knives are produced is
P (450) = 5 −
450
100
= 0,5.
The sale of one additional knife will yield an additional profit of R0,50.
The marginal profit when 750 knives are produced, is
750
100
= −2,5.
P (750) = 5 −
The marginal profit is negative, which indicates a decrease in profit. Selling one additional knife
will yield a decrease in profit of R2,50.
Exercise 6.3:2
The profit function is
P (x) = R(x) − C(x).
Thus
x2
− (7 000 + 2x)
1 000
x2
− 7 000 − 2x
= 10x −
1 000
x2
− 7 000.
= 8x −
1 000
P (x) = 10x −
The derivative of P (x) is
P (x) = 8 −
x
.
500
439
The marginal profit when 2 000 scissors are produced is
P (2 000) = 8 −
2 000
500
= 4.
The marginal profit when 4 000 scissors are produced is
P (4 000) = 8 −
4 000
500
= 0.
The marginal profit when 5 000 scissors are produced is
P (5 000) = 8 −
= −2.
5 000
500
When 2 000 scissors are produced, an increase in production will yield an increase of R4,00 per
pair of scissors in profit.
When 4 000 scissors are produced, the marginal profit is 0, which indicates that 4 000 is the
number of scissors that should be produced to maximise profit.
When 5 000 scissors are produced, the marginal profit is negative. This indicates that the profit
will decrease by R2,00 per pair of scissors.
440