The Open University of Tanzania
Institute of Continuing Education
Basic Mathematics for Diploma in
Primary Teacher Education
ODC 055
First Edition: 2013
Copyright © 2013
All Rights Reserved
Published by
THE OPEN UNIVERSITY OF TANZANIA
Kawawa Road,
P. O. Box 23409,
Dar es Salaam,
TANZANIA
2 Basic Mathematics for Diploma in Primary Teacher Education
Contents
PART I: ARITHMETIC
Lecture 1 Some Basic Concepts
Lecture 2 Sequences and Series of Numbers
7
23
PART II: ALGEBRA
Lecture 3 Algebraic Expressions
37
Lecture 4 Quadratic Equations
59
PART III: SET THEORY
Lecture 5 Set Presentations and Notations
73
Lecture 6 Operations with Sets
81
PART IV: ELEMENTARY STATISTICS
Lecture 7 Data Collection and Presentation
97
Lecture 8 Measures of Central Tendency
117
Lecture 9 Measures of Dispersion
125
Basic Mathematics for Diploma in Primary Teacher Education 3
ART I
Arithmetic
Arithmetic is the branch of Mathematics that deals with numbers, operations
on numbers, and computation. It is one of the very basic branches of
Mathematics taught right from elementary levels. The usual numbers of
arithmetic are whole numbers, fractions, decimals, and percents. Beyond
these numbers there are negative numbers, rational numbers, and irrational
numbers. The rational and irrational numbers together constitute what is
known as the real numbers.
At this level you must have already mastered the elementary operations of
addition, subtraction, multiplication and division of fractions and decimals, as
well as their applications in solving problems. This section is divided into two
lectures, in the first lecture we will look at some basic concepts of rounding
off of numbers, scientific notations of numbers, and significant figures.
Furthermore, we will discuss the concepts of ratios, proportions and
percentages and how to apply them in solving practical problems. In the
second lecture of this section we will discuss number sequences and series,
arithmetic and geometric progressions, and solve some related problems.
4 Basic Mathematics for Diploma in Primary Teacher Education
LECTURE 1
Some Basic Concepts
1.1 Introduction
Arithmetic – commonly considered a separate branch but in actuality a part of
algebra – is a branch of mathematics concerned with operations on sets of numbers
or other elements that are often represented by symbols. Algebra is a
generalization of arithmetic and gains much of its power from dealing symbolically
with elements and operations. Conventionally the term has been most widely
applied to simple teaching of the skills of dealing with Numbers for practical
purposes, e.g., computation of areas, proportions, costs, and the like. The four
fundamental operations of this study are addition, subtraction, multiplication, and
division. In advanced study the concept of number is greatly generalized to include
not only complex numbers, but also quaternions, tensors, and abstract entities
with no other meaning than that they obey certain laws (see algebra).
Learning Objectives
At the end of this lecture, you will be able to:

Write numbers in scientific notations, correctly rounding off numbers in
specified number of decimal places and writing correct number of
significant figures;

Apply basic arithmetic operations concepts in solving practical problems
related to fractions and decimals;

Write proportions and ratios and solve problems involving the concept of
ratio and proportions;

Solve problems involving percentage increase or decrease, profit, loss and
percentage discount.
1.2 Programme Overview
To start with, let us quickly remind ourselves of some basic definitions and do
some practical examples to refresh;
Basic Mathematics for Diploma in Primary Teacher Education 5
Fractions are rational numbers of the form
a
where b ≠ 0
b
a is called the numerator
b is the denominator and
a
is undefined for any value of a, meaning that no denominator can be 0.
0
Decimals are extended notations of numbers to include numbers smaller than 1
unit, e.g. 1.5, 35. 68, 90.375 etc.
Example:
1
1
1
of her salary in food,
in house rent and in bus fare to and
3
6
4
from her workplace, she then keeps the remaining part in her serving account.
What fraction of the salary is kept in the account?
A woman uses
Solution:
To get the fraction of the salary that is kept in the account, we have to subtract the
total amount that is spent in different items from the whole salary, assuming the
whole salary to be 1 unit.
Fraction of the salary that is spent:
=
1
1
1
+ +
4
3
6
=
4 2 3
12
=
9
12
=
3
4
Fraction of the salary that is kept:
=1–
=
3
4
1
4
Therefore, the woman keeps
Example:
1
of the salary in her serving account.
4
6 Basic Mathematics for Diploma in Primary Teacher Education
How many pieces of
1
1
metres chord can be cut from a metre piece of chord.
8
2
Solution:
1
1
metres chords cut from a metre piece of chord, we
8
2
will have to divide the two fractions
To obtain the number of
=
1
1
÷
2
8
=
1
8
×
2
1
=
8
2
= 4 pieces of chords
Example:
3
1
of a 300 hectors farm is grown maize,
is grown beans and the remaining
5
6
portion is grown coffee plants. How many hectors of the farm is grown coffee
plants?
If
Solution:
The whole farm, i.e. 300 hectors is considered as 1 unit.
First, we have to find the fraction of the farm that is grown coffee plants. This will
3
1
be obtained by subtracting and from 1
5
6
3 1

= 1 – 


5 6
18  5

= 1 – 

30
=1–
=
23
30
7
30
Second, we have to convert it into hectors
Basic Mathematics for Diploma in Primary Teacher Education 7
This means that
7
of a 300 hectors farm is grown coffee plants
30
This fraction is equal to
7
× 300
30
= 70 hectors
In your previous level, you must have worked with some problems that require
you to convert a decimal number to a fraction and vice versa. Here are some
highlights to remind you.
In converting a decimal number to a fraction:
1.
Write the decimal number as a fraction with denominator 1.
2.
Multiply the numerator by a number which will make it a whole
number, then multiply the numerator 1 by the same number.
3.
Simplify if possible.
Activity 1
How can you perform operations involving numbers in both fraction
and decimal forms?
1.3 Estimations and Approximations
In estimation, we are interested in giving a rough idea of the actual amount. There
are many occasions when one needs to estimate. For instance, when cooking, one
needs to estimate the amount of salt, or when taking tea one has to estimate the
amount of sugar. In mathematics, such quantities as weights, heights, distances and
so on are sometimes estimated.
Approximations or approximate figures are amounts given as estimations.
For example, 499 may be estimated as 500
may be estimated as 750
7
3
may be estimated as 8
5
Rounding off of Numbers to given Decimal Places
Rounding off of a number means to find another number closer to the given
number with the desired number of decimal places.
8 Basic Mathematics for Diploma in Primary Teacher Education
For example, you may be carrying out an operation which involves decimals, and
you get such an answer as 0.836, then you are required to round it off to 2 places of
decimal.
In doing so, you will have to critically look at the second and the third decimal
places. The third place, which is 6, is bigger than half of 10, which is 5; therefore it
is big enough to influence the size of the next number, so the second number will
be changed from 3 to 4. Therefore, the number will become 0.84.
0.836 = 0.84
Suppose now you have 0.832, and are required to write it in two decimal places.
Again, critically look at the second and the third decimal places. The third place,
which is 2, is less than 5; therefore it is not big enough to influence the size of the
next number. So the number will remain 0.83
0.832 ≈ 0.82
Rules for rounding off decimal numbers:
1.
Look at the single digit just to the right of the place of desired accuracy.
2.
If the digit is 5 or greater, make the digit one larger, if it is less than 5,
leave it as it is.
3.
Remove all other digits to its right.
Examples:

0.895
to 2 decimal places is 0.90

0.4532 to 3 decimal places is 0.453

0.5278 to 3 decimal places is 0.528

0.5278 to 2 decimal places is 0.53
Examples:
1.
Convert 0.75 to a fraction of a rational form
0.75
100
75
3
×
=
=
100
1
100
4
2.
Convert 0.637 to a fraction form
0.637
1000
637
×
=
1
1000
1000
Example:
Basic Mathematics for Diploma in Primary Teacher Education 9
1
marks in her assignment and 29.54
2
marks in the annual examination. What is the total mark acquired in all three
correct to one decimal place?
A student achieved 15.3 marks in her test, 6
Solution:
1
To get the total mark, add the three, but before that, we will have to convert 6 to
2
the decimal form.
6
1
= 6.5
2
Then;
15.3
6.5
+29.54
51.34
You are required to give the answer correct to one decimal place. Following the
rules given above, this will be 51.3.
Example:
A foreign currency exchange rate is 1:1240 TSh. How much is 13450 TSh worth of
the currency? Give the answer in two decimal places.
Solution:
This will require dividing 13450 by 1240
13450
= 10.8467 ≈ 10.85 units of foreign currency.
1240
1.4 Scientific Notations
When writing numbers, especially those involving many zeros before and after the
decimal point, it is convenient to employ the scientific notation using powers of
10.
It is written as A × 10n where A lies between a and 10 and n is an integer.
For example,
10 Basic Mathematics for Diploma in Primary Teacher Education
1000 = 10 × 10 × 10 = 103
100 000 =10 × 10 × 10 × 10 × 10 =105
100 = 1
10–1 = 0.1
10–2 = 0.01
10–5 = 0.00001
725 000 000 = 7.25 × 108
0.000 034 16 = 3.416 × 10–5
Note that multiplying a number by 108 for example, has the effect of moving the
decimal point of the number 8 places to the right. Multiplying a number by 10–8
has the effect of moving the decimal point of the number 8 places to the left.
The scientific notation is often useful in computation, especially in locating
decimal points.
Activity 2
You have to remember the following very important rules:
(10a) × (10b) = 10a+b
10a = 10(a-b)
10b
Where a and b are any numbers.
Examples:
1.
(103) × (102) = 105
2.
(4 000 000) × (0.000 000 000 2) = (4 × 106) × (2 × 10–10) = (4 × 2) × 106–10
=
8 × 10–4
1.5 Significant Figures
Significant figures of a number are the accurate digits, apart from zeros needed to
locate the decimal point.
For example,
65.4
has 3 significant figures
4.5300
has 5 significant figures. The two zeros to the right of 3 are significant.
Basic Mathematics for Diploma in Primary Teacher Education 11
0.0028
has 2 significant figures. The two zeros to the left of 2 are space fillers.
0.001800 has 4 significant figures. The two zeros to the left of 1 are space fillers.
1.6 Ratios, Proportions and Percents
1.6.1 Ratios
Definition
A ratio is a comparison of two quantities. The ratio of two numbers a and b can be
a
written as or a : b
b
To avoid confusion, the units in a ratio should be written down or otherwise
explained in the problem. Generally, the ratio should also be reduced to the
simplest form.
For example, if 2cm in a drawing represents 100m in the actual map, then the ratio
is 2cm:100m, which is 2cm:10000cm = 1cm:5000cm or 1cm:50m
Example:
In one shop a 5kg package of sugar was priced at 6500 TSh. In another shop a 3kg
package was priced at 4500 TSh. Which one was a better buy?
Solution:
To determine this, we have to calculate the price per unit in each shop and then
compare.
In the first shop, the ratio is
6500TSh
= 1300 TSh/Kg
5Kg
In the second shop, the ratio is
4500TSh
= 1500TSh/Kg
3Kg
This means that the first shop is cheaper, hence a better buy.
Example:
A mapmaker uses a scale of 2cm to represent 150m. Write the ratio in a simpler
form?
Solution:
12 Basic Mathematics for Diploma in Primary Teacher Education
The ratio in this case is 2cm:150m, in simpler form 1cm:75m
Example:
A parent left 6,000TSh to be divided amongst her three kids, Ann, Joe and Suzy in
the ratio 1:2:3. Find how much each one will receive.
Solution:
Suppose Ann receives 1x, Joe 2x and Suzy 3x
Then x +2x +3x = 6000
6x = 6000,
therefore x = 1000
So, Ann will receive 1000 TSh, Joe will receive 2000 TSh and Suzy will receive
3000 TSh
1.6.2 Proportions
A proportion is a statement that two ratios are equal.
75 3
 is an equation that says that the two fractions
100 4
are equal. This is therefore a proportion.
For example, the statement
a c
has four terms, a, b, c and d. Terms a and d are called the

b d
extremes, whereas b and c are called the means.
A proportion
A proportion
the means.
a c
 is true if the product of the extremes equal to the product of
b d
Note that the terms can be decimals, fractions, mixed numbers or whole numbers.
Examples:
Determine whether the following proportions are true or false:
1.
5 7

8 9
Solution:
For this to be true, 5 × 9 should be equal to 8 × 7
But 5 × 9 = 45 and 8 × 7 = 56
45 ≠ 56 therefore the proportion is not true.
2.
11 55

12 60
Basic Mathematics for Diploma in Primary Teacher Education 13
Solution:
Again, we compare 11 × 60 and 12 × 55
11 × 60 = 660 12 × 55 = 660
The two products are equal therefore the proportion is true.
Proportions are useful in solving certain types of word problems. In these
applications, there will be some unknown quantity that can be represented by a
letter, say x, y etc. Finding the value of this unknown term is called solving the
proportion.
For example, find x if
4 12

11 x
4x should be equal to 12 × 11
4x = 132
x=
132
= 33
4
x = 33
Direct Proportion
Two quantities are said to be direct proportional to each other if when one is
increased by a certain factor, the other also increases by the same factor.
Example:
If x is direct proportional to y and x = 10 when y = 8, find x when y = 15
Solution:
The ratio x:y is always equal to 10:8 respectively.
x 10
10 x

………. which implies that,

y 8
8 15
8x = 10 × 15
x=
150
8
x = 18
3
4
Inverse Proportion
In inverse proportion, two quantities vary such that if one increases the other
decreases,
14 Basic Mathematics for Diploma in Primary Teacher Education
Examples of quantities that varies inversely in everyday life:
1.
The time taken to do a job is inversely proportional to the number of
people. i.e. when the number of people increases the time taken decreases.
2.
The time taken by a car to travel a certain distance is inversely
proportional to its speed, i.e. when speed increases, the time spent
decreases.
Activity 3
Give examples of quantities that vary directly in everyday life.
Example:
If a varies inversely as b and a = 6 when b = 3, find a when b = 8.
Solution:
Generally, if a is inversely proportional to b, this means that a is directly
proportional to
1
1
k
i.e. a 
this is equivalent to the equation a = or ab = k
b
b
b
where k is a constant to be found
Substituting the given values of a and b to obtain k we have
k = ab
k = 6 × 3 = 18
Since a =
1
k
18
then when b = 8, a =
=2
8
4
b
1.6.3 Percent
The word percent comes from a Latin word per centum meaning ‘per hundred’, so
percent means hundredth or the ratio of a number to 100. The symbol % is called
the percent sign.
For example,
50
 50%;
100
25
= 25%
100
Percent is commonly used in connection with taxes, discount, commissions, profit
and loss.
Example:
Basic Mathematics for Diploma in Primary Teacher Education 15
An item is on sale at 30% discount. What is the discount if the original amount was
6750 shs? What is the sale price?
Solution:
The discount is 30% of 6750TSh =
30
× 6750TSh = 825TSh
100
The sale price will be,
6750 – 825 = 5925TSh
To change decimals to percent and fractions and vice versa;
Examples:

0.67
=
67
= 67%
100

0.342
=
342
= 34.2%
100

0.005
= 0.5%

1.65
= 165%

0.0045 = 0.45%

3
4
= 75%

5
8
= 0.625 = 62.5%
Move two decimal places to the right and put %
Summary
Fractions are rational numbers of the form
a
where b ≠ 0
b
a is called the numerator
b is the denominator and
a
is undefined for any value of a, meaning that no denominator can be 0.
0
Decimals are extended notations of numbers to include numbers smaller than 1
unit, e.g. 1.5, 35. 68, 90.375 etc.
16 Basic Mathematics for Diploma in Primary Teacher Education
In estimation, we are interested in giving a rough idea of the actual amount. There
are many occasions when one needs to estimate.
Rounding off of a number means to find another number closer to the given
number with the desired number of decimal places.
When writing numbers, especially those involving many zeros before and after the
decimal point, it is convenient to employ the scientific notation using powers of
10.
Significant figures of a number are the accurate digits, apart from zeros needed to
locate the decimal point.
A ratio is a comparison of two quantities. The ratio of two numbers a and b can be
written as
a
or a : b
b
A proportion is a statement that two ratios are equal.
Two quantities are said to be direct proportional to each other if when one is
increased by a certain factor, the other also increases by the same factor.
In inverse proportion, two quantities vary such that if one increases the other
decreases.
The word percent comes from a Latin word per centum meaning ‘per hundred’, so
percent means hundredth or the ratio of a number to 100.
Exercise
1.
2.
Simplify the following expressions:
(i)
12 34
5 15


+ 
×




17 60
6 2
(ii)
11 7
12 5


÷ 





13 16
20 9
(i)
Change the following numbers to mixed numbers
(a)
(ii)
53
91
234
(b)
(c)
13
8
100
Change the following fractions into the rational forms
(a) 5
2
11
1
(b) 6
(c) 14
13
20
3
Basic Mathematics for Diploma in Primary Teacher Education 17
3.
7
13
1
4
2
3


Simplify (i) 
+ 
(ii) 3  5
4 2
4 3


6
5
7
5
8
16
4.
If
5.
If John takes 1
6.
The sum of 12
7.
8.
3 th
of a number is 105, what is the number?
7
3
2
hours to clean the farm and Andrew takes 2 hours to
4
3
clean the same farm. How much longer does Andrew take?
5
1
and another number is 20 , what is the other number?
8
4
3
A board is 32m long. If a piece of 5 m length is cut off, how long is the
4
remaining piece?
A telephone pole is 32metres long. If
5
of the pole must be underground
16
11
of the pole aboveground. How much pole is underground and how
16
much is aboveground?
and
1
of the time in
6
1
2
introduction and
in presentation. He then spent
of the remaining
3
3
time in giving exercises and the rest of the time was spent in marking.
What fraction of the lesson time was spent in marking?
9.
During a Mathematics lesson, the teacher used
10.
Find the average of numbers 86.7, 49.2 and 75.4 correct to one decimal
place.
11.
A cyclist rode 250.6Km in 15.3 hours, what was her average speed to the
nearest 1 decimal place?
12.
An architect makes a drawing to scale so that, 1cm represent 6.75m. What
distance will represent 19.25m?
13.
In a certain school 0.25 of the students are in form one,
14.
Write the following number in scientific notations;
1
in form two and
6
0.375 in form three. If the rest of the students are in form four, what
fraction of the total number of students are in form four?
(i)
0.0000567
18 Basic Mathematics for Diploma in Primary Teacher Education
15.
16.
17.
(ii)
456.75 × 103
(iii)
8950000
(iv)
0.0035 × 10–2
Simplify the following:
(i)
(3.4 × 105) (2.3 × 10–2)
(ii)
(1.21 × 106) ÷ 11000
(iii)
(56.7 × 10–3) ÷ (0.80 × 10–4)
How many significant figures are in each of the following, assuming the
numbers are accurately recorded?
(i)
2.54 mm
(ii)
3.51 million litres
(iii)
0.005400 m
(iv)
10.00100 m
Write 500.3272
(i)
Correct to four significant figures
(ii)
Correct to three decimal places.
18.
An architect draws the plans for a building using a scale 0.5cm to represent
10m. How many metres will be represented by 6cm?
19.
A recommended mixture of weed killer is 3 cupfuls for 1 bucket of water.
How many cupfuls should be mixed with 6 buckets of water?
20.
If fuel sells for 1500 shs per litre, how many litres will be bought by 10000
shs?
21.
An investor thinks that he would make 300TSh for every 5000TSh
invested. How much would he expect to make on an 100 000TSh
investment?
22.
If 40kg of fertilizer are used on a 250 hectors farm, how much will be
needed for a 125 hectors farm?
23.
An employee earns a salary of 550,000 shs. If she is then given a salary
increase of 7%, what will be her new salary?
24.
A sales tax is figured at 10% for each item in the shop. What will be the
total cost of three books sold at shs. 15,000; 20,000 and 12,000?
Basic Mathematics for Diploma in Primary Teacher Education 19
25.
In one month Mr. J. earned 50,000Shs. He spent 32% of it on food, 0.2 of it
on rent
3
of it on taxes and the rest was spent on school fees for his
25
daughter. How much was spent on each item?
26.
27.
(a)
If x is inversely proportional to the square of y and x = 4 when y
= 6, find y when x = 16.
(b)
16 men can complete a certain piece of work in 15 days. How long
would it take for 24 men to complete the same piece of work?
Given that y is inversely proportional to x and that x = 5 when y = 8.
Find the value of y when x = 4.
References
Ascott, Justin (2011), Learn Basic Math NOW: Math for the Person Who Has Never
Understood Math! Minute Help Press.
Laing, Laura (2011), Math for Grownups, F+W Media.
Ross, Debra Anne (2009), Master Math: Basic Math and Pre-algebra, Course
Technology PTR.
Zakon, Elias (2001), Basic Concepts of Mathematics, F+W Media.
20 Basic Mathematics for Diploma in Primary Teacher Education
Basic Mathematics for Diploma in Primary Teacher Education 21
LECTURE 2
Sequences and Series of Numbers
2.1 Introduction
A sequence is an ordered list of numbers. The sum of the terms of a sequence is
called a series. While some sequences are simply random values, other sequences
have a definite pattern that is used to arrive at the sequence's terms. In this lecture,
we will discuss about both sequences and series.
Learning Objectives
At the end of this lecture, you will be able to:

Define the terms sequence and series;

Find the general form of any given simple series;

Differentiate between an arithmetic progression and a geometric
progression;

Find missing terms in arithmetic progressions and geometric progression;

Find sums of terms in arithmetic progression and geometric progressions.
2.2 Sequences and Series
2.2.1 Sequences
Definition
Sequences are sets of numbers that show simple patterns.
Some of these sets are sets of natural numbers 1, 2, 3, …, the set of even numbers 2,
4, 6, …, the set of odd numbers 1, 3, 5, …. Given any of these patterns, it is easy to
determine the next number.
For example, the pattern 1, 4, 7, 10, 13 shows that the difference between
consecutive numbers is 3, hence the sixth number is 17.
22 Basic Mathematics for Diploma in Primary Teacher Education
Every member of a particular pattern is called a term. Each term has a particular
position value. Using such naming, a term corresponding to the last position is
called the nth term.
A sequence is therefore a set of terms written in a definite order with a defined
rule by which the terms are obtained.
The order in which terms appear in a certain pattern is important. For example,
10, 20, 30, 40, … is a completely different pattern from 40, 30, 20, 10, …
The pattern can be discovered by examining three or more consecutive terms to
see how they are related.
Example:
Find the fifth term of the sequence 5, 10, 15, 20, 25, …
Solution:
The pattern governing the given terms is such that the difference between
consecutive terms is 5.
Adding 5 to the fifth term therefore generates the fifth term.
25 + 5 = 30
The sequence is therefore
5, 10, 15, 20, 25, 30, . . .
2.2.2 Series
When the terms of a sequence are considered as a sum, the resulting expression is
known as a series progression.
Examples of a series are:
1.
1, 2, 3, 4, … , 50
2.
2, 4, 6, 8, 10, …
3.
3, 6, 9, 12, …, 99
4.
5, 10, 15, 20, …
If a series ends after a finite number of terms, like 1 and 3 in the example above, it
is said to be a finite series. However, a series is infinite if it does not have an end,
like series 2 and 4 in the above example.
Consider a series with n terms. The sum of all the n terms of the series is denoted
by Sn
Thus, for a series 2, 4, 6, 8, 10, …+ n, the sum is;
Basic Mathematics for Diploma in Primary Teacher Education 23
Sn = 2 + 4 + 6 + 8 + 10 + …+ n
The general form of a simple series can be obtained by studying the terms.
For example,
for the sequence 2, 4, 6, 8, 10, … the general term is 2n
for the sequence 5, 10, 15, 20, … the general term is 5n
for the sequence 1, 3, 5, 7, …. The general term is 2n + 1.
2.3 Arithmetic Progression
An arithmetic progression is a sequence in which each term after the first is
obtained by adding a fixed number to the preceding term. The fixed number,
which is the difference between any two consecutive terms, is called the common
difference, denoted by d.
For example, 2, 4, 6, 8, 10, … is an arithmetic series with a common difference of 2.
If n is the number of terms of an arithmetic progression, then the nth terms is
denoted by ‘An’.
Therefore, for an arithmetic progression beginning with A1 with a common
difference d, the second term, A2 = A1 + d
A3 = A2 + d
the third term,
A4 = A3 + d, and so on.
the fourth term,
We will then have a general form,
An+1 = An + d
for all natural numbers n
Considering that A2 = A1 + d
A3 = A2 + d
(but A2 = A1 + d)
Substituting it we get
= (A1 + d) + d
= A1 + 2d
Likewise,
A4 = A3 + d
= (A1 + 2d) + d
= A1 + 3d
We will then end up with a general form,
An = A1 +(n – 1)d
(but A3 = A1 + 2d)
24 Basic Mathematics for Diploma in Primary Teacher Education
Example:
The first term of an arithmetic progression is 1 and the common difference is 2.
Find, the 23rd term.
Solution:
Given: A1 = 1
d=2
A23 = ?
From the general formula,
An = A1 +(n – 1)d
We substitute the given values
A23 = 1 + (23 – 1)2
= 1 + (22)2
= 1 + 44
= 45
Example:
The first term of an arithmetic progression is 6 and the common difference is 5.
Find;
(i) The 3rd term
(ii) The nth term
Solution:
Given: A1 = 6
d=5
A3 = ? An = ?
From the general formula,
An = A1 +(n – 1)d
We substitute the given values
A3 = 6 + (3 – 1)5
= 6 + (2)5
= 6 + 10
= 16
An = 6 + (n – 1)5
Basic Mathematics for Diploma in Primary Teacher Education 25
= 6 +5n – 5
= 5n + 1
Example:
The 2nd term of an arithmetic progression is 15 and the 5th term is 21. Find;
(i) The common difference
(ii) The first term
Solution:
Given: A2 = 15
A5 = 21
d=?
A23 = ?
Substituting the given values in the general formula,
A2 = A1 + d ……… 15 = A1 + d
…(1)
A5 = A1 + 4d …….. 21 = A1 + 4d
…(2)
Substituting equation (1) in equation (2), we get
21 = (A1 + d) + 3d
(but A1 + d = 15)
21 = 15 + 3d
21 – 15 = 3d
6 = 3d
d=2
Substitute d in one of the equations to get A1
A2 = A1 + d
15 = A1 + 2
A1 = 15 – 2
A1 = 13
Sum of the First n Terms of an Arithmetic Progression
For any arithmetic progression with the first term A1, the last term An and the
common difference d, the sum Sn can be obtained by the formula:
Sn =
n
(A1 + An)
2
26 Basic Mathematics for Diploma in Primary Teacher Education
The formula can be further extended by replacing An by A1 +(n – 1)d
We thus get, Sn =
n
[2A1 + (n – 1)d]
2
The sum of the first n terms of an arithmetic progression is obtained by the
general formula:
Sn =
n
[2A1 + (n – 1)d]
2
Example:
Find the sum of the first 16 terms of the following series:
3, 10, 17, ….
Solution:
Given: A1 = 3
n = 16
d=7
From the general formula,
Sn =
n
[2A1 + (n – 1)d]
2
S16 =
16
[(2 × 3)+ (16 – 1)7]
2
= 8[6 + (15 × 7)]
= 8[6 + 105]
= 8 × 111
= 888
2.4 Geometric Progression
A geometric progression is a sequence in which each term after the first is obtained
by multiplying the preceding term by a fixed number called the common ratio,
denoted by r.
For example, 3, 9, 27, 81, … is a geometric progression with a common ratio of 3.
For a geometric progression whose first term is G1 and common ratio r, the
preceding terms will by as follows:
G2 = rG1
Basic Mathematics for Diploma in Primary Teacher Education 27
G3 = rG2 = r(rG1) = r2(G1)
G4 = rG3 = r(G3) = r(r2G1) = r3G1
Following this trend, we will end up with a general form Gn = G1rn–1
For a geometric progression with a common ratio of r, and the first term G1, the
nth term, Gn, is obtained by the expression Gn = G1rn–1
Example:
The 4th term of a geometric progression is 9 and the 6th term is 81. Find:
(a) The common ratio
(b) The 1st term
Solution:
Given: G4 = 9
G6 = 81
From the general formula,
G4 = G 1 r 3
…(1)
G 6 = G1 r 5
… (2)
Substitute the given values in the two equations
9 = G1r 3
… (1)
81= G1r5
… (2)
Substitute equation (1) into equation (2)
81 = (G1r3)r2
= 9 × r2
from which we get r = 3
Substitute r in equation (1) to get the value of G1
9 = G1 × (33)
9 = G1 × 27
G1 =
9
27
G1 =
1
3
The Sum of the First n Terms of a Geometric Progression
28 Basic Mathematics for Diploma in Primary Teacher Education
The sum of the first n terms of a geometric progression is obtained by adding the
terms,
Sn = G1 + G2 + G3 + … + Gn
But since the common ratio is r, then we have,
Sn = G1 + G1r + G1r2 + G1r3… + G1rn–1
… (1)
Multiplying eqn. (1) by r we have,
rSn = G1r + G1r2 + G1r3 + G1r4… + G1rn
… (2)
Subtracting equation (1) from equation (2) we have,
rSn – Sn = G1rn – G1
Sn(r – 1) = G1(rn – 1)
Sn =
G1 r n  1
r1
The formula holds for r ≠ 1
When r < 1
then Sn =
G1 r n  1
1 r
When r = 1 then Sn = nG1
The sum of the first n terms of a geometric progression is given by:
n
Sn = G1 r  1 for r ≠ 1
r1
n
Sn = G1 r  1 ………… for r < 1
1 r
Sn = nG1….….………… for r = 1
Example:
Find the sum of the first 6 terms of the following geometric progression
–81 –27 –9 ….
Solution:
Given:
G1 = –81,
n=6
First we evaluate the value of r, this is obtained by dividing two consecutive terms
Basic Mathematics for Diploma in Primary Teacher Education 29
r=
G2  27 1

G1  81 3
Since r < 1 we apply the formula
Sn =
G1 r n  1
1 r
Substituting the given values,
S6 =
6

1


81 
1 

3

1
1
3
2
1

÷
= 
81 
1

729
3
1
= –121 .
3
Activity
1.
A plant grows 1.67cm in its first week. Each week it grows by
4% more than it did the week before. By how much does it
grow in nine weeks, including the first week?
2.
After how many complete years will a starting capital of 5000
first exceed 10,000 if it grows at 6% per annum?
Summary
In this lecture, we have discussed some basic concepts such as estimations and
approximations and learnt how to correctly round off figures to the desired
number of decimal places. Rounding off of a number means to find another
number closer to the given number with the desired number of decimal places.
We have also discussed how to use Scientific Notations in writing numbers,
especially those involving many zeros before and after the decimal point. We
discussed the concept significant figures of number as the accurate digits, apart from
zeros, needed to locate the decimal point. Significant figures of a number are the
accurate digits, apart from zeros needed to locate the decimal point.
30 Basic Mathematics for Diploma in Primary Teacher Education
We also discussed the concepts Ratios, Proportions and Percents, where we learned
a
or a : b; and a
b
a c
proportion as a statement that two ratios are equal. A proportion  has four
b d
that a ratio is a comparison of two quantities a and b, written as
terms, a, b, c and d; where the terms a and d are called the extremes, and b and c
are called the means. A proportion
equal to the product of the means.
a c
is true if the product of the extremes is

b d
Two quantities are said to be directly proportional to each other if when one is
increased by a certain factor, the other also increases by the same factor. Two
quantities are said to be inversely proportional to each other if when one is
increased by a certain factor, the other decreases by the same factor.
In this lecture, we discussed about sequences and series, looked at different types of
sequences and series and used the general forms to find missing numbers. A
sequence is a set of terms written in a definite order with a defined rule by which
the terms are obtained.
We have also differentiated between an arithmetic progression and a geometric
progression. An arithmetic progression is a sequence in which each term after the
first is obtained by adding a fixed number to the preceding term. The fixed
number, which is the difference between any two consecutive terms, is called the
common difference, denoted by d. The nth term of an arithmetic progression,
denoted by An, is given by a general expression An = A1 +(n – 1)d; where A1 is the
first term and d is the common difference.
A geometric progression is a sequence in which each term after the first is obtained
by multiplying the preceding term by a fixed number called the common ratio,
denoted by r. The nth term of a geometric progression, denoted by Gn, is obtained
by
the
expression
n–1
Gn = G1r where r is a common ratio and G1 is the first term.
Exercise
1.
2.
(i)
The nth term of the sequence is given by 2n+1. Write down the
first five terms.
(ii)
The nth term of a certain sequence is 2n–1. Find the sum of the first
three terms.
If the general term of a certain sequence is 3n +4, write down the first four
terms and find their sum.
Basic Mathematics for Diploma in Primary Teacher Education 31
22n 2
. Find the first five terms.
2
3.
Given the general form of a series as
4.
The fourth and the fifth terms of an arithmetic progression are 47 and 52
respectively. Find;
(i)
the common difference
(ii)
the first term
(iii)
the 30th term
5.
The third term of an arithmetic progression is 9 and the common
difference is 2. Find (a) The 1st term (b) The 200th term (c) The nth term.
6.
Find the nth term of an arithmetic progression whose first term is x + 2
and the common difference is 3.
7.
Find the sum of the eighth and the tenth terms of the geometric
progression 2, 6, 18, …
8.
If the sum of n terms of a geometric progression with the 1st term 1 and the
31
common ratio of is
find the number of terms
16
9.
(i)
If a geometric progression is given by Gn = 2n, find the sum of the
first 5 terms.
(ii)
In a geometric progression, Sn = 1248, r = 5 and n = 4. Find G1.
The second, fifth and seventh term of an arithmetic progression form three
consecutive terms of a geometric progression. Find;
(i)
the common ration of the geometric progression
(ii)
the sum of the first five terms of the geometric progression.
10.
The fourth, fifth and sixth terms of a series are (2x + 10), (4x – 4) and (8x
+ 40) respectively. Calculate the value of x and the sum of the first ten
terms if the series is an arithmetic progression.
11.
If it is known that x:y = 5:1 find the value of
12.
If a = 2.432 × 104, b = 7.42 × 10–2 and c = 3.0324 × 10–2, find the value of
ab
.
R in standard form correct to two decimal places given that R =
c
13.
Given that y varies inversely as the square root of x, and y = 2 when x =
25; find the value of x when y = 4.
x y
.
3x  4 y
32 Basic Mathematics for Diploma in Primary Teacher Education
14.
If 18000 Shs was divided among four people in the ratio 1:2:3:3. Find out
the amount of each of the portions.
15.
Find the first five terms of a series with a general formula
16.
A man weighed 80kg. He then lost 5kg in three months. What percent did
he lose?
4n2
.
3
References
Wright, F. D. (1987), Arithmetic for College Students, 5th Ed., Heath and Company,
USA.
Dressler, I. (1981), Preliminary Mathematics, AMSCO School Publications Inc.
USA.
Setek, W. M. Jr. (1989), Fundamentals of Mathematics, 5th Ed., MacMillan
Publishing Company, New York.
Nelson, B. (2001), Mathematics for Elementary Teachers: A Conceptual Approach,
MacGraw-Hill Companies Inc., New York.
Musser, G. L. and Burger, W. F. (1988), Mathematics for Elementary Teachers: A
Contemporary Approach, MacMillan Publishing Company, New York.
Tanzania Institute of Education (2000), Secondary Basic Mathematics Book Three,
TIE Dar es Salaam.
Basic Mathematics for Diploma in Primary Teacher Education 33
34 Basic Mathematics for Diploma in Primary Teacher Education
PART II
Algebra
Algebra is that branch of mathematics in which symbols, usually letters of the
alphabet, are used to represent numbers or members of a specified set or
certain quantities. It also involves studying their rules of functioning using the
elementary operations of addition, subtraction, multiplication and division.
Algebra is required in many fields, ranging from business, natural sciences to
social sciences. For example, equations, graphs and functions occur again and
again in many different areas such as predictions of population growth,
investigation of costs versus benefit of projects etc.
This section is divided into two lectures. In the first lecture, Algebraic
Expression we will discuss various basic concepts in algebra such as
translating word problems into mathematical equations and solving equations
involving both numbers and variables. We will also discuss the important
concepts of linear equations and inequalities, simplification and factorization
of expressions, graphing of equations and solving quadratic equations and
simultaneous equations by using different methods.
Basic Mathematics for Diploma in Primary Teacher Education 35
36 Basic Mathematics for Diploma in Primary Teacher Education
LECTURE 3
Algebraic Expressions
3.1 Introduction
As we have already seen, the operations in algebra are similar to those in
arithmetic, but in algebra, letters are used to denote numbers or a certain set of
numbers. In arithmetic we encounter an expression such as 2 + 3 = 5; whereas in
algebra we encounter expressions such as x + 2 = 5 and so on.
Learning Objectives
At the end of this lecture, you will be able to:

Construct simple algebraic expressions;

Simplify and factorize algebraic expressions;

Translate word problems into mathematical equations and solve them;

Find solution sets of linear equations and linear inequalities;

Solve simultaneous equations by using different methods.
3.2 Simplifying and Factorizing Algebraic
Expressions
Simplifying Expressions
When simplifying algebraic expressions, it will be important to remember the
following:

When using letters to stand for numbers, omit the multiplication sign.
For example x + x + x = 3x (not 3 × x). The number 3 is called the
coefficient of x
Basic Mathematics for Diploma in Primary Teacher Education 37

Parts of an expression connected by a plus or minus sign are called the
terms. Terms containing the same algebraic letter are called like terms. For
example, in the expression 6y + 5x – 4 + 4y + 10 – 3x; 6y and 4y are like
terms, 5x and 3x are like terms; 4 and 10 are like terms.

Like terms are collected or written next to each other. The coefficients of
the like terms are then added or subtracted. For example, 5x + 4 + 3x + 7
= 8x + 11. Note that we cannot combine 5x and 4 because 5x is a term
associated with a variable x whereas 4 is a constant. They are not like
terms.
Simple Factorization of Algebraic Expressions
When factorizing algebraic expressions, it is important to remember that:

The order of the factors in a product does not matter, so xy is the same as
yx

We can write a common factor outside the bracket
For example, 9x + 9y = 9(x + y)

Algebraic fractions can be simplified if there is a common factor to the
numerator and the denominator.
For example,
1
3 2
1
1
xy +
xy can be simplified as
xy(3x + 1) since
and xy are
4
4
4
4
common in both two terms.
Example:
Write down an expression for the perimeter of a triangle whose sides are (x +
12)cm,
(2x – 7)cm and (10 + 4x)cm.
Solution:
We know that the perimeter of a triangle is obtained by adding the values of its
sides. In this case we will add;

x  12 2x  7 10  4x

Combining the like terms we get
x + 2x + 4x + 12 – 7 + 10
= 7x + 15
The expression for the perimeter is (7x + 15)cm.
38 Basic Mathematics for Diploma in Primary Teacher Education
3.3 Translating Word Problems into
Mathematical Expressions
In many applications of algebra, we are faced with problems that are stated in
words. Such problems are called word problems. It is possible to solve some
simple word problems by applying arithmetic as we have seen in the previous
lecture, but most word problems require the use of algebra in order to find the
solution in a systematic manner. In such cases, we can introduce variables for the
unknowns and transform the given information into an equation that we can
solve.
Equations are mathematical statements which compare two expressions using the
equality symbol ‘=’
Variables are letters used in equations. They are symbols used to represent
unknown numbers.
The following are some of the translations of word phrases into mathematical
phrases:
Word Phrase
1.
2.
3.
Addition:
The sum of x and y
x plus y
A number added to 5
A number increased by 5
5 more than a number
x+y
x+y
5+x
x+5
x+5
Subtraction
A number decreased by 5
5 less than a number
The difference between x and y
x minus y
x–5
n–5
x–y
x–y
Multiplication
The product of a and b
A number multiplied by 5
3
of a number ‘a’
4
30 per cent of a number ‘a’
4.
Mathematical Phrase
Division
The ratio of a and b
The quotient of a and b
a divided by b
a × b = ab
a × 5 = 5a
3
3
×a= a
4
4
30
×a
100
a
b
a
b
a
b
Basic Mathematics for Diploma in Primary Teacher Education 39
In solving word problems, the following strategy may be useful:

Read the problem carefully, getting well in mind what are the ‘knowns’
and what are the ‘unknowns’.

Draw a picture if possible.

Introduce a variable to represent the unknown.

If there are several unknowns, use the information in the problem to write
each unknown in terms of one variable only.

From the information given, write down the equation(s) that relates the
known and unknown numbers.

Solve the equation.

Check to see if the solution you have obtained gives a solution to the
original problem.
Example:
The sum of the present ages of Harriet and Sam is 50. In 5 years Harriet will be
twice as old as Sam. Find their present ages.
Solution:
Let x be the present age of Harriet
Since the sum of Harriet’s and Sam’s age is 50, then Sam’s age will be 50 – x
In 5 years time Harriet’s age will be
and Sam’s age will be
x+5
(50 – x) + 5 = 55 – x
From the information given, in 5 years time Harriet will be twice as old as Sam,
Then x + 5 = 2(55 – x)
x + 5 = 110 – 2x
x + 2x = 110 – 5
3x = 105
x = 35
Harriet’s present age is 35
Sam’s present age is 15
Check: 35 + 15 = 50.
Example:
The length of a rectangle is 3cm less than four times its width. Find the dimensions
of this rectangle if its perimeter is 64cm.
40 Basic Mathematics for Diploma in Primary Teacher Education
Solution:
Let x be the width of the rectangle. Then its length will be 4x – 3
From the description, we get the following figure:
(4x –3)cm
x cm
x cm
(4x –3)cm
The perimeter of the rectangle is given by [x + x + (4x – 3) + (4x – 3)]
Therefore, [x + x + (4x – 3) + (4x – 3)] = 64
[2x + 2(4x –3)] = 64
2x + 8x – 6 = 64
10x – 6 = 64
10x = 64 + 6
10x = 70
x = 7cm
Hence the width of the rectangle is 7cm
And its length will be 4(7) – 3 = 25cm
Check: 7cm + 25cm + 7cm + 25cm = 64cm
Example:
A teacher drove his car from his home to school at an average speed of 50Km per
hour. On the return trip, he drove at an average speed of 60 Km per hour, and
made the trip 1 hour less. Find the distance between his home and the school.
Solution:
Let the distance between his home and the school be x. We know that from the
general formula,
Time =
Distance
Speed
Thus, time taken from home to school at 50 Km per hour =
x
hours
50
Basic Mathematics for Diploma in Primary Teacher Education 41
And time taken from school to home at 60 Km per hour =
x
hours
60
Since he spent 1 hour less from school to home, we have the equation
Multiply each term by 300
x
x
=
+1
60
50
x
x
(300) = 300(
+ 1)
50
60
6x = 5x + 300
6x – 5x = 300
x = 300
Check: 50Km per hour from home to school will take
60Km per hour from school to home will take
300
hours = 6 hours
50
300
hours = 5 hours
60
The difference is 1 hour
You can always check a factorization by expanding the expression.
3.4 Linear Equations
Definition
Generally, equations are statements of the form A = B, where A is called the lefthand side of the equation and B is called the right-hand side of the equation. When
we put an algebraic expression equal to another expression or a number we form
an equation.
An algebraic equation in the variable x is a statement that expresses the relation of
equality between two algebraic expressions in x.
For example, the following statements are algebraic equations;

x2 – 4 = 0

3x + 4 = 7

(x + 1)2 = x2 + 2x +1

6x2 + 12x + 4 = 8

8x = 2x + 12
42 Basic Mathematics for Diploma in Primary Teacher Education
We note that the equation has two sides separated by an equal sign, ‘=’
Solution Set of an Equation
When a variable in an equation is replaced by a specific numeral, the resulting
statement may be either true or false.
For example, given an equation x2 – 4 = 0, if we replace x by 3 we get 9 – 4 = 0
which is false. However, if we replace x by 2 we get 4 – 4 = 0 which is true. The
equation is also true when x is –2. Thus (2, –2) is the solution set of the equation.
Two equations are said to be equivalent if they have exactly the same solution sets.
A solution to an equation in one variable is a real number which when substituted
for the variable results in a true statement.
Definition
A linear equation is an equation that can be expressed in the form ax = b where x
is the unknown and ‘a’ and ‘b’ are known. Note that, such equations do not
contain x2 or any higher power of x. Such equations are called linear equations
because their graphs are straight lines.
Solving Linear Equations in one Variable
The approach to solving these equations involves application of a series of algebraic
operations to both sides of the given equation in order to produce equivalent
equations that are easier to solve.
Example:
Solve the following linear equations
5x – 1 = 3x + 11
Solution:
Collect the like terms 5x – 1 = 3x + 11
5x – 3x = 11 + 1
2x = 12
x=6
Linear Equations in two Variables
Any equation of the form ax + by = c, where a, b and c are real numbers is called
a linear equation in two variables. These equations take as their solution, ordered
pairs. To find an ordered pair that satisfies a given linear equation we apply the
following procedure:
Basic Mathematics for Diploma in Primary Teacher Education 43

Select a number for either x or y

Substitute the number chosen into the equation for the remaining variable.
Example:
Find three solutions for the equation 2x + y = –6.
Solution:
Let us select any number for x. Suppose we choose to let x = 0; 2; 5
We now substitute those numbers for x into the given equation and then find the
corresponding number of y.
If x = 0, then
If x = 2, then
If x = 5, then
2x + y = –6
2x + y = –6
2x + y = –6
2(0) + y = –6
2(2) + y = –6
2(5) + y = –6
0 + y = –6
4 + y = –6
10 + y = –6
y = –6
y = –10
y = –16
Thus the ordered pairs are (0, –6), (2, –10) and (5, –16).
Applications involving Linear Equations
Now let us put our knowledge to some practical applications and see how we can
solve problems.
For example, if you invest P shillings in a savings account that pays simple interest
at an annual rate equal to r, then after t years you will have amount A of money
equal to;
A = P (1 + rt)
This equation gives us a relationship between the amount A, the principal P, the
interest r and the time t. Given any three of these quantities, we can solve this
equation to find the fourth quantity. (Please note that the interest rate r must be
expressed as a decimal in this formula. For example, an interest rate of 12% will be
r = 0.12.
Suppose for example we want to express the interest rate r, in terms of A, P and t,
we will have to change the equations to be in terms of r i.e. making r the subject of
our equation.
A = P (1 + rt)
Divide by P both sides we get,
44 Basic Mathematics for Diploma in Primary Teacher Education
A
P
= 1  rt 
A
1 rt

P
AP
r
Pt
r=
AP
Pt
Graphing Linear Equations
Generally, graphs of equations, inequalities, functions and relations are geometrical
representation of algebraic ideas. Studying the graphs gives a better understanding
of algebra. On the other hand, algebra gives further insight into geometric concepts
and relations. Arithmetic and Algebra are used to describe geometrical figures and
to derive some of their properties. This is dealt with in another branch of
mathematics called Analytical Geometry, which will not be covered in this unit.
Solutions of linear and quadratic equations can be presented graphically by the use
of Cartesian coordinate system, named in honour of the great French
mathematician and philosopher, Rene Descartes (1596–1650) who developed the
idea. The Cartesian coordinate consists of two lines, one vertical and one
horizontal, that cross at their respective zero points. The vertical line is called the
y-axis and the horizontal line is called the x-axis, and their point of intersection is
called the origin.
y-axis
Origin
x-axis
In this lecture, we will deal with graphs of linear equations only and see how to
solve related problems.
Example:
Basic Mathematics for Diploma in Primary Teacher Education 45
Draw the graph of the following equation:
y = 2x + 4
Note that this is a linear equation, i.e. when plotted it will give a straight line.
The first step is to prepare the table of values as follows;
Select some values of x and calculate the corresponding values of y
y = 2(–6) + 4

2  6 4
 12  4
 8
Likewise for the rest of the x values
x
–6
–5
–4
–3
–2
–1
0
1
2
y
–8
–6
–4
–2
0
6
4
6
8
To get the value of y when x = –6, substitute –6 into the equation y = 2x + 4.
Then plot the points (–6, –8); (–5, –6); (–4, –4); (–3, –2); (–2, 0); (–1, 6); (0, 4); (1, 6)
and (2, 8) in the Cartesian coordinates as shown below.
46 Basic Mathematics for Diploma in Primary Teacher Education
8
7
6
5
4
3
2
1
0
-8
-7
-6
-5
-4
-3
-2
-1 -1 0
1
2
3
-2
-3
-4
-5
-6
-7
-8
3.5 Linear Inequalities
An inequality is an expression involving >, <, ≤ or ≥
For any two real numbers a and b, we say a is greater than b denoted by a > b if a
– b is positive. Further, we say a is less than b, a < b, if a – b is negative. Similarly,
a ≤ b means ‘a < b or a = b’ and a ≥ b means ‘a > b or a = b’
For each pair of real numbers a and b exactly one of the following relations holds:
a > b, a < b or a = b
Just as in equations, an inequality in the variable x may be true for a certain values
of x and false for other values of x.
Example:
For an inequality
2x + 5 < x + 8
Suppose we put x = 1 we get;
2+5<1+8
7<9
which is true
But if we put x = 6, it will become;
12 + 5 < 6 + 8
17 < 14
which is false
Basic Mathematics for Diploma in Primary Teacher Education 47
The value of x that holds the inequality true is called the solution set of the
inequality.
Double Inequalities
To denote that a number is between two other numbers, we combine the two
separate inequalities to form a double inequality.
For example, since 5 is between 2 and 7, we can combine the two inequalities
2 < 5 and 5 < 7 to get
2<5<7
Solutions of Linear Inequalities
When solving inequalities, we use the same rules as those for equations with only
two exceptions:

When multiplying or dividing an inequality by a negative number, we
must reverse the sign of the inequality.

When interchanging the left hand side and the right hand side, the
inequality sign must be reversed. Otherwise, remember to do exactly the
same to both sides. There will be several solution sets for one inequality.
Examples:
Solve the following inequalities.
1.
2x + 5 < 11
Solution:
2x + 5 < 11
Subtract 5 from both sides
2x + 5 – 5 < 11 – 5
2x < 6
Divide both sides by 2
x<3
2.
3x – 1 > x + 7
Solution:
3x – 1 > x + 7
Add 1 to both sides
48 Basic Mathematics for Diploma in Primary Teacher Education
3x – 1 + 1 > x + 7 + 1
3x > x + 8
Subtract x from both sides to get
3x – x > x – x + 8
Divide both sides by 2
2x > 8
x>4
3.
–11 < 5x – 1 < 14
Solution:
–11 < 5x – 1 < 14
Add 1 to all parts of the inequality
–11 + 1 < 5x – 1 + 1 < 14 + 1
–10 < 5x < 15
Divide by 5
–2 < x < 3
Note that this is a double inequality but it is solved using the same rules and
operations.
Using Linear Inequalities to Solve Problems
Example:
A student needs at least a total of 250 marks in her five examinations to be able to
move to the second year. There are 5 subjects. What is the average mark that she
should score in each subject?
Solution:
Let the average mark for each subject be x
Then 5x ≥ 250
x ≥ 50.
3.6 Simultaneous Equations
Basic Mathematics for Diploma in Primary Teacher Education 49
Given two linear equations in two variables, solving them simultaneously is to find
a solution set that satisfies both of them. This may be done by the method of
elimination.
In this method, we multiply one or each of the two equations by a suitable number
or numbers, which enables us to write two equivalent equations where the
coefficients of one of the unknowns are numerically equal. Then by addition, we
are able to eliminate the unknown x and find y, or vice versa.
When solving simultaneous equations:

Consider each term of the equations carefully.

Find suitable number(s) which when multiplied to one or both of the
equations will make the coefficients of one of the unknowns numerically
equal, then eliminate one of the unknowns and find the other.

Remember when multiplying or dividing an equation by a certain factor,
each term in the equation must be multiplied or divided by the same factor.

Substitute the values you have obtained in each equation to confirm your
solution.
Examples:
Solve the following systems of simultaneous equations by elimination:
1.
y 1
2x 

3x  3y  6
Solution:
Let 2x – y = 1
… (i)
3x + 3y = 6
… (ii)
Multiplying equation (i) by 3 and equation (ii) by (2), we get
6x – 3y = 3
… (iii)
6x + 6y = 12
… (iv)
Subtract equation (i) from equation (ii)
6x + 6y = 12
6x – 3y = 3
0 + 9y = 9
9y = 9
y =1
50 Basic Mathematics for Diploma in Primary Teacher Education
Substitute the value of y in equation (i) to obtain the value of x
2x – y = 1
2x – 1 = 1
2x = 1+1
2x = 2
x =1
Thus, the solution to this system of simultaneous equation is x = 1 and y = 1.
Check: Substitute these values into the two equations to confirm your answer
2(1) – 1 = 1
2 – 1 = 1√
3(1) +3(1) = 6
3 + 3 = 6√
x  y  1

3y 3
x 
2.
Solution:
Let
x+y=1
…(i)
x – 3y = 3
…(ii)
Subtract equation (ii) from equation (i)
x – 3y = 3
x+y=1
0 – 4y = 2
– 4y = 2
y=–
1
2
Substitute the value of y in equation (i) to get the value of x
x+y=1
x–
1
=1
2
x=1
1
2
Basic Mathematics for Diploma in Primary Teacher Education 51
1
1
The solution for this system of simultaneous equations is x = 1 , y = –
2
2
Check: Substitute these values into the two equations to confirm your answer.
1
1 1
– =1√
2 2
1
1
1
3 3 6
– 3(– ) =   3 √
2 2 2
2
2
x  2 y  4

2y 6
3x 
3.
Solution:
Add the two equations
x + 2y = 4
… (i)
3x – 2y = 6
… (ii)
4x + 0 = 10
4x = 10
x= 2
1
2
Substitute the value of x in equation (i)
2
1
+ 2y = 4
2
2y = 4 – 2
y=
1
2
3
4
1
3
The solution is x = 2 , y =
2
4
Check: Substitute these values into the two equations to confirm your answer
2
1
3
+ 2( ) = 4
2
4
5 3
8
 =  4√
2 2
2
52 Basic Mathematics for Diploma in Primary Teacher Education
1
3
3( 2 ) – 2( ) = 6
2
4
15 3
12
 =  6√
2 2
2
Applications of Simultaneous Equations
Some problems require the application of the concept of simultaneous equations to
solve them.
Example:
Ali paid 750 shillings for 7oranges and 4 mangoes. Juma went to the same fruit
vendor and bought 12 oranges and 8 mangoes and paid 1400 shillings. What is the
price for a mango and for an orange?
Solution:
Let
Oranges be x
Mangoes be y
From the information, we get the following simultaneous equations;
7x  4y  750

12x  8y  1400
…(i)
…(ii)
Multiply equation (i) by –2 to get the following equations;
14x 8y 1500


12x  8y  1400
Add the two equations;
–14x – 8y = –1500
12x + 8y = 1400
–2x + 0 = –100
x = 50
Substitute x in equation (i) to solve for y;
7(50) + 4y = 750
350 + 4y = 750
4y = 400
y = 100
…(iii)
…(iv)
Basic Mathematics for Diploma in Primary Teacher Education 53
Check:
7(50) + 4(100) = 750
350 + 400 = 750√
12(50) + 8(100) = 1400
600 + 800 = 1400√
We can apply the concept of simultaneous equations to determine the point of
intersection of two lines without having to actually draw them. This can be done
by solving the two linear equations simultaneously to get a solution which satisfy
them both. This will be their point of intersection.
Example:
Determine where lines y = 2x + 1 and y = –x + 4 cross each other.
Solution:
Let
y = 2x + 1
…(i)
y = –x + 4
...(ii)
Re-writing the two equations we get;
2x – y = –1
x+y=4
which is a system of simultaneous equations
y 1
2x 

x  y  4
We may solve this system of simultaneous equations by elimination.
Adding the two equations we get;
3x = 3
Therefore x = 1
Substituting the value of x to solve for y;
2(1) – y = –1
2 – y = –1
–y = –3
y=3
Solution set of the two equations is;
x = 1 and y = 3,
54 Basic Mathematics for Diploma in Primary Teacher Education
Thus the point of intersection of the two lines will be (1, 3)
This concept can also be used to determine the points of intersection between
graphs of quadratic equations and straight lines without drawing them.
This is the point of intersection of the two graphs which can be obtained
geometrically by drawing the two graphs in the same plane and then read the
points of intersection. But by applying algebra, we can find the points without
drawing the graphs by solving the two equations simultaneously. We will do this
in the next lecture after discussing quadratic equations.
Activity
1.
Describe a real-world situation in which you might use the
expression 50 + 2x.
2.
Explain what you need to write an algebraic expression.
Summary
When simplifying algebraic expressions, it will be important to remember a few
things. In many applications of algebra, we are faced with problems that are stated
in words. Such problems are called word problems.
Generally, equations are statements of the form A = B, where A is called the lefthand side of the equation and B is called the right-hand side of the equation. When
we put an algebraic expression equal to another expression or a number we form
an equation.
An algebraic equation in the variable x is a statement that expresses the relation of
equality between two algebraic expressions in x.
A linear equation is an equation that can be expressed in the form ax = b where x
is the unknown and ‘a’ and ‘b’ are known.
Generally, graphs of equations, inequalities, functions and relations are geometrical
representation of algebraic ideas.
An inequality is an expression involving >, <, ≤ or ≥
To denote that a number is between two other numbers, we combine the two
separate inequalities to form a double inequality.
When solving inequalities, we use the same rules as those for equations with only
two exceptions.
Basic Mathematics for Diploma in Primary Teacher Education 55
Given two linear equations in two variables, solving them simultaneously is to find
a solution set that satisfies both of them. This may be done by the method of
elimination.
Some problems require the application of the concept of simultaneous equations to
solve them.
Exercise
1.
Factorize (x + 2)2 – (x – 4)2
2.
Simplify the following expressions:
3.
4.
(i)
8x + (5x +4) – 7
(ii)
5(2x – 15) – 7x + 8
(iii)
3m2n + 2mn2
(iv)
10x – 6 + x + 4(2x – 2)
(v)
4a2b – 6ab2 + 2ab
Compute the following expressions:
(i)
3xy(2x – 5y + 7) when x = 3, y = 2
(ii)
5x(2x + 6) + 3y(5x – 8) when x = 4, y = 5
Factorize the following expressions:
(i)
9x3y2 – 3x2y3 + 6x2y2
(ii)
4a2b + 2 ab2
2
1
of one pile is equal to
of
3
4
the other. How many oranges are in each pile?
5.
44 oranges are divided into two piles so that
6.
Grace is 2 years older than Ton and Sam is half of Grace’s age. The sum of
their age is 23. Form an equation using the given information and use it to
find how old each one of them is.
7.
1
Mary spent 1 hours doing her homework. She spent 2n minutes doing
2
Mathematics, n minutes doing English and (n + 14) minutes doing
56 Basic Mathematics for Diploma in Primary Teacher Education
Kiswahili. Write down an equation for n and find the time she spent doing
each subject.
8.
Two sides of a square are (2x – 5)cm and (27 – 2x)cm. What is the value of
its area?
9.
Suppose that there are a total of 5 equally weighted papers in your annual
examination, and you have scored 68, 72, 44 and 59 in four of them. What
score must you get in the fifth paper in order to get an average of 60 in all
the five papers?
10.
Find three consecutive numbers whose sum is 39.
11.
A woman is 40 years old and her daughter is 8 years old. In how many
years will the woman be twice as old as her daughter.
12.
Solve the following equations:
(i)
(4x  5) (3x  1)
1
–
=
4
6
3
(ii)
2
1
(x + 2) – (x + 1) = 1
3
2
13.
Determine whether the given values are solutions to the following
equations:
14.
(i)
3x + 5 = 2x – 4x = –9
(ii)
x2 – 2 = x x = 1
(iii)
x2 – 4x – 5 = 0 x = 5, x = –1
(iv)
xx – x = 24 x = 0, x = 2.
Suppose that you deposited 100 000 Shs in an account 5 years ago, and that
it has grown to an amount of 120 000 Shs. What is the interest rate of this
account? Given that:
A = P (1 + rt) where A = final amount, P = deposited amount, t = time r
= interest rate.
15.
The total energy of an object is given by the formula
1
mv2 + mgh where m = mass of the object, v = its velocity, h = its
2
height above ground and g = acceleration due to gravity.
E=
Make h the subject of the formula.
Basic Mathematics for Diploma in Primary Teacher Education 57
1
1
16.
Make p the subject of the formula tp 2  q p  r 2 .
17.
Solve the following inequalities:
18.
19.
20.
(i)
4x + 5 > 2x – 1
(ii)
3x – 8 < 7x + 16
(iii)
–9 < 2x + 3 < 15
Write an inequality for each of the following information and use it to
solve the problem.
(i)
When 6 is added to three times a certain number, the result is less
than 63. What is the number?
(ii)
Multiplying a number by 4 and then adding 5 gives a greater result
than multiplying it by 5 then adding 4. What is the number?
Solve the following inequalities:
(i)
5 – (1 + x) > 3 – (2 + 3x)
(ii)
3(1 – 4x ) < 9 – 6x
Write down the solution of the inequality;
x 3
>2
x 2
21.
Solve the following simultaneous equations;
(i)
3y 19
4x 

6x  7y  17
(ii)
3x  4y  3

2x  5y  5  0
(iii)
7x  3y  12  0

5x  2y  2  0
22.
The sum of the ages of a father and his son is 52 years. Eight years ago the
father was exactly eight times as old as his son. How old are they now?
23.
Find the point where the lines y = 2x + 4 and y = 3x + 6 intersect.
24.
During a trip between two towns, one family had 2 adults and 3 children
and paid a total fare of 2100 shillings. Another family had 3 adults and 4
children and paid a total fare of 2900 shillings. How much was the fare for
one child and how much was it for one adult?
58 Basic Mathematics for Diploma in Primary Teacher Education
x y
 and 2x  3y  18 .
3 4
25.
Find the values of x and y given that
26.
I am thinking of two numbers. Twice the first number plus thrice the
second number gives twelve, and thrice the first number minus the second
number gives seven. What are the numbers?
References
Auvil, D. L. (1996), Algebra for College Students, McGraw-Hill Companies Inc.,
USA.
Blyth T. S. and Robertson E. F (1984), Algebra through Practice Cambridge
University Press, Great Britain.
Hirsch, L. & Goodman, A. (1987), Understanding Intermediate Algebra, West
Publishing Company, St. Paul, Minesota, USA.
Lial, M. L. & Miller, C. D. (1986), Algebra and Trigonometry, 4th Ed., Scott,
Foresman & Company, USA.
Sullivan, M. College (1990), Algebra with Review, Dellen Publishing Company, San
Francisco, California.
Tanzania Institute of Education, (2000), Secondary Basic Mathematics Book Two,
TIE, Dar es Salaam.
Willis, A. T. & Johnston, C. L. (1981), Intermediate Algebra, Wadsworth
Publishing Company, California.
Basic Mathematics for Diploma in Primary Teacher Education 59
60 Basic Mathematics for Diploma in Primary Teacher Education
LECTURE 4
Quadratic Equations
4.1 Introduction
A quadratic equation is an equation of the form ax2 + bx + c = 0, where a ≠ 0
a is called the coefficient of x2; b is called the coefficient of x and c is called the
constant term.
For example, equation 2x2 + 3x + 1 is a quadratic equation with a = 2, b = 3 and c
= 1.
In this lecture, we will discuss four methods of solving quadratic equations that are
factoring, completing the square, using the quadratic formula and graphical
method.
Learning Objectives
At the end of this lecture, you will be able to:

Explain the meaning of a quadratic equation;

Solve quadratic equations by using different methods;

Apply quadratic equations to solve different types of practical problems.
4.2 Solving Quadratic Equations by Factoring
This method relies on the fact that if the product of two real numbers is 0, then
one or both of them must be 0.
Examples:
Solve the following equations by the method of factoring
1.
6x2 – 3x – 3 = 0
Solution:
The first step is to factor the left hand side of the equation as follows:
6x2 – 3x – 3 = 3(2x +1)(x – 1) = 0
Basic Mathematics for Diploma in Primary Teacher Education 61
3(2x +1)(x – 1) = 0
Dividing both sides by 3 we get;
(2x +1)(x – 1) = 0
For a real number to satisfy this equation, it must make either
2x – 1 = 0 or (x – 1) = 0
2x – 1 = 0
2x = 1
x=
x–1=0
x – 1 + 1= 1
1
2
x=1
So the solution of the original equation is x =
2.
1
or x = 1
2
(x + 6)(x + 2) = –4
Solution:
In order to use the method of factoring to solve a quadratic equation, one side of
the equation must be 0. So the first step in this case will be to make the equation in
the form ax2 + bx + c = 0
(x + 6)(x + 2) = –4
x2 + 2x + 6x + 12= –4
x2 + 8x + 12 = –4
x2 + 8x + 12 + 4 = 0
x2 + 8x + 16 = 0
After expressing the equation in the quadratic form, we move ahead to factorize it:
x2 + 8x + 16 = 0
x2 + 4x + 4x + 16 = 0
(x + 4)(x + 4) = 0
x+4=0
or x + 4 =0
So the solution of the original equation is x = –4
(Note that in this case the solution is one real number)
62 Basic Mathematics for Diploma in Primary Teacher Education
4.3 Solving
Quadratic
Completing the Square
Equations
by
This method involves the concept of a perfect square. A quadratic equation is said
to be a perfect square if it is a square of a linear equation.
For example, x2 + 8x + 16 = 0 in example 1 above is a perfect square since it can
be written as (x + 4)2
Now, if a is a constant, then the equation x2 + bx can be made into a perfect
2
2
2
b
b
b



x
square by adding the quantity  since x2 + bx +  = 



2
2
2
2
5
5


For example to make x + 5x +  which is equal to 
x


2
2
2
2
b

The process of making x + bx into a perfect square by adding the quantity 

2
is called completing the square.
2
Note that this process applies only to quadratic equations of the form x2 + bx,
where the coefficient of x2 is 1 and the constant term c is 0.
Examples:
Solve the following equations by completing the square
1.
x2 – 6x + 2= 0
Solution:
We first write the equation in the form x2 + bx, it becomes
x2 – 6x = –2
To make the left side of the equation a perfect square, we add the square of half the
coefficient of x to both sides.
2
6

In this case, the coefficient of x, i.e. b = –6, so we add  = (–3)2 = 9

2
Adding 9 to both sides we have,
x2 – 6x + 9 = –2 + 9
(but x2 – 6x + 9 is the same as (x – 3)2)
(x – 3)2 = 7
The equation is now in the form (x + d) = k. Solving by using square root
property we get,
Basic Mathematics for Diploma in Primary Teacher Education 63
x–3= ±
x=3+
2.
7
7 or × = 3 –
7
x2 + 5x + 6 = 0
Solution:
Writing the equation in the form x2 + bx we have,
x2 + 5x = –6
2
5
25


In this case, b = 5, so we add  =  to both sides of the equation


2
4
25
25


x2 + 5x +  = –6 + 


4
4
25

(but x2 + 5x +  =

4
2
5

x
)


2
2
5
 24  25

x
=


4
2
2
5
1

x
=


2
4
x
x=
5
=
2
1
4
5 1
5 1
 or x =

2 2
2 2
x = –2 or x = –3
3.
x2 – 6x – 16 = 0
Solution:
x2 – 6x = 16
2
6

In this case b = –6, so we add  = 9 to both sides of the equation:

2
x2 – 6x + 9= 16 + 9
x2 – 6x + 9 = 25
(x – 3)2 = 25
x – 3 =  25
x – 3 = ±5
(but x2 – 6x + 9 = (x – 3)2)
64 Basic Mathematics for Diploma in Primary Teacher Education
x = 3 + 5 or x = 3 – 5
x = 8 or x = –2
4.4 Solving Quadratic Equations by using the
Quadratic Formula
This is the general formula x =

b
b2 4 ac
known as the quadratic formula.
2a
Please note that here we will not go into details as to how this formula is derived.
Instead you are required to look into references provided at the end of this lecture
to see how this was done.
This formula gives solutions to the general quadratic equation ax2 + bx + c = 0,
expressed in the coefficients a, b, and c. We can find the solutions to any specific
quadratic equation simply by substituting the appropriate values of a, b, and c.
The solutions to a quadratic equation ax2 + bx + c = 0 where a ≠ 0 are given by a
b
b2 4 ac
quadratic formula x = 
2a
If b2 – 4ac > 0, then the equation has two distinct solutions.
If b2 – 4ac = 0, then the equation has exactly one solution.
If b2 – 4ac < 0, then the equation has no real roots i.e. has complex roots.
Examples:
Solve the following quadratic equations by using the formula
1.
3x2 – 4x + 1 = 0
Solution:
In this case, a = 3, b = – 4 and c = 1
Substituting these values into the formula we get;

b
b2 4 ac
x=
2a
x=

( 4)
(-4)2 4(3)(1)
2(3)
x=
4 
16 12
6
Basic Mathematics for Diploma in Primary Teacher Education 65
x=
4 4
6
x=
42
6
x=
6
2
or x =
6
6
x = 1 or x =
2.
1
3
x(x + 6) = 2x – 6
Solution:
The first step in this problem is to put the equation in the quadratic form.
x(x + 6) = 2x – 6
x2 + 6x = 2x – 6
x2 + 6x – 2x + 6 = 0
x2 + 4x + 6 = 0
From this quadratic form, a = 1, b = 4 and c = 6
We substitute these values into the quadratic formula
x=
b
b2 4 ac

2a

4
4 2 4(1)(6)
x=
2(1)
x=

4
16 24
2
x=

4
8
2
At this point we observe that the number under the square root sign is negative.
Since we cannot take the square root of a negative number, we conclude that this
quadratic equation has no real solution.
There are four methods of solving quadratic equations:

The first method is Factoring.
66 Basic Mathematics for Diploma in Primary Teacher Education
This is a simple method, but does not always work, because we cannot
always factor a given equation.

The second method is Completing the Square.
This method always works, but it may involve more steps than the other
methods.

The third method is using the Quadratic Formula.
This method always works, is simple to use, and has the advantage that it
tells us easily by means of discriminant, what type of and how many
solutions does the equation have.

The fourth method is using graph i.e. Graphical Method.
4.5 Graphing Quadratic Equations
Refer to our discussion of lecture 3 part 3.4 where we plotted the graph of the
linear equation y  2x  4 . The same process is applied when graphic quadratic
equations. The difference here is that instead of obtaining a straight line, with
quadratic equations we get parabola shapes.
In solving quadratic equations graphically, we will first draw the graph of the
equation and then use it to obtain the solution to the equation.
Example:
Solve the quadratic equation x2 – 5x + 4 = 0 by using graphical method.
Solution:
Select some values of x and calculate the corresponding values of y
For x = –2,
Substituting –2 in the equation y = x2 – 5x + 4 we get
(–2)2 – 5(–2) + 4
= 4 + 10 + 4
= 18
Thereby obtaining a point (–2, 18)
Likewise, substituting –1 in the equation y = x2 – 5x + 4 we get
(–1)2 – 5(–1) + 4
=1+5+4
Basic Mathematics for Diploma in Primary Teacher Education 67
= 10
Thereby obtaining a point (–1, 10)
And also Substituting 0 in the equation y = x2 – 5x + 4 we get
(0)2 – 5(0) + 4
=0+0+4
=4
Thereby obtaining a point (0, 4)
We do the same procedure for the rest of the x values and get our table of values as;
x
–2
–1
0
1
2
3
4
5
6
y
18
10
4
0
–2
–2
0
4
10
The next step is to plot the graph using the obtained table of values
11
10
9
8
7
6
5
4
3
2
1
-2
0
-1 -1 0
1
2
3
4
5
6
7
-2
-3
The solution of the quadratic equation x2 – 5x + 4 = 0 can be obtained from the
graph
y = x2 – 5x + 4 at the point where y = 0. In this case y = 0 when x = 1 and when
x=0
Therefore the solution of our equation is x = 1 or x = 4.
Solve the quadratic equation in the above example by factoring and compare the
result.
Example:
Without drawing, find the points where the line y = –2x – 4 touches the graph
y = x2 + 3x – 10
68 Basic Mathematics for Diploma in Primary Teacher Education
Solution:
Let
y = x2 + 3x – 10
…(i)
y = –2x – 4
…(ii)
To solve the two equations simultaneously, substitute equation (i) into equation
(ii), this means we replace y by –2x – 4
We get,
–2x – 4 = x2 + 3x – 10
x2 + 3x – 10 + 2x + 4 = 0
(combine like terms)
x2 + 5x – 6 = 0
We have ended up with a quadratic equation, which we can solve by using any
method:
By factoring:
x2 + 5x – 6 = 0
x2 – x + 6x– 6 = 0
(x – 1)(x + 6) = 0
x = 1 or x = –6then for the two values of x find the corresponding values of y
Find y when x = 1
Find y when x = –6
y = –2x – 4
y = –2x – 4
y = –2(1) – 4
y = –2(–6) – 4
y = –2 – 4
y = 12 – 4
y = –6
y=8
So there are two points of intersection between the graph y = x2 + 3x – 10 and y
=
–2x – 4 which are (1, –6) and (–6, 8).
Summary
A quadratic equation is an equation of the form ax2 + bx + c = 0, where a ≠ 0.
a is called the coefficient of x2; b is called the coefficient of x and c is called the
constant term. A quadratic equations can be solved by factoring, completing the
square, using the quadratic formula or graphically.
Basic Mathematics for Diploma in Primary Teacher Education 69
Exercise
1.
Solve the following equations
(i)
x2 + x – 6 = 0
(ii)
2x  1 +
x1 = 5
(iii)
x 2 +
2x  4 = 6
(iv)
x4 – 16 = 0.
2.
Determine whether the solutions of 5x2 + 2x – 4 = 0 are real or not real.
3.
Solve the following by using the quadratic formula
(i)
x2 –3x – 10 = 0
(ii)
x2 – 2 = 2x
(iii)
6x2 = 1 – x
4.
The sides of a square are given by (x – 2) cm and its area is 64cm2. Find x.
5.
Solve the following equations by completing the square
(i)
x2 – 6x + 34 = 0
(ii)
x2 – 4x – 7 = 0
(iii)
3x2 – 9x + 12 = 0
6.
Find the points where the graph of y = x2 + 10x +5 and the line y = 4x –
3 intersect.
7.
The sides of a right angled triangle are (2x + 1)cm, (2x)cm and (x – 1) cm.
8.
(i)
Find the value of x
(ii)
Find the area of the triangle (numerically in cm)
(i)
Factorize completely the expression pq  qr 
q 2 qr
(ii)
Find the value of the expression given in (a) above if p = 11.1; q =
7.1 and r = 2.9
x
write y in terms of A and x.
y
9.
If A  1 
10.
Solve the following simultaneous equations given that z = 2
70 Basic Mathematics for Diploma in Primary Teacher Education
2
2
x  2y 

2x  y  2z  9
11.
Find the points of intersection of the parabola y = x2 – 3x + 7 and the line
y = x + 4.
12.
Solve the following inequalities;
13.
14.
(i)
1
1
x–2< x+1
4
2
(ii)
1
x > 2.
2
Solve the following quadratic equations by factorization
(i)
18x2 – 9x – 2 = 0
(ii)
8x2 + 14x + 3 = 0
Draw the graph of the equation y = x2 – x – 6 and use it to find the
solution of the equation x2 – x – 6 = 2.
References
Hirsch, L. & Goodman, A. (1987), Understanding Intermediate Algebra, West
Publishing Company, St. Paul, Minesota, USA.
Auvil, D. L. (1996), Algebra for College Students, McGraw-Hill Companies Inc.,
USA.
Muller, F. J. (1969), Elements of Algebra, Prentice-Hall Inc., Englewood Cliffs, New
Jersey, USA.
Willis, A. T. & Johnston, C. L. (1981), Intermediate Algebra, Wadsworth
Publishing Company, California.
Blyth T. S. and Robertson E. F. (1984), Algebra through Practice, Cambridge
University Press, Great Britain.
Sullivan, M. College (1990), Algebra with Review, Dellen Publishing Company, San
Francisco, California.
Gardner K. L (1966), Discovering Modern Algebra, Oxford University Press,
London.
Rees, P. K. (1975), Algebra, Trigonometry and Analytical Geometry, 2nd Ed.,
McGraw-Hill, Inc. USA.
Basic Mathematics for Diploma in Primary Teacher Education 71
Lial, M. L. & Miller, C. D. (1986), Algebra and Trigonometry, 4th Ed., Scott,
Foresman & Company, USA.
Tanzania Institute of Education, (2000) Secondary Basic Mathematics Book Two,
TIE, Dar es Salaam.
72 Basic Mathematics for Diploma in Primary Teacher Education
Basic Mathematics for Diploma in Primary Teacher Education 73
PART III
Set Theory
In mathematics, a set can be defined as any collection of objects. Sets are one
of the most important and fundamental concepts in modern mathematics.
Having only been invented in the 19th century, set theory is now an
important part of mathematics, being introduced from primary schools in
many countries. Set theory can be viewed as the foundation upon which
nearly all mathematics can be derived.
The concept of sets is commonly used in everyday life. For example, we have
used terms such as dinner set, mathematical set etc. This section is divided
into two lectures, the first one deals with the basic concepts of set
presentations and notations, and the second one deals with operations with
sets.
74 Basic Mathematics for Diploma in Primary Teacher Education
Basic Mathematics for Diploma in Primary Teacher Education 75
LECTURE 5
Set Presentations and Notations
5.1 Introduction
This lecture aims to introduce you to some basic concepts in the set theory,
including definitions of some key terms, set presentations and notations.
Learning Objectives
At the end of this lecture, you will be able to:

Explain what sets are and different ways of describing them;

Define some key concepts underlying the set theory;

Determine the number of subsets of a given set.
5.2 Basic Concepts
Definition of Set
A set may be defined as any well-defined list or collection of objects. Objects of a
set are referred to as elements of that set. Sets are denoted by capital letters e.g. A,
B, C, D, …. Elements of a set are, if alphabets, denoted by small letters.
We can tell whether an element is a member of a given set or not. For example if
we say A is a set of all English vowels, and B is a set of all counting numbers less
that 10, then a is a member of set A but not B and 2 is a member of set B but not
A. The symbol (is used to denote ‘a member of.’ Thus we write a (A, 2 (B whereas
2 (A means 2 is not a member of set A.
Description of Sets
Sets can be described by the following main ways;
76 Basic Mathematics for Diploma in Primary Teacher Education
1.
Listing all the members or element of a set in a roster form
Examples:
If A is a set of English vowels, then it can be presented as
A = {a, e, i, o, u}
If B is a set of all counting numbers less than 10, then it can be presented as
B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
But if we were for example to give a set of all saloon cars in Dar es Salaam
city by registration numbers, how many pages of pages of paper and time
we would need to accomplish the task? Therefore, the other way is by
stating properties that characterize the elements of the set. For example,
C = {All saloon cars in Dar es Salaam city}
If a set contains many elements but continuous in a pattern, we often use
three dots …, called ellipsis, to indicate that they are elements of that set, in
the same pattern, but have not been listed down.
Examples:
N = {1, 2, 3, …}
N = {1, 2, 3, …, 98, 99, 100}
The use of ellipsis in the first case indicates that the pattern has to be
continued indefinitely, while in the second case it indicate that some
elements in the same pattern have been left out.
The list or description of a particular set is usually separated by commas and
enclosed in braces { } and not parenthesis ( ) or square brackets [ ]
Listing elements of a set in different orders does not change the set
In using ellipsis some elements must be listed so that the pattern can be
determined.
2.
The second way of describing a set is the use of set builder notation. It
combines the use of braces and the concept of variables.
Examples:
A = {x: x is an English vowel}
B = {x: x is an integer, x>0}
C = {x: 0 < x <20}
The colon: reads “such that”. Sometimes a vertical line | is used.
Basic Mathematics for Diploma in Primary Teacher Education 77
3.
Another way of representing a set is by using Venn diagrams. This will be
discussed further in the following lecture.
Finite and Infinite Sets
A set can be finite or infinite. A finite set is the one that consists of countable
number of elements, whereas an infinite set consists of uncountable number of
elements. Recall our previous examples,
N = {1, 2, 3, …}
N = {1, 2, 3, …, 98, 99, 100}
In the first case, the pattern continues indefinitely. The set has unlimited
(uncountable) number of elements. It is thus an infinite set. In the second case, the
pattern continues definitely, meaning that we know the last element of the set and
we can count the number of elements in the set. This set is a finite set.
Cardinality of Sets
Cardinality of a set is simply the number of elements contained in that set. This is
denoted as n(A) or A. For example, set A = {a, e, i, o, u}, then n(A) = 5. It is
only possible to determine the cardinality of a set for finite sets.
Example:
Given set P = {x: x all odd numbers between 1 and 25 inclusive}. Find n(P)
Solution:
To do this, first we will have to list all the elements of set P. The term inclusive
here implies that we have to take all the odd numbers between 1 and 25, including
1 and 25 themselves. It is mentioned here so as to avoid confusion as to whether
the minimum and maximum numbers are in the set or not.
P = {x: x all odd numbers between 1 and 25}
P = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25}
Empty, Equal, Equivalent Sets
Empty Sets
We have so far gone through sets that contain finite and infinite elements. It is also
true that a set may contain no element. If a set contains no elements it is still a
finite set with zero number of elements. This is known as an empty set and is
denoted by { } or ∅
For example, A = {all three eyed students at the Open University of Tanzania}
A={}
78 Basic Mathematics for Diploma in Primary Teacher Education
Equal Sets
Two sets that contain exactly the same number of elements are said to be equal,
denoted by A = B. Consider the following sets
A = {1, 2, 3, 4, 5} and B = {5, 4, 3, 2, 1} then A = B because each member of A
belongs to B and vice versa. If two sets are not equal, we write A ≠ B
Equivalent Sets
Equivalent sets are two sets that contain exactly the same number of elements.
For example, A = {!, @, #, $, %} and B = {6, &, a, 9, 10} then we say A is
equivalent to B because they both contain five elements.
Subsets
Given any two sets A and B, set A is said to be a subset of set B if and only if all
elements of set A are also in set B, denoted by A ⊆ B. In other words we can say
set A is contained in B.
For example, Let A = {2, 4, 6, 8, 10, 12, 14, 16, 18}
B = {1, 2, 3, … , 20}
Then A ⊆ B
A subset of a given set which is not the set itself is called a proper subset.
Number of Subsets
Consider a set A such that A = {a, b}. If we are to write the subsets of this simple
set using the definition we have already discussed above we will have first the
empty set{ }, then taking one element of the set at a time we will have {a} and then
{b}and finally taking two at a time we will have {a, b} and {b, a}. But since {b, a}
is the same as {a, b} we will list it only once to avoid repetition. Thus we have four
subsets. Following this trend, consider how cumbersome it would be if we were to
give the subsets of a set say with ten elements. We have an already derived general
rule that the number of subsets of a given set is equal to 2n where n is equal to the
number of elements in that set.
Example:
Given A = {1, 2, 3, 4, 5} how many subsets does set A have?
Using the general rule, number of subsets of any set is given as 2n, where n is the
number of elements of that set
Set A has five elements; therefore number of its subsets will be 25 = 32.
Basic Mathematics for Diploma in Primary Teacher Education 79
An empty set is only denoted as { } or ∅, a set denoted as {∅} is not an empty set
but a set with one element which is ∅
An empty set is also a subset of any set
Every set is a subset of itself
The number of subset of any given set can be obtained by using the general rule 2n
where n is the number of elements of that set.
5.3 Universal Sets
A universal set, denoted by U is a general set from which elements of all other sets
under consideration are drawn. The universal set can change from one problem to
another depending on the nature of the problem under consideration.
For example, consider the following sets:
A = {All students in the Open University of Tanzania}
B = {All students in the Faculty of Science}
C = {All students in the Faculty of Arts}
D = {All students in the Faculty of Law}
E = {All students in the Faculty of Education}
F = {All students in the Faculty of Business Management}
G = {All students in the Institute of Continuing Education}
It is clear that sets B, C, D, E, F and G are all subsets of set A. Since set A contains
all the other sets listed, it is the universal set. Thus in all problems that relate to
students in different faculties at this university we would find ourselves referring to
sets whose elements are members of set A, the universal set.
Example:
Let A = {All boys in x primary school}
B = {All girls in x primary school}
C = {All pupils in x primary school}
Then set C in this case is the universal set.
Activity
1.
Is A a subset of B, where A = {1, 3, 4} and B = {1, 4, 3, 2}?
2.
Let A be all multiples of 4 and B be all multiples of 2. Is A a
subset of B? And is B a subset of A?
80 Basic Mathematics for Diploma in Primary Teacher Education
5.4 Complementary Sets
Given a set A and a universal set U, the complement of set A denoted by A' is a set
of all elements that are in the universal set but are not in set A. It implies therefore
that to be able to give a complement of any set one must first know the universal
set.
For example, consider the sets listed above,
U = {All students in the Open University of Tanzania}
B = {All students in the Faculty of Science}
B' = {All students who are not in the Faculty of Science}
Summary
A set may be defined as any well-defined list or collection of objects. Objects of a
set are referred to as elements of that set.
A set can be finite or infinite. A finite set is the one that consists of countable
number of elements, whereas an infinite set consists of uncountable number of
elements.
Cardinality of a set is simply the number of elements contained in that set.
If a set contains no elements it is still a finite set with zero number of elements.
This is known as an empty set and is denoted by { } or ∅
Two sets that contain exactly the same number of elements are said to be equal,
denoted by A = B.
Equivalent sets are two sets that contain exactly the same number of elements.
Given any two sets A and B, set A is said to be a subset of set B if and only if all
elements of set A are also in set B, denoted by A ⊆ B. In other words we can say
set A is contained in B.
A universal set, denoted by U is a general set from which elements of all other sets
under consideration are drawn.
Given a set A and a universal set U, the complement of set A denoted by A' is a set
of all elements that are in the universal set but are not in set A. It implies therefore
that to be able to give a complement of any set one must first know the universal
set.
Exercise
Basic Mathematics for Diploma in Primary Teacher Education 81
1.
2.
Given A = {All odd numbers less than 20}. Present set A in;
(i)
Roster form
(ii)
Set builder notation
List the member s of the following sets
A = {All counting number between 10 and 15}
B = {All even numbers between 0 and 25}
C = {All countries in East Africa}
3.
Which of the following sets if finite and which is infinite and why?
A = {x: x all numbers greater than 0}
B = {x: x is a region in Tanzania}
C = {All people in the world}
D = {0, 1, 2, 3, …, 1000}
4.
Given the following sets:
P = {1, 2, 3, …20}
Q = {All even numbers less than 30}
R = {All primes between 10 and 15}
5.
Which of the following is true and which is false?
(i) 6 ( P (ii) 7 (Q (iii) 9 ( Q
(iv) 15( R (v) 23 ( P (vi) 20 ( Q
(vii) 20 ( R (viii) 1( Q (ix) 19 ( P
6.
Give three examples of equal sets.
7.
Let U = {3, 4, 5, 6, 7, 8, 9, 10}
A = {3, 4, 5, 6}
B = {7, 8, 9, 10}
8.
(i)
Find A' and B'
(ii)
Comment on the relationship between A' and B'
Suppose U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {2, 3, 7, 8}
B = {0, 4, 5, 7, 8, 9}
Find (i) (A)' (ii) (B)'
82 Basic Mathematics for Diploma in Primary Teacher Education
9.
Given that P = {all positive real numbers x: x ≤ 20}
Q = {all odd numbers between 0 and 20}
R = {all even numbers between 0 and 20}
Find (i) (Q)' (ii) (R)'.
References
Henle, J. M. (1986), An Outline of Set Theory, R. R. Donnelly & Sons,
Harrisonburg, Virginia, USA.
Meserve, B. E. & Sobel, M. A. (1977), Contemporary Mathematics, Prentice-Hall
Inc., New Jersey
Moshe, M.(1996), Set Theory, Logic and their Limitations, Cambridge University
Press, Great Britain.
Nathan, R. & Moshi, A. M. (1993), Set Theory Advanced Level Mathematics, Vol. 1,
IVST Book Series, Dar Es Salaam University Press, Dar es Salaam
Basic Mathematics for Diploma in Primary Teacher Education 83
LECTURE 6
Operations with Sets
6.1 Introduction
Just as in arithmetic where we have additions and subtractions that enable us to
combine numbers, in sets we have similar mechanisms that enable us to combine
sets. These are intersections and unions of sets.
Learning Objectives
At the end of this lecture, you will be able to:

Define unions and intersections of sets;

Define disjoint sets;

Evaluating simple set expressions;

State different laws of algebra of sets;

Apply the laws of algebra of sets in simplifying set expressions;

Draw venn diagrams and apply them in solving set problems.
6.2 Unions and Intersections of Sets
Union
Given any two sets A and B, the union of A and B, denoted by AB, is the set of
all elements that are members of set A or set B or both. If we are to list down the
elements of the new set, AB, we should remember that elements contained in
both sets are listed only once. This means that the union of sets A and B is the set
of elements that are members of at least one of the sets.
Example:
Let
A = {a, e, i, o, u}
B = {a, b, c, d, e}
Then AB = {a, e, i, o, u, b, c, d}
84 Basic Mathematics for Diploma in Primary Teacher Education
Intersection
Given any two sets A and B, the intersection of A and B, denoted by A
∩B, is a set
of all elements that are members of both A and B. This means that the intersection
of two sets results into another set, A
∩B, containing a
A and B.
Example:
Let
A = {a, e, i, o, u}
B = {a, b, c, d, e}
Then A
∩B = { a, e}
Disjoint Sets
Two sets are said to be disjoint if none of the members of one set belong to the
other. This means that they have no common element, i.e. A
∩B
}. = {
Example:
Let
A = {All even numbers less than 10}
B = {All odd numbers less than 10}
The two sets are disjoint since there is never a number which is both odd and even,
which means that A
∩B is an empty set.

The intersection of two sets is the set whose elements belong to both sets.
In finding the union of two sets, the members of both sets are written only
once.


Two sets A and B are said to be disjoint in A
∩B = { }
6.3 Evaluating Set Expressions
In many occasions we may find set expressions that may involve more than two
sets
Example:
Consider the following sets:
U = {1, 2, 3, …, 10}
A = {2, 4, 6, 8, 10}
B = {1, 3, 5, 7, 9}
Basic Mathematics for Diploma in Primary Teacher Education 85
C = {3, 4, 5, 6, 7, 8, 9}
D = {3, 5, 6, 7}
Find:
(i)
(BCD)
(ii)
(B
∩C∩
D)
Solution:
To evaluate (BCD), list all members that are contained in one, two or all of the
three sets B, C and D.
Thus, (BCD) = {1, 3, 4, 5, 6, 7, 8, 9}
Likewise, (B
D ) is a set of all members that are
∩C∩
Thus, (B
∩C∩
D ) = { 3, 5, 7}
There are some expressions in which you may be required to do several operations
at the same time. Suppose you are given any four sets U, A, B and C and asked to
find say
(A'  B)
∩
C).
(AHow would you proceed?
Such cases will require you to follow the following order;

Work out the expression from left to right by evaluating first the
expression within the parenthesis.

Evaluate all the complements if there are any.

Evaluate unions and intersections.

If the expression contains nested grouping symbols, begin with the
innermost and work out towards the outermost.
Example:
Consider sets U, A, B, C and D given in the example above, U = {1, 2, 3, …, 10};
A = {2, 4, 6, 8, 10}; B = {1, 3, 5, 7, 9}; C = {3, 4, 5, 6, 7, 8, 9} and D = {3, 5, 6, 7}
Find the following;
(i)
(A'∩D')
(ii)
B∩(AC)
(iii)
[(B∩C')D]
(iv)
[(AC)'∩B]'
Solution:
(i)
In solving for A'
∩ D ', w e start by evaluating the compleme
86 Basic Mathematics for Diploma in Primary Teacher Education
A' is a set of all elements which are in the universal set but not in A.
Therefore A' = {1, 3, 5, 7, 9}, likewise D' = {1, 2, 4, 8, 9, 10}
then A'
(ii)
∩ D ' = { 1, 9}
C) is evaluated by first determining (AC), then finding the
B 
∩(A
intersection between the resulting set and set B
(AC) = {2, 3, 4, 5, 6, 7, 8, 9, 10}
We have set B = {1, 3, 5, 7, 9}
Therefore, B
(iii)

C) =
∩(A
{3, 5, 7, 9}
For [(B
∩C')

D], we start by determining the complement of set C, and
then find its intersection with set B. Finally we find the union of the
resulting set and set D
C' = {1, 2, 10}
(B
∩C') = { 1}
therefore, [(B
(iv)
∩C')

D]=
{1, 3, 5, 6, 7}
In evaluating [(AC)' ∩B
]', we start with the innermost bracketed group
(AC), then find its complement. We then intersect the resulting set with
set B. Finally we determine the complement of the resulting set.
(AC) = {2, 3, 4, 5, 6, 7, 8, 9, 10}
(AC)' = {1}
(AC)'
∩B = { 1}
Therefore, [(AC)' ∩B
]' = {2, 3, 4, 5, 6, 7, 8, 9, 10}
6.4 Laws of Algebra of Sets
Since it is possible to have a more complex set expression, the evaluation exercise
may be made considerably easy by applying some of the laws of algebra of sets
which exists. These when applied, can reduce the most complex operations down
to simple workable operations.
The following are the laws of algebra of sets;
1.
Idempotent laws
AA = A
A
2.
∩A = A
Commutative laws
AB = BA
A
3.
∩B = B∩
A
Associative laws
Basic Mathematics for Diploma in Primary Teacher Education 87
(AB)C = A(BC)
(A
4.
∩B)∩
C = A∩
(B∩
C)
Distributive laws
(AB)

∩C
(B
(A
5.
= ∩C)
(A ∩
C)
∩B)U
= (A

C) C

∩(B
C)
Identity laws
AØ = A
A
∩Ø = Ø
AU = U
A
6.
∩U = A
Complementary laws
AA' = U
A
∩A ' = Ø
(A')'= A
Ø' = U
U' = Ø
7.
De Morgan’s laws
(AB)' = A'
(A
∩B'
∩B)'
A'B'=
Applying the Laws of Algebra of Set to Work-out Set Expressions
Examples:
Prove the following expressions by using the laws of algebra of set;
1.
[A
∩=(AA'
]'
∩
B)'
B
Solution:
[A
∩(A=∩
]'
B)'
[A

B')]''
∩(A
= [(A
∩A ') U(Distributive
]'
(A ∩
B')
law)
= [ Ø  (A
= [(A
(De Morgan’s law)
]'
∩B')
]'
∩B')
(Complementary law)
(Identity law)
= A'(B')'
(De Morgan’s law)
= A'B
(Complementary law)
88 Basic Mathematics for Diploma in Primary Teacher Education
2.
A(BA)' = AB'
Solution:
A (BA)'
= A(B'
(De Morgan’s law)
∩A ')
= (AUB')
= (AB')
∩(A U A(Distributive
')
law)
(Identity law)
∩U
= AB'
3.
[(B
(Identity law)
∩A
(A ') ∩B')
(A ∩B)
] = AB
Solution:
[(B
∩A
(A ')
]∩B')U
= (B
(A ∩
B)
∩A
((A')
= (B

((A')
∩A
= (B

(A ')
∩A
= (B

A ')
∩A
= (BA)
= (BA)
∩B')U
(Associative
(A ∩
B)) law)

∩(B'
B))
∩U )

∩(A
A) '
∩U
(Distributive law)
(Complementary law)
(Identity law)
(Distributive law)
(Complementary law)
= (BA)
(Identity)
= AB
(Commutative law)
6.5 Venn Diagrams
Set and relationships between sets may be represented by means of pictures and
diagrams. These diagrams are referred to as Venn diagrams.
A Venn diagram may therefore be defined as a pictorial presentation of the
relationship between sets. It will consist of the following:
1.
A rectangle showing the universal set U.
2.
Circles showing sets that are subsets of U drawn within the rectangle.
3.
Members of the sets distributed in the regions of the diagram according to
the relationships that exists between the sets.
With such presentation, operations with sets become more simplified.
Examples:
Let
U = {a, b, c, d, e, f, g, h, i, j, k}
A = {b, c, d, e, j}
B = {d, e, f, c, g, h}
Basic Mathematics for Diploma in Primary Teacher Education 89
C = {a, b, d, e, h}
Sets U, A, B and C can be represented in a Venn diagram as follows:
b
j
C
a
d, e
c
h
g, f
A
i, k
B
From this diagram we find that:
(A
∩B∩
=C){d, e}
(A
∩B) = {c, d, e}
(A
∩C) = {b, d, e}
(B
∩C) = {h, d, e}
(ABC)' = {i, k}
(AB)
= {c, d, e, f, g, b, h, j}
It is also possible to show the relationships that exist between sets in the Venn
diagram without having to allocate the elements. This may be done by shading or
labelling the regions.

Shading
Prepare a well-labelled two or more sets Venn diagram depending on the
problem under consideration.
Consider and perform all complements, unions and intersections.
Shade the region(s) that correspond to the relations of the sets.

Numbering the disjoint subsets
Number all the regions then perform operations as per given relationship.
90 Basic Mathematics for Diploma in Primary Teacher Education

Venn diagram is a very useful and efficient approach of solving
various set problems as it shows all relations that exist between the
sets.

The shading and labelling approaches work more efficiently with
simple expressions, in complex expressions the results may be more
difficult to interpret.
Number of Elements
The concept of cardinality of sets that we discussed in the last lecture can be
extended to more than one set by determining the number of elements in each
region of the Venn diagram.
We may need to follow the following steps:

Determine the regions that make up the Venn diagram.

Identify the number of elements contained in each of the regions of the
Venn diagram.

Find the sum of these elements in each set of the Venn diagram.
General Rules
Let x, y and z be the number of elements of each region in the diagram below:
B
x
y
z
A
n(A) = x + y
n(B) = y + z
n(AB) = (x + y) + (y + z)
so n(AB) would have (x + y) + (y + z) elements.
But since y elements are contained in both sets, they must be considered only once.
Thus y elements are deducted once from the expression, giving:
n(AB) = (x + y) + (y + z) – y
(But, y = n(A∩B)
Basic Mathematics for Diploma in Primary Teacher Education 91
n(AB) = n(A) + n(B) – n(A∩B)
This result can be extended to three sets A, B, and C, in such a way that we get the
expression
n(ABC) = n(A) + n(B) + n(C) – n(A∩B) – n(A∩C) – n(B∩C) + n(A∩B∩C)

Given any two sets A and B;

n(AUB) = n(A) + n(B) – n(A∩B)

Given any three sets A, B and C;

n(AUBUC) = n(A) + n(B) + n(C) – n(A∩B) – n(A∩C) – n(B∩C) +
n(A∩B∩C)
Example:
In a class of 100 students, 66 like Mathematics, 57 like English and 35 like both
subjects.
Draw a Venn diagram showing the number of students in each region and then
find the number of students who like neither subject.
Solution:
Let
U be the universal set,
M be students who like Mathematics
E be students who like English
n(U) = 100
n(M) = 66
n(E) = 57
n(M∩E) = 35
Since 35 students like both subjects, these are placed in region II of the Venn
diagram. To get the number of students in region I which represent number of
students who like Mathematics only we take 66 – 35 = 31
Likewise, region III represent those who like English only, and are obtained from
57 – 35 = 22
Then we fill the numbers in their respective regions as shown in the diagram.
92 Basic Mathematics for Diploma in Primary Teacher Education
I
31
II
35
E
III
22
12
M
Regions I, II and III together represent set (ME) and the area outside the circles
represent set (ME)'
n(ME)= 31 + 35 + 22
= 88
n(ME)' = n(U) – n(ME)
= 100 – 88
= 12 Number of students who like neither subject is therefore 12
Example:
200 second year students were required to join at least one of the three science
clubs in their college. 85 students joined the Physics club, 81 students joined the
Chemistry club, 21 students joined both the Physics and Mathematics clubs, 27
joined both the Mathematics and Chemistry clubs and 5 students joined all three
clubs.
How many joined:
(i)
Mathematics club only
(ii)
Only one club
Solution:
Since each student joined at least one club, then n(U) = n(PCM)
Given: n(PCM) = 200
n(P)
= 85
n(C)
= 81
n(P∩M)
= 21
n(M∩C)
= 27
n(P∩C)
= 13
n(P∩C∩M) = 5
Basic Mathematics for Diploma in Primary Teacher Education 93
Filling in the amounts in the respective regions we have,
M
56
16
X
5
8
22
46
P
C
To find the number of students joined the Mathematics club only, we apply the
formula
n(PCM) = n(P) + n(C) + n(M) – n(P∩M) – n(M∩C) – n(P∩C) +
n(P∩M∩C)
200 = 85 + 81 + (16 +5 +22 +x) – 21 – 27 – 13 + 5
200 = 166 + 43 + x – 61 + 5
200 = 153 + x
x = 47
Students with only one club: Physics only
= 56
Chemistry only = 46
Mathematics only = 47
Total
= 149
(i)
47 students joined the Mathematics club only
(ii)
149 students joined only one club.
Activity
In a particular insurance life office, employees Smith, Jones, Williams
and Brown have 'A’ levels, with Smith and Brown also having a
degree. Smith, Melville, Williams, Tyler, Moore and Knight are
associate members of the Chartered Insurance Institute (ACII) with
Tyler, and Moore having 'A’ levels. Identifying set A as those
employees with 'A' levels, set C as those employees who are ACII and
set D as graduates:
(a)
Specify the elements of sets A, C and D.
(b)
Draw a Venn diagram representing sets A, C and D, together
with their known elements.
94 Basic Mathematics for Diploma in Primary Teacher Education
(c)
What special relationship exists between sets A and D?
(d)
Specify the elements of the following sets and for each set,
state in words what information is being conveyed.
(i) A
(e)
∩C
(ii) D
D
C (iii)
∩C
What would be a suitable universal set for this situation?
Summary
Given any two sets A and B, the union of A and B, denoted by AUB, is the set of
all elements that are members of set A or set B or both. Given any two sets A and
B, the intersection of A and B, denoted by A∩B, is a set of all elements that are
members of both A and B. Two sets are said to be disjoint if none of the members
of one set belong to the other. This means that they have no common element.
Given any two sets A and B;
n(AB) = n(A) + n(B) – n(A∩B)
Given any three sets A, B and C;
n(ABC) = n(A) + n(B) + n(C) – n(A∩B) – n(A∩C) – n(B∩C) + n(A∩B∩C)
Exercise
1.
Given three sets P, Q and R such that;
P = {3, 2, 6, 8}
Q = {4, 5, 7}
R = {1, 3, 6, 9}
Find the following sets;
(i) PR (ii) P
2.
∩Q
(iii) R
∩Q
(iv) P
∩R∩
(v) PQ
RQ
Given the following sets;
X = {3, 6, 9, 12}
Y = {2, 4, 8, 12}
Z = {2, 3, 4, 5}
Find (i) (X
Y (ii) (XZ)
∩Z)
∩(Y ∩
Z)
Basic Mathematics for Diploma in Primary Teacher Education 95
3.
Given the following sets;
U = {х: 0
≤ х≤10}
A = {х: x is an odd number less than ten}
B = {х: x is an even number less than ten}
C = {х: x is a prime number less than ten}
Evaluate the following:
(i) A' ∩B
(ii) A'C' (iii) [(B
(iv) B(A
4.
5.
∩C)
(v) [C'

∩C)'
A']
∩(B
A')]'
Simplify the following expressions;
(i)
[(AB)'
(ii)
[(AB)
∩(A ∩
]'
B)'
∩(A
B') 
' (A
]
∩B)
Prove by using the laws of algebra of sets that;

(A'B')
∩B')
(i)
(A
(ii)
(AUB)
= AB'
∩A ' = B∩
A'
6.
A certain school has 140 students in Std. V. Each student takes part in at
least one sport out of football and basketball. If 90 play football and 85
play basketball, how many play both games?
7.
When the Headmaster sent for the students in his school who were either
Prefects or members of the School Council, 20 students came to his office.
He then asked the Prefects to stay and the rest to leave. He remained with
14 students in his office. If no Prefect is a member of the School Council,
find how many students were:
8.
(i)
Members of the School Council but not Prefect
(ii)
Prefects but not members of the School Council
A group of high school leavers was surveyed, and the following
information was obtained. 95 students had studied Mathematics, 76 had
studied Chemistry, 133 had studied History, 25 had studied all three
subjects, 22 had studied none, 38 had studied Mathematics and Chemistry,
46 had studied Chemistry and History and 54 had studied Mathematics and
History.
(i)
What percent of the students surveyed had studied none of the
three subjects?
96 Basic Mathematics for Diploma in Primary Teacher Education
(ii)
What percent of the students surveyed took exactly one of the
three subjects?
9.
Give three examples of equivalent sets.
10.
Given set A = {a, b, c}
11.
12.
13.
(i)
How many subsets does set A have?
(ii)
List all subsets of set A.
An institute that specializes in language tutoring lists the following
information concerning its students. 23 took English classes, 25 took
Swahili classes, 31 took French classes, 13 took both French and Swahili
classes, 10 took Swahili and English classes, 10 took only English classes,
and 8 took all English, Swahili and French classes.
(i)
How many students took at least two of the classes?
(ii)
How many took only one class?
(i)
In a group of 35 boys 25 play soccer and 15 play basketball. If each
boy takes part in at least one game, how many boys play both
soccer and basketball?
(ii)
Given the universal set U = 1,2,3,4,5,6,7,8,9 and the sets
A = {multiples of 2} and B = {Prime numbers}; find (i) AB' (ii)
A∩B.
If E = {integers between 1 and 11}
A = {x: 2 < x ≤ 9}
B = {x: 1 ≤ x ≤ 10}
(i)
Draw a Venn diagram to illustrate these sets
(ii)
List the elements belonging to AB and A
.
∩B'
References
Nathan, R. & Moshi, A. M. (1993), Set Theory Advanced Level Mathematics, Vol. 1,
IVST Book Series, Dar Es Salaam University Press, Dar es Salaam.
Henle, J. M. (1986), An Outline of Set Theory, R. R. Donnelly & Sons,
Harrisonburg, Virginia, USA.
Moshe, M. (1996), Set Theory, Logic and their Limitations, Cambridge University
Press, Great Britain.
Basic Mathematics for Diploma in Primary Teacher Education 97
Meserve, B. E. & Sobel, M. A. (1977), Contemporary Mathematics, Prentice-Hall
Inc., New Jersey.
98 Basic Mathematics for Diploma in Primary Teacher Education
PART IV
Elementary Statistics
The word statistics has two dimensions. The first meaning refers to a body of
data such as may be found in some technical reports. For instance you may
encounter phrases such as employment statistics, accident statistics etc. The
second meaning of Statistics refers to the methods and techniques employed
in collecting, organizing, summarizing, processing, analyzing, presenting and
interpreting of any kind of data. In the later sense, statistics is therefore a
branch of applied mathematics with the following main objectives:

To present facts in a definite form

To simplify and classify large masses of facts

To furnish methods of comparison

To indicate trends and tendencies
This section is divided into three lectures. The first lecture deals with data
collection and presentation methods, the second lecture deals with measures
of central tendency and the last one is on measures of dispersion.
Basic Mathematics for Diploma in Primary Teacher Education 99
100 Basic Mathematics for Diploma in Primary Teacher Education
LECTURE 7
Data Collection and Presentation
7.1 Introduction
The study of statistics can be divided into two major segments:
1.
Descriptive Statistics: This is concerned with the methods and techniques
of data collection, organization and presentation, as well as the description
of some of their important features.
2.
Inferential Statistics: This is concerned with techniques of using features
resulting from descriptive statistics in making generalizations. It includes
such concepts as probability theory, distributions, theory of estimation
and hypothesis testing.
For the scope of this unit, we will deal with the first segment i.e. descriptive
statistics. This lecture introduces you briefly to the basic concepts of data
collection. We will then discuss some basic data presentation methods such as
different kinds of charts. Furthermore, we will discuss about grouping data from
raw information and presenting them graphically.
Learning Objectives
At the end of this lecture, you will be able to:

Describe and explain different methods of collecting data;

Represent data by using pie charts and bar charts;

Group statistical data from raw information;

Construct a frequency distribution table;

Describe and apply various methods of representing grouped data from a
frequency distribution table, i.e. histograms, frequency polygons and
cumulative frequency curves.
7.2 Data Collection Methods
Basic Mathematics for Diploma in Primary Teacher Education 101
Before we move on to the different methods of data collection, there are two very
basic concepts that are of significant importance in statistics worth mentioning
here; population and sample.
Population
Population in statistics refers to the collection of individual items such as people,
places or things about which information is required. For example, we want to
know the diversity of occupation of the Open University of Tanzania students,
then all students enrolled in its different programs would form a population.
Sample
A sample refers to a group of individual items selected from the population for a
study. That is the group of n observations drawn from a population of N size
where n < N.
From our example, it may be difficult to study occupations of all students at the
Open University of Tanzania, and instead we may decide to cover only students in
one of the programs, one intake etc to represent the population. This selected
group forms a study sample. Results from a study sample are then used in making
general conclusion about the observed variables over the whole population.
In statistics, we usually study a sample instead of a population. This is due to the
following reasons:

A population may be relatively and even infinitely large. For example, if a
study aims at observing behaviour of a certain species of animals, it will be
impossible to cover all of them.

Resource constraints may prevent persons involved from studying the
entire population.

Inaccessibility to the entire population may also be a hindering factor.
Types of Data
In statistics, there are two basic types of data:
Primary data: Which refer to the original information gathered for the specific
purpose at hand. For instance if a company gathers data from its prospective
customers about the suitability of a certain product, the resulting data would be
primary data.
Secondary data: Refers to the information that already exists somewhere having
been collected for some other purposes. This includes data from such sources as
publications by various organizations found in libraries etc.
102 Basic Mathematics for Diploma in Primary Teacher Education
There are organizations that collect primary data on various issues. These data
serve as secondary data to various users in handling numerous problems of study.
Examples of these organizations in Tanzania are:

The Central Bureau of Statistics

Bank of Tanzania

Board of External Trade

Tanzania Chambers of Commerce Industry and Agriculture

Economic Research Bureau etc.
These data are present in government publications, journals, reports, bulletins,
trade and specific magazines etc.
Data Collection Methods
We have so far gone through some concepts of statistics including the types of data.
Statistical data can be collected through various methods:
1.
Questionnaire: These contain a list of questions compiled in a way that the
required information can be obtained from the respondents. They need to
be carefully and skilfully designed for the objective to be achieved. These
must contain specific, clear, understandable and free from bias questions.
They must also be designed to fit a particular sample of the population
from which information is sought.
2.
Observation: This is done by observing and recording the actual events or
behaviour of elements of the sample. For example, observing customers
entering and coming out of a certain supermarket at a given period of
time., one could record the number of customers, male or females, young
or old, what they have bought etc. This method is preferred in situations
where the issues under the study are not the type that one could easily ask
questions.
3.
Experiments: Experiments can be carried out to collect primary data. For
example, verifying certain principles or laws in say Physics, Chemistry,
Commerce, etc. This method is reliable as the researcher gets the first hand
information.
7.3 Presentation of Statistical Data
For numerical data to be able to give meaningful interpretation, they need to be
summarized and logically presented. In this part, we shall explore various ways of
summarizing, tabulating and presenting statistical data.
Basic Mathematics for Diploma in Primary Teacher Education 103
1.
Charts: One method of presenting statistical data is by using charts. There
are several types of charts, some of which are pie charts and bar charts.
(i)
Pie charts: The pie chart can be used to display the percentage of
the total number of observations (measurements) that fall into each
of the data categories. This is done by partitioning a circle into
sectors whose size reflects their proportion of the whole.
To present data in a pie chart, we will go through the following
steps:
(a)
Sum up the items
(b)
Find the proportion of each individual item in relation to
the total sum of the items.
(c)
Multiply the relatives by 100 to convert each relative into a
percentage of the total sum.
(d)
Multiply each result by 3.6 (a full circle is 360°) to obtain
proportion of each relative in degrees.
(e)
Draw the circle and partition it basing on the degrees
obtained in (d) above.
Example:
Annual collection of fees in TShs. from students of four classes in a certain
secondary school was as follows:
Class
Amount (TShs)
Form I
4,500,000
Form II
2,000,000
Form III
2,600,000
Form IV
1,000,000
Present this data in a pie chart
Solution:
From these records, we can work out the following:
Class
Amount
Percentage
% of 360
Degrees
o
160.56 o
Form I
4,500,000
4500000

100 44.6
10100000
0.446 × 360
Form II
2,000,000
2000000

100 19.8
10100000
0.198 × 360o
71.28o
Form III
2,600,000
2600000

100 25.7
10100000
0.257 × 360o
92.52o
104 Basic Mathematics for Diploma in Primary Teacher Education
Form IV
Total
1,000,000
10,100,000
1000000

100 9.9
10100000
0 099 × 360o
35.6 o
360o
100%
Then from these results, a pie chart would be drawn as follows:
Form IV
Form I
Form III
Form II
From this pie chart, we can quickly make certain inferences about fees
collection. For example, we can easily see that more fees were collected
from Form I students than from other classes.
Please note that pie charts are useful when there is a small number of
variables, for too many categories make the pie chart difficult to interpret.
A maximum of 5 to 6 is preferable.
2.
Bar charts: These can be vertical or horizontal, single or multiple. They
can be used to display data across time and sometimes display percentage
rather than the number of observations in each category.
The following are important steps in the construction of bar charts:
Label frequencies along one axis and categories of the variables along the
other. Horizontal axis acts as the base of the bars.
Construct a rectangle at each category of variable with the height equal to
the frequency of the observation in that category.
Leave space between each category to connote distinct, separate categories
and to clarify the presentation. The vertical axis starts at zero.
Examples of Bar Charts
Multiple Bar Charts
This form of presentation facilitates comparison of data overtime or space.
Basic Mathematics for Diploma in Primary Teacher Education 105
Example:
The values of tobacco and tea produce in metric tones in Tanzania for the period
of four years from 1971 were as follows:
Year
Tobacco
Tea
1971
14,200
10,500
1972
17,300
12,700
1973
13,000
12,600
1974
18,200
12,900
If we present this data on a multiple chart we get the following results:
20000
18000
16000
VOLUME
14000
12000
Tobacco
10000
Tea
8000
6000
4000
2000
0
1971
1972
1973
1974
YEARS
Percentage Bar Charts
In this case, instead of using observations or measurements taken on variables, we
have converted them into percentages of the total. The vertical axis is therefore
marked with the percentages.
Example:
Contribution of three products namely, cotton, sisal and coffee to the total
country exports in three consecutive years was as follows:
Year
Cotton
Tonnes
Sisal
%
Tonnes
Coffee
%
Tonnes
Total
%
1984
400
25.0
650
40.6
550
34.4
1600
1985
500
27.8
700
38.9
600
33.3
1800
1986
550
27.1
780
38.4
700
34.5
Total
1450
2130
1850
2030
5430
In each of the three years there is a contribution percentage of the three crops to
the total exports.
106 Basic Mathematics for Diploma in Primary Teacher Education
Presenting these data in a bar chart we have:
50
45
% VOLUME
40
35
30
Cotton
25
Coffee
20
Sisal
15
10
5
0
1984
1985
1986
YEARS
Contribution of Three Crops to the Total Exports (1984–1986)
(For all these examples the data are hypothetical)
Please note that all charts must:

Have titles.

Have axis labelled properly.

Indicate the source of data where required.
7.4 Frequency Distribution
In this section, we will discuss about grouped data, frequency table of grouped
data, class limits and class boundaries.
Grouped Data
Consider for example the following set of 30 scores for an examination in a statistic
class:
52
54
58
57
52
55
56
61
63
59
59
60
52
58
55
56
57
50
54
64
56
51
50
60
56
59
60
65
55
62
These scores do not give us much information because they are presented in the
order of their occurrence. This set of data is known as raw data. If for example, we
want to find out the highest score we will have to list the scores in a specified
order. For example, we may list them from the highest to the lowest or vice versa.
The set of raw data arranged in ascending or descending order of magnitude is
known as array data.
Basic Mathematics for Diploma in Primary Teacher Education 107
When raw data is very large, for instance 30 or more observations, it becomes
difficult to handle them. Even presenting them in an array becomes cumbersome.
To summarize the data we then have to group them in classes, and to determine
the number of observations belonging to each class.
For example, we can group the scores of the 30 students in the above example as
follows:
Table 7.1
Scores
Number of Students
50 – 53
6
54 – 57
12
58 – 61
8
62 – 65
4
Total
30
This is an example of grouped data. Although the grouping process generally
destroys much of the original detail of the data, an important advantage is gained
in the clear overall picture that is obtained in vital relationships, which are hereby
made evident.
The difference between the largest and the smallest number is called the range of
the data. For example in this case, the highest score is 65 and the lowest is 50, so
the range will be 65 – 50 = 15.
A Frequency distribution is therefore a tabular arrangement of data by classes
together with the corresponding class frequencies.
A frequency distribution examines the frequency of occurrence of different values
of a variable at a given point of time, and that the values of a variable are combined
together into classes of a predetermined size. The number of times a variable
occurs is the frequency of that variable.
Table 7.1 is an example of a frequency distribution table. The first class or
category in the table, for example, consists of scores 50 to 53. Since 6 students have
score belonging to this class, the corresponding class frequency is 6.
Class Intervals and Class Limits
A symbol defining a class such as 50 – 53 in table 7.1 is called a class interval. The
end numbers 50 and 53 are called class limits. The smaller number, 50 is called the
lower class limit and the larger number 53 is the upper class limit. The terms class
and class interval are often used interchangeably, although the class interval is
actually a symbol for the class.
Guidelines to Choice of Class Interval

Find the range of the data.
108 Basic Mathematics for Diploma in Primary Teacher Education

Decide on the number of classes you wish to use.

Divide the range by the number of classes chosen. The result is an
approximation of the class size.

Avoid having classes like 20 – 30, 30 – 40,40 – 50 etc. Classes like these
make it difficult to place values such as 20, 30, 40 etc. For example where
will 30 be placed? Will it be placed in 20 – 30 or in 30 – 40?
Raw data – data taken direct from observation which has not been organized
Array data – data arranged in ascending or descending order
Grouped data – data organized and summarized in classes or categories
How to prepare frequency distribution table:

Find the lowest value and the highest value

Construct classes or intervals for the set of data provided. The first class
starts from the lowest value and the last class ends with the highest value

Tally each observation to its corresponding class. Once any observation is
tallied, it should be marked to avoid repetition

Count the tallies in each class and write the total in the frequency column.

Sum the frequencies to check whether the total agrees with the total
number of observations.
When constructing a frequency distribution, avoid having too many or too few
classes. Having too many classes means that the table is doing very little to
summarize the data. Having too few classes means that a lot of important
information is obscured. Good number of classes is advisable to be between 6 and
12.
Class Boundaries
The class boundaries are the numbers that fall halfway between the upper class
limit of one class and the lower class limit of the next class.
In practice, class boundaries are obtained by adding the upper limit of one class
interval to the lower limit of the next higher class interval and dividing by 2.
Example:
Referring to the example in table 7.1, the classes were as follows:
50 – 53
54 – 57
58 – 61
Basic Mathematics for Diploma in Primary Teacher Education 109
62 – 65
We shall find the class boundaries for the first class 50 – 53
Solution:
Using the same interval, assume a class just before the lower class limit for the class
50 – 53, which would be 46 – 49.
49  50
 49.5
2
Then the lower class boundary is:
Similarly, the upper class boundary is:
53  54
 53.5
2
Therefore we will have 49.5 – 53.5 as the class boundaries for the 50 – 53 class
Please note that some references do not use these class boundaries and only write
about class limits to mean class boundaries.
Modal Class
Modal class is the class containing the highest frequency. For example, in table 7.1,
54 – 57 is the modal class.
Class Mark
The class mark is the midpoint of the class interval and is obtained by adding the
lower and upper class limits and dividing by two. Thus the class mark of the
interval 50 – 53 is
1
103
50  53 51.5
2
2
The class mark is also called the class midpoint.
Example:
The following marks were scored by 40 students in a mathematics test:
54
42
73
54
58
85
52
60
58
48
70
52
53
53
45
60
55
50
53
75
58
63
58
57
82
30
35
49
48
53
52
57
60
25
65
72
54
55
68
28
110 Basic Mathematics for Diploma in Primary Teacher Education
Use the marks to:
(i)
Find the range.
(ii)
Group the data into classes.
(iii)
Construct a frequency distribution table.
(iv)
Find class boundaries for the second class.
(v)
Find the modal class.
(vi)
Find the class mark for the last class.
Solution:
(i)
From this data, the highest mark is 85 and the lowest mark is 25. The range
is therefore 85 – 25 = 60
(ii)
In grouping the data into classes, first we have to decide on the number of
classes we are going to use, from which we will get an approximation of
the class interval.
60
=5
For 12 classes, the interval will be
12
The classes may then be:
25 – 29
60 – 64
30 – 34
65 – 69
35 – 39
70 – 74
40 – 44
75 – 79
45 – 49
80 – 84
50 – 54
85 – 89
55 – 59
(iii)
Constructing a frequency distribution table.
Class
Tally
Frequency
Class
Tally
Frequency
25 – 29
//
2
60 – 64
////
4
30 – 34
/
1
65 – 69
//
2
35 – 39
/
1
70 – 74
///
3
40 – 44
/
1
75 – 79
/
1
45 – 49
////
4
80 – 84
/
1
50 – 54
/////
///// /
11
85 – 89
/
1
55 – 59
///// ///
8
Basic Mathematics for Diploma in Primary Teacher Education 111
(iv)
Class boundaries for the second class
From the table, the second class is the one containing 30 – 34
The lower class boundary is:
The upper class boundary is:
29 + 30 = 29.5
2
34 + 35 = 34.5
2
(v)
From the frequency distribution table we have constructed, class 50–54
contains the highest frequency therefore this is the modal class.
(vi)
The last class contains 85–89, the class mark for this class will be
85  89 174
 87
2
2
7.5 Graphical Presentation of Grouped Data
In the last lecture, we studied frequency distribution of grouped data. In this
lecture, we are going to study the various graphical methods of representing
grouped data from a frequency table.
Histogram
A histogram is one way by which a frequency distribution can be represented. It is
a form of a bar chart in which the areas of the rectangles are proportional to the
frequencies. The bars are the same width and attached to one another.
To draw a histogram, frequencies are marked on the y-axis (vertical) and class
marks are plotted on the x-axis (horizontal). Please note that in some references
class boundaries and not class marks are plotted on the x-axis.
Example:
Using the frequency table for grouped data in Section 7.3, draw a histogram.
Solution:
Classes
Class Marks
Frequency
25 – 29
27
2
30 – 34
32
1
35 – 39
37
1
40 – 44
42
1
45 – 49
47
4
50 – 54
52
11
55 – 59
57
8
112 Basic Mathematics for Diploma in Primary Teacher Education
60 – 64
62
4
65 – 69
67
2
70 – 74
72
3
75 – 79
77
1
80 – 84
82
1
85 – 89
87
1
Histogram: Scores of 40 students in Mathematics test.
12
11
10
9
FREQUENCY
8
7
6
5
4
3
2
1
0
27
32
37
42
47
52
57
62
67
72
77
82
87
CLASS MARKS
Frequency Polygon
This is another method of representing grouped data graphically. It is also used to
graph a frequency distribution. It is constructed in a much more the same manner
as a histogram, using the same kind of vertical and horizontal scales. Frequency
polygon can be obtained when the midpoints of the classes in the histogram are
linked with straight lines.
Example:
The following are mathematics assignment results of 45 University students.
Construct a frequency polygon to represent the data:
2 students scored 10 – 19; 3 students scored 20 – 29; 4 students scored 30 – 39; 5
scored 40 – 49; 10 scored 50 – 59; 11 scored 60 – 69; 6 scored 70 – 79 and 4 scored
80 – 89
Solution:
Basic Mathematics for Diploma in Primary Teacher Education 113
The scores are already grouped into classes with their corresponding number of
students. To construct a frequency polygon, we need to add another column with
the class mark for each class as follows:
Class
Frequency
Class mark
10 – 19
2
14.5
20 – 29
3
24.5
30 – 39
4
34.5
40 – 49
5
44.5
50 – 59
10
54.5
60 – 69
11
64.5
70 – 79
6
74.5
80 – 89
4
84.5
Total
45
12
11
10
9
FREQUENCY
8
7
6
5
4
3
2
1
0
14.5
24.5
34.5
44.5
54.5
64.5
74.5
84.5
CLASS MARKS
Frequency Polygon: Mathematics assignment scores for 45 University students.
Cumulative Frequency Curve
A cumulative frequency table is also called ‘ogive’. This is constructed from a
cumulative frequency table. Cumulative frequency is a total obtained after adding a
given frequency to the previous total. Cumulative frequency table can be
constructed in two ways, that is ‘less than’ or ‘more than’.
114 Basic Mathematics for Diploma in Primary Teacher Education
Consider the example above of assignment scores for 45 University students. In the
first class i.e. 10 – 19 there are two students, this means there are only two students
who scored less or equal to 19. In the second class there are 3 students who scored
less or equal to 29. Now adding the two students in the 10 – 19 class we have 5
students who scored less or equal to 29. Likewise we have 4 students who scored
less or equal to 39, adding the 5 students in the previous two classes we have 9
students who scored less or equal to 39; and so on. We can thus construct a ‘less
than’ cumulative frequency table as follows:
Class
Frequency
Cumulative frequency
10 – 19
2
2
20 – 29
3
5
30 – 39
4
9
40 – 49
5
14
50 – 59
10
24
60 – 69
11
35
70 – 79
6
41
80 – 89
4
45
Total
45
We can construct the above table on a more than basis. In this case, we are doing
the opposite of the first case. Here the cumulative frequency for the first class, i.e.
10 – 19 is 45, meaning that 45 students scored 10 or more. Likewise, for the second
class we have 42 students who scored 20 or more, and so on. The ‘more than’
cumulative frequency table will be as follows:
Class
Frequency
Cumulative frequency
10 – 19
2
45
20 – 29
3
42
30 – 39
4
38
40 – 49
5
33
50 – 59
10
23
60 – 69
11
12
70 – 79
6
6
80 – 89
4
2
Total
45
For the above two cumulative frequency tables we can construct frequency curves.
Please, note that instead of class marks, here we use the class boundaries. The
ogives will be as follows:
‘Less than’ Ogive
CUMULATIVE FREQUENCY
Basic Mathematics for Diploma in Primary Teacher Education 115
45
40
35
30
25
20
15
10
5
0
19.5
29.5
39.5
49.9
59.5
69.5
79.5
89.5
CLASS BOUNDARIES
CUMULATIVE FREQUENCY
‘More than’ Ogive
45
40
35
30
25
20
15
10
5
0
19.5
29.5
39.5
49.9
59.5
69.5
CLASS BOUNDARIES
Activity
Which of the following would give:
(a)
Qualitative data
(b)
Discrete quantitative data
(c)
Continuous quantitative data
(i)
Mass
(ii)
Number of cars
(iii)
Favourite football team
(iv)
Colour of car
(v)
Price of chocolate bars
(vi)
Amount of pocket money
(vii)
Distance from home to school
(viii)
Number of pets
(ix)
Number of sweets in a jar
(x)
Mass of crisps in a packet
79.5
89.5
116 Basic Mathematics for Diploma in Primary Teacher Education
Summary
There are two very basic concepts that are of significant importance in statistics
worth mentioning in this lecture: population and sample. Population in statistics
refers to the collection of individual items such as people, places or things about
which information is required. A sample refers to a group of individual items
selected from the population for a study. In statistics, there are two basic types of
data: Primary data and Secondary data.
Statistical data can be collected through various methods. For numerical data to be
able to give meaningful interpretation, they need to be summarized and logically
presented. There are various graphical methods of representing grouped data from
a frequency table. Grouped data, frequency table of grouped data, class limits and
class boundaries have also been discussed in this lecture.
Exercise
1.
Define the term statistics.
2.
Explain briefly the two types of statistics.
3.
Briefly explain the following terms as used in statistics:
(i)
Sample
(ii)
Population
(iii)
Primary data
(iv)
Secondary data
4.
Briefly describe three methods of collecting data.
5.
Production in metric tones of three crops in one region of Tanzania for
five years was as follows:
Year
Cotton
Coffee
Sisal
1988
76
49
202
1989
65
54
180
1990
79
51
157
1991
65
54
155
1992
71
44
124
Present these data using:
Basic Mathematics for Diploma in Primary Teacher Education 117
6.
(i)
Pie charts, each for the first two years
(ii)
Multiple bar charts
(iii)
Percentage bar charts
The following are the areas in millions of square kilometres of oceans of
the world.
Ocean
2
Area (million Km )
Pacific
Atlantic
Indian
Antarctic
Arctic
183.4
106.7
73.8
19.7
12.4
Graph the data using:
(i) a bar chart (ii) a pie chart
7.
Mr. R. had totalled his son’s expenses for the last school year as follows.
Food 20%; Rent 20%; Clothing 10%; Fees and Books 25%; Entertainment
10%; Others 15%.
Draw a pie chart to represent this data.
8.
array
9.
(a)
Arrange the numbers 12, 56, 21, 5, 18, 10, 3, 61, 34, 65, 24 in an
(b)
determine their range
The final marks in mathematics of 80 students at the Open University are
recorded in the accompanying table.
68
84
75
82
68
90
62
88
76
93
73
79
88
73
60
93
71
59
85
75
61
65
75
87
74
62
95
78
63
72
66
78
82
75
94
77
69
74
68
60
96
78
89
61
75
95
60
79
83
71
79
62
67
97
78
85
76
65
71
75
65
80
73
57
88
78
62
76
53
74
86
67
73
81
72
63
76
75
85
77
With reference to this table:
(i)
Find, (a) The highest mark (b) the lowest mark (c) the range
(ii)
Construct a grouped frequency distribution table
(iii)
Find the modal class
(iv)
Find class marks for the first five classes.
118 Basic Mathematics for Diploma in Primary Teacher Education
10.
11.
12.
13.
In the following table, lengths of 40 pieces of timber were recorded to the
nearest centimetre. Construct a frequency distribution table and determine
the modal class.
138
164
150
132
144
125
149
157
146
158
140
147
136
148
152
144
168
126
138
176
163
119
154
165
146
173
142
147
135
153
140
135
161
145
135
142
150
156
145
128
The number of patients attending a certain clinic were recorded for a
period of 40 days as follows:
12
15
11
21
23
19
30
36
29
35
25
18
38
25
31
32
34
36
23
22
14
13
22
24
25
30
37
21
10
12
22
23
21
23
36
22
37
16
39
16
(i)
Construct a grouped frequency table
(ii)
Write down the modal class.
Below are the marks obtained by fifty Diploma students at the Open
University of Tanzania in Basic Mathematics test;
38
54
50
35
59
52
61
50
47
71
65
55
57
35
54
69
58
70
49
45
67
64
60
59
50
63
40
50
57
57
45
57
62
50
59
60
64
67
50
51
49
36
48
51
58
60
62
63
61
60
(i)
Using the class interval of 5, construct a frequency distribution
table for the data and draw a histogram.
(ii)
Draw a frequency polygon representing the data.
Draw a frequency polygon for the following data of ages of 40 people in a
meeting.
Age
No. of
people
14.
21-25
2
26-30
4
31-35
5
36-40
41-45
8
12
46-50
4
51-55
3
56-60
2
Given the following data, construct cumulative frequency tables on a ‘less
than’ and on a ‘more than’ basis and then draw their respective cumulative
frequency curves.
Basic Mathematics for Diploma in Primary Teacher Education 119
Class
Frequency
30 – 39
7
40 – 49
8
50 – 59
15
60 – 69
4
70 – 79
28
80 – 89
10
90 – 99
12
Total
84
References
Brookes, B. C. (1965), The Teaching of Statistics in Schools, Heinemann Educational
Books Ltd., London.
Mosteller, F. & Fienberg, S. (1983), Beginning Statistics, Addison-Wesley Publishing
Company Inc., USA.
Murray, R. S. and Boxer, R. W. (1972), Schaum’s Outline Series: Theory and
Problems of Statistics, McGraw-Hill Book Company UK.
Ott, L. and Mendenhall, W. (1990), Understanding Statistics, 5th Ed., PWS-KENT
Publishing Company, Boston.
120 Basic Mathematics for Diploma in Primary Teacher Education
Basic Mathematics for Diploma in Primary Teacher Education 121
LECTURE 8
Measures of Central Tendency
8.1 Introduction
A measure of central tendency describes a set of data by locating the middle region
of the set. The common measures of central tendency are the arithmetic mean (or
simply the mean), the median and the mode. In this lecture, we are going to discuss
methods of calculating mean, median and mode for both grouped and ungrouped
data.
Learning Objectives
At the end of this lecture, you will be able to:

Define the terms mean, median and mode;

Calculate mean, median and mode of ungrouped data;

Calculate mean, median and mode of grouped data.
8.2 Mean
This is the most popular and best understood measure of central tendency for a
quantitative data set.
Arithmetic Mean of Ungrouped Data
Suppose we have a set of n numbers, x1, x2, x3, x4, …..xn, The arithmetic mean, or
simply the mean of this set of number, denoted by x , is given by dividing the sum
of all the individual numbers or measurements by the total number of
measurements.
Thus x =
Where,
x1  x2  x3  x4  ......  xn
x
symbolically written as x 
n
n
n = the number of measurements
x denotes the arithmetic mean
Example:
122 Basic Mathematics for Diploma in Primary Teacher Education
The mean of the numbers 10, 12, 14, 23, 16, 19, 12 is:
x =
10  12  14  23  16  19  12  14 120
 15
8
8
Thus the mean is 15.
Activity 1
Find the arithmetic mean of the following sets of numbers:
1.
39, 45, 67, 13, 24, 34, 16, 42, 25, 51
2.
15, 35, 60, 75, 90, 25, 30, 45, 55, 80
Arithmetic Mean of Grouped Data
We can expand the concept of arithmetic mean to include any number of
measurements. Suppose now we have grouped data presented in a frequency
distribution, we cannot reconstruct the actual sample measurement from this data,
instead we represent all values in a given class interval by the midpoint (or class
mark) of the interval. If we let x be the midpoint of the interval and let f denote
the frequency in that interval, then the sum of measurements in that interval will
be fx. Following this, the sum of measurements across all n classes, denoted by will
be obtained by adding fx1  fx2  fx3  .....  fxn
Thus, the mean of grouped data will be:
fx
fx

x 
n
f
Example:
Find the mean of the following grouped data.
Solution:
Class
Frequency
10 – 19
2
20 – 29
3
30 – 39
3
40 – 49
5
50 – 59
5
60 – 69
11
70 – 79
10
80 – 89
6
Total
45
Basic Mathematics for Diploma in Primary Teacher Education 123

The first step is to determine the midpoint, x, of each class. Referring to
the previous lecture, this is obtained by adding the lower and upper class
limits and dividing by two. In doing so, we add another column in our
table for the midpoints, x.

The second step is to multiply each class midpoint with their respective
frequencies, f. We again add another column for fx in our table.

Lastly, we substitute the values in our formula.
Class
Frequency (f)
Midpoint (x)
fx
10 – 19
2
14.5
29.0
20 – 29
3
24.5
73.5
30 – 39
3
34.5
103.5
40 – 49
5
44.5
222.5
50 – 59
5
54.5
272.5
60 – 69
11
64.5
709.5
70 – 79
10
74.5
745.0
80 – 89
6
84.5
507.0
Total
 f  45
 f x  2662.5
Substituting the results in the formula:
 fx 2662.5
x  59.17
45
 f
Thus the mean is 59.17
8.3 Median
Median of a set of data is obtained by first, ranking the data from the lowest to the
highest, and then find the middle number. Median is the most useful measure in
describing large set of data.
Median of Ungrouped Data
In determining the range of ungrouped set of data, arrange the numbers in a
sequential order and pick the middle number. If n is odd, then the median is the
middle number, however, if n is even, the median is the mean of the two middle
numbers.
Example:
Find the median of the following sets of scores:
1.
84, 71, 72, 75, 66, 60, 62
124 Basic Mathematics for Diploma in Primary Teacher Education
2.
10, 12, 13, 14, 16, 18, 19, 20
Solution:
1.
Arranging the numbers in a sequential order, we have:
60, 62, 66, 71, 72, 75, 84
71
Note that there are 7 data (odd number), so the median is the fourth data,
Thus the median is 71
2.
Arranging the numbers in a sequential order, we have:
10, 12, 13, 14, 16, 18, 19, 20
Note that we there are 8 data (even number), so we take the mean of the
fourth and the fifth data
14  16 30
 15
2
2
Thus the median is 15.
Median of Grouped Data
Since the actual measurements or observations in grouped data are unknown, we
can only determine the class containing the median, but not where the median is
located in that interval. This problem is eased by the application of the general
formula
n

 f1

2
Median = L  
c
fm


Where,
L = Lower class limit of the class that contains the median (median class)
n = Total number of measurements or items in the data
f1 = The cumulative frequency of the class just before the median class
fm= Frequency of the median class
c = Size of median class interval
The class containing the median can be identified by looking at the class for which
the cumulative frequency exceeds 50% of all the observations.
Example:
Find the median of the data from the previous example.
Basic Mathematics for Diploma in Primary Teacher Education 125
Solution:
To the original table, we add a column of cumulative frequency, c.
Class
Frequency
cf
10 – 19
2
2
20 – 29
3
5
30 – 39
3
8
40 – 49
5
13
50 – 59
5
18
60 – 69
11
29
70 – 79
10
39
80 – 89
6
45
Total
45
The class containing the median will be 60 – 69 since its cumulative frequency is 29
which is more than half of 45.
Applying the formula we have:
Median =
n

 f1

2
L  
c
fm


45

 18

2
Median = 59.5 + 
10
11


22.5  18

11
Median = 59.5 + 10

4.5

11
Median = 59.5 + 10

Median = 59.5 + 4.1
Median = 63.6
When calculating the median, you have to make sure that the final number you
get as the median falls within the class you had previously identified as the one
containing the median.
In the above example, our median 63.6 falls within 60 – 69, which was earlier
identified as the class containing the median.
8.4 Mode
The mode of a set of data or measurement is that number which occurs most often
in the set. Although mode is the least common of the three measure of central
126 Basic Mathematics for Diploma in Primary Teacher Education
tendency we have discussed, it is very useful as a measure of popularity that reflects
central tendency or opinion. For example, in planning purchases in retail business,
mode can be used to give the most popular brand of goods preferred by customers.
Mode of Ungrouped Data
In finding the mode of ungrouped data we look at individual numbers, count each
items and the one that has appeared mostly is the mode. Sometimes all
observations appear the same number of times. In this case the mode gives no
information for locating the central tendency. We therefore say that the set
possesses no mode. In other cases the set of data may possess more than one mode
(bi-modal for two or tri-modal for three etc).
Examples:
Find the mode of each of the following sets of numbers:
1.
20, 28, 12, 15, 16, 10, 28, 21, 14, 20, 28
The mode of this set of data is 28 since it appeared 3 times
2.
1, 2, 2, 5, 7, 7, 9, 9, 9, 10, 10, 10
This set has two modes 9 and 10. It is bimodal.
3.
5, 8, 9, 11, 2, 13, 9
This set has no mode.
Activity 2
Find the mode of the following sets of numbers:
1.
13, 12, 11, 13, 14, 15, 16, 13, 10, 14
2.
23, 26, 22, 20, 22, 27, 28, 21, 24, 29, 24, 25
Mode of Grouped Data
When dealing with grouped data, we define mode to be the midpoint of the class
with the highest frequency. This gives an approximation to the actual mode of the
measurements.
A more precise formula for computing mode from grouped data is given as:


1
c
Mode = L + 
 

1
2
Where,
L = Lower class boundary of the modal class
Basic Mathematics for Diploma in Primary Teacher Education 127
∆1 = Excess of modal frequency over the frequency of the preceding
class
class
∆ 2 = Excess of modal frequency over the frequency of the next higher
c = Class size
Example:
Taking the data from the previous examples:
Class
Frequency
10 – 19
2
20 – 29
3
30 – 39
3
40 – 49
5
50 – 59
5
60 – 69
11
70 – 79
10
80 – 89
6
Total
45
The modal class is 60 – 69 since it is the one with the highest frequency i.e. 11.
The preceding class, 50 – 59 has a frequency of 5 and the next higher class has a
frequency of 10.
Therefore, ∆1 = 11 – 5 = 6
∆ 2 = 11 – 10 = 1
c = 10
Substituting these values in our formula we have:


1  2

1
c
Mode = L + 
6
61

Mode = 59.5  10

Mode = 59.5  0.85710
Mode = 59.5  8.57
Mode = 68.07.
128 Basic Mathematics for Diploma in Primary Teacher Education
Summary
Mean is the most popular and best understood measure of central tendency for a
quantitative data set. The arithmetic mean, or simply the mean of this set of
number, denoted by x , is given by dividing the sum of all the individual numbers
or measurements by the total number of measurements. We can expand the
concept of arithmetic mean to include any number of measurements.
Median of a set of data is obtained by first, ranking the data from the lowest to the
highest, and then find the middle number. In determining the range of ungrouped
set of data, arrange the numbers in a sequential order and pick the middle number.
Since the actual measurements or observations in grouped data are unknown, we
can only determine the class containing the median, but not where the median is
located in that interval.
The mode of a set of data or measurement is that number which occurs most often
in the set. In finding the mode of ungrouped data we look at individual numbers,
count each items and the one that has appeared mostly is the mode. Sometimes all
observations appear the same number of times. When dealing with grouped data,
we define mode to be the midpoint of the class with the highest frequency.
Exercise
1.
Four groups of students consisting of 15, 20, 10 and 18 individuals reported
mean heights of 1.62, 1.48, 1.53 and 1.40 metres respectively.
2.
Find the mean height of all the students.
3.
Find the mean, median and mode of each of the following sets of numbers
4.
(i)
5, 3, 6, 5, 4, 5, 2, 8, 6, 4, 5, 3, 8, 9
(ii)
12, 17, 25, 23 15 17, 15, 16, 18
The table below shows the weights in Kg of 100 male students in a certain
University.
Weight (kg)
Frequency
60 – 62
5
63 – 65
18
66 – 68
42
69 – 71
27
Basic Mathematics for Diploma in Primary Teacher Education 129
72 – 74
8
Calculate:
(i)
Mean
(ii)
Median
(iii)
Mode
References
Brookes, B. C. (1965), The Teaching of Statistics in Schools, Heinemann Educational
Books Ltd., London.
Mosteller, F. & Fienberg, S. (1983), Beginning Statistics, Addison-Wesley Publishing
Company Inc., USA.
Murray, R. S. and Boxer, R. W. (1972), Schaum’s Outline Series: Theory and
Problems of Statistics, McGraw-Hill Book Company, UK.
Ott, L. and Mendenhall, W. (1990), Understanding Statistics, 5th Ed., PWS-KENT
Publishing Company, Boston.
LECTURE 9
Measures of Dispersion
9.1 Introduction
In the last lecture, we discussed about the measures of central tendency. These were
given as the mean, median and mode. In this lecture, we are going to discuss the
measures of dispersion or variation of a given set of data. Dispersion or variation of
the data is the degree to which that data tend to spread about an average. The
measures we are going to discuss are Range, Standard deviation and Variance.
These measures tell us how observations are spread, particularly how they are
spread about the middle value.
130 Basic Mathematics for Diploma in Primary Teacher Education
Learning Objectives
At the end of this lecture, you will be able to:

Explain the meaning of the terms range, standard deviation and variance;

Calculate range, standard deviation and variance of ungrouped data;

Calculate range, standard deviation and variance of grouped data.
9.2 Range
The range of a set of data is used extensively as a measure of variability in
summaries of data that are made available to the general public, for example in
rating quality of goods etc. As we have already mentioned in the previous lecture,
the range of a set of data is given by the difference between the largest and the
smallest measurements.
Example:
The range of the set 2, 3, 3, 5, 5, 5, 8, 10, 12 is 12 – 12 = 10.
Sometimes the range is given by simply quoting the smallest and the largest
numbers in the set, for example in this set the range could be indicated as 2 to 12,
or simply 2 – 12.
Example:
The largest of 50 measurements is 8.34kg. If the range is 0.46kg, find the smallest
measurement.
Solution:
Range = Largest measurement – Smallest measurement
Let the smallest measurement be x.
Then,
0.46kg = 8.34kg – xkg
x = 8.34kg – 0.46kg
= 7.88kg
The smallest measurement is 7.88kg.
Activity 1
Find the range of each sets of numbers:
Basic Mathematics for Diploma in Primary Teacher Education 131
1.
12, 6, 7, 3, 15, 10, 18, 5
2.
9, 3, 8, 8, 9, 8, 8, 9, 18
Range of a Grouped Data
For a grouped data, the range will be obtained by calculating the difference
between the largest and the smallest midpoints.
Example:
Determine the range from the following frequency distribution table:
Weight (kg)
Frequency
57 – 59
3
60 – 62
7
63 – 65
10
66 – 68
22
69 – 71
25
72 – 74
9
Here the smallest midpoint is
72  74
57  59
 73
 58 and the largest is
2
2
Therefore the range will be 73 – 58 = 15.
9.3 Standard Deviation
The standard deviation is the root mean square of the deviations of each of the
numbers xi from the mean x . It is a very important measure of dispersion.
Standard Deviation of Nngrouped Data
For ungrouped data, the standard deviation is calculated through the following
steps:
Given n number of measurements, say x1, x2, x3, …..xn
1.
Find the mean of the data, x
2.
For each measurement, determine its difference with the mean i.e. xi – x
3.
Square these deviations to obtain (xi – x )2
4.
Add all squared deviations to obtain  xi  x
5.
2

xi  x
Divide by n to get the mean of values in step 4 i.e.  
n


2


132 Basic Mathematics for Diploma in Primary Teacher Education
6.
Take the square root in step 5 to end up with

 xi  x

2
n
What we now have is referred to as the Standard Deviation (SD) and is commonly
denoted by σ.
Hence, for ungrouped data, standard deviation is obtained from the expression;
σ=

 xi  x

2
n
Please note that the values (xi – x ) in step 2 always add up to 0. To avoid this we
have squared them in step 3. We then find the square root in step 6 so as to get the
average deviations that are free from the effect of squatting.
Example:
Find the standard deviation for the following set of numbers: 12, 6, 7, 3, 15, 10, 18,
5.
Solution:
The first step is to find the mean;
12  6  7  3  15  10  18  5
 9.5
8
There are 8 values, so n = 8
We can now organize our work in a tabular form to work out steps 2 and 3
xi
xi – x
(xi – x )2
12
12 – 9.5 = 2.5
6.25
6
6 – 9.5 = –3.5
12.25
7
7 – 9.5 = –2.5
6.25
3
3 – 9.5 = –6.5
42.25
15
15 – 9.5 = 5.5
30.25
10
10 – 9.5 = 0.5
0.25
18
18 – 9.5 = 8.5
72.25
5
5 – 9.5 = –4.5
20.25
 xi – x =0
190.00


Now we shall substitute our values in the formula,
σ=

 xi  x
n
σ
=
190
8
σ
=
23.75

2
Basic Mathematics for Diploma in Primary Teacher Education 133
σ
= 4.870.
Activity 2
Find the standard deviation of the following set of numbers:
2, 5, 8, 11, 14
Standard Deviation of Grouped Data
When working with grouped data, the same standard deviation formula is extended
by using  f instead of n, and the midpoints of each group as the corresponding ×
values. Here also the mean of the data x , will be obtained by using the method for
grouped data, discussed in the previous lecture.
Therefore instead of x , our formula will include
 fx
and instead of n it will be
 f
 f
Substituting these expressions in the formula σ =
f x 
fx

2
the formula σ =

 xi  x

2
n
we will end up with
2


f
f


Hence, for grouped data, standard deviation is obtained from the expression;
fx 
fx

2
2


f
f


Example:
Find the standard deviation for the following data:
Class
Frequency
30 – 34
4
35 – 39
9
40 – 44
14
45 – 49
8
50 – 54
5
Solution:
We shall organize our work in a tabular form as follows:
Class
Midpoint (x)
Frequency (f)
fx
x2
fx2
30 – 34
32
4
128
1024
4096
134 Basic Mathematics for Diploma in Primary Teacher Education
35 – 39
37
9
333
1369
12321
40 – 44
42
14
588
1764
24696
45 – 49
47
8
376
2209
17672
50 – 54
52
5
260
2704
13520
 f =40
 fx  1685
2
 fx  72305
Then, substituting the values in the formula we get:
σ
=
fx 
fx

2
2


f
f


2
σ
=
72305 
1685
 

40
40
σ
=
72305 2839225

40
1600
σ
=
33.11
σ
= 5.75
We can observe that the dispersion of a set of data is small if the values are bunched
closely about their mean and that it is large if the values are scattered widely about
their mean. Consequently, the standard deviation of a set of data is small if the
values are concentrated near the mean and is large if the values are widely scattered
about the mean.
9.4 Variance
This is another measure of dispersion.
The variance of a set of data is defined as the square of the standard deviation, and
thus given by σ 2.
Variance = σ 2
Activity 3
Variance is the square of standard deviation.
Thus, taking the formula for standard deviation,
Basic Mathematics for Diploma in Primary Teacher Education 135
σ
=

fx
fx

2
2
the square of σ i.e. variance will be:


f
f


Variance (σ 2) =
2

fx
fx

2
 
f
f


Example:
Find the variance for the data in the previous page.
Variance (σ2) = 5.752
= 33.11.
Summary
A measure of dispersion of a set of data is the degree to which that data tend to
spread about the average. The measures of dispersion discussed in this section are
range, standard deviation and variance.
The range of a set of data is given by the difference between the largest and the
smallest measurements. For a grouped data, the range will be obtained by
calculating the difference between the largest and the smallest midpoints.
The standard deviation is the root mean square of the deviations of each of the
numbers, xi, from the mean, x .
For ungrouped data, the standard deviation is calculated by using the expression

 xi  x
σ=

2
n
Where xi are the individual values, x is the mean of the data and n is the number
of values
For grouped data, standard deviation is obtained by the formula
fx 
fx

2
2


f
f


The variance of a set of data is defined as the square of the standard deviation, and
thus given by σ2.
Variance = σ2.
136 Basic Mathematics for Diploma in Primary Teacher Education
Exercise
1.
Find (a) the range (b) the standard deviation (c) the variance for the
following set of numbers: 9, 3, 8, 8, 9, 8, 9, 18
2.
From the table of weights of 100 male University students
Weight (kg)
Frequency
60 – 62
5
63 – 65
18
66 – 68
42
69 – 71
27
72 – 74
8
Find:
3.
(i)
The range
(ii)
The standard deviation
(iii)
The variance
The following table shows the temperature (oF) recordings of twenty days
in a certain town by the meteorology department:
Temperature (oF)
Frequency
76 – 79
2
80 – 83
3
84 – 87
8
88 – 91
4
92 – 95
1
96 – 99
1
Compute (a) Standard deviation, (b) Variance.
4.
Find the mode and the median of 54, 30, 36, 50, x, 40 if their mean is 40.
5.
Given below are data on primary school enrolment for girls and boys in
one region during six years from 2000–2005.
Year
% of Girls
% of Boys
2000
53
47
2001
49
51
2002
54
46
2003
44
56
2004
57
43
Basic Mathematics for Diploma in Primary Teacher Education 137
2005
41
59
Represent these data by using a bar chart.
6.
7.
Determine the mean of the following frequency distribution:
x
18
19
22
21
22
23
f
3
5
4
6
3
2
The table below gives the diameters in mm of sample tubes produces in a
certain industry:
Diameter
(mm)
58 – 60
61 – 63
64 – 66
67 – 69
70 – 72
Frequency
3
4
12
9
2
(i)
Draw a frequency distribution table which will include mid-points
of the intervals and the cumulative frequencies
(ii)
Draw a histogram of the distribution
(iii)
Plot the cumulative frequency curve (Ogive)
(iv)
Find from the distribution table, the number of tubes with
diameters greater that 66mm.
References
Murray, R. S. and Boxer, R. W. (1972), Schaum’s Outline Series: Theory and
Problems of Statistics, McGraw-Hill Book Company, UK.
Mosteller, F. & Fienberg, S. (1983), Beginning Statistics, Addison-Wesley Publishing
Company Inc., USA.
Ott, L. and Mendenhall, W. (1990), Understanding Statistics, 5th Ed., PWS-KENT
Publishing Company, Boston.
Brookes, B. C. (1965), The Teaching of Statistics in Schools, Heinemann Educational
Books Ltd., London.