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The real Number system
What is the real number system?
All the numbers we use for counting, measuring and calculating are called real numbers. All the
numbers you will come across at this stage are real numbers. They are made up of a collection of
different types of numbers.
Types of Numbers
Natural numbers (counting numbers)
These are the numbers 1, 2, 3, 4, etc. The smallest is 1. We also call these numbers COUNTING
NUMBERS. There is no largest natural number. This set of numbers is denoted by the symbol N.
We may therefore write this set as follows:
N  1; 2; 3; 4;...
Whole numbers (W)
Whole numbers include all natural numbers and zero. The numbers 0, 1, 2, 3,… are called whole
numbers. Remember that every natural number is a whole number. Zero is not a natural number
therefore not every whole number is a natural number.
Integers (Z)
All the positive and negative whole numbers including zero are integers. This set is denoted by
the symbol Z. We may write this set as follows:
Z  ..., 4,  3,  2, 1, 0, 1, 2, 3, 4,...
You can represent these numbers on a number line as shown below:
Rational numbers (Q)
These are numbers of the form
a where a and b are integers and b  0 . This means that
b
the numbers we can write down as one integer divided by another integer so long as the bottom
integer is not zero. These integers can be positive or negative. The list of rational numbers is also
1 ,  3 , 11, 201, 3 3 is also a rational number,
2 5 6 49
a as 3 .
because we can express 3 in terms of
b 1
endless. Some rational numbers are
This collection is denoted by the symbol Q. Thus we can state that:
Q   a such that a and b areintegersand b  0 
b

Irrational numbers ( Q )
Any number which is not rational is referred to as an IRRATIONAL NUMBER.
Examples:
 (which has the value 3.141592654...), 2, 5, 2.7182818284590452358
and so on. These numbers cannot be written as a quotient of two integers. More generally, the
square roots of all prime numbers are irrational numbers. We shall denote the collection of all
irrational numbers by
Q.
Primes and Composites
x 1
is said to be prime if and only if it is divisible by 1 and itself.
A natural number
Meaning, a prime number has only two natural factors. Otherwise, a number which is not a
prime is called a composite - i.e. it is composed of more than two natural factors.
EXAMPLES

Primes:

Composites:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, etc.
4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, etc.
NOTE: Zero and number 1 are not prime numbers.
Factors
The factors of a number are those numbers which
divide exactly into a given number.
E.G. The factors of 24 are 1, 2, 3, 4, 6, 12, 24
The pairs of factors of 30 are:
1 x 30
2 x 15
3 x 10 and
5x6
Therefore, the factors of 30 are 1, 2, 3, 5, 6, 10, 15, 30
Multiples
A multiple of a number
n is k  n where k is a counting
number.
Examples:
Some multiples of 5 are 5,10,15, …
Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, …
Factors and Multiples (open slides)
Prime Factors
Remember that the factors of 30 are:
1, 2, 3, 5, 6, 10, 15 and 30
The prime factors of 30 are:
2, 3, and 5
We can find the prime factors by expressing a number as a product of a primes.
How to express a number as a product of primes or product of prime factors:
E.G.
Express 44 as a product of primes (prime factors)
1.
Write down the 1st few prime numbers e.g.
2, 3, 5, 7, 11, 13
Divide 44 by the 1st prime number (2) as many times as possible until it can no
longer divide exactly into that number.
2.
3.
4.
Divide 44 by the next prime number (3) as many times as possible and so on until
you get 1.
Write down the product of all the prime numbers you divided in.
44 : 22 : 11 : 1
2
2
11
e.g.
or
2 44
2 22
11 11
1
Therefore
5.
44  2  2  11
Write any repeated prime number as powers
(using index form) e.g.
44  22 11
Exercises
(a)
(b)
Or
Express 6960 as a product of primes
Express each of the following numbers as a product of prime numbers.
Decompose the following numbers into prime factors




1200
4464
8 000
2464
Highest Common Factors ( HCF) or Highest Common Divisor(HCD)
To find the HCF we can apply the following:
Method 1: List all factors of the given numbers and
Find the highest common factor of 8 and 12
We know that factors of 12 are: 1, 2, 3, 4, 6, 12
identify the HCF.
and factors of 8 are: 1, 2, 4, 8
The common factors of 8 and 12 are 1, 2, and 4.
But the highest common factor is 4, hence HCF is 4.
Method 2: Factorise the given numbers into their prime factors respectively. Select those
common factors (with lowest power) and multiply them together.
e.g.
for 12:
12 : 6 : 3 :1
2 2 3
12  22 3
and
For 8:
8 : 4 : 2 :1
2 2 2
3
8  2 2  2  2
Hence, the HCF of 12 and 8 is
22  4
Exercises:
Find the HCF of
(a)
(b)
(c)
(d)
(e)
24, 72, 96 and 300
25, 455, 1050
9 and 45
12, 26 and 36
255 and 75
Multiples
A multiple of a number n is k x n where k is a counting
number.
Examples:
Multiples of 5 are 5,10,15, …
Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, …
Lowest Common Multiple (LCM)
Method 1: List the multiples of the given numbers and
LCM.
identify the
Find the LCM of 12 and 8
8 = 8, 16, 24, 32
12 = 12, 24, 36
24 is the LCM of 12 and 8
Method 2: Factorise the given numbers into their prime factors respectively. Select every
number (prime factors) with highest power which occur in any of the decompositions
(prime factors) of each of the given numbers and multiply them together.
The LCM of 12 and 8
For 12:
12 : 6 : 3 :1
2 2 3
12  22 3
And
For 8:
8 : 4 : 2 :1
2 2 2
8  2 2 2  23
Hence, the LCM of 8 and 12 is =
Exercises:
Find the LCM of
23 3  24
(a)
24, 72, 96 and 300
(b)
25, 455, 1050
(c)
9 and 45
(d)
(e)
12, 26 and 36
255 and 75
Problem sums on HCF and LCM can be sometimes tricky as they are not easy to identify. The
main focus here is how to determine when to find the HCF and when to find the LCM of the
numbers involved in the problem sums.
First let’s take a look at a problem involving the HCF.
3 strings of different lengths, 240 cm, 318 cm and 426 cm are to be cut into equal
lengths. What is the greatest possible length of each piece?6
LCM problem:
Two lighthouses flash their lights every 20s and 30s respectively. Given that they
flashed together at 7pm, when will they next flash together?
One method to finding the next time the lighthouses flash together is:
20, 40, 60
30, 60, 90
60 is a multiple common to 20 and 30, and thus the lighthouses will flash together
in 60s’ time, i.e. at 7:01pm.
This is the same as finding the lowest common multiple, or LCM:
More Examples:
1.
As a humanitarian effort, food ration is distributed to each refugee in a refugee camp. If
a day’s ration is
284 packets of biscuits, 426 packets of instant
noodles and 710
bottles of water, what is the greatest possible number of refugees are there in the
camp?
[142 refugees]
2.
294 blue balls, 252 pink balls and 210 yellow balls are distributed equally among some
students with none left over. What is the biggest possible number of students? [42 students]
3.
A group of girls bought 72 rainbow hairbands, 144 brown and black hairbands, and 216
bright-coloured
hairbands. What is the largest possible number of girls in the group?
[72 girls]
4.
A man has a garden measuring 84 m by 56 m. He wants to divide them equally into the
minimum
number of square plots. What is the length of each square plot? [28 m]
5.
Leonard wants to cut identical square as big as he can from a piece of paper 168 mm by
196 mm. What
is the length of each square? []
6.
Candice, Gerald and Johnny were jumping up a flight
of stairs. Candice did 2 steps
at a time, Gerald 3 steps
at time while Johnny 4 steps at a time. If they started
on the
bottom step at the same, on which step will all 3 land together the first time? []
7.
Heidi helps out at her mum’s stall every 9 days while her sister every 3 days. When will
they be together if they last helped out on June 16, 2008?
8.
A group of students can be further separated into groups of 5, 13 and 17. What is the
smallest possible total number of students?
9.
Jesslyn goes to the market every 64 days. Christine goes to the same market every 72
days. They met each other one day. How many days later will they meet each other
again? []
10.
A Polytechnic choir coordinator wants to
divide the choir into smaller
groups. There are
24 sopranos, 60 altos and 36 tenors. Each group
will have the same number of each type of
voice.
10.1 What is the greatest number of
can be formed?
groups that
10.2 How many sopranos, altos and
in each group?
tenors will be
FUNDAMENTAL OPERATIONS ON WHOLE NUMBERS
Directed Numbers
To add two directed numbers with the same sign, find the sum of the numbers and give the
answer the same sign.
Examples
3 (5)  8
7  (3)  10
9.1 (3.1)  12.2
2  (1)  (5)  8
To add two directed numbers with different signs, find the difference between the numbers
and give the answer the sign of the larger number.
Examples
7  (3)  7  3  4
9  (12)  9 12  3
8  (4)  8  4  4
To subtract a directed number, change its sign and add.
Examples
7  (5)  7  5  2
7  (5)  7  5  12
8  (4)  8  4  12
9  (11)  9 11 2
MULTIPLICATION
(+75) x (-88) =
(-97) x (-93) =
(-27) x (+49) =
(-78) x (-33) =
(+31) x (-52) =
(-44) x (+22) =
Integer Division
(+4437) ÷ (-87) = -
(-7644) ÷ (-98) = +
(-2560) ÷ (+64) = -
(+2376) ÷ (+54) =
(+360) ÷ (+36) =
+
(0) ÷ (-6) =
(-240) ÷ (-3) =
(+15) ÷ (+1) =
Rules for multiplications
Pos. number x pos. number = pos. number
Neg. number x neg. number = pos. number
Neg. number x pos. number = neg. number
Pos. number x neg. number = neg. number
BASIC ARITHMETIC
Rules of Arithmetic
BEDMAS
(brackets, exponents, division, multiplication, addition and subtraction)
BODMAS
(brackets, powers, division, multiplication, addition and subtraction)
1.
Work out brackets
2.
work out powers (exponents)
3.
Divide and multiply
4.
Add and subtract
1. 3(4 1)  35 15
2. 4  3 2  4  6
3. 3 4(5  2)  8  2  5
4. 412  6  6
5. 12  22  (4  23)
6. 32 3 3 22   7  411


7. 2 4  23 8  4  (7)
2
8. 2  4    2  21  4
Simplify each of the following
1. 2(4  23) 8  4
2.



4  6   3 4  2   6
3. (2  3)3   6  2 4  7 
4. 2 3 7  2   5  6   2  3  
5.  4  2   6  4    4  12 10 


6. 25  3 2   7  (8)  (5  6  44)
7.



2  5   2  5  3  8
8. 5  3  4  7   6   6  9  


9.   813 209   54
10. 72  2(2  4)  24   24  (3 5) 17 
3 7  2   5 × 6   2  3 

11. 2  
 



12.
13.

14. 

4  2   6  4   + 4 ÷ 12 10 



3 9   6  3   4  12  7 

5  3  4  7    6   6  9 
Vulgar Fractions – Concepts and operations
In a fraction
7,
8
7 is a numerator and 8 is a denominator.
7 is referred to as a proper fraction and
8
8 is referred to as an improper.
7
2 7 is referred to as a mixed number.
8
Equivalent fractions:
246 8
3 6 9 12
OPERATIONS
Evaluate and simplify your answer.
1.
1  2  1  1   2 2  2  1
5 5  2 4  3 3 2
2.
1  2  1  2   4 3  1
3 5  4 3  4 2
3.


113 2  2 1   1  2 5  1 
3 3 2 5 3
2
Work out and simplify.
1.
31
8 5
7 2
10 3
3.
2  1  3 1
3 5 4 3
2. 3  1  7  2
8 5 10 3


4. 11  2 2   4 11 
2 3 
3
5. 11  2 2  7  7
2 3
9
6. 2 2 11  5 3
3 2


7. 2 3  2 3   7  5 
7 4 8 8
5.
From a group of athletes,
1 of the athletes are chosen for long jump and 1 of the
8
4
remaining athletes were chosen for javelin. One hundred and five athletes remained and they
were all chosen for relay race.
5.1 How many athletes were chosen for long
A. 75
B. 105
jump?
C. 20
D. 35
5.2 How many athletes were chosen for
A. 160
6.
B.
40
javelin?
C. 35
Jane earns a salary every month. She spends
accommodation and
for other
D.
30
N $6800 which is 1 of her salary on
5
N $3 400 on food. What
purposes?
fraction of her salary is left
A.
7.
1
3
B.
2
5
In 2007, a number of
auctioneer sold
number of
C.
7
10
D.
5 200 vehicles were sold
3 of the vehicles. In the next
5
3
5
at an auction. In the first 3 hours the
2 hours, he sold
vehicles. In the last hour the auctioneer sold
1 of the remaining
5
1 of the original number of
5
vehicles.
7.1
How many vehicles were sold during the first
A.
7.2
749
B.
1456
C.
333
1
4
5
12
C.
3120
hours?
D.
1040
decide to buy a car. Alex pays
1 of the cost and Charles pays
3
B.
D.
7
12
1 of the
4
the rest.
D.
2
7
N $12 000 more than Alex. Calculate the cost of the car.
B. N $144 000
A. N $58 000
Brenda pays
C.
9.
416
C.
What fraction of the cost does Charles Pay?
A.
8.2
4160
Three friends, Alex, Brenda and Charles
cost, Brenda pays
8.1
B.
How many vehicles were sold in the last three
A.
8.
3536
five hours?
N $102 000
D.
N $60 000
Frieda earns a salary every month. She spends
accommodation and
for other purposes?
N $3 400 which is 2 on
5
N $1 700 on food. What fraction of her salary
is left
A.
10.
1
3
B.
2
3
In 2007, a number of
auctioneer sold
number of
C.
1
5
5 200 vehicles were sold
2 of the vehicles. In the next
5
vehicles. In the last hour the auctioneer sold
D.
2
5
at an auction. In the first 3 hours the
2 hours, he sold
1 of the remaining
4
1 of the original number of
4
vehicles.
11.1
How many vehicles were sold during the
A.
11.2
B.
780
C.
How many vehicles were sold in total?
A.
12.
2 860
4160
B.
585
C.
3 380
4680
Mr. Kakololo accumulated a number of shares. He sold
cousin and 5 340
which is
best friend, Lucas.
2 of
15
first five hours?
D.
D.
2 080
1 300
1
5
1 of his shares to his brother, 1 to
5
3
the original number of shares to his
12.1
What fraction of shares remained with Mr. Kakololo?
12.2
How many shares did Mr. Kakololo sell in total?
Work out each of the following and simplify
13.
Mr. Titus had 640 shares. He sold out one third of them to a trading company and
2 of
5
the remainder to another company. How many shares remained with Titus?
14.
Andrew sold half of his cows; gave his younger
cows. How
15.
many cows
brother
1 of the original number of
4
does Andrew now have if he had 88 originally?
The Simon’s family spends
2 of their income on
5
rent,
1 on food, and 1 on clothes.
5
4
If they are left with N$390.00 each month, find:
16.
(a)
the fraction which is left.
(b)
their monthly income
A man spends
left with
42.
2 of their income on rent, 1 on
5
4
N$390.00
food, and
1 on clothes. If they are
5
each month, find:
(a)
The total fraction, spent.
(b)
The amount, which is left
There are 60 000 soccer supporters at a game
and the police estimated that
5 of
8
them support
the
43.
home team. Estimate the number who
A man saves N$240 every month. This is
supports the away team.
4 of his monthly salary. Calculate his
25
monthly salary.
44.
The company decided to donate
went to the orphans association,
25000 shares to some institutions. 2 of the shares
5
1 went to the cancer association and the remaining part was
4
donated to a church ‘Praise The Lord’
45.
(a)
How many shares did church ‘Praise The
(b)
What fraction of the shares did ‘Praise The Lord’ receive?
A farmer takes 250 chickens to be sold at a market. In the first hour he sells
chickens. In the second hour he sells
46.
Lord’ receive?
3 of those he has left.
5
(a)
How many chickens were sold in the second
(b)
How many chickens has he sold in total?
Frieda earns a salary every month. She spends
accommodation and
2 of his
5
hour?
N $3400 which is 2 on
5
N $1700 on food. What fraction of her salary is left for other
purposes?
Madam Ecka shares her monthly salary with her children as follows: Maria receives
mother’s salary and Tom receives
2 of her
7
1 . If Tom receives N $950 , how much does Maria
5
receive?
47.
Three friends, Alex, Brenda and Charles decide to buy a car. Alex pays
Brenda pays
1 of the cost and Charles pays the rest.
3
47.1
What fraction of the cost does Charles Pay?
47.2
Brenda pays
cost of the car.
N $12 000 more than Alex.
1 of the cost,
4
Calculate the
Decimal Fractions
Converting vulgar Fractions to Decimal Fractions
1. 7  7 8  0.875
8
2. 1  0.3333
3
3. 9  0.9
10
4. 1 7 1.7
10
5. 2 2  2.285714285714
7
6. 3 19  3.19
100
Change decimals to Vulgar fractions and simplify
1. 0.35  35  7
100 20
2. 0.7  7
10
3. 3.2  3 2  31
10 5
4. 0.007  7
1000
5. 0.00011
11
100 000
6. 2.35  2 35  2 7
100 20
7. 0.625  625  5
1000 8
Types of Decimals
1.
e.g.
2.
Terminating Decimals
7 1.4 or
5
1  0.5
2
Recurring or repeating decimals
e.g.
3.
e.g.
2  0.66666 or
3
Non-recurring – non terminating decimals
12  0.214285714
56
Use your calculator to evaluate
1.
2.
3.
4.
0.6
0.27140.00589
2.41670.000717
6.51 0.1114
7.24 1.653
4.7  1.6
11.4 3.61 9.7




9.6  1.5
2.4 0.74




2


4.2  1  1 
 5.5 7.6 
5.
APPROXIMATIONS
Decimal Places and Significant Figures
A significant figure is a first non-zero digit.
(a)
7.8126  7.81 to 3 sf
(b)
0.078126  0.0781 to 3 sf
(c)
3596  3600 to 2 sf
and 0.078 to 3 dp
SIGNIFICANT FIGURES AND DECIMAL PLACES
1.
(a)
(b)
(c)
Write the following numbers correct to:
Three significant figures
Two significant figures
Two decimal places
1. 8.174
2.
3. 20.041
4. 0.814 52
5. 311.14
7. 0.007 47
19.617
6.
8.
15.62
0.275
9. 900.12
10.
3.555
11. 5.454
12.
20.961
13. 0.0851 14.
0.5151
15. 3.071
2. Express the following numbers correct to the indicated number of significant figures.
40,283 (2 s.f.)
a)
b) 0,0275 (2 s.f.)
c)
4090,01249 (3 s.f.)
d) 20,17 (3 s.f.)
e)
38 (4 s.f.)
f)
1017,2 (3 s.f.)
4099,789 (3 s.f.)
g)
h)
3.4104 (1 s.f.)
i)
4.2578103
j)
4.2578104
( 3 s.f.)
(3 s.f.)
STANDARD FORM
The number
integer.
To write
a10n is in standard form when 1 a 10 and n is a positive or negative
3 000 in standard form 3 1000  3103
150 1.5100 1.5102
0.0004  4 1  4104
10 000
Write the following numbers in standard form
1.
46 000
3.
46
900 000
4.
0.007
5.
0.421
6.
0.000 055
7.
564 000
8.
0.0004
9.
19 millions
10.
A hydrogen atom weighs 0.000 000 000 000 000 000 001 67 grams. Write this weight in
standard form.
11.
The population of China is estimated at 1 100
000 000. Write this in standard form.
12.
to 2
The population of China is estimated at 1 100
significant figures.
000 000. Write this in standard form
2.
POWERS AND ROOTS. REAL NUMBERS
xn signifies that x is multiplied by itself n times.
x is referred to as the base and n is termed as an exponent or index.
Given a positive integer,
By convention an exponent of 1 is not expressed.
Square numbers are 1, 4, 9, 16, 36 etc. They are called perfect squares
The square of
a is a2 which is a  a .
A cube number is the result of multiplying a number by itself three times.
222  23  8
and
38  2
Cube numbers are e.g. 1, 8, 27, 64 etc
64 means 6 multiplied itself four times
64  66666 1296
Remember:
and
(7)2  77  49
4 1296  6
but
72  (77)  49
Taking a square root of a positive number gives two possible answers, - or +
Work out:
1.
0.36
2.
25
3.
0.09
4.
5 32
5.
3 27
81
6.
5 167.9421
leave answer to 3dp


 5.67193 


7.


3 2.312  0.92
2.319.81
(2sf)
1
5103
8.
4
9.
812 802
10.
11.



2.311.01
(3sf)
3
2
 0.79  0.041




6 64
1331
(standard form)
(4sf)
ALGEBRAIC EXPRESSIONS
Terms, Constants, Coefficients and Variables
An algebraic expression is made up of the signs and symbols of algebra. These symbols include
the Arabic numerals, literal numbers, the signs of operation, and so forth.
TERMS AND COEFFICIENTS
The terms of an algebraic expression are the parts of the expression that are connected by plus
and minus signs.
An expression
150 px  25 py  80 pz , is an algebraic expression.
Let us consider the algebraic expression below.
2 x 2  3 x  6 xy  4 y  5
The expression above has 5 terms, namely,
2x2 ,
3 x,
 6 xy ,
 4 y , and 5
An expression containing only one term, such as 3ab, is called a monomial (mono means one).
A binomial contains two terms; e.g. 2r + by.
A trinomial consists of three terms.
Any expression containing two or more terms may also be called by the general name,
polynomial (poly means many).
See the following term:
Examples
Given the two algebraic expressions below, identify each variable and its coefficient. Also state
the constants.
(a)
(b)
3b  5  4 xy  x3
x2  4x 12
BASIC ALGEBRAIC EXPRESSIONS AND OPERATIONS
1.
Terms, Constants, Variables and Coefficients
2.
Simplification (Addition, Subtraction)
3.
Expansion of algebraic expressions
Brackets and simplifying
Two brackets (Expansion)
4.
Factorization
5.
Addition, Subtraction, Multiplication and
Division of Algebraic expressions
Simplification:
‘Like terms’ are terms that contain the same variables raised to the same power.
When we add or subtract algebraic expressions, we simply collect the like terms. The process of
collecting the like terms is called simplification.
4 x  7 y  5x , simply identify the like terms. In this
case 4 x and 5x are like terms hence we can add the two terms to become 1x which
we normally write as x since the coefficient is one. There is only one term with y variable
which is 7 y . Since there is no other like term, we keep the term as it is. If we then simplify the
whole expression then we have x  7 y
If we are to simplify the expression
Algebraic Manipulation
Addition and Subtraction of polynomials
A monomial is the product of non-negative integer powers of variables. Consequently, a monomial has
NO variable in its denominator. It has one term. (mono implies one)
13, 3x, -57, x², 4y², -2xy, or 520x²y² (note: no negative exponents, no fractional exponents)
A binomial is the sum of two monomials. It has two unlike terms.
(bi implies two)
3x + 1, x² - 4x, 2x + y, or y - y²
A trinomial is the sum of three monomials. It has three unlike terms. (tri implies three)
x2 + 2x + 1, 3x² - 4x + 10,
2x + 3y + 2
A polynomial is the sum of one or more terms. (poly implies many)
x2 + 2x, 3x3 + x² + 5x + 6,
4x - 6y + 8
Polynomials are in simplest form when they contain no like terms.
x2 + 2x + 1 + 3x² - 4x when simplified becomes 4x2 - 2x + 1
Polynomials are generally written in descending order.
Descending: 4x2 - 2x + 1 exponents of variables decrease from left to right
Polynomials
A polynomial can be one monomial or a bunch of monomials hooked together with plus/minus signs.
Examples of Polynomials are:
2x3  5x2  8x 1,
y 2  7 y  6,
4z  3
4x2 is a monomial because it has one term.
An expression 4 x2  3x is a binomial because it has two terms.
An expression 4x2  3x  5 is a trinomial because it has three terms.
An expression
All three expressions can be called polynomials.
The degree of the polynomial is the highest power in the variable. An expression
of degree one and it is called a linear expression.
2x1 is a polynomial
2x2 3x 1 is a polynomial of degree two and it is called a quadratic expression.
An expression 2 x3  3x is a polynomial of degree three and it is called a cubic expression.
An expression 4 x4  2 x3 1is a polynomial of degree four and it is called a quartic expression.
An expression
All monomials and polynomials are algebraic expressions.
FOR ADDITION AND SUBTRACTION:





LIKE TERMS CONTAIN THE SAME VARIABLES WITH THE SAME EXPONENT
REMEMBER ONLY LIKE TERMS MAY BE ADDED OR SUBTRACTED
GROUP LIKE TERMS, THEN PERFORM ADDITION OR SUBTRACTION OPERATION BY:
ADDING / SUBTRACTING COEFFICIENT/S OF LIKE TERMS. POWER/S REMAINS
UNCHANGED
HINT: TO AVOID ERRORS, STRIKE OUT TERMS WHICH YOU HAVE ADDED OR
SUBTRACTED.
Simplify each of the following expressions as much as possible.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
11x 12 y  7 x  5 y
21a2 12a  2a  a2  4
7 x3 115x2 9x3
22xy2 17 x2 y 17 yx2  23 y2 x  5
ad  2x  x 11ad 2
5xt  5x2t  (10tx2)  (7 xt 2)  8
35xy  20xy2 19xy 19xy2 10xy 1
55x y  60x y  x y  x y  x y
11
21 11 2 2 2 2
25x5 y3z2  xyz2  44x5 y3z  xyz2  26x5 y3z2
 xy  4ws  xy  2ws  2ws  7
571
x x 2
47 1 2
x y x y
nmnm
4 3 2 3
(3 y)2  x2  (2 y)2
15.
16.
3  5 6
x2 x 2 x
1 xy2  2 y2x  5xy  5 5xy
5
3
1.
Simplify:
(a)
(b)
(c)
(d )
12x  3 y  4x
2cd 2  5cd 2  6d 2c  4c2d  7dc2  dc2
x2 y  2xy2  3x2 y
15a2  2ab  7a2 11bc 8a2  2ba 1
(e) 7a2  2a  8b  c  4a2 12c  2a2  9a3  5c  2b  6a2  8b2 13a3  5ab
( f ) 22a2 17a2  5,7a  2,8b  5,2a2  6b  0,4a
( g ) 4a2  2,3b3 8a3  3ab 1,8b3  3,6a2 12,5ab 14a3  2a2
(h) 34a  22b 12b2  6c  2abc
Expansion of Algebraic Expressions
Expand and simplify
1. 3x  2( x 1)
2. 9  2(3x 1)
3. 3ab  2a(b  2)
4. 7 x  ( x  3)
5. 7(2x  2)  3(2x  2)
6. 3( x  2)  3( x  2)
7. 5(6a  8)  4(2a  4)
8. x( x 1)  x(2  3x)
9. 5n(4n  2n2  6)  3(4  n2)
10. xy(2x  2)  5(2x  xy)
11. 3x  4(2x  3)
12. x( x  2)  3x( x  4)
13. 4x3 y2( xy)  2x 8x4 y3  4( x  y)  3x
14. 1 x2(2  y)  2 x2 y  1 x2
3
3
4
15. 23 xy  3 xy  x( y  4)  1 x
8
4
TWO BRACKETS
Remove the brackets and simplify:
1. ( x  y) ( x  y)
2.  x  y 2
3. ( x  y) ( x2  2x  y2)
4. (a  7)2  a  b 
5.  a  b  c   a  b  c 
6.



k 12   k 12 
7. 4 2 y 1  3 y  2 
8. 3 y  y  2   y  3
2
9. 3( x  2)2   x  4 
2
2
10. 2 x 1   x  2   x( x  3)
11. 4  ( x 1)2
12. (2x 1)2  ( x  3)2
13. 3( x  2)2  ( x  4)2
14. ( y  3)2  ( y  2)2
Real problems in science or in business occur in ordinary language. To do such problems, we
typically have to translate them into algebraic language.
Problem 7. Write an algebraic expression that will symbolize each of the following.
a) Six times a certain number. 6n, or 6x, or 6m. Any letter will do.
b) Six more than a certain number. x + 6
c) Six less than a certain number. x − 6
d) A certain number less than 6. 6 − x
e) A number repeated as a factor three times. x· x· x = x3
f) A number repeated as a term three times. x + x + x
g) The sum of three consecutive whole numbers. The idea, for example,
g) of 6 + 7 + 8. [Hint: Let x be the first number.]
g) x + (x + 1) + (x + 2)
h) Eight less than twice a certain number. 2x − 8
i) One more than three times a certain number. 3x + 1
Now an algebraic expression is not a sentence, it does not have a verb, which is typically the
equal sign =. An algebraic statement has an equal sign.
FACTORIZATION OF ALGEBRAIC EXPRESSIONS
In the previous section we expanded expressions such as
reverse of this process is called factorizing.
To factorize linear expressions:
x(3x1) to give 3x2  x.The
We can check the answer by multiplying out the brackets: 2(2x+3) = 4x+6
Example
Factorise 4x² + 6x.
In this case 2x is the highest factor of both 4x² and 6x, so 2x will go outside the brackets.
The remaining factors of each term are left inside the brackets, where they are recombined.
We can check the answer by multiplying out the brackets: 2x(2x+3) = 4x²+6x
Example
Factorise 3xy² + 12x²y.
In this case 3xy is the highest factor of both 3xy² and 12x²y, so 3xy will go outside the brackets.
The remaining factors of each term are left inside the brackets, where they are recombined.
Hence,
To factorise a polynomial:
-
identify the HCF of the coefficients
identify any variable(s) which appear commonly (with lowest power) in given
terms.
Take out the common factor
We can also see it in this way:
4 xy  2 xz , The HCF of the coefficients is 2 and the common variable is x
therefore we can factor out 2x . Now from the term 4xy if 2x is a factor then
4xy  2 y and 2xz  z
2x
2x
then 4 xy  2 xz can be factorized as 2 x(2 y  z)
In expression,
12ax 18x2  42bx The HCF of the coefficients is 6 and the
common variable is x therefore we can factor out 6x . Now from the term 12ax if 6x is a
12ax  2a and from 18x2  3x and 42bx  7b
factor then
6x
6x
6x
thus 12ax 18x2  42bx can be factorized as 6 x(2a  3x  7b)
In expression,
Factorize the following expressions:
1. 21a  7
2. 3xy  6x
3. xy2  7 xy3
4.12x3 y3  x4 y2  4x2 y
5. aby  2aby  aby2
6. ax  bx  2cx
7. x2 y  y3  z3 y
8. 3a2b  2ab2
9. ax2  ay  2ab
10. ax2 y  2ax2z
11. abx  6ky  4kz
12. x2 y  y3  z2 y
13. 3a2b  2ab2
14. 6a2  4ab  2ac
15. 2a2e  5ae2
16. 2abx  2ab2  2a2b
17. ayx  yx3  2 y2 x2
To factorise algebraic expressions involved four terms:
Factorise
-
ah  ak bh  bk
Divide into pairs (in each pair must have a
variable in common)
e.g.
ah  ak  bh  bk
here a is common to the first pair and b is common
to the second pair, therefore, we factorise each factor as follows:
a(h  k ) b(h  k ) . Since (h  k ) is common to both terms, thus we have
(h  k ) (a  b) .
We refer to the process as factorization by grouping.
Factorise the following expressions
1. 6mx  3nx  2my  ny
2. xh  xk  yh  yk
3. ay  az  by  bz
4. as  ay  xs  xy
5. 2ax  6ay  bx  3by
6. 6ax  2bx  3ay  by
7. 2ax  2ay  bx  by
8. ms  2mt 2  ns  2nt 2
9. am  bm  an  bn
10. xs  xt  ys  yt
6. km  4m  kn  4n
7. 4x2  6xy  6xk  9 yk
8. 20x2 y3 8xp2  6 p2 15xy3
(9) ax  3x  2a  6
(10) xa  2xb  ya  2 yb
(11) ab2  b3  ad 2  bd 2
(12) 6a2  ab  2b2  2a  b
ARITHMETIC OF FRACTIONAL ALGEBRAIC EXPRESSIONS
Addition and Subtraction of algebraic fractions
To simplify, write as a single fraction.
e.g.
1.
2 3 8
9 17
 


3 4 12 12 12
2.
2 3
 the LCM of x and y is xy
x y
 2x  3y
 2xyy  3xyx
 3x
 2 yxy
3.
2  x
x 1 2 x  3
 2(2 x  3)  x( x 1)
( x 1) (2 x  3)
2
 4x  6  x  x
 x 1  2x  3
2
 x  5x  6
( x 1) (2 x  3)
4.
3
4

x2 x
5.
2
5

x  3 x 1
Simplify the following algebraic expressions:
1.
5
3

x  2 2x  1
answer
13x  1
( x  2)(2 x  1)
 x 2  x  10
(2 x  5)(5  x)
2.
x
2

2x  5 5  x
answer
3.
5
2
3


2
x 1 x 1 x  2
 x2  7x  9
answer
( x  1)( x  1)( x  2)
4.
2
5

x3 x4
answer
5.
x 1 x 1

x 1 x 1
answer 
6.
2x 7
1


7 2x x  1
answer
4 x 3  4 x 2  35 x  49
14 x( x  1)
7.
4
3

x 1 x 1
answer
 3x  7
x2 1
8.
x  x6
x 2  2x  3
answer
x2
x 1
9.
x  3x  10
x2  4
answer
x5
x2
10.
3x 2  9 x
x 2  4x  3
answer
3x
x 1
11.
6x 2  2x
12 x 2  4 x
answer
1
2
12.
x 2  4 x  21
x 2  5 x  14
answer
x3
x2
2
7x  7
( x  3)( x  4)
4x
x 1
2
Multiplication of Algebraic Fractions
To multiply two algebraic fractions, we simply multiply the numerator to get the numerator of
the product, and multiply the denominators to get the denominator of the product.
e.g. To multiply:
x
3
x3
3x



2 x  1 x  1 2 x  1 x  1 2 x  1 x  1
Linear Equations
1.
Solving linear equations in one variable.
2.
Simple word problems involving linear
equations
An equation has to have an equals sign, as in 3x + 5 = 11 .
bx  c  0 where b  0
bx  c  0 has the solution x   c
b
A linear equation has the general form;
 c
b
  c 0
c
Checking the solution by substituting  for x in the equation:  b 
b
c  c  0
Example:
4  3x  2
4  2  3x
4  2  3x
2  3x
2x
3
x terms on both sides, collect them on one side.
2 x  7  5  3x
2 x  3x  5  7
5x 12
x  12  2 2
5
5
If there is a fraction in the x term, multiply out to simplify the equation.
If there are
2x 10
3
2x  30
x  30 15
2
Solve the following equations:
1.
2x 5 11
2.
3x  7  20
3.
2 x  6  20
4.
5 x  10  60
5.
6.
3x 7  10
7 x
2
7.
x 1

2 3
8.
3  2x
4 3
10.
 3  3x
4 5
2x  4  x  3
11.
2y 1  4  3y
12.
7  3x  5  2 x
13.
x  16  16  2 x
9.
14.
15.
 x 1 1
2
4
 3  x  1  x
5 10 5 5
16.
x  2(x 1) 1 4(x 1)
17.
x  3( x  1)  2 x
18.
4(1  2 x )  3( 2  x )
19.
7  ( x  1)  9  ( 2 x  1)
20.
3( 2 x  1)  2( x  1)  23
21.
4( y 1)  3( y  2)  5( y  4)
22.
10(2x  3) 8(3x 5)  5(2x 8)  0
23.
10 ( x  4)  9( x  3)  1  8( x  3)
24.
6(3 x  4)  10 ( x  3)  10( 2 x  3)
25.
1

6 x  30( x  12)  2 x  1 
2

26.


6(2x 1)  9( x 1)  8 x 11 
4

27.
10 ( 2.3  x )  0.1(5 x  30 )  0
28.


8 2 1 x  3   1 (1 x)  1
4 4
2
 2
29.
(6  x )  ( x  5)  ( 4  x )  
30.
101 x   (10  x)  1 (10  x)  0.05
10 
100

x
2
Example:
2
2
x  3    x  2   32
( x  3)( x  3)  ( x  2)( x  2)  9
x2  6 x  9  x2  4 x  4  9
6 x  9  4 x 13
2x  4
x2



Solve the following equations
1.
x 2  4  ( x  1)( x  3)
2.
x  3x  1  x 2  5
3.
x  2x  3  x  7x  7
4.
x 2   x  1  2 x  1 x  4
5.
x  1x  3  x  12  2 xx  4
6.



7.
2
2
3
x  2    x  3  3x 11
2
2
2 x 1   x  2   x  x  3
When solving equations involving fractions, multiply both sides of the equation by a suitable
number to eliminate the fractions.
x  4 2x  1

4
3
x  4  12 2 x  1
12
4
3
3 x  4  42 x  1
3 x  12  8 x  4
16  5 x
16
x
5
1
x3
5
Solve the following equations:
1.
x3 x4

2
5
2.
x  2 3x  6

7
5
3.
x
x
 2
3
4
4.
5
10

x 1 x
5.
5
15

x5 x7
6.
4
7

x  1 3x  2
7.
x 1 x 1 1


2
3
6
8.
1
 x  2   1 (3 x  2 )
3
5
9.
1
x  1  1 x  1  0
2
6
10.
1
 x  5  2 x  0
4
3
11.
x  1 2x  3 1


4
5
20
12.
4
3

1 x 1 x
Problems solved by linear equations:
1.
The sum of three consecutive whole numbers is 78. Find the numbers.
2.
The sum of four consecutive numbers is 90. Find the numbers.
3.
Find three consecutive even numbers which add up to 1524.
4.
When a number is doubled and then added to 13, the result is 38.
5.
When 7 is subtracted from three times a certain number, the result is 28.
What is the number?
6.
The sum of two numbers is 50. The second number is five times the first.
Find the numbers.
7.
The difference between two numbers is 9. Find the numbers, if their sum is 46.
8.
The product of two consecutive even numbers is 12 more than the square
of the smaller number. Find the numbers.
9.
The sum of three numbers is 66. The second number is twice the first and
six less than the third. Find the numbers.
10.
David weighs 5kg less than John. John weighs 8kg less than Paul.
If their total weight is 197kg, how heavy is each person?
11.
Brian is 2 years older than bob who is 7 years older than mark.
If their combined age is 61 years, find the age of each person.
12.
Richard has four times as many marbles as John. If Richard gave 18 to John
they would have the same number. How many marbles has each?
13.
Stella has five times as many books as Tina. If Stella gave 16 books to Tina,
they would each have the same number. How many books did each girl have?
14.
A tennis racket costs N$12 more than a hockey stick. If the price of the two
is N$31, find the cost of the tennis racket. Answ.N$21.50
1.
One half of Mari’s age two years from now plus one-third of her
age three years ago is twenty years. If we let Mari’s age be x,
which of the equations below give the correct mathematical
translation of the statement?
1
1
A. x  2  x  3  20
2
3
1
1
B. x  2  ( x  3)  20
2
3
1
1
C. ( x  2)  ( x  3)  20
2
3
1
1
D. ( x  2)  x  3  20
2
3
How old was maria three years ago?
How old will maria be in 2yrs from now?
How old will maria be in 10 years time?
2.
During the class period, the number of girls is 10 less than 2 times
the number of boys.
2.1
Formulate a mathematical equation to express the number
of girls in terms of boys, given that g represent the number
of girls and b represent number of boys.
2.2
If the total number of learners in that class period were 80,
use the equation to formulate the above statement to determine
the number of boys and girls in that class period.
3.
During the Global Leadership Convention, it is discovered that
the number of men who are attending the convention is nine
hundred and forty less than four times the number of women in
attendance.
3.1
From the statement above, formulate a mathematical equation
expressing the number of men in terms of women, given that
represent the number of women and
3.2
represent the number of men.
Given that the total number of people who are attending the
Global Leadership Convention
are twenty thousand five hundred
and sixty, use the equation you formulated in 3.1 to
determine the number of men who are attending the convention.
3.3
Fruit & Veg. shop in Windhoek sells 5l of water bottles.
3.3.1
On Wednesday Fruit & Veg shop received N $2 530 from selling 5l
bottles of water at N $11.50 . How many bottles of water were sold?
3.3.2
On Thursday, the shop received N$ x by selling bottles of
water at N $11.50 each. In terms of x, how many bottles of water
were sold?
3.3.3
On Friday the shop received N $( x  20) by selling bottles of
water at N $9 each. In terms of x, how many bottles of water
were sold?
3.4
If the length of a rectangular play field is double the width and the area of the play field is
24 200 square metres, calculate the perimeter of the play field .
3.5 I am 41 years old and my son is 5 years old. After x years, my son’s age will be half my
age. What is the value of x?
3.6 You had a sum of money. Two hundred dollars have just been added on to it. What you
now have is four hundred dollars more than half of what you originally had. How
much
did you originally have?
3.7
John has N$6000 to invest. He invests part of it at 5% and the rest at 8%. How much
should be invested at each rate to yield 6% on the total amount?
3.8
A retailer incurs a fixed cost of N$330 when purchasing sugar for his stock. He pays
N$15 per packet which he resells at N$18 per packet. How many packets should he purchase
and
sell in order to break even?
3.9
The sum of four consecutive numbers is 20 more than the sum of the second and the forth
numbers. Find the consecutive numbers.
SETS and SET THEORY
A set is a collection of distinct objects, e.g.symbols, numbers, names etc. considered as an object
in its own right. “A collection of well-defined objects".
The objects in a set are called the members of the set or the elements of the set.
A set should satisfy the following:
1) The members of the set should be distinct.(not be repeated)
2) The members of the set should be well-defined.(well-explained)
Sets are one of the most fundamental concepts in mathematics.
There are two ways of describing, or specifying the members of, a set.
One way is by intensional definition, using a rule or semantic description:
For Example
A is the set whose members are the first four positive integers.
B is the set of colors of the French flag.
The second way is by extension – that is, listing each member of the set. An extensional
definition is denoted by enclosing the list of members in brackets:
C = {4, 2, 1, 3}
D = {blue, white, red}
Unlike a multiset, every element of a set must be unique; no two members may be identical.
NOTATIONS:
1.
 intersection
Intersection of the sets A and B, denoted A ∩ B, is the set of all objects that are members
of both A and B. The intersection of {1, 2, 3} and {2, 3, 4} is the set {2, 3} .
If set Set A  1, 2, 3, 4, 5, 7, 9, 10 and Set B  2, 3, 4, 6, 8, 11, 12then
A  B is 2, 3, 4
A  B is shaded
2.
 union
Union of the sets A and B, denoted A ∪ B, is the set of all objects that are a member of A,
or B, or both. The union of {1, 2, 3} and {2, 3, 4} is the set {1, 2, 3, 4} .
If set Set A  1, 2, 3, 4, 5, 7, 9, 10 and Set B  2, 3, 4, 6, 8, 11, 12then
A  B is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
A  B is shaded
3.
 is a subset of
A set is a subset of another set when all the elements in the first set are also a
member of the second set.
By definition, all sets are subsets of themselves and by convention, the null set is a
subset of all sets.
If A  1, 2, 3 and
4.
B  1, 2, 3, 4, 5 then A is a subset of B. A  B
 is a member of or belongs to.
If A  1, 2, 3, 4 then 3 is a member of A. 3  A and 5  A
5.
 universal set or S
The totality of all sets. The universe (usually represented as ) is a set containing all
possible elements
If set Set A  1, 2, 3, 4, 5, 7, 9, 10 and Set B  2, 3, 4, 6, 8, 11, 12then
  1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14
6.
A
complement of or not in A
Complement of set A relative to set U, denoted Ac, is the set of all members of U that are
not members of A.
The complement of a set is the set containing all elements of the universe which are not
elements of the original set.
This terminology is most commonly employed when U is a universal set, as in the study
of Venn diagrams.
This operation is also called the set difference of U and A, denoted U \ A. The
complement of {1,2,3} relative to {2,3,4} is {4} , while, conversely, the complement of
{2,3,4} relative to {1,2,3} is {1} .
If A  1, 2, 3, 4 and   1, 2, 3, 4, 5, 6,7, 8 then A  5, 6, 7, 8
This compliment contains all those elements of  that are not in A.
7.
n( A) the number of elements in set A
If A  1, 2, 3, 4 then n ( A)  4
8.
A  x : x is an int eger, 2  x  9
A is the set of elements x such that x is an integer and 2  x  9.
The set A is 2, 3, 4, 5, 6, 7, 8, 9 This is an example of property definition method
9.
  empty set
Note an
10.
 
for any set A.
Difference and Symmetric difference
Difference: If A  1, 2, 3, 4, 5, 6, 7 and B  1, 5, 6, 7, 8, 9, 10 then
A difference B which is denoted by A-B or A\B, is the set of all those elements of A which are
not in B. A  B  2, 3, 4
Symmetric difference:
Symmetric difference of sets A and B is the set of all objects that are a member of exactly one of
A and B (elements which are in one of the sets, but not in both). For instance, for the sets {1,2,3}
and {2,3,4} , the symmetric difference set is {1,4} . It is the set difference of the union and the
intersection, (A ∪ B) \ (A ∩ B).
The symmetric difference between two sets A and B is defined as the set of all those elements
that belong to A or to B but NOT to both A and B.
A  B  2, 3, 4, 8, 9, 10
A  B  ( A  B)  ( A  B)
11.
Power set of a set A is the set whose members are all possible subsets of
example, the powerset of {1, 2} is { {}, {1}, {2}, {1,2} } .
A. For
Example:
  1, 2, 3...,12,
(a )
(b)
(c )
A  2, 3, 4, 5, 6 and B  2, 4, 6, 8, 10
A  B  2, 3, 4, 5, 6, 8, 10
A  B  2, 4, 6
A  1, 7, 8, 9, 10, 11, 12
( d ) n A  B   7
(e) B   A  3, 5
1.
If X  1, 2, 3, ..., 10, Y  2, 4, 6, ..., 20 and Z  x : x is an int eger ,15  x  25
Find: (a)
(d)
2.
X Y
(b)
Y Z
(c)
X Z
n( X  Y )
(e)
n(Z )
(f)
n( X  Z )
If A  a, b, c, d , e B  a, b, d , f , g  C  b, c, e, g , h D  d , e, f , g , h
(a) A  ( B  D)
(b) ( A  D )  B
(e) ( A  D )  C
( f ) (C  A)  D
(c ) B  C  D
(d ) B  (C  D )
(g) ( A  C)  B
( h) p ( A  C )
Find:
From the Venn diagram above find:
(a) M   N
(b) N   M
( f ) M  N
(c ) ( M  N ) 
(d ) M  N 
( e) N  M 
In a school with a student population of 204 it was found that the number of girls in that school is
105. It was also discovered that there are 117 students who can swim, 97 students who are lefthanded, 80 girls who can swim, 65 girls who are left-handed, 62 left-handed students who can
swim and 50 left-handed girls who can swim.
Draw a Venn diagram and present the information given on that Venn diagram and answer the
following questions.
(a)
How many left-handed children are there?
(b)
How many girls cannot swim?
©
How many boys can swim?
(d)
How many girls are left-handed?
(e)
How many boys are left-handed?
(f)
How many left-handed girls can swim?
(g)
How many boys are there in the school?
SHADING Venn diagrams
1.
Draw Venn diagrams and shade the following areas.
(a) A  ( B  C )
(b) ( A  B )  C
(e) A  ( B  C ) 
( f ) (B  C)  A
(i ) ( A  C )  ( B  C )
(c ) A  B 
(d ) B  ( A  C )
( g ) C   ( A  B)
( h) ( A  C )  B 
( j ) A  ( B  C )
Application of Venn Diagram
n( A  B )  n( A)  n( B )  n( A  B )
n( M  N  Q )  n( M )  n( N )  n(Q )  n( M  N )  n( M  Q )  n( N  Q )  n( M  N  Q )
1.
A survey on regular payment of municipal bills was carried out on 140 house owners. It
was found that 60 pay electricity (E) bills regularly and 45 pay water (W) bills regularly.
Further, 20 pay both bills regularly. Use a Venn diagram to find the number of
house owners who
(a)
(b)
(c)
2.
pay at least one of the bills regularly.
pay exactly one of the two bills regularly
do not pay either bill regularly.
In a class of 30 girls, 18 play netball and 14 play hockey, whilst 5 play neither.
Find the number who play both netball and hockey.
Let
  girls in the class
N  girls who play netball 
H  girls who play hockey
x  the number of girls who play both netball and hockey
The number of girls in each portion of the universal set is shown in the Venn diagram.
n()  30
18  x  x  14  x  5  30
Since
37  x  30
x7
7 girls play both netball and hockey
3.
In the Venn diagram n ( A)  10, n( B )  13, n( A  B )  x and n( A  B )  29.
(a)
Write in terms of x the number of elements in A but not in B.
(b)
Write in terms of x the number of elements in B but not in A.
©
Add together the number of elements in the three parts of the diagram to obtain
the equation 10  x  x  13  x  18
(d)
4.
Hence find the number of elements in both A and B.
The sets M and N intersect such that n ( M )  31, n( N )  18 and n( M  N )  35. How
many elements are in both M and N?
5.
In a school, students must take at least on of these subjects: Maths, Physics or Chemistry.
In a group of 50 students, 7 take all three subjects, 9 take physics and Chemistry only, 8
take Maths and Physics only and 5 take Maths and Chemistry only. Of these 50 students,
x take Math only, x take physics only and x  3 take Chemistry only. Draw a Venn
diagram, find x, and hence find the number taking Maths.
6.
All of 60 different vitamin pills contain at least one of the vitamins A, B and C. Twelve
have A only, 7 have B only, and 11 have C only. If 6 have all three vitamins and there are
x having A and B only, B and C only and A and C only, how many pills contain vitamin
A?
7.
In a street of 150 houses, three different newspapers are delivered. T, G, and M. Of these,
40 receive T, 35 receive G, and 60 receive M, 7 receive T and G, 10 receive G and M and
4 receive T and M, 34 receive no paper at all. How many receive all three?
8.
In a survey conducted on 2000 officers in an establishment, 48% prefer coffee ©, 54%
like tea (T), and 64% do smoke (S). Further, 28% use C and T, 32 use T and S, and 39%
use C and S. Only 6% use none of these.
Find: How many use all three
How many use T and S but not C
How many use C only
9.
In a survey of 60 people, it was found that 25 read the Namibian, 26 read the Republikein
and 23 read the New Era, Also 9 read both the Namibian and the New Era, 11
read the Namibian and the Republikein, 8 read the Republikein and the New Era. All
three papers are read by 3 people.
Draw a Venn diagram to represent the given information
Find the number of people in the survey who read:
only the Namibian
only the Republikein
only the New Era
the Namibian and the Republikein but not the new Era
only one of the paper
none of the papers
1.1
Given S  1, 2, 3, 4, 5, 6
1.1.1
Find  A  B 
A. 2, 3, 4, 6
1.2
B. 1, 5
A  1, 3, 4, 5,
C. 1, 3, 4, 6
B  1, 2, 5,
D. 2, 3, 4,
In a survey of 200 households regarding the ownership of desktop and laptop
computers,
the following information was obtained:
120 households own only desktop computers, 10 households own only laptop computers
and 40 households own neither desktop nor laptop computers.
How many households own both desktop and laptop computers?
A. 70
B. 30
C. 40
Which of the following statements is false?
D. 170
A. a, b, c  c, a, b
2.2
B.    
C. a, b  a, b, c D. A  A
Given that A  x : x is a whole number between 0 and 4 and
B  x : x is a negative int eger greater than  4 , find A  B.
A. 0
B. 
C. 0, 4,  4
D.  
The values of p, q and r in the Venn diagram below are:
2.6
A. p  160, q  200 and r  200
B. p  130, q  200 and r  320
C. p  90, q  110 and r  220
D. p  90, q  110 and r  320
From the Venn diagram below, describe the region shaded.
A. A  B
3.3
B. ( A  B )  C
C. ( A  B )  C 
D. ( A  B )  C
Out of 240 students interviewed, it was found that 120 students speak Spanish
(S), 60
students speak neither Spanish nor Portuguese. Further
more ( x  10 ) students speak
Portuguese (P) only and x speak both languages.
3.3.1
Draw a Venn diagram and show the information as given above on the Venn diagram.
3.3.2
Solve for x .
3.3.3
Find the number of students who speak Spanish only.
1.5
The values of p, q and r in the Venn diagram below are:
1.3
The values of p, q and r in the Venn diagram below are:
A. p  160, q  200 and r  200
B. p  130, q  200 and r  320
C. p  90, q  110 and r  220
D. p  90, q  110 and r  320
In a survey of 240 households regarding the ownership of desktop and laptop
following information was obtained:
computers, the
130 households own only desktop computers, 25 households own only laptop computers and
36 households own neither desktop nor laptop computers.
How many households own both desktop and laptop computers?
A. 179
B. 74
C. 49
D. 104
The values of p, q and r in the Venn diagram below are:
A. p  50, q  270 and r  500
B. p  500, q  50 and r  230
C. p  320, q  50 and r  592
D. p  320, q  50 and r  642
3.3 Out of 360 students interviewed, it was found that 185 students speak Spanish (S), 55
students speak neither Spanish nor Portuguese. Further more ( x  7 ) students speak
Portuguese (P) only and x speak both languages.
3.3.1
Draw a Venn diagram and show the information as given above on the Venn
diagram.
3.3.2
Solve for x .
3.3.3
Find the number of students who speak Spanish only.
,
If Ω
, , , , , , , , , , ,
, , , , ,
′
∩
′
′
∪
. .
,
, , , , ,
, , , ,
,
,
, work out the following sets:
∩
∩
1.2
In a survey conducted on 3400 officers in an establishment, 48% prefer coke (C), 54%
like juice (J) while 64% like milk (M), Furthermore 28% drink coke and juice, 32% drink juice
and milk and 30% drink
coke and milk. Only 6% use none of these.
1.2.1
Draw a Venn diagram to represent this information.
1.2.2
How many officers use neither coke nor juice?
1.2.2
How many drink milk only?
1.3
Draw a Venn diagram and shade the region,
∪
′
∩
EXERCISES
1.
Given that of 380 soccer players, 210 drink tea and coffee, 260 drink coffee and 60
drink neither tea nor coffee. How many golf players drink tea only?
A. 210
B. 60
C. 120
D. 270
E. 200
2.
Out of 120 students interviewed, it was found that 60 students speak Spanish (S),
30
students speak neither Spanish nor Portuguese. Further more x  8 students speak Portuguese
(P) only and x speak both languages.
2.1
Draw a Venn diagram and show the information as given above on the Venn diagram.
2.2
Solve for x .
2.3
Find the number of students who speak Spanish only.
3.
A survey on regular payment of municipal bills was carried out on 140 house owners. It
was found that 60 pay electricity (E) bills regularly and 45 pay water (W) bills regularly.
Further,
20 pay both bills regularly. How many house owners pay at least one of the bills
regularly?
A. 40
B.20
C. 65
D. 85
4.
The values of p, q and r in the Venn diagram below are:
5.
A. p  160, q  200 and r  200
B. p  130, q  200 and r  320
C. p  90, q  110 and r  220
D. p  90, q  110 and r  320
In a group of 155 students, it was discovered that 70 students are male (M ) , 90
students are first year students (Y1 ) and 15 are neither male nor first year
students.
5.1
Present this information in a Venn diagram.
5.2
How many female students were first year?
5.3
How many male students were first year?
6.
(H).
A team of athletes was selected to compete in long jump (L), javelin (J) and high jump
The Venn diagram is a complete representation of the distribution of the selected athletes.
From the above Venn diagram find the total number of athletes in:
6.1
( L  H )  J
A. 51
6.2
8.
C. 102
D. 131
B. 51
C. 18
D. 21
(L  H )  J
A. 29
7.
B. 22
From the Venn diagram below, describe the region shaded.
In a group of 255 students, it was discovered that 140 students are
male (M ) , 110
students are first year students (Y1 ) and 35 are neither male nor first year students.
8.1
Present this information in a Venn diagram.
8.2
How many female students were first year?
8.3
How many male students were first year?
9.
The values of p, q and r in the Venn diagram below are:
A. p  160, q  200 and r  200
B. p  130, q  200 and r  320
C. p  90, q  110 and r  220
D. p  90, q  110 and r  320
10.
Out of 360 students interviewed, it was found that 185 students speak Spanish (S),
55 students speak neither Spanish nor Portuguese. Further more ( x  7 ) students
speak Portuguese (P) only and x speak both languages.
10.1
Draw a Venn diagram and show the information as given above on the Venn
diagram.
10.2
Solve for x .
10.3
Find the number of students who speak Spanish only.
11.
A survey shows that 71% of Indians like to watch cricket, whereas 64% like to watch
hockey. What percentage of Indians like to watch both cricket and hockey? (Assuming
every Indian watches at least one of these games)
that
A. 135%%
B. 36%
C. 7%
D. 35%
12.
In a class of 85 boys, there are 60 boys who play chess and 35 play table tennis.
12.1
How many boys play chess only?
A. 45
B. 100
How many play table tennis only?
C. 15
2.7.2
A. 50
C. 15
B. 35
D. 25
D. 25
MATRIX ALGEBRA
1.
VECTORS
1.1
ROW AND COLUMN VECTOR
A vector is a special type of matrix that has only one row (called a row vector) or one column
(called a column vector).


A vector of the general form 
 is called an n-component row vector.


a , a ,..., an
1 2
a 
 1 


a 
 2


.

A vector of the general form 
 is called an n-component column vector.
.





.



a 
n


EXAMPLES:
A vector
 3;  4;  5 is a 3 – component row vector.
A vector
 5; 0; 1; 4; 9  is a 5 – component row vector.
 1
 3
 is a 4 – component column vector.
A vector 
 10 


5


A vector
 0
 0  is a 2 – component column vector. It is called a null / zero vector since
 
all the components are zeros.
The zero vector has zero magnitude and no direction.
1.2
ADDITION AND SUBTRACTION OF VECTORS
Two or more vectors can be added or subtracted if they are both row vectors with the
same number of components or if they are both column vectors with the same number of
components.
It is customary to denote vectors by bold, lower case
EXAMPLES
letters (e.g.,
a)
Given:
a   2; 1;  3 ,
Work out:

3 
2 

b   2;  3; 9  , c  1  , d   0 



 5 
4 









1. a  b
2. b  a
3. b  c
4. d  c
1.3
SCALAR MULTIPLES OF VECTORS
If
is a vector and k is a constant (number) then, we refer to the vector (k ) as a scalar
multiple of
where k is the scalar. It is customary to denote scalars by italicized, lower case
letters (e.g., k).
a
a
a
If
a   a1, a2,...,an , then ka   ka1, ka2,...,kan .
If
k  0, k only changes the magnitude (size) of the vector a .
If
k  0, k
changes both the magnitude (size) and direction of the vector
a.
EXAMPLES
a   2; 1;  3 ,
Work out:
1. 2a
2.  5c
3. 1 b
3
 2
3 




b   2;  3; 9  , c  1  , d   0 



 5 
4 









4.  2a  4b
1.
MATRICES
In this unit, we shall focus on how to:


Express data correctly in matrix format
Perform matrix operations (addition, subtraction and multiplications)
In our discussion, we are limited to
2 2 matrices.
A matrix is a collection of numbers ordered by rows and columns. It is a rectangular array of
numbers arranged in rows and columns. It is customary to enclose the elements of a matrix in
parentheses, brackets, or braces, hence the definition of a matrix.
The array of numbers below is an example of a matrix.
21 62
44 95
The number of rows and columns that a matrix has is called its dimension or its order. By
convention, rows are listed first; and columns, second. Thus, we would say that the dimension
(or order) of the above matrix is 2 x 2, meaning that it has 2 rows and 2 columns.
Numbers that appear in the rows and columns of a matrix are called elements of the matrix.
For example, the following is a matrix:








A
5
9
8
7
3
2
The matrix A has two rows and three columns, so it is referred to as a “2 by 3” matrix. The order
(size) of a matrix depends on firstly the number of rows it has and secondly the number of
columns it has. The matrix
A above has order 23 .
1.1
Matrix Notation
Statisticians use symbols to identify matrix elements and matrices.
Matrix elements.
Consider the matrix below, in which matrix elements are represented entirely by symbols. This is
what we refer to as general representation of matrices.
a
ij
a a
11 12
A
a a
21 22










By convention, first subscript refers to the row number; and the second subscript, to the
column number.
Thus, the first element in the first row is represented by
a
11
The second element in the first row is represented by
a
12
There are several ways to represent a matrix symbolically. The simplest is to use a boldface
letter, such as A, B, or C.
EXAMPLE
Given:









2
1
2
1

0 1 
A  1
2 11
3
Entries
4
9
1 5







a 2 , a  ,
11
12
a  ,
32
a 
23
1.3
Matrix Addition and Subtraction
Addition and subtraction of a matrix of order 2 x 2.
If
a
11
A
a
21






a 
12 

a 
22 
(a ; a ) and (a ; a ) It also has two
11 12
21 22
a

 12 
and 

a

 22 


The matrix has two rows namely


columns namely 



a 
11 

a 
21 
Matrices of the same order are added (or subtracted) by adding (or subtracting) the
corresponding elements in each matrix.
To add two matrices, they both must have the same number of rows and they both must have the
same number of columns. The elements of the two matrices are simply added together, element
by element (corresponding elements), to produce the results.

So we can add matrices 


1 4 
5 3









1 4 
7 9


 to get 




2
0 
12 6
.


1.4
Scalar Multiple of a matrix
To multiply matrix A  a by a scalar,
Thus,
 
ij
k , we multiply each and every element of A by k .
kA  k aij  ka .
ij
EXAMPLE
Given:
3 4 
B 
,
0
5


Find:
 3 4   6  8 
2B   2



 0 5   0 10 
Multiplication by another matrix
For 2 x 2 matrices





b   w
d   y
a
c
x   aw  by

z   cw  dy

ax  bz 
cx  dz 
The same process is used for matrices of other orders.
To perform the following multiplication:
(a)





3
4
2 
2


1 
1

1  (3 2)  (21)

5   (4 2)  (11)

(31)  (25)   6  2

 41  (15)   8 1

3 10   8

4  5   9
13 
9 
Matrices may be multiplied only if they are compatible. The number of columns in the left-hand
matrix must equal the number of rows in the-hand matrix. Matrix multiplication is not
commutative, i.e. for square matrices A and B, the product AB does not necessarily equal the
product BA.
Exercises
2
A
3
 1
0
 B
4
1
5
4
 c
 2
1
1. A  B
2. B  c
3. 2 B
4. 3 A  B
5. 2C  3 A
6. 2 A  B
7. C  B  A
8. AB
9. 2( BC )
10. C 2
Find the value of the letters.
2
1. 
y
x  4

7    3
y  x

2   z
9

9 
3

 2
x

2.  1
w

a
3. 
c
 x
4. 
 2
2
5. 
0
p
6. 
q
2  x
y  8
 
 
 2   y
3    x
3   v
5   w
b  2
5
1

  2
0   3
d 
b
3  2   5 
    
y 1   0 
0  m  10 
    
 3  n  1 
 2
2
1    5 
 1 

2
2q     10 
3
0  y
z 6
 3



x  4
0  8
w
3
7. 
2
3z   6
 3
 3y
8. 


2z   8
w
 2 y  4x
1   8
e
2
3
9. 
k

3
 2   3
a
0
4
10. 
1
z 

w
8 
a

1
6

 1
p   20
0  n
12 



m  2
q 
0   1
0
0
1
x
11. if A  
, B  
 , and AB  BA, find x
2
3
3
1
3
 3
13. B  

 1  1
(a) Find k if B 2  kB
SIMPLE INTERESTS
P = Money borrowed or invested
i = Interest on P
r = annual interest rate
t = time in years
A = the amount due after t year
Simple Interest is calculated on a on-time investment at the end of the investment period. It does
not generate any interest itself.
Formulas to be used in calculating simple interests:
I  prt
A  P 1 rt 
I
r  pt
and
I
p  rt
or
or
A P I
I
t  pr
p  A I
or
p A
1 rt
EXAMPLES
1.
Find the simple interest payable on a loan of N$2 500 at
years.
25% p.a. at the end of 3
I  prt
I  2500 25%3
I  N $1875
2.
Find the simple interest payable on a loan of N$2 500 at
18 months.
12 1 % p.a. at the end of
2
I  prt
I  250012.51.5
100
I  N $468.75
3.
For how long should an amount of N$5000 be invested at 5% p.a. to generate an
interestN$750?
t ?
p  5000
r  5%
I  750
I
t  pr
750
50000.05
t 3
t
4.
John wants to buy a car after 10 years. He wants to have N$75 000 at the time of
purchase. How much should he
invest in a savings account that pays simple interest at
12%
t = 10yrs,
A = 75 00
r = 12% = 0.12
p=?
A  p(1 rt )
or
A
p
 N $34 090.91
(1 0.1210)
5.
Andrew invested N$12 550 for 5 years. After 5 yrs he
received a total amount of
N$22 500 from his investment. Calculate the annual rate at which interest was paid.
r=?
p = 12 550
A = 22 500
t = 5yrs
A  p 1 rt 
22500 125501 r 5 
22500 1 5r
12550
22500 1 5r
12550
r  0.158565737
r  0.16 16%
6.
Find the simple interest on N$8 500 loan at an annual
p = 8 500
r = 12%
interest rate of 12% for 2yrs.
t = 2yrs
I  prt
I  85000.12 2  N $2040
7.
Calculate the maturity value of an investment of N$30 000 due in 5yrs when the annual
simple interest rate is 16%.
r = 0.16
t=5
p = 30 000
A=?
A  p 1 rt 
A  30000(1 50.16)
A  N $54000
8.
Benson wishes to take a loan at an annual simple interest rate of 14.5% for 7 months. He
is told that he will have to pay back the sum of N$5422.92 at the end of the 7th month.
Calculate the loan Benson wishes to take.
r = 0.145
t=
7
12
A = 5422.92
p=?
A  p(1 rt)
5422.92  p(1 7 0.145)
12
5422.92  p(1.084583333)
p  N $5000
9.
The maturity value of a loan of N$30 000.00 is N$54 000.00.
(a)
Calculate the annual simple interest if the loan takes 5 yrs to mature.
(b)
Calculate the time the loan takes to mature if the annual simple interest rate is
16%
(a)
I  prt (there is no r, therefore find r first)
A  p(1 rt )
54000  30000(1 5r)
54000  30000 150000r
r  54000  30000
150000
r  24000  .16 16%
150000
I  prt
I  300000.165
I  N $24000.00
(b)
I
t  pr
24000
30000.16
t  24000  5 yrs
4800
t
MORE EXERCISES
1.
How much would you have to invest for nine years at a simple interest rate of
17.25% per
annum in order to receive
N $250 840.00 at the end of the ninth
year?
1.2million in her estate account. This amount is to be
invested in the estate for 3 years at simple interest rate of 12.5% per annum. After 3 years
2.
Madam Henk left N$
the maturity value will be distributed amongst her 3 sons in the ratio of their age. Mark will be
24 years
old, Paul will be 36 years old and Cyril will be 60 years old.
2.1
The maturity value after 3 years will be;
A.
N $1708 593.75
C.
N $1650 000
D.
B.
N $1200 000
N $450 000
3.
Dora invested N$40 000 for 10 years. After 10 years she received a total amount of N$52
000 from her investment. Calculate the annual simple interest rate at which interest
was paid.
4.
Find the simple interest payable on a loan of
the end of 9
N $170 000 at 6.75% p.a. at
years.
5.
Benson wishes to take a loan at an annual simple interest rate of 14.5% for 7 months. He
is told that he will have to pay back the sum of N$5422.92 at the end of the 7th month.
Calculate the loan Benson wishes to take?
6.
The maturity value of a loan of N$30 000.00 is N$54 000.00.
(a)
Calculate the annual simple interest if the loan takes 5 yrs to mature.
(b)
is 12.75%
7.
Calculate the time the loan takes to mature if
the annual simple interest rate
Dora invested N$40 000 for 10 years. After 10 years she received a total amount of N$52
000 from her investment. Calculate the annual rate at which interest was paid.
COMPOUND INTEREST
SI is calculated once on a once-off investment at the end of the
investment period. Compound Interest is calculated periodically
(within the investment period).
p = capital or investment
A = amount at the end of investment period
i = interest rate per compounding period
n = number of compounding periods
Formulas:
A  p(1 i)n
1.
and
P
A
n
1 i 




Calculate the amount payable for a loan of N$1000 for 3yrs at the rate of 10% p.a.
compounding annually.
p = 1000
r = 0.1
n=3
A  p 1 i 
n
A 1000(1 0.1)3
A  N $1331
2.
Calculate the amount payable for a loan of N$1000 for 3yrs at the rate of 10% p.a.
compounded quarterly.
p = 1000
i=
0.1
4
n=12
A  p(1 i)n
A 1000(1 0.025)12
A 1000(1.025)12
A  N $1344.89
3. Jane inherited a sum of money from her father. She wants to invest part of the inherited
money so that after 10 years, she could get N$250 000 from the investment. The bank has
accepted to pay interest at 7
(a)
(b)
1 % p.a. compounded semi-annually.
2
How much should Jane invest?
How much interest would her investment generate?
(a) A=250 000
A  p(1  i) n
A
p
(1  i ) n
250000
P
20
 0.075 
1 

2 

p  N $119723.09
I=
7.5%  0.0375
2
n = 20
(b)
I  A p
250000  119723.09  N $130276.91
4.
A trust fund is expected to grow from 360 000 to N$500 000 in 4 years when the interest
rate is compounded monthly. At what annual interest rate is the trust expected to grow?
p = 360 000
A= 500 000
n = 12 x 4yrs = 48
i=?
A  p(1  i ) n
500000
 (1  i ) 48
360000
48
1.388888889  1  i 
1.006867307  1  i
i  0.00687
annual int erest rate is 0.00687  12  0.0824  8.24%
5.
Determine the compound amount if N$5000 is invested for 10 years at 5%p.a.
compounded annually.
p = 5000
n = 10
i = 5%
A  p (1  i ) n
A  5000(1  0.05)10
A  N $8144.47
6.
Tony invested a sum of money for 2 years at 8%p.a. compounded annually. At the end of
the 2 years
he received a total amount of N$1166.40. How much did Tony invest?
t = 2yrs
N$1166.40
r = 8%
n=2
A=
A
(1  i) n
1166.40
p
(1  0.08) 4
1166.40
p
 N $1000
1.1664
p
7.
Determine the sum to be invested for 4 yrs at *% p.a. compounded semi-annually to
amount to N$3 500 at the end of the investment period.
p=?
p
p
A = 3 500
i=
8% 0.08

2
2
n=4x2=8
A
(1  i ) n
3500
1  0.048
P  N $2557.42
8.
If N$750 amounts to N$1200 in 3years, determine the nominal rate converted monthly.
A = 1200
p = 750
n = 3 x12
I=?
A  p (1  i ) n
1200
36
 1  i 
750
1.6  (1  i ) 36
36
1. 6  1  i
36
1. 6  1  i
0.013141253  i
i  0.013141253  12  0.157695036  15.8%
9.
Fifty-five years old Tate Paul invested N$80000 in a savings account that paid 10% p.a.
compounded semi-annually. After 5 years, the interest rate increased by 2%. The
compounding period also changed to quarterly. Tate Paul made no withdrawal from this
savings account until he was seventy years old. How much was in Tate Paul’s savings
account
at the age of seventy?
For the 1st part, at the end of the 1st five years:
p = 80000
I=
10%
 0.05
2
n = 5 x 2 = 10
A  p (1  i ) n
A  800001  0.05
A  N $130311.57
10
For the 2nd part, at the end of the next 10 years:
p = 130311.57
i=
12%
 0.03
4
n = 10 x 4 = 40
A  p (1  i ) n
A  130311.571  0.03
A  N $425081.27
40
10.
Miss Ndapandula wishes to save for her wedding day, which comes up exactly two and a
half years
from now. She has N$6000 to invest in a savings account that pays interest
at 10% p.a. compounded
every two months. How much will she have to borrow to add to
her investment amount if her wedding
budget stands at N$12500 on the day of her
wedding?
p = 6000
i=
10%
 0.016666666
6
1
n = 2 6
2
A  p (1  i ) n
 0 .1 
A  60001 

6 

A  N $7688.29
15
N $12500  N $7688.29  N $4811.71
MORE EXERCISES
1.
Leon left N$ 800 000 in his estate account. This amount is to be invested in the estate for
6
years at the interest rate of 12.75% p.a. compounded
monthly. After 6 years the
maturity
value will be distributed amongst his 4
daughters in the ratio of their age.
Maria will be 15
years old, Jolene will be 22 years old, Rolna will be 28 years old and Tina
will be 8 years old.
1.1
The maturity value after 6 years will be:
A. N $1 643 574 .11
N $1 712 271 .66
B. N $1 412 000
C. N $4 523 573 243 D.
2.
to
Determine the sum to be invested for 4 years at 7.5% per annum compounded quarterly
amount to N$45 000 at the end of the investment.
3.
The sum to be invested for four years at 8% p.a. compounded semi-annually to
to
N $3 500 at the end of the investment period is:
A. N $2 651 .52
B. N $4 761 .71
C. N $2 572 .60
amount
D. N $2 557.42
4.
Determine the sum to be invested for 4 years at 4.5% per annum compounded
monthly to
amount to N$25 000 at the end of the investment.
5.
Kavita has N $30 000 .00 to invest in an account that pays interest at
12.75% p.a. for five years. He has two options:
Option A:
Investment at simple interest.
Option B:
Investment with interest compounded quarterly.
By showing full calculations, determine which interest option is better for
Kavita
6.
Determine the sum to be invested for 4 years at 7% per annum compounded semiannually to amount to N$55 000 at the end of the investment.
7.
A trust fund is expected to grow from 360 000 to N$500 000 in 4 years when the interest
rate is compounded monthly. At what annual interest rate is the trust expected to grow?
8.
Fifty-five years old Tate Paul invested N$80000 in a savings account that paid 10% p.a.
compounded semi-annually. After 5 years, the interest rate increased by 2%. The
compounding period also changed to quarterly. Tate Paul made no withdrawal
from
savings account until he was seventy years old. How much was in Tate Paul’s savings
account at the age of seventy?
this
9.
to
Determine the sum to be invested for 4 years at 12.5% per annum compounded monthly
amount to N$65 000 at the end of the investment.
10.
Determine the sum to be invested for 4 years at 7.5% per annum compounded
quarterly to amount to N$45 000 at the end of the investment.