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Transcript
1.
BASIC ARITHMETIC OPERATIONS
• Differentiate between a factor and multiple
• Define a prime factor
• Prime Factorization/ Decomposing into Prime Factors
• Define Highest Common factor (HCF) and Lowest Common Multiple
(LCM)
• Determine HCF and LCM
• Applications of HCF and LCM
Factors
The factors of a number are those numbers which divide exactly into the
number.
Factor – a number that is multiplied by another to give a product.
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 5 x 6 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.
Multiple – An answer to a multiplication problem, e.g 7 8 56 , 56 is a
multiple
Examples:
Some multiples of 5 are 5,10,15, …
Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, …
1|Page
Prime Numbers
A natural number x 1 is said to be prime if and only if it is divisible by 1 and
itself. Meaning, a prime number has only two natural factors, itself and 1. Zero
and 1 are not prime numbers. 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: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, etc.
Composites: 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, etc.
Example: 7 is prime because the only numbers that will divide into it evenly are
1 and 7.
Few prime numbers are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 etc.
Prime Factors
Remember the factors of 30 are: 1, 2, 3, 5, 6, 10, 15, 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 360 as a product of primes
1.
2.
3.
4.
Write down the 1st few prime numbers e.g. 2,3,5,7,11,13
Divide 360 by the 1st prime number (2) as many times as possible
until it can no longer divide exactly into that number.
Divide 360 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.
One way of doing this is:
360 180 90 45 15 5
:
: : : : :1 This means that
2
2 2 3 3 5
5.
360 2 2 2 3 3 5
We can write any repeated prime number as powers (using index
form) as: 360 23 32 5
2|Page
Exercise
Express 378 and 18522 as a product of prime numbers.
Highest Common Factors ( HCF) and Highest Common Divisor (HCD)
To find the HCF:
Method 1: List all factors of the given numbers and identify the HCF.
For example, to the highest common factor of 8 and 12
Factors of 12 are: 1, 2, 3, 4, 6, 12
Factors of 8 are: 1, 2, 4, 8
Those bolded factors are common factors but the biggest common factor is 4.
Hence, the HCF is 4.
Method 2: Factorise the given numbers into their prime factors respectively.
Select those numbers (with lowest power) which occur commonly in all of the
decompositions and multiply them together.
For twelve
For eight
12 6 3
: : :1 12 22 3
2 2 3
8 4 2
: : :1 8 23 1
2 2 2
HCF 22 4
Exercise
Find the HCF of:
(a)
(b)
24, 72, 96 and 300
255 and 75
Lowest Common Multiple (LCM)
Method 1: List the multiples of the given numbers and identify the LCM.
3|Page
Find the LCM of 12 and 8
Multiples of 8 are 8, 16, 24, 32, 40,...............
12 = 12, 24, 36, 48,.............
We see that the very first (lowest) number appears in both lists is 24, hence we
say that 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 its highest power which occur in any
of the decompositions (prime factors) of each of the given numbers and
multiply them together.
To find the LCM of 12 and 8, we decompose as follows:
For 12 we have
For 8 we have
12 6 3
: : :1whichis 12 22 3
2 2 3
8 4 2
: : :1whichgives 8 23
2 2 2
Hence, LCM = 23 3 24
Exercise:
Find the LCM of
(a)
72, 96 and 300
(b)
455, 1050
Problem sums on HCF and LCM can be really tricky as they are not easy to
identify. Thus in this case the main focus is not on going through how to find
HCF and LCM (please refer to your notes on those), but more importantly to go
through how to determine when to find the HCF and when to find the LCM of
the numbers involved in the problem sums.
Let’s take a look at a typical problem involving the HCF.
4|Page
HCF – Example
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?
If you notice, finding the HCF is crucial here because you are trying to find what
the 3 numbers have in common, i.e. a common factor. All 3 numbers must be
able to be divided by the same number in order for all 3 strings to be cut into
equal lengths. HCF is needed here because you are asked to find the greatest
possible length.
Therefore,
LCM – Example
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 to list
the seconds:
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:
5|Page
There are other different types of problems involving LCM, but
just remember that such questions involve you trying to find a
multiple that is common to the numbers involved.
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, how many 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]
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]
3.
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]
6|Page
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? [28 cm]
6.
A small bus interchange has 2 feeder services that start
simultaneously at 9am. Bus number 801 leaves the
interchange at 15-min intervals, while bus number 802
leaves at 20-min intervals. On a particular day, how many
times did both services leave together from 9 am to 12 noon
inclusive? [4 times]
7.
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? [12th step]
8.
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? [June 25 2008]
9.
A group of students can be further separated into groups of
5, 13 and 17. What is the smallest possible total number of
students? [1105 students]
10.
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?
[576 days]
7|Page
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) 3 5 8
7 (3) 7 3 10
9.1 (3.1) 9.1 3.1 12.2
2 (1) (5) (2 1) 5 3 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
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
Same goes for division.
8|Page
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. 32 3 3 22 7 4 1 1
2. 2 4 2 3 8 4 (7)
2
3. 2 4 2 2 1 4
Simplify each of the following
1. 2 3 7 2 5 6 2 3
2. 4 2 18 6 4 4 12 10
3. 25 3 2 7 (8) (52 6) 4 4
4. 5 3 4 7 6 6 9
5. 81 3 209 54
6. 72 2(2 4) 24 24 (3 5) 17
9|Page
Vulgar Fractions – Concepts and operations
In a fraction
7
8
7 is a numerator and 8 is a denominator.
to as a proper fraction and
7
is referred
8
8
is referred to as a improper.
7
7
2 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.
3 4
9 9
2.
1 2
4 3
3.
3 4
9 9
4.
1 2
4 3
5.
1 2
4 3
6.
1 2
4 3
7.
1
2
1 2
3 5
8.
2 1
2
3 4
9.
2 1
3 5
3 1
4 3
10.
3 1 7 2
8 5 10 3
12.
2 1
2 1 5 3
3 2
13.
7 2 3 2
8 7 5 3
11.
14.
1 2
1
1 2 4 1
2 3
3
3
3 7 5
2 2
7
4 8 8
Fractions Word Problems
1.
Johannes had 640 shares. He sold out one third of them to a trading
company and
2
of the remainder to another company. How many
5
shares remained with Johannes?
10 | P a g e
2.
Andrew sold half of his cows; gave his younger brother
1
of the original
4
number of cows. How many cows does Andrew now have if he had 88
originally?
3.
The Martin’s family spends
2
1
1
of their income on rent, on food, and
5
4
5
on clothes. If they are left with N$390.00 each month, find:
(a) the fraction which is left.
(b) their monthly income
4.
A man spends
2
1
1
of their income on rent, on food, and on clothes. If
5
4
5
they are left with N$390.00 each month, find:
5.
(a)
The total fraction, spent.
(b)
The amount, which is left
A man saves N$240 every month. This is
4
of his monthly salary.
25
Calculate his monthly salary.
6.
The company decided to donate 25 000 shares to some institutions.
the shares went to the orphans association,
2
of
5
1
went to the cancer
4
association and the remaining part was donated to a church.
7.
(a)
How many shares received by the church?
(b)
What fraction of the shares did the church receive?
Three friends, Alex, Brenda and Charles decide to buy a car. Alex pays
of the cost, Brenda pays
7.1
7.2
1
4
1
of the cost and Charles pays the rest.
3
What fraction of the cost does Charles Pay?
Brenda pays N $12 000 more than Alex. Calculate the cost of the car.
11 | P a g e
Decimal Fractions
Conversion:
Changing vulgar fractions to decimals fractions and simplify
1.
2.
3.
4.
5.
6.
7
7 8 0.875
8
1
0.3333
3
9
0.9
10
7
1 1.7
10
2
2 2.285714285714
7
19
3
3.19
100
Change decimals to fractions and simplify
35 7
100 20
4 2
0.8
10 5
2
1
3.2 3 3
10
5
6
3
0.006
1000 500
44
11
0.00044
100 000 25000
1. 0.35
2.
3.
4.
5.
Types of Decimals
1.
e.g.
2.
e.g.
Terminating Decimals
7
1 .4
5
or
1
0.5
2
Recurring or repeating decimals
2
0.66666 or
3
0. 6
12 | P a g e
3.
e.g.
Non-recurring – non terminating decimals
12
0.214 285 714
56
Use your calculator to evaluate
1.
0.2714 0.00589
2.4167 0.000717
4.
9.6 1.5
2.4 0.74
2.
6.51 0.1114
7.24 1.653
5.
4.2
2
1
3.
4.7
1.6
11.4 3.61 9.7
1
5.5 7.6
POWERS AND ROOTS
Given a positive integer, x n 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.
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 a 2 which is a a
A cube number is the result of multiplying a number by itself three times.
2 2 2 23 8
and
3
82
Cube numbers are e.g. 1, 8, 27, 64 etc
6 4 means 6 multiplied itself four times
6 4 6 6 6 6 6 1296
and
4
1296 6
Remember: (7) 2 7 7 49 but 7 2 (7 7) 49
Taking a square root of a positive number gives two possible answers, - or +
13 | P a g e
Use your calculator to work out:
1.
0.36
2.
0.850.2
3.
0.98
1.09
4.
3
5
5.
3
27
81
167.9421
5.6719 3
6.
3
2.312 0.9 2
2.31 9.81
7.
4
1
5 10 3
8.
2.31 1.013
0.79 0.0412
9.
6
64
1331
INDICES
The arithmetic operation of raising a number to a power is devised from
repetitive multiplication. Exponents represent shortcuts in multiplication
The use of powers (also called indices or exponents provides a convenient form
of algebraic shorthand. Repeated factors of the same base, for example
b b b b b can be written as b5 where the number 5 is called the
index or exponent or power and b is called the base.
Given a positive integer m, x m implies that x is multiplied by itself m number
of times. For example, the expression 2 2 2 2 2 2 2 2 can be written as
28 in exponential format. You would much rather write the expression (5)100
14 | P a g e
than write out 5 one hundred times. We can also use exponents with
variables for example the expression x y x y x y x y x y is more
Conveniently written as x y 5 .
Note:
Any number raised to the power one equals itself.
By definition any nonzero number or variable raised to exponent zero is equal
0
to 1. For example 80 1 or 2 c 1.
If a, b, m, and n are real numbers with a and b positive, then the following
rules apply.
(a)
am an amn
(c)
am
am an or n amn
a
n
am amn
(d)
m
ab ambm
(e)
a
b
(f)
1
a m m
a
(g)
a
b
(b)
m
am
m
b
m
b
a
m
a m b m
or m m
b
a
Note: From rule f;
1
1
a m m and bm m then it follows that
a
b
1
a m a m
1
1
1 bm bm
1
bm
a m bm a m 1 a m
bm
(h)
a0 1 for a 0 since 00 is undefined
15 | P a g e
It can easily be shown why any variable or nonzero number raised to the zero
power is one.
x3 33 0
x
x , but
x3
x3
1, therefore x0 1
3
x
1 1
32
32 9
3
5
4
1
4
4
3
1 1
1 54 625
3
4
1
4
4
4
1
81
5
3 5
3
4
5
32 1
1 1
1
1
43
43 32
32 43 9 64 576
Worked examples
Simplify each of the following expressions as much as possible.
5m
n
4
1.
2n 2 4
2.
3.
x y x y
53m
5
2
4.
2s
Solutions
1.
2n 24 2n4
4.
2s2
5
2.
n
4
n4
x y x y x y
3.
5m
5m3m 54m
53m
25 s 10 32s 10
Simplify the following expressions
(a) 3n 32n
(c)
y(ab)
y(ab)
(b)
(d)
x2
b3
y
y
2
ab
ab
16 | P a g e
2
(e)
85
84
(g)
1
3
5
x y z2
x3 y 6 z
3
3x a yb
(i)
2
3x a yb
(f)
1
1
3
m m4
(h)
a x xa
t
t
4
2
SURDS
A number b is said to be a square root of y if b 2 y. Thus, 4 is a square root of
2
16 , because 4 2 16, and 4 is also a square root of 16, because 4 9.
Similarly, 6 is a third root (called a cube root) of 216, because 6 3 216. The
number 216 has no other real-number cube roots.
The symbol b denotes the nonnegative square root of b and the symbol 3 b
denotes the real-number cube root of b. Then the symbol n z denotes the nth
root of z. From the symbol n z , the symbol is called a radical and the
expression under the radical symbol is called the radicand and n is called index.
By convention 2 16 is written as 16 with index 2 omitted. Hence keep in mind
that when we write a square root of a given number then the index is two.
Examples of roots where n 3, 5 and 2 respectively are;
3
125 5 because 5 3 125,
5
32 2 because 2 5 32 and
121 11 because 112 121.
We should also be aware that a 2 a1 which can be expressed in exponential
1
2
1
n
format as a and a a .
n
3
5
3
a can be expressed as a in exponential format.
5
Properties of Radicals
Let a and b be any real nonnegative numbers and m and n are positive integers,
the rules of radicals are given as:
17 | P a g e
a a
If n is odd, a a
(a)
If n is even,
(b)
n
n
n
n
(c)
n
a n b n ab
(d)
n
a na
, where b 0
b nb
(e)
n
am
(f)
m n
a
m
n
a mn a
Worked Examples
Simplify each of the following surds.
(a)
3
(b)
3
4
(c)
(d)
(e)
(f)
(g)
27
3
64
1782
4
22
8 18
50
3
85
36 x 4 y 6
Solutions
(a)
27
(b)
3
4
(c)
3
3
27 27
64 32 64 6 64 2
1782
4
4
1782 4
81 3
22
22
8 18 144 12
50 25 2 25 2 5 2
(d)
(e)
(f)
3
3
3
85
8
3
5
2 5 32
4
(g)
6
36 x 4 y 6 36 x 4 y 6 6 x 2 y 2 6 x 2 y 3
18 | P a g e
Simplify the following expressions
(a)
169 x 6 y 8
(b)
81x 2 y 3 z 6
294
6
(c)
27 x 6 y 9
3
8 16
4
(d) 3 x y
3
(e)
432 3 4
1
4
(f)
m
4
1
5
m 5
Sometimes we can use the properties of radicals to solve equations.
Let us look at the following two examples.
Example 1.
Given 3 y 6 x 5 , solve for y.
Solution
If we square both sides of the equation, we have;
3 y 6 x
2
5 2
3 y 36 x10
y 12 x10
Example 2.
Given 7 y 42x 3
Solution
7 y 42 x 3
7 y 42x
2
3 2
7 y 1764 x 6
y 252 x 6
19 | P a g e
Logarithms
Indicial and Logarithmic Functions
When we dealt with powers and exponents, we learnt that any real number
can be written as another number raised to a power. For example:
16 4 2 , 27 33 and 64 4 3
A logarithm is an exponent. It is the exponent to which the base must be raised
to produce a given number.
If p, q and r are real numbers where:
p q r and q 0 , the power r is called the logarithm of the number p to base q
and it is written r log q p, read as r is the log of p to base q.
Therefore having b x y which implies that x log b y , we say that x is the
logarithm of y with base b if and only if b to the power x equals y.
For example, because 36 6 2 the power 2 is the logarithm of 36 to the base 6.
That is: 2 log 6 36.
Let’s work on the following examples. Try to work them out yourself before
you look at the solutions.
Worked Examples
1.
2.
In each of the following what is the value of x, remembering that if
p q r then r log q p.
(a)
x log 3 81
(b)
4 log x 16
(c)
2 log 5 x
Express the following logarithms in exponential format.
(a)
(b)
(c)
(d)
(e)
log 3 81 4
log c x z
log 2 32 5
2 log 7 49
x log 5 25
20 | P a g e
Since Logarithms are powers, the rules that govern the manipulation of
logarithms closely follow the rules of powers.
(a)
M ba so that a log M and
b
and N bcso that c log N then :
b
MN babc bac hence log MN a c log M log N .
b
b
b
That is log MN log M log N
b
b
b
The log of a product equals the sum of the log s
(b)
Similary, M N ba bc bac so that
M N ba bc bac so that
log (M N ) a c log M log N
b
b
b
That is :
log (M N ) log M log N
b
b
b
The log of a quotient equals the difference of the log s
(c)
Because M q (ba )q baq , log
b
M q aq q logb M .
log (M q ) q log M
b
b
The log of a number raised to a power is the product of
the power and the log of the number.
(d)
(e)
(f)
(g)
log 1 0 because, from the laws of powers a0 1.
b
Therefore, from the definition of the logarithm log 1 0.
b
log b 1 because b1 b so that log b 1.
b
b
log bc c because log bc c log b c 1
b
b
b
log q
b b q because if we take the log of the left-hand side of this
equation:
21 | P a g e
log q
log q
log b b log q log b log q so that b b q.
b
b
b
b
(h)
1
log a
because, if loga b c then ac b and so
b
loga b
1
1
1
b c b b c . Hence, log a
b
c loga b
Considering the manipulation of logarithms above,
The Rules / Laws of logarithms
For all b 0, b 1, and M , N 0(b, M , N , are real numbers) ,
We have:
(a)
(b)
(c)
log MN Log M log N
b
b
b
M
log
log M log N
bN
b
b
log M n n loga M
b
(d)
log 1 0
b
(e)
log b 1
b
log bn n
b
(f)
(g)
log n
b b n
(h)
1
log a
also,
b
log a b
logc a
log a
(change of base if c 0 and c 1)
b
logc b
Note: There are no rules for the log (M N ) or log (M N )
b
b
22 | P a g e
Base 10 and base e
Logarithms to base 10 are called common logarithms and are written without
indicating the base. For example log a is simply written as log a without
10
indicating base 10 and log 1000 is simply written as log1000.
10
To find the value of the common logarithm, log1000, on the calculator simply
press the key, log, the value 1000 and then the key.
The logarithms to base e are called natural logarithms . They are written as ln x .
Instead of writing loge 10, we write ln10. You can also find the value of loge 10
by simply pressing ln10 on the calculator.
The value of log10 1and the value of ln10 2.302585093
Note that for any base the:
Logarithm of 1 is zero
Logarithm of 0 is not defined
Logarithm of a number greater than 1 is positive
Logarithm of a number between 0 and 1 is negative
Logarithm of a negative number cannot be evaluated as a real number
Let us now try some of the examples below keeping in mind the logarithms
rules.
Exercise
Evaluate the following logarithms
(a) log 3 9
1
8
(c) 4 log x 81
(b) log 2
(d) log 2 8 log 3 9 log 5 25
(e) log 4 8 2
(f) log 2
(g)
2
8
log e 7
log e 3
1
(h) log 3 81 log 6 36 log 2 8
1
3
23 | P a g e
(i) Simplify log a y 3 5 log a y 3 log a 4 y
ALGEBRAIC EXPRESSIONS
In this section, we will discuss:
1.
Terms, Variables and Coefficients
2.
Simplification
3.
Expansion of algebraic expressions
Brackets and simplifying
Two brackets (Expansion)
4.
Factorization
5.
Addition, Subtraction, Multiplication and Division of Algebraic
expressions
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. Such an expression represents one number or one quantity. Thus,
just as the sum of 4 and 2 is one quantity, that is, 6, the sum of c and d is one
quantity, that is, c + d.
TERMS, VARAIBLES AND COEFFICIENTS
The terms of an algebraic expression are the parts of the expression that are
connected by plus and minus signs. In the expression 3abx cy k for example,
3abx, cy and k are the terms of the expression.
An expression containing only one term, such as 3ab, is called a monomial
(mono means one).
A binomial contains two terms; for example, 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).
24 | P a g e
Usually special names are not given to polynomials of more than three times.
The expression x3 3x2 7 x 1 is a polynomial of four terms.
The trinomial x2 + 2x + 1 is an example of a polynomial which has a special
name (quadratic)
In general, a COEFFICIENT of a term is any factor or group of factors of a term
by which the remainder of the term is to be multiplied. Thus in the term 2axy,
2ax is the coefficient of y, 2a is the coefficient of xy, and 2 is the coefficient of
axy. The word "coefficient" is usually used in reference to that factor which is
expressed in Arabic numerals. This factor is sometimes called the NUMERICAL
COEFFICIENT. The numerical coefficient is customarily written as the first
factor of the term. In 4x , 4 is the numerical coefficient, or simply the
coefficient, of x . When no numerical coefficient is written it is understood to
be 1. Thus in the term xy, the coefficient is 1.
Practice problems. Combine like terms in the following expression:
In 5x 2 x 3xy 1 there are 4 terms. x and y are variables. In 5 x, 5 is the
coefficient of x , 3 is the coefficient of xy and -1 is a constant.
5x and 2x are like terms. They contained the same variable.
2x2 y 4xy2 xy2 x2 y ,
2x2 y and x2 y are like terms while 4xy2 and xy2
are like terms.
Simplify each of the following:
1.
11x 12 y 7 x 5 y x y
2.
21a2 ax 12a 2a a 2 4ax
3.
5xt 5x2t (10tx2) (7 xt 2) 8
25 | P a g e
4.
5 7 1
x x 2
5.
4 7 1 2
x y x y
6.
n m n m
4 3 2 3
7.
(3 y)2 y 2 y2 x2 (2 y)2
8.
3 4 6 5
x x2 x x2
Expansion of Algebraic Expressions
Brackets and simplifying
Expand and simplify
3x 2( x 1)
9 2(3x 1)
3ab 2a(b 2)
7 x ( x 3)
7(2 x 2) 3(2 x 2)
3( x 2) 3( x 2)
5(6a 8) 4(2a 4)
x( x 1) x(2 3x)
9. 5n(4n 2n2 6) 3(4 n2 )
10. xy(2 x 2) 5(2 x xy )
11. 3x 4(2 x 3)
12. x( x 2) 3x( x 4)
13. 4 x3 y 2 ( xy) 2 x 8 x 4 y3 4( x y) 3x
1.
2.
3.
4.
5.
6.
7.
8.
1 2
2
1
x (2 y) x2 y x2
3
3
4
3
1
15. 23 xy xy x( y 4) x
8
4
14.
26 | P a g e
TWO BRACKETS
Remove the brackets and simplify:
1. ( x y ) ( x y)
2
2. x y
3. ( x y ) ( x 2 2 x y 2 )
4. (a 7)2 a b
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
5.
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. (2 x 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 in to algebraic language.
For example to 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
27 | P a g e
g) The sum of three consecutive whole numbers. The idea, for example,
is 6 + 7 + 8. [Hint: Let x be the first number.] then we have 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
The sum of a number and 5
5 more than a number
5 less than a number
5 less a number
Twice the sum of a number and 3. 2(x + 3)
3 more than twice a number
2x + 3
10 less five times a number
10 - 5x
divide the sum of a number and 5
(x + 5)/2
by 2
2 more than the quotient of 3 and
3/x + 2
a number
3 more than four times the sum
of a number and 2
4(x + 2) + 3
28 | P a g e
FACTORIZATION OF ALGEBRAIC EXPRESSIONS
In the previous section we expanded expressions such as x(3x 1) to give
3x2 x. The reverse of this process is called factorizing.
To factorize linear expressions:
-
identify the HCF of the coefficients
identify any variable(s) which is in common
In expression, 4 xy 2xz , 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
4 xy
2xz
2 y and
z
2x
2x
then 4 xy 2xz 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
18x2
42bx
2a and from
3x and
7b
12ax if 6x is a factor then
6x
6x
6x
thus 12ax 18x2 42bx can be factorized as 6x(2a 3x 7b)
29 | P a g e
Factorize the following expressions:
1. 21a 7
2. 3xy 6 x
3. xy 2 7 xy3
4. 12 x3 y3 x 4 y 2 4 x 2 y
5. aby 2aby aby 2
6. ax bx 2cx
7. x2 y y3 z3 y
8. 3a 2b 2ab2
9. ax2 ay 2ab
10. ax2 y 2ax2 z
11. abx 6ky 4kz
12. x2 y y3 z 2 y
13. 3a 2b 2ab2
14. 6a 2 4ab 2ac
15. 2a 2e 5ae2
16. 2abx 2ab2 2a 2b
17. ayx yx3 2 y 2 x 2
To factorise algebraic expressions involved four terms: (Factorization by
Grouping)
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:
30 | P a g e
a(h k ) b(h k ) . Since (h k ) is common to both terms, thus we have
(h k ) (a b)
Factorise the following expressions
6mx 3nx 2my ny
xh xk yh yk
ay az by bz
as ay xs xy
2ax 6ay bx 3by
6ax 2bx 3ay by
2ax 2ay bx by
8. ms 2mt 2 ns 2nt 2
9. am bm an bn
10. xs xt ys yt
1.
2.
3.
4.
5.
6.
7.
Factorisation of difference between two squares: (Difference of two squares)
This expression is called a difference of two
squares.
(Notice the subtraction sign between the
terms.)
You may remember seeing expressions like this one when you worked with
multiplying algebraic expressions. Do you remember ...
If you remember this fact, then you already know that:
The factors of
are
and
Do not get confused with expression like
( x 6)2,which gives x 6 x 6 when wefactorise . These are called Perfect
Square Trinomials.
31 | P a g e
Remember:
An algebraic term is a perfect square when the numerical
coefficient (the number in front of the variables) is a
perfect square and the exponents of each of the variables
are even numbers
Example:
Factor: x2 - 9
Both x2 and 9 are perfect squares. Since subtraction is occurring between these
squares, this expression is the difference of two squares.
What times itself will give x2 ? The answer is x.
What times itself will give 9 ? The answer is 3.
These answers could also be negative values, but positive values will make our
work easier.
The factors are (x + 3) and (x - 3).
Answer: (x + 3) (x - 3) or (x - 3) (x + 3) (order is not important)
Factorising Quadratic Expressions
An expression of the form ax b where a and b are numbers is called a linear
expression in x.
When two linear expressions in x are multiplied together, the result usually
contains three terms: a term in x2 , a term in x and a number.
Expressions of the form i.e. ax2 bx c where a, b, and c are numbers and
a 0 are called quadratic expressions.
32 | P a g e
Number 8 can be expressed as 4 2 , whereby 4 and 2 are factors of 8. So,
x2 4x 12 can be expressed as x 2 x 6 , therefore x 2 x 6 are
both factors of the expression x2 4x 12 .
To factorise x2 6x 8
Find two numbers which multiply to give 8 and add up to 6. In this case the
numbers are 4 and 2.
Put these numbers into brackets.
So x2 6x 8 = ( x 4) ( x 2)
To factorise x2 2x 15
Two numbers which multiply to give -15 and add up to +2 are -3 and 5
Hence: x2 2x 15 = ( x 3) ( x 5)
For x2 6x 8 , two numbers which multiply to give +8 and add up to -6 are -2
and-4.
x2 6x 8 ( x 2) ( x 4)
Factorise the following:
33 | P a g e
1. x 2 7 x 10
2. x 2 7 x 12
3. x 2 8 x 15
4. x 2 10 x 21
5. y 2 10 y 25
6. y 2 15 y 36
7. a 2 3a 10
8. a 2 a 12
9. z 2 z 6
10. x 2 2 x 35
11. x 2 5 x 24
12. x 2 8 x 16
13. x 2 8 x 240
14. x 2 26 x 165
15. y 2 3 y 108
16. x 2 49
17. x 2 144
18. y 2 1
To factorise 3x2 13x 4
Find two numbers which multiply to give 12 and add up to 13. In this case the
numbers are 1 and 12.
Split the 13x term, 3x2 x 12x 4
Factorise in pairs
x(3x 1) 4(3x 1)
(3x 1) is common
(3x 1) ( x 4)
Factorise the following:
34 | P a g e
1. 2 x 2 5 x 3
2. 2 x 2 7 x 3
3. 3x 2 8 x 4
4. 3x 2 5 x 2
5. 2 x 2 x 15
6. 2 x 2 x 21
7. 3x 2 17 x 28
8. 6 x 2 7 x 2
9. 6 x 2 19 x 3
10. 8 x 2 10 x 3
11. 12 x 2 4 x 5
12. 4a 2 4a 1
13. 12 x 2 17 x 14
14. 15 x 2 44 x 3
15. 120 x 2 67 x 5
16. 1 4 z 2
17. 16a 2 4
18. 16a 2 b2
35 | P a g e
Linear Equations
1.
Solving linear equations in one variable.
2.
Solve simultaneous linear equations in 2 variables
3.
Simple word problems involving linear equations
A linear equation has the general form; bx c 0 where b 0
bx c 0 has the solution x
c
b
c
b c 0(b)
c
Checking the solution by substituting for x in the equation: b
b
cc 0
Solving Linear Equations
An equation has to have an equals sign, as in 3x + 5 = 11 .
A solution to an equation is a number that can be plugged in for the variable to
make a true number statement.
For example, putting 2 in for x above in 3x + 5 = 11 gives
3(2) + 5 = 11 , which says 6 + 5 = 11 ; that's true! So 2 is a solution.
The thing that makes this equation linear is that the highest power of x is x1
(not x 2 or other powers; for those "quadratic equations" go to intermediate
algebra).
To solve the following equations:
1.
4 3x 2
4 2 3x
2 3x
2
x
3
36 | P a g e
2.
2 x 7 5 3x
2 x 3x 5 7
5 x 12
12
2
x 2
5
5
If there is a fraction in the x term, multiply out to simplify the equation.
3.
2x
10
3
2 x 30
30
x 15
2
Solve the following equations:
1.
2x 5 11
2.
3x 7 20
3.
2x 6 20
4.
5x 10 60
5.
3x 7 10
6.
7
7.
x 1
2 3
8.
3 2x
4
3
9.
10.
2x 4 x 3
11.
2 y 1 4 3y
12.
7 3x 5 2x
13.
x 16 16 2x
x
2
3 3x
4
5
37 | P a g e
14.
x
1
1
2
4
15.
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(2 x 3) 8(3x 5) 5(2 x 8) 0
23.
10( x 4) 9( x 3) 1 8( x 3)
24.
6(3x 4) 10( x 3) 10(2 x 3)
25.
1
6 x 30( x 12) 2 x 1
2
26.
1
6(2 x 1) 9( x 1) 8 x 1
4
27.
10(2.3 x) 0.1(5 x 30) 0
28.
3 1
1
1
8 2 x (1 x)
4 4
2
2
29.
(6 x) ( x 5) (4 x)
30.
x
1
101 (10 x)
(10 x) 0.05
100
10
x
2
38 | P a g e
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
x2
Solve the following equations
1.
x 2 4 ( x 1)( x 3)
2.
x 3x 1 x 2 5
3.
x 2x 3 x 7x 7
4.
x 2 x 1 2 x 1x 4
5.
x 1x 3 x 12
6.
x 22 x 33 3x 11
7.
2 x 1 x 2 xx 3
2
2
2 x x 4
2
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 42 x 1
3 x 12 8 x 4
16 5 x
16
x
5
1
x3
5
39 | P a g e
Solve the following equations:
1.
x3 x4
2
5
2.
x 2 3x 6
7
5
3.
x
x
2
3
4
4.
5
10
x 1 x
5.
5
15
x5 x7
6.
4
7
x 1 3x 2
7.
x 1 x 1 1
2
3
6
8.
1
x 2 1 (3x 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.
Let the smaller number be x ; then the other numbers are x 1 and
x 2
2.
The sum of four consecutive numbers is 90. Find the numbers. Answ.
21, 22 23, 24
40 | P a g e
3.
Find three consecutive even numbers which add up to 1524. answ. 506,
508, 510
4.
When a number is doubled and then added to 13, the result is 38. answ.
12
5.
1
2
When 7 is subtracted from three times a certain number, the result is
28. What is the number? Answ. 11
6.
The sum of two numbers is 50. The second number is five times the first.
1
3
Find the numbers. Answ. 8 and 41
7.
2
3
2
3
The difference between two numbers is 9. Find the numbers, if their
sum is 46.
1
2
Answ. 18 , 27
1
2
8.
The product of two consecutive even numbers is 12 more than the
square of the smaller number. Find the numbers. Answ. 6, 8
9.
The sum of three numbers is 66. The second number is twice the first
and six less than the third. Find the numbers. Answ. 12, 24, 30
10.
David weighs 5kg less than John, who in turn is 8kg lighter than Paul. If
their total weight is 197kg, how heavy is each person? Answ.
2
2
2
59 kg, 64 kg, 72 kg
3
3
3
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. Answ. 24, 22, 15
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?
Answ. 48, 12
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? Answ.40, 8
41 | P a g e
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 𝑤
42 | P a g e
represent the number of women and 𝑚 represent the number of men
3.2
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?
SOLUTION
After x years, my age will be 41 + x and my son’s age will be 5 + x.
At that time the equation is 5 x
1
41 x .
2
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?
43 | P a g e
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.
Simultaneous Linear Equations
To find the value of two unknown in a problem, two different equations must
be given that relate the unknown to each other. These two equations are
called simultaneous equations.
Two methods we use at this stage:
Elimination method and Substitution method
Solve the following equations simultaneously.
1.
3x 2 y 0
2x y 7
2.
4 x y 14
x 5 y 13
3.
2 x y 13
5 x 4 y 13
4.
2w 3x 13 0
x 5w 13 0
5.
x 2( y 6) 0
3 x 4 y 30
6.
5c d 11 0
4d 3c 5
7.
2x y 5
x y
2
4 3
44 | P a g e
x 2y 4
8.
3x y 9
1
2
9.
x 3y 7 0
2y x 3 0
10.
2 x 11 y
x y
1
5 4
11.
3x 2 y 5
2x y
7
3 2
9
12.
0.4 3 y 2.6
x 2 y 4.6
PROBLEMS SOLVED BY SIMULTANEOUS EQUATIONS
1.
A motoris buys 24 litres of petrol and 5 litres of oil for N$10.70, while
another motorist buys 18 litres of petrol and 10 litres of oil for N$12.40.
Find the cost of 1 litre of petrol and 1 litre of oil at this garage.
24 x 5 y 1070
18 x 10 y 1240
2.
Find two numbers with a sum of 15 and a difference of 4.
3.
Twice one number added to three times another number gives 21. Find
the numbers if the difference between them is 3.
4.
The average of two numbers is 7, and three times the difference
between them is 18. Find the numbers.
5.
A fishing enthusiast buys fifty maggots and twenty worms for N$1.10
and her mother buys thirty maggots and forty worms for N$1.50. Find
the cost of one maggot and one worm. Answ:
M=1c and w=3c
6.
A television addict can buy either two televisions and three videorecorders for N$1 750 or four televisions and video-recorder for N$1
250. Find the cost of one of each. TV200,V 450
45 | P a g e
7.
Half the difference between two numbers is 2. The sum of the greater
number and twice the smaller number is 13. Find the numbers. Answer 7
and 3
1
( x y ) 2 and x 2 y 13
2
8.
A snake can lay either white or brown eggs. Three white eggs and two
brown eggs weigh 13 grams, while five white eggs and four brown eggs
weigh 24 grams. Find the weight of a brown egg and of a white egg.
9.
A bag contains forty coins, all of them either 2 cent or 5 cent coins. If the
value of the money in the bag is N$1.55, find the number of each kind.
15 two cents and 25 five cents.
10.
A slot machine takes only 10cent and 50cent coins and contains a total
of twenty-one coins altogether. If the value of the coins is N$4.90, find
the number of coins of each value. 14 ten cents and 7 fifty cents.
11.
Thirty tickets were sold for a concert, some at 60 cents and the rest at
N$1. If the total raised was N$22, how many had the cheaper tickets? 20
12.
The wage bill for five men and six woman workers is N$6 700, while the
bill for eight men and three women is N$6 100. Find the wage for a
man and for a woman. Man N$500, woman N$700.
13.
If the numerator and denominator of a fraction are both decreased by
2
. If the numerator and denominator are
3
3
both increased by 1 he fraction becomes . Find the original fraction.
4
5
answer .
7
one the fraction becomes
14.
The denominator of a fraction is 2 more than the numerator. If both
denominator and numerator are increased by 1 the fraction becomes
2
3
. Find the original fraction. =
3
5
15.
In three years’ time a pet mouse will be as old as his owner was four
years ago. Their present ages total 13 years. Find the age of each now.
Boy 10, mouse 3
16.
Find two numbers where three times the smaller number exceeds the
larger by 5 and the sum of the numbers is 11. 4 and 7
46 | P a g e
17.
A wallet containing N$40 has three times as many N$1 notes as N$5
notes. Find the number of each kind. N$1x15, N$5x5.
18.
At the present time a man is four times as old as his son. Six years ago he
was 10 times as old. Find their present ages. 36, 9.
QUADRATIC EQUATIONS
Quadratic equations always have a x2 term, and often an x term and a
number term, and generally have two different solutions.
The general form: ax2 bc c 0, where a 0 and b and c are numbers. Not
all quadratic equations can be solved to get real number as the solutions.
When solve, sometimes you get 2 real number solutions, sometimes you get
one real number solution which is a repeated solution.
DISCRIMINANT:
The type of solution that a quadratic equation has depends on its discriminant.
The discriminant of the quadratic equation ax2 bx c 0 is given by the
expression, b2 4ac .
The discriminant of the quadratic equation can be positive, zero or negative.
e.g.
2x2 5x 3 0 the 49
4 x 2 12 x 9 0
the 0
2x2 x 2 2x the 15
If 0 the equation has two distinct real number solutions (distinct roots)
If 0 the equation has a repeated real number solution.
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If 0 then the equation does not have real number solutions because we
can't do a real square root of a negative number. However, there are two
complex solutions.
Three methods which help us to find the correct value of x:
Factorising
Completion of the square
Using quadratic formula
For these course, we will apply the first two methods.
Solving Quadratic Equations
Recall a linear equation is one that looks like ax + b = cx + d, and our strategy
was to get all x terms on the left, all constants on the right, then divide by the
coefficient on x to solve.
A quadratic equation has an x2 (x-squared) term; "quadrat" is Latin for square.
The general quadratic equation looks like ax2 bx c
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Solve the following Equations:
1. x 2 7 x 12
2. x 2 8 x 12
3. x 2 4 x 5 0
4. 3 x 2 10 x 8 0
5. 2 y 2 3 2 y y 2 5
6. 2 x 2 7 x 15 0
7. y 2 15 y 56 0
8. z 2 8 z 65
9. 36 x 2 x 2 0
10. 6a 2 a 1
11. x 2 3 x 0
12. 3 x 2 x 0
13. 6a 2 9a
14. 4 x 2 1
15. 12 x 5 x 2
x
16. 2 x 2
3
2
17. x 49
x
18. x 2
4
SOLUTION BY FORMULA
Quadratic formula:
x
b b 2 4ac
2a
Solve the following, giving answers to two decimal places where necessary:
1.
2 x 2 11x 5 0
2.
3x 2 11x 6 0
3.
x 2 8x 2 0
4.
1
y 3y 1 0
2
5.
5x 2 7 x 2 0
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6.
3x 2 7 x 20 0
7.
6 x 2 11x 7 0
8.
3x 2 1 4 x 0
9.
5x 2 x 2 2 0
10.
2 x( x 1) ( x 1) 2 5
11.
x
12.
10 x 1
13.
3
3
4
x 1 x 1
14.
2
4
3
x 2 x 1
15.
16
16.
4( y 1)
17.
9 x 2 16
18.
x 2 6x
15
22
x
3
x
1
x2
3
1 y
Some word problems solved by quadratic equations
1.
The length of a rectangle is 6 inches more than its width. The area of
the rectangle is 91 square inches. Find the dimensions of the rectangle.
2.
The product of two consecutive odd integers is 1 less than four times
their sum. Find the two integers
3.
The product of two consecutive odd integers is 1 less than twice
their sum. Find the integers.
4.
The length of a rectangle is 4 times its width. The area of the rectangle is
144 square inches. Find the dimensions of the rectangle.
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SETS and SET THEORY
A set is a collection of distinct objects, 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, e.g.
A is the set whose members are the first four positive integers.
B is the set of colours of the French flag.
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.
This is an example of property definition method. This means set A contains
elements 2, 3, 4, 5, 6, 7, 8, 9
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.
The curly braces are the customary notation for sets.
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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, 12then
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, 12then
A B is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
3.
(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 B 1, 2, 3, 4, 5 then A is a subset of B. A B
4.
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
The totality of all sets. The universe (usually represented as or ) 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, 12then
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14
52 | P a g e
6.
A
complement of or not in A
Complement of set A relative to set U, denoted Ac or A 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} .
OR
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.
empty set
Note an for any set.
Example:
Given That:
1, 2, 3...,12 , A 2, 3, 4, 5, 6 and B 2, 4, 6, 8,10
(a) A B 2, 3, 4, 5, 6, 8,10
(b) A B 2, 4, 6
(c) A 1, 7, 8, 9,10,11,12
(d ) n A B 7
(e) B A 3, 5
53 | P a g e
Exercise
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)
2.
X Y
(b)
n( X Y )
(c)
X Z
(d)
n( X Z )
Given that:
A a, b, c, d , e B a, b, d , f , g C b, c, e, g, h D d , e, f , g, h
and a, b, c, d ,..., j , find:
(a) A ( B D)
(b) ( A D) B
(c) n( B C D)
(d ) B B
VENN DIAGRAMS
In this section we introduce the ideas of sets and Venn diagrams. A set is a list
of objects in no particular order; they could be numbers, letters or even words.
A Venn diagram is a way of representing sets visually.
The formal definition is: A Venn diagram or set diagram is a diagram that
shows all possible logical relations between a finite collection of sets. They
were conceived around 1880 by John Venn. In this course we use them to
illustrate relationships in sets
The intersection of set A and set B can be displayed showing a Venn diagram
as:
54 | P a g e
A B
For A
We can also display the following information on a Venn diagram.
A 1, 2, 3, 4
B 3, 4, 5, 6 and 1, 2, 3, 4, 5, 6, 7, 8
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APPLICATION OF VENN DIAGRAMS
EXERCISES
1.
In the Venn diagram,
U people in the hotel
B people who like bacon
E people who like eggs
(a)
How many people like bacon?
(b)
How many people like eggs but not bacon?
(c)
How many people like bacon and eggs?
(d)
How many people are in the hotel?
(e)
How many people like neither bacon nor egg?
56 | P a g e
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.
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.
(c)
Add together the number of elements in the three parts of the
diagram to obtain the equation.
(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 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?
6.
The values of p, q and r in the Venn diagram below are:
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7.
The values of p, q and r in the Venn diagram below are:
8. 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.
8.1 Draw a Venn diagram and show the information as given above on the
Venn diagram.
8.2 Solve for x .
8.3 Find the number of students who speak Spanish only.
9. 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.
9.1
Present this information in a Venn diagram.
9.2
How many female students were first year?
9.3
How many male students were first year?
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10.
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 that every Indian watches at least one of
these games)
A. 135%%
B. 36%
C. 7%
D. 35%
Venn Diagrams of three sets
1.
In the Venn diagram,
U cars in a street
B blue cars
L cars with left hand drive
F cars with four doors
(a)
(b)
How many cars are blue?
How many blue cars have four doors?
(c)
How many cars with left-hand drive have four doors?
(d)
How many blue cars have left-hand drive?
(e)
How many cars are in the street?
(f)
How many blue cars with left-hand drive do not have four doors?
2.
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 left-handed, 80
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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?
(c)
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?
You may need to make use of the two formulas below at some point hence it is
wise that you recognise them.
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)
3.
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.
4.
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?
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5.
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?
6.
From the Venn diagram below, describe the region shaded.
A. A B
7.
B. ( A B) C
C. ( A B) C
D. ( A B) C
A team of athletes was selected to compete in long jump (L), javelin (J)
and high jump (H). 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:
7.1
( L H ) J
A. 51
7.2
B. 22
C. 102
D. 131
B. 51
C. 18
D. 21
(L H ) J
A. 29
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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:
8
3
5
A
7 2
9
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
23 .
62 | P a g e
Matrix Notation
Statisticians / mathematicians 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
A 11 12
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:
1
2
2
A 1
1
2
11
3
0 1
4
9
1 5
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Entries a 2 , a ,
11
12
a ,
32
a
23
Matrix Addition and Subtraction
Addition and subtraction of a matrix of order 2 x 2.
a
If A 11
a
21
a
12
a
22
The matrix has two rows namely (a ; a ) and (a ; a ) It also has two
11 12
21 22
a
a
11
columns namely
and 12
a
a
21
22
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.
4 1 4
to get
5 3 7 9
1
So we can add matrices
2
12
0
.
6
Scalar Multiple of a matrix
To multiply matrix A aij by a scalar, k , we multiply each and every element
of A by k . Thus, kA k aij kaij .
EXAMPLE
Given:
3 4
B
0 5
,
Find: 2 B 2
3 4
0
5
10
6 8
0
64 | P a g e
Multiplication by another matrix
For 2 x 2 matrices
a
c
b w
d y
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:
3
(a)
4
2 2
1
1
1 (3 2) (2 1)
5 (4 2) (11)
(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
3
2
A B
Bc
2B
3 A B
2C 3 A
2A B
CB A
AB
2( BC )
10. C 2
1.
2.
3.
4.
5.
6.
7.
8.
9.
Find the value of the letters.
2
1.
y
x 4
7 3
y x
2 z
9
9
65 | P a g e
x
2. 1
w
a
3.
c
x
4.
2
2 x
2 y
3 v
b 2
0 3
y 8
3 x
5 w
z
w
8
5
a
1
1
2
d b
3 2 5
y
1
0
2
5.
0
0 m 10
3 n 1
p
6.
q
2
2
3
7.
2
0 y
x 4
3y
8.
2 y 4x
2
1 5
1
2q 10
3
3z 6
2 z 8
2
9.
a
e 3
k
3 0
4
10.
1
0 n
m
2
1
11. if A
3
3
13. B
1
3
w
z 6
0 8
3
w
1 8
2 3
p 20
0 1
0
x
, B
2
1
6
1
12
q
0
, and AB BA, find x
3
3
1
(a) Find k if B 2 kB
66 | P a g e
THE INVERSE OF A MATRIX
The inverse of a matrix A is written A 1 , and the inverse exists if AA1 A1 A I
where I is called the identity matrix.
Only square matrices possess an inverse.
1
0
1
For 2x2 matrices, I
0
1
For 3x3 matrices, I 0
0
If
a
A
c
0
1
0
0
0 , etc
1
b
d
1
1
1
, the inverse A is given by A
(ad cb) c
d
b
a
Here, the number ad cb is called the determinant of the matrix and is
written A or det (A).
If A 0 , then the matrix has no inverse.
To find the inverse of A if
3
A
1
4
2
2
1
A
(3 2) (1 4) 1
1
1 2
A 1 A
2 1
Check:
0
1
1
0
4
1 2
3
2 1
4 3
3 1
1
4
1
3
2
4
1 2
2
2 0
2
3
2
0
2
Multilying by the inverse of a matrix gives the same result as dividing by the
matrix: the effect is similar to ordinary algebraic operations.
67 | P a g e
e.g. if AB C
A 1 AB A 1C
B A 1C
Find the inverse of the matrix:
4
1.
3
1
1
1
2.
2
2
5.
1
2
2
4
6.
1
1
9.
1
2
3
2
10.
1
2
13.
1
3
4
7
14.
5
2
If B
1
2
2
0
0 1
1 0
3
2
4
2
0
1
4
2
5
4.
1
2
7.
2
1
3
3
11.
1
2
4
2
15.
2
0
8.
2
3
12.
2
2
1
3
4
1
1
1
4
0
1
3
1
X
4
0
0
, find X
1
2
1
2
4
and AB
3
0
3
1
X
5
0
0
, find X .
1
Find B if A =
If
3
2
3
3.
1
4
and AB I , find A.
3
Find Y if Y
3
If
2
5
2
7
68 | P a g e
1
2
Find M if
1
1
M 2
1
0
3
1
and B
1
1
2
0
A=
Find: (a)
(b)
A 1
©
B 1
0
1
1
3
AB
Show That ( AB) 1 B 1 A1
3
If M= 2
.
2
1
1
7
and MN
1
2
1
11
; C . If B is a (2 1) matrix such that AB C , find B.
1
7
A=
x
1
Find x if the determinant of
(a)
9
, find N .
6
5
(b)
1
x
If the matrix
-1
3
is
2
(c)
0
2
has no inverse, what is the value of x ?
4
69 | P a g e
APPLICATION OF MATRICES-CRAMER’S RULE
Consider the simultaneous equations:
2x y 3
x 2y 4
This system of pair of equations can be written in matrix form as AX b, where
1
is the matrix of coefficients,
2
2
A
1
x
3
X and b
y
4
2
Find det
1
3
Find x det
4
1
(2 2) (1 1) 5
2
1
(3 2) (1 4) 10.
2
Note that the first column (or x-column) has been replaced by b.
2
Find y det
1
3
(2 4) (3 1) 5
4
Note that the second column (or y-column) has been replaced by
b.
x
y
x 10
2
5
y
5
1
5
Hence x 2 and
y 1
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Use Cramer’s rule to solve the following pairs of simultaneous linear equations:
1 . 2 x y 1
4x 3y 8
2. 3 x 4 y 7
x 5 y 4
3. 3 p 4 q
2q p 1
4. 4 x 2 y 2
2x 3y 5
5. 2 p 3q 6
3 p q 2
6. m 4 2 n
2n m 4
EXERCISES
1.
2
Given 𝐴 = [
3
1
2 −2
] 𝑎𝑛𝑑 𝐵 = [
]
−1
4
0
Work out:
1
− 𝐵
2
𝐴+𝐵
𝐴𝐵
2.
x
8
B. 1
A. 0
3.
1
has no inverse, what is the value of x ?
1
If the matrix
2
3
If matrix A
C. 8
2
, find AA1
1
D. 8
(show A 1 )
71 | P a g e
4.
x
For what values of x will the matrix
3
A. x 5 and x 3 B. x 17
1
1
if matrix C 2
1
5.
Find CC
6.
If
7.
Given matrix C
2
1
3 2
a
1 0
1 8
4 c
3
1
5
have no inverse?
x 2
D. x 0
C. x 5 and x 3
1
1
2 .(show C )
2
b
, find the values of a, b, c and d .
d
3
, calculate the value of x if C 2 xC . (Show
1
matrices C 2 and xC. )
8.
Find A2 if the matrix A is given as:
2
A
2
9.
3
.
0
x
2
Find the values of the letters in the matrices,
A. x 3 and y 0
B. x 0 and y 2
3 2 5
y 1 0
C. x 4 and y 2
D. x 1 and y 4
10.
3
2
Given the matrices A
0
5
1
B
0
2
8
Find:
3A
Det (2 B )
A2
72 | P a g e
11.
2
ak
Given matrices
e 3
k
3 0
1 8
2 6
6a
1
, the values of
a, e, and k are:
A. k 2, e 20 and a 3 B. k 2, e 4 and a 3
C. k 2, e 20 and a 3 D. k 2, e 20 and a 3
12.
2
2
Given the matrices A
3
0
1
B
3
0
6
Find:
A B
2A B
AB
73 | P a g e
ARITHMETIC AND GEOMETRIC PROGRESSIONS
AP is a progression with a common difference.
If given a progression:
2, 5, 8, 11, 14, 17
The 1st term is 2, The second term is 5 and the 3rd term is 8
T1
T2
T3
2
5
8
T4
11
T5
14
T6
15
In this progression, we find the next number by adding 3 to the previous one.
3 = Common Difference
While in the progression8; 2; -4; -10; -16; -22 we find the next number by
adding – 6 or subtracting 6.
The last term is referred to as Tn therefore the term next to Tn is Tn 1 and the
term before that is Tn 1 .
To find the Common Difference: (d)
2nd term – 1st term;
3rd term – 2nd term; 4th term – 3rd term
d = Tn1 Tn
74 | P a g e
To find the common difference of the progression:
2; 9; 16; 23; 30
D=7
9 – 2 = 7;
16 – 9 = 7; 23 – 16 = 7 and 30 – 23 = 7
In a progression 2; 9; 16; 23; 30
T1 2; T2 9; T3 16; T4 23; T5 30
We can find that T11 72 and that T101 702
using the formula:
Tn T1 (n 1)d
T11 2 (11 1)7 72 and
T101 2 (101 1)7 702
Therefore to find Tn we use the above formula, Tn T1 (n 1)d
In AP 6; 11; 16; 21;…
What are the values of T1 , T2 , and T5 ?
2.
Find the common difference.
3.
Find
d=5
75 | P a g e
T12
61
T30
151
T72
361
Tn
5n 1
In AP 8; 4; 0; -4; -8; -12;…
Find
T101
ans 392
Tn 1
ans 4t 8
Tn
ans 4t 12
For a certain AP, T1 5 and d 3
Find T3
T8
ans 1
ans 16
HW: AP T1 12 and d
1
2
Find T15
To find the 1st term and the common difference given any two terms of the
progression: e.g.
The 4th term of an AP is 3 and the 7th term is -6. Find the 1st term and the d
(common difference)
76 | P a g e
Tn T1 (n 1)d
Tn T1 (n 1)d
T4 T1 3d
T7 T1 6d
3 T1 3d
6 T1 6d
3 T1 3d
6 T1 6d
9 3d
d 3
3 T1 9
T1 12
Finding The Term’s Position:
Sometimes you are given an AP of 4; 7; 10; 13;…;100
What is n? or what position does the term 100 occupy?
Tn T1 (n 1)d
100 4 (n 1)3
100 1 3n
n 33
Therefore T33 100
EXAMPLES
Find the 100th term of the following progressions:
2; 5; 8; 11;…
ans. 299
16; 12; 8; 4;…
ans. -380
Find the 50th term of the following progressions:
1.
8; 5; 2; 1;...
answ. -139
77 | P a g e
2.
1 1
3
2 ; 3 ; 4; 4 ;...
2 4
4
answ. 39
1
4
The common difference of an AP is 3, the 24th term is 74. What is the 1st term? Ans. T1 5
The first term of an AP is 3. There are 25 terms is and the last term is 195. Find common difference.
Answ. d = 8
The 3rd term of an AP is 18 and the 50th term is 347. Find the common difference and the 1st term.
Answ. d = 7
T1 4
The 5th term of an AP is 3 and the 9th term is 5. Find the 1st term and the common difference.
Answ. D =
2 1
and T1 1
4 2
The 1st term of an AP is 1 and the d is 2. The last term is 61. How many terms are there in this
progression.
61 1 (n 1)d
n 31
There are 31 terms in the progression.
THE SUM OF AN ARITHMETIC PROGRESSION
How to add the terms of a progression:
Having the AP 1; 3; 5; 7; 9; 11…
We add the terms as 1+3+5+7+9+11. This is called a series and this series has the sum 36. If the series is large
or some of terms are not listed then we use S n
n
a1 l
2
Sum of a series is denoted by S
78 | P a g e
S n means the sum of the 1st n terms of the series or progression.
The progression 1; 3; 5; 7; 9; 11
S 1 3 5 7 9 11
adding the two together we get
S 11 9 7 5 3 1
2S 12 12 12 12 12 12
2S 6 12
2S 6(1 11)
6
Thus S (1 11)
2
From
6
1 11, 6 is the number of terms, 1 is the 1st term and 11 is the last term.
2
Sn
n
T1 l , where l is the last term of the series .
2
Example:
Find the sum of the progression: 5; 9; 13; 17; 21; 25
S 5 9 13 17 21 25
T1 5
l 25
n6
n
T1 l
2
6
S 6 5 25
2
S 6 3 30 90
Sn
If we do not have the last term (the last term not given) then the last term is n.
Tn T1 (n 1)d
Therefore S n
Sn
n
T1 l will be
2
Sn
n
T1 T1 n 1d
2
n
2T1 n 1d
2
79 | P a g e
Find the sum of the first 120 terms of the series 5 + 9 + 13 +…
S
S120
S120
n
2T1 (n 1)d
2
120
2 5 120 14
2
29160
Find the AP if the sum of the first ten terms is 210 and the sum of the next ten terms of an AP is 610.
S
n
2T1 (n 1)d
2
20
2T1 (20 1)d 210 610
2
20T1 190d 820 equ 1
S 20
10
2T1 (10 1)d 210
2
10T1 45d 210 equ. 2
S10
20T1 190d 820
10T1 45d 210
20T1 90d 820
20T1190(4) 820
20T1 90d 420
T1 3
100d 400
d 4
Hence the progression is 3; 7; 11; 15
Find the sum of the series 3 + 9 + 15 +… as far as the 50th.
S
n
2T1 (n 1)d
2
50
2(3) (50 1)6
2
7500
S 50
Find the sum of the series 3 + 5 + 7 +…+ 103
1st we have to find the position of 103 (to find n)
80 | P a g e
a n T1 (n 1)d
103 3 (n 1)2
103 3 2n 2
51 n
Then the sum of the series will be:
S
n
2T1 (n 1)d
2
51
2(3) (51 1)2
2
51
(106) 2703
2
S
Find the sum of the series 12 + 8 + 4 + 0 + … + (-32)
a n 32
a n T1 (n 1)d
32 12 (n 1) 4
32 12 4n 4
n 12
S
n
2T1 (n 1)d
2
12
2(12) (12 1) 4
2
S 120
The sum of the first 20 terms is 420 and the sum of the next 10 terms is 510. Find the 1st term and the
common difference.
S
n
2T1 (n 1)d
2
S
n
2T1 (n 1)d
2
81 | P a g e
20
2T1 (20 1)d 420
2
20T1 190d 420 equ.1
30
2T1 30 1d 930
2
30T1 435d 930 equ.2
S 20
60T1 570d 1260
60T1 870d 1860
S 30
subst . 30T1 435(2) 930
T1 2
300d 600
d 2
4.
n
2T1 (n 1)d
2
27
2T1 (27 1)d
S 27
2
702 13.52T1 26d
S
27T1 351d 702
12
2T1 (12 1)d
2
177 12T1 66d
S12
27T1 351d 702
12T1 66d 177
d 1.5 and T1 6.5
GEOMETRIC PROGRESSIONS
We recognize a geometric progression by the presence of a common ratio.
The progression 1; 2; 4; 8; 16;…
We noticed that a term needs to be multiplied by 2 to get the next term. Hence the next three terms
in this progression will be 32, 64 and 128.
The common ratios of some GP are not easily recognizable, therefore:
To obtain the common ratio (r) of a GP use the formula:
r
(n 1)thterm
nth term
82 | P a g e
We must make sure that the r is indeed common
e.g. r
2nd term 3rd term 4th term
etc.
1st term
2nd term 3rd term
Find the common ratio and write down the next three terms of each progression.
1.
2; 6; 18; 54;….
6
3
2
18
3
6
The next term is 54 3 162 ,
2.
5; -10; 20; -40; …
3.
16; 8; 4; 2; …
54
3
18
then 162 3 486 , and next is 486 3 1458
answ. Next is 80, then -160 and 320
r
1
2
next three are 1;
1
1
and
2
4
Terms of a progression
The nth term of a GP is given by the formula:
nth term T1 r n1
Tn T1 r n1
In the GP 2; 6; 18; 54;… find the 8th term
Tn T1 r n 1
T8 2 381
T8 4374
T1 of a GP is 5 and the third term is 45, find r.
T1 5
T3 T1 r
and
T3 45
n 1
45 5 r n 1
45 5r 2
9 r2
3 r
The r is 3 or 3 .
When T1 and r are not given, we need two equations.
83 | P a g e
Find the 1st term and the common ratio of a GP with the 2nd term is 10 and the 4th term is 250. Also write
down the 1st 4 terms of the GP.
T2 10
T4 250
Tn T1 r n 1
Tn T1 r n 1
10 T1 r 21
250 T1 r 41
10 T1 r equ.1
250 T1 r 3
equ.2
Dividing equ 2 by equ. 1
T1 r 3 250
T1 r
10
r 2 25
r 5
If r = 5 then T1 T1 5 T2 : T2 T1 r 10 T1 5 therefore the 1st term is 2.
If r = -5 then T1 T1 5 T2 therefore the 1st term is -2.
For r = 5 then the GP is 2; 10; 50; 250;…
For r = -5 then the GP is -2; 10; -50; 250;…
To find the number of terms in a given GP.
What term of the GP 128; 64; 32; 16;… has the value
1
?
4
84 | P a g e
64
0.5
128
32
0 .5
64
16
0.5
32
1
2
Tn T1 r n 1
r
n 1
1
1
128
4
2
n 1
1
1
512 2
512 2 n 1
2 9 2 n 1
n 10
1
is the 10th term.
4
If T6 3888
T10 5038848
Tn T1 r n 1
Tn T1 r n 1
3888 T 1 r 6 1
5038848 T 1 r 10 1
3888 T 1 r 5
5038848 T 1 r 9
T1 r 9
T1 r 5
5038848
3888
r 4 1296
4
r4
r 6
4
1296
If r 6 then Tn T1 r n 1
T6 3888
3888 T1 6 5
T1
T1
3888 1
7776 2
1
6 T2 3 6 T3 18 etc.
2
85 | P a g e
ACTIVITY 11.3
1.
For the progression 6; 30; 150; 750; …
What is r? r=5
What are the values of T5 , T6 and T10 ? T5 3750 T6 18750 T10 11718750
Write down an expression for Tn .
Tn T1r n 1
6 5 n1
2.
Find the common ratio of each of the following GP’s
(a)
1
2
13
5; 3 ; 2 ; 1
2
7
27
(b)
3; 3; 3; 3;.....
ans. 1
©
2; 4; 8; 16;.......
ans. -2
ans.
The 1st term of a GP is 2 and the fifth term is 162. Find two possible values for the common ratio. For each
common ratio, write down the first three terms of the GP.
T1 2 and T5 162
T5 T1 r n 1
162 2 r 51
r 4 81
r 3
if r 3 then T1 2; T2 2 3 6; T3 6 3 18
if r 3 then T1 2; T2 2 3 6; T3 6 3 18
The fourth term of a GP is 27 and the sixth term is 243. Find the 1st term and the common ratio. Also write down
the first four terms of the progression.
86 | P a g e
T4 27
T6 243
and
Tn T1 r n 1
27 T1 r 3
243 T1 r 5
243 T1 r 5
27 T1 r 3
9 r2
r 3
If r 3 then T1 1 then the GP is 1; 3; 9; 27;....
If r 3 then T1 1 then the GP is 1; 3; 9; 27;....
How many terms are there in the progression 3; 6; 12;…; 12288
Tn T1r n 1
12288 3 2 n 1
4096 2 n 1
212 2 n 1
12 n 1
n 13
There are 13 terms in the progression
THE SUM OF A GEOMETRIC PROGRESSION
The GP 3; 6; 12; 24; 48; 96;…
The sum of the 1st 5 terms is S 5 3 6 12 24 48
We can write this as:
S 5 3 2 0 3 21 3 2 2 3 2 3 3 2 4
where T1 3 and
r2
*
S 5 3 2 0 3 21 3 2 2 3 2 3 3 2 4
**
2S 5 3 21 3 2 2 3 2 3 3 2 4 3 2 5
Subtracting * from * * we have:
87 | P a g e
2S 5 S 5 3 2 5 3 2 0
S 5 3 2 2 3(32 1)
5
0
3(31)
93
If given a GP with the 1st term and the common ratio, we can find the sum by using the formulas:
T1 r n 1
Sn
r 1
*** IF r > 1 we use:
T1 1 r n
Sn
1 r
*** IF r < 1 we use:
If r = 0 then the formula breaks down. (Why)
The 2nd formula is obtainable from the 1st formula using the fact that x y ( y x) or
a b ba
cd d c
Find the sum of the 1st ten terms of the GP 3; 9; 37; 81;….
r is 3 which is more than one hence:
T1 r n 1
r 1
3 310 1
S10
88572
3 1
Sn
Hence the sum is 88572
Find the sum of the GP 8; -4; 2; -1; …;
T1 8
and
r
1
32
1
2
88 | P a g e
First we have to find n. (The position of
1
)
32
Tn T1 r n 1
n 1
1
1
8
32
2
n 1
1
1
256
2
1
1
n 1
8
2
2
8 n 1
n9
n = 9 and r is less than 1 then The sum will be:
T1 1 r 9
1 r
1
81 ( ) 9
2
S9
1
1 ( )
2
11
S9 5
32
S9
Find the 1st term of GP with r 5 and the sum of the 1st 6 terms is 2 604, write down the GP.
T1 1 r n
1 r
T 1 (5) 6
2604 1
1 (5)
T (15624)
2604 1
6
2604 2604T1
Sn
T1 1
GP 1; 5; 25; 125; 625
Find the sum of the first ten terms of the GP 3; 9; 27; 81
89 | P a g e
T1 3
r 3
T1 r n 1
r 1
3 310 1
S10
88572
3 1
Sn
Find the sum of the GP 1;
r
1
2
Sn
1
;
2
1 1
;
;...
4 8
which is less than one.
T1 1 r n
1 r
1 10
11
2
511
S10
1
or 1.998
1
512
1
2
Find the sum of the GP -32; 64; -128; 256
r 2
T1 1 r n
1 r
32 1 (2)10
S10
10912
1 (2)
Sn
Find the sum of the GP 3; 6; 12; 24;…;1536
90 | P a g e
Tn T1 r n 1
1536 3 2 n 1
512 2 n 1
2 9 2 n 1
n 10
S10
3 210 1
3069
2 1
The GP 1; 4; 16; 64;….is given. The sum of the 1st n terms is greater than 8500. What is the value of n?
S n 8500
T1 r n 1 1 4 n 1 4 n 1
r 1
4 1
3
therefore :
Sn
4n 1
8500
3
4 n 1 25500
4 n 25501
n log 4 25501
Changing the basis
log 25501
n
log 4
n 7.319
Find the sum of the GP 1 4 16 ... 1024
91 | P a g e
Tn T1 r n 1
1024 1 4 n 1
1024 4 n 1
4 5 4 n 1
n6
The Sum :
T1 r n 1
r 1
1 4 6 1 4096 1
S6
1365
4 1
3
Sn
The Sigma Notation
“Sigma” one of the 24 letters of the greek alphabet.
6
Having
a
i 1
i
a1 a 2 a3 a 4 a5 a 6
EXAMPLES
5
1.
2n 2(1) 2(2) 2(3) 2(4) 2(5)
n 1
2 4 6 8 10 30
6
2.
(1 2k ) 1 2 3 1 2 4 1 2 5 1 2 6 40
k 3
4
3.
2
n
2 0 21 2 2 2 3 2 4 31
n 0
7
4.
1
1
1
1
1
319
i 4 5 6 7 420
i 4
92 | P a g e
EVALUATE:
3
1.
n 1
4
1
1
n
4
2.
4
1
1
1
4
1
1
2
4
1
1
3
4
4
4 4 8 12
2
7
1
1 2 3 4
11
3
1
1
2
3
i 4 2 4 3 4 4 4
3
3
3
3
9
i 2
FIND THE FOLLOWING SUMMATIONS
5
1.
i
12 2 2 3 2 4 2 5 2 55
2
i 1
4
2.
1
2
i 0
i
1
1
1
1
1
15
1 2 3 4 1
0
16
2
2
2
2
2
2k k 2 2 2 2 3 3 2 4 4 47
4
3.
2
2
2
2
k 2
5
4.
2
21 2 2 2 3 2 4 2 5 62
r
r 1
Evaluate
4
1.
n
n 0
2
n
0
1
2
2
1 0 1
1 1
2
2 1
2
3
3 1
2
4
37
1
85
4 1
2
i 2 1 12 1 2 2 1 32 1
5
7
i
1
2
3
6
i 1
3
2.
5
3.
3 p
2
3 4 2 3 5 2 123
p 4
93 | P a g e
x
2
4.
2
2 x 3 0 2 2 0 3 12 2 1 3 2 2 2 2 3 2
s
11 2 2 33 32
x 0
3
5.
S
s 1
100
6.
n 1 2 3 4 5 ... 100
n 1
Tn T1 (n 1)d
100 1 (n 1)1
100 1 n 1
100 n
n
2T1 (n 1)d
2
100
2 1 (100 1)1
S
2
S 5050
S
94 | P a g e
PERCENTAGES:
Percentages are simply a convenient way of expressing fractions or decimals. 50% 0f N$60 means
N$60. or more simply
50
of
100
1
of N$60. Percentage are used very frequently in everyday life.
2
Example
Change 80% to a fraction.
80%
80 4
100 5
Change
3
to a percentage
8
3 3 100
1
% 37 %
8 8 1
2
Change 8% to a decimal
8
0.08
100
Exercise:
Change to fractions
1. 60%
2. 24%
3. 35%
4. 2%
Change to percentages
1.
1
4
2.
1
10
3.
7
8
4.
1
3
5. 0.72
6. 0.31
Change to decimals
1. 36%
2. 28%
3. 7%
4. 13.4%
5.
3
%
5
6.
7
%
8
The following are marks obtained in various tests. Convert them to percentages.
(a)
17 out of 20
(b) 31 out of 40
95 | P a g e
(b)
19 out of 80
(c) 112 out of 200
(c)
1
2 out of 25
2
(d) 7
1
out of 20
2
A car costing N$400 is reduced in price by 10%. Find the new price.
10
2400 240
100
new price of car (2400 240) N $2160
10% of N $400
After a price increase of 10% a television set cost N$286. What was the price before the increase?
x
10
x 286
100
The price before the increase is 100%.
Or
110% of old price = N$286
100x 10x 286000
100% of old price=
110
28600
x
then x N $260
110
110
286
110
10% of old price =
286 100
110 1
Old price of TV =N4260
Exercise
1.
Calculate:
(a) 30% of N$50
(b) 45% of 200kg (c) 4% of N$70
(d) 2.5% of 5000 people
2. In a sale, a jacket costing N$40 is reduced by 20%. What is the sale price?
3. The charge for a telephone cal costing 12 cents is increased by 10%. What is the new charge?
4. In peeling potatoes 4% of the mass of the potatoes is lost as ‘peel’. How much is left for use
From a bag containing 55 kg?
5. Work out to the nearest cent:
(a) 6.4% of N$15.95
© 8.6 of N$25.84
(b) 11.2% of N$192.66
(d) 2.9% of N$18.18
6. Find the total bill:
(a) 5 golf clubs at N$18.65 each
(b) 60 golf balls at N$16.50 per dozen
© 1 bag at N$35.80
96 | P a g e
Sales tax at 15% is added to the total cost.
7. In 2000 a club has 250 members who each pay N$95 annual subscription. In 2001 the membership increases
by 4% and the annual subscription is increased by 6%. What is the total income from subscriptions in 2001
8. In 1999 the prison population was 48 700 men and 1600 women. What percentage of the total prison
population were men?
9. In 1999 there were 21 280 000 licensed vehicles on the road. Of these, 16 486 000 were private cars. What
percentage of the licensed vehicles were private cars?
10. A quarterly telephone bill consists of N$19.15 rental plus 4.7 cents for each dialed unit. Sales tax is added at
15%. What is the total bill for Mrs. Jones who used 915 dialed units?
11. Hassan thinks his goldfish got chickenpox. He lost 70% of his collection of goldfish. If he has 60 survivors,
how many did he have originally?
12. The average attendance at Parma football club fell by 7% in 1999. If 2030 fewer people went to matches in
1999, how many went in 1998?
13. In the last two weeks of a sale, prices are reduced first by 30% and then by a further 40% of the new price.
What is the final sale price of a shirt which originally cost N$15?
14. Over a period of 6 months, a colony of rabbits increases in number by 25% and then by a further 30%. If
there were originally 200 rabbits in the colony how many were there at the end?
15. A television cost N$ N$270.25 including 15% sales tax. How much of the cost is tax?
16. The cash price for a car was N$7640. Mr Khan bought the car on the following hire purchase terms:’ A
deposit of 20% of the sash price and 36 monthly payment of N$191.60’. Calculate the amount Mr. Khan paid.
1.
Mr. Hansen’s annual salary was N $282 000 in the year 2004. In 2005 his salary was
increased by 12.5% and in 2006 his monthly salary increased to N $28 645.03.
From the information above, determine:
1.1
Mr. Hansen’s monthly salary in 2005.
A. N $2 643.50
1.2
C. N $317 250.00
D. N $26 437.50
The percentage increase for 2006.
A. 8.34%
1.3
B. N $35 250.00
B. 1.835%
C. 0.0835%
D. 8.35%
His total income for the three years.
A. N $78 582.53
B. N $942 990.36 C. N $942 970.00 D. N $660 990.38
97 | P a g e
2.
In 2006 an association had 10 000 members who each paid N$500 monthly
membership fees. In 2007 the monthly membership fees increased by 2% and the
members increased by 40%. In 2008, 8% of the members cancelled their
membership and no payment was received from them. At the same time 2000 members
signed up for membership and joined the association.
From the information above, determine:
2.1
The total income from membership fees for 2007.
A. N $60 000 000
2.2
B. N $5 000 000
B. 12 880
C. 14 880
D. 9 200
The difference between the income of 2007 and 2008.
A. N $5 385 600
2.4
D. N $7 140 000
The number of members in 2008.
A. 14 000
2.3
C. N $85 680 000
B. N $25 680 000
C. N $31 065 600 D. N $448 800
Find the total bill of the items given below.
(a) 5 golf clubs at N$18.65 each
(b) 60 golf balls at N$16.50 per dozen
(c) 1 bag at N$35.80
Sales tax at 15% is added to the total cost.
A. N $211.55
2.5
B. N $1 286.91
C. N $167.86
D. N $243.28
After a price increase of 10% , the price of a car is N $105 000 . What was the
price before
the increase?
2.6
Mr. Goagoseb has 24 goats and Mrs. Namises has 30 goats. They decide to share 648kg of
animal feed between them, in the ratio of the numbers of their animals. How many
kilogram of animal feed does each get?
2.7
Convert 0.75% to a vulgar fraction in its simplest form.
98 | P a g e
RATIOS
How to Determine a Ratio
Ratios represent how one quantity is related to another quantity. A ratio is a comparison between
two or more quantities
A ratio may be written as A:B or A/B or by the phrase "A to B".
A ratio of 1:5 says that the second quantity is five times as large as the first.
The following steps will allow a ratio to be determination if two numbers are known.
Example: Determine the ratio of 24 to 40.
Divide both terms of the ratio by the greatest common factor (24/8 = 3, 40/8=5)
State the ratio. (The ratio of 24 to 40 is 3:5)
SIMPLIFYING A RATIO
24 to 40 can be simplified to 3:5
Simplify the following ratios to their simplest forms:
1.
75mm : 10cm
2.
45minutes : 3hrs
3.
N$1.50 : 25c
Suppose there are thirty-five people, fifteen of whom are men. Then the ratio of men to
women is 15 to 20.
Notice that, in the expression "the ratio of men to women", "men" came first. This order is
very important, and must be respected: whichever word came first, its number must come
first. If the expression had been "the ratio of women to men", then the numbers would have
been "20 to 15".
Expressing the ratio of men to women as "15 to 20" is expressing the ratio in words. There
are two other notations for this "15 to 20" ratio:
odds notation: 15 : 20
fractional notation:
15/
20
99 | P a g e
You should be able to recognize all three notations; you will probably be expected to know
them for your test.
Given a pair of numbers, you should be able to write down the ratios. For example:
There are 16 ducks and 9 geese in a certain park. Express the ratio of ducks to
geese in all three formats.
Consider the above park. Express the ratio of geese to ducks in all three
formats.
If Sarah and Tom are to share N$15 with Sara taking one two third and
Tom taking one third, express their share in ratios.
Express each of the following in the form 1:n
1.
25 : 75 =1 : 3
2.
42 :8 = 1: 4/21
3.
N$0.80 : 25c = 1: 5/16
The numbers were the same in each of the above exercises, but the order in
which they were listed differed, varying according to the order in which the
elements of the ratio were expressed. In ratios, order is very important.
Let's return to the 15 men and 20 women in our original group. I had expressed
the ratio as a fraction, namely, 15/20. This fraction reduces to 3/4. This means
that you can also express the ratio of men to women as 3/4, 3 : 4, or "3 to 4".
Dividing a quantity is a given ration:
1.
Divide 105 into a ratio of 2:3
100 | P a g e
2.
A piece of wire is 105 m long. Paul cuts this piece into two parts, making
one part twice as long as the other. What are the lengths of the two
pieces?
1:2
1+2=30
1/3 times 105 and 2/3 times 105
3.
Divide 8484 in the ration of 2:3:7
EXERCISES
1.1
Madam Henk left N$ 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 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.
1.2.1
The maturity value after 3 years will be;
A. N $1 708 593.75
1.1.2
C. N $1 650 000 D. N $450 000
Paul will receive an amount of;
A. N $240 000
1.2.3
B. N $1 200 000
(3)
B. N $360 000
(2)
C. N $341 718.75 D. N $495 000
How much will Mark and Cyril receive altogether?
A. N $155 000
B. N $825 000
(3)
C. N $1 495 000 D. N $1 155 000
101 | P a g e
2.
A ratio of boys to girls in a class room is 16 to 13. If there are 174 boys and girls in the
classroom altogether then how many boys are there? ANS 96
3.
The ratio of boys to girls in a class room is 8 to 5. If there are 72 boys then how many girls
are there? Answer 45
4.
Jane and Tom spend N$5 between them on their lottery tickets each week. Jane puts in N$1
and Tom puts in N$4. They have agreed to share the winnings according to the amount they
put in. If one week they win N$45 000, how much will each get? Jane 9000 and Tom 36 000
PROPORTIONS
A proportion is simply a statement that two ratios are equal. It can be written in two ways:
As two equal fractions
a c
or
b d
Using a colon a : b c : d
Given a proportion,
20 4
we can read this as twenty is to twenty-five as four is to five.
25 5
We can use cross products to test whether two ratios are equal and form a proportion.
To find the cross products of a proportion, we multiply the outer terms called the extremes and the
middle terms called the means.
For example: in the proportion
20 4
which can be expressed as 20 : 25 4 : 5 ,
25 5
20 and 5 are the extremes, and 25 and 4 are the means.
Since the cross products are both equal to 100 then the ratios are equal and that is true proportion.
We can also use cross products to find a missing term in a proportion.
20 30
50
x
x 75
DIRECT PROPOTION
Direct proportion is when an increase in one quantity causes increase in other quantity or decrease
in one quantity causes decrease in other quantity. We say that the two quantities are related
directly.
102 | P a g e
PRINCIPLE OF DIRECT PROPORTION
EXAMPLE:
If 30 dozen of eggs cost N$300. Find the cost of 5 dozen of eggs
Solution:
Let x be the required price of 5 dozens of eggs
EGGS (dozens) COST(N$)
30
5
300
x
OR
Since quantities are in direct proportion, so we use the above principle
30 300
5
x
x N $50
EXAMPLE
A car travel 81 km in 4.5 liters. How far will it go by 20 liters of petrol?
Solution:
Let x be required distance travelled by car in 20 litres.
PETROL
DISTANCE
4.5
81
20
x
x 360km
EXAMPLE
4 brick layers could lay 240 bricks in one day, working at the same rate:
(a)
How many bricks could 12 brick layers lay in one day?
(b)
How many bricklayers could lay 180 bricks in one day?
SOLUTION
(a) Let x be the number of bricks that 12 brick layers can lay in 1 day.
103 | P a g e
Brick Layers
Bricks
4
240
12
x
4 x 12 240
x 720
12 brick layers can lay 720 in one day.
(b)
Brick Layers
Bricks
4
240
x
180
240 x 4 180
x3
EXERCISES:
1.
The cost of a particular quality of cloth is N$210. Calculate the cost of 2, 4, 10 and
13 metres of cloth of the same type.
2.
If the weight of 12 sheets of thick paper is 40 grams, how many sheets of the
1
same paper would weigh 2 kg.
2
3.
A train is moving at a uniform speed of 75km/hour
How far will it travel in 20 minutes? Ans.25
Find the time required to cover a distance of 250km. Ans. 200min or 3hrs 20min
4.
A loaded truck travels 14km in 25 minutes. If the speed remains the same, how far
can it travel in 5 hours?
INVERSE PROPORTION (INDIRECT PROPOTION)
Two quantities may change in such a manner that if one quantity increases, the other
quantity decreases and vice versa. For example, if we increase the speed, the time taken to
cover a given distance decreases.
EXAMPLE:
If it takes 4 days for 10 men to dig a trench, how long will it take 8 men?
104 | P a g e
MEN
DAYS
10
1
10 4 40
8
40
5
8
4
It takes 5 days for 8 men to dig the trench.
Or
Vedic Method
The first by the first and the last by the last
4 : 10
8: x
4 10 8 x
40 8 x
x5
8 men take 5 days
Example
If 15 workers can build a wall in 48 hours, how many workers will be required to do the same
work in 30 hours?
WORKER
HOURS
15
48
x
30
15 48 30 x
720 30 x
x 24
To finish work in 30 hours, 24 workers are required.
Exercises
1.
A farmer has enough food to feed 20 animals in his cattle for 6 days. How long would
the food last if there were 10 more animals in his cattle?
105 | P a g e
2.
A factory requires 42 machines to produce a given number of articles in 63
days. How many machines would be required to produce the same number of
articles in 54days?
3.
A school has 8 periods a day each of 45 minutes duration. How long would
each period be, if the school has 9 periods a day, assuming the number of school
hours to be the same?
106 | P a g e
INTERESTS AND DISCOUNT
INTEREST:
Money paid for the use of capital borrowed or invested.
RATE or INTEREST RATE:
The percentage of the principal payable as interest on the capital
TIME:
The period for which the principal is borrowed or invested.
AMOUNT:
The sum of the principal and all the interests paid during the investment
period.
NOMINAL RATE of INTEREST: The interest rate is often given per year (p.a.) “per annum”
EFECTIVE INTEREST RATE:
The rate that actually determines the interest earned on the capital
i = nominal rate
r = effective rate
Relationship between r and i;
r 1 i 1
m
If 10% pa on investment
n=4 (4 quarters in a year)
i
10
% because interest is to be calculated 4 times in a year.
4
4
10
r 1 1
4
4
10
r 1
1
400
r 1.1038 1
r 0.1038
r 10.38%
107 | P a g e
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 P1 rt
I
r
and
pt
I
p
or
rt
A PI
or
t
I
pr
p A I
or
p
A
1 rt
EXAMPLES
1.
Find the simple interest payable on a loan of N$2 500 at 25% p.a. at the end of 3 years.
I prt
I 2500 25% 3
I N $1 875
2.
Find the simple interest payable on a loan of N$2 500 at 12
1
% p.a. at the end of 18
2
months.
I prt
12.5
1 .5
100
I N $468.75
I 2500
3.
For how long should an amount of N$5000 be invested at 5% p.a. to generate an interestN$750?
108 | P a g e
t ?
p 5000
r 5%
I 750
t
I
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 p1 rt
22500 125501 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 interest rate of 12% for 2yrs.
p = 8 500
r = 12%
t = 2yrs
I prt
I 8500 0.12 2 N $2040
109 | P a g e
7.
Calculate the maturity value of an investment of N$30 000 due in 5yrs when the annual
interest rate is 16%.
r = 0.16
t=5
p = 30 000
simple
A=?
A p1 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 )
7
0.145)
12
5422.92 p(1.084583333)
p N $5000
5422.92 p(1
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
54000 30000
r
150000
24000
r
.16 16%
150000
I prt
I 30000 0.16 5
I N $24000.00
(b)
110 | P a g e
I
pr
24000
t
30000 .16
24000
t
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 $250 840.00 at the end of the ninth year?
2.
Madam Henk left N$ 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 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 $1 708 593.75
B. N $1 200 000
C. N $1 650 000 D. 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 N $170 000 at 6.75% p.a. at the end of 9
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.
7.
(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
12.75%
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.
111 | P a g e
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
1 i n
Calculate the amount payable for a loan of N$1000 for 3yrs at the rate of 10% p.a. compounding
annually.
p = 1000
A p 1 i
r = 0.1
n=3
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?
112 | P a g e
(a) A=250 000
I=
7.5%
0.0375
2
n = 20
A p (1 i ) n
A
p
(1 i ) n
250000
P
20
0.075
1
2
p N $119723.09
(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
r = 8%
n=2
A = N$1166.40
113 | P a g e
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 0.048
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
1.6 1 i
0.013141253 i
i 0.013141253 12 0.157695036 15.8%
36
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
114 | P a g e
A 800001 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 130311.571 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
2
n = 2 6
A p(1 i) n
0.1
A 60001
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
B. N $1 412 000
C. N $4 523 573 243
D.
N $1 712 271.66
115 | P a g e
2.
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.
3.
The sum to be invested for four years at 8% p.a. compounded semi-annually to amount 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
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 this
savings account until he was seventy years old. How much was in Tate Paul’s savings
account at the age of seventy?
9.
Determine the sum to be invested for 4 years at 12.5% per annum compounded monthly to
amount to N$65 000 at the end of the investment.
116 | P a g e
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.
117 | P a g e
