ENG1C03 - sdeuoc.ac.in
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19655
BASIC NUMERICAL METHODS (BCM3 A11/BBA3 A11)
STUDY MATERIAL
III SEMESTER
CORE COURSE
Bcom/BBA (2019 ADMISSION ONWARDS)
UNIVERSITY OF CALICUT SCHOOL OF DISTANCE EDUCATION
CALICUT UNIVERSITY P.O., MALAPPURAM - 673 635, KERALA
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SCHOOL OF DISTANCE EDUCATION UNIVERSITY OF CALICUT
STUDY MATERIAL THIRD SEMESTER
Bcom/BBA (2019 ADMISSION ONWARDS)
COMMON COURSE BCM3 A11/BBA3 A11: BASIC NUMERICAL METHODS
Prepared by:
1. Dr P Siddeeque Melmuri Assistant Professor
School of Distance Education
University of Calicut.
2, Sri. Udayakumar O.K, Associate Professor,
Govt College Madappally
Scrutinised by:
Prof.P.BAIJUMON
Assistant Professor,
Department of Commerce,
Govt. College, Malappuram
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CONTENTS
Modules Contents Page No.
1 Numerical Expressions and Equations
2 Matrices
3 Progressions
4 Interest and Time Value
5 Descriptive Statistics
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Syllabus
Common Course
BBA & BCom โ BASIC NUMERICAL METHODS
Objectives:
To enable the students to acquire knowledge of numerical equations,
matrices progressions, financial mathematics and descriptive statistics.
Learning Outcome: On completing the course ,the students will be able
to understand, numerical equations, matrix, progression, financial
mathematics, descriptive statistics and their applications.
Module I: Numerical expressions and Equations:
Simultaneous linear equations (upto three variables), Quadratic equations
in one variable-factorization and quadratic formula
(10 Hours)
Module II Matrices: introduction - type of matrices โ - trace and
transpose and determinants - matrix operations โadjoint and inverse โ
rank- solving equations by matrices: Cramerโs Rule( not more than Three
variables). (15 Hours)
Module III Sequence, Series and Progression :Concepts
and differences - Arithmetic progression- n th term and sum of n
terms of an AP - Insertion of Arithmetic means in AP - Geometric
progression- โnโth term and sum of n terms of an GP - Insertion of
Geometric Mean in GP - Harmonic progression. (20 Hours)
Module IV Interest and Time value : Concept of interest-Types of
interest: Simple interest and compound interest โ nominal, real and
effective rate of interest. Future value and Present Value; Annuity and
Perpetuity . Computing future and present values of annuity ( regular and
immediate) - multi and growing period perpetuity. Compound annual
growth rate- computation of Equated Monthly Instalments (EMI).
(15 Hours)
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Module V: Descriptive Statistics: Measures of Central Tendency โ
Mean : Arithmetic mean , Geometric mean and Harmonic Mean-
Median ,Mode and other position values. Measures of Dispersion: mean
deviation, quartile deviation, standard deviation and coefficient of
variation. Measures of Skewness and Kurtosis.
(20 Hours)
Reference Books
1 Business Mathematics and Statistics- N G Das & J K Das (Tata
McGraw Hill)
2 Basic Mathematics and its Application in Economics โ S. Baruah
(Macmillan )
3 Mathematics for Economics and Business โ R. S. Bhardwaj (Excel
Books)
4 Business Statistics โ G. C. Beri (Tata McGraw Hill)
5 Fundamentals of Statistics โ S.C.Gupta (Himalaya Publishing House
6 SP Gupta ,Statistical Methods, Sultan Chand
7 Dinesh Khattar-The Pearson guide to quantitative aptitude for
competitive examinations.
8 Dr. Agarwal.R.S โ Quantitative Aptitude for Competitive
Examinations, S.Chand and Company Limited. 9.. Abhijit Guha,
Quantitative Aptitude for Competitive Examinations, Tata Mcgraw
Hill,
(Theory and problems may be in the ratio of 20% and 80% respectively.
An over view of the topics is expected and only simple problems shall be
given)
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Module I
NUMERICAL EXPRESSIONS AND EQUATIONS
THEORY OF EQUATIONS
An equation is a statement of equality between two expressions.
For eg:โ x +2 = 5. An equation contains one or more unknowns.
Types of Equations
1)Linear Equation
It is an equation when one variable is unknown.
For example 2x + 3 = 7
Practical Problems
1. Solve 2x + 3 = 7
Ans: 2x = 7 โ 3
2x = 4, x = 4
2 = 2
2. Solve 3x + 4x = 35
Ans: 7x = 35, x = 35
7 = 5
3. Solve 4 ( x โ 2 ) + 5 ( x โ 3 ) โ 25 = x + 8
Ans: = 4x โ 8 + 5x โ 15 โ 25 = x + 8
= 4x + 5x โ x = 8 + 8 + 15 + 25
8x = 56
x = 56/8 = 7
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4. 7x โ 21 โ 3x + 13 = 7 + 6x โ 19
Ans: 7x โ 3x โ 6x
=7 โ 19 + 21 โ 13
= โ 2x = โ 4
2x = 4
x = 4
2 = 2
5. โ23x + 14 โ 7x + 16 = 10x โ 17 + 3x + 4
Ans: โ23x โ 7x โ 10x โ 3x = 17 + 4 โ 14 โ 16
โ23x = โ23 23x = 23
x = 23/23 =1
6. Find two numbers whose sum is 30 and difference is 4
Ans: Let one number = x
Then other number = 30 โ x
Numbers = (30 โx) โ x = 4
โ2x = 4 โ 30
โ2x = โ26 2x = 26
x= 26
2 = 13
โด numbers are 13, 17
7. Two third of a number decreased by 2 equals 4. Find the number
Ans: Let the number = x
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Then 2/3(x) โ 2 = 4
2x โ 6 = 12
2x = 12+ 6
2x = 18
x = 9
8. Solve 7๐ฅ+4
๐ฅ+2 =
โ4
3
Ans: = 3(7x + 4) = -4 (x + 2 )
= (21x + 12) = โ 4x + โ8
21x + 4x = โ8 โ 12
25x = โ20
x = 20
โ25 =
4
โ5
9. The ages of Hari and Hani are in the ratio of 4 : 5. Eight years from
now, the ratio of their ages will be 5:6. Find their present age?
Ans: Let present age = 4x and 5 x
After 8 years = 4๐ฅ+8
5๐ฅ+8 =
5
6
= 6(4x + 8 ) = 5 (5x + 8)
= 24x + 48 = 25x + 40
= 24x โ 25x = 40 โ 48
= โ1x =โ8
= x = 8
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Present ages of Hari and Hani are Hari = 4x = 4 ร 8 = 32 years
Hani = 5x = 5 ร 8 = 40 years
2) Simultaneous equations in two unknowns
For solving the equations, firstly arrange the equations. For
eliminating one unknown variable, multiply the equation 1 or 2 or
both of them with certain amount and then deduct or add some
equation with another, we get the value of one variable. Then
substitute the value in the equation, we get the values of
corresponding variable.
PRACTICAL PROBLEMS
1. Solve 3x + 4y = 7
4x โ 7 = 3
Ans: 3x + 4y = 7 โโโโโโ (1)
4x โ y = 3 โโโโโโโ (2)
Multiply the equation 2 by 4, then
3x + 4y = 7 โโโโโโโโโ (1)
16x โ 4y = 12
Add 19 x = 19
x = 19
19 = 1
Substitute to value of x
3x + 4y = 7
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3 ร 1 + 4y = 7
3 + 4y = 7
4y = 7 โ 3 = 4
4y= 4
y 4
4 = 1
2. 4x + 2y = 6
5x + y = 6
Ans: 4x + 2y = 6 โโโโโโโโโโโ (1)
5x + y = 6 โโโโโโโโโโโ (2)
Multiply the equation 2 by 2, then
4x + 2y = 6
10x + 2y = 12
โ6x = โ6 (Deduct 1 โ 2)
6x = 6
x= 6
6 =1
5x+ y = 6
5 ร 1 + y = 6
5 + y = 6, y = 6 โ 5 = 1
3. Solve y = 3(x + 1)
4x = 4 + 1
Ans: y = 3x +1
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4x = 4 +1
Arrange the equation
ยญ3x + y = 3 ยญยญยญยญยญ-- (1)
4x โ y = 1 ยญยญยญยญยญยญยญยญยญยญ (2)
1x = 4 Add
x = 4
Substituting the value of x
4x โ y = 1
16 โ y = 1
Y = 16 โ 1 = 15
X = 4, y = 15
4. Solve 8x + 7y = 10 11x = 10(1โy)
Ans: 8x + 7y = 10 โโโโโโ (1)
11x = 10 โ 10 y
11x + 10 y = 10 โโโโโโโโโ (2)
Multiply equation (1) by 11 and (2) by 8
88x + 77y = 110
88x + 80 y =80
(1ยญ2) ยญ3y = 30
y= 30
โ3 = ยญ10
Substituting the value of y
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8x + 7y = 10
8x + 7 ร ยญ10 = 10
8x + ยญ70 = 10
8x = 10 + 70
8x = 80, x =80
8 =10
x = 10, y = ยญ10
5. Solve ๐ฅโ๐ฆ
2=
๐ฆโ1
3 ๐๐๐
3๐ฅโ4๐ฆ
5 = ๐ฅ โ 10
Ans: ๐ฅโ๐ฆ
2=
๐ฆโ1
3
3(xยญy) = 2 (yยญ1)
3x โ 3y = 2y โ 2
3x โ 3y โ 2y = ยญ2
3x โ 5y = ยญ2 ยญยญยญยญยญยญยญยญยญยญ (1)
3๐ฅโ4๐ฆ
5 = ๐ฅ โ 10
3x โ 4 y = 5 (xยญ10)
3x โ 4y = 5x โ 50
3x ยญ5x โ 4y = ยญ50
2x + 4y = 50
= x + 2y = 25 ยญยญยญยญยญยญยญยญ (2)
Multiply equation (2) by 3
3x โ 5y = ยญ2
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3x + 6y = 75
(1 -2 ยญ11y = ยญ77
y = โ77
โ11 = 7
Substituting the value
x + 2y = 25
x + 2ร7 = 25
x = 11
x = 11, y = 7
6. A man sells 7 horses and 8 cows at Rs. 2940/โ and 5 horses
and 6 cows at Rs. 2150/โ. What is selling price of each?
Ans: Let the selling price of horse = x
Cow = y
7x + 8y = 2940 ยญยญยญยญยญยญยญ (1)
5x + 6y = 2150 ยญยญยญยญยญยญยญยญ(2)
Multiply equation (1) by 5 and 2 by 7
Then 35x + 40 y = 14700
35x + 42y = 15050
(1โ2) โ2y = โ350
y =โ350
โ2 = 175
Substituting the value of y
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7x + 8y = 2940
7x + 8 ร175= 2940
7x = 2940 โ 1400
7x = 1540
x = 1540
7
Selling price of horse = 220
Selling price of cow = 175
3. Simultaneous Equations in three unknowns
Firstly, eliminate one of the unknown from first two equations. Then
eliminate the same unknown from second and third equations. Then
we get two equations. Solve such equations, we get the values of x,
y and z.
1) Solve 4x + 2y โ 32 = 2
3x + 4yโ2z = 10
2x โ 5y = 5
Ans: First consider first two equations and eliminate one unknown
4x + 2y โ 3z = 2
3x + 4y โ 2z = 10
For eliminating 2 multiply equation in 1 by 2 and 2 by 3, then
8x + 4y โ62 = 4
9x + 12y = 30
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(2โ1) x + 8y = 26 (1)
Consider equation 2 and 3
3x + 4y โ 2z = 10
2x โ 5y +4z = 5
On multiply xy equals 2 by 2
6x โ 8y โ 42 = 20
2x โ 5y + 42 = 5
add 8x + 3y = 25 (2)
Solve the new equation 1 and 2
x + 8y = 26 (1)
8x + 3y = 25 (2)
Multiply equation 1 by 8, then
8x + 64y = 208
8x + 3y = 25
(1โ2) 61y = 183
Substitute value of Y
x + 8y = 26
x + 8 x 3 = 26
x + 24 = 26
x = 26 โ 24 = 2
Substitute the value of x, y,
4x + 2y โ 3z = 2
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4 x 2 + 2 x โ3z = 2
8 + 6 โ 3z = 2
14 โ 3z = 2
3z = 14 โ z
3z = 12
z = 12/ 3 = 4
x = 2, y=3, z=4
4. Quadratic equations
The equation of the form ax2 + bx+ c = 0 in which a, b, c are constant is
called a quadratic equation in x. Here x is the unknown.
Solution of quadratic equations
There are three methods to solve a quadratic equation.
1. Method by formula
2. Method of factorization
3. Method of completing the squre
Quadratic formula method
One general quadratic equation is ax2 + bx + c = 0
Then x = โ๐ยฑโ๐2โ4๐๐
2๐
1. Solve the equation x2 โx โ 12 = 0
Ans: a = 1, b= โ1, c=โ12
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x = โ๐ยฑโ๐2โ4๐๐
2๐ =
โโ1ยฑโ12โ4ร1รโ12
2ร1
= 1 ยฑ โ49
2 = 1ยฑ
7
2
= 8/2 or -6/2
= 4 or โ3
2. Solve the equation 2x +5
๐ฅ = 7
Ans: Multiply the equation by x Then
2x2 +5 = 7x
2x2 โ 7x + 5 = 0
a = 2, b= โ7, c=5
= โ๐ยฑโ๐2โ4๐๐
2๐
= โโ7ยฑโโ72โ4ร2ร5
2ร2
= 7ยฑโ9
4
= 10
4 or
4
4
= 2/5 or 1
3. Solve the equation ( x + 1) (x +2) โ 3 = 0
Ans: x2 + 2x + x + 2 โ 3 = 0
x2 +3x + 2 โ 3 = 0
x2 + 3xโ1 = 0
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a = 1, b = 3, c= โ1
x= โ๐ยฑโ๐2โ4๐๐
2๐
= โ3ยฑโ32โ4ร1รโ1
2ร1
= โ3ยฑโ9โโ4
2
= -3ยฑโ13
2
4) Solve x4 โ 10 x2 + 9 = 0
Ans: Let x2 = y
Then equation
= y2 โ 10y + 9 = 0
y = โ๐ยฑโ๐2โ4๐๐
2๐
= โโ10ยฑโโ102โ4ร1รโ9
2ร1
= 10ยฑโ64
2
= 10 ยฑ 8
2
Y = 9, 1
x2 =y , then x = โy
Y = 1, x= โ1 = ยฑ1
Y = 9, x =โ9 = ยฑ 3
X = โ1, 1, 3, โ3
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5) 2xโ7โx + 5 = 0
Answer = Letโx = y, then equation
2y2 โ 7y + 5 = 0
Y= โ๐ยฑโ๐2โ4๐๐
2๐
= โ7ยฑโ72โ4ร2ร5
2ร2
=โ7ยฑโ49โ40
4
= 7ยฑ3
4 =
10
4 ๐๐
4
4
y= 1, x = 12 = 1
y= 10
4, x=
100
16 =
25
4
x = 1, 25
4
6) Solve x 10 โ 33x5 + 32 = 0
Ans: Let y = x5, Then equation
= y2 โ 33y + 32 = 0
Use quadratic formula
Y = 32, 1
Y = 32 then x2 = 32
= 25=32
โด x = 2
y = 1 then x5 = 1
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= 15 = 1, x = 1
X = 2, 1
7. Solve x + y = 10
xy = 24
Ans: change to equation in the form of quadratic
x + y = 10
x= 10โy
Substitute the value in second equation
xy = 24
(10โy) y = 24
=10y โy2 = 24
y2โ 10 + 24 = 0
Use quadratic formula
y = โ๐ยฑโ๐2โ4๐๐
2๐
= โ10ยฑโ102โ4ร1ร24
2ร1
= โ10ยฑโ102โ96
2
= 10 ยฑ2
2 = 6, 4
when y = 6, x = 4 y= 4, x = 6
Simultaneous equations of two unknowns when one of them is
quadratic and the other is linear
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1) x + y = 7
x2 + y2 = 25
Ans:
x + y = 7
y = 7 โ x
Substistue the value of y, y in the second equation, then
x2 + (7โ x)2 = 25
We know (aโ b)2 = a2 โ2ab+b2
x2 + 72โ 2 x 7 x x + x2 = 25
x2 + 49 โ 14x + x2 = 25
x2 + x2 โ 14x + 49 โ 25
2x2โ14x + 24 = 0
Use quadratic formula
Y= โ๐ยฑโ๐2โ4๐๐
2๐
= โโ14ยฑโโ142โ4ร2ร24
2ร2
14ยฑโ4
4
= 14ยฑ 2
4 =4, 3
When y= 4, x = 3
Y=3, x=4
2. Solve x + y = 5
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2 x2 โ y2 โ 10x โ 2xy โ 28 = 0
Ans: y = 5โ x
Substitute the value of y in equation (2)
2 x2 โ (5- x)2 โ 10x โ 2x(5 - x)+ 28 = 0
= 3x2 -10x+ 3 = 0
Use quadratic formula
X = 3 or 1
3
When x = 3, y =2
When x = 1
3, y =
14
3
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Module II
MATRICES
Matrices
A matrix is an ordered rectangular array of numbers or functions. It
is a rectangular presentation of numbers arranged systematically in
rows and columns one number or functions are called the elements
of the matrix. The horizontal lines of elements of the matrix are
called rows and vertical lines of elements of matrix are called
columns.
Order of Matrix
A matrix having โmโ rows โnโ columns are called a matrix of order
โm x nโ or simply โm x nโ matrix (read as an โmโ by โnโ matrix)
Types of Matrices
(i) Rectangular matrix : Any matrix with โmโ rows and โnโ column
is called a rectangular matrix. It is a matrix of Order m x n. For
example,
A = 1 2 32 1 23 2 1
4 1 2
is a 3 x 4 matrix
(ii) Square matrix : A matrix by which the number of rows are
equal to the number of columns, is said to be a square matrix. Thus
an m x n matrix is said to be square matrix if m= n and is known as
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a square matrix of order โnโ. For example,
A = [2 6 11 8 45 4 0
] is a square matrix of order 3
(iii) Row matrix : A matrix having only one row is called a row
matrix. For example,
A = [1 2 3 ]is a row matrix.
(iv) Column matrix : A matrix having only column is called
column matrix. For example,
A = [123
] is a column matrix.
(v) Diagonal matrix: A square matrix is said to be diagonal it all
elements except leading diagonal are zero. Elements a11, a22, a33
etc. termed as leading diagonal of a matrix. Example of Diagonal
matrix is
A = [2 6 1
1 8 4
5 4 0
] is a diagonal matrix. Leading diagonal elements
are 2, 8, 0.
(vi) Scalar Matrix : A diagonal matrix is said to be scalar matrix,
if its diagonal elements are equal. For example.
A = [2 6 1
1 2 4
5 4 2]
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(vii) Unit matrix of identity matrix : A diagonal matrix in which
diagonal elements are 1 and rest are zero is called Unit Matrix or
identity matrix. It is denoted by 1.
A = [1 0 00 1 00 0 1
] is a Unit matrix or Identity matrix.
(viii) Null Matrix or Zero matrix: A matrix is said to be zero or
null matrix if all its elements are zero. For example
A = [0 0 00 0 00 0 0
] is a Null matrix or Zero matrix
(ix) Triangular matrix: If every element above or below the
leading diagonal is zero, the matrix is called Triangular matrix. It
may be upper triangular or lower triangular. In upper triangular all
elements below the leading diagonal are zero and in the lower
triangular all elements above the leading diagonal are zero. For
example,
A = [1 6 10 8 40 4 2
] is a matrix of upper triangular.
A = [2 0 01 8 05 4 1
] is matrix of lower triangular
(x) Symmetric matrix : Any square matrix is said to be symmetric
if it is equal to transpose. That is, A = At
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Transpose of a matrix as a matrix obtained by interchanging its
rows and columns. It is denoted by At or A1. Example of symmetric
matrix
A = [2 11 4
] , = [2 11 4
]
(xi) Skew Symmetric Matrix : Any square matrix is said to be
skew symmetric if it is equal to its negative transpose. That is A =
โAt
For example A = [0 2 3
โ2 0 โ4โ3 4 0
] = At
At =[0 โ2 โ32 0 43 โ4 0
]
-At = [0 2 3
โ2 0 โ4โ3 4 0
]
OPERATION OF MATRICES
Operation of matrices relate to the addition of matrices, difference,
multiplication of matrix by a scalar and multiplication of matrices.
Addition of matrices: If A and B are any two matrices of the same
order, their sum is obtained by the elements of A with the
corresponding elements of B.
For example :
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A = [8 โ6 2
โ7 3 2โ4 3 2
] B = A = [โ5 2 3โ3 โ2 13 โ2 2
]
Then A + B = A = [3 โ4 5
โ10 1 3โ1 1 4
]
Difference of Matrices : if A and B are, two matrices of the same
order, then the difference is obtained by deducting the element of B
from A.
If A = [ 1 2 32 3 1
] B = [ 3 โ1 3
โ1 0 2]
Then A- B = [ โ2 3 03 3 โ1
]
Multiplication of a Matrix by a Scalar
The elements of Matrix A is multiplied by any value (ie. K) and
matrix obtained is denoted byK
For example : A = [1 2 32 3 12 2 1
]
Then 5A = [5 10 15
10 15 510 10 5
]
Practical Problems
1) If A = [ 0 2 32 1 4
] , ๐ต = [ 7 6 31 4 5
] find A-B?
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Ans: 3A = [ 0 6 96 3 12
]
3A โB = [ 0 6 96 3 12
] - [ 7 6 31 4 5
]
= [ โ7 0 65 โ1 7
]
(2) Solve the equation:
2 [๐ฅ ๐ฆ๐ง ๐ก
] +3[1 โ10 2
] = 5 [3 54 6
]
Ans: 2 [๐ฅ ๐ฆ๐ง ๐ก
] = [2๐ฅ 2๐ฆ2๐ง 2๐ก
]
3[2 11 4
] = [3 โ30 6
]
5 [3 54 6
] = [15 2520 30
]
โด [2๐ฅ 2๐ฆ2๐ง 2๐ก
] + [3 โ30 6
] = [15 2520 30
]
2x+3 = 15, 2x = 15 -3 =12, x= 12
2 = 6
2y+ -3 = 25, 2y = 25+3 = 28, y 28
2 = 14
2z+0 = 20, 2z = 20 , z = 20
2 = 10
2t+6 = 30, 2t = 30 -6 =24, t = 24
2 = 12
(3) Find the value of a, b if
2 ร [๐ 57 ๐ โ 3
] + [3 โ41 2
] = [7 6
15 14]
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Ans: 2 ร [๐ 57 ๐ โ 3
] = [2๐ 1014 2๐ โ 6
]
[2๐ 1014 2๐ โ 6
]+ [3 โ41 2
] = [7 6
15 14]
2a+3 = 7, 2a = 7 - 3 =4, a = 4
2 = 2
2b-6+2 = 14, 2b = 14+ 6 -2 =18, b = 18
2 = 9
2t+6 = 30, 2t = 30 -6 =24, t = 24
2 = 12
Multiplication of two matrices
For multiplication, take each row of the left hand side matrix with
all colums of the right hand side matrix.
For example
A = [๐ ๐๐ ๐
] ๐ต = [๐ ๐๐ โ
] ๐กโ๐๐ ๐ด๐ต = [๐๐ + ๐๐ ๐๐ + ๐โ๐๐ + ๐๐ ๐๐ + ๐โ
]
Practical Problems
(1) Let A = [ 1 2 3
โ2 1 4] , ๐ต = [
2 3 15 4 21 5 3
] Compute AB
Ans: AB =
[ 1 ร 2 + 2 ร 5 + 3 ร 1 1 ร 3 + 2 ร 4 + 3 ร 5 1 ร 1 + 2 ร 2 + 3 ร 3
โ2 ร 2 + 1 ร 5 + 4 ร 1 โ2 ร 3 + 1 ร 4 + 4 ร 5 โ2 ร 1 + 1 ร 2 + 4 ร 3]
AB =[ 2 + 10 + 3 3 + 8 + 15 1 + 4 + 9โ4 + 5 + 4 โ6 + 4 + 20 โ2 + 2 + 12
]
AB = [ 15 26 145 18 12
]
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(2) Let A = [1 23 4
] ๐๐๐ ๐ต = [โ2 21 โ1
] Find AB and BA and
hence show that ABโ BA.
Ans: AB = [1 ร โ2 + 2 ร 1 1 ร 2 + 2 ร โ13 ร โ2 + 4 ร 1 3 ร 2 + 4 ร โ1
]
= [โ2 + 2 2 + โ2โ6 + 4 6 + โ4
]
= [0 0
โ2 2]
BA = [โ2 ร 1 + 2 ร 3 โ2 ร 2 + 2 ร 41 ร 1 + โ1 ร 3 1 ร 2 + โ1 ร 4
]
= [โ2 + 6 โ4 + 81 + โ3 2 + โ4
]
= [4 42 โ2
]
Therefore, ABโ BA
(3) Let A = [0 6 7
โ6 0 87 โ8 0
] , B = [0 1 11 0 21 2 0
] , C = [2
โ23
]
Calculate AC, BC and (A+B)C and verify that (A+B)C = AC +
BC.
Ans: AC = [0 โ 12 + 21โ12 + 0 + 2414 + 16 + 0
] = [9
1230
]
BC = [0 โ 2 + 32 + 0 + 62 โ 4 + 0
] = [18
โ2]
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A+B = [0 7 8โ5 0 108 โ6 0
]
(A+B)C = [0 โ 14 + 24โ10 + 0 + 3016 + 12 + 0
] = [102028
]
AC+BC = [102028
]
โด (A+B)C = AC +BC
(4) Let A = [2 5
โ3 1] B =[
4 โ53 ๐
] What is the value of โkโ if any
make AB = BA
Ans: AB = [23 โ10 + 5๐โ9 15 + ๐
]
BA = [23 โ10 + 5๐
6 โ 3๐ 15 + ๐]
AB = BA
-10+5k =15
5k = 15+10 =25
โด k = ๐๐
๐ = ๐
(5) Two shops have the stock of large, medium and small size of a
tooth paste. The number of each size stocked is given by the matrix
A, where
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A = [๐๐๐๐๐ ๐๐๐๐๐ข๐ ๐ ๐๐๐๐150 240 120 90 300 210
] ๐ โ๐๐ ๐๐.1
๐ โ๐๐ ๐๐.2
are cost matrix 1 of the different size of the tooth paste is given by
cost (Rs.)
B = [14 ๐๐๐๐๐10 ๐๐๐๐๐ข๐
6 ๐ ๐๐๐๐
]
Find the investment in the toothpaste by each shop.
Ans : Investment = AB
AB = [150 240 12090 300 210
] ร [14106
]
= [2100 + 2400 + 720
1260 + 3000 +1260]
= [52205220
]
Investment in toothpaste by
Shop 1 = 5220
Shop 2 = 5520
(6) In a large legislative Assembly electron, a political group hired
a public relations firm to promote its candidate in three ways;
telephonic, housecalls, and letters. The cost per contract (in paise)
is given in matrix A as.
Cost per contract
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A = [40 ๐ก๐๐๐๐โ๐๐๐100 โ๐๐ข๐ ๐ ๐๐๐๐
50 ๐๐๐ก๐ก๐๐
]
The number of contract of each type made in two cities X and Y is
given by
B = [๐ก๐๐๐๐โ๐๐๐ โ๐๐ข๐ ๐ ๐๐๐๐๐ ๐๐๐ก๐ก๐๐
1000 500 5000 3000 1000 10000
] ๐ฅ๐ฆ
Find the total amount spent by the group in the two cities x and y ?
Amount spent = BA
BA = [1000 500 50003000 1000 10000
] ร [40
10050
]
= [40000 + 50000 + 250000
120000 + 100000 +500000]
= [340000720000
]
Amount spent by
City X = 3,40,000 paise i.e. 3400/โ
City Y = 7,20,000 paise i.e. 7200/โ
Determinants
A determinant is a compact form showing a set of numbers
arranged in rows and columns, the number of rows and the number
of columns being equal. The number in a determinant are known as
the elements of the determinant.
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Matrics which are not square do not have determinants.
Determinant of Square matrix of order 1
The determinants of 1 x 1 matrix A [a] is denoted by |A| or det. A
(i.e. determinant of A) and its value is a.
Determinant of Square matrix of order 2
Let A = [๐ ๐๐ ๐
] be a matrix of order 2 x 2
Then the determinant A is defined as
|A|= [๐ ๐๐ ๐
] = ad โ bc
Determinant with 3 rows and columns
Let A = [ ๐ ๐ ๐๐ ๐ ๐๐ โ ๐
] be a matrix of order 3 x 3.
Then the determinant A is defined as
|A| =๐ |๐ ๐โ ๐
|- b|๐ ๐๐ ๐
| ๐ |๐ ๐๐ โ
|
ie. a(ei - hf) - b(di - gf) + c (dh - ge)
Practical Problems
1) Evaluate the determinant
|2 โ34 9
|
Ans: |2 โ34 9
| = 2ร 9 โ 4 ร โ3
= 18 + 12 = 30
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2) Find the value of the determinant
| 1 2 โ32 โ1 23 2 4
|
Ans: | 1 2 โ32 โ1 23 2 4
| = 1 |โ1 22 4
| -2 |2 23 4
| โ 3|2 โ13 2
|
= 1(-4 -4) โ 2 (8 - 6) โ 3 (4- -3)
= 1 (โ8) - 2 (2) โ 3 (7)
= โ8 โ 4 โ 21 = โ33
Singular and Non singular matrices โ A square matrix โAโ is said
to be singular if its determinant value is zero. If |A| โ 0, then A is
called nonโsingular.
Minor elements of a matrix:
Minor element is the determinant obtained by deleting its rows and
the column in which element lies.
Example โ (1) Find the Minor of element 6 in the
determinant A =| 1 2 โ32 โ1 23 2 4
|
Ans : Minor of 6 = |1 27 8
|
= 1 x 8 โ 2 x 7
= 8 โ 14 = โ 6
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2) If A = | 3 1 22 1 04 2 0
| Find the minor of 3
Ans: Minor of 3 =|1 02 2
|
= 1 x 2 โ 0 x 2 = 2 โ 0 = 2 Cofactor of an element
Coโfactor of an element is obtained by multiplying the minor of
that element with (โ1) ( i + j) .
Where i = the row in which the element belongs, s = the column in
which the element belongs.
Coโfactpr of an element = Minor of an element X (โ1)i+j
Example 1. Find the Coโfactors of all the element of the
determinant|1 โ24 3
|
Ans : Minor element
1 = 3, โ2 = 4
4 = โ2, 3= 1
Coโfactors 1 = 3 x โ11+1 = 3 x โ12 = 3
โ2 = 4 x โ1 1 + 2 = 4 x โ13 = โ4
4 = โ2 x โ12+1 = โ2 x โ13 = 2
3 = 1 x โ1 2+2 = 1x โ14 = 1
2) Find the coโfactors of the elements of the determinant
| 2 โ3 56 0 41 5 โ7
| and verify that a11 A31 + a12 A32 + a13 A33 = 0
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Ans : Minor of an element :
2 = |0 45 โ7
| = (0 ร โ7) โ (4 ร 5)= 0 โ 20 = โ20
-3 = |6 41 โ7
| = (6 ร โ7) โ (4 ร 1)= -42 โ 4 = โ46
5 = |6 01 5
| = (6 ร 5) โ (0 ร 15)= 30 โ 0 = 30
6 = |โ3 55 โ7
| = (-3 ร โ7) โ (5 ร 5)= 21 โ 25 = -4
0 = |2 51 โ7
| = (2 ร โ7) โ (5 ร 1)= -14 โ 5 = โ19
4 = |2 โ31 5
| = (2 ร 5) โ (5 ร 1)= 10 โ -3 = 13
1 = |โ3 50 4
| = (-3 ร 4) โ (5 ร 0)= -12 โ 0 = โ12
5 = |2 56 4
| = (2 ร 4) โ (5 ร 6)= 8 โ 30 = โ22
-7 = |2 โ36 0
| = (2 ร 0) โ (-3 ร 6)= 0 โ -18 = 18
Coโfactors:
2 = โ20 x โ11+1 = โ20 x โ12 = โ20
โ3= โ46 x โ1 1+2 = โ46 x โ13 = 46
5 = 30 x โ11+3 = 30 x โ14 = 30
6 = โ4 x โ12+1 = โ4 x โ13 = 4
0 = โ19 x โ1 2+2 = โ19 x โ14 = โ19
4 = 13 x โ12+3 = 13 x โ15 = โ13
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1 = โ12 x โ13+1 = โ12 x โ14 = โ12
5 = โ22 x โ13+2 = โ22 x โ15 = 22
โ7 = 18 x โ13+3 = 18 x โ16 = 18
a11 = 2, a12= โ3, a13= 5
A31 = โ12, A32 = 22, A33 = 18
a11 A31 + a12 A32 + a13 A33 = 0
i.e., = 2 x โ12 + โ3 x 22 + 5 x 18
= โ24 + โ66 + 90
= โ90 +90 = 0
Adjoint Matrix
Adjoint of a given matrix is the transpose of the matrix formed by
coโfactors of the elements. It is denoted by Adj A.
Let A= [
๐11 ๐12 ๐13
๐21 ๐22 ๐23
๐31 ๐32 ๐33
]
then Adj A= Transpose [
๐ด11 ๐ด12 ๐ด13
๐ด21 ๐ด22 ๐ด23
๐ด31 ๐ด32 ๐ด33
]
= [
๐ด11 ๐ด21 ๐ด31
๐ด12 ๐ด22 ๐ด32
๐ด13 ๐ด23 ๐ด33
]
Practical Problems
1) Find adj A for A = |2 31 4
|
Ans: Minor element:
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2 = 4, 3 = 1, 1 = 3, 4 = 2
Coโfactors:
2 = 4 x โ11+1 = 4, 3 = 1 x โ11+2 = โ1
1 = 3 x โ12+1 = โ3, 4 = 2 x โ12+2 = 2
adj A = Transpose = |4 โ1
โ3 2|
= |4 โ3
โ1 2|
2) Find adj A for A = [ 2 1 33 1 21 2 3
]
Ans: Minor element:
2 = โ1, 1 = 7, 3 = 5
3 = โ3, 1 = 3, 2 = 3
1 = โ1,
Coโfactor elements
2 = โ5, 3 = โ1
2 = โ1 x โ11+1 = โ1, 1 = 7 x โ11+2 = โ7
3 = 5 x โ11+3 = 5
3 = โ3 x โ12+1 = 3, 1 = 3 x โ12+2 =3
2 = 3 x โ12+3 = โ3
1 = โ1 x โ13+1 = โ1, 2 = โ5 x โ13+2 = 5
3 = โ1 x โ13+3 = โ1
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adj A = Transpose = [โ1 โ7 53 3 โ3
โ1 5 โ1]
= [โ1 3 โ1โ7 3 55 โ3 โ1
]
Invertible Matrix and Inverse of a Matrix
Let A be a square matrix of order n, if there exist a square matrix B of
order n, such that AB = BA = I
Then A is said to be convertible and B is called on inverse of A and A is
called inverse of B
Where I = Identity Matrix
Inverse of A is denoted by Aโ1
Aโ1 = 1
|A| adj A or
Aโ1 = adj A
|A|
1) Find the inverse matrix A = [2 โ11 3
]
Ans: |A| = (2 x 3 โ 1 x โ1) = 6 โ โ1 = 7
Minor element:
2 = 3, โ1 = 1, 1 = โ1, 3 = 2
Coโfactors element
2 = 3 x โ11+1 = 3, โ1 = 1 x โ1 1+2 = โ1
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1 = โ1 x โ12+1 = 1, 3= 2 x โ12+2 = 2
adj A = Transpose [3 โ11 2
]
adj A = [3 1
โ1 2]
Aโ1 = 1
|A| adj A
Aโ1 = 1
7 [3 โ11 2
]
= [๐
๐๐
๐
โ๐
๐๐
๐
]
2. Compute the inverse of [1 2 52 3 1
โ1 1 1]
Ans. |A| = 1(3โ1) โ2(2โ โ1) + 5 (2โโ3)
= 1(2) โ 2(3) + 5(5)
= 2 โ 6 + 25 = 21
Minor element:
1 = 2, 2 = 3, 5 = 5
2 = โ3, 3 = 6, 1 = 3
โ1= โ13, 1 = โ9, 1 = โ1
Coโfactors element
1 = 2 x โ1 1+1 = 2, 2 = 3 x โ1
1+2 = โ3, 5 = 5 x โ1 1+3 = 5
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2 = โ3 x โ11+2 = 3, 3 = 6 x โ1
2+2 = 6, 1 = 3 x โ12+3 = โ3
โ1= โ13 x โ13+1 = โ13, 1 = โ9 x โ1
3+2 = 9, 1 = โ1 x โ13+3 = โ1
Adj A = Transpose [2 โ3 53 6 โ3
โ13 9 โ1]
Aโ1 = 1
|A| adj A
= 1
21[
2 3 โ13โ3 6 95 โ3 โ1
]
= [
2
21โ3
21
3
216
21
โ13
219
21
5
21
โ3
21
โ1
21
]
Solving simultaneous equations with the help of Matrices
Firstly, express the equation in the form of AX = B
Then possibilities When |A| โ 0
Then X = Aโ1B i.e., the system has a unique solution.
therefore the system is consistant.
Aโ1 = 1
|A| adj A
When |A| = 0
Then we calculate (adj A)B
If (adj A)B = 0, then the system will have infinite solution were
the system is consistent.
[[
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If (adj A)B โ 0, then the system will have no solution.
Problem
1) Solve the linear equation by using matrix
5x + 2y = 4
7x + 3y = 5
Ans: AX = B
Let A= [5 27 3
], X = [๐ฅ๐ฆ]
B = [45
]
|A| = (15 โ 14) = 1
i.e., 1 โ 0
Then X = Aโ1B
Aโ1 = 1
|A| adj A
๐๐๐ฃ ๐ :
Minor element 5 = 3, 2 = 7, 7 = 2, 3 = 5
Coโfactors element 5 = 3, 2 = โ7, 7 = โ2, 3 = 5
adj A = Transpose [3 โ7
โ2 5] = [
3 โ2โ7 5
]
Aโ1 = 1
|A| adj A =
1
1[3 โ2
โ7 5]
= [3 โ2
โ7 5]
X = Aโ1 B = [3 โ2
โ7 5] ร [
45
]
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X = [12 โ10
โ28 25] = [
2โ3
]
X= [2
โ3] = [
๐ฅ๐ฆ]
x= 2 y = -3
2) Solve the equation by using matrix
x โ y + z = 4
2x + y โ 3z = 0
x + y + z = 2
Ans: AX = B
Let A = [1 โ1 12 1 โ31 1 1
] , X = [๐ฅ๐ฆ๐ง
]
B = [402
]
|A| = 1(1+3) โ (โ1) (2+3) +1(2โ1)
= 1(4) +1(5) +1(1)
= 4 + 5 + 1 = 10 ie โ 0
Then X = Aโ1B
Aโ1 = 1
|A| adj A
Factor elements:
1 = 4, โ1 = โ5, 1 = 1
2 = 2, 1 = 0 โ3 = โ2
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1 = 2, 1 = 5, 1 =3
Adj A = Transpose [4 โ5 12 0 โ22 5 3
] = [4 2 2
โ5 0 51 โ2 3
]
X = Aโ1B
Aโ1 = 1
|A| adj A
= 1
10[
4 2 2โ5 0 51 โ2 3
]
X = Aโ1B = 1
10[
4 2 2โ5 0 51 โ2 3
] ร [402
]
= 1
10[
16 + 0 + 4โ20 + 0 + 104 + 0 + 6
]
= 1
10[
20โ1010
] =[2
โ11
]
x = [2
โ11
]
ie., x = 2, y= -1, z = 1
3) Solve the following equation by using matrix
5x โ 6y + 4 z = 15
7x + 4y โ 32 = 19
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2x + y + 6 z = 46
Ans: AX = B
Let A = [5 โ6 47 4 โ32 1 6
] , X = [๐ฅ๐ฆ๐ง
]
B = [151946
]
|A| = 5(24 โ โ3) โ โ6 (42 โ โ6) + 4 (7 โ 8)
= 5(27) + 6(48) + 4(โ1)
= 135 +288 โ 4 = 419
Then X = Aโ1B
Then X = Aโ1B
Aโ1 = 1
|A| adj A
Coโfactor elements:
5 = 27, โ6 = โ48, 4 = โ1
7 = 40, 4 = 22, โ 3 = โ17
2 = 2, 1 = 43, 6 = 62
Adj A = Transpose [27 โ48 โ140 22 โ172 43 62
] = [27 40 2
โ48 22 43โ1 โ17 62
]
X = Aโ1B
Aโ1 = 1
|A| adj A
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=1
419 [
27 40 2โ48 22 43โ1 โ17 62
] ร [151946
]
=1
419 [
125716762514
]
= [364
]
X = [โ364
]
โด x = 3, y = 4, z = 6
Cramerโs rule
Cramerโs rule is a method for solving linear simultaneous
equations. It makes use of determinants and so knowledge of these
is necessary before proceeding.
Cramerโs Rule - two equations
If we are given a pair of simultaneous equations
a1x + b1y = d1
a2x + b2y = d2
then x, and y can be found from
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x = |
๐1๐2
๐1๐2
๐1๐2
๐1๐2
| y = |
๐1๐2
๐1๐2
๐1๐2
๐1๐2
|
Example
Solve the equations
3x + 4y = โ14
โ2x โ 3y = 11
Ans:
Using Cramerโs rule we can write the solution as the ratio of two
determinants.
x = |โ14
114
โ33
โ2
4โ3
| = โ2
โ1 = 2
y = |3 โ2
โ14 11
3 โ2
4
โ3
| = 5
โ1 = โ5
The solution of the simultaneous equations is then x = 2, y = โ5.
Cramerโs Rule for solving 3ร ๐ ๐ฌ๐ฒ๐ฌ๐ญ๐๐ฆ๐ฌ
Consider the system
a1x + b1y + c1z = d1
a2x + b2y + c1z = d2
a3x + b3y + c3z = d3
Let the four determinants D, Dx, Dy and Dz be defined as
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D= |
๐1
๐2
๐3
๐1 ๐2
๐3
๐1
๐2
๐3
| Dx= |
๐1
๐2
๐3
๐1 ๐2
๐3
๐1
๐2
๐3
|
Dy= |
๐1
๐2
๐3
๐1
๐2
๐3
๐1
๐2
๐3
| Dz = |
๐1
๐2
๐3
๐1 ๐2
๐3
๐1
๐2
๐3
|
Then, if D โ 0, the system has a unique solution of
x = Dx
D x =
Dy
D x =
Dz
D
Exercise
Use Cramerโs rule to solve the following sets of simultaneous
equations.
Solve the system by using Cramerโs Rule
x โ y + 2z = -4
3x + y -4z = -6
2x + 3y โ 4z =4
Solution:
The first thing we need to do here is to evaluate all of the
corresponding determinants. We start with the coefficient
determinant.
D= |
๐1
๐2
๐3
๐1 ๐2
๐3
๐1
๐2
๐3
| = |13
2
โ1 13
2โ4โ4
| = 1|1 โ43 โ4
| - (-1) |3 โ42 โ4
|
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+2|3 12 3
|
= (-4 + 12) + (-12+8) + 2(9-2)
= 8 โ 4 +14 = 18
Now we compute the other determinants
Dx= |
๐1
๐2
๐3
๐1 ๐2
๐3
๐1
๐2
๐3
|
= |โ4โ64
โ1 13
2โ4โ4
| = -4|1 โ43 โ4
| - (-1) |โ6 โ44 โ4
| +2|โ6 14 3
|
= -4(-4 + 12) + (24+16) + 2(-18-4)
= 32 +40-44 = -36
Dy= |
๐1
๐2
๐3
๐1
๐2
๐3
๐1
๐2
๐3
|
= |132
โ4 โ64
2โ4โ4
| = 1 |โ6 โ44 โ4
| - (-4)|3 โ42 โ4
| +2|3 โ62 4
|
= (24 + 16) + 4(-12+8) + 2(12+12)
= 40 โ 16+ 48 = 72
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Dz = |
๐1
๐2
๐3
๐1 ๐2
๐3
๐1
๐2
๐3
|
= |132
โ1 13
โ4โ6โ4
| = 1 |1 โ63 4
| - (-1)|3 โ62 4
| +(-4)|3 12 3
|
= (4 + 18) + (12+12) - 4(9-2)
= 22 +24 - 28 = 18
Now that we have that, we simply need to determine the solutions
by the formulas
We get
x = Dx
D x =
Dy
D x =
Dz
D
x = โ36
18 x =
72
18 x =
18
18
So the solution to the system is (-2, 4, 1).
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MODULE III
SEQUENCE, SERIES AND PROGRESSION
ARITHMETIC PROGRESSION
A series is said to be in Arithmetic Progression, it its terms
continuously increase or decrease by a constant number . It is a
series, in which each term is obtained by adding or deducting a
constant number to the preceding term. The constant number is
called common difference of the progression and is denoted by โdโ.
It is the difference between the two term of the series i.e., the
difference between second term and first term or third term and
second term and so on.
The first term of an A.P. is usually denoted by โaโ. One general
form of an A.P is a, a+ d, a+2d, a+ 3d, โฆโฆโฆ
For example
1) The sequence 1, 3, 5, 7,is an A.P whose first term is 1 and d = 2
2) The sequence โ5, โ2, 1, 4, 7,โฆ.., whose โaโ = โ5, d = 3
General term of an AP or nth term
Ler โaโ be the first term and โdโ be the common difference of an
A.P, then an denotes the nth term of the A.P.
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an = a + (nยญ1)d
n = number of term in a series.
Practical Problems
1) Find the 12th term of an A.P 6, 2, โ2
Ans: an = a + (nยญ1)d
a = 6, n = 12, d = โ4
= 6 + (12โ1) โ 4
= 6 + (11) โ 4
= 6 + โ 44 = โ38
12th term is โ38
2) Find the 8th term of the series 6, 5ยฝ, 5, 4ยฝ, . . . . .
Ans: a = 6, d = โยฝ,n = 8
an = a + (nยญ1)d
= 6 + (8โ1)โยฝ
= 6 + (7) โยฝ
= 6 + โ3.5 = 2.5
3) Which term of the A.P 21, 18, 15, โ81 ?
Ans: a = 21 , d = โ3, an = โ81 n = ?
an = a + (nยญ1)d
โ81 = 21 + (nโ1)โ3
โ81 = 21 + โ3n +3
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โ81 = 24 โ 3n
โ81 โ 24 = โ3n 3n = 105
n = 105/3 = 35
Therefore the 35th term of the given A.P = โ81
4) Which term of the A.P 21,18,15, 0 ?
Ans: a = 21, d = โ3, an = 0, n = ?
an = a + (nยญ1)d
0 = 21 + (n โ 1)โ3
0 = 21 + โ3n + 3
0 = 24 โ 3n
3n = 24, n = 8
Therefore, the 8th term = 0
5) If the 9th term of an A.P is 99 and 99th term is 9. Fine 108th
term?
Ans: an = a + (nยญ1)d
n = 9, an = 99
= a + (9โ1)d = 99
= a + 8d = 99 โโโโโโโโโโโโโโโโ(1)
n = 99, an = 9
= a + (99 โ 1)d =9
= a + 98d = 9 โโโโโโโโโโโโโโโ(2)
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Solve the equations
a + 8d = 99 โโโโโโโโโโโโโโโโ(1)
a + 98d = 9 โโโโโโโโโโโโโโโโ(2)
Then (1) โ (2) โ 90d = 90
d = 90/โ90 = โ1
Substitute the value of โdโ
a + 8d = 99
a + 8 x โ1 = 99 a + โ8 = 99
a = 99 + 8 = 107
108th term = a + (nโ1)d
= 107 + (108 โ 1)โ1
= 107 + (107)โ1
= 107 โ 107 = 0
108th term = 0
6) Determine the A.P whose 3rd term is 5 and the 6th term is 8
Ans: a + 2d = 5 โโโโโโโโโโโโโโโโโโโโโ (1)
a + 5d = 8 โโโโโโโโโโโโโโโโโโโโโ(2)
Then (1) โ (2) = โ3d = โ3
๐ =3
3 = 1
A.P = 3,4,5,6,7,8โฆโฆ.
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7) Find many two digit numbers are divisible by 3 ?
Ans: Numbers = 12, 15, 18, โ โ โ โ โ โ 99
a = 12, d = 3, an = 99
an = a + (nยญ1)d
99 = 12 + (nโ1)3
99 = 12 + 3n โ 3
99 = 12 โ 3 + 3n
99 = 9 + 3n
3n = 99 โ 9,
3n = 90
n = 90
3 = 30
โด Two digit numbers are divisible by 3 = 30 number
8) Determine the 25th term of the A.P, whose 9th term is โ6 and the
common difference is 5/4.
Ans: d = 5/4, a9 = โ6
a9 = a + (nยญ1)d
โ6 = a + 8 x 5
4
โ6 = a + 10
a = โ10 โ 6 = โ16
a25 = a + (n โ 1)d
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= โ16 + (25 โ 1) 5
4
= โ16 + 24 x5
4
= โ16 + 30 = 14
25th term = 14
Sum of n terms of an A.P
Let Sn denotes the sum of โnโ terms of an A.P, whose first term is
โaโ and common difference is โdโ.
Sn = n/2 [2a + (nโ 1)d]
2a = a + a or 2 x a
Practical Problems
1) Find the sum of the first 20 terms of 1 + 4 + 7 + 10 . . . . . . .
Ans: Sn = n/2 [2a + (nโ 1)d]
n = 20, a = 1, d = 3
Sn = 20
2 [2 x 1 + (20 โ 1 )3]
= 10 (2+ 19 x 3)
= 10(2 + 57), 10 x 59 =590
Sum of the first 20 terms = 590
2) Find the sum of the series 5, 3, 1, โ1,โฆโฆโฆ. โ23
Ans: a= 5, d = โ2, n = ?, an =โ23
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Sn = n/2 [2a + (nโ 1)d]
We know, an = a + (n โ 1)d
โ23 = 5 + (n โ 1)โ2
โ23 = 5 + โ2n + 2
โ23 = 5 + 2 โ 2n
โ23 = 7 โ 2n
2n = โ23 โ 7
2n =30
n = 30
2= 15
Sn = 15
2 [2 x 5 + (15 โ 1 )-2]
= 15
2 [10 + - 28]
= 15
2ร -18
= 15 x โ9 = โ135
Sum of the series = โ135
3) How many terms of the sequence 54, 51, 48, โฆโฆโฆโฆ be taken
so that their sum is 513. Explain the double answer.
Ans: Sn = 513, a = 54, d = โ3
Sn = n/2 [2a + (nโ 1)d]
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513 = ๐
2 (2 x 54 + (n โ 1)โ3)
513 = ๐
2 (108 โ 3n + 3)
513 = ๐
2 (111 โ 3n)
1026 = n(111 โ 3n)
1026 = 111n โ 3n2
3n2 โ 111n = โ1026
3n2 โ 111n + 1026 = 0
n2 โ 37n + 342 = 0
Solve by using quadratic formula
= โ๐ยฑโ๐2โ4๐๐
2๐
a = 1, b = โ37, c = 342
= 37ยฑโ372โ4ร1ร342
2ร1
=
37ยฑโ1369โ1368
2
= 37ยฑโ1
2 =
37ยฑ1
2
=37+1
2 or
37โ1
2
= 19 or 18
N = 18 or 19
4) Find the sum of all natural numbers between 500 and 1000 which
are divisible by 13.
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Ans: Number between 500 and 1000 which are divisible by 13
507, 520, 533, 988
a = 507, d = 13, an = 988
an = a + (nยญ1)d
988 = 507 + (nโ1)13
988 = 507 + 13n โ 13
988 = 507 โ 13 + 13n
988 = 494 + 13n
13n = 988 โ 494 = 494
13n = 494
n = 494
13 =38
Sn = n/2 [2a + (nโ 1)d]
= 19(1014 +37 x 13)
= 19(1014 + 481)
= 19 1495 = 28405
5) Find the sum of all natural numbers from 1 to 200 excluding those
divisible by 5
Ans: Natural number from 1 to 200 = 1, 2, 3, 4, โฆโฆ. 200
Divisible by 5 = 5, 10, 15, 20 200
โด Natural numbers from 1 to 200, excluding divisible by 5
= (1, 2, 3, 4 . . . . 200) โ (5, 10, 15โฆโฆ200)
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Sum of (1, 2, 3, 4, 200)
Sn = n/2 [2a + (nโ 1)d]
= 200
2
[2x1 + (200 โ 1)1]
= 100 (2+199)
=100 x 201 = 20,100
Sum of (5, 10, 15,20,โฆโฆโฆ200)
= 40
2 (2 x5 +(40โ1)5)
=20 (10 + 39 x 5)
=20(10 + 195)
= 20 x 205 = 4100
Sum by natural numbers from 1 to 200 excluding divisible by 5
= 20100 โ 4100 = 16000
6) The sum of the first 3 terms of an A.P is 30 and the sum of first
7 terms is 140. Find the sum of the first 10 terms.
Ans: S3 = 30, s7 = 30,
Sn = n/2 [2a + (nโ 1)d]
= ๐
2
[2a + (3 โ 1)d] = 30
= 2a +2d = 30 x 3
2
= 2a + 2d = 20
= a + d = 10 โโโโโโโโโโโโโโ(1)
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= 7
2
[2a + 6d] = 140
= 2a + 6d = 140ร 7
2
= 2a + 6d = 40
= a + 3d = 20 โโโโโโโโโโโโโโโ(2)
Solving the equation (1) and (2) d = 5
Then a = 5
S10 = 10
2
[2ร 5 + 9ร 5] = 275
7) Find three numbers in A. P whose sum is 9 and the product is โ165.
Ans: Let the numbers be aโd, a, a+d
(aโd) + a + (a +d) = 9
3a = 9, a = 3
(aโd) x a x (a+d) = โ165
= (3 โ d) x 3 x (3 +d) = โ165
= 9 โ d2 = โ165
3
= 9 โ d2 = โ55
= โd2 = โ55 โ 9 = โ64
= d2 = 64, d = 8
a = 3, d = 8
Numbers = (a โ d), a, (a+d)
= โ5, 3, 11
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8) Find four numbers of A.P whose sum is 20 and the sum of whose
square is 120
Ans: Let numbers be (aโ3d), (aโd) (a+d) (a+3d)
Given (aโ3d)+(aโd)+(a+d)+(a+3d) = 20
4a = 20, a = 20
4 = 5
(aโ3d)2 x (aโd)2 x (a+d)2 x (a+3d)2 = 120
= (5โ3d)2 x (5โd)2 x (5+d)2 x (5+3d)2 = 120
We know (aโb)2 = a2 โ 2ab + b2
= 25 โ30d+9d2+25 โ10d+d2+25+10d + d2+25+30d+9d2 = 120
= 100 + 20d2 = 120
20d2 = 120 = 100
20d2 = 20, d2 = 20/20 = 1, d = 1 a = 5, d = 1
Numbers are = (a โ 3d), (aโd), (a + d), (a + 3d)
= (5โ3). (5โ1), (5+1), (5+3)
` = 2, 4, 6, 8
9) A manufacturing of radio sets produced 600 units in the third
year and 700 units in the seventh year. Assuming that the
production uniformly increases by a fixed number every year.
Find
1) One production in the first year
2) The production in the 10th year.
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3) The total production in 7 year.
Ans: Since the production increases uniformly by a fixed number
in every year, it form an A.P.
Let a3 = 600, a7 = 700
an = a + (n โ 1)d
600 = a + (3โ1)d
600 = a + 2dโฆโฆโฆ..(1)
700 = a + 6dโฆโฆโฆ..(2)
a + 2d = 600โฆโฆโฆโฆ. (1)
a + 6d = 700โฆโฆโฆโฆ.(2)
โ4d = โ100
d = 100
4 = 25
1) Production in the first year
a + 2d = 600
a + 50 = 600
a = 550
2) Production in the 10th year
i.e., an = a + (nโ1)d
= 550 + (10 โ 1) 25
= 550 + 9 x 25
= 550 + 225 = 775
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3) Total production in 7th year
Sn = n/2 [2a + (nโ 1)d]
= 7
2 [2 x 550 + (7 โ 1)25]
= 7
2 (1100 + 6 x 25)
= 7
2 (1100 + 150)
= 7
2 (1250)
= 7 x 625 = 4375 units
Arithmetic Mean (A.M)
Given two numbers a and b, we can insert a number A between them,
so that a, A, b is an A.P. Such a number A is called the Arithmetic
Mean of the number a and b.
We can insert as many numbers as we like between them. Let A,
A2, A3 An be โnโ numbers between a and b, Then
A1 = a + d
A2 = a + 2d
A3 = a +3 d
An = a + nd
Example
1) Find A.M between 2 and 6
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Ans: A.M between 2 and 6 = 2+6
2 = 4
Then A.P. = 2, 4, 6
2) Insert 4 Arithmetic means between 5 and 20
a = 5, n = 6, an =20, d = ?
an = a + (n โ 1)d
20 = 5 + (6โ1)d
20 = 5 + 5d
20 = 5 + 5d
5d = 20 โ 5 = 15
d = 15/5 = 3
A1 = a + d i.e., 5 +3 = 8
A2 = a + 2d i.e., 5 +6 = 11
A3 = a +3d i.e., 5 +9 = 14
A4 = a + 4d i.e., 5 +12 = 17
Arithmetic means are 8, 11, 14, 17
A.P. = 5, 8, 11, 14, 17, 20
3) Insert six numbers between 3 and 24 such that the resulting
sequence is an A.P.
Ans: a = 3, n = 8, an = 24, d = ?
an = a + (n โ 1)d
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24 = 3 + 7d
7d = 21, d = 3
A1 = 3 +3 = 6
A2 = 3 + 6 = 9
A3 = 3 + 9 = 12
A4 = 3 + 12 = 15
A5 = 3 + 15 = 18
A6 = 3 + 18 = 21
A.M. = 6, 9, 12, 15, 18, 21
A.P. = 3, 6, 9, 12, 15, 18, 21, 24
Geometric Progression
A series is said to be in G.P if every term of it is obtained by
multiplying the previous term by a constant number. This constant
number is called common ratio, denoted by โrโ. r = ๐ ๐๐๐๐๐ ๐ก๐๐๐
๐๐๐๐ ๐ก ๐ก๐๐๐ or
third term by second term etc.
The first term of a G.P is usually denoted by a. The general form of
a G.P is usually denoted by a. The general form of a G.P is a, ar,
ar2, ar3 โฆ.. If the number of terms of a G.P is finite, it is called a
finite G.P, otherwise it is called an infinite G.P. For example.
(i) 1, ยฝ, ยผ , 1 8โ โฆโฆโฆ is a G.P, whose first term is 1 and r = ยฝ
(ii) 3, โ6, 12, โ24 โฆโฆโฆโฆ is a G.P whose a = 3 , r = โ2
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General term of a G.P or nth term of a G.P
Let โaโ be the first term and โrโ be the common ratio of a G.P, then
an = arn-1
1) Find 10th term of series 9, 6, 4โฆโฆโฆ
Ans: a=9 r = 6
9 =
2
3 ๐ = 10
an = arn-1
= 9ร(2
3)10-1
= 9ร(2
3)9
= 9(๐
๐)9
2) Find the 12th term of 2, 6, 18, 54 โฆโฆโฆ
a = 2, r = 6/2 =3, n = 12
an = arnยญ1 = 2 ร 312โ1
= 2 ร 311 = 2 x 177147 = 3, 54,294
3) Which term of the G.P 2, 8, 32 up to n terms is 131072 ?
a = 2, r = 4, an = 1,31,072
an = arnยญ1
1,31,072 = 2 ร 4nโ1
4nโ1 =
131072
2
4nโ1 = 65536
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i.e., 48 = 65536
i.e. nโ1 = 8
n = 8 + 1 = 9
Hence 1,31,072 is the 9th term of the G.P.
4) In a G.P the third term is 24 and 6th term is 192. Find the 10th term
Ans: a3 = 24, a6 = 192
an = arnยญ1
a3 = ar2 = 24
a6 = ar5 = 192
i.e., ar2 = 24 โโโโโโโโโ (1)
ar2 = 192 โโโโโโโ(2)
Divide (2) by (1),
= ar5
ar2 = 192
24
r3 = 8 ie 23
r = 2
Substituting r = 2 in (1)
ar2 = 24,
a x 22 = 24
a x 4 = 24,
a = 24/4 = 6
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a10 = arnยญ1 = 6(2)9 = 3072
Sum of โnโ terms of a G.P
Let โaโ be the first term and โrโ be the common ratio and Sn the sum
of the โnโ terms of G.P.
Then Sn = ๐(1โ๐๐)
(1โ๐) =
๐(๐๐โ1)
(rโ1)
When r is less than 1, we can apply first formula.
1) Find the sum of the series.
1024 + 512 + 256 โฆโฆโฆto 15 terms
Asn: a = 1024, n = 15, r = ยฝ
Sn = ๐(1โ๐๐)
(1โ๐)
=
1024(1โ1
2
15)
(1โ1
2)
=
1024(1
2
15)
(1โ1
2)
=
1024(1
2
15)
(1โ1
2)
= 1024 ร2
1ร (
1
2
15)
= ๐๐๐๐ ร (๐
๐
๐๐)
2) Find the sum of 1 + 3+ 9 + 27 to 10 terms.
a = 1, r = 3, n = 10
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Sn = ๐(๐๐โ1)
(rโ1)
= 1(310โ1)
(3โ1)
= 59049โ1
2 = 29524
3) How many terms of the G.P 3, 3/2, 3/4 , โฆโฆโฆโฆโฆ.. are needed to
give the sum
a = 3, r = ยฝ, Sn= 3069
512
Sn = ๐(1โ๐๐)
(1โ๐)
= 3069
512 =
3(1
2
n)
(1โ1
2)
= 3069
512 =
3(1
2
n)
1
2
= 3069
512 = 3 ร
2
1(1 โ
1
2
n)
= 3069
512 = 3 ร
2
1(1 โ
1
2
n)
= 3069
512 = 6(1 โ
1
2
n)
= 3069
512ร6 = (1 โ
1
2
n)
= 3069
3072 = 1 โ
1
2๐)
1
2๐= 1 - 3069
3072 =
3
3072 =
1
1024
2n = 1024
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210 = 1024, n = 10
Geometric Mean
One geometric mean of two positive numbers a and b is the number
โab. Therefore, the geometric mean of 2 and 8 is 4. We can insert as
many numbers as we like between a and b to make the sequence in a
G.P. Let G1, G2, G3, Gn be โnโ number between a and b, then
G1 = ar, G2 = ar2, G3= ar3, Gn = arn
1) Insert three G.M. between 1 and 256
Ans. a = 1, an = 256, n = 5, r = ?
an = arnโ1
256 = 1 rnโ1
256 = rnโ1
256 = r5โ1
256 = r4
256 = 44, r = 4
G.M. are ar, ar2, ar3
1 x 4, 1 x 42, 1 x 43 = 4, 16, 64
G.P = 1, 4, 16, 64, 256
2) Find the G.M between 4 is 16
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Ans: G.M = โ4 ร 16= โ64 = 8
3) Insert 5 geometric means between 2 and 1458
Ans: a = 2, n = 7, an = 1458
an = arnโ1
1458 = 2 r7โ1
1458 = 2r6
2r6 = 1458
r6 = 1458/2 = 729
r6 = 36
G.M. = ar, ar2, ar3, ar4, ar5
= 2 x 3, 2 x 32, 2 x 33, 2 x 34, 2 x 35
= 6, 18, 54, 162, 486, 486
G.P. = 2, 6, 18, 54, 162, 486, 1458
4) Find the three numbers in G.P whose sum is 26 and product is 216.
Ans: Let the number is G.P be
a/r, a, ar
a/r, a/ ar = 216
i.e. a3 = 216, 63 = 216
โด a = 6
a/r + a + ar = 6/r + 6 + 6r = 26
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= 6/r + 6r = 26 โ 6
= 6/r + 6r = 20
Multiply by r
= 6 + 62 = 20r
= 62 โ 20r + 6
= 62 โ 20 r + 6 = 0
Solving by using quadratic formula
Then r = 1/3 or 3
Required numbers a/r, a, ar
r = 3
6/3, 6, 6 x 3 = 2, 6, 18
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MODULE IV
INTEREST AND TIME VALUE
Simple interest
It is the interest calculated on principal amount at the fixed rate .
Simple Interest = ๐๐๐
100
Where P = Principal amount, n = number of year,
r = rate of interest per annum
Amount at the end of nth year = P + ๐๐๐
100 or
= P(1 + ๐๐
100)
or principal amount + interest
1) What is the simple interest for Rs. 10, 000 at the rate of 15% per
annum for 2 years?
Ans: P = 10,000, n = 2 years, r = 15
Interest = ๐๐๐
100 =
10000ร2ร15
100
= Rs. 3, 000
2) Find the total interest and amount of the end of 5th year for as
10,000 at 10% per annum, simple interest.
Ans: P = 10,000, n = 5 years, r = 10%
Interest = ๐๐๐
100 =
10000ร2ร10
100
= Rs. 5, 000
Amount at the end
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5th year = P(1 + ๐๐
100)
= 10000(1 + 5ร10
100)
= 10000(1 + 50
100)
= 10000( 150
100)
= 10000 ร 1.5
= 15000
3) Find the simple interest and amount for Rs. 25,000 at 10% p. a
for 26 weeks. Ans: P = 25,000 n = 26/52, r = 10%
Ans: Interest = ๐๐๐
100
=25000ร
26
52ร10
100
=25000ร
1
2ร10
100
=25000ร5
100 = 1250
Amount at the end
= P(1 + ๐๐
100)
= 25000(1 +
26
52ร10
100)
= 25000(1 + 5
100)
= 25000( 105
100)
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= 25000ร 1.05 = 26250
4) Find the simple interest and amount for Rs. 50,000 at 7.5% p. a
for 4 months.
Ans: P = 50,000, n = 4/12, r = 7.5%
Interest = ๐๐๐
100
=50000ร
4
12ร7.5
100
=50000ร
1
3ร7.5
100
=50000ร2.5
100 = 1250
Amount
= P(1 + ๐๐
100)
= 50000(1 +
4
12ร7.5
100)
= 50000(1 + 2.5
100)
= 50000( 102.5
100)
= 50000ร 1.025 = 51250
5) Find the number of years in which a sum of money will double
itself at 25% p. a, simple interest.
Ans: P = p, Amount = 2P, r = 25, n = ?
Amount = P(1 + ๐๐
100)
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2P= P(1 + ๐๐
100)
i.e., 2= (1 + ๐๐
100)
= 2 -1 = ๐๐
100
=1 = ๐๐
100
nr = 100
r= 25 โด ๐ = 4
number of years = 4
6) Find the rate of interest at which an amount of Rs. 12000 will
become Rs. 15000 at the end of 10th year.
Ans: A = 15000, P = 12000, n = 1, r = ?
Total interest 15000 โ 12000 = 3000
Interest = ๐๐๐
100
3000 = 12000ร10ร๐
100
3000ร 100 =12000 ร 10 ร ๐
300000 = 120000r
r = 300000
120000 = 2.5
Rate of interest = 2.5%
7) A certain sum amounts to Rs. 678 in 2 years and to Rs. 736.50 in
3โ5 years find the rate of interest and principal amount.
Ans: Amount for 2 years = 678
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โ 3โ5 years = 736.50
Amount = P(1 + ๐๐
100)
678 = P(1 + 2๐
100) โฆโฆโฆ(1)
736.50 = P (1 + 3.5๐
100) โฆโฆโฆ(2)
Divide (1) by (2)
= 678
736.50 =
1+2๐
100
1+ 3.5๐
100
= 678
736.50 =
100+2r
100+ 3.5r
= 678 (100+3.5r) = 736.50 (100 + 2r)
= 67800 + 2373r = 73650 + 1473r
= 2373r โ 1473r = 73650 โ 67800
= 900r = 5850
= r = 5850/900 = 6.5
Substituting the value of r
P (1 + 2๐
100) = 678
P (1 + 2ร6.5
100) = 678
P (1 + 13
100) = 678
P ( 113
100) = 678
P(1.13) = 678
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P = 678/1.13 = 600
Rate of interest = 6.5%
Principal amount at the begining = 600
Compound Interest
Compound interest means interest calculated on principal amount
plus interest. Let โpโ be the principal โrโ be the rate of interest
(compound) p.a., โnโ be the number of years then
Amount = P(1 + ๐
100)๐
Total interest = A โ P
1) Find CI on Rs. 25200 for 2 years at 10% p.a compounded
annually?
Ans: P = 25200, r = 10, n = 2
A = P(1 + ๐
100)๐
= 25200(1 + 10
100)2
= 25200( 110
100)2
= 25200ร (1.10)2
= 25200ร 1.21
= 30492
CI = 30492 โ 25200 = 5292
2) Find the Compound Interest Rs.10,000/โ for 2ยฝ years at 10%
p.a..
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)
Ans: P = 10,000 n = 2ยฝ r = 10
Amount for 2 years = P(1 + ๐
100)๐
= 10000(1 + 10
100)2
= 10000( 110
100)2
= 10000ร (1.10)2
= 10000ร 1.21
= 12100
Interest for 2 years = 2100
Interest for 6 months = 12100 ร10
100ร
6
12
= 605
Total interest for 2ยฝ years = 2100 + 605
= 2,705
3) Mr. A borrowed Rs.20,000/โ from a person, but he could
not repay any amount in a period of 4 years. So the lender
demanded as 26500 which is the rate of interest charged.
Ans: Here interest charged on compound
P = 20,000 n = 4 A = 26500 r = ?
A = P(1 + ๐
100)๐
26500 = 20000 (1 + ๐
100)4
26500
20000 = (1 +
๐
100)4
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1.325 = (1 + ๐
100)4
Log 1.325 = (1 + ๐
100)4
Log 1.325 = 4 log (1 + ๐
100)
0.1222= 4 log (1 + ๐
100)
log (1 + ๐
100) =
0.1222
4
log (1 + ๐
100) = 0.03055
Antilog 0.03055 = 1.073
(1 + ๐
100) = 1.073
๐
100 = 1.073 โ 1
๐
100 = 0.073
r = 100 x 0.073 = 7.3%
4) The population of a country increases every year by 2.4%
of the population at the beginning of first year. In what time
will be population double itself? Answer to the nearest year?
Ans: p = p A = 2p r = 2.4 n = ?
A = P (1 + ๐
100)๐
2P = P (1 + 2.4
100)๐
2P = P (1 + 102.4
100)๐
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2P = p(1.024)n
P = (1.024)n
log 2 = n log 1.024
0.3010 = n x 0.0103
n = 0.3010
0.0103 = 29.22 = 30
5) The population of a city increases every year by 1.8% of the
population at the beginning of that year, in how many years will
the total increase of population be 30%?
Ans: p = p A = 1.3p r = 1.8 n = ?
A = P (1 + ๐
100)๐
1.3p = P (1 + 1.8
100)๐
1.3p = P (1 + 101.8
100)๐
1.3p = p (1.018)n
1.3 = (1.018)n
log 1.3 = n log 1.018
0.1139 = n x 0.0076
n = 0.1139
0.0076 = 14.987
= 15
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COMPOUNDING HALF YEARLY OR QUARTERLY
When interest is compounded half yearly, then r = r/2 , n = 2n.
When interest is compounded quarterly, then r = r/4 , n = 4n.
When interest is compounded monthly, then r = r/12 , n = 12n.
1) Find the compound interest on Rs.50,000/โ for 2 ยฝ years at 6%
p.a. interest being compounded half yearly.
Ans: p = 50,000 n = 2 ยฝ x 2 = 5
r = 6/2 = 3
Amount = 50000 (1 + 3
100)5
= 50000 ( 103
100)5
= 50000 ร (1.03)5
= 57964
CI = 7964
2) Find the compound interest on Rs.60,000/โ for 4 years, if interest
is payable half yearly for due first 3 years at the rate of 8% p.a. and
for the fourth year, the interest is being payable quarterly at the rate
of 6% p.a.
Ans: Amount at in end of 3 years
n = 3 x 2 = 6, r = 8
2 = 4
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p = 6000
= 6000 (1 + 4
100)6
= 6000 ( 104
100)6
= 6000 (1.04)6
= 6,000 x 1.2653 = 7592
For last year
n = 1 x 4 = 4,
r = 6
2 =1.5
p = 7,592
Amount at the end of 4th year
= 7592 (1 + 1.5
100)4
= 7592 (1.015 )4
= 7592 x 1.0613 = 8057
Interest = 8057 โ 6000 = 2057
3) Find the effective rate of interest if interest is calculated at 10%
p.a. half yearly?
Ans: Let p = 100, n = 1 x 2 = 2 r =10
2 = 5
= 1 00 (1 + 5
100)2
= 1 00 ( 105
100)2
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= 100 x 1.1025 = 110.25
C 1 = 110.25 โ 100 = 10.25
Effective rate = 10.25% p.a.
ANNUITY
An annuity is a fixed sum paid at regular intervals under certain
conditions. The interval may be either a year or a half-year or, a
quarter year or a month.
Amount of an annuity :
Amount of an annuity is the total of all the instalments left unpaid
together with the compound interest of each payment for the period
it remains unpaid.
Formula : A = P
i {(1 + ๐)2 โ 1}
(i) Where A = total(s) amount after n years,
i = rate of interest per rupee per annum.
p = yearly annuity
(ii) If an annuity is payable half-yearly and interest is also
compounded half-yearly, then amount A is given by
A = 2๐
๐ 1+
๐2๐
2 - 1
(iii) If an annuity is payable quarterly and interest is also
compounded quarterly, then amount A is given by
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A = 4๐
๐ 1+
๐4๐
2 - 1
Present value of an annuity :
Definition : Present value of an annuity is the sum of the present
values of all payments (or instalments) made at successive annuity
periods.
Formula :
(i) The present value V of an annuity P to continue for n years is
given by
V = ๐
๐{1 โ (1 + ๐)โ๐}
Where i = interest per rupee per annum.
(ii) The Present value V of an annuity P payable half-yearly, then
V = 2๐
๐{1 โ (1 +
๐
2
โ2๐}
(iii) The Present value V of an annuity P payable quarterly, then
V = 4๐
๐{1 โ (1 +
๐
4
โ4๐}
Example:
A man decides to deposit 20,000 at the end of each year in a bank
which pays 10% p.a. compound interest. If the instalments are
allowed to accumulate, what will be the total accumulation at the
end of 9 years?
Solution :
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Let ` A be the total accumulation at the end of 9 years. Then we
have
Example :
A truck is purchased on instalment basis, such that ` 10,000 is to be
paid on the signing of the contract and five yearly instalments of `
5,000 each payable at the end of 1st, 2nd, 3rd, 4th and 5th years. If
interest is charged at 10% per annum what would be the cash down
price?
Solution :
Let V be the present value of the annuity of ` 5,000 for 5 years at
10% p.a. compound interest, then cash down price of the truck is `
(10,000 + V).
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Hence the required cash down price of the truck
= ` (18,953.95 + 10,000)
= ` 28,953.95
Example :
A man purchased a house valued at ` 3,00,000. He paid ` 2,00,000
at the time of purchase and agreed tob pay the balance with interest
of 12% per annum compounded half yearly in 20 equal half yearly
instalments.
If the first instalment is paid after six months from the date of
purchase, find the amount of each instalment.
[Given log 10.6 = 1.0253 and log 31.19 = 1.494]
Solution :
Since ` 2,00,000 has been paid at the time of purchase when cost of
house was ` 3,00,000, we have to consider 20 equated half yearly
annuity payment ` P when 12% is rate of annual interest
compounded half yearly for present value of ` 1,00,000.
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PERPETUITY
Perpetuity can be well defined as an annuity without any end, or it
can be said that perpetuity features a stream of cash payments
continuing forever. To describe in detail, perpetuity is an annuity
wherein the periodic payments commence on a specific date and
continue to an indefinite time. Sometimes, it is even referred to as
perpetual annuity. Some of the prime examples perpetuities include
fixed coupon payments on permanently invested money, or consol
(the British issued bond). The concept of perpetuity is, very often,
used in financial theory, like the Dividend Discount Model (DDM).
Calculation (formula) of perpetuity
The value of perpetuity or a perpetual annuity is calculated by a
simple formula: PV = ๐ด
๐
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where,
PV = represents the present value of the perpetuity,
A = represents the amount of periodic payments, and
r = represents the discount rate, yield, or interest rate.
Besides, the present value of perpetuity can also be determined by
the following steps:
Find out the annual payment, rate of interest, and growth rate of the
perpetuity.
Put the apt numbers into the formula:
Present Value of a growing perpetuity = P / (i โ g),
Where โPโ represents the annual payment, โiโ represents the interest
or discount rate, and โgโ is the growth rate.
You can apply the following example as a point of reference so as
to crosscheck the calculations. Let us assume that a XYZ company
pays a dividend of Rs.100 per annum forever at an interest rate of
5% and a growth rate of 1%. To estimate the present value of this
perpetuity, you can apply the following formula:
PV of preferred stock in XYZ = 100 / (Rs.0.05-0.01)
PV of preferred stock in XYZ = Rs.100 / 0.04
PV of preferred stock in XYZ = Rs.2,500
Therefore, the present value of a share of XYZโs preferred stock is
expected to be Rs.2,500.
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Although the payments made on perpetuity are assumed to be
indefinite, there is still a finite value for the perpetuityโs present
value which is accrued to the expected returns in future featuring a
low present value.
The value of perpetuity can, however, change over the time period,
in spite of same amount of payments. This might be due to changes
in discount rate. The value of perpetuity increases with a decrease
in the discount rate and vice versa.
Equated monthly installment
An equated monthly installment (EMI) is defined as "A fixed
payment amount made by a borrower to a lender at a specified date
each calendar month. Equated monthly installments are used to pay
off both interest and principal each month, so that over a specified
number of years, the loan is fully paid off along with interest."
It further explains that, with most common types of loans, such as
real estate mortgages, the borrower makes fixed periodic payments
to the lender over the course of several years with the goal of
retiring the loan. EMIs differ from variable payment plans, in
which the borrower is able to pay higher payment amounts at his or
her discretion. In EMI plans, borrowers are mostly only allowed
one fixed payment amount each month.
The benefit of an EMI for borrowers is that they know precisely
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how much money they will need to pay toward their loan each
month, making the personal budgeting process easier.
The formula for EMI (in arrears) is:
P = A .1โ(1+๐)โ๐
๐
Or P = P .๐(1+๐)๐
(1+๐)๐โ1
where: P is the principal amount borrowed, A is the
periodic amortization payment, r is the annual interest rate divided
by 100 (annual interest rate also divided by 12 in case of monthly
installments), and n is the total number of payments (for a 30-year
loan with monthly payments n = 30 ร 12 = 360).
For example, if you borrow 10,000,000 units of a currency from
the bank at 10.5% annual interest for a period of 10 years (i.e., 120
months), then EMI = units of currency
10,000,000 ยท 0.00875 ยท (1 + 0.00875)120/((1 + 0.00875)120 โ 1)
= units of currency 134,935. i.e., you will have to pay total
currency units 134,935 for 120 months to repay the entire loan
amount.
The total amount payable will be 134,935 ร 120 = 16,192,200
currency units that includes currency units 6,192,200 as interest
toward the loan.
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MODULE V
DESCRIPTIVE STATISTICS
Meaning and Definitions of Statistics
The word statistics is derived from the Latin word โStatusโ or Italian
word โStatistaโ or German word โStatistikโ which means a Political
State. It is termed as political state, since in early years, statics
indicates a collection of facts about the people in the state for
administration or political purpose.
Statistics has been defined either as a singular non or as a plural
noun.
Definition of Statistics as Plural noun or as numerical facts:โ
According to Horace Secrist, โStatistics are aggregates of facts
affected to a marked extent by multiplicity of causes numerically
expressed, enumerated or estimated according to a reasonable
standard of accuracy, collected in a systematic manner for a
predetermined purpose and placed in relation to each otherโ.
Definition of Statistics as a singular noun or as a method:โ
According to Seliman, โ Statistics is the science which deals with the
methods of collecting classifying, comparing and interpreting
numerical data collected, to know some light on any sphere of
enquiryโ.
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Characteristics of Statistics
1. Statistics show be aggregates of facts
2. They should be affected to a marked extent by multiplicity of
causes.
3. They must be numerically expressed.
4. They should be enumerated or estimated according to a
reasonable standard of accuracy.
5. They should be collected in a systematic manner.
6. They should be collected for a predetermined purpose.
7. They should be placed in relation to each other.
Function of Statistics
The following are the important functions of statistics:
1. It simplifies complexity:โ Statistical methods make facts
and figures easily understandable form. For this purpose
Graphs and Diagrams, classification, averages etc are used.
2. It presents facts in a proper form:โ Statistics presents facts
in a precise and definite form.
3. It facilitates for comparison:โ When date are presented in a
simplified form, it is easy to compare date.
4. It facilitates for formulating policies:โ Statistics helps for
formulating policies for the companies, individuals, Govt.
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etc. it is possible only with the help of date presented in a
suitable form.
5. It tests hypothesis:โ Hypothesis is an important concept in
research studies. Statistics provides various methods for
testing the hypothesis. The important tests are Chi โ square,
Zโtest, Tโtest and Fโtest.
6. It helps prediction or forecasting:โ Statistical methods
provide helpful means of forecasting future events.
7. It enlarges individualโs knowledge:โWhen data are presented
in a form of comparison, the individuals try to find out the
reasons for the variations of two or more figures. It thereby
helps to enlarge the individualโs knowledge.
8. It measures the trend behavior:โ Statistics helps for predicting
the future with the help of present and past data. Hence
plans, programs, and policies are formulated in advance
with the help of statistical techniques.
Scope of Statistics or importance or utility of statistics.
The Scope of Statistics in various field are:
1. Statistics in Business:โ Statistics is most commonly used in
business. It helps to take decision making of the business.
The statistical data regarding the demand and supply of
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product can be collected and analyzed to take decisions.
The company can also calculate the cost of production and
then the selling price. The existing firms can also make a
comparative study about their performance with the
performance of others through statistical analysis.
2. Statistics in Management:โ Most of the managerial
decisions are taken with the help of statistics. The important
managerial activities like planning, directing and controlling
are properly executed with the help of statistical data and
statistical analysis. Statistical techniques can also be used
for the payment of wages to the employees of the
organization.
3. Statistics in Economics:โ Statistical data and methods of
statistical analysis render valuable assistance in the proper
understanding of the economic problems and the
formulation of economic policy.
4. Statistics in Banking and Finance:โ Banking and financial
activities use statistics most commonly.
5. Statistics in Administration:โ The govt. frames polices on
the basis of statistical information.
6. Statistics in Research:โ Research work are undertaken with
the help of statistics.
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Limitation of statistics
1. Statistics studies only numerical data
2. Statistics does not study individual cases.
3. Statistical result are true only an average.
4. Statistics does not reveal the entire story of the problem.
5. Statistics in only one of the methods of study a problem.
6. Statistics can be misused.
Statistical Enquires or Investigation
Statistical Investigation is concerned with investigation of some
problem with the help of statistical methods. It implies search for
knowledge about some problems through statistical device.
Different stages in statistical enquiry are:
1. Planning the enquiry
2. Collection of data.
3. Organization of data.
4. Presentation of data.
5. Analysis of data.
6. Interpretation of data.
Planning the enquiry:โ The first step in statistical investigation is
planning. The investigator should determine the objective and
scope of the investigation. He should decide in advance about the
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type of enquiry to be conducted, source of information and the unit
of measurement.
Object and scope:โ The objective of the Statistical enquiry
must be clearly defined. Once the objective of enquiry has been
determined, the next step is to decide the scope of enquiry. It refers to
the coverage of the enquiry.
Source of information:โ After the purpose and scope have
been defined, the next step is to decide about the sources of data.
The sources of information may be either primary or secondary.
Types of enquiry:โ Selection of type of enquiry depends on a
number of factors like object and scope of enquiries, availability of
time, money and facilities. Enquiries may be (1) census or sample
(2) original or repetitive (3) direct or indirect (4) open or
confidential (5) General or special purpose.
Statistical unit:โ The unit of measurements which are
applied in the collected data is called statistical unit. For example
ton, gram, meter, hour etc.
Degree of accuracy:โ The investigator has to decide about
the degree of accuracy that he wants to attain. Degree of accuracy
desired primarily depends up on the object of an enquiry.
Cost of plan:โ An estimate of the cost of the enquiry must
be prepaid before the commencement of enquiry.
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Collection of data:โ Collection of data implies accounting and
systematic recoding of the information gathered in a statistical
investigation. Depending on the source, the collected statistical data
are classified under two categories namely primary data and
secondary data.
Organization of data:โ Organization of data implies the
arrangement and presentation of data in such a way that it becomes
easy and convenient to use them. Classification and tabulation are
the two stages of organizing data.
Presentation of data:โ They are numerous ways in which statistical
data may be displayed. Graphs and diagrams are used for presenting
the statistical data.
Analysis data:โ Analysis of data means critical examination of the
data for studying characteristics of the object under study and for
determining the pattern of relationship among the variables.
Interpretation of data:โ Interpretation refers to the technique of
drawing inference from the collected facts and explaining the
significance.
MEASURES OF CENTRAL TENDENCY OR AVERAGES
An average is a single value that represents a group of values. It
represents the whole series and conveys general idea of the whole
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group. Characteristics of a good average or Requisites or Essentials
properties of average
1. Clearly defined
2. Easy to understand
3. Simple to compute
4. Based on all items
5. Not be unduly affected by extreme observations.
6. Capable of further algebraic treatment
7. Sampling stability.
Types of averages
1. Arithmetic Mean
2. Median
3. Mode
4. Geometric mean
5. Harmonic Mean
ARITHMETIC MEAN (AM)
It is the value obtained by adding together all the items and by
dividing the total number of items.
Arithmetic mean may either be
1. Simple arithmetic Mean or
2. Weighted arithmetic Mean
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Simple Arithmetic Mean
It is the mean of items which give equal importance to all items.
It is denoted by xฬ
xฬ = ฮฃ๐ฅ
๐
Where = Sum of given variables
N = Number of items
Calculation of Arithmetic Mean
a) Individual Series
i) Direct Method
xฬ = ฮฃ๐ฅ
๐
ii) Short Cut Method
xฬ =A+ ฮฃ๐ฅ
๐
A = Assumed mean
D = X โ A
n = total number of items
b) Descrete Series
i) Direct Method
xฬ = ฮฃf๐ฅ
๐
ii) Short Cut Method
xฬ =A+ ฮฃf๐ฅ
๐
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d = X โ A
iii) Step deviation method
xฬ =A+ ฮฃf๐1
๐ร ๐
๐1 = ๐โ๐ด
๐
c = common factor
c) Continuous Series
i) Direct Method
xฬ = ฮฃfm
๐
m = midpoint of X
N = Total frequency
ii) Short Cut Method
xฬ =A+ ฮฃfd
๐
d = m โ A
iii) Step deviation method
xฬ =A+ ฮฃf๐1
๐ร ๐
๐1 = ๐โ๐ด
๐
C = Common factor or class interval
Practical Problems
1) Calculate A.M. of the weight of 10 students in a Class
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Sl. No. 1 2 3 4 5 6 7 8 9 10
Weight in
Kg
42 56 49 50 49 53 52 48 47 54
Ans: This is an individual series.
xฬ = ฮฃ๐ฅ
๐
ฮฃ๐ฅ = 42+56+49+50+49+53+52+48+47+54
= 500
n = 10
xฬ = 500
10 = 50kg
2) Calculate mean from the following data.
Marks 25 30 35 40 45 50 55 60 65 70
No. of students 3 8 12 9 4 7 15 5 10 7
Ans:
Marks
x
No. of
students f
d
(x โ 55)
dโ fdโ
25 3 โ30 โ6 โ18
30 8 โ25 โ5 โ40
35 12 โ20 โ4 โ48
40 9 โ15 โ3 โ27
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45 4 โ10 โ2 โ8
50 7 โ5 โ1 โ7
55 15 0 0 0
60 5 5 1 5
65 10 10 2 20
70 7 15 3 21
80 โ102
xฬ =A+ ฮฃf๐1
๐ร ๐
xฬ =55+ โ102
80ร 5
xฬ =55+ โ510
80ร 5
= 55 + โ6.375
= 48.625
3) Calculate Arithmatic Mean
Production in tons No. of factories
10 โ 20 5
20 โ 30 4
30 โ 40 7
40 โ 50 12
50 โ 60 10
60 โ 70 8
70 โ 80 4
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Ans:
X f m fm
10 โ 20 5 15 75
20 โ 30 4 25 100
30 โ 40 7 35 245
40 โ 50 12 45 540
50 โ 60 10 55 550
60 โ 70 8 65 520
70 โ 80 4 75 300
80 2330
xฬ = ฮฃfm
๐
xฬ = 2330
80= 46.6
4) Following are the data related with the production of a product
during January in 100 factories
Production in tons No. of factories
0 โ 100 7
100 โ 200 15
200 โ 300 10
300 โ 400 9
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400 โ 500 10
500 โ 600 12
600 โ 700 8
700 โ 800 13
800 โ 900 9
900 โ 1000 7
Ans:
x f m d (m โ A) dโ fdโ
0 โ 100 7 50 โ500 โ5 โ35
100 โ 200 15 100 โ400 โ4 โ60
200 โ 300 10 250 โ300 โ3 โ30
300 โ 400 9 350 โ200 โ2 โ18
400 โ 500 10 450 โ100 โ1 โ10
500 โ 600 12 550 0 0 0
600 โ 700 8 650 100 1 8
700 โ 800 13 750 200 2 26
800 โ 900 9 850 300 3 27
900 โ 1000 7 950 400 4 28
100 โ64
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xฬ =A+ ฮฃf๐1
๐ร ๐
xฬ =550+ โ64
100ร 100
= 486
Calculation of Arithmatic Mean for open end classes
If the lower limit of the first class and upper limit of the last class are
not known, it is called open end classes.
Calculate A.M.
Below 10 5
10 โ 20 12
20 โ 30 14
30 โ 40 10
Above 40 8
Ans:
X f m fm
0 โ 10 5 5 25
10 โ 20 12 15 180
20 โ 30 14 25 350
30 โ 40 10 35 350
40 โ 50 8 45 360
49 1265
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๏ฟฝฬ ๏ฟฝ = ฮฃfm
๐ =
1265
49
= 25.82
Weighted Mean
Weighted means are obtained by taking in to account of weights.
Each value is multiplied by its weight and total is divided by the
total weight to get weighted mean.
๏ฟฝฬ ๏ฟฝ๐ค= ฮฃ๐ค๐ฅ
ฮฃ๐ค
๏ฟฝฬ ๏ฟฝ๐ค = weighted A.M
w = weight
x = given variable
MEDIAN
Median is the middle value of the series. When the series are
arranged in the ascending order
or descending order Median is a positional average.
Calculation of Median
Individual series
Firstly arrange the series.
Median = Size of (๐+1
2)th item
Discrete series
Median = Size of (๐+1
2)th item
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Continuous series
Median Class =๐
2
Median = L1+
๐
2โ ๐๐
๐ร C
L1 = Lowerlimit of median class
c.f = culmulative frequency of preceding median class
f = frequency of median class
C = Class interval
1) Find the median for the following data
4, 25, 45, 15, 26, 35, 55, 28, 48
Ans :
4, 15, 21, 25, 26, 28, 35, 45, 48, 55
Median = Size of (๐+1
2)th item
= (5+1
2)th item = 5th item
Median = 28
2) Calculate median
25, 35, 15, 18, 17, 36, 28, 24, 22, 26
Ans :
15, 17, 18, 22, 24, 25, 26, 28, 35, 36
Median = Size of (๐+1
2)th item
= (10+1
2)th item
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= 5.5th item
Median = 5๐กโ ๐๐ก๐๐+6๐กโ ๐๐ก๐๐
2
= 24+25
2 = 24.5
3) Calculate median
Size : 5 8 10 15 20 25
Frequency : 3 12 8 7 5 4
Ans:
Size
Frequency
Cf
5 3 3
8 12 15
10 8 23
15 7 30
20 5 35
25 4 39
Median = Size of (๐+1
2)th item
= (39+1
2)th item = 20th item
Median = 10
4) Find median from the following
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Marks No. of students
0โ5 29
10โ15 195
15โ20 241
20โ25 117
25โ30 52
30โ35 10
35โ40 6
40โ45 2
Ans:
Marks f c.f
0โ5 29 29
5โ10 195 227
10โ15 241 465
15โ20 117 582
20โ25 52 634
25โ30 10 644
30โ35 6 650
35โ40 3 653
40โ45 3 656
656
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Median Class = N/2 = 656
2 = 328th item
Median = L1+
๐
2โ ๐๐
๐ร C
= 10+328โ 224
241ร 5
= 12.2
MODE
Mode is the value of item of series which occurs most frequently.
Mode in individual series
In the case of individual series, the value which occurs more
number of times is mode.
When no items appear more number of times than others, then mode
is the ill defined. In this case :
Mode = 3 median โ 2 mean
Mode in discrete series
In the case of discrete series, the value having highest frequency is
taken as mode.
Mode in continuous series
Mode lies in the class having the highest frequency.
Mode = l1+(๐1โ ๐0)ร๐
2๐1โ ๐0โ๐2
l1= lower limit of the model class
f1= frequency of the model class
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f0 , f1= frequency of class preceding and succeeding modal class.
1) Find mode
1, 2, 5, 6, 7, 3, 4, 8, 2, 5, 4, 5
Ans:
Mode = 5
2) Find mode
4, 2, 6, 3, 8, 7, 9, 1
Ans:
Mode is ill defined
Mode = 3 median โ 2 mean
๏ฟฝฬ ๏ฟฝ = ฮฃx
๐ =
40
8 = 5
Median : 1, 2, 3, 4, 6, 7, 9
Median = Size of (๐+1
2)th item
= 8+1
2 = 4.5
= 4๐กโ ๐๐ก๐๐+5๐กโ ๐๐ก๐๐
2
= 4+6
2 = 5
Mode = 3 ร 5 โ 2 ร 5 = 5
3) Find mode
Size : 5 8 10 12 15 20 25
Frequency: 3 7 2 9 5 6 2
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Mode = 2, since 12 has the highest frequency
4) Calculate mode
Size : 0โ
5
5โ1
0
10โ1
5
15โ2
0
20โ2
5
25โ3
0
Frequenc
y:
20 24 32 28 20 26
Ans:
Size Frequency
0โ5 20
5โ10 24
10โ15 32 โโโโ โModels class
15โ20 28
20โ25 20
25โ30 26
Mode = l1+(๐1โ ๐0)ร๐
2๐1โ ๐0โ๐2
= 10+(32โ 24)ร5
2ร32โ 24โ28
= 10 + 40
12
= 13.3
Geometric Mean
Geometric mean is defined as the nthroot of the product of those in
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values.
GM = Antilog (ฮฃ log ๐ฅ
๐)
GM in individual series
GM = Antilog (ฮฃ log ๐ฅ
๐)
GM in descrete series
GM = Antilog (ฮฃ f log ๐ฅ
๐)
GM in continous series
GM = Antilog (ฮฃ f log ๐ฅ
๐)
X = midpoint of x
1) Find Geometric mean of the following
57.5, 87.75, 53.5, 73.5, 81.75
Ans:
X logx
57.5 1.7597
87.75 1.9432
53.5 1.7284
73.5 1.8663
81.75 1.9125
9.2101
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GM = Antilog (ฮฃ log ๐ฅ
๐)
= Antilog (9.2101
5)
= Antilog (1.84202)
= 69.51
2) Find G.M from the following data
Size : 5 8 10 12
Frequency: 2 3 4 1
Ans:
X f logX f logX
5 2 .6990 1.3980
8 3 .9031 2.7093
10 4 1.0000 4.0000
12 1 1.0792 1.0792
10 9.1865
GM = Antilog (ฮฃ f log ๐ฅ
๐)
= Antilog (9.1865
10)
= Antilog (.91865)
= 8.292
3) Calculate G.M.
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Daily Income 0 -20 20 - 40 40 - 60 60 โ 80 80 - 100
Frequency: 5 7 12 8 4
Ans:
X f x( logx f logx
0โ20 5 10 1.0000 5.0000
20โ40 7 20 1.4771 10.3397
40โ60 12 30 1.6990 20.3880
60โ80 8 40 1.8451 14.7608
80โ100 4 50 1.9542 7.8168
36 58.3053
GM = Antilog (ฮฃ f log ๐ฅ
๐)
= Antilog (58.3053
36)
= Antilog 1.6195916
= 41.65
Harmonic Mean
Harmonic mean is defined as the reciprocal of the mean of the
reciprocals of those values. It applied in averaging rates, times etc.
HM = ๐
โ1
๐ฅ
H.M in Discrete series
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HM = ๐
โ๐(1
๐ฅ)
H.M in Continous series
HM = ๐
โ๐(1
๐ฅ)
X = midpoint of x
1)Find the H.M.
2, 3, 4, 5
Ans:
x 1
x
2 0.5
3 0.33
4 0.25
5 0.20
1.28
HM = ๐
โ(1
๐ฅ)
= 4
1.28 = 3.125
2) Find the H.M.
Size : 6 10 14 18
Frequency: 20 40 30 10
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Ans:
Size f 1
x
f(1/X)
6 20 0.1667 3.334
10 40 0.1000 4.000
14 30 0.0714 2.142
18 10 0.0556 0.556
100 10.032
HM = ๐
โ๐(1
๐ฅ)
= 100
10.032
= 9.97
3) From the following data, calculate the value of HM?
Income ( ) No. of persons
10 โ 20 4
20 โ 30 6
30 โ 40 10
40 โ 50 7
50 โ 60 3
Ans:
Income ( )
f
x in m
1
x ๐(
1
๐ฅ)
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10 โ 20 4 15 0.667 0.2666
20 โ 30 6 25 0.0400 0.2400
30 โ 40 10 35 0.0286 0.2857
40 โ 50 7 45 0.0222 0.1556
50 โ 60 3 55 0.0182 0.0545
30 1.0023
HM = ๐
โ๐(1
๐ฅ)
= 30
1.0023
= 29.93
MEASURES OF DISPERSION OR VARIABILITY
Dispersion means a measure of the degree of deviation of data from
the central value.
Measures of Dispersion are classified into
(1) Absolute Measures
(2) Relative Measures.
Absolute Measures of dispersion are expressed in the same units in
which data are collected. They measure variability of series.
Various absolute measures are:
(i) Range
(ii) Quartile Deviation
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(iii) Mean Deviation
(iv) Standard Deviation
Relative measure is also called coefficient of dispersion. They are
useful for comparing two series for their variability. Various
relative measures are:
(i) Coefficient Range
(ii) Coefficient of Quartile Deviation
(iii) Coefficient of Mean Deviation
(iv) Coefficient of Variation
RANGE
The range of any series is the difference between the highest and
the lowest values
in the series.
Range = H โ L
H = Highest variable
L = Lowest variable
Coefficient of Range =๐ปโ๐ฟ
๐ป+๐ฟ
1) Find the Range and Coefficient of Range. 75, 29, 96, 15,7,8,11, 7, 49 Ans:
Range = H โ L = 96 โ 74 = 92
Coefficient of Range = =๐ปโ๐ฟ
๐ป+๐ฟ = =
96โ4
96+4 =
92
100 = 0.92
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2) Find Range and Coefficient of Range
Wages 5 10 15 20 25 30
No. of
employees
2 5 6 7 4 6
Range = H โ L
= 30 โ 5 = 25
Coefficient of Range = =๐ปโ๐ฟ
๐ป+๐ฟ = =
30โ5
30+5 =
25
35 = 0.71
3) Find out Range and Coefficient of Range
Marks 20 โ 29 30 โ 39 40 โ 49 50 โ 59 60 โ 69
No. of Students 8 12 20 7 3
Ans:
Marks f
19.5 โ 29.5 8
29.5 โ 39.5 12
39.5 โ 49.5 20
49.5 โ 59.5 7
59.5 โ 69.5 3
Range = H โ L
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= 69.5 โ 19.5 = 50
Coefficient of Range = =๐ปโ๐ฟ
๐ป+๐ฟ = =
69.5โ19.5
69.5+19.5 =
50
89 = 0.56
QUARTILE DEVIATION
Quartile Deviation is defined as the half distance between the third
and first quartiles.
Quartile Deviation = ๐3โ ๐1
2
Coefficient of Quartile Deviation = ๐3โ ๐1
๐3+ ๐1
Quartile Deviation in Individual Series
Quartile Deviation= ๐3โ ๐1
2
๐1= size of ๐+1
4 th item
๐3= size of 3( ๐+1
4) th item
Quartile Deviation in Discrete Series
Quartile Deviation= ๐3โ ๐1
2
๐1= size of ๐+1
4 th item
๐3= size of 3( ๐+1
4) th item
Quartile Deviation in Continous Series
Quartile Deviation= ๐3โ ๐1
2
๐1= size of ๐
4 th item
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๐3= size of 3( ๐
4) th item
Then ๐1 = L1+
๐
4โ ๐๐
๐ร C
๐3 = L1+3(
๐
4)โ ๐๐
๐ร C
1) Calculate Quartile Deviation from the following:
25, 15, 30, 45, 40, 20, 50
Also find coefficient of quartile deviation.
Ans: Arrange the series, then
15, 20, 25, 30, 40, 45, 50
๐1= size of ๐+1
4 th item =
8
4 = 2nd item
๐3= size of 3( ๐+1
4) th item = 3 ร 2 = 6th item
Quartile Deviation = ๐3โ ๐1
2 =
45โ 20
2 = 12.5
Coefficient of Quartile Deviation = ๐3โ ๐1
๐3+ ๐1
= 25
45+ 20
= 25
65 = 0.385
2) Find the value of Quartile Deviation and coefficient of Quartile
Deviation?
Marks 25 30 40 50 60 70 80 90
No. of Students 4 7 12 8 9 15 7 3
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Ans:
x f c.f.
25 4 4
30 7 11
40 12 23
50 8 31
60 9 40
70 15 55
80 7 62
90 3 65
65
๐1= size of ๐+1
4 th item =
65+1
4 = 16.5th item
๐3= size of 3( ๐+1
4) th item = 3 ร 16.5 =49.5th item
๐1= 40
๐3= 70
Quartile Deviation = ๐3โ ๐1
2 =
70โ 40
2 = 15 marks
Coefficient of Quartile Deviation = ๐3โ ๐1
๐3+ ๐1
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= 70โ40
70+ 40
= 0.27
3) Compute Quartile Deviation and coefficient of Quartile
Deviation?
x 0 โ 10
10 โ 20
20 โ 30
30 โ 40
40 โ 50
50 โ 60
f 5 12 15 9 10 3
Ans:
x f c.f.
0 โ 10 5 5
10 โ 20 12 17
20 โ 30 15 32
30 โ 40 9 41
40 โ 50 10 51
50 โ 60 3 54
54
๐1= size of ๐
4 th item
= 54
4 th item = 13.5th item
Which lies in 10 โ 20 , then
Then ๐1 = L1+
๐
4โ ๐๐
๐ร C
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= 10+13.5โ 5
12ร 10
= 10+8.5
12ร 10
= 10+85
12 = 17.08
๐3= size of 3( ๐
4) th item
3ร13.5 = 40.5th item
Which lies in 30 โ 40 , then
๐3 = L1+3(
๐
4)โ ๐๐
๐ร C
= 30+40.5โ 32
9ร 10
= 30+8.5
9ร 10
= 30+85
9 = 39.44
Quartile Deviation = ๐3โ ๐1
2 =
39.44โ 17.08
2
= 11.18 marks
Coefficient of Quartile Deviation = ๐3โ ๐1
๐3+ ๐1
= 39.44 โ 17.08
39.44 + 17.08 =
22.36
56.52
= 0.396
MEAN DEVIATION
Mean Deviation is defined as the arithmetic mean of deviations of
all the values in a series from their average. The average may be
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mean, median or mode.
Mean Deviation = ฮฃ|๐|
๐
Where |d| = deviation from an average without sign
Mean Deviation in Individual Series
Mean Deviation = ฮฃ|๐|
๐
Coefficient of Mean Deviation = Mean Deviation
๐ด๐ฃ๐๐๐๐๐
Average = Mean, Median or Mode from which the deviation is
taken
Mean Deviation in Discrete Series
Mean Deviation =ฮฃf|๐|
๐
Coefficient of Mean Deviation = Mean Deviation
๐ด๐ฃ๐๐๐๐๐
Mean Deviation in Continous Series
Mean Deviation =ฮฃf|๐|
๐
1) Calculate Mean Deviation from the following.
14, 15, 23, 20, 10, 30, 19, 18, 16, 25, 12
Ans:
Arrange the data
10, 12, 14, 15, 16, 18, 19, 20, 23, 25, 30
Median size of 11+1
2 item
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= 6th Item = 18
X |d| ie. X โ median
14 4
15 3
23 5
20 2
10 8
30 12
19 1
18 0
16 2
25 7
12 6 50
Mean Deviation = ฮฃ|d|
n =
50
11 = 4.54 marks
2) Calculate Mean Deviation from the following data:
Size of item 6 7 8 9 10 11 12
Freequency 3 6 9 13 8 5 4
Ans:
Size f c.f |d| f |d|
6 3 3 3 9
7 6 9 2 12
8 9 18 1 9
9 13 31 0 0
10 8 39 1 8
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11 5 44 2 10
12 4 48 3 12
48 60
Median size of 48+1
2 item = 24.5
Median = 9
= 18
Mean Deviation = ฮฃf|d|
n =
60
48 = 1.25
3) Calculate the Mean Deviation from the following data: Marks 0 โ 10 10 โ 20 20โ 30 30โ 40 40 โ 50 50 โ 60 60 โ 70 70 โ
80
Freequency 18 16 15 12 10 5 2 2
Ans:
x f m c.f. |d| ie. X โ median f|d|
0 โ 10 18 5 18 19 342
10 โ 20 16 15 34 9 144
20 โ 30 15 25 49 1 15
30 โ 40 12 35 61 11 132
40 โ 50 10 45 71 21 210
50 โ 60 5 55 76 31 155
60 โ 70 2 65 78 41 82
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70 โ 80 2 75 80 51 102
80 1182
Median = ๐
2th item
=80
2th item = 40th item
Which lies on 20 โ 30
Median = L1+
๐
2โ ๐๐
๐ร C
= 20+40โ 34
15ร 10
= 20+6
15ร 10
= 24
Mean Deviation = ฮฃf|d|
n =
1182
80 = 14.775
STANDARD DEVIATION
Standard Deviation is defined as the square root of the mean of
the squares of the deviations of individual items from their
arithmetic mean. It is denoted by ฯ (sigma).
ฯ =โฮฃ(๐ฅโ๏ฟฝฬ ๏ฟฝ)2
n
Standard Deviation in Individual Series
ฯ =โฮฃ(๐ฅโ๏ฟฝฬ ๏ฟฝ)2
n or ฯ =โ
ฮฃx2
nโ (
ฮฃx
n)2
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Coefficient of variation = ฯ
๏ฟฝฬ ๏ฟฝร 100
Standard Deviation in Discrete Series
ฯ =โฮฃfx2
Nโ (
ฮฃfx
N)2
Shortcut method:
ฯ =โฮฃfd2
Nโ (
ฮฃfd
N)2
d = x โ A
Standard Deviation in Continuous Series
(i) Direct Method:
ฯ =โฮฃfx2
Nโ (
ฮฃfx
N)2
x = mid point of X
(ii) Shortcut method:
ฯ =โฮฃfd2
Nโ (
ฮฃfd
N)2
d = m โ A or x โ A
(iii) Step Deviation method:
ฯ =โฮฃfdโฒ 2
Nโ (
ฮฃfdโฒ
N)2 ร c
dโ= ๐
๐
c = class interval.
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VARIANCE
Variance is defined as the mean of the squares of the deviations
of all the values in the series from their mean. It is the sqare
root of the Standard Deviation.
Variance = ฯ
1) Compute S.D
4, 8, 10, 12, 15, 9, 7, 7
Ans:
x x2
4 16
8 64
10 100
12 144
15 225
8 81
7 49
7 49
72
728
ฯ =โฮฃx2
nโ (
ฮฃx
n)2
=โ728
8โ (
72
8)2
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=โ91 โ 92
=โ91 โ 81
= โ10
= 3.16
2) Find the S.D and C.V
10, 12, 80, 70, 60, 100, 0, 4
Coefficient of variation = ฯ
๏ฟฝฬ ๏ฟฝร 100
Ans:
x x2
10 100
12 144
80 6400
70 4900
60 3600
100 10000
0 0
4 16
336 25160
ฯ =โฮฃx2
nโ (
ฮฃx
n)2
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=โ25160
8โ (
336
8)2
= โ3145 โ 422
= โ3145 โ 1764
= โ1381
= 37.16
C.V.
= ฯ
๏ฟฝฬ ๏ฟฝร 100
๏ฟฝฬ ๏ฟฝ = 336
8= 42
C.V = 37.16
42ร 100 = 88.48
3) Find out S.D
Production in tones : 50 100 125 150 200 250 300
No. of factories: 2 5 7 12 9 5 3
Ans:
X f d(xโA) d1 d12 fd1 fd12
50 2 โ100 โ4 16 โ8 32
100 5 โ50 โ2 4 โ10 20
125 7 โ25 โ1 1 โ7 7
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150 12 0 0 0 0 0
200 9 50 2 4 18 36
250 5 100 4 16 20 80
300 3 150 6 36 18 108
43 31 283
A= 150
dโ = ๐
25
ฯ =โฮฃfdโฒ 2
Nโ (
ฮฃfdโฒ
N)2 ร c
=โ283
43โ (
31
43)2 ร 25
=โ6.58 โ 0.52 ร 25
=โ6.06 ร 25
= 2.46ร 25
= 61.5
4) The scores of the batsmen A and B the six innings during a
certain match are as follows.
Expenditure
(Rs): 100โ200 200โ300 300โ400 400โ500 500โ600
No. of families 30 20 40 5 10
(i) Find which of the two batsman is more consistant in scoring.
(ii) Find who is more efficient batchman.
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Ans:
Batsman A Batsman B
X X2 X X2
10 100 8 64
12 144 9 81
80 6400 7 49
70 4900 10 100
60 3600 5 25
100 10000 9 81
0 0 10 100
4 16 8 64
336 25160 66 564
(i) For finding consistant, C.V is calculated
C.V = = ฯ
๏ฟฝฬ ๏ฟฝร 100
Batsman A
๏ฟฝฬ ๏ฟฝ =336
8 = 42
ฯ =โฮฃx2
nโ (
ฮฃx
n)2
=โ25160
8โ (
236
8)2
= 37.16
C.V = 37.16
42ร 100
= 88.48
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Batsman B
๏ฟฝฬ ๏ฟฝ = 66
8 = 8.25
ฯ =โฮฃx2
nโ (
ฮฃx
n)2
=โ564
8โ (
66
8)2
= 1.562
C.V = 1.562
8.25ร 100
= 18.93
B is more consistent since C.V. is less.
(ii) For finding more efficient, average is taken
A = 42 B = 8.25
Batsman A is more consistent since he has greater
average.
Merits of S.D
1. S.D. is based on all the values of a series.
2. It is rigidly defined
3. It is capable of further mathematical treatment.
4. It is not much affected by sampling fluctuations.
Demerits
1. It is difficult to calculate.
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2. Signs of the deviations are not ignored.
Measures of skewness
Skewness means lack of symmetry when a frequency distribution
is not symmetrical, it is said t be asymmetrical or skewed. In the
case of a skewed distribution, the mean, median and mode are not
equal. Similarly for a skewed distribution Q1 and Q3 will not be
equidistant from median. It is an asymmetrical distribution. It has a
long tail on one side and a start tail on the other side.
A distribution is said to be skewed when:
(1) Mean, media and mode are not equal.
(2) Q1 and Q3 are not equidistant from median.
(3) Frequencies on either side of mode are not equal.
(4) The frequency curve has longer tail on the left side or on the
right side.
Positive and Negative skewness
Skewnwss may be either positive or negative. Skewness is said to
be positive when the mean is greater than the median and median is
greater than mode. More than half area falls to right side of the
highest ordinate.
Swewness is said to be negative when the mean is less than median
and the median is less than mode. In this case curve is skewed to
the left more than half the area falls to the left of the highest
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ordinate.
Measures of skewness
Karl Pearsonโs measure of skewness
Skewness = ๐๐๐๐โ๐๐๐๐๐๐
๐
Bowleyโs measure of skewness
Skewness = ๐3+๐1โ2๐๐๐๐๐๐
๐3โ๐1
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Kelleyโs measure of skewness
Skewness = ๐90+๐10โ2๐๐๐๐๐๐
๐90โ๐10
Measure of skewness Based on Moments
Skewness = ๐3
โ๐23
KURTOSIS
Kurtosis is a measure of peakdness. It refers a distribution
which is relatively fetker than the normal curve.
When a frequency curve is more peaked than the normal curve, it is
called lepto kurtic and when it is more flat topped than the normal
curve it is called platy kurtic. When a curve is neither peaked nor plat
topped, it is called meso kurtic normal.
Lorenz Curve
Lorenz curve is a graphical method of studying dispersion. It is
used in business to study the disparities of the distribution of
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wages, sales, production etc. In Economics it is useful to measure
inequalities in the distribution of income.
It is a graph down to a frequency distribution. While drawing the
graph, cumulative percentage values of frequencies on X axis and
cumulative percentage values of the variable on Y axis.