mathematics - Kopykitab

As per the New Revised Syllabus of Maharashtra State Board of Secondary

and Higher Secondary Education

PRAGATI

MATHEMATICS PART −−−− I

ALGEBRA

STANDARD X

SALIENT FEATURES

✒ Written strictly as per the New Revised Syllabus and upgraded Question Paper Pattern effective

from the academic year commencing from June 2018 onwards.

✒ Perfect presentation from the exam point of view.

✒ Chapterwise important formulae and theory for revision at a glance.

✒ Complete step-by-step solution of the problems along with markings of important steps and

important problems.

− : By : − A. A. Tathe

Price : `̀̀̀ 150.00

P0P0P0P0083083083083

STD. X : MATHEMATICS (PART − I) : ALGEBRA

First Edition : June 2018

© : Authors

The text of this publication, or any part thereof, should not be reproduced or transmitted in any form or stored in any computer storage system or device for distribution including photocopy, recording, taping or information retrieval system or reproduced on any disc, tape, perforated media or other information storage device etc., without the written permission of Authors with whom the rights are reserved. Breach of this condition is liable for legal action. Every effort has been made to avoid errors or omissions in this publication. In spite of this, errors may have crept in. Any mistake, error or discrepancy so noted and shall be brought to our notice shall be taken care of in the next edition. It is notified that neither the publisher nor the authors or seller shall be responsible for any damage or loss of action to any one, of any kind, in any manner, therefrom.

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PREFACE

It gives great pleasure in presenting 'Pragati Mathematics Part - I (Algebra)' to the

students of Standard X in view of the New Revised Syllabus and Upgraded Evaluation

Plan of Maharashtra State Board of Secondary and Higher Secondary Education, effective

from the academic year 2018-19.

While writing this book, I have kept in mind the general level of understanding of the

students and my every sincere effort to make this book error-free and to increase the utility

of the book. The problem analysis, mathematical logic and its sequencing is so framed that

it will boost-up the interest and confidence of the users in the subject. I am sure, it will

definitely enhance the performance of the students in the Board Examination.

This book is really speaking, a joint effort. I wish to take this opportunity to acknowledge

my deep gratitude to the publisher especially Shri. Dineshbhai Furia and Shri.

Jigneshbhai Furia and the members of his team for their deep interest and quick

response given by them to my efforts without which this the book would not have taken up

shape.

This preface would be incomplete without the mention of active encouragement given

by all my family members for their strong moral support.

I am indebted to all my students and colleagues whose love, faith and affection is a real

driving force behind this book. I am eagerly waiting for the response of the students and

teachers. The difficulties and the constructive suggestions of the users for the improvement

and utility of this book are most welcome.

June 2018

−−−− A. A. Tathe

✤ ✤ ✤

SCHEME OF EVALUATION

Std. X : Mathematics

The revised syllabus of std. X will be implemented from the academic year 2018 - 2019. Two

text books, Mathematics - Part I and Mathematics - Part - II are prepared for std. X. In all 100

marks are stipulated for the evaluation of Mathematics for std. X.

The division of 100 marks for S. S.C. Examination is as given below.

S.S.C. Examination :

Mathematics : Part I 40 Marks Written examination Time 2 hours

Mathematics : Part II 40 Marks Written examination Time 2 hours

80 Marks

+ 20 Marks Internal Evaluation

100 Marks Total

The scheme of Internal Evaluation will be as follows :

Aggregate of marks of tests of

Maths - Part I and Maths - Part II, converted 10

Marks of Practicals on Maths - Part I 5

Marks of Practicals on Maths - Part II 5

Total marks of internal evaluation 20

In the academic year, two unit tests (Mathematics - Part I, 20 + 20 marks, Mathematics - Part

− II, 20 + 20 marks) should be conducted. The aggregate marks of the two tests should be

converted into 10 marks.

The above scheme of evaluation will be implemented for the S. S. C. Examination from the

academic year 2018-2019.

In the written examination, questions carrying 80% marks will be based on all units learnt in

std. X. The remaining questions carrying 20% marks will be based on previous knowledge means

units learnt in std. IX.

S.S.C. Examination - Std. X : Mathematics

The division of marks in question papers as per objectives will be as follows :

Mathematics - Part I Mathematics - Part II

Objectives Percentage of

Marks

Objectives Percentage of

Marks

Previous knowledge 20 Previous knowledge 20

Knowledge and

understanding

30 Knowledge and

understanding

30

Application 40 Application 30

Skill 10 Skill 20

OUTLINE OF QUESTION PAPER FOR MATHEMATICS PART - I AND

MATHEMATICS PART - II

Marks Marks with

option

Q. 1 : Based on Previous knowledge (Syllabus of std. IX)

Q. 1 : (A) Solve 4 out of 6 sub-questions. [1 Mark each] 4 (6)

(B) Solve 2 out of 3 sub-questions. [2 Marks each] 4 (6)

Q. 2 to 6 will be based on the syllabus of std. X.

Q. 2 : (A) Solve 4 out of 4 MCQ. [ 1 Mark each] 4 (4)

(B) Solve 2 out of 3 sub-questions. [2 Marks each] 4 (6)

Q. 3 : (A) Solve activity based 2 sub-questions out of 3 [2 Marks each] 4 (6)

(B) Solve 2 out of 3 sub-questions [2 Marks each] 4 (6)

Q. 4 : Solve 3 out of 4 sub-questions. [3 Marks each] 9 (12)



Total Marks 40 (60)

NOTE FOR STD. IX

• In the written examination of first semester, 80% marks should be based on all units

studied in the first term of std. IX. The remaining questions of 20% marks should be based

on the previous knowledge of units studied in std. VIII.

• In the written examination of semester II, 80% marks should be based on all units studied

is semester II of std. IX. The remaining questions of 20% marks should be based on the

previous knowledge of unit studied in std. VIII and also on the units studied in the first

semester of std. IX.

• The paper pattern of std. IX and X will remain same.

Note : In this book we covered all textbook questions thoroughly. All solved questions are

from textbook only and other questions are given for practice which would be expected to solve

by students only.

✤ ✤ ✤

CONTENTS

1. Linear Equations in Two Variables ................................................................. 1.1 − 1.31

2. Quadratic Equations ...................................................................................... 2.1 − 2.41

3. Arithmetic Progression ................................................................................... 3.1 − 3.22

4. Financial Planning ......................................................................................... 4.1 − 4.25

5. Probability ..................................................................................................... 5.1 − 5.16

6. Statistics ........................................................................................................ 6.1 − 6.46

✡ Sample Question Paper..................................................................................S.1 − S.3

✤ ✤ ✤

(1.1)

ChapterChapterChapterChapter 1

Linear Equations in Two Variables

INTRODUCTION

1. General form of a linear equation in 2

variables x and y is ax + by + c = 0 where a,

b and c are real numbers or constants and a ≠

0, b ≠ 0.

2. Solution of the linear equation is a pair of

values of x and y which satisfies the equation.

A single linear equation in 2 variables has

infinite number of solutions.

For example, consider the equation 2x + y = 5.

The (x, y) pairs (0, 5), (− 1, 7), (1, 3) (2, 1) …

are all the solutions of this equation.

3. Linear equation

Graph of the linear equation

(i) x = a, a ≠ 0 Straight line parallel to Y-axis,

at a distance a units from it.

If a > 0 (positive), the line is on

the R.H.S. of Y axis. If a < 0 (a

is negative), the line lies on the

L.H.S. of Y axis.

(ii) x = 0 The line is the Y axis itself. In

other words, the equation of Y

axis is x = 0.

(iii) y = b, b ≠ 0 Straight line parallel to X axis,

at a distance b units from it.

If b > 0 (positive), the line is

above the X axis. If b < 0

(negative), then the line is

below the X axis.

(iv) y = 0 The line is the X axis itself. i.e.

The equation of X axis is y =

0.

4. Methods of Solution of simultaneous linear

equations.

(1) Graphical method. (2) Cramer’s Rule

using Determinants.

5. Determinant of second order : It is the

arrangement of 4 numbers in 2 rows and 2 columns and enclosed by two vertical bars.

Thus if a, b, c and d are 4 numbers, then

a b

c d = ad −−−− bc

↑ ↑

Determinant of

order 2

Value or the

expansion of

the determinant

6. Solution of the simultaneous linear

equations by Cramer's rule.

For using this rule, the constant term must be

on the R.H.S. If ax + by = e

cx + dy = f … (1) … (2)

a‚ b‚ c‚ d‚ e‚ f

: constants

then, by Cramer's rule x

Dx

= y

Dy

= 1

D

Where,

D = a b

c d =

the “Determinant

of the coefficients

of x and y

Dx = e b

f d =

D1 : Determinant

obtained by replacing

the 1st column of D

Dy = a e

c f =

D2 : Determinant

obtained by replacing the 2nd column of D

by the constants column

on R.H.S.

Std. X – Mathematics (Part −−−− I) 1.2 Linear Equations of Two Variables

7. Consistency in simultaneous Equations :

The simultaneous equations are said to be consistent if they have a unique (i.e. one and only one)

solution or a common solution. Thus the equations are satisfied simultaneously by the same values of the

variables (or the unknowns of the equation).

The simultaneous equations are said to be inconsistent if they do not have a solution or infinite

number of solutions.

Given : Simultaneous linear equations in x and y namely, a1 x + b1 y = c1; a2 x + b2 y = c2 The constant term must be on the R.H.S. To examine whether the given simultaneous linear equations are consistent or not.

Comparison of ratios Algebraic

Interpretation :

Equations have

Graphical Interpretation i.e.

Graph of Equations

a1

a2 ≠

b1

b2

Unique Solution Intersecting lines

Fig. 1.1 (a)

a1

a2 =

b1

b2 ≠

c1

c2

No Solution Parallel lines

Fig. 1.1 (b)

a1

a2 =

b1

b2 =

c1

c2

Infinitely many

Solutions

Overlapping lines or coincident

lines

Fig. 1.1 (c)

Practice Set 1.1

1. Complete the following activity to solve the simultaneous equations.

Ans. : 5x + 3y = 9 …(I)

2x − 3y = 12 …(II)

Let's add equations (I) and (II)

5x + 3y = 9

+ 2x − 3y = 12

7 x = 21

x = 21

7 x = 3

Place x = 3 in equation (I).

5 × 3 + 3y = 9

3y = 9 − 15

3y = −−−− 6

y = −−−−6

3

y = −−−−2

∴ Solution is (x, y) = ( 3 , −−−−2 )

2. Solve the following simultaneous equations.

(1) 3a + 5b = 26; a + 5b = 22

Ans. : 3a + 5b = 26 …(i)

a + 5b = 22 …(ii)

Subtracting equations (i) and (ii), we get

3a + 5b = 26

− a + 5b = 22

− − −

2a = 4

∴ a = 2

Substituting a = 2 in eq. (ii) we get

2 + 5b = 22

∴ 5b = 22 – 2 = 20

∴ b = 4

Solution is a = 2, b = 4

(x, y) = (2, 4)


(2) x + 7y = 10; 3x −−−− 2y = 7

Ans. : x + 7y = 10 …(i)

3x − 2y = 7 …(ii)

Multiplying both sides of equation (i) by − 3

−3x − 21y = − 30 …(iii)

Now, add equations (ii) and (iii)

3x − 2y = 7

+ − 3x − 21y = − 30

− 23y = − 23

23y = 23

y = 1

Substituting y =1 in equation (i)

x + 7 (1) = 10

x + 7 = 10

∴ x = 3

Solution is x = 3, y = 1

(x, y) = (3, 1)

(3) 2x −−−− 3y = 9; 2x + y = 13

Ans. : 2x − 3y = 9 …(i)

2x + y = 13 …(ii)

Subtracting equation (ii) from equation (i)

2x − 3y = 9

− 2x − y = −13

− 4y = − 4

∴ 4y = 4

∴ y = 1

Substituting y = 1 in equation (i)

2x − 3y = 9

∴ 2x − 3 (1) = 9

2x − 3 = 9

2x = 12

∴ x = 6


(x, y) = (6, 1)

(4) 5m −−−− 3n = 19; m −−−− 6n = −−−− 7

Ans. : 5m − 3n = 19 …(i)

m − 6n = − 7 …(ii)

Multiplying both sides of equation (i) by − 2.

We get, − 10m + 6n = − 38 …(iii)

Now add equations (iii) and (ii)

m − 6n = − 7

+ − 10m + 6n = − 38

− 9m = − 45

∴ 9m = 45

∴ m = 5

Substituting m = 5 in equation (ii)

m − 6n = − 7

∴ 5 − 6n = − 7

− 6n = − 12

∴ n = 2

Solution is m = 5, n = 2

∴ (m, n) = (5, 2)

(5) 5x + 2y = −−−−3; x + 5y = 4

Ans. : 5x + 2y = −3 …(i)

x + 5y = 4 …(ii)

Multiplying both sides of equation (ii) by − 5

−5x − 25 y = − 20 …(iii)

Add equations (i) and (iii)

5x + 2y = − 3

+ − 5x − 25y = − 20

− 23y = − 23

∴ 23y = 23

∴ y = 1

Substituting y = 1 in equation (ii)

x + 5 (1) = 4

x + 5 = 4

∴ x = − 1

Solution is x = − 1, y = 1

∴ (x, y) = (−1, 1)

(6) 1

3 x + y =

10

3 ; 2x +

1

4 y =

11

4

Ans. : 1

3 x + y =

10

3

∴ x + 3y

3 =

10

3

∴ x + 3y = 10 …(i)

2x + 1

4 y =

11

4

∴ 8x + y

4 =

11

4

∴ 8x + y = 11 …(ii)

Multiplying both sides of equation (i) by 8

8x + 24y = 80 …(iii)

Substract equation (ii) from (iii)

8x + 24y = 80

− − 8x − y = − 11

23y = 69

∴ y = 3


Substituting y = 3 in equation (i)

∴ x + 3 × 3 = 10

∴ x = 10 − 9

∴ x = 1


∴ (x, y) = (1, 3)

(7) 99x + 101y = 499; 101x + 99y = 501

Ans. : 99x + 101y = 499 …(i)

101x + 99y = 501 …(ii)

In the two equations above, the co-efficients

of x and y are interchanged. For solving such

equations, we will find two simple equations

by adding and substracting the given

equations. After solving these equations we

can find the solution.

Add the two given equations

99x + 101 y = 499

+ 101 x + 99y = 501

200x + 200y = 1000

Dividing both sides of the equation by 200.

x + y = 5 …(iii)

Now, subtract equations (ii) from (i)

99x + 101y = 499

− 101x − 99y = − 501

−2x + 2y = − 2

Dividing the equation by 2

− x + y = − 1 …(iv)

Now add equations (iii) and (iv)

x + y = 5

+ − x + y = − 1

2y = 4

∴ y = 2

Place this value in equation (iii)

x + y = 5

∴ x + 2 = 5

∴ x = 5 − 2 = 3

Hence, (x, y) = (3, 2) is the solution

(8) 49x −−−− 57y = 172; 57x −−−− 49y = 252

Ans. : 49x − 57y = 172 …(i)

57x − 49 y = 252 …(ii)

In the above two equations, the co-efficients

of x and y are interchanged. While solving

such equations we will find two simple

equations by adding and substracting the given

equations. After solving these equations, we

can easily find the solution.

Add the two given equations.

49x − 57y = 172

+ 57x − 49y = 252

106x − 106y = 424

Dividing both the sides of equation by 106

x − y = 4 …(iii)

Now, substract equation (ii) from (i)

49x − 57y = 172

− 57x + 49y = − 252

− 8x − 8y = − 80

Dividing the equation by 8

− x − y = − 10

∴ x + y = 10 …(iv)

Now, add equations (iii) and (iv)

x − y = 4

+ x + y = 10

2x = 14

∴ x = 7

Place this value in equation (iii)

x − y = 4

∴ 7 − y = 4

∴ y = 7 − 4

∴ y = 3

(x, y) = (7, 3) is the solution.

Practice Set 1.2

1. Complete the following table to draw graph of the equations :

(I) x + y = 3, (II) x −−−− y = 4

x + y = 3 x −−−− y = 4

x 3 −2 0 x 4 − 1 0

y 0 5 3 y 0 −5 − 4

(x, y) (3, 0) −2, 5 (0, 3) (x, y) 4, 0 −1, 5 (0, −4)


2. Solve the following simultaneous equations graphically.

(1) x + y = 6; x −−−− y = 4

Ans. : x + y = 6 x −−−− y = 4

x 5 0 3 6 x 4 2 5 0

y 1 6 3 0 y 0 − 2 1 − 4

(x, y) (5, 1) (0, 6) (3, 3) (6, 0) (x, y) (4, 0) (2, −2) (5, 1) (0, − 4)

x' x1 2 3 4 5 6 7 8 9 100 11 12 131-2-3-

1-

2-

3-

4-

5-

1

2

3

4

5

6

7

y

y'

(0, )-4

(4,0)

(6,0)

(5,1)

(3,3)

(0,6)

xy

=4

-

(2, 2)-

x+

y=

6

Scale:1cm = 1unitfor both the axes

Fig. 1.2

(2) x + y = 5; x −−−− y = 3

Ans. : x + y = 5

x 5 0 4 2

y 0 5 1 3

(x, y) (5, 0) (0, 5) (4, 1) (2, 3)

x −−−− y = 3

x 0 3 4 1

y − 3 0 1 − 2

(x, y) (0, − 3) (3, 0) (4, 1) (1, − 2)


x' x1 2 3 4 5 6 7 8 9 100 11 121-2-3-

1-

2-

3-

4-

5-

1

2

3

4

5

6

7

y

(3,0) (5,0)

(0,5)

xy

=3

-

4-

x + y = 5

y'

(0, )-3

(4,1)


Fig. 1.3

(3) x + y = 0; 2x −−−− y = 9

Ans. : x + y = 0 2x −−−− y = 9

x 1 3 5 2 x 0 2 3 4

y − 1 − 3 − 5 − 2 y −9 −5 −3 −1

(x, y) (1, −1) (3, −3) (5, −5) (2, −2) (x, y) (0, − 9) (2, − 5) (3, − 3) (4, − 1)

1

2

3

1 2 3 4 5 6 7 8 9 xx' 1-2-3-4-5-6-7-

1-

2-

3-

4-

5-

6-

7-

8-

9-

10-

0

(0,0)

(1, )-1

(3, )-3

(2, )-5(5, )-5

(0, )-9

y'

y2x y = 9-x + y = 0

7


Fig. 1.4


(4) 3x −−−− y = 2; 2x −−−− y = 3 x 0 −1 2 1 x 3 1 −1 0 y −2 −5 4 1 y 3 − 1 −5 − 3

(x, y) (0, − 2) (−1, −5) (2, 4) (1, 1) (x, y) (3, 3) (1, − 1) (−1, − 5) (0, − 3)

3 4 5 6 7 8 92-3- 1-

5-

1

2

3

4

104-5-6-x' x

y

y'

6-

7-

8-

102

2-

3-

4-

1-

(3, )3

(2, )42x y = 3-

3x

y = 2

-

(1, )1

(1, )-1

(0, )-3

( , )- -1 5

(0, )-2


Fig. 1.5

(5) 3x −−−− 4y = −−−− 7; 5x −−−− 2y = 0 x 3 1 −1 −5 x 2 −2 0 −1 y 4 2.5 1 −2 y 5 −5 0 − 2.5

(x, y) (3, 4) (1, 2.5) (−1, 1) (−5, −2) (x, y) (2, 5) (−2, −5) (0, 0) (−1, −2.5)

2 3 4 5 6 7 8 92- 1-

2-

3-

4-

5-

1

2

3

4

5

6

7

( , )- -2 5

104-5-6-7-x' x

y

y'

(2, )5

(3,4)

(1, 2. )5

5x

2y

=0

-

3x4y =

-

7-

(0,0)


3-

( , )- - .1 2 5

( , )-1 1

5x

2y

=0

-

( , )- -5 2

Fig. 1.6


(6) 2x −−−− 3y = 4; 3y −−−− x = 4

Ans. : 2x −−−− 3y = 4 3y −−−− x = 4

x 5 2 8 −2.5 x 8 2 −4 −1

y 2 0 4 −3 y 4 2 0 1

(x, y) (5, 2) (2, 0) (8, 4) (−2.5, − 3) (x, y) (8, 4) (2, 2) (−4, 0) (−1, 1)

1

2

3

4

5

6

0

y

2-

3-

4-

1-

5-

2 3 4 5 6 7 8 91 x10 11 12

6-

2-3- 1-4-

x'

(2, )0

(5,2)

(8,4)

(2,2)

2x3y = 4

-

3yx = 4

-

y'


(-1, 1)

( 2.5, 3)- -

(-4,0)

Fig. 1.7

Exercise 1.2

Q. Solve the following simultaneous equations using Graphical method.

(i) x + y = 8; x −−−− y = 2

Ans. :

(ii) 3x + 4y + 5 = 0; y = x + 4

Ans. :

Mathematics Part - I Algebra

Publisher : Nirali Prakashan Author : A. A. Tathe

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