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      Class 9 Maths

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      • Class 09
      • Class 9 Maths
      CoursesClass 09MathsClass 9 Maths
      • 1.Number System
        14
        • Lecture1.1
          Numbers and its Classification, Rational Number and its Representation on the Number Line 50 min
        • Lecture1.2
          Decimal Representation of Rational Number and Conversion of Decimal Number into Rational Number 01 hour
        • Lecture1.3
          Representation of Decimal Number on Number Line, Finding Irrational Numbers Between Given Two Numbers 46 min
        • Lecture1.4
          Representing Irrational Number on the Number Line 41 min
        • Lecture1.5
          Rationalisation 54 min
        • Lecture1.6
          Introduction to Exponents, Laws of Exponents, BODMAS Rule 41 min
        • Lecture1.7
          Integral Exponents of Real Numbers 57 min
        • Lecture1.8
          Integral Exponents of Real Numbers and Solving Equation for x 01 hour
        • Lecture1.9
          Rational Exponents of Real Numbers 52 min
        • Lecture1.10
          Examples Based on Rational Exponents of Real Numbers 30 min
        • Lecture1.11
          Chapter Notes – Number System
        • Lecture1.12
          NCERT Solutions – Number System Exercise 1.1 – 1.6
        • Lecture1.13
          R D Sharma Solutions Number System
        • Lecture1.14
          Revision Notes – Number System
      • 2.Polynomials
        10
        • Lecture2.1
          Introduction to Polynomials and Some Basics Questions 58 min
        • Lecture2.2
          Value of Polynomial, Zeroes of Polynomial and Relationship between Zeroes and Coefficient of Linear Equation 01 hour
        • Lecture2.3
          Division of Polynomials and Division Algorithm and Remainder Theorem 51 min
        • Lecture2.4
          Factorising by Spliting Method 01 hour
        • Lecture2.5
          Factorisation By Different Methods and Factor Theorem 27 min
        • Lecture2.6
          Factorisation of Polynomial by Using Algebric Identities 01 hour
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.5
        • Lecture2.9
          R S Aggarwal Solutions Polynomials
        • Lecture2.10
          Revision Notes Polynomials
      • 3.Coordinate Geometry
        8
        • Lecture3.1
          Representation of Position on Number Line, Finding the location by Using the Axis 01 hour
        • Lecture3.2
          Quadrant, Coordinates of Point on Axis, Distance from Axis 01 hour
        • Lecture3.3
          Finding the Area of Triangle and Rectangle and Mirror Image of Point On Axis 57 min
        • Lecture3.4
          Chapter Notes – Coordinate Geometry
        • Lecture3.5
          NCERT Solutions – Coordinate Geometry Exercise 3.1 – 3.3
        • Lecture3.6
          R D Sharma Solutions Coordinate Geometry
        • Lecture3.7
          R S Aggarwal Solutions Coordinate Geometry
        • Lecture3.8
          Revision Notes Coordinate Geometry
      • 4.Linear Equations
        8
        • Lecture4.1
          Introduction of Linear Equation in Two Variables and Graphical Representation 01 hour
        • Lecture4.2
          Example to find the value of Constant and Graph of Linear Equation in Two Variables 57 min
        • Lecture4.3
          Equations of lines Parallel to the x-axis and y-axis 46 min
        • Lecture4.4
          Chapter Notes – Linear Equations
        • Lecture4.5
          NCERT Solutions – Linear Equations Exercise 4.1 – 4.4
        • Lecture4.6
          R D Sharma Solutions Linear Equations
        • Lecture4.7
          R S Aggarwal Solutions Linear Equations
        • Lecture4.8
          Revision Notes Linear Equations
      • 5.Euclid's Geometry
        7
        • Lecture5.1
          Euclid’s Definitions and Euclid’s Axioms 32 min
        • Lecture5.2
          Eulcid’s Postulates and Playfair’s Axiom 37 min
        • Lecture5.3
          Chapter Notes – Euclid’s Geometry
        • Lecture5.4
          NCERT Solutions – Euclid’s Geometry Exercise 5.1 5.2
        • Lecture5.5
          R D Sharma Solutions Euclid’s Geometry
        • Lecture5.6
          R S Aggarwal Solutions Euclid’s Geometry
        • Lecture5.7
          Revision Notes Euclid’s Geometry
      • 6.Lines and Angles
        10
        • Lecture6.1
          Introduction to Lines and Angles, Types of Angle and Axioms 01 hour
        • Lecture6.2
          Vertically opposite Angles and Angle bisector 46 min
        • Lecture6.3
          Angle made by a transversal with two Lines and Parallel Lines 01 hour
        • Lecture6.4
          Lines Parallel to same Line; Exterior angle of a triangle; Angle Sum property of a Triangle 52 min
        • Lecture6.5
          Angle Sum Property of Triangle Including Interior and Exterior Angles 01 hour
        • Lecture6.6
          Chapter Notes – Lines and Angles
        • Lecture6.7
          NCERT Solutions – Lines and Angles Exercise 6.1 – 6.3
        • Lecture6.8
          R D Sharma Solutions Lines and Angles
        • Lecture6.9
          R S Aggarwal Solutions Lines and Angles
        • Lecture6.10
          Revision Notes Lines and Angles
      • 7.Triangles
        11
        • Lecture7.1
          Introduction, Types of Triangle, Altitude & Median of Triangle, Congruences, SAS criterion of Congruence 42 min
        • Lecture7.2
          Questions Based on SAS Criterion 01 hour
        • Lecture7.3
          ASA Criterion of Congruence and Its based Questions 42 min
        • Lecture7.4
          Questions Based on AAS and SSS Criterion 29 min
        • Lecture7.5
          RHS Criterion of Congruence and Its based Questions, Questions Based on Multiple Criterion 01 hour
        • Lecture7.6
          Inequalities of Triangle 42 min
        • Lecture7.7
          Inequalities of Triangle cont., Distance Between a line and The Point, Miscellaneous Questions 48 min
        • Lecture7.8
          Chapter Notes – Triangles
        • Lecture7.9
          NCERT Solutions – Triangles Exercise 7.1 – 7.5
        • Lecture7.10
          R D Sharma Solutions Triangles
        • Lecture7.11
          Revision Notes Triangles
      • 8.Quadrilaterals
        13
        • Lecture8.1
          Congruence of Triangles- SAS Congruence Criteria, ASA Congruence Criteria, SSS Congruence Criteria, RHS Congruence Criteria; Types of Quadrilateral- Parallelogram, Rectangle, Rhombus, Square, Trapezium, Kite 01 hour
        • Lecture8.2
          Theorem 1; Properties of parallelogram- Theorem 2, Theorem 3; 01 hour
        • Lecture8.3
          Theorem 4; Theorem 5; 24 min
        • Lecture8.4
          Conditions for a Quadrilateral to be a parallelogram- Theorem 6 01 hour
        • Lecture8.5
          Theorem 8 01 hour
        • Lecture8.6
          Theorem 9 51 min
        • Lecture8.7
          Theorem 9(Mid-point Theorem); 01 hour
        • Lecture8.8
          Theorem 10 27 min
        • Lecture8.9
          Chapter Notes – Quadrilaterals
        • Lecture8.10
          NCERT Solutions – Quadrilaterals Exercise 8.1 8.2
        • Lecture8.11
          R D Sharma Solutions Quadrilaterals
        • Lecture8.12
          R S Aggarwal Solutions Quadrilaterals
        • Lecture8.13
          Revision Notes Quadrilaterals
      • 9.Area of Parallelogram
        11
        • Lecture9.1
          Parallelograms on the Same Base and between the Same Parallels; A Diagonal of a parallelogram divides it into two triangles of Equal Area 57 min
        • Lecture9.2
          Questions based on parallelogram and Triangle are on same base & parallel lines and Area of Triangle 01 hour
        • Lecture9.3
          Triangles on the same Base and between the same Parallels 56 min
        • Lecture9.4
          Miscellaneous Questions Based on Triangles on the same Base and between the same Parallels 54 min
        • Lecture9.5
          Sums Based on Median of a Triangle 52 min
        • Lecture9.6
          Two Triangles on the same Base and equal Area lie b/w the same parallel 01 hour
        • Lecture9.7
          Summary of All Concepts and Miscellaneous Questions 56 min
        • Lecture9.8
          Chapter Notes – Area of Parallelogram
        • Lecture9.9
          NCERT Solutions – Area of Parallelogram Exercise 9.1 – 9.4
        • Lecture9.10
          R D Sharma Solutions Area of Parallelogram
        • Lecture9.11
          Revision Notes Area of Parallelogram
      • 10.Constructions
        7
        • Lecture10.1
          Construction of Perpendicular Bisector of a Line Segment, Different Angles and Angle bisector 42 min
        • Lecture10.2
          Construction of Different types of Triangle 43 min
        • Lecture10.3
          Chapter Notes – Constructions
        • Lecture10.4
          NCERT Solutions – Constructions Exercise
        • Lecture10.5
          R D Sharma Solutions Constructions
        • Lecture10.6
          R S Aggarwal Solutions Constructions
        • Lecture10.7
          Revision Notes Constructions
      • 11.Circles
        11
        • Lecture11.1
          Introduction, Important Terms related to circle, Theorem-1, 2, 3 and 4 01 hour
        • Lecture11.2
          Problems Solving 48 min
        • Lecture11.3
          Problems Based on Parallel Chords, Common Chord of Circles 01 hour
        • Lecture11.4
          Problems Based on Concentric Circles, Triangle in Circle 41 min
        • Lecture11.5
          Equal Chords and Their Distance, Theorem-5, 6, 7, 8 01 hour
        • Lecture11.6
          Circle Through Points, Theorem-9, Arc of a Circle, Theorem-10 01 hour
        • Lecture11.7
          Chapter Notes – Circles
        • Lecture11.8
          NCERT Solutions – Circles Exercise
        • Lecture11.9
          R D Sharma Solutions Circles
        • Lecture11.10
          R S Aggarwal Solutions Circles
        • Lecture11.11
          Revision Notes Circles
      • 12.Heron's Formula
        8
        • Lecture12.1
          Introduction of Heron’s Formula; Area of Different Triangles 01 hour
        • Lecture12.2
          Area of Triangles Using Heron’s Formula (Type-1 level-1 and 2) 53 min
        • Lecture12.3
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-1) 32 min
        • Lecture12.4
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-2) 38 min
        • Lecture12.5
          Chapter Notes – Heron’s Formula
        • Lecture12.6
          NCERT Solutions – Heron’s Formula Exercise 12.1 12.2
        • Lecture12.7
          R D Sharma Solutions Heron’s Formula
        • Lecture12.8
          Revision Notes Heron’s Formula
      • 13.Surface Area and Volume
        16
        • Lecture13.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere 40 min
        • Lecture13.2
          Type-1: Surface Area and Volume of Cube and Cuboid 27 min
        • Lecture13.3
          Level-2, Question based on Cube and Cuboid, Level-1 Questions based on Cylinder 28 min
        • Lecture13.4
          Level-2 Questions based on Cylinder, Type-3 Surface Area and Volume of Right Circular Cone (Level-1) 25 min
        • Lecture13.5
          Level-2 Questions based on Right Circular Cone, Level-1& 2 Questions Based on Sphere & Hemisphere 27 min
        • Lecture13.6
          Type-5 Short Answer Types Questions, 12 min
        • Lecture13.7
          Chapter Notes – Surface Area and Volume of a Cuboid and Cube
        • Lecture13.8
          Chapter Notes – Surface Area and Volume of A Right Circular Cylinder
        • Lecture13.9
          Chapter Notes – Surface Area ad Volume of A Right Circular Cone
        • Lecture13.10
          Chapter Notes – Surface Area ad Volume of A Sphere
        • Lecture13.11
          NCERT Solutions – Surface Area and Volume Exercise 13.1 – 13.9
        • Lecture13.12
          R D Sharma Solutions Surface Areas And Volumes of A cuboid and A cube
        • Lecture13.13
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cylinder
        • Lecture13.14
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cone
        • Lecture13.15
          R D Sharma Solutions Surface Areas And Volumes of A Sphere
        • Lecture13.16
          Revision Notes Surface Areas And Volumes
      • 14.Statistics
        15
        • Lecture14.1
          Data; Presentation of Data; Understanding Basic terms; 01 hour
        • Lecture14.2
          Questions; 52 min
        • Lecture14.3
          Bar Graph, Histogram, Frequency Polygon; 01 hour
        • Lecture14.4
          Histogram for Grouped Frequency Distribution (Class interval in Inclusive form), Frequency Polygon, Measures of Central Tendency, 01 hour
        • Lecture14.5
          Examples Based on Mean of Ungrouped Data 56 min
        • Lecture14.6
          Direct Method of Finding Mean of Ungrouped Data, Median of Cont. Freq. Distribution (Grouped Data) 01 hour
        • Lecture14.7
          Mean; Mean for an ungrouped freq. Dist. Table 25 min
        • Lecture14.8
          Chapter Notes – Statistics – Tabular Representation of Statistical Data
        • Lecture14.9
          Chapter Notes – Statistics – Graphical Representation of Statistical Data
        • Lecture14.10
          Chapter Notes – Statistics – Measures of Central Tendency
        • Lecture14.11
          NCERT Solutions – Statistics Exercise 14.1 – 14.4
        • Lecture14.12
          R D Sharma Solutions Tabular Representation of Statistical Data
        • Lecture14.13
          R D Sharma Solutions Graphical Representation of Statistical Data
        • Lecture14.14
          R S Aggarwal Solutions Statistics
        • Lecture14.15
          Revision Notes Statistics
      • 15.Probability
        8
        • Lecture15.1
          Activities; Understanding term Probability; Experiments; 44 min
        • Lecture15.2
          Event; Probability; Empirical Probability- Tossing a coin; 31 min
        • Lecture15.3
          Rolling a dice; Miscellaneous Example; 27 min
        • Lecture15.4
          Chapter Notes – Probability
        • Lecture15.5
          NCERT Solutions – Probability Exercise 15.1
        • Lecture15.6
          R D Sharma Solutions Probability
        • Lecture15.7
          R S Aggarwal Solutions Probability
        • Lecture15.8
          Revision Notes Probability

        NCERT Solutions – Number System Exercise 1.1 – 1.6

        Exercise 1.1

        Q.1     Is zero a rational number ? Can you write it in the form pq, where p and q are integers and q≠0?
        Sol.

        Yes, zero is a rational number.
        Zero can be written in any of the following forms :
        01,0−1,02,0−2 and so on.
        Thus, 0 can be written as pq, where p = 0 and q is any non- zero integer.
        Hence , 0 is a rational number.


        Q.2      Find six rational numbers between 3 and 4.
        Sol.

        We know that between two rational numbers x and y, such that x < y, there is a rational number x+y2. That is , x<x+y2<y
        There are two rational numbers 3 and 4 , such that 3 < 4 , a rational number between 3 and 4 is 12(3+4)i.e.,72
        Therefore 3<72<4
        Now ,two rational numbers 3 and72 such that, a rational number between 3 and 72is
        12(3+72)=12×6+72=134
        There are two rational numbers72 and 4 , a rational number between 72 and 4 is
        12(72+4)=12×7+82=154
        Therefore 3<134<72<154<4
        There are two rational numbers 3 and 134  ,a rational number between 3 and 134is
        12(3+134)=12×12+134=258
        There are two rational numbers154 and 4 , a rational number between 154 and 4 is
        12(154+4)=12×15+164=318
        There are two rational numbers 318 and 4 , a rational number between 318 and 4 is
        12(318+4)=12×31+328=6316
        Therefore 3<258<134<72<154<318<6316<4
        Hence, six rational numbers between 3 and 4 are :
        258,134,72,154,318and6316
        ALTERNATIVE METHOD
        Since we want 6 rational number between 3 and 4, so we write
        3=3×71×7=217 and4=4×71×7=287
        We know that 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28
        ⇒ 217<227<237<247 <257<267<277<287
        Hence, six rational numbers between 3=217and4=287are
        227,237,247,257,267 and277


        Q.3      Find five rational numbers between 35and45.
        Sol.

        Since we want 5 rational numbers between 35and45, so we write.
        35=3×65×6=1830 and 45=4×65×6=2430
        We know that 18 < 19 < 20 < 21 < 22 < 23 < 24
        ⇒    1830<1930<2030<2130<2230 <2330<2430
        Hence , 5 rational numbers between 35=1830and45=2430are:
        1930,2030,2130,2230,2330and2430


        Q.4     State whether the following statements are true or false. Give reasons for your answers.
                   
        (i) Every natural number is a whole number.
                        (ii) Every integer is a whole number.
                        (iii) Every rational number is a whole number.
        Sol.

        (i) True : Every natural number lies in the collection of whole numbers.
        (ii) False : -3 is not a whole number.
        (iii) False 35 is not a whole number.

        Exercise 1.2

        Q.1      State whether the following statements are true or false. Justify your answers.
                    (i) Every irrational number is a real number.
                    (ii) Every point on the number line is of the form m−−√, where m is a natural number.
                    (iii) Every real number is an irrational number.

        Sol.

        (i)     Every irrational number is a real number – True
                 Justification : – A real number is either rational or irrational.
        (ii)   Every point on the number line is of the form m−−√  – False
                 Justification : – Numbers  of other types also lie on the number line.
        (iii)  Every real number is an irrational number – False
                  Justification : – Rational numbers are also real numbers.


        Q.2     Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational numbers.
        Sol.

        No, the square roots of all positive integers are not irrational.
        Example : –  4 is a positive integer but 4–√=2 is a  natural number.


        Q.3       Show how 5–√ can be represented on the number line.
        Sol.

        We shall now show how to represent 5–√ on the number line.
        We first represent 5–√ on the number line l. We construct a right – angled Δ OAB, right – angled at A such that OA = 2 and AB = 1  unit (see figure)
        ns1
        Then,  OB=OA2+AB2−−−−−−−−−−√=4+1−−−−√=5–√
        Now, we cut off a length OC=OB=5–√ on the number line.
        Then the point C represents the irrational number 5–√.


        Q.4      Classroom activity (Constructing the ‘square root spiral’) : Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1P2 perpendicular to OP1 of unit  length (see figure). Now draw a line segment P2P3 perpendicular to OP2. Then draw a line segment P3P4 perpendicular to OP3. Continuing in the manner, you can get the line segment Pn−1Pn by drawing a line segment of unit length perpendicular to OPn−1. In this manner, you will have created the points P2,P3....Pn,.... and joined them to create a beautiful spiral depicting 2–√,3–√,4–√,.....

        ns2Sol.        Classroom activity – Do as directed.

        Exercise 1.3

        Q.1     Write the following in decimal form and say what kind of decimal expansion each has :
                   
        (i) 36100                           
                   (ii) 111            
                   
        (iii) 418
                   
        (iv) 313                          
                   
        (v) 211             
                   (vi) 329400
        Sol. 

        (i)    36100 in decimal form –
                 36100=0.36,, terminating decimal.
        (ii)   By long division,we have

        1.3
                                    Therefore 111=0.090909.....=0.09¯¯¯¯¯, non- terminating and repeating decimal.
                          (iii)   418=4×8+18=338 By long division , we have
        1.4
                                       Therefore 338=4.125, terminating decimal.
                               (iv)  By long division, we have

        1.5
                                           Therefore 313=0.23076923..=0.230769¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯,
                                           non- terminating and repeating decimal.  
                                   (v)   By long division we have1.6

                                            Therefore 211=0.181818..=0.18¯¯¯¯¯,
                                            non- terminating and repeating decimal.                 
                                    (vi)  By long division, we have

        1.7
                                                Therefore 329400=0.8225, terminating decimal.


        Q.2         You know that 17=0.142857¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ . Can you predict what the decimal expansions of 27,37,47,57,67 are , without actually doing the long division? If so, how?
        Sol.

        Yes. All of the above will have repeating decimals which are permutations of 1, 4, 2, 8, 5, 7.
        For example, here

        1.8

        To find 27, locate when the remainder becomes 2 and respective quotient (here it is 2), then write the new quotient beginning from there (the arrows in the above division) using the repeating digits 1, 4, 2, 8, 5, 7. Therefore 27=0.285714¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
        Similarly,      37=0.428571¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯,47=0.571428¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
        57=0.714285¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯,67=0.857142¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯


        Q.3      Express the following in the form pq, where p and q are integers and q≠0..
                    
        (i) 0.6¯¯¯                
                    (ii) 0.47¯¯¯              
                    (iii) 0.001¯¯¯¯¯¯¯¯
        Sol.

        (i)    Let x=0.6¯¯¯
                Then, x = 0.666 … … (1) Here , we have only one repeating digit.
                So, we multiply both sides of (1) by 10 to get
                10x = 6.66…. … (2)
                Subtracting (1) from (2) we get
                10x – x = (6.66…) – (0.66…) ⇒ 9x = 6 ⇒ x=23
                 Hence, 0.6¯¯¯=23
        (ii)   Let x=0.47¯¯¯
                 Clearly, there is just one digit
                 on the right side of the decimal point which is without bar.
                 So, we multiply both sides by 10.
                 So that only the repeating decimal is left on the right side of the decimal point.
                 Therefore 10x=4.7¯¯¯
                 ⇒ 10x=4+0.7¯¯¯
                 ⇒ 10x=4+79
                 For example : 0.7¯¯¯=79,0.35¯¯¯¯¯=3599 etc.
                 ⇒ 10x=4×9+79 ⇒ 10x=439
                 ⇒ x=4390
                 Hence 0.47¯¯¯=4390
                 ALITER
                 Let x=0.47¯¯¯=0.4777…
                 Therefore 10x = 4.777… … (1)
                  and 100x = 47.777… … (2)
                  Subtracting (1) from (2) we get
                  100 x – 10 x = (47.777 … ) – (4.777… )
                  ⇒ 90x = 43 ⇒ x=4390
                  Hence, 0.47¯¯¯=4390
        (iii)   Let x=0.001¯¯¯¯¯¯¯¯
                  ⇒ x = 0.001001001 …. … (1)
                  Here , we have three repeating digits after the decimal point. So we multiply (1) by 103=1000,
                  we get
                  1000x = 1.001001… … (2)
                  Subtracting (1) from (2) we get –
                  1000 x – x = (1.001001…) – (0.001001…)
                  ⇒ 999x = 1 ⇒ x=1999
                  Hence, 0.001¯¯¯¯¯¯¯¯=1999


        Q.4      Express 0.99999… in the form pq. Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
        Sol.

        Let x = 0.9999 … … (1)
        Here , we have only one repeating digit. So, we multiply both sides of (1) by 10 to get.
        10x = 9.999… … (2)
        Subtracting (1) from (2) we get  
        10 x – x = (9.999…) – (0.999…)
        ⇒ 9x = 9 ⇒ x = 1
        Hence, 0.9999 … = 1
        Since , 0.9999… goes on forever. So, there is no gap between 1 and 0.9999… and hence they are equal.


        Q.5     What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 117? Perform the division to check your answer.
        Sol.         We have

        Q.5_

        Thus, 117=0.588235294117647¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
        Therefore the maximum number of digits in the quotient while computing 117 are 15.


        Q.6       Look at several examples of rational numbers in the form pq(q≠0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
        Sol. 

        Consider several rational numbers in the form pq(q≠0), where p and q are integers with common factors other than 1 and having terminating decimal representation.
        Let the various such rational numbers be 12,14,78,3725,
        8125,1720,3116etc.
        In all cases, we think of the natural number which when multiplied  by their respective denominators gives 10 or a power of 10.
        12=1×52×5=510=0.5 [Since 2 × 5 = 10]
        14=1×254×25=25100=0.25 [Since 4 ×25 = 100]
        78=7×1258×125=8751000=0.875 [Since 8 × 125 = 1000]
        3725=37×425×4=128100=1.28 [Since 25 × 4 = 100]
        8125=8×8125×8=641000=0.064 [Since 125 × 8 = 1000]
        1720=17×520×5=85100=0.85 [Since 20 × 5 = 100]
        3116=31×62516×625=1937510000=1.9375 [Since 16 × 625 = 10000]
        We have seen that those rational numbers whose denominators when multiplied by a suitable integer produce a power of 10 are expressible in the finite decimal form. But this can always be done only when the denominator of the given rational number has either 2 or 5 or both of them as the only prime factors. Thus, we obtain the following property :
        If the denominator of a rational number in standard form has no prime factors other than 2 or 5, then and only then it can be represented as a terminating decimal.


        Q.7      Write three numbers whose decimal expansions are non- terminating non- recurring.
        Sol.

        Three numbers whose decimal representations are non- terminating non- recurring are
        2–√,3–√and5–√ or we can say 0.100100010001…, 0.20200200020002… and .003000300003.


        Q.8       Find three different irrational numbers between the rational numbers 57and911.
        Sol.          We have

        1.10

        Therefore 57=0.714285¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯and911=0.81¯¯¯¯¯
        Thus , three different irrational numbers between 57and911 are.
        0.75075007500075000075…. , 0.767076700767000… and 0.80800800080000…

         


        Q.9          Classify the following numbers as rational or irrational :
                        
        (i) 23−−√                      
                        (ii) 225−−−√                            
                        (iii) 0.3796
                        
        (iv) 7.478478 …                          
                        (v) 1.101001000100001…
        Sol.

        (i) 23−−√ is an irrational number as 23 is not a perfect square.
        (ii) 225−−−√=3×3×5×5−−−−−−−−−−−√ ns3
        =3×5
        =15
        Thus, 15 is a rational number.
        (iii) 0.3796 is a rational number as it is terminating decimal.
        (iv) 7.478478 …. is non- terminating but repeating , so, it is a rational number.
        (v) 1.101001000100001 … is non – terminating and non- repeating so, it is an irrational number.

        Exercise 1.4

        Q.1       Visualise 3.765 on the number line, using successive magnification.
        Sol.

        We know that 3.765 lies between 3 and 4, that is, in the interval [3, 4] to have a rough idea where it is. Now, we divide the interval [3,4] into 10 equal parts and look at [3.7, 3.8] through a magnifying glass and realize that 3.765 lies between 3.7 and 3.8 (see figure (i)). Now, we imagine that each of the new  intervals [3.1, 3.2], [3.2, 3.3], … [3.9, 4] has been sub divided into 10 equal parts. As before , we can now visualize through the magnify glass that 3.765  lies in the interval [3.76, 3.77] (see figure (ii)).

        ns4

        So, we have seen that it is possible by sufficient successive magnifications to visualize the position (or representation of a real number with a terminating decimal expansion on the number line. Let us now try and visualize the position (or representation) of a real number with a non- terminating recurring  decimal expansion on the number line. We can look at appropriate intervals through a magnifying glass and by successive magnifications visualize the  position of the number on the number line.


        Q.2       Visualise 4.26¯¯¯¯¯ on the number line, upto 4 decimal places.
        Sol.

        We proceed by successive magnifications, and successively decrease the lengths of the intervals in which 4.26¯¯¯¯¯ is located. 4.26¯¯¯¯¯  is located in the interval [4, 5] of length 1. We further locate 4.26¯¯¯¯¯ in the interval [4.2, 4.3] of length 0.1. To get more accurate visualization of the representation, we divide even this interval into 10 equal parts and use a magnifying glass to visualize that 4.26¯¯¯¯¯ lies in the interval [4.26, 4.27] of length 0.01. To visualize 4.26¯¯¯¯¯ in an interval of length 0.001, we again divide each of the new intervals  into 10 equal parts and visualize the representation of 4.26¯¯¯¯¯  in the interval [4.262, 4.263] of length 0.001. Notice that 4.26¯¯¯¯¯ is located closer to 4.263 than to 4.262.
        Note : – We can proceed endlessly in this manner and simultaneously imagining the decrease in the length of the interval in which  4.26¯¯¯¯¯ is located.

        ns5

        Exercise 1.5

        Q.1      Classify the following numbers as rational or irrational :
                       (i) 2−5–√                                      
                    (ii) (3+23−−√)−23−−√

                       (iii) 27√77√
                    (iv) 12√

                       (v) 2π
        Sol.

        (i) 2−5–√ is an irrational number being a difference between a rational and an irrational.
        (ii) (3+23−−√)−23−−√=3+23−−√−23−−√=3, which is a rational number.
        (iii) 27√77√=27, which is a rational number.
        (iv) −12√ is an irrational number being the quotient of a rational and an irrational number .
        (v) 2π is irrational being the product of rational and irrational number .


        Q.2      Simplify each of the following expressions :
                     
        (i) (3+3–√)(2+2–√)
                     
        (ii) (3+3–√)(3−3–√)
                     
        (iii) (5–√+2–√)2
                     
        (iv) (5–√−2–√)(5–√+2–√)
        Sol.

        (i)   (3+3–√)(2+2–√)
                =3×2+32–√+23–√+2–√×3–√
                =6+32–√+23–√+6–√
        (ii)  (3+3–√)(3−3–√)
                = (3)2−(3–√)2
                =9−3=6
        (iii) (5–√+2–√)2
                 (5–√)2+25–√2–√+(2–√)2
                 5+210−−√+2=7+210−−√
        (iv)   (5–√−2–√)(5–√+2–√)
                 = (5–√)2−(2–√)2
                 =5−2=3


        Q.3      Recall , π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is π=cd. This seems to  contradict the fact that π is irrational. How will you resolve this contradiction?
        Sol.         There is no contradiction as either c or d irrational and hence π is an irrational number.


        Q.4      Represent 9.3−−−√ on the number line.
        Sol.

        Mark the distance 9.3 units from a fixed point A on a given line to obtain a point B such that AB = 9.3 units. From B, mark a distance of 1 unit and mark the new point as C. Find the mid- point of AC and mark that point as O. Draw a semi-circle with centre O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semicircle at D.

        3
        Then BD=9.3−−−√. To represent 9.3−−−√ on the number line. Let us treat the line BC as the number line, with B as zero, C as 1 and so on. Draw an arc with centre B and radius BD, which intersects the number line in E. Then E represent 9.3−−−√


         

        Q.5      Rationalise the denominators of the following :
                    (i) 17√
                    (ii) 17√−6√

                       (iii) 15√+2√
                    (iv) 17√−2

        Sol.

        (i)    17√=1×7√7√×7√=7√7
        (ii)   17√−6√ =17√−6√×7√+6√7√+6√ =7√+6√(7√)2−(6√)2
                 =7√+6√7−6=7–√+6–√
        (iii)  15√+2√ =15√+2√×5√−2√5√−2√ =5√−2√(5√)2−(2√)2
                  =5√−2√5−2=5√−2√3
        (iv) 17√−2 =17√−2×7√+27√+2
                =7√+2(7√)2−(2)2
                =7√+27−4 =7√+23

        Exercise 1.6

        Q.1     Find
                      (i) 6412   (ii) 3215   (iii)12513
        Sol.

        (i)    6412 =(8×8)12 =(82)12 =82×12=81=8
        (ii)   3215 =(2×2×2×2×2)15=(25)15 =25×15=21=2
        (iii)  12513 =(5×5×5)13=(53)13 =53×13=51=5


        Q.2       Find :
                     
        (i) 932   (ii) 3225   (iii) 1634   (iv) 125−13
        Sol.

        (i)    932 =(3×3)32=(32)32=32×32 =33=3×3×3=27
        (ii)   3225=(2×2×2×2×2)25=(25)25 25×25=22=2×2=4
        (iii)  1634=(2×2×2×2)34=(24)34 =24×34=23=2×2×2=8
        (iv)    125−13=112513 =1(5×5×5)13=15


        Q.3      Simplify :
                    
        (i) 223.215    
                    
        (ii) (133)7
                    
        (iii) 11121114
                    
        (iv) 712.812
        Sol.

        (i)    223.215=223+15=210+315=21315
        (ii)   (132)7=(1)732×7=1321
        (iii)  11121114=1112−14=1114
        (iv)   712.812=(7×8)12=5612 = (4×14)12=(56)12

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