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      Class 10 Maths

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      • Class 10 Maths
      CoursesClass 10MathsClass 10 Maths
      • 01. Real Numbers
        9
        • Lecture1.1
          Real Numbers and Fundamental Theorem of Arithmetic 59 min
        • Lecture1.2
          Divisibility and Euclid’s Division Lemma 49 min
        • Lecture1.3
          Finding HCF and LCM Using Fundamental Theorem of Arithmetic 02 hour
        • Lecture1.4
          Miscellaneous Questions 31 min
        • Lecture1.5
          Chapter Notes – Real Numbers
        • Lecture1.6
          NCERT Solutions – Real Numbers Exercise 1.1 – 1.4
        • Lecture1.7
          Revision Notes Real Numbers
        • Lecture1.8
          R S Aggarwal Real Numbers
        • Lecture1.9
          R D Sharma Real Numbers
      • 02. Polynomials
        11
        • Lecture2.1
          Introduction to Polynomials and Its Zeroes 44 min
        • Lecture2.2
          Finding Zeroes of Quadratic Polynomial 01 hour
        • Lecture2.3
          Relationship b/w the Zeroes and coefficients of a Polynomial 01 hour
        • Lecture2.4
          Cubic Polynomial 35 min
        • Lecture2.5
          Division and Division Algorithm for Polynomials 01 hour
        • Lecture2.6
          Geometrical Meaning of Zeroes of Polynomial 15 min
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.4
        • Lecture2.9
          Revision Notes Polynomials
        • Lecture2.10
          R S Aggarwal Polynomials
        • Lecture2.11
          R D Sharma Polynomials
      • 03. Linear Equation
        9
        • Lecture3.1
          Solution of linear Equation in Two Variable by Graphical and Algebraic Methods 01 hour
        • Lecture3.2
          Some Basic Questions on Solving Pair of Linear Equations 01 hour
        • Lecture3.3
          Some Basics Problems on Numbers and Ages 40 min
        • Lecture3.4
          Problems on Money Matters, Time, Distance, Speed, Time and Work 01 hour
        • Lecture3.5
          Chapter Notes – Linear Equation
        • Lecture3.6
          NCERT Solutions – Linear Equation Exercise 3.1 – 3.7
        • Lecture3.7
          Revision Notes Linear Equation
        • Lecture3.8
          R S Aggarwal Linear Equation
        • Lecture3.9
          R D Sharma Linear Equation
      • 04. Quadratic Equation
        8
        • Lecture4.1
          Introduction and Finding The Roots of a Quadratic Equation 37 min
        • Lecture4.2
          Miscellaneous Questions 1 02 hour
        • Lecture4.3
          Miscellaneous Questions 2 02 hour
        • Lecture4.4
          Chapter Notes – Quadratic Equation
        • Lecture4.5
          NCERT Solutions – Quadratic Equation Exercise 4.1 – 4.4
        • Lecture4.6
          Revision Notes Quadratic Equation
        • Lecture4.7
          R S Aggarwal Quadratic Equation
        • Lecture4.8
          R D Sharma Quadratic Equation
      • 05. Arithmetic Progressions
        11
        • Lecture5.1
          Introduction to Arithmetic Progression 59 min
        • Lecture5.2
          Miscellaneous Questions Based on nth Term Formula of A.P. 55 min
        • Lecture5.3
          Middle term (s) of Finite A.P. and Arithmetic Mean 01 hour
        • Lecture5.4
          Selection of the Terms and Sum of nth Terms of A.P. 01 hour
        • Lecture5.5
          Miscellaneous Questions Based on Sum of nth Term of A.P. 01 hour
        • Lecture5.6
          Word Problems related to nth term and sum of nth terms of A.P. 01 hour
        • Lecture5.7
          Chapter Notes – Arithmetic Progressions
        • Lecture5.8
          NCERT Solutions – Arithmetic Progressions Exercise 5.1 – 5.4
        • Lecture5.9
          Revision Notes Arithmetic Progressions
        • Lecture5.10
          R S Aggarwal Arithmetic Progressions
        • Lecture5.11
          R D Sharma Arithmetic Progressions
      • 06. Some Applications of Trigonometry
        7
        • Lecture6.1
          Angle of Elevation and Depression; Problems involving Elevation and Double Elevation 01 hour
        • Lecture6.2
          Miscellaneous Problems Involving Double Elevation, Depression and Elevation 01 hour
        • Lecture6.3
          Miscellaneous Questions (Level-1, 2, 3, 4) 49 min
        • Lecture6.4
          Chapter Notes – Some Applications of Trigonometry
        • Lecture6.5
          NCERT Solutions – Some Applications of Trigonometry
        • Lecture6.6
          Revision Notes Some Applications of Trigonometry
        • Lecture6.7
          R D Sharma Some Applications of Trigonometry
      • 07. Coordinate Geometry
        17
        • Lecture7.1
          Introduction to Terms related to Coordinate Geometry 49 min
        • Lecture7.2
          Locating Coordinates of Point on the Axes 37 min
        • Lecture7.3
          Finding vertices of a Geometrical Fig. and Distance b/w Two Points 58 min
        • Lecture7.4
          Distance b/w Two Points Formula Based Examples 44 min
        • Lecture7.5
          Finding the type of Triangle and Quadrilateral 01 hour
        • Lecture7.6
          Find the Collinear or Non-collinear points and Missing Vertex of a Geometrical Figure 42 min
        • Lecture7.7
          Section Formula and Corollary 01 hour
        • Lecture7.8
          Finding the Section Ratio and Involving Equation of Line 53 min
        • Lecture7.9
          Proof Related to Mid-points and Centroid of a Triangle 59 min
        • Lecture7.10
          Finding Area of Triangle and Quadrilateral 53 min
        • Lecture7.11
          Collinearity 34 min
        • Lecture7.12
          Examples Based on If Areas are Given 34 min
        • Lecture7.13
          Chapter Notes – Coordinate Geometry
        • Lecture7.14
          NCERT Solutions – Coordinate Geometry Exercise 7.1 – 7.4
        • Lecture7.15
          Revision Notes Coordinate Geometry
        • Lecture7.16
          R S Aggarwal Coordinate Geometry
        • Lecture7.17
          R D Sharma Coordinate Geometry
      • 08. Triangles
        15
        • Lecture8.1
          Introduction, Similarity of Triangles, Properties of Triangles, Thales’s Theorem 52 min
        • Lecture8.2
          Questions Based on Thales’s Theorem, Converse Thales’s Theorem 01 hour
        • Lecture8.3
          Questions Based on Converse Thales’s Theorem, Questions Based On Mid-points and its Converse 59 min
        • Lecture8.4
          Questions Based on Internal Bisectors of an Angle, Similar Triangles, AAA criterion for Similarity 52 min
        • Lecture8.5
          SSS & SAS Criterion of Similarity, Question Based on Criteria for Similarity 02 hour
        • Lecture8.6
          Questions Based on Criteria for Similarity Conti. 01 hour
        • Lecture8.7
          Questions Based On Area of Similar Triangles 02 hour
        • Lecture8.8
          Questions Based On Area of Similar Triangles cont., Pythagoras Theorem Level-1 01 hour
        • Lecture8.9
          Pythagoras Theorem Level-1 cont. 01 hour
        • Lecture8.10
          Miscellaneous Questions 01 hour
        • Lecture8.11
          Chapter Notes – Triangles
        • Lecture8.12
          NCERT Solutions – Triangles Exercise 8.1 – 8.6
        • Lecture8.13
          Revision Notes Triangles
        • Lecture8.14
          R S Aggarwal Triangles
        • Lecture8.15
          R D Sharma Triangles
      • 09. Circles
        8
        • Lecture9.1
          Introduction to Circle and Tangent and Theorems-1, 2 01 hour
        • Lecture9.2
          Property of Chord of Circle and Theorem-3 52 min
        • Lecture9.3
          Theorem 4 and Angle in Semicircle 01 hour
        • Lecture9.4
          Chapter Notes – Circles
        • Lecture9.5
          NCERT Solutions – Circles Exercise
        • Lecture9.6
          Revision Notes Circles
        • Lecture9.7
          R S Aggarwal Circles
        • Lecture9.8
          R D Sharma Circles
      • 10. Areas Related to Circles
        10
        • Lecture10.1
          Area and Perimeter of Circle, Semicircle and Quarter circle 56 min
        • Lecture10.2
          Touching Circles and Example on Bending Wire, Wheel Rotation 58 min
        • Lecture10.3
          Sector of A Circle 51 min
        • Lecture10.4
          Segment of Circle and Area of Combination of Plane Figures 01 hour
        • Lecture10.5
          Questions Based on Area of Combination of Plane Figures 01 hour
        • Lecture10.6
          Chapter Notes – Areas Related to Circles
        • Lecture10.7
          NCERT Solutions – Areas Related to Circles
        • Lecture10.8
          Revision Notes Areas Related to Circles
        • Lecture10.9
          R S Aggarwal Areas Related to Circles
        • Lecture10.10
          R D Sharma Areas Related to Circles
      • 11. Introduction to Trigonometry
        7
        • Lecture11.1
          Introduction, Trigonometric Ratios, Sums when Sides of Triangle are given, Miscellaneous Questions 01 hour
        • Lecture11.2
          Trigonometric Ratios for Some Specific Angles, Trigonometric Ratio for 30 & 60, Value Table, Word Problems 48 min
        • Lecture11.3
          Complimentary angles in Inverse Multiplication, Miscellaneous Problems 01 hour
        • Lecture11.4
          Miscellaneous Questions 01 hour
        • Lecture11.5
          Chapter Notes – Introduction to Trigonometry
        • Lecture11.6
          NCERT Solutions – Introduction to Trigonometry
        • Lecture11.7
          Revision Notes Introduction to Trigonometry
      • 12. Surface Areas and Volumes
        9
        • Lecture12.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere, Type-1 : Surface Area and Volume of A Solid 01 hour
        • Lecture12.2
          Volume Related Questions, Based On Embarkment 48 min
        • Lecture12.3
          Based on the Rate of Flowing, Frustum of Cone (A) Surface Area (B) Volume 52 min
        • Lecture12.4
          Quantity, Percentage, Very Short Answer Types Questions 44 min
        • Lecture12.5
          Chapter Notes – Surface Areas and Volumes
        • Lecture12.6
          NCERT Solutions – Surface Areas and Volumes
        • Lecture12.7
          Revision Notes Surface Areas and Volumes
        • Lecture12.8
          R S Aggarwal Surface Areas and Volumes
        • Lecture12.9
          R D Sharma Surface Areas and Volumes
      • 13. Statistics
        12
        • Lecture13.1
          Introduction and Mean of Ungrouped Data 46 min
        • Lecture13.2
          Mean of Grouped Data 58 min
        • Lecture13.3
          Median of Ungrouped and grouped Data 01 hour
        • Lecture13.4
          Mode of Ungrouped and Grouped Data 49 min
        • Lecture13.5
          Mean, Median and Mode & Cumulative Frequency Curve-Less Than Type 53 min
        • Lecture13.6
          Cumulative Frequency Curve (Ogive)-More Than Type 52 min
        • Lecture13.7
          Miscellaneous Questions 38 min
        • Lecture13.8
          Chapter Notes – Statistics
        • Lecture13.9
          NCERT Solutions – Statistics
        • Lecture13.10
          Revision Notes Statistics
        • Lecture13.11
          R S Aggarwal Statistics
        • Lecture13.12
          R D Sharma Statistics
      • 14. Probability
        9
        • Lecture14.1
          Activities, Understanding term Probability, Experiments 54 min
        • Lecture14.2
          Calculations 43 min
        • Lecture14.3
          Event, Favorable Outcomes, Probability, Some important experiments- Tossing a coin, Rolling dice, Drawing a card 01 hour
        • Lecture14.4
          Complement of an event, Elementary Events, Equally Likely events, Some problems of Probability 58 min
        • Lecture14.5
          Chapter Notes – Probability
        • Lecture14.6
          NCERT Solutions – Probability
        • Lecture14.7
          Revision Notes Probability
        • Lecture14.8
          R S Aggarwal Probability
        • Lecture14.9
          R D Sharma Probability
      • 15. Construction
        7
        • Lecture15.1
          Introduction, Construction to Divide a Line Segment in Certain Ratio, Construction of Similar Triangles 44 min
        • Lecture15.2
          Construction of Different Angles, Construction of Angle Bisector 41 min
        • Lecture15.3
          Construction of Tangent of Circle at a Given Point 35 min
        • Lecture15.4
          Chapter Notes – Construction
        • Lecture15.5
          NCERT Solutions – Construction
        • Lecture15.6
          R S Aggarwal Construction
        • Lecture15.7
          R D Sharma Construction

        NCERT Solutions – Real Numbers Exercise 1.1 – 1.4

        Q.1.      Use Euclid’s division algorithm to find the HCF of :
        (i) 135 and 225        (ii) 196 and 38220        (iii) 867 and 255
        Sol.     (i)  135 and 225

        Given integers are 135 and 225 clearly 225 > 135 applying Euclid’s division lemma on 135 and 225.

        We get

        225 = 135 × + 90……………… (i)

        Here remainder  So we again apply EDL on divisor 135 and remainder 90

        135 = 90 × 1 + 45………………(ii)

        Here, remainder  , so we apply  Euclid’s division lemma on divisor 90 and remainder 45

        90 = 45 × 2 + 0 ………………….. (iii)

        From equation (iii), remainder = 0. So the divisor at this stage and remainder of previous stage

        i.e. 45 is HCF ( 135, 225) = 45

        (ii) 196 and 38220

        Given positive integers are 196 and 38220 and 38220 > 196 so applying EDL,

        we get

        38220 = 196 × 195 + 0 …………… (i)

        Remainder at this stage is zero. So, the divisor of this stage i.e 196 is HCF of 38220 and 196

        HCF ( 196 , 38220) = 196

        (iii) 867 and 255

        Given positive integers are 867 and 255 and 867 > 255 So, applying Euclid’s division algorithm

        We get

        867 = 255 × 3 + 102  ………………. (i)

        Here, remainder . So, we again apply Euclid’s division algorithm on division 255 and remainder 102 .


        Q.2     Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.
        Sol.       Let a be any positive integer and b = 6.Then, by Euclid’s algorithm a = 6q + r, for some integer q≥0 and where 0≤r<6 the possible remainders are 0, 1, 2, 3, 4, 5 i.e, a can be 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5,where q is the quotient. If a = 6q or 6q + 2 or 6q + 4, then a is an even integer. Also, an integer can be either even or odd.Therefore, any odd integer is of the form 6a + 1 or 6q + 3 or 6q + 5, where q is some integer.


        Q.3     An army contingent of 616 members is to march behind an army band of 32 members in parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ?
        Sol.       To find the maximum number of columns, we have to find the HCF of 616 and 32.

                    100                                                                                                 5

        ⇒616=32×19+8  
                     ⇒32=8×4+0
                    Therefore,the HCF of 616 and 32 is 8.Hence, maximum number of columns is 8.


        Q.4     Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m +1 for some integer m.
        Sol.       Let x be any positive integer, then it is of the form 3q, 3q + 1 or 3q +2. Now, we have to prove that the squre of each of these can be written in the form 3m or 3m +1.
                     Now, (3q)2=9q2=3(3q2)=3m, where m=3q2
                     (3q+1)2=9q2+6q+1
                      =3(3q2+2q)+1
                     = 3m + 1, where m=3q2+2q
                     and, (3q+2q)2=9q2+12q+4
                      =3(3q2+4q+1)+1
                      = 3m + 1, where m=3q2+4q+1
                     Hence, the result.


        Q.5     Use Euclid’s division lemma to show that cube of any positive integer is either of the form 9q, 9q + 1 or 9q + 8.
        Sol.      Let x be any positive integer, then it is of  the form 3m, 3m + 1 or 3m +2. Now, we have prove that the cube of each of these can be rewritten in the form 9q + 1 or 9q + 8.
                    Now, (3m)3=27m3=9(3m3)
                    = 9q, where q=3m3
                    (3m+1)3=(3m)3+3(3m)2.1+3(3m).12+1
                    =27m3+27m2+9m+1
                    =9(3m3+3m2+m)+1
                    = 9q + 1, where q=3m3+3m2+m
                    and (3m+2)3=(3m)3+3(3m)2.2+3(3m).22+8
                    =27m3+54m2+36m+8
                    =9(3m3+6m2+4m)+8
                    = 9q + 8, where q=3m3+6m2+4m

         

        Q.1     Express each number as product of its prime factors:
        (i) 140      (ii) 156      (iii) 3825      (iv) 5005      (vi) 7429

        Sol.       (i) We use the division method as shown below :
        6
        Therefore, 140 = 2 × 2 × 5 × 7 =22×5×7

        (ii) We use the division method as shown below:
        7
        Therefore, 156 = 2 × 2 × 3 × 13  =22×3×13

        (iii) We use the division method as shown below :
        8

        Therefore, 3825 = 3 × 3 × 5 × 5 × 17 =32×52×17

        (iv) We use the division method as shown below :

        9

        Therefore, 5005 = 5 × 7 × 11 × 13

        (v) We use the division method as shown below :

        10

        Therefore, 7429 = 17 × 19 × 23


        Q.2     Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
        (i) 26 and 91     (ii) 510 and 92     (iii) 336 and 54
        Sol.       (i) 26 and 91

        11                             12

        26 = 2 × 13 and 91 = 7 × 13
        Therefore, LCM of 26 and 91 = 2 × 7 × 13 = 182
                       and HCF of 26 and 91 = 13
        Now, 182 × 13 = 2366 and 26 × 91 = 2366
        Since, 182 × 13 = 26 × 91
        Hence verified.

        (ii) 510 and 92

        13                              14


                      510 = 2 × 3 × 5 × 17 and 92 = 2 × 2 × 23
        Therefore, LCM of 510 and 92 = 2 × 2 × 3 × 5 × 17 × 23 = 23460
                      and HCF of 510 and 92 = 2
                      Now, 23460 × 2 = 46920 and 510 × 92 = 46920
        Since 23460 × 2 = 510 × 92
        Hence verified.

        (iii) 336 and 54

        15                                 16


                   336 = 2 × 2 × 2 × 2 × 3 × 7
        and 54 = 2 × 3 × 3 × 3
        Therefore, LCM of 336 and 54 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
        and HCF of 336 and 54 = 2 × 3 = 6
        Now, 3024 × 6 = 18144
        and 336 × 54 = 18144
        Since, 3024 × 6 = 336 × 54
        Hence verified.


        Q.3       Find the LCM and HCF of the following integers by applying the prime factorisation method
        (i) 12,15 and 21      (ii) 17, 23 and 29      (iii) 8, 9 and 25
        Sol.         (i) First we write the prime factorisation of each of the given numbers.
        12 = 2 × 2 × 3 = 22 × 3, 15 = 3 × 5 and 21 = 3 × 7
        Therefore, LCM = 22 × 3 × 5 × 7 = 420
        and, HCF = 3

        (ii) First we write the prime factorisation of each of the given numbers.
        17 = 17, 23 = 23 and 29 = 29
        Therefore, LCM = 17 × 23 × 29 = 11339
        and HCF = 1

        (iii)  First we write the prime factorisation of each of the given numbers.
                          8 = 2 × 2 × 2 =23,9=3×3=32, 25 = 5 × 5 = 52
        Therefore, LCM =23×32×52=8×9×25=1800
        and HCF = 1


        Q.4     Given that HCF (306, 657) = 9, find LCM (306, 657).
        Sol.       We know that the product of the HCF and the LCM of two numbers is equal to the product of the given numbers.
        Therefor, HCF (306,657) × LCM (306,657) = 306 × 657
        ⇒ 9 × LCM (306 × 657) = 306 × 657
        ⇒ LCM (306,657) =306×6579  = 22338


        Q.5.     Check whether 6n can end with the digit 0 for any natureal number n.
        Sol.        If the number 6n, for any n ends with the digit zero, then it is divisible by 5. That is, the prime factorisation of 6n contains the prime 5. That is, not possible as the only prime in the factorisation of 6n is 2 and 3 and the uniqueness of the Fundamental Theorem of Arihmetic guarantees that there are no other primes in the factorisation of 6n. So, there is no n∈N for which 6n ends with the digit zero.


        Q.6    Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
        Sol.     Since, 7 × 11 × 13 + 13 = 13 × (7 × 11 × 1 + 1)
        = 13 × (77 + 1)
        = 13 × 78
        ⇒ It is a composite number.
        Again, 7 × 6 × 5 × 4 × 3 × 1 × 1 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 1 × 1 + 1)
        ⇒ It is a composite number.


        Q.7     There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the path, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they again at the starting point ?
        Sol.      To find the LCM of 18 and 12, we have

        17                      18

        18 = 2 × 3 × 3 and 12 = 2 × 2 × 3
        LCM of 18 and 12 = 2 × 2 × 3 × 3 = 36
        So, Sonia and Ravi will meet again at the starting point after 36 minutes.

         

        Q.1     Prove that 5–√ is irrational.
        Sol.       Let us assume, to the contrary, that 5–√ is rational.
        Now, let 5–√=ab , where a and b are coprime and b≠0. Squaring on both side, we get
        5=a2b2⇒5b2=a2                                                          …(1)
        This shows that a2 is divisible by 5
                     It follows that a is divisible by 5                                              …(2)
        ⇒a=5m for some integer m.
        Substituting a = 5m in (1), we get
        5b2=(5m)2=25m2
        or b2=5m2
        ⇒b2 is divisible by 5
        and hence b is divisible by 5                                                   …(3)
        From (2) and (3), we can conclude that 5 is a common factor of both a and b.
        But this contradicts our supposition that a and b are coprime.
        Hence, 5–√ is irrational.


        Q.2     Prove that 3 + 25–√ is irrational.
        Sol.       Let us assume, to the contrary, that 3+25–√, is a rational number.
                     Now, let  3+25–√=ab, where a and b are coprime and b≠0
        ⇒25–√=ab−3 or 5–√=a2b−32
        Since, a and b are integers.
        Therefore, a2b−32 is a rational number
        ⇒ 5–√ is a rational number.
        But 5–√ is an irrational number.
        This shows that our assumption is incorrect.
        So, 3+25–√ is an irrational number .


        Q.3     Prove that the following are irrationals :
        (i) 12√     (ii) 75–√     (iii) 6+2–√
        Sol.       (i) Let us assume, to the contrary, that 12√ is rational. That is, we can find co-prime integers p and q(≠0) such that
        12√=pq⇒1×2√2√×2√=pq⇒2√2=pq
        ⇒2–√=2pq
        Since p and q are integers, 2pq is rational, and so 2–√ is rational.
        But this contradicts the fact that 2–√ is irrational.So, we conclude that 12√ is irrational.

                     (ii) Let us assume, to the contrary, that 75–√ is rational. 
                     That is, we can find co-prime integers p and q(≠0) such that 75–√=pq
        Since p and q are integers, p7q is rational and so is 5–√ 
                     But this contradicts the fact that 5–√ is irrational . So, we conclude that 75–√ is irrational.

        (iii) Let us assume, to the contrary, that 2–√ is rational. That is, we can find integers p and q(≠0) such that
        6+2–√=pq⇒6−pq=2–√
        ⇒2–√=6−pq
        Since p and q are integers, we get 6−pq is rational, and so 2–√ is rational.
        But this contradicts the fact that 2–√ is irrational.
        So, we conclude that 6 + 2–√ is irrational

         

        Q.1     Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non – terminating repeating decimal expansion :
                  (i) 133125                                   (ii) 178
                  (iii) 64455                                   (iv) 151600
                  (v) 29343                                    (iv) 232352
                 (vii) 129225775             (viii) 615
                 (ix) 3550                                      (x) 77210

        Sol.     We know that if the denominator of a rational number has no prime factors other than 2 or 5, then it is expressible as a terminating, otherwise it has non – terminating  repeating decimal representation. Thus, we will have to check the prime factors of the denominators of each of the given rational numbers.  
                  (i) In 133125, the denominator is 3125.
        19
        We have, 3125 = 5 × 5 × 5 × 5 × 5.
        Thus, 3125 has 5 as the only prime factor.
        Hence, 133125 must have a terminating decimal representation.

         (ii) In 178, the denominator is 8.
        20
        We have, 8 = 2 × 2 × 2
        Thus, 8 has 2 as the only prime factor.
        Hence, 178 must have a terminating decimal representation.

        (iii) In 64455, denominator is 455. We have, 455 = 5 × 7 × 13
        Clearly, 455 had prime factors other than 2 and 5. So, it will not have a terminating decimal representation.

        (iv) In 151600, the denominator is 1600.
        22
        We have, 1600
        = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
        Thus, 1600 has only 2 and 5 as prime factors.
        Hence, 151600 must have a terminating decimal representation.

         (v) In 29343, the denominator is 343.
        23
        We have, 343 = 7 × 7 × 7
        Clearly, 343 has prime factors other than 2 and 5.
        So, it will not have terminating decimal representation.

        (vi) In 2323.52 Clearly, the denominator 23.52 has only 2 and 5 as prime factors.
               Hence, 2323.52 must have a terminating decimal representation.
               (vii) In 12922.57.75 Clearly, the denominator 22.57.75 has prime factors other than 2 and 5.So, it will not have terminating decimal representation.

         (viii) In 615, we have 15 = 3 × 5
        Clearly, 15 has prime factors other than 2 and 5. So, it will not have terminating decimal representation.

          (ix) In 3550 , we have 50 = 2 × 5 × 5 The denominator has only 2 and 5 as prime factors. Hence, 3550 must have a terminating decimal representation.

        (x) In 77210, the denominator is 210.
        24
        We have, 210 = 2 × 3 × 5 × 7
        Clearly, 210 has prime factors other than 2 and 5.
        So, it will not have terminating decimal representation.


        Q.2     Write down the decimal expansion of those rational numbers in Question 1 above which have terminating decimal expansions.
        Sol.      (i) 133125=135×5×5×5×5
        =13×2×2×2×2×25×2×5×2×5×2×5×2×5×2
        =13×3210×10×10×10×10=416100000 = 0.00416

        (ii) 178=17×5323×53=17×53103=17×125103
        =21251000=2.125

         (iii) Non – terminating.

        (iv) 151600=1526×52=1524×22×52
        =1524×102=15×5424×54×102
        =15×625104×102=93751000000=0.009375

           (v) Non – terminating.

         (vi) 2323.52=232.22.52=232.102=23×52×5×102
        =11510×102=1151000=0.115

         (vii) Non – terminating.

         (viii) 615=25=410=0.4

          (ix) 3550=35×250×2=70100=0.70

          (x) Non – terminating.


        Q.3     The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form pq, what can you say about the prime factors of q ?
        (i) 43.123456789      (ii) 0.120120012000120000…….      (iii) 43.123456789¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

        Sol.     (i) 43.123456789 is terminating.
        So, it represents a rational number.
        Thus, 43.123456789 = pq, where q=109.

        (ii) 0.12012001200012000… is non – terminating and non-repeating. So, it is irrational.

         (iii) 43.123456789¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ is non – terminating but repeating. So, it is rational.
        Thus, 43.123456789¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ =pq, where q = 999999999.

        Prev Chapter Notes – Real Numbers
        Next Revision Notes Real Numbers

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          6 Comments

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          September 15, 2021

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          December 17, 2022

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