• Home
  • Courses
  • Online Test
  • Contact
    Have any question?
    +91-8287971571
    contact@dronstudy.com
    Login
    DronStudy
    • Home
    • Courses
    • Online Test
    • Contact

      Class 10 Maths

      • Home
      • All courses
      • Class 10
      • 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 – Statistics

        Q.1     A survey was conducted by a group of students as a part of their environment awareness 
        programmer, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

        108

        Which method did you use for finding the mean, and why ?
        Sol.      Calculation of Mean

        109
                  Hence, mean x¯=1nΣfixi
                                           = 120×162=8.1 plants
        We have used direct method because numerical values of xi and fi are small.


        Q.2     Consider the following distribution of daily wages of 50 workers of a factory.

        3
                  Find the mean daily wages of the workers of the factory by using an appropriate method.
        Sol.      Let the assumed mean, A =150; class-interval, h = 20
        so         ui=xi−Ah=xi−15020
        We construct the table :

        100
                       x¯=A+hΣfiuiΣfi=150+20×−1250 = 150 – 4.8 = 145.2               
                       Hence, mean = Rs 145.20 .


        Q.3     The following distribution shows the daily pocket allowance of children locality mean. The mean pocket allowance is Rs 18. Find the missing frequency f.

        110

        Sol.       Let the assumed mean, A = 16, class interval h = 2
                     So       ui=xi−Ah=xi−162

        111
        We have, x¯ = 18, A = 16 and h = 2
        Therefore, x¯=A+h(1NΣfiui)
        ⇒18=16+2(2f+24f+44)
        ⇒2=2(2f+24f+44)
        ⇒f+44=2f+24
        ⇒2f – f = 44 – 24
                     ⇒f=20
                     Hence, the missing frequency is 20.


        Q.4      Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

        7.
        Sol.       Let the assumed mean, A = 75.5, class interval
        h = 3, so     ui=xi−Ah=xi−75.53

        8
        x¯=A+hΣfiuiΣfi=75.5+3×430
                        = 75.5 + 0.4 = 75.9
        Hence, the mean heart beats per minute for these women is 75.9.


        Q.5     In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes
        contained varying number of mangoes. The following was the distribution of mangoes according to
        the number of boxes.

        table of statistics


        Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose ?

        Sol.      Here, the class intervals are formed by the exclusive method. If we make the series an inclusive one the mid-values remain  same. So there is no need to convert the series.
        Let the assumed mean be A = 60 and h = 3
        so      ui=xi−Ah=xi−603
        Calculation of Mean
        104-640x375 22222
                  Therefore x¯=A+hΣfiuiΣfi=60+3×−375400
                                      = 60 – 2.8125 = 57.1875 = 57.19(nearly)
        Hence, average number of mangoes per box = 57.19 (nearly)


        Q.6      The table below shows the daily expenditure on food of 25 households in a locality.

        11

                   Find the mean daily expenditure on food by a suitable method.
        Sol.       Let the assumed mean, A = 225, class interval, h = 50
        so      ui=xi−Ah=xi−22550
        We construct the following table :
        113                x¯=A+hΣfiuiΣfi
        = 225 +50×−725
        = 225 – 14 = 211
        Hence, the mean daily expenditure of food is Rs 211.


        Q.7     To find the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below :

        14

                   Find the mean concentration of SO2 in the air.
        Sol.       Calculation of mean by direct method

        205
                       Hence, x¯=∑fixi∑fi = 2.9630 = 0.099 ppm

        Q.8     A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

         

        16
        Sol.      Here, the class size varies, and xi ‘s are small. Let us apply the direct method here.

        211           
                     Therefore, mean, x¯=ΣfixiΣfi  = 49940 = 12.475
                     Hence,      mean = 12.48 days


        Q.9     The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

        116
        Sol.
               Let the assumed mean be A = 70 and h = 10
        so      ui=xi−7010

        117

        Therefore,  x¯=A+h×ΣfiuiΣfi
                                            = 70 + 10 ×−235
                                            = 70 – 0.57 = 69.43
        Thus, mean literacy rate is 69.43% .

         

        Q.1      The following table shows the ages of the patients admitted in a hospital during a year :

        20

                     
                  Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

        Sol.      The class 35 – 45 has the maximum frequency, therefore, this is the modal class.

        Here ℓ = 35, h = 10, f1 = 23, f0 = 21, f2 =14
        Now, let us substitute these values in the formula
        Mode =ℓ+(f1−f02f1−f0−f2)×h
                              = 35 + 23−212×23−21−14×10
                              =35+246−21−14×10
                               = 35 + 211×10
                              = 35 + 1.8 = 36.8
        Calculation of Mean
        Let assumed mean, A = 30, class interval h = 10
        so      ui=xi−Ah=xi−3010

        206
        Therefore, x¯=A+h×ΣfiuiΣfi = 30 + 10 ×4380
                                       = 30 + 5.37 = 35.37
        Hence, Mode = 36.8 years, Means = 35.37 years.
        Maximum number of patients admitted in the hospital are of age group 36.8 years (approx.) while on an
        average the age of a patient admitted to the hospital is 35.37 years.


        Q.2      The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :

        222-607x640

                    Determine the modal lifetimes of the components.
        Sol.        The class 60 – 80 has the maximum frequency, therefore, this is the modal class.
        Here ℓ = 60 , h = 20, f1 = 61, f0 = 52 and f2 = 38 .
        Now, let us substitute these values in the formula
        Mode = ℓ+(f1−f02f1−f0−f2)×h
                                =60+61−522×61−52−38×20
                                 = 60+(61−52122−52−38)×20
                                 = 60+932×20= 60 + 5.625 = 65.625
        Thus, the modal lifetimes of the components is 65.625 hours.


        Q.3      The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean
        monthly expenditure :
        23
        Sol.      
        The class 1500 – 2000 has the maximum frequency,therefore, this is the modal class.
        Here ℓ = 1500, h = 500, f1=40,f0=24 and f2=33
        Now, let us substitute these values in the formula
        Mode = ℓ+(f1−f02f1−f0−f2)×h
                              =1500+(40−242×40−24−33)×500
                              = 1500+40−2480−24−33×500
                              = 1500+1623×500
                              = 1500 + 347.83 = 1847.83
        Therefore, modal monthly expenditure = Rs. 1847.83
        Let the assumed mean be A = 3250 and h = 500.

        120
                     We have, N = 200, A = 3250, h = 500 and Σfiui=−235
        Therefore, mean = A+h(1NΣfiui)
                                             = 3250 + 500 ×−235200
                                             = 3250 – 587.5 = 2662.5 
                    Hence, the average expenditure is Rs 2662.50.


        Q.4      The following distribution gives the state – wise teacher- student ratio in higher secondary schools of India. Find the mode and mean of this data. interpret the two measures.

        216

        Sol.      The class 30 – 35 has the maximum frequency, therefore, this is the modal class.
        Here ℓ = 30, h = 5, f1 =10, f0 = 9 and f2 =3
        Now, let us substitute these values in the formula
        Mode = ℓ+(f1−f02f1−f0−f2)×h
                              =30+(10−92×10−9−3)×5
                              = 30 + 10−920−9−3×5
                              = 30 +18×5
                              = 30 + 0.625
                              = 30.625 = 30.6(approx.)
        Calculation of mean :

        207               x¯=ΣfiuiΣfi
                           = 1022.535 = 29.2
        Therefore, mean = 29.2
        Thus, most states/U.T. have a student teacher ratio of 30.6 and on an average, the ration is 29.2.


         

        Q.1      The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

        29

        Sol.      Calculation of median
        First, we prepare the following table to compute the median :

        30
        We have n = 68 ⇒n2=682=34
                       The cumulative frequency just greater than n2, is 42 and the corresponding class is 125 – 145.
                       Thus, 125 – 145is the median class such that n2 = 34, ℓ = 125, c.f. = 22, f  = 20, h = 20 .
                       Substituting these values in the formula
                       Median = ℓ+(n2−cff)×h, 
                       Median = 125 + (34−2220)×20 = 125 – 12 = 137 units
                       Calculation of mean
                       Let the assumed mean, A = 135,class interval, h = 20
                       so ui=xi−Ah=xi−13520

        208

        x¯=A+h×ΣfiuiΣfi
                      = 135 + 20 ×768
        135 + 2.05 = 137. 05
        Therefore, Mean = 137.05 units
        Calculation of mode
        The class 125 – 145 has the maximum frequency,therefore, this is the modal class.
        Here ℓ =125, h =20, f1 =20, f0 = 13 and f2 =14
        Now, let us substitute these values in the formula
        Mode = ℓ+(f1−f02f1−f0−f2)×h
                               =125+20−1340−13−14×20
                               =125+713×20
                               = 125 + 10.76 = 135.76
        Therefore, mode = 135.76 units
        Clearly, the three measures are approximately the same in this case.


        Q.2       If the median of the distribution given below is 28.5, find the values of x and y.

        106
        Sol.       Here it is given that median is 28.5 and n = Σfi = 60
        We now prepare the following cumulative frequency table :

        107
        Here n = 60          ⇒n2=30
        Since the median is given to be 28.5, thus the median class is 20 – 30.
        Therefore, ℓ = 20, h = 10, f = 20 and cf = 5 + x
        Therefore, median = ℓ+(n2−cff)×h
        ⇒ 28.5 = 20+30−(5+x)20×10
        ⇒ 28.5 = 20 + 25−x2
        ⇒ 57 = 40 + 25 – x
        ⇒ x = 65 – 57 = 8
        ⇒ Also, 45 + x + y = 60
        ⇒ 45 + 8 + y = 60                   [As x = 8]
        ⇒ y = 60 – 53 = 7
        Hence, x = 8, y = 7


        Q.3     A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
        34
        Sol.          We are given the cumulative frequency distribution. So, we first construct a frequency table from the given cumulative frequency distribution and then we will make necessary computations to compute median .

        35
        Here n = Σfi=100⇒n2=50
        We see that the cumulative frequency just greater than n2, i.e., 50 is 78 and the corresponding class is 35 – 40. So, 35 – 40 is the median class.
        Therefore, n2=50,ℓ=35, c.f = 45 , f = 33, h = 5
        Now, let us substitute these values in the formula
        Median = ℓ+(n2−cff)×h
                                   = 35+50−4533×5 =35+533×5
                                   = 35 + 0.76 = 35.76
        Hence, the median age = 35.76 years .


        Q.4      The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

        36
                      Find the median length of the leaves.

        Sol.           Here the frequency table is given in the inclusive form. So, we first convert it into exclusive form by subtracting and adding h2 to the lower and upper limits respectively of each class, where h denotes the difference of lower limit of a class and the upper limit of the previous class. Converting the given table into exclusive form and preparing the cumulative frequency table, we get

        37
        We have n = 40 ⇒n2=20
        The cumulative frequency just greater than n2 is 29 and the corresponding class is 144.5 – 153.5.
        So, 144.5 – 153.5 is the median class.
        Here n2=20,ℓ = 144.5, h = 9, f = 12, cf =17
        Substituting these values in the formula
        Median = ℓ+(n2−cff)×h, we have
        Median = 144.5+(20−1712)×9
                                   = 144.5 +312×9
                                   = 144.5 + 2.25
                                   = 146.75 mm


        Q.5     The following table gives the distribution of the life time of 400 neon lamps:

        221
                  Find the median life time of a lamp.

        Sol.       First, we prepare the following table to compute the median:

        121121209121             
                       We have n = 400 ⇒n2=200
        The cumulative frequency just greater than n2 is 216 and the corresponding class is 3000 – 3500.
        Thus, 3000 – 3500 is the median class such that n2=200,ℓ = 3000, cf = 130, f = 86 and h = 500.
        Substituting these values in the formula
        Median = ℓ+(n2−cff)×h, we have
        Median = 3000+(200−13086)×500
                                 = 3000+7086×500
                                 = 3000 + 406.98 = 3406.98
        Hence, median life = 3406.98 hours


        Q.6     100 surnames were randomly picked up from a local telephone directory and the frequency
        distribution of the number of letters in English alphabets in the surnames was obtained as follows:

        43
                  Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? 
        Also, find the modal size of the surnames.
        Sol.      Calculation of median
        First, we prepare the following table to compute the median

        210             
                      We have n = 100 ⇒n2=50
        The cumulative frequency just greater than n2 is 76 and the corresponding class is 7 – 10 . Thus 7 – 10 is the median class such that
        n2=50,ℓ = 7, cf = 36, f = 40 and h = 3
        Substitute these values in the formula
        Median = ℓ+(n2−cff)×h, we have
        Median =  7+(50−3640)×3
                                  =7+1440×3
                                  = 7 + 1.05 = 8.05
                     Calculation of mean

        45
        Therefore, Mean, x¯=ΣfixiΣfi=832100 = 8.32
        Calculation of mode
        The class 7 – 10 has the maximum frequency therefore, this is the modal class.
        Here ℓ = 7 , h = 3, f1 = 40, f0 = 30 and f2 = 16
        Now, let us substitute these values in the formula
        Mode=ℓ+(f1−f02f1−f0−f2)×h
                               = 7+40−3080−30−16×3
                               = 7+1034×3 = 7 + 0.88 = 7.88
        Hence, median = 8.05, mean = 8.32 and mode = 7.88

        Q.1     The following distribution gives the daily income of 50 workers if a factory.

        48

        Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
        Sol.       Converting the given distribution to a less than type cumulative frequency distribution, we get

        49
        Let us now plot the points corresponding to the ordered pairs (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and join them by a free hand smooth curve.

        213               The curve thus obtained is  less than type ogive.


        Q.2     During the medical check-up of 35 students of a class, their weights were recorded as follows :

        50
        Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
        Sol.       Here the values 38, 40, 42, 44, 46, 48, 50 and 52 are the upper limits of the respective class-intervals. To represent the data in the table graphically, we mark the upper limits on the class intervals on x-axis and
        their corresponding cumulative frequencies on the y-axis, choosing a convenient scale. Let us now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper an join them by a free hand smooth curve.

        215
        The curve thus obtained is  less than type ogive To locate n2=352= 17.5 on the y-axis. From this point draw a line parallel to the x-axis cutting the curve at a point. From this point, draw a perpendicular to the x-axis. The point of intersection of this perpendicular with the x-axis gives the median of the data. In this case it is 46.95. Let us make the following table in order to find mode by using the formula

        51

        The class 46 – 48 has the maximum frequency, therefore, this is modal class
        Here, ℓ = 46, h = 2, f1 = 14, f0 = 5 and f2 = 4
              Now, Mode = ℓ+(f1−f02f1−f0−f2)×h
                                = 46+14−528−5−4×2 = 46+919×2
                                = 46 + 0.95 = 46.95
        Thus, mode is verified.


        Q.3      The following tables gives production yield per hectare of wheat of 100 farms of a village.

        52
        Change the distribution to a more than type distribution and draw its ogive.
        Sol.      Converting the given distribution to a more than type distribution, we get

        53.
        Now, draw the ogive by plotting the points (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on
        this graph paper and join them by a free hand smooth curve.

        56
        The curve thus obtained is  more than type ogive.

        Prev Chapter Notes – Statistics
        Next Revision Notes Statistics

        Leave A Reply Cancel reply

        Your email address will not be published. Required fields are marked *

          6 Comments

        1. Priyanshu Kumar
          May 28, 2021

          How we access live tutorial classes please tell me

          • Dronstudy
            May 31, 2021

            For any information regarding live class please call us at 8287971571

        2. Rohan Yedla
          September 15, 2021

          I am not able to access any test of coordinate geometry, can you pls resolve the issue?

          • Dronstudy
            September 20, 2021

            We have fixed the issue. Please feel free to call us at 8287971571 if you face such type of issues.

        3. Ananyaa
          November 12, 2021

          Hi!
          I am a student in class 10th and ive noticed that you dont give full courses on your youtube channel. I mean, its understandable. But we request you to at least put full course of only one chapter on youtube so that we can refer to that. Only one from the book Because your videos are very good and they enrich our learning.
          Please give this feedback a chance and consider our request.

          Dronstudy lover,
          Ananyaa

        4. Nucleon IIT JEE KOTA
          December 17, 2022

          Great article, the information provided to us. Thank you

        All Courses

        • Backend
        • Chemistry
        • Chemistry
        • Chemistry
        • Class 08
          • Maths
          • Science
        • Class 09
          • Maths
          • Science
          • Social Studies
        • Class 10
          • Maths
          • Science
          • Social Studies
        • Class 11
          • Chemistry
          • English
          • Maths
          • Physics
        • Class 12
          • Chemistry
          • English
          • Maths
          • Physics
        • CSS
        • English
        • English
        • Frontend
        • General
        • IT & Software
        • JEE Foundation (Class 9 & 10)
          • Chemistry
          • Physics
        • Maths
        • Maths
        • Maths
        • Maths
        • Maths
        • Photography
        • Physics
        • Physics
        • Physics
        • Programming Language
        • Science
        • Science
        • Science
        • Social Studies
        • Social Studies
        • Technology

        Latest Courses

        Class 8 Science

        Class 8 Science

        ₹8,000.00
        Class 8 Maths

        Class 8 Maths

        ₹8,000.00
        Class 9 Science

        Class 9 Science

        ₹10,000.00

        Contact Us

        +91-8287971571

        contact@dronstudy.com

        Company

        • About Us
        • Contact
        • Privacy Policy

        Links

        • Courses
        • Test Series

        Copyright © 2021 DronStudy Pvt. Ltd.

        Login with your site account

        Lost your password?

        Modal title

        Message modal