NCERT Exemplar Solution for Class 10 Mathematics: Statistics, Exercise 13.1

Sep 8, 2017 12:17 IST

Class 10 Maths NCERT Exemplar Solutions
Class 10 Maths NCERT Exemplar Solutions

Here you get the CBSE Class 10 Mathematics chapter 10, Constructions: NCERT Exemplar Problems and Solutions. This part includes solutions for questions 1 to 11 that based on Statistics from Exercise 13.1.  All these are the Multiple Choice Questions framed from various important topics in the chapter. Each question is provided with a detailed solution.

CBSE Class 10 Mathematics Syllabus 2017-2018

NCERT Exemplar problems are a very good resource for preparing the critical questions like Higher Order Thinking Skill (HOTS) questions. All these questions are very important to prepare for CBSE Class 10 Mathematics Board Examination 2017-2018 as well as other competitive exams.

Find below the NCERT Exemplar problems and their solutions for Class 10 Mathematics Chapter, Statistics:

Exercise 13.1

Multiple Choice Questions:

(a) lower limits of the classes

(b) upper limits of the classes

(c) mid-points of the classes

(d) frequencies of the class marks

Answer. (c)          

Explanation:

Here, a is assumed mean from class marks (xi) and di is the difference between a and each of the xi’s (mid-value).

Thus, di’s are the deviation from a of mid-points of the classes.

Question. 2 While computing mean of grouped data, we assume that the frequencies are

(a) evenly distributed over all the classes

(b) centred at the class marks of the classes

(c) centred at the upper limits of the class

(d) centred at the lower limits of the class

Answer. (b)          

Expalnation:

While grouping the data all the observations between lower and upper limits of class marks are taken in one group then mid value is taken for any further calculation. In computing the mean of grouped data, it is assumed that the frequency of each class interval is centred around its mid value or class mark.

Question. 5 The abscissa of the point of intersection of the Less than type and of the more than type cumulative frequency curves of a grouped data gives its

(a) mean                                               

(b) median           

(c) mode                                               

(d) All of these

Answer. (b)

Explanation:

Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa because on X-axis we take the upper or lower limits respectively and on Y-axis we take cumulative frequency.

Answer. (b)          

Explanation:

Commulative frequency for the given data is given below:

Class

Frequency

Cumulative frequency

0-5

10

10

5-10

15

25

10-15

12

37

15-20

20

57

20-25

9

66

Now, the modal class is the class having the maximum frequency.

The maximum frequency 20 belongs to class (15 - 20).

So, lower limit of modal class is 15.

And, the class whose cumulative frequency is greater than (and nearest to) n/2 is called the median class.

Here, n/2 = 66/2 = 33, which lies in the interval 10-15.

So, lower limit of the median class is 10.

Hence, required sum of lower limits of the median class and modal class is 10 + 15 = 25.

(a) 17                                    

(b) 17.5                                 

(c) 18                                     

(d) 18.5

Answer. (b)

Explanation:

First make the classes continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.

Then find the Commulative frequency for the given data as below:

Class

Frequency

Cumulative frequency

0.5-5.5

13

13

5.5-11.5

10

23

11.5-17.5

15

38

17.5-23.5

8

46

23.5-29.5

11

57

Now, the class whose cumulative frequency is greater than (and nearest to) n/2 is called the median class.

Here, n/2 = 57/2 = 28.5 which lies in the interval 11.5 - 17.5.

Hence, the upper limit is 17.5.

Question. 8 For the following distribution,

Marks

Number of students

Below 10

3

Below 20

12

Below 30

27

Below 40

57

Below 50

75

Below 60

80

the modal class is

(a) 10-20                                     

(b) 20-30                                     

(c) 30-40                                     

(d) 50-60

Answer. (c)

Explanation:

Commulative frequency for the given data is calculated as below:

Marks

Number of students

Cumulative frequency

Below 10

 3

3

10-20

 9

12

20-30

15

27

30-40

30

57

40-50

18

75

50-60

5

80

Now the class whose frequency is maximum is called the modal class. Here, we see the highest frequency is 30, which lies in the interval 30-40. Therefore, 30-40 is the modal class.

Question.  9 Consider the data

The difference of the upper limit of the median class and the lower limit of the modal class is

(a) 0                                       

(b) 19                                    

(c) 20                                     

(d) 38

Answer.  (c)

Explanation:

Commulative frequency for the given data is calculated as below:

Class

Frequency

Cumulative frequency

65-85

4

4

85-105

5

9

105-125

13

22

125-145

20

42

145-165

14

56

165-185

7

63

185-205

4

67

Now, the class whose cumulative frequency is greater than (and nearest to) n/2 is called the median class.

Here, n/2 = 67/2 = 33.5, which lies in the interval 125 - 145.

Hence, upper limit of median class is 145.

Again, the class whose frequency is maximum is called the modal class.

Here, 20 is the highest frequency which lies in the interval 125 –145.

Hence, the lower limit of modal class is 125.

Thus, the required difference = Upper limit of median class - Lower limit of modal class

                                                = 145 –125 = 20.

Question. 10 The times (in seconds) taken by 150 atheletes to run a 110 m hurdle race are tabulated below

The number of atheletes who completed the race in less than 14.6 s is

(a) 11    

(b) 71    

(c) 82     

(d) 130

Answer. (c)

Exlanation:

 The number of atheletes who completed the race in less than 14.6

= 2 + 4 + 5 + 71 = 82.

Question.  11 Consider the following distribution

Marks obtained

Number of students

More than or equal to 0

63

More than or equal to 10

58

More than or equal to 20

55

More than or equal to 30

51

More than or equal to 40

48

More than or equal to 50

42

the frequency of the class 30-40 is

(a) 3       

(b) 4       

(c) 48     

(d) 51

Answer. (a)

Explanation:

The given frequency data can be represented in the form of class intervals as follows:

Marks obtained

Number of students

Frequency

0-10

(63 – 58) = 5

5

10-20

(58 – 55) = 3

3

20-30

(55 – 51) = 4

4

30-40

(51 – 48) = 3

3

40-50

(48 – 42) = 6

6

50…

42 − 0 = 42

42

Hence the frequency of 30 – 40 class interval is 3.

CBSE Class 10 NCERT Textbooks & NCERT Solutions

NCERT Solutions for CBSE Class 10 Maths

NCERT Exemplar Problems and Solutions Class 10 Science: All Chapters

Show Full Article

Post Comment

Register to get FREE updates

All Fields Mandatory
  • Please Select Your Interest
  • By clicking on Submit button, you agree to our terms of use
  • A verifcation code has been sent to
    your mobile number

    Please enter the verification code below

X

Register to view Complete PDF

Newsletter Signup

Copyright 2017 Jagran Prakashan Limited.