RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions for Class 7 Mathematics

Created by: Abhishek Kapoor

Class 7 : RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

The document RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev is a part of the Class 7 Course RD Sharma Solutions for Class 7 Mathematics.
All you need of Class 7 at this link: Class 7

Question 1:

Find the mode and median of the data: 13, 16, 12, 14, 19, 12, 14, 13, 14
 By using the empirical relation also find the mean.

Answer 1:

Arranging the data in ascending order such that same numbers are put together, we get:
12,12,13,13, 14,14,14, 16, 19
Here, n = 9.
∴ Median = Value of  RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev  observation = Value of the 5th observation = 14.
Here, 14 occurs the maximum number of times, i.e., three times. Therefore, 14 is the mode of the data.

Now,
Mode = 3 Median - 2 Mean
⇒ 14 = 3 x 14 - 2 Mean
⇒2 Mean  = 42 - 14 = 28
⇒ Mean = 28 ÷ 2 = 14.

 

Question 2:

Find the median and mode of the data: 35, 32, 35, 42, 38, 32, 34

Answer 2:

Arranging the data in ascending order such that same numbers are put together, we get:

32, 32, 34,35,35, 38,42.

Here, n = 7
∴ Median = Value of RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev observation = Value of the 4th observation = 35.
Here, 32 and 35, both occur twice. Therefore, 32 and 35 are the two modes.

 

Question 3:

Find the mode of the data: 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4

Answer 3:

Arranging the data in ascending order such that same values are put together, we get:

0, 2, 2, 2, 3, 3,3,4,4,4,5,5,6.

Here, 2,3 and 4 occur three times each. Therefore, 2 ,3 and 4 are the three modes.

Alternate Solution
Arranging the data in the form of a frequency table, we have:

ValuesTally BarsFrequency
01
2∣∣∣3
3∣∣∣3
4∣∣∣3
5∣∣2
61
Total 13

 

Clearly, the values 2,3 and 4 occur the maximum number of times, i.e., three times.
Hence, the mode is 2,3 and 4.

 

Question 4:

The runs scored in a cricket match by 11 players are as follows:
 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 10
 Find the mean, mode and median of this data.

Answer 4:

Arranging the data in ascending order such that same values are put together, we get:

6,8,10,10,10,15,15,50,80,100,120.

Here, n = 11
∴ Median = Value of RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev observation = Value of the 6th observation = 15.
Here, 10 occurs three times. Therefore, 10 is the mode of the given data.
Now, 
Mode = 3 Median - 2 Mean
⇒ 10 = 3 x 15 - 2 Mean
⇒2 Mean  = 45 - 10 = 35
⇒ Mean = 35 ÷ 2 = 17.5.

 

Question 5:

Find the mode of the following data:
 12, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14

Answer 5:

Arranging the data in ascending order such that same values are put together, we get:

10,12,12,14,14,14,14,14,14, 16, 16, 18.

Here, clearly, 14 occurs the most number of times.
Therefore, 14 is the mode of the given data.

Alternate solution:
Arranging the data in the form of a frequency table, we get:

  ValuesTally BarsFrequency
101
12∣∣2
14RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev6
16∣∣2
181
Total 12


Clearly, 14 has maximum frequency. So, the mode of the given data is 14. 

 

Question 6:

Heights of 25 children (in cm) in a school are as given below:
 168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 163, 165, 163, 162, 163, 164, 163, 160, 165, 163, 162
 What is the mode of heights?
 Also, find the mean and median.

Answer 6:

Arranging the data in tabular form, we get:

Height of Children (cm) Tally BarsFrequency
160∣∣∣3
1611
162∣∣∣∣4
163RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev10
164∣∣∣3
165∣∣∣3
1681
Total 25


Here, n = 25
∴ Median = Value of RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev observation = Value of the 13th observation = 163 cm.
Here, clearly, 163 cm occurs the most number of times. Therefore, the mode of the given data is 163 cm.
Now, 
Mode = 3 Median - 2 Mean
⇒ 163 = 3 x 163 - 2 Mean
⇒2 Mean  =  326
⇒ Mean = 326 ÷ 2 = 163 cm.

Question 7:

The scores in mathematics test (out of 25) of 15 students are as follows:
 19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20
 Find the mode and median of this data. Are they same?

Answer 7:

Arranging the data in ascending order such that same values are put together, we get:

5,9,10,12,15,16, 19, 20, 20, 20, 20,  23, 24, 25, 25.
Here, n = 15
∴ Median = Value of  RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev observation = Value of the 8th observation = 20.
Here, clearly, 20 occurs the most number of times, i.e., 4 times. Therefore, the mode of the given data is 20.
Yes, the median and mode of the given data are the same.

 

Question 8:

Calculate the mean and median for the folllowing data:

Marks:1011121314161920
Number of students:35452321

Using empirical formula, find its mode.

Answer 8:

 

                               Calculation of Mean

Marks (xi)1011121314161920Total
Number of Students (fi)35452321RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev
fixi3055486528483820 RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

 

Mean  =    RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

Here, n = 25, which is an odd number. Therefore, 
Median = Value of RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev observation = the 13th observation = 13.
Now,
Mode = 3 Median - 2 Mean

⇒Mode = 3 x 13 - 2 x (13.28) 
⇒Mode = 39 - 26.56
⇒Mode = 12.44.

 

Question 9:

The following table shows the weights of 12 persons.

Weight (in kg):4850525458
Number of persons:43221

Find the median and mean weights. Using empirical relation, calculate its mode.

Answer 9:

Calculation of Mean

Weight (xi)4850525458Total
Number of Persons (fi)43221  RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev
fixi19215010410858RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

Here, n = 12

RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

RD Sharma Solutions - Ex-23.4, Data Handling II Central Values, Class 7, Math Class 7 Notes | EduRev

Now, 
     Mode = 3 Median - 2 Mean
⇒ Mode = 3 x 50 - 2 x 51
⇒Mode  = 150 - 102 
⇒ Mode = 48 kg.
Thus, Mean = 51 kg, Median = 50 kg and Mode = 48 kg.

Complete Syllabus of Class 7

Dynamic Test

Content Category

Related Searches

Extra Questions

,

RD Sharma Solutions - Ex-23.4

,

Sample Paper

,

pdf

,

Math Class 7 Notes | EduRev

,

practice quizzes

,

study material

,

RD Sharma Solutions - Ex-23.4

,

ppt

,

Data Handling II Central Values

,

shortcuts and tricks

,

mock tests for examination

,

Previous Year Questions with Solutions

,

Viva Questions

,

past year papers

,

Exam

,

Summary

,

Important questions

,

Objective type Questions

,

Data Handling II Central Values

,

video lectures

,

Semester Notes

,

Math Class 7 Notes | EduRev

,

Class 7

,

Class 7

,

Free

,

RD Sharma Solutions - Ex-23.4

,

Math Class 7 Notes | EduRev

,

MCQs

,

Data Handling II Central Values

,

Class 7

;