RD Sharma Solutions for Class 7 Maths Chapter 23 Data Handling III (Construction of Bar Graphs)
Chapter 22 of RD Sharma's Class 7 Maths book, titled "Data Handling II (Central Values)," introduces students to the concept of central values in statistics. Data handling is an essential part of mathematics as it helps us understand how to organize, analyze, and interpret information in a meaningful way. In this chapter, students will learn about various ways to summarize and represent data, especially focusing on central values like the mean, median, and mode.
The concept of central values helps us find the "central" or "average" point of a set of numbers. By calculating these central values, we can easily understand the overall pattern or trend of the data, which is often helpful in making decisions or conclusions. For example, if you want to know the average score of students in a class or the most common number in a list of exam results, central values provide a simple way to summarize large amounts of data.
In this chapter, students will first learn about the mean, which is calculated by adding up all the numbers in a data set and dividing the total by the number of values. Next, students will explore the median, which is the middle value when the data is arranged in order. Finally, the mode is introduced, which is the value that occurs most frequently in a data set. Understanding these concepts is essential for solving real-life problems that involve data analysis.
Do Check: RD Sharma Solutions for Class 7 Maths
The RD Sharma Solutions for Class 7 Maths Chapter 22 offer detailed explanations and step-by-step methods to solve problems related to central values. Each solution is designed to help students build a strong foundation in understanding data handling, allowing them to tackle a variety of questions with ease. These solutions not only help in grasping the theoretical aspects but also ensure that students can apply these concepts to practical situations.
With the help of this chapter, students will be able to improve their skills in handling and interpreting data, which will be useful not only in exams but also in day-to-day life where data plays an important role.
Download RD Sharma Solutions for Class 7 Maths Chapter 23 Data Handling II PDF Here
RD Sharma Class 7 Chapter 23 PDF includes detailed solutions, examples, and extra questions to help you master real numbers and other topics. Click here to download the RD Sharma Class 7 Chapter 23 PDF.
Access Answers to RD Sharma Solutions for Class 7 Maths Chapter 23 Data Handling II
In this chapter, students will learn about decimals and how to perform basic operations with them. The solutions provided here are detailed and easy to follow, helping students understand each concept thoroughly.
Q1. Ashish studies for 4 hours, 5 hours, and 3 hours on three consecutive days. How many hours does he study daily on average?
Solution:
Given that Ashish studies for 4 hours, 5 hours, and 3 hours on three consecutive days.
Average study hours = Total hours ÷ Number of days
Average study hours = (4 + 5 + 3) ÷ 3
= 12 ÷ 3
= 4 hours
Thus, Ashish studies for 4 hours on average.
Q2. A cricketer scores the following runs in 8 innings: 58, 76, 40, 35, 48, 45, 0, 100. Find the mean score.
Solution:
Given the runs in 8 innings: 58, 76, 40, 35, 48, 45, 0, 100
Mean score = Total runs ÷ Number of innings
Mean score = (58 + 76 + 40 + 35 + 48 + 45 + 0 + 100) ÷ 8
= 402 ÷ 8
= 50.25 runs
Q3. The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 84, 39, 48, 56, 95, 81, and 75. Find the:
(i) Highest and the lowest marks obtained by the students.
(ii) Range of marks obtained.
(iii) Mean marks obtained by the group.
Solution:
To find the highest and lowest marks, we arrange the marks in ascending order:
39, 48, 56, 75, 76, 81, 84, 85, 90, 95
(i) The highest mark is 95, and the lowest mark is 39.
(ii) The range of marks = 95 – 39 = 56.
(iii) Mean marks = Sum of marks ÷ Total number of students
Mean marks = (39 + 48 + 56 + 75 + 76 + 81 + 84 + 85 + 90 + 95) ÷ 10
= 729 ÷ 10
= 72.9 marks
Thus, the mean marks of the students is 72.9.
Q4. The enrolment of a school during six consecutive years was as follows: 1555, 1670, 1750, 2019, 2540, and 2820. Find the mean enrolment of the school for this period.
Solution:
Given the enrolment of the school during six consecutive years: 1555, 1670, 1750, 2019, 2540, and 2820
The mean enrolment = Sum of enrolments ÷ Total number of years
The mean enrolment = (1555 + 1670 + 1750 + 2019 + 2540 + 2820) ÷ 6
= 12354 ÷ 6
= 2059
Thus, the mean enrolment of the school for the given period is 2059.
Q5. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:
| Day | Mon | Tue | Wed | Thu | Fri | Sat | Sun |
| Rainfall (in mm) | 0.0 | 12.2 | 2.1 | 0.0 | 20.5 | 5.3 | 1.0 |
(i) Find the range of the rainfall from the above data.
(ii) Find the mean rainfall for the week.
(iii) On how many days was the rainfall less than the mean rainfall?
Solution:
(i) The range of rainfall = Maximum rainfall – Minimum rainfall
= 20.5 – 0.0
= 20.5 mm
(ii) The mean rainfall = (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.3 + 1.0) ÷ 7
= 41.1 ÷ 7
= 5.87 mm
(iii) There are 5 days (Mon, Wed, Thu, Sat, and Sun) where the rainfall was less than the mean rainfall (5.87 mm).
Q6. If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm, and 161 cm respectively, find the mean height.
Solution:
Mean height = Sum of heights ÷ Total number of persons
Mean height = (140 + 150 + 152 + 158 + 161) ÷ 5
= 761 ÷ 5
= 152.2 cm
Q7. Find the mean of 994, 996, 998, 1002, and 1000.
Solution:
Mean = Sum of the given numbers ÷ Total number of numbers
Mean = (994 + 996 + 998 + 1002 + 1000) ÷ 5
= 4990 ÷ 5
= 998
Q8. Find the mean of the first five natural numbers.
Solution:
The first five natural numbers are 1, 2, 3, 4, and 5.
Mean of the first five natural numbers = (1 + 2 + 3 + 4 + 5) ÷ 5
= 15 ÷ 5
= 3
Q9. Find the mean of all factors of 10.
Solution:
Factors of 10 are 1, 2, 5, and 10.
Arithmetic mean of all factors of 10 = (1 + 2 + 5 + 10) ÷ 4
= 18 ÷ 4
= 4.5
Q10. Find the mean of the first 10 even natural numbers.
Solution:
The first 10 even natural numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20.
Mean of the first 10 even natural numbers = (2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20) ÷ 10
= 110 ÷ 10
= 11
Q11. Find the mean of x, x + 2, x + 4, x + 6, x + 8.
Solution:
Mean = Sum of observations ÷ Number of observations
Mean = (x + x + 2 + x + 4 + x + 6 + x + 8) ÷ 5
Mean = (5x + 20) ÷ 5
Mean = 5 (x + 4) ÷ 5
Mean = x + 4
Q12. Find the mean of the first five multiples of 3.
Solution:
The first five multiples of 3 are 3, 6, 9, 12, and 15.
Mean of the first five multiples of 3 = (3 + 6 + 9 + 12 + 15) ÷ 5
= 45 ÷ 5
= 9
Q13. The following are the weights (in kg) of 10 newborn babies in a hospital on a particular day: 3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6. Find the mean weight.
Solution:
Mean weight = Sum of weights ÷ Number of babies
Mean weight = (3.4 + 3.6 + 4.2 + 4.5 + 3.9 + 4.1 + 3.8 + 4.5 + 4.4 + 3.6) ÷ 10
= 40 ÷ 10
= 4 kg
Q14. The percentage of marks obtained by students of a class in mathematics are: 64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1. Find their mean.
Solution:
Mean = Sum of the marks obtained ÷ Total number of students
Mean = (64 + 36 + 47 + 23 + 0 + 19 + 81 + 93 + 72 + 35 + 3 + 1) ÷ 12
= 474 ÷ 12
= 39.5%
Q15. The number of children in 10 families of a locality are: 2, 4, 3, 4, 2, 3, 5, 1, 1, 5. Find the mean number of children per family.
Solution:
Mean number of children per family = Sum of total number of children ÷ Total number of families
Mean number of children = (2 + 4 + 3 + 4 + 2 + 3 + 5 + 1 + 1 + 5) ÷ 10
= 30 ÷ 10
= 3
Thus, on average, there are 3 children per family in the locality.
FAQs on RD Sharma Solutions for Class 7 Maths Chapter 23 Data Handling II (Central Values)
What are central values in statistics, and why are they important?
Central values, such as the mean, median, and mode, help us summarize large sets of data by finding a typical or average value. These values allow us to understand the overall trend or pattern in a dataset. They are essential for making sense of data and drawing conclusions from it.
What is the difference between the median and the mode?
Median is the middle value when the data is arranged in ascending or descending order. If there is an even number of values, the median is the average of the two middle values.
Mode is the value that appears most frequently in the data set. A dataset can have one mode, more than one mode, or no mode at all.
How do RD Sharma Solutions help in understanding central values?
The RD Sharma Solutions provide step-by-step methods and explanations for calculating central values like the mean, median, and mode. The solutions also include examples and practice problems to ensure students grasp the concepts thoroughly and can apply them in different scenarios.
Are there any real-life applications of central values taught in this chapter?
Yes, central values are used in various real-life situations, such as finding the average score of students in a class, determining the most common item purchased in a store, or analyzing the median income in a neighborhood. Understanding central values helps in making data-driven decisions and conclusions.
What should I do if I find the problems in Chapter 23 difficult to solve?
If you find the problems challenging, you can refer to the RD Sharma Solutions for detailed explanations. Practicing multiple problems from the chapter will also help you become more familiar with the methods. Additionally, seeking help from a teacher or tutor can provide personalized guidance.