Physics Wallah

RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3 PDF

RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3 Statistics has been provided here. Students can refer to these questions before their examinations for better preparation.
authorImageNeha Tanna15 Nov, 2024
Share

Share

RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3

RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3: Chapter 7 of RD Sharma's Class 10 Maths book, which focuses on "Statistics," introduces methods to systematically organize and interpret data. Exercise 7.3 specifically covers the computation of cumulative frequency distributions, cumulative frequency curves, and their applications in analyzing grouped data.

Students learn how to convert data into cumulative frequencies, construct graphs, and utilize these visuals to estimate median values. This exercise strengthens skills in understanding data distribution patterns and prepares students for practical applications of statistics. It emphasizes visualizing data trends, a fundamental aspect of real-world data analysis and interpretation.

RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3 Overview

Chapter 7 of RD Sharma’s Class 10 Maths book covers Statistics, which is crucial for developing data interpretation skills. Exercise 7.3, in particular, focuses on calculating measures of central tendency, such as mean, median, and mode, which help summarize data into a single representative value. These concepts are foundational for higher-level statistics and practical data analysis in various fields, including economics, science, and social studies. Understanding and practicing this exercise equips students with the skills to organize, interpret, and make informed conclusions from data, an essential competency in our data-driven world.

RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3 PDF

Below, we have provided a PDF for RD Sharma Solutions for Class 10 Maths Chapter 7, Exercise 7.3, focusing on Statistics. This exercise is essential for students aiming to understand data interpretation, mean, median, and mode calculations, and graphical representation techniques. These solutions are comprehensive, making it easier to grasp concepts and solve problems accurately. Perfect for revising before exams, this PDF guide will help you tackle all statistics-related questions confidently. Download the PDF to access step-by-step solutions for every problem in Exercise 7.3.

RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3 PDF

RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3 Statistics

Below is the RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3 Statistics -

1. The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.

Expenditure (in rupees) (x) Frequency (f i ) Expenditure (in rupees) (x i ) Frequency (f i )
100 – 150 24 300 – 350 30
150 – 200 40 350 – 400 22
200 – 250 33 400 – 450 16
250 – 300 28 450 – 500 7

Find the average expenditure (in rupees) per household.

Solution:

Let the assumed mean (A) = 275
Class interval Mid value (x i ) d i = x i – 275 u i = (x i – 275)/50 Frequency f i f i u i
100 – 150 125 -150 -3 24 -72
150 – 200 175 -100 -2 40 -80
200 – 250 225 -50 -1 33 -33
250 – 300 275 0 0 28 0
300 – 350 325 50 1 30 30
350 – 400 375 100 2 22 44
400 – 450 425 150 3 16 48
450 – 500 475 200 4 7 28
N = 200 Σ f i u i = -35
It’s seen that A = 275 and h = 50 So, Mean = A + h x (Σf i u i /N) = 275 + 50 (-35/200) = 275 – 8.75 = 266.25

2. A survey was conducted by a group of students as a part of their environmental awareness program, in which they collected the following data regarding the number of plants in 200 houses in a locality. Find the mean number of plants per house.

Number of plants: 0 – 2 2 – 4 4 – 6 6 – 8 8 – 10 10 – 12 12 – 14
Number of house: 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why?

Solution:

From the given data, To find the class interval, we know that, Class marks (x i ) = (upper class limit + lower class limit)/2 Now, let’s compute x i and f i x i by the following
Number of plants Number of house (f i ) x i f i x i
0 – 2 1 1 1
2 – 4 2 3 6
4 – 6 1 5 5
6 – 8 5 7 35
8 – 10 6 9 54
10 – 12 2 11 22
12 – 14 3 13 39
Total N = 20 Σ f i u i = 162
Here, Mean = Σ f i u i / N = 162/ 20 = 8.1 Thus, the mean number of plants in a house is 8.1 We have used the direct method as the values of class mark x i and f i are very small.

3. Consider the following distribution of daily wages of workers of a factory

Daily wages (in ₹) 100 – 120 120 – 140 140 – 160 160 – 180 180 – 200
Number of workers: 12 14 8 6 10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

Let the assume mean (A) = 150
Class interval Mid value x i d i = x i – 150 u i = (x i – 150)/20 Frequency f i f i u i
100 – 120 110 -40 -2 12 -24
120 – 140 130 -20 -1 14 -14
140 – 160 150 0 0 8 0
160 – 180 170 20 1 6 6
180 – 200 190 40 2 10 20
N= 50 Σ f i u i = -12
It’s seen that, A = 150 and h = 20 So, Mean = A + h x (Σf i u i /N) = 150 + 20 x (-12/50) = 150 – 24/5 = 150 = 4.8 = 145.20

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

Number of heart beats per minute: 65 – 68 68 – 71 71 – 74 74 – 77 77 – 80 80 – 83 83 – 86
Number of women: 2 4 3 8 7 4 2

Solution:

Using the relation (x i ) = (upper class limit + lower class limit)/ 2 And, class size of this data = 3 Let the assumed mean (A) = 75.5 So, let’s calculate d i , u i , f i u i as follows:
Number of heart beats per minute Number of women (f i ) x i d i = x i – 75.5 u i = (x i – 755)/h f i u i
65 – 68 2 66.5 -9 -3 -6
68 – 71 4 69.5 -6 -2 -8
71 – 74 3 72.5 -3 -1 -3
74 – 77 8 75.5 0 0 0
77 – 80 7 78.5 3 1 7
80 – 83 4 81.5 6 2 8
83 – 86 2 84.5 9 3 6
N = 30 Σ f i u i = 4
From the table, it’s seen that N = 30 and h = 3 So, the mean = A + h x (Σf i u i /N) = 75.5 + 3 x (4/30 = 75.5 + 2/5 = 75.9 Therefore, the mean heartbeats per minute for those women are 75.9 beats per minute.

Find the mean of each of the following frequency distributions: (5 – 14)

5.

Class interval: 0 – 6 6 – 12 12 – 18 18 – 24 24 – 30
Frequency: 6 8 10 9 7

Solution:

Let’s consider the assumed mean (A) = 15
Class interval Mid – value x i d i = x i – 15 u i = (x i – 15)/6 f i f i u i
0 – 6 3 -12 -2 6 -12
6 – 12 9 -6 -1 8 -8
12 – 18 15 0 0 10 0
18 – 24 21 6 1 9 9
24 – 30 27 12 2 7 14
N = 40 Σ f i u i = 3
From the table, it’s seen that, A = 15 and h = 6 Mean = A + h x (Σf i u i /N) = 15 + 6 x (3/40) = 15 + 0.45 = 15.45

6.

Class interval: 50 – 70 70 – 90 90 – 110 110 – 130 130 – 150 150 – 170
Frequency: 18 12 13 27 8 22

Solution:

Let’s consider the assumed mean (A) = 100

Class interval Mid – value x i d i = x i – 100 u i = (x i – 100)/20 f i f i u i
50 – 70 60 -40 -2 18 -36
70 – 90 80 -20 -1 12 -12
90 – 110 100 0 0 13 0
110 – 130 120 20 1 27 27
130 – 150 140 40 2 8 16
150 – 170 160 60 3 22 66
N= 100 Σ f i u i = 61
From the table, it’s seen that, A = 100 and h = 20 Mean = A + h x (Σf i u i /N) = 100 + 20 x (61/100) = 100 + 12.2 = 112.2

7.

Class interval: 0 – 8 8 – 16 16 – 24 24 – 32 32 – 40
Frequency: 6 7 10 8 9

Solution:

Let’s consider the assumed mean (A) = 20
Class interval Mid – value x i d i = x i – 20 u i = (x i – 20)/8 f i f i u i
0 – 8 4 -16 -2 6 -12
8 – 16 12 -8 -1 7 -7
16 – 24 20 0 0 10 0
24 – 32 28 8 1 8 8
32 – 40 36 16 2 9 18
N = 40 Σ f i u i = 7
From the table, it’s seen that, A = 20 and h = 8 Mean = A + h x (Σf i u i /N) = 20 + 8 x (7/40) = 20 + 1.4 = 21.4

8.

Class interval: 0 – 6 6 – 12 12 – 18 18 – 24 24 – 30
Frequency: 7 5 10 12 6

Solution:

Let’s consider the assumed mean (A) = 15
Class interval Mid – value x i d i = x i – 15 u i = (x i – 15)/6 f i f i u i
0 – 6 3 -12 -2 7 -14
6 – 12 9 -6 -1 5 -5
12 – 18 15 0 0 10 0
18 – 24 21 6 1 12 12
24 – 30 27 12 2 6 12
N = 40 Σ f i u i = 5
From the table, it’s seen that, A = 15 and h = 6 Mean = A + h x (Σf i u i /N) = 15 + 6 x (5/40) = 15 + 0.75 = 15.75

9.

Class interval: 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50
Frequency: 9 12 15 10 14

Solution:

Let’s consider the assumed mean (A) = 25
Class interval Mid – value x i d i = x i – 25 u i = (x i – 25)/10 f i f i u i
0 – 10 5 -20 -2 9 -18
10 – 20 15 -10 -1 12 -12
20 – 30 25 0 0 15 0
30 – 40 35 10 1 10 10
40 – 50 45 20 2 14 28
N = 60 Σ f i u i = 8
From the table, it’s seen that, A = 25 and h = 10 Mean = A + h x (Σf i u i /N) = 25 + 10 x (8/60) = 25 + 4/3 = 79/3 = 26.333

10.

Class interval: 0 – 8 8 – 16 16 – 24 24 – 32 32 – 40
Frequency: 5 9 10 8 8

Solution:

Let’s consider the assumed mean (A) = 20
Class interval Mid – value x i d i = x i – 20 u i = (x i – 20)/8 f i f i u i
0 – 8 4 -16 -2 5 -10
8 – 16 12 -4 -1 9 -9
16 – 24 20 0 0 10 0
24 – 32 28 4 1 8 8
32 – 40 36 16 2 8 16
N = 40 Σ f i u i = 5
From the table, it’s seen that, A = 20 and h = 8 Mean = A + h x (Σf i u i /N) = 20 + 8 x (5/40) = 20 + 1 = 21

11.

Class interval: 0 – 8 8 – 16 16 – 24 24 – 32 32 – 40
Frequency: 5 6 4 3 2

Solution:

Let’s consider the assumed mean (A) = 20
Class interval Mid – value x i d i = x i – 20 u i = (x i – 20)/8 f i f i u i
0 – 8 4 -16 -2 5 -12
8 – 16 12 -8 -1 6 -8
16 – 24 20 0 0 4 0
24 – 32 28 8 1 3 9
32 – 40 36 16 2 2 14
N = 20 Σ f i u i = -9
From the table, it’s seen that, A = 20 and h = 8 Mean = A + h x (Σf i u i /N) = 20 + 6 x (-9/20) = 20 – 72/20 = 20 – 3.6 = 16.4

12.

Class interval: 10 – 30 30 – 50 50 – 70 70 – 90 90 – 110 110 – 130
Frequency: 5 8 12 20 3 2

Solution:

Let’s consider the assumed mean (A) = 60
Class interval Mid – value x i d i = x i –60 u i = (x i – 60)/20 f i f i u i
10 – 30 20 -40 -2 5 -10
30 – 50 40 -20 -1 8 -8
50 – 70 60 0 0 12 0
70 – 90 80 20 1 20 20
90 – 110 100 40 2 3 6
110 – 130 120 60 3 2 6
N = 50 Σ f i u i = 14
From the table, it’s seen that, A = 60 and h = 20 Mean = A + h x (Σf i u i /N) = 60 + 20 x (14/50) = 60 + 28/5 = 60 + 5.6 = 65.6

13.

Class interval: 25 – 35 35 – 45 45 – 55 55 – 65 65 – 75
Frequency: 6 10 8 12 4

Solution:

Let’s consider the assumed mean (A) = 50
Class interval Mid – value x i d i = x i – 50 u i = (x i – 50)/10 f i f i u i
25 – 35 30 -20 -2 6 -12
35 – 45 40 -10 -1 10 -10
45 – 55 50 0 0 8 0
55 – 65 60 10 1 12 12
65 – 75 70 20 2 4 8
N = 40 Σ f i u i = -2
From the table, it’s seen that, A = 50 and h = 10 Mean = A + h x (Σf i u i /N) = 50 + 10 x (-2/40) = 50 – 0.5 = 49.5

14.

Class interval: 25 – 29 30 – 34 35 – 39 40 – 44 45 – 49 50 – 54 55 – 59
Frequency: 14 22 16 6 5 3 4

Solution:

Let’s consider the assumed mean (A) = 42
Class interval Mid – value x i d i = x i – 42 u i = (x i – 42)/5 f i f i u i
25 – 29 27 -15 -3 14 -42
30 – 34 32 -10 -2 22 -44
35 – 39 37 -5 -1 16 -16
40 – 44 42 0 0 6 0
45 – 49 47 5 1 5 5
50 – 54 52 10 2 3 6
55 – 59 57 15 3 4 12
N = 70 Σ f i u i = -79
From the table, it’s seen that, A = 42 and h = 5 Mean = A + h x (Σf i u i /N) = 42 + 5 x (-79/70) = 42 – 79/14 = 42 – 5.643 = 36.357

Benefits of Solving RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3

Solving RD Sharma Solutions for Class 10 Maths Chapter 7 Exercise 7.3 on Statistics offers several benefits for students preparing for their exams. Here’s how it helps: 1. Strengthens Conceptual Understanding Exercise 7.3 focuses on the application of statistical measures, such as mean, median, and mode. Working through these solutions deepens the understanding of these fundamental statistical concepts and helps students remember the formulas and methods. 2. Enhances Problem-Solving Skills The exercise includes various types of questions, allowing students to practice and strengthen their problem-solving abilities. By attempting diverse problems, students learn to apply the correct methods in a structured way. 3. Builds Analytical Skills Solving statistics problems involves analyzing data and choosing appropriate techniques. This analytical approach to solving questions can benefit students not just in mathematics but in other subjects involving data interpretation as well. 4. Improves Accuracy and Speed Regular practice with these solutions improves calculation accuracy and speeds up the time taken to solve problems, which is crucial in a timed exam setting. Consistency in solving RD Sharma exercises helps students become faster and more precise. 5. Prepares for Competitive Exams This chapter is fundamental in many entrance exams and competitive tests. Understanding the basic statistical concepts here helps students prepare for exams beyond school, as they form the foundation for more advanced topics in statistics.

RD Sharma Solutions Class 10 Maths Chapter 7 Exercise 7.3 FAQs

What are the 2 main methods in statistics?

Two main statistical methods are used in data analysis: descriptive statistics, which summarizes data using indexes such as mean and median and another is inferential statistics, which draw conclusions from data using statistical tests such as student's t-test.

What is the basic concept of statistics?

In general, statistics is a study of data: describing properties of the data, which is called descriptive statistics, and drawing conclusions about a population of interest from information extracted from a sample, which is called inferential statistics.

What does ∑ mean in statistics?

summation

What is the N symbol in statistics?

N = population size. ND = normal distribution, whose graph is a bell-shaped curve; also “normally distributed”. p = probability value. The specific meaning depends on context.
Popup Close ImagePopup Open Image
Talk to a counsellorHave doubts? Our support team will be happy to assist you!
Popup Image
CTA Image
008448982616
Join 15 Million students on the app today!
Point IconLive & recorded classes available at ease
Point IconDashboard for progress tracking
Point IconMillions of practice questions at your fingertips
Download ButtonDownload Button
Banner Image
Banner Image
Free Learning Resources
Know about Physics Wallah
Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.
We Stand Out because
We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.
Our Key Focus Areas
Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.
What Makes Us Different
Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others
Privacy Policy
Terms of use
Copyright © 2025 Physicswallah Limited All rights reserved.