NCERT Solutions Class 7 Maths Chapter 3
NCERT Solutions Class 7 Mathematics Chapter 3
NCERT Solutions for Class 7 Mathematics Chapter 3 Data Handling
In this world, data is the fuel that allows us to explore and interpret a wide range of possibilities. The concept of data and its importance are introduced in the NCERT Class 7 Mathematics Chapter 3 data handling. This chapter covers everything from data collection to data storage and display, as well as how to determine the central tendency of the data collected. In Chapter 3, the relevance of the average, median, and mode to various data sets is also discussed, with examples. It also introduces the concept of probability and its theorem, which can be used to determine the likelihood of an event occurring.
To understand the chapter in a better way, students should practise in-text and end-text exercises.They can solve these questions on their own or refer to NCERT Solutions for Class 7 Mathematics Chapter 3 to get answers to all the questions. Extramarks provides these solutions to help students to strengthen their preparation for the tests, exams or Olympiad .
NCERT Solutions for Class 7 Mathematics Chapter 3 Data Handling
Chapter 3 – Data Handling Exercises | |
Exercise 3.1 | Questions & Solutions |
Exercise 3.2 | Questions & Solutions |
Exercise 3.3 | Questions & Solutions |
Exercise 3.4 | Questions & Solutions |
NCERT Solutions for Class 7 Mathematics Chapter 3
Chapter 3 Data Handling
Chapter 3 of Class 7 Mathematics Data handling is a chapter that describes the procedures for working with data and storing it in various formats. According to the CBSE board, this is an important chapter because it teaches us how to analyse data from various fields.
NCERT Solutions Class 7 Mathematics Chapter 3
The answers given in NCERT Solutions for Class 7 Mathematics Chapter 3 cover a variety of important concepts that are necessary for understanding higher-level mathematics. The concept of calculating mean, median, and mode is thoroughly explained, and students will be able to recall it with ease. The average value is referred to as ‘mean,’ the midpoint value is referred to as ‘median,’ and the most common value is referred to as ‘mode.’
All the exercises from 3.1 to 3.9 are included in the NCERT Solutions for Class 7 Mathematics Chapter 3. In this article, you will find detailed solutions prepared by Extramarks experienced subject matter experts. These solutions are provided here in accordance with the NCERT textbook and CBSE guidelines.
3.1 Introduction
Students can learn what data is in the introduction to the data handling chapter. Data is simply well-organised information. Students will learn about the various steps in handling specific data, such as collecting the data, properly analysing it, evaluating it to obtain an average, and finally, how to represent the data.
3.2 Collecting Data
This section explains how students can learn about data collection from various sources. It also explains how to extract data from it and how to find answers by correlating with the situation. Consider some of the following scenarios:
- Gujarat literacy rate-year?
- Bangalore current weather report
- The number of females in a school
3.3 Organisation of Data
Students will learn how to properly organise data collected in this section. Students can learn how to organise and record data in a specific format. NCERT Solution for Class 7 Mathematics Chapter 3 explains all of the different types of data that are used to represent forms. The tabular format is a widely used and basic format for accurately organising data. Proper organisation of data is very important since it ensures that data is easy to understand and interpret.
3.4 Representative Values
Students become aware of these values, which are referred to as representative values, in this lesson. These are used to explain how the data is organised. In the data handling chapter, students must remember the word ‘average.’ The term can be applied to a variety of situations, such as average temperature, average time, average distance, average number of boys, and so on. As a result, students should bear in mind the approximate value rather than an exact value. In NCERT Solutions for Class 7 Mathematics Chapter 3, it is explained with examples and tables.
3.5 Arithmetic Mean
Students will learn about an important aspect of data management in this section. The average value of a group of units is known as the arithmetic mean or mean. It is computed using the formula:
The sum of a unit’s total values/The total number of units
The NCERT Solutions contains examples with solutions applying the formula. If they ever get stuck on a question, they can always refer to the solutions prepared by subject matter experts.
3.5.1 Range
Students are taught the sub-topic of arithmetic mean. They must take an average for the mean, but must subtract the minimum from the maximum to obtain the range of complete data or observation.
3.6 Mode
Students can learn about another important topic known as a mode in this section. The mode, like the mean, is another type of symbolic value. Students must use various types of values depending on the type of data. The mode is a value that appears repeatedly in a set of observations. It means that the value that appears the most times can be considered the mode.
3.6.1 Mode of Large Data
We tabulate the large data to find the mode, and the frequency of the data helps us find the mean of the data. Students may be sceptical of the mode. How can the problem be solved if the data is too large? The problem is addressed in this part.
3.7 Median
Students will learn about yet another type of representative value in this section. The ascending order should be remembered here. Find the exact middle value by writing all the observations in ascending order. It is capable of accurately classifying them into two groups. It’s known as the median because it’s located in the centre. In other words, the Median is the mid-value that divides a unit into two equal halves.
3.8 Use of Bar Graphs with a Different Purpose
The use of bar graphs explores the various applications of the bar graph, including how it can be used to present data on frequency of mode on a graph. Students will learn about the concept of a bar graph and its variations in this lesson. It’s a simple way to represent data in a structured way.
3.8.1 Choosing a Scale
After learning about bar graphs, students will be aware of using the scale to draw the bar graph. Students can choose a larger scale if the data is too large, and vice versa. Students can choose one unit that equals 1,10,100, and so on, depending on the values and size. The choosing scale may vary, but the plotting must be on point.
Drawing Double Bar Graphs
Students can learn more about bar graphs in this section. NCERT Mathematics Class 7 Chapter 3 clearly explains the concepts when data is given for two or more people to compare. Students must create bar Graphs for two groups, but must differentiate them by shading or colouring.
3.9 Chance and Probability
Students must estimate and expect chance and probability in the final section. Simply put, it is the possibility of something occurring. Probability is the term used when chance is defined mathematically. The ratio of favourable cases to the total number of possible cases determines the probability of an event occurring. Each topic is explained separately in NCERT Solutions for Class 7 Mathematics Chapter 3 so that the concept is clearly understood.
3.9.1 Chance
Chance is nothing more than a forecast. Students will be able to predict and express the data’s outcome. It is necessary to predict the outcome, whether it is correct or incorrect. This concept is explained clearly with solved examples in these NCERT Solutions.
3.9.2. Probability
Probability is the final topic that students can learn in this chapter. Probability is the estimation of the chances of something happening, such as tossing a coin or rolling dice.
Key Factors of NCERT Solutions for Class 7 Mathematics Chapter 3
While solving these NCERT Solutions, students gain a variety of benefits, some of them are listed below:
- The solutions were created and compiled by subject matter experts , hence their authenticity and accuracy is guaranteed. They are useful for both the teachers and the students, and they can be accessed anywhere without much hassle.
- The solutions are written in clear, simple language and include easily understandable charts and diagrams to ensure that students fully comprehend the subject.
- The solutions have been created with the goal of enhancing the students’ potential and achieving better results.
NCERT Solutions for Class 7 Mathematics
All textbook sums based on triangles, area, and perimeter of different shapes, data handling, integers, and other topics are covered in NCERT Solutions Class 7 Mathematics. As a result, CBSE students will be able to develop Mathematical knowledge and analytical skills.
These solutions allow students of grade 7 to develop an in-depth understanding of Mathematics. Students from other boards refer to the NCERT Solutions for Class 7 Mathematics. They are reliable and accurate. You can access these chapter-wise solutions from the link given below.
NCERT Solutions for Class 7
Students who have a good grasp on the answers to all of the questions in the NCERT textbook can ace the exams easily. We have provided NCERT Solutions Class 7 to assist students in finding the most suitable solutions to their problems. Subject experts have written the answers based on the CBSE curriculum. Every question in the most recent NCERT books has 100% accurate step-by-step solutions. NCERT Solutions will help you imbibe your concepts to the core, ensuring that you retain them in the long run. It encourages students to practise daily without getting stressed and to improve their performance significantly.
Q.1 Find the range of heights of any ten students of your class.
Ans.
$\begin{array}{l}\text{Let the heights (in cm) of any ten students be}\\ \text{124, 126, 133,135, 136, 139, 140, 142, 144, 147}\\ \text{Highest value among these observations}=\text{147}\text{\hspace{0.17em}}\text{cm}\\ \text{Lowest value among these observations}=\text{124 cm}\\ \text{Range}=\text{Highest value-Lowest value}\\ =(\text{147}-\text{124})\text{cm}\\ =\overline{)\text{23 cm}}\text{}\end{array}$
Q.2
4 | 6 | 7 | 5 | 3 | 5 | 4 | 5 | 2 | 6 |
2 | 5 | 1 | 9 | 6 | 5 | 8 | 4 | 6 | 7 |
$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{\hspace{0.17em}}\mathrm{Which}\mathrm{number}\mathrm{is}\mathrm{the}\mathrm{greatest}?\\ \left(\mathrm{ii}\right)\mathrm{Which}\mathrm{number}\mathrm{is}\mathrm{the}\mathrm{lowest}?\\ \left(\mathrm{iii}\right)\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{range}\mathrm{of}\mathrm{the}\mathrm{data}?\\ \left(\mathrm{iv}\right)\mathrm{Find}\mathrm{the}\mathrm{arithmetic}\mathrm{mean}.\end{array}$
Ans.
\begin{array}{l}\text{(i)}\text{\hspace{0.17em}}\text{9}\\ \text{(ii)}\text{\hspace{0.17em}}1\\ (iii)\text{\hspace{0.17em}}\text{Range=9-1}=\overline{)\text{8}}\\ (iv)\text{\hspace{0.17em}}\text{Arithmetic}\text{\hspace{0.17em}}\text{mean=}\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{100}{20}=\overline{)5}\end{array}
Q.3 Find the mean of the first five whole numbers.
Ans.
$\begin{array}{l}\text{First five whole numbers are 0, 1, 2, 3 and 4}\\ \text{Since Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ \text{So, Mean =}\frac{0+1+2+3+4}{5}=\frac{10}{2}=\overline{)5}\\ \text{Thus, mean of first five whole number is 5.}\end{array}$
Q.4 A cricket scores the following runs in eight innings:
58, 76, 40, 35, 46, 45, 0, 100.
Find the mean score.
Ans.
$\begin{array}{l}\text{Since Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\text{}\\ \text{So},\text{\hspace{0.17em}}\text{Mean}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{58+76+40+35+46+45+0+100}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{400}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\overline{)50}\\ \text{Thus, the mean score is 50}\end{array}$
Q.5 Following table shows the points of each player scored in four game.
Player | Game | Game | Game | Game |
A | 14 | 16 | 10 | 10 |
B | 0 | 8 | 6 | 4 |
C | 8 | 11 | Did not Play | 13 |
Now answer the following questions:
(i) find the mean to determine A’ save range number of points scored per game.
(ii) To find mean number of points per game of C,
would you divide the total points by 3 or by 4? Why?
(iii) B played in all four games. How would you find the mean?
(iv) Who is the best performer?
Ans.
\begin{array}{l}\text{(i)}\\ \text{Since Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\text{}\\ \text{So,}\text{\hspace{0.17em}}\text{Mean}=\frac{14+16+10+10}{4}\\ =\frac{50}{4}=\overline{)12.5}\\ \text{Thus, A}\u2019\text{s average number of points scored per game is 12}\text{.5}\\ \text{(ii)}\\ \text{Since C played only 3 games, so to find the mean number of}\\ \text{points per game for C},\text{divide the total points by 3}\text{.}\\ \text{(iii)}\\ \text{Mean of B =}\frac{0+8+6+4}{4}=\frac{18}{4}=4.5\\ \text{(iv)}\\ \text{Since, the mean of A}\text{}\text{\hspace{0.17em}}\text{is better and bigger than B and C, so A is}\\ \text{the best performer}\text{.}\end{array}
Q.6 The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75.
Find the :
(i) Highest and the lowest marks obtained by the students.
(ii) Range of the marks obtained.
(iii) Mean marks obtained by the group.
Ans.
\begin{array}{l}\text{(i) Clearly, the from the given data, the highest and the lowest}\\ \text{marks are 95 and 39 respectively}\text{.}\\ \text{(ii) Range}=\text{Highest}-\text{Lowest}\\ =95-39=56\\ \text{(iii)}\text{}\text{\hspace{0.17em}}\text{Mean}=\text{\hspace{0.17em}}\frac{\text{Sum of all observations}}{\text{Total number of observations}}\text{}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{85+76+90+58+39+48+56+95+81+75}{10}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\overline{)70.3}\end{array}
Q.7 The enrolment in a school during six consecutive years was as follows:
1555, 1670, 1750, 2013, 2540, 2820
Find the mean enrolment of the school for this period.
Ans.
\begin{array}{c}\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ =\frac{1555+1670+1750+2013+2540+2820}{6}\\ =\frac{12348}{6}\\ =\overline{)2058}\end{array}
Q.8 The enrolment in a school during six consecutive years was as follows:
1555, 1670, 1750, 2013, 2540, 2820
Find the mean enrolment of the school for this period.
Ans.
\begin{array}{c}\text{Mean}=\frac{\text{Sum of all observations}}{\text{Total number of observations}}\\ =\frac{1555+1670+1750+2013+2540+2820}{6}\\ =\frac{12348}{6}\\ =\overline{)2058}\end{array}
Q.9 The rain fall (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 | 5.5 | 1.0 |
(
Ans.
\begin{array}{l}\text{(i) Range}=\text{Highest-Lowest}\\ =20.\text{5 mm}\u20130.\text{0 mm}=\overline{)20.5}\text{\hspace{0.17em}}\text{mm}\\ \text{(ii)}\text{\hspace{0.17em}}\text{Mean rainfall}=\text{}\frac{0.0+12.2+2.1+0.0+20.5+5.5+1.0}{7}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{41.3}{7}=5.9\text{\hspace{0.17em}}\text{mm}\\ \text{(iii)}\text{\hspace{0.17em}}\text{For five days}\text{\hspace{0.17em}}\text{(Monday, Wednesday, Thursday, Saturday}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and Sunday), rainfall was less than the mean rainfall}\text{.}\end{array}
Q.10
Ans.
\begin{array}{l}\text{(i) The height of the tallest girl is 151 cm}\text{.}\\ \text{(ii) The height of the shortest girl is 128}\text{.}\\ \text{(iii) Range}=\text{Highest}-\text{Lowest}=\left(151-128\right)\text{cm}=\overline{)23\text{\hspace{0.17em}}\text{cm}}\\ \text{(iv) Mean height}\\ =\frac{135+150+139+128+151+132+146+149+143+141}{10}\\ =\frac{1414}{10}\\ =\overline{)141.4\text{\hspace{0.17em}}\text{cm}}\\ \text{Thus, five girls have more height than the mean height}\text{.}\end{array}
Q.11 The scores in mathmatics test (out of 25) of 15 students is 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?
Ans.
$\begin{array}{l}\text{Arranging the score in ascending order.}\\ \text{5,9,10,12,15,16,19,20,20,20,20,23,24,25,25}\\ \text{Since, mode is that value of observation which occurs for}\\ \text{the most number of time and median is the middle observation.}\\ \text{There are 15 values, so median would be the 8th observation}\\ \text{and 20 occurs 4 times.}\\ \text{So,}\overline{)\text{Mode}=\text{20\hspace{0.17em}and Median}=\text{20}}\\ \text{Yes both are same.}\end{array}$
Q.12
Ans.
\begin{array}{l}\text{Arranging in ascending order to get}\\ \text{6,8,10,10,15,15,15,50,80,100,120}\\ \text{Mean}=\frac{6+8+10+10+15+15+15+50+80+100+120}{11}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{429}{11}=\overline{)39}\\ \text{Mode is that value of observation which occurs for the most}\\ \text{number of time and median is the middle observation}\text{.}\\ \text{There are 11 values, so median would be the 6th observation}\\ \text{and 15 occurs 3 times}\text{.}\\ \text{So,}\overline{)\text{Mode}=\text{15}\text{\hspace{0.17em}}\text{and Median}=\text{15}}\\ \text{No, these three are not same}\text{.}\end{array}
Q.13
Ans.
\begin{array}{l}\text{Arranging the weight in ascending order}\text{.}\\ \text{32,35,36,37,38,38,38,40,42,43,43,43,45,47,50}\\ \text{Mode of given data is that value of observation which occurs for}\\ \text{the most number of time and median is the middle observation}\text{.}\\ \text{There are 15 values, so median would be the 8th observation}\\ \text{So, Median}=\overline{)\text{40}}\\ \text{Also 38 and 43 both occurs 3 times}\text{.}\\ \text{So,}\overline{)\text{Mode}=\text{38}\text{\hspace{0.17em}}\text{and 43}}\\ \text{Yes there are two mode for given data}\text{.}\end{array}
Q.14 Find the mode and median of the data:
13, 16, 12, 14, 19, 12, 14, 13, 14
Ans.
\begin{array}{l}\text{Arrange in ascending order to get:}\\ \text{12,12,13,13,14,14,1416,19}\\ \text{Mode of given data is that value of observation which occurs for}\\ \text{the most number of time and median is the middle observation}\text{.}\\ \text{There are 9 values, so median would be the 5th observation}\\ \text{So, Median}=\overline{)\text{14}}\\ \text{Also 14 occurs 3 times}\text{.}\\ \text{So,}\overline{)\text{Mode}=\text{14}}\end{array}
Q.15
\begin{array}{l}Tellwhetherthestatementistrueorfalse:\\ (i)Themodeisalwaysoneofthenumbersinadata.\\ \left(ii\right)Themeancanbeoneofthenumbersinadata.\\ \left(iii\right)Themedianisalwaysoneofthenumbersinadata.\\ \left(iv\right)Adataalwayshasamode.\\ \left(v\right)Thedata6,4,3,8,9,12,13,9hasmean9.\end{array}
Ans.
\begin{array}{l}\text{(i)}\overline{)\text{True}}\text{, because mode is the value that occurs most of}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{the time and it belongs to the given data}\text{.}\\ \text{(ii)}\overline{)\text{False}}\text{, Mean may or may not belong to the data}\text{.}\\ \text{(iii)}\overline{)\text{True}}\text{, because median is the middle observation}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{of given data}\text{.}\\ \text{(iv)}\overline{)\text{False}}\\ \text{The given data is: 6,4,3,8,9,12,13,9}\text{.}\\ \text{So, Mean}=\frac{6+4+3+8+9+12+13+9}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{64}{8}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8\end{array}
Q.16 Use the bar graph (Fig below) to a answer the following question
(a) Which is the most popular pet?
(b) How many children have dog as a pet?
Ans.
(a) The bar showing cats is the tallest, so cat is the most popular pet.
(b) From the bar graph, the number of children having dog as a pet is 8.
Q.17
$\text{Read the bar graph}\left(\text{Figbelow}\right)\text{and answer the questions}\phantom{\rule{0ex}{0ex}}\text{hat follow:}$ $\text{Number of books sold by a book store during five}\phantom{\rule{0ex}{0ex}}\mathrm{consecutive}\mathrm{years}.$\begin{array}{l}\left(\text{i}\right)\text{About how many books were sold in 1989? 1990? 1992?}\\ \left(\text{ii}\right)\text{Inwhich year were about 475 books sold?}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{About 225 books sold?}\\ \left(\text{iii}\right)\text{In which years were fewer than 25}0\text{books sold}?\\ \left(\text{iv}\right)\text{Can you explain how you would estimate the number}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{of books sold in 1989?}\end{array}
Ans.
\begin{array}{l}\text{(i) In 1989, 175 books were sold in 1990, 475 books were sold}\text{.}\\ \text{In 1992, 225 books were sold}\text{.}\\ \text{(ii) From the graph, it can be concluded that 475 books were}\\ \text{sold in the year 1990 and 225 books were sold in the year 1992}\text{.}\\ \text{(iii) From the graph, it can be concluded that in the years}\\ \text{1989 and 1992, the number of books sold were less than 250}\text{.}\\ \text{(iv) From the graph, it can be concluded that the number of}\\ \text{books sold in the year 1989 is about 1 and}\frac{3}{4}\text{th}\text{\hspace{0.17em}}\text{part of 1 cm}\\ \text{The scale is taken as 1 cm = 100 books}\\ \text{So, 100+}\frac{3}{4}\times 100=100+75=175\\ \text{Therefore, about 175 books were sold in the year 1989}\text{.}\end{array}
Q.18 Number of children in six different classes are given below.
Represent the data on a bar graph.
Class | Fifth | Sixth | Seventh | Eighth | Ninth | Tenth |
Number of Children | 135 | 120 | 95 | 100 | 90 | 80 |
(
Ans.
\begin{array}{l}\text{(a)}\text{\hspace{0.17em}}\text{Choose a scale as 1 unit = 10 children, because we can}\\ \text{show a clear difference between the number of students of}\\ \text{class 7th and class 9th}\text{.}\\ \text{(b)}\\ \text{(i)}\\ \text{The bar representing the number of children of class fifth is}\\ \text{tallest, there are maximum number of children in class fifth}\\ \text{and the bar representing the number of children of class tenth is}\\ \text{smallest, there are minimum number of children in class tenth}\text{.}\\ \text{(ii)}\\ \text{The number of student in class sixth is 120 and the number of}\\ \text{student in class eight is 100}\text{.}\\ \text{So, the ratio ratio is =}\frac{120}{100}=\frac{20\times 6}{20\times 5}=\overline{)\frac{6}{5}}\end{array}
Q.19 The performance of student sin 1^{st} Term and 2^{nd} Term is given.
Draw a double bar graph choosing appropriate
s
Subject | English | Hindi | Maths | Science | S.Science |
1st Term
(M.M. 100) |
67 | 72 | 88 | 81 | 73 |
2nd Term
(M. M. 100) |
70 | 65 | 95 | 85 | 75 |
(i) In which subject, has the child improved h is performance the most?
(ii) in which subject is the improvement the least?
(iii) Has the performance gone down in any subject?
Ans.
\begin{array}{l}\text{(i) There was a maximum increase in the marks obtained in}\\ \text{Maths}\text{. Thus, the child improved his performance the most}\\ \text{in Maths}\text{.}\\ \text{(ii) From the graph, we observe that the improvment was least}\\ \text{in S}\text{. Science}\text{.}\\ \text{(iii) From the graph, we observe that the performance in Hindi}\\ \text{has grown down}\text{.}\end{array}
Q.20 Consider this data collected from a survey of a colony.
Favorite Sport | Cricket | Basket Ball | Swimming | Hockey | Athletics |
Watching | 1240 | 470 | 510 | 423 | 250 |
Participating | 620 | 320 | 320 | 250 | 105 |
(
(
Ans.
(i)
\begin{array}{l}\text{(ii) From the bar graph, the bar representing the number}\\ \text{of people who like watching and participating in cricket}\\ \text{is the tallest among all the bars}\text{.Thus, cricket is the most}\\ \text{popular sport}\text{.}\\ \text{(iii) The bars showing watchig spot are longer than the bars}\\ \text{showing participating in spot}\text{. Thus, watching different types}\\ \text{of sports is more preffered than participating in the sports}\text{.}\end{array}
Q.21
City | Max. | Min |
Ahemdabad | 38°C | 29°C |
Amritsar | 37°C | 26°C |
Banglore | 28°C | 21°C |
Chennai | 36°C | 27°C |
Delhi | 38°C | 28°C |
Jaipur | 39°C | 29°C |
Jammu | 41°C | 26°C |
Mumbai | 32°C | 27°C |
(
(
(
(
Ans.
\text{A double bar graph for given data is shown as below:}
\begin{array}{l}\text{(i) From the graph, we observe that Jammu has the largest}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{difference in its minimum and maximum temprature on the}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{given date}\text{.}\\ \text{(ii) From graph, we observe that Jammu is the hottest city}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and Banglore is the coldest city}\text{.}\\ \text{(iii) Banglore and Jaipur, Banglore and Ahemdabad}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{For Banglore, the maximum temprature was 28}\xb0C\text{, while}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{temprature of both cities was 29}\xb0C.\\ \text{(iv) From graph, we observe that the city which has the}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{least difference between its minimum and maximum}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{temprature is Mumbai}\text{.}\end{array}
Q.22 Tell whether the following is cretain to happen, impossible, can happed but not certain.
(i) You are older today than yesterday.
(ii) Atossed coin will land heads up.
(iii) A die when tossed shall land up with 8 on top.
(iv) The next traffic light seen will be green.
(v) Tomorrowl will be a coludy day.
Ans.
(i) Certain
(ii) Can happen but not certain
(iii) Impossible
(iv) Can happen but not certain
(v) Can happen but not certain
Q.23 There are 6 marbles in a box with numbers from 1 to 6 marked on each o0f them.
(i) What is the probability of drawing a marble with number 2?
(ii) What is the probability of drawing a marble with umber 5?
Ans.
\begin{array}{l}\text{(i)}\text{\hspace{0.17em}}\text{Since,}\text{\hspace{0.17em}}\overline{)\text{Probability}=\text{}\frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{So,P(apperance of 2)}=\overline{)\frac{1}{6}}\\ \text{(ii) P(apperance of 2)}=\overline{)\frac{1}{6}}\end{array}
Q.24 A coin is flipped to decide which team starts the game.
what is the probability that your team will start?
Ans.
\begin{array}{l}\text{A coin has two faces namely}-\text{Head and Tail}\text{.}\\ \text{One team can opt either Head or Tail}\text{.}\\ \text{Since,}\overline{)\text{Probability}=\frac{\text{Number of favourable outcomes}}{\text{Number of favourable outcomes}}}\\ \text{So, Probability(our team start first)}=\overline{)\text{}\frac{1}{2}}\end{array}
Q.25 A box contains pairs of socks of two colours (black and white) I have picked out a white sock.
I pick out one more with my eyes closed. What is the probability that it will make a pair?
Ans.
\begin{array}{l}\text{While closing the eyes, one can draw either a black sock}\\ \text{or a white sock}\text{.}\\ \text{Therefore, there are two possible cases}\text{.}\\ \text{Since,}\overline{)\text{Probability=}\frac{\text{Number of favourable outcomes}}{\text{Number of favourable outcomes}}}\\ \text{So, P(a pair of white socks will be formed) =}\overline{)\frac{1}{2}}\end{array}
FAQs (Frequently Asked Questions)
1. How many exercises are there in Chapter 3 data handling of the NCERT Solutions for Class 7 Mathematics?
This chapter has four exercises. Exercise 3.1 has five short and four long answer questions. Exercise 3.2 has two short and three long answer questions. Exercise 3.3 has three short and three long answer questions and Exercise 3.4 has two short and two long answer questions.
2. What are the key formulas in Chapter 3 of the NCERT textbook for Class 7 Mathematics?
Calculation of arithmetic mean, mean, median, and mode, probability, and other formulas are described in these solutions. These are the foundations for learning data handling techniques. These formulas, as well as the key definitions, can be found at the end of the chapter for a quick review.
3. Why is Chapter 3 of the NCERT Solutions for Class 7 Mathematics so important?
The concepts covered in NCERT Solutions for Class 7 Mathematics Chapter 3 are excellent for honing study skills because the examples and exercises in the chapter are based on experiential learning. Students can access hundreds of practise problems on mean, median, mode, bar graphs, and probability, questions. Besides this they can use the problem-solving techniques explained in the examples to stay ahead in the competition.
4. Is it mandatory for me to practise all of the questions in NCERT Solutions Class 7 Mathematics Data Handling?
To improve their data handling skills, students should complete all of the examples and practise questions. The sample examples and exercise questions will ensure that students have a strong foundation to deal with higher mathematical concepts and even Mathematics for competitive exams in the future.