NCERT Solutions For Class 7 Maths Chapter 3 Data Handling are given here in a simple and detailed way. These NCERT Solutions for chapter 3 of class 7 maths can be extremely helpful for the students to clear all their doubts easily and understand the basics of this chapter in a better and detailed way. NCERT class 7 maths chapter 3 Data Handling solutions given here are very easily understandable so that students does not face any difficulties regarding any of the solutions. The NCERT solutions for class 7 maths Data Handling PDF is also available here that the students can download and study. These solutions present are useful for helping students understand the format and type of problems. It is important to understand the various types of problems to figure out the solutions for them. You can find the NCERT solutions of class 7 where you can understand how to solve the different problems.
NCERT Solutions For Class 7 Maths Chapter 3 Exercises
- NCERT Solutions For Class 7 Maths Chapter 3 Data Handling Exercise 3.1
- NCERT Solutions For Class 7 Maths Chapter 3 Data Handling Exercise 3.2
- NCERT Solutions For Class 7 Maths Chapter 3 Data Handling Exercise 3.3
- NCERT Solutions For Class 7 Maths Chapter 3 Data Handling Exercise 3.4
Exercise 3.1
Question 1:
The following table represents the heights of ten students. Calculate the range of heights.
Answer:
S. No. | Name | Height (in feet) |
1. | Aditi | 4.6 |
2. | Vipul | 4.8 |
3. | Sangam | 4.5 |
4. | Manish | 4.8 |
5. | Sakhshi | 4.1 |
6. | Parul | 4.7 |
7. | Ayushi | 4.9 |
8. | Manoj | 4.8 |
9. | Akash | 5.2 |
10. | Rahul | 5.4 |
Range = Highest height – Lowest height = 5.4 – 4.1 = 1.3 feet.
Question 2:
Organize the following marks in a class test, in a tabular form:
5, 4, 7, 2, 6, 4, 7, 8, 9, 2, 1, 4, 2, 9, 1, 3, 4, 8, 7, 6.
a) Find the number which is the highest.
b) Find the number which is the lowest.
c) Find the range of marks.
d) Find the arithmetic mean.
Answer:
S. No. | Marks | Frequency
(No. of students) |
1. | 1 | 2 |
2. | 2 | 3 |
3. | 3 | 1 |
4. | 4 | 4 |
5. | 5 | 1 |
6. | 6 | 2 |
7. | 7 | 3 |
8. | 8 | 2 |
9. | 9 | 2 |
a) Highest number = 9.
b) Lowest number = 1.
c) Range = 9 – 1 = 8.
d) Arithmetic mean = \(\frac{5+4+7+2+6+4+7+8+9+2+1+4+2+9+1+3+4+8+7+6}{22}=\frac{99}{22}=4.5\)
Question 3:
Calculate the mean of the first five whole numbers.
Answer:
The first five whole numbers are 0, 1, 2, 3, 4.
Therfore,
Mean = Sum of number/Total numbers = (0 + 1 + 2 + 3 + 4)/5 = 10/5 = 2.
Thus, the mean of first five whole numbers is 5.
Question 4:
A cricketer scores the following runs in eight innings: 58, 56, 89, 69, 76, 78, 44, 90. Calculate the mean score.
Answer:
Number of innings = 8
Mean 0f score = Sum of score/Number of innings = (58 + 56 + 89 + 69 + 76 + 78 + 44 + 90)/8
= 560/8 = 70.
Thus, the mean score is 70.
Question 5:
The score of each player in four games is shown in the following table:
Player | Game 1 | Game 2 | Game 3 | Game 4 |
X | 15 | 25 | 12 | 8 |
Y | 0 | 8 | 4 | 6 |
Z | 9 | 11 | Did not play | 10 |
Now,
a) Calculate the mean to determine X’s average number of points scored per game.
b) To calculate the mean of points per game for Z, we will divide by 3 or 4? Why?
c) Calculate the mean of Y.
d) Who performed the best?
Answer:
a) Mean of player X = Sum of scores by X/No. of games played by X
= (15 + 25 + 12 + 8)/4 = 60/4 = 15
b) We will divide by 3 as player Z played only three games.
c) Mean of player Y = Sum of scores by Y/No. of games played by Y
= (0 + 8 + 4 + 6)/4 = 18/4 = 4.5
d) Mean of player X = 15
Mean of player Y = 4.5
Mean of player Z = Sum of scores by Z/No. of games played by Z
= (9 + 11 + 10)/3 = 30/3 = 10
Player A has the highest mean. Therefore, A performed the best.
Question 6:
A group of students scored 45, 48, 49, 50, 47, 35, 41, 32, 22 and 42 (out of 50) in a science test. Find:
a) the highest and lowest marks obtained by the students.
b) range of the marks obtained.
c) mean marks obtained by the group.
Answer:
a) Highest marks obtained by the student = 50
Lowest marks obtained by the student = 22
b) Range of the marks obtained = 50 – 22 = 28
c) Mean of obtained marks = Sum of marks/Total number of marks
= (45 + 48 + 49 + 50 + 47 + 35 + 41 + 32 + 22 + 42)/10
= 411/10 = 41.1
Thus, the mean mark obtained by the group of student is 41.1.
Question 7:
During the six consecutive years the enrolment in a school was as follows:
1449, 1454, 1419, 2539, 2509, 2630
Find the mean.
Answer:
Mean enrolment = Sum of numbers of enrolment/Total number of enrolment
= (1449 + 1454 + 1419 + 2539 + 2509 + 2630)/6
= 12000/6 = 2000
Therefore, the mean enrolment of the school is 2,000.
Question 8:
The following table shows the rainfall (in mm) in a city of 7 days:
Day | Mon | Tue | Wed | Thurs | Fri | Sat | Sun |
Rainfall (in mm) | 0.0 | 12.3 | 2.5 | 0.0 | 19.5 | 5.5 | 4.5 |
Find:
a) range of the rainfall.
b) mean rainfall for the week
c) on how many days was the rainfall less than the mean rainfall?
Answer:
a) Range of the rainfall = Highest rainfall – Lowest rainfall = 19.5 – 0.0 = 19.5 mm
b) Mean rainfall = Sum of rainfall recorded/Total number of days
= (0.0 + 12.3 + 2.5 + 0.0 + 19.5 + 5.5 + 4.5)/7
= 6.32 mm
c) 5 days, i.e., Monday, Wednesday, Thursday, Saturday, Sunday rainfalls were less than the mean rainfall.
Question 9:
The height of 10 boys (in cm) was as follows:
130, 150, 145, 165, 150, 143, 146, 123, 128, 136
a) Find the height of the tallest boy.
b) Find the height of the shortest boy.
c) Calculate the range of data.
d) Calculate the mean height of the boys.
e) How many boys have heights more than the mean height?
Answer:
a) Height of the tallest boy = 165 cm
b) Height of the shortest boy = 123 cm
c) Range = 165 – 123 = 42 cm
d) Mean height = Sum of the heights of the boys/Total number of boys
= (130 + 150 + 145 + 165 + 150 + 143 + 146 + 123 + 128 + 136)/10
= 1416/10 = 141.6 cm
e) Six boys have heights more than the mean height.
Exercise 3.2
Q1 : The scores of 15 students (out of 35) of 15 students in Science test is as follows:
20, 26, 24, 21, 10, 21, 16, 11, 6, 17, 26, 21, 25, 13, 21
Find median and mode of this data and check if they are same.
Answer:
Scores of 15 students in mathematics test are
20, 26, 24, 21, 10, 21, 16, 11, 6, 17, 26, 21, 25, 13, 21
Arranging these scores in an ascending order,
6, 10, 11, 13, 16, 17, 20, 21, 21, 21, 21, 24, 25, 26, 26
Mode of a certain data is the value of observation which repeats itself for the most number of times.
Median of a certain data is the observation which occurs in the middle when the data is sorted in an ascending order or descending order.
There are totally 15 values given in the question. Therefore 8^{th} observation is the median of the given data.
Hence, median = 21
For mode, we observe that 21 repeat 4 times (i.e., maximum number of times).
So, mode of the given data = 21
Yes, both mode and median are same.
Q2. The runs scored by 11 players in a cricket match are as follows:
7, 16, 121, 51, 101, 81, 11, 16, 9, 11, 16
Find the median, mean and mode of the above data and check if the three are same.
Answer:
The runs scored by the 11 players are
7, 16, 121, 51, 101, 81, 11, 16, 9, 11, 16
Arranging the scores in an ascending order,
7, 9, 11, 11, 16, 16, 16, 51, 81, 101, 121
Mean = \(\frac{7+9+11+11+16+16+16+81+101+121}{11}=35.36\)
Mode of a certain data is the value of the observation which repeats itself for the most number of times.
Median of a certain data is the observation which occurs in the middle when the data is sorted in an ascending order or descending order.
There are totally 11 values given in the question. Therefore, the 6^{th} observation is the median of this data.
Median = 16
For mode, we observe that 16 repeat 3 times (i.e., maximum number of times).
So, mode of the given data = 16
We can understand from the above results that median, mode and mean are not the same.
Q3. 15 students in a class weigh (in kg):
39, 43, 36, 38, 46, 51, 33, 44, 44, 41, 37, 39, 44, 39, 48
(i) Determine the median and mode of the above data.
(ii) How many modes are there?
Answer:
The weights given (in kg) are
39, 43, 36, 38, 46, 51, 33, 44, 44, 41, 37, 39, 44, 39, 48
Arranging the given weights in ascending order,
33, 36, 37, 38, 39, 39, 39, 41, 43, 44, 44, 44, 46, 48, 51
Mode of a certain data is the value of observation which repeats itself for the most number of times.
Median of a certain data is the observation which occurs in the middle when the data is sorted in an ascending order or descending order.
There are totally 15 values given in the question. Therefore 8^{th} observation is the median of this data.
Hence, median = 41
For mode, we observe that both 44 and 39 repeat 3 times.
So, mode of the given data = 44 and 39.
Therefore, there are 2 modes in the above given question.
Q4: Calculate the mode and median of the following data:
14, 16, 11, 13, 13, 14, 19, 13, 12.
Answer:
Rearranging the given data in ascending order:
11, 12, 13, 13, 13, 14, 14, 16, 19
Mode = 13
Median = 13
Q5. Which of the below-given statements are true? Justify your answer:
(i) One of the numbers is always the mode of the data.
(ii) One of the numbers is the mean of the data.
(iii) One of the numbers is always the median of the data.
(iv) The data 7, 5, 4, 9, 10, 13, 14, 10 has its mean value as 9.
Ans:
(i) True
Mode of a certain data is the value of the observation, which repeats itself for the most number of times. Therefore, it is one of the observations given in the data.
(ii) False
Mean may or may not be one of the numbers in the data.
(iii) True
Median of a certain data is the observation which occurs in the middle when the data is sorted in an ascending order or descending order.
(iv) True
Mean = \(\frac{7+5+4+9+10+13+14+10}{8}=9\)
Exercise 3.3
Q1. Make use of the given bar graph to answer the questions below:
(a) Which is the least popular pet?
(b) How many students own hamsters?
Answer:
(a) Rabbits are the least popular pet.
(b) 5 students own hamsters .
Q2.Study the bar graph which represents the number of books sold by a store in 5 years. Then answer the given questions.
(i) How many books were sold in 1993, 1991 and 1989?
(ii) In which years were approximately 225 and 475 books sold?
(iii) In which year was the least amount of books sold?
(iv) Explain how you estimated the number of books sold in 1991.
Answer:
(i) Number of books sold in : 1993 = 600, 1991 = 300 and 1989 =180.
(ii) Approximately 225 book were sold in the year 1992 and 475 books were sold in 1990.
(iii) In 1989 the least amount of books were sold.
(iv) By studying the graph we can see that in 1991 300 books were sold.
Q3. Given below is a data of the number of children in six different classes. Represent the data on a bar graph.
Class | Fifth | Sixth | Seventh | Eighth | Ninth | Tenth |
No. of children | 135 | 120 | 95 | 100 | 90 | 80 |
(a) How is the scale chosen ?
(b) Which classes have the largest and the smallest number of children?
(c) Calculate the ratio of class ten to class six.
Answer:
(a) Scale : 1unit = 25 children.
(b) Class 5 have the largest number of students, while class 10 has the smallest number of students.
(c) Ratio = (No. of class 10 students)/ (no. of class 6 students)
= 100/120 = 5/6
Question 4:
The performance of a student in two terms is given. Draw a double bar graph by choosing appropriate scale and answer the following question:
Subject | English | Hindi | Maths | Science | S. Science |
1^{st} term (M.M. = 100 ) | 67 | 72 | 88 | 81 | 73 |
2^{nd} term (M.M. = 100) | 70 | 65 | 95 | 85 | 75 |
a) In which subject, the child improved his performance the most?
b) In which subject, the child’s improvement is least?
c) Has the performance gone down in any subject?
Answer:
The given data is represented in a bar graph:
Difference of marks of first term and second term
English = 3
Hindi = -7
Maths = 7
Science = 4
- Science = 2
- a) Maths.
- b) S. Science.
- c) Yes, Hindi.
Question 5:
A survey of a colony was done. The data are as following:
Favourite Sport | Cricket | Basket Ball | Swimming | Hockey | Athletics |
Watching | 1240 | 470 | 510 | 423 | 250 |
Participating | 620 | 320 | 320 | 250 | 105 |
a) Draw a double bar graph choosing an appropriate scale. What do you get from the bar graph?
b) Which sport is most popular?
c) Which is more preferred, watching or participating in sports?
Answer:
The given data is represented in a bar graph:
a) The bar graph represents the number of persons who are watching and participating in their favourite sports.
b) Cricket is more popular.
c) Watching sports is more preferred.
Question 6:
Minimum and maximum temperatures of various cities are calculated. The data are as following:
City | Ahmedabad | Amritsar | Bangalore | Chennai | Delhi | Jaipur | Jammu | Mumbai |
Max. | 38^{o}C | 37^{o}C | 28^{o}C | 36^{o}C | 38^{o}C | 39^{o}C | 41^{o}C | 32^{o}C |
Min. | 29^{o}C | 26^{o}C | 21^{o}C | 27^{o}C | 28^{o}C | 29^{o}C | 26^{o}C | 27^{o}C |
Plot a double bar graph using the date and answer the following:
a) Which city has the largest difference in the minimum and maximum temperature?
b) Which are the hottest city and the coldest city?
c) Name two cities where maximum temperature of one was less than the minimum temperature of the order.
d) Name the city which has the least difference between its minimum and the maximum temperature.
Answer:
The given data is represented in a bar graph:
a) Jammu has the largest difference in temperature i.e.,
Maximum temperature = 41^{o}C and
Minimum temperature = 26^{o}C.
Difference = 41^{o}C – 26^{o}C = 15^{o}C
b) Jammu is the hottest city due to maximum temperature is high and Bangalore is the coldest city due to maximum temperature is low.
c) Maximum temperature of Bangalore is 28^{o}C
Minimum temperature of two cities whose minimum temperature is higher than the maximum temperature of Bangalore are Ahmedabad and Jaipur where the minimum temperature is 29^{o}C
d) Mumbai has the least difference in temperature i.e.,
Maximum temperature = 32^{o}C and
Minimum temperature = 27^{o}C
Difference = 32^{o}C – 27^{o}C = 5^{o}C
Exercise 3.4
Q1. State whether the following events are : certain to happen, can happen but not certain or impossible.
(i) Tomorrow will be sunny.
(ii)A tossed coin will land and stand on its edge.
(iii) Monday will follow Sunday.
(iv ) Sun will rise from the south.
(v) You will read this line again.
Answer:
(i) Can happen but not certain.
(ii) Can happen but not certain.
(iii) Certain to happen.
(iv) Impossible.
(v) Certain to happen.
Q2. There are 7 balls in a bag with numbers from 1 to 7 marked on them.
(i)What is the probability of drawing a ball numbered 2?
(ii) What is the probability of drawing a ball numbered 8?
Total number of balls = 7
(i) Number of balls numbered 2 = 1
Therefore, the probability of drawing a ball numbered 2 = 1/7
(ii) Number of balls numbered 8 = 0
Therefore, the probability of drawing a ball numbered 8 = 0/7
= 0.
Q3. In a football match, a coin is tossed to decide which team starts. What is the probability that your team won’t start? Assuming head starts and your team has chosen tails.
A coin has two sides; head and tail.
Thus, the probability of getting tails = 1/2
The mentioned above is the NCERT solutions for class 7 maths chapter 3. It is important to have a good understanding of the different solutions for the different problems. Data Handling is an important concept that needs to be studied in the field of mathematics. A student can read through the solutions to have a much better understanding of the difference between the types of problems. Referring to the NCERT books for class 7 maths can give a much better viewpoint and reference for understanding exercises and solutions much better. This is much better in solving for problems as a student can refer to the main problems and figure out the best solutions in solving for problems. It is important to have a clear well thought out understanding of the chapters, especially data handling. As a student prepares for the upcoming math exams it is important to have a clear understanding. Thus, this describes the major NCERT solutions for class 7 maths.