NCERT Class 7 Maths Chapter 3 Data Handling Exercise 3.1, 3.2, 3.3 Solutions
In this page we have provided solutions of the Exercises 3.1, 3.2, 3.3 of NCERT Class 7 Maths Chapter 3 Data Handling. These solutions are made by our team of expert teachers. Practice these solutions carefully for a better understanding of the topic which will help in scoring good marks in the examination.
Publishing Organisation 
NCERT 
Class 
7 
Subject 
Mathematics 
Chapter 
3: Data Handling 
Exercise 
3.1, 3.2, 3.3 
Exercise – 3.1
(1) Please do Yourself
(2)
Marks 
Tally Marks  Frequency 
1  I 
1 
2 
II  2 
3  I 
1 
4 
III  3 
5 
5 

6 
IIII  4 
7  II 
2 
8 
I  1 
9  I 
1 
(i) Highest number is 9.
(ii) Lowest number is 1.
(iii) Range of the data = Highest observation – Lowest observation = 9 – 1 = 8.
(iv) Arithmetic mean
= Sum all observations/ number of observations
4+6+7+5+3+5+4+5+2+6+2+5+1+1+9+6+5+8+4+6+7/20.
= 10/20 = 5.
(3) First five whole numbers are 0, 1, 2, 3 and 14.
Mean = 0 + 1 + 2 + 3 + 4/5 = 10/5 = 2.
(4) Runs Scored by the cricketer are 58, 76, 40, 35, 46, 45, 0, and 100.
Mean Score = Total runs scored in all the innings/ Total numbers of the innings.
Therefore, mean Score is 50.
(5) (i) Mean = Sum of all observation/number of observation
= 14 + 16 + 10 + 10/4
= 50/4 = 12.5
So, As average number of points scored per game is 12.5.
(ii) To find the mean number of points per game for c. we shall divide the total points by 3 because the number of games under consideration is 4 but ‘c’ did not play game3.
(iii) Mean of B’s Score = 0 + 8 + 6 + 4/4 = 18/4 = 4.5
(iv) C’s average number of points scored per game = Sum of all observations/number of observations
= 8 + 11 + 13/3 = 32/3 = 10.6
[∵ ‘c’ did not play Game 3]
So, the best performer is A.
(6) (i) Highest marks obtained by the students = 95
Lowest marks obtained by the students = 39
(ii) Range of the marks obtained = Highest marks – Lowest marks = 95 – 39 = 56
(iii) Mean marks obtained = Sum of all observations/ by the group number of observations
= 85+76+90+85+39+48+56+95+81+75/10
= 739/10 = 73.
(7) Mean enrolment = (1555 + 1670 + 1750 + 2013 + 2540 + 2820)/6
= 12348/6 = 2058
(8) (i) Range of the rainfall.
= Heights rainfall – Lowest rainfall = 20.5 mm – 0.0 mm = 20.5m.
(ii) Mean rainfall for the week
= Sum of all observation/number of observation
= 0.0+12.2+2.1+0.0+20.5+5.5+1.0/7
= 41.3/7 = 5.9 mm
(iii) The rainfall was less than the mean rainfall on 5 days.
(9) Arranging the heights of 10 girls in an ascending order.
128, 132, 135, 139, 141, 143, 146, 149, 150, 151.
(i) Highest of the tallest girl = 151 cm
(ii) Highest of the Shortest girl = 128 cm
(iii) Range = (151 – 128) cm = 23 cm
(iv) Mean highest = (135+150+139+128+151+132+146+143 + 141)/10
= 1414/10 = 141.4 cm
(v) The highest of 5 girls are greater than the mean highest are 143, 146, 149, 150, and 151 cm.
Exercise – 3.2
(1) For Median we arrange the data in ascending order. We get 5, 9, 10, 12, 15, 16, 19,20,20,20,20, 23, 24, 25, 25
Median is the middle observation
Therefore, 20 is the median
For mode.
Mode = Observation with highest frequency = 20
Yes! They are same.
(2) The runs scored by 11 players are 6, 15, 120, 50,100,80,10,15,8,10,15
Arranging these scored in an ascending order 6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120
Mean = 6+8+10+10+15+15+15+50+80+100+120/11
= 429/11 = 39
More of a given data is that value of observation which occurs for the most number of times and the median of the given data is the middle observation when the data is arranged in an ascending or descending order.
As there are 11 terms in the given data, therefore, the median of this data will be the 6^{th} observation.
Median = 15
Also, it can be observed that 15 occurs 3 times (i.e., maximum number of times).
Therefore, made of this data = 15.
No these three are not same.
(3) (i) For median we arrange the data in ascending order, we get 32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50
Median is the middle observation.
Therefore, 40 kg is the median.
For mode . mode = Observation with highest frequency = 38kg and 43 kg.
(ii) Yes! There are 2 (more than one) modes.
(4) For Median we arrange the data in ascending order we get 12, 12, 13, 13, 14, 14, 14, 16, 19.
Median is the middle observation.
Therefore 14 is the median.
For Mode. Mode = Observation with highest frequency = 14.
(5) (i) True (ii) False (iii) True (iv) False
Therefore, mode of this data = 15.
No, these three are not same.
(3) (i) For median we arrange the data in ascending order, we get 32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50
Median is the middle observation.
Therefore, 40 kg is the median.
For mode. Mode = Observation with highest frequency = 38kg and 43kg.
(ii) Yes! There are 2 (more than one) modes.
(4) For Median we arrange the data in ascending order we get 12, 12, 13, 13, 14, 14, 16, 19.
Median is the middle observation.
Therefore 14 is the median.
For mode. Mode = Observation with highest frequency = 14.
(5) (i) True (ii) False (iii) True (iv) False.
Exercise – 3.3
Solution
(1)
(a) Since the bar representing cats is the tallest, Cat is the most popular pet.
(b) The number of children having dog as a pet are 8.
(2)
(i) Number of books in 1989 = 175 (approx.)
Number of books sold in 1990 = 475 (approx.)
Number of books sold in 1992 = 225 (approx.)
(ii) About 475 were sold in the year 1990. About 225 books were sold in the year 1992.
(iii) Fewer than 250 books were sold in the years 1989 and 1992.
(iv) Length of the bar Corresponding to 1989 is about 1.75 cm. The scale is 1 cm = 100 book, So, the number of books estimated to be sold in 1989 = 1.75 × 100 = 175.
(3)
(a) Start scale at 0. The greatest value in the data is 135. So end the scale at a value greater than 135. Such as 140. Use equal divisions along the axes, such as increments of 10. We know that all the bars would lie between 0 and 140.
(b) (i) Fifth class has the maximum number of children. Tenth class has the minimum number of children.
(ii) Ratio of students of class sixth to eight.
= 120:100 = 120/100 = 6/5 = 6:5.
(4)
(i) The child improved his performance the most in the subject of Math’s.
(ii) The improvements is the least in the Subject of S, Science.
(5)
(i) The double bar graph represents the number of people who like watching and participating in different sports. It can be observed that most of the people like watching and participating in Cricket while the least number of people like watching and participating in athletics.
(ii) From the bar graph, it can be observed that the bar representing the number of people who like watching and participating in cricket is the tallest among all the bars. Hence, cricket is the most popular sport.
(iii) The bars representing watching sport are longer than the bars representing participating in sport. Hence, watching different types of sports is more preferred than participating in the sports.
(6) A double bar graph for the given data is constructed as follows –
Scale: 1cm = 10c
Max. temperature
Min: temperature.
(i) From the graph, it can be concluded that Jammu has the largest difference in its minimum and maximum temperatures on 20.6.2006.
(ii) From the graph, it can be concluded that Jammu is the hottest city and Bangalore in the coldest city.
(iii) Bangalore and Jaipur, Bangalore and Ahmedabad for Bangalore, the maximum temperature was 28° c. While minimum temperature of both cities, Ahmedabad and Jaipur, was 29°c.
(iv) From the graph, it can be concluded that the city which has the least difference between its minimum and maximum temperatures is Mumbai.
See Also NCERT Class 7 Math Solution