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NCERT Solutions for Class 7 Maths: Chapter 3 - Data Handling

Get NCERT Solutions for Class 7 Maths, Chapter 3 - Data Handling. With this article, you can also download the PDF of the Class 7 Maths NCERT textbook in PDF format.

Jul 25, 2019 16:27 IST
NCERT Solutions for Class 7 Maths: Chapter 3 - Data Handling

NCERT Solutions for Class 7 Maths, Chapter 3 - Data Handling is available here. Here you will get solutions to every question of this chapter. The PDF of this chapter is also available here for download in PDF format.

NCERT Solutions for Class 7 Maths, Chapter 3 - Data Handling:

1. Find the range of heights of any ten students of your class.

Solution1:

Heights of 10 students of my class are 130, 140, 150, 160, 145, 135, 165, 132, 142, 152

Here, the highest value among these observations = 152

The lowest value among these observations = 130

Now, Range = Highest value − Lowest value = (152 − 130) cm = 22 cm.

2. Organise the following marks in a class assessment, in a tabular form.

4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7

(i) Which number is the highest?

(ii) Which number is the lowest?

(iii) What is the range of the data?

(iv) Find the arithmetic mean.

Solution 2:

(i) Highest number = 9

(ii) Lowest number = 1

(iii) Range of data = (9 − 1) = 8

(iv) As, arithmetic mean = (Sum of all the observations)/(Total number of observations)

Here, Sum of all the observations = 4 + 6 + 7 + 5 + 3 + 5 + 4 + 5 + 2 + 6 + 2 + 5 + 1 + 9 + 6 + 5 + 8 + 4 + 6 + 7 = 100

Total number of observations = 20

So, arithmetic mean = 100/20 = 5.

3. Find the mean of the first five whole numbers.

Solution 3:

As the first five whole numbers are 0, 1, 2, 3, and 4.

Mean = (Sum of observations)/(Number of observations) = (0 + 1+ 2 + 3 +4)/5 = 2.

4. A cricketer scores the following runs in eight innings: 58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.

Solution 4:

Mean Score = (Sum of runs of all the innings)/(Number of innings) = (58 + 76 + 40 + 35 + 46 + 45 + 0 + 100)/8 = 2 = 50.

5. Following table shows the points of each player scored in four games:

Player

Game 1

Game 2

Game 3

Game 4

A

14

16

10

10

B

8

6

4

C

8

11

Did not Play

13

Now answer the following questions:

(i) Find the mean to determine A’s average number of points scored per game.

(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?

(iii) B played in all the four games. How would you find the mean?

(iv) Who is the best performer?

Solution 5:

(i) As, A’s average number of points scored per game = (Total number of points of A in all the games)/(Number of games A played)

So , A’s average number of points scored per game = (14 + 16 +10 +10)/4 = 50/4 = 12.5

(ii) To find the mean number of points per game for C, we will divide it by because C played 3 matches.

(iii) Mean of B = Total number of points of A in all the games)/(Number of games B played) = (0 + 8 + 6 + 4)/4 = 4.5

(iv) The player with the greatest average among all in the best performer. Here, A has an average of 12.5 which is greater than B and C. So, A is the best performer.

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.

Solution 6:

Marks in ascending order are given below: 39, 48, 56, 75, 76, 81, 85, 85, 90, 95

(i) Highest marks = 95, lowest marks = 39

(ii) Range = Highest marks - Lowest marks = 95 - 39 = 56.

(iii) Mean of the marks = (Total marks of all the students)/(Number of students) = 730/1= 73.

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.

Solution 7:

 The mean enrolment of the school for this period = (Sum of enrolments)/(Number of years) = (1555 + 1670 + 1750 + 2013 + 2540 + 2820)/6 = 12348/6 = 2058.

8. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

Day

Mon

Tue

Wed

Thurs

Fri

Sat

Sun

Rainfall (in mm)

0.0

12.2

2.1

0.0

20.5

5.5

1.0

(i) Find the range of the rainfall in 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 8:

(i) Range of the rainfall for the week = (20.5 - 0.00) = 20.5 mm

(ii) Mean rainfall = (0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0)/7 = 41.3/7 = 5.9 mm

(iii) 5 days, i.e. Monday, Wednesday, Thursday, Saturday, Sunday

9. The heights of 10 girls were measured in cm and the results are as follows:

135, 150, 139, 128, 151, 132, 146, 149, 143, 141.

(i) What is the height of the tallest girl?

(ii) What is the height of the shortest girl?

(iii) What is the range of the data?

(iv) What is the mean height of the girls?

(v) How many girls have heights more than the mean height.

Solution 9:

Heights of 10 girls in ascending order are given below:

128, 132, 135, 139, 141, 143, 146, 149, 150, 151

(i) The height of the tallest girl = 151 cm

(ii) The height of the shortest girl = 128 cm

(iii) Range of the data = (151 - 128) cm = 23 cm

(iv) Mean height of the girls = (135 + 150 + 139 + 128 + 151 + 132 + 146 + 149 + 143 + 141)/10 = 1414/10 = 14.14

(v) 5 girls have more heights than the mean height.

Solutions to rest of the questions of this chapter will be available here shortly.

Download Chapter 3 (Data Handling) of Class 7 Maths NCERT Textbook