Ads Blocker Image Powered by Code Help Pro

Ads Blocker Detected!!!

We have detected that you are using extensions to block ads. Please support us by disabling these ads blocker.

Mathematics Class 7 Chapter 3 – Data Handling – Exercise 3.1 – NCERT Exercise Solution

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

Solution:

Let us taken the height of 10 students in cm

150, 122, 135, 166, 117, 118, 113, 100, 112, 132

Now, By observing this data we find the highest value =166

By observing this data we find the lowest value = 100

Then,

Range of heights = Highest value – Lowest value

= 166 – 100

= 66 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

Note: First we have to arrange the given marks in ascending order:

i.e. 1, 2, 2, 3, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 8, 9

Let us form a frequency table of the given data.

Marks (xi)Tally MarksFrequency (fi)fi xi
1I11
2II24
3I13
4III312
5IIII525
6IIII424
7II214
8I18
9I19
    
Total – 20100

(i) Which number is the highest?

Solution:

By observing the given data, the highest number is = 9

(ii) Which number is the lowest?

Solution:

By observing the given data, the lowest number is = 1

(iii) What is the range of the data?

Solution:

Range = Highest value – Lowest value

= 9 – 1

= 8

Hence, the range of the data is 8.

(iv) Find the arithmetic mean.

Solution:

= 100/20

= 5

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

Solutions:

Mathematics Class 7 Chapter 3 - Data Handling - Exercise 3.1

4. A cricketer scores the following runs in eight innings:

58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.

Solution:

Mathematics Class 7 Chapter 3 - Data Handling - Exercise 3.1

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

PlayerGame 1Game 2Game 3Game 4
A14161010
B0864
C811Did not Play13

Now answer the following questions:

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

Solution:

A’s average number of points scored per day = total points scored by a in 4 games/Total number of games

= (14+16+10+10)/4

= 50/4 = 12.5 points

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

Solution:

we divide by 3 to find the mean number of points per game for C because C played only 3 games.

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

Solution:

Total number of games = 4

Given, B played in all the four games

So, To find out the mean we will divide the total points by 4

Now,

Mean of B’s score = Total point scored by B in 4 games/Total number of games

= (0+8+6+4)/4

= 18/4 = 4.5 points

(iv) Who is the best performer?

Solution:

To find the best performer among 3 player

First, we have to find the average point of C = (8+11+13)/3

= 32/3 = 10.66

Now, we have average points scored by A is 12.5 which is more than B and C.

Hence, it is clear that A is the best performer among the 3 player.

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:

First of all, we have to arrange the marks in ascending order:

= 39,48,56,75,76,81,85,85,90,95

(i) Highest and the lowest marks obtained by the students.

Solution:

By observing the given marks:

The highest marks obtained by the student = 95

The lowest marks obtained by the student=39

(ii) Range of the marks obtained.

Solution:

Range of marks = Highest marks – Lowest marks

= 95-39 = 56

(iii) Mean marks obtained by the group.

Solution:

Mean of marks obtained by group = (sum of all marks obtained by the group of students)/(Total number of marks)

= (39+48+56+75+76+81+85+85+90+95)/10

= 730/10 = 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:

Mean enrolment = (Sum of all observation )/ (Number of Observation)

= ( 1555 + 1670 + 1750 + 2013 + 2540 + 2820)/6

= 12348/6 = 2058

Hence, the required mean enrolment of a school for the given period is 2058

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

DayMonTueWedThursFriSatSun
Rainfall (in mm)0.012.22.10.020.55.51.0

(i) Find the range of the rainfall in the above data.

Solution:

Range of rainfall for the given data = highest rainfall – lowest rainfall

= 20.5 – 0.0 = 20.5 mm

(ii) Find the mean rainfall for the week.

Solution:

Mean of rainfall for the given data = (Sum of all observation) / (Total 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) On how many days was the rainfall less than the mean rainfall.

Solution:

After observing the average rainfall,

We may find the rainfall of 5 days was less than the average rainfall i.e Monday, Wednesday, Thursday, Saturday, and 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.

We have to arrange the given data in ascending order:
= 128, 132, 135, 139, 141, 143, 146, 149, 150, 151

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

Solution:

By observing the given data the height of the tallest girl = 151 cm

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

Solution:

By observing the given data the height of the shortest girl = 128 cm

(iii) What is the range of the data?

Solution:

Range of the given data = Tallest height – Shortest height

= 151 – 128 = 23 cm

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

Solution:

Mean height of the girl = (Sum of the height of all-girl) / Total number of girls)

= (128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150 + 151) / 10

= 1414 / 10 = 141.4 cm

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

Solution:

The mean height is 141.4 cm

Hence, 5 girls have a height more than the mean height.

👍👍👍

1 thought on “Mathematics Class 7 Chapter 3 – Data Handling – Exercise 3.1 – NCERT Exercise Solution”

Leave a Comment

error: