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.

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