Questions and Solutions for NCERT Class 7th Mathematics , Detailed solutions for class 7th NCERT Mathematics, Solutions for Maths Chapter 3 -Data Handling Exercise 3.1 for Class 7th, NCERT Maths solutions for class 7th, Questions and answers of NCERT class 7th Mathematics, detailed solution Notes for NCERT Class 7th Mathematics, Class 7th Maths solutions wikipedia, Solutions for class 7th NCERT Mathematics.
Please find the detailed Solutions for class 7th Maths Chapter 3 -Data Handling Exercise 3.1 Solutions, you can check all the solutions from here chapter wise
NCERT Class 7th Mathematics Chapter 3 -Data Handling Exercise 3.1 Solutions
1.Find the range of heights of any ten students of your class.
Sol: Let the height (in cm) of ten students of class be
129, 126, 127, 125, 124, 131, 136, 132, 128, 137
Highest value among all the observations = 137 cm
Lowest value among all the observations = 124 cm
Range = Highest value – Lowest value
= (137-124) cm
Therefore range = 13 cm
2. Organise the following marks in a class assessment, in a tabular form.
4 6 7 5 3 5 4 5 26
2 5 1 9 6 5 8 4 67
(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.
Sol: Marks Frequency
1 1
2 2
3 1
4 3
5 5
6 4
7 2
8 1
9 1
Sol: (i) Highest number is 9
(ii) Lowest number is 1
(iii) Range of data = Highest value – Lowest value
= 9-1
= 8
(iv) Arithmetic mean = Sum of all observations / No. of observations
= (4+6+7++5+3+5+4+5+2+6+2+5+1+9+6+5+8+4+6+7 ) / 20
= 100/20
= 5
3. Find the mean of the first five whole numbers.
Sol: First five whole numbers are 0,1,2,3,4
Mean = Sum of all observations / no. of observations
= (0+1+2+3+4) / 5
= 10/5
= 2
Therefore mean of first five whole numbers is 2
4. A cricketer scores the following runs in eight innings:
58, 76, 40, 35, 46, 45, 0, 100.
Find the mean score.
Sol: Mean = Sum of all observations / No. of observations
Sum of all observations = 58+76+40+35+46+45+0+100
= 400
Total observations = 8
Mean = 400/8
= 50
Therefore mean = 50
5. Following table shows the points of each player scored in four games:
Player Game Game Game Game
1 2 3 4
A 14 16 10 10
B 0 8 6 4
C 8 11 Did not 13
Play
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?
Sol: (i) A’s average number of points = (14+16+10+10) / 4
= 50/4
= 12.5
(ii) To find the mean number of points per game for C, we would divide the total
points by 3 because C played 3 games.
(iii) Mean = Sum of all observations / No. of observations
= (0+8+6+4) / 4
= 18 / 4
= 4.5
(iv) The best performer will have the highest average among all. Here the average of A is more than B and C both. 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.
Sol: (i) Let us first arrange the data in increasing order = 39, 48, 56, 75, 76, 81, 85, 85, 90 ,95
Highest marks obtained by the students = 95
Lowest marks obtained by the students = 39
(ii) Range = Highest value – Lowest Value
= 95 – 39
= 56
(iii) Mean = Sum of all observations / No. of observations
Sum of all observations = 39+48+56+75+76+81+85+85+90+95
= 730
Mean = 730/ 10
= 73
So the mean marks obtained by the group = 73
7. The enrollment in a school during six consecutive years was as follows:
1555, 1670, 1750, 2013, 2540, 2820
Find the mean enrollment of the school for this period.
Sol: Mean = Sum of all observations / No. of observations
Sum of all observations = 1555+1670+1750+2013+2540+2820
= 12348
Mean = 12348 / 6
= 2058
The mean enrollment of the school for six years = 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 0.0 12.2 2.1 0.0 20.5 5.5 1.0
(in mm)
(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.
Sol: (i) Range = Highest value – Lowest value
= 20.5 – 0.0
= 20.5
(ii) Mean = Sum of all observations / No. of observations
Sum of all observations = 0.0+12.2+2.1+0.0+20.5+5.5+1.0
= 41.3
Mean = 41.3 / 7
= 5.9
The mean rainfall for the week = 5.9 mm
(iii) For 5 days the rainfall less than the mean rainfall. The days are 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.
(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.
Sol: Let us first arrange the heights in increasing order = 128, 132, 135, 139, 141, 143, 146, 149, 150, 151
(i) Height of tallest girl = 151 cm
(ii) The height of the shortest girl = 128 cm
(iii) The range of the data = Highest value – Lowest value = 151 -128 = 23 cm
(iv) Mean = Sum of all observations / No. of observations
Sum of all observations = 128+132+135+139+141+143+146+149+150+151
= 1414
Mean = 1414 / 10
= 141.4 cm
So the mean height of the girls = 141.4 cm
(v) Five girls have the height more than mean height.
You can find the solutions for previous exercise from here
- Class 7th Maths – Exercise 2.7 Solutions
- Class 7th Maths – Exercise 2.6 Solutions
- Class 7th Maths – Exercise 2.5 Solutions
- Class 7th Maths – Exercise 2.4 Solutions
- Class 7th Maths – Exercise 2.3 Solutions
- Class 7th Maths – Exercise 2.2 Solutions
- Class 7th Maths – Exercise 2.1 Solutions
- Class 7th Maths – Exercise 1.4 Solutions
- Class 7th Maths – Exercise 1.3 Solutions
