(i) The range of the rainfall = Highest rainfall – Lowest rainfall = 20.5 – 0.0 = 20.5 mm (ii) Main rainfall = Sum of rainfall recorded/Total number of days = 0.0+12.2+2.1+2.2+20.5+5.5+1.0/7 = 41.3/7 = 5.9 mm (iii) 5 days. i.e., Monday, Wednesday, Thursday, Saturday and Sunday rainfalls were less thRead more
(i) The range of the rainfall = Highest rainfall – Lowest rainfall
= 20.5 – 0.0 = 20.5 mm
(ii) Main rainfall = Sum of rainfall recorded/Total number of days
= 0.0+12.2+2.1+2.2+20.5+5.5+1.0/7 = 41.3/7 = 5.9 mm
(iii) 5 days. i.e., Monday, Wednesday, Thursday, Saturday and Sunday rainfalls
were less than the mean rainfall.
Mean enrolment = Sum of numbers of enrolment/Total number of enrolment = 1555+1670+1750+2013+2540+2820/6 = 12348/6 = 2058 Thus, the mean enrolment of the school is 2,058. Class 7 Maths Chapter 3 Exercise 3.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapteRead more
Mean enrolment = Sum of numbers of enrolment/Total number of enrolment
= 1555+1670+1750+2013+2540+2820/6
= 12348/6 = 2058
Thus, the mean enrolment of the school is 2,058.
(i) Highest marks obtained by the student = 95 Lowest marks obtained by the student = 39 (ii) Range of marks = Highest marks – Lowest marks = 95 – 39 = 56 (iii) Mean of obtained marks = Sum of marks/Total number of marks = 85+76+90+85+39+48+56+95+81+75/10 = 730/10 =73 Thus, the mean marks obtained bRead more
(i) Highest marks obtained by the student = 95
Lowest marks obtained by the student = 39
(ii) Range of marks = Highest marks – Lowest marks = 95 – 39 = 56
(iii) Mean of obtained marks = Sum of marks/Total number of marks
= 85+76+90+85+39+48+56+95+81+75/10
= 730/10 =73
Thus, the mean marks obtained by the group of students is 73.
(i) Mean of player A = Sum of scores by A / No. of games played by A = 14+16+10+10/4 = 50/4 = 12.5 (ii) We should divide the total points by 3 because player C played only three games. (iii) Player B played in all the four games. ∴ Mean of player B = Sum of scores by B/No. of games played by B = 0+8Read more
(i) Mean of player A = Sum of scores by A / No. of games played by A
= 14+16+10+10/4 = 50/4 = 12.5
(ii) We should divide the total points by 3 because player C played only three
games.
(iii) Player B played in all the four games.
∴ Mean of player B = Sum of scores by B/No. of games played by B
= 0+8+6+4/4 = 18/4 = 4.5
(iv) To find the best performer, we should know the mean of all players.
Mena of player A = 12.5
Mean of player B = 4.5
Mean of player C = 8+11+3/3 = 32/3 = 10.67
Therefore, on comparing means of all players, player A is the best performer.
Number of innings = 8 Mean of score = Sum of scores/Number of innings = 58+76+40+35+46+45+0+100/8 = 400/8 = 50 = Thus, the mean score is 50. Class 7 Maths Chapter 3 Exercise 3.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-3/
Number of innings = 8
Mean of score = Sum of scores/Number of innings
= 58+76+40+35+46+45+0+100/8
= 400/8 = 50
= Thus, the mean score is 50.
The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows: (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?
(i) The range of the rainfall = Highest rainfall – Lowest rainfall = 20.5 – 0.0 = 20.5 mm (ii) Main rainfall = Sum of rainfall recorded/Total number of days = 0.0+12.2+2.1+2.2+20.5+5.5+1.0/7 = 41.3/7 = 5.9 mm (iii) 5 days. i.e., Monday, Wednesday, Thursday, Saturday and Sunday rainfalls were less thRead more
(i) The range of the rainfall = Highest rainfall – Lowest rainfall
= 20.5 – 0.0 = 20.5 mm
(ii) Main rainfall = Sum of rainfall recorded/Total number of days
= 0.0+12.2+2.1+2.2+20.5+5.5+1.0/7 = 41.3/7 = 5.9 mm
(iii) 5 days. i.e., Monday, Wednesday, Thursday, Saturday and Sunday rainfalls
were less than the mean rainfall.
Class 7 Maths Chapter 3 Exercise 3.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-3/
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.
Mean enrolment = Sum of numbers of enrolment/Total number of enrolment = 1555+1670+1750+2013+2540+2820/6 = 12348/6 = 2058 Thus, the mean enrolment of the school is 2,058. Class 7 Maths Chapter 3 Exercise 3.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapteRead more
Mean enrolment = Sum of numbers of enrolment/Total number of enrolment
= 1555+1670+1750+2013+2540+2820/6
= 12348/6 = 2058
Thus, the mean enrolment of the school is 2,058.
Class 7 Maths Chapter 3 Exercise 3.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-3/
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) The highest and the lowest marks obtained by the students. (ii) Range of the marks obtained. (iii) Mean marks obtained by the group.
(i) Highest marks obtained by the student = 95 Lowest marks obtained by the student = 39 (ii) Range of marks = Highest marks – Lowest marks = 95 – 39 = 56 (iii) Mean of obtained marks = Sum of marks/Total number of marks = 85+76+90+85+39+48+56+95+81+75/10 = 730/10 =73 Thus, the mean marks obtained bRead more
(i) Highest marks obtained by the student = 95
Lowest marks obtained by the student = 39
(ii) Range of marks = Highest marks – Lowest marks = 95 – 39 = 56
(iii) Mean of obtained marks = Sum of marks/Total number of marks
= 85+76+90+85+39+48+56+95+81+75/10
= 730/10 =73
Thus, the mean marks obtained by the group of students is 73.
Class 7 Maths Chapter 3 Exercise 3.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-3/
Following table shows the points of each player scored in four games: 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 47? Why? (iii) B played in all the four games. How would you find the mean? (iv) Who is the best performer?
(i) Mean of player A = Sum of scores by A / No. of games played by A = 14+16+10+10/4 = 50/4 = 12.5 (ii) We should divide the total points by 3 because player C played only three games. (iii) Player B played in all the four games. ∴ Mean of player B = Sum of scores by B/No. of games played by B = 0+8Read more
(i) Mean of player A = Sum of scores by A / No. of games played by A
= 14+16+10+10/4 = 50/4 = 12.5
(ii) We should divide the total points by 3 because player C played only three
games.
(iii) Player B played in all the four games.
∴ Mean of player B = Sum of scores by B/No. of games played by B
= 0+8+6+4/4 = 18/4 = 4.5
(iv) To find the best performer, we should know the mean of all players.
Mena of player A = 12.5
Mean of player B = 4.5
Mean of player C = 8+11+3/3 = 32/3 = 10.67
Therefore, on comparing means of all players, player A is the best performer.
Class 7 Maths Chapter 3 Exercise 3.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-3/
A cricketer scores the following runs in eight innings: 58, 76, 40, 35, 46, 45, 0, 100 Find the mean score.
Number of innings = 8 Mean of score = Sum of scores/Number of innings = 58+76+40+35+46+45+0+100/8 = 400/8 = 50 = Thus, the mean score is 50. Class 7 Maths Chapter 3 Exercise 3.1 for more answers vist to: https://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-3/
Number of innings = 8
Mean of score = Sum of scores/Number of innings
= 58+76+40+35+46+45+0+100/8
= 400/8 = 50
= Thus, the mean score is 50.
Class 7 Maths Chapter 3 Exercise 3.1
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-7/maths/chapter-3/