Class 9 RD Sharma Solutions – Chapter 24 Measures of Central Tendency – Exercise 24.1 | Set 1
Question 1. If the heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm respectively. Find the mean height.
Solution:
Given: The heights of 5 persons are 140 cm, 150 cm, 152 cm, 158 cm and 161 cm
Mean height = (Sum of heights) / (Total number of persons)
Sum of heights = 140 + 150 + 152 + 158 + 161 = 761
Total number of persons = 5
So, Mean height = 761/5 =152.2
Question 2. Find the mean of 994, 996, 998, 1002, 1000.
Solution:
Given numbers are: 994, 996, 998, 1002, 1000
Sum of numbers = 994+996+998+1000+100 = 4990
Total count = 5
Therefore, Mean = (Sum of numbers)/(Total Count)
= 4990/5
= 998
Therefore, Mean = 998
Question 3. Find the mean of the first five natural numbers.
Solution:
First five natural numbers are 1, 2, 3, 4, 5.
Sum of all the numbers = 1+2+3+4+5 = 15
Total Numbers = 5
Therefore,, Mean = (Sum of numbers)/(Total Numbers)
= 15/5
= 3
Therefore, Mean = 3
Question 4. Find the mean of all factors of 10.
Solution:
Factors of 10 are 1, 2, 5, 10.
Sum of all the factors = 1+2+5+10 = 18
Total Numbers = 4
Therefore, Mean = (Sum of factors)/(Total Numbers)
= 18/4
= 4.5
Therefore, Mean = 4.5
Question 5. Find the mean of first 10 even natural numbers.
Solution:
First 10 even natural numbers = 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Sum of numbers = 2+4+6+8+10+12+14+16+18+20 = 110
Total Numbers = 10
Mean = (Sum of numbers) / (Total Numbers)
= 110/10
Therefore, Mean = 11
Question 6. Find the mean of x, x + 2, x + 4, x + 6, x + 8.
Solution:
Given numbers are: x, x + 2, x + 4, x + 6, x + 8.
Sum of numbers = x+(x+2) + (x+4) + (x+6) + (x+8) = 5x+20
Total Numbers = 5
Mean = (Sum of numbers) / (Total Numbers)
= (5x+20)/5
= 5(x + 4)/5
= x + 4
Therefore, Mean = x + 4
Question 7. Find the mean of first five multiples of 3.
Solution:
First five multiples of 3 are 3, 6, 9, 12, 15.
Sum of numbers = 3+6+9+12+15 = 45
Total Numbers = 5
Mean = (Sum of numbers) / (Total Numbers)
= 45/5
=9
Therefore, Mean = 9
Question 8. Following are the weights (in kg) of 10 new born babies in a hospital on a particular day: 3.4, 3 .6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6. Find the mean.
Solution:
Given: The weights of 10 new born babies (in kg): 3.4, 3 .6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6
Sum of weights = 3.4+3.6+4.2+4.5+3.9+4.1+3.8+4.5+4.4+3.6 = 40
Total number of babies = 10
No, Mean = (Sum of weights) / (Total number of babies)
= 40/10
= 4
Therefore, Mean weight = 4 kg
Question 9. The percentage marks obtained by students of a class in mathematics are : 64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1. Find their mean.
Solution:
Given: The percentage marks obtained by students: 64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1
Sum of marks = 64+36+47+23+0+19+81+93+72+35+3+1 = 474
Total students = 12
Mean marks = (Sum of marks) / (Total students)
=474/12
= 39.5
Therefore,Mean Marks = 39.5
Question 10. The numbers of children in 10 families of a locality are:
2, 4, 3, 4, 2, 3, 5, 1, 1, 5. Find the number of children per family.
Solution:
Given: The numbers of children in 10 families: 2, 4, 3, 4, 2, 3, 5, 1, 1, 5
Total number of children = 2+4+3+4+2+3+5+1+1+5 = 30
Total Families = 10
Number of children per family = Mean = (Total number of children) / (Total Families) = 30/10
= 3
Therefore, Number of children per family is 3.
Question 11. Explain, by taking a suitable example, how the arithmetic mean alters by (i) adding a constant k to each term, (ii) subtracting a constant k from each them, (iii) multiplying each term by a constant k and (iv) dividing each term by a non-zero constant k.
Solution:
Let say numbers are 7,8,9
Therefore, Mean=(sum o f numbers)/ (total numbers)
=(7+8+9)/(3)=8
(i) Adding constant term k = 4 in each term.
New numbers are = 11,12,13
Therefore, Mean=(sum o f numbers)/ (total numbers)
=(11+12+13)/(3)=12
Therefore, new mean will be 4 more than the original mean.
(ii) Subtracting constant term k = 4 in each term.
New numbers are = 3,4,5
Therefore, Mean=(sum o f numbers)/ (total numbers)
=(3+4+5)/(3)=4
Therefore, new mean will be 4 less than the original mean.
(iii) Multiplying by constant term k = 4 in each term.
New numbers are = 28,32,36
Therefore, Mean=(sum o f numbers)/ (total numbers)
=(28+32+36)/(3)=32
Therefore, new mean will be 4 times of the original mean.
(iv) Divide the constant term k =4 in each term.
New numbers are = 1.75,2,2.25
Therefore, Mean=(sum o f numbers)/ (total numbers)
=(1.75+2+2.25)/(3)=2
Therefore, new mean will be one-fourth of the original mean.
Question 12. The mean of marks scored by 100 students was found to be 40. Later on its was discovered that a score of 53 was misread as 83. Find the correct mean.
Solution:
Mean marks of 100 students = 40
Sum of marks of 100 students = 100 * 40
= 4000
Correct value = 53
Incorrect value = 83
Correct sum = 4000 – 83 + 53 = 3970
Mean=(sum of numbers)/ (total numbers)
Therefore, correct mean = = 3970/100=39.7
Please Login to comment...