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