Chapter 24 Measures of Central Tendency RD Sharma Solutions Exercise 24.1 Class 9 Maths
Chapter Name  RD Sharma Chapter 24 Measures of Central Tendency Exercise 24.1 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 24.1 Solutions
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
It is given that,
The heights of 5 persons are  140 cm, 150 cm, 152 cm, 158 cm and 161 cm.
∴ Mean height = Sum of heights/Total No. of persons
= (140 + 150 + 152 + 158 + 161)/5
= 761/5
= 152.2
2. Find the mean of 994, 996, 998, 1002 and 1000.
Solution
Given numbers are 994, 996, 998, 1000 and 1002.
∴ Mean = Sum of Numbers/Total Numbers
= (994 + 996 + 998 + 1000 + 1002)/5
= 4990/5
= 998
3. Find the mean of first five natural numbers.
Solution
Given that,
The first five natural numbers are 1, 2, 3, 4, 5
∴ Mean = Sum of Numbers/Total Numbers
= (1 + 2 + 3 + 4 + 5)/5
= 15/5
Mean = 3
4. Find the mean of all factors of 10 .
Solution
All factors of 10 are 1, 2, 5, 10
∴ Mean = Sum of factors/Total factors
= (1 + 2 + 5 + 12)/4
= 18/4
= 9/2 = 4.5
∴ Mean = 4.5
5. Find the mean of first 10 even natural numbers.
Solution
Given that,
The first 10 natural numbers be 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
∴ Mean = Sum of all Numbers/Total numbers
= (2+4+6+8+10+12+14+16+18+20)= 110/10
= 110/10 = 11
∴ Mean = 11
6. Find the mean of x, x+2, x + 4, x + 6, x + 8.
Solution
Numbers be x, x+2, x + 4, x + 6 and x + 8.
∴ Mean = Sum of numbers/Total numbers
= (x + x + 2 + x + 4 + x + 6 + x + 8)/5
= (5x + 20)/5
= 5(x + 4)/5
= x + 4
7. Find the mean of first five multiples of 3.
Solution
First five multiple of 3 : 3,6,9,12,15
∴ Mean = Sum of Numbers/Total Numbers
= (3 + 6 + 9 + 12 + 15)/5
= 45/9
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 X .
Solution
The weight (in kg) of 10 new born babies
3.4, 3.6, 4.2, 4.5, 3.9, 4.1, 3.8, 4.5, 4.4, 3.6
∴ Mean (x ) = Sum of weights/Total babies
= 3.4+ 3.6 + 4.2 + 4.5 + 3.9 + 4.1 + 3.8 + 4.5 + 4.4 + 3.6)/10
= 40/10 = 4 kg
9. The percentage of 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
The percentage marks obtained by students are = 64, 36, 47, 23, 0, 19, 81, 93, 72, 35, 3, 1
∴ Mean marks = (64+ 36 + 47 + 23 + 0 + 19 + 81 + 93 + 72 + 35 + 3 + 1)/12
= 474/12 = 39.5
∴ Mean marks = 39.5
10. The numbers of children in 10 families of a locality are :
2, 4, 3, 4, 2, 0, 3, 5, 1, 1, 5. Find the mean number of children per family.
Solution
The number of children in 10 families is 2,4,3,4,2,3,5,1,1,5.
∴ Mean number of children per family
= Total no. of children /Total families
= (2 + 4 + 3 + 4 +2 + 3 + 5 + 1 + 1 + 5)/10
= 30/10
= 3
11. If M is the mean of x_{1} , x_{2} , x_{3} ,x_{4} ,x_{5} and x_{6} prove that
(x_{1}  M) + (x_{2}  M) + (x_{3}  M) + (x_{4}  M) + (x_{5}  M) + (x_{6}  M) = 0.
Solution
(ii) Verify that
= 56/10 = 5.6
∴ Mean = Sum of number/Total number
= 12/3 = 4
(i) Adding constant term k = 2 in each term
∴ New mean = (5 + 6 + 7)/3
= 18/3 = 6 = 4 + 2
∴ New mean will be 2 more than the original mean.
∴ New mean = (1 + 2 +3)/3 = 6/3 = 2 = 4  2
∴ New mean will be 2 less than the original mean.
(iii) Multiplying by constant term k = 2 in each term
New mean = (6 + 8 + 10)/3
= 24/3
= 8
= 4 × 2
∴ New mean will be 2 times of the original mean.
∴ New mean = (1.5 + 2 + 2.5)/3
= 6/3 = 2 = 4/2
∴ New mean will be half of the original mean.
⇒ Sum of marks of 100 students = 100 × 40 = 4000
Correct value = 53.
Incorrect value = 83.
Correct sum = 4000  83 + 53 = 3970
∴ Correct mean = 3970/100 = 39.7
= 552/10
The, sum of five numbers = 5 × 27 = 135.
If one number is excluded, then the new mean is 25
∴ Sum of numbers = 4 × 25 = 100
∴ Excluded number = 135  100 = 35
Let weight of 7th student = x kg
⇒ 385 = 324 + x
⇒ x = 385  324
⇒ x = 61 kg
∴ Weight of 7th student = 61 kg
∴ New mean = 240/8 = 30.
Then, the sum of 5 numbers = 5 × 18 = 90
If the one number is excluded
∴ Sum of 4 numbers = 4 × 16 = 64
Excluded number = 90  64 = 26
Then the sum of 200 items = 200 × 50 = 10000
Correct values = 192 and 88
Incorrect values = 92 = 8
∴ Correct sum = 10000  92  8 + 192 + 88 = 10180
∴ Correct mean = 10180/200 = 50.9
= 101.8/2 = 50.9
∴ Mean = Sum of numbers/Total number
= 42/6 = 7
∴ Sum of deviation of values from their mean
⇒ (4) + (3) + (1) + (0) + (1) + (7)
= 8 + 8 = 0