RD Sharma Solutions Class 9 Chapter 24 are given here which consist of questions and answers related to Measure of Central Tendency. Associated with statistics, measures of central tendency is the typical or central value of a probability distribution. The measure of Central Tendency makes an attempt to signify a group of data by a single value, it is also called measures of central location. We are all familiar with at least one of the measures of central tendency, wherein Mean is the most commonly used one. The mean of data by the simplest of words means the average of the given data. Other measures of central tendency are median and mode. Practicing form RD Sharma class 9 solutions will help students to score good marks in exams.
Download PDF of RD Sharma Solutions for Class 9 Maths Chapter 24 Measure of Central Tendency
Access Answers to Maths RD Sharma Solutions for Class 9 Chapter 24 Measure of Central Tendency
Exercise 24.1 Page No: 24.9
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:
The heights of 5 persons are 140 cm , 150 cm , 152 cm , 158 cm and 161 cm (Given)
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:
Sum of numbers = 994+996+998+1000+100 = 4990
Total counts = 5
Therefore, Mean = (Sum of numbers)/(Total Counts)
= 4990/5
= 998
Mean = 998
Question 3: Find the mean of 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
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
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
Now,
Mean = (Sum of numbers) / (Total Numbers)
= 110/10
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
Now,
Mean = (Sum of numbers) / (Total Numbers)
= (5x+20)/5
= 5(x + 4)/5
= x + 4
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
Now,
Mean = (Sum of numbers) / (Total Numbers)
= 45/5
=9
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:
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
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:
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
Now, Mean marks = (Sum of marks ) / (Total students )
=474/12
= 39.5
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:
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.
Exercise 24.2 Page No: 24.14
Question 1: Calculate the mean for the following distribution:
Solution:
Formula to calculate mean:
= 281/40
= 7.025
⇒ Mean for the given distribution is 7.025.
Question 2: Find the mean of the following data:
Solution:
Formula to calculate mean:
= 2650/106
= 25
⇒ Mean for the given data is 25.
Question 3: The mean of the following data is 20.6 .Find the value of p.
Solution:
Formula to calculate mean:
= (25p + 530)/50
Mean = 20.6 (Given)
So,
20.6 = (25p + 530)/50
25p + 530 = 1030
25p = 1030 − 530 = 500
or p = 20
⇒ The value of p is 20.
Question 4: If the mean of the following data is 15, find p.
Solution:
Formula to calculate mean:
= (10p + 445)/(p + 27)
Mean = 15 (Given)
So, (10p + 445)/(p + 27) = 15
10p + 445 = 15(p + 27)
10p – 15p = 405 – 445 = -40
-5p = -40
or p = 8
⇒ The value of p is 8.
Question 5: Find the value of p for the following distribution whose mean is 16.6.
Solution:
Formula to calculate mean:
= (24p + 1228)/100
Mean = 16.6 (given)
So, (24p + 1228)/100 = 16.6
24p + 1228 = 1660
24p = 1660 – 1228 = 432
p = 432/24 = 18
⇒ The value of p is 18.
Question 6: Find the missing value of p for the following distribution whose mean is 12.58.
Solution:
Formula to calculate mean:
= (7p + 524)/50
Mean = 12.58 (given)
So, (7p + 524)/50 = 12.58
7p + 524 = 12.58 x 50
7p + 524 = 629
7p = 629 – 524 = 105
p = 105/7 = 15
⇒ The value of p is 15.
Question 7: Find the missing frequency (p) for the following distribution whose mean is 7.68.
Solution:
Formula to calculate mean:
= (9p + 303)/(p+41)
Mean = 7.68 (given)
So, (9p + 303)/(p+41) = 7.68
9p + 303 = 7.68 (p + 41)
9p + 303 = 7.68p + 314.88
9p − 7.68p = 314.88 − 303
1.32p = 11.88
or p = (11.881)/(1.32) = 9
⇒ The value of p is 9.
Exercise 24.3 Page No: 24.18
Question 1: Find the median of the following data:
83 , 37 , 70 , 29 , 45 , 63 , 41 , 70 , 34 , 54
Solution:
Arranging given numbers in ascending order:
29 , 34 , 37 , 41 , 45 , 54 , 63 , 70 , 70 , 83
Here, Total number of terms = n = 10 (even)
Question 2: Find the median of the following data:
133 , 73 , 89 , 108 , 94 , 104 , 94 , 85 , 100 , 120
Solution:
Arranging given numbers in ascending order:
73 , 85 , 89 ,94 , 94 , 100 , 104 , 108 , 120 , 133
Here, total number of terms = n = 10 (even)
Question 3: Find the median of the following data:
31 , 38 , 27 , 28 , 36 , 25 , 35 , 40
Solution:
Arranging given numbers in ascending order
25 , 27 , 28 , 31 , 35 , 36 , 38 , 40
Here, total number of terms = n = 8 (even)
Question 4: Find the median of the following data:
15 , 6 , 16 , 8 , 22 , 21 , 9 , 18 , 25
Solution:
Arranging given numbers in ascending order
6 , 8 , 9 , 15 , 16 , 18, 21 , 22 , 25
Here, total number of terms = n = 9 (odd)
Question 5: Find the median of the following data:
41 , 43 , 127 , 99 , 71 , 92 , 71 , 58 , 57
Solution:
Arranging given numbers in ascending order
41 , 43 , 57 , 58 , 71 , 71 , 92 , 99 , 127
Here, total number of terms = n = 9 (odd)
Question 6: Find the median of the following data:
25 , 34 , 31 , 23 , 22 , 26 , 35 , 29 , 20 , 32
Solution:
Arranging given numbers in ascending order
20 , 22 , 23 , 25 , 26 , 29 , 31 , 32 , 34 , 35
Here, total number of terms = n = 10 (even)
Question 7: Find the median of the following data:
12 , 17 , 3 , 14 , 5 , 8 , 7 , 15
Solution:
Arranging given numbers in ascending order
3 , 5 , 7 , 8 , 12 , 14 , 15 , 17
Here, total number of terms = n = 8(even)
Question 8: Find the median of the following data:
92 , 35 , 67 , 85 , 72 , 81 , 56 , 51 , 42 , 69
Solution:
Arranging given numbers in ascending order
35 , 42 , 51 , 56 , 67 , 69 , 72 , 81 , 85 , 92
Here, total number of terms = n = 10(even)
Exercise 24.4 Page No: 24.20
Question 1: Find out the mode of the following marks obtained by 15 students in a class:
Marks : 4 , 6 , 5 , 7 , 9 , 8 , 10 , 4 , 7 , 6 , 5 , 9 , 8 , 7 , 7.
Solution:
Mode is the value which occurs most frequently in a set of observations.
Frequency of given set of observations are:
Here, we can see that 7 occurred most frequently.
So, Mode = 7
Question 2: Find out the mode from the following data :
125 , 175 , 225 , 125 , 225 , 175 , 325 , 125 , 375 , 225 , 125
Solution:
Find the frequency of given set of observations:
125 occurred for 4 times than any other values.
So, Mode = 125
Question 3: Find the mode for the following series:
7.5 , 7.3 , 7.2 , 7.2 , 7.4 , 7.7 , 7.7 , 7.5 , 7.3 , 7.2 , 7.6 , 7.2
Solution:
Find the frequency:
Maximum frequency 4 corresponds to the value 7.2.
So, mode = 7.2
Exercise VSAQs Page No: 24.21
Question 1: If the ratio of mean and median of a certain data is 2:3, then find the ratio of its mode and mean.
Solution:
Empirical formula: Mode = 3 median – 2 mean
Since, ratio of mean and median of a certain data is 2:3, then mean = 2x and median = 3x
Mode = 3(3x) – 2(2x)
= 9x – 4x
= 5x
Therefore,
Mode: Mean = 5x:2x or 5: 2
Question 2: If the ratio of mode and median of a certain data is 6 : 5, then find the ratio of its mean and median.
Solution: We know, Empirical formula: Mode = 3 Median – 2 Mean
Since, ratio of mode and median of a certain data is 6:5.
⇒ Mode/Median = 6/5
or Mode = (6 Median)/5
Now,
(6 Median)/5 = 3 Median – 2 Mean
(6 Median)/5 – 3 Median = – 2 Mean
or 9/10 (Median) = Mean
or Mean/ Median = 9/10 or 9:10.
Question 3: If the mean of x+2, 2x+3, 3x+4, 4x+5 is x+2, find x.
Solution:
Given: Mean of x+2, 2x+3, 3x+4, 4x+5 is x+2
We know, Mean = (Sum of all the observations) / (Total number of observations)
Sum of all the observations = x+2 + 2x+3 + 3x+4 + 4x+5 = 10x + 14
Total number of observations = 4
⇒ Mean = (10x + 14)/4
or (x + 2) = (10x + 14)/4 (using given)
4x + 8 = 10x + 14
x = -1
Question 4: The arithmetic mean and mode of a data are 24 and 12 respectively, then find the median of the data.
Solution:
Given: The arithmetic mean and mode of a data are 24 and 12 respectively
We know, Empirical formula: Mode = 3 Median – 2 Mean
or 3 Median = Mode + 2 Mean
Using given values, we get
3 Median = 12 + 2(24) = 60
or Median = 20
Question 5: If the difference of mode and median of a data is 24, then find the difference of median and mean.
Solution:
Given: difference of mode and median of a data is 24.
That is, Mode – Median = 24
or Mode = 24 + Median …(1)
We know, Empirical formula: Mode = 3 Median – 2 Mean
24 + Median = 3 Median – 2 Mean
(Using (1))
24 = 2 Median – 2 Mean
or 12 = Median – Mean
Therefore, the difference of median and mean is 12.
RD Sharma Solutions for Class 9 Maths Chapter 24 Measures of Central Tendency
In the 24th chapter of RD Sharma Solutions class 9 students will study important concepts listed below:
- Measure of Central Tendency Introduction
- Arithmetic mean of individual observations or ungrouped data
- Properties of arithmetic mean
- Median and its Properties
- Mode and its Properties
