RD Sharma Solutions Class 9 Maths Chapter 24 – Measure of Central Tendency (Updated for 2021-22)

RD Sharma Solutions Class 9 Maths Chapter 24 – RD Sharma Solutions Class 9 Maths Chapter 24 has a very high weightage in your maths exams and the student needs to understand the concepts of this unit in detail. RD Sharma Class 9 Maths Chapter 24 focuses on the calculation of mean, median, and mode of the data by using both traditional and formula methods. To know more, read the whole blog.

Download RD Sharma Solutions Class 9 Maths Chapter 24 PDF

RD Sharma Class 9 Solutions Chapter 24

Exercise-wise RD Sharma Solutions Class 9 Maths Chapter 24 PDF

RD Sharma Solutions Class 9 Maths Chapter 24 Exercise 24.1

RD Sharma Solutions Class 9 Maths Chapter 24 Exercise 24.2

RD Sharma Solutions Class 9 Maths Chapter 24 Exercise 24.3

RD Sharma Solutions Class 9 Maths Chapter 24 Exercise 24.4

Acces answers of RD Sharma Solutions Class 9 Maths Chapter 24

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.

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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.

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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)

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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

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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.

Detailed Exercise-wise Explanation with Listing of Important Topics in the Exercise

Before understanding the detailed explanation for the exercise, we should understand some basic terms-

1. Mean- It is the method of calculating the average of all the numbers present in the data.
2. Median- The middle number of an ascended or descended series is called a median.
3. Mode- The number that repeats the most times in a series is called the mode of the series.

RD Sharma Solutions Class 9 Maths Chapter 24 Ex 24.1

This exercise contains 10 questions that talk about calculating the mean by using the average method. It is not a very complex exercise but the student should know the concept well to avoid any mistakes in their maths examination.

RD Sharma solution PDF contains answers in a very detailed manner so that a student can understand the concept very well. It is a high-quality study material, which includes a step-wise method for answering questions and the student can incorporate the same pattern in their maths examination.

RD Sharma Solutions Class 9 Maths Chapter 24 Ex 24.2

There are a total of 7 questions in this exercise which also focuses on calculating mean but by using a formula. The data given is in a tabular form that contains frequency and to handle such types of questions you need to use a mean formula. The formula is a little complex but the student can easily learn through regular practice.

RD Sharma Solutions Class 9 Maths Chapter 24 PDF will provide a stepwise answer so that you can easily understand the concept very well. These types of questions contain high weightage in the maths exams so every student needs to be extra careful while solving this exercise.

RD Sharma Solutions Class 9 Maths Chapter 24 Ex 24.3

The following exercise contains 8 questions which are based on the calculation of Median. The students have to calculate the median by using the formula method and there are two formulas one is- if the total number in the series is even then the other one will be used if they are odd.

RD Sharma solutions class 9-chapter 24 24.2 will help you in solving all the questions accurately. Our pdf is made following the latest CBSE pattern so that you easily use the methods and steps we use in your maths exams or while doing homework.

RD Sharma Solutions Class 9 Maths Chapter 24 Ex 24.4

 There are a total of 3 questions in this exercise that are dedicated to calculating the mode of the series. It is an easy exercise and can be solved by a student who has thoroughly learned all the concepts of mode.

The pdf solution of RD Sharma class 9 chapter 23 will give an insight on how to solve these types of questions and we also mention tips and tricks so that you can solve the questions more efficiently.

Important Topics From RD Sharma Solutions Class 9 Maths Chapter 24

Before solving the questions of this unit, you need to know all the important topics that will help you score high grades in your maths exam.

Here, is the list of all the topics that you should study before solving this chapter-

• Calculation of mean and its formulas
• Calculation of mode and its formulas
• Calculation of mode and its formulas

his is the complete blog on RD Sharma Solutions Class 9 Maths Chapter 24. To know more about the CBSE Class 9 Maths exams, ask in the comments. 

RD Sharma Solutions for Class 9 Maths Chapter 24