CBSE Class 10 Mathematics Chapter 29 NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 10 PDF

This chapter provides comprehensive NCERT Solutions for Class 10 Mathematics, focusing on Chapter 29, Statistics. It covers essential concepts related to data representation and analysis, including calculating the mean for grouped data using different methods. The solutions offer clear, step-by-step guidance for solving problems from the exercises, ensuring students grasp the underlying principles. Key topics include understanding frequency distributions, calculating class marks, and applying the direct method and assumed mean method for finding the mean. These solutions are designed to aid students in understanding complex statistical concepts and preparing effectively for their board examinations by reinforcing problem-solving skills and conceptual clarity.

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Quick info

Board	CBSE
Class	Class 10
Subject	Mathematics
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 29

Chapter summary

Chapter 29 of the Class 10 Mathematics NCERT Solutions delves into the core concepts of Statistics. This chapter focuses on understanding and calculating the mean for grouped frequency distributions. It introduces students to methods like the direct method and the assumed mean method, explaining when and why to use each. The solutions provide detailed walkthroughs for exercises, helping students interpret data tables and apply the correct formulas to find the average value of a dataset, which is crucial for data analysis.

Learning outcomes

Understand the concept of mean for grouped data.
Calculate the class mark for a given class interval.
Apply the direct method to find the mean of a frequency distribution.
Apply the assumed mean method to find the mean of a frequency distribution.
Determine the appropriate method for calculating the mean based on the data.
Interpret statistical data presented in frequency tables.

Topics covered

Paper topics

Statistics
Mean of Grouped Data
Frequency Distribution
Class Interval
Class Mark
Direct Method for Mean
Assumed Mean Method for Mean
Data Interpretation

Important topics

Mean of Grouped Data
Direct Method for Mean
Assumed Mean Method for Mean
Class Mark Calculation
Frequency Distribution Analysis

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Questions and Solutions

Question 1

A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants:

0 - 2	2 - 4	4 - 6	6 - 8	8 - 10	10 - 12	12 - 14
Number of houses	1	2	1	5	6	2	3

Which method did you use for finding the mean, and why?

Solution:

To find the mean number of plants per house, we first need to calculate the class mark (x_i) for each class interval. The class mark is the midpoint of the interval and is calculated using the formula:

x_i = \frac{\text{Upper limit} + \text{Lower limit}}{2}

Next, we calculate the product of the frequency (number of houses, f_i) and the class mark (f_i x_i) for each interval. The total number of houses is given as 20, which is Σf_i.

Here is the table with the calculated values:

Number of plants (Class Interval)	Number of houses (f_i)	Class Mark (x_i)	f_i x_i
0 - 2	1	$(0+2)/2 = 1$	$1 \times 1 = 1$
2 - 4	2	$(2+4)/2 = 3$	$2 \times 3 = 6$
4 - 6	1	$(4+6)/2 = 5$	$1 \times 5 = 5$
6 - 8	5	$(6+8)/2 = 7$	$5 \times 7 = 35$
8 - 10	6	$(8+10)/2 = 9$	$6 \times 9 = 54$
10 - 12	2	$(10+12)/2 = 11$	$2 \times 11 = 22$
12 - 14	3	$(12+14)/2 = 13$	$3 \times 13 = 39$
Total	$\sum f_i = 20$		$\sum f_i x_i = 162$

From the table, we have:

\sum f_i = 20

\sum f_i x_i = 162

We use the Direct Method to calculate the mean ( $\bar{x}$ ):

\bar{x} = \frac{\sum f_i x_i}{\sum f_i}

Substituting the values:

\bar{x} = \frac{162}{20} = 8.1

Therefore, the mean number of plants per house is 8.1.

Method Used and Why: The Direct Method was used. This method was chosen because the class marks (x_i) and the frequencies (f_i) are relatively small numbers. Calculating the product f_i x_i and their sum (Σf_i x_i) results in manageable values, making the direct computation of the mean straightforward and less prone to calculation errors compared to other methods like the Assumed Mean Method in this specific case.

Question 2

Consider the following distribution of daily wages of 50 workers of a factory.

Daily wages (in Rs):

100 - 120	120 - 140	140 - 160	160 - 180	180 - 200
Number of workers	12	14	8	6	10

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

To find the mean daily wages, we first determine the class mark (x_i) for each interval. The formula for the class mark is:

x_i = \frac{\text{Upper limit} + \text{Lower limit}}{2}

The class size (h) for all intervals is $120 - 100 = 20$ .

Since the class marks and frequencies involve larger numbers, the Assumed Mean Method is an appropriate and efficient method to use. Let's assume a mean (a) from the class marks. A common choice is the middle class mark. Let's choose $a = 150$ (the class mark for the 140-160 interval).

Now, we calculate the deviation $d_i = x_i - a$ for each class, and then the product $f_i d_i$ .

Daily Wages (Class Interval)	Number of Workers (f_i)	Class Mark (x_i)	Deviation $(d_i = x_i - a)$	$f_i d_i$
100 - 120	12	$(100+120)/2 = 110$	$110 - 150 = -40$	$12 \times (-40) = -480$
120 - 140	14	$(120+140)/2 = 130$	$130 - 150 = -20$	$14 \times (-20) = -280$
140 - 160	8	$(140+160)/2 = 150$	$150 - 150 = 0$	$8 \times 0 = 0$
160 - 180	6	$(160+180)/2 = 170$	$170 - 150 = 20$	$6 \times 20 = 120$
180 - 200	10	$(180+200)/2 = 190$	$190 - 150 = 40$	$10 \times 40 = 400$
Total	$\sum f_i = 50$			$\sum f_i d_i = -480 - 280 + 0 + 120 + 400 = -240$

We have:

\sum f_i = 50

\sum f_i d_i = -240

The formula for the mean using the Assumed Mean Method is:

\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}

Substituting the values:

\bar{x} = 150 + \frac{-240}{50}

\bar{x} = 150 - 4.8

\bar{x} = 145.2

Therefore, the mean daily wages of the workers of the factory is Rs 145.20.

Common mistakes

Errors in calculating the class mark (midpoint) of intervals.
Incorrectly applying the formula for the mean (direct or assumed mean method).
Calculation errors in summing up frequencies (Σf_i) or products (Σf_i x_i or Σf_i u_i).
Choosing an inappropriate assumed mean, leading to complex calculations.
Misinterpreting the class limits or frequencies from the given data.

Revision tips

Practice calculating the mean using both the direct and assumed mean methods for various datasets.
Pay close attention to the class intervals and ensure correct class marks are calculated.
Review the formulas for each method carefully before attempting problems.
Understand the rationale behind choosing the assumed mean method for larger values.
Verify your calculations, especially the summation steps, to avoid arithmetic errors.

Practice MCQs

Q1. What is the formula for calculating the class mark (x_i) of a class interval?

Q2. Which method is generally preferred when the class marks (x_i) and frequencies (f_i) are small?

Q3. In the Assumed Mean Method for finding the mean, what does 'd_i' represent?

Q4. What is the primary goal of calculating the mean in statistics?

Q5. If the class intervals in a frequency distribution are large, which method is often more efficient for calculating the mean?

Frequently asked questions

What is Chapter 29 of Class 10 Maths NCERT about?

Chapter 29 of the Class 10 Maths NCERT Solutions focuses on Statistics, specifically on calculating the mean for grouped frequency distributions using methods like the Direct Method and the Assumed Mean Method.

How do I calculate the class mark?

The class mark (or midpoint) of a class interval is calculated by adding the lower limit and the upper limit of the interval and then dividing the sum by 2. The formula is: Class Mark (x_i) = (Lower Limit + Upper Limit) / 2.

When should I use the Direct Method to find the mean?

The Direct Method is generally suitable when the class marks (x_i) and the frequencies (f_i) are relatively small numbers. This makes the calculation of Σf_i x_i manageable.

What is the advantage of using the Assumed Mean Method?

The Assumed Mean Method is advantageous when dealing with large class marks or frequencies. By assuming a mean ('a') within the data range, it simplifies calculations by working with smaller deviation values (d_i = x_i - a), reducing the risk of arithmetic errors.

How do these NCERT solutions help in exam preparation?

These solutions provide clear, step-by-step explanations for solving problems related to calculating the mean. Practicing these helps in understanding the concepts, mastering the formulas, and improving problem-solving accuracy for exams.

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