NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers All Exercises

NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers

When students enter the sixth grade, they are exposed to the CBSE syllabus which includes different subjects. NCERT Solutions for Class 6 Maths Chapter 3 is quite diverse and covers a wide range of topics. Playing with Numbers, the third chapter of the Class 6 Maths syllabus focuses on teaching pupils about multiples and divisors.

Students are introduced to subjects such as common factors and multiples, divisibility laws, greatest common factors, lowest common factors, and so on as the chapter proceeds. Students will be able to understand prime and composite numbers, as well as their differences, provided they have a thorough understanding of the chapter.

You may also get NCERT Class 6 Maths solutions and NCERT Class 6 Science solutions to help you revise the entire syllabus and get better grades in your exams.

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NCERT Solutions for Class 6 Maths Chapter 3 PDF Download Updated 2024-25

3. Playing with numbers

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.1 to 3.7

NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers is a crucial chapter for students to understand the concepts of multiples, divisors, factors, and how to identify factors and multiples, along with HCF and LCM. The chapter consists of 7 exercises, each focusing on specific topics related to numbers.
Exercise 3.1 starts with finding the factors of numbers, including 24, 15, 21, 27, 12, 20, 18, 23, 36. Students are introduced to the concept of factors and how to find them.
Exercise 3.2 focuses on finding multiples of numbers, including 5, 7, 8, 9, 11, 12, 15, 16, 18, 19, 20, and identifying multiples of the given numbers.
Exercise 3.3 introduces the concept of prime and composite numbers. Students learn to identify prime and composite numbers and understand the difference between them.
Exercise 3.4 focuses on the tests for divisibility of numbers. Students learn to identify whether a number is divisible by 2, 3, 4, 5, 9, or 10.
Exercise 3.5 covers common factors and common multiples. Students learn to find the common factors and common multiples of two numbers.
Exercise 3.6 focuses on prime factorisation, HCF, and LCM. Students learn to find the prime factors of a number, HCF, and LCM of two or more numbers.
Exercise 3.7 covers some problems related to LCM and HCF. Students learn to solve problems related to LCM and HCF.

Class 6 CBSE Foundation Course

NCERT Solutions for Class 6 Maths Chapter 3 All Exercise

Below are the class 6 chapter 3 exercise-wise solve NCERT questions

NCERT Solutions for Class 6 Maths Chapter 3 Exercise 3.7

1. Renu purchases two bags of fertiliser of weights 75 kg and 69 kg. Find the maximum value of weight, which can measure the weight of the fertiliser exact number of times.

Solution:

1. List the factors of each number:
   • Factors of 75: 1, 3, 5, 15, 25, 75
   • Factors of 69: 1, 3, 23, 69
2. Find the highest common factor:
   • The common factors of 75 and 69 are 1 and 3.
3. Choose the greatest common factor:
   • The greatest common factor is 3.

2. Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm, and 77 cm, respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

Solution:

• Step 1: The problem requires finding the Least Common Multiple (LCM) of the step lengths.
• Step 2: Use prime factorization:
  • 63 = 7 × 9 = 7 × 3^2
  • 70 = 2 × 35 = 2 × 5 × 7
  • 77 = 7 × 11
• Step 3: The LCM is the product of the highest powers of all primes appearing in these factorizations:
  • LCM = 2^1 × 3^2 × 5^1 × 7^1 × 11^1 = 13860 cm

Answer: The minimum distance they all should cover in complete steps is 13860 cm.

3. The length, breadth, and height of a room are 825 cm, 675 cm, and 450 cm, respectively. Find the longest tape that can measure the room’s three dimensions exactly.

Solution:

• Step 1: This problem involves finding the GCD of three numbers (length, breadth, height).
• Step 2: Apply the Euclidean algorithm sequentially:
  • GCD(825, 675) = GCD(675, 825 % 675) = GCD(675, 150)
  • GCD(675, 150) = GCD(150, 675 % 150) = GCD(150, 75)
  • GCD(150, 75) = 75
  • Now, use this GCD with the third dimension: GCD(75, 450) = 75

Answer: The longest tape that can measure all dimensions exactly is 75 cm.

4. Determine the smallest 3-digit number, which is exactly divisible by 6, 8, and 12.

Solution:

• Step 1: Identify the requirement for the LCM of the numbers.
• Step 2: Calculate the LCM:
  • LCM(6, 8) = LCM(2 × 3, 2^3) = 2^3 × 3 = 24
  • LCM(24, 12) = LCM(2^3 × 3, 2^2 × 3) = 24
• Step 3: The smallest three-digit number divisible by 24 is 120 (24 × 5).

Answer: The smallest three-digit number divisible by 6, 8, and 12 is 120.

5. Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.

Solution:

• Step 1: Calculate the LCM of 8, 10, and 12:
  • LCM(8, 10) = LCM(2^3, 2 × 5) = 2^3 × 5 = 40
  • LCM(40, 12) = LCM(2^3 × 5, 2^2 × 3) = 2^3 × 3 × 5 = 120
• Step 2: Identify the greatest three-digit number divisible by 120 by dividing 999 by 120 and multiplying the integer quotient by 120:
  • 999 ÷ 120 ≈ 8.325 → Integer part is 8
  • Greatest number = 120 × 8 = 960

Answer: The greatest three-digit number divisible by 8, 10, and 12 is 960.

6. The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds, respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?

Solution:

1. Identify the Requirement: Find the least common multiple (LCM) of the intervals at which the lights change (48, 72, 108 seconds) to determine when they will next change simultaneously.
2. Calculate the LCM:
   • Prime factorization:
     • 48 = 2^4 × 3^1
     • 72 = 2^3 × 3^2
     • 108 = 2^2 × 3^3
   • Take the highest power of each prime factor:
     • LCM = 2^4 × 3^3 = 16 × 27 = 432 seconds
3. Convert Seconds to Time: 432 seconds is 7 minutes and 12 seconds.
4. Add to Initial Time: Starting at 7:00 a.m., add 7 minutes and 12 seconds.

Answer: The lights will change simultaneously again at 7:07:12 a.m.

7. Three tankers contain 403 litres, 434 litres and 465 litres of diesel, respectively. Find the maximum capacity of a container that can measure the diesel of the three containers the exact number of times.

Solutions:

Identify the common factor: The number 31 is the only factor that appears in all three breakdowns.

Determine the HCF: Since 31 is common in all three, the highest common factor is 31 liters.

8. Find the least number, which, when divided by 6, 15, and 18 leave the remainder 5 in each case.

Solution:

Formulate the Problem: The number can be expressed as

𝑛 = 6𝑎 + 5 = 15𝑏 + 5 = 18𝑐 + 5.

This means 𝑛−5 is divisible by 6, 15, and 18.

Calculate the LCM: LCM(6, 15, 18) = 90

Add the Remainder:

𝑛 = 90 + 5 = 95

Answer: The least number that leaves a remainder of 5 when divided by 6, 15, and 18 is 95.

9. Find the smallest 4-digit number, which is divisible by 18, 24 and 32.

Solution:

Calculate the LCM:

Prime factorization:

• 18 = 2 × 3^2
• 24 = 2^3 × 3
• 32 = 2^5

Take the highest power of each prime factor:

LCM = 2^5 × 3^2 = 32 × 9 = 288

Find the Smallest 4-Digit Number: The smallest 4-digit number is 1000, so we divide and find the nearest multiple of 288.

1000 ÷ 288 ≈ 3.47, thus the nearest multiple is

288 × 4 = 1152

Answer: The smallest 4-digit number divisible by 18, 24, and 32 is 1152.

10. Find the LCM of the following numbers:

1. 9 and 4
2. 12 and 5
3. 6 and 5
4. 15 and 4

Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?

(a) LCM of 9 and 4

Factors: 9 is 3 × 3 and 4 is 2 × 2.

LCM Calculation: Take all prime factors at their highest powers, so 2² × 3² = 36.

Result: The LCM of 9 and 4 is 36.

(b) LCM of 12 and 5

Factors: 12 breaks down to 2² × 3 and 5 is a prime number.

LCM Calculation: Combine the highest powers of all primes, 2² × 3 × 5 = 60.

Result: The LCM of 12 and 5 is 60.

(c) LCM of 6 and 5

Factors: 6 is 2 × 3 and 5 is a prime.

LCM Calculation: Multiply the highest powers, 2 × 3 × 5 = 30.

Result: The LCM of 6 and 5 is 30.

(d) LCM of 15 and 4

Factors: 15 is 3 × 5 and 4 is 2 × 2.

LCM Calculation: Combine these, 2² × 3 × 5 = 60.

Result: The LCM of 15 and 4 is 60.

11. Find the LCM of the following numbers in which one number is the factor of the other.

1. 5, 20
2. 6, 18
3. 12, 48
4. 9, 45

What do you observe in the results obtained?

Solutions

(a) For numbers 5 and 20:

• LCM calculation: The largest number, 20, already includes 5 as a factor.
• Result: The LCM of 5 and 20 is 20.

(b) For numbers 6 and 18:

• LCM calculation: Since 18 includes 6 as a factor (18 = 2 × 3 × 3 and 6 = 2 × 3), the larger number is the LCM.
• Result: The LCM of 6 and 18 is 18.

(c) For numbers 12 and 48:

• LCM calculation: 48 already includes 12 as a factor (48 = 2^4 × 3 and 12 = 2^2 × 3).
• Result: The LCM of 12 and 48 is 48.

(d) For numbers 9 and 45:

• LCM calculation: 45 includes 9 as a factor (45 = 3^2 × 5 and 9 = 3^2).
• Result: The LCM of 9 and 45 is 45.

NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Overview

Students may easily study for their examinations with the help of NCERT answers for Class 6 Maths Chapter 3. Various solved examples are provided here to assist students in understanding how to calculate these sums. They also allow kids to acquire this fundamental knowledge, which will be useful in the future.

On this website, you can get a free PDF of Class 6 Maths Chapter 3 for free. It assists you in comprehending the importance of different numbers and their multiples in solving a problem. The formulas utilized and used in this chapter are primary equations that will help the student answer more difficult equations in the future.

NCERT Solutions for Class 6 Maths All Chapters

• Chapter 1 Knowing Our Numbers
• Chapter 2 Whole Numbers
• Chapter 3 Playing With Numbers
• Chapter 4 Basic Geometrical Ideas
• Chapter 5 Understanding Elementary Shapes
• Chapter 6 Integers
• Chapter 7 Fractions
• Chapter 8 Decimals
• Chapter 9 Data Handling
• Chapter 10 Mensuration
• Chapter 11 Algebra
• Chapter 12 Ratio and Proportion
• Chapter 13 Symmetry
• Chapter 14 Practical Geometry

NCERT Class 6 Maths Chapter 3 Exercises Overview

Exercise 3.1 – Introduction

The goal of this introduction is to provide students with a fundamental understanding of what constitutes a divisor and what constitutes a factor. This section serves as the foundation for the rest of the chapter.

Exercise 3.2 – Factors and Multiples

The first set of tasks in NCERT answers for Class 6th Maths Chapter 3 – Playing with Numbers encourages students to recognize multiples and factors. Every integer is also a multiple and factor of itself, according to the rule.

Exercise 3.3 – Prime and Composite Numbers

This portion of the textbook explains how to determine which numbers are prime and which are composite. The segment also explains why 2 is the smallest prime number and why 1 cannot be both a composite and a prime number.

Exercise 3.4 – Tests for Divisibility of Numbers

The purpose of this section of Chapter 3 in the NCERT Maths book for Class 6 is to teach pupils which integers are divisible by 10, 5, 2, 3, 6, 4, 8, 9, and 11. Find out how this divisibility test can help you.

Exercise 3.5 – Common Factors and Common Multiples

Understanding common multiples and common factors are essential for understanding the subject of numbers. The goal of this activity is to demonstrate how some numbers are multiples and factors of other numbers.

Exercise 3.6 – Some More Divisibility Rules

To obtain a thorough understanding of the subject of factors and multiples, the learner will need to know some divisibility rules. For a more in-depth understanding of the issue, consult the NCERT Chapter 3 answers.

Exercise 3.7 – Prime Factorization

The prime factors are discussed in this portion of NCERT Class 6 Maths Chapter 3 and how they are multiplied to form the original number. In this segment, you’ll learn about prime factor number play.

Exercise 3.8 – Highest Common Factor

The HCF, or the highest common factor, is an important component on which students must focus. The reason for this is that most examination questions are generated from this part.

Exercise 3.9 – Lowest Common Factor

Teachers prefer LCF (Lowest Common Factor) to be included in tests. Once students grasp the equations and how they are framed, the LCF calculation method is relatively simple.

Exercise 3.10 – Problems of HCF and LCM

The learner must tackle both theoretical and practical problems in this phase. Students are given a brief review of the entire topic at the end of the chapter.

Key Features of NCERT Solutions for Class 6 Maths Chapter 3

Students may learn a lot from NCERT Solutions on INFINITY learn’s official website. It’s similar to an assorted back with a wide range of requirements. The following are some of the advantages of NCERT solutions:

• The explanation will be simple to grasp and tailored to the student’s level.
• It includes a test purpose and enough unresolved problems to become proficient in that subject for each chapter.
• The pupils will profit from the free PDF in a variety of ways, and parents will no longer have to worry about the cost.
• Students’ doubts can be cleared by participating in a live session on the official website.

FAQs on NCERT Solutions for Class 6 Maths Chapter 3

What is the name of the chapter 3 of maths class 6?

The name of the chapter is Playing with Numbers and it is the third chapter of Class 6 Maths.

How many questions are there in class 6 maths chapter 3 exercise 3.7?

Exercise 3.7 of Class 6 Maths Chapter 3 has 11 questions.

How many Exercise are in class 6 maths chapter 3?

There are 7 exercises in Class 6 Maths Chapter 3.

According to the latest CBSE syllabus, what types of numbers are explored in Chapter 3 of NCERT Solutions for Class 6 Maths?

Prime numbers, even numbers, odd numbers, a set of whole numbers, natural numbers, composite numbers are discussed in Chapter 3 of NCERT Solutions for Class 6 Maths.

What are the benefits of NCERT Solutions for Class 6 Maths Chapter 3 for CBSE exams?

NCERT Solutions for Class 6 Maths Chapter 3 provides answers with thorough descriptions as per the period limit specified by the board, helping students gain valuable experience and prepare effectively for exams.

In Chapter 3 of NCERT Solutions for Class 6 Maths, what will I learn about factor properties?

Factors of a number have properties such as every number has a factor of one, every digit is a factor in and of itself, every divisor of an integer is also a factor of that number, every factor has a value that is less than or equal to the provided value, a given number has a finite number of factors.