NCERT Solutions Class 6 Maths Chapter 3 Number Play Exercise 3.7

Exercise 3.7 of Class 6 Maths Chapter 3 - Playing with Numbers teaches you to find common multiples and common factors of numbers. This is also helpful when you are solving bigger problems in higher classes. The questions are simple and help you to understand the basis of numbers and how they relate to each other. 

Solving this exercise will help to strengthen your basic number concepts and also help you to achieve better marks in school exams. Regular practice may also help you in maths olympiads and other competitive exams. The questions in this exercise are made as per the latest NCERT Syllabus adhering to the Class 6 curriculum.

You can download the NCERT Solutions for this exercise in PDF format from this page. These solutions are written in a sequential step by step form to help learn faster and better conceptual understanding. 

1.0Download NCERT Solutions Class 6 Maths Chapter 3 Number Play Exercise 3.7: Free PDF

Exercise 3.7 focuses on identifying common factors and multiples, which are key concepts in number theory. The NCERT Solutions for Class 6 Maths Chapter 3 offer clear, step-by-step answers as per the latest syllabus. Download the free PDF from below to strengthen your understanding and improve your exam preparation.

2.0NCERT Solutions Class 6 Chapter 3 Number Play: All Exercises

3.0NCERT Class 6 Maths Chapter 3 Number Play Exercise 3.7: Detailed Solutions

1. Pratibha uses the digits ' 4 ', ' 7 ', ' 3 ' and ' 2 ', and makes the smallest and largest 4 -digit numbers with them: 2347 and 7432. The difference between these two numbers is 7432−2347= 5085. The sum of these two numbers is 9779 . Choose 4 -digits to make:

a. The difference between the largest and smallest numbers greater than 5085.

b. The difference between the largest and smallest numbers less than 5085.

c. The sum of the largest and smallest numbers greater than 9779 .

d. The sum of the largest and smallest numbers less than 9779.

Sol.

a. Let's use the digits 2, 3, 1, 8:

Largest number: 8321

Smallest number: 1238

Difference =8321−1238=7083

Since 7083 is greater than 5085, this works.

b. Let's use the digits 4,5,6,8:

Largest number: 8654

Smallest number: 4568

Difference =8654−4568=4086

Since 4086 is less than 5085, this works.

c. Let's use the digits 5,9,2,3 :

Largest number: 9532

Smallest number: 2359

Sum =9532+2359=11891

Since 11891 is greater than 9779 , this works.

d. Let's use the digits 7,2,5,1 :

Largest number =7521

Smallest number =1257

Sum =7521+1257=8778

Since 8778 is less than 9779 , this works.

2. What is the sum of the smallest and largest 5 -digit palindrome? What is their difference?

Sol. Smallest 5-Digit Palindrome: 10001.

Largest 5-Digit Palindrome: 99999.

Sum: 10001+99999=110000.

Difference: 99999−10001=89998.

3. The time now is 10:01. How many minutes until the clock shows the next palindromic time? What about the one after that?

Sol. The next palindromic time after 10:01 is 11:11.

The difference between 10:01 and 11:11 is 1 hour and 10 minutes, which equals to 70 minutes. The next palindromic time after 11:11 is 12:21.

The difference between 11:11 and 12:21 is also 1 hour and 10 minutes, or 70 minutes.

4. How many rounds does the number 5683 take to reach the Kaprekar constant?

Sol.

It takes 7 rounds for the number 5683 to reach the Kaprekar constant.

4.0Key Features and benefits for Class 6 Maths Chapter 3 Exercise 3.7

• This exercise has straightforward questions on common factors and common multiples.
• All of these problems are based on the new NCERT syllabus of Class 6 Maths.
• It will also help to build a strong foundation in number concepts when it comes to later classes.
• Practicing these NCERT solutions, students will increase their accuracy and speed when solving questions in the CBSE exam.
• Students can use these solutions to prepare for competitive exams like the maths olympiads.
• The step-by-step solutions also give students the clarity and confidence they need to learn problem-solving skills.