RD Sharma Solutions for Class 6 Maths Chapter 2 Exercise 2.7: RD Sharma Solutions for Class 6 Maths Chapter 2 Exercise 2.7 Playing with Numbers – is a vital resource for students aiming to strengthen their basic number concepts.
This exercise covers divisibility rules, common multiples, and factors, which are important for building a strong mathematical foundation.
Understanding these concepts is not only crucial for school exams but also aligns with the CBSE exam pattern and syllabus. Practicing these solutions regularly helps improve problem-solving speed and accuracy. Referring to RD Sharma along with previous year papers allows students to gain confidence and prepare effectively for their exams.
What is Covered in RD Sharma Solutions for Class 6 Chapter 2 Exercise 2.7?
Exercise 2.7 of Chapter 2, Playing with Numbers from RD Sharma Class 6 Maths focuses on the concept of prime factorization using the factor tree method. Here are the detailed pointers on what is covered:
Topics Covered in Exercise 2.7:
-
Prime Factorization:
-
Factor Tree Method:
-
Identifying Prime and Composite Numbers:
-
Verification of Prime Factorization:
-
Practice with Different Types of Numbers:
-
Application in Real Problems:
-
Builds a base for advanced topics like LCM, HCF, fractions, and algebra.
-
Indirectly helps in solving questions from competitive exams and higher classes.
RD Sharma Solutions for Class 6 Chapter 2 Playing with Numbers Exercise 2.7 Question Answers
RD Sharma Solutions for Class 6 Chapter 2 Playing with Numbers Exercise 2.7 help students master prime factorization using the factor tree method. These solutions are accurate and exam-oriented. Below, we have provided the complete solutions for reference.
1. Determine the HCF of the following numbers by using Euclid’s algorithm (i – x):
(i) 300, 450
(ii) 399, 437
(iii) 1045, 1520
Solution:
(i) 300, 450
Taking 450 as dividend and 300 as divisor
We know that the last divisor is 150
Therefore, HCF of 300, 450 is 150.
(ii) 399, 437
Taking 437 as dividend and 399 as divisor
We know that the last divisor is 19
Therefore, HCF of 399, 437 is 19.
(iii) 1045, 1520
Taking 1520 as dividend and 1045 as divisor
We know that the last divisor is 95
Therefore, HCF of 1045, 1520 is 95.
2. Show that the following pairs are co-prime:
(i) 59, 97
(ii) 875, 1859
(iii) 288, 1375
Solution:
(i) 59, 97
Taking 97 as dividend and 59 as divisor
We know that the last divisor is 1.
Therefore, the numbers 59, 97 are co-prime.
(ii) 875, 1859
Taking 1859 as dividend and 875 as divisor
We know that the last divisor is 1.
Therefore, the numbers 875, 1859 are co-prime.
(iii) 288, 1375
Taking 1375 as dividend and 288 as divisor
We know that the last divisor is 1.
Therefore, the numbers 288, 1375 are co-prime.
3. What is the HCF of two consecutive numbers?
Solution:
We know that the HCF of two consecutive numbers is 1.
For example consider 4 and 5 as two consecutive numbers
Taking 5 as dividend and 4 as divisor
We know that the last divisor is 1.
Therefore, HCF of 4 and 5 is 1.
4. Write true (T) or false (F) for each of the following statements:
(i) The HCF of two distinct prime numbers is 1.
(ii) The HCF of two co-prime number is 1.
(iii) The HCF of an even and an odd number is 1.
(iv) The HCF of two consecutive even numbers is 2.
(v) The HCF of two consecutive odd numbers is 2.
Solution:
(i) True.
(ii) True.
(iii) False. The HCF of even number 6 and odd number 9 is 3.
(iv) True.
(v) False. The HCF of numbers 25 and 27 is 1.
RD Sharma Solutions for Class 6 Maths Chapter 2 Exercise 2.7 PDF Download
RD Sharma Solutions for Class 6 Chapter 2 Playing with Numbers – is an essential tool for mastering prime factorization through the factor tree method. These solutions are prepared as per the latest CBSE syllabus and exam pattern, making them perfect for exam preparation and concept clarity.
Practicing these problems also helps in understanding the types of questions asked in previous year papers. Below, we have provided the PDF download link for easy access and offline study.
