RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers

RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers provides comprehensive answers to the textbook exercise questions, crafted by subject experts at Infinity Learn to help students effectively learn the concepts. 

These solutions are designed to offer the best study material for students to prepare well for the board exams. Students can easily download the RD Sharma Class 10th Chapter 1 PDF from the links provided here and study without any time constraints.

The RD Sharma Class 10 Chapter 1 PDF includes precise answers to all questions found in the chapter. It serves as an excellent resource for clearing doubts students might have while revising problems from RD Sharma Class 10 Chapter 1. Additionally, students can access RD Sharma Class 10 Real Numbers solutions and enhance their understanding with Class 10 Maths Chapter 1 Extra Questions from RD Sharma.

The solutions are designed to help students grasp the essential concepts from Real Numbers Class 10. These concepts are crucial for those considering a future in Mathematics.

The RD Sharma Chapter 1 Class 10 solutions are aligned with the latest CBSE syllabus 2025-26, ensuring that students are well-prepared for their exams. Whether you're looking for RD Sharma Class 10 Chapter 1 PDF Questions or Real Numbers Class 10 Extra Questions, these solutions will provide all the support you need to excel academically.

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Access Answers to D Sharma Solutions for Class 10 Maths Chapter 1 – Real Numbers

Q. Prove that the products of two consecutive positive integers is divisible by 2.

Solutions: Let the numbers are a and a-1

Product of these number: 𝑎(𝑎 −1) = 𝑎² −𝑎

Case 1: When a is even:

𝑎 =2𝑝

then (2𝑝)² − 2𝑝 ⇒ 4𝑝²−2𝑝

2𝑝(2𝑝 −1).......... it is divisible by 2

Case 2: When a is odd:

𝑎 =2𝑝+1

then (2𝑝 + 1)² −(2𝑝+1) ⇒ 4𝑝² +4𝑝+1−2𝑝−1

=4𝑝²+2𝑝⇒2𝑝(2𝑝+1) it is divisible by 2

Hence, we conclude that the product of two consecutive integers is always divisible by 2.

Q. Prove that the product of three consecutive positive integers is divisible by 6

Solution: Let 𝑎 be any positive integer and 𝑏 = 6. By division lemma there exists integers 𝑞 and 𝑟 such that

𝑎 =6𝑞+𝑟 where 0≤𝑟 <6

𝑎 =6𝑞, or 𝑎 =6𝑞+1 or, 𝑎 =6𝑞+2 or, 𝑎 =6𝑞+3 or 𝑎 =6𝑞+4 or 𝑎 =6𝑞+5

Let 𝑛 is any positive integer.

Since any positive integer is of the form 6𝑞 or, 6𝑞 + 1 or, 6𝑞 + 2 or, 6𝑞 +3 or, 6𝑞 +4 or, 6𝑞 +5.

Case 1:

If 𝑛 = 6𝑞 then

𝑛(𝑛 +1)(𝑛 +2) = 6𝑞(6𝑞+1)(6𝑞+2), which is divisible by 6 .

Case 2:

If 𝑛 = 6𝑞 +1, then

𝑛(𝑛 +1)(𝑛 +2) = (6𝑞+1)(6𝑞+2)(6𝑞+3)

=(6𝑞+1)2(3𝑞+1)3(2𝑞+1)

=6(6𝑞+1)(3𝑞+1)(2𝑞+1), which is divisible by 6 .

Case 3:

If 𝑛 = 6𝑞 +2, then

𝑛(𝑛 +1)(𝑛 +2) = (6𝑞+2)(6𝑞+3)(6𝑞+4)

=2(3𝑞+1)3(2𝑞+1)(6𝑞+4)

=6(3𝑞+1)(2𝑞+1)(6𝑞+4), which is divisible by 6

Case 4:

If 𝑛 = 6𝑞 +3, then

𝑛(𝑛 +1)(𝑛 +2) = (6𝑞+3)(6𝑞+4)(6𝑞+5)

=3(2𝑞+1)2(3𝑞+2)(6𝑞+5)

=6(2𝑞+1)(3𝑞+2)(6𝑞+5), which is divisible by 6.

Case 5 :

If 𝑛 = 6𝑞 +4, then

𝑛(𝑛 +1)(𝑛 +2) = (6𝑞+4)(6𝑞+5)(6𝑞+6)

=(6𝑞+4)(6𝑞+5)6(𝑞+1)

=6(6𝑞+4)(6𝑞+5)(𝑞+1), which is divisible by 6.

Case 6:

If 𝑛 = 6𝑞 +5, then

𝑛(𝑛 +1)(𝑛 +2) = (6𝑞+5)(6𝑞+6)(6𝑞+7)

=(6𝑞+5)6(𝑞+1)(6𝑞+7)

=6(6𝑞+5)(𝑞 +1)(6𝑞+7), which is divisible by 6 .

Hence, 𝑛(𝑛 +1)(𝑛 +2) is divisible by 6.

Q. For any positive integer, prove that 𝑛3 − 𝑛 divisible by 6 .

Solutions:

n³ −n=n(n²−1)=n(n−1)(n+1)

For a number to be divisible by 6, it should be divisible by 2 and 3 both,

Divisibility by 3:

n −1,n and n+1 are three consecutive whole numbers.

By Euclid's division lemma

n +1=3𝑞+𝑟, for some integer 𝑘 and 𝑟 < 3

As, 𝑟 < 3 possible values of ' 𝑟 ' are 0,1 and 2.

If 𝑟 = 0

n +1=3q

⇒n+1 is divisible by 3

⇒n(n−1)(n+1) is divisible by 3

⇒(n3 −n) is divisible by 3

If 𝑟 = 1

⇒n+1=3q+1

⇒n=3q

⇒n is divisible by 3

⇒n(n−1)(n+1) is divisible by 3

⇒(n3 −n) is divisible by 3

If 𝑟 = 2

⇒n+1=3q+2

⇒n+1−2=3q

⇒n−1=3q

⇒n−1 is divisible by 3

⇒n(n−1)(n+1) is divisible by 3

⇒(𝑛³−𝑛) is divisible by 3

Divisibility by 2:

If 𝑛 is even

Clearly, 𝑛(𝑛 − 1)(𝑛 + 1) is divisible by 2

If 𝑛 is odd

⇒n+1 is even

⇒n+1 is divisible by 2

⇒n(n−1)(n+1) is divisible by 2

Hence, for any value of 𝑛,𝑛³ − 𝑛 is divisible by 2 and 3 both, therefore 𝑛³ − 𝑛 is divisible by 6 .

Q. Prove that if a positive integer is of the form 6𝑞 + 5, then it is of the form 3𝑞 + 2 for some integer 𝑞, but not conversely.

Solution: let 𝐴 = 6𝑞 +5, be any number, where 𝑞 is any positive integer.

Part 1: To show 𝐴 is in the form of 3𝑞 + 2, where 𝑞 is another integer

𝐴 =6𝑞+5=6𝑞+3+2=3(2𝑞 +1)+2=3𝑞′+2

As,𝑞 is any positive integer, 𝑞′ = 3𝑞 + 2 is also a positive integer and hence 6𝑞 + 5 is in form of 3𝑞′ + 5

Part 2: To show converse is not true, i.e. if a number is in the form of 3𝑞 + 2, then it may or may not be in the form of 6𝑞 +5

For example, consider: 8 = 3(2) + 2 is in the 3𝑞 + 2 form, but it can't be expand in 6𝑞 +5 form.

Advantages of Solving RD Sharma Class 10 Chapter 1 Real Numbers

1. Strengthens Key Concepts: Helps students develop a strong understanding of essential topics like rational and irrational numbers, HCF, LCM, and Euclid's Division Lemma, which form the foundation for higher-level mathematics.
2. Improves Problem-Solving Skills: Provides a variety of problems, including extra questions from RD Sharma, enhancing students' analytical and problem-solving abilities.
3. Board Exam Preparation: Prepares students for Class 10 Maths Chapter 1 board exams by offering practice questions aligned with the latest CBSE syllabus.
4. Boosts Speed and Accuracy: Regular practice with RD Sharma Class 10 Chapter 1 PDF Questions helps improve speed and accuracy in solving problems under exam conditions.
5. Comprehensive Solutions: Offers detailed solutions to all exercises, allowing students to understand each step and clarify any doubts they encounter during practice.
6. Prepares for Competitive Exams: Lays the groundwork for entrance exams like JEE and NEET, where a solid understanding of real numbers is crucial for success.