Class 10 students struggle to balance many subjects and assignments as they advance in their education. For this reason, ALLEN has developed comprehensive NCERT Solutions for Maths Class 10 Chapter 1 Real Numbers. The answers adhere to the most recent CBSE curriculum. ALLEN's subject matter experts have meticulously researched NCERT Class 10 Maths - Chapter 1 to ensure ease of comprehension. You can master complex problem-solving skills by understanding real numbers at the fundamental level by going through these NCERT class 10 Maths chapter 1 pdf solutions.
Mathematics introduces us to a wide range of numbers, including real numbers. To comprehend higher mathematics, it makes sense to start by thoroughly examining the number system, and this is exactly what this book has accomplished. The NCERT book has an intriguing feature that breaks up the monotony of study material by providing some historical notes.
Additionally, the solutions include some pointers and advice to help students make learning easier. The following are links to the NCERT Solutions Class 10 Maths PDF downloads, which are available for free:
Real Numbers is one of the important chapters covered in class 10. It has four exercises. Below is a summary of the significant topics addressed in the chapter. Students are advised to carefully review these topics to understand, apply, and become proficient with this chapter and the problems based on real numbers.
Brief Overview Of Class 10 Chapter 1 Real Numbers
1. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational, or not. If they are rational, and of the form p/q, what can you say about the prime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000…...
(iii) 43.123456789 (with a bar over 123456789 indicating repetition)
Solution:
(i) 43.123456789
Since the decimal expansion terminates, the given real number is rational.
It can be expressed in the form p/q as:
43.123456789 = 43123456789 / 1000000000
43.123456789 = 43123456789 / 10⁹
43.123456789 = 43123456789 / (2 × 5)⁹
43.123456789 = 43123456789 / (2⁹ × 5⁹)
Therefore, q = 2⁹ × 5⁹.
The prime factorization of q is of the form 2ⁿ ⋅ 5ᵐ, where n = 9 and m = 9.
(ii) 0.120120012000120000....
Since the decimal expansion is neither terminating nor non-terminating, the given real number is irrational.
(iii) 43.123456789 (with a bar over 123456789)
Since the decimal expansion is non-terminating and repeating, the given real number is rational.
However, since the decimal is non-terminating repeating, q is not of the form 2ᵐ × 5ⁿ.
The prime factors of q will contain factors other than 2 and 5.
2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
Solution:
(i) 13/3125 = 13/5⁵ = (13 × 2⁵)/(5⁵ × 2⁵) = 416/10⁵ = 0.00416
(ii) 17/8 = 17/2³ = (17 × 5³)/(2³ × 5³) = 2125/10³ = 2.125
(iv) 15/1600 = 3/(2⁶ × 5) = (3 × 5⁵)/(2⁶ × 5⁶) = 9375/10⁶ = 0.009375
(vi) 23/(2³ × 5²) = (23 × 5)/(2³ × 5³) = 115/10³ = 0.115
(viii) 6/15 = 2/5 = 4/10 = 0.4
(ix) 35/50 = 7/10 = 0.7
3. Prove that 3 + 2√5 is irrational.
Solution:
Let us assume, to the contrary, that 3 + 2√5 is rational. So, we can find coprime integers a and b (b ≠ 0) such that 3 + 2√5 = a/b, b ≠ 0, a, b ∈ I.
Therefore, a/b - 3 = 2√5 ⇒ (a - 3b)/b = 2√5 ⇒ (a - 3b)/(2b) = √5.
Since a and b are integers, we get (a - 3b)/(2b) is rational, and so √5 is rational.
But this contradicts the fact that √5 is irrational. This contradiction has arisen because of our incorrect assumption that 3 + 2√5 is rational.
So, we conclude that 3 + 2√5 is irrational.
4. Given that HCF(306, 657) = 9, find LCM(306, 657).
Solution:
LCM × HCF = product of two numbers.
LCM(306, 657) = (306 × 657) / HCF(306, 657)
LCM(306, 657) = (306 × 657) / 9 = 22338
5. Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solution:
If the number 6ⁿ, for any natural number n, ends with digit 0, then it would be divisible by 5. That is, the prime factorisation of 6ⁿ would contain the prime number 5. This is not possible because 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ; so the only primes in the factorisation of 6ⁿ are 2 and 3 and the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of 6ⁿ. So, there is no natural number n for which 6ⁿ ends with the digit zero.
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