REAL NUMBER NCERT EXCERCISE 1.1 and 1.2 SOLUTION

CLASS 10 MATHS NCERT SOLUTION

NCERT Exercise 1.1 & 1.2 Solutions

NCERT EXERCISE 1.1
[A] Write the answer of the following questions.
1.	Express number as a product of its prime factors: 140 Click to show answer Solution: Using prime factorization: 140 ÷ 2 = 70 70 ÷ 2 = 35 35 ÷ 5 = 7 7 ÷ 7 = 1 ∴ 140 = 2 × 2 × 5 × 7 = \(2^2 \times 5 \times 7\)
2.	Express number as a product of its prime factors: 156 Click to show answer Solution: 156 ÷ 2 = 78 78 ÷ 2 = 39 39 ÷ 3 = 13 13 ÷ 13 = 1 ∴ 156 = 2 × 2 × 3 × 13 = \(2^2 \times 3 \times 13\)
3.	Express number as a product of its prime factors: 3825 Click to show answer Solution: 3825 ÷ 3 = 1275 1275 ÷ 3 = 425 425 ÷ 5 = 85 85 ÷ 5 = 17 17 ÷ 17 = 1 ∴ 3825 = 3 × 3 × 5 × 5 × 17 = \(3^2 \times 5^2 \times 17\)
4.	Express number as a product of its prime factors: 5005 Click to show answer Solution: 5005 ÷ 5 = 1001 1001 ÷ 7 = 143 143 ÷ 11 = 13 13 ÷ 13 = 1 ∴ 5005 = 5 × 7 × 11 × 13
5.	Express number as a product of its prime factors: 7429 Click to show answer Solution: 7429 ÷ 17 = 437 437 ÷ 19 = 23 23 ÷ 23 = 1 ∴ 7429 = 17 × 19 × 23
6.	Find the LCM and HCF of 12, 15 and 21 using prime factorization Click to show answer Solution: 12 = \(2^2 \times 3\) 15 = 3 × 5 21 = 3 × 7 HCF = Common factor with lowest power = 3¹ = 3 LCM = All prime factors with highest powers = \(2^2 \times 3 \times 5 \times 7 = 4 \times 3 \times 5 \times 7 = 420\)
7.	Find the LCM and HCF of 17, 23 and 29 Click to show answer Solution: 17 = 17 (prime) 23 = 23 (prime) 29 = 29 (prime) HCF = 1 (no common factors except 1) LCM = 17 × 23 × 29 = 11339
8.	Find the LCM and HCF of 8, 9 and 25 Click to show answer Solution: 8 = \(2^3\) 9 = \(3^2\) 25 = \(5^2\) HCF = 1 (no common factors) LCM = \(2^3 \times 3^2 \times 5^2 = 8 \times 9 \times 25 = 1800\)
9.	Check whether \(6^n\) can end with digit 0 for any natural number \(n\) Click to show answer Solution: For a number to end with digit 0, its prime factorization must contain factors 2 and 5. \(6^n = (2 \times 3)^n = 2^n \times 3^n\) Prime factors of \(6^n\) are 2 and 3 only (no factor of 5). ∴ \(6^n\) cannot end with digit 0 for any natural number \(n\).
[B] Write the answer of the following questions.
10.	Find LCM & HCF of 26 and 91, verify LCM × HCF = product Click to show answer Solution: 26 = 2 × 13 91 = 7 × 13 HCF = 13 LCM = 2 × 7 × 13 = 182 Verification: LCM × HCF = 182 × 13 = 2366 Product of numbers = 26 × 91 = 2366 ∴ LCM × HCF = Product of numbers (Verified)
11.	Find LCM & HCF of 510 and 92, verify LCM × HCF = product Click to show answer Solution: 510 = 2 × 3 × 5 × 17 92 = \(2^2 \times 23\) HCF = 2 LCM = \(2^2 \times 3 \times 5 \times 17 \times 23 = 4 \times 3 \times 5 \times 17 \times 23 = 23460\) Verification: LCM × HCF = 23460 × 2 = 46920 Product of numbers = 510 × 92 = 46920 ∴ LCM × HCF = Product of numbers (Verified)
12.	Find LCM & HCF of 336 and 54, verify LCM × HCF = product Click to show answer Solution: 336 = \(2^4 \times 3 \times 7\) 54 = \(2 \times 3^3\) HCF = 2 × 3 = 6 LCM = \(2^4 \times 3^3 \times 7 = 16 \times 27 \times 7 = 3024\) Verification: LCM × HCF = 3024 × 6 = 18144 Product of numbers = 336 × 54 = 18144 ∴ LCM × HCF = Product of numbers (Verified)
13.	Given HCF(306, 657) = 9, find LCM(306, 657) Click to show answer Solution: Using formula: HCF × LCM = Product of numbers 9 × LCM = 306 × 657 LCM = \(\frac{306 \times 657}{9}\) = 34 × 657 = 22338 ∴ LCM(306, 657) = 22338
14.	Explain why these are composite numbers: (i) \(7 \times 11 \times 13 + 13\) (ii) \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\) Click to show answer Solution: (i) \(7 \times 11 \times 13 + 13\) = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78 = 13 × 2 × 3 × 13 = \(2 \times 3 \times 13^2\) Has factors other than 1 and itself, so it’s composite. (ii) \(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5\) = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009 Has factors 5 and 1009, so it’s composite.
15.	Sonia takes 18 min, Ravi takes 12 min for one round. When will they meet again? Click to show answer Solution: Time for Sonia = 18 minutes Time for Ravi = 12 minutes They will meet again at LCM of their times 18 = 2 × \(3^2\) 12 = \(2^2 \times 3\) LCM = \(2^2 \times 3^2 = 4 \times 9 = 36\) ∴ They will meet again after 36 minutes at starting point.
NCERT EXERCISE 1.2
16.	Prove that \(7\sqrt{5}\) is irrational Click to show answer Proof: Assume \(7\sqrt{5}\) is rational. Then \(7\sqrt{5} = \frac{p}{q}\), where p, q are integers, q ≠ 0, HCF(p,q)=1 ∴ \(\sqrt{5} = \frac{p}{7q}\) Since p and 7q are integers, RHS is rational ⇒ \(\sqrt{5}\) is rational. But we know \(\sqrt{5}\) is irrational (contradiction). ∴ Our assumption is wrong ⇒ \(7\sqrt{5}\) is irrational.
17.	Prove that \(6 + \sqrt{2}\) is irrational Click to show answer Proof: Assume \(6 + \sqrt{2}\) is rational. Then \(6 + \sqrt{2} = \frac{p}{q}\), where p, q are integers, q ≠ 0 ∴ \(\sqrt{2} = \frac{p}{q} – 6 = \frac{p – 6q}{q}\) Since p-6q and q are integers, RHS is rational ⇒ \(\sqrt{2}\) is rational. But we know \(\sqrt{2}\) is irrational (contradiction). ∴ Our assumption is wrong ⇒ \(6 + \sqrt{2}\) is irrational.
[C] Write the answer of the following questions.
18.	Prove that \(\sqrt{5}\) is irrational Click to show answer Proof: Assume \(\sqrt{5}\) is rational. Then \(\sqrt{5} = \frac{p}{q}\), where p, q are integers, q ≠ 0, HCF(p,q)=1 Squaring: \(5 = \frac{p^2}{q^2}\) ⇒ \(p^2 = 5q^2\) ∴ 5 divides \(p^2\) ⇒ 5 divides p Let p = 5k, then \((5k)^2 = 5q^2\) ⇒ \(25k^2 = 5q^2\) ⇒ \(5k^2 = q^2\) ∴ 5 divides \(q^2\) ⇒ 5 divides q Thus 5 divides both p and q, contradicting HCF(p,q)=1 ∴ \(\sqrt{5}\) is irrational.
19.	Prove that \(3 + 2\sqrt{5}\) is irrational Click to show answer Proof: Assume \(3 + 2\sqrt{5}\) is rational. Then \(3 + 2\sqrt{5} = \frac{p}{q}\), where p, q are integers, q ≠ 0 ∴ \(2\sqrt{5} = \frac{p}{q} – 3 = \frac{p – 3q}{q}\) ∴ \(\sqrt{5} = \frac{p – 3q}{2q}\) Since p-3q and 2q are integers, RHS is rational ⇒ \(\sqrt{5}\) is rational. But we know \(\sqrt{5}\) is irrational (contradiction). ∴ \(3 + 2\sqrt{5}\) is irrational.
20.	Prove that \(\frac{1}{\sqrt{2}}\) is irrational Click to show answer Proof: Assume \(\frac{1}{\sqrt{2}}\) is rational. Then \(\frac{1}{\sqrt{2}} = \frac{p}{q}\), where p, q are integers, q ≠ 0, HCF(p,q)=1 ∴ \(\sqrt{2} = \frac{q}{p}\) Since q and p are integers, RHS is rational ⇒ \(\sqrt{2}\) is rational. But we know \(\sqrt{2}\) is irrational (contradiction). ∴ \(\frac{1}{\sqrt{2}}\) is irrational.
[D] Write the answer of the following questions.
21.	Prove that the following are irrationals: (i) \(\frac{1}{\sqrt{2}}\) – Same as Q20 (ii) \(7\sqrt{5}\) – Same as Q16 (iii) \(6 + \sqrt{2}\) – Same as Q17