Ch 1: Real Numbers (NCERT Solutions)
Based on the latest rationalized CBSE syllabus. Master the Fundamental Theorem of Arithmetic and Irrationality proofs.
Exercise 1.1 Solutions
Q1. Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429
Solution:
- (i) 140: 140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7 = 2² × 5 × 7
- (ii) 156: 156 = 2 × 78 = 2 × 2 × 39 = 2 × 2 × 3 × 13 = 2² × 3 × 13
- (iii) 3825: 3825 = 3 × 1275 = 3 × 3 × 425 = 3 × 3 × 5 × 85 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17
- (iv) 5005: 5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13
- (v) 7429: 7429 = 17 × 437 = 17 × 19 × 23
Q2. Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers: (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
Solution for (i) 26 and 91:
26 = 2 × 13
91 = 7 × 13
HCF = 13 | LCM = 2 × 7 × 13 = 182
Verification: LCM × HCF = 182 × 13 = 2366. Product = 26 × 91 = 2366. Verified.
Solution for (ii) 510 and 92:
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23 = 2² × 23
HCF = 2 | LCM = 2² × 3 × 5 × 17 × 23 = 23460
Verification: LCM × HCF = 23460 × 2 = 46920. Product = 510 × 92 = 46920. Verified.
Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
We know that: HCF × LCM = Product of the two numbers
9 × LCM = 306 × 657
LCM = (306 × 657) / 9
LCM = 34 × 657 = 22338
Q5. Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solution:
For any number to end with the digit 0, its prime factorization must contain at least one pair of 2 and 5.
Prime factorization of 6 = (2 × 3). Therefore, 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ
Since the prime factor 5 is missing, 6ⁿ can never end with the digit 0.
Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
Part 1: 7 × 11 × 13 + 13
Take 13 common: 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78.
Since it has factors other than 1 and itself (13 and 78), it is a composite number.
Part 2: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Take 5 common: 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × (1008 + 1) = 5 × 1009.
Since it has factors other than 1 and itself, it is a composite number.
Exercise 1.2 Solutions
⚠️ Board Exam Tip: This exercise guarantees a 3-mark question in the board exam.
Q1. Prove that √5 is irrational.
Solution (Proof by Contradiction):
Let us assume to the contrary that √5 is a rational number.
Then, we can find two co-prime integers a and b (b ≠ 0) such that: √5 = a / b
Squaring both sides:
5 = a² / b²
⇒ 5b² = a² --- (Equation 1)
This means a² is divisible by 5. By theorem, if a prime number divides a², it also divides a. So, a is divisible by 5.
Let a = 5c for some integer c. Substituting this in Eq 1:
5b² = (5c)²
5b² = 25c²
⇒ b² = 5c²
This means b² is divisible by 5, hence b is also divisible by 5.
Conclusion: Since both a and b are divisible by 5, they are not co-prime. This contradicts our initial assumption. Therefore, √5 is irrational. ∎
Q2. Prove that 3 + 2√5 is irrational.
Solution:
Let us assume that 3 + 2√5 is rational. Then we can write it as a/b where a and b are co-prime integers (b ≠ 0).
3 + 2√5 = a / b
Rearranging the equation:
2√5 = (a / b) - 3
2√5 = (a - 3b) / b
√5 = (a - 3b) / 2b
Since a and b are integers, the expression (a - 3b) / 2b is rational. This would mean that √5 is also rational.
But we know that √5 is an irrational number. This contradiction proves that 3 + 2√5 is irrational. ∎
Q3. Prove that the following are irrational: (i) 1/√2 (ii) 7√5 (iii) 6 + √2
Solution for (i) 1/√2:
Assume 1/√2 is rational. So, 1/√2 = a/b (a, b are co-prime).
Reversing both sides: √2 = b/a.
Since a and b are integers, b/a is rational, which means √2 is rational. This contradicts the fact that √2 is irrational. Hence, 1/√2 is irrational.
Solution for (ii) 7√5:
Assume 7√5 is rational = a/b.
Then √5 = a / 7b.
Since a, 7, b are integers, a/7b is rational. This contradicts the fact that √5 is irrational. Hence, 7√5 is irrational.
Previous Year Questions (PYQs)
Q. The HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, find the other number.
Solution:
Formula: HCF × LCM = Product of numbers
27 × 162 = 54 × x
x = (27 × 162) / 54 = 81