NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers — Exercise 1.1 & 1.2

How to Use This Page

These are complete NCERT Solutions for CBSE Class 10 Maths Chapter 1 — Real Numbers, covering every question of Exercise 1.1 and Exercise 1.2. Try each problem yourself first, then check the full method below. Chapter 1 is based on the Fundamental Theorem of Arithmetic (prime factorisation, HCF/LCM) and irrationality proofs.

Key Concepts (Quick Recap)

• Fundamental Theorem of Arithmetic: Every composite number is a unique product of primes (order apart).
• HCF = product of the smallest power of each common prime factor.
• LCM = product of the greatest power of every prime factor that appears.
• For two numbers a and b: HCF(a,b) × LCM(a,b) = a × b.
• A number ends in 0 only if it is divisible by both 2 and 5.

Exercise 1.1

Q1. Express each number as a product of its prime factors.

• (i) 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7
• (ii) 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13
• (iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17
• (iv) 5005 = 5 × 7 × 11 × 13 = 5 × 7 × 11 × 13
• (v) 7429 = 17 × 19 × 23 = 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 26 = 2 × 13; 91 = 7 × 13. HCF = 13; LCM = 2 × 7 × 13 = 182. Check: HCF × LCM = 13 × 182 = 2366 = 26 × 91. ✓

(ii) 510 and 92 510 = 2 × 3 × 5 × 17; 92 = 2² × 23. HCF = 2; LCM = 2² × 3 × 5 × 17 × 23 = 23460. Check: 2 × 23460 = 46920 = 510 × 92. ✓

(iii) 336 and 54 336 = 2⁴ × 3 × 7; 54 = 2 × 3³. HCF = 2 × 3 = 6; LCM = 2⁴ × 3³ × 7 = 3024. Check: 6 × 3024 = 18144 = 336 × 54. ✓

Q3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21 12 = 2² × 3; 15 = 3 × 5; 21 = 3 × 7. HCF = 3; LCM = 2² × 3 × 5 × 7 = 420.

(ii) 17, 23 and 29 All are prime. HCF = 1; LCM = 17 × 23 × 29 = 11339.

(iii) 8, 9 and 25 8 = 2³; 9 = 3²; 25 = 5². HCF = 1; LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800.

Q4. Given that HCF(306, 657) = 9, find LCM(306, 657).

HCF × LCM = product of the numbers. LCM = (306 × 657) ÷ 9 = 201042 ÷ 9 = 22338.

Q5. Check whether 6ⁿ can end with the digit 0 for any natural number n.

6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ. For a number to end in 0 it must have 5 as a factor. Since 6ⁿ has only the primes 2 and 3 (no 5), 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.

First: 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × 78. It has a factor 13 (other than 1 and itself), so it is composite.

Second: 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × 1009. It has a factor 5, so it is composite.

Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round, while Ravi takes 12 minutes. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

They meet again at the starting point after a time equal to the LCM of 18 and 12. 18 = 2 × 3²; 12 = 2² × 3 → LCM = 2² × 3² = 36 minutes.

Exercise 1.2

Q1. Prove that √5 is irrational.

Proof (by contradiction). Assume √5 is rational, so √5 = p/q where p and q are integers, q ≠ 0, and p, q are coprime (no common factor other than 1).

Then √5 = p/q ⟹ 5 = p²/q² ⟹ 5q² = p². So 5 divides p², which means 5 divides p. Write p = 5m. Then 5q² = (5m)² = 25m² ⟹ q² = 5m², so 5 divides q², hence 5 divides q.

Now 5 divides both p and q — contradicting that they are coprime. Hence our assumption is wrong, and √5 is irrational.

Q2. Prove that 3 + 2√5 is irrational.

Proof (by contradiction). Assume 3 + 2√5 is rational, equal to a. Then 2√5 = a − 3 ⟹ √5 = (a − 3)/2. Since a is rational, (a − 3)/2 is rational, which would make √5 rational. But √5 is irrational (proved above). This is a contradiction. Hence 3 + 2√5 is irrational.

Q3. Prove that the following are irrational.

(i) 1/√2 Assume 1/√2 is rational = a. Then √2 = 1/a, which is rational. But √2 is irrational — contradiction. Hence 1/√2 is irrational.

(ii) 7√5 Assume 7√5 is rational = a. Then √5 = a/7, which is rational. But √5 is irrational — contradiction. Hence 7√5 is irrational.

(iii) 6 + √2 Assume 6 + √2 is rational = a. Then √2 = a − 6, which is rational. But √2 is irrational — contradiction. Hence 6 + √2 is irrational.

Common Mistakes

• Taking HCF as the product of the greatest powers (that is the LCM) — HCF uses the smallest power of common primes only.
• Forgetting that HCF × LCM = product works only for two numbers, not three.
• In irrationality proofs, not stating that p and q are coprime — that assumption is what produces the contradiction.

Exam Tips

• Prime-factorise neatly; show the factor tree or division steps for full marks.
• Irrationality proofs are 3-mark questions — write the contradiction clearly: assume rational → derive a common factor → contradiction.
• LCM/HCF word problems (like Q7) almost always use the LCM for "meet again / ring together" and the HCF for "largest size / maximum number of groups."