Math chapter 1 real number

Class 10 Maths Chapter 1 (Real Numbers), Exercise 1.1 – Full Solution in English (all together)

📘 Exercise 1.1 – Full Solution

✏️ Q1. Use Euclid’s Division Algorithm to find the HCF:

(i) 135 and 225

225 = 135 × 1 + 90

135 = 90 × 1 + 45

90 = 45 × 2 + 0

👉 HCF = 45

(ii) 196 and 38220

38220 = 196 × 195 + 0

👉 HCF = 196

(iii) 867 and 255

867 = 255 × 3 + 102

255 = 102 × 2 + 51

102 = 51 × 2 + 0

👉 HCF = 51

✏️ Q2. Prove that any positive odd integer is of the form 6q + 1, 6q + 3, or 6q + 5

Let any integer be written as:

👉 6q + r, where r = 0, 1, 2, 3, 4, 5

Odd integers occur when r = 1, 3, 5

👉 So, any odd integer can be written as:

✔️ 6q + 1

✔️ 6q + 3

✔️ 6q + 5

👉 Hence proved

✏️ Q3. Prove that the square of any positive integer is of the form 3m or 3m + 1

Let any integer be:

👉 3q, 3q + 1, or 3q + 2

Now squaring:

(3q)² = 9q² = 3m

(3q + 1)² = 9q² + 6q + 1 = 3m + 1

(3q + 2)² = 9q² + 12q + 4 = 3m + 1

👉 In all cases, square is of form 3m or 3m + 1

👉 Hence proved

✏️ Q4. Use Euclid’s Algorithm to find HCF:

(i) 306 and 657

657 = 306 × 2 + 45

306 = 45 × 6 + 36

45 = 36 × 1 + 9

36 = 9 × 4 + 0

👉 HCF = 9

(ii) 510 and 92

510 = 92 × 5 + 50

92 = 50 × 1 + 42

50 = 42 × 1 + 8

42 = 8 × 5 + 2

8 = 2 × 4 + 0

👉 HCF = 2

✅ Final Summary

Euclid’s Division Algorithm → used to find HCF

Odd integers → 6q + 1, 6q + 3, 6q + 5

Squares → always of form 3m or 3m + 1

Great 👍 Now here is Class 10 Maths Chapter 1 (Real Numbers), Exercise 1.2 – Full Solution in English

📘 Exercise 1.2 – Full Solution

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

(i) 140

140 = 2 × 2 × 5 × 7

👉 Prime factorization = 2² × 5 × 7

(ii) 156

156 = 2 × 2 × 3 × 13

👉 = 2² × 3 × 13

(iii) 3825

3825 = 3 × 3 × 5 × 5 × 17

👉 = 3² × 5² × 17

(iv) 5005

5005 = 5 × 7 × 11 × 13

👉 Prime factors = 5 × 7 × 11 × 13

(v) 7429

7429 = 17 × 19 × 23

👉 Prime factors = 17 × 19 × 23

✏️ Q2. Find LCM and HCF using prime factorization

(i) 26 and 91

26 = 2 × 13

91 = 7 × 13

👉 HCF = 13

👉 LCM = 2 × 7 × 13 = 182

(ii) 510 and 92

510 = 2 × 3 × 5 × 17

92 = 2 × 2 × 23

👉 HCF = 2

👉 LCM = 2² × 3 × 5 × 17 × 23 = 23460

(iii) 336 and 54

336 = 2⁴ × 3 × 7

54 = 2 × 3³

👉 HCF = 2 × 3 = 6

👉 LCM = 2⁴ × 3³ × 7 = 3024

✏️ Q3. Find the smallest number divisible by 6, 15 and 20

👉 LCM of 6, 15, 20

6 = 2 × 3

15 = 3 × 5

20 = 2² × 5

👉 LCM = 2² × 3 × 5 = 60

✔️ Answer = 60

✏️ Q4. Find the LCM and HCF of 12, 15 and 21

12 = 2² × 3

15 = 3 × 5

21 = 3 × 7

👉 HCF = 3

👉 LCM = 2² × 3 × 5 × 7 = 420

✅ Final Summary

Prime factorization → numbers into primes

HCF → common smallest powers

LCM → highest powers of primes

✏️ 1. Prove that √3 is irrational

Assume √3 = p/q (in lowest form)

Squaring:

3 = p² / q²

⇒ p² = 3q²

👉 p divisible by 3 ⇒ p = 3k

Substitute:

(3k)² = 3q²

9k² = 3q²

⇒ q² = 3k²

👉 q also divisible by 3

❌ Contradiction

✔️ Hence, √3 is irrational 

Some important question 

1- Prove that √7 + √3 is irrational .

2-Prove that (√5 − 2) is irrational

3- Prove that √11 is irrational

4- Prove that 2√3 is irrational

5- Prove that 5 + √7 is irrational

6- Prove that 2√3 is irrational

7- find composite number

Find all composite numbers less than 15