NCERT Class 10 Maths Chapter 1 Real Numbers Solutions & Formulae

Class 10 Mathematics Chapter 1: Real Numbers - Complete Solutions

Welcome students! Below are the complete step-by-step solutions for all exercises in Class 10 NCERT Chapter 1: Real Numbers.

EXERCISE 1.1

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

(i) 140
140 ÷ 2 = 70
70 ÷ 2 = 35
35 ÷ 5 = 7
7 ÷ 7 = 1
Prime factorisation of 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7

(ii) 156
156 ÷ 2 = 78
78 ÷ 2 = 39
39 ÷ 3 = 13
13 ÷ 13 = 1
Prime factorisation of 156 = 2² × 3 × 13

(iii) 3825
3825 ÷ 3 = 1275
1275 ÷ 3 = 425
425 ÷ 5 = 85
85 ÷ 5 = 17
17 ÷ 17 = 1
Prime factorisation of 3825 = 3² × 5² × 17

(iv) 5005
5005 ÷ 5 = 1001
1001 ÷ 7 = 143
143 ÷ 11 = 13
13 ÷ 13 = 1
Prime factorisation of 5005 = 5 × 7 × 11 × 13

(v) 7429
7429 ÷ 17 = 437
437 ÷ 19 = 23
23 ÷ 23 = 1
Prime factorisation of 7429 = 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
Verification: HCF × LCM = 13 × 182 = 2366
Product of numbers = 26 × 91 = 2366
Hence, HCF × LCM = Product of two numbers. Verified.

(ii) 510 and 92
510 = 2 × 3 × 5 × 17
92 = 2² × 23
HCF = 2
LCM = 2² × 3 × 5 × 17 × 23 = 23460
Verification: HCF × LCM = 2 × 23460 = 46920
Product of numbers = 510 × 92 = 46920
Hence, HCF × LCM = Product of two numbers. Verified.

(iii) 336 and 54
336 = 2⁴ × 3 × 7
54 = 2 × 3³
HCF = 2 × 3 = 6
LCM = 2⁴ × 3³ × 7 = 3024
Verification: HCF × LCM = 6 × 3024 = 18144
Product of numbers = 336 × 54 = 18144
Hence, HCF × LCM = Product of two numbers. Verified.

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
17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 17 × 23 × 29 = 11339

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

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

We know that: HCF × LCM = Product of two numbers
9 × LCM(306, 657) = 306 × 657
LCM(306, 657) = (306 × 657) / 9 = 34 × 657 = 22338

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

If a number ends with 0, its prime factorisation must contain both 2 and 5.
Prime factorisation of 6ⁿ = (2 × 3)ⁿ = 2ⁿ × 3ⁿ.
Since 5 is not in the prime factorisation of 6ⁿ, by the uniqueness of the Fundamental Theorem of Arithmetic, 6ⁿ cannot end with the digit 0 for any natural number n.

Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

First Expression:
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1) = 13 × 78.
Since it has factors other than 1 and itself, it is a composite number.

Second Expression:
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 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.

Q7. Sonia takes 18 minutes and Ravi takes 12 minutes to drive one round. After how many minutes will they meet again at the starting point?

To find the time when they meet again, we need to find LCM(18, 12).
18 = 2 × 3²
12 = 2² × 3
LCM(18, 12) = 2² × 3² = 4 × 9 = 36.
They will meet again at the starting point after 36 minutes.

EXERCISE 1.2

Q1. Prove that √5 is irrational.

Let us assume to the contrary that √5 is rational.
So, √5 = a/b where a and b are coprime integers and b ≠ 0.
Squaring both sides: 5 = a²/b² ⇒ a² = 5b² --- (Equation 1)
This means 5 divides a², so 5 divides a.
Let a = 5c for some integer c.
Substituting a = 5c in Equation 1:
(5c)² = 5b² ⇒ 25c² = 5b² ⇒ b² = 5c²
This means 5 divides b², so 5 divides b.
Thus, 5 is a common factor of both a and b, which contradicts our assumption that a and b are coprime.
Therefore, √5 is irrational.

Q2. Prove that 3 + 2√5 is irrational.

Let us assume that 3 + 2√5 is rational.
Then, 3 + 2√5 = a/b (where a and b are coprime integers, b ≠ 0).
Rearranging: 2√5 = (a/b) - 3 ⇒ 2√5 = (a - 3b)/b
√5 = (a - 3b) / (2b)
Since a and b are integers, (a - 3b)/(2b) is rational, which implies √5 is rational.
This contradicts the fact that √5 is irrational.
Hence, 3 + 2√5 is irrational.

Q3. Prove that the following are irrationals:

(i) 1/√2
Assume 1/√2 is rational = a/b.
√2 = b/a.
Since a and b are integers, b/a is rational, implying √2 is rational. This is a contradiction.
Hence, 1/√2 is irrational.

(ii) 7√5
Assume 7√5 is rational = a/b.
√5 = a / (7b).
Since a, b, and 7 are integers, a/(7b) is rational, implying √5 is rational. This is a contradiction.
Hence, 7√5 is irrational.

(iii) 6 + √2
Assume 6 + √2 is rational = a/b.
√2 = (a/b) - 6 = (a - 6b) / b.
Since a and b are integers, (a - 6b)/b is rational, implying √2 is rational. This is a contradiction.
Hence, 6 + √2 is irrational.