NCERT Solutions For Class 10 Maths Chapter 1 Real Number, Download PDF

NCERT Solutions for Class 10 Maths Chapter 1, Real Numbers:   NCERT solutions play a crucial role in helping students clear concepts and build a strong foundation. By mastering these solutions, students can gain confidence in their subject knowledge and perform better in their exams. With this article, students will be able to access the NCERT solutions for Chapter 1 of Class 10 Mathematics. NCERT Solutions for Real Numbers have been prepared as per the CBSE marking scheme and special care has been taken to keep them simple yet inclusive. 

EXERCISE 1.1

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

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

(i) 140

Divide by 2: 140÷2=70

Divide by 2 again: 70÷2=35

Divide by 5: 35÷5=7

7 is a prime number.

So, the prime factorization of 140 is: 140=2²×5×7

(ii) 156

Divide by 2: 156÷2=78

Divide by 2 again: 78÷2=39 

Divide by 3: 39÷3=13

13 is a prime number.

So, the prime factorization of 156 is: 156=2²×3×13

(iii) 3825

Divide by 3: 3825÷3=1275

Divide by 3 again: 1275÷3=425

Divide by 5: 425÷5=85

Divide by 5 again: 85÷5=17

17 is a prime number.

So, the prime factorization of 3825 is: 3825=3²×5²×17

(iv) 5005

Divide by 5: 5005÷5=1001

Divide by 7: 1001÷7=143

Divide by 11: 143÷11=13

13 is a prime number.

So, the prime factorization of 5005 is: 5005=5×7×11×13

(v) 7429

Check for divisibility by primes up to √7429 (approximately 86).

It is divisible by 29: 7429÷29=257

257 is a prime number.

So, the prime factorization of 7429 is: 7429=29×257

2. 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

(i) 26 and 91

Prime Factorization:

• 26=2×1326 = 2
• 91=7×1391 = 7

HCF:

• Common prime factor: 13

LCM:

Highest powers of all prime factors: 2¹, 7¹, 13¹

So, LCM=2¹×7¹×13¹

=2×7×13=182

Verification:

LCM×HCF=182×13=2366

Product of the two numbers=26×91=2366

(ii) 510 and 92

Prime Factorization:

• 510=2×3×5×17
• 92=22×23

HCF:

• Common prime factor: 212^121

LCM:

Highest powers of all prime factors: 2²,3¹,5¹,17¹,23¹

So,

LCM=2²×3¹×5¹×17¹×23¹=4×3×5×17×23

Calculating step-by-step:

• 4×3=124
• 12×5=6012 
• 60×17=102060 
• 1020×23=23460

Verification:

LCM×HCF=23460×2=46920 

Product of the two numbers=510×92=46920

(iii) 336 and 54

Prime Factorization:

• 336=24×3×7
• 54=2×33

HCF:

• Common prime factors: 2¹ and 3¹, 

 HCF=21×31=6

LCM:

• Highest powers of all prime factors: 2⁴,3³,7¹

Calculating step-by-step:

• 2⁴=16
• 3³=27
• 7¹=7

LCM=16×27×7

Calculating 16×27=432 

Now, 432×7=3024

So, LCM=3024

Verification:

LCM×HCF=3024×6=18144

Product of the two numbers=336×54=18144

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

method.

(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

(i) 12, 15, and 21

Prime factorization of 12: 2 x 2 x 3

Prime factorization of 15: 3 x 5

Prime factorization of 21: 3 x 7

HCF: The common prime factors are 3. So, HCF(12, 15, 21) = 3.

LCM: Take the highest powers of each prime factor that appears in any of the numbers. LCM(12, 15, 21) = 2 x 2 x 3 x 5 x 7 = 420.

(ii) 17, 23, and 29

These numbers are all prime numbers, so they have no common factors other than 1.

HCF: HCF(17, 23, 29) = 1.

LCM: Since they are all prime, their LCM is simply their product. LCM(17, 23, 29) = 17 x 23 x 29 = 11,339.

(iii) 8, 9, and 25

Prime factorization of 8: 2 x 2 x 2

Prime factorization of 9: 3 x 3

Prime factorization of 25: 5 x 5

HCF: There are no common prime factors. So, HCF(8, 9, 25) = 1.

LCM: Take the highest powers of each prime factor that appears in any of the numbers. LCM(8, 9, 25) = 2 x 2 x 2 x 3 x 3 x 5 x 5 = 1800.

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

There is a fundamental relationship between the HCF (Highest Common Factor) and LCM (Least Common Multiple) of two numbers:

Product of HCF and LCM = Product of the two numbers

• LCM(306, 657)

Using the formula:

• 9 x LCM(306, 657) = 306 x 657

Now, solve for LCM(306, 657):

• LCM(306, 657) = (306 x 657) / 9

Calculating the LCM:

• LCM(306, 657) = 200982 / 9
• LCM(306, 657) = 22338

Therefore, the LCM of 306 and 657 is 22338.

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

A number ends with 0 if it is divisible by 10. For a number to be divisible by 10, it must be divisible by both 2 and 5.

• Divisibility by 2: Since 6 is divisible by 2, any multiple of 6 (6n) will also be divisible by 2.
• Divisibility by 5: However, 6 is not divisible by 5. Therefore, no multiple of 6 (6n) can be divisible by 5.

Conclusion:

Because 6n cannot be divisible by both 2 and 5, it cannot be divisible by 10. As a result, 6n cannot end with the digit 0 for any natural number n.

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

Here's the breakdown of why both numbers are composite:

7 × 11 × 13 + 13

Factorization:

• The expression can be factored as 13 x (7 x 11 + 1).

Composite Nature:

• The expression is a product of two factors: 13 and (7 x 11 + 1). Since it has factors other than 1 and itself, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Factorization:

• The expression can be factored as 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1).

Composite Nature:

• The expression is a product of two factors: 5 and (7 x 6 x 4 x 3 x 2 x 1 + 1). Since it has factors other than 1 and itself, it is a composite number.

In summary, both expressions are composite because they can be factored into two smaller factors other than 1 and themselves.

7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. 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?

To find when Sonia and Ravi will meet again at the starting point, we need to find the least common multiple (LCM) of their round times.

• Sonia takes 18 minutes per round.
• Ravi takes 12 minutes per round.

LCM of 18 and 12:

• Prime factorization of 18: 2 x 3 x 3
  
  
• Prime factorization of 12: 2 x 2 x 3
  
  
• LCM(18, 12) = 2 x 2 x 3 x 3 = 36
  

Therefore, Sonia and Ravi will meet again at the starting point after 36 minutes.

EXERCISE 1.2

1. Prove that √5 is irrational.

Let a = 5k

(5k)² = 5b²

b² = 5b²

Thus,  a and b have 5 as a common factor.

Hence, √5 is an irrational no.

2. Prove that 3+2√5  is irrational.

Let 3+2√5  is a rational number.

Thus,

3+2√5 = a/b

√5=1/2 = (a/b-3)

1/2 (a/b-3) is rational

Hence, 3+2√5  is irrational no.

3. Prove that the following are irrationals :

(i) 1/ √2

Let 1/ √2 irrational number

Thus, 

1/ √2 = a/b

√2 = b/a

b/a is rational 

Hence, 1/ √2 is a irrational number

(ii) 7√5 

Let 7√5 irrational number

Thus, 

7√5 = a/b

√5=b/a

a/7b is rational 

Hence, 7√5 irrational no. 

(iii) 6 + √2

Let 6 + √2 irrational number

Thus,

6 + √2 = a/b 

√2 = a/b-6

(a/b-6) is a rational 

Hence, 6 + √2 irrational number

For all the questions given in CBSE Class 10 Mathematics NCERT book, you will find detailed and accurate answers at Jagran Josh. These solutions will help you find the right approach to solve different questions accurately and secure high marks in exams.

Check NCERT Solutions for Class 10 Maths chapter 1 Real Numbers (PDF)

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