Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 NCERT Solutions PDF Link Here

Class 10 Maths Chapter 1 Real Numbers Exercise 1.1: The NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 help students understand how the Fundamental Theorem of Arithmetic is used in prime factorisation. Through clear explanations, these real numbers class 10 exercise 1.1 solutions teach students how to express composite numbers as a product of prime numbers and apply the concept to solve different questions.

Each answer is explained step by step, making the concepts easy to follow and strengthening problem-solving skills. These class 10 maths real numbers exercise 1.1 solutions build a strong foundation for advanced topics and support effective exam preparation.

Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 Questions and Solutions

The NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 provide simple and clear explanations to help students understand the Fundamental Theorem of Arithmetic.

This exercise mainly covers prime factorisation, along with concepts related to HCF and LCM, enabling students to express composite numbers in their unique prime factor forms.

These real numbers class 10 exercise 1.1 solutions strengthen conceptual clarity and improve problem-solving skills. Students preparing for exams will find these Class 10 Maths Real Numbers Exercise 1.1 resources extremely helpful for quick revision and accuracy.

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

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Solutions:

(i) 140 By taking the LCM of 140, we will get the product of its prime factors. Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 2 2 ×5×7 

(ii) 156 By taking the LCM of 156, we will get the product of its prime factors. Hence, 156 = 2 × 2 × 13 × 3 × 1 = 2 2 × 13 × 3 

(iii) 3825 By taking the LCM of 3825, we will get the product of its prime factors. Hence, 3825 = 3 × 3 × 5 × 5 × 17 × 1 = 3 2 ×5 2 ×17 

(iv) 5005 By taking the LCM of 5005, we will get the product of its prime factors. Hence, 5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13 

(v) 7429. By taking the LCM of 7429, we will get the product of its prime factors. Hence, 7429 = 17 × 19 × 23 × 1 = 17 × 19 × 23

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

Solutions:

(i) 26 and 91

Expressing 26 and 91 as the product of its prime factors, we get 26 = 2 × 13 × 1 91 = 7 × 13 × 1. Therefore, LCM (26, 91) = 2 × 7 × 13 × 1 = 182 And HCF (26, 91) = 13 Verification 

 Now, product of 26 and 91 = 26 × 91 = 2366 And product of LCM and HCF = 182 × 13 = 2366 Hence, LCM × HCF = product of the 26 and 91

(ii) 510 and 92

Expressing 510 and 92 as the product of its prime factors, we get 510 = 2 × 3 × 17 × 5 × 1 92 = 2 × 2 × 23 × 1 Therefore, LCM (510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460 And HCF (510, 92) = 2 Verification 

Now, product of 510 and 92 = 510 × 92 = 46920 And product of LCM and HCF = 23460 × 2 = 46920 Hence, LCM × HCF = product of the 510 and 92

(iii) 336 and 54

Expressing 336 and 54 as the product of its prime factors, we get 336 = 2 × 2 × 2 × 2 × 7 × 3 × 1 54 = 2 × 3 × 3 × 3 × 1 Therefore, LCM (336, 54) = = 3024 And HCF (336, 54) = 2×3 = 6 Verification 

Now, product of 336 and 54 = 336 × 54 = 18,144 And product of LCM and HCF = 3024 × 6 = 18,144 Hence, LCM × HCF = product of the 336 and 54

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

Solutions: 

(i) 12, 15 and 21 Writing the product of prime factors for all the three numbers, we get 12=2×2×3 15=5×3 21=7×3 Therefore, HCF(12,15,21) = 3 LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23 and 29 Writing the product of prime factors for all the three numbers, we get 17=17×1 23=23×1 29=29×1 Therefore, HCF(17,23,29) = 1 LCM(17,23,29) = 17 × 23 × 29 = 11339 

(iii) 8, 9 and 25 Writing the product of prime factors for all the three numbers, we get 8=2×2×2×1 9=3×3×1 25=5×5×1 Therefore, HCF(8,9,25)=1 LCM(8,9,25) = 2×2×2×3×3×5×5 = 1800

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

Solution:

As we know, HCF×LCM=Product of the two given numbers Therefore, 9 × LCM = 306 × 657 LCM = (306×657)/9 = 22338 Hence, LCM(306,657) = 22338

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

Solution:

If the number 6 n ends with the digit zero (0), then it should be divisible by 5, as we know any number with the unit place as 0 or 5 is divisible by 5.

Prime factorisation of 6 n = (2×3) n. Therefore, the prime factorisation of 6 n doesn’t contain the prime number 5. Hence, it is clear that for any natural number n, 6 n is not divisible by 5, and thus it proves that 6 n 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.

Solution:

By the definition of a composite number, we know if a number is composite, then it means it has factors other than 1 and itself. Therefore, for the given expression 7 × 11 × 13 + 13

Taking 13 as a common factor, we get =13(7×11×1+1) = 13(77+1) = 13×78 = 13×3×2×13 Hence, 7 × 11 × 13 + 13 is a composite number. Now let’s take the other number, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Taking 5 as a common factor, we get =5(7×6×4×3×2×1+1) = 5(1008+1) = 5×1009 Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

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?

Solution: 

Since Sonia and Ravi move in the same direction and at the same time, the method to find the time when they will be meeting again at the starting point is LCM of 18 and 12.

Therefore, LCM(18,12) = 2×3×3×2×1=36. Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.

Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 Download PDF 

Students can access the Real Numbers Class 10 Exercise 1.1 PDF to revise concepts like prime factorisation, HCF, and LCM in an easy, step-by-step format. This PDF is helpful for quick revision, homework support, and strengthening basics for board exams.

NCERT Solutions for Class 10 Maths PDF