--- title: "NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1" url: https://www.swavid.com/maths/class/10/chapter/real-numbers/ncert-solutions/exercise-1-1 dateModified: 2026-10-07T14:41:11+00:00 --- # NCERT Solutions for Class 10 Maths Chapter 1 Exercise 1.1 Chapter 1: Real Numbers. Every question from Exercise 1.1, with full working and the final answer. Free PDF (7 pages): https://www.swavid.com/api/seo/pdf/ncert/maths/class-10/swavid-ncert-solutions-class-10-maths-chapter-1-real-numbers-a484766ea2.pdf ## EXERCISE 1.1 ### Question 1 *3 marks · Short answer* Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 **Part (i)** 1. Divide 140 by the smallest prime factor 2 to get 70. 2. Divide 70 by 2 to get 35, divide 35 by 5 to get 7, and 7 by 7 to get 1. 3. Express 140 as $2 \times 2 \times 5 \times 7 = 2^2 \times 5 \times 7$. Answer (i): $2^2 \times 5 \times 7$ **Part (ii)** 1. Divide 156 by the smallest prime factor 2 to get 78. 2. Divide 78 by 2 to get 39, divide 39 by 3 to get 13, and 13 by 13 to get 1. 3. Express 156 as $2 \times 2 \times 3 \times 13 = 2^2 \times 3 \times 13$. Answer (ii): $2^2 \times 3 \times 13$ **Part (iii)** 1. Divide 3825 by 3 to get 1275, and again by 3 to get 425. 2. Divide 425 by 5 to get 85, divide 85 by 5 to get 17, and 17 by 17 to get 1. 3. Express 3825 as $3 \times 3 \times 5 \times 5 \times 17 = 3^2 \times 5^2 \times 17$. Answer (iii): $3^2 \times 5^2 \times 17$ **Part (iv)** 1. Divide 5005 by the prime factor 5 to get 1001. 2. Divide 1001 by 7 to get 143, divide 143 by 11 to get 13, and 13 by 13 to get 1. 3. Express 5005 as $5 \times 7 \times 11 \times 13$. Answer (iv): $5 \times 7 \times 11 \times 13$ **Part (v)** 1. Divide 7429 by 17 to get 437. 2. Divide 437 by 19 to get 23, and 23 by 23 to get 1. 3. Express 7429 as $17 \times 19 \times 23$. Answer (v): $17 \times 19 \times 23$ **Answer:** (i) $2^2 \times 5 \times 7$, (ii) $2^2 \times 3 \times 13$, (iii) $3^2 \times 5^2 \times 17$, (iv) $5 \times 7 \times 11 \times 13$, (v) $17 \times 19 \times 23$ > Common mistake: Stopping the prime factorisation before all factors are prime numbers, or writing composite numbers instead of primes. ### Question 2 *3 marks · Short answer* Find the LCM and HCF of the following pairs of integers and verify that $\text{LCM} \times \text{HCF} = \text{product of the two numbers.}$ (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54 **Part (i)** 1. $26 = 2 \times 13$ and $91 = 7 \times 13$ 2. $\text{HCF}(26, 91) = 13$ and $\text{LCM}(26, 91) = 2 \times 7 \times 13 = 182$ 3. $\text{HCF} \times \text{LCM} = 13 \times 182 = 2366$ and product $= 26 \times 91 = 2366$, hence verified. Answer (i): HCF = 13, LCM = 182, Product = 2366 **Part (ii)** 1. $510 = 2 \times 3 \times 5 \times 17$ and $92 = 2^2 \times 23$ 2. $\text{HCF}(510, 92) = 2$ and $\text{LCM}(510, 92) = 2^2 \times 3 \times 5 \times 17 \times 23 = 23460$ 3. $\text{HCF} \times \text{LCM} = 2 \times 23460 = 46920$ and product $= 510 \times 92 = 46920$, hence verified. Answer (ii): HCF = 2, LCM = 23460, Product = 46920 **Part (iii)** 1. $336 = 2^4 \times 3 \times 7$ and $54 = 2 \times 3^3$ 2. $\text{HCF}(336, 54) = 2 \times 3 = 6$ and $\text{LCM}(336, 54) = 2^4 \times 3^3 \times 7 = 3024$ 3. $\text{HCF} \times \text{LCM} = 6 \times 3024 = 18144$ and product $= 336 \times 54 = 18144$, hence verified. Answer (iii): HCF = 6, LCM = 3024, Product = 18144 **Answer:** LCM and HCF found and verified for all pairs. > Common mistake: Confusing the powers of common and non-common prime factors when calculating LCM and HCF. ### Question 3 *3 marks · Short answer* 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 **Part (i)** 1. $12 = 2^2 \times 3$, $15 = 3 \times 5$, $21 = 3 \times 7$ 2. $\text{HCF}(12, 15, 21) = 3$ 3. $\text{LCM}(12, 15, 21) = 2^2 \times 3 \times 5 \times 7 = 420$ Answer (i): HCF = 3, LCM = 420 **Part (ii)** 1. $17 = 17$, $23 = 23$, $29 = 29$ 2. $\text{HCF}(17, 23, 29) = 1$ 3. $\text{LCM}(17, 23, 29) = 17 \times 23 \times 29 = 11339$ Answer (ii): HCF = 1, LCM = 11339 **Part (iii)** 1. $8 = 2^3$, $9 = 3^2$, $25 = 5^2$ 2. $\text{HCF}(8, 9, 25) = 1$ 3. $\text{LCM}(8, 9, 25) = 2^3 \times 3^2 \times 5^2 = 1800$ Answer (iii): HCF = 1, LCM = 1800 **Answer:** Found HCF and LCM for the given sets of integers. > Common mistake: Taking common factors when none exist, resulting in an HCF other than 1. ### Question 4 *3 marks · Short answer* Given that $\text{HCF} (306, 657) = 9$, find $\text{LCM} (306, 657)$. **Solution** 1. We know that for two positive integers $a$ and $b$, $\text{HCF}(a, b) \times \text{LCM}(a, b) = a \times b$. 2. Substituting the given values, $9 \times \text{LCM}(306, 657) = 306 \times 657$. 3. $\text{LCM}(306, 657) = \frac{306 \times 657}{9}$. 4. $\text{LCM}(306, 657) = 34 \times 657 = 22338$. **Answer:** 22338 > Common mistake: Multiplying the numbers first and making calculation errors while dividing by HCF. ### Question 5 *3 marks · Short answer* Check whether $6^n$ can end with the digit 0 for any natural number $n$. **Solution** 1. If the number $6^n$, for any $n$, were to end with the digit zero, then it would be divisible by 5. 2. That is, the prime factorisation of $6^n$ would contain the prime 5. 3. However, $6^n = (2 \times 3)^n = 2^n \times 3^n$; so the only primes in the factorisation of $6^n$ are 2 and 3. 4. The uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorisation of $6^n$. 5. Therefore, there is no natural number $n$ for which $6^n$ ends with the digit zero. **Answer:** No, $6^n$ cannot end with the digit 0 for any natural number $n$. > Common mistake: Stating that $6^n$ can end with zero because 6 is a multiple of 2 and 3 without mentioning the necessity of factor 5. ### Question 6 *3 marks · Short answer* Explain why $7 \times 11 \times 13 + 13$ and $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$ are composite numbers. **Solution** 1. Given numbers can be simplified by taking out common factors as per the distributive property. 2. For the first number: $7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 \times 1 + 1) = 13 \times (77 + 1) = 13 \times 78$. 3. Since $13 \times 78$ has factors other than 1 and itself, it is a composite number. 4. For the second number: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5 = 5 \times (7 \times 6 \times 4 \times 3 \times 2 \times 1 + 1) = 5 \times (1008 + 1) = 5 \times 1009$. 5. Since $5 \times 1009$ has factors other than 1 and itself, it is a composite number. **Answer:** Both numbers are composite because they can be expressed as a product of factors other than 1 and themselves. > Common mistake: Multiplying out the entire expression fully instead of factoring out the common term. ### Question 7 *3 marks · Short answer* 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** 1. Time taken by Sonia to drive one round = 18 minutes 2. Time taken by Ravi to drive one round = 12 minutes 3. The time when they meet again at the starting point is given by the LCM of 18 and 12 4. Find the prime factorisation of 18 and 12: $18 = 2 \times 3^2$ and $12 = 2^2 \times 3$ 5. Calculate the LCM of 18 and 12: $\text{LCM}(18, 12) = 2^2 \times 3^2 = 4 \times 9 = 36$ 6. Sonia and Ravi will meet again at the starting point after 36 minutes **Answer:** 36 minutes > Common mistake: Finding the HCF instead of the LCM of the two numbers. ## Related pages - [All Chapter 1 NCERT solutions](https://www.swavid.com/maths/class/10/chapter/real-numbers/ncert-solutions) - [Exercise 1.2](https://www.swavid.com/maths/class/10/chapter/real-numbers/ncert-solutions/exercise-1-2) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student