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NCERT Solutions 10th Maths Chapter 1 Real Number Exercise 1.2
NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.2 we will restart our exploration of the world of real numbers. We will study Euclid’s division algorithm and the Fundamental Theorem of Arithmetic. Also, we will see decimal representation of real numbers.
Exercise 1.2 1. Express each number as product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429 Answer (i) 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7 (ii) 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13 (iii) 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17 (iv) 5005 = 5 × 7 × 11 × 13 (v) 7429 = 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 Answer (i) 26 = 2 × 13 91 =7 × 13 HCF = 13 LCM =2 × 7 × 13 =182 Product of two numbers 26 × 91 = 2366 Product of HCF and LCM 13 × 182 = 2366 Hence, product of two numbers = product of HCF × LCM (ii) 510 = 2 × 3 × 5 × 17 92 =2 × 2 × 23 HCF = 2 LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460 Product of two numbers 510 × 92 = 46920 Product of HCF and LCM 2 × 23460 = 46920 Hence, product of two numbers = product of HCF × LCM (iii) 336 = 2 × 2 × 2 × 2 × 3 × 7 54 = 2 × 3 × 3 × 3 HCF = 2 × 3 = 6 LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 =3024 Product of two numbers 336 × 54 =18144 Product of HCF and LCM 6 × 3024 = 18144 Hence, product of two numbers = product of HCF × LCM. 3. Find the LCM and HCF of the following integers by applying the prime factorization method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25 Answer (i) 12 = 2 × 2 × 3 15 =3 × 5 21 =3 × 7 HCF = 3 LCM = 2 × 2 × 3 × 5 × 7 = 420 (ii) 17 = 1 × 17 23 = 1 × 23 29 = 1 × 29 HCF = 1 LCM = 1 × 17 × 19 × 23 = 11339 (iii) 8 =1 × 2 × 2 × 2 9 =1 × 3 × 3 25 =1 × 5 × 5 HCF =1 LCM = 1 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800 4. Given that HCF (306, 657) = 9, find LCM (306, 657). Answer We have the formula that Product of LCM and HCF = product of number LCM × 9 = 306 × 657 Divide both side by 9 we get LCM = (306 × 657) / 9 = 22338 5. Check whether 6n can end with the digit 0 for any natural number n. Answer If any digit has last digit 10 that means it is divisible by 10 and the factors of 10 = 2 × 5. So value 6n should be divisible by 2 and 5 both 6n is divisible by 2 but not divisible by 5 So it can not end with 0. 6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. Answer 7 × 11 × 13 + 13 Taking 13 common, we get 13 (7 x 11 +1 ) 13(77 + 1 ) 13 (78) It is product of two numbers and both numbers are more than 1 so it is a composite number. 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 Taking 5 common, we get 5(7 × 6 × 4 × 3 × 2 × 1 +1) 5(1008 + 1) 5(1009) It is product of two numbers and both numbers are more than 1 so it 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? Answer They will be meet again after LCM of both values at the starting point. 18 = 2 × 3 × 3 12 = 2 × 2 × 3 LCM = 2 × 2 × 3 × 3 = 36 Therefore, they will meet together at the starting point after 36 minutes.Post your comments
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