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Page 1 of 5, Mathematics, (Chapter-1) (Real Numbers), (Class X), Exercise 1.2, Question 1:, Express each number as product of its prime factors:, (i) 140, (ii) 156, (iii) 3825, (iv) 5005, (v) 7429, Answer 1:, (i), 140 = 2x 2x5x7=22 x5x7, (ii), 156 = 2x2x3x13 = 2' x3x13, (ii), 3825 = 3x3x 5x5x17 = 3 x5 ×17, (iv) 5005 = 5x7x11x13, (v), 7429 17x19×23, Question 2:, Find the LCM and HCF of the following pairs of integers and verify that, LCM x HCF, product of the two numbers., %3D, (i) 26 and 91, (ii) 510 and 92, (iii) 336 and 54, Answer 2:, (i), 26 and 91, 26 = 2x13, 91 = 7x13, HCF = 13, LCM = 2x7x13 = 182, Product of the two numbers = 26x91 = 2366, HCFX LCM =13x182 2366, Hence, product of two numbers HCF x LCM, (ii), 510 and 92, 510 = 2x3x5x17, 92 = 2x 2x 23, HCF = 2, LCM = 2x2x3×5x17x23 23460, Product of the two numbers = 510x92 = 46920, HCFX LCM = 2x 23460, = 46920, Hence, product of two numbers = HCF x LCM, 1| Page
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Page 3 of 5, (iii) 8,9 and 25, 8 = 2x 2x2, 9 = 3x3, 25 = 5x5, HCF = 1, LCM = 2x2x 2x3x3x5x5%31800, Question 4:, Given that HCF (306, 657) = 9, find LCM (306, 657)., Answer 4:, HCF(306, 657) = 9, We know that, LCM x HCF = Product of two numbers, .. LCM ×IICF = 306x 657, 306x 657 306x 657, LCM =, HCF, LCM = 22338, Question 5:, Check whether 6" can end with the digit 0 for any natural number n., Answer 5:, If any number ends with the digit 0, it should be divisible by 10 or in, other words, it will also be divisible by 2 and 5 as 10 = 2 x 5, Prime factorization of 6° =, (2 x3)", It can be observed that 5 is not in the prime factorisation of 6"., Hence, for any value of n, 6" will not be divisible by 5., Therefore, 6" cannot end with the digit 0 for any natural number n., 3 | Page
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Page 4 of 5, Question 6:, Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are, composite numbers., Answer 6:, Numbers are of two types - prime and composite. Prime numbers can be, divided by 1 and only itself, whereas composite numbers have factors, other than 1 and itself., It can be observed that, 7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) = 13 x (77 + 1), %3D, = 13 x 78, = 13 x13 x 6, The given expression has 6 and 13 as its factors. Therefore, it is a, composite number., 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x(7 x 6 x 4 x 3 x 2 x 1 + 1), = 5 x (1008 + 1), = 5 x1009, 1009 cannot be factorised further. Therefore, the given expression has 5, and 1009 as its factors. Hence, it is a composite number., Question 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 7:, It can be observed that Ravi takes lesser time than Sonia for completing, 1 round of the circular path. As they are going in the same direction, they, will meet again at the same time when Ravi will have completed 1 round, 4 | Page
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Page 5 of 5, of that circular path with respect to Sonia. And the total time taken for, completing this 1 round of circular path will be the LCM of time taken by, Sonia and Ravi for completing 1 round of circular path respectively i.e.,, LCM of 18 minutes and 12 minutes., 18 = 2 x 3 x 3, And, 12 = 2 x 2 x 3, LCM of 12 and 18 = 2 x 2 x 3 x 3 = 36, Therefore, Ravi and Sonia will meet together at the starting point after, 36 minutes., 5 | Page
