{"id":24920,"date":"2022-06-05T16:00:08","date_gmt":"2022-06-05T10:30:08","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=24920"},"modified":"2022-05-23T15:58:48","modified_gmt":"2022-05-23T10:28:48","slug":"ncert-solutions-for-class-10-maths-chapter-1-ex-1-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-1-ex-1-2\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 1 Real Numbers Ex 1.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers<\/h2>\n

Exercise 1.2<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nExpress each number as a product of its prime factors:
\n(i) 140
\n(ii) 156
\n(iii) 3825
\n(iv) 5005
\n(v) 7429
\nSolution:
\n(i) 140
\n140 = 2 x 2 x 7 x 5
\n140 = 22 x 7 x 5<\/p>\n

(ii) 156 = 2 x 2 x 3 x 13 = 22 x 3 x 13<\/p>\n

(iii) 3825 = 5 x 5 x 3 x 3 x 17
\n= 32 x 52 x 17<\/p>\n

(iv) 5005 = 5 x 7 x 11 x 13<\/p>\n

(v) 7429 = 17 x 19 x 23<\/p>\n

Question 2.
\nFind the LCM and HCF of the following pairs of integers and verify that LCM x HCF = Product of the two numbers.
\n(i) 26 and 91
\n(ii) 510 and 92
\n(iii) 336 and 54
\nSolution:
\nThe Prime factorisation of 26 and 91 is
\n26 = 2 x 13
\n91 = 7 x 13
\nTherefore, the LCM is 2 x 7 x 13 = 182.
\nAlso HCF is 13.
\nBy using the formula LCM x HCF = Product of two numbers
\n182 x 13 = 26 x 91
\n2366 = 2366
\n\u2234 LMC x HCF = Product of two numbers. [Verified]<\/p>\n

(ii) 510 and 92
\nThe prime factorisation of 510 and 92 is
\n510 = 2 x 3 x 5 x 17 92 = 2 x 2 x 23
\nTherefore, the LCM is 2 x 2 x 3 x 5 x 17 x 23 = 23,460 and HCF is 2.
\nNow, LCM x HCF = Product of two numbers
\nL.H.S. = LCM x HCF = 23640 x 2 = 46920
\nR.H.S. = Product of two numbers
\n= 510 x 92 = 46920
\nL.H.S. = R.H.S. [Verified]<\/p>\n

(iii) 336 and 54
\nThe prime factorisation of 336 and 54 is
\n336 = 24<\/sup> x 3 x 7 54
\n= 2 x 3\u00b3
\nTherefore, the LCM is 24<\/sup> x 3\u00b3 x 7 = 3024
\nand HCF is 2 x 3 = 6
\nNow, LCM x HCF = Product of two numbers
\nL.H.S. = LCM x HCF = 3024 x 6 = 18144
\nR.H.S. = Product of two numbers = 336 x 54 = 18144
\nNow, L.H.S.= R.H.S. [Verified]<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nFind the LCM and HCF of the following integers by applying the prime factorisation method.
\n(i) 12,15 and 21
\n(ii) 17, 23 and 29
\n(iii) 8, 9 and 25
\nSolution:
\n(i) 12,15 and 21
\n12 = 2 x 2 x 3
\n15 = 3 x 5
\n21 = 3 x 7
\nTherefore the LCM is 2\u00b2 x 3 x 7 = 420
\nAnd the HCF is 3! = 3<\/p>\n

(ii) 17, 23 and 29
\n17 = 1 x 17
\n23 = 1 x 23
\n29 = 1 x 29
\nTherefore tht LCM is 1 x 17 x 23 x 29 = 11339
\nAnd the HCF = 1<\/p>\n

(iii) 8,9 and 25
\n8 = 2 x 2 x 2 x 1
\n9= 3 x 3 x 1
\n25 = 5 x 5 x 1
\nTherefore the LCM is 1 x 2\u00b3 x 3\u00b2 x 5\u00b2 = 1800
\nAnd HCF is = 1<\/p>\n

Question 4.
\nGiven that HCF (306,657) = 9, find LCM (306, 657).
\nSolution:
\nHCF (306, 657) = 9 We know that
\nHCF x LCM = Product of two numbers
\n9 x LCM = 306 x 657
\nLCM = \\(\\frac { 201042 }{ 9 }\\)
\nLCM = 22338<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nCheck wether 6n <\/sup>can end with the digit 0 for any natural number n.
\nSolution:
\nIf the number 6n<\/sup> for any n, were to end with digit zero, then it will be divisible by 5. That is, the prime factorisation of 6n<\/sup> would contain the prime 5. This is not possible because 6n<\/sup> = (2 x 3)n<\/sup> = 2n<\/sup>. 3n<\/sup>, so the only prime in the factorisation of 6n<\/sup> is 2. So the uniqnessess of the fundamental theorem of arithemetic gurantees that, there are no other primes in the factorisation of 6n<\/sup>.<\/p>\n

Question 6.
\nExplain 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.
\nSolution:
\nWe know by fundamental theorem of arithmetic, “every composite number can be factorised as a product of primes”. So we can explain 7 x 11 x 13 + 13 and 7 x 3 x 2 x 5 x 2 x 2 x 3 x 2 x 1 + 5 are composite numbers or 7 x 11 x 13 + 13 and 7 x 32 x 24 + 5 are composite numbers.<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nThere is a circular path around a sports field. Sonia takes 18 minutes to derive one round of the field, while Ravi takes 12 minutes for 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.
\nSolution:
\nL.C.M (12,18)
\n12 = 2 x 2 x 3
\n18 = 2 x 3 x 3
\nSo the L.C.M of 12 and 18 is 2 x 3 x 2 x 3 = 36 After 36 minutes they will will meet at the starting point.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.2 Question 1. Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[2],"tags":[],"yoast_head":"\nNCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2 - MCQ Questions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-1-ex-1-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2 Questions and Answers are prepared by our highly skilled subject experts. 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