{"id":6003,"date":"2023-07-14T04:45:33","date_gmt":"2023-07-14T04:45:33","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6003"},"modified":"2023-07-23T05:41:36","modified_gmt":"2023-07-23T05:41:36","slug":"ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-2\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.2"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 1 (Real Numbers)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 1 Real Numbers\u00a0 <\/strong>Exercise 1.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Chapter : 1 Real Numbers<\/strong><\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-1\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 1.1<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 1.3<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 1.4<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 1.2\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Express each number as a product of its prime factors:<br \/>\n<\/strong><strong>(i) 140<br \/>\n<\/strong><strong>(ii) 156<br \/>\n<\/strong><strong>(iii) 3825<br \/>\n<\/strong><strong>(iv) 5005<br \/>\n<\/strong><strong>(v) 7429<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211;<br \/>\n<\/strong><strong>(i) 140<br \/>\n<\/strong>By taking the LCM of 140, we will get the product of its prime factor.<br \/>\n140 = 2 \u00d7 2 \u00d7 5 \u00d7 7 \u00d7 1 = 2<sup>2 <\/sup>\u00d7 5 \u00d7 7<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 156<br \/>\n<\/strong>By Taking the\u00a0LCM of 156, we will get the product of its prime factor.<br \/>\n156 = 2 \u00d7 2 \u00d7 13 \u00d7 3 \u00d7 1 = 2<sup>2 <\/sup>\u00d7 13 \u00d7 3<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 3825<br \/>\n<\/strong>By taking the LCM of 3825, we will get the product of its prime factor.<br \/>\n3825 = 3 \u00d7 3 \u00d7 5 \u00d7 5 \u00d7 17 \u00d7 1 = 3<sup>2 <\/sup>\u00d7 5<sup>2 <\/sup>\u00d7 17<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iv) 5005<br \/>\n<\/strong>By Taking the\u00a0LCM of 5005, we will get the product of its prime factor.<br \/>\n5005 = 5 \u00d7 7 \u00d7 11 \u00d7 13 \u00d7 1 = 5 \u00d7 7 \u00d7 11 \u00d7 13<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(v) 7429<br \/>\n<\/strong>By taking the LCM of 7429, we will get the product of its prime factor.<br \/>\n7429 = 17 \u00d7 19 \u00d7 23 \u00d7 1 = 17 \u00d7 19 \u00d7 23<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Find the LCM and HCF of the following pairs of integers and verify that LCM \u00d7 HCF = product of the two numbers.<br \/>\n<\/strong><strong>(i) 26 and 91<br \/>\n<\/strong><strong>(ii) 510 and 92<br \/>\n<\/strong><strong>(iii) 336 and 54<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211; <\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(i) 26 and 91<br \/>\n<\/strong>Expressing 26 and 91 as product of its prime factors, we get,<br \/>\n26 = 2 \u00d7 13 \u00d7 1<br \/>\n91 = 7 \u00d7 13 \u00d7 1<br \/>\nTherefore, LCM (26, 91) = 2 \u00d7 7 \u00d7 13 \u00d7 1 = 182<br \/>\nAnd HCF (26, 91) = 13<br \/>\nVerification Now, product of 26 and 91 = 26 \u00d7 91 = 2366<br \/>\nAnd product of LCM and HCF = 182 \u00d7 13 = 2366<br \/>\nHence, LCM \u00d7 HCF = product of the 26 and 91.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 510 and 92<br \/>\n<\/strong>Expressing 510 and 92 as product of its prime factors, we get,<br \/>\n510 = 2 \u00d7 3 \u00d7 17 \u00d7 5 \u00d7 1<br \/>\n92 = 2 \u00d7 2 \u00d7 23 \u00d7 1<br \/>\nTherefore, LCM(510, 92) = 2 \u00d7 2 \u00d7 3 \u00d7 5 \u00d7 17 \u00d7 23 = 23460<br \/>\nAnd HCF (510, 92) = 2<br \/>\nVerification<br \/>\nNow, product of 510 and 92 = 510 \u00d7 92 = 46920<br \/>\nAnd Product of LCM and HCF = 23460 \u00d7 2 = 46920<br \/>\nHence, LCM \u00d7 HCF = product of the 510 and 92.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 336 and 54<br \/>\n<\/strong>Expressing 336 and 54 as product of its prime factors, we get,<br \/>\n336 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 7 \u00d7 3 \u00d7 1<br \/>\n54 = 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 1<br \/>\nTherefore, LCM(336, 54) = 3024<br \/>\nAnd HCF(336, 54) = 2\u00d73 = 6<br \/>\nVerification<br \/>\nNow, product of 336 and 54 = 336 \u00d7 54 = 18,144<br \/>\nAnd product of LCM and HCF = 3024 \u00d7 6 = 18,144<br \/>\nHence, LCM \u00d7 HCF = product of the 336 and 54.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>3. Find the LCM and HCF of the following integers by applying the prime factorisation method.<br \/>\n<\/strong><strong>(i) 12, 15 and 21<br \/>\n<\/strong><strong>(ii) 17, 23 and 29<br \/>\n<\/strong><strong>(iii) 8, 9 and 25<\/strong><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Solutions &#8211;<br \/>\n<\/strong><strong>(i) 12, 15 and 21<br \/>\n<\/strong>Writing the product of prime factors for all the three numbers, we get,<br \/>\n12 = 2 \u00d7 2 \u00d7 3<br \/>\n15 = 5 \u00d7 3<br \/>\n21 = 7 \u00d7 3<br \/>\nTherefore,<br \/>\nHCF (12, 15, 21) = 3<br \/>\nLCM (12, 15, 21) = 2 \u00d7 2 \u00d7 3 \u00d7 5 \u00d7 7 = 420<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 17, 23 and 29<br \/>\n<\/strong>Writing the product of prime factors for all the three numbers, we get,<br \/>\n17 = 17 \u00d7 1<br \/>\n23 = 23 \u00d7 1<br \/>\n29 = 29 \u00d7 1<br \/>\nTherefore,<br \/>\nHCF (17, 23, 29) = 1<br \/>\nLCM (17, 23, 29) = 17 \u00d7 23 \u00d7 29 = 11339<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 8, 9 and 25<br \/>\n<\/strong>Writing the product of prime factors for all the three numbers, we get,<br \/>\n8 = 2 \u00d7 2 \u00d7 2 \u00d7 1<br \/>\n9 = 3 \u00d7 3 \u00d7 1<br \/>\n25 = 5 \u00d7 5 \u00d7 1<br \/>\nTherefore,<br \/>\nHCF (8, 9, 25) = 1<br \/>\nLCM (8, 9, 25) = 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 5 = 1800<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Given that HCF (306, 657) = 9, find LCM (306, 657).<br \/>\n<\/strong><strong>Solution &#8211;\u00a0 <\/strong>As we know that,<br \/>\nHCF \u00d7 LCM = Product of the two given numbers<br \/>\nTherefore,<br \/>\n9 \u00d7 LCM = 306 \u00d7 657<br \/>\nLCM = (306 \u00d7 657)\/9 = 22338<br \/>\nHence, LCM (306, 657) = 22338<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>5. Check whether 6<sup>n<\/sup> can end with the digit 0 for any natural number n.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>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 \u00d7 5<\/span><br \/>\n<span style=\"color: #000000;\">Prime factorisation of 6<sup><i>n <\/i><\/sup><i>=<\/i> (2 \u00d73)<sup>n<br \/>\n<\/sup>It can be observed that 5 is not in the prime factorisation of 6<sup><i>n<\/i><\/sup>.<\/span><br \/>\n<span style=\"color: #000000;\">Hence, for any value of <i>n<\/i>, 6<sup><i>n<\/i><\/sup> will not be divisible by 5.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, 6<sup><i>n<\/i><\/sup> cannot end with the digit 0 for any natural number <i>n<\/i>.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>6. Explain why 7 \u00d7 11 \u00d7 13 + 13 and 7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 5 are composite numbers.<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Numbers are of two types &#8211; prime and composite. Prime numbers can be divided by 1 and only itself, whereas composite numbers have factors other than 1 and itself.<\/span><br \/>\n<span style=\"color: #000000;\">It can be observed that <\/span><br \/>\n<span style=\"color: #000000;\">7 \u00d7 11 \u00d7 13 + 13 = 13 \u00d7 (7 \u00d7 11 + 1) = 13 \u00d7 (77 + 1)<\/span><br \/>\n<span style=\"color: #000000;\">= 13 \u00d7 78<\/span><br \/>\n<span style=\"color: #000000;\">= 13 \u00d713 \u00d7 6<\/span><br \/>\n<span style=\"color: #000000;\">The given expression has 6 and 13 as its factors. Therefore, it is a composite number. <\/span><br \/>\n<span style=\"color: #000000;\">7 \u00d7 6 \u00d7 5 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 5 = 5 \u00d7 (7 \u00d7 6 \u00d7 4 \u00d7 3 \u00d7 2 \u00d7 1 + 1) <\/span><br \/>\n<span style=\"color: #000000;\">= 5 \u00d7 (1008 + 1) <\/span><br \/>\n<span style=\"color: #000000;\">= 5 \u00d7 1009 <\/span><br \/>\n<span style=\"color: #000000;\">1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>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?<br \/>\n<\/strong><strong>Solution &#8211; <\/strong>Since, Both 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.<br \/>\nTherefore, LCM (18, 12) = 2 \u00d7 3 \u00d7 3 \u00d7 2 \u00d7 1 = 36<br \/>\nHence, Sonia and Ravi will meet again at the starting point after 36 minutes.<\/span><\/p>\n<p><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\"><b><i>Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 1 (Real Numbers)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 1 Real Numbers\u00a0 Exercise 1.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 1 Real Numbers NCERT Class 10 Maths Solution Ex &#8211; 1.1 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1046,1050,1044,1049,1048],"class_list":["post-6003","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-1-real-numbers-solutions","tag-ncert-class-10-mathematics-exercise-1-2-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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