{"id":24847,"date":"2022-06-05T04:00:18","date_gmt":"2022-06-04T22:30:18","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=24847"},"modified":"2022-05-23T15:41:13","modified_gmt":"2022-05-23T10:11:13","slug":"ncert-solutions-for-class-10-maths-chapter-1-ex-1-1","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-10-maths-chapter-1-ex-1-1\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1"},"content":{"rendered":"

In this Page, you will learn how to solve questions of Ex 1.1 Class 10 Maths<\/a> NCERT Solutions recommended in CBSE 10 Board Maths Exam syllabus.<\/p>\n

These NCERT Solutions for Class 10 Maths<\/a> Chapter 1 Real Numbers Ex 1.1 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Exercise 1.1<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nEuclid’s division algoritgm to find the HCF of:
\n(i) 135 and 225
\n(ii) 196 and 38220
\n(iii) 867 and 225
\nSolution:
\n(i) 135 and 225
\nSince 225 > 135, we apply the division leema to 225 and 135, to get
\n225 = 135 x 1 + 90
\nSince the remainder 90 \u2260 0, we can apply the division lemma to 135 and 90, to get 135 = 90 x 1 + 45
\nWe consider the new divisor 90 and new remainder 45, and apply division lemmat to get 90 = 45 x 2 + 0
\nThe remainder has now become zero, so the HCF of 135 and 225 is 45.<\/p>\n

(ii) 196 and 38220
\nSince 38220 > 196, we apply the division leema to 38220 and 196, to get
\n38220 = 196 x 195 + 0
\nThe remainder has now become zero, so the HCF of 38220 and 196 is 196<\/p>\n

(iii) 867 and 255
\nSince 867 > 255, we apply the division lemma to 867 and 225, to get
\n867 = 255 x 3 + 102
\nSince the remainder 102 = 0, we apply the division lemma to 255 and 102, to get 255 = 102 x 2 + 51
\nWe consider the new divisor 102 and new remainder 51, and apply division lemma to get 102 = 51 x 2 + 0
\nThe remainder has now become zero, so the HCF of 867 and 255 is 51.<\/p>\n

Question 2.
\nShow that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
\nSolution:
\nLet a be any positive odd integer. We apply the division algorithm with a and b = 6.
\nSince 0 < r < 6, the positive remainders are 0,1, 2, 3, 4, 5.
\nNow putting the values r = 0,1, 5 we get
\n\u21d2 a = bq + r
\n\u21d2 a = 6q + 0 \u21d2 a = 6q [r = 0]
\n\u21d2 a = 6q + 1 \u21d2 a = 6q +1 [r = 1]
\nSimilarly integers of the form 6q, 6q + 1,6q + 2, 6q + 3, 6q + 4, 6q + 5.<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nAn army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
\nSolution:
\nThe maximum number of columns is HCF of (616,32).
\nNow, let us apply Euclid’s division algorithm to find their HCF
\n616 = 32 x 16 + 8
\nSince the remainder 8 \u2260 0, we apply the division lemma, to get
\n32 = 8 x 4 + 0
\nSo, the HCF of 616 and 32 is 8.
\nTherefore, the maximum number of columns is 8.<\/p>\n

Question 4.
\nUse Euclid’s division lemma to show, that the square of any positive integer is either of the form 3m or 3m +1 for some integer m.
\nSolution:
\n[Hint : Let x be any positive integer then it is of the form 3q, 3q +1, or 3q + 2. Now square each of these and show that they can be written in the form 3m or 3m+ 1]<\/p>\n

\"NCERT<\/p>\n","protected":false},"excerpt":{"rendered":"

In this Page, you will learn how to solve questions of Ex 1.1 Class 10 Maths NCERT Solutions recommended in CBSE 10 Board Maths Exam syllabus. These NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 …<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.1<\/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 1.1 - 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-1-1\/\" \/>\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 1.1 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"In this Page, you will learn how to solve questions of Ex 1.1 Class 10 Maths NCERT Solutions recommended in CBSE 10 Board Maths Exam syllabus. 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