NCERT Solutions for Class 9 Maths<\/a> Chapter 1 Number Systems Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\nNCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.3<\/h2>\n Question 1. \nWrite the following in decimal form and say what kind of decimal expansion each has: \n(i) \\(\\frac{36}{100}\\) \n(ii) \\(\\frac{1}{11}\\) \n(iii) \\(4 \\frac{1}{8}\\) \n(iv) \\(\\frac{3}{13}\\) \n(v) \\(\\frac{2}{11}\\) \n(vi) \\(\\frac{329}{400}\\) \nSolution: \n(i) By actual division we have \n \nThe decimal form of \\(\\frac{36}{100}\\) is 0.36 \nHere, the remainder becomes 0, therefore it has a terminating decimal expansion.<\/p>\n
(ii) By actual division we have \n \nThe decimal form of \\(\\frac{1}{11}\\) is 0.090909….. of \\(0 . \\overline{09}\\). \nHence remainder never becomes zero but repeats. Therefore, it has non terminating but repeating decimal expansion.<\/p>\n
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(iii) \\(4 \\frac{1}{8}\\) \nWe can also write \\(\\frac{33}{8}\\) \nBy actual division we have, \n \nThe decimal form of \\(4 \\frac{1}{8}\\) is 4.125. \nHere remainder becomes zero after certain steps. Therefore it has a terminating decimal expansion.<\/p>\n
(iv) By actual division we have \n \nThe decimal form of \\(\\frac{3}{13}\\) is 0.2307692307….. or \\(0 . \\overline{230769}\\). \nHere, the remainder never becomes zero but repeats after some steps. \nTherefore, it has a non terminating but repeating decimal expansion.<\/p>\n
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(v) By actual division we have \n \nThe decimal form of \\(\\frac{2}{11}\\) is 0.181818…. or \\(0 . \\overline{18}\\). Here remainder never becomes zero but repeat after some steps. \nTherefore, it has non terminating but repeating decimal expansion.<\/p>\n
(vi) By actual division we have, \n \nThe decimal form of \\(\\frac{329}{400}\\) is 0.8225. Here remainder becomes zero after some steps. Therefore, it has terminating decimal expansion.<\/p>\n
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Question 2. \nYou know that \\(\\frac{1}{7}=0 . \\overline{142857}\\). Can you predict what the decimal expansions of \\(\\frac{2}{7}, \\frac{3}{7}, \\frac{4}{7}, \\frac{5}{7}, \\frac{6}{7}\\) are, without actually doing the long devision? If so how? \nSolution: \nWe have given that \n\\(\\frac{1}{7}=0 . \\overline{142857}\\) \n <\/p>\n
Question 3. \nExpress the following in the form \\(\\frac {p}{q}\\), where p and q are integers and q \u2260 0. \n(i) \\(0 . \\overline{6}\\) \n(ii) \\(0 . \\overline{47}\\) \n(iii) \\(0 . \\overline{001}\\) \nSolution: \nLet x = \\(0 . \\overline{6}\\) \nor, x = 0.66666…… (i) \nNow, add 6 both side in equation (i) \nx + 6 = 0.66666…… + 6 \nor x + 6 = 6.6666……. (ii) \nAgain, multiply 10 both side in equation (i) \n10x = 0.66666…. \u00d7 10 \nor, 10x = 6.6666…. (iii) \nNow, from equation (ii) and (iii) \nx + 6 = 10x \n9x = 6 \n\u2234 x = \\(\\frac{6}{9}\\) \nTherefore, \\(0 . \\overline{6}\\) = x = \\(\\frac{2}{3}\\)<\/p>\n
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(ii) Let x = \\(0.4 \\overline{7}\\) \nor x = 0.47777……. (i) \nNow, multiply 10 both side in equation (i) \n10x = 0.47777……. \u00d7 10 \nor, 10x = 4.77777…… (ii) \nAgain, multiply 100 both side in equation (i) \n100x = 0.47777 \u00d7 100 \nor, 100x = 4.77777…….. (iii) \nSubtract equation (ii) from equation (iii) \n90x = 43.0000 \nor 90x = 43 \n\u2234 x = \\(\\frac{43}{90}\\) \nTherefore, \\(0.4 \\overline{7}\\) = x = \\(\\frac{43}{90}\\) \nwhich is in the form of \\(\\frac {p}{q}\\)<\/p>\n
(iii) Let x = \\(0 . \\overline{001}\\) \nor, x = 0.001001001……. (i) \nNow, add 1 both side in equation (i) \nx + 1 = 0.001001001…… + 1 \nor x + 1 = 1.001001001……. (ii) \nAgain, multiply 1000 both side in equation (i) \n1000 \u00d7 x = 1000 \u00d7 0.001001001…….. \nor, 1000x = 1.001001001….. (iii) \nFrom equation (ii) and (iii) we get \nx +1 = 1000x \nor 999x = 1 \nor x = \\(\\frac{1}{999}\\) \nTherefore, \\(0 . \\overline{001}=x \\times \\frac{1}{999}\\) \nwhich is in the form of \\(\\frac {p}{q}\\)<\/p>\n
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Question 4. \nExpress 0.99999…… in the form \\(\\frac {p}{q}\\). Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense. \nSolution: \nLet x = 0.9999…… (i) \nAdd 9 both side in equation (i) \nx + 9 = 9 + 0.99999……… \nor, x + 9 = 9.99999…… (ii) \nAgain, multiply 10 both side in equation (i) \n10x = 9.9999……. (iii) \nFrom equation (ii) and (iii) we get \n10x = x + 9 \nor, 9x = 9 \nor, x = 1 \nTherefore, 0.9999…… = 1 \nIt is because there is infinite 9 comes after the point; which is very-very close to 1.<\/p>\n
Question 5. \nWhat can the maximum number of digits be in the repeating block of digits in the decimal expansion of \\(\\frac{1}{17}\\)? Perform the division to check your answer. \nSolution: \nBy actual division we have \nIt is clear that, \\(\\frac{1}{17}=0 . \\overline{0588235294117647}\\) \n \nIt is clear that, \\(\\frac{1}{17}=0 . \\overline{0588235294117647}\\) \nSo, the maximum number of digits be in the repeating block of digits in die decimal expansion of \\(\\frac{1}{17}\\) is 16.<\/p>\n
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Question 6. \nLook at several examples of rational numbers in the form \\(\\frac{p}{q}\\) (q \u2260 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy? \nSolution: \nSome examples of a rationed number having terminating decimal representations are: \n(i) \\(\\frac{1}{2}\\) = 0.5 \n(ii) \\(\\frac{7}{4}\\) = 1.75 \n(iii) \\(\\frac{7}{8}\\) = 0.875 \n(iv) \\(\\frac{2}{5}\\) = 0.4 \nIt is clear that the prime factorization of q has only power of 2 or power of 5 or both.<\/p>\n
Question 7. \nWrite three numbers whose decimal expansions are non-terminating non-recurring. \nSolution: \nWe knew that the decimal expansions of an irrational number are non-terminating non-recurring. Three examples of such numbers are: \n\u221a2 = 1.4142135……. \n\u221a3 = 1.732050807……… \n\u03c0 = 3.1415926535………<\/p>\n
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Question 8. \nFind three different irrational numbers between the rational numbers \\(\\frac{5}{7}\\) and \\(\\frac{9}{11}\\). \nSolution: \nTo find an irrational numbers between \\(\\frac{5}{7}\\) and \\(\\frac{9}{11}\\) is non terminating non recurring lying between them. \nHere, \\(\\frac{5}{7}=0 . \\overline{714285}\\) and \\(\\frac{9}{11}=0 . \\overline{81}\\) \nTherefore, the required three different irrational number which is lying between \\(\\frac{5}{7}\\) and \\(\\frac{9}{11}\\) are 0.720720072000…….., 0.730730073000……., and 0.740740074000………<\/p>\n
Question 9. \nClassify the following number as rational or irrational: \n(i) \u221a23 \n(ii) \u221a225 \n(iii) 0.3796 \n(iv) 7.478478……. \n(v) 1.101001000100001……. \nSolution: \n(i) We have \n\u221a23 = 4.795831523…….. \nIt is non-terminating non-recurring. \nSo, \u221a23 is an irrational number<\/p>\n
(ii) We have \n\u221a225 = 15 \nor \u221a225 = \\(\\frac{15}{1}\\) \nwhich is a rational number<\/p>\n
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(iii) We have, \n0.3796 = \\(\\frac{3796}{10000}\\) \nwhich is in the form of \\(\\frac{p}{q}\\) \nSo, 0.3796 is a rational number.<\/p>\n
(iv) We have \n7.478478……. = \\(7 . \\overline{478}\\) which is non terminating recurring. \nTherefore, 7.478478…….. is an irrational number.<\/p>\n
(v) We have, 1.101001000100001…….. \nwhich is non-terminating non-recurring. \nTherefore, 1.101001000100001……… is an irrational number.<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.3 Question 1. Write the following in decimal form and say what kind of decimal expansion each has: (i) (ii) …<\/p>\n
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3<\/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":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n