{"id":24562,"date":"2022-12-28T11:28:39","date_gmt":"2022-12-28T05:58:39","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=24562"},"modified":"2022-12-28T14:49:00","modified_gmt":"2022-12-28T09:19:00","slug":"ncert-solutions-for-class-9-maths-chapter-1-ex-1-3","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-1-ex-1-3\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3"},"content":{"rendered":"

These 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>\n

NCERT 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\"NCERT
\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\"NCERT
\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

\"NCERT<\/p>\n

(iii) \\(4 \\frac{1}{8}\\)
\nWe can also write \\(\\frac{33}{8}\\)
\nBy actual division we have,
\n\"NCERT
\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\"NCERT
\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

\"NCERT<\/p>\n

(v) By actual division we have
\n\"NCERT
\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\"NCERT
\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

\"NCERT<\/p>\n

5\/9 as a decimal<\/a>\u00a0is 0.56<\/p>\n

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\"NCERT<\/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

\"NCERT<\/p>\n

(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

\"NCERT<\/p>\n

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\"NCERT
\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

\"NCERT<\/p>\n

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

\"NCERT<\/p>\n

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

\"NCERT<\/p>\n

(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":{"spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.3 - NCERT Solutions<\/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:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-1-ex-1-3\/\" \/>\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 9 Maths Chapter 1 Number Systems Ex 1.3 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"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. 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NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.3 Question 1. 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