{"id":24592,"date":"2021-06-19T12:40:59","date_gmt":"2021-06-19T07:10:59","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=24592"},"modified":"2022-03-02T10:38:16","modified_gmt":"2022-03-02T05:08:16","slug":"ncert-solutions-for-class-9-maths-chapter-1-ex-1-5","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-9-maths-chapter-1-ex-1-5\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 1 Number Systems Ex 1.5 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.5<\/h2>\n

Question 1.
\nClassify the following numbers as rational or irrational:
\n(i) 2 – \u221a5
\n(ii) (3 + \u221a23) – \u221a23
\n(iii) \\(\\frac{2 \\sqrt{7}}{7 \\sqrt{7}}\\)
\n(iv) \\(\\frac{1}{\\sqrt{2}}\\)
\n(v) 2\u03c0
\nSolution:
\n(i) We have, 2 – \u221a5
\nor, 2 – 2.236067977…….
\nor, -0.236067977……..
\nwhich is non terminating non recurring.
\nSo, it is an irrational number.<\/p>\n

(ii) We have, (3 + \u221a23) – \u221a23
\nor, 3 + \u221a23 – \u221a23
\nor, 3
\nwhich is a rational number.<\/p>\n

\"NCERT<\/p>\n

(iii) We have, \\(\\frac{2 \\sqrt{7}}{7 \\sqrt{7}}\\)
\nor \\(\\frac{2}{7}\\)
\nwhich is a rational number.<\/p>\n

(iv) We have, \\(\\frac{1}{\\sqrt{2}}\\)
\nor, \\(\\frac{1}{1.414213562 \\ldots}\\)
\nTherefore the quotient of this number is irrational.<\/p>\n

(v) We have,
\n2\u03c0 = 2 \u00d7 3.1415926535……. = 6.2831853070……..
\nwhich is non-terminating non-recurring.
\nTherefore, it is an irrational number.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nSimplify each of the following expressions:
\n(i) (3 + \u221a3) (2 + \u221a2)
\n(ii) (3+ \u221a3) (3 – \u221a3)
\n(iii) (\u221a5 + \u221a2)2<\/sup>
\n(iv) (\u221a5 – \u221a2) (\u221a5 + \u221a2)
\nSolution:
\n(i) We have, (3 + \u221a3) (2 + \u221a2)
\nor 6 + 3\u221a2 + 2\u221a3 + \u221a6<\/p>\n

(ii) We have, (3 + \u221a3) (3 – \u221a3)
\nWe know that (a + b) (a – b) = a2<\/sup> – b2<\/sup>
\n\u2234 (3 + \u221a3) (3 – \u221a3) = (3)2<\/sup> – (\u221a3)2<\/sup>
\n= 9 – 3
\n= 6
\nSo, (3 + \u221a3) (3 – \u221a3) = 6<\/p>\n

\"NCERT<\/p>\n

(iii) We have, (\u221a5 + \u221a2)2<\/sup>
\nWe know that (a + b)2<\/sup> = a2<\/sup> + b2<\/sup> + 2ab
\n\u2234 (\u221a5 + \u221a2)2<\/sup> = (\u221a5)2<\/sup> + (\u221a2)2<\/sup> + 2(\u221a5)(\u221a2) = 5 + 2 + 2\u221a10
\nor, (\u221a5 + \u221a2)2<\/sup> = 7 + 2\u221a10<\/p>\n

(iv) We have,
\n(\u221a5 – \u221a2) (\u221a5 + \u221a2) = (\u221a5)2<\/sup> – (\u221a2)2<\/sup>
\n[\u2234 (a + b) (a – b) = a2<\/sup> – b2<\/sup>]
\n= 5 – 2
\n= 3
\n\u2234 (\u221a5 – \u221a2) (\u221a5 + \u221a2) = 3<\/p>\n

Question 3.
\nRecall, \u03c0 is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is \u03c0 = \\(\\frac{c}{d}\\). This seems to contradict the fact that \u03c0 is irrational. How will you resolve this contradiction?
\nSolution:
\nCircumference = 2\u03c0R (Irrational number) of the circle (c)
\nR \u2192 Radius of the circle
\nDiameter of circle (D) = 2R (Rational number)
\n\\(\\frac{C}{D}=\\frac{2 \\pi R}{2 R}=\\pi=\\frac{\\text { Irrational number }}{\\text { Rational number }}\\) = Irrational number
\nAs we know irrational number divided by a rational number results in an irrational number.<\/p>\n

\"NCERT<\/p>\n

Question 4.
\nRepresent \\(\\sqrt{9.3}\\) on the number line.
\nSolution:
\n\"NCERT
\nTo represent \\(\\sqrt{9.3}\\) on the number line.
\nWe mark a point B on the number line so that AB = 9.3 units.
\nAgain mark a point C so that BC = 1 unit.
\nNow take the midpoint of AC and mark that point as O.
\nDraw a semicircle with centre O and radius OC.
\nDraw the line perpendicular to AC passing through B and intersecting the semicircle at D. Then, BD = \\(\\sqrt{9.3}\\)
\nAgain taking BD as a radius and draw an arc which intersecting the number line at E.
\nThen E represents \\(\\sqrt{9.3}\\) on number line.
\nTake b at zero on number line, then point E represent \\(\\sqrt{9.3}\\).<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nRationalise the denominators of the following:
\n(i) \\(\\frac{1}{\\sqrt{7}}\\)
\n(ii) \\(\\frac{1}{\\sqrt{7}-\\sqrt{6}}\\)
\n(iii) \\(\\frac{1}{\\sqrt{5}+\\sqrt{2}}\\)
\n(iv) \\(\\frac{1}{\\sqrt{7}-2}\\)
\nSolution:
\n(i) We have, \\(\\frac{1}{\\sqrt{7}}\\)
\nMultiply \u221a7 both numerator and denominator.
\n\\(\\frac{1}{\\sqrt{7}} \\times \\frac{\\sqrt{7}}{\\sqrt{7}}=\\frac{\\sqrt{7}}{7}\\)<\/p>\n

(ii) We have, \\(\\frac{1}{\\sqrt{7}-\\sqrt{6}}\\)
\nMultiply \u221a7 + \u221a6 both numerator and denominator
\n\\(\\frac{1}{\\sqrt{7}-\\sqrt{6}} \\times \\frac{\\sqrt{7}+\\sqrt{6}}{\\sqrt{7}+\\sqrt{6}}=\\frac{\\sqrt{7}+\\sqrt{6}}{7-6}\\)
\n= \\(\\frac{\\sqrt{7}+\\sqrt{6}}{1}\\)
\n= \u221a7 + \u221a6<\/p>\n

(iii) We have, \\(\\frac{1}{\\sqrt{5}+\\sqrt{2}}\\)
\nMultiply \u221a5 – \u221a2 both numerator and denominator
\n\\(\\frac{1}{\\sqrt{5}+\\sqrt{2}} \\times \\frac{\\sqrt{5}-\\sqrt{2}}{\\sqrt{5}-\\sqrt{2}}=\\frac{\\sqrt{5}-\\sqrt{2}}{5-2}=\\frac{\\sqrt{5}-\\sqrt{2}}{3}\\)<\/p>\n

\"NCERT<\/p>\n

(iv) We have
\n\\(\\frac{1}{\\sqrt{7}-2}=\\frac{1}{\\sqrt{7}-2} \\times \\frac{\\sqrt{7}+2}{\\sqrt{7}+2}\\)
\n= \\(\\frac{\\sqrt{7}+2}{7-4}\\)
\n= \\(\\frac{\\sqrt{7}+2}{3}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.5 Question 1. Classify the following numbers as rational or irrational: (i) 2 – \u221a5 (ii) (3 + \u221a23) – …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5<\/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 1.5 - 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-9-maths-chapter-1-ex-1-5\/\" \/>\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 1.5 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 Questions and Answers are prepared by our highly skilled subject experts. 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