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>\nNCERT 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
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(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
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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 <\/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>\nQuestion 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
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Question 4. \nRepresent \\(\\sqrt{9.3}\\) on the number line. \nSolution: \n \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
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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
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(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 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n