5<\/sup><\/p>\n <\/p>\n
(vi) 2\u00b0 + 3\u00b0+ 4\u00b0 = 1 + 1 + 1 = 3 (a\u00b0 = 1) \n(vii) 2\u00b0 x 3\u00b0 \u00d7 4\u00b0 = 1 \u00d7 1 \u00d7 1 = 3 (a\u00b0 = 1) \n(viii) (3\u00b0 + 2\u00b0) \u00d7 5\u00b0 = (1 + 1) \u00d7 1 = 3 (a\u00b0 = 1) \n= 2 \u00d7 1 = 2<\/p>\n
<\/p>\n
(ix) \\(\\frac{2^{8} \\times a^{5}}{4^{3} \\times a^{3}}=\\frac{2^{8} \\times a^{5}}{\\left(2^{2}\\right)^{3} \\times a^{3}}\\) \n= \\(\\frac{2^{8} \\times a^{5}}{2^{6} \\times a^{3}}\\) [(am<\/sup>)n<\/sup> = amn<\/sup>] \n= 28-6<\/sup> \u00d7 a 5-3<\/sup> (\\(\\frac{\\mathrm{a}^{\\mathrm{m}}}{\\mathrm{a}^{\\mathrm{n}}}\\) = am-n<\/sup>) \n= 22<\/sup> \u00d7 a2<\/sup> \n= 2a2<\/sup><\/p>\n(x) \\(\\left(\\frac{a^{5}}{a^{3}}\\right)\\) x a8<\/sup> = (a5-3<\/sup>) x a8<\/sup> \n[(am<\/sup>)n<\/sup> = amn<\/sup>] \n= a2<\/sup> \u00d7 a8<\/sup> (am<\/sup> x an<\/sup> = am+n<\/sup>) \n= a2+8<\/sup> \n= a10<\/sup><\/p>\n(xi) \\(\\frac{4^{5} a^{8} b^{3}}{4^{5} a^{5} b^{2}}=\\frac{4^{5-5} \\times a^{8} \\times b^{3}}{a^{5} \\times b^{2}}\\) \n= 4\u00b0 \u00d7 a8-5<\/sup> \u00d7 b3-2<\/sup> ( \\(\\frac{\\mathrm{a}^{\\mathrm{m}}}{\\mathrm{a}^{\\mathrm{n}}}\\) = am-n<\/sup>) \n= 1 \u00d7 a3<\/sup> \u00d7 b \n= a3<\/sup>b (a0<\/sup> = 1)<\/p>\n(xii) (23<\/sup> x 2)2<\/sup> = (23+1<\/sup>)2<\/sup> \n(am<\/sup> x an<\/sup> = am+n<\/sup>) \n= (24<\/sup>)2<\/sup> \n= 28<\/sup> (am<\/sup>)n<\/sup> = amn<\/sup><\/p>\nQuestion 3. \nSay true or false and justify your answer: \n(i) 10 \u00d7 1011<\/sup> = 10011<\/sup> \n(ii) 23<\/sup> > 52<\/sup> \n(iii) 23<\/sup> \u00d7 32<\/sup> = 65<\/sup> \n(iv) 3\u00b0 = (1000)\u00b0 \nAnswer: \n(i) 10 \u00d7 1011<\/sup> = 10011<\/sup> \nL.H.S = 10 x 1011<\/sup> \n= 10(1+11)<\/sup>= 1012<\/sup> \n(am<\/sup> \u00d7 an<\/sup> = am+n<\/sup>) \nR.H.S = 10011<\/sup> \n= (102<\/sup>)H= 1022<\/sup> \n(am<\/sup>)n<\/sup> = amn<\/sup> \nL.H.S \u2260 R.H.S. \n\u2234 10 \u00d7 1011<\/sup> \u2260 10011<\/sup> \nThis statement is false.<\/p>\n <\/p>\n
(ii) 23<\/sup> > 52<\/sup> \n23<\/sup> = 2 \u00d7 2 \u00d7 2 = 8 \n52<\/sup> = 5 \u00d7 5 = 25 \n8 < 25 \ni. e. 23<\/sup> > 52<\/sup> \nThis statement is false.<\/p>\n(iii) 23<\/sup> \u00d7 32<\/sup> = 65<\/sup> \nL.H.S = 23<\/sup> \u00d7 32<\/sup> \n= 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 = 72 \nR.H.S = 65<\/sup> = 6 \u00d7 6 \u00d7 6 \u00d7 6 \u00d7 6 \n= 36 \u00d7 36 \u00d7 6 \n= 7776 \nL.H.S \u2260 R.H.S. \n72 \u2260 7776 \n\u2234 23<\/sup> \u00d7 32<\/sup> \u2260 65<\/sup> \nThis statement is false.<\/p>\n(iv) 3\u00b0 = (1000)\u00b0 \nL.H.S = 3\u00b0 = 1 \nR.H.S = (1000)\u00b0= 1 \n3\u00b0 = (1000)\u00b0 \nL.H.S = R.H.S \n3\u00b0 = (1000)\u00b0 \n\u2234 This statement is true.<\/p>\n
<\/p>\n
Question 4. \nExpress each of the following as a product of prime factors only in exponential form: \n(i) 108 \u00d7 192 \n(ii) 270 \n(iii) 729 \u00d7 64 \n(iv) 768 \nAnswer: \n(i) 108 \u00d7 192 \n= 2 x 2 x 3 x 3 x 3 x 2 x 2 \n \n= 28<\/sup> \u00d7 34<\/sup><\/p>\n(ii) 270 = 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 5 \n \n= 2 \u00d7 33<\/sup> \u00d7 5<\/p>\n(iii) 729 \u00d7 64 \n= 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \n \n= 36<\/sup> \u00d7 26<\/sup><\/p>\n(iv) 768 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 = 28<\/sup> \u00d7 3 \n <\/p>\nQuestion 5. \nSimplify: \n(i) \\(\\frac{\\left(2^{5}\\right)^{2} \\times 7^{3}}{8^{3} \\times 7}\\) \n(ii) \\(\\frac{25 \\times 5^{2} \\times t^{8}}{10^{3} \\times t^{4}}\\) \n(iii) \\(\\frac{3^{5} \\times 10^{5} \\times 25}{5^{7} \\times 6^{5}}\\) \nAnswer: \n(i) \\(\\frac{\\left(2^{5}\\right)^{2} \\times 7^{3}}{8^{3} \\times 7}=\\frac{2^{10} \\times 7^{3}}{\\left(2^{3}\\right)^{3} \\times 7}\\) \n[(am<\/sup>)n<\/sup> = amn<\/sup> ] \n= \\(\\frac{2^{10} \\times 7^{3}}{2^{9} \\times 7}\\) \n= 210-9<\/sup> \u00d7 73-1<\/sup> [\\(\\frac{a^{m}}{a^{n}}\\) = am-n<\/sup>] \n= 21<\/sup> \u00d7 72<\/sup> \n= 2 \u00d7 49 = 98 \nThus, \\(\\frac{\\left(2^{5}\\right)^{2} \\times 7^{3}}{8^{3} \\times 7}\\) = 98<\/p>\n <\/p>\n
\n \n <\/p>\n
= 35-5<\/sup> \u00d7 25-5<\/sup> \u00d7 57-7<\/sup> \n= 3\u00b0 \u00d7 2\u00b0 \u00d7 5\u00b0 \n= 1 \u00d7 1 \u00d7 1 = 1<\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.2 Question 1. Using laws of exponents, simplify and write the answer in exponential form: (i) 32 \u00d7 …<\/p>\n
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2<\/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":[9],"tags":[],"yoast_head":"\nNCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2 - 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