{"id":26688,"date":"2021-06-26T16:25:35","date_gmt":"2021-06-26T10:55:35","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=26688"},"modified":"2022-03-02T10:47:30","modified_gmt":"2022-03-02T05:17:30","slug":"ncert-solutions-for-class-7-maths-chapter-13-ex-13-2","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-7-maths-chapter-13-ex-13-2\/","title":{"rendered":"NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Ex 13.2"},"content":{"rendered":"

These NCERT Solutions for Class 7 Maths<\/a> Chapter 13 Exponents and Powers Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.2<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nUsing laws of exponents, simplify and write the answer in exponential form:
\n(i) 32<\/sup> \u00d7 34<\/sup> \u00d7 38<\/sup>
\n(ii) 615<\/sup> \u00f7 610<\/sup>
\n(iii) a3<\/sup> \u00d7 a2<\/sup>
\n(iv) 7x<\/sup> \u00d7 72<\/sup>
\n(v) (52<\/sup>)3<\/sup> \u00f7 53<\/sup>
\n(vi) 25<\/sup> \u00d7 55<\/sup>
\n(vii) a4<\/sup> \u00d7 b4<\/sup>
\n(viii) (34<\/sup>)3<\/sup>
\n(ix) (220<\/sup> \u00f7 215<\/sup>) \u00d7 23<\/sup>
\n(x) 8t<\/sup> \u00f7 82<\/sup>
\nAnswer:
\n(i) 32<\/sup> \u00d7 34<\/sup> \u00d7 38<\/sup> = 32 + 4 + 8<\/sup>
\n(am<\/sup> \u00d7 an<\/sup> \u00d7 r<\/sup> = am+n+r<\/sup>) = 314<\/sup>
\n(ii) 615<\/sup> \u00f7 610<\/sup> = \\(\\frac{6^{15}}{6^{10}}\\) = 615-10<\/sup>
\n(\\(\\frac{a^{\\mathrm{m}}}{\\mathrm{a}^{\\mathrm{n}}}\\) = am-n<\/sup>) = 65<\/sup>
\n(iii) a3<\/sup> \u00d7 a2<\/sup> = a3+2<\/sup>
\n(am<\/sup> \u00d7 an<\/sup> = am+n<\/sup> ) = a5<\/sup>
\n(iv) 7x<\/sup> \u00d7 72<\/sup> = 7x+2<\/sup> (am<\/sup> \u00d7 an<\/sup> = am+n<\/sup> )
\n(v) (52<\/sup>)3<\/sup> \u00f7 53<\/sup>
\n\"NCERT
\n(vi) 25<\/sup> \u00d7 55<\/sup> = (2 \u00d7 5)5<\/sup>
\n[am<\/sup> \u00d7 bm<\/sup> = (ab)m<\/sup>]
\n= (ab)4<\/sup>
\n(vii) a4<\/sup> x b4<\/sup>
\n(viii) (34<\/sup>)3<\/sup> = 312<\/sup>
\n[(am<\/sup>)n<\/sup> = amn<\/sup>]
\n(ix) (220<\/sup> \u00f7 215<\/sup>) x 23<\/sup> = (\\(\\frac{2^{20}}{2^{15}}\\)) x 23<\/sup>
\n= (220-15<\/sup>) \u00d7 23<\/sup> [ \\(\\frac{a^{\\mathrm{m}}}{\\mathrm{a}^{\\mathrm{n}}}\\) = am-n<\/sup>]
\n= 25<\/sup> \u00d7 23<\/sup>
\n= 25 + 3<\/sup> (am<\/sup> x an<\/sup> = am+n<\/sup>) = 28<\/sup>
\n(x) 8t<\/sup> \u00f7 82<\/sup> = \\(\\frac{8^{t}}{8^{2}}=8^{t-2}\\left(\\frac{\\mathrm{a}^{\\mathrm{m}}}{\\mathrm{a}^{\\mathrm{n}}}=\\mathrm{a}^{\\mathrm{m}-\\mathrm{n}}\\right)\\)<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nSimplify and express each of the following in exponential form:
\n(i) \\(\\)
\n(ii) [(52<\/sup>)3<\/sup> \u00d7 54<\/sup>)] \u00f7 57<\/sup>
\n(iii) 254<\/sup> \u00f7 53<\/sup>
\n(iv) \\(\\frac{3 \\times 7^{2} \\times 11^{8}}{21 \\times 11^{3}}\\)
\n(v) \\(\\frac{3^{7}}{3^{4} \\times 3^{3}}\\)
\n(vi) 2\u00b0+ 3\u00b0+ 4\u00b0
\n(vii) 2\u00b0 \u00d7 3\u00b0 \u00d7 4\u00b0
\n(viii) (3\u00b0 + 2\u00b0) \u00d7 5\u00b0
\n(ix) \\(\\frac{2^{8} \\times \\mathrm{a}^{5}}{4^{3} \\times \\mathrm{a}^{3}}\\)
\n(x) \\(\\left(\\frac{a^{5}}{a^{3}}\\right) \\times a^{8}\\)
\n(xi) \\(\\frac{4^{5} \\times a^{8} b^{3}}{4^{5} \\times a^{5} b^{2}}\\)
\n(xii) (23<\/sup> \u00d7 2)2<\/sup>
\nAnswer:
\n\"NCERT
\n= 2\u00b0 \u00d7 33<\/sup>
\n= 1 \u00d7 33<\/sup> = 33<\/sup>
\n(a\u00b0 = 1)<\/p>\n

\"NCERT
\n\"NCERT
\n\"NCERT<\/p>\n

(iv) \\(\\frac{3 \\times 7^{2} \\times 11^{8}}{21 \\times 11^{3}}=\\frac{3 \\times 7^{2} \\times 11^{8}}{3 \\times 7 \\times 11^{3}}\\)
\n= 31-1<\/sup> \u00d7 72-1<\/sup> \u00d7 118-3<\/sup>
\n(\\(\\frac{\\mathrm{a}^{\\mathrm{m}}}{\\mathrm{a}^{\\mathrm{n}}}\\) = am-n<\/sup>)
\n= 3\u00b0 \u00d7 71<\/sup> \u00d7 115<\/sup>
\n= 1 \u00d7 7 \u00d7 115<\/sup>
\n= 7 \u00d7115<\/sup><\/p>\n

\"NCERT<\/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

\"NCERT<\/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>\n

Question 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

\"NCERT<\/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

\"NCERT<\/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\"NCERT
\n= 28<\/sup> \u00d7 34<\/sup><\/p>\n

(ii) 270 = 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 5
\n\"NCERT
\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\"NCERT
\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\"NCERT<\/p>\n

Question 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

\"NCERT<\/p>\n

\"NCERT
\n\"NCERT
\n\"NCERT<\/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<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-7-maths-chapter-13-ex-13-2\/\" \/>\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 7 Maths Chapter 13 Exponents and Powers Ex 13.2 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"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. 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NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.2 Question 1. 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