{"id":22303,"date":"2022-01-14T17:46:06","date_gmt":"2022-01-14T12:16:06","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=22303"},"modified":"2022-01-14T17:46:06","modified_gmt":"2022-01-14T12:16:06","slug":"ap-board-7th-class-maths-solutions-chapter-8-unit-exercise","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-7th-class-maths-solutions-chapter-8-unit-exercise\/","title":{"rendered":"AP Board 7th Class Maths Solutions Chapter 8 Exponents and Powers Unit Exercise"},"content":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf<\/a> Chapter 8 Exponents and Powers Unit Exercise Questions and Answers.<\/p>\n

AP State Syllabus 7th Class Maths Solutions 8th Lesson\u00a0Exponents and Powers Unit Exercise<\/h2>\n

Question 1.
\nAnswer the following.
\n(i) The exponential form 149<\/sup> should read as
\nAnswer:
\n14 is raised to the power of 9.<\/p>\n

(ii) When base is 12 and exponent is 17, it\u2019s exponential form is _________
\nAnswer:
\n1217<\/sup>.<\/p>\n

(iii) The value of (14 \u00d7 21)0<\/sup> is
\nAnswer:
\nWe know a0<\/sup> = 1
\nSo, (14 \u00d7 21)0<\/sup> = 1<\/p>\n

Question 2.
\nExpress the following numbers as a product of powers of prime factors :
\n(i) 648
\nAnswer:
\n\"AP
\nGiven 648 = 2 \u00d7 324
\n= 2 \u00d7 2 \u00d7 162
\n= 2 \u00d7 2 \u00d7 2 \u00d7 81
\n= 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 27
\n= 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 9
\n= 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3
\n\u2234 648 = 23<\/sup> \u00d7 34<\/sup><\/p>\n

(ii) 1600
\nAnswer:
\nGiven 1600 = 2 \u00d7 800
\n= 2 \u00d7 2 \u00d7 400
\n= 2 \u00d7 2 \u00d7 2 \u00d7 200
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 100
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 25
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 5 \u00d7 5
\n\"AP
\n\u2234 1600 = 26<\/sup> \u00d7 52<\/sup><\/p>\n

(iii) 3600
\nAnswer:
\n\"AP
\nGiven 3600 = 2 \u00d7 1800
\n= 2 \u00d7 2 \u00d7 900
\n= 2 \u00d7 2 \u00d7 2 \u00d7 450
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 225
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 75
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 25
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 \u00d7 5
\n\u2234 3600 = 24<\/sup> \u00d7 32<\/sup> \u00d7 52<\/sup><\/p>\n

\"AP<\/p>\n

Question 3.
\nSimplify the following using laws of exponents.
\n(i) a4<\/sup> \u00d7 a10<\/sup>
\nAnswer:
\na4<\/sup> \u00d7 a10<\/sup>
\nWe know am<\/sup> \u00d7 an<\/sup> = am+n<\/sup>
\n= a4+10<\/sup>
\n\u2234 a4<\/sup> \u00d7 a10<\/sup> = a14<\/sup><\/p>\n

(ii) 1818<\/sup> \u00f7 1814<\/sup>
\nAnswer:
\n1818<\/sup> \u00f7 1814<\/sup>
\nWe Know am<\/sup> \u00f7 an<\/sup> = am-n<\/sup>
\n= 1818-14<\/sup>
\n\u2234 1818<\/sup> \u00f7 1814<\/sup> = 184<\/sup><\/p>\n

(iii) (xm<\/sup>)0<\/sup>
\nAnswer:
\n(xm<\/sup>)0<\/sup>
\nWe Know (am<\/sup>)n<\/sup> = am.n<\/sup>
\n= xm\u00d7n<\/sup> = x0<\/sup>(\u2235 a0<\/sup> = 1)
\n\u2234 (xm<\/sup>)0<\/sup> = 1<\/p>\n

(iv) (62<\/sup> \u00d7 64<\/sup>) \u00f7 63<\/sup>
\nAnswer:
\n(62<\/sup> X 64<\/sup>) \u00f7 63<\/sup>
\nWe Know am<\/sup> \u00d7 an<\/sup> = am+n<\/sup>
\n= (62+4<\/sup>) \u00f7 63<\/sup>
\n= 66<\/sup> \u00f7 63<\/sup>
\nWe Know am<\/sup> \u00f7 an<\/sup> = am-n<\/sup>
\n= 66-3<\/sup>
\n\u2234 (62<\/sup> \u00d7 64<\/sup>) \u00f7 63<\/sup> = 63<\/sup><\/p>\n

(v) \\(\\left(\\frac{2}{3}\\right)^{p}\\)
\nAnswer:
\n\\(\\left(\\frac{2}{3}\\right)^{p}\\)
\nWe Know \\(\\left(\\frac{a}{b}\\right)^{m}=\\frac{a^{m}}{b^{m}}=\\frac{2^{p}}{3^{p}}\\)
\n\u2234 \\(\\left(\\frac{2}{3}\\right)^{\\mathrm{p}}=\\frac{2^{\\mathrm{p}}}{3^{\\mathrm{p}}}\\)<\/p>\n

Question 4.
\nIdentify the greater number in each of the following andjustify your answer.
\n(i) 210<\/sup> or 102<\/sup>
\nAnswer:
\n210<\/sup> = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2
\n210<\/sup> = 1024
\n102<\/sup> = 10 \u00d7 10
\n102<\/sup> = 100
\n1024 > 100
\nSo, 210<\/sup> > 102<\/sup>
\n\u2234 210<\/sup> is greater.<\/p>\n

(ii) 54<\/sup> or 45<\/sup>
\nAnswer:
\n54<\/sup> = 5 \u00d7 5 \u00d7 5 \u00d7 5 = 625
\n45<\/sup> =4 \u00d7 4 \u00d7 4 \u00d7 4 \u00d7 4 = 1024
\n1024 > 625
\nSo, 45<\/sup> > 54<\/sup>
\n\u2234 45<\/sup> is greater number.<\/p>\n

Question 5.
\nIf \\(\\left(\\frac{4}{5}\\right)^{2} \\times\\left(\\frac{4}{5}\\right)^{5}=\\left(\\frac{4}{5}\\right)^{k}\\), then find the value of ‘k’
\nAnswer:
\nGiven \\(\\left(\\frac{4}{5}\\right)^{2} \\times\\left(\\frac{4}{5}\\right)^{5}=\\left(\\frac{4}{5}\\right)^{k}\\)
\nWe Know am<\/sup> \u00d7 an<\/sup> = am+n<\/sup>
\n\u21d2 \\(\\left(\\frac{4}{5}\\right)^{2+5}=\\left(\\frac{4}{5}\\right)^{\\mathrm{k}}\\)
\n\u21d2 \\(\\left(\\frac{4}{5}\\right)^{7}=\\left(\\frac{4}{5}\\right)^{\\mathrm{k}}\\)
\nIf the bases are equal powers should be equal
\n\u21d2 7 = k
\n\u2234 k = 7<\/p>\n

\"AP<\/p>\n

Question 6.
\nIf 52p+1<\/sup> \u00f7 52<\/sup> = 125, then find the value of ‘p’.
\nAnswer:
\nGiven 52p+1<\/sup> \u00f7 52<\/sup> = 125
\nWe Know am<\/sup> \u00f7 an<\/sup> = am-n<\/sup>
\n\u21d2 52p+1-2<\/sup> = 5 \u00d7 5 \u00d7 5
\n\u21d2 52p-1<\/sup> = 53<\/sup>
\nIf the bases are equal, powers should be equal.
\n\u21d2 2p – 1 =3
\n\u21d2 2p = 3 + 1
\n\u21d2 2p = 4
\n\u21d2 \\(\\frac{2 \\mathrm{p}}{2}=\\frac{4}{2}\\)
\n\u2234 p = 2<\/p>\n

Question 7.
\nProve that \\(\\left(\\frac{x^{b}}{x^{c}}\\right)^{a} \\times\\left(\\frac{x^{c}}{x^{a}}\\right)^{b} \\times\\left(\\frac{x^{a}}{x^{b}}\\right)^{c}\\) = 1
\nAnswer:
\nGiven \\(\\left(\\frac{x^{b}}{x^{c}}\\right)^{a} \\times\\left(\\frac{x^{c}}{x^{a}}\\right)^{b} \\times\\left(\\frac{x^{a}}{x^{b}}\\right)^{c}\\) = 1
\n\"AP<\/p>\n

Question 8.
\nExpress the following numbers in the expanded form.
\n(i) 20068
\nAnswer:
\n20068 = (2 \u00d7 10,000) + (0 \u00d7 1000) + (0 \u00d7 100) + (6 \u00d7 10) + (8 \u00d7 1)
\n\u2234 20068 = (2 \u00d7 104) + (6 \u00d7 101) + (8 \u00d7 1)<\/p>\n

(ii) 120718
\nAnswer:
\n120718 = (1 \u00d7 1,00,000) + (2 \u00d7 10,000) + (0 \u00d7 1000) + (7 \u00d7 100) + (1 \u00d7 10) + (8 \u00d7 1)
\n\u2234 120718 = (1 \u00d7 105) + (2 \u00d7 104) + (7 \u00d7 102) + (1 \u00d7 101) + (8 \u00d7 1)<\/p>\n

Question 9.
\nExpress the number appearing in the following statements in standard form :
\n(i) The Moon is 384467000 meters away from the Earth approximately.
\nAnswer:
\nDistance of Moon from the Earth = 384467000 metres
\n= 3.84467000 \u00d7 100000000<\/p>\n

Decimal is shifted eight places to the left.
\n= 3.84 4 67 \u00d7 108<\/sup> m<\/p>\n

Distance of Moon from the earth = 3.84 4 67 \u00d7 108<\/sup> m<\/p>\n

(ii) Mass of the Sun is 1 ,989,000,000,000,000,000,00000,000000 kg.
\nAnswer:
\nMass of Sun = 1 ,989,000,000,000,000,000,00000,000000 kg
\n= 1.989 \u00d7 1 ,000,000,000,000,000,00000,000000
\n= 1.989 \u00d7 1030<\/sup> kg
\n\u2234 Mass of Sun = 1.989 \u00d7 1030<\/sup> kg.<\/p>\n

Question 10.
\nLasya solved some problems of exponents and powers in the following way. Do you agree with the solution ? If not why? Justify your answer.
\nAnswer:
\n(i) x3<\/sup> \u00d7 x2<\/sup> = x6<\/sup>
\nAnswer:
\nNo. I won\u2019t agree with this solution.
\nGiven, x3<\/sup> \u00d7 x2<\/sup>
\nWe know am<\/sup> \u00d7 an<\/sup> = am+n<\/sup>
\n= a3+2<\/sup>
\n= x5<\/sup> which is \u2260 x6<\/sup>
\nso, x3<\/sup> \u00d7 x2<\/sup> \u2260 x6<\/sup>
\nWe have to add to powers. But, Lasya multiplied the powers. That\u2019s why Lasya\u2019s solution is wrong.<\/p>\n

(ii) (63<\/sup>)10<\/sup> = 613<\/sup>
\nAnswer:
\nNo, (63<\/sup>)10<\/sup> is not equal to 6.
\nWe know (am<\/sup>)n<\/sup> = amn<\/sup>
\n= 63\u00d710<\/sup>
\n= 630<\/sup> which is \u2260 613<\/sup>
\nso, (63<\/sup>)10<\/sup> \u2260 613<\/sup>
\nWe have to multiply. the powers. But, Lasya added the powers. That\u2019s why Lasya\u2019s solution is wrong.<\/p>\n

(iii) \\(\\frac{4 x^{6}}{2 x^{2}}\\) = 2x3<\/sup>
\nAnswer:
\nNo, I won\u2019t agree with this solution.
\n\\(\\frac{4 x^{6}}{2 x^{2}}=\\frac{2^{2}}{2^{1}} \\times \\frac{x^{6}}{x^{2}}\\)
\nWe know am<\/sup> \u00f7 an<\/sup> = am-n<\/sup>
\n= 22-1<\/sup> \u00d7 x6-2<\/sup>
\n= 21<\/sup>.x4<\/sup>
\n= 2x4<\/sup> which is \u2260 2x3<\/sup>
\nso, \\(\\frac{4 x^{6}}{2 x^{2}}\\) \u2260 2x3<\/sup>
\nWe have to subtract the powers. But, Lasya divided the powers. That\u2019s why Lasya\u2019s solution is wrong.<\/p>\n

(iv) \\(\\frac{3^{5}}{9^{5}}=\\frac{1}{3}\\)
\nAnswer:
\nNo. I won\u2019t agree with this solution.
\n\\(\\frac{3^{5}}{9^{5}}=\\frac{3^{5}}{\\left(3^{2}\\right)^{5}}\\)<\/p>\n

We Know (am<\/sup>)n<\/sup> = amn<\/sup>
\n\"AP
\nWe have to subtract the powers. But, Lasya divided the powers. That\u2019s why Lasya\u2019s solution is wrong.<\/p>\n

\"AP<\/p>\n

Question 11.
\nIs – 22<\/sup> is equal to 4? Justify your answer.
\nAnswer:
\n– 22<\/sup> = – (2 \u00d7 2) = – 4 \u2260 4
\n\u2234 – 22<\/sup> = – 4<\/p>\n

Question 12.
\nBeulah computed 25<\/sup> \u00d7 210<\/sup> = 250<\/sup>. Has she done it correctly? Give the reason.
\nAnswer:
\nGiven 25<\/sup> \u00d7 210<\/sup>
\nWe know am<\/sup> \u00d7 an<\/sup> = am+n<\/sup>
\n= 25+10<\/sup> = 215<\/sup> \u2260 250<\/sup>
\n\u2234 25<\/sup> \u00d7 210<\/sup> \u2260 250<\/sup>
\nBeulah did wrong.
\nHere we have to add the powers.
\nBut, he multiplied the powers.<\/p>\n

Question 13.
\nRafi computed \\(\\frac{3^{9}}{3^{3}}\\) as 33<\/sup>. Has he done
\nAnswer:
\nGiven \\(\\frac{3^{9}}{3^{3}}\\)
\nWe know \\(\\frac{a^{m}}{a^{n}}\\) = am-n<\/sup>
\n= 39-3<\/sup> = 36<\/sup> \u2260 33<\/sup>
\nSo, \\(\\frac{3^{9}}{3^{3}}\\) \u2260 33<\/sup>
\nRafi computed wrong.
\nHere we have to subtract the powers. But, he divided the powers.<\/p>\n

Question 14.
\nIs (a2<\/sup>)3<\/sup> equal to a8<\/sup>? Give the reason.
\nAnswer:
\nGiven (a2<\/sup>)3<\/sup> equal to a8<\/sup>
\nWe Know (am<\/sup>)n<\/sup> = amn<\/sup>
\n= (a2<\/sup>)3<\/sup> = a2\u00d73<\/sup> = a6<\/sup> \u2260 a8<\/sup>
\n\u2234 (a2<\/sup>)3<\/sup> = a6<\/sup><\/p>\n

We have to do 2 \u00d7 3 = 6, But not
\n23<\/sup> = 2 \u00d7 2 \u00d7 2 = 8<\/p>\n","protected":false},"excerpt":{"rendered":"

SCERT\u00a0AP 7th Class Maths Solutions Pdf Chapter 8 Exponents and Powers Unit Exercise Questions and Answers. AP State Syllabus 7th Class Maths Solutions 8th Lesson\u00a0Exponents and Powers Unit Exercise Question 1. Answer the following. (i) The exponential form 149 should read as Answer: 14 is raised to the power of 9. (ii) When base is …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[9],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/22303"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=22303"}],"version-history":[{"count":3,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/22303\/revisions"}],"predecessor-version":[{"id":22311,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/22303\/revisions\/22311"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=22303"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=22303"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=22303"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}