{"id":27440,"date":"2020-07-01T11:03:01","date_gmt":"2020-07-01T05:33:01","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=27440"},"modified":"2021-07-02T15:07:09","modified_gmt":"2021-07-02T09:37:09","slug":"kseeb-solutions-for-class-8-maths-chapter-5-additional-questions","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-8-maths-chapter-5-additional-questions\/","title":{"rendered":"KSEEB Solutions for Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions"},"content":{"rendered":"

Students can Download Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka State Syllabus Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions<\/h2>\n

Question 1.
\nMatch the following numbers \u00a1n the column A with their squares \u00a1n the column B.
\n\"KSEEB
\nSolution:
\n(1) (a) 25
\n(2) (e) 64
\n(3) (f) 4
\n(4) (c) 36
\n(5) (d) 484
\n(6) (b) 144<\/p>\n

Question 2.
\nChoose the correct option:
\n(a) The number of perfect squares from 1 to 500 is
\nA. 1
\nB. 16
\nC. 22
\nD. 25
\nSolution:
\nC. 22<\/p>\n

\"KSEEB<\/p>\n

(b) The last digit of a perfect square can never be
\nA. 1
\nB. 3
\nC. 5
\nD. 9
\nSolution:
\nB. 3<\/p>\n

(c) If a number ends in 5 zeros, its square ends in
\nA. 5 zeros
\nB. 8 zeros
\nC. lo zeros
\nD. 12 zeros
\nSolution:
\nC. 10 zeros<\/p>\n

(d) Which could be the remainder among the following when a perfect square is devided by 8?
\nA. 1
\nB. 3
\nC. 5
\nD. 7
\nSolution:
\nA. 1<\/p>\n

(e) The 6th triangular number is
\nA. 6
\nB. 10
\nC. 21
\nD. 28
\nSolution:
\nC. 21<\/p>\n

\"KSEEB<\/p>\n

Question 3.
\nConsider all integers froni – 10 to 5 and square each of them. How many distinct numbers you get?
\nSolution:
\n(-10)2<\/sup> = 100, (-9)2<\/sup> = 81, (-8)2<\/sup>
\n= 64,(-7)2<\/sup> = 49
\n(-6)2<\/sup> = 36, (5)2<\/sup> = 25, (-4)2<\/sup> 16, (-3)2<\/sup> =9,
\n(2)2<\/sup> =4, (-1)2<\/sup> = 1, 02<\/sup> = 0, 12<\/sup> = 1, 22<\/sup> = 4, 32<\/sup> = 9, 42<\/sup> = 16, 52<\/sup> = 25
\nThere are 11 distinct numbers.<\/p>\n

Question 4.
\nWrite the digit in units place when the following numbers are squared.
\nSolution:
\n\"KSEEB<\/p>\n

Question 5.
\nWrite all the numbers from 400 to 425 which end in 2, 3, 7 or 8. Check if any of these in a perfect square.
\nSolution:
\n402, 403, 407, 408, 412, 413, 417, 418, 422, 423
\nNone of these is a perfect square.<\/p>\n

Question 6.
\nFind the sum of the digits of (1111111111)2<\/sup>.
\nSolution:
\nObserve the pattern
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nSuppose x2<\/sup> + y2<\/sup> = Z2<\/sup>
\n(i) If x = 4 and y = 3 find z
\n(ii) If x = 5 and z = 13 find y
\n(iii) If y = 15 and z = 17 find x.
\nSolution:
\n(i) x = 4, y = 3
\nx2<\/sup> + y2<\/sup> = z2
\n42<\/sup> + 32<\/sup> = z2<\/sup>
\n16 + 9 = z2<\/sup>
\n25 = z2<\/sup>
\n\u2234 z = \u221a25 = 5<\/p>\n

(ii) x = 5, z = 13
\nx2<\/sup> + y2<\/sup> = z2<\/sup>
\n52<\/sup> + y2<\/sup> = 132<\/sup>
\n25+ y2<\/sup> =169
\ny2<\/sup> =169 – 25
\ny2<\/sup> = 144
\ny = \u221a144 = 12<\/p>\n

(iii) y = 15, z = 17
\nx2<\/sup>+ y2<\/sup> = z2<\/sup>
\nx2<\/sup> + 152<\/sup> = 172<\/sup>
\nx2<\/sup> + 225 = 289
\nx2<\/sup> = 289 – 225
\nx2<\/sup> = 64
\nx = \u221a 64 = 8<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nA sum of Rs. 2304 is equally distributed among several people. Each gets as many rupees as the number of persons. How much does each one get?
\nSolution:
\n\"KSEEB
\nLet the number of persons be x. Each person get rupees x.
\n\u2234 x \u00d7 x = 2304
\nx2<\/sup> = 2304
\nx = \u221a2304
\n2304 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3
\n= 16 \u00d7 3 \u00d7 16 \u00d7 3
\n2304 = 48 \u00d7 48
\n\u221a2304 = 48 Each person gets Rs. 48<\/p>\n

Question 9.
\nDefine a new addition e on (be set of all natural numbers by m * n = m2<\/sup> + n2<\/sup>
\n(i) Is N closed under e?
\n(ii) Is * commutative on N?
\n(iii) Is * associative on N?
\n(iv) Is there an identity element in N with respect * ?
\nSolution:
\n(i) Let m 2 and n 5
\nm * n = m2<\/sup> + n2<\/sup>
\n2 * 5 = 22<\/sup> + 52<\/sup> = 4 + 25 = 29 \u2208 N
\n\u2200 m, n \u2208 N, m2<\/sup> + n2<\/sup> \u2208 N
\n\u2234 * is closed under N.<\/p>\n

(ii) Let m = 4 and n = 7
\nm * n = m2<\/sup> + n2<\/sup> = 42<\/sup> + 72<\/sup> = 16 + 49 = 65
\nn * m = n2<\/sup> + m2<\/sup> = 72<\/sup> + 42<\/sup> = 49 + 16 = 65
\n\u2234 m * n = n * m \u2200 m, n \u2208 N
\nHence * is commutative on N.<\/p>\n

\"KSEEB<\/p>\n

(iii) Lei m = 2, n = 3 and p = 4.
\nm * (n * P) = 2 * (3*4)
\n= 2*(32<\/sup> + 42<\/sup>) = 2*(9 + 16)
\n= 2*25 = 22<\/sup> + 252<\/sup> = 4 + 625
\n= 629 ….(i)
\n(m * n) * p = (2 * 3)*4
\n=(22<\/sup> + 32<\/sup>) * 4 = (4 * 9) * 4
\n= 13 * 4 = 132<\/sup> + 42<\/sup>.
\n= 169 + 16 = 185 …(ii)
\nFrom (i) and (ii) m*(n*p) \u2260 (m*n)*p
\n\u2234 * is not associative on N.<\/p>\n

(iv) Let K be the identity element then m * k = m, m2<\/sup> + k2<\/sup> = m2<\/sup> which means
\nk2<\/sup> = m2<\/sup> – m2<\/sup> = O
\nK = 0, O does not belong to N.
\nThere is no identity element in N with respect to *.<\/p>\n

\"KSEEB<\/p>\n

Question 10.
\n(Exploration) Find all the perfect squares from 1 to 500 each of which is a sum of two perfect squares.
\nSolution:
\nPerfect squares from 1 to 500 are 1, 4, 9, 1, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289. 324, 361, 400, 441, 484.
\n25 = 9 + 16
\n100 =36 + 64
\n169 = 25 + 144
\n289 = 225 + 64<\/p>\n

Question 11.
\nSupposes the area of a square field is 7396 find its perimeter.
\nSolution:
\nLet each side of the square be
\nl2<\/sup> = 7396 = 2 \u00d7 2 \u00d7 43 \u00d7 43 = 2 \u00d7 43 \u00d7 2 \u00d7 43
\nl2<\/sup> =7396 = 86 \u00d7 86.<\/p>\n

\"KSEEB
\n\u2234 \u221a7396 = 86 = l
\nPerimeter = 4l = 4 \u00d7 86 = 344 m.<\/p>\n

Question 12.
\nCan 1010 be written as a difference of two perfect squares?
\nSolution:
\nIf a2<\/sup> – b2<\/sup> = 1010 for any two integers a and b then either both \u2018a\u2019 and \u2018b\u2019 are odd or both even. Hence a2<\/sup> – b2<\/sup> is divisible by 4. But 1010 is not divisible by 4. Hence 1010 is not the difference of two perfect squares.<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nWhat are the remainders when a perfect cube is divided by 7?
\nSolution:
\nWhen the perfect cubes 8, 27, 64, 125, 216, 343, 512, 729, 1000 are divided by 7 the remainders respectively are 1, 6, 1, 6, 6, 0, 1, 1, 6. Hence the remainders are 0. 1 or 6.<\/p>\n

Question 14.
\nWhat is the least perfect square which leaves the remainder 1 when divided by 7 as welt as by 11?
\nSolution:
\nThe least number divisible by both 7 and 11 is 7 \u00d7 11 = 77. When 1 is added to 77 we get 78. But 78 is not a perfect square. (77 \u00d7 2) + 1 = 154 + 1 = 155 is not a perfect square. In the same way, continue we find that
\n(77 \u00d7 15) + 1 = 1155 + 1 = 1156 in a perfect square.
\n\u2234 The number required is 1156 = 342<\/p>\n

Question 15.
\nFind two smallest perfect squares whose product is a perfect cube.
\nSolution:
\n4 and 16 are perfect squares. 4 \u00d7 16 = 64 is a perfect cube.<\/p>\n

Question 16.
\nFind a proper positive factor of 48 and a proper positive multiple of 48 which add up to a perfect square. Can you prove that there are infinitely many such pairs? By considering 16 as a factor of 48 and 48 as a factor of 240, we can write 16 + 240 = 162<\/sup>.
\nSolution:
\nMultiple of 48 is 48 and considering it as 48l where l = m (3m + 2), m – 1, 2, 3 …. we can get infinite numbers.<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka State Syllabus Class 8 Maths Chapter 5 Squares, Square Roots, Cubes, Cube Roots Additional …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27440"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=27440"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/27440\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=27440"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=27440"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=27440"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}