{"id":27286,"date":"2021-06-30T12:04:50","date_gmt":"2021-06-30T06:34:50","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=27286"},"modified":"2022-03-02T10:31:30","modified_gmt":"2022-03-02T05:01:30","slug":"ncert-solutions-for-class-8-maths-chapter-6-ex-6-3","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/ncert-solutions-for-class-8-maths-chapter-6-ex-6-3\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.3"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 6 Square and Square Roots Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.3<\/h2>\n

Question 1.
\nWhat could be the possible \u2018ones\u2019 digits of the square root of each of the following numbers?
\n(i) 9801
\n(ii) 99856
\n(iii) 998001
\n(iv) 657666025
\nSolution:
\n(i) The unit\u2019s digit of the square root of the number 9801 could be 1 or 9 [1 \u00d7 1 = 1, 9 \u00d7 9 = 81]
\n(ii) The units digit of the square root of the number 99856 could be 4 or 6 [4 \u00d7 4 = 16 or 6 \u00d7 6 = 36]
\n(iii) The unit\u2019s digit of the square root of the number 998001 could be 1 or 9 [1 \u00d7 1 = 1 or 9 \u00d7 9 = 81]
\n(iv) The units digit of the square root of the number 657666025 could be 5 [5 \u00d7 5 = 25]<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nWithout doing any calculation, find the numbers which are surely not perfect squares.
\n(i) 153
\n(ii) 257
\n(iii) 408
\n(iv) 441
\nSolution:
\nWe know that the ending digit of a perfect square is 0, 1, 4, 5, 6 and 9
\nA number ending in 2, 3, 7, or 8 can never be a perfect square
\n(i) 153 cannot be a perfect square.
\n(ii) 257 cannot be a perfect square.
\n(iii) 408 cannot be a perfect square.
\n(iv) 441 can be a perfect square.<\/p>\n

Question 3.
\nFind the square roots of 100 and 169 by the method of repeated subtraction.
\nSolution:
\n(i) \u221a100
\n100 – 1 = 99
\n99 – 3 = 96
\n96 – 5 = 91
\n91 – 7 = 84
\n84 – 9 = 75
\n75 – 11 = 64
\n64 – 13 = 51
\n51 – 15 = 36
\n36 – 17 = 19
\n19 – 19 = 0
\n\u2234 We reach at 0 by successive subtraction of 10 odd numbers
\n\u2234 \u221a100 = 10<\/p>\n

\"NCERT<\/p>\n

(ii) \u221a169
\n169 – 1 = 168
\n168 – 3 = 165
\n165 – 5 = 160
\n160 – 7 = 153
\n153 – 9 = 144
\n144 – 11 = 133
\n133 – 13 = 120
\n120 – 15 = 105
\n105 – 17 = 88
\n88 – 19 = 69
\n69 – 21 = 48
\n48 – 23 = 25
\n25 – 25 = 0
\n\u2234 We reach at 0 by successive subtraction of 13 odd numbers
\n\u2234 \u221a169 = 13<\/p>\n

Question 4.
\nFind the square roots of the following numbers by the Prime Factorisation Method.
\n(i) 729
\n(ii) 400
\n(iii) 1764
\n(iv) 4096
\n(v) 7744
\n(vi) 9604
\n(vii) 5929
\n(viii) 9216
\n(ix) 529
\n(x) 8100
\nSolution:
\n(i) \u221a729
\n= \\(\\sqrt{3^{2} \\times 3^{2} \\times 3^{2}}\\)
\n= 3 \u00d7 3 \u00d7 3
\n= 27
\n\u2234 \u221a729 = 27
\n\"NCERT<\/p>\n

(ii) \u221a400
\n= \\(\\sqrt{2^{4} \\times 5^{2}}\\)
\n= 22<\/sup> \u00d7 5
\n= 20
\n\u2234 \u221a400 = 20
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

(iii) \u221a1764
\n= \\(\\sqrt{2^{2} \\times 3^{2} \\times 7^{2}}\\)
\n= 2 \u00d7 3 \u00d7 7
\n= 42
\n\u2234 \u221a1764 = 42
\n\"NCERT<\/p>\n

(iv) \u221a4096
\n= \\(\\sqrt{2^{12}}\\)
\n= 26<\/sup>
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2
\n= 64
\n\u2234 \u221a4096 = 64
\n\"NCERT<\/p>\n

(v) \u221a7744
\n= \\(\\sqrt{2^{6} \\times 11^{2}}\\)
\n= 23<\/sup> \u00d7 11
\n= 88
\n\u2234 \u221a7744 = 88
\n\"NCERT<\/p>\n

(vi) \u221a9604
\n= \\(\\sqrt{2^{2} \\times 7^{4}}\\)
\n= 2 \u00d7 72<\/sup>
\n= 2 \u00d7 49
\n= 98
\n\u2234 \u221a9604 = 98
\n\"NCERT<\/p>\n

(vii) \u221a5929
\n= \\(\\sqrt{7^{2} \\times 11^{2}}\\)
\n= 7 \u00d7 11
\n= 77
\n\u2234 \u221a5929 = 77
\n\"NCERT<\/p>\n

(viii) \u221a9216
\n= \\(\\sqrt{2^{5} \\times 3^{2}}\\)
\n= 25<\/sup> \u00d7 3
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3
\n= 32 \u00d7 3
\n= 96
\n\u2234 \u221a9216 = 96
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

(ix) \u221a529
\n= \\(\\sqrt{23^{2}}\\)
\n= 23
\n\u2234 \u221a529 = 23
\n\"NCERT<\/p>\n

(x) \u221a8100
\n= \\(\\sqrt{2^{2} \\times 3^{4} \\times 5^{2}}\\)
\n= 2 \u00d7 32<\/sup> \u00d7 5
\n= 2 \u00d7 9 \u00d7 5
\n= 90
\n\u2234 \u221a8100 = 90
\n\"NCERT<\/p>\n

Question 5.
\nFor each of the following numbers, find the smallest whole number by which it should be multiple so as to get a perfect square number. Also, find the square root of the square number so obtained.
\n(i) 252
\n(ii) 180
\n(iii) 1008
\n(iv) 2028
\n(v) 1458
\n(vi) 768
\nSolution:
\n(i) 252 = 22<\/sup> \u00d7 32<\/sup> \u00d7 7
\nThe prime factor 7 has no pair
\n\u2234 The number 252 should be multiply by 7 to get a perfect square number 252 \u00d7 7 = 1764
\n\u221a1764
\n= \\(\\sqrt{2^{2} \\times 3^{2} \\times 7^{2}}\\)
\n= 2 \u00d7 3 \u00d7 7
\n= 42
\n\"NCERT<\/p>\n

(ii) 180 = 22<\/sup> \u00d7 32<\/sup> \u00d7 5
\nThe prime factor 5 has no pair
\n\u2234 The number 180 should be multiplied by 5 to get a perfect square number 180 \u00d7 5 = 900
\n\u221a900
\n= \\(\\sqrt{2^{2} \\times 3^{2} \\times 5^{2}}\\)
\n= 2 \u00d7 3 \u00d7 5
\n= 30
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

(iii) 1008 = 24<\/sup> \u00d7 32<\/sup> \u00d7 7
\nThe prime factor 7 has no pair
\n\u2234 The number 1008 should be multiplied by 7 to get a perfect square number 1008 \u00d7 7 = 7056
\n\u221a7056
\n= \\(\\sqrt{2^{4} \\times 3^{2} \\times 7^{2}}\\)
\n= 4 \u00d7 3 \u00d7 7
\n= 84
\n\"NCERT<\/p>\n

(iv) 2028 = 22<\/sup> \u00d7 33<\/sup> \u00d7 132<\/sup>
\nThe Prime factor 3 has no pair
\n\u2234 The number 2028 should be multiplied by 3 to get a perfect square number 2028 \u00d7 3 = 6084
\n\u221a6084
\n= \\(\\sqrt{2^{2} \\times 3^{2} \\times 13^{2}}\\)
\n= 2 \u00d7 3 \u00d7 13
\n= 78
\n\"NCERT<\/p>\n

(v) 1458 = 2 \u00d7 36<\/sup>
\nThe prime factor 2 has no pair
\n\u2234 The number 1458 should be multiplied by 2 to get perfect square number 1458 \u00d7 2 = 2916
\n\u221a2916
\n= \\(\\sqrt{2^{2} \\times 3^{6}}\\)
\n= 2 \u00d7 33<\/sup>
\n= 2 \u00d7 3 \u00d7 3 \u00d7 3
\n= 54
\n\"NCERT<\/p>\n

(vi) 768 = 28<\/sup> \u00d7 3
\n\u2234 The prime factor 3 has no pair
\n\u2234 The number 768 should be multiply be 3 to make it a perfect square.
\n768 \u00d7 3 = 2304
\n\u221a2304
\n= \\(\\sqrt{2^{8} \\times 3^{2}}\\)
\n= 24<\/sup> \u00d7 3
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3
\n= 48
\n\"NCERT<\/p>\n

Question 6.
\nFor each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also, find the square root of the square number so obtained
\n(i) 252
\n(ii) 2925
\n(iii) 396
\n(iv) 2645
\n(v) 2800
\n(vi) 1620
\nSolution:
\n(i) 252 = 22<\/sup> \u00d7 32<\/sup> \u00d7 7
\nThe prime factor 7 has no pair
\n\u2234 The given number should be divided by 7 to get a perfect square
\n\\(\\frac{252}{7}=\\frac{2^{2} \\times 3^{2} \\times 7}{7}\\) = 36
\n\u221a36
\n= \\(\\sqrt{2^{2} \\times 3^{2}}\\)
\n= 2 \u00d7 3
\n= 6
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

(ii) 2925 = 32<\/sup> \u00d7 52<\/sup> \u00d7 13
\nThe prime factor 13 has no pair
\n\u2234 The given number should be divided by 13 to get a perfect square
\n\\(\\frac{2925}{13}=\\frac{3^{2} \\times 5^{2} \\times 13}{13}\\)
\n= 9 \u00d7 25
\n= 225
\n\u221a225
\n= \\(\\sqrt{3^{2} \\times 5^{2}}\\)
\n= 3 \u00d7 5
\n= 15
\n\"NCERT<\/p>\n

(iii) 396 = 22<\/sup> \u00d7 32<\/sup> \u00d7 11
\nThe prime factor 11 has no pair.
\n\u2234 The given no. should be divided by 11 to get a perfect square
\n\\(\\frac{396}{11}=\\frac{2^{2} \\times 3^{2} \\times 11}{11}\\)
\n= 4 \u00d7 9
\n= 36
\n\u221a36
\n= \\(\\sqrt{2^{2} \\times 3^{2}}\\)
\n= 2 \u00d7 3
\n= 6
\n\"NCERT<\/p>\n

(iv) 2645 = 5 \u00d7 232<\/sup>
\nThe prime factor 5 has no pair.
\n\u2234 The given number should be divided by 5 to get a perfect square
\n\\(\\frac{2645}{5}=\\frac{5 \\times 23^{2}}{5}\\) = 232<\/sup> = 529
\n\u221a529 = \\(\\sqrt{23^{2}}\\) = 23
\n\"NCERT<\/p>\n

(v) 2800 = 24<\/sup> \u00d7 52<\/sup> \u00d7 7
\nThe prime factor 7 has no pair
\n\u2234 The given number should be divided by 7 to get a perfect square
\n\\(\\frac{2800}{7}=\\frac{2^{4} \\times 5^{2} \\times 7}{7}\\)
\n= 24<\/sup> \u00d7 52<\/sup>
\n= 16 \u00d7 25
\n= 400
\n\u221a400 = 20
\n\"NCERT<\/p>\n

(vi) 1620 = 22<\/sup> \u00d7 34<\/sup> \u00d7 5
\nThe prime factor 5 has no pair.
\nThe given number should be divided by 5 to get a perfect square
\n\\(\\frac{1620}{5}=\\frac{2^{2} \\times 3^{4} \\times 5}{5}\\)
\n= 22<\/sup> \u00d7 34<\/sup>
\n= 4 \u00d7 81
\n= 324
\n\u221a324
\n= \\(\\sqrt{2^{2} \\times 3^{4}}\\)
\n= 2 \u00d7 32<\/sup>
\n= 2 \u00d7 9
\n= 18
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nThe students of class VIII of a school donated \u20b9 2401 in all, for Prime Minister National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.
\nSolution:
\nLet the number of students be \u2018x\u2019
\nAmount donated by each student = x
\nTotal amount donated by the class = \u20b9 x \u00d7 x = \u20b9 x2<\/sup>
\nGiven, x2<\/sup> = 2401
\nx = \u221a2401
\n= \\(\\sqrt{7^{2} \\times 7^{2}}\\)
\n= 7 \u00d7 7
\n= 49
\nNumber of students in the class = 49.
\n\"NCERT<\/p>\n

Question 8.
\n2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.
\nSolution:
\nLet the number of rows be x\u2019
\nNumber of plants in a row = x
\nNumber of plants to be planted = x \u00d7 x = x2<\/sup>
\nGiven, x2<\/sup> = 2025
\nx = \u221a2025
\n= \\(\\sqrt{3^{4} \\times 5^{2}}\\)
\n= 32 \u00d7 5
\n= 9 \u00d7 5
\n= 45
\n\u2234 The required number of rows = 45
\nAlso, no. of plants in a row = 45
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nFind the smallest square number that is divisible by each of the numbers 4, 9, and 10.
\nSolution:
\nThe least number divisible by each one of 4, 9, and 10 is their L.C.M.
\n\"NCERT
\n\u2234 L.C.M = 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 = 180
\nPrime factors of 180 = 22<\/sup> \u00d7 32<\/sup> \u00d7 5
\nThe Prime factor 5 is not in pair
\n180 is not a perfect square
\nIn order to get a perfect square 180 should be multiplied by 5
\nThe required square number = 180 \u00d7 5 = 900<\/p>\n

Question 10.
\nFind the smallest square number that is divisible by each of the numbers 8, 15, and 20.
\nSolution:
\nThe least number divisible by each one of 8, 15, and 20 is their L.C.M.
\n\"NCERT
\n\u2234 L.C.M = 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 5 = 120
\nThe prime factors 2, 3, and 5 are not in pairs.
\n\u2234 120 is not a perfect square.
\nTo get a perfect square 120 should be multiplied by 2, 3, and 5 i.e. 30.
\n\u2234 The required smallest square number = 120 \u00d7 30 = 3600<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.3 Question 1. What could be the possible \u2018ones\u2019 digits of the square root of each …<\/p>\n

NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.3<\/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":[7],"tags":[],"yoast_head":"\nNCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.3 - 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-8-maths-chapter-6-ex-6-3\/\" \/>\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 8 Maths Chapter 6 Square and Square Roots Ex 6.3 - MCQ Questions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Ex 6.3 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots Exercise 6.3 Question 1. 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