\nQuestion 1.
\nFill in the blanks:
\n(i) A number ending in ……… is never a perfect square.
\n(ii) On combining two consecutive triangular number, we get a ………
\n(iii) If a number has digits ……… in the unit’s place, then its square ends in 1.
\n(iv) Sum of first 10 odd natural numbers is ………
\n(v) Number of non-square numbers between 112<\/sup> and 122<\/sup> is ………
\n(vi) Number of zeros in the end of the square of 400 is ………
\n(vii) Square of any ……… number can be expressed as the sum of two consecutive natural numbers.
\n(viii)For a natural number m > 1, (2m, m2<\/sup> – 1, m2<\/sup> + 1) is called ………
\nSolution:
\n(i) A number ending in 2,3,7 and 8 is never a perfect square.
\n(ii) On combining two consecutive triangular number,
\nwe get a square number.
\n(iii) If a number has digits 1 or 9 in the unit’s place,
\nthen its square ends in 1.
\n(iv) Sum of first 10 odd natural numbers is (10)2<\/sup> = 100.
\n(v) Number of non-square numbers between
\n112<\/sup> and 122<\/sup> is (122<\/sup> – 112<\/sup>) – 1 = 12 + 11 – 1 =22.
\n(vi) Number of zeros in the end of the square of 400 is 0000 (four zeros).
\n(vii) Square of any odd number can be expressed as
\nthe sum of two consecutive natural numbers.
\n(viii) For a natural number m > 1,
\n(2m, m2<\/sup> – 1, m2<\/sup> + 1) is called Pythagorean triplet.<\/p>\n
Question 2. Multiple Choice Questions<\/strong> Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Question 12. Question 13. Question 14. Question 15. Value Based Questions<\/strong> Question 2. Higher Order Thinking Skills (Hots)<\/strong> Question 2.
\nState whether the following statements are true (T) or false (F):
\n(i) All natural numbers are not perfect squares.
\n(ii) A perfect square can never be expressed as the product of pairs of equal prime factors.
\n(iii) A number having 2,3,7 or 8 at its unit place is never a square number.
\n(iv) A number having 0, 1, 4, 5, 6 or 9 at its unit place is always a square number.
\n(v) A number ending in an even number of zeros is always a perfect square.
\n(vi) Square of an odd number is always an odd number.
\n(vii) 1, 3, 6, 10, 15, are called triangular numbers.
\n(viii)There are 2n non-square numbers between the squares of consecutive numbers n and (n + 1).
\n(ix) (4, 6, 8) is a Pythagorean triplet.
\nSolution:
\n(i) All natural numbers are not perfect squares. (True)
\n(ii) A perfect square can never be expressed as
\nthe product of pairs of equal prime factors. (True)
\n(iii) A number having 2, 3, 7 or 8 at its unit place is never a square number. (Time)
\n(iv) A number having 0, 1, 4, 5, 6 or 9 at its unit place is always a square number. (False)
\nCorrect:
\nAs 24, 14, 34, 26, 19, 50, 61, 35, etc are not a perfect square
\n(v) A number ending in an even number of zeros is always a perfect square. (False)
\nCorrect:
\nAs 200, 500, 8000, etc. are not perfect squares.
\n(vi) Square of an odd number is always an odd number. (True)
\n(vii) 1, 3, 6, 10, 15, are called triangular numbers. (True)
\n(viii) There are 2n non-square numbers between
\nthe squares of consecutive numbers n and (n + 1). (True)
\n(ix) (4, 6, 8) is a Pythagorean triplet. (False)
\nCorrect:
\nAs 42<\/sup> + 62<\/sup> \u2260 82<\/sup> \u21d2 16 + 36 \u2260 64
\n\u21d2 52 \u2260 64<\/p>\n
\nChoose the correct answer from the given four options (3 to 15):<\/strong>
\nQuestion 3.
\nHow many natural numbers lie between 252<\/sup> and 262<\/sup>?
\n(a) 49
\n(b) 50
\n(c) 51
\n(d) 52
\nSolution:
\nNatural numbers between 252<\/sup> and 262<\/sup>
\n25 + 26 – 1 = 50 (b)<\/p>\n
\nSquare of an even number is always
\n(a) even
\n(b) odd
\n(c) even or odd
\n(d) none of these
\nSolution:
\nSquare of an even number is always even. (a)<\/p>\n
\n1+ 3 + 5 + 7 + ……….. up to n terms is equal to
\n(a) n2<\/sup> – 1
\n(b) (n + 1)2<\/sup>
\n(c) n2<\/sup> + 1
\n(d) n2<\/sup>
\nSolution:
\n1 + 3 + 5 + 7 + …….. up to n terms is equal to\u00a0n2<\/sup> + 1\u00a0(c)<\/p>\n
\n\\(\\sqrt{208+\\sqrt{2304}}\\) is equal to
\n(a) 18
\n(b) 16
\n(c) 14
\n(d) 22
\nSolution:
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-3-Squares-and-Square-Roots-Objective-Type-Questions-Q6.1.png\")
<\/p>\n
\n\\(\\sqrt{0.0016}\\) is equal to
\n(a) 0.04
\n(b) 0.004
\n(c) 0.4
\n(d) none of these
\nSolution:
\n\\(\\sqrt{0.0016}\\) = 0.04
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-3-Squares-and-Square-Roots-Objective-Type-Questions-Q7.1.png\")
<\/p>\n
\nThe smallest number by which 75 should be divided to make it a perfect square is
\n(a) 1
\n(b) 2
\n(c) 3
\n(d) 4
\nSolution:
\n75 = 3 \u00d7 5 \u00d7 5
\nFactor 3 is unpaired
\n\u2234 By dividing 75 by 3, we get a perfect square of 5.<\/p>\n
\n\\(\\sqrt{3 \\frac{6}{25}}\\) is equal to
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-3-Squares-and-Square-Roots-Objective-Type-Questions-Q9.1.png\")
\nSolution:
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-3-Squares-and-Square-Roots-Objective-Type-Questions-Q9.2.png\")
<\/p>\n
\nThe smallest number by which 162 should be multiplied to make it a perfect square is
\n(a) 4
\n(b) 3
\n(c) 2
\n(d) 1
\nSolution:
\n162 = 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 3
\nFor 2 is left unpairs. So, by multiplying 162 by 2,
\nwe get a perfect square.
\n\u2234 Required least number to be multiplied = 2 (c)<\/p>\n
\nIf the area of a square field is 961 unit2<\/sup>, then the length of its side is
\n(a) 29 units
\n(b) 41 units
\n(c) 31 untis
\n(d) 39 units
\nSolution:
\nArea of a square = 961 unit2<\/sup>
\n\u2234 It’s side = \\(\\sqrt{961}\\) unit = 31 unit (c)
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-3-Squares-and-Square-Roots-Objective-Type-Questions-Q11.1.png\")
<\/p>\n
\nThe smallest number that should be subtracted from 300 to make it a perfect square is
\n(a) 11
\n(b) 12
\n(c) 13
\n(d) 14
\nSolution:
\n300
\nTaking the square root of 300,
\nwe see that 11 is left unpaired.
\n\u2234 11 be subtracted. (a)
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-3-Squares-and-Square-Roots-Objective-Type-Questions-Q12.1.png\")
<\/p>\n
\nIf one number of Pythagoream triplet is 6, then the triplet is
\n(a) (4, 5, 6)
\n(b) (5, 6, 7)
\n(c) (6, 7, 8)
\n(d) (6, 8, 10)
\nSolution:
\nOne number of a Pythagorean triplet is 6
\nLet 2n = 6 \u21d2 n = 3
\nn2<\/sup> – 1 = 32<\/sup> – 1 = 8 and n2<\/sup> + 1 = 32<\/sup> + 1 = 10
\n\u2234 Triplet is (6, 8, 10) (d)<\/p>\n
\nnth triangular number is
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-3-Squares-and-Square-Roots-Objective-Type-Questions-Q14.1.png\")
\nSolution:
\nnth triangular number = \\(\\frac{n(n+1)}{2}\\) (a)<\/p>\n
\nGiven that \\(\\sqrt{1521}\\) = 39, the value of \\(\\sqrt{0.1521}+\\sqrt{15.21}\\) is
\n(a) 42.9
\n(b) 4.29
\n(c) 3.51
\n(d) 35.1
\nSolution:
\n\\(\\sqrt{1521}\\) = 39, then value of \\(\\sqrt{0.1521}+\\sqrt{15.21}\\)
\n= 0.39 + 3.9 = 4.29 (b)<\/p>\n
\nQuestion 1.
\nIn a school, students of class VIII collected \u20b99216 to give a donation to an NGO working for the education of poor children. If each student donated as many rupees as the number of students in class VIII. Find the number of students in class VIII.
\nWhy should we donate money for the education of poor children? What values are being promoted?
\nSolution:
\nAmount collected by students = \u20b9 9216
\nand each donated amount equal to the number of students
\nNumber of students = \\(\\sqrt{9216}\\) = \u20b9 96
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-3-Squares-and-Square-Roots-Objective-Type-Questions-value-Q1.1.png\")
\nDonation to an NGO, who is working for the education
\nof poor children is a noble cause.<\/p>\n
\nA person wants to plant 2704 medicinal plants with a board depicting the diseases in which that can be used. He planted these in the form of rows. If each row contains as many plants as the number of rows, then find the number of rows.
\nWhy should we plant medicinal plants? What values are being promoted?
\nSolution:
\nTotal number of plants = 2704
\nThese are planted in such a way that
\nNumber of rows = number of plants in each row
\n\u2234 Number of rows = \\(\\sqrt{2704}\\)
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-3-Squares-and-Square-Roots-Objective-Type-Questions-value-Q2.1.png\")
\nMedicinal plants = 52
\nThese plants give us many kinds of medicines
\nwhich are useful for different diseases.
\nSo, it is a noble cause for society.<\/p>\n
\nQuestion 1.
\nA square field is to be ploughed. Ramu get it ploughed in \u20b934560 at the rate of \u20b915 per sq. m. Find the length of side of square field.
\nSolution:
\nTotal expenditure for ploughing the field = \u20b934560
\nThe rate of ploughing = \u20b915 per sq. m
\n\u2234 Total area of the square field = \\(\\frac{34560}{15}\\)
\n= 2304 m2<\/sup>
\nSide of the square field = \\(\\sqrt{2304}\\) m
\n= 48 m
\n![\"ML](\"https:\/\/cbselibrary.com\/wp-content\/uploads\/2019\/07\/ML-Aggarwal-Class-8-Solutions-for-ICSE-Maths-Chapter-3-Squares-and-Square-Roots-Objective-Type-Questions-hots-Q1.1.png\")
<\/p>\n
\nLalit has some chocolates. He distributed these chocolates among 13 children in such a way that he gave one chocolate to first child, 3 chocolates to second child, 5 chocolates to third and so on. Find the number of chocolates Lalit had.
\nSolution:
\nNumber of children = 13
\nLalit gave chocolates to the children in such a way that
\nhe gives one chocolate to first child, 3 chocolates to the second child
\n5 chocolates to a third child and so on
\n\u2234 1 + 3 + 5 + up to 13 terms
\n(\u2235 1 + 3 + 5 + 7 + ……… n terms = n2<\/sup>)
\n= 169 chocolates<\/p>\nML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"