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Exercise 6.3, Question 1:, What could be the possible ‘one’s’ digits of the square root of each of the following, numbers:, , (i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025, €. Answer 1:, Since, Unit's digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible, unit's digits of the given numbers are:, , (i)1 (ii) 6 (iii) 1 (iv) 5, Question 2:, Without doing any calculation, find the numbers which are surely not perfect squares:, (i) 153 (ii) 257 (iii) 408 (iv) 441, €. Answer 2:, , Since, all perfect square numbers contain their unit's place digits 0, 1, 4, 5, 6 and 9., (i) But given number 153 has its unit digit 3. So itis nota perfect square number., (ii) Given number 257 has its unit digit 7. So it is not a perfect square number., (iii) | Given number 408 has its unit digit 8. So it is not a perfect square number., (iv) Given number 441 has its unit digit 1. So it would be a perfect square number, , Question 3:, , Find the square roots of 100 and 169 by the method of repeated subtraction., €. Answer 3:, , By successive subtracting odd natural numbers from 100,, , 100-1=99 99-3=96 96-5=91 91-7=84, 84-9=75 75-11=64 64-13=51 51-15=36, 36-17=19 19-19=0, , This successive subtraction is completed in 10 steps. Therefore V100 =10, By successive subtracting odd natural numbers from 169,, , 169 - 1= 168 168 -3=165 165 -5=160 160 -7=153, 153-9=144 144-11=133 133 -13=120 120-15=105, 105-17=88 88-19 =69 69-21=48 48 -23=25, 25-25=0, , This successive subtraction is completed in 13 steps. Therefore Vi69 =13, , Question 4:, , Find the square roots of the following numbers by the Prime Factorization method:, , (i) 729 (ii) 400 (iii) 1764 (iv) 4096 (v) 7744, , (vi) 9604 (vii) 5929 (viii) 9216 (ix)529 —(x) 8100, €. Answer 4:, , 3 729, , 10) 729 3 243, 729 = \3x3x3x3x3x3 3 81, =3x3x3 3 27, , =27 3 9, , 3 3

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(Class - VIII), , (ix) 529 23 | 529, 529 = 23x23 23 | 23, , =23 1, , 2 8100, , 2 4050, , (x) 8100 3 2025, 8100 = J2x2x3x3x3x3x5x5 3 675, =2x3x3x5 3 225, , =90 3 75, , 5 25, , 5 5, , Question 5: 1, , For each of the following numbers, find the smallest whole number by which it should be, multiplied so as to get a perfect square number. Also, find the square root of the square, number so obtained:, , (i 252 (ii) 180, (iii) 1008 (iv) 2028, (v) 1458 (vi) 768, €. Answer 5:, (i) 252=2x2x3x3x7, Here, prime factor 7 has no pair. Therefore 252 must be 2 252, multiplied by 7 to make it a perfect square. 2 126, And 1764 =2x3x7=42 3 21, 7 7, 1, (ii) 180 =2x2x3x3x5, Here, prime factor 5 has no pair. Therefore 180 must be 2 180, multiplied by 5 to make it a perfect square. 2 90, “ 180x5=900 3 45, And 900 =2x3x5=30 3 15, 5 | 5, 1, 2 1008, (iii) 1008 =2x2x2x2x3x3x7 2 504, Here, prime factor 7 has no pair. Therefore 1008 must be 2 252, multiplied by 7 to make it a perfect square. 2 126, i x7 = 7056 3 63, And V1056 =2x2x3x7=84 3 21, tu 7

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(iv) 2028=2x2x3x13x13, , Here, prime factor 3 has no pair. Therefore 2028 mustbe 2 2028, multiplied by 3 to make it a perfect square. 2 1014, , i 2028 x 3 = 6084 3. 507, And 6084 =2x2x3x3x13x13=78 13-169, 13 13, , 1, , (v) 1458 =2x3x3x3x3x3x3 2, Here, prime factor 2 has no pair. Therefore 1458 must be 3, multiplied by 2 to make it a perfect square. 3, , - 1458x2=2916 3 81, And 2916 =2x3x3x3=54 3, 3, 3, , , , 27, , 9, , 3, , i, , 2 768, , (vi) 768 =2x2x2x2x2x2x2x2x3 z t 33, , Here, prime factor 3 has no pair. Therefore 768 must be 21 9%, , multiplied by 3 to make it a perfect square. 2 a, 768 x 3 = 2304, , And J2304 =2x2x2x2x3=48 2 4, , 2 6, , 3 3, , Question 6: 1, , For 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:, , (i) 252 (ii) 2925, (iii) 396 (iv) 2645, (v) 2800 (vi) 1620, €. Answer 6: 2 | 252, {fi 252=2x2x3x3x7, Here, prime factor 7 has no pair. Therefore 252 must be 2 126, divided by 7 to make it a perfect square. 3 63, “© 252 + 7=36 3 2, And 36 =2x3=6 TIES :, (ii) 2925=3x3x5x5x13 3 975, Here, prime factor 13 has no pair. 5 325, Therefore 2925 must be divided by 13 5 65, to make it a perfect square. 13 13, 2925 + 13 =225 1, , fal 225 =3x5=15

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(iii) 396=2x2x3x3x1l1, , Here, prime factor 11 has no pair. Therefore 396 must be 2 396, divided by 11 to make it a perfect square. 2 198, oe 396 + 11=36 3 99, And 36 =2x3=6 3 33, 11 11, i, , (iv) 2645=5x23x23 5 2645—, Here, prime factor 5 has no pair. Therefore 2645 must be 23 529, divided by 5 to make it a perfect square. 23. 23, is 2645 + 5=529 1, , And 529 = 23x 23=23, , 2 2800, I) 2800 = 2x2x2x2x5x5x7 2 1400, Here, prime factor 7 has no pair. Therefore 2800 must be — 2 700, divided by 7 to make it a perfect square. 2 350, i 2800 + 7=400 5 175, And 400 =2x2x5=20 5 35, 7 7, 1, 2 1620, (vi) 1620=2x2x3x3x3x3x5 2810, Here, prime factor 5 has no pair. Therefore 1620 must be 3 405, divided by 5 to make it a perfect square. 3 135, 1620 + 5=324 3 45, And 324 =2x3x3=18 2 )__ts, 5 5, 1, , Question 7:, The students of Class VIII of a school donated %2401 in all, for Prime Minister's 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., €. Answer 7:, , Here, Donated money = % 2401, , Let the number of students be x. aE!, Therefore donated money = xxx 7 | 343, According to question, 7 49, , ? = 2401 7 d, , = x= 240 = f7x7x7x7, , > x=7x7=49, , Hence, the number of students is 49.