{"id":27426,"date":"2023-02-28T15:40:50","date_gmt":"2023-02-28T10:10:50","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=27426"},"modified":"2023-02-28T16:55:40","modified_gmt":"2023-02-28T11:25:40","slug":"ncert-solutions-for-class-8-maths-chapter-6-intext-questions","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-8-maths-chapter-6-intext-questions\/","title":{"rendered":"NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots InText Questions"},"content":{"rendered":"

These NCERT Solutions for Class 8 Maths<\/a> Chapter 6 Square and Square Roots InText 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 InText Questions<\/h2>\n

Try These (Page No. 90)<\/span><\/p>\n

Question 1.
\nFind the perfect square numbers between
\n(i) 30 and 40 and
\n(ii) 50 and 60.
\nSolution:
\n(i) Since,
\n1 \u00d7 1 = 1
\n2 \u00d7 2 = 4
\n3 \u00d7 3 = 9
\n4 \u00d7 4 = 16
\n5 \u00d7 5 = 25
\n6 \u00d7 6 = 4
\n7 \u00d7 7 = 49
\nThus 36 is a perfect square number between 30 and 40.<\/p>\n

(ii) Since, 7 \u00d7 7 = 49 and 8 \u00d7 8 = 64.
\nIt means there is no perfect number between 49 and 64 and thus there is no perfect number between 50 and 60.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nCan we say whether the following numbers are perfect squares? How do we know?
\n(i) 1057
\n(ii) 23453
\n(iii) 7928
\n(iv) 222222
\n(v) 1069
\n(vi) 2061
\nWrite five numbers that you can decide by looking at their ones digit that they are not square numbers.
\nSolution:
\n(i) 1057
\n\u2235 The ending digit is 7 (which is not one of 0, 1, 4, 5, 6, or 9).
\n\u2234 1057 cannot be a square number.<\/p>\n

(ii) 23453
\n\u2235 The ending digit is 3 (which is not one of 0, 1, 4, 5, 6, or 9).
\n\u2234 23453 cannot be a square number.<\/p>\n

(iii) 7928
\n\u2235 The ending digit is 8 (which is not one of 0, 1, 4, 5, 6, or 9).
\n\u2234 7928 cannot be a square number.<\/p>\n

(iv) 222222
\n\u2235 The ending digit is 2 (which is not one of 0, 1, 4, 5, 6, or 9).
\n\u2234 222222 cannot be a square number.<\/p>\n

(v) 1069
\n\u2235 The ending digit is 9
\n\u2234 It may or may not be a square number.
\nAlso, 30 \u00d7 30 = 900
\n31 \u00d7 31 = 961
\n32 \u00d7 32 = 1024
\n33 \u00d7 33 = 1089
\ni.e., No natural number between 1024 and 1089 which is a square number.
\n1069 cannot be a square number.<\/p>\n

\"NCERT<\/p>\n

(iv) 2061
\n\u2235 The ending digit is 1
\n\u2234 It may or may not be a square number.
\n\u2235 45 \u00d7 45 = 2025 and 46 \u00d7 46 = 2116
\ni.e., No natural number between 2025 and 2116 which is a square number.
\n\u2234 2061 is not a square number.
\nWe can write many numbers which do not end with 0, 1, 4, 5, 6, or 9. (i.e., which are not square numbers)
\nFive such numbers can be: 1234, 4312, 5678, 87543, 1002007<\/p>\n

Question 3.
\nWrite five numbers that you cannot decide just by looking at their units digit (or one’s place) whether they are square numbers or not.
\nSolution:
\nProperty 1.
\nAny natural number ending in 1, 4, 5, 6, or 9 can be or cannot be a square number.
\nFive such numbers are: 56790, 3671, 2454, 76555, 69209
\nProperty 2.
\nIf a number has 1 or 9 in the unit’s place, then its square ends in 1.
\nFor example: (1)2<\/span><\/sup><\/span> = 1, (9)2<\/span><\/sup><\/span> = 81, (11)2<\/span><\/sup><\/span> = 121, (9)2<\/span><\/sup><\/span> = 361, (21)2<\/span><\/sup><\/span> = 441<\/p>\n

Try These (Page No. 91)<\/span><\/p>\n

Question 4.
\nWhich of 1232<\/span><\/sup><\/span> , 772<\/span><\/sup><\/span>, 822<\/span><\/sup><\/span>, 1612<\/span><\/sup><\/span>, 1092<\/span><\/sup><\/span> would end with the digit 1?
\nSolution:
\nThe squares of those numbers end in 1 which end in either 1 or 9.
\n\u2234 The squares of 161 and 109 would end in 1.<\/p>\n

Question 5.
\nWhich of the following numbers would have digit 6 at the unit place.
\n(i) 192<\/span><\/sup><\/span>
\n(ii) 242<\/span><\/sup><\/span>
\n(iii) 262<\/span><\/sup><\/span>
\n(iv) 362<\/span><\/sup><\/span>
\n(v) 342<\/span><\/sup><\/span>
\nSolution:
\n(i) 192<\/span><\/sup><\/span>: Unit place digit = 9.
\n192<\/span><\/sup><\/span> would not have unit digit as 6.<\/p>\n

(ii) 242<\/span><\/sup><\/span>: Unit place digit = 4.
\n242<\/span><\/sup><\/span> would not have unit digit as 6.<\/p>\n

(iii) 262<\/span><\/sup><\/span>: Unit place digit = 6.
\n262<\/span><\/sup><\/span> would have 6 as unit place.<\/p>\n

(iv) 362<\/span><\/sup><\/span>: Unit place digit = 6.
\n362<\/span><\/sup><\/span> would end in 6<\/p>\n

\"NCERT<\/p>\n

(v) 342<\/span><\/sup><\/span>: Since the unit place digit is 4
\n342<\/span><\/sup><\/span> would have unit place digit as 6.<\/p>\n

Try These (Page No. 92)<\/span><\/p>\n

Question 6.
\nWhat will be the \u201cone’s digit\u201d in the square of the following numbers?
\n(i) 1234
\n(ii) 26387
\n(iii) 52698
\n(iv) 99880
\n(v) 21222
\n(vi) 9106
\nSolution:
\n(i) 192<\/span><\/sup><\/span>: Ending digit = 4 and 42<\/span><\/sup><\/span> = 16.
\n(1234)2<\/span><\/sup><\/span> will have 6 as the ones digit.<\/p>\n

(ii) 242<\/span><\/sup><\/span>: Ending digit is 7 and 72<\/span><\/sup><\/span> = 49.
\n(26387)2<\/span><\/sup><\/span> will have 9 as the ones digit.<\/p>\n

(iii) 262<\/span><\/sup><\/span>: Ending digit is 8, and 82<\/span><\/sup><\/span> = 64.
\n(52698)2<\/span><\/sup><\/span> will end in 4.<\/p>\n

(iv) 362<\/span><\/sup><\/span>: Ending digit is 0.
\n(99880)2<\/span><\/sup><\/span> will end in 0.<\/p>\n

(v) 22<\/span><\/sup><\/span> = 4
\nEnding digit of (21222)2<\/span><\/sup><\/span> is 4.<\/p>\n

(v) 62<\/span><\/sup><\/span> = 36
\nEnding digit of (9106)2<\/span><\/sup><\/span> is 6.<\/p>\n

Question 7.
\nThe square of which of the following numbers would be an odd number\/an even number? Why?
\n(i) 727
\n(ii) 158
\n(iii) 269
\n(iv) 1980
\nSolution:
\n(i) 727
\nSince 727 is an odd number.
\nIts square is also an odd number.<\/p>\n

(ii) 158
\nSince 158 is an even number.
\nIts square is also an even number.<\/p>\n

\"NCERT<\/p>\n

(iii) 269
\nSince 269 is an even number.
\nIts square is also an odd number.<\/p>\n

(iv) 1980
\nSince 1980 is an even number.
\nIts square is also an even number.<\/p>\n

Question 8.
\nWhat will be the number of zeros in the square of the following numbers?
\n(i) 60
\n(ii) 400
\nSolution:
\n(i) In 60, a number of zero is 1. Its square will have 2 zeros.
\n(ii) There are 2 zeros in 400. Its square will have 4 zeros.<\/p>\n

Try These (Page No. 94)<\/span><\/p>\n

Question 9.
\nHow many natural numbers lie between 92 and 102 Between 112 and 122?
\nSolution:
\n(a) Between 92 and 102
\nHere, n = 9 and n + 1 = 10
\nNatural numbers between 92 and 102 are (2 \u00d7 n) or 2 \u00d7 9, i.e. 18.<\/p>\n

(b) Between 112 and 122
\nHere, n = 11 and n + 1 = 12
\nNatural numbers between 112<\/span><\/sup><\/span> and 122<\/span><\/sup><\/span> are (2 \u00d7 n) or 2 \u00d7 11, i.e. 22.<\/p>\n

Question 10.
\nHow many non-square numbers lie between the following pairs of numbers:
\n(i) 1002<\/span><\/sup><\/span> and 1012<\/span><\/sup><\/span>
\n(ii) 902<\/span><\/sup><\/span> and 912<\/span><\/sup><\/span>
\n(iii) 10002<\/span><\/sup><\/span> and 10012<\/span><\/sup><\/span>
\nSolution:
\n(i) Between 1002<\/span><\/sup><\/span> and 1012<\/span><\/sup><\/span>
\nHere, n = 100
\nn \u00d7 2 = 100 \u00d7 2 = 200
\n200 non-square numbers lie between 1002<\/span><\/sup><\/span> and 1012<\/span><\/sup><\/span><\/p>\n

\"NCERT<\/p>\n

(ii) Between 902<\/span><\/sup><\/span> and 912<\/span><\/sup><\/span>
\nHere, n = 90
\n2 \u00d7 n = 2 \u00d7 90 = 180
\n180 non-square numbers lie between 902<\/span><\/sup><\/span> and 912<\/span><\/sup><\/span>.<\/p>\n

(iii) Between 10002<\/span><\/sup><\/span> and 10012<\/span><\/sup><\/span>
\nHere, n = 1000
\n2 \u00d7 n = 2 \u00d7 1000 = 2000
\n2000 non-square numbers lie between 10002<\/span><\/sup><\/span> and 10012<\/span><\/sup><\/span>.<\/p>\n

Question 11.
\nFind whether each of the following numbers is a perfect square or not?
\n(i) 121
\n(ii) 55
\n(iii) 81
\n(iv) 49
\n(v) 69
\nSolution:
\nIf a natural number cannit be expressed as a sum of successive off natural numbers starting from 1, then it is not a perfect square.
\n(i) 121
\nSince, 121 – 1 = 120
\n120 – 3 = 117
\n117 – 5 = 112
\n112 – 7 = 105
\n105 – 9 = 96
\n96 – 11 = 85
\n85 – 13 = 72
\n72 – 15 = 57
\n57 – 17 = 40
\n40 – 19 = 21
\n21 – 21 = 0
\ni.e., 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21.
\nThus, 121 is a square.<\/p>\n

(ii) 55
\nSince, 55 – 1 = 54
\n54 – 3 = 51
\n51 – 5 = 46
\n46 – 7 = 39
\n39 – 9 = 30
\n30 – 11 = 19
\n19 – 13 = 6
\n6 – 15 = -9
\nSince 55 cannot be expressed as the sum of successive odd numbers starting from 1.
\n55 is not a perfect square.<\/p>\n

\"NCERT<\/p>\n

(iii) 81
\nSince, 81 – 1 = 80
\n80 – 3 = 77
\n77 – 5 = 72
\n72 – 7 = 65
\n65 – 9 = 56
\n56 – 11 = 45
\n45 – 13 = 32
\n32 – 15 = 17
\n17 – 17 = 0
\n81 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17
\nThus, 81 is a perfect square.<\/p>\n

(iv) 49
\nSince, 49 – 1 = 48
\n48 – 3 = 45
\n45 – 5 = 40
\n40 – 7 = 33
\n33 – 9 = 24
\n24 – 11 = 13
\n13 – 13 = 0
\n49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
\nThus, 49 is a perfect square.<\/p>\n

(iv) 69
\nSince, 69 – 1 = 68
\n68 – 3 = 65
\n65 – 5 = 60
\n60 – 7 = 53
\n53 – 9 = 44
\n44 – 11 = 33
\n33 – 13 = 20
\n20 – 15 = 5
\n5 – 17 = -12
\n49 = 1 + 3 + 5 + 7 + 9 + 11 + 13
\n69 cannot be expressed as the sum of consecutive odd numbers starting from 1.
\nThus, 69 is not a perfect square.<\/p>\n

Try These (Page No. 95)<\/span><\/p>\n

Question 12.
\nExpress the following as the sum of twp consecutive integers.
\n(i) 212<\/span><\/sup><\/span>
\n(ii) 132<\/span><\/sup><\/span>
\n(iii) 112<\/span><\/sup><\/span>
\n(iv) 192<\/span><\/sup><\/span>
\nSolution:
\n(i) n = 21
\n(n2<\/span><\/sup><\/span> – 1)\/2 = (441 – 1)\/2 = 440\/2 = 220
\n(n2<\/span><\/sup><\/span> + 1)\/2 = (441 + 1)\/2 = 442\/2 = 221
\nn2<\/span><\/sup><\/span> = (n2 – 1)\/2 + (n2 + 1)\/2
\n212<\/span><\/sup><\/span> = 220 + 221 = 441<\/p>\n

\"NCERT<\/p>\n

(ii) n = 13
\n(n2<\/span><\/sup><\/span> + 1)\/2 = (132 + 1)\/2 = (169 + 1)\/2 = 85
\n(n2<\/span><\/sup><\/span> – 1)\/2 = (132 – 1)\/2 = (169 – 1)\/2 = 84
\n132<\/span><\/sup><\/span> = 85 + 84 = 169<\/p>\n

(iii) n = 11
\n(n2<\/span><\/sup><\/span> + 1)\/2 = (112 + 1)\/2 = 61
\n(n2<\/span><\/sup><\/span> – 1)\/2 = (112 – 1)\/2 = 60
\n112<\/span><\/sup><\/span> = 60 + 61 = 121<\/p>\n

(iv) n = 19
\n(n2<\/span><\/sup><\/span> – 1)\/2 = (192 – 1)\/2 = 180
\n(n2<\/span><\/sup><\/span> + 1)\/2 = (192 + 1)\/2 = 181
\n192<\/span><\/sup><\/span> = 180 + 181 = 361<\/p>\n

Question 13.
\nDo you think the reverse is also true, i.e., is the sum of any two consecutive positive integers is a perfect square of a number? Give example to support your answer.
\nSolution:
\nNo, it is not always true.
\nFor example:
\n(i) 5 + 6 = 11, 11 is not a perfect square.
\n(ii) 21 + 22 = 43, 43 is not a perfect square.
\nExamples:
\n92<\/span><\/sup><\/span> – 82<\/span><\/sup><\/span> = 81 – 64 = 17 = 9 + 8
\n102<\/span><\/sup><\/span> – 92<\/span><\/sup><\/span> = 100 – 81 = 19 = 10 + 9
\n152<\/span><\/sup><\/span> – 142<\/span><\/sup><\/span> = 225 – 196 = 29 = 15 + 14
\n1012<\/span><\/sup><\/span> – 1002<\/span><\/sup><\/span> = 10201 – 10000 = 201 = 101 + 100
\nFor Example,
\n10 \u00d7 12 = (11 – 1) \u00d7 (11 + 1) = 112<\/span><\/sup><\/span> – 1
\n11 \u00d7 12 = (12 – 1) \u00d7 (12 + 1) = 112<\/span><\/sup><\/span> – 1
\n25 \u00d7 27 = (26 – 1) \u00d7 (26 + 1) = 262<\/span><\/sup><\/span> – 1<\/p>\n

\"NCERT<\/p>\n

Question 14.
\nWrite the square, making use of the above pattern.
\n(i) 1111112<\/span><\/sup><\/span>
\n(ii) 11111112<\/span><\/sup><\/span>
\nSolution:
\nusing the above pattern we can write
\n(i) (111111)2<\/span><\/sup><\/span> = 12345654321
\n(ii) (1111111)2<\/span><\/sup><\/span> = 1234567654321<\/p>\n

Question 15.
\nCan you find the square of the following numbers using the above pattern?
\n(i) 66666672<\/span><\/sup><\/span>
\n(ii) 666666672<\/span><\/sup><\/span>
\nSolution:
\nusing the above pattern we can write
\n(i) (6666667)2<\/span><\/sup><\/span> = 44444448888889
\n(ii) (66666667)2<\/span><\/sup><\/span> = 4444444488888889<\/p>\n

Try These (Page No. 97)<\/span><\/p>\n

Question 16.
\nFind the square of the following numbers containing 5 in unit place.
\n(i) 15
\n(ii) 95
\n(iii) 105
\n(iv) 205
\nSolution:
\n(i) (15)2<\/span><\/sup><\/span> = 1 \u00d7 (1 + 1) \u00d7 100 + 25
\n= 1 \u00d7 2 \u00d7 100 + 25
\n= 200 + 25
\n= 225<\/p>\n

(ii) (95)2<\/span><\/sup><\/span> = 9 \u00d7 (1 + 1) \u00d7 100 + 25
\n= 9 \u00d7 10 \u00d7 100 + 25
\n= 9000 + 25
\n= 9025<\/p>\n

(iii) (105)2<\/span><\/sup><\/span> = 10 \u00d7 (10 + 1) \u00d7 100 + 25
\n= 10 \u00d7 11 \u00d7 100 + 25
\n= 1100 + 25
\n= 11025<\/p>\n

\"NCERT<\/p>\n

(iv) (205)2<\/span><\/sup><\/span> = 20 \u00d7 (20 + 1) \u00d7 100 + 25
\n= 20 \u00d7 21 \u00d7 100 + 25
\n= 4200 + 25
\n= 42025<\/p>\n

Try These (Page No. 99)<\/span><\/p>\n

Question 17.
\n(i) 112<\/span><\/sup><\/span> = 121. What are the square roots of 121?
\n(ii) 142<\/span><\/sup><\/span> = 196. What are the square roots of 196?
\nSolution:
\n(i) The square root of 121 is 11.
\n(ii) The square root of 196 is 14.<\/p>\n

Try These (Page No. 100)<\/span><\/p>\n

Question 18.
\nBy repeated subtraction of numbers starting from 1, find whether the following numbers are perfect squares the number is a perfect square, then square root.
\n(i) 121
\n(ii) 55
\n(iii) 36
\n(iv) 49
\n(v) 90
\nSolution:
\n(i) Subtracting the successive odd number from 121,
\nWe have 121 – 1 = 120
\n120 – 3 = 117
\n117 – 5 = 112
\n112 – 7 = 105
\n105 – 9 = 96
\n96 – 11 = 85
\n85 – 13 = 72
\n72 – 15 = 57
\n57 – 17 = 40
\n40 – 19 = 21
\n21 – 21 = 0
\n\u221a121 = 11 [we had to subtract the first 11 odd numbers.]<\/p>\n

(ii) 55 – 1 = 54
\n54 – 3 = 51
\n51 – 5 = 46
\n46 – 7 = 39
\n39 – 9 = 30
\n30 – 11 = 19
\n19 – 13 = 6
\n6 – 15 = -9
\nand we do not reach to 0.
\n55 is not a perfect square.<\/p>\n

(iii) 36 – 1 = 35
\n35 – 3 = 32
\n32 – 5 = 27
\n27 – 7 = 20
\n20 – 9 = 11
\n11 – 11 = 0
\nand we do have obtained 0 after subtracting 6 successive odd numbers.
\n36 is a perfect square.
\nThus, \u221a36 = 6.<\/p>\n

\"NCERT<\/p>\n

(iv) We have
\n49 – 1 = 48
\n48 – 3 = 45
\n45 – 5 = 40
\n40 – 7 = 33
\n33 – 9 = 24
\n24 – 11 = 13
\n13 – 13 = 0
\nWe have obtained 0 after successive subtraction of 7 odd numbers.
\n49 is a perfect square Thus, \u221a49 = 7.<\/p>\n

(v) We have
\n90 – 1 = 89
\n89 – 3 = 86
\n86 – 5 = 81
\n81 – 7 = 74
\n74 – 9 = 65
\n65 – 11 = 54
\n54 – 13 = 41
\n41 – 15 = 26
\n26 – 17 = 9
\n9 – 19 = -10
\nSince we can not reach 0 after subtracting successive odd numbers.
\n90 is not a perfect square.<\/p>\n

Try These (Page No. 105)<\/span><\/p>\n

Question 19.
\nWithout calculating square roots, find the number of digits in the square root of the following numbers.
\n(i) 25600
\n(ii) 100000000
\n(iii) 36864
\nSolution:
\n(i) 25600
\nn = 5 [an odd number]
\nIts square root will have \\(\\frac{n+1}{2}\\)
\ni.e. \\(\\frac{5+1}{2}=\\frac{6}{2}\\) = 3 digits<\/p>\n

(ii) 100000000
\nn = 9 (odd number)
\nNumber of digits of its square root
\n\\(\\frac{n+1}{2}=\\frac{9+1}{2}=\\frac{10}{2}=5\\) digits<\/p>\n

\"NCERT<\/p>\n

(iii) 36864
\nn = 5 (odd number)
\nNumber of digits in its square root
\n\\(\\frac{n+1}{2}=\\frac{5+1}{2}=\\frac{6}{2}=3\\) digits<\/p>\n

Try These (Page No. 107)<\/span><\/p>\n

Question 20.
\nEstimate the value of the following the nearest whole number.
\n(i) \u221a180
\n(ii) \u221a100
\n(iii) \u221a350
\n(iv) \u221a500
\nSolution:
\n(i) \u221a80
\n102<\/span><\/sup><\/span> = 100, 92<\/span><\/sup><\/span> = 81, 82<\/span><\/sup><\/span> = 64
\nand 80 is between 64 and 81.
\ni.e., 64 < 80 < 81
\nor 82<\/span><\/sup><\/span> < 80 < 92<\/span><\/sup><\/span>
\nor 8 < \u221a80 < 9
\nThus, \u221a80 lies between 8 and 9.<\/p>\n

(ii) \u221a1000
\nWe know that
\n302<\/span><\/sup><\/span> = 900, 312<\/span><\/sup><\/span> = 961, 322<\/span><\/sup><\/span> = 1024
\n1000 lies between 961 and 1024.
\ni.e., 916 < 1000 < 1024
\nor 312<\/span><\/sup><\/span> < 1000 < 322<\/span><\/sup><\/span>
\nor 312<\/span><\/sup><\/span> < \u221a1000 < 32.
\nThus, \u221a1000 lies between 31 and 32.<\/p>\n

(iii) \u221a350
\nWe have 182<\/span><\/sup><\/span> = 324, 192<\/span><\/sup><\/span> = 361
\nSince 350 lies between 324 and 316.
\ni.e., 324 < 350 < 361
\nor 182<\/span><\/sup><\/span> < 350 < 192<\/span><\/sup><\/span>
\nor 18 < \u221a350 < 19.
\nThus, \u221a350 lies between 18 and 19.<\/p>\n

\"NCERT<\/p>\n

(iv) \u221a500
\n222<\/span><\/sup><\/span> = 484 and 232<\/span><\/sup><\/span> = 529
\nSince, 500 lies between 484 and 529.
\nor 484 < 500 < 529
\nor 222<\/span><\/sup><\/span> < 500 < 232<\/span><\/sup><\/span>
\nor 22 < \u221a350 < 23.
\nThus, \u221a500 lies between 22 and 23.<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots InText Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 8 Maths Chapter 6 Square and Square Roots InText Questions Try These (Page No. 90) Question 1. Find the perfect square numbers between (i) 30 and …<\/p>\n

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