# Squares And Square Roots : Exercise 6.1 (Mathematics NCERT Class 8th) Q.1 What will be the unit digit of the squares of the following numbers?
(i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 26387 (vii) 52698 (viii) 99880
(ix) 12796 (x) 55555
Sol. (i) 81
The digit 81 ends with ‘1’. And square of 1 is 1.
Hence, the unit digit of square of 81 is 1.

(ii) 272
The digit 272 ends with ‘2’. And square of 2 is 4.
Hence, the unit digit of square of 272 is 4.

(iii) 799
The digit 799 ends with ‘9’. And square of 9 is 81.
Hence, the unit digit of square of 799 is 1.

(iv) 3853
The digit 3853 ends with ‘3’. And square of 3 is 9.
Hence, the unit digit of square of 3583 is 9.

(v) 1234
The digit 1234 ends with ‘4’. And square of 4 is 16.
Hence, the unit digit of square of 1234 is 6.

(vi) 26387
The digit 26387 ends with ‘7’. And square of 7 is 49.
Hence, the unit digit of square of 26387 is 9.

(vii) 52698
The digit 52698 ends with ‘8’. And square of 8 is 64.
Hence, the unit digit of square of 52698 is 4.

(viii) 99880
The digit 99880 ends with ‘0’. And square of 0 is 0.
Hence, the unit digit of square of 99880 is 0.

(ix) 12796
The digit 12796 ends with ‘6’. And square of 6 is 36.
Hence, the unit digit of square of 12796 is 6.

(x) 55555
The digit 55555 ends with ‘5’. And square of 5 is 25.
Hence, the unit digit of square of 55555 is 5.

Q.2 The following numbers are obviously not perfect squares. Give reason.
(i) 1057 (ii) 23453 (iii) 7928 (iv) 222222
(v) 64000 (vi) 89722 (vii) 222000 (viii) 505050
Sol. (i) 1057
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(ii) 23453
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(iii) 7928
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(iv) 222222
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(v) 64000
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(vi) 89722
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(vii) 222000
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

(viii) 505050
It is not a perfect square since its unit’s digit does not end with 1, 4, 5, 6, 9 or even numbers of 0.

Q.3 The squares of which of the following would be odd numbers?
(i) 431 (ii) 2826 (iii) 7779 (iv) 82004
Sol. (i) 431
The digit 431 ends with ‘1’. And square of 1 is 1.
Hence, the square of 431 would be an odd number.

(ii)2826
The digit 2826 ends with ‘6’. And square of 6 is 36.
Hence, the square of 2826 would not be an odd number.

(iii)7779
The digit 7779 ends with ‘9’. And square of 9 is 81.
Hence, the square of 7779 would be an odd number.

(iv)82004
The digit 82004 ends with ‘4’. And square of 4 is 16.
Hence, the square of 82004 would not be an odd number.

Q.4 Observe the following pattern and find the missing digits.
112 = 121
1012 = 10201
10012 = 1002001
1000012 = 1 ......... 2 ......... 1
100000012 = ...........................
Sol.
1000012 = 10000200001
100000012 = 100000020000001

Q.5 Observe the following pattern and supply the missing numbers.
112 = 1 2 1
1012 = 1 0 2 0 1
101012 = 102030201
10101012 = ...........................
............ 2 = 10203040504030201
Sol.
10101012 = 1020304030201
1010101012 = 10203040504030201

Q.6 Using the given pattern, find the missing numbers.
12 + 22 + 22 = 32
22 + 32 + 62 = 72
32 + 42 + 122 = 132
42 + 52 + _2 = 212
52 + _2 + 302 = 312
62 + 72 + _2 = __2
Sol.
42 + 52 + 202 = 212
52 + 62 + 302 = 312
62 + 72 +422 = 432

Q.7 Without adding, find the sum.
(i) 1 + 3 + 5 + 7 + 9
(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19
(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Sol. We know that the sum of first n odd natural numbers is n2.
(i) 1 + 3 + 5 + 7 + 9
Here, there are five odd numbers.
Therefore, 1 + 3 + 5 + 7 + 9 = 52 = 25.

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19
Here, there are ten odd numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 +19 = 102 = 100

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
Here, there are twelve odd numbers.
Therefore, 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23= 122 = 144

Q.8 (i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.
Sol. We know that the sum of first n odd natural numbers is n2.

(i) 49 = 72. Therefore, it is the sum of 7 odd numbers.
Hence, 49 = 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) 121 = 112. Therefore, it is the sum of 11 odd numbers.
Hence, 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

Q.9 How many numbers lie between squares of the following numbers?
(i) 12 and 13 (ii) 25 and 26 (iii) 99 and 100
Sol. We know that there will be 2n numbers in between the squares of the numbers n and (n+1).
(i) 12 and 13
Here, n = 12. Therefore, there will 2n = 2 x 12 = 24 square numbers between 12 and 13.

(ii) 25 and 26
Here, n = 25. Therefore, there will 2n = 2 x 25 = 50 square numbers between 25 and 26.

(iii) 99 and 100
Here, n = 99. Therefore, there will 2n = 2 x 99 = 198 square numbers between 99 and 100.

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