# Class 8 Maths NCERT Solutions for Chapter – 6 Squares and Square Roots Ex – 6.1

## Squares and Square Roots

Question 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

Solution:
The unit digit of the squares of the given numbers is shown against the numbers in the following table : Question 2.
The following numbers are not perfect squares. Give reason,
(i)
1057
(ii) 23453
(iii) 7928
(iv) 222222
(v) 64000
(vi) 89722
(vii) 222000
(viii) 505050

Solution:
A number that ends either with 2, 3, 7 or 8 cannot be a perfect square. Also, a number that ends with odd number of zero(s) cannot be a perfect square.
(i) Since the given number 1057 ends with 7, so it cannot be a perfect square.
(ii) Since the given number 23453 ends with 3, so it cannot be a perfect square.
(iii) Since the given number 7928 ends with 8, so it cannot be a perfect square.
(iv) Since the given number 222222 ends with 2, so it cannot be a perfect square.
(v) Since the number 64000 ends in an odd number of zeros, so it cannot be a perfect square.
(vi) Since the number 89722 ends in 2, so it cannot be a perfect square.
(vii) Since the number 222000 ends in an odd number of zeros, so it cannot be a perfect square.
(viii) Since the number 505050 ends in an odd number of zeros, so it cannot be a perfect square.

Question 3.
The squares of which of the following would be odd numbers?
(i)
431
(ii) 2826
(iii) 7779
(iv) 82004

Solution:
(i) The given number 431 being odd, so its square must be odd.
(ii) The given number 2826 being even, so its square must be even.
(iii) The given number 7779 being odd, so its square must be odd.
(iv) The given number 82004 being even, so its square must be even.
Hence, the numbers 431 and 7779 will have squares as odd numbers.

Question 4.
Observe the following pattern and find the missing digits :
112 =121
1012 =10201
10012 =1002001
1000012 = 1 ………….. 2 ……….. 1
100000012 = ………………..

Solution:
The missing digits are as under :
1000012 = 10000200001
100000012 = 100000020000001

Question 5.
Observe the following pattern and supply the missing numbers:
112 = 121
1012 =10201
101012 =102030201
10101012 = ………..
……………. 2 =10203040504030201

Solution:
The missing numbers are as under :
10101012 = 1020304030201
1010101012 = 10203040504030201

Question 6.
Using the given pattern, find the missing
12 + 22 + 22 = 32
22+ 32 + 62 = 72
32 + 42 + 122 = 132
422 + 52 + _2 =212
52+ _2 + 302 = 312
62 + 72 + _2 = _2

Solution:
The missing numbers are as under :
42 + 52 + 202 =212
52 + 62 +302 =312
62 +72 + 422 = 432.

Question 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

Solution:
(i) 1+3 + 5 + 7 + 9
= Sum of first 5 odd numbers
= 52 = 25

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
= Sum of first 10 odd numbers
= 102 =100

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23
= Sum of first 12 odd numbers
= 122 =144

Question 8.
(i) Express 49 as the sum of 7 odd numbers.
(ii) Express 121 as the sum of 11 odd numbers.

Solution:
(i) 49 =72 =1 + 3 + 5 + 7 + 9 +11 +13
(ii) 121 =112 =1 + 3 + 5 + 7 + 9 + 11 +13 + 15 + 17 + 19 + 21

Question 9.
How many numbers lie between squares of the following numbers?
(i)
12 and 13
(ii) 25 and 26
(iii) 99 and 100

Solution:
(i) Between 122 and 132 there are twenty four (i.e., 2 x 12) numbers.
(ii) Between 252 and 262 there are fifty (i.e., 2 x 25) numbers.
(iii) Between 992 and 1002 there are one hundred ninety-eight (i. e., 99 x 2) numbers.