# Mathematics – Class 8 – Chapter 6 – Squares and Square root – Exercise 6.1 – NCERT Exercise Solution

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:

i.Unit digit of given number 81 = 1

Hence, unit digit of square of 1 = 1

ii. Unit digit of given number 272 = 2

Hence, unit digit square of 2 = 4

iii. Unit digit of given number 799 = 9

square of 9 = 81

Hence, its unit digit = 1

iv. Unit digit of given number 3853 =3

Hence, unit digit of suare of 3 = 9

v. Unit digit of given number 1234 =4

square of 4 = 16

Hence, unit digit = 6

vi. Unit digit of given number 26387 =7

square of 7 = 49

Hence, unit digit number = 9

vii. Unit digit of given number 52698 = 8

square of 8 = 64

Hence, its unit digit = 4

viii. The unit digit of given number 99880 = 0

Hence, Unit digit of the square of number 99880 is equal to 0.

ix. The unit digit of given number 12796 = 6

Square of 6 = 36

Hence, its unit digit = 6

x. The unit digit of given number 55555 = 5

square of 5 = 25

Hence, its unit digit = 5

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

Solution:

We know that natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect squares.

i. 1057 =Not a perfect square because it ends with 7

ii. 23453 = Not a perfect square because it ends with 3

iii. 7928 = Not a perfect square because it ends with 8

iv. 222222 = Not a perfect square because it ends with 2

v. 64000 = Not a perfect square because it ends with 0

vi. 89722 = Not a perfect square because it ends with 2

vii. 222000 = Not a perfect square because it ends with 0

viii. 505050 = Not a perfect square because it ends with 0

3. The squares of which of the following would be odd numbers?

i. 431

ii. 2826

iii. 7779

iv. 82004

Solution:

We know that the square of an odd number is odd and the square of an even number is even.

i. The square of 431 is an odd number.

ii. The square of 2826 is an even number.

iii. The square of 7779 is an odd number.

iv. The square of 82004 is an even number.

4. Observe the following pattern and find the missing numbers.

112 = 121

1012 = 10201

10012 = 1002001

1000012 = 1 …….2……1

100000012 = ………………

Solution:

According to the above pattern:

1000012 = 10000200001

100000012 = 100000020000001

5. Observe the following pattern and supply the missing numbers.

112 = 121

1012 = 10201

101012 = 102030201

10101012 = ………………………

…………2 = 10203040504030201

Solution:

According to the above pattern:

10101012 =1020304030201

1010101012 =10203040505030201

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

Solution:

According to the given pattern, we have

42 + 52 + 202 = 212

52 + 62 + 302 = 312

62 + 72 + 422 = 432

7. Without adding, find the sum.

We know that the sum of n odd numbers = n2

i. 1 + 3 + 5 + 7 + 9

Solution:

Sum of first five odd number = (5)2 = 25

ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19

Solution:

Sum of first ten odd number = (10) 2 = 100

iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Solution:

Sum of first thirteen odd number = (12) 2 = 144

8. (i) Express 49 as the sum of 7 odd numbers.

Solution:

We know, sum of first n odd natural numbers = n2

Since,49 = 72

Here n = 7

Hence, 49 = sum of first 7 odd natural numbers

= 1 + 3 + 5 + 7 + 9 + 11 + 13

(ii) Express 121 as the sum of 11 odd numbers.

Solution:

Since, 121 = 112

Here n = 11

Hence, 121 = sum of first 11 odd natural numbers

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

9. How many numbers lie between squares of the following numbers?

i. 12 and 13

ii. 25 and 26

iii. 99 and 100

Solution:

We know that between n2 and (n+1)2, there are 2n non–perfect square numbers.

Now,

i. 122 and 132 there are 2×12 = 24 natural numbers.

ii. 252 and 262 there are 2×25 = 50 natural numbers.

iii. 992 and 1002 there are 2×99 =198 natural numbers.

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