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**Squares and Square Roots Class 8 Extra Questions**

**NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots Exercise 6.1**

**Ex 6.1 Class 8 Maths 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) 20387

(vii) 52698

(viii) 99880

(ix) 12796

(x) 55555**Solution**:

(i) Unit digit of 812 = 1

(ii) Unit digit of 2722 = 4

(iii) Unit digit of 7992 = 1

(iv) Unit digit of 38532 = 9

(v) Unit digit of 12342 = 6

(vi) Unit digit of 263872 = 9

(vii) Unit digit of 526982 = 4

(viii) Unit digit of 998802 = 0

(ix) Unit digit of 127962 = 6

(x) Unit digit of 555552 = 5

**Ex 6.1 Class 8 Maths 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**:

(i) 1057 ends with 7 at unit place. So it is not a perfect square number.

(ii) 23453 ends with 3 at unit place. So it is not a perfect square number.

(iii) 7928 ends with 8 at unit place. So it is not a perfect square number.

(iv) 222222 ends with 2 at unit place. So it is not a perfect square number.

(v) 64000 ends with 3 zeros. So it cannot a perfect square number.

(vi) 89722 ends with 2 at unit place. So it is not a perfect square number.

(vii) 22000 ends with 3 zeros. So it can not be a perfect square number.

(viii) 505050 ends with 1 zero. So it is not a perfect square number.

**Ex 6.1 Class 8 Maths Question 3.**

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

(i) 431

(ii) 2826

(iii) 7779

(iv) 82004**Solution**:

(i) 4312 is an odd number.

(ii) 28262 is an even number.

(iii) 77792 is an odd number.

(iv) 820042 is an even number.

**Ex 6.1 Class 8 Maths Question 4.**

Observe the following pattern and find the missing digits.

112 = 121

1012 = 10201

10012 = 1002001

1000012 = 1…2…1

100000012 = ………**Solution**:

According to the above pattern, we have

1000012 = 10000200001

100000012 = 100000020000001

**Ex 6.1 Class 8 Maths Question 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, we have

10101012 = 1020304030201

1010101012 = 10203040504030201

**Ex 6.1 Class 8 Maths Question 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

**Ex 6.1 Class 8 Maths 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**:

We know that the sum of n odd numbers = n2

(i) 1 + 3 + 5 + 7 + 9 = (5)2 = 25 [∵ n = 5]

(ii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = (10)2 = 100 [∵ n = 10]

(iii) 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = (12)2 = 144 [∵ n = 12]

**Ex 6.1 Class 8 Maths 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 = 1 + 3 + 5 + 7 + 9 + 11 + 13 (n = 7)

(ii) 121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 (n = 11)

**Ex 6.1 Class 8 Maths 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) We know that numbers between n2 and (n + 1)2 = 2n

Numbers between 122 and 132 = (2n) = 2 × 12 = 24

(ii) Numbers between 252 and 262 = 2 × 25 = 50 (∵ n = 25)

(iii) Numbers between 992 and 1002 = 2 × 99 = 198 (∵ n = 99)

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