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

NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square
Roots Ex 6.1 are the part of NCERT Solutions for Class 8 Maths. Here you can
find the NCERT Solutions for Class 8 Maths Chapter 6 Squares and Square Roots
Ex 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)**26387

**(vii)**52698

**(viii)**99880

**(ix)**12796

**(x)**55555

**Solution:**

The unit
digits of the squares of the given numbers are shown against the numbers in the
table given below:

**Ex 6.1 Class 8
Maths Question 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:**

A number which ends with 2, 3, 7 or 8 cannot be a perfect square. Again, a
number which ends with odd number of zero(s) cannot be a perfect square.

**(i)** Since
the number 1057 ends with 7, so it cannot be a perfect square.

**(ii)** Since
the number 23453 ends with 3, so it cannot be a perfect square.

**(iii)** Since
the number 7928 ends with 8, so it cannot be a perfect square.

**(iv)** Since
the number 222222 ends with 2, so it cannot be a perfect square.

**(v)** Since
the number 64000 ends with an odd number of zeros, so it cannot be a perfect
square.

**(vi)** Since
the number 89722 ends with 2, so it cannot be a perfect square.

**(vii)** Since
the number 222000 ends with an odd number of zeros, so it cannot be a perfect
square.

**(viii)** Since
the number 505050 ends with an odd number of zeros, so it cannot be a perfect
square.

**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)** The
number 431 is an odd number, so its square would be odd.

**(ii)** The
number 2826 is an even number, so its square would be even.

**(iii)** The
number 7779 is an odd number, so its square would be odd.

**(iv)** The
number 82004 is an even number, so its square would be even.

Hence, the squares of the numbers 431 and 7779 would be odd numbers.

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

Observe the following pattern and find the missing digits:11

^{2}= 121

101

^{2}= 10201

1001

^{2}= 1002001

100001

^{2}= 1 ………….. 2 ……….. 1

10000001

^{2}= ………………..

**Solution:**

The missing digits are as follows:

100001^{2} = __10000200001__

10000001^{2} = __100000020000001__

**Ex 6.1 Class 8
Maths Question 5.**

Observe the following pattern and supply the missing numbers:11

^{2}= 121

101

^{2}= 10201

10101

^{2}= 102030201

1010101

^{2}= ………..

…………….

^{2}= 10203040504030201

**Solution:**

The missing numbers are as follows:

1010101^{2} = __1020304030201__

__101010101__^{2} = 10203040504030201

**Ex 6.1 Class 8
Maths Question 6.**

Using the given pattern, find the missing numbers.1

^{2}+ 2

^{2}+ 2

^{2}= 3

^{2}

2

^{2}+ 3

^{2}+ 6

^{2}= 7

^{2}

3

^{2}+ 4

^{2}+ 12

^{2}= 13

^{2}

4

^{2}+ 5

^{2}+ ____

^{2}= 21

^{2}

5

^{2 }+ ____

^{2}+ 30

^{2}= 31

^{2}

6

^{2}+ 7

^{2}+ ____

^{2}= ____

^{2}

**Solution:**

The missing numbers are as follows:

4^{2} + 5^{2} + 20^{2} = 21^{2}

5^{2} + 6^{2} + 30^{2} = 31^{2}

6^{2} + 7^{2} + 42^{2} = 43^{2}

**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:**

**(i)** 1
+ 3 + 5 + 7 + 9 = Sum of first 5 odd numbers

= 5^{2} =
25

**(ii)** 1
+ 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

= Sum
of first 10 odd numbers

= 10^{2} =
100

**(iii)** 1
+ 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

= Sum
of first 12 odd numbers

= 12^{2} =
144

**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
= 7^{2} = 1 + 3 + 5 + 7 + 9 + 11 + 13

**(ii)** 121
= 11^{2} = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21

**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:**

**We** know that there are 2n numbers lie
between n^{2} and (n + 1)^{2}.

**(i)** Therefore,
between 12^{2} and 13^{2}, there are 24 (i.e., 2 × 12)
numbers.

**(ii)** Therefore,
between 25^{2} and 26^{2}, there are 50 (i.e., 2 × 25)
numbers.

**(iii)** Therefore,
between 99^{2} and 100^{2}, there are 198 (i.e., 99 × 2)
numbers.

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