In this page we have *NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.1 *.This exercise has questions about Pattern of numbers , properties of the square number, how to find the unit digit of the square. Hope you like them and do not forget to like , social share
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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

The following numbers are obviously not perfect squares. Give reason.

- 1057
- 23453
- 7928
- 222222
- 64000
- 89722
- 222000
- 505050

So (i), (ii), (iii), (iv), (vi) don’t have any of the 0, 1, 4, 5, 6, or 9 at unit’s place, so they are not be perfect squares.

So (v), (vii) and (viii) don’t have even number of zeroes at the end so they are not perfect squares.

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

- 431
- 2826
- 7779
- 82004

- 431 square will end in 1,So odd number
- 2826 square will end in 6 ,so even number
- 779 square will end in 1,So odd number
- 82004 square will end in 6 ,so even number

Observe the following pattern and find the missing digits.

$11^2= 121$

$101^2= 10201$

$1001^2= 1002001$

$100001^2$= 1.........2.......1

$10000001^2$= ...............

$100001^2= 10000200001$

$10000001^2= 100000020000001$

Observe the following pattern and supply the missing numbers.

$11^2= 121$

$101^2= 10201$

$10101^2= 102030201$

$1010101^2$= ..................

..............

$1010101^2= 1020304030201$

$101010101^2=10203040504030201$

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

5

6

Relation among first, second and third number - Third number is the product of first and second number

Relation between third and fourth number - Fourth number is 1 more than the third number

$4^2+ 5^2+ 20^2= 21^2$

$5^2+ 6^2+ 30^2= 31^2$

$6^2+ 7^2+ 42^2= 43^2$

Without adding, find the sum.

(i) $1 + 3 + 5 + 7 + 9$

(ii) $1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19$

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

Explanation:

$1 + 3 = 2^2= 4$

$1 + 3 + 5 = 3^2= 9$

$1 + 3 + 5 + 7 = 4^2=16$

$1 + 3 + 5 + 7 + 9 = 5^2= 25$

So Sum of n odd numbers starting from 1 = $n^2$

From the above derivation we can answer the above questions

- Since, there are 5 consecutive odd numbers, Thus, their sum = $5^2= 25$
- Since, there are 10 consecutive odd numbers, Thus, their sum = $10^2= 100$
- Since, there are 12 consecutive odd numbers, Thus, their sum = $12^2= 144$

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

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

Explanation:

$1 + 3 = 2^2= 4$

$1 + 3 + 5 = 3^2= 9$

$1 + 3 + 5 + 7 = 4^2=16$

$1 + 3 + 5 + 7 + 9 = 5^2= 25$

So Sum of n odd numbers starting from 1 = $n^2$

(1) Since, $49 = 7^2$

So, $7^2$ can be expressed as follows:

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

(2) Since, $121 = 11^2$

Therefore, $121 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21$

How many numbers lie between squares of the following numbers?

(i) 12 and 13

(ii) 25 and 26

(iii) 99 and 100

i. $12^2= 144$

$13^2= 169$

Now, 169 - 144 = 25

So, there are 25 - 1 = 24 numbers lying between $12^2$ and $13^2$

ii. We know that, $25^2= 625$

And, $26^2= 676$

Now, 676 - 625 = 51

So, there are 51 - 1 = 50 numbers lying between $25^2$ and $26^2$

iii. We know that, $99^2= 9801$

And, $100^2= 10000$

Now, $10000 - 9801 = 199$

So, there are 199 - 1 = 198 numbers lying between $99^2$ and $100^2$

- NCERT Solutions for Class 8 Maths Chapter 6 Exercise 6.1 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also

**Download Class 8 Maths Chapter 6 Exercise 6.1 as pdf** - This chapter 6 has total 4 Exercise 6.1 ,6.2,6.3 and 6.4. This is the First exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below

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