## MP Board Class 8th Maths Solutions Chapter 6 Square and Square Roots Ex 6.1

**MP Board Class 8 Maths Chapter 6 Exercise 6.1 Question 1.**

What will be the unit digit of the squares of the following numbers?

Solution:

(i) The unit digit of (81)^{2} is 1. Because when we multiply digit 1 by itself, we get 1.

(ii) The unit digit of (272)^{2} is 4. Because when we multiply digit 2 by itself, we get 4.

(iii) The unit digit of (799)^{2} is 1. Because when we multiply digit 9 by itself, we get 81.

(iv) The unit digit of (3853)^{2} is 9. Because when we multiply digit 3 by itself, we get 9.

(v) The unit digit of (1234)^{2} is 6. Because when we multiply digit 4 by itself, we get 16.

(vi) The unit digit of (26387)^{2} is 9. Because when we multiply digit 7 by itself, we get 49.

(vii) The unit digit of (52698)^{2} is 4. Because when we multiply digit 8 by itself, we get 64.

(viii) The unit digit of (99880)^{2} is 0. Because when we multiply digit 0 by itself, we get 0.

(ix) The unit digit of (12796)^{2} is 6. Because when we multiply the unit digit 6 by itself, we get 36.

(x) The unit digit of (55555)^{2} is 5. Because when we multiply the unit digit 5 by itself, we get 25.

**MP Board Class 8 Maths Chapter 5 Square And Square Roots 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:

We know that number ending in 2, 3, 7 or 8 are not perfect squares.

∴ (i) 1057,

(ii) 23453,

(iii) 7928,

(iv) 222222

and

(vi) 89722 are not perfect squares.

Since, for perfect squares, there should be even number of zeroes at the end.

∴ (v) 64000,

(vii) 222000 and

(viii) 505050. Are not perfect squares.

**Class 8 Maths MP Board Chapter 6 Question 3.**

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

(i) 431

(ii) 2826

(iii) 7779

(iv) 82004

Solution:

(i) When we multiply the unit digit 1 by itself, we get 1 at the end, which shows that the square of 431 is an odd number.

(ii) When we multiply the unit digit 6 by itself we get 36, i.e., we get 6 at the end, which shows that the square of 2826 is an even number.

(iii) When we multiply the unit digit 9 by itself we get 81, i.e., we get 1 at the end, which shows that the square of 7779 is an odd number.

(iv) When we multiply the unit digit 4 by itself we get 16, i.e., we get 6 at the end, which shows that the square of 82004 is an even number.

**MP Board Class 8 Maths Chapter 6 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:

100001^{2} = 10000200001

10000001^{2} = 100000020000001.

**Class 8 Maths Chapter 6 MP Board 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:

1010101^{2} = 1020304030201

101010101^{2} = 10203040504030201.

**Class 8 MP Board Maths Chapter 6 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:

In the pattern,

Third number = first number × second number and fourth number = third number + 1

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

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

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

**Class 8 Maths Chapter 6 Exercise 6.1 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 have to find the sum of first 5 odd numbers,

1 + 3 + 5 + 7 + 9 = 5^{2} = 25.

(ii) We have to find the sum of the first 10 odd numbers,

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 10^{2} = 100.

(iii) We have to find the sum of first 12 odd numbers,

1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 12^{2} = 144.

**Class 8 Maths Exercise 6.1 Solutions 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.

**8th Class Maths Chapter 6 Exercise 6.1 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 can find the number of terms between the squares of 12 and 13, by doubling the first term from 12 and 13. i.e., 2 × 12 = 24:

∴ Total number of terms = 24.

(ii) We can find the number of terms between the squares of 25 and 26, by doubling the first term from 25 and 26. i.e., 2 × 25 = 50.

∴ Total number of terms = 50.

(iii) We can find the number of terms between the squares of 99 and 100, by doubling the first term from 99 and 100. i.e., 2 × 99 = 198.

∴ Total number of terms = 198.