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

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.

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.

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.

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.

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}.

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.

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.

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.