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Question 1. What will be the unit digit of the squares of the following numbers?

1. 81

2. 272

3. 799

4. 3853

5. 1234

6. 26387

7. 52698

8. 99880

9. 12796

10. 55555

Answer.

i. 81 

The unit digit (1) of the square of the following number is 1 x 1 = 1

ii. 272

The unit digit is 2 of the square of the following number is 2 x 2 = 4

iii. 799

The unit digit is 9 of the square of the following number is 9 x 9 = 81

iv. 3853

The unit digit is 3 of the square of the following number is 3 x 3 = 9

v. 1234

The unit digit is 4 of the square of the following number is 4 x 4 = 16

vi. 26387

The unit digit is 7 of the square of the following number is 7 x 7 = 49

vii. 52698

The unit digit is 8 of the square of the following number is 8 x 8 = 64

viii. 99880

The unit digit is 0 of the square of the following number is 0 x 0 = 0

ix. 12796

The unit digit is 6 of the square of the following number is 6 x 6 = 36

x. 55555

The unit digit is 5 of the square of the following number is 5 x 5 = 25

 

Question 2. The following numbers are not perfect squares. Give a reason.

1. 1057

2. 23453

3. 7928

4. 222222

5. 64000

6. 89722

7. 222000

8. 505050

Answer. 

i. 1057

Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.

The following number is 1057, and the unit digit is 7. Thus, the following numbers are not perfect squares.

ii. 23453

Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.

The following number is 23453, and the unit digit is 3. Thus, the following numbers are not perfect squares.

iii. 7928

Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.

The following number is 1057, and the unit digit is 7. Thus, the following numbers are not perfect squares.

iv. 222222

Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.

The following number is 7928, and the unit digit is 2. Thus, the following numbers are not perfect squares.

v. 64000

Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.

The following number is 64000, and the unit digit is 0. Thus, the following numbers are not perfect squares.

vi. 89722

Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.

The following number is 89722, and the unit digit is 0. Thus, the following numbers are not perfect squares.

vii. 222000

Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.

The following number is 222000, and the unit digit is 0. Thus, the following number is not perfect squares.

viii. 505050

Natural numbers with the digits 0, 2, 3, 7, and 8 are not perfect squares, as we know.

The following number is 505050, and the unit digit is 0. Thus, the following number is not perfect squares.

 

Question 3. The squares of which of the following would be odd numbers?

1. 431

2. 2826

3. 7779

4. 82004

Answer. 

i. 431

We know that an odd number's square is odd, while an even number's square is even. The square of 431 is an odd number. 

ii. 2826

We know that an odd number's square is odd, while an even number's square is even. The square of 2826 is an even number. 

iii. 7779

We know that an odd number's square is odd, while an even number's square is even. The square of 7779 is an odd number. 

iv. 82004 

We know that an odd number's square is odd, while an even number's square is even. The square of 82004 is an even number.

 

Question 4. Observe the following pattern and find the missing numbers. 112 = 121

1012 = 10201

10012 = 1002001

1000012 = 1 …….2………1

100000012 = ……………………..

Answer. 

The square on the R.H.S of the equality contains an odd number of digits, with the first and last digits being 1 and the middle digit being 2. And the number of zeros between the left and middle digits 1 and 2 and the right and middle digits 1 and 2 is the same as the number of zeros in the supplied integer.

∴, 1000012 = 100020001

100000012 = 100000020000001

 

Question 5. Observe the following pattern and supply the missing numbers. 112 = 121

1012 = 10201

101012 = 102030201

10101012 = ………………………

…………2 = 10203040504030201

Answer. 

1012 = 10201

101012 = 102030201

10101012 = ………………………

…………2 = 10203040504030201

We see that the square of the R.H.S. number of the equality has an odd number of digits, with the first and last digits both being 1. Furthermore, the square is symmetrical around the central digit. If the middle digit is 4, the squared number is 10101, and the squared number is 102030201.

∴, 10101012 = 1020304434

1010101012 = 10203040504030201

 

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 

Answer.

Given, 12 + 22 + 22 = 32

i.e 12 + 22 + (1×2 )2 = ( 12 + 22 -1 × 2 )2

22 + 32 + 62 = 72

∴, 22 + 32 + (2 x 3)2 = 72 

= (22 + 32 -2 x 3)2

32 + 42 + 122 = 132 

∴, 32 + 42 + (3 x 4)2 = 132 

= (32 + 42 -3 x 4)2

42 + 52 + _2 = 212 

∴, 42 + 52 + (4 x 5)2 = 212 

= (42 + 52 -4 x 5)2

52 + _ 2 + 302 = 312 

∴, 52 + 62 + (5 x 6)2 = 312 

= (52 + 62 -5 x 6)2

6 + 7 + _ 2 = _ 2 

∴, 62 + 72 + (6 x 7)2 = 432

= (62 + 72 -6 x 7)2

 

Question 7. Without adding, find the sum.

1. 1 + 3 + 5 + 7 + 9

2. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19

3. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

Answer. 

i. 1 + 3 + 5 + 7 + 9

(5)2 = 25 as the sum of the first five odd numbers.

ii. 1 + 3 + 5 + 7 + 9 + I1 + 13 + 15 + 17 +19

(10)2 = 100 as the sum of the first ten odd numbers added together.

iii. 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23

(12)2 = 144, which is the total of the first thirteen odd numbers.

 

Question 8. (i) Express 49 as the sum of 7 odd numbers.

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

Answer. 

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

We already know that the sum of the first n odd natural numbers is n2. 

Because 49 is 72, 1+3+5+7+9+11+13 = sum of the first 7 odd natural numbers

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

We already know that the sum of the first n odd natural numbers is n2. 

Because 121 is 112, 1+3+5+7+9+11+13+15+17+19+21 = sum of the first 11 odd natural numbers

 

Question 9. How many numbers lie between squares of the following numbers?

1. 12 and 13

2. 25 and 26

3. 99 and 100

Answer.

i. 12 and 13

Between n2 and (n+1)2, there are 2n non–perfect square numbers. There are 212 = 24 natural numbers) between 122 and 132. 

ii. 25 and 26

Between n2 and (n+1)2, there are 2n non–perfect square numbers. There are 225 = 50 natural numbers between 252 and 262. 

iii. 99 and 100

Between n2 and (n+1)2, there are 2n non–perfect square numbers. There are 299 =198 natural numbers between 992 and 1002.