**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:**

**i**.Unit digit of given number 81 = 1

Hence, unit digit of square of 1 = 1

**ii.** Unit digit of given number 272 = 2

Hence, unit digit square of 2 = 4

**iii.** Unit digit of given number 799 = 9

square of 9 = 81

Hence, its unit digit = 1

**iv**. Unit digit of given number 3853 =3

Hence, unit digit of suare of 3 = 9

**v**. Unit digit of given number 1234 =4

square of 4 = 16

Hence, unit digit = 6

**vi**. Unit digit of given number 26387 =7

square of 7 = 49

Hence, unit digit number = 9

**vii. **Unit digit of given number 52698 = 8

square of 8 = 64

Hence, its unit digit = 4

**viii**. The unit digit of given number 99880 = 0

Hence, Unit digit of the square of number 99880 is equal to 0.

**ix.** The unit digit of given number 12796 = 6

Square of 6 = 36

Hence, its unit digit = 6

**x.** The unit digit of given number 55555 = 5

square of 5 = 25

Hence, its unit digit = 5

**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 natural numbers ending in the digits 0, 2, 3, 7 and 8 are not perfect squares.

**i.** 1057 =Not a perfect square because it ends with 7

**ii.** 23453 = Not a perfect square because it ends with 3

**iii.** 7928 = Not a perfect square because it ends with 8

**iv.** 222222 = Not a perfect square because it ends with 2

**v.** 64000 = Not a perfect square because it ends with 0

**vi.** 89722 = Not a perfect square because it ends with 2

**vii.** 222000 = Not a perfect square because it ends with 0

**viii.** 505050 = Not a perfect square because it ends with 0

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

**i. 431**

**ii. 2826**

**iii. 7779**

**iv. 82004**

**Solution:**

We know that the square of an odd number is odd and the square of an even number is even.

**i**. The square of 431 is an odd number.

**ii.** The square of 2826 is an even number.

**iii.** The square of 7779 is an odd number.

**iv.** The square of 82004 is an even number.

**4. Observe the following pattern and find the missing numbers.**

**11 ^{2} = 121**

**101 ^{2} = 10201**

**1001 ^{2} = 1002001**

**100001 ^{2} = 1 …….2……1**

**10000001 ^{2} = ………………**

**Solution:**

According to the above pattern:

100001^{2} = 10000200001

10000001^{2} = 100000020000001

**5. Observe the following pattern and supply the missing numbers.**

**11 ^{2} = 121**

**101 ^{2} = 10201**

**10101 ^{2} = 102030201**

**1010101 ^{2} = ………………………**

**………… ^{2 }= 10203040504030201**

**Solution:**

According to the above pattern:

1010101^{2} =1020304030201

101010101^{2} =10203040505030201

**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:**

According to the given pattern, we have

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

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

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

**7. Without adding, find the sum.**

**We know that the sum of n odd numbers = n**^{2}

**i. 1 + 3 + 5 + 7 + 9**

**Solution:**

Sum of first five odd number = (5)^{2} = 25

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

**Solution:**

Sum of first ten odd number = (10)^{ 2} = 100

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

**Solution:**

Sum of first thirteen odd number = (12)^{ 2} = 144

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

Solution:

We know, sum of first n odd natural numbers = n^{2}

Since,49 = 7^{2}

Here n = 7

Hence, 49 = sum of first 7 odd natural numbers

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

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

Solution:

Since, 121 = 11^{2}

Here n = 11

Hence, 121 = sum of first 11 odd natural numbers

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

**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 between n^{2} and (n+1)^{2}, there are 2n non–perfect square numbers.

Now,

i. 12^{2} and 13^{2} there are 2×12 = 24 natural numbers.

ii. 25^{2} and 26^{2} there are 2×25 = 50 natural numbers.

iii. 99^{2} and 100^{2} there are 2×99 =198 natural numbers.

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