**Exercise 3 (A)**

**Question 1**

**Find the square of the following numbers :**

**(i) 15**

**(ii) 48**

**(iii) **

**(iv) **

**(v) **

**(vi) 0.9**

**(vii) 1.1**

**(viii) 0.018**

Sol :

**Question **

**Determine whether square of the following numbers will be even or odd.(Do not find the square)**

Note : Square of even number are even

Square of odd number is odd

**(i) 529**

Sol :

529 is odd and we know square of odd number is odd

**(ii) 30976**

Sol :

30976 is even and we know square of even number is even

**(iii) 893025**

Sol :

893025 is odd and we know square of odd number is odd

**(iv) 6990736**

Sol :

6990736 is even and we know square of even number is even

**Question 3**

**Just by looking at the following numbers, give reason why they are not perfect squares.**

Note : Any number ending with 2,3,7,8 and having odd number of zeros are not perfect square

**(i) 4893**

Sol : Not a perfect square

**(ii) 65000**

Sol : Not a perfect square

**(iii) 4422**

Sol : Perfect square

**(iv) 89138**

Sol : Perfect square

**(v) 150087**

Sol : Not a perfect square

**Question 4**

**Using prime factorization method, find which of the following are perfect square numbers.**

Note : For a number to be a perfect square it’s prime factors should be to a power of 2 or any even number.

**(i) 100**

Sol :

2 | 100 |

2 | 50 |

5 | 25 |

5 | 5 |

1 |

100=2×2×5×5 =2^{2}×5^{2}

Since the powers of the prime factors are even therefore 100 is a perfect square.

**(ii) 284**

Sol :

2 | 284 |

2 | 142 |

71 | 71 |

1 |

281=2×2×71 =2^{2}×71^{1}^{ }

Since the powers of the prime factors are not even (i.e. 71^{1}) therefore 284 is not a perfect square.

**(iii) 784**

Sol :

2 | 784 |

2 | 392 |

2 | 196 |

2 | 98 |

7 | 49 |

7 | 7 |

1 |

784=2×2×2×2×7×7 =2^{4}×7^{2}

Since the powers of the prime factors are even therefore 784 is a perfect square.

**(iv) 1444**

Sol :

2 | 1444 |

2 | 722 |

19 | 361 |

19 | 19 |

1 |

1444=2×2×19×19 =2^{2}×19^{2}

Since the powers of the prime factors are even therefore 1444 is a perfect square.

**(v) 768**

Sol :

2 | 768 |

2 | 384 |

2 | 192 |

2 | 96 |

2 | 48 |

2 | 24 |

2 | 12 |

2 | 6 |

3 | 3 |

1 |

768=2×2×2×2×2×2×2×2×3 =2^{8}×3^{1}

Since the powers of the prime factors are not even therefore 768 is not a perfect square.

**(vi) 4225**

Sol :

5 | 4225 |

5 | 845 |

13 | 169 |

13 | 13 |

1 |

4225=5×5×13×13 =5^{2}×13^{2}

Since the powers of the prime factors are even therefore 4225 is a perfect square.

**(vii) 3375**

Sol :

3 | 3375 |

3 | 1125 |

3 | 375 |

5 | 125 |

5 | 25 |

5 | 5 |

1 |

3375=3×3×3×5×5×5 =3^{3}×5^{3}

Since the powers of the prime factors are not even therefore 3375 is not a perfect square.

**(viii) 15625**

Sol :

5 | 15625 |

5 | 3125 |

5 | 625 |

5 | 125 |

5 | 25 |

5 | 5 |

1 |

15625=5×5×5×5×5×5 =5^{6}

Since the powers of the prime factors are even therefore 15625 is a perfect square.

**Question 5**

**Using prime factorization method, find the square root of the following numbers.**

**(i) 256**

Sol :

2 | 256 |

2 | 128 |

2 | 64 |

2 | 32 |

2 | 16 |

2 | 8 |

2 | 4 |

2 | 2 |

1 |

√15625=√2×2×2×2×2×2×2×2 =2^{4}=16

**(ii) 1936**

Sol :

2 | 1936 |

2 | 968 |

2 | 484 |

2 | 242 |

11 | 121 |

11 | 11 |

1 |

√1936=√2×2×2×2×11×11=2×2×11=44

**(iii) 3364**

Sol :

2 | 3364 |

2 | 1682 |

29 | 841 |

29 | 29 |

1 |

√3364=√2×2×29×29=2×29=58

**(iv) 5625**

Sol :

3 | 5625 |

3 | 1875 |

5 | 625 |

5 | 125 |

5 | 25 |

5 | 5 |

1 |

√5625=√3×3×5×5×5×5=3×5×5=75

**(v) 7744**

Sol :

2 | 7744 |

2 | 3872 |

2 | 1936 |

2 | 968 |

2 | 484 |

2 | 242 |

11 | 121 |

11 | 11 |

1 |

√7744=√2×2×2×2×2×2×11×11=2×2×2×11=88

**(vi) 12544**

Sol :

2 | 12544 |

2 | 6272 |

2 | 3136 |

2 | 1568 |

2 | 784 |

2 | 392 |

2 | 196 |

2 | 98 |

7 | 49 |

7 | 7 |

1 |

√12544=√2×2×2×2×2×2×2×2×7×7=2×2×2×2×7=112

**Question 6**

**Find the square root of the following fractions.**

**(i) **

**(ii) **

**(iii) **

**(iv) 0.01**

**(v) 0.1764**

**(vi) 0.00003136**

Sol :

**Question 7**

**Simplify:**

**(i) **

**(ii) **

**(iii) **

Sol :

**Question 8**

**Find the square root of 1024. Hence find the value of .**

Sol :

**Question 9**

**Find the smallest number by which each of the given number s ,ust be multiplied so that the product is a perfect square. Also find the square root of this new number.**

**(i) 175**

Sol :

**(ii) 325**

Sol :

**(iii) 720**

Sol :

2 | 720 |

2 | 360 |

2 | 180 |

2 | 90 |

3 | 45 |

3 | 15 |

5 | 5 |

1 |

Resolving Prime factors

720= (2×2)×(2×2)×(3×3)×5

Above ,product of 2 and 3 occur in pairs whereas 5 does not exist in pair so we have to multiply 720 by 5 to make it a perfect square

Perfect square number=3600

Square root=√(2×2)×(2×2)×(3×3)×(5×5)

=2×2×3×5 =60

**(iv) 3150**

Sol :

2 | 3150 |

3 | 1575 |

3 | 525 |

5 | 175 |

5 | 35 |

7 | 7 |

1 |

Resolving Prime factors

3150= 2×(3×3)×(5×5)×7

Above ,product of 3 and 5 occur in pairs whereas 2 and 7 does not exist in pair so we have to multiply 3150 by 2 and 7 (i.e. 14) to make it a perfect square

Perfect square number=44100

Square root=√(2×2)×(3×3)×(5×5)×(7×7)

=2×3×5×7 =210

**Question 10**

**Find the smallest number by which each of the given numbers must be divided so that the quotient is a perfect square.Also find the square root of this quotient.**

**(i) 2700**

Sol :

2 | 2700 |

2 | 1350 |

3 | 675 |

3 | 225 |

3 | 75 |

5 | 25 |

5 | 5 |

1 |

Resolving Prime factors

2700= (2×2)×(3×3)×3×(5×5)

Above ,product of 2 , 3 and 5 occur in pairs but their is a single 3 extra so we have to divide 2700 by 3 to make it a perfect square

Perfect square number=900

Square root=√(2×2)×(3×3)×(5×5)

=2×3×5 =30

**(ii) 5488**

Sol :

2 | 5488 |

2 | 2744 |

2 | 1372 |

2 | 686 |

7 | 343 |

7 | 49 |

7 | 7 |

1 |

Resolving Prime factors

5488= (2×2)×(2×2)×(7×7)×7

Above ,product of 2 and 7 occur in pairs but their is a single 7 extra so we have to divide 5488 by 7 to make it a perfect square

Perfect square number=784

Square root=√(2×2)×(2×2)×(7×7)

=2×2×7 =28

**(iii) 7203**

Sol :

3 | 7203 |

7 | 2401 |

7 | 343 |

7 | 49 |

7 | 7 |

1 |

Resolving Prime factors

7203= 3×(7×7)×(7×7)

Above ,product of 7 occur in pairs but their is a single 3 extra so we have to divide 7203 by 3 to make it a perfect square

Perfect square number=2401

Square root=√(7×7)×(7×7)

=7×7 =49

**(iv) 20886**

Sol :

2 | 20886 |

3 | 10443 |

59 | 3481 |

59 | 59 |

1 |

Resolving Prime factors

7203= 3×(7×7)×(7×7)

Above ,product of 7 occur in pairs but their is a single 3 extra so we have to divide 7203 by 3 to make it a perfect square

Perfect square number=2401

Square root=√(7×7)×(7×7)

=7×7 =49

**Question 11**

**Find the smallest number which is completely divisible by each of the numbers 10 , 16 and 24.**

Sol :

Here, we need to find the perfect number divisible by square of L.C.M of 10 , 16 and 24

L.C.M of 10 ,16 , 24 is =2×2×2×2×3×5 =240

2 | 10 , 16 , 24 |

2 | 5 , 8 , 12 |

2 | 5 , 4 , 6 |

2 | 5 , 2 , 3 |

3 | 5 , 1 , 3 |

5 | 5 , 1 , 1 |

1 , 1 , 1 |

Resolving Prime factors

240= (2×2)×(2×2)×3×5

To make it into a perfect square is should be multiplied by 3×5=15

∴Required square number =240×15=3600

**Question 12**

**The children of class VIII of a school contributed ₹ 3025 for the Prime Minister’s National Relief Fund, contributing as much money as there are children in the class.How many children are there in the class ?**

Sol :

Let the number of children be x

Then, according to question,

Money donated by individual = number of children = x

Now,

Total money donated = money donated by individual × number of children

⇒x×x=3025

⇒x^{2}=3025

⇒x=√3025

⇒x=√5×5×11×11

⇒x=5×11=55

Hence there are 55 children in the class and all of them donated 55 ₹ each