Exercise 5.3

**1. What could be the possible ‘one’s’ digits of the square root of each of the following numbers?**

**i. 9801**

**ii. 99856**

**iii. 998001**

**iv. 657666025**

Solution:

i. We know that the unit’s digit of the square of a number having digit as unit’s

place 1 is 1 and also 9 is 1[9^{2}=81 whose unit place is 1].

∴ Unit’s digit of the square root of number 9801 is equal to 1 or 9.

ii. We know that the unit’s digit of the square of a number having digit as unit’s

place 6 is 6 and also 4 is 6 [6^{2}=36 and 4^{2}=16, both the squares have unit digit 6].

∴ Unit’s digit of the square root of number 99856 is equal to 6.

iii. We know that the unit’s digit of the square of a number having digit as unit’s

place 1 is 1 and also 9 is 1[9^{2}=81 whose unit place is 1].

∴ Unit’s digit of the square root of number 998001 is equal to 1 or 9.

iv. We know that the unit’s digit of the square of a number having digit as unit’s

place 5 is 5.

∴ Unit’s digit of the square root of number 657666025 is equal to 5.

**2. Without doing any calculation, find the numbers which are surely not perfect squares.**

**i. 153**

**ii. 257**

**iii. 408**

**iv. 441**

Solution:

We know that natural numbers ending with the digits 0, 2, 3, 7 and 8 are not perfect square.

i. 153⟹ Ends with 3.

∴, 153 is not a perfect square

ii. 257⟹ Ends with 7

∴, 257 is not a perfect square

iii. 408⟹ Ends with 8

∴, 408 is not a perfect square

iv. 441⟹ Ends with 1

∴, 441 is a perfect square.

**3. Find the square roots of 100 and 169 by the method of repeated subtraction**.

Solution:

100

100 – 1 = 99

99 – 3 = 96

96 – 5 = 91

91 – 7 = 84

84 – 9 = 75

75 – 11 = 64

64 – 13 = 51

51 – 15 = 36

36 – 17 = 19

19 – 19 = 0

Here, we have performed subtraction ten times.

∴ √100 = 10

169

169 – 1 = 168

168 – 3 = 165

165 – 5 = 160

160 – 7 = 153

153 – 9 = 144

144 – 11 = 133

133 – 13 = 120

120 – 15 = 105

105 – 17 = 88

88 – 19 = 69

69 – 21 = 48

48 – 23 = 25

25 – 25 = 0

Here, we have performed subtraction thirteen times.

∴ √169 = 13

**4. Find the square roots of the following numbers by the Prime Factorisation Method.**

**i. 729**

**ii. 400**

**iii. 1764**

**iv. 4096**

**v. 7744**

**vi. 9604**

**vii. 5929**

**viii. 9216**

**ix. 529**

**x. 8100**

Solution:

i )

4096 = 2×2×2×2×2×2×2×2×2×2×2×2

⇒ 4096 = (2×2)×(2×2)×(2×2)×(2×2)×(2×2)×(2×2)

⇒ 4096 = (2×2×2×2×2×2)×(2×2×2×2×2×2)

⇒ 4096 = (2×2×2×2×2×2)^{2}

⇒ √4096 = 2×2×2 ×2×2×2 = 64

vi.

5929 = 7×7×11×11

⇒ 5929 = (7×7)×(11×11)

⇒ 5929 = (7×11)×(7×11)

⇒ 5929 = (7×11)^{2}

⇒ √5929 = 7×11 = 77

viii.

8100 = 2×2×3×3×3×3×5×5×1

⇒ 8100 = (2×2) ×(3×3)×(3×3)×(5×5)

⇒ 8100 = (2×3×3×5)×(2×3×3×5)

⇒ 8100 = 90×90

⇒ 8100 = (90)^{2}

⇒ √8100 = 90

**5. For each of the following numbers, find the smallest whole number by which it should be multiplied so as to get a perfect square number. Also find the square root of the square number so obtained.**

**i. 252**

**ii. 180**

**iii. 1008**

**iv. 2028**

**v. 1458**

**vi. 768**

Solution:

i.

1764 = 2×2×3×3×7×7

⇒ 1764 = (2×2)×(3×3)×(7×7)

⇒ 1764 = 2^{2}×3^{2}×7^{2}

⇒ 1764 = (2×3×7)^{2}

⇒ √1764 = 2×3×7 = 42

ii.

900 = 2×2×3×3×5×5×1

⇒ 900 = (2×2)×(3×3)×(5×5)

⇒ 900 = 2^{2}×3^{2}×5^{2}

⇒ 900 = (2×3×5)^{2}

⇒ √900 = 2×3×5 = 30

iii.

1008 = 2×2×2×2×3×3×7

= (2×2)×(2×2)×(3×3)×7

Here, 7 cannot be paired.

∴ We will multiply 1008 by 7 to get perfect square.

New number = 1008×7 = 7056

7056 = 2×2×2×2×3×3×7×7

⇒ 7056 = (2×2)×(2×2)×(3×3)×(7×7)

⇒ 7056 = 2^{2}×2^{2}×3^{2}×7^{2}

⇒ 7056 = (2×2×3×7)^{2}

⇒ √7056 = 2×2×3×7 = 84

iv.

2028 = 2×2×3×13×13

= (2×2)×(13×13)×3

Here, 3 cannot be paired.

∴ We will multiply 2028 by 3 to get perfect square. New number = 2028×3 = 6084

6084 = 2×2×3×3×13×13

⇒ 6084 = (2×2)×(3×3)×(13×13)

⇒ 6084 = 2^{2}×3^{2}×13^{2}

⇒ 6084 = (2×3×13)^{2}

⇒ √6084 = 2×3×13 = 78

v.

1458 = 2×3×3×3×3×3×3

= (3×3)×(3×3)×(3×3)×2

Here, 2 cannot be paired.

∴ We will multiply 1458 by 2 to get perfect square. New number = 1458 × 2 = 2916

2916 = 2×2×3×3×3×3×3×3

⇒ 2916 = (3×3)×(3×3)×(3×3)×(2×2)

⇒ 2916 = 3^{2}×3^{2}×3^{2}×2^{2}

⇒ 2916 = (3×3×3×2)^{2}

⇒ √2916 = 3×3×3×2 = 54

vi.

768 = 2×2×2×2×2×2×2×2×3

= (2×2)×(2×2)×(2×2)×(2×2)×3

Here, 3 cannot be paired.

∴ We will multiply 768 by 3 to get perfect square.

New number = 768×3 = 2304

2304 = 2×2×2×2×2×2×2×2×3×3

⇒ 2304 = (2×2)×(2×2)×(2×2)×(2×2)×(3×3)

⇒ 2304 = 2^{2}×2^{2}×2^{2}×2^{2}×3^{2}

⇒ 2304 = (2×2×2×2×3)^{2}

⇒ √2304 = 2×2×2×2×3 = 48

**6. For each of the following numbers, find the smallest whole number by which it should be divided so as to get a perfect square. Also find the square root of the square number so obtained.**

**i. 252**

**ii. 2925**

**iii. 396**

**iv. 2645**

**v. 2800**

**vi. 1620**

Solution:

i.

252 = 2×2×3×3×7

= (2×2)×(3×3)×7

Here, 7 cannot be paired.

∴ We will divide 252 by 7 to get perfect square. New number = 252 ÷ 7 = 36

36 = 2×2×3×3

⇒ 36 = (2×2)×(3×3)

⇒ 36 = 2^{2}×3^{2}

⇒ 36 = (2×3)^{2}

⇒ √36 = 2×3 = 6

ii.

2925 = 3×3×5×5×13

= (3×3)×(5×5)×13

Here, 13 cannot be paired.

∴ We will divide 2925 by 13 to get perfect square. New number = 2925 ÷ 13 = 225

225 = 3×3×5×5

⇒ 225 = (3×3)×(5×5)

⇒ 225 = 3^{2}×5^{2}

⇒ 225 = (3×5)^{2}

⇒ √36 = 3×5 = 15

iii.

396 = 2×2×3×3×11

= (2×2)×(3×3)×11

Here, 11 cannot be paired.

∴ We will divide 396 by 11 to get perfect square. New number = 396 ÷ 11 = 36

36 = 2×2×3×3

⇒ 36 = (2×2)×(3×3)

⇒ 36 = 2^{2}×3^{2}

⇒ 36 = (2×3)^{2}

⇒ √36 = 2×3 = 6

iv.

2645 = 5×23×23

⇒ 2645 = (23×23)×5

Here, 5 cannot be paired.

∴ We will divide 2645 by 5 to get perfect square.

New number = 2645 ÷ 5 = 529

529 = 23×23

⇒ 529 = (23)^{2}

⇒ √529 = 23

v.

2800 = 2×2×2×2×5×5×7

= (2×2)×(2×2)×(5×5)×7

Here, 7 cannot be paired.

∴ We will divide 2800 by 7 to get perfect square. New number = 2800 ÷ 7 = 400

x^{2} = 7×7×7×7

⇒ x^{2} = (7×7)×(7×7)

⇒ x^{2 }= 49×49

⇒ x = √(49×49)

⇒ x = 49

∴ The number of students = 49

**8. 2025 plants are to be planted in a garden in such a way that each row contains as many plants as the number of rows. Find the number of rows and the number of plants in each row.**

Solution

Let the number of rows be, x.

∴ the number of plants in each rows = x.

Total many contributed by all the students = x × x =x^{2}

Given,

x_{2} = Rs.2025

**10. Find the smallest square number that is divisible by each of the numbers 8, 15 and 20.**

Solution:

L.C.M of 8, 15 and 20 is (2×2×5×2×3) 120.

120 = 2×2×3×5×2

= (2×2)×3×5×2

Here, 3, 5 and 2 cannot be paired.

∴ We will multiply 120 by (3×5×2) 30 to get perfect square.

Hence, the smallest square number divisible by 8, 15 and 20 =120×30 = 3600