# NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots Exercise 6.3

NCERT Solutions have been carefully compiled and developed as per the latest CBSE syllabus. NCERT solutions for Class 8 maths, Chapter 6 Exercise 6.3, exercise questions and answers helps students to understand the topics and concepts related to Squares and Square Roots.

NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots Exercise 6.3 are prepared using step by step approach, with an aim to improving problem-solving skills.

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### Access other exercise solutions of Class 8 Maths Chapter 6- Squares and Square Roots

Exercise 6.1 Solutions 9 Questions

Exercise 6.2 Solutions 2 Questions

Exercise 6.4 Solutions 9 Questions

### Access Answers of Maths NCERT Class 8 Chapter 6- Squares and Square Roots Exercise 6.3 Page Number 102

1. What could be the possible â€˜oneâ€™sâ€™ 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 unitâ€™s digit of the square of a number having digit as unitâ€™s

place 1 is 1 and also 9 is 1[92=81 whose unit place is 1].

âˆ´ Unitâ€™s digit of the square root of number 9801 is equal to 1 or 9.

ii. We know that the unitâ€™s digit of the square of a number having digit as unitâ€™s

place 6 is 6 and also 4 is 6 [62=36 and 42=16, both the squares have unit digit 6].

âˆ´ Unitâ€™s digit of the square root of number 99856 is equal to 6.

iii. We know that the unitâ€™s digit of the square of a number having digit as unitâ€™s

place 1 is 1 and also 9 is 1[92=81 whose unit place is 1].

âˆ´ Unitâ€™s digit of the square root of number 998001 is equal to 1 or 9.

iv. We know that the unitâ€™s digit of the square of a number having digit as unitâ€™s

place 5 is 5.

âˆ´ Unitâ€™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. 153âŸ¹ Ends with 3.

âˆ´, 153 is not a perfect square

ii. 257âŸ¹ Ends with 7

âˆ´, 257 is not a perfect square

iii. 408âŸ¹ Ends with 8

âˆ´, 408 is not a perfect square

iv. 441âŸ¹ Ends with 1

âˆ´, 441 is a perfect square.

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

Solution:

100

100 â€“ 1 = 99

99 â€“ 3 = 96

96 â€“ 5 = 91

91 â€“ 7 = 84

84 â€“ 9 = 75

75 â€“ 11 = 64

64 â€“ 13 = 51

51 â€“ 15 = 36

36 â€“ 17 = 19

19 â€“ 19 = 0

Here, we have performed subtraction ten times.

âˆ´ âˆš100 = 10

169

169 â€“ 1 = 168

168 â€“ 3 = 165

165 â€“ 5 = 160

160 â€“ 7 = 153

153 â€“ 9 = 144

144 â€“ 11 = 133

133 â€“ 13 = 120

120 â€“ 15 = 105

105 â€“ 17 = 88

88 â€“ 19 = 69

69 â€“ 21 = 48

48 â€“ 23 = 25

25 â€“ 25 = 0

Here, we have performed subtraction thirteen times.

âˆ´ âˆš169 = 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.

729 = 3Ã—3Ã—3Ã—3Ã—3Ã—3Ã—1

â‡’ 729 = (3Ã—3)Ã—(3Ã—3)Ã—(3Ã—3)

â‡’ 729 = (3Ã—3Ã—3)Ã—(3Ã—3Ã—3)

â‡’ 729 = (3Ã—3Ã—3)2

â‡’ âˆš729 = 3Ã—3Ã—3 = 27

ii.

400 = 2Ã—2Ã—2Ã—2Ã—5Ã—5Ã—1

â‡’ 400 = (2Ã—2)Ã—(2Ã—2)Ã—(5Ã—5)

â‡’ 400 = (2Ã—2Ã—5)Ã—(2Ã—2Ã—5)

â‡’ 400 = (2Ã—2Ã—5)2

â‡’ âˆš400 = 2Ã—2Ã—5 = 20

iii.

1764 = 2Ã—2Ã—3Ã—3Ã—7Ã—7

â‡’ 1764 = (2Ã—2)Ã—(3Ã—3)Ã—(7Ã—7)

â‡’ 1764 = (2Ã—3Ã—7)Ã—(2Ã—3Ã—7)

â‡’ 1764 = (2Ã—3Ã—7)2

â‡’ âˆš1764 = 2 Ã—3Ã—7 = 42

iv.

4096 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2

â‡’ 4096 = (2Ã—2)Ã—(2Ã—2)Ã—(2Ã—2)Ã—(2Ã—2)Ã—(2Ã—2)Ã—(2Ã—2)

â‡’ 4096 = (2Ã—2Ã—2Ã—2Ã—2Ã—2)Ã—(2Ã—2Ã—2Ã—2Ã—2Ã—2)

â‡’ 4096 = (2Ã—2Ã—2Ã—2Ã—2Ã—2)2

â‡’ âˆš4096 = 2Ã—2Ã—2 Ã—2Ã—2Ã—2 = 64

v.

7744 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—11Ã—11Ã—1

â‡’ 7744 = (2Ã—2)Ã—(2Ã—2)Ã—(2Ã—2)Ã—(11Ã—11)

â‡’ 7744 = (2Ã—2Ã—2Ã—11)Ã—(2Ã—2Ã—2Ã—11)

â‡’ 7744 = (2Ã—2Ã—2Ã—11)2

â‡’ âˆš7744 = 2Ã—2Ã—2Ã—11 = 88

vi.

9604 = 62 Ã— 2 Ã— 7 Ã— 7 Ã— 7 Ã— 7

â‡’ 9604 = ( 2 Ã— 2 ) Ã— ( 7 Ã— 7 ) Ã— ( 7 Ã— 7 )

â‡’ 9604 = ( 2 Ã— 7 Ã—7 ) Ã— ( 2 Ã— 7 Ã—7 )

â‡’ 9604 = ( 2Ã—7Ã—7 )2

â‡’ âˆš9604 = 2Ã—7Ã—7 = 98

vii.

5929 = 7Ã—7Ã—11Ã—11

â‡’ 5929 = (7Ã—7)Ã—(11Ã—11)

â‡’ 5929 = (7Ã—11)Ã—(7Ã—11)

â‡’ 5929 = (7Ã—11)2

â‡’ âˆš5929 = 7Ã—11 = 77

viii.

9216 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—1

â‡’ 9216 = (2Ã—2)Ã—(2Ã—2) Ã— ( 2 Ã— 2 ) Ã— ( 2 Ã— 2 ) Ã— ( 2 Ã— 2 ) Ã— ( 3 Ã— 3 )

â‡’ 9216 = ( 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3) Ã— ( 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3)

â‡’ 9216 = 96 Ã— 96

â‡’ 9216 = ( 96 )2

â‡’ âˆš9216 = 96

ix.

529 = 23Ã—23

529 = (23)2

âˆš529 = 23

x.

8100 = 2Ã—2Ã—3Ã—3Ã—3Ã—3Ã—5Ã—5Ã—1

â‡’ 8100 = (2Ã—2) Ã—(3Ã—3)Ã—(3Ã—3)Ã—(5Ã—5)

â‡’ 8100 = (2Ã—3Ã—3Ã—5)Ã—(2Ã—3Ã—3Ã—5)

â‡’ 8100 = 90Ã—90

â‡’ 8100 = (90)2

â‡’ âˆš8100 = 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.

252 = 2Ã—2Ã—3Ã—3Ã—7

= (2Ã—2)Ã—(3Ã—3)Ã—7

Here, 7 cannot be paired.

âˆ´ We will multiply 252 by 7 to get perfect square.

New number = 252 Ã— 7 = 1764

1764 = 2Ã—2Ã—3Ã—3Ã—7Ã—7

â‡’ 1764 = (2Ã—2)Ã—(3Ã—3)Ã—(7Ã—7)

â‡’ 1764 = 22Ã—32Ã—72

â‡’ 1764 = (2Ã—3Ã—7)2

â‡’ âˆš1764 = 2Ã—3Ã—7 = 42

ii.

180 = 2Ã—2Ã—3Ã—3Ã—5

= (2Ã—2)Ã—(3Ã—3)Ã—5

Here, 5 cannot be paired.

âˆ´ We will multiply 180 by 5 to get perfect square.

New number = 180 Ã— 5 = 900

900 = 2Ã—2Ã—3Ã—3Ã—5Ã—5Ã—1

â‡’ 900 = (2Ã—2)Ã—(3Ã—3)Ã—(5Ã—5)

â‡’ 900 = 22Ã—32Ã—52

â‡’ 900 = (2Ã—3Ã—5)2

â‡’ âˆš900 = 2Ã—3Ã—5 = 30

iii.

1008 = 2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—7

= (2Ã—2)Ã—(2Ã—2)Ã—(3Ã—3)Ã—7

Here, 7 cannot be paired.

âˆ´ We will multiply 1008 by 7 to get perfect square.

New number = 1008Ã—7 = 7056

7056 = 2Ã—2Ã—2Ã—2Ã—3Ã—3Ã—7Ã—7

â‡’ 7056 = (2Ã—2)Ã—(2Ã—2)Ã—(3Ã—3)Ã—(7Ã—7)

â‡’ 7056 = 22Ã—22Ã—32Ã—72

â‡’ 7056 = (2Ã—2Ã—3Ã—7)2

â‡’ âˆš7056 = 2Ã—2Ã—3Ã—7 = 84

iv.

2028 = 2Ã—2Ã—3Ã—13Ã—13

= (2Ã—2)Ã—(13Ã—13)Ã—3

Here, 3 cannot be paired.

âˆ´ We will multiply 2028 by 3 to get perfect square. New number = 2028Ã—3 = 6084

6084 = 2Ã—2Ã—3Ã—3Ã—13Ã—13

â‡’ 6084 = (2Ã—2)Ã—(3Ã—3)Ã—(13Ã—13)

â‡’ 6084 = 22Ã—32Ã—132

â‡’ 6084 = (2Ã—3Ã—13)2

â‡’ âˆš6084 = 2Ã—3Ã—13 = 78

v.

1458 = 2Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3

= (3Ã—3)Ã—(3Ã—3)Ã—(3Ã—3)Ã—2

Here, 2 cannot be paired.

âˆ´ We will multiply 1458 by 2 to get perfect square. New number = 1458 Ã— 2 = 2916

2916 = 2Ã—2Ã—3Ã—3Ã—3Ã—3Ã—3Ã—3

â‡’ 2916 = (3Ã—3)Ã—(3Ã—3)Ã—(3Ã—3)Ã—(2Ã—2)

â‡’ 2916 = 32Ã—32Ã—32Ã—22

â‡’ 2916 = (3Ã—3Ã—3Ã—2)2

â‡’ âˆš2916 = 3Ã—3Ã—3Ã—2 = 54

vi.

768 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3

= (2Ã—2)Ã—(2Ã—2)Ã—(2Ã—2)Ã—(2Ã—2)Ã—3

Here, 3 cannot be paired.

âˆ´ We will multiply 768 by 3 to get perfect square.

New number = 768Ã—3 = 2304

2304 = 2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—2Ã—3Ã—3

â‡’ 2304 = (2Ã—2)Ã—(2Ã—2)Ã—(2Ã—2)Ã—(2Ã—2)Ã—(3Ã—3)

â‡’ 2304 = 22Ã—22Ã—22Ã—22Ã—32

â‡’ 2304 = (2Ã—2Ã—2Ã—2Ã—3)2

â‡’ âˆš2304 = 2Ã—2Ã—2Ã—2Ã—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 = 2Ã—2Ã—3Ã—3Ã—7

= (2Ã—2)Ã—(3Ã—3)Ã—7

Here, 7 cannot be paired.

âˆ´ We will divide 252 by 7 to get perfect square. New number = 252 Ã· 7 = 36

36 = 2Ã—2Ã—3Ã—3

â‡’ 36 = (2Ã—2)Ã—(3Ã—3)

â‡’ 36 = 22Ã—32

â‡’ 36 = (2Ã—3)2

â‡’ âˆš36 = 2Ã—3 = 6

ii.

2925 = 3Ã—3Ã—5Ã—5Ã—13

= (3Ã—3)Ã—(5Ã—5)Ã—13

Here, 13 cannot be paired.

âˆ´ We will divide 2925 by 13 to get perfect square. New number = 2925 Ã· 13 = 225

225 = 3Ã—3Ã—5Ã—5

â‡’ 225 = (3Ã—3)Ã—(5Ã—5)

â‡’ 225 = 32Ã—52

â‡’ 225 = (3Ã—5)2

â‡’ âˆš36 = 3Ã—5 = 15

iii.

396 = 2Ã—2Ã—3Ã—3Ã—11

= (2Ã—2)Ã—(3Ã—3)Ã—11

Here, 11 cannot be paired.

âˆ´ We will divide 396 by 11 to get perfect square. New number = 396 Ã· 11 = 36

36 = 2Ã—2Ã—3Ã—3

â‡’ 36 = (2Ã—2)Ã—(3Ã—3)

â‡’ 36 = 22Ã—32

â‡’ 36 = (2Ã—3)2

â‡’ âˆš36 = 2Ã—3 = 6

iv.

2645 = 5Ã—23Ã—23

â‡’ 2645 = (23Ã—23)Ã—5

Here, 5 cannot be paired.

âˆ´ We will divide 2645 by 5 to get perfect square.

New number = 2645 Ã· 5 = 529

529 = 23Ã—23

â‡’ 529 = (23)2

â‡’ âˆš529 = 23

v.

2800 = 2Ã—2Ã—2Ã—2Ã—5Ã—5Ã—7

= (2Ã—2)Ã—(2Ã—2)Ã—(5Ã—5)Ã—7

Here, 7 cannot be paired.

âˆ´ We will divide 2800 by 7 to get perfect square. New number = 2800 Ã· 7 = 400

400 = 2Ã—2Ã—2Ã—2Ã—5Ã—5

â‡’ 400 = (2Ã—2)Ã—(2Ã—2)Ã—(5Ã—5)

â‡’ 400 = (2Ã—2Ã—5)2

â‡’ âˆš400 = 20

vi.

1620 = 2Ã—2Ã—3Ã—3Ã—3Ã—3Ã—5

= (2Ã—2)Ã—(3Ã—3)Ã—(3Ã—3)Ã—5

Here, 5 cannot be paired.

âˆ´ We will divide 1620 by 5 to get perfect square. New number = 1620 Ã· 5 = 324

324 = 2Ã—2Ã—3Ã—3Ã—3Ã—3

â‡’ 324 = (2Ã—2)Ã—(3Ã—3)Ã—(3Ã—3)

â‡’ 324 = (2Ã—3Ã—3)2

â‡’ âˆš324 = 18

7. The students of Class VIII of a school donated Rs 2401 in all, for Prime Ministerâ€™s National Relief Fund. Each student donated as many rupees as the number of students in the class. Find the number of students in the class.

Solution:

Let the number of students in the school be, x.

âˆ´ Each student donate Rs.x .

Total many contributed by all the students= xÃ—x=x2 Given, x2 = Rs.2401

x2 = 7Ã—7Ã—7Ã—7

â‡’ x2 = (7Ã—7)Ã—(7Ã—7)

â‡’ x2 = 49Ã—49

â‡’ x = âˆš(49Ã—49)

â‡’ x = 49

âˆ´ 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.

âˆ´ the number of plants in each rows = x.

Total many contributed by all the students = x Ã— x =x2

Given,

x2 = Rs.2025

x2 = 3Ã—3Ã—3Ã—3Ã—5Ã—5

â‡’ x2 = (3Ã—3)Ã—(3Ã—3)Ã—(5Ã—5)

â‡’ x2 = (3Ã—3Ã—5)Ã—(3Ã—3Ã—5)

â‡’ x2 = 45Ã—45

â‡’ x = âˆš45Ã—45

â‡’ x = 45

âˆ´ The number of rows = 45 and the number of plants in each rows = 45.

9. Find the smallest square number that is divisible by each of the numbers 4, 9 and 10.

Solution:

L.C.M of 4, 9 and 10 is (2Ã—2Ã—9Ã—5) 180.

180 = 2Ã—2Ã—9Ã—5

= (2Ã—2)Ã—3Ã—3Ã—5

= (2Ã—2)Ã—(3Ã—3)Ã—5

Here, 5 cannot be paired.

âˆ´ we will multiply 180 by 5 to get perfect square.

Hence, the smallest square number divisible by 4, 9 and 10 = 180Ã—5 = 900

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 (2Ã—2Ã—5Ã—2Ã—3) 120.

120 = 2Ã—2Ã—3Ã—5Ã—2

= (2Ã—2)Ã—3Ã—5Ã—2

Here, 3, 5 and 2 cannot be paired.

âˆ´ We will multiply 120 by (3Ã—5Ã—2) 30 to get perfect square.

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

Exercise 6.3 of NCERT Solutions for Class 8 Maths Chapter 6- Squares and Square Roots is based on the following topics:

1. Square Roots
• Finding square roots
• Finding square root through repeated subtraction
• Finding square root through prime factorisation