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# Class 8 NCERT Solutions – Chapter 6 Squares and Square Roots – Exercise 6.3

• Last Updated : 13 Nov, 2020

### Question 1. What could be the possible oneâ€™s digits of the square root of each of the following numbers?

i. 9801

Solution:

Unit place digit of the number is 1

And we all know 12 = 1 & 92 = 81 whose unit place is 1

Therefore, oneâ€™s digit of the square root of 9801 should equal to 1 or 9.

ii. 99856

Solution:

Unit place digit of the number is 6

And we all know 62 = 36 & 42 = 16, both the squares have unit place 6.

Therefore, oneâ€™s digit of the square root of 99856 is equal to 6 or 4.

iii. 998001

Solution:

Unit place digit of the number is 1

And we all know 12 = 1 & 92 = 81 whose unit place is 1

Therefore, oneâ€™s digit of the square root of 998001 should equal to 1 or 9.

iv. 657666025

Solution:

Unit place digit of the number is 5

And we all know 52 = 25 whose unit place is 5

Therefore, oneâ€™s digit of the square root of 657666025 should equal to 5.

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

i. 153

Solution:

Unit place digit of the number is 3.

Therefore, 153 is not a perfect square [As natural numbers having Unit place digits as 0, 2, 3, 7 and 8 are not perfect square].

ii. 257

Solution:

Unit place digit of the number is 7.

Therefore, 257 is not a perfect square [As natural numbers having Unit place digits as 0, 2, 3, 7 and 8 are not perfect square].

iii. 408

Solution:

Unit place digit of the number is 8.

Therefore, 408 is not a perfect square [As natural numbers having Unit place digits as 0, 2, 3, 7 and 8 are not perfect square].

iv. 441

Solution:

Unit place digit of the number is 1.

Therefore, 441 is a perfect square

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

Solution:

For 100

100 – 1 = 99                 [1]

99 – 3 = 96                   [2]

96 – 5 = 91                   [3]

91 – 7 = 84                   [4]

84 – 9 = 75                   [5]

75 – 11 = 64                 [6]

64 – 13 = 51                 [7]

51 – 15 = 36                 [8]

36 – 17 = 19                 [9]

19 -19 = 0                 [10]

Here, subtraction has been performed for ten times.

Therefore,  âˆš100 = 10

For 169

169 – 1 = 168                 [1]

168 – 3 = 165                 [2]

165 – 5 = 160                 [3]

160 – 7 = 153                 [4]

153 – 9 = 144                 [5]

144 – 11 = 133                 [6]

133 – 13 = 120                 [7]

120 – 15 = 105                 [8]

105 – 17 = 88                 [9]

88 – 19 = 69                 [10]

69 – 21 = 48                 [11]

48 – 23 = 25                 [12]

25 – 25 = 0                 [13]

Here, subtraction has been performed for thirteen times.

Therefore, âˆš169 = 13

### Question 4. Find the square roots of the following numbers by the Prime Factorization Method.

i. 729

Solution:

729 = 1 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3

729 = (3 Ã— 3) Ã— (3 Ã— 3) Ã— (3 Ã— 3)

729 = (3 Ã— 3 Ã— 3) Ã— (3 Ã— 3 Ã— 3)

729 = (3 Ã— 3 Ã— 3)2

Therefore, âˆš729 = 3 Ã— 3 Ã— 3  = 27

ii. 400

Solution:

400 = 1 Ã— 5 Ã— 5 Ã— 2 Ã— 2 Ã— 2 Ã— 2

400 = (2 Ã— 2) Ã— (2 Ã— 2) Ã— (5 Ã— 5)

400 = (2 Ã— 2 Ã— 5) Ã— (2 Ã— 2 Ã— 5)

400 = (2 Ã— 2 Ã— 5)2

Therefore, âˆš400 = 2 Ã— 2 Ã— 5 = 20

iii. 1764

Solution:

1764 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 7 Ã— 7 Ã— 1

1764 = (2 Ã— 2) Ã— (3 Ã— 3) Ã— (7 Ã— 7)

1764 = (2 Ã— 3 Ã— 7) Ã— (2 Ã— 3 Ã— 7)

1764 = (2 Ã— 3 Ã— 7)2

Therefore, âˆš1764 = 2 Ã— 3 Ã— 7 = 42

iv. 4096

Solution:

4096 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 1

4096 = (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2)

4096 = (2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2) Ã— (2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã—2)

4096 = (2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2)2

Therefore, âˆš4096 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 = 64

v. 7744

Solution:

7744 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 11 Ã— 11 Ã— 1

7744 = (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— (11 Ã— 11)

7744 = (2 Ã— 2 Ã— 2 Ã— 11) Ã—( 2 Ã— 2 Ã— 2 Ã— 11)

7744 = (2 Ã— 2 Ã— 2 Ã— 11)2

Therefore, âˆš7744 = 2 Ã— 2 Ã— 2 Ã— 11 = 88

vi. 9604

Solution:

9604 = 2 Ã— 2 Ã— 7 Ã— 7 Ã— 7 Ã— 7Ã— 1

9604 = (2 Ã— 2) Ã— (7 Ã— 7) Ã— (7 Ã— 7)

9604 = (2 Ã— 7 Ã— 7) Ã— (2 Ã— 7 Ã—7)

9604 = (2 Ã— 7 Ã— 7)2

Therefore, âˆš9604 = 2 Ã— 7 Ã— 7 = 98

vii. 5929

Solution:

5929 = 7 Ã— 7 Ã— 11 Ã— 11

5929 = (7 Ã— 7) Ã— (11 Ã— 11)

5929 = (7 Ã— 11) Ã— (7 Ã— 11)

5929 = (7 Ã— 11)2

Therefore, âˆš5929 = 7 Ã— 11 = 77

viii. 9216

Solution:

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

9216 = (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— (3 Ã— 3)

9216 = (2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3) Ã— (2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3)

9216 = 96 Ã— 96

9216 = (96)2

Therefore, âˆš9216 = 96

ix. 529

Solution:

529 = 23 Ã— 23 Ã— 1

529 = (23)2

Therefore, âˆš529 = 23

x. 8100

Solution:

8100 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 5 Ã— 5 Ã— 1

8100 = (2 Ã— 2) Ã— (3 Ã— 3) Ã— (3 Ã— 3) Ã— (5 Ã— 5)

8100 = (2 Ã— 3 Ã— 3 Ã— 5) Ã— (2 Ã— 3 Ã— 3 Ã— 5)

8100 = 90 Ã— 90

8100 = (90)2

Therefore, âˆš8100 = 90

### Question 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

Solution:

252 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 7

= (2 Ã— 2) Ã— (3 Ã— 3) Ã— 7

7 cannot be paired.

Therefore, multiply by 7 to get perfect square.

New number obtained = 252 Ã— 7 = 1764

1764 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 7 Ã— 7

1764 = (2 Ã— 2) Ã— (3 Ã— 3) Ã— (7 Ã— 7)

1764 = (2 Ã— 3 Ã— 7)2

Therefore, âˆš1764 = 2Ã—3Ã—7 = 42

ii. 180

Solution:

180 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5

= (2 Ã— 2) Ã— (3 Ã— 3) Ã— 5

5 cannot be paired.

Therefore, multiply by 5 to get perfect square.

New number obtained = 180 Ã— 5 = 900

900 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5 Ã— 5 Ã— 1

900 = (2 Ã— 2) Ã— (3 Ã— 3) Ã— (5 Ã— 5)

900 = (2 Ã— 3 Ã— 5)2

Therefore, âˆš900 = 2 Ã— 3 Ã— 5 = 30

iii. 1008

Solution:

1008 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 7

= (2 Ã— 2) Ã— (2 Ã— 2) Ã— (3 Ã— 3) Ã— 7

7 cannot be paired.

Therefore, multiply by 7 to get perfect square.

New number obtained = 1008 Ã— 7 = 7056

7056 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 7 Ã— 7

7056 = (2 Ã— 2) Ã— (2 Ã— 2) Ã— (3 Ã— 3) Ã— (7 Ã— 7)

7056 = (2 Ã— 2 Ã— 3 Ã— 7)2

Therefore, âˆš7056 = 2 Ã— 2 Ã— 3 Ã— 7 = 84

iv. 2028

Solution:

2028 = 2 Ã— 2 Ã— 3 Ã— 13 Ã— 13

= (2 Ã— 2) Ã— (13 Ã— 13) Ã— 3

3 cannot be paired.

Therefore, multiply by 3 to get perfect square.

New number obtained = 2028 Ã— 3 = 6084

6084 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 13 Ã—13

6084 = (2 Ã— 2) Ã— (3 Ã— 3) Ã— (13 Ã— 13)

6084 = (2 Ã— 3 Ã— 13)2

Therefore, âˆš6084 = 2Ã—3Ã—13 = 78

v. 1458

Solution:

1458 = 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3

= (3 Ã— 3) Ã— (3 Ã— 3) Ã— (3 Ã— 3) Ã— 2

2 cannot be paired.

Therefore, multiply by 2 to get perfect square.

New number obtained = 1458 Ã— 2 = 2916

2916 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 3

2916 = (3 Ã— 3) Ã— (3 Ã— 3) Ã— (3 Ã— 3) Ã— (2 Ã— 2)

2916 = (3Ã—3Ã—3Ã—2)2

Therefore, âˆš2916 = 3Ã—3Ã—3Ã—2 = 54

vi. 768

Solution:

768 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3

= (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— 3

3 cannot be paired.

Therefore, multiply 768 by 3 to get perfect square.

New number obtained  = 768Ã—3 = 2304

2304 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3

2304 = (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— (2 Ã— 2) Ã— (3 Ã— 3)

2304 = (2 Ã— 2 Ã— 2 Ã— 2 Ã— 3)2

âˆš2304 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 = 48

### Question 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

Solution:

252 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 7

= (2 Ã— 2) Ã— (3 Ã— 3) Ã— 7

7 cannot be paired.

Divide 252 by 7 to get perfect square.

Therefore, New number obtained = 252 Ã· 7 = 36

36 = 2 Ã— 2 Ã— 3 Ã— 3

36 = (2 Ã— 2) Ã— (3 Ã— 3)

36 = (2 Ã— 3)2

Therefore, âˆš36 = 2 Ã— 3 = 6

ii. 2925

Solution:

252 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 7

= (2 Ã— 2) Ã— (3 Ã— 3) Ã— 7

7 cannot be paired.

Divide by 7 to get perfect square.

Therefore, New number obtained = 252 Ã· 7 = 36

36 = 2 Ã— 2 Ã— 3 Ã— 3

36 = (2 Ã— 2) Ã— (3 Ã— 3)

36 = (2 Ã— 3)2

Therefore, âˆš36 = 2 Ã— 3 = 6

iii. 396

Solution:

396 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 11

= (2 Ã— 2) Ã— (3 Ã— 3) Ã— 11

11 cannot be paired.

Divide by 11 to get perfect square.

Therefore, New number obtained = 396 Ã· 11 = 36

36 = 2 Ã— 2 Ã— 3 Ã— 3

36 = (2 Ã— 2) Ã— (3 Ã— 3)

36 = (2 Ã— 3)2

Therefore, âˆš36 = 2 Ã— 3 = 6

iv. 2645

Solution:

2645 = 5 Ã— 23 Ã— 23

2645 = (23 Ã— 23) Ã— 5

5 cannot be paired.

Divide  by 5 to get perfect square.

Therefore, New number obtained = 2645 Ã· 5 = 529

529 = 23 Ã— 23

529 = (23)2

Therefore, âˆš529 = 23

v. 2800

Solution:

2800 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 5 Ã— 5 Ã— 7

= (2 Ã— 2) Ã— (2 Ã— 2) Ã— (5 Ã— 5) Ã— 7

7 cannot be paired.

Divide by 7 to get perfect square.

Therefore, New number obtained = 2800 Ã· 7 = 400

400 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 5 Ã— 5

400 = (2 Ã— 2) Ã— (2 Ã— 2) Ã— (5 Ã— 5)

400 = (2 Ã— 2 Ã— 5)2

Therefore, âˆš400 = 20

vi. 1620

Solution:

1620 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3 Ã— 5

= (2 Ã— 2) Ã— (3 Ã— 3) Ã— (3 Ã— 3) Ã— 5

5 cannot be paired.

Divide by 5 to get perfect square.

Therefore, New number obtained = 1620 Ã· 5 = 324

324 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3 Ã— 3

324 = (2 Ã— 2) Ã— (3 Ã— 3) Ã— (3 Ã— 3)

324 = (2 Ã— 3 Ã— 3)2

âˆš324 = 18

### Question 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 as assume number of students be, a

So, Each Student has donated Rs a.

Therefore, Total amount donated = a x a

That mean’s a x a = 2401

a2 = 2401

a2 = 7 Ã— 7 Ã— 7 Ã— 7

a2 = (7 Ã— 7) Ã— (7 Ã— 7)

a2 = 49 Ã— 49

a = âˆš(49 Ã— 49)

a = 49

Therefore, The number of students = 49

### Question 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 as assume number of rows be, a

So, Each row has number of plants = a.

Therefore, Total number of plants = a x a

That mean’s a x a = 2025

a2 = 3 Ã— 3 Ã— 3 Ã— 3 Ã— 5 Ã— 5

a2 = (3 Ã— 3) Ã— (3 Ã— 3) Ã— (5 Ã— 5)

a2 = (3 Ã— 3 Ã— 5) Ã— (3 Ã— 3 Ã— 5)

a2 = 45 Ã— 45

a = âˆš(45 Ã— 45)

a = 45

Therefore, The number of rows = 45 and also number of plants in each rows = 45.

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

Solution:

First, we have to find L.C.M of 4, 9 and 10

4 = 2 x 2 x 1

9 = 3 x 3 x 1

5 = 1 x 5

Therefore,  L.C.M = (2 Ã— 2 Ã— 3 x 3 Ã— 5) = 180.

Now we have to find  the smallest whole number divisible by 180

180 = 2 Ã— 2 Ã— 9 Ã— 5

= (2 Ã— 2)Ã— 3 Ã— 3 Ã— 5

= (2 Ã— 2) Ã— (3 Ã— 3) Ã— 5

5 cannot be paired.

Therefore,  multiply 180 by 5 to get perfect square.

The smallest square number divisible by 180 and also by  4, 9 and 10 = 180 Ã— 5

= 900

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

Solution:

First, we have to find L.C.M of 8, 15 and 20

8 = 1 x 2 x 2 x 2

15 = 1 x 5 x 3

20 = 1 x 2 x 5 x 2

Therefore,  L.C.M = (2 Ã— 2 Ã— 5 Ã— 2 Ã— 3) = 120.

Now we have to find  the smallest whole number divisible by 120

120 = 2 Ã— 2 Ã— 3 Ã— 5 Ã— 2

= (2 Ã— 2) Ã— 3 Ã— 5 Ã— 2

3, 5 and 2 cannot be paired.

Therefore, multiply 120 by (3 Ã— 5 Ã— 2) i.e 30 to get perfect square.

The smallest square number divisible by 120 and also by 8, 15 and 20 = 120 Ã— 30

= 3600

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