# NCERT Solutions for Class 8 Maths Chapter 6 (Ex 6.3) Squares and Square Roots

## NCERT Solutions for Class 8 Chapter 6 Squares and Square Roots -Free PDF Download

Free PDF download of NCERT Solutions Maths Class 8 Solutions Chapter 6 – Squares and Square Roots solved by Expert Maths Teachers on CoolGyan.Org. All Chapter 6 – Squares and Square Roots Questions with Solutions for NCERT to help you to revise complete Syllabus and Score More marks.

Maths Revision Notes for Class 8

 Chapter Name Squares and Square Roots Chapter Chapter 6 Exercise Exercise 6.3 Class Class 8 Subject Maths NCERT Solutions Board CBSE TEXTBOOK CBSE NCERT Category NCERT Solutions

# NCERT SOLVED

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

Ans. Since, Unit’s digits of square of numbers are 0, 1, 4, 5, 6 and 9. Therefore, the possible unit’s digits of the given numbers are:

(i) 1 (ii) 6  (iii) 1  (iv) 5

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

(i) 153     (ii) 257     (iii) 408       (iv) 441

Ans. Since, all perfect square numbers contain their unit’s place digits 0, 1, 4, 5, 6 and 9.

(i) But given number 153 has its unit digit 3. So it is not a perfect square number.

(ii) Given number 257 has its unit digit 7. So it is not a perfect square number.

(iii) Given number 408 has its unit digit 8. So it is not a perfect square number.

(iv) Given number 441 has its unit digit 1. So it would be a perfect square number

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

Ans. By successive subtracting odd natural numbers from 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

This successive subtraction is completed in 10 steps.

Therefore

By successive subtracting odd natural numbers from 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

This successive subtraction is completed in 13 steps.

Therefore

4. Find the square roots of the following numbers by the Prime Factorization method:

(i) 729     (ii) 400    (iii) 1764       (iv) 4096   (v) 7744

(vi) 9604   (vii) 5929    (viii) 9216    (ix) 529      (x) 8100

Ans. (i) 729

= 27

(ii) 400

= 20

(iii) 1764

= 42

(iv) 4096

= 64

(v) 7744

= 88

(vi) 9604

= 98

(vii) 5929

=  = 77

(viii) 9216

= 96

(ix) 529

= 23

(x) 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

Ans. (i)

252 =

Here, prime factor 7 has no pair. Therefore 252 must be multiplied by 7 to make it a perfect square.

= 1764

= = 42

(ii)

180 =

Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.

180 x 5 = 900

= = 30

(iii)

1008 =

Here, prime factor 7 has no pair. Therefore 1008 must be multiplied by 7 to make it a perfect square.

= 7056

And  = = 84

(iv)

2028 =

Here, prime factor 3 has no pair. Therefore 2028 must be multiplied by 3 to make it a perfect square.

= 6084

And  = = 78

(v)

1458 =

Here, prime factor 2 has no pair. Therefore 1458 must be multiplied by 2 to make it a perfect square.

= 2916

And  = = 54

(vi)

768 =

Here, prime factor 3 has no pair. Therefore 768 must be multiplied by 3 to make it a perfect square.

= 2304

And  = = 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

Ans. (i)

252 =

Here, prime factor 7 has no pair. Therefore 252 must be divided by 7 to make it a perfect square.

252  7 = 36

And  =  = 6

(ii)

2925 =

Here, prime factor 13 has no pair. Therefore 2925 must be divided by 13 to make it a perfect square.

2925  13 = 225

And  = = 15

(iii)

396 =

Here, prime factor 11 has no pair. Therefore 396 must be divided by 11 to make it a perfect square.

396  11 = 36

And  = = 6

(iv)

2645 =

Here, prime factor 5 has no pair. Therefore 2645 must be divided by 5 to make it a perfect square.

2645  5 = 529

And  = = 23

(v)

2800 =

Here, prime factor 7 has no pair. Therefore 2800 must be divided by 7 to make it a perfect square.

2800  7 = 400

And  =  = 20

(vi)

1620 =

Here, prime factor 5 has no pair. Therefore 1620 must be divided by 5 to make it a perfect square.

16205 = 324

And  = = 18

7. The students of Class VIII of a school donated Rs. 2401 in all, for Prime Minister’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.

Ans. Here, Donated money = Rs 2401

Let the number of students be

Therefore donated money =

According to question,

= 2401

=

= = 49

Hence, the number of students is 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.

Ans. Here, Number of plants = 2025

Let the number of rows of planted plants be

And each row contains number of plants =

According to question,

= 2025

= = 45

Hence, each row contains 45 plants.

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

Ans. L.C.M. of 4, 9 and 10 is 180.

Prime factors of 180 =

Here, prime factor 5 has no pair. Therefore 180 must be multiplied by 5 to make it a perfect square.

= 900

Hence, the smallest square number which is divisible by 4, 9 and 10 is 900.

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

Ans. L.C.M. of 8, 15 and 20 is 120.

Prime factors of 120 =

Here, prime factor 2, 3 and 5 has no pair. Therefore 120 must be multiplied by

to make it a perfect square.

120  2  3  5 = 3600

Hence, the smallest square number which is divisible by 8, 15 and 20 is 3600.