NCERT Solutions for class 8 Maths chapter 6 Square and square roots

Exercise 6.1  Exercise 6.2  Exercise 6.3  Exercise 6.4

Exercise 6.3

Question 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

Sol :

(i) If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 9801 is either 1 or 9.

(ii) If the number ends with 6, then the one’s digit of the square root of that number may be 4 or 6. Therefore, one’s digit of the square root of 99856 is either 4 or 6.

(iii) If the number ends with 1, then the one’s digit of the square root of that number may be 1 or 9. Therefore, one’s digit of the square root of 998001 is either 1 or 9.

(iv) If the number ends with 5, then the one’s digit of the square root of that number will be 5. Therefore, the one’s digit of the square root of 657666025 is 5.

 

Question 2

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

(i) 153 (ii) 257

(iii) 408 (iv) 441

Sol :

The perfect squares of a number can end with any of the digits 0, 1, 4, 5, 6, or 9 at unit’s place. Also, a perfect square will end with even number of zeroes, if any.

(i) Since the number 153 has its unit’s place digit as 3, it is not a perfect square.

(ii) Since the number 257 has its unit’s place digit as 7, it is not a perfect square.

(iii) Since the number 408 has its unit’s place digit as 8, it is not a perfect square.

(iv) Since the number 441 has its unit’s place digit as 1, it is a perfect square.

 

Question 3

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

Sol :

We know that the sum of the first n odd natural numbers is n2.

Consider.

(i) 100 − 1 = 99 (ii) 99 − 3 = 96 (iii) 96 − 5 = 91

(iv) 91 − 7 = 84 (v) 84 − 9 = 75 (vi) 75 − 11= 64

(vii) 64 − 13 = 51 (viii) 51 − 15 = 36 (ix) 36 − 17 = 19

(x) 19 − 19 = 0

We have subtracted successive odd numbers starting from 1 to 100, and obtained 0 at 10th step.

Therefore,

The square root of 169 can be obtained by the method of repeated subtraction as follows.

(i) 169 − 1 = 168 (ii) 168 − 3 = 165 (iii) 165 − 5 = 160

(iv) 160 − 7 = 153 (v) 153 − 9 = 144 (vi) 144 − 11 = 133

(vii) 133 − 13 = 120 (viii) 120 − 15 = 105 (ix) 105 − 17 = 88

(x) 88 − 19 = 69 (xi) 69 − 21 = 48 (xii) 48 − 23 = 25

(xiii)25 − 25 = 0

We have subtracted successive odd numbers starting from 1 to 169, and obtained 0 at 13th step.

Therefore,

Question 4

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

(i) 729

Sol :

(i) 729 can be factorised as follows.

3

729

3

243

3

81

3

27

3

9

3

3
1

729 = 3 × 3 × 3 × 3 × 3 × 3

= 27

 

(ii) 400

Sol :

(ii) 400 can be factorised as follows.

2

400

2

200

2

100

2

50

5

25

5

5
1

400 = 2 × 2 × 2 × 2 × 5 × 5

= 20

 

 

(iii) 1764

Sol :

(iii) 1764 can be factorised as follows.

2

1764

2

882

3

441

3

147

7

49

7

7
1

1764 = 2 × 2 × 3 × 3 × 7 × 7

= 42

 

(iv) 4096

Sol :

(iv) 4096 can be factorised as follows.

2

4096

2

2048

2

1024

2

512

2

256

2

128

2

64

2

32

2

16

2

8

2

4

2

2
1

4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 64

 

(v) 7744

Sol :

(v) 7744 can be factorised as follows.

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

= 88

 

(vi) 9604

Sol :

(vi) 9604 can be factorised as follows.

2

9604

2

4802

7

2401

7

343

7

49

7

7
1

9604 = 2 × 2 × 7 × 7 × 7 × 7

= 98

 

 

(vii) 5929

Sol :

(vii) 5929 can be factorised as follows.

7

5929

7

847

11

121

11

11
1

5929 = 7 × 7 × 11 × 11

= 77

 

(viii) 9216

Sol :

(viii) 9216 can be factorised as follows.

2

9216

2

4608

2

2304

2

1152

2

576

2

288

2

144

2

72

2

36

2

18

3

9

3

3
1

9216 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3× 3

= 96

 

 

(ix) 529

Sol :

(ix) 529 can be factorised as follows.

23

529

23

23
1

529 = 23 × 23

 

(x) 8100

Sol :

(x) 8100 can be factorised as follows.

2

8100

2

4050

3

2025

3

675

3

225

3

75

5

25

5

5

1

8100 = 2 × 2 × 3 × 3 × 3 × 3 × 5 × 5

= 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

Sol :

(i)252 can be factorised as follows.

2

252

2

126

3

63

3

21

7

7
1

252 = 2 × 2 × 3 × 3 × 7

Here, prime factor 7 does not have its pair.

If 7 gets a pair, then the number will become a perfect square. Therefore, 252 has to be multiplied with 7 to obtain a perfect square.

252 × 7 = 2 × 2 × 3 × 3 × 7 × 7

Therefore, 252 × 7 = 1764 is a perfect square.

 

(ii) 180

Sol :

(ii)180 can be factorised as follows.

2

180

2

90

3

45

3

15

5

5

1

180 = 2 × 2 × 3 × 3 × 5

Here, prime factor 5 does not have its pair. If 5 gets a pair, then the number will become a perfect square. Therefore, 180 has to be multiplied with 5 to obtain a perfect square.

180 × 5 = 900 = 2 × 2 × 3 × 3 × 5 × 5

Therefore, 180 × 5 = 900 is a perfect square.

= 30

 

 

(iii) 1008

Sol :

(iii)1008 can be factorised as follows.

2

1008

2

504

2

252

2

126

3

63

3

21

7

7

1

1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7

Here, prime factor 7 does not have its pair. If 7 gets a pair, then the number will become a perfect square. Therefore, 1008 can be multiplied with 7 to obtain a perfect square.

1008 × 7 = 7056 = 2 × 2 ×2 × 2 × 3 × 3 × 7 × 7

Therefore, 1008 × 7 = 7056 is a perfect square.

= 84

 

(iv) 2028

Sol :

(iv) 2028 can be factorised as follows.

2

2028

2

1014

3

507

13

169

13

13

1

2028 = 2 × 2 × 3 × 13 × 13

Here, prime factor 3 does not have its pair. If 3 gets a pair, then the number will become a perfect square. Therefore, 2028 has to be multiplied with 3 to obtain a perfect square.

Therefore, 2028 × 3 = 6084 is a perfect square.

2028 × 3 = 6084 = 2 × 2 × 3 × 3 × 13 × 13

= 78

 

 

(v) 1458

Sol :

(v) 1458 can be factorised as follows.

2

1458

3

729

3

243

3

81

3

27

3

9

3

3
1

1458 = 2 × 3 × 3 × 3 × 3 × 3 × 3

Here, prime factor 2 does not have its pair. If 2 gets a pair, then the number will become a perfect square. Therefore, 1458 has to be multiplied with 2 to obtain a perfect square.

Therefore, 1458 × 2 = 2916 is a perfect square.

1458 × 2 = 2916 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3

= 54

 

(vi) 768

Sol :

(vi) 768 can be factorised as follows.

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

Here, prime factor 3 does not have its pair. If 3 gets a pair, then the number will become a perfect square. Therefore, 768 has to be multiplied with 3 to obtain a perfect square.

Therefore, 768 × 3 = 2304 is a perfect square.

768 × 3 = 2304 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 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 number. Also find the square root of the square number so obtained.

(i) 252

Sol :

(i) 252 can be factorised as follows.

2

252

2

126

3

63

3

21

7

7

1

Here, prime factor 7 does not have its pair.

If we divide this number by 7, then the number will become a perfect square. Therefore, 252 has to be divided by 7 to obtain a perfect square.

252 ÷7 = 36 is a perfect square.

 

(ii) 2925

Sol :

(ii) 2925 can be factorised as follows.

3

2925

3

975

5

325

5

65

13

13

1

Here, prime factor 13 does not have its pair.

If we divide this number by 13, then the number will become a perfect square. Therefore, 2925 has to be divided by 13 to obtain a perfect square.

2925 ÷13 = 225 is a perfect square.

 

 

(iii) 396

Sol :

(iii)396 can be factorised as follows.

2

396

2

198

3

99

3

33

11

11

1

Here, prime factor 11 does not have its pair.

If we divide this number by 11, then the number will become a perfect square. Therefore, 396 has to be divided by 11 to obtain a perfect square.

396 ÷11 = 36 is a perfect square.

 

(iv) 2645

Sol :

(iv) 2645 can be factorised as follows.

5

2645

23

529

23

23

1

Here, prime factor 5 does not have its pair.

If we divide this number by 5, then the number will become a perfect square.

Therefore, 2645 has to be divided by 5 to obtain a perfect square.

2645 ÷5 = 529 is a perfect square.

 

 

(v) 2800

Sol :

(v)2800 can be factorised as follows.

2

2800

2

1400

2

700

2

350

5

175

5

35

7

7

1

Here, prime factor 7 does not have its pair.

If we divide this number by 7, then the number will become a perfect square.

Therefore, 2800 has to be divided by 7 to obtain a perfect square.

2800 ÷7 = 400 is a perfect square.

 

(vi) 1620

Sol :

(vi)1620 can be factorised as follows.

2

1620

2

810

3

405

3

135

3

45

3

15

5

5

1

Here, prime factor 5 does not have its pair.

If we divide this number by 5, then the number will become a perfect square.

Therefore, 1620 has to be divided by 5 to obtain a perfect square.

1620 ÷5 = 324 is a perfect square.

 

Question 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.

Sol :

It is given that each student donated as many rupees as the number of students of the class. Number of students in the class will be the square root of the amount donated by the students of the class.

The total amount of donation is Rs 2401.

Number of students in the class =

Hence, the number of students in the class is 49.

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