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Question 1

The following numbers are not perfect squares. Give reason.
       (i) 1057 (ii) 23453 (iii) 7928 (iv) 222222

Solution:
(i) The units digit of the number 1057 is 7. 1057 is not a perfect square since no square number ends in the digit 7.

(ii) The units digit of the number 23453 is 3. 23453 is not a perfect square since no square number ends in the digit 3.

(iii) The units digit of the number 7928 is 8. 7928 is not a perfect square since no square number ends in the digit 8.

(iv) The units digit of the number 222222 is 2. 222222 is not a perfect square since no square number ends in the
digit 2.

 

Question 2

What will be the units digit of the squares of the following numbers?

       (i) 81 (ii) 272 (iii) 799 (iv) 3853 (v) 1234 (vi) 26387 (vii) 52698 (viii) 99880 (ix) 12796 (x) 55555

Solution:
(i) 1 is the units digit of the square of 81, since 1 × 1 =1.

(ii) 4 is the units digit of the square of 272, since 2 × 2 =4.

(iii) 1 is the units digit of the square of 799, since 9 × 9 = 81.

(iv) 9 is the units digit of the square of 3853, since 3 × 3 =9.

(v) 6 is the units digit of the square of 1234, since 4 × 4 =16.

(vi) 9 is the units digit of the square of 26387, since 7 × 7 = 49.

(vii) 4 is the units digit of the square of 52698, since 8 × 8 =64.

(viii) 0 is the units digit of the square of 99880, since 0 × 0 = 0.

(ix) 6 is the units digit  of the square of 12796, since 6 × 6 = 36.

(x) 5 is the units digit of the square of 55555, since 5 × 5 =25.

Question 3

The following numbers are not square numbers. Give reason.

       (i) 64000 (ii) 89722 (iii) 222000 (iv) 505050

Solution:
(i) The number of zeros at the end of the number 64000 is 3 which is odd. A number ending with odd number of zeros is never a perfect square.

(ii) The number 89722 ends in the digit 2. No square number ends in 2.

(iii) The number of zeros at the end of the number 222000 is 3 which is odd. A number ending with odd number of zeros is never a perfect square.

(iv) The number of zeros at the end of the number 505050 is 1 which is odd. A number ending with odd number of zeros is never a perfect square.

Question 4

The squares of which of the following would be odd numbers?

       (i) 431 (ii) 2826 (iii) 7779 (iv) 82004

Solution:
(i) The square of 431 is an odd number, since 431 is odd.

(ii) The square of 2826 is not odd, since 2826 is even whose square is also even.

(iii) The square of 7779 is odd, since 7779 is odd.

(iv) The square of 82004 is not odd, since 82004 is even whose square is also even.

Question 5

Show that the following numbers are not perfect squares:

       (i) 7927 (ii) 1058 (iii) 33453 (iv) 22222

Solution:
(i) The unit digit of the number 7927 is 7. No perfect square ends in 7.

∴ 7927 is not a perfect square.

(ii) The unit digit of the number 1058 is 8. No perfect square ends in 8.

∴ 1058 is not a perfect square.

(iii) The unit digit of the number 33453 is 3. No perfect square ends in 3.

∴ 33453 is not a perfect square.

(iv) The unit digit of the number 22222 is 2. No perfect square ends in 2.

∴ 22222 is not a perfect square.        

Question 6

Observe the following pattern and find the missing digits:

       112 = 121
       1012 = 10201
       10012 =1002001
       1000012 = 1_______2_____1
       100000012 = ____________

Solution:
112 = 121

1012 = 10201

10012 =1002001

1000012 = 10000200001

100000012 = 100000020000001

Question 7

Observe the following pattern and supply the missing numbers:

       112 = 121
       1012 = 10201
       101012 = 102030201
       10101012     = _____________
       _________2 =10203040504030201    

Solution:
112 = 121

1012 = 10201

101012 = 102030201

10101012 = 1020304030201

1010101012 =10203040504030201

Question 8

Using the given pattern, find the missing numbers:

       12 + 22 + 22 = 32
      
22 + 32 + 62 = 72
      
32 + 42 + 122 = 132
      
42 + 52 + ___2 = 212
      
52 + ___2 + 302 = 312
      
62 + 72 + ____2 = ___2

Solution:

12 + 22 + 22 = 32

22 + 32 + 62 = 72

32 + 42 + 122 = 132

42 + 52 + 202 = 212

52 + 62 + 302 = 312

62 + 72 + 422 = 432

Question 9

Using suitable pattern, complete the following:

       (i) = _____  (ii) = ______

Solution:
(i)  121    = 
             
     12321 =

  =1+2+3+2+1=9.

  ∴ = 9.  

(ii)  1234321 =

   123454321 =

12345654321 =

=1+2+3+4+5+6+5+4+3+2+1=36.
                                          
       ∴ =36.       


Question 10

Write true (T) or false (F) for the following statements:

         (i) The numbers of digits in a square number is even.
         (ii) The square of a prime number is prime.
         (iii) The sum of two square numbers is a square number.
         (iv) The difference of two square numbers is a square number.
         (v) The product of two square numbers is a square number.
         (vi) No square number is negative.
         (vii) There is no square number between 50 and 60.
         (viii) There are fourteen square numbers upto 200.

Solution:
(i) False, since all square numbers will not have even number of digits even.

(ii) False, since square of a prime number is never prime.

(iii) False, since the sum of two square numbers is not always a square number.

(iv) False, since the difference of two square numbers is not always a square number.

(v) True.

(vi) True.

(vii) True.

(viii) True.

Question 11

Consider the following pattern:

         252 = 2 × (2 + 1) hundreds + 25 = 625
         452 = 4 × (4 + 1) hundreds + 25 = 2025
         1152 = 11 × (11 + 1) hundreds + 25 = 13225
         Using this pattern, find the square of:
         (i) 35 (ii) 75 (iii) 95 (iv) 105 (v) 205

Solution:
(i) 352 = 3 × (3 + 1) hundreds + 25

          = (3 × 4) hundreds + 25

          = 12 × 100 + 25

          = 1200 + 25

         =1225

(ii) 752 = 7 × (7 + 1) hundreds + 25

           = (7 × 8) hundreds + 25

           = 56 × 100 + 25

           = 5600 + 25

           = 5625

(iii) 952 = 9 × (9 + 1) hundreds + 25

            =(9 × 10) hundreds + 25

            = 90× 100 + 25

            = 9000 + 25

            = 9025

(iv) 1052 = 10 × (10 + 1) hundreds +25

              = (10 × 11) hundreds + 25

              = 110× 100 + 25

              = 11000 + 25

             = 11025

(v) 2052 = 20 × (20+ 1) hundreds + 25

             = (20 × 21) hundreds + 25

             = 420 × 100 + 25

             = 42000 + 25

             = 42025

Question 12

Consider the following pattern:

         522 = (52+ 2) hundreds + 22 =2704
         572 = (52+ 7) hundreds + 72 =3249
         Using this pattern, find the squares of:
         (i) 51 (ii) 54 (iii) 56 (iv) 58 (v) 59

Solution:
(i) 512 = (52+ 1)hundreds + 12

          = (25 + 1) × 100 + 1

          = 26 × 100 + 1

          = 2601.

(ii) 542 = (52+ 4) hundreds + 42

            = (25 + 4) × 100 + 16

           = 29 × 100 + 16

           = 2900 + 16

           = 2916.

(iii) 562 = (52+ 6) hundreds + 62

            = (25 + 6) × 100 + 36

            = 31 × 100 + 36

            = 3100 + 36

            =3136.

(iv) 582 = (52+ 8) hundreds + 82

            = (25 + 8) × 100 + 64

            = 33 × 100 + 64

            = 3300 + 64

            =3364.

(v) 592 = (52+ 9) hundreds + 92

           = (25 + 9) × 100 + 81

           = 34 × 100 + 81

           = 3400 + 81

          = 3481.

Question 13

Consider the following pattern:

         5112 = (250 + 11) thousands + 112 = 261121
         5902 = (250 + 90) thousands + 902 = 348100
         Using this pattern find the squares of:
         (i) 509 (ii) 515 (iii) 525 (iv) 580 (v) 534

Solution:
(i) 5092 = (250 + 9) thousands + 92

            = 259 × 1000 + 81

            = 259000 + 81

            = 259081.

(ii) 5152 = (250 + 15) thousands + 152

             = 265 × 1000 + 225

             = 265000 + 225

             = 265225.

(iii) 5252 = (250 + 25) thousands + 252

              = 275 × 1000 + 625

              = 275000 + 625

              = 275625.

(iv) 5802 = (250 + 80) thousands + 802

              = 330 × 1000 + 6400

              = 330000 + 6400

              = 336400.

(v) 5342 = (250 + 34) thousands + 342

              = 284 × 1000 + 81

              = 284000 + 81

              = 284081.

Question 14

Find the squares of following numbers using the identity: (a +b)2 = a2 + 2ab + b2

         (i) 509 (ii) 211 (iii) 625

Solution:
(i) 509 = (500 + 9)2

          = (500)2 + 2 × 500 × 9 + (9)2

[(a +b)2 = a2 + 2ab + b2, where a = 500 and b= 9.]

             = 250000 + 9000 + 81

             = 259081.

(ii)   211 = (200 + 11)2

             = (200)2 + 2 × 200 × 11 + (11)2

[(a +b)2 = a2 + 2ab + b2, where a = 200 and b= 11.]

             = 40000 + 4400 + 121

             = 44521.

(iii)  625 = (600 + 25)2

             = (600)2 + 2 × 600 × 25 + (25)2

[(a +b)2 = a2 + 2ab + b2, where a = 600 and b= 25.]

             = 360000 + 30000 + 625

             = 390625.

Question 15

Find the squares of the following numbers using the identity: (a -b)2 = a2 - 2ab + b2

         (i) 491 (ii) 189 (iii) 575

Solution:

(i) 491 = (500 - 9)2

           = (500)2 - 2 × 500 × 9 + (9)2

[(a - b)2 = a2 - 2ab + b2, where a = 500 and b= 9.]

             = 250000 - 9000 + 81

             = 241081.

(ii) 189 = (200 - 11)2

           = (200)2 - 2 × 200 × 11 + (11)2

[(a - b)2 = a2 - 2ab + b2, where a = 200 and b= 11.]

             = 40000 - 4400 + 121

             = 35721.

(iii) 575 = (600 - 25)2

            = (600)2 - 2 × 600 × 25 + (25)2

[(a - b)2 = a2 - 2ab + b2, where a = 600 and b= 25.]

             = 360000 - 30000 + 625

             = 330625

Question 16

Write the possible units digits of the square root of the following numbers. Which of these numbers have odd square roots?
         (i) 9801 (ii) 99856 (iii) 998001 (iv) 657666025

Solution:
(i) The possible units digit of 9801 is either 1 or 9 and it will have an odd square root.

(ii) The possible units digit of 99856 is either 4 or 6 and it will have even square root.

(iii) The possible units digit of 998001 is either 1 or 9 and it will have an odd square root.

(iv) The possible units digit of 657666025 is 5 and it will have an odd square root.

Question 17

Find the square roots of 121 and 169 by the method of repeated subtraction.


Solution:
(i) 121 – 1 = 120,
(ii) 120 – 3 = 117,
(iii) 117 – 5 = 112,
(iv) 112 – 7 = 105,
(v) 105 – 9 = 96,
(vi) 96 – 11 = 85,
(vii) 85 – 13 = 72,
(viii) 72 – 15 = 57,
(ix) 57 – 17 = 40,
(x) 40 – 19 = 21,
(xi) 21 – 21 = 0.

From 121 we have subtracted successive odd numbers starting from 1 and obtained 0 at 11th step.

Therefore  = 11.

(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.

From 169 we have subtracted successive odd numbers starting from 1 and obtained 0 at 13th step.

Therefore =13

Question 18

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

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

Solution:
(i)


729 = 9× 9 × 3 × 3  =92 × 3=(9 × 3)2 (By pairing the prime factors )

... The square root of 729 = 9 × 3= 27.
    
(ii)

400 = 2 × 2 × 10 × 10  =22 × 10=(2 × 10)2 (By pairing the prime factors )
   
... The square root of 400 = 2 × 10= 20  

(iii)   

1764 = 2 × 2 × 3 × 3 × 7 × 7 =22 × 32  × 72   =(2 × 3 × 7 )2 (By pairing the prime factors )
                 
... The square root of 1764 =2 × 3 × 7= 42.      
  
(iv)


4096 = 2
×2×2×2×2×2×2×2×2×2×2×2=22 × 22 × 22 × 22 × 22 × 22  

             = (2×2×2×2×2×2)2(By pairing the prime factors )
   
  ...
The square root of 4096=2×2×2×2×2×2= 64

Question 19

Write the prime factorization of the following numbers and hence find their square roots:

         (i) 7744 (ii) 9604 (iii) 5929 (iv) 7056

Solution:
(i) 7744

7744 = 2 × 2 × 2 × 2 × 2 × 2 × 11 × 11 == 22 × 22 × 22 × 112

 =(2 × 2 × 2 × 11)2 (By pairing the prime factors)

∴ The square root of 7744 = 2 × 2 × 2 × 11 = 88.

(ii) 9604

  9604 = 2 × 2 × 7 × 7 × 7 × 7 = 22  × 72  × 72

∴ The square root of 9604 = 2 × 7 × 7 = 98.

(iii) 5929
5929 = 7 × 7 × 11 × 11  = 72 × 112
= (7 × 11)2  (By pairing the prime factors)

∴ The square root of 5929 = 7 × 11 = 77.

(iv) 7056
∴ 7056= 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7  = 22 × 22 × 32 ×  72

  = (2 × 2 × 3×  7 )2 (By pairing the prime factors)

∴ The square root of 7056 = 2 × 2 × 3 × 7 = 84.

Question 20

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

         (i) 180 (ii) 1458 (iii) 1200 (iv) 1008 (v) 2028

Solution:
(i) 180

180 = 2 × 2 × 3 × 3 × 5 = (2)2 × (3)2 × 5

∴ 180 should be multiplied by 5 to make it a perfect square.

Also the square root of the square number so obtained = 2 × 3 × 5 = 30.

(ii) 1458

∴ 1458 = 2 × 3 × 3 × 9 × 9 = (2)1 × (3)2 × (9)2

∴ 1458 is to be multiplied by 2 to make it a perfect square.

Also the square root of the square number so obtained = 2 × 3 × 9 = 54.

(iii) 1200

∴ 1200 = 2 × 2 × 2 × 2 × 3 × 5 × 5 = (2)4 × (3)1 × (5)2

∴ 1200 should be multiplied by 3 to make it a perfect square.

Also the square root of the square number so obtained = 2 × 2 × 3 × 5 = 60.

(iv) 1008

∴ 1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7 = (2)4 × (3)2 × (7)1

∴ 1008 should be multiplied by 7 to make it a perfect square.

Also the square root of the square number so obtained = 2 × 2 × 3 × 7 = 84.

(v) 2028

2028 = 2 × 2 × 3 × 13 × 13 = (2)2 × (3)1 × (13)2

∴ 2028 should be multiplied by 3 to make it a perfect square.

Also the square root of the square number so obtained = 2 × 13 × 3 = 78.

Question 21

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) 180 (ii) 3645 (iii) 2800 (iv) 45056

Solution:
(i) 180

180 = 2 × 2 × 3 × 3 × 5 =(2)2 × (3)2 × (5)1

∴ 180 should be divided by 5 to make it a perfect square.

Also the square root of the square number so obtained = 2 × 3 = 6.

 

(ii) 3645

3645 = 3 × 3 × 3 × 3 × 3 × 3 × 5

∴ 3645 = (3)4 × (5)1

∴ 3645 should be divided by 5 to make it a perfect square.

Also the square root of the square number so obtained = 3 × 3 × 3 = 27.

(iii) 2800

2800 = 2 × 2 × 2 × 2 × 5 × 5 × 7

∴ 2800 = (2)4 × (5)2 × (7)1

∴ 2800 should be divided by 7 to make it a perfect square.

Also the square root of the square number so obtained = 2 × 2 × 5 = 20.


(iv) 45056

∴ 45056 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 x 11

             = (2)12 × 11

∴ 45056 should be divided by 11 to make it a perfect square.

Also the square root of the square number so obtained = 2 × 2 × 2 × 2 × 2 × 2 = 64.

Question 22

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.


Solution:
Amount donated by the students of Class VIII of a school= Rs. 2401

Amount each student donated= Number of students in the class = Square root of 2401

      2401= 7 × 7 × 7 × 7 = (7)4

= 7 × 7 = 49.

∴ The number of students in the class = 49.

Question 23

A PT teacher wants to arrange maximum possible number of 6000 students in a field such that the number of rows is equal to the number of columns. Find the number of rows if 71 students were left out  after the arrangement.


Solution:
Number of students            = 6000

Number of students left out = 71

Number of students arranged in rows and columns = 6000 – 71 = 5929

The number of rows = The number of columns

                                      =
                  

5929 = 7 x  7 x 11 x 11 = (7 x 11)2

Therefore the number of rows is 77

Question 24

Find the number of digits in the square roots of the following numbers:

         (i) 64 (ii) 144 (iii) 4489 (iv) 27225 (v) 390625

Solution:
(i) The number of digits in the square root of 64 is 1.

(ii) We know that 100 < 144 < 225 and and

So 10 < <15

 Therefore the number of digits in the square root of 144 is 2.

(iii) We know that 4225 < 4489 < 5625 and and

So 65 < < 75

Therefore the number of digits in the square root of 4489 is 2.

(iv)  We know that 10000 < 27225 < 40000 and and .
So 100 < < 200
Therefore the number of digits in the square root of 27225 is 3.

(v) We know that 250000 < 390625 < 490000 and and .
So 500 < < 700

Therefore the number of digits in the square root of 390625 is 3.  


Question 25

Put a dot on the units digit of a number. Now put a dot over every alternate digit. The number of dots gives the number of digits in the square root of the given number. Find the number of digits in the square root of:

         (i) 1234321 (ii) 2124449 (iii) 3915380329

Solution:
(i) The number of dots = 4

The number of digits in the square root of 1234321 = 4.

(ii) The number of dots = 4

∴ The number of digits in the square root of 2124449 = 4.

(iii) The number of dots = 5

∴ The number of digits in the square root of 3915380329 = 5.


Question 26

Using the division method, find the square root of the following numbers:
         (i) 44100 (ii) 27225 (iii) 54756 (iv) 49284 (v) 99856

Solution:
(i) 44100

∴ The square root of 44100 is 210.

(ii) 27225

∴ The square root of 27225 is 165.

(iii) 54756

∴ The square root of 54756 is 234.

(iv) 49284

The square root of 49284 is 222.

(v) 99856

∴ The square root of 99856 is 316.

Question 27

Using the division method, find the square roots of the following:

         (i) 390625 (ii) 119025 (iii) 193600

Solution:
(i) 390625

∴ The square root of 390625 is 625.

(ii) 119025

The square root of 119025 is 345.

(iii) 193600

The square root of 193600 is 440.

Question 28

Find the square roots of the following numbers:

         (i) 1444 (ii) 1849 (iii) 5776 (iv) 7921

Solution:
(i) 1444

= 38

(ii) 1849

= 43

(iii) 5776

= 76

(iv) 7921

=89

Question 29


 

Find the least numbers which must be subtracted from the following numbers so as to leave a perfect square:
         (i) 2361 (ii) 4931 (iii) 18265 (iv) 390700

Solution:
(i) 2361 < 2500

< 50

1600 < 2361

40 <

⇒ 40 < < 50

2025 < 2361

⇒ 45 <

2116 < 2361

⇒ 46 <

2209 < 2361

⇒ 47 <

2304 < 2361

⇒ 48 <

∴ 2401 > 2361

⇒ 48 < <49

The nearest square number lesser than 2361 is 2304.

⇒ The number to be subtracted from 2316 to make it a perfect square is 2361 – 2304 = 57.

∴ 57 is the least number to be subtracted from 2361 to make it a perfect square.

(ii) 4931

4900 < 4931

⇒ 70 <

5041 > 4931

⇒ 71 >

⇒ The nearest square number lesser than 4931 is 4900.

⇒ The number to be subtracted from 4931 to make it a perfect square is 4931 – 4900 = 31.

∴ 31 is the least number to be subtracted from 4931 to make it a perfect square.

(iii) 18265

16900 < 18265

⇒ 130 <

17161 < 18265

⇒ 131 <

17424 < 18265

⇒ 132 <

17689 < 18265

⇒ 133 <

17956 < 18265

⇒ 134 <

18225 < 18265

⇒ 135 <

18496 > 18265

⇒ 136 >

⇒ The nearest square number lesser than 18265 is 18225.

⇒ The number to be subtracted from 18225 to make it a perfect square is 18265 – 18225 = 40.

∴ 40 is the least number to be subtracted from 18265 to make it a perfect square.

(iv) 390700

390625 < 390700

⇒ 625 <

391876 > 390700

⇒ 625 >

⇒  The nearest square which is lesser than 390700 is 390625.

⇒  The number to be subtracted from 390700 to make it a perfect square is 390700 – 390625 = 75.

∴ 75 is the least number to be subtracted from 390700 to make it a perfect square.

Question 30

Find the smallest number that must be added to get a perfect square.

         (i) 2361 (ii) 4931 (iii) 18265 (iv) 390700

Solution:

(i) 2361

2500 > 2361

⇒ 50 >

2401 >2361

⇒ 49 >

2304 < 2361

⇒ 48 <

∴ 2401 is the perfect square nearest to 2361.

∴ 2401 – 2361 = 40 is the number to be added.

∴ 40 is the least number to be added to 2361 to make it a perfect square.

(ii) 4931

4900 < 4931

⇒ 70 <

5041 > 4931

⇒ 71 >

⇒ 5041 is the perfect square nearest to 4931.

⇒ 5041 – 4931 = 110 is the number to be added.

∴ 110 is the least number to be added to 4931 to make it a perfect square.

(iii) 18265

18225 < 18265

⇒ 135 <

18496 > 18265

⇒ 136 >

⇒ 18496 is the perfect square nearest to 18265.

⇒ 18496 – 18265 = 231 is the number to be added to 18265.

∴ 231 is the least number to be added to 18265 to make it a perfect square.

(iv) 390700

390625 < 390700

⇒ 625 <

391876 > 390700

⇒ 626 >

⇒ 391876 is the perfect square nearest to 390700.

⇒ 391876 – 390700 = 1176 is the number to be added to 390700.

∴ 1176 is the least number to be added to 390700 to make it a perfect square.

Question 31

Find the square roots of the following decimal numbers:

         (i) 7.29 (ii) 16.81 (iii) 9.3025 (iv) 84.8241

Solution:
(i) 7.29

=2.7.

(ii) 16.81

=4.1.

(iii) 9.3025

=3.05.

(iv) 84.8241

=9.21.


Question 32

Find the square roots of the following numbers correct to two places of decimal:
         (i) 1.7 ii) 23.1 (iii) 5 (iv) 20 (v) 0.1

Solution:

(i) 1.7

= 1.303 to three decimal places.
      = 1.30 correct to two decimal places.

(ii) 23.1

= 4.806 to three decimal places.
        = 4.81 correct to two decimal places.

(iii) 5

= 2.236 to three decimal places.
    = 2.24 correct to two decimal places.

(iv) 20

= 4.472 to three decimal places.
     = 4.47 correct to two decimal places.


Question 33

Write true (T) or false (F) for the following statements:
         (i) = 0.3
         (ii) If a is a natural number, then is a rational number
         (iii) If a is negative, then a2 is also negative.
         (iv) If p and q are perfect squares, then is a rational number.
         (v) The square root of a prime number may be obtained approximately, but never exactly.

Solution:
(i) False, since the square of a number with one decimal will have two decimal places.

(ii) False, since square root of a natural numbers is not a rational numbers.

(iii) False, since the square of a negative number is a positive number.

(iv) True.

(v) True.





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