PRACTICE PROBLEMS ON SQUARES AND SQUARE ROOTS

Example 1 :

196 is the square of

(A)  11     (B)  12     (C)  14     (D)  16

Solution :

By using the property of square numbers,

If a number has 4 or 6 in the unit's place then its square ends in 6.

Here 196 is the square number of 14.

That is,

196  =  14²

So, the answer is 14.

Example 2 :

Which of the following is a square of an even number ?

(A)  144     (B)  169     (C)  441     (D)  625

Solution :

By using the property of square numbers,

Square of even numbers is always even. 

144 is an even square number of 12.

So, the answer is 144.

Example 3 :

The square root of 1521 is 31 ?

(A)  True     (B)  False     (C)  None of these

Solution :

The square root of 1521  =  √(3 × 3 × 13 × 13)

Inside the radical sign, if the same number is repeated twice, take one number out of the radical sign.

  =  3 × 13

  =  39

The square root of 1521 is 39.

So, it is false.

Example 4 :

The square of 2.8 is 78.4 ?

(A)  True     (B)  False     (C)  None of these

Solution :

Square number of 2.8  =  2.8 × 2.8

=  7.84

The square number of 2.8 is 7.84

So, it is false.

Example 5 :

1000 is a perfect square ?

(A)  True     (B)  False     (C)  None of these

Solution :

√1000  =  √(5 × 5 × 5 × 2 × 2 × 2)

Since there are not grouped in identical pairs, 1000 is not a perfect square.

So, it is false.

Example 6 :

Which of the following is a square of an odd number ?

(A)  256     (B)  361     (C)  144     (D)  400

Solution :

By using the property of square numbers ,

Square of odd numbers is always odd. 

361 is an odd square number of 19.

So, the answer is 361.

Example 7 :

Which of the following is not a perfect square ?

(A)  361     (B)  1156     (C)  1128     (D)  1681

Solution :

By using the property of square numbers,

The number having 2, 3, 8, 7 are its end place are not perfect square numbers.

Here 1128 is not a perfect square number.

So, the answer is 1128.

Example 8 :

The value of √(176 + √2401)

(A)  14     (B)  15     (C)  16     (D)  17

Solution :

Find the square root of 2401

√2401  =  √(7 × 7 × 7 × 7)

Inside the radical sign, if the same number is repeated twice, take one number out of the radical sign.

=  7 × 7

√2401  =  49

Then,

The value of √(176 + √2401)  =   √(176 + 49)

=  √225

=  √(15 × 15)

=  15

So, the answer is 15.

Example 9 :

A perfect square can never have the following digit at ones place.

(A)  1     (B)  6     (C)  5     (D)  3

Solution :

By using the property of square numbers,

In perfect square numbers, the digits at the unit place are always 0, 1, 4, 5, 6 or 9

The number having 2, 3, 8, 7 are its end place are not perfect square numbers.

So, we have only one option (D) that has a digit 3 and if 3 is at unit place then the given number is not a perfect number.

Example 10 :

Which of the following will have 1 at its unit place ? 

(A)  19²     (B)  17²     (C)  18²    (D)  16²

Solution :

By using the property of square numbers,

If a number has 1 or 9 in the unit's place then its square ends in 1.

Then,

19²  =  19 × 19

19²  =  361

So, the answer is 19².

Apart from the stuff given above,  if you need any other stuff in math, please use our google custom search here.

Kindly mail your feedback to v4formath@gmail.com

We always appreciate your feedback.