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.
Example 11 :
If 19x2 = 1002 - 902 , then find the value of x
A.10 B. 9 C. 11 D. 12
Solution :
Given that, 19x2 = 1002 - 902
19x2 = (100 + 90)(100 - 90)
19x2 = (190)(10)
x2 = (190)(10) / 19
x2 = 10 (10)
x = 10
Example 12 :
√(248 + √(51+√169))
Solution :
Given that, √(248 + √(51 + √169))
First we have simplify the numbers inside the most interior bracket.
= √(248 + √(51 + √13⋅13))
Since we have two 13's, we can factor one 13 out of the square root sign.
= √(248 + √(51 + 13))
= √(248 + √64)
= √(248 + √8⋅8)
= √(248 + 8)
= √256
= 16
Example 13 :
a * b * c = [√(a + 2)(b + 3)] / (c + 1)
then find the value of 6 * 15 * 3
Solution :
a * b * c = [√(a + 2)(b + 3)] / (c + 1)
Here a = 6, b = 15 and c = 3
Applying these values in the given expression, we get
= [√(6 + 2)(15 + 3)] / (3 + 1)
= [√8 x 18] / 4
= √144 / 4
= √(12⋅12) / 4
= 12 /4
= 3
So, the answer is 3.
Example 14 :
What will come in place of question mark in each of the following questions ?
√(32.4 / ?) = 2
Solution :
√(32.4 / ?) = 2
Let ? = x
√(32.4 / x) = 2
Squaring on both sides.
32.4/x = 4
Multiplying by x on both sides
32.4 = 4x
Dividing by 4,
32.4/4 = x
x = 8.1
So, the answer is 8.1
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