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Bunuel
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Statement 1: x = t^n, where t is a positive integer and n is odd.
If x = 2^1 , then x is not a perfect square
If x = 4^1 , then x is a perfect square
Not Sufficient

Statement 2: sqrt(x) = k
x = k^2.
Sufficient

Correct option : B
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Great Question
Here we need to check if x is a perfect square of not
Statement 1
x=5^3 => no
x=9^3=> 3^6 => yes
Hence insufficient
Statement 2
x^1/2 is basically √x => √x=k => x=k^2 hence sufficient
hence B
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The question checks the basics of perfect squares.

1. x= t^n. n is odd. Although no other odd no. in the power will give a perfect square, but any perfect square raised to the power of 1(which is an odd number) will obviously remain a perfect square. But any perfect square raised to another odd numbers(3, 5, 7,etc.) will not result in a perfect square. Insufficient. AD is out.

2. x^(1/2) is an integer. A square root of any number can only be an integer if that number is a perfect square. Sufficient. B is the correct choice.

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