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Bhavikasingla
Is x < y ?

(1) \(\sqrt{x} > y\)
(2) \(x^2 > y\)
­There are certain properties associated with where a number is placed on a numberline.

Between 0 and 1, the value of x decreases as power increases =>\( \sqrt{x}>A>x>x^2>B\).
If y is at A's position, the answer is Yes, but if at B, the answer is NO for statement 1.
Greater than 1, the value of x increases as power increases => \(x^2>B>x> \sqrt{x}>A\)
If y is at A's position, the answer is Yes, but if at B, the answer is NO for statement 2.

When you combine....
x is always between \(\sqrt{x} \)and \(x^2\), so if these both are greater than y, then surely all values between \(\sqrt{x} \)and \(x^2\), will also be greater than y or x>y
Sufficient

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