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abhaypratapsingh
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abhijit_sen
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pmenon
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abhijit_sen
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pmenon
i think its C.

all you know from 1 is that x > y^2, i.e. that x is a positive number. we can have x=1/2 and y=1/2, but y^2=1/4 and so even though x=y, x >y^2 ... so this statement alone cant be sufficient.
Thanks missed that.
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rao
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abhijit_sen
1) 2 > 1 also 4 > 1
5 > 3 also 25 > 1
0 > -1
This statement will always hold true. So question can be answer hence answer should be A or D.

2) x^3 > y.
(4)^3 > 5 but 4 4 also 5 > 4 (supporting statement 2)
Thereby Maybe case so on basis of B question cannot be answered.

So answer is A.

your are only taking interger values. If you take value of x between 0 and 1. then A is not suff.


Is x > y?

(1) SQRT(X)> y

(2) x^3 > y


1) x > y^2
2) x^3 > y


from 1, X can be 'integere/improper fraction ( X can be any value more than 1)' more than Y or X can be a proper fraction greater than Y
from 2, X can be 'integere/improper fraction ( X can be any value more than 1)' greater than Y or X can be a proper fraction less than Y.

adding together, X is a 'integere/improper fraction ( X can be any value more than 1)' more than Y

IMO C



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