[quote="venmic"]Is x > y?

(1) x^2 > y

(2) √x < y

Can anyone explain a simple method to this could not follovv statement B

(1) For \(x=-2 > y=-3, \, x^2=4>-3.\) But \(x=2 < y=3\), although \(x^2=4>y=3.\)

Not sufficient.

(2) From the given inequality it follows that \(x\) must be non-negative (because of the square root) and since \(y>\sqrt{x}\geq0\), necessarily \(y\) is positive.

Therefore, we can square the given inequality and get \(x<y^2.\)

For \(x=4, y=3, \,x=4<y^2=9,\) and \(x>y.\)

But if \(y=1,\) we cannot have simultaneously \(x<y^2=1\) and \(x>y=1.\)

Not sufficient.

(1) and (2) together:

Consider the two cases: \(x=2, y=3\) and \(x=4,y=3.\)

Again, not sufficient.

Answer E.

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