so it means that:

case 1: if x = 9 (perfect sq) and z = cannot be any integer as it will not satisfy the equation sqrt xz = i... it has to be a perfect sq it self, such as 4... sqrt 9*4 = 6

Case 2: but if it x - not a perfect square say 3, then z will have to be 3 so satisfy the equation sqrt xz = Integer, but in that case sqrt z will not be an integer..

Case 3: if x is a reduced fraction: 5/2 then z will have to be 2 or 8 to satisfy the equation sqrt xz = Integer and even in this case sqrt z is not an integer

so the fact that \sqrt{z}is an integer depends on value of x.. therefore, Insufficient.

I hope i have finally understood.

st 2... in this x= z3 means nothing as z= cuberoot x... x can be 27 in this case z= 3 but \sqrt{z}= irrational number

or x can be 64.. making z = 4 and \sqrt{z} and integer... Insufficient again

together (st 1 & st2): \sqrt{xz} = \sqrt{z^4} = z^2 = Integer.... but \sqrt{z} can be an integer or irrational number

i think i have got it now. Am i?

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Hope to clear it this time!!

GMAT 1: 540

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