A list contains 11 consecutive integers. What is the greatest integer on the list?
I. If x is the smallest integer on the list, then $$(x + 72)^{\frac{1}{3}} = 4$$.
II. If x is the smallest integer on the list, then $$\frac{1}{64} = x^{-2}$$.

Statement I : If x is the smallest integer on the list, then $$(x + 72)^{\frac{1}{3}} = 4$$

=> $$(x+72)=4^3=64$$

=> $$x=64-72=-8$$

Now, 11 consecutive integers starting from -8 : -8,-7,-6,-5,-4,-3,-2,-1,0,1,2

Thus, greatest integer is 2 and statement I alone is sufficient.

Statement II : If x is the smallest integer on the list, then $$\frac{1}{64} = x^{-2}$$

=> $$(x)^{2}=64$$

=> $$x=\pm8$$

Since, there are two values of $$x$$, we cannot find the greatest integer. statement II alone is insufficient.

=> Ans - (A)