Is x > y ?
(1) √x > y
(2) x³ > y
Statement 1: √x > y
It's possible that x=4 and y=1, in which case x>y.
It's possible that x=1/4 and y=1/3, in which case x<y.
INSUFFICIENT.
Statement 2: x³ > y
It's possible that x=2 and y=1, in which case x>y.
It's possible that x=-1/2 and y=-1/4, in which case x<y.
INSUFFICIENT.
Statements 1 and 2 combined:
Since we can't take the square root of a negative, statement 1 implies that x≥0.
Thus, when we combine the two statements, if y<0, we know that y<x.
Our concern is what happens when y≥0.
One approach is to memorize the shapes of some basic graphs:
Only in the yellow region is y<√x and y<x³.
The entire yellow region is below the graph of y=x, implying that y<x throughout the entire region.
Thus, combining the two statements, we know that y<x.
SUFFICIENT.
The correct answer is
C.
An alternate approach would be to use algebra to test the 3 cases: y=x, y>x, and y<x.
Case 1: y=x.
Statement 1: If y=x and y<√x, then x < √x.
Statement 2: If y=x and y<x³, then x < x³.
No value for x will work here: a number cannot be less than both its root and its cube.
Thus, y≠x.
Case 2: y>x.
Statement 1: If x<y and y<√x, then x < √x.
Statement 2: If x<y and y<x³, then x < x³.
No value for x will work here: a number cannot be less than both its root and its cube.
Thus, it is not possible that y>x.
Since it is not possible that y=x or that y>x, we know that y<x.
SUFFICIENT.
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