First, since we know x and y are both positive, we don't have to worry about 0 or negative numbers.Abhijit K wrote:If x and y are positive, is x³ > y?
(1) √x > y
(2) x > y
Statement 1 is tricky. If √x > y, then it seems that x³ would also be greater than y, but things are not so simple, because fractions can work differently from whole numbers.
If x = 1/4 and y = 1/3, then √x = 1/2 and 1/2 > 1/3.
However (1/4)³ is way less than 1/3.
So Statement 1 is insufficient.
Statement 2 also seems to indicate that x³ > y. For instance, if x = 2 and y = 1, then x³ > y.
But once again given fractional values of x and y, x could be greater than y and x³ could be less than y. For instance if x = 1/2 and y = 1/3, then x³ = (1/2)³ = 1/8 which is less than 1/3.
So Statement 2 is insufficient.
Combining the statements we can still find examples that make x³ > y and fractions that fit both statements and result in x³ < y.
For example if x = 1/2 and y = 1/3, then x > y and √x > y, but (1/2)³ = 1/8 which is less than 1/3.
So the statements in combination are still insufficient.
Choose E.
