shrouded1 wrote:
I thought this was a really tough question !
Is \(x > y\) ?
(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)
One of those gorgeous questions that seem so simple at first but surprise you later...
Best way to work on these is to fall back on your drawing skills (Yes, I love diagrams!)
Statement (1): If I can say that \(x >= \sqrt{x}\) for all values of x, then I can say that x > y. The green line shows me the region where \(x >= \sqrt{x}\) but the red line shows me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient.
I also understand from this statement that x >= 0.
Statement (2): If I can say that \(x >= x^3\) for all values of x, then I can say that x > y. The green lines show me the region where \(x >= x^3\) but the red lines show me the region where it isn't. Then, for the red line region, x MAY NOT be greater than y. Not Sufficient.
Attachment:
Ques.jpg [ 9.04 KiB | Viewed 139596 times ]
Using both together, I know x >= 0.
If 0 <= x <= 1, then we know \(x >= x^3\). Since statement (2) says that \(x^3 > y\), I can say that x > y.
If x > 1, then we know \(x > \sqrt{x}\). Since statement (1) says that \(\sqrt{x} > y\), I can deduce that x > y.
For all possible values of x, we can say x > y. Sufficient. Answer (C).