shrouded1 wrote:
I thought this was a really tough question !
Is \(x > y\) ?
(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. In the area of inequality, the number of equations means the number of inequalities, since the number of equations means the number of conditions.
In the original question, the number of variables is 2 and the number of equation is 0.
Hence, C is most likely the answer.
For the condition (1), we have two cases.
\(x = 1/4\) and \(y = 1/3\) satisfies the condition (1), but \(x > y\) is not true.
\(x = 4\) and \(y = 1\) satisfies the condition (1) and \(x > y\) is true.
Thus, the condition (1) is not sufficient.
For the condition (2), we also have two cases.
\(x = 2\) and \(y = 3\) satisfies the condition (2), but \(x > y\) is not true.
\(x = 3\) and \(y = 2\) satisfies the condition (2) and \(x > y\) is true.
Thus, the condition (2) is not sufficient, either.
For both the conditions (1) and (2) together, we should consider two cases.
Case 1: \(0 \leq x \leq 1\).
Since \(x^3 \leq x\), if \(x^3 > y\) by the condition (2), \(x \geq x^3 > y\), that is, \(x >y\).
Case 2: \(x > 1\)
Since \(x > \sqrt{x}\), if \(\sqrt{x} > y\) by the condition (1), \(x > \sqrt{x} >y\), that is, \(x > y\).
Hence, both the conditions (1) and (2) together are sufficient, as expected earlier.