Is x > y ? (1) x^(1/2) > 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.

Hey VeritasKarishma,
great approach, thanks a lot.
But here I have a doubt (excuse me if silly ) -
When we are combining both graphs - its clear that X>Y when 0>x>1 AND 1>x>+.
But what about when the value of x is between "-1 to 0".....as shown with red color in second graph.

please explain.

regards,
Tamal

Stmnt 1 tells you that x >= 0 (since \(\sqrt{x}\) is not defined for negative values of x). Hence, when you use both stmnts together, negative values of x are to be ignored.

(1) \(\sqrt{x} > y\)
We can say that x is positive, as we deal with square roots of positive numbers only.
\(x\geq \sqrt{x} > y\)......When \(x\geq 1\)
But when 0<x<1, \(\sqrt{x}>x\) and we can not say anything about placement of y.
\(\sqrt{x}>x>y\).......\(\sqrt{\frac{1}{4}}>\frac{1}{4}>0\)
\(\sqrt{x}>y>x\).......\(\sqrt{\frac{1}{4}}>\frac{1}{3}>\frac{1}{4}\)
Insufficient

(2) \(x^3 > y\)
\(x\geq x^3 > y\)......When \(x< 1\)
But when x>1, \(x^3>x\) and we can not say anything about placement of y.
\(x^3>x>y\).......\(2^3>2>0\)
\(x^3>y>x\).......\(2^3>5>2\)
Insufficient


Combined
x is positive and both \(x^3\) and \(\sqrt{x}\) are greater than y.
Say x>1......\(x^3>x>\sqrt{x}>y\)
But if x<1....\(\sqrt{x}>x>x^3>y\)
x=1.....\(\sqrt{x}=x=x^3>y\)
Hence, in all cases, x>y


Sufficient


C

Hi Bunuel can y have negative value

Posted from my mobile device

Good question. When we consider the two cases together, y can absolutely be some negative number, say -10. If y is negative and since we concluded that x must be positive, then in this case (y = negative) < (x = positive). So, we'd get the same YES answer to the question: x is greater than y.

Does this make sense?

We can use the concept of a number line to solve this question. Attached video solution to the question -

I still didn't understand can you explain. I can't seem to grasp the idea of red and green regions