Hi Pugal:

We are asked to find whether x > 0.
Let's look at the statements.

(i)
x^2 = 2x => x(x-2)=0. So either x=0 or x=2. If x=0, x is not greater than 0, if x=2, x is greater than 0. Insufficient.

(2)
x^3 = 3x => x(x^2-3)=0 => x=0 or x^2=3 => x=0 or x=-sqrt(3) or x=+sqrt(3). Since x can either be 0 or positive or negative, we can't know whether x is greater than 0. Insufficient.

Combining the two.
The only common value that satisfies both equations is x=0. Since x is equal to 0, x is not greater than 0. Sufficient.

Let me know if this helps

eagleeye wrote:Hi Pugal:

We are asked to find whether x > 0.
Let's look at the statements.

(i)
x^2 = 2x => x(x-2)=0. So either x=0 or x=2. If x=0, x is not greater than 0, if x=2, x is greater than 0. Insufficient.

(2)
x^3 = 3x => x(x^2-3)=0 => x=0 or x^2=3 => x=0 or x=-sqrt(3) or x=+sqrt(3). Since x can either be 0 or positive or negative, we can't know whether x is greater than 0. Insufficient.

Combining the two.
The only common value that satisfies both equations is x=0. Since x is equal to 0, x is not greater than 0. Sufficient.

Let me know if this helps

Great summary! This ia a good, thorough way to solve this. In particular, the key is spotting that 0 is a possible solution. The "trap" choice some test-takers may fall into is trying to simplify by dividing out x's. But unless you know a variable is non-zero, you can't divide by it!

Interestingly, we could use that very trap to get to answer (C) by a roundabout method. If we assume x is non-zero and divide each side of statement 1) by x, we get x = 2, so x ^ 2 = 4. If we divide each side of statement 2) by x, we'll get x ^ 2 = 3. But it already equalled 4! Our assumption that x was non-zero led to a contradiction, and the two statements on in a DS problem will never contradict. Therefore, the only logical conclusion is that our assumption was wrong, x = 0, and the solution is (C).