rommysingh wrote:Is xy > x²y²?
(1) 14x² = 3
(2) y² = 1

xy > x²y² only if xy≠0, with the result that x²y² > 0 (since the square of a nonzero value must be positive).
Thus, we can safely divide each side by x²y², which must be a POSITIVE value:
(xy)/(x²y²) > (x²y²)/(x²y²)
1/xy > 1.

1/xy > 1 only if xy is a POSITIVE FRACTION between 0 and 1.
Question stem, rephrased:
Is xy a positive fraction between 0 and 1?

Statement 1: x² = 3/14, implying that x = ±√(3/14).
Statement 2: y² = 1, implying that y = ±1.

Both statements are satisfied if x=√(3/14) and y=1.
In this case, xy = √(3/14), which is a positive fraction between 0 and 1.
Both statements are satisfied if x=√(3/14) and y=-1.
In this case, xy = -√(3/14), which is NOT a positive fraction between 0 and 1.
Thus, the two statements combined are INSUFFICIENT.

The correct answer is E.

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.

Is xy > x^2y^2?
(1) 14x^2 = 3
(2) y^2 = 1

If we convert the original condition, we get (xy)^2-xy<0, xy(xy-1)<0 and ultimately we want to know whether 0<xy<1.
We have 2 variables, and hence we need 2 equations to match the number of variables; the conditions provide 2 equations, so the answer is likely to be (C).
Looking at the conditions together, x^2=3/14, x=-sqrt(3/14),sqrt(3/14) and y^2=1, y=-1,1, which means
The question is answered 'no' for xy=-sqrt(3/14) but 'yes' for xy=sqrt(3/14). No unique answer is given, and therefore the conditions are insufficient, making the answer (E).

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