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rommysingh
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xy > x²y² only if xy≠0, with the result that x²y² > 0 (since the square of a nonzero value must be positive).rommysingh wrote:Is xy > x²y²?
(1) 14x² = 3
(2) y² = 1
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.
