sohrabkalra wrote:is x^2+y^2 > 100?
1) 2xy<50
2) (X+Y)^2>200
OA and Doubt after a few replies !
Statement 1: 2xy < 100.
Thus, xy < 50.
If x=1 and y=1, then x²+y² < 100.
If x=2 and y=10, then x²+y² > 100.
Insufficient.
Statement 2: (x+y)² > 200.
x² + 2xy + y² > 200.
Since the question stem asks about x²+y², we want to eliminate 2xy from statement 2.
One approach: (x+y)² + (x-y)² = (x² + 2xy + y²) + (x² - 2xy + y²) = 2x² + 2y² = 2(x²+y²).
Since the square of a value cannot be negative, (x-y)² ≥ 0.
Thus, we can add together (x+y)² > 200 and (x-y)² ≥ 0:
(x+y)² + (x-y)² > 200+0.
2(x²+y²) > 200.
x² + y² > 100.
Sufficient.
The correct answer is
B.
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