ronaldramlan wrote:Is x^2 + y^2 > 100?
(1) 2xy < 100
(2) (x+y)^2 > 200
Statement 2: (x+y)^2 > 200
Since the square of any value cannot be negative, (x-y)^2 ≥ 0.
Adding the two inequalities, we get:
(x+y)^2 + (x-y)^2 > 200+0.
x^2 + 2xy + y^2 + x^2 - 2xy + y^2 > 200.
2x^2 + 2y^2 > 200.
x^2 + y^2 > 100.
Sufficient.
The correct answer is
B.
Last edited by
GMATGuruNY on Sun Aug 14, 2011 11:57 am, edited 1 time in total.
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