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<50
xy<25
x=4, y=4 x^2+y^2=32 so x^2+y^2<100

x=10, y=1 x^2+y^2=101 so x^2+y^2>100
INSUFFICIENT.

Statement 2:

(x+y)^2>200
The expression we're concerned with is x^2+y^2, so we can restrict our attention to positive values of the inequality:

(x+y)>10√2

First, note that if x+y=c, x^2+y^2 is minimized when x=y=c/2. Assume for a second that this is an equation instead of an inequality, and x+y=10√2. That means that x=5√2 and y=5√2 minimizes x^2+y^2, and plugging in we get x^2+y^2=100. But since we need x+y to be GREATER than 10√2, the values of x and y that minimize x^2+y^2 would have to be at least slightly larger than 5√2, in which case x^2+y^2 would always be greater than 100. SUFFICIENT.

Ans: B

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.

Yes, even i solved for B but OA was given as C!
This comes from the free veritas prep CAT, guess just a mistake in OA!

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