Data Sufficiency

confuse mind wrote:Is x^2 + y^2 > 100

1. 2xy < 100
2. (x + y)^2 > 200

[spoiler]OA - C
IMO - B will post my complete solution later[/spoiler]

(1) 2xy < 100
If x = y = 0, then x² + y² = 0 < 100
If x = 10, y = -10, then x² + y² = 100 + 100 = 200 > 100
No definite answer; NOT sufficient.

(2) (x + y)² > 200
x² + 2xy + y² > 200
Since the square of any number is more than or equal to zero, so (x - y)² more than or equal to zero.
So, x² + y² more than or equal to 2xy implies x² + (x² + y²) + y² > 200
2(x² + y²) > 200
x² + y² > 100; SUFFICIENT.

The correct answer is B.

confuse mind wrote:Is x^2 + y^2 > 100

1. 2xy < 100
2. (x + y)^2 > 200

Target Question: Is x^2 + y^2 > 100?

Statement 1: 2xy < 100
The only way to show that this statement is not sufficient is by counter-example.
There are several pairs of values that satisfy the condition that 2xy < 100
case a: x=0 y=0, in which case x^2 + y^2 is less than 100
case b: x=0 y=100, in which case x^2 + y^2 is greater than 100
So, statement 1 is NOT SUFFICIENT

Statement 2: (x+y)^2 > 200
Expand to get: x^2 + 2xy + y^2 > 200
Rearrange to get: x^2 + y^2 + 2xy > 200

Aside: (this is the tricky part)
Notice that for any values of x and y, it must be true that (x-y)^2 > 0
Expand to get x^2 - 2xy + y^2 > 0
Rearrange to get x^2 + y^2 > 2xy
Now add x^2 + y^2 to both sides to get: x^2 + y^2 + x^2 + y^2 > x^2 + y^2 + 2xy
Simplify to get: 2(x^2 + y^2) > x^2 + y^2 + 2xy

Now notice that the right-hand side of this inequality matches the left-hand side of the inequality we derived from statement 2.
So, we can write: 2(x^2 + y^2) > x^2 + y^2 + 2xy > 200
Now ignore the middle part to get: 2(x^2 + y^2) > 200
Divide both sides by 2 to get x^2 + y^2 > 100
Since we can answer the target question with certainty, statement 2 is SUFFICIENT and the answer is B

Cheers,
Brent