shrey2287 wrote:If x^2 > y^2, is x>y?
1) x > |y|
2) |x| > y

Let's begin by exploring what conclusions we can make about x and y if x^2 > y^2.
This tells us that the magnitude of x is greater than the magnitude of y.
In other words, if we examine x and y on the number line, the distance from x to 0 will be greater than the distance from y to 0.

So, if x^2 > y^2, there are 6 possible cases to consider (all shown on the number line).
case a: ......x...y...0.............
case b: ......x.......0...y..........
case c: ......x.......0............. (and y=0)
case d: ..........y...0...........x..
case e: ..............0..y...x.......
case f: ...............0.......x...... (and y=0)

Now let's tackle the question.

Target question: Is x > y?

Statement 1: x > |y|
Since |y| is greater than or equal to 0, statement 1 tells us that x must be positive.
When we check our 6 possible cases, we can see that this eliminates cases a, b and c (where x is negative), leaving us with cases d, e and f.
In cases d, e and f, x is definitely greater that y.
As such, statement 1 is SUFFICIENT

Statement 2: |x| > y
If |x| > y, which of our 6 possible cases can we eliminate?
We cannot eliminate any of the cases.
In one case, (case a) x is not greater that y.
In one case, (case c) x is greater that y.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent