Hi All,

This DS question is rooted in Classic Quadratics and Radical rules. There's a heavy "logic" component to this prompt, and you'll have to do a certain amount of math to get to the solution...

We're given a number of facts to work with:

1) A = X+Y

2) B = X-Y

3) A^2 = B^2

We're asked for the value of Y.

Before dealing with the two Facts, I'm going to take a moment to simplify the information in the prompt.

A^2 = (X+Y)^2 = X^2 + 2XY + Y^2

B^2 = (X-Y)^2 = X^2 - 2XY + Y^2

A^2 = B^2.....so......

X^2 + 2XY + Y^2 = X^2 - 2XY + Y^2

If we cancel out the terms, we're left with...

2XY = -2XY

There are only 3 ways for this equation to exist....

X = 0 OR Y = 0 OR BOTH = 0

This is an important restriction that will impact the answer to this question.

Fact 1: √X + √Y > 0

Since radicals are either POSITIVE or ZERO, either X or Y could = 0 (but not both).

IF....X=1, Y=0 the answer to the question is 0

IF....X=0, Y=1 the answer to the question is 1

Fact 1 is INSUFFICIENT

Fact 2: √X − √Y > 0

Here, the DIFFERENCE is > 0, so the only way for that difference to be POSITIVE, given ALL of the restrictions we've already discussed, is for the X to be POSITIVE and the Y to be 0. Here, the answer to the question is ALWAYS 0.

Fact 2 is SUFFICIENT.

Final Answer:

GMAT assassins aren't born, they're made,

Rich

Nice explanation and powerful 'TESt It' methodology. I reached the same result: X = 0 OR Y = 0 OR BOTH = 0 but I was afraid that I'm mistaken because of zeros in the result so I applied 'Testing it' in the equation but it got so complicated with many variables and situations.

However, I want to ask about Fact 2. If I choose X=0 & Y= 9 for easy roots example. So The result will be the following:

√X − √Y > 0 then Fact 2: √0 − √9 > 0------> .... √9 will be either 3 or -3. If it is 3, then 0-3>0 so not satisfying the condition but if Y=-3, then 0- (-3)= 3>0 so it is OK with condition. Therefore, Y will always any positive number. So Fact 2 should be insufficient as there many values for Y.