mbaprep2016 wrote:

Bunuel I have one doubt here

|x−z−y|>x−z+y? in normal case open the mode by multiplying + and another by multiplying with negative

x−z−y>x−z+y --(1) and -x+z+y>x−z+y --(2)

solving 1 we y<0

solving second we get x<z

now if we see option A it gives us hint that basically

|x-z-y| = -(x-z-y) = y + z - x. The question becomes: is y + z - x > x - z + y?--- so I thought Its sufficient

Now in case of B we already know that mode is negative ... so why cant it be sufficient

Please help me understand what M i missing here

I think the key here is to note that statement one DOES NOT satisfy the condition of y < 0 (the statement says that 0<x<y<z) and thus, we need to only take care of the other case and can ignore the y < 0 case which makes statement 1 sufficient.

In other words, if we solve the |x-z-y|, we can have two cases - one positive and one negative as you specified above.

Now, if we talk about statement 2, it says that (x–z–y) is negative (less than zero). So, this basically means that our expression in the absolute expression (in the question stem is negative) or in other words, |x−z−y| = - (x-z-y). BUT, this does not answer the main question we have in the question stem.

Knowing that the absolute value of an expression is negative does NOT answer the main question at hand which is whether |x−z−y| > x−z+y?

Hope the information helps. All the best to you !!

_________________

Best Regards,

RS