Bunuel wrote:
If w, x and y are integers such that w < x < y, are w, x and y consecutive integers?
(1) y - w = 2 --> y = w + 2. Thus we have that w < x < (w + 2). There is only one integer between w and w + 2, which is w + 1. Thus the three integers are w, x = w + 1, and y = w + 2 --> w, x and y are constitutive integers. Sufficient.
(2) The average (arithmetic mean) of w, x and y is x. If w=0, x=5 and y=10, then the answer is NO but if w=0, x=1 and y=2, then the answer is YES. Not sufficient.
Answer: A.
Hope it's clear.
Hi
Bunuel,
I understand the no. approach for St.2, but I don't understand what I am doing wrong.
I assumed that the nos. are consecutive. Let w=n, so x=n+1 and y=n+2
The avg of three no. would be [n+n+1+n+2]/[3]= n+1, which is equal to x, same as St.2, so the assumption is correct.
Kindly explain.