Official Answer: С
Given that |x| = |y+1|
both have same signs (either positive or negative), then x = y+1
, which makes x
both do not have the same signs (one is negative and the other is positive), then x = -(y+1)
or (x + y + 1) = 0
. In that case, either one could be greater in absolute value. S1 is not sufficient by itself.
If x^y = x! + |y|
both have to be positive because x! + |y|
is always a positive integer. If y
is negative, x^y
can never be an integer. So for x^y
to be an integer, y
has to be positive. Let's see what values satisfy the equation:
If x = 1
, and y = 0
, x^y = x! + |y| = 1
If x = 2
, and y = 2
, x^y = x! + |y| = 4
have at least two possibilities. S2 is not sufficient by itself either.
Statement (1)+Statement (2):
Because of Statement (2), x
have to be positive, in which case, by Statement (1) x > y
. Sufficient. Therefore, the answer is C.
I don't get it..Statement 2 gives you 2 solutions 1,0 and 2,2 so x and y are non-negative .. even combining with statement 2 you get 2 solutions.. so answer is E..