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in fact, st2 is sufficient as well so answer is D.

st1 is sufficient as noted above:
if x<0 then |x|=-x
so y=|x|+x = -x+x = 0
suff.

st2 is sufficient as well:
from stem only we can deduce that y is non-negative:
if x<0 then y=0 as we showed above,
if x=0 then obviously y=0.
if x>0 then |x|=x and y = 2x > 0

also note that y is an integer (!!!!!!!)

st2 says y<1. the only non-negative integer less than 1 is 0.
so y must be 0.

Given: y is integer, y = |x| + x
asked : does y = 0 ?

from the question, 0=< y =< infinity; that is, y can range between zero and any positive integer.

because |x| is always positive.
if x is positive then y = 2x
if x is negative then y = 0

(1) x<0
--------------
x is negative then y = x - x = 0 .. YES y = 0
statement 1 is sufficient (2) y<1
--------------
the only possible value for y < 1 is zero because y can never be negative and y is an integer so it can't be some fraction less than 1 like 1/2 or so.

So y in this case has to equal zero [ y = 0 ]
statement 2 is sufficient

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