from 1:
|x-3|>= y, a non-negative number. Since, y it could be any number, x could take various values.
Not Sufficient

from 2:
|x-3|<= -y
since absolute value can never be negative, this is only possible, if y=0.
therefore x-3 = 0, x=3
Sufficient

IMO answer B

iamseer wrote:from 1:
|x-3|>= y, a non-negative number. Since, y it could be any number, x could take various values.
Not Sufficient

from 2:
|x-3|<= -y
since absolute value can never be negative, this is only possible, if y=0.
therefore x-3 = 0, x=3
Sufficient

IMO answer B

IMO B

oops it has to be B. Seer explanation does all!

Ans: B

Since, | x -3 | >= 0 always true hence not sufficient.
|x -3| <= -y and Y >= 0
LHS is always +ve and rhs is -ve hence it is only satisfied when Y = 0
and hence X = 3

neoreaves wrote:If y > = 0, what is the value of x?
1. |x - 3| >= y
2. |x - 3| <= - y

st1: |x - 3| >= y

as y is a positive quantity and |x-3| > y .

implies that x-3 itself is a positive quantity and we can safely write it as x-3 > y . However, this doesn't give us an solid value for X. Hence, insufficient.

st2: |x - 3| <= - y

LHS of this is a modulus, which can be either 0 or positive number. However, on the RHS we have -y. Suppose, y is a non-zero number. In that case, RHS would become a negative quantity and cannot be equated to a modulus . Hence, y MUST be zero.

which will give us |X-3| = 0 or X = 3.

IMO B