1) Let y = 0. x can be anything.

2) 0 ≤ │x – 3│ ≤ –y
y ≤ 0

Combine that with y ≥ 0, then y = 0
Then x = 3

I would choose B as well. I want to see what other people think on this one

GMATQuantCoach wrote:1) Let y = 0. x can be anything.

2) 0 ≤ │x – 3│ ≤ –y
y ≤ 0

Combine that with y ≥ 0, then y = 0
Then x = 3

I would choose B as well. I want to see what other people think on this one

St2 can be valid only if y=0, as LHS |x-3| always positive which is less than equal to a number, y, greater than equal to zero

hence |x-3|=0, x=3

Ans B

why don't you test statement (1) with x<0 and x>0 for the statement (2) ?
I have some difficulties to follow

Xbond wrote:why don't you test statement (1) with x<0 and x>0 for the statement (2) ?
I have some difficulties to follow

If y &#8805; 0, what is the value of x ?

(1) &#9474;x – 3&#9474; &#8805; y

(2) &#9474;x – 3&#9474; &#8804; –y

Oh..i haven't mentioned anything about statement 1 because it's quite easy to understand that you won't get any single value of x simply by the equation &#9474;x – 3&#9474; &#8805; y when y &#8805; 0

Consider y=1 or y=2, we can have so many values of x...it is insufficient..
But statement 2 alone is sufficient, considering we've been given that y &#8805; 0 in the question.

|x-a| is the distance from x to a
1) X can be anything depending upon y. NOT SUFF
2) Distance cannot be negative so only possible value is y=0. so x=3. SUFF

B

Many thks guys