ghoward199 wrote:From a GMAT CAT Practice Test:

If y>=0, what is the value of x?

I. lx-3l>=y
II. lx-3l<=-y

I am not sure why the correct answer is B.

y>= 0, x = ?

Statement 1:
|x-3| >=y , clearly x can have any value based on y's value - Insufficient

Statement 2:
|x-3| <= -y
Since mod of any value has non-negative value, hence |x-3| >=0
Also, y >=0 , hence -y <= 0
Therefore, |x-3| = 0 => x= 3 --- Sufficient, B

That explains it! Thanks for the help!

@anshu, that |x-3|=<-y and y<=0 doesn't mean that we have fixed condition for |x-3|=<0 - and even if we have there are still two solutions, one is interval x<3 and the other is x=3.

I approached this question differently.
st(1) |x-3|>=y, since y can be 0 OR +ve we get at least two values for x --> x=3 when x-3=0, and x>3 when when x-3>0;
st(2) |x-3|<= -y, -y can be -ve or 0 only, because y>=0. BUT if we square both parts we get (x-3)^2<=0 WHICH means (x-3) CAN be only 0 BUT not -ve, as the squared value cannot be -ve. Hence x-3=0 and x=3 Sufficient.

Choice B should be relevant here.

anshumishra wrote:

ghoward199 wrote:From a GMAT CAT Practice Test:

If y>=0, what is the value of x?

I. lx-3l>=y
II. lx-3l<=-y

I am not sure why the correct answer is B.

y>= 0, x = ?

Statement 1:
|x-3| >=y , clearly x can have any value based on y's value - Insufficient

Statement 2:
|x-3| <= -y
Since mod of any value has non-negative value, hence |x-3| >=0
Also, y >=0 , hence -y <= 0
Therefore, |x-3| = 0 => x= 3 --- Sufficient, B

Night reader wrote:@anshu, that |x-3|=<-y and y<=0 doesn't mean that we have fixed condition for |x-3|=<0 - and even if we have there are still two solutions, one is interval x<3 and the other is x=3.

|x-3| <= -y and y>=0
Minimum value of |x-3| = 0 >= - y ( -y =0 only if y=0, else -y = -ve, as y has non-negative value)
That means the inequality |x-3| < -y is impossible, so the equality (|x-3| = -y) must hold for some values of x and y.
The only possible solution is |x-3| = 0 (which the max. value for -y ).

yes, you are right - i did overlook modes going straight to squaring both sides. |x-3| cannot be -ve either, so squaring both sides is additional check on (perhaps, time consuming in GMAT). Your solution is adequate
//Thanks

anshumishra wrote:

Night reader wrote:@anshu, that |x-3|=<-y and y<=0 doesn't mean that we have fixed condition for |x-3|=<0 - and even if we have there are still two solutions, one is interval x<3 and the other is x=3.

|x-3| <= -y and y>=0
Minimum value of |x-3| = 0 >= - y ( -y =0 only if y=0, else -y = -ve, as y has non-negative value)
That means the inequality |x-3| < -y is impossible, so the equality (|x-3| = -y) must hold for some values of x and y.
The only possible solution is |x-3| = 0 (which the max. value for -y ).