let me try:

sandipchowdhury wrote:

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

A. !X-3!>=Y

B. !X-3!<=-Y

From Stmt 1--->

Lets take Y as 2 (as y >= 0)

we get, |x-3|>=2

Here X can have any values like 5,6,7....etc

Hence not sufficient

From Stmt 2---->

As per the question stem, the possible values of Y can be 0,1,2,3....etc

Now a modular value cannot be anything but 0.

Hence, |x-3|=0 and x = -3

Sufficient

Answer is B

Sandeep's first part is clear that x can h ave multiple values. Gurmeet , you are correct that y can be 0 in which case x=3. But still, y can be 1 also, in which case x can have multiple values. Hence A is not sufficient.

Now coming to (2),

Since, y >=0, it means that -y <=0 which implies |x-3|<=0 (since |x-3|<=y). Now the left hand (|x-3|)is a modulus which can never be negative, it can only be greater than or equal to 0.

Which implies |x-3| = 0 --> x=3. Hence (B) is sufficient.

I hope this was bit clear.

let me try:

sandipchowdhury wrote:

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

A. !X-3!>=Y

B. !X-3!<=-Y

From Stmt 1--->

Lets take Y as 2 (as y >= 0)

we get, |x-3|>=2

Here X can have any values like 5,6,7....etc

Hence not sufficient

From Stmt 2---->

As per the question stem, the possible values of Y can be 0,1,2,3....etc

Now a modular value cannot be anything but 0.

Hence, |x-3|=0 and x = -3

Sufficient

Answer is B

Sandeep's first part is clear that x can h ave multiple values. Gurmeet , you are correct that y can be 0 in which case x=3. But still, y can be 1 also, in which case x can have multiple values. Hence A is not sufficient.

Now coming to (2),

Since, y >=0, it means that -y <=0 which implies |x-3|<=0 (since |x-3|<=y). Now the left hand (|x-3|)is a modulus which can never be negative, it can only be greater than or equal to 0.

Which implies |x-3| = 0 --> x=3. Hence (B) is sufficient.

I hope this was bit clear.

i think you want to write since |x-3|<=-y

if it is then a very good explaination.

thanks

Shobuj