Statement give Sqrt((x-3)^2)
i.e positive root of (x-3)^2.

Now statement 1 tells us that X is not 3.
So X can be anything except 3. If X = 2 then the main statement holds true, i.e
Sqrt (2-3)^2 = 1 = 3-2
When X = 4, this doesn't hold good.
Sqrt (4-3)^2 = 1 not equal to 3-4
So statement 1 is insufficient.

From statement two, we get X is negative value. Then only -x * (abs value of x) can be > 0.
for eq, let x = -2 then
Sqrt (-2-3)^2 = 5
3 - (-2) = 5.

This will hold true for any negative value of X.

Hence B is sufficient.

Have small doubt why is it not D ?

Is sqrt((x-3)^2) = 3 - x

Squaring both sides


(x-3)^2=(3-x)^2---cant we use this ??

Anand,

I am not sure whether this is the correct reason, but just trying.

The question Is sqrt((x-3)^2) = 3 - x ? So before starting we are not sure, whether the equality stands good, so how can we square it and proceed?

Suppose we are asked Is -5 = 5? If we square this we get 25 = 25 and we say the equation satisfies, but that is not the actual case right? -5 is not equal to 5 at the first place.

In the way i have solved, i found the values of the two sides of the equation separately and then tried to equate it.

Hope this makes sense.