Since, it is given that
|a-b| = b-a, we can infer following from Absolute Value -
|a-b| = -(a-b) = b-a => a-b must be negative or 0
i.e. a-b < 0 (a=B, so a-b cannot be 0)
=> a<b Hence, only option III satisfies.
Now, let us see other options -
I.a < 0 even without plugging numbers, we can see this cannot be true always. No info ob B, and depending upon b the equation may vary.
II.a + b < 0 B=Not neccessarily tru, since we already know that only a<b is sufficient. It foes
...
|a-b| = b-a, we can infer following from Absolute Value -
|a-b| = -(a-b) = b-a => a-b must be negative or 0
i.e. a-b < 0 (a=B, so a-b cannot be 0)
=> a<b Hence, only option III satisfies.
Now, let us see other options -
I.a < 0 even without plugging numbers, we can see this cannot be true always. No info ob B, and depending upon b the equation may vary.
II.a + b < 0 B=Not neccessarily tru, since we already know that only a<b is sufficient. It foes
...
