Hi,

in a data sufficiency question involving inequalities, start thinking about signs (pos vs neg).

The question is:

Is 1/(a-b)<b-a?

We should note that (a-b) and (b-a) will have opposite signs (or if they are both zero, then no sign).

(1) tells us that a<b. When we subtract something large from something smaller, the result is negative. Thus, 1/(a-b) is negative. And when we subtract something small from something larger, the result is positive. Thus, (b-a) is positive. Thus, the question becomes:

Is (-ve)<(+ve)?

The answer is "definitely yes", and so (1) is sufficient. Eliminate B, C, and E.

(2) tells us that |a-b|>1. "|a-b|" means the distance between "a" and "b" on the number line. So, the distance between a and b on the number line is greater than 1. So, for example, a could be 4 and b could be 2, in which case 1/(a-b) is positive and (b-a) is negative, and the question becomes:

Is (+ve)<(-ve)? and the answer is "no".

However, it could also be the other way around: a could be 2 and b could be 4 (the distance between them is still greater than 1, so we are satisfying the statement). But in this case the answer to the question is "yes".

Because we can get both a "yes" and "no" answer, (2) is not sufficient.

(1) is sufficient by itself; (2) is not; choose A.

Rahul@gurome wrote:The question is Is 1/(a-b) < (b-a)?
Consider first (1) alone.
It says a<b.
Or a-b<0.
Or a-b is negative.
So even 1/(a-b) is negative.
Also b-a>0.
Or b-a is positive.
So obviously, since a negative quantity is less than a positive quantity, 1/(a-b) < (b-a).
So (1) alone is sufficient.
Next consider (2) alone.
It says 1 < la-bl.
It means a-b > 1 or a-b < -1.
So a-b can be either positive or negative and we cannot say definitely whether 1/(a-b) < (b-a) or not.
Or (2) alone is not sufficient.
The correct answer is (A).

my speed is so bad for math... i am dumb these days... but rahul, can i solve the problem like this:
1/(a-b)<(b-a) <=> 1/(a-b)+(a-b) <0 <=> [1+(a-b)^2]/(a-b)<0
now i will have to see if [1+(a-b)^2]/(a-b)<0
1. a<b
because (a-b)^2>0 with all a, b so 1+(a-b)^2>0 with all a and b
and a-b <0 so sufficient
2. 1<|a-b|
a-b>1 or a-b<-1
no conclusion about this, thus sufficient