Is 1/(a-b) < b-a?
1. a<b
2. 1<|a-b|
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inequalities
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(1) if a<b, then a-b is negative, meaning that 1/(a-b) is negative.beater wrote:Is 1/(a-b) < b-a?
1. a<b
2. 1<|a-b|
Conversely, if a<b, then b-a is positive.
So, it must be the case that 1/(a-b) < b-a (SUFFICIENT)
(2) Knowing that 1<|a-b|doesn't help us here. For example, a-b can equal either 2 (positive) or -2 (negative). Each case yields a different answer to the question "Is 1/(a-b) < b-a?" (INSUFFICIENT)
Brent Hanneson - Creator of GMATPrepNow.com
I would first consider the different cases for a and b:
1. a>0, b>0
2. a<0, b<0
3. a<0, b>0
4. a>0, b<0
Statement (1) tells us a<b, therefore it can only be case 1., 2. or 3.
In all cases a-b<0 and b-a>0, and consequently 1/(a-b)<b-a. SUFFICIENT
Statement (2) tells us 1<|a-b|, so there are two cases: a-b>1 and a-b<-1.
Let's first consider a-b>1 which means that its case 1., 2. or 4. In any of these cases b-a<0 and consequently 1/(a-b)>b-a.
Now let's consider a-b<-1 which means that its case 1., 2. or 3. In any of these cases b-a>0 and consequently 1/(a-b)<b-a. INSUFFICIENT
So its answer choice A.
1. a>0, b>0
2. a<0, b<0
3. a<0, b>0
4. a>0, b<0
Statement (1) tells us a<b, therefore it can only be case 1., 2. or 3.
In all cases a-b<0 and b-a>0, and consequently 1/(a-b)<b-a. SUFFICIENT
Statement (2) tells us 1<|a-b|, so there are two cases: a-b>1 and a-b<-1.
Let's first consider a-b>1 which means that its case 1., 2. or 4. In any of these cases b-a<0 and consequently 1/(a-b)>b-a.
Now let's consider a-b<-1 which means that its case 1., 2. or 3. In any of these cases b-a>0 and consequently 1/(a-b)<b-a. INSUFFICIENT
So its answer choice A.
