Going through the solutions posted above, I realized that this question is a good illustration of
the importance of first analyzing the question statement before moving to St. 1 and 2 in a DS question.
Most students dived straight into Statement 1 and then tried to draw inferences from St. 1 to determine whether the inequality given in the question statement was true or not.
Imagine if you had done this instead before going to St. 1:
The question is asking if:
\(\frac{1}{a-b} < b-a\)
Case 1: a - b is positive (that is, a > b)Multiplying both sides of an inequality with with the positive number (a-b) will not change the sign of inequality.So, the question simplifies to: Is 1 < (b-a)(a-b)
Now, b - a will be negative.
So, the question simplifies to: Is \(1 < -(a-b)^2\) . . . (1)
\((a-b)^2\), being a square term, will be >0 (Note: a - b cannot be equal to zero because then the fraction given in the question: 1/a-b becomes undefined)
So, \(-(a-b)^2\) will be < 0
Therefore, the question simplifies to: Is 1 < (a negative number?)
And the answer is
NO.
Case 2: a - b is negative (that is, a < b)Multiplying both sides of an inequality with the negative number (a-b) will change the sign of inequality.So, the question simplifies to: Is 1 > (b-a)(a-b)
Now, b - a will be positive.
So, the question simplifies to: Is \(1 > -(a-b)^2\) . . . (2)
Again, by the same logic as above, we see that the question simplifies to: Is 1 > (a negative number?)
And the answer is
YESThus, from the question statement itself, we've inferred that:
If a > b, the answer to the question asked is
NOIf a < b, the answer to the question asked is
YESThus, the only thing we need to find now is whether a > b or a < b.
Please note how we are going to Statement 1 now with a much simpler 'To Find' task now. One look at St. 1 and we know that it will be sufficient.
One look at St. 2 and we know that it doesn't give us a clear idea of which is greater between a and b, and so, is not sufficient.
To sum up this discussion,
spending time on analyzing the question statement before going to the two statements usually simplifies DS questions a good deal.Hope this helped!
Japinder