faraday88 wrote:Is 1/m < 1/n?
1) m-n=3
2) 3n=m+2

One look at the question should tell us that whether 1/m is less than 1/n or not depends upon the signs of m and n, and also on whether m > n or not. So we will plug number accordingly to prove insufficiency.

Statement 1: m = (n + 3)
Consider the following two examples,

• m = 4, n = 1 ---> 1/m < 1/n
  m = 1, n = -2 ---> 1/m > 1/n

Not sufficient

Statement 2: m = (3n - 2)
Consider the following two examples,

• m = 1, n = 1 ---> 1/m = 1/n
  m = 4, n = 2 ---> 1/m < 1/n

Not sufficient

1 & 2 Together: m = (n + 3) and m = (3n - 2)
As we have two equations, we can solve for both m and n to get definite values for them. With definite values for m and n, we can definitely say whether 1/m is less than 1/n or not.

Sufficient

The correct answer is C.

dhonu121 wrote:Plugging numbers is fairly easy, just that even you have plugged a few of them, you aren't sure whether you have tried all possible scenarios.
Any approach to ensure that please ?

In DS, plug numbers only to ensure insufficiency.
It's more or less intuition (experience, I prefer) when it comes to "How do you know when to plug numbers?". For example, in this case as I mentioned before doing analysis that "whether 1/m is less than 1/n or not depends upon the signs of m and n, and also on whether m > n or not." Now from both statements we can easily conclude that m > n. But what about the signs? We cannot conclude anything about that from the statements. Hence, I had a hunch they are insufficient individually. So I plugged numbers to prove that. The same could've been shown algebraically but it's a bit time consuming to explain all the logic and follow them.

However, you can never prove sufficiency by plugging numbers (unless a finite range is defined and you have checked for each of the members in that range). Even if it seems that by plugging ten sets of numbers the statement is sufficient, you may be missing some.