kashishh wrote: Is 0<n<1
St.1.) n^2 < n
St.2.) n^3>0

Plug in the values
Statement 1: SUFFICIENT
We will consider 3 cases here
1) Assume n<0
We know that n^2 is always positive
So when n<0 ,n^2 will always be greater than 0.
=> n^2>n But the statement says the opposite . Hence our assumption is wrong.

2)Assume n>1
For every value of n>1, n^2 will be greater than n
=> n^2> n Again our assumption is wrong.

3)Assume 0<n<1
take n = 1/2, then n^2 = 1/4
=> n^2<n. Hence our assumption is correct and we can answer a definite Yes to our question.

Note : n^2< n only when 0<n<1 . So from next time you won't need to take all the cases mentioned above, as they were just for the explanation purpose.

Statement 2:INSUFFICIENT
This statement only tells us that n>0 . Hence we don't know whether 0<n<1 or n>1

The correct answer is A

kashishh wrote:St.1.) n^2 < n
St.2.) n^3>0

Please tell how to solve this DS problem.
I got B as answer. [/list]

The answer is A.

For statement 1, we know that n can NOT be negative since a negative number squared will always be positive, hence greater than n. The only numbers that meet this criteria are positive numbers between 0 and 1. It is thus sufficient.

Statement 2, we know that n can NOT be negative because a negative number cubed will always be less than 0. Any positive number cubed will always be positive, so this statement is not sufficient.

Statement 1
Think about when squaring a number makes it smaller. (Drawing graphs of y=x and y=x^2 can help).
The answer is only when 0<n<1 -- SUFFICIENT.
(Further explanation: if n<0, n^2 will be positive and hence greater. If n>1, n^2 will be larger than n).

Statement 2
What does this tell us? Cubing a number retains the sign, so this simply tells us that n is positive.
Since we don't know that n<1, this is INSUFFICIENT.

Answer: A