nidhis.1408 wrote:If n is not equal to 0, is |n| < 4 ?
(1) n^2 > 16
(2) 1/|n| > n
Can somebody please explain the 2nd set of data- 1/|n| > n
Hi, there. I'm happy to help.
It sounds like you understand statement #1, which of course is sufficient by itself, because if n^2 > 16, then either n > + 4 or n < -4, and in either case, |n| > 4, giving a definitive "no" answer to the prompt question. Statement #1, alone and by itself, is sufficient.
What is going on with that expression in statement #2?
1/|n| > n
Well, as is often the case when we have an algebraic expression involving absolutely value, we will consider two separate cases, one in which n is positive and one in which n is negative.
Case I: n > 0
Then, the inequality becomes
1/n > n
1 > n^2
That would be true for any positive fraction less than one. So, if we assume n > 0, we get a solution of 1 > n > 0.
Case II: n < 0
Then, the inequality becomes
-1/n > n
Multiply by n --- BUT notice that, in doing so, we are multiplying by a negative, so the order of the inequality reverses.
-1 < n^2
Well, the right side, the square of a negative number, will be positive. Any positive is greater than -1, so this works for all values of x in the region. From the n < 0 region, every member works.
Combined solution
Now, we take the union of the individual solution sets.
If 1/|n| > n
then 1 > n > 0 or n < 0 --- that's the complete solution region
Obviously, that region includes some value for which |n| < 4 and others for which |n| > 4. We can find values to satisfy the inequality both ways, so we are unable to determine a definitive answer to the prompt question. This statement, alone and by itself, is insufficient.
Statement #1 = sufficient
Statement #2 = insufficient
Answer =
A
Does all that make sense?
Here's another challenging absolute value question:
https://gmat.magoosh.com/questions/126
When you submit your answer to that question, the very next screen will have a video explanation. Each one of our 800+ GMAT practice questions has it's own video explanation, for accelerated learning.
Please let me know if you have any further questions.
Mike
