apoorva.srivastva wrote:If 1/(n+1)<1/31+1/32+1/33<1/n, n=?
(A) 9
(B) 10
(C) 11
(D) 12
(E) 13
please solve it algebraically!!

I don't believe an algebraic solution is the way to go here.
First, recognize that 1/33 < 1/32 (and 1/33 < 1/31)
So, we know that 1/33 + 1/33 + 1/33 < 1/31 + 1/32 + 1/33
We should also see that 1/33 and 1/32 are very close in value so we might say that (1/33 + 1/33 + 1/33) is just a little bit less than (1/31 + 1/32 + 1/33)

Also recognize that 1/33 + 1/33 + 1/33 = 1/11
So 1/33 + 1/33 + 1/33 < 1/31 + 1/32 + 1/33 can be rewritten as 1/11 < 1/31 + 1/32 + 1/33

Compare this to the inequality in our question: 1/(n+1)<1/31+1/32+1/33
We can see that n+1 = 11, which means n=10

zonda12 wrote:Just a little concern: how in the hell is an average person supposed to come up with that approach in under minutes? I mean, my math is decent, but no way near the level of thinking of that on the spot. I don't wish to belittle your answer, because you are indeed a genuius; I'm merely curious.

Is this a normal (easy/mid) question for gmat, or is this considered at a hard one (750+ score) one?

That's a valid point. Sometimes when explaining a solution to a problem it's easy to make it sound like one just steps up and . . . voila . . . everything just falls into place. The actual process is much more complex, and seldom so straightforward.

Here's another way to tackle the question:
First, I'd love to know the sum of 1/31 + 1/32 + 1/33 without actually doing all of the calculations.
One option would be to conclude that, since 1/32 falls between the other two fractions, then 1/31 + 1/32 + 1/33 should be approximately equal to 1/32 + 1/32 + 1/32 = 3/32

At this point I ask how 3/32 compares with 1/n and 1/(n+1)

I know that 3/32 is a little less than 1/10 since 3.2/32 would be equal to 1/10
I also know that 3/32 is greater than 1/11 since 3/33 = 1/11
So, from here, I could conclude that 1/11 < 1/31 + 1/32 + 1/33 < 1/10

The original question features the inequality 1/(n+1) < 1/31 + 1/32 + 1/33 < 1/n
So, we can see that 1/11 = 1/(n+1) . . . which means n=10

I'm finding it hard to predict the degree of difficulty on this one. I'm going to go out on a limb and say it falls somewhere around 650 (even though I should be predicting an approximate raw score )